TITLE: Products and sum of cubes in Fibonacci QUESTION [6 upvotes]: Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$. Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence, QUESTION. Is there a combinatorial or more conceptual reason for this "pretty" identity? $$F_nF_{n-1}F_{n-2}=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}3.$$ Caveat. I'm open to as many alternative replies, of course. Remark. The motivation comes as follows. Define $F_n!=F_1\cdots F_n$ and $F_0!=1$. Further, $\binom{n}k_F:=\frac{F_n!}{F_k!\cdot F_{n-k}!}$. Then, I was studying these coefficients and was lead to $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$ REPLY [15 votes]: $F_n$ is the number of compositions (ordered partitions) of $n-1$ into parts equal to 1 or 2. The number of triples $(a,b,c)$ of such compositions is $F_n^3$. The number such that $a,b,c$ all begin with 1 is $F_{n-1}^3$. The number such that $a,b,c$ all begin with 2 is $F_{n-2}^3$. Otherwise either one of $a,b,c$ begins with 1 and the others begin with 2, or vice versa. There are $3F_{n-1}F_{n-2}^2$ such triples of the first type. Similarly there are $3F_{n-1}^2F_{n-2}$ of the second type, i.e., one of $a,b,c$ begins with 2 and the others begin with 1. Hence \begin{eqnarray*} F_n^3 & = & F_{n-1}^3 + F_{n-2}^3 +3(F_{n-1}^2F_{n-2}+F_{n-1}F_{n-2}^2)\\ & = & F_{n-1}^3 + F_{n-2}^3 +3F_{n-1}F_{n-2}(F_{n-1}+F_{n-2})\\ & = & F_{n-1}^3 + F_{n-2}^3 + 3F_nF_{n-1}F_{n-2}. \end{eqnarray*}<|endoftext|> TITLE: Turning simplicial complexes into simplicial sets without ordering the vertices QUESTION [14 upvotes]: Given an abstract simplicial complex $K$, one can make a simplicial set $X(K)$ with $n$-simplices given by sequences $(x_0, \ldots, x_n)$ such that $\{x_0, x_1, \ldots, x_n\}$ is a simplex of $K$. The face maps delete entries and the degeneracy maps repeat entries. I'd like a reference for the fact that the geometric realization of $X(K)$ is homotopy equivalent to the geometric realization of $K$ itself. (Note that $|X(K)|$ is typically very big: for $K$ a single edge, $|X(K)|$ is the infinite-dimensional sphere $S^\infty$.) I've sketched a proof of this fact here, but hope there is a reference I can just cite since, as I expected, every algebraic topologist I've asked in person already knew the fact. :) Also, does this $X(K)$ have a standard name or notation? Or if not, can someone think of a catchy name or nice notation? REPLY [5 votes]: Let $K$ be a simplicial complex with vertex set $V$. Let $S_\bullet (K)$ be the simplicial set whose p-simplices are the maps $f:[p]\to V$ such that $f([p])$ is a simplex of $K$, or alternatively the set of maps $\Delta^p \to K$ of simplicial complexes. There is an obvious map $$\pi_K:|S_\bullet (K)| \to |K|$$ which you ask to be a homotopy equivalence. Here is an argument. I find it easier to work with the fat geometric realization $||S_\bullet (K)||$ instead, but the difference is minimal, since the quotient map to the ordinary geometric realization is a homotopy equivalence. Step 1. Consider first the case $K=\Delta^n$ (rather, it is the full simplicial complex with vertex set $[n]$). I claim that $||S_\bullet \Delta^n||$ is contractible. For sake of notational clarity, let me write $\nabla^p$ for the topological $p$-simplex. Consider the map $$H_p: S_p (\Delta^n) \times \nabla^p \times [0,1] \to S_{p+1}(\Delta^n) \times \nabla^{p+1} $$ which is given by the formula $$H(f,v,t):= (f \ast n,((1-t)v,t)). $$ Explanation: $f \ast n: [p+1] \to [n]$ is the map whose restriction to $[p]$ is $f$ and which has $f(p+1)=n$. Furthermore $((1-t)v,t) \in \mathbb{R}^{p+1} \times \mathbb{R}$ is a point of $\nabla^{p+1}$. It is easily checked that the different $H_p$ glue together to a map $H:|| S_\bullet (\Delta^n)|| \times [0,1] \to ||S_{\bullet}(\Delta^n) ||$ (use that products and quotients commute in this setting, as the interval is compact, or work in the context of compactly generated spaces). It is clear that $H(0,\_)$ is the identity, and $H(1,\_)$ is the constant map to the vertex $n$. So we are done in this case. Step 2. Now we prove the claim for finite complexes, by induction over both, the dimension and the number of top-dimensional simplices. The induction beginning $K=\emptyset$ is trivial. For the induction step, let $K$ be $n$-dimensional and let $L$ be obtained from $K$ by deleting one $n$-simplex. Then $|K| \cong |L| \cup_{|\partial \Delta^n|} |\Delta^n|$ and $||S_\bullet (K)|| \cong ||S_\bullet (L)|| \cup_{||S_\bullet (\partial \Delta^n)||} ||S_\bullet (\Delta^n)||$. The map $\pi_K$ is the pushout of the maps $\pi_{\Delta^n}$ and $\pi_L$, along $\pi_{\partial \Delta^n}$. These maps are homotopy equivalences, by step 1 and by induction hypothesis, respectively. The maps $|\partial \Delta^n| \to |\Delta^n|$ and $||S_\bullet (\partial \Delta^n)|| \to ||S_\bullet (\Delta^n)||$ are cofibrations, and so the gluing lemma implies that $\pi_K$ is a homotopy equivalence. Step 3. Having shown the claim for finite complexes, it follows by a colimit argument that $\pi_K$ is a weak homotopy equivalence for arbitrary $K$, and hence a homotopy equivalence, by Whiteheads theorem.<|endoftext|> TITLE: Interpretations for higher Tor functors QUESTION [19 upvotes]: Let's work in the category $R$-${\sf mod}$, for concreteness. I know that one can see the modules ${\rm Ext}^n_R(M,N)$ as modules of equivalence classes of $n$-extensions of $M$ by $N$ (Yoneda extensions), namely, exact sequences of the form $$0 \to N \to E_1 \to \cdots \to E_n \to M \to 0,$$with certain operations (more precisely, if one denotes such collection by ${\rm E}^n(M,N)$, there are natural isomorphisms ${\rm E}^n(M,N)\cong {\rm Ext}_R^n(M,N)$ for each $n$). Is there anything similar for ${\rm Tor}^R_n(M,N)$? I expect the answer to be highly non-trivial, for the following analogy: this business about $n$-extensions effectively gives us a way to describe the elements of ${\rm Ext}^n_R(M,N)$, but why should we expect any simple explanation for ${\rm Tor}$ if we cannot even describe the elements of ${\rm Tor}_0^R(M,N) = M\otimes_RN$ in general? Apologies if by any chance this is a repeated question, a quick search on the website didn't show up anything here. REPLY [3 votes]: I don't know an answer for general rings/modules but the best approach is probably to write Tor in terms of Ext when you want a similar interpretation for Tor with exact sequences. For Artin algebras $R$ with duality $D$ (which include for example all quiver algebras $KQ/I$) and finitely generated modules $Y$ and $Z$ one then has $$\operatorname{Tor}_n^R(Y,Z)=D\operatorname{Ext}_R^n(Y,D(Z)).$$ So in this case Tor for $Y$ and $Z$ has an interpretation as a dual space of extensions between $Y$ and $D(Z)$. I think the same should work for general rings/modules with a duality having good properties.<|endoftext|> TITLE: "Oddity" of Fibonacci-Catalan numbers QUESTION [5 upvotes]: As a follow up to my previous two MO questions, here and here, let's consider the below inquiry. Define the Fibonacci-Catalan numbers by $FC_n=\frac1{F_{n+1}}\binom{2n}n_F$ where $F_0=0, F_1=1, F_0!=1$, $$F_n=F_{n-1}+F_{n-2} \qquad F_n!=F_1\cdot F_2\cdots F_n, \qquad \binom{n}k_F=\frac{F_n!}{F_k!\cdot F_{n-k}!}.$$ Recall the property that the Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$ satisfy: $C_n$ is odd iff $n=2^j-1$ for some $j\in\mathbb{N}$. In the same vain, QUESTION. For $n\geq2$, is this true? $$\text{$FC_n$ is odd iff $n=3\cdot2^j-1$ for some $j\in\mathbb{N}$}.$$ POSTSCRIPT. In response to Alexander Burstein question below, I suppose the following should hold. If we write $F$ for $F(s,t)$ and $FC_n(s,t)=\frac1{F_{n+1}(s,t)}\binom{2n}n_F$ then $$\text{$FC_n(2s-1,2t-1)$ is odd iff $n=3\cdot2^j-1$} \qquad \text{and}$$ $$\text{$FC_n(2s,2t-1)$ is odd iff $n=2^j-1$}.$$ REPLY [11 votes]: Let $\alpha=(1+\sqrt{5})/2,\beta=(1-\sqrt{5})/2$, then by Binet formula for Fibonacci numbers we have $F_n=(\alpha^n-\beta^n)/(\alpha-\beta)=:P_n(\alpha,\beta)$. Factorize our Catalan-like expression onto cyclotomics: $$ \frac{\prod_{j=1}^{2n} P_j(x,y)}{\prod_{i=1}^{n+1}P_i(x,y)\cdot \prod_{i=1}^{n}P_i(x,y)}=\prod_{s\geqslant 2} (\Phi_s(x,y))^{\eta(n,s)},\quad (\star)\\ \text{where}\quad\eta(n,s)=\left[\frac{2n}s\right]-\left[\frac{n}s\right]-\left[\frac{n+1}s\right].\quad (\bullet) $$ Here $\Phi_s(x,y)$ are homogeneous cyclotomic polynomials, and $(\star)$ immediately follows from $P_j=\prod_{d|j,d>1} \Phi_d$. Therefore we get $$ FC_n=\prod_{s>2} (\Phi_s(\alpha,\beta))^{\eta(n,s)}. $$ Now let us find out which numbers $g_s:=\Phi_s(\alpha,\beta)$ are even. Recall that $F_n$ is even if and only if $n$ is divisible by 3. Since $$ F_n=P_n(\alpha,\beta)=\prod_{d|n} \Phi_d(\alpha,\beta)=\prod_{d|n} g_d, $$ we conclude that $g_s$ are odd when $s$ is not divisible by 3. Next, if $n=3kl$ for odd $l>1$, then $g_{3kl}$ divides $$\frac{F_{3kl}}{F_{3k}}=\frac{\alpha^{3kl}-\beta^{3kl}}{\alpha^{3k}-\beta^{3k}}= \alpha^{3k(l-1)}+\ldots+\beta^{3k(l-1)},$$ and substituting $\alpha^3=2+\sqrt{5},\beta^3=2-\sqrt{5}$ and expanding the brackets we see that it is odd. Therefore all $g_s$ for $s\ne 3\cdot 2^m,m=0,1,\ldots$, are odd. $g_3=2$ and if $s=3\cdot 2^m$, $m>0$, we have $$g_{3\cdot 2^m}g_{2^m}F_{3\cdot 2^{m-1}}=F_{3\cdot 2^m}.$$ Using the formula $F_{2k}=F_k(F_{k-1}+F_{k+1})$ for $k=3\cdot 2^{m-1}$ we finally conclude that $g_{3\cdot 2^m}$ is even. So your claim is equivalent to the following: if $n\geqslant 2$, then all exponents of the form $\eta(n,3\cdot 2^m)$ are equal to 0 if and only if $n=3\cdot 2^j-1$ for some $j=0,1,\ldots$. If $n=3m$ or $n=3m+1$ for $m\geqslant 1$, $k=3\cdot 2^s$, where $2^s$ is the maximal power of 2 which divides $2m$. We get from $(\bullet)$ that $\eta(n,k)=[2m/2^s]-2[m/2^s]=1$. If $n=3m+2$ and $k=3t$, we get $\eta(3m+2,3t)=[(2m+1)/t]-[m/t]-[(m+1)/t]$. For $t=1$ this is always 0, for $t=2^s$, $s>0$, this is the same as $[(2m)/t]-[m/t]-[(m+1)/t]$, and this expression is already studied in the answer to your $q$-Catalan question. Namely, it is zero for all positive integer $s$ if and only if $m=2^j-1$ which means $n=3\cdot 2^j-1$.<|endoftext|> TITLE: Does "every" first-order theory have a finitely axiomatizable conservative extension? QUESTION [32 upvotes]: I originally asked this question on math.stackexchange.com here. There's a famous theorem (due to Montague) that states that if $\sf ZFC$ is consistent then it cannot be finitely axiomatized. However $\sf NBG$ set theory is a conservative extension of $\sf ZFC$ that can be finitely axiomatized. Similarly, if $\sf PA$ is consistent then it is not finitely axiomatizable (Ryll-Nardzewski) but it has a conservative extension, $\sf ACA_0$, which is finitely axiomatizable. (Usually $\sf ACA_0$ is considered a second-order theory, but my understanding is that there isn't really a difference between first and second-order from the syntactic point of view.) I'm wondering if this happens in general. At first I thought it might be that every theory had a finitely axiomatizable conservative extension. But then I realised that every theory with a finitely axiomatized conservative extension must be effectively axiomatizable. So we shouldn't hope for theories to have finitely axiomatizable conservative extensions unless they're already effectively axiomatizable. Similarly if we add countably many logical constants to the language and demand that they're all true then that's effectively axiomatizable but doesn't have a finitely axiomatizable conservative extension, so we should restrict our attention to finite languages. Does every effectively axiomatizable first-order theory over a finite language have a finitely axiomatizable conservative extension? REPLY [27 votes]: Essentially, yes. An old result of Kleene [1], later strengthened by Craig and Vaught [2], shows that every recursively axiomatizable theory in first-order logic without identity, and every recursively axiomatizable theory in first-order logic with identity that has only infinite models, has a finitely axiomatized conservative extension. See also Mihály Makkai’s review, Richard Zach’s summary, and a related paper by Pakhomov and Visser [3]. Let me stress that the results above apply to the literal definition of conservative extension, i.e., we extend the language of $T$ by additional predicate or function symbols, and we demand that any sentence in the original language is provable in the extension iff it is provable in $T$. If we loosen the definition so as to allow additional sorts, or extension by means of a relative interpretation, then every recursively axiomatizable first-order theory has a finitely axiomatized conservative extension. But back to the standard definition. I’m assuming logic with identity from now on. What happens for theories that may also have finite models? First, [2] give the following characterization (they call condition 1 “f.a.${}^+$”, and condition 2 “s.f.a.${}^+$”): Theorem. For any theory $T$ in a finite language, the following are equivalent: $T$ has a finitely axiomatized conservative extension. There exists a finitely axiomatized extension $T'\supseteq T$ such that every model of $T$ expands to a model of $T'$. $T$ is equivalent to a $\Sigma^1_1$ second-order sentence. Following Dmytro Taranovsky’s comments, we have the following necessary condition (which is actually also mentioned in [2], referring to Scholz’s notion of spectrum instead of NP, which was only defined over a decade later): Theorem. For any theory $T$ in a finite language, 1 implies 2: $T$ has a finitely axiomatized conservative extension. $T$ is recursively axiomatizable, and the set of its finite models is recognizable in NP. Indeed, the truth of a fixed $\Sigma^1_1$ sentence in finite models can be checked in NP. It is not known if this is a complete characterization: if $T$ is a r.e. theory whose finite models are NP, then $T$ is equivalent to a $\Sigma^1_1$ sentence on infinite models by [1,2], and $T$ is equivalent to a $\Sigma^1_1$ sentence on finite models by Fagin’s theorem, but it is unclear how to combine the two to a single $\Sigma^1_1$ sentence that works for all models. This is mentioned as an open problem in the recent paper [3]. References: [1] Stephen Cole Kleene: Finite axiomatizability of theories in the predicate calculus using additional predicate symbols, in: Two papers on the predicate calculus. Memoirs of the American Mathematical Society, no. 10, Providence, 1952 (reprinted 1967), pp. 27–68. [2] William Craig and Robert L. Vaught: Finite axiomatizability using additional predicates, Journal of Symbolic Logic 23 (1958), no. 3, pp. 289–308. [3] Fedor Pakhomov and Albert Visser, On a question of Krajewski’s, Journal of Symbolic Logic 84 (2019), no. 1, pp. 343–358. arXiv: 1712.01713 [math.LO]<|endoftext|> TITLE: Kolmogorov superposition on the Hilbert Cube QUESTION [7 upvotes]: A result of Kolmogorov and Arnold says that continuous functions on $\mathbb{R}^n$ can be represented as sums of the form $$ f(x_1,\dots,x_n)=\sum_{q=0}^{2n}\Phi_q\left(\sum_{p=1}^n\phi_{p,q}(x_p)\right),$$ where $\Phi_p$ and $\phi_{p,q}$ are unary continuous functions. I'm curious about analogous results for functions on the Hilbert cube. Suppose I have a continuous function $f:[0,1]^\omega \rightarrow [0,1]$. Is it always possible to write this in the form $$ f(x_0,x_1,\dots) = \sum_{q<\omega} \Phi_q\left( \sum_{p<\omega}\phi_{p,q}(x_p) \right) ,$$ where $\Phi_p$ and $\phi_{p,q}$ are continuous functions and all of the sums converge uniformly? If this fails are there known analogous results? EDIT: What we do get immediately is this. For each $n>\omega$ let $$f_n(x_0,\dots,x_{n-1})=\inf_{\overline{y}\in[0,1]^\omega} f(x_0,\dots,x_{n-1},\overline{y}),$$ i.e. $f_n$ is an approximation of $f$ from below by a continuous function on $[0,1]^n$. Let $f_{0}=0$ and $g_{n}(x_0,\dots,x_{n-1})=f_n(x_0,\dots,x_{n-1})-f_{n-1}(x_0,\dots,x_{n-2})$, so that in particular $$f(x_0,x_1,\dots)=\sum_{1\leq n < \omega}g_{n}(x_0,\dots,x_{n-1}).$$ Now we can use Kolmogorov superposition on each $g_{n}$, $$g_n(x_0,\dots,x_{n-1}) = \sum_{q=1}^{2n} \Phi_{n,q}\left( \sum_{p=0}^{n-1} \phi_{n,p,q}(x_p) \right).$$ So then we have $$f(x_0,x_1,\dots)=\sum_{1\leq n < \omega} \sum_{q=1}^{2n} \Phi_{n,q}\left( \sum_{p=0}^{n-1} \phi_{n,p,q}(x_p) \right) $$. By Dini's theorem the outermost sum is uniformly convergent and we can rearrange this sum to be one of the desired form, but it's unclear if the rearranged sum converges uniformly, in particular it's not clear to me that it is monotonic. For a related posts see: Is there any continuous ternary function which can not be represented by composition of continuous binary functions? Kolmogorov superposition for smooth functions Kolmogorov-Arnold theorem for (just-)functions Continuous functions of three variables as superpositions of two variable functions REPLY [2 votes]: After finally getting around to learning the proof of the Kolmogorov–Arnold representation theorem (thanks to this recorded talk by Bar-Natan) I now know that the functions can be chosen so that the rearranged sequence in my edit converges uniformly (in particular the terms are all positive, so the sum is monotonic and therefore uniformly convergent by Dini's theorem). By shifting the functions around you can remove the requirement on the range of $f$. The precise statement can be taken to be this: There is a fixed sequence $\{\phi_p\}_{p<\omega}$ of continuous unary functions $\mathbb{R}\rightarrow\mathbb{R}$ and a fixed sequences $\{a_n\}_{n<\omega}$ and $\{b_n\}_{n<\omega}$ of intergers such that for any continuous function $f:[0,1]^\omega \rightarrow \mathbb{R}$ there exists a sequence $\{\Phi_n\}_{n<\omega}$ of continuous unary functions $\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(x_0,x_1,\dots)=\sum_{n<\omega}\Phi_n\left(\sum_{p = a_n}^{b_n} \phi_p(x_p)\right),$$ with the outermost sum converging monotonically (after the first term) and uniformly. The key being that (after shifting so that $f$ is non-negative), each of the terms in the sum are non-negative. I suspect that we can choose $a_n=0$ and $b_n = n$, but I haven't proven that. It also seems likely that this result or something similar to it is actually easier to prove directly than via the Kolmogorov–Arnold representation theorem. In the broadest sense we already know that every function on the Hilbert cube can be represented by an infinite sum of compositions of unary functions and addition, since this is sufficient to give polynomials in the coordinates and polynomials can uniformly approximate functions on the Hilbert cube, so this isn't really as exciting as the finite dimensional case.<|endoftext|> TITLE: The union of all coreflective Cartesian closed subcategories of $\mathbf{Top}$ QUESTION [7 upvotes]: Very often, in topology, one restricts to a coreflective Cartesian closed subcategory of $\mathbf{Top}$ in order to freely use exponential laws for mapping spaces, which imply things like "the internal categorical product of quotient maps is a quotient map." I have a situation where an argument works regardless of the particular "convenient" subcategory that I pick. Hence, as long as a space lies in some coreflective Cartesian closed subcategory, I may apply my argument. What is the class of topological spaces, that lie in some coreflective Cartesian closed subcategory of $\mathbf{Top}$? Update: It is well-known that every class of spaces generates a Cartesian closed subcategory of $\mathbf{Top}$ (See Booth-Tillotson) but it may not be coreflective. On the other hand, many coreflective categories like the category of locally path-connected spaces are not Cartesian closed. As pointed out by David White below, it is a result of Juraj Cincura that there is no largest coreflective Cartesian closed subcategory of $\mathbf{Top}$. I'd be content to just to know that there is a space that does not lie in any coreflective Cartesian closed subcategory of $\mathbf{Top}$. REPLY [5 votes]: I contacted Juraj Činčura and he kindly wrote back and directed me to the following observation that is a consequence of results in the paper that David White noted in his answer. Cartesian closed coreflective subcategories of the category of topological spaces, Topology and its Applications 41 (1991) 205-212. For an infinite cardinal $a$, let $C(a)$ be the space defined on set $a\cup \{a\}$ where $V\subseteq C(a)$ is open if and only if $V\subseteq a$ or if $a\in V$ and $|a\backslash V|b$, then any coreflective subcategory in $\mathbf{Top}$ or $\mathbf{Haus}$ containing the space $C(a)\times C(b)$ is not Cartesian closed. Hence, there is no coreflective Cartesian closed subcategory of $\mathbf{Top}$ or $\mathbf{Haus}$ containing $C(a)\times C(b)$ if $a\neq b$. This is a specific example showing that there is no largest coreflective Cartesian subcategory of $\mathbf{Top}$ or $\mathbf{Haus}$.<|endoftext|> TITLE: Two monoidal structures and copowering QUESTION [6 upvotes]: Let $(\mathbf{M},\otimes,1)$ be a closed monoidal category and $(\mathbf{C},\oplus,0)$ an $\mathbf{M}$-enriched monoidal category. Furthermore, assume that we have a copowering $\odot:\mathbf{M}\times\mathbf{C}\to \mathbf{C}$. Is there a canonical morphism $$(A\odot X)\oplus (B\odot Y)\to (A\otimes B)\odot (X\oplus Y)$$ The question came to my mind because in order to spell out the axioms (in one of the definitions) for an algebra over an operad $\mathcal{O}$ in the above setting, we need for the associativity axiom a morphism $$\mathcal{O}(r)\odot \left(\bigoplus_i (\mathcal{O}(k_i)\odot X^{\oplus k_i})\right)\to\left(\mathcal{O}(r)\otimes\bigotimes_{i}\mathcal{O}(k_i)\right)\odot \left(\bigoplus_iX^{\oplus k_i}\right)$$ Or the other direction. If $\mathbf{M}$ is considered to be enriched over itself, everything is fine because then $\otimes=\odot=\oplus$, but in general? REPLY [9 votes]: No. Consider the case where $(M,\otimes,1)$ is $(\mathbf{Set},\times,1)$, so the enrichment is vacuous, and $(C,\oplus,0)$ is $(\mathbf{Set},+,0)$, with copowering $\odot$ given by $\times$. Then the morphism you ask for would give a map $$(A \times X) + (B \times Y) \longrightarrow (A \times B) \times (X + Y) $$ which doesn’t exist in general: consider $A = X = Y = 1$, $B = 0$. However, there is a natural map in the other direction. There are natural maps $A \to C(X,A \odot X)$ and $B \to C(Y,B \odot Y)$, the structure maps of the copowering. Also, the definition of enriched monoidal category includes the condition that $\oplus$ is an enriched bifunctor, so there’s a general map $C(X,X') \otimes C(Y,Y') \to C(X \oplus Y, X' \oplus Y')$. Putting these together, we get a map $$ A \otimes B \longrightarrow C(X, A \odot X) \otimes C(Y, B \odot Y) \longrightarrow C(X \oplus Y, (A \odot X) \oplus (B \odot Y)) $$ which corresponds under copowering to a map $(A \otimes B) \odot (X \oplus Y) \to (A \odot X) \oplus (B \odot Y)$.<|endoftext|> TITLE: $p$-adic equivalence of spectra with $G$-action QUESTION [6 upvotes]: In Krause and Nikolaus' "Lectures on topological Hochschild homology and cyclotomic spectra" (which can be found here), there is a lemma I'm trying to understand, but one line in the proof eludes me. The lemma and proof are as follows: Lemma A.4. Let $X\to Y$ be a $G$-equivariant map of connective spectra where $G$ is a discrete group. Assume that $G$ acts trivially on $\Bbb{F}_p$-homology of $X$ and $Y$ and that the induced map $$X_{hG}\to Y_{hG}$$ is a $p$-adic equivalence. Then also the initial map $X\to Y$ is a $p$-adic equivalence. Proof: Let $Z$ be the cofibre of the map $X\to Y.$ Then it also carries a $G$-action and the homotopy orbits $Z_{hG}$ are $p$-adically trivial. We want to show that $Z$ is $p$-adically trivial, i.e. that the $\Bbb{F}_p$-homology of $Z$ vanishes. Assume that it has a lowest non-trivial homology group $H_n(Z;\Bbb{F}_p).$ We get that $$H_n(Z_{hG},\Bbb{F}_p) = H_n(Z,\Bbb{F}_p)/G.$$ The action of $G$ on $H_n(Z,\Bbb{F}_p)$ is not necessarily trivial, but still $2$-stage nilpotent by the long exact sequence. Thus the coinvariants are non-trivial which gives a contradiction. Question 1: I'm assuming that $H_n(Z,\Bbb{F}_p)/G$ refers to the coinvariants of the $G$ action on $H_n(Z,\Bbb{F}_p),$ although I'm not 100% sure about this. It's the only thing that seems to make sense in the context, but I'm not familiar with this notation being used to mean coinvariants. Is this indeed what is meant? (Also, where can I find a proof of the stated isomorphism? I think I have a spectral sequence argument that proves this, but it would be nice to find a reference as well.) Question 2: Aside from the coinvariant notation, I follow the argument until the penultimate sentence. I have never heard of the notion of a "$2$-stage nilpotent group action," and I did not find any explanation when I did a search. I only find references to nilpotent groups, which does not seem to be what is meant here -- all abelian groups, trivial or not, are nilpotent, so this doesn't seem like it would be particularly useful in the argument. What does the "$2$-stage nilpotence" here mean, and why does it prove the result? Although I could not figure out the meaning of $2$-stage nilpotence, I played around with long exact sequences in homology associated to $X\to Y\to Z$ to try to show that $H_n(Z,\Bbb{F}_p)_G$ is nontrivial, and I think I've found a proof (which may or may not be what Krause and Nikolaus had in mind); I've included it for completeness. Proof: Consider the long exact sequence in homology $$ \cdots\xrightarrow{\partial_{n+1}}H_n(X,\Bbb{F}_p)\xrightarrow{f_n}H_n(Y,\Bbb{F}_p)\xrightarrow{g_n}H_n(Z,\Bbb{F}_p)\xrightarrow{\partial_n}H_{n-1}(X,\Bbb{F}_p)\xrightarrow{f_{n-1}}H_{n-1}(Y,\Bbb{F}_p)\to 0. $$ Set $K := \ker f_{n-1} = \operatorname{im}\partial_n.$ Observe that $K$ fits into an exact sequence $$ 0\to K\xrightarrow{i_{n-1}}H_{n-1}(X,\Bbb{F}_p)\xrightarrow{f_{n-1}}H_{n-1}(Y,\Bbb{F}_p)\to 0 $$ and that the map $\partial_n$ factors as $H_n(Z,\Bbb{F}_p)\twoheadrightarrow K\xrightarrow{i_{n-1}} H_{n-1}(X,\Bbb{F}_p).$ Then there are two cases to consider: $K = 0.$ In this case, it follows that $\partial_n = 0,$ and so we get a short exact sequence $$ 0\to K'\xrightarrow{i_n} H_n(Y,\Bbb{F}_p)\xrightarrow{g_n}H_n(Z,\Bbb{F}_p)\to 0, $$ where $K' =\ker g_n\subseteq H_n(Y,\Bbb{F}_p).$ Taking coinvariants, we get an exact sequence $$ (K')_G\xrightarrow{(i_n)_G} H_n(Y,\Bbb{F}_p)_G\xrightarrow{(g_n)_G}H_n(Z,\Bbb{F}_p)_G\to 0. $$ However, the $G$-action on $H_n(Y,\Bbb{F}_p)$ is assumed to be trivial, which implies that the $G$-action on $K'$ is also trivial and that the map $(i_n)_G = i_n.$ Thus, $$ \ker\left((g_n)_G\right) = \operatorname{im}\left((i_n)_G\right) = \operatorname{im}i_n = \ker g_n, $$ so the $G$-action on $H_n(Z,\Bbb{F}_p)$ is trivial. This implies that $H_n(Z,\Bbb{F}_p)_G = H_n(Z,\Bbb{F}_p)\neq 0,$ and we get the desired contradiction. $K\neq 0.$ Because $H_n(Z,\Bbb{F}_p)$ surjects onto $K$ and taking coinvariants is right exact, it suffices to prove that $K_G\neq 0.$ Taking coinvariants of the short exact sequence $$ 0\to K\xrightarrow{i_{n-1}} H_{n-1}(X,\Bbb{F}_p)\xrightarrow{f_{n-1}}H_{n-1}(Y,\Bbb{F}_p)\to 0 $$ gives the exact sequence $$ K_G\xrightarrow{(i_{n-1})_G} H_{n-1}(X,\Bbb{F}_p)_G\xrightarrow{(f_{n-1})_G}H_{n-1}(Y,\Bbb{F}_p)_G\to 0, $$ which simplifies to $$ K_G\xrightarrow{(i_{n-1})_G} H_{n-1}(X,\Bbb{F}_p)\xrightarrow{f_{n-1}}H_{n-1}(Y,\Bbb{F}_p)\to 0 $$ because the action of $G$ is trivial on both $H_{n-1}(X,\Bbb{F}_p)$ and $H_{n-1}(Y,\Bbb{F}_p).$ But now $K_G$ surjects onto $\operatorname{im}\left((i_{n-1})_G\right),$ and we have $$ \operatorname{im}\left((i_{n-1})_G\right) = \ker((f_{n-1})_G)= \ker f_{n-1}= K\neq 0, $$ so we are finished. REPLY [9 votes]: I don't know if "$2$-stage nilpotent" is a standard term for this, but I'm sure that what the author means is this: a group $A$ that $G$ is acting on has a subgroup $B$ such that the action of $G$ fixes every element of $B$ and such that also the resulting action of $G$ on $A/B$ is trivial. This is true for $A=H_n(Z;F_p)$, with $B$ being the image of $H_n(Y;F_p)$. That implies that the group of coinvariants $A_G$ is nontrivial if $A$ is nontrivial, because either (1) $B\neq A$, so that $A/B$ is nontrivial and $A_G$ maps onto the nontrivial group $(A/B)_G$, or (2) $B=A$ so that $A_G=A$ is nontrivial. When $G$ acts on a spectrum $Z$ and $H_j(Z)=0$ for $j TITLE: Properties of categories that can not be proven by abstract nonsense QUESTION [18 upvotes]: What are examples of properties of particular categories that can be formulated in categorical language and "feel" like they ought to be provable formally but they actually are not? I think that this question is inherently imprecise but one can not seriously say that every question on this site is precise. To clarify, the existence of fiber products in the category of schemes would not count IMHO because I personally do not expect a random category to have fiber products. A non-example is the fact that for a morphism of schemes $X\rightarrow Y$, the diagonal morphism $X\rightarrow X\times_Y X$ is a monomorphism. At first I thought that this could only proven non-formally by noticing that the diagonal morphism is an immersion and immersions are monomorphisms but it turns out that in any category with fiber products the diagonal morphism is a monomorphism. REPLY [8 votes]: It's unclear to me both what counts as "abstract nonsense" and what counts as "looks like it should be provable by abstract nonsense", so I think the question is rather subjective. Nonetheless, I think it has merit, so here's a possible example under my own subjective interpretation. Taking inspiration from Aknazar Kazhymurat's comment, my own take is that one's understanding of "what is likely to be provable by abstract nonsense" should be constantly evolving based on new evidence, so maybe the question is really more like "what are some interesting examples of subtleties in the scope of abstract nonsense?". Consider the category $Top_\Delta$ of Delta-generated spaces. This is the smallest full subcategory of the category $Top$ of topological spaces which is closed under colimits and contains the unit interval. Since it is generated under colimits by a small amount of data, it is natural to think that $Top_\Delta$ should be a locally presentable category (which is indeed the case). Indeed, until recently it was beleived that any cocomplete category with a small colimit-generator was locally presentable under the set-theoretical hypothesis known as Vopenka's Principle. So it was thought that, modulo set-theoretical subtleties, the local presentability of $Top_\Delta$ should be viewed as following directly from its definition. Unfortunately, there turned out to be an error in the proof of this consequence of Vopenka's principle (and it turns out that $Set^{op}$ is a counterexample), so this line of reasoning is actually not valid. Even before this was realized, though, it was desirable to have a proof that worked without strong set-theoretical assumptions. Perhaps surprisingly it turns out that the proof in the literature that $Top_\Delta$ is locally presentable relies on notions of infinitary logic. Now, in some sense, this just means that the proof is "doubly-abstract nonsense", but in another sense, it's a type of abstract nonsense that even sophisticated category theorists are not generally familiar with. An unwinding of this proof avoids explicitly using infinitary logic, but still requires one to think about topological spaces in terms of ultrafilters rather than open sets. What is the moral of the story? I'm not quite sure. Maybe it's simply "accessibility questions sometimes require you to think hard and creatively about the specifics of your situation", even though they are often treated as an afterthought (I've certainly been guilty of this attitude before!).<|endoftext|> TITLE: Hyperelliptic Jacobians with (or without) CM QUESTION [6 upvotes]: Let $C$ be a hyperelliptic curve $y^2 = f(x) $ defined over $\mathbb{Q}$, where $f(x) \in \mathbb{Q} [x]$ is a polynomial of degree $n=5$ or $6$, and $J = Jac(C)$ its Jacobian. I know Zarhin's result [Hyperelliptic Jacobians without complex multiplication], which states that if the Galois group $G= Gal(f)$ of $f$ is either the Symmetric group $S_n$ or the alternating group $A_n$, then the endomorphism ring of $J$ is $\mathbb{Z}$ (that is, $J$ has no complex multiplication). My question is that whether there is a condition on $G$ under which $J$ has CM. The reason why I ask this is the follwing; In [Wamelen, Examples of genus two CM curves defined over the rationals], Wamelen found 19 curve (of gunus 2) whose Jacobian has CM. In all of these examples, the Galois group of $f$ is either the cyclic group $C_4$ of order $4$ or the Frobenius group $F_5$ of order $20$. So my question is that; Is there any example of $C$ without CM and with $G = C_4$ ? REPLY [2 votes]: The curve $$y^2 = x^5 + 2x^4 + 2x^3 +2x^2 + 2x +1$$ has $G= C_4$ and its Jacobian $J$ has endomorphism algebra ${\rm End}^0(J) \cong \mathbb{Q}$. For the curve $$y^2 = x^5 + x^4 + x^3 + x^2 + x$$ we have $G= C_4$ and its Jacobian $J$ has endomorphism algebra ${\rm End}^0(J) \cong \mathbb{Q} \times \mathbb{Q}$. As explained in Jedrzej's answer, it is possible to impose conditions on $G$ which restrict the possible endomorphism algebras. A relevant example here is: if $C:y^2=f(x)$ is a hyperelliptic curve with $f(x) \in \mathbb{Q}[x]$ of degree 5 or 6, and $|G|$ is odd, then the Jacobian of $C$ does not have CM. In fact, a stronger statement is true: if $f(x)$ has degree 5 or 6 with coefficients in some number field $K$, Galois group $G$ of odd order, and $J$ has CM, then $K$ must contain a real quadratic field.<|endoftext|> TITLE: Ordered union of Borel sets QUESTION [5 upvotes]: Let $\mathfrak{A}$ be an uncountable collection of Borel sets in $\mathbb{R}^d$ such that for any $A,B\in\mathfrak{A}$, either $A\subset B$ or $A\supset B$. Then is it necessarily true that the union $\bigcup_{A\in\mathfrak{A}}A$ is Borel measurable? REPLY [2 votes]: Every coanalytic set can be written as a union of $\aleph_1$-many Borel sets. (For example, see Theorem 4.3.17 of Srivastava.) Take a coanalytic non-Borel set $B \subseteq \mathbb{R}^d$. Then $B = \bigcup_{\alpha < \omega_1} B_{\alpha}$ for some Borel sets $B_{\alpha}$. For each $\alpha < \omega_1$, set $C_{\alpha} = \bigcup_{\delta < \alpha} B_{\delta}$. Clearly each $C_{\alpha}$ is Borel since it is a countable union of Borel sets. On the other hand, $\bigcup_{\alpha < \omega_1} C_{\alpha}=\bigcup_{\alpha < \omega_1} B_{\alpha}=B$ is non-Borel.<|endoftext|> TITLE: Isomorphisms in enriched categories QUESTION [6 upvotes]: Let $(M,\otimes,1)$ be closed monoidal category and $C$ an $M$-enriched category. Assume we have $C$-objects $X$ and $X'$ and a morphism $f:1\to C(X,X')$ in $M$. We call $f$ an isomorphism if there is a $g:1\to C(X',X)$ in $M$ such that $$1\cong 1\otimes 1\stackrel{g\otimes f}{\to} C(X',X)\otimes C(X,X')\to C(X,X)$$ equals $\mathrm{id}:1\to C(X,X)$ and the same for $C(X',X')$. Now for each $Y$, $f$ induces a morphism $$f^*_Y:C(X',Y)\cong C(X',Y)\otimes 1\stackrel{\mathrm{id}\otimes f}{\to} C(X',Y)\otimes C(X,X')\to C(X,Y).$$ Assume that $f_Y^*$ is an $M$-isomorphism for all $Y$. As in the basic case ($C$ enriched over $\mathbf{Set}$), I want to conclude that $f$ is an isomorphism. To construct an inverse $g$, we should use $$g:1\stackrel{\mathrm{id}}{\to} C(X,X)\stackrel{(f_X^*)^{-1}}{\to} C(X',X).$$ First of all: Is this correct? At some point, it seems that I have to use the (enriched) naturality of $f^*$, now as a map $f^*_Y:1\to C(X,Y)^{C(X',Y)}$. REPLY [6 votes]: Unless I've misunderstood something, I think this is all fine. You would need the enriched naturality of $f^*$ in order to conclude that the collection of $f^*_Y$ was induced by some $f$, but in your setup you already assumed this. It's hard to gauge from your question how much you know about enriched categories, so let me recommend Kelly's book. He uses $\mathcal{V}$ where you use $M$, and then $\mathcal{V}$-naturality in general is (1.7), and in the context of your question is (1.21). That's where he shows that $\mathcal{V}$-naturality can be checked variable-by-variable, and what I stated about $f_Y^*$ inducing $f$. I would caution against using the name isomorphism for your $f:1\to C(X,X')$, because this is a morphism in $M$, so there's already a notion of isomorphism. If you call this an isomorphism, you should prove it really is one in $M$. Perhaps you could instead say "$f$ induces an isomorphism" if this condition is met, and then you check that the corresponding morphism $X \to X'$ in $C$ is an isomorphism. To get that corresponding morphism, look at (1.33) in Kelly's book. He painstakingly defines a 2-functor $(-)_0: \mathcal{V}-CAT \to CAT$, which would take $C$ to an actual category $C_0$. Under this 2-functor the $\mathcal{V}$-object $C(X,X')$ goes to a set $C_0(X,X')$, and $f$ picks out an element of this set. Using this approach, it's easy to check that the composites in your setup are the identities on $X$ and $X'$, in $C_0$, and hence are isomorphic to the identities on $X$ and $X'$ in $C$. Again, I refer you to Kelly's book.<|endoftext|> TITLE: Model category structure on spectra QUESTION [7 upvotes]: I have a concrete question for the algebraic category of spectra, but if there is an answer for its topological analogue I would be interested in it. Let $S$ be a finite dimensional Noetherian scheme and $\mathbf{Spt}(S)$ the category of spectra over $S$. After inverting $\mathbb{A}^1$-stable equivalences we obtain Voevodsky's stable homotopy category $\mathbf{SH}(S)$. My question is: Is there a model structure on $\mathbf{Spt}(S)$, having $\mathbf{SH}(S)$ as homotopy category, such that every object is fibrant? If so, could you provide a reference? For example, does the obvious candidate, given by the class of $\mathbb{A}^1$-stable equivalences as weak equivalences, surjective morphisms as fibrations, and cofibrations defined via the left lifting property, define a model structure on $\mathbf{Spt}(S)$? REPLY [4 votes]: Is there a model structure on Spt(S), having SH(S) as homotopy category, such that every object is fibrant? If so, could you provide a reference? No, if the given model category of spectra Spt(S) (for which there are many different, but Quillen equivalent, definitions) is not right proper, because if all objects are fibrant, the model category is necessarily right proper. Yes, if we are allowed to pick a specific Spt(S). According to Corollary 2.21 in a paper by Nikolaus (Algebraic models for higher categories), every cofibrantly generated model category C where all trivial cofibrations are monic is Quillen equivalent to a model category where all objects are fibrant. Thus, one can take the projective or injective model structure on simplicial presheaves, localize it, pass to spectra (which gives one possible choice for Spt(S)), and then apply the cited result to get a desired model category.<|endoftext|> TITLE: Why only some del Pezzo are toric? QUESTION [7 upvotes]: Let us define smooth del Pezzo surfaces $dP_r$ as the blowup of $r$ generic points in $\mathbb{CP}_2$. One can show that if we request $dP_r$ to be Fano, then $r=0,...,8$. In theoretical physics literature is common to define also $dP_9$, which is the blowup of $9$ points in $\mathbb{CP}_2$ and refer to this as well as a del Pezzo surface, while as far as I understand mathematicians would call this a generic rational elliptically fibered surface. Now, the question is why only $dP_r$ for $r=0,1,2,3$ are toric varieties? This statement is given for example in the appendix A.3 of http://inspirehep.net/record/1707635 For example, consider the fan of $\mathbb{CP}_2$. In order to perform the $i$-th blow-up, at the level of the fan, we add a new 1-dimensional cone which is representing the exceptional divisor $E_i$, associated to the new $\mathbb{CP}_1$. Which is the obstruction from keeping adding cones also for $r>3$? Why does the toric description fail? REPLY [3 votes]: You have no problem for adding cones from a toric surface $\mathbb{P}^2$ or any surface obtained by blowing-up points on it. However, there are only three toric points on $\mathbb{P}^2$, so once you have blown-up these $3$ points, the toric points that you see on your surface lie on $(-1)$-curves, and thus do not correspond anymore to points of $\mathbb{P}^2$. You can still blow-up them, and it corresponds to what you expect on the fan point of view, but you no longer get a del Pezzo surface, as there are $(-2)$-curves. In general, if you blow-up more than $8$ points of $\mathbb{P}^2$ in (very) general points, you get a non-toric surface with only curve of self-intersection $\ge -1$ and such that negative curves should be $(-1)$-curves (it is in fact a conjecture, see https://arxiv.org/pdf/math/0512631.pdf ), but this is no longer del Pezzo as the anti-canonical divisor is no longer ample, only nef and not big.<|endoftext|> TITLE: Explicit solution of a Hamiltonian system QUESTION [6 upvotes]: It is well-known that the following Hamiltonian system \begin{eqnarray} \left\{\begin{array}{rcl} \frac{dx}{dt}&=&y,\\ \frac{dy}{dt}&=&x(-1+x^2), \end{array}\right. \end{eqnarray} with $$ H(x,y)=\frac{x^2}2+\frac{y^2}{2}-\frac{x^4}{4}$$ has the solution $$ x=\tanh\left(\frac{t}{\sqrt2}\right),\qquad y=\frac1{\sqrt2}\text{sech}^2\left(\frac{t}{\sqrt2}\right) $$ such that $H(x,y)=\frac14$. Now consider the following Hamiltonian system \begin{eqnarray} \left\{\begin{array}{rcl} \frac{dx}{dt}&=&y,\\ \frac{dy}{dt}&=&x(-1+x^4), \end{array}\right.\tag{1} \end{eqnarray} with $$ H(x,y)=\frac{x^2}2+\frac{y^2}{2}-\frac{x^6}{6}$$ and I want to find the explicit solution such that $H(x,y)=\frac{1}{3}$. I tried many ways to get the solution but failed. Any help is appreciated. REPLY [12 votes]: Edited. Surprisingly, there is an elementary solution, if I made no mistake in the following computation. Your equation is equivalent to $$\left(\frac{dx}{dt}\right)^2=\frac{1}{3}(x^6-3x^2+2),$$ (I just plugged $y=dx/dt$ to your Hamiltonian, and used its value $H=1/3$.) This equation is separable, $$\frac{t}{\sqrt{3}}=\int\frac{dx}{\sqrt{x^6-3x^2+2}}=:I$$ and requires inversion of the integral. To reduce it to a standard integral, change $x^2=1/(u+1), \; dx=-(1/2)(u+1)^{-3/2}du,$ and the integral becomes $$I=-\frac{1}{2}\int\frac{du}{u\sqrt{2u+3}}.$$ The inverse function to this integral is elementary. Integrating and returning to the original variables, I obtained the general solution $$x(t)=\frac{2\sinh(t+c)}{\sqrt{6+4\sinh^2(t+c)}}.$$ Can you check this computation? Of course, your value $1/3$ for the Hamiltonian is crucial here. With some other value you obtain a much more complicated integral which cannot be expressed in elementary functions. Where did this $1/3$ come from?<|endoftext|> TITLE: Example of a Mathematician/Physicist whose Other Publications during their PhD eclipsed their PhD Thesis QUESTION [30 upvotes]: I am wondering if there is some example of a mathematician or physicist who published other papers at the same time as their PhD work and independently of it which actually eclipsed the content of the PhD thesis. The only semi-example I can think of immediately is Einstein, whose other publications in 1905 (especially on special relativity and the photoelectric effect) eclipsed his PhD thesis which was published in the same year. Although it contained important insights, it was somewhat forgotten to the point where he felt that he had to remind people about it. Although this is a soft question, I didn't ask in Academia as I didn't want examples outside of mathematics and physics. REPLY [5 votes]: Well, Art Garfunkel (Simon and Garfunkel) certainly has his other "publications" being more famous than his (never-completed) PhD in math. Then there's Brian May from Queen, with a PhD in astrophysics, who is more famous for other publications.<|endoftext|> TITLE: Does this "mixable" property have a standard name in constructive mathematics? QUESTION [12 upvotes]: While thinking about constructive mathematics, I stumbled on the following notion, and I would like to know if it has a standard name, a simpler equivalent, or has appeared in the literature: Say that a set $X$ is mixable (temporary term, for lack of a better one) when it satisfies the following property: $$\forall S\subseteq X.(\forall u\in S.\forall v\in S.(u=v) \Rightarrow \exists x\in X.\forall y\in S.(y=x))$$ In other words, every subset $S$ of $X$ all of whose elements are equal is a subset of the singleton $\{x\}$ for a certain $x\in X$. Classically, it is trivial that every (edit:) nonempty set is mixable. Example: Every power set object $\mathscr{P}(Z)$ is mixable. (Indeed, given $S \subseteq \mathscr{P}(Z)$ as assumed, consider $x := \bigcup_{t\in S} t$, that is, $\{z\in Z : \exists t\in S.(z\in t)\}$. Given $y\in S$, since we have $S = \{y\}$, we have $x = \bigcup\{y\} = y$, as announced.) For the same reason, in CZF, we have: $\forall S.(\forall u\in S.\forall v\in S.(u=v) \Rightarrow \exists x.\forall y\in S.(y=x))$. Counterexample: If $\{0,1\}$ is mixable then the Weak Law of Excluded Middle holds. (Indeed, given $p$ a truth value, consider $S := \{0 : p\} \cup \{1 : \neg p\}$. Clearly all elements of $S$ are equal. But if there is $x\in\{0,1\}$ such that $S \subseteq \{x\}$ then either $x=0$ or $x=1$, and in the first case $\neg\neg p$ while in the second $\neg p$ holds, so on the whole, $\neg p \lor \neg\neg p$.) In sheaf semantics, saying that a sheaf $X$ on a topological space $E$ is mixable means that (†) for every section $s$ of $X$ on an open set $U$ of $E$, there is a covering of $E$ by open sets $V_i$ such that $X$ has a section $x_i$ on each $V_i$ which coincides with $s$ on $V_i \cap U$. This is the case in particular if $X$ is flabby, but I don't know if this property has a name in the sheaf context. Edit (2019-03-29): As Mike Shulman points out in his answer, the above property of sheaves is actually equivalent to being flabby, so “flabby” is a good name for the property. Since the proof (by a “typical argument”) is not actually written out on nLab, here it is for the completeness of MathOverflow: assume the sheaf $X$ on $E$ has the property (†) described in the previous paragraph, and let $s$ be a section on $U$. Consider sections of $X$ extending $s$, partially ordered by extension: since chains of sections give rise to a single section on the union by the seaf property, Zorn's lemma implies that there is a maximal extension $s' \in X(U')$ of $s$; now apply (†) to $s'$: for any $i$, the sections $s'$ and $x_i$ glue to give a section of $X$ on $U' \cup V_i$, extending $s$, so by maximality of $s'$ we have $V_i\subseteq U'$, and since this holds for every $i$, in fact $U' = E$. Thus, $X$ is flabby. Question: Does the above “mixable” property have a standard name or a nicer equivalent, and where might I learn more about it? REPLY [8 votes]: According to this proposition, the property in question is (by "a typical argument with Zorn's lemma") equivalent to flabbiness for sheaves on topological spaces. Thus, the term "flabby" is sometimes used more generally inside constructive mathematics for sets with this property.<|endoftext|> TITLE: Existence of a projective small resolution QUESTION [7 upvotes]: It is known that three-dimensional ordinary double points, that is singular points which complete locally have the equation $xy - zw = 0$ are resolved by a single blow up, with exceptional divisor being a smooth quadric surface $\mathbf{P^1 \times P^1}$. It is also known that sometimes one of the two rulings of the exceptional divisor can be contracted to form a so-called small resolution, that is the one where only a curve is blown down. So let $X$ be a threefold with isolated ordinary double points $P_1, \dots, P_r$. Let $\pi: Y \to X$ be the blow up of the singular points. Let $E_i$ be the exceptional divisors. I would like to form a small resolution by contracting one of the two rulings on each of the exceptional divisors. My question is what is the numerical condition for existence of a projective small resolution in terms of divisors on $Y$? I see that a necessary condition is existence of a divisor $D$ on $Y$ which restricts "nondiagonally" (that is has bidegree $(a,b)$ for $a \ne b$) on each exceptional divisor. Indeed, otherwise, we have no chance to contract one set of rulings of $E_i$ without contracting the other. Is this also a sufficient condition? Thank you! REPLY [4 votes]: The necessary condition that you write is also sufficient. By construction, $Y$ is projective. Thus, if there exists any "nondiagonal" divisor class, then (after twisting by a sufficient powers of an ample invertible sheaf) there exists a very ample invertible sheaf $\mathcal{L}$ on $Y$ whose restriction to each exceptional divisor $E_i$ is ample and "nondiagonal". Denote by $N_i$ the normal bundle of $E_i$ in $Y$. This is an antiample invertible sheaf that is diagonal. Thus, there exist integers $d_i\geq 0$ such that the invertible sheaf $$\mathcal{M}:=\mathcal{L}\left( \sum_i d_i \underline{E}_i \right),$$ restricts on every $E_i$ as a globally generated invertible sheaf that is not big, i.e., it contracts one of the two rulings. By Michael Artin's theorems in "Algebraization of Formal Moduli, II" (ed.I will add a reference soon) there exists a proper, birational morphism of normal, proper algebraic spaces, $$\nu:Y\to Z,$$ and a nef invertible sheaf $\mathcal{N}$ on $Z$ such that $\nu^*\mathcal{N}$ equals $\mathcal{M}$ and such that $\nu$ contracts a curve in $E_i$ if and only if it has zero intersection number with $\mathcal{M}$. To prove that $\mathcal{N}$ is ample, it suffices to use the version of the Nakai-Moishezon Criterion for algebraic spaces proved by János Kollár, Theorem 3.11, p. 248 of the following. MR1064874 (92e:14008) Kollár, János Projectivity of complete moduli. J. Differential Geom. 32 (1990), no. 1, 235–268. https://projecteuclid.org/download/pdf_1/euclid.jdg/1214445046 Nakai-Moishezon hypothesis for curves. To check the hypotheses of the Nakai-Moishezon criterion, begin with irreducible curves. Every irreducible curve in $Z$ is the finite image of an irreducible curve in $Y$. If that curve is not contained in any divisor $E_i$, then its intersection with each of these is nonnegative. Since also the intersection of the curve with $c_1(\mathcal{L})$ is positive, the intersection of the irreducible curve with $c_1(\mathcal{M})$ is also positive. Also, the restriction of $\mathcal{M}$ to each $E_i$ is globally generated and has zero intersection number only on those curves contracted by $\nu$. Thus, for every irreducible curve in $E_i$ that maps finitely to its image in $Z$, the intersection number with $c_1(\mathcal{M})$ is positive. Nakai-Moishezon hypothesis for surfaces. Next consider surfaces. As above, it suffices to check for every noncontracted irreducible surface in $Y$ that the intersection number with $c_1(\mathcal{M})\smile c_1(\mathcal{M})$ is positive. For an irreducible surface that is contained in none of the divisors $E_i$, the intersection of every $E_i$ with the surface is an effective curve. Also the intersection of the surface with a general member of the linear system of the very ample invertible sheaf $\mathcal{L}$ is an effective curve, at least one component of which is contained in no $E_i$. Thus, the intersection class of the surface with $c_1(\mathcal{M})$ is rationally equivalent to a nontrivial effective curve, at least one component of which is contained in no $E_i$. Now apply the previous paragraph to deduce that intersection number of this curve with $c_1(\mathcal{M})$ is positive. Since each surface $E_i$ is contracted by $\nu$, we do not need to check the criterion for these surfaces. Nakai-Moishezon hypothesis for threefolds. Finally, $c_1(\mathcal{M})$ restricts to a divisor class on each $E_i$ that is effective. Thus, since the restriction of the effective divisor class $c_1(\mathcal{M})$ to a general member of the complete linear system of $\mathcal{L}$ is also effective, the intersection class $c_1(\mathcal{M})\smile c_1(\mathcal{M})$ is rationally equivalent to an effective curve, not all of whose components are contained in exceptional divisors $E_i$. Thus, by the first paragraph once more, the intersection number of this curve with $c_1(\mathcal{M})$ is positive, i.e., $c_1(\mathcal{M})\smile c_1(\mathcal{M})\smile c_1(\mathcal{M})$ has positive degree. Altogether, this confirms the Nakai-Moishezon criterion for ampleness of $\mathcal{N}$ on $Z$.<|endoftext|> TITLE: Guessing each other's coins QUESTION [59 upvotes]: I recently thought about the following game (has it been considered before?). Alice and Bob collaborate. Alice observes a sequence of independent unbiased random bits $(A_n)$, and then chooses an integer $a$. Similarly, Bob observes a sequence of independent unbiased random bits $(B_n)$, independent from $(A_n)$, and then chooses an integer $b$. Alice and Bob are not allowed to communicate. They win the game if $A_b=B_a=1$. What is the optimal winning probability $p_{opt}$? A strategy for each player is a (Borel) function $f : \{0,1\}^{\mathbf{N}} \to \mathbf{N}$, and we want to maximize the winning probability over pairs of strategies $(f_A,f_B)$. Constant strategies win with probability $1/4$, and it is perhaps counterintuitive that you can do better. Choosing $f$ to be the index of the first $1$ wins with probability $1/3$. This is not optimal though, by running a little program trying randomly modified strategies on a finite window I could find that $p_{opt} \geq 358/1023 \approx 0.3499$, with some pair (with $f_A=f_B$) lacking any apparent pattern. But a more interesting question is: can you prove any upper bound on $p_{opt}$, besides the trivial $p_{opt} \leq 1/2$? Edit. As has been pointed out by Édouard Maurel-Segala, the problem has been studied in this paper, where it is proved (as is also proved in the present thread) that $0.35 \leq p_{opt} \leq 0.375$, stated without proof that $p_{opt} \leq \frac{81}{224} \approx 0.3616$, and conjectured that $p_{opt} = 0.35$. Edit (clarifying what I said in the comments). You can ask the same question for the finite version of the game, with strings $(A_1,\dots,A_N)$ and $(B_1,\dots,B_N)$, giving optimal winning probability $p_N$. It can be checked than $(p_N)$ is non-decreasing with limit $p_{opt}$. Moreover the inequality $p_{opt} \geq \frac{4^N}{4^N-1} p_N$ holds, because in the infinite game, when a player sees a string of $N$ $0$s, he may discard them and apply the strategy to the next $N$ bits. We have $p_1=1/4$, $p_2=5/16$, $p_3=22/64 > p_2$. It seems that $p_4=89/256$ (therefore $p_4 > p_3$, but $\frac{256}{255} p_4 < \frac{64}{63} p_3$, so $4$-bit strategies are worse than $3$-bit for the infinite game), and I know that $p_5 \geq 358/1024$ and $p_6 \geq 1433/4096$. For $p_3$ and $p_4$ one strategy achieving the value is: when the observed string contains a single block of $1$s, Alice (resp. Bob) picks the index of the $0$ immediately after (resp. before) that block; what they do in the remaining cases is irrelevant. REPLY [22 votes]: I discussed this with Arvind Singh a while ago and I think we can show the non trivial inequality $p_{opt}\leqslant\frac{3}{8}$ with simple arguments. The proof relies on the symmetry of the problem and the intuition is that one can not find a strategy wich is good simultaneously for a configuration and its inverse. It will be simpler to work with the sets of indices such that the coin is on $1$: $$A=\{1\leqslant i\leqslant n : A_i=1\},\quad B=\{1\leqslant i\leqslant n : B_i=1\}.$$ We want to bound $$G=\mathbb P (f_a(B)\in A, f_b(A) \in B).$$ Introducing the function $$g(A,B)=\frac{1}{4}\left(1_{f_a(B)\in A, f_b(A) \in B}+1_{f_a(B^c)\in A^c, f_b(A^c) \in B^c}+1_{f_a(B)\in A^c, f_b(A^c) \in B}+1_{f_a(B^c)\in A, f_b(A) \in B^c}\right),$$ we get by symmetry (since for example $A^c,B$ has the same law than $A,B$): $$G=\mathbb E [g(A,B)].$$ But there are some incompatibilities in $g$: the first term and the third term can not be both equal to $1$ since one contains $f_a(B)\in A$ and the other $f_a(B)\in A^c$. The same thing applies for the second and the fourth. Thus $g(A,B)\in\{0;\frac{1}{4};\frac{1}{2}\}$ almost surely. On the event $E_1=\{f_b(A)\in B, f_b(A^c)\in B\}$, only the first and the third term can be non vanishing and since they are incompatible $g(A,B)$ is at most $1/4$ (in fact it is equal to $1/4$). Besides, by first conditionning on $A$ we see that $E_1$ is of probability at least $1/4$ (the probability that $B$ contains (one or) two elements). The same applies to $E_2=\{f_b(A)\in B^c, f_b(A^c)\in B^c\}$. If we consider the event $$E=\{f_b(A)\in B, f_b(A^c)\in B\}\cup\{f_b(A)\in B^c, f_b(A^c)\in B^c\}.$$ we have built an event such that $g1_E\leqslant \frac{1}{4}$ and $\mathbb P(E)\geqslant \frac{1}{2}$ (the union is disjoint). Thus since $g\leqslant \frac{1}{2}$, $$G=\mathbb E[g(A,B)]\leqslant \mathbb E[g1_E]+\mathbb E[g1_{E^c}]\leqslant \frac{1}{4}\mathbb P(E)+\frac{1}{2}(1-\mathbb P(E))\leqslant \frac{1}{4}\frac{1}{2}+\frac{1}{2}(1-\frac{1}{2})=\frac{3}{8}.$$<|endoftext|> TITLE: What does convergence in distribution "in the Gromov–Hausdorff" sense mean? QUESTION [9 upvotes]: I am trying to understand this survey article by Le Gall on Brownian geometry, especially the statement of Theorem 1. The basic statement of the theorem is $$(m_n,d_n) \to (m_{\infty}, d_{\infty})$$ "in the Gromov–Hausdorff sense" as $n \to \infty$, where the convergence is in distribution. Here $(m_n,d_n)$ and $(m_{\infty},d_{\infty})$ are both random compact metric spaces. So how do we interpret this? We might hope for a statement along the lines of the following. For every compact metric space $(X,d)$ and $R > 0$, we have $$(*) \, \, \, \mathbb{P} \left[ d_{GH}[ (m_n,d_n), (X,d) ] < R \right] \to \mathbb{P} \left[ d_{GH}[ (m_{\infty},d_{\infty}), (X,d) ] < R \right]$$ as $n \to \infty$. But even when we talk about convergence in distribution for real random variables (instead of compact-metric-space-valued random variables), we have to be careful to restrict our attention to points where the cumulative distribution function is continuous. So I wonder if (*) is too strong? REPLY [11 votes]: Following the notation of the paper, let $\mathbb{K}$ be the metric space of all compact metric spaces, equipped with the Gromov-Hausdorff metric $\mathrm{d_{GH}}$. Then we can express convergence in distribution in the usual way: for every bounded continuous $F : \mathbb{K} \to \mathbb{R}$, we have $\mathbb{E}[F((m_n, d_n))] \to \mathbb{E}[F((m_\infty, d_\infty))]$. The portmanteau theorem gives you several other equivalent statements. In other words, this is just the usual notion of convergence in distribution for random variables taking their values in a metric space $S$, where that metric space happens to be $S = (\mathbb{K}, \mathrm{d_{GH}})$, the metric space of all compact metric spaces. In particular, if $(X,d)$ is a fixed compact metric space, the function $\mathrm{d_{GH}}(\cdot, (X,d)) : \mathbb{K} \to \mathbb{R}$ is a continuous function. So if we let $Y_n = \mathrm{d_{GH}}((m_n, d_n), (X, d))$, then the scalar-valued random variables $Y_n$ converge in distribution to $Y$. So your formula (*) holds, but as you say, only for values of $R$ at which the function $R \mapsto \mathbb{P}[\mathrm{d_{GH}}((m_\infty, d_\infty), (X,d)) < R]$ is continuous.<|endoftext|> TITLE: Is there an analogue of projective spaces for proper schemes? QUESTION [13 upvotes]: Does there exist a countable set of connected proper smooth $\mathbb{C}$-schemes such that any connected proper smooth $\mathbb{C}$-scheme admits a $\mathbb{C}$-immersion into one of them? REPLY [12 votes]: I am just posting my comment as one answer. So long as you are only asking about schemes (rather than complex analytic spaces), you can avoid the hard analysis from the previous MathOverflow answer. The argument below sketches this. The main additional detail beyond Hilbert scheme techniques is a strong variant of Chow's Lemma. Let $k$ be an algebraically closed field (soon we will assume it has characteristic $0$). Chow's Lemma. Let $X$ be a separated, finitely presented scheme over a field $k$, and let $U\subset X$ be a dense open subscheme that is a quasi-projective $k$-scheme. There exists a strongly projective morphism $\nu:\widetilde{X}\to X$ such that $\widetilde{X}$ is a quasi-projective $k$-scheme and such that the restriction of $\nu$ over $U$ is an isomorphism. There may be an earlier source, but the source that I know is the following article. MR0308104 (46 #7219) Raynaud, Michel; Gruson, Laurent Critères de platitude et de projectivité. Techniques de "platification'' d'un module. Invent. Math. 13 (1971), 1–89. Finally, the very last step of the argument requires Nagata compactification. Nagata compactification Every separated, finite type $k$-scheme is isomorphic to a dense open subscheme of a proper $k$-scheme. Definition. For every separated morphism that restricts as an isomorphism over a dense open in the target, the isomorphism locus is the maximal open subscheme of the target over which the morphism is an isomorphism. For a separated, flat, finite type morphism of schemes, $$\rho:\mathcal{X}\to B,$$ a Chow covering is an ordered $n$-tuple (for some integer $n\geq 0$) of pairs of $B$-morphisms, $$(\nu_\ell:\widetilde{\mathcal{X}}_\ell \to \mathcal{X}, e_\ell:\widetilde{\mathcal{X}}_\ell \to \mathbb{P}^m_B),$$ where each $e_\ell$ is a locally closed immersion of flat $B$-schemes and where the morphisms $\nu_\ell$ are strongly projective morphisms whose isomorphism loci give an open covering of $\mathcal{X}$. Corollary. Every separated, finite type $k$-scheme has a Chow covering. Proof. Since $X$ is quasi-compact, there exists a finite covering of $X$ by open affine subschemes. By Chow's Lemma, for each open affine, there exists a strongly projective $k$-morphism from a quasi-projective $k$-scheme to $X$ whose isomorphism locus contains that open affine. QED Hypothesis. The field $k$ has characteristic $0$. Definition. A smooth Chow covering is a Chow covering such that every $B$-scheme $\widetilde{\mathcal{X}}_\ell$ is smooth over $B$. Corollary. Every smooth, separated, quasi-compact $k$-scheme has a smooth Chow covering. Proof. This follows from the previous corollary and Hironaka's Theorem. QED Notation. For every smooth Chow covering over $B$, denote by $\widetilde{\mathcal{X}}$ the disjoint union of the $B$-schemes $\widetilde{\mathcal{X}}_\ell$. Denote by $\nu:\widetilde{\mathcal{X}}\to \mathcal{X}$ the unique $B$-morphism whose restriction to each component $\widetilde{\mathcal{X}}_\ell$ equals $\nu_\ell$. Denote by $\mathcal{Y}$ the disjoint union over all ordered pairs $(j,\ell)$ with $1\leq j,\ell\leq n$ of the closed subscheme of the product, $$\mathcal{Y}_{j,\ell}:= \widetilde{\mathcal{X}}_j\times_{\mathcal{X}} \widetilde{\mathcal{X}}_\ell \subseteq \widetilde{\mathcal{X}}_j\times_{B} \widetilde{\mathcal{X}}_\ell,$$ together with its two projections, $$\text{pr}_{1,(j,\ell)}:\mathcal{Y}_{j,\ell}\to \widetilde{\mathcal{X}}_j, \ \ \text{pr}_{2,(j,\ell)}:\mathcal{Y}_{j,\ell}\to \widetilde{\mathcal{X}}_\ell.$$ Denote the disjoint union of these morphisms by $$\text{pr}_1:\mathcal{Y} \to \widetilde{\mathcal{X}}, \ \ \text{pr}_2:\mathcal{Y}\to \widetilde{\mathcal{X}}.$$ For every $\ell=1,\dots,n$, denote the diagonal morphism by $$\delta_\ell:\widetilde{\mathcal{X}}_\ell \to \mathcal{Y}_{\ell,\ell}.$$ Denote the disjoint union of these morphisms by $$\delta:\widetilde{\mathcal{X}}\to \mathcal{Y}.$$ For every $(j,\ell)$, denote the involution of $B$-schemes that transposes the factors by $$\sigma_{j,\ell}:\mathcal{Y}_{j,\ell}\to \mathcal{Y}_{\ell,j}.$$ Denote the disjoint union of these involutions by the involution, $$\sigma:\mathcal{Y}\to \mathcal{Y}, \ \ \text{pr}_2\circ \sigma = \text{pr}_1, \ \ \text{pr}_1\circ \sigma = \text{pr}_2.$$ For every ordered triple $(j,\ell,r)$ with $1\leq j,\ell,r\leq n$, denote by $$c_{j,\ell,r}:\mathcal{Y}_{j,\ell}\times_{\widetilde{\mathcal{X}}_\ell} \mathcal{Y}_{\ell,r} \to \mathcal{Y}_{j,r},$$ the morphism induced by the first and final projections. Denote the disjoint union of these morphisms by $$c:\mathcal{Y}\times_{\text{pr}_2,\widetilde{\mathcal{X}},\text{pr}_1} \mathcal{Y} \to \mathcal{Y}.$$ Definition. For every $k$-scheme $B$, a Chow descent predatum over $B$ is a collection of projective $B$-schemes and morphisms of $B$-schemes, $$((\pi_\ell:\widetilde{\mathcal{X}}_\ell \to B, ((\text{pr}_{1,(j,\ell)},\text{pr}_{2,(j,\ell)}):\mathcal{Y}_{j,\ell}\hookrightarrow \widetilde{\mathcal{X}}_j\times_{B}\widetilde{\mathcal{X}}_\ell)_{j,\ell}, $$ $$(\delta_\ell:\widetilde{\mathcal{X}}_\ell\to Y_{\ell,\ell})_\ell, (\sigma_{j,\ell}:Y_{j,\ell}\to Y_{\ell,j})_{j,\ell}, (c_{j,\ell,r}:Y_{j,\ell}\times_{\text{pr}_2,\widetilde{X}_\ell,\text{pr}_1} Y_{\ell,r}\to Y_{j,r})_{j,\ell,r})$$ together with a collection of locally closed immersions of $B$-schemes, $$e_\ell:\widetilde{\mathcal{X}}_\ell\hookrightarrow \mathbb{P}^m_B.$$ The isomorphism locus is the maximal open subscheme $U_\ell$ of $\widetilde{X}_\ell$ on which the closed immersion $\delta_\ell$ is an open immersion. Constraint axioms for a Chow datum. As the $1$-truncation of the simplicial scheme induced by $\nu$, there are many constraints satisfied by the Chow descent predatum of a Chow covering. Taken together, these constraints define a Chow descent datum. Constraint 1. For every $(j,\ell)$, the $\text{pr}_1$-inverse image in $Y_{j,\ell}$ of $U_j$ is an open subscheme whose closed complement in $\mathcal{Y}_{j,\ell}$ equals its total inverse image under $\text{pr}_2$ of its closed image in $\widetilde{\mathcal{X}}_\ell$. Denote by $\widetilde{\mathcal{X}}_{\ell,j}$ the open complement in $\widetilde{\mathcal{X}}_\ell$ of this closed image. Constraint 2. For each $\ell=1,\dots,n$, the collection of open subschemes $$(\widetilde{\mathcal{X}}_{\ell,j})_{j=1,\dots,n},$$ form an open covering of $\widetilde{\mathcal{X}}_\ell$. Stated differently, the $n$-fold intersection of the closed complements is empty. Constraint 3. The isomorphism locus of each projection $$\text{pr}_{2,(j,\ell)}:\mathcal{Y}_{j,\ell} \to \widetilde{\mathcal{X}}_\ell,$$ contains $\widetilde{\mathcal{X}}_{\ell,j}$. Thus, the inverse image of $\widetilde{\mathcal{X}}_{\ell,j}$ under this isomorphism equals the graph of a unique $k$-morphism, $$\nu_{\ell,j}:\widetilde{\mathcal{X}}_{\ell,j} \to U_j.$$ Constraint 4. Denote by $U_{\ell,j}$ the intersection of $U_\ell$ and the $\nu_{\ell,j}$-inverse image of $U_j$ in $\widetilde{\mathcal{X}}_{\ell,j}$. For each triple $(j,\ell,r)$, the inverse image in $U_{\ell,j}$ under $\nu_{\ell,j}$ of $U_{j,r}$ equals $U_{\ell,j}\cap U_{\ell,r}$. Denote this open by $U_{\ell,j,r}$. Also, the inverse image in $\widetilde{\mathcal{X}}_{\ell,j}$ under $\nu_{\ell,j}$ of $U_j\cap \widetilde{\mathcal{X}}_{j,r}$ equals the inverse image in $\widetilde{\mathcal{X}}_{\ell,r}$ under $\nu_{\ell,r}$ of $U_r\cap \widetilde{\mathcal{X}}_{r,j}$. Denote this open by $\widetilde{\mathcal{X}}_{\ell,j,r}$. Constraint 5. On the open $U_{\ell,j,r}$, the composition $\nu_{j,r}\circ \nu_{\ell,j}$ equals $\nu_{\ell,r}$. Similarly, on the open $\widetilde{\mathcal{X}}_{\ell,j,r}$, the composition $\nu_{j,r}\circ \nu_{\ell,j}$ equals $\nu_{\ell,r}$. Effectivity Proposition. Every Chow descent datum is effective, i.e., it is isomorphic to the Chow descent datum of a Chow covering. Proof Altogether, these constraints are the descent constraints for a Zariski descent datum of $B$-schemes, $$((U_\ell)_\ell, (U_{\ell,j}\subset U_\ell)_{\ell,j}, (\nu_{\ell,j}:U_{\ell,j}\to U_j)_{\ell,j})$$ together with the descent constraints for each Zariski descent datum of $B$-morphisms, $$(\widetilde{X}_{\ell,j}\to U_j)_j.$$ By Zariski gluing for schemes, this descent datum for a $B$-scheme is effective. By Zariski gluing for morphisms, each of these descent data for a $B$-morphism is effective. For each $U_\ell$, each $\widetilde{\mathcal{X}}_{\ell,j}\to U_\ell$ is proper and surjective. Thus, the induced morphism $\nu_j$ is proper and surjective. Since also $\widetilde{\mathcal{X}}_j$ is proper over $B$, the $B$-scheme $\mathcal{X}$ is proper. Since $\mathcal{X}$ is covered by open $U_\ell$ that are smooth $B$-schemes, also $\mathcal{X}$ is a smooth $B$-scheme. QED Constraint 6. The morphisms $\sigma$ and $c$ equal the natural morphisms induced by this effective descent datum. Nota bene. Since we have already incorporated the glueing constraints above, Constraint 6, and even the data of the morphisms $\sigma$ and $c$, are extraneous. However, from the point of view of a "groupoid scheme" associated to the Chow covering, it seems best to include this data as part of the definition. Constructibility Proposition. For every separated, finite type $k$-scheme $B$, for every Chow descent predatum over $B$, there exists a finite collection of locally closed subschemes of $B$ (with the induced reduced structure) whose set of geometric points are precisely those geometric points of $B$ where the Chow descent predatum is a Chow descent datum. Proof sketch. Each constraint condition involves equalities or inclusions of closed subsets of $\widetilde{X}_\ell$, or, equivalently, equalities or inclusions of open subsets, it involves a morphism being an isomorphism, or it involves certain associated morphisms to a self-fiber product factoring through the diagonal. Pass to flattening stratifications of the base $B$ for the relevant closed subschemes with their reduced structure. By separatedness of the Hilbert scheme, the locus in $B$ where two $B$-flat closed subschemes of a projective $B$-scheme are equal is a closed subset of $B$. Similarly, using properness of the Hilbert scheme and quasi-compactness of $B$, the locus where one $B$-flat closed subscheme factors through a second $B$-flat closed subscheme is also a closed subset of $B$. The isomorphism locus is open. By taking set complements, inclusion or equality of open subsets is equivalent to inclusion or equality of the closed complements. This defines finitely many locally closed subsets of $B$ by the previous paragraph. Finally, since all $B$-schemes in this setup are separated, factorization of a morphism through the diagonal is equivalent to equality of the domain of the morphism with the closed inverse image of the diagonal. Up to taking closures of each of these in the appropriate ambient projective $B$-scheme, this again reduces to an inclusion of closed subschemes. Thus each of these conditions defines a finite union of locally closed subscheme of $B$. QED By the proposition, we can form a countable collection of families of Chow descent data such that every Chow descent datum occurs as a fiber of one of these families.. For every Chow descent predatum,$$(([\widetilde{X}_\ell])_{\ell},([Y_{j,\ell}])_{j,\ell},([\delta_\ell])_{\ell},([\sigma_{j,\ell}])_{j,\ell}, ([c_{j,\ell,r}])_{j,\ell,r}),$$ the first two parts of the datum give points in appropriate Hilbert schemes. The last three parts give points in appropriate Hom schemes. Altogether, there are countably many ordered pairs $(n,m)\in \mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$. For each such pair, there are countably many ordered $n$-tuples of Hilbert polynomials for the closed subschemes $\widetilde{X}_\ell$ of $\mathbb{P}^m_k$. For each such ordered $n$-tuple of Hilbert polynomials, for the associated $n$-fold product of Hilbert schemes parameterizing $(\widetilde{X}_\ell)_{\ell=1,\dots,n}$, there are countably many $n^2$-tuples of Hilbert polynomials for the closed subschemes $Y_{j,\ell}$. For each such tuple of Hilbert polynomials, there is a further relative Hilbert scheme parameterizing closed subschemes $Y_{j,\ell}$ in $\widetilde{X}_j\times_{\text{Spec}\ k}\widetilde{X}_\ell$. Then there are countably many components of each Hom scheme for the morphisms $\delta_\ell$, $\sigma_{j,\ell}$ and $c_{j,\ell,r}$. For each of these countably many quasi-projective $k$-schemes parameterizing a Chow descent predatum, by the Constructibility Proposition, there are finitely many locally closed subschemes (with reduced structures) in the base parameterizing those Chow descent predata that are Chow descent data. Over this countable disjoint union of quasi-projective $k$-schemes, we glue together the isomorphism loci to form a proper flat morphism $\mathcal{X}\to B$ and the strongly projective $B$-morphisms $\nu_\ell$. By the existence of Chow coverings, every proper $k$-scheme occurs as a fiber over a $k$-point of one of the schemes $B$ in the countable collection. Denote by $I$ the countable set of irreducible components for all bases of all of these countably many families. For each $i\in I$, the Effectivity Proposition gives a smooth Chow covering. Denote this by $$(\rho:\mathcal{X}_i \to B_i, ((\nu_i,e_i):\widetilde{\mathcal{X}}_{i,\ell} \hookrightarrow \mathcal{X}_i\times_{\text{Spec}\ k} \mathbb{P}^m_{\mathbb{C}})_{\ell}).$$ Finally, by applying Hironaka's Theorem to each irreducible, quasi-projective $k$-scheme $B_i$, we can assume that each $B_i$ is a smooth, connected quasi-projective $k$-scheme. The effect is that every $\mathcal{X}_i$ is itself a separated, quasi-compact, smooth $k$-scheme. Certainly $\mathcal{X}_i$ can be non-proper over $k$. However, we can apply Nagata compactification and Hironaka's Theorem (for a third time) to realize $\mathcal{X}_i$ as a dense open subscheme of a proper, smooth $k$-scheme $\overline{\mathcal{X}}_i$. Since every smooth proper $k$-scheme arises as a fiber of some $\pi_i$, every smooth proper $k$-scheme admits a closed immersion in one of the countably many smooth, proper $k$-schemes $\overline{\mathcal{X}}_i$.<|endoftext|> TITLE: Dynamics of Riemann zeta function QUESTION [8 upvotes]: Has the dynamics of the Riemann zeta function been studied? By dynamics I mean the limiting behavior of the sequence of iterates $s, \zeta (s), \zeta (\zeta (s)), \zeta (\zeta (\zeta (s)))\dots $ for various starting values of $s$ in the (extended) complex plane. In particular, for which $s \in \mathbb{C}$ do the iterates $\zeta^{(n)}(s)$ converge to 0? REPLY [3 votes]: There is also a fixed point near $s=1.83377$. The Mathematica graphic below shows the argument of $\zeta(s)-s$, for $-10\le\sigma\le 5$, $-30\le t\le 30$,interpreted as a color. We see the two real fixed points and the pole at $s=1$, as well as a number of complex conjugate pairs of fixed points.<|endoftext|> TITLE: Circle $x^2 + y^2 = n!$ doesn't hit any lattice points for any $n$ except for $0$, $1$, $2$ and $6$ or does it? QUESTION [20 upvotes]: I stumbled across the following problem in high school:$$ x^2 + y^2 = n! $$ I tested it within my laptop capabilities, watched a 3b1b video Pi in prime regularities, where he explains how to find the number of integer solutions based on prime factors. There doesn't seem to be any above $30!$. Maybe I'm wrong and there are infinitely many exceptions like $2$ and $6$, maybe the proof is too difficult for me to grasp or... I hope I'm just too blind to see the obvious. REPLY [55 votes]: For $n\geq 7$, Erdős proved in 1932 that there is a prime $n/2 TITLE: Is a finite group given by its character table if its Sylow subgroups are so? QUESTION [6 upvotes]: As pointed out by Mikko Korhonen in this answer, Özdem Çelik proved (in 1976 here) that a finite group whose Sylow subgroups are cyclic (called a Z-group) is determined by its character table. Now there are many results and conjectures relating character tables and Sylow subgroups (see this paper of Gabriel Navarro), the most famous being perhaps the McKay conjecture. This leads to wonder whether Çelik's theorem can be extended*. Question 1: Is a finite group determined by its character table only if its Sylow subgroups are so? Answer (Alex B.): No. Question 2: Is a finite group not in a Brauer pair only if its Sylow subgroups are so? (negative answer suspected by Alex B.) *Question 3: Is a finite group determined by its character table if its Sylow subgroups are so? (it is this question which wonders whether Çelik's theorem can be extended) Question 4: Is a finite group not in a Brauer pair if its Sylow subgroups are so? REPLY [6 votes]: The answer to the first question is negative. The group ${\rm SL}_2(\mathbb{F}_3)$ has a $2$-Sylow subgroup isomorphic to $Q_8$, which is not determined by its character table, but ${\rm SL}_2(\mathbb{F}_3)$ is the only group with its character table. I strongly suspect that the answer to the second question is also negative, and it should be not too hard to check this computationally, using the following strategy, but I have not attempted this. One way of seeing how the counterexample above worked is this: the group $Q_8$ has an outer automoprhism of order $3$ (cyclicly permuting $i$, $j$, and $k$), and ${\rm SL}_2(\mathbb{F}_3)$ is a semidirect product of $Q_8$ and $C_3$, with the action given by this automorphism. On the other hand, $D_8$ has no automorphisms of order $3$, so one cannot produce an analogous construction with this cousin of $Q_8$. That makes it "unlikely" (in no precise sense) that there is another group with the same character table as ${\rm SL}_2(\mathbb{F}_3)$. I have checked that the group that Lux and Pahlings call $G_{3378}$ in Theorem 2.6.2 has an automorphism $\alpha$ of order $3$, while the other group in the Brauer pair, $G_{3380}$, does not. So I would conjecture that the group $G_{3378}\rtimes\langle \alpha\rangle$ is determined by its character table. This is a group of order $768$, so magma has the complete list of all groups of this order, and the conjecture should not be too hard to check, but I have not done that.<|endoftext|> TITLE: Reference request: Grassmannian and Plucker coordinates in type B, C, D QUESTION [16 upvotes]: Grassmannian $Gr(k,n)$ is the set of $k$-dimensional subspace of an $n$-dimensional vector space. What are the Grassmannian in types B, C, D? What are the analog of Plucker coordinates and Plucker relations in these cases? Are there some references about this? Thank you very much. REPLY [2 votes]: For the orthogonal case (B and D), this was developed in great detail in Elie Cartan's "Theory of Spinors" (Dover Publications Inc, Mineola N.Y., 1981); see also: Claude Chevalley, "The Algebraic Theory of Spinors and Clifford Algebras" (Springer, 1996, Vol. 2 of collecteed works). Of special importance are the "maximal" isotropic Grassmannians, which are related, in Cartan's approach, to "pure spinors". The analog of the Plucker relations are the "Cartan relations", or "Pfaffian Plucker relations". Further details and developments may be found in: J. Harnad and S.Shnider, "Isotropic geometry and twistors in higher dimensions I. The generalized Klein correspondence and spinor flags in even dimensions," J. Math. Phys. 33, 3191-3208 (1992) and J. Harnad and S.Shnider, "Isotropic geometry and twistors in higher dimensions II. Odd dimensions, reality conditions and twistor superspaces", J. Math. Phys. 36 1945-1970 (1995). For the symplectic case (C), again, the "maximal isotropic" case is of special importance, or "Lagrangian Grassmannians".<|endoftext|> TITLE: Existence of an isotopy in Riemannian manifold QUESTION [5 upvotes]: Let $(M,g)$ be a Riemannian manifold, and $p,q\in M$ be two fixed points. We assume $p,q$ are close enough. Say, we assume $p$ and $q$ are in the same normal coordinate chart. It is clear that there is some diffeomorphism $F:M\to M$ so that $F(p)=q$ and $F$ equals the identity outside a small neighborhood of $p$ and $q$. For example, we take $F=\exp(X)$ for some vector field $X$ which flows from $p$ to $q$ and which is supported near $p$ and $q$. Question: Now, suppose $F'$ is another such diffeomorphism ($F'(p)=q$ and equal to the identity outside a small neighborhood of $p$ and $q$ too). Can we find an isotopy of diffeomorphisms $F_s,s\in [0,1]$ so that $F_0=F$, $F_1=F'$ and for each $s$ we have $F_s(p)=q$? Moreover, can we require $F_s=\exp(X_s)$ for a smooth family of vector fields $X_s,s\in[0,1]$? I guess the answer should be positive. For example, in the special case $p=q$ and $F'=\mathrm{id}$; write $F=\exp(X)$ and then $X_p=0$. Thus, $F_s=\exp(sX)$ suffices. My attempt in general is that if $F'=\exp(X')$ then we may first take $G_s=\exp((1-s)X+sX')$. But we cannot guarantee $G_s(p)=q$. Maybe we can do some modification. REPLY [8 votes]: For the first question, which concerns just the smooth category without reference to metrics, the answer depends on the dimension of $M$. Before explaining this a small clarification is needed. By "a small neighborhood of $p$ and $q$" you probably mean a ball containing $p$ and $q$, otherwise one could take the neighborhood to consist of disjoint balls about $p$ and $q$, and then there would be no chance of finding a diffeomorphism $F$ with $F(p)=q$ and with $F$ the identity outside the neighborhood. Let $Diff(D^n)$ be the group of diffeomorphisms $F:D^n\to D^n$ which are the identity in a small neighborhood $N$ of the boundary of $D^n$, with the $C^\infty$ topology on $Diff(D^n)$. The set $\pi_0Diff(D^n)$ of path components of $Diff(D^n)$ is the same as the set of isotopy classes of diffeomorphisms in $Diff(D^n)$ with the understanding that isotopies $F_s$ restrict to the identity on $N$ for all $s$. You are interested in the subspace $Diff(D^n;p,q)$ consisting of diffeomorphisms taking $p$ to $q$. It is not hard to see that every $F\in Diff(D^n)$ can be isotoped to take $p$ to $q$, and also that any isotopy $F_s$ between diffeomorphisms $F_0,F_1$ taking $p$ to $q$ can be deformed, fixing $F_0$ and $F_1$, to an isotopy with $F_s(p)=q$ for all $s$. In other words the natural map $\pi_0Diff(D^n;p,q)\to \pi_0Diff(D^n)$ is a bijection. (In fact this holds for all higher homotopy groups as well.) If $\pi_0Diff(D^n)=0$, i.e., there is a single isotopy class, then the the answer to the first question is Yes. It is known that $\pi_0Diff(D^n)=0$ for $n=1$ (elementary), $n=2$ (Smale), and $n=3$ (Cerf). For $n=4$ it is still unknown whether $\pi_0Diff(D^n)$ is $0$ or not. For $n\geq 5$ it is known that $\pi_0Diff(D^n)$ is isomorphic to the group of exotic spheres of dimension $n+1$ (via the h-cobordism theorem and Cerf's pseudoisotopy theorem). This group is known to be $0$ for $n=5,11,55,60$ and it is known to be nonzero for all other $n\leq 125$ and all other even $n$, using hard calculations in the stable homotopy groups of spheres. An embedding $D^n \subset M$ induces maps $\pi_0Diff(D^n)\to\pi_0Diff(M)$ and $\pi_0Diff(D^n;p,q)\to\pi_0Diff(M;p,q)$ by extending diffeomorphsms via the identity outside $D^n$. In these terms the question becomes whether the image of the map $\pi_0Diff(D^n;p,q)\to\pi_0Diff(M;p,q)$ is $0$. If $M$ is simply connected then it is not hard to show that the map $\pi_0Diff(M;p,q)\to \pi_0Diff(M)$ is a bijection, so in these cases the question boils down to whether the image of the map $\pi_0Diff(D^n)\to\pi_0Diff(M)$ is $0$. The answer is certainly Yes if $\pi_0Diff(D^n)=0$, so the answer is Yes for $M$ simply connected of dimension $n=1,2,3,5,11,55,60$. In the special case that $M$ is the sphere $S^n$ there is a classical elementary argument showing that the map $\pi_0Diff(D^n)\to\pi_0Diff(S^n)$ is injective (also on all higher homotopy groups), so this provides a No answer for $M=S^n$ for all other $n\leq 125$ and all even $n\geq 126$. For other manifolds $M$ the question of injectivity of $\pi_0Diff(D^n)\to\pi_0Diff(M)$ was studied in the heyday of differential topology but I can't recall any specific results or references at the moment.<|endoftext|> TITLE: Is there always a complete, orthogonal set of unitary matrices? QUESTION [6 upvotes]: The set of size-$n$ unitary matrices span $\Bbb C^{n \times n}$ (this can be proven nicely using polar decomposition). If we select a maximal linear subset of unitary matrices, then we have a basis of $\Bbb C^{n \times n}$ consisting of $n^2$ unitary matrices. My question is whether we can find a basis that satisfies the additional constraint of orthogonality. That is: Does there exist a basis $\mathcal B$ of $\Bbb C^{n \times n}$ such that every $P \in \mathcal B$ is unitary (that is, $P^*P = I$) and for all distinct $P,Q \in \mathcal B$, we have $\langle P,Q \rangle = 0$? Here, $\langle \cdot , \cdot \rangle$ refers to the Frobenius (AKA Hilbert-Schmidt) inner product, namely $\langle P, Q \rangle = \operatorname{trace}(PQ^*)$. When $n = 2$, the Pauli matrices provide a convenient solution. That is, we can take $$ \mathcal B = \{I,\sigma_1,\sigma_2,\sigma_3\} \subset \Bbb C^{2 \times 2}. $$ We can use this to produce a solution whenever $n = 2^k$. In particular, if we define $\sigma_0 = I$ for convenience, we can take $$ \mathcal B = \{\sigma_{m_1} \otimes \cdots \otimes \sigma_{m_k} : 0 \leq m_j \leq 3\} \subset \Bbb C^{2^k \times 2^k}. $$ Could we come up with a basis for any other $n$? Could we do so for every $n$? Some observations so far: Without loss of generality, we can assume that $\mathcal B$ contains the $n \times n$ identity matrix $I$. If $I$ is an element of the basis, it follows that the remaining matrices form a basis for the subspace of all trace-$0$ matrices. A commuting set of matrices spans at most an $n$-dimensional subset, so there must be elements of $\mathcal B$ that fail to commute REPLY [10 votes]: Yes, consider the group of $n^2$ matrices generated by the shift $e_i \mapsto e_{i+1}$ and the diagonal matrix with entries $(1,\omega,\omega^2,\cdots,\omega^{n-1})$ where $\omega$ is a primitive $n$th root of unit.<|endoftext|> TITLE: Maximum dimension of space of matrices with a real eigenvalue QUESTION [15 upvotes]: Let $M_n(\mathbb{R})$ denote the space of all $n\times n$ real matrices. What is the maximum dimension $f(n)$ of a subspace $V$ of $M_n(\mathbb{R})$ such that every matrix in $V$ has at least one real eigenvalue? It is easy to see that if $n$ is odd then $f(n)=n^2$, while if $n$ is even then $n^2-n+1\leq f(n)\leq n^2-1$. There is also the ``complementary'' problem: what is the maximum dimension $g(n)$ of a subspace $W$ of $M_n(\mathbb{R})$ such that every nonzero matrix in $W$ has no real eigenvalues? Clearly if $n$ is odd then $g(n)=0$. When $n$ is even, can one have $g(n)>1$? REPLY [8 votes]: The "complementary" problem is strongly related to the problem of independent vector fields on $S^{n-1}$. Indeed, if $W$ is a space of matrices none of them having a real eigen-value and $(A_1,...,A_k)$ is its basis, it means that for every $v$, the vectors $(v,A_1(v),...,A_k(v))$ are linearly independent. Hence, the projections of $(A_1v,...,A_kv)$ to the plane orthogonal to $v$ give rise to $k$ independent vector fields on the sphere $S^{n-1}$. This problem was solved by Adams using non-trivial algebraic topology, and the exact possible number is $\rho(n)-1$ where $\rho((2k+1)2^{4a+b})=2^b+8a$. Note that this is consistent with the previous answer by Noam, since $S^1,S^3,S^7$ are really the only parallelizable spheres. Now, to see that this is the answer in our case, it suffices to note that the construction of the examples in Adams work actually arrises from a solution to the question you ask: they all come from construction of linear examples. Namely, consider a Clifford algebra $Cl_k$ having a representation of dimension $n$. Then the imaginary clifford elements $(e_1,...,e_k)$ give example to a space of matrices as you wish, and it is known that this is a maximal example for independent vector fields on on the $n-1$-sphere as well.<|endoftext|> TITLE: Voisin examples in $p$-adic geometry QUESTION [8 upvotes]: Let $K$ be an algebraic closure of p-adic rationals. Does there exist a proper smooth rigid-analytic variety over $K$ whose etale homotopy type is not isomorphic to etale homotopy type of a proper smooth scheme over $K$? REPLY [3 votes]: Welcome new contributor. There is an analogue of the Hopf surface in rigid analytic geometry. Here is one link to an article about these. MR1149806 (93g:32048) Voskuil, Harm Non-Archimedean Hopf surfaces. Sém. Théor. Nombres Bordeaux (2) 3 (1991), no. 2, 405–466. http://www.numdam.org/article/JTNB_1991__3_2_405_0.pdf In particular, since the (first) homotopy group of these rigid analytic spaces is Abelian of odd rank, they cannot be "homotopic" to a proper smooth scheme over $K$. This was not asked, but I will relate the following anyway. Over the rigid analytic space of the projective line, there exists a proper smooth morphism to the projective line whose fibers are these rigid Hopf surfaces such that the total space is "rationally connected" in the sense that any pair of points are contained in a rigid analytic subspace that is isomorphic to the projective line. Yet the total space admits a free action of a finite cyclic group. Via "twisting", this gives a "rationally connected" fibration over a rigid analytic curve that has no section (not even topologically). Thus, just as the usual Hopf surfaces show that there can be no complex-analytic analogue of the "Rationally Connected Fibration Theorem", also these rigid analytic Hopf surfaces prove that also there is no rigid-analytic analogue.<|endoftext|> TITLE: What numbers can simulate 1/2? QUESTION [7 upvotes]: Given two numbers $p,q\in(0,1)$, we say that $p$ can simulate $q$ if, given a biased coin with probability $p$, we can toss it a bounded number of times and use the results to simuate a biased coin with probability $q$. In this math.SE question the following results were proved: No rational number except $1/2$ can simulate $1/2$; Some irrational numbers can simulate $1/2$, for example: $1/2 \pm 1/\sqrt{12}$, or $1/2^{1/n}$ for any integer $n$. Q1. What is the set of all numbers $p\in (0,1)$ which can simulate $1/2$? Q2. What is the set of all numbers $p\in (0,1)$ which can simulate $1/3$? REPLY [4 votes]: Depends on what kind of "description" you want, but consider the question of the probabilities $p$ that can simulate $1/2$ with exactly $n$ flips. The events you have are: One event with probability $p^n$ $n$ events with probability $p^{n-1}(1-p)$ $\binom{n}{2}$ events with probability $p^{n-2}(1-p)^2$ ... One event with probability $(1-p)^n$ So if you pick any integers $a_0, a_1, \ldots a_n$ with $0 \leq a_i \leq \binom{n}{i}$ then you get a polynomial that $p$ has to satisfy. Of course such a polynomial may or may not have a root on $(0, 1)$, but we can at least enumerate the solutions in this manner. Clearly transcendental values for $p$ are right out. Moreover we have an explicit description of what the minimal polynomial of a solution has to look like, and any number $p \in (0, 1)$ satisfying such a polynomial is indeed a solution. In some sense that could be considered a complete answer, although maybe we want to know which potential minimal polynomials correspond to actual solutions. Notice that if we have two symmetries here. First, a biased coin that comes up heads with probability $p$ is just as good as a biased coin that comes up heads with probability $1 - p$; we can just relabel "heads" and "tails" on the coin. Second, a scheme that assigns a set of outcomes to "heads" and the complement to "tails" is just as good as the scheme making the opposite assignment. In terms of the choices of $a_i$ this amounts to (1) replacing $a_i$ with $\binom{n}{i} - a_i$ or (2) replacing $a_i$ with $a_{n-i}$. We can also rule out any case where swapping the roles of $p$ and $1-p$ results in larger coefficients across the board. For instance, if we have $n = 2$ then clearly we can't have ${HT}$ having an equal probability to ${HH, TH, TT}$ since $p(HT) = p(TH)$. Let's do the first few values of $n$. For $n = 0$ there are of course no solutions and for $n = 1$ there is only the solution $p = 1/2$. For $n = 2$ there are the solutions $p = \sqrt{2}/2$ as well as $p = 1/2$ again. Okay, now let's try $n = 3$. If I've enumerated right, the possible equivalence classes are: (1, 0, 0, 0); (0, 3, 3, 1); (0, 0, 0, 1); (1, 3, 3, 0) (0, 2, 0, 0); (1, 1, 3, 1); (0, 0, 2, 0); (1, 3, 1, 1) (1, 2, 0, 0); (0, 1, 3, 1); (0, 0, 1, 2); (1, 3, 1, 0) (0, 3, 0, 0); (1, 0, 3, 1); (0, 0, 3, 0); (1, 3, 0, 1) (1, 3, 0, 0); (0, 0, 3, 1) (1, 0, 1, 0); (0, 3, 2, 1); (0, 1, 0, 1); (1, 2, 3, 0) (1, 1, 1, 0); (0, 2, 2, 1); (0, 1, 1, 1); (1, 2, 2, 0) (0, 2, 1, 0); (1, 1, 2, 1); (0, 1, 2, 0); (1, 2, 1, 1) (1, 2, 1, 0); (0, 1, 2, 1); (0, 3, 1, 0); (1, 0, 2, 0); (0, 1, 3, 0); (0, 2, 0, 1) (1, 0, 2, 0); (0, 3, 1, 1); (0, 2, 0, 1); (1, 1, 3, 0) (1, 1, 2, 0); (0, 2, 1, 1) (0, 2, 2, 0); (1, 1, 1, 1) (0, 3, 2, 0); (1, 0, 1, 1); (0, 2, 3, 0); (1, 1, 0, 1) (1, 0, 3, 0); (0, 3, 0, 1) Let's try one representative from each class. $p^3 = 1/2$. This gives us the solution $p = \sqrt[3]{4}/2$ $2p^2(1-p) = 1/2$. The real root here is negative. $p^3 + 2p^2(1-p) = 1/2$. This polynomial has three real roots, one of which is a valid probability. $3p^2(1-p) = 1/2$. The real root here is negative. $p^3 + 3p^2(1-p) = 1/2$. This has the solution $p = 1/2$, and corresponds to taking whichever of heads or tails we got more of. $p^3 + p(1-p)^2 = 1/2$. The real solution is a valid probability. Okay, I lied, I'm not going to try every single case. But we did come up with a couple of new solutions we didn't previously know about. There are some things we could systematically do here, like eliminating the cases that correspond to ignoring some number of the coin flips or employing Descartes's rule of signs to skip some cases where there are no real roots. I don't know if this would reach a bijective solution or not, but it's worth a try.<|endoftext|> TITLE: hook-length formula: "Fibonaccized" Part I QUESTION [15 upvotes]: Consider the Young diagram of a partition $\lambda = (\lambda_1,\ldots,\lambda_k)$. For a square $(i,j) \in \lambda$, define the hook numbers $h_{(i,j)} = \lambda_i + \lambda_j' -i - j +1$ where $\lambda'$ is the conjugate of $\lambda$. The hook-length formula shows, in particular, that if $\lambda\vdash n$ then $$\text{$n!\prod_{\square\,\in\,\lambda}\frac1{h_{\square}}$} \qquad \text{is an integer}.$$ Recall the Fibonacci numbers $F(0)=0, \, F(1)=1$ with $F(n)=F(n-1)+F(n-2)$. Define $[0]!_F=1$ and $[n]!_F=F(1)\cdot F(2)\cdots F(n)$ for $n\geq1$. QUESTION. Is it true that $$\text{$[n]!_F\prod_{\square\,\in\,\lambda}\frac1{F(h_{\square})}$} \qquad \text{is an integer}?$$ REPLY [6 votes]: This is a less elementary but maybe more conceptual proof, also giving some combinatorial meaning: Use the formulas $F(n) = \frac{\varphi^n -\psi^n}{\sqrt{5}}$, $\varphi =\frac{1+\sqrt{5}}{2}, \psi = \frac{1-\sqrt{5}}{2}$. Let $q=\frac{\psi}{\varphi} = \frac{\sqrt{5}-3}{2}$, so that $F(n) = \frac{\varphi^n}{\sqrt{5}} (1-q^n)$ Then the Fibonacci hook-length formula becomes: \begin{align*} f^{\lambda}_F:= \frac{[n]!_F}{\prod_{u\in \lambda}F(h(u))} = \frac{ \varphi^{ \binom{n+1}{2} } [n]!_q }{ \varphi^{\sum_{u \in \lambda} h(u)} \prod_{u \in \lambda} (1-q^{h(u)})} \end{align*} So we have an ordinary $q$-analogue of the hook-length formula. Note that $$\sum_{u \in \lambda} h(u) = \sum_{i} \binom{\lambda_i}{2} + \binom{\lambda'_j}{2} + |\lambda| = b(\lambda) +b(\lambda') +n$$ Using the $q-$analogue hook-length formula via major index (EC2, Chapter 21) we have \begin{align*} f^\lambda_F = \varphi^{ \binom{n}{2} -b(\lambda)-b(\lambda')} q^{-b(\lambda)} \sum_{T\in SYT(\lambda)} q^{maj(T)} = (-q)^{\frac12( -\binom{n}{2} +b(\lambda') -b(\lambda))}\sum_T q^{maj(T)} \end{align*} Now, it is clear from the q-HLF formula that $q^{maj(T)}$ is a symmetric polynomial, with lowest degree term $b(\lambda)$ and maximal degree $b(\lambda) + \binom{n+1}{2} - n -b(\lambda) -b(\lambda') =\binom{n}{2} - b(\lambda')$ so the median degree term is $$M=\frac12 \left(b(\lambda) +\binom{n}{2} - b(\lambda')\right)$$ which cancels with the factor of $q$ in $f^{\lambda}_F$, so the resulting polynomial is of the form \begin{align*} f^{\lambda}_F = (-1)^{M} \sum_{T: maj(T) \leq M } (q^{M-maj(T)} + q^{maj(T)-M}) \\ = (-1)^{M} \sum_{T} (-1)^{M-maj(T)}( \varphi^{2(M-maj(T))} + \psi^{2(M-maj(T)}) = \sum_T (-1)^{maj(T)} L(2(M-maj(T))) \end{align*} where $L$ are the Lucas numbers. **byproduct of collaborations with A. Morales and I. Pak.<|endoftext|> TITLE: Frattini subgroup is normal-monotone QUESTION [5 upvotes]: On page 199 of Dummit and Foote's Abstract Algebra (Here $\Phi(G)$ is the Frattini subgroup of a group $G$, not necessarily finite): If $N\unlhd G$, then $\Phi(N)\subseteq\Phi(G)$. First, When every proper subgroup of $N$ is contained in a maximal subgroup of $N$, I know how to prove the statement. (By taking $M$ as a maximal subgroup of $G$ that fails to contain $\Phi(N)$, deriving $N=\Phi(N)(N\cap M)$ and taking the maximal subgroup $H$ of $N$ containing $N\cap M$, then $\Phi(N)\subset H$, a contradiction.) But in the general case, as there may not exist a maximal subgroup of $N$ containing $N\cap M$, the case is different. The process of deriving $N=\Phi(N)(N\cap M)$ is the same, but I find no way to proceed after that. Hence I have several questions: (1) If the statement still holds in the case when not all proper subgroup of $N$ is contained in some maximal subgroup of $N$? Or does there exist an counterexample? (2) Moreover, I'm wondering that if a group $G$ satisfies the condition that every proper subgroup is contained in a maximal subgroup, could it be possible that the condition does not apply to its normal subgroup $N$? I've posted the question on StackExchange, and just get a partial answer for the question(1). (Actually, by the definition of Frattini subgroup being the set of all non-generators, the statement is proved in the case of $\phi(N)$ being finitely generated). Hence I hope for an answer for both questions! Thanks in advance! REPLY [7 votes]: Indeed the result is false. Consider the affine group $G=\mathbf{Q}^*\ltimes\mathbf{Q}$ and $N$ the normal subgroup $\mathbf{Q}$. Since $N$ has no maximal proper subgroup $\Phi(N)=N$. Since $\mathbf{Q}^*$ is a maximal proper subgroup of $G$ and since the intersection of its conjugates is trivial, we have $\Phi(G)=\{1\}$. So $\Phi(N)\nsubseteq \Phi(G)$. The problem is exactly what you're pointing out. The result however holds whenever $N$ has the property that every proper subgroup of $N$ is contained in a maximal proper subgroup of $N$.<|endoftext|> TITLE: Cardinal characteristics of amorphous sets QUESTION [7 upvotes]: In a universe where the continuum hypothesis ($CH$) fails we can ask about combinatorial cardinal characteristics of the continuum, but in a universe where $CH$ is true no such cardinals exist so this study becomes vacuous. Does a similar phenomenon occur at the countable level in a universe without choice? Specifically, are there properties which are true for finite sets but false for $\omega$ which are still true for the cardinal of an amorphous set, like divisibility as suggested here by François G. Dorais? In a universe without choice we have the existence of amorphous sets and we can ask about their 'amorphous cardinals' which are incomparable with $\omega$ (thank you Asaf for the correction) and may satisfy nice theorems, but in a universe with choice there are no infinite sets whose cardinality is incomparable with $\omega$ so this study becomes vacuous in similar fashion to the uncountable case. A possible candidate for characteristics smaller than $\omega$ could come from theorems in finite group theory that become false for countable groups, since it is possible to have a group structure on an unbounded amorphous cardinal as constructed by Asaf Karagila here. There is an article behind a paywall published in 2010 that appears to touch on these matters but I can't access it; if anyone is familiar with its contents and willing to give a brief exposition it would be greatly appreciated. REPLY [8 votes]: Any cardinal smaller than $\aleph_0$ is finite. Amorphous sets are not "smaller", they are just incomparable with. They are very small, in some sense, for example we cannot even divide them into two infinite sets, but they are still infinite. With respect to the article you linked, let me point out that amorphous sets cannot even be mapped onto $\omega$, so they are definitely not the countable union of pairs. Now, there are some combinatorial characteristics one can assign to general sets, which may be of interest in the case of amorphous sets. For example, if $A$ is amorphous, then any partition of $A$ is up to finitely many parts constant in size (i.e. all but finitely many parts are singletons, or pairs, or so on). We call this size the gauge of the partition, and we can ask what is the supremum of the gauges of possible partitions. This can be $1$, or some finite $n$, or it can be "unbounded". We can prove, for example, that if $A$ is an amorphous set which can be made into a group, then it is unbounded. So it gives us some information. But in general, this is not something too similar to cardinal characteristics in the traditional sense, and it is not something too helpful, since $\omega$ is a very unique and a very concrete set, whereas amorphous sets can come in many different flavors, sizes, and support different structures.<|endoftext|> TITLE: Is the Perron-Frobenius dimension of a G-Set given by its cardinality? QUESTION [5 upvotes]: Given a ring $R$ with finite additive basis $\{e_i\}_{i=1}^{n}$, such that $e_i e_j=\sum c_{ijk}e_k$ with $c_{ijk}\in \mathbb{N}$, we define the Perron-Frobenius dimension $FPDim(e_i)$ of a basis element $e_i$ to be the maximal positive real eigenvalue of matrix $M_{e_i}$, multiplication by $e_i$. This exists by the Perron-Frobenius theorem, and we extend by linearity to all of $R$. Now take $B(G)$ to be the burnside ring of a finite group $G$, with basis given by isomorphism classes of transitive actions of $G$. One can directly check for $G\cong C_p$ that $FPDim(X)=|X|$. Does this hold in general for finite $G$-sets? Note that our ring $B(G)$ is not necessarily transitive in the sense of Etingof's Tensor Categories (Definition 3.3.1), so this doesn't seem to follow immediately from the results in there. REPLY [7 votes]: For $H \subset G$, write $e_H$ for the basis element of $B(G)$ corresponding to the $G$-set $G/H$. Recall that for $K \subset G$, $$ e_H\cdot e_K = \sum\limits_{H g K \in H \backslash G / K} e_{H \cap g K g^{-1}}. $$ Say that $K$ is subconjugate to $H$ if $K$ is conjugate to a subgroup of $H$. This gives rise to a partial order on the set of conjugacy classes of subgroups of $G$, and hence on the basis elements $e_H$. With respect to this partial ordering, $M_{e_H}$ is lower-triangular for each $H \subset G$. The eigenvalues are then the diagonal elements. The diagonal element corresponding to $e_{\{\textrm{id}_G\}}$ is $|G/H|$. The diagonal element corresponding to $e_K$ for $K \subset G$ is bounded above by the number of $(H, K)$-double cosets in $G$, which is bounded above by $|G/H|$.<|endoftext|> TITLE: Finitely generated matrix groups whose eigenvalues are all algebraic QUESTION [15 upvotes]: Let $G$ be a finitely generated subgroup of $GL(n,\mathbb{C})$. Assume that there exists a number field $k$ (i.e. a finite extension of $\mathbb{Q}$) such that for all $g \in G$, the eigenvalues of $g$ all lie in $k$. This implies that $g$ is conjugate to an element of $GL(n,k)$. Question: must it be the case that some conjugate of $G$ lies in $GL(n,k)$? Or at least $GL(n,k')$ for some finite extension $k'$ of $k$? If this is not true, what kinds of assumptions can I put on $G$ to ensure that it is? REPLY [2 votes]: Re-edited following YCor's comments: A nice theorem of Schur, building on earlier work of Jordan and Burnside, states that any finitely generated periodic subgroup $G$ of ${\rm GL}(n,\mathbb{C})$ is finite ( this is Theorem 36.2 of the 1962 edition of Curtis and Reiner)-and hence is completely reducible. Hence the answer to your question is "yes" , if every eigenvalue of every element of $G$ is a root of unity and every element of $G$ is semisimple. In that case, once we know that $G$ is finite, then a Theorem of Brauer (which makes use of his induction theorem) asserts that every finite subgroup $X$ of ${\rm GL}(n,\mathbb{C})$ is conjugate within ${\rm GL}(n,\mathbb{C})$ to a subgroup of ${\rm GL}(n,\mathbb{Q}[\omega]),$ where $\omega$ is a primitive complex $|G|$-th root of unity.<|endoftext|> TITLE: Is this a new Fibonacci Identity? QUESTION [9 upvotes]: I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity \begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} \end{equation} where $F_1 = F_2 = 1$, $r \leq n$, $h \leq g$, and $n, g, k \in \mathbb{N}$. It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input! REPLY [13 votes]: This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4\times 4$ matrix. Concrete mathematics gives the following reference: As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_{m,n}=s_{m+n}s_{m-n}$ $(m,n\in \mathbb{Z})$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.<|endoftext|> TITLE: A question about the (motivic) integral cohomology of the Eilenberg-MacLane spectrum QUESTION [6 upvotes]: Let $H\mathbb{Z}$ be the Eilenberg-MacLane spectrum. Let $n\geq 0$ be any integer. Is it known the structure of the group $[H\mathbb{Z},\Sigma^{n}H\mathbb{Z}]$? Is there any reference in this direction? What about of its motivic version, i.e. if $M\mathbb{Z}$ denotes the motivic Eilenberg-MacLane spectrum, What $[M\mathbb{Z},\Sigma^{2n,n}M\mathbb{Z}]$ is? I would appreciate any help. REPLY [10 votes]: For the topological question, this thread and this thread have interesting answers and references about integral (co)homology of Eilenberg-MacLane spaces and spectra. Let me add a few comments. I'll focus on integral homology. Let $A$ be an abelian group and consider the stable homology group $$H\mathbb{Z}_n HA = \pi_n(H\mathbb{Z} \wedge HA) \cong H_{m+n}(K(A,m);\mathbb{Z})$$ for $m > n$ (the stable range). Eilenberg and MacLane give us: $$H\mathbb{Z}_n HA = \begin{cases} A &n = 0 \\ 0 &n = 1 \\ A/2 &n = 2 \\ \mbox{}_2 A &n = 3 \\ A/2 \oplus A/3 &n = 4 \\ \mbox{}_2 A \oplus \mbox{}_3 A &n = 5 \\ \end{cases}$$ respectively in Theorems 20.3, 20.5, 23.1, 24.1, 25.1, and 25.3. (Ok, the cases $n=0,1$ are rather an exercise.) Here, $\mbox{}_2 A$ denotes the $2$-torsion subgroup of $A$. Sections 26-27 address the cohomology $H^*(K(A,m); \mathbb{Z})$ for small $m$. Section 5 of Cartan and Section 6 of this Séminaire Henri Cartan provide a way to compute $H\mathbb{Z}_n HA$ entirely. It is a torsion group whose $p$-primary part is a finite direct sum of copies of $A/p$ and $\mbox{}_p A$ indexed by some combinatorial formulas. For the motivic question, I'll defer to the algebraic geometers. Here are a few ideas. For a nice base scheme $S$ (say, essentially smooth over a field), we have$$M\mathbb{Z}^{0,0}M\mathbb{Z} = \mathbb{Z}^{\pi_0(S)},$$where $\pi_0(S)$ denotes the set of connected components of the base scheme $S$. See Lemma 6.14 in this preprint and the references therein. In Lemma 4.10, we had an $M\mathbb{Z}$-module map $M\mathbb{Z} \to \Sigma^{2,1} M\mathbb{Z}$, which corresponds to a map $\mathbf{1} \to \Sigma^{2,1} M\mathbb{Z}$ in $SH(S)$. Those are given by $$[\mathbf{1}, \Sigma^{2,1} M\mathbb{Z}] = H^{2,1}(S;\mathbb{Z}) = \mathrm{Pic}(S),$$the Picard group of the base scheme $S$ (see Lemma 4.8). I'm not sure what maps $[M\mathbb{Z}, \Sigma^{2,1} M\mathbb{Z}]$ in $SH(S)$ look like. These lecture notes by Spitzweck might be helpful, especially Section 3. This book by Mazza, Voevodsky, and Weibel is also a nice reference.<|endoftext|> TITLE: Volume of balls in homogeneous manifolds QUESTION [5 upvotes]: Let $X=G/H$ be a homogeneous manifold, where $G$ and $H$ are connected Lie groups and assume there is given a $G$-invariant Riemannian metric on $X$. Let $B(R)$ be the closed ball of radius $R>0$ around the base point $eH$ and let $b(R)$ denote its volume. Is it rue that $$ \lim_{\varepsilon\to 0}\ \limsup_{R\to\infty}\ \frac{b(R+\varepsilon)}{b(R)}=1?\qquad(\#) $$ The idea somehow being that volume growth is largest with constant negative curvature in which case it is exponential and thus satisfies our claim. REPLY [7 votes]: The idea somehow being that volume growth is largest with constant negative curvature is essentially the content of the Bishop–Cheeger–Gromov comparison theorem. See for instance Lemma 36 of Peter Petersen's Riemannian Geometry, 2ed. It states, in his notation, that in any complete Riemannian manifold $(M,g)$ of dimension $n$ with Ricci curvature bounded below by $(n-1)k$, the function $$r \mapsto \frac{\operatorname{vol} B(p,r)}{v(n,k,r)}$$ is nonincreasing, where $B(p,r)$ is the ball in $(M,g)$ centered at $p$ with radius $r$, and $v(n,k,r)$ is the volume of the ball of radius $r$ in the space form of dimension $n$ and constant sectional curvature $k$. A homogeneous space certainly has bounded Ricci curvature, and the interesting case for us is when $k$ is negative, so that $v(n,k,r)$ is the volume in hyperbolic space, which as you say grows exponentially. So this result tells us that $$\frac{\operatorname{vol} B(p,r)}{v(n,k,r)} \ge \frac{\operatorname{vol} B(p,r+\epsilon)}{v(n,k,r+\epsilon)}$$ or in other words $$1 \le \frac{\operatorname{vol} B(p,r+\epsilon)}{\operatorname{vol} B(p,r)} \le \frac{v(n,k,r+\epsilon)}{v(n,k,r)}.$$ As $r \to \infty$ the right side converges to something of the form $e^{c \epsilon}$, which in turn goes to $1$ as $\epsilon \to 0$.<|endoftext|> TITLE: Sums of two squares in arithmetic progressions QUESTION [9 upvotes]: Let $r(n)$ denote the number of representations of $n$ as the sum of two squares. Are there any known results on $$\sum _{n\leq x\atop {n\equiv a(q)}}r(n)$$ and in particular is there an asymptotic formula? REPLY [17 votes]: The first result in this direction seems to be due to R. A. Smith, ``The Circle Problem in an Arithmetic Progression,'' Can. Math. Bull. 11 (2), 175–184 (1968). He showed that if we write $$\sum _{n\leq x\atop {n\equiv a(q)}}r(n) =\pi x \cdot \frac{\eta_{a}(q)}{q^2}+ R_{q,a}(x)$$ where $\eta_{a}(q) := \{ (x_1,x_2) \in (\mathbb{Z}/q\mathbb{Z)}^2 : x_1^2 +x_2^2 \equiv a \bmod q\}$, then $$R_{q,a}(x) = O\left( x^{\frac{2}{3} + \xi} q^{-\frac{1}{2}(1+3\xi)}\gcd(a,q)^{1/2}\tau(q) \right)$$ for any $\xi \in (0,1/3)$. This is non-trivial for $q \le x^{\frac{2}{3}-\varepsilon}$. In particular, as $x$ tends to infinity, your expression is asymptotic to $\pi x$ times the probability that $x_1^2+x_2^2 \equiv a \bmod q$ (for uniformly drawn $x_1,x_2$ mod $q$). If you consider $a$ and $q$ as fixed, this answers your question. The state-of-the-art result is due to D. I. Tolev, ``On the remainder term in the circle problem in an arithmetic progression,'' Tr. Mat. Inst. Steklova 276 (2012), Teoriya Chisel, Algebra i Analiz, 266--279; translation in Proc. Steklov Inst. Math. 276 (2012), no. 1, 261–274. He showed that $$R_{q,a}(x) = O\left( (q^{\frac{1}{2}}+x^{\frac{1}{3}}) \gcd(a,q)^{1/2}\tau^4(q)\log^4 x \right).$$ Interestingly, for $a=1$, there is a result which is superior in certain ranges of $x$ and $q$, see P. D. Varbanets, “Lattice Points in a Circle Whose Distances from the Center Are in an Arithmetic Progression,” Mat. Zametki 8 (6), 787–798 (1970) [Math. Notes 8, 917–923 (1970)]. All three results are explained in Tolev's paper. In the 2002 PhD thesis of Michael J. Dancs under R. Vaughan, titled "On a Variance arising in the Gauss Circle Problem", he proves (independently of the above) that $$f(q,a):=\lim_{x \to \infty} \frac{\sum _{n\leq x\atop {n\equiv a(q)}}r(n) }{\pi x}$$ exists and can be written as $$f(q,a)=q^{-3} \sum_{k=1}^{q} \exp\left( 2\pi i \frac{-ak}{q} \right) S(q,k)^2,$$ where $S(q,a)$ is a quadratic Gauss sum mod $q$. This gives a different expression for $\eta_a(q)/q^2$, which was better suited for the applications given by Dancs. This is done in Theorem 2.1. He then proceeds to prove, in Theorem 2.2, the estimate $$R_{q,a}(x) = O\left( (\sqrt{x}+q) \log q\right).$$ For small $q$ this is better than Smith's work. Still, it is superseded by Tolev's estimates.<|endoftext|> TITLE: how do we prove that a sum of two periods is still a period? QUESTION [12 upvotes]: Kontsevich and Zagier define periods as the values of absolutely convergent integrals $\int_\sigma f$ where $f$ is a rational function with rational coefficients and $\sigma$ is a semi-algebraic subset of $\mathbb{R}^n$. How do we prove that the sum of two such numbers is still of this form? I've tried a few things but they don't seem to work... REPLY [23 votes]: Let $\alpha$ and $\beta$ be two periods corresponding respectively to two absolutely convergent integrals $\int_\sigma f(x)dx$ and $\int_\tau g(y)dy$, where $f$ (resp. $g$) is a rational function on $\Bbb Q$ with $r$ (resp. $s$) variables and $\sigma$ (resp. $\tau$) is a semi-algebraic subset of $\Bbb R^r$ (resp. $\Bbb R^s$). Setting $\omega:=\sigma\times(0,1)\times(0,1)^s\coprod(0,1)^r\times(1,2)\times\tau$, one immediately gets that $$\alpha+\beta=\int_\omega \left[f(x)+g(y)\right]dxdydt$$which is again an absolutely convergent integral, so that $\alpha+\beta$ is a period.<|endoftext|> TITLE: Unipotent completion of free group QUESTION [5 upvotes]: Whilst I am reading articles on unipotent completion to understand its basic construction, I found something confusing. Let $F$ be a free group of rank 2 whose generating letters are $x$ and $y$ and let $K$ be a field of characteristic 0. Then the unipotent completion of $F$ over $K$ would be the set of group-like elements in the completed Hopf algebra $K[F]^\wedge=\varprojlim K[F]/I^n$ where $I=\ker(K[F]\to K)$ is the kernel of the augmentation map. Note that $K[F]^\wedge$ is isomorphic to the ring of non-commutative formal power series in two variables $K\langle\!\langle X,Y\rangle\!\rangle$, i.e. $K[F]^\wedge\simeq K\langle\!\langle X,Y\rangle\!\rangle$. Now, here is the point I am confusing: why is the natural embedding $F\to K[F]^\wedge\simeq K\langle\!\langle X,Y\rangle\!\rangle$ given explicitly by $x\mapsto e^X$ and $y\mapsto e^Y$? Is this a chosen one? More precisely, I guess that the natural embedding $F\to K[F]^\wedge$ would be $x\mapsto x$ and $y\mapsto y$ (up to the abuse of notations). So, I guess that the explicit form is given by the isomorphism $K[F]^\wedge\simeq K\langle\!\langle X,Y\rangle\!\rangle$. But I think that the map $x\mapsto 1+X$, $y\mapsto 1+Y$ also gives an isomorphism, is it wrong? REPLY [3 votes]: You're absolutely right. The last map you mention is called the Magnus expansion. One of the reasons it's interesting is that this formula is valid over $\mathbb{Z}$ and induces an injective map $$F\rightarrow \mathbb{Z}\langle \langle X,Y\rangle\rangle$$ which you can use to prove that $F$ is residually torsion-free-nilpotent (I believe this was Magnus motivation). On the other hand, the map $x\mapsto e^X, y\mapsto e^Y$ is an Hopf algebra morphism, hence it identifies the unipotent completion of $F$, with the exponential group of the degree completion of the free Lie algebra on $X,Y$. This is also very useful, and implies for example that $F$ is 1-formal which is an important property. This is a somewhat distinguished choice for such an isomorphism, at least once you've chosen generators, but by no means is it canonical and there are in fact many of those. For example, another important one is obtained from the monodromy of the flat regular connection on $P^1(\mathbb{C})-\{0,1,\infty\}$ given by $$d- (\frac xz + \frac y{1-z}).$$ This one is defined only for $K=\mathbb{C}$ but have some extra nice properties. It has a version for any field of characteristic 0 but those are much less explicit (they are constructed using so-called Drinfeld associators).<|endoftext|> TITLE: Rigorous proof of the good regulator theorem QUESTION [8 upvotes]: As an applied mathematician with an interest in control theory, I have read several research papers that explicitly use the good regulator theorem of Conant and Ashby 1 which states that: Every good regulator of a system must be a model of that system. As pointed out by the authors of [3], the importance and generality of this theorem in control theory makes it comparable in importance to Einstein's $E=mc^2$ for physics. However, as John C. Baez carefully argues in a blog post titled The Internal Model Principle it's not clear that Conant and Ashby's paper demonstrates what it sets out to prove. I'd like to add that many other researchers, besides myself, share John C. Baez' perspective. At present, I think an information-theoretic approach may be used to demonstrate a general version of the good regulator theorem and in the near future I will probably attempt such a demonstration. Meanwhile, might there be control theorists on the MathOverflow that know of a proof of the good regulator theorem that is both clear and rigorous? Such a publication, if it exists, would be of interest to applied mathematicians working on control-theoretic problems, researchers in the area of behavioural neuroscience as well as artificial intelligence researchers. References: Roger C. Conant and W. Ross Ashby, Every good regulator of a system must be a model of that system), International Journal of Systems Science 1 (1970), 89–97. B. A. Francis and W. M. Wonham, The internal model principle of control theory, Automatica 12 (1976) 457–465 Daniel L. Scholten, Every good key must be a model of the lock it opens (the Conant & Ashby Theorem revisited), 2010. REPLY [2 votes]: There are two principled information-theoretic approaches to identifying a good regulator for an organism's environment. Within the context of reinforcement learning, both approaches are unsupervised in the sense that organisms don't need an explicit reward function. Instead of maximising a reward function, the organism's objective is to construct a suitable model of their environment for decision-making and planning. Empowerment: The empowerment approach, developed by Daniel Polani and his collaborators consists of approximating the channel capacity in the perception-action loop: \begin{equation} C = \max_{p(x)} I(X;Y) \end{equation} where $X$ represents the space of possible actions and $Y$ represents the sensory space(i.e. percepts). One practical challenge facing this approach is that it generally requires identifying an environment's state-transition function which is intractable for large and complex state-spaces. That said, scientists at Deep Mind have recently demonstrated that variational approximations of the channel capacity $C$ address certain scalability issues [2]. Free Energy Principle: The Free Energy Principle, developed by Karl Friston rests upon the ergodic assumption that self-organising biological agents minimise the expected surprise or the entropy of their sensory states: \begin{equation} H(y) = - \int p(y|m) \ln p(y|m) dy = \lim_{T \to \infty} \frac{1}{T} \int_{0}^T - \ln p(y|m) dt \end{equation} where $m$ represents the model used by the agent and $y$ represents its sensory input. Some scientists have pointed out that Friston's theory is a theory of cognitive dissonance minimisation which leaves it vulnerable to the dark room problem. However, scientists at Google Brain and Google DeepMind recently argued that the Free Energy Principle is a theory of niche construction which forces organisms' to find the ecological niche they are most suited for [5]. Challenges facing a definitive proof: An important challenge facing any mathematical formalism for good regulators is that the practical consequences of such a formalism in complex environments requires experimentation which escapes the analytical process. References: Christoph Salge, Cornelius Glackin and Daniel Polani. Empowerment — An Introduction. 2013. Shakir Mohamed and Danilo J. Rezende. Variational Information Maximisation for Intrinsically Motivated Reinforcement Learning. 2015. Karl Friston. The free-energy principle: a rough guide to the brain? Cell press. 2009. Zekun Sun & Chaz Firestone. The Dark Room Problem. Cell Press. 2020. Danijar Hafner et al. Action and Perception as Divergence Minimization. 2020.<|endoftext|> TITLE: hook-length formula: "Fibonaccized": Part II QUESTION [6 upvotes]: This is a natural follow-up to my previous MO question, which I share with Brian Hopkins. Consider the Young diagram of a partition $\lambda = (\lambda_1,\ldots,\lambda_k)$. For a square $(i,j) \in \lambda$, define the hook numbers $h_{(i,j)} = \lambda_i + \lambda_j' -i - j +1$ where $\lambda'$ is the conjugate of $\lambda$. The hook-length formula shows that if $\lambda\vdash n$ then $$n!\prod_{\square\,\in\,\lambda}\frac1{h_{\square}}$$ counts standard Young tableaux whose shape is the Young diagram of $\lambda$. Recall the Fibonacci numbers $F(0)=0, \, F(1)=1$ with $F(n)=F(n-1)+F(n-2)$. Define $[0]!_F=1$ and $[n]!_F=F(1)\cdot F(2)\cdots F(n)$ for $n\geq1$. QUESTION. What do these integers count? $$[n]!_F\prod_{\square\,\in\,\lambda}\frac1{F(h_{\square})}.$$ REPLY [10 votes]: This is my answer to the original question (https://mathoverflow.net/a/327022/50244) whether these numbers are integers to begin with, it gives some combinatorial meaning as well: Use the formulas $F(n) = \frac{\varphi^n -\psi^n}{\sqrt{5}}$, $\varphi =\frac{1+\sqrt{5}}{2}, \psi = \frac{1-\sqrt{5}}{2}$. Let $q=\frac{\psi}{\varphi} = \frac{\sqrt{5}-3}{2}$, so that $F(n) = \frac{\varphi^n}{\sqrt{5}} (1-q^n)$ Then the Fibonacci hook-length formula becomes: \begin{align*} f^{\lambda}_F:= \frac{[n]!_F}{\prod_{u\in \lambda}F(h(u))} = \frac{ \varphi^{ \binom{n+1}{2} } [n]!_q }{ \varphi^{\sum_{u \in \lambda} h(u)} \prod_{u \in \lambda} (1-q^{h(u)})} \end{align*} So we have an ordinary $q$-analogue of the hook-length formula. Note that $$\sum_{u \in \lambda} h(u) = \sum_{i} \binom{\lambda_i}{2} + \binom{\lambda'_j}{2} + |\lambda| = b(\lambda) +b(\lambda') +n$$ Using the $q-$analogue hook-length formula via major index (EC2, Chapter 21) we have \begin{align*} f^\lambda_F = \varphi^{ \binom{n}{2} -b(\lambda)-b(\lambda')} q^{-b(\lambda)} \sum_{T\in SYT(\lambda)} q^{maj(T)} = (-q)^{\frac12( -\binom{n}{2} +b(\lambda') -b(\lambda))}\sum_T q^{maj(T)} \end{align*} Now, it is clear from the q-HLF formula that $q^{maj(T)}$ is a symmetric polynomial, with lowest degree term $b(\lambda)$ and maximal degree $b(\lambda) + \binom{n+1}{2} - n -b(\lambda) -b(\lambda') =\binom{n}{2} - b(\lambda')$ so the median degree term is $$M=\frac12 \left(b(\lambda) +\binom{n}{2} - b(\lambda')\right)$$ which cancels with the factor of $q$ in $f^{\lambda}_F$, so the resulting polynomial is of the form \begin{align*} f^{\lambda}_F = (-1)^{M} \sum_{T: maj(T) \leq M } (q^{M-maj(T)} + q^{maj(T)-M}) \\ = (-1)^{M} \sum_{T} (-1)^{M-maj(T)}( \varphi^{2(M-maj(T))} + \psi^{2(M-maj(T)}) = \sum_T (-1)^{maj(T)} L(2(M-maj(T))) \end{align*} where $L$ are the Lucas numbers. Remark. This is a byproduct of collaboration with A. Morales and I. Pak.<|endoftext|> TITLE: Conjecture: $a^n+b^n+c^n\ge x^n+y^n+z^n$ QUESTION [7 upvotes]: let $a,b,c,x,y,z>0$ such $x+y+z=a+b+c,abc=xyz$,and $a>\max\{x,y,z\}$, I conjecture $$a^n+b^n+c^n\ge x^n+y^n+z^n,\forall n\in N^{+}$$ Maybe this kind of thing has been studied, like I found something relevant, but I didn't find the same one.[Schur convexity and Schur multiplicative convexity for a class of symmetric function] I found sometime maybe The sum of squared logarithms conjecture REPLY [7 votes]: A triple $(a,b,c)$ with $a+b+c=s$, $abc=p$, and $ab+bc+ca=t$ is the triple of roots of $X^3-sX^2+tX-p=0$, i.e., of $-X^2+sX+p/X=t$. Fix $s$ and $p$; the left-hand part has just two local extrema on the right semiaxis. Let $a(t)>b(t)>c(t)$ be the three roots (defined for all values of $t$ when they exist and are sdistinct). Looking at the graph, one can easily see that $a(t)$ and $c(t)$ decrease as $t$ grows, while $b(t)$ grows as well. Now it suffices to prove that $(a^n+b^n+c^n)'=n(a^{n-1}a'+b^{n-1}b'+c^{n-1}c')\leq 0$ (the derivative is taken with respect to $t$). We know that $a'+b'+c'=a'bc+b'ca+c'ab=0$, hence $a'=\lambda a(b-c)$, $b'=\lambda b(c-a)$, and $c'=\lambda c(a-b)$, where $\lambda<0$ as $a'<0$. Hence the required inequality reads $a^n(b-c)+b^n(c-a)+c^n(a-b)\geq0$, or $$ a(b-c)(a^{n-1}-b^{n-1})\geq c(a-b)(b^{n-1}-c^{n-1}), $$ which follows from $$ \frac{a^{n-1}-b^{n-1}}{a-b}\geq\frac{b^{n-1}-c^{n-1}}{b-c}. $$ The last inequality holds for all (not neccessarily integer) $n\geq 2$, e.g., by Lagrange's mean value theorem. REPLY [5 votes]: Yes, it is true. Without loss of generality, $a\geq b \geq c$. Let $a+b+c=s,\ abc=p$ then $b=\frac{(s \ - \ a) \ + \ d}{2}, \ c=\frac{(s \ - \ a) \ - \ d}{2}$ with $d=b-c=\sqrt{(s-a)^2- \frac{4p}{a}}$. If we consider $b,c$ as variables in $a,s,p$, we find that $$ \begin{eqnarray} \frac{\partial (a^n+b^n+c^n)}{\partial a} &=& na^{n-1}+nb^{n-1}\frac{1}{2}\left(-1+\frac{\partial d}{\partial a}\right)+nc^{n-1}\frac{1}{2}\left(-1-\frac{\partial d}{\partial a}\right)\\ &=& \frac{n}{2}\left(\left(2a^{n-1}-b^{n-1}-c^{n-1}\right)+\frac{\partial d}{\partial a}\left(b^{n-1}-c^{n-1}\right) \right). \end{eqnarray} $$ Hence it is enough to prove the following: $$ \frac{\partial d}{\partial a}+\frac{2a^{n-1}-b^{n-1}-c^{n-1}}{b^{n-1}-c^{n-1}}\geq 0 $$ We have $$ \frac{\partial d}{\partial a}=\frac{a-s+2p/a^2}{d}=\frac{-b-c+2bc/a}{b-c}=\frac{2bc(b^{n-2}+b^{n-3}c+\dots)/a-(b+c)(b^{n-2}+b^{n-3}c+\dots)}{b^{n-1}-c^{n-1}} \ $$ Hence, $$ \frac{\partial d}{\partial a}+\frac{2a^{n-1}-b^{n-1}-c^{n-1}}{b^{n-1}-c^{n-1}} =\frac{2bc(b^{n-2}+b^{n-3}c+\cdots)/a+2a^{n-1}-(b+c)(b^{n-2}+b^{n-3}c+\dots)-(b^{n-1}+c^{n-1})}{b^{n-1}-c^{n-1}} $$ Taking the derivative with respect to $a$ yields $$\frac{\partial^2 d}{\partial a^2} = \frac{2(n-1)a^{n-2} - \frac{2bc}{a^2(b^{n-2} \ + \ b^{n-3}c \ + \ \cdots)}}{b^{n-1} - c^{n-1}}$$ which is greater than or equal to zero. Hence the expression above is minimal if we choose $a$ to be minimal, i.e. $a=b$. In this case, we have: $$\frac{\partial d}{\partial a}=-1=-\frac{2a^{n-1}-b^{n-1}-c^{n-1}}{b^{n-1}-c^{n-1}}$$.<|endoftext|> TITLE: A few questions while reading Higher Topos Theory QUESTION [6 upvotes]: I am now reading Higher Topos Theory. I have recently met the following questions that I am not able to figure out and I am here to look for some answer or help. First, in Lemma 2.2.3.6, while proving $(a)\implies (c)$, Lurie constructed a class $\scr U$ of simplicial sets which contains all simplicial sets $A$ satisfying $(c)$. The author claimed that using Lemma 2.2.3.4, it can be shown that the condition $(e)$ of Lemma 2.2.3.5 is satisfied. But I have no idea how Lemma 2.2.3.4 can be used here, nor can I figure out any other proof. How should I use Lemma 2.2.3.4 to prove this? Second, in Corollary 2.2.3.12, Lurie claimed that for the simplicial equivalence $f:X\to Y$ in ${\sf SSet}_{/S}$ constructed there exists a map $g:Y\to X$ in ${\sf SSet}_{/S}$ and a homotopy $h:X\times\Delta[1]\to X$ from $\mathbb1_X$ to $g\circ f$. Such claim is true in a topological category, but in my thoughts two maps $f_1,f_2:A\to B$ in a simplicial category coincides in the homotopy category does not mean that there exists a homotopy $h:A\times\Delta[1]\to B$ connecting $f_1,f_2$. Is my thought incorrect, or the addition assumptions to $X,Y$ ($X\to S$ is a right fibration and $Y$ is contravariantly fibrant) may ensure that such $h$ does exist? REPLY [6 votes]: $\newcommand{\SSet}{\mathsf{SSet}}\DeclareMathOperator{\Map}{Map}$First, let's record the fact that for any $A$ in $\SSet_{/S}$ and any right fibration $p : X \to S$, the simplicial set $\Map_{\SSet_{/S}}(A, X)$ is a Kan complex. This follows from Lemma 2.2.3.4 by applying it to the inclusion $\emptyset \subset S$. As far as I can tell, this is the only way we'll actually use Lemma 2.2.3.4. Here is another lemma, which is an exercise in diagram chasing: suppose given morphisms $A \to B$ and $X \to S$ and equip $B$ with some map $B \to S$. Then the induced map $\Map_{\SSet_{/S}}(B, X) \to \Map_{\SSet_{/S}}(A, X)$ is a pullback of the induced map $X^B \to X^A \times_{S^A} S^B$. When $A \to B$ is a monomorphism and $X \to S$ is a right fibration, the latter map is again a right fibration by Corollary 2.1.2.9, so $\Map_{\SSet_{/S}}(B, X) \to \Map_{\SSet_{/S}}(A, X)$ is also a right fibration. But we know $\Map_{\SSet_{/S}}(A, X)$ is a Kan complex, so by Lemma 2.1.3.3 this map is actually a Kan fibration. For your first question, suppose $A_0 \to A_1 \to A_2 \to \cdots$ is a sequence of monomorphisms between objects belonging to $\mathcal{U}$ and let $A$ be the colimit of the $A_i$ and suppose $A$ is equipped with some map to $S$. Then the $A_i$ inherit maps to $S$ by composition and $A$ is also the colimit of the $A_i$ in $\SSet_{/S}$. Then the mapping space $\Map_{\SSet_{/S}}(A, X)$ is the inverse limit of the spaces $\Map_{\SSet_{/S}}(A_i, X)$. We showed the that all the objects involved are Kan complexes and that the transition maps are fibrations. Therefore taking the inverse limit is a homotopy-invariant thing to do, so the induced map $\Map_{\SSet_{/S}}(A, X) \to \Map_{\SSet_{/S}}(A, Y)$ is a weak homotopy equivalence in the Kan model structure, hence a homotopy equivalence since both objects are Kan complexes. A similar argument applies to condition (iv). For your second question, the existence of a genuine homotopy inverse follows from the fact that $\Map_{\SSet_{/S}}(A, X)$ and $\Map_{\SSet_{/S}}(A, Y)$ are Kan complexes for each $A$. For $X$ this follows from our first observation, and for $Y$ it follows from the fact that $Y$ is assumed to be fibrant in the simplicial model category $\SSet_{/S}$ (and every object is cofibrant).<|endoftext|> TITLE: Examples of smooth manifolds admitting inbetween one and a continuum of complex structures QUESTION [15 upvotes]: For smooth manifolds it is known that they can admit a unique, finitely many, or a continuum of distinct smooth structures (I don't know whether there are any examples admitting precisely a countably infinite number). For complex manifolds there are examples of smooth manifolds admitting a unique complex structure ($\mathbb{CP}^1$) or a continuum (compact Riemann surfaces, K3 surfaces, etc.) Q. Are there examples admitting only finitely many or a countably infinite number? By deformation theory the tangent space to the moduli space of complex structures on $X$ should be given by $H^1(X, TX)$ (at least morally) so it must be necessary for this to vanish for every possible complex structure on $X$ to have any hope. REPLY [17 votes]: There are countably many complex structures on $S^2 \times S^2$ up to isomorphism. Specifically, the even Hirzebruch surfaces $F_{2k}$ are the only options. This is the main result of Qin, Zhenbo, Complex structures on certain differentiable 4-manifolds, Topology 32, No. 3, 551-566 (1993). ZBL0796.57010. The author also proves that the odd Hirzebruch surfaces are the only complex structures on the $4$-fold $\mathbb{CP}^2 \# \overline{\mathbb{CP}^2}$.<|endoftext|> TITLE: Radin generics from iterated ultrapowers QUESTION [5 upvotes]: Let $u$ be a measure sequence derived from some embedding $j:V\rightarrow M$. I heard that from iterating $j$ is possible to define a generic filter for some Radin forcing $\mathbb{R}_w$. Specifically, $\mathbb{R}_w$ is the Radin forcing defined using the measure sequence $w=j_{\alpha}(u)$, where $j_\alpha$ is the $\alpha^{th}$-iterated of $j$. My question is the following: Which are the measures that we use to carry out this process? Are the second coordinates (i.e. $u_\gamma(1)$) of the corresponding measure sequence $u_\gamma$? Are the whole measure sequences $u_\gamma$? In this latter case, how looks like the definition of a ultrapower given by a measure sequence $u_\gamma$? REPLY [4 votes]: See section 5 (called "Generating generic sequences by iterating j") of Radin's paper Adding closed cofinal sequences to large cardinals. More details: The iteration is the usual one, using $j$ (see my comment below). Now the point is that is $u$ is the measure sequence in $V$, then there is $\sigma: ORD \to V$ so that $\sigma(\alpha)=j_{0, \alpha}(u)\restriction \beta_\alpha,$ for some suitable $\beta_\alpha$ and such that for any $\alpha, \sigma \restriction \alpha+1$ is $\mathbb{R}_{\sigma(\alpha)}$-generic over $M_\alpha$.<|endoftext|> TITLE: Classification of the Extraspecial 2-groups $H_n$ QUESTION [7 upvotes]: I have a sequence of groups $H_n$ which I know to be extraspecial 2-groups of order $2^{2n+1}$. I also know the number of order 4 elements I have in $H_n$ for every $n$. Precisely, the number of order 4 elements is given by $2\sum_{i=0}^{4i \leq 2n-1} {2n+1 \choose 2+4i}$. Is there a slick way of determining which extraspecial 2-group I have in general. For $H_1$, the explicit paper calculations gave me $Q_8$ and for $H_2$ I got $Q_8*D_8$ (here $*$ denotes the $\mathbb{Z}_2$ central product). It appears knowing the number of order 4 elements should be sufficient to decide which one I have, yet I do not know a nice way of computing the number of order 4 elements for arbitrary $n$ copies of central products of Quaternion or Dihedral groups. REPLY [10 votes]: This may be answered as follows: any extra-special $2$ group of order $2^{2n+1}$ is either the central product of $n$-copies of $D_{8}$ or else is the central product of $n-1$ copies of $D_{8}$ with one copy of $Q_{8}.$ The first type has all its complex irreducible representations realizable over $\mathbb{R}$, while the second type has $2^{2n}$ linear characters and one irreducible representation of degree $2^{n}$ which has Frobenius-Schur indicator $-1$ (and which is not realizable over $\mathbb{R}$). The number of solutions of $x^{2} = 1_{G}$ in $G$ of the first type is $\sum_{ \chi \in {\rm Irr}(G)} \nu(\chi)\chi(1) = 2^{2n}+2^{n},$ so a group of the first type has $2^{2n}-2^{n}$ elements of order $4$, while the number of solutions of $x^{2} = 1_{G}$ in $G$ of the second type is $\sum_{ \chi \in {\rm Irr}(G)} \nu(\chi)\chi(1) = 2^{2n}-2^{n},$ so a group of the second type has $2^{2n}+2^{n}$ elements of order $4.$<|endoftext|> TITLE: Do matrices with only elements along the main and anti-diagonals have a name? QUESTION [5 upvotes]: To expand upon the title, I am wondering if there is a specific name for square matrices of the form: $$M = \begin{bmatrix} a_{11} & 0 & \cdots & 0 & \cdots & & 0 & b_{1n} \\ 0 & a_{22} & 0 & \cdots & & b_{2 \ i} & & 0 \\ 0 & \ddots & \ddots & & \cdot^{\textstyle \cdot^{\textstyle \cdot}} & 0 & & 0 \\ & & & a_{} & & \vdots & & \\ \vdots & 0 & \cdot^{\textstyle \cdot^{\textstyle \cdot}} & 0 & \ddots & 0 & & \vdots\\ & \cdots & b_{i \ n-(i+1)} & \cdots & & a_{i \ i} &\\ 0 & \cdot^{{\textstyle \cdot}^{\textstyle \cdot}} & & & & \ddots & \ddots & 0\\ b_{n 1} & 0 & \cdots & 0 & & \cdots & & a_{nn} \end{bmatrix} \\$$ where, $ M = A_{\text{diagonal}} + B_{\text{anti-diagonal}}$ Essentially, M is a square matrix which is zero everywhere except the main and anti-diagonals. In other words, it is the sum of diagonal and anti-diagonal square matrices with the same dimension. Apologies for the formatting, the normal LaTeX notation for up-right dots is not supported on this site. REPLY [6 votes]: I am continuing in the answer box, to get this out of the "unanswered" queue. The OP asks "for a more standard terminology that is perhaps present in the literature." The name "X-matrices" or "X-form matrices" has also been used in the published literature, for example, Properties of Central Symmetric X-Form Matrices (2011) and The exponential functions of central-symmetric X-form matrices (2016). The reason, I guess, why this name is not used more extensively, is that it seems more natural to reorder the basis vectors and work with $2\times 2$ block-diagonal matrices.<|endoftext|> TITLE: Gauge invariance of Chern-Simons functional integral for a 3-manifold with boundary QUESTION [5 upvotes]: I am trying to understand how the functional integral for Chern-Simons theory for a possibly non-compact 3-manifold with boundary is made gauge invariant. For a compact 3-manifold, $M$, without boundary, it is well known (see, for example, section 2 of this reference), that for a compact simple Lie group $G$ and trivial principal G-bundle $P\rightarrow M$, one may define the Chern-Simons action \begin{equation} S[A]=\frac{k}{4\pi}\int_M\textrm{Tr}\bigg(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\bigg). \end{equation} Here, the group, $\mathcal{G}$, of gauge transformations of $P$, is isomorphic to the group of smooth maps from $M$ to $G$. Under a gauge transformation $g\in \mathcal{G}$, the action changes by the sum of a boundary term and $\textrm{deg}(g)$, which labels the corresponding component of $\mathcal{G}$. The group, $\pi_0(\mathcal{G})$, of components is isomorphic to the group of homotopy classes of maps from $M$ to $G$, which for simply connected $G$ is isomorphic to $\pi_3(G)$. Since $G$ is simple, $\pi_3(G)\cong\mathbb{Z}$. Thus, upon requiring that $k$ is quantized, we find that the integrand of the functional integral, $e^{iS}$, is invariant under gauge transformations. My question is, how does one extend this to the case where $M$ has a boundary, and is possibly noncompact? The example I have in mind is $M=D\times \mathbb{R}$, where $D$ is the disk. In this paper, it is explained that for gauge invariance, one first chooses one of the boundary components of the connection, $A$, to be zero, and with such a boundary condition the functional integral is invariant only under gauge transformations which are one at the boundary. This requirement is also alluded to below equation 3.18 of this paper by Dijkgraaf and Witten. It is clear to me that the aforementioned boundary term that arises via gauge transformation will vanish via the boundary condition. However, it is not clear to me why we require that $g$ be 1 at the boundary for gauge invariance. Firstly, why do we need to impose another boundary condition on $g$, i.e., in addition to the constraint required to preserve the boundary condition on $A$ under gauge transformations? Secondly, why would another constant value for $g$ at the boundary not suffice? I would think that any common value for $g$ along the boundary would imply that we are effectively studying $M$ with boundary points identified, which is a closed 3-manifold, for which we can apply the arguments of the second paragraph above. REPLY [5 votes]: It's not necessary to consider only gauge transformations that are constant on the boundary. However, you won't get that the CS invariant is well-defined, even modulo integers (which is what I understand the word quantized to signify). The usual proof of this invariance shows that the change in CS under a gauge transformation is the integral of a certain closed 3-form over the 3-manifold. With the correct normalization, this is an integral for closed 3-manifolds, but there is no reason for it to be so when the boundary is non-empty. The fix for this issue was sorted out in the late 1980s. See Ramadas, T., Singer, I. and Weitsman, J., Some comments on Chern-Simons gauge theory, Comm. Math. Phys. 126 (1989), 409-420. The CS invariant becomes a section of a certain U(1) bundle over the moduli space of connections mod gauge on the boundary. This makes precise the observation that the failure for the CS invariant to change by an integer just depends on the restriction of the gauge transformation to the boundary. As far as my limited understanding goes, this is viewed as the right notion of quantized in this situation, and there is quite a bit of literature on the subject. A nice treatment (maybe helpful for mathematicians) of the geometry is in C. Herald, Legendrian Cobordism and Chern-Simons Theory on 3-Manifolds with Boundary, Comm. Anal. Geom. 2, Number 3, 337-413, 1994.<|endoftext|> TITLE: A category-like structure without composition? QUESTION [9 upvotes]: Is there a name for the 'category-like' structure which satisfies the axioms for a category except for composition, i.e. identities exist for every object, if $f\in Hom(A,B)$ and $g \in Hom(B,C)$ then $g\circ f$ may not exist in $Hom(A,C)$, but when the relevant compositions do exist, then composition is associative. 'Category-like' structures derived from directed graphs with at most one edge in each direction, where the vertices are the objects and the edges are the morphisms, provide plentiful examples, as do (equivalently) not-necessarily-transitive relations on a set $X$. Could anyone provide references which discuss this from a categorical perspective? Thanks in advance! REPLY [12 votes]: As Qiaochu says, one way to talk about categories with partially defined composition is to talk about categories enriched over the monoidal category $Par$ of sets and partial functions with the cartesian product (that is, the cartesian product in $Set$, which is not the cartesian product in $Par$). Since $Par$ is equivalent to the category of pointed sets with its monoidal smash product, where the basepoint in a pointed set is a formal way to represent "not defined", it is equivalent to talk about categories enriched over the latter. A different notion of "category with partially defined composition" is called a paracategory. This has $n$-ary partial composition functions for all $n$, which are associative insofar as defined in an "unbiased" way. It was apparently defined by Peter Freyd in unpublished work, and studied further by Hermida and Mateus; see the references at the link. REPLY [4 votes]: Jørgen Ellegaard Andersen calls this a "categroid". I'm not particularly fond of that term.<|endoftext|> TITLE: What does Rosenlicht mean by a "point"? By $k(v_1,v_2)$? QUESTION [5 upvotes]: This is cross-posted from Math.SE at the recommendation of a commenter. I'm reading M. Rosenlicht's 1956 paper, "Some Basic Theorems on Algebraic Groups" [link], and having trouble with some of the language and notation. In particular, I have two questions: 1) Rosenlicht defines an algebraic group as a variety with a group structure on the points. He pointedly does not assume the ground field $k$ is algebraically closed. As he was writing prior to Grothendieck's development of scheme theory, I am not sure how to interpret "points". In this context, does he mean the $k$-points? The $\overline k$-points i.e. the geometric points? Or what? 2) With $G$ an algebraic group, and $V$ a variety with an action of $G$, all defined over a field $k$, Rosenlicht frequently writes expressions like "$k(g,v)$" with $g\in G, v\in V$; for example, on p. 403, we have We say $G$ operates on $V$ (or that $V$ is a pre-transformation space for $G$) if for each component $G_\alpha$ of $G$ we are given a rational map $g\times v\to g(v)$ of $G_\alpha\times V\rightarrow V$ such that if $k$ is a field of definition for $G$, $V$, and each of these rational maps and if $g_1\times g_2 \times v$ is a generic point over $k$ of $G_\alpha \times G_\beta \times V$ ($G_\alpha$, $G_\beta$ being any components of $V$) then $$(1)\:\:\:\: g_1(g_2(v)) = g_1g_2(v).$$ $$(2)\:\:\:\: k(g_1,g_1(v)) = k(g_1,v).$$ What's meant by $k(g_1,v)$ here? I want it to be a residue field but it has 2 arguments. Is it the composite of the two residue fields? (In which case, the answer to question 1 cannot be "$k$-points"?) Aside: I am particularly interested in Theorem 2 on p. 407 ("Rosenlicht's theorem"). However, my goal is to be able to read Rosenlicht's paper, so I would prefer a gloss of his usages over a reference to a more modern statement and proof. Addendum: The accepted answer contains a link to another question/answer which links to this very helpful piece by Raynaud in the Sept. 1999 Notices, discussing Weil's foundational approach (which turns out to be the context for Rosenlicht's paper). I include the link here for the sake of self-containedness of the question/answer pair. REPLY [11 votes]: Weil's foundations (which Rosenlicht uses) use a universal domain, which is an algebraically closed extension of $k$ of infinite transcendence degree. There is a discussion of universal domains here. In this language, "point" means a point with coordinates in a universal domain. The expression $k(g,v)$ means the subfield of the universal domain generated over $k$ by the coordinates of the points $g$ and $v$.<|endoftext|> TITLE: Is there a proof of quadratic reciprocity using $p$-adic numbers? QUESTION [11 upvotes]: I asked same question on MSE before, but I didn't get any answer yet. I know that the quadratic reciprocity can be regarded as a special case of Artin reciprocity (class field theory), and we can get it by considering the cyclotomic extension of $\mathbb{Q}_{p}$. However, I want to know if there's any proof of quadratic reciprocity that doesn't use any stronger result, but only uses some properties of $p$-adic numbers. Thanks in advance. More precisely, we can do analysis on $\mathbb{Q}_{p}$. We have an exponential and logarithm function on $\mathbb{Q}_{p}$ (at least for $p>2$), and we understand unit group $\mathbb{Z}_{p}^{\times}$ well, etc. But I don't know how to prove it in a purely local and analytic way. REPLY [11 votes]: The quadratic reciprocity law is a special case of the product formula for Hilbert's symbol: for all $(a,b)\in\mathbb{Q}^{\times}$ \begin{equation} \underset{v}{\prod}(a,b)_{v}=1 \end{equation} where $v$ ranges in the product other all prime numbers and $\infty$. The definition of the symbol $(a,b)_{p}$ can be given in purely $p$-adic terms, but of the course the proof of the product formula must involve global arguments, so is global in nature (as any proof of the quadratic reciprocity law must be at least in the sense that the statement of this theorem involves two primes). When done through the structure of the Brauer group and cyclic algebras, the proof of the product formula is however minimally global in some sense. Is that a satisfying answer for you?<|endoftext|> TITLE: A reference to a well-known characterization of scattered compact spaces QUESTION [8 upvotes]: It is well-known that a compact Hausdorff $X$ space is scattered if and only if admits no continuous maps onto the unit interval $[0,1]$. Surprisingly, but I cannot find a good reference to this well-known fact (desirably some textbook). In the survey paper "Scattered spaces" in Encyclopedia of General Topology this fact is not mentioned, unfortunately. REPLY [13 votes]: The proof in the direction that there is no continuous surjection from a compact scattered $X$ onto $[0,1]$ may be found here (Theorem 1): W. Rudin, Continuous functions on compact spaces without perfect subsets, Proc. Amer. Math. Soc. 8 (1957), 39-42. And the full characterisation may be found in (Theorem 8.5.4, p. 148): Z. Semadeni, Banach spaces of continuous functions, vol. 1, PWN - Polish Scientific Publishers, Warsaw 1971.<|endoftext|> TITLE: Graded commutativity of the $n$th Browder bracket QUESTION [5 upvotes]: Let $\mathcal{O}$ be a topological operad and $X$ an algebra over $\mathcal{O}$. Then $H_*(X)$ is an algebra (in the category of $\mathbb{Z}$-graded $R$-modules) over $H_*(\mathcal{O})$. Each $e\in H_n(\mathcal{O}(2))$ gives us a graded product $$H_p(X)\otimes H_q(X)\to H_{p+q+n}(X),a\otimes b\mapsto e(a\otimes b).$$ Now $\mathfrak{S}_2$ acts on $H_*(\mathcal{O}(2))$ on the right and on $H_*(X)^{\otimes 2}$ by $\tau_*(a\otimes b) = (-1)^{ab}(b\otimes a)$. By equivariance, we get $$e(b\otimes a) = (-1)^{ab} (\tau^*e)(a\otimes b).$$ Now consider the special operad $\mathcal{C}_{n+1}$ of little $(n+1)$-cubes. Here, $\mathcal{C}_{n+1}(2)\simeq \mathbb{S}^n$, so its $n$th homology is generated by the fundamental class $s$ of the submanifold $$\mathbb{S}^n\to \mathrm{Conf}_2(\mathbb{R}^{n+1}), x\mapsto (0,x).$$ Now it is clear that $\tau^*s=(-1)^{n+1}s$ as $x\mapsto (x,0)$ is homotopic to $x\mapsto (0,-x)$. Cohen defines the Browder bracket by $[a,b]:=s(a\otimes b)$. By the above argumentation, we would get $$[b,a]=s(b\otimes a)=(-1)^{ab}(\tau^*s)(a\otimes b)= (-1)^{ab+n+1}[a,b].$$ However, in both Cohen’s paper The homology of $\mathcal{C}_{n+1}$-spaces (1976) and in May’s Operads, algebras and modules, the sign is different, namely $$[b,a] = (-1)^{ab+1+n(a+b+1)}[a,b].$$ It is clear that they don’t coincide for $|a|+|b|$ odd. What am I missing? REPLY [3 votes]: Both papers choose a normalization that is different from yours: they define $$ [a,b] = (-1)^{na+1} s(a \otimes b). $$ This is definition 5.7 in Cohen's paper that you mentioned. The reason for this convention is that it is more consistent with writing the Browder bracket as a binary operation, rather than a function applying on the left: we should think of this as the image of $a \otimes s \otimes b$ under the composite which first twists the two factors and then acts. If you don't do this, then you start to get into sign mishaps when you verify things like the Jacobi identity because the triple bracket $[a,[b,c]]$ involves $s (a \otimes s(b\otimes c))$ and introduces an unexpected sign when moving $s$ across $a$. Similar remarks apply to the derivation property for the bracket acting on products.. See also this paper of Xianglong Ni.<|endoftext|> TITLE: Homology of the free loop space of a Grassmanian QUESTION [10 upvotes]: Is there any reference for calculation of the rational homology of the free loop space $H_*(\mathcal{L}Gr(k,n),\mathbb{Q})$ of a complex Grassmanian? More precisely, I am interested in computing ranks $$r_i=rk(H_i(\mathcal{L}Gr(2,4),\mathbb{Q}))$$ (Here $Gr(k,n)$ stands for k-planes in n-space). REPLY [17 votes]: The complex Grassmannian $Gr(2,4)$ can be realized up to homotopy as the homotopy fiber of the map $BU(2) \times BU(2) \rightarrow BU(4)$ which corresponds to the Whitney sum of two complex rank 2 bundles. To obtain the rational cohomology of it (and then of its free loop space), we'll calculate its minimal model as the cofiber of the map this topological map induces on minimal models. This lengthy and perhaps unnecessarily detailed calculation ends with "end of calculation" below; so skip there to just get the minimal model. The minimal model (in the sense of rational homotopy theory) of $BU(2)$ is given by $\Lambda(a_1,a_2)$, and so the model of $BU(2)\times BU(2)$ is given by $\Lambda(a_1,a_2)\otimes \Lambda(b_1,b_2) = \Lambda(a_1,a_2,b_1,b_2)$. Let us denote the minimal model of $BU(4)$ by $\Lambda(c_1,c_2,c_3,c_4).$ The differential is zero in all of these algebras, and the variables $a_i,b_i,c_i$ are of degree equal to twice their index (they are universal Chern classes). The notation $\Lambda$ is used for the free exterior (graded-commutative) algebra on a given set of generators in prescribed degrees. On the level of cohomology (or equivalently, minimal models, since everything is formal), the map $BU(2)\times BU(2) \rightarrow BU(4)$ is given by $H^*(BU(4)) \rightarrow H^*(BU(2))\otimes H^*(BU(2))$, that is, $$\Lambda(c_1,c_2,c_3,c_4) \rightarrow \Lambda(a_1,a_2,b_1,b_2).$$ To figure out the image of $c_i$, observe that if a complex rank 4 bundle $E$ splits into the sum of two complex rank 2 bundles, $C = A\oplus B$, then on Chern classes we have $$1+c_1(C) + c_2(C) + c_3(C) + c_4(C) = (1+c_1(A)+c_2(A))(1+c_1(B)+c_2(B)).$$ Denoting by $a_i,b_i,c_i$ the $i$th Chern class of $A$,$B$,or $C$ respectively, we obtain the equations \begin{align*} c_1 &= a_1+b_1, \\ c_2 &= a_2+b_2+a_1b_1, \\ c_3 &= a_1b_2 + a_2b_1, \\ c_4 &= a_2b_2. \end{align*} These relations tell us how the map of minimal models $\Lambda(c_1,c_2,c_3,c_4) \rightarrow \Lambda(a_1,a_2,b_1,b_2)$ is prescribed. Now we turn this map $\Lambda(c_1,c_2,c_3,c_4) \rightarrow \Lambda(a_1,a_2,b_1,b_2)$ into a fibration. That is, we will find a differential graded algebra $E$ containing $\Lambda(c_1,c_2,c_3,c_4)$ with a quasi-isomorphism to $\Lambda(a_1,a_2,b_1,b_2)$ such that the map $\Lambda(c_1,c_2,c_3,c_4) \to \Lambda(a_1,a_2,b_1,b_2)$ factors into the inclusion followed by the quasi-isomorphism. This should be thought of as dual to the mapping path space construction, turning any map $X\to Y$ into a fibration. Just as the homotopy fiber is the fiber of the map from the mapping path space to $Y$, here the algebraic model of the homotopy fiber is the fiber of the inclusion $\Lambda(c_1,c_2,c_3,c_4) \hookrightarrow E$, which is defined to be $E/\textrm{ideal}(c_1,c_2,c_3,c_4)$. Now we have to construct this $E$ to contain $\Lambda(c_1,c_2,c_3,c_4)$ and to be quasi-isomorphic via some map $f$ to $\Lambda(a_1,a_2,b_1,b_2)$. In order to do this, prescribe $f(c_i)$ to be the corresponding polynomial in $a_i,b_i$ given above. Introduce a variable $\overline{a_i}$ in degree 2 so that $f(\overline{a_1}) = a_1$. Note that now we have $a_1$ and $b_1$ in the image of $f$, since we have $a_1+b_1$ thanks to $c_1$ and $a_1$ is the image of $\overline{a_1}$. Next, introduce a variable $\overline{a_2}$ such that $f(\overline{a_2}) = a_2$. This gets us $a_1,b_1,a_2,b_2$ in the image of $f$, due to $c_1$,$c_2$,$\overline{a_1}$, $\overline{a_2}$. So, we have made $f$ surjective. Now we look at its kernel, and introduce variables to kill these elements in cohomology. Since $c_3$ has to be mapped to $a_1b_2 + a_2b_1$, we see that we necessarily have $$f(c_3 - \overline{a_1}c_2 + 2\overline{a_1}\overline{a_2} + \overline{a_1}^2c_1 - \overline{a_1}^3 + \overline{a_2}c_1) = 0.$$ Similarly, since $f(c_4) = a_2b_2$, we get that $$f(c_4 - \overline{a_2}c_2 + \overline{a_2}^2 + \overline{a_1}\overline{a_2}c_1 - \overline{a_1}^2\overline{a_2}) = 0.$$ To get rid of this non-injectivity on cohomology, we introduce variables $\eta_5$ and $\eta_7$ in $E$ (in degrees 5 and 7, respectively) with the prescription that \begin{align*} d\eta_5 &= c_3 - \overline{a_1}c_2 + 2\overline{a_1}\overline{a_2} + \overline{a_1}^2c_1 - \overline{a_1}^3 + \overline{a_2}c_1, \\ d\eta_7 &= c_4 - \overline{a_2}c_2 + \overline{a_2}^2 + \overline{a_1}\overline{a_2}c_1 - \overline{a_1}^2\overline{a_2}. \end{align*} Setting $f(\eta_5) = f(\eta_7) = 0$, we have made $f$ injective as well on cohomology, and thus a quasi-isomorphism. So, our $E$ is given by $\Lambda(c_1,c_2,c_3,c_4, \overline{a_1}, \overline{a_2} , \eta_5, \eta_7)$ with the first six variables closed, and the differential on $\eta_5$ and $\eta_7$ given as above. Now we take the cofiber of this map, that is, we take $\Lambda(c_1,c_2,c_3,c_4, \overline{a_1}, \overline{a_2} , \eta_5, \eta_7)$ mod the elements in the differential ideal generated by the $c_i$. Our cofiber is given by $\Lambda(\overline{a_1}, \overline{a_2}, \eta_5, \eta_7)$, where $d\overline{a_1} = d\overline{a_2} = 0$ and \begin{align*}d\eta_5 &= 2\overline{a_1}\overline{a_2} - \overline{a_1}^3, \\ d\eta_7 &= \overline{a_2}^2 - \overline{a_1}^2\overline{a_2}.\end{align*} For notational convenience in what follows, let us rename the variables to $c_1 = \overline{a_1}, c_2 = \overline{a_2}, u = \eta_5, v = \eta_7$. END OF CALCULATION So we have that the minimal model of $Gr(2,4)$ is given by $$\Lambda(c_1,c_2, u,v), \textrm{ where }dc_1 = dc_2 = 0, du = -c_1(c_1^2 - 2c_2), dv = c_2^2 - c_1^2c_2.$$ As for the free loop space, we use the first algebraic construction in Section 11 of Sullivan's Infinitesimal Computation in Topology. To form the minimal model of $LGr(2,4)$, we take the minimal model of $Gr(2,4)$, introduce a new copy of all the generators but with degree shifted down by one, define a derivation $i$ on this new free algebra given by declaring it sends a generator from the original minimal model to its shifted-down-by-one counterpart, and sends these new shifted generators to 0 (and extending so that it is a derivation). The differential on this new larger algebra extends the old differential and satisfies $di+id=0$. Concretely, in our case, the minimal model of $LGr(2,4)$ will have underlying algebra $$\Lambda(c_1,c_2, u,v, ic_1, ic_2, iu, iv),$$ where $deg(ic_1) = 1, deg(ic_2) = 3, deg(iu) = 4, deg(iv) = 6$. The condition $di+id = 0$ gives us $d(ic_1) = 0$ and $d(ic_2) = 0$, along with $d(iu) = (ic_1)\cdot (c_1^2-2c_2) + c_1 \cdot (2c_1\cdot ic_1 - 2ic_2)$ and $d(iv) = -2c_2 \cdot ic_2 + 2c_1 \cdot ic_1 \cdot c_2 + c_1^2 \cdot ic_2$. From here we can now calculate the ranks of the rational (co)homology groups, certainly with a computer algebra system, and maybe with a clever closed-form expression (though I don't know it, sorry). The first few ranks are $r_1 = 1, r_2 = 1, r_3 = 2, r_4 = 3, r_5 = 2$. By a theorem of Sullivan and Vigué-Poirrier, this sequence of ranks will be unbounded. A similar calculation will give you the rational homology of the free loop space of other Grassmannians; just consider $BU(k) \times BU(n-k) \rightarrow BU(n)$ instead.<|endoftext|> TITLE: Theorems that impeded progress QUESTION [82 upvotes]: It may be that certain theorems, when proved true, counterintuitively retard progress in certain domains. Lloyd Trefethen provides two examples: Faber's Theorem on polynomial interpolation: Interpreted as saying that polynomial interpolants are useless, but they are quite useful if the function is Lipschitz-continuous. Squire's Theorem on hydrodynamic instability: Applies in the limit $t \to \infty$ but (nondimensional) $t$ is rarely more than $100$. Trefethen, Lloyd N. "Inverse Yogiisms." Notices of the American Mathematical Society 63, no. 11 (2016). Also: The Best Writing on Mathematics 2017 6 (2017): 28. Google books link. In my own experience, I have witnessed the several negative-result theorems proved in Marvin Minsky and Seymour A. Papert. Perceptrons: An Introduction to Computational Geometry, 1969. MIT Press. impede progress in neural-net research for more than a decade.1 Q. What are other examples of theorems whose (correct) proofs (possibly temporarily) suppressed research advancement in mathematical subfields? 1 Olazaran, Mikel. "A sociological study of the official history of the perceptrons controversy." Social Studies of Science 26, no. 3 (1996): 611-659. Abstract: "[...]I devote particular attention to the proofs and arguments of Minsky and Papert, which were interpreted as showing that further progress in neural nets was not possible, and that this approach to AI had to be abandoned.[...]" RG link. REPLY [5 votes]: The Mermin-Wagner theorem (1966) proved that for two-dimensional models with a continuous symmetry there was no finite temperature transition to a phase with long-range order via spontaneous breaking of this symmetry. My understanding is that this was generally interpreted as proving that "two-dimensional models with continuous symmetries do not have phase transitions". I do not know for sure that this impeded progress, but I suspect that as it appeared to be apparently a no-go theorem this would have prevented researchers from looking for examples of phase transitions. Remarkably, not too long after the theorem was published, Kosterlitz and Thouless (1973) showed that the two-dimensional XY model does indeed have a phase transition with diverging correlation length, but no long-range order. The mechanism was completely different (due to topological effects), and so managed to avoid the apparently hard constraints imposed by the theorem.<|endoftext|> TITLE: Number of ultrafilters in an extender QUESTION [5 upvotes]: Assume GCH. Suppose $j: V\to M$ is an elementary embedding such that $\mathrm{crit}(j)=\kappa$, ${}^\kappa M \subset M$, $M\supset V_{\kappa+2}$. We can assume $j$ is defined from some extender of length $\kappa^{++}$. Hence, in particular, we know $\kappa^{++} TITLE: Infinitely many initial ideals for non-Artinian monomial orders? QUESTION [5 upvotes]: Consider the polynomial ring $R=\mathbb Z[x_1,\ldots,x_n]$ and an ideal $I\subset R$. Let $<$ be a monomial order, i.e. a total order on the set of monomials in $R$ such that for any monomials $a$, $b$ and $c$ we have $ab TITLE: Hyperbolic PDE in mathematics QUESTION [8 upvotes]: Hyperbolic PDE (like the wave equation) are roughly speaking, PDE that satisfy the “finite propagation speed of information” property. They are ubiquitous in mathematical physics (essentially, most fundamental laws of nature are hyperbolic). However, do hyperbolic PDE occur in any other areas of mathematics that do not have ties to the real world ? REPLY [6 votes]: Hyperbolic PDEs arise unexpectedly in some differential geometric questions involving prescribed data. What's weird in these cases is that there is no natural time coordinate in the PDEs. Here are some examples: Bryant; Griffiths; Yang. Characteristics and existence of isometric embeddings. Duke Math. J. 50 (1983), no. 4, 893–994. DeTurck, Yang. Existence of elastic deformations with prescribed principal strains and triply orthogonal systems. Duke Math. J. 51 (1984), no. 2, 243–260. As an aside, the triply orthogonal system result implies the local existence of coordinates on a Riemannian 3-manifold for which the metric tensor is diagonal. This generalizes isothermal coordinates on a Riemannian 2-manifolds.<|endoftext|> TITLE: Disjoint paths between four vertices QUESTION [6 upvotes]: Consider the following property of an undirected graph: For any four distinct vertices $a,b,c,d$, there is a path from $a$ to $b$ and a path from $c$ to $d$ such that the two paths do not share any vertex. Is there a name for this property, or has it been studied? It looks related to vertex connectivity but not quite the same. Also it seems that any graph satisfying this property must have vertex connectivity at least $2$. REPLY [8 votes]: The property that you are describing is called $2$-linked. More generally, we say that a graph is $k$-linked if it has at least $2k$ vertices and for all distinct vertices $s_1, \dots s_k, t_1, \dots, t_k$ there are $k$-vertex disjoint paths connecting $s_i$ to $t_i$ for all $i \in [k]$. Note that every $k$-linked graph is $k$-connected. The converse is not true. For example, a cycle is $2$-connected but is not $2$-linked. However, it is a classic theorem that there is a function $f(k)$ such that every $f(k)$-connected graph is $k$-linked. The current best bound for $f(k)$ is due to Thomas and Wollan, where they prove that every $10k$-connected graph is $k$-linked. The paper is titled An improved linear edge bound for graph linkages, and is available here.<|endoftext|> TITLE: Can the number of solutions to a system of PDEs be bounded using the characteristic variety? QUESTION [6 upvotes]: I've recently come across a system of PDEs which I'd like to understand better. The particular system I'm interested in locally solves for a 2-dimensional Riemannian metric as the Hessian of a potential function (which was addressed in a separate question). In that question, Robert Bryant noted that the characteristic variety consists of three points. Concretely, I'm wondering if this means that there are at most three solutions to this system of equations (perhaps modulo some affine transformations). More generally, does the number of points in the characteristic variety bound the number of real analytic solutions for a given system of PDEs? My understanding is that the characteristic variety gives something like the formal power series solutions to the system, so there shouldn't be more real analytic solutions than that. However, my knowledge of $D$-modules is really lacking, so I was wondering if anyone could either correct me on this or else point me to a good reference to learn more. I've been trying to read through Bryant's Exterior Differential Systems, and I apologize if this question is obvious for those who understand the theory. REPLY [6 votes]: The wave equation in the plane is $\partial^2_x-\partial^2_y=(\partial_x+\partial_y)(\partial_x-\partial_y)$, so two points in the characteristic variety, but infinite dimensional family of solutions. Each isolated point in the characteristic variety represents a hypersurface foliation inside every sufficiently smooth solution, by a theorem of Ofer Gabber. For the wave equation, this is the foliation by the two directions spanned by the vector fields $\partial_x+\partial_y, \partial_x-\partial_y$.<|endoftext|> TITLE: Examples of Kreisel-Putnam topological spaces QUESTION [10 upvotes]: Let us say that a topological space $X$ is a Kreisel-Putnam space when it satisfies the following property: For all open sets $V_1, V_2$ and regular open set $W$ of $X$, if a point $x\in X$ has a neighborhood $N$ such that $N \cap W \subseteq V_1 \cup V_2$ then in fact it has a neighborhood $N'$ such that either $N' \cap W \subseteq V_1$ or $N' \cap W \subseteq V_2$. (Equivalently, $\operatorname{int}(V_1 \cup V_2 \cup (X\setminus W))$ is contained in the union of $\operatorname{int}(V_1 \cup (X\setminus W))$ and $\operatorname{int}(V_2 \cup (X\setminus W))$.) (There may be clearer ways to phrase this. Perhaps passing to the closed complements is more palatable.) The reason for this name and condition is that the above is equivalent to saying that the Heyting algebra of open sets of $X$ satisfies the Kreisel-Putnam axiom $(\neg u \Rightarrow (v_1\lor v_2)) \Rightarrow ((\neg u \Rightarrow v_1) \lor (\neg u \Rightarrow v_2))$ of interest in the study of intermediate logics. But of course, is this property has a different, more classical name, this is part of my question. A counterexample is provided by $\mathbb{R}^2$: this does not satisfy the Kreisel-Putnam property, as shown by taking $V_1 = \{x_1 > 0\}$ and $V_2 = \{x_2 > 0\}$ and $W = V_1 \cup V_2$, which is indeed regular open, and the point $x = (0,0)$. (A slighly more complicated counterexample works for $\mathbb{R}$.) Note the requirement that $W$ be regular open (or equivalently, be the pseudocomplement $\operatorname{int}(X\setminus U)$ of an open set $U$). If we drop this requirement, we get a (presumably!) stronger condition on $X$ which I might call a Gödel-Dummett space, because its Heyting algebra of open sets satisfies the axiom $(w \Rightarrow (v_1\lor v_2)) \Rightarrow ((w \Rightarrow v_1) \lor (w \Rightarrow v_2))$, which turns out to be equivalent to $(v_1 \Rightarrow v_2) \lor (v_2 \Rightarrow v_1)$, the Gödel-Dummett axiom. As I find this property a little hard to visualize, I would like to ask: Question: What are some interesting examples of Kreisel-Putnam spaces? I don't have a precise definition of “interesting”, of course (I am trying to gain an intuitive grasp on the notion), but for example, discrete spaces (which are indeed Kreisel-Putnam) are definitely not interesting. Ideally, I would like something which is “somewhat like $\mathbb{R}^n$”, but some criteria which would help make a space interesting might be: being regular, connected (or at the very least, not extremally disconnected) and not satisfying the Gödel-Dummett condition. It is probably worth pointing out that, as shown in this answer, the topos of simplicial sets satisfies the Kreisel-Putnam axiom. The corresponding condition for the topos of sheaves of sets on $X$ would be that every open set in $X$ satisfies the Kreisel-Putnam condition (maybe this is follows from merely $X$ satisfying it, this is one of the many things unclear to me). REPLY [6 votes]: Some observations (hopefully correct): It is enough to consider $x ∈ ∂W$ and $V_1, V_2 ⊆ W$. So for fixed $x, W$ we want that $\mathcal{N}_{x, W} := \{N ∩ W: N$ open neighborhood of $x\}$ is a prime filter (in the distributive lattice of open subsets of $W$). Extremally disconnected spaces are exactly those where every regular open set is clopen, so such spaces are Kreisel–Putnam simply because of lack of suitable regular open sets. (That is probably why you are not interested in such examples). The space $\{0, 1, 2\}$ with the topology generated by $\{1\}, \{2\}$ is a simple example of a connected Kreisel–Putnam space that is neither Gödel–Dummet nor extremally disconnected. The space $\{0, 1, 2, 3\}$ with the topology generated by $\{1, 3\}, \{2, 3\}$ is a simple example of a connected extremally disconnected (and so Kreisel–Putnam) space that is not Gödel–Dummet. A $T_1$ connected Gödel–Dummet space has no cut-points, i.e. $X \setminus \{x\}$ is connected for every $x$. Proof: If $X \setminus \{x\} = V_1 ∪ V_2$ for some disjoint nonempty open sets, we put $W := V_1 ∪ V_2$ and have a contradiction since $x ∈ \overline{V_1} ∩ \overline{V_2}$. No Hausdorff Kreisel–Putnam space contains a nontrivial convergent sequence, so it is quite far from being “somewhat like $ℝ^n$”. Proof: Let $x_n \to x$ be a nontrivial convergent sequence. There are pairwise disjoint respective open neighborhoods $U_n$ not nontaining $x$. Now $W := \operatorname{int} \overline{⋃_{n ∈ ω} U_{2n}}$ is a regular open set with $x ∈ ∂W$ and $V_1 := W \setminus \{x_{4n}: n ∈ ω\}$ and $V_2 := W \setminus \{x_{4n + 2}: n ∈ ω\}$ are its open subsets such that $W = V_1 ∪ V_2$, yet $x$ is in the closure of their complements. It seems that a maximal connected expansion of the real line is a Hausdorff connected Kreisel–Putnam space that is not Gödel–Dummet. So let $X$ be such expansion, which exists by [1], [2]. For every $x$, $\mathcal{N}_{x, (x, ∞)}$ is a maximal filter of open subsets of $(x, ∞)$. Hence, for every open $W ⊆ (x, ∞)$ such that $x ∈ ∂W$ we have $V ∈ \mathcal{N}_{x, W}$ iff $x ∈ \overline{V}$. The same holds for $(-∞, x)$. So $X$ is even Gödel–Dummet at pairs $x ∈ ∂W$ such that $W$ is only on one side. On the other hand, $X$ is not really Gödel–Dummet by 5. since every point of $X$ is a cut-point: $X = (-∞, x) ∪ \{x\} ∪ (x, ∞)$. By an almost clopen set at $x$ we mean an open set $U ≠ ∅$ such that $U ∪ \{x\}$ is closed. Note that in our $X$ there are only few almost clopen sets. If $U ⊆ X$ is almost clopen at $x$, then $U ∩ (x, ∞)$ is clopen in $(x, ∞)$. But $(x, ∞)$ is connected in $X$ since every connected expansion of $ℝ$ has the same connected sets as $ℝ$ [3]. So $U ∩ (x, ∞) ∈ \{∅, (x, ∞)\}$. The same holds for $U ∩ (-∞, x)$. To show that $X$ is Kreisel–Putnam let $x ∈ ∂W$ for some regular open set $W ⊆ X$. Since $X$ is nearly maximal connected [4], there is $U$ a neighborhood of $x$ and an almost clopen set $V$ at $x$ such that $U ∩ V ∩ W = ∅$. Since by 9. we have $V = (-∞, x)$ or $(x, ∞)$, $W$ is essentially only on one side of $x$, and so we may apply 7. Therefore, being Kreisel–Putnam look like quite exotic property, which may be connected with the notions of maximal connectedness and near maximal connectedness. Existence of a regular maximal connected space is an open problem. Recently [5], it was shown that there is no regular maximal connected expansion of $ℝ$. [1] Simon, Petr, An example of maximal connected Hausdorff space, Fundam. Math. 100, 157-163 (1978). ZBL0435.54017. [2] Guthrie, J. A.; Stone, H. E.; Wage, M. L., Maximal connected expansions of the reals, Proc. Am. Math. Soc. 69, 159-165 (1978). ZBL0396.54001. [3] Hildebrand, S. K., A connected topology for the unit interval, Fundam. Math. 61, 133-140 (1967). ZBL0185.26101. [4] Clark, Bradd; Schneider, Victor, A characterization of maximal connected spaces and maximal arcwise connected spaces, Proc. Am. Math. Soc. 104, No. 4, 1256-1260 (1988). ZBL0692.54010. [5] Kalapodi, A.; Tzannes, V., The non-existence of a regular maximal connected expansion of the reals, Topology Appl. 232, 34-38 (2017). ZBL1377.54026.<|endoftext|> TITLE: Homology of spectra vs homology of infinite loop spaces QUESTION [16 upvotes]: Let $X$ be a CW complex and let $\Sigma^\infty X$ denote its suspension spectrum. By definition, the $n$th singular homology group of $\Sigma^\infty X$ with coefficients in $\mathbb{Z}$ is $\pi_n(\Sigma^\infty X \wedge H\mathbb{Z})$. Now, connective spectra are in bijection with infinite loop spaces. The infinite loop space corresponding to $\Sigma^\infty X$ is $QX := \Omega^\infty \Sigma^ \infty X = \lim_{n \to \infty} \Omega^n \Sigma^n X$. Is the homology of $\Sigma^\infty X$ (as a spectrum) isomorphic to the homology of $QX$ (as a CW complex)? I would expect this. But what confuses me is a theorem is Rudyak's On Thom spectra, orientability, and cobordism (Theorem 7.11.(i)). Rudyak states that for every rational spectrum $E_\mathbb{Q}$ the Hurewicz map $\pi_\ast(E_\mathbb{Q}) \rightarrow H_\ast(E_\mathbb{Q})$ is an isomorphism. This seems to prove that the homotopy/homology groups of spectra do not agree with homotopy/homology groups of their corresponding infinite loop spaces. First, $\pi_\ast(Q X) \otimes \mathbb{Q}$ is isomorphic to the stable rational homotopy $\pi_\ast^S(X) \otimes \mathbb{Q}$ of $X$. Then, by the Milnor-Moore theorem, $H_\ast(QX,\mathbb{Q}) \cong \mathbb{Q}[\pi_\ast^S(X) \otimes \mathbb{Q}]$. So we'd have that $H_\ast(X,\mathbb{Q})$ is the free commutative-graded algebra on the rational stable homotopy groups of $X$ for any CW complex $X$. But it is easy to write down examples of CW complexes whose rational homology is not free e.g. projective spaces. So what, then, is the relationship between $\pi_\ast(\Sigma^\infty X \wedge H\mathbb{Z})$ and $H_\ast(QX, \mathbb{Z})$? REPLY [2 votes]: First of all, I think the concern of the OP is the fact that the homology of the infinite loop space doesn't agree with that of the spectra. So let's start with this. Let $Y$ be a spectrum, then it is homology is related to its infinite loop spaces' homology by $$H_*(Y)=\lim _i H_{*+i}(\underline{Y}_{i+j})$$ where $Y_k$ is the $k$-th infinite loop space associated to $Y$. Notably we have $Y_{k-1}=\Omega Y_k$, so it should be more or less clear that the homology of a spectrum can almost never agree with that of the associated infinite loop spaces. Now, let's get to $H_*(QX)$. As has been noted, its rational homology is easy. So basically we need to know $p$-complete homology for all prime numbers $p$ to recover integral homology. In other words, we need to know the Bockstein spectral sequence for all prime $p$. The input of the Bockstein spectral sequence $H_*(QX;Z/p)$ was computed by Kudo-Araki for $p=2$, Dyer-Lashof for odd prime. The rest was worked out by May in Chapter 3, Theorem 3.12 of Frederick R. Cohen, Thomas J. Lada, J. Peter May "The homology of iterated loop space" Springer Lecture Notes in Mathematics 533.<|endoftext|> TITLE: Sobolev functions on $\mathbb{R}^N$ cannot be discontinuous on a $(N-1)$-dimensional submanifold QUESTION [5 upvotes]: How can one prove (or where can I find a proof) that if $u \in W^{1,p}(\Omega)$, where $\Omega \subset \mathbb{R}^N$, then $u$ cannot have a $(N-1)$-manifold of discontinuity points? REPLY [8 votes]: Any function in $W^{1,p}$, $p>N$, has a continuous representative by the Sobolev embedding theorem so there is no issue here. However: Proposition. There is a function $f\in W^{1,p}$, $p\leq N$, that is essentially discontinuous everywhere. In fact you can find a function such that the essential supremum on every open set is $+\infty$ and the essential infimum on every open set is $-\infty$. See Example 2.26 in [2]. For a similar construction see also: A function in $W^{1,p}(\Omega)$ for $1 < p < n$ which is not differentiable a.e. Thus the answer to the question the way it is stated is no, the function can be essentially discontinuous everywhere. There is however, a different point of view which shows that, in fact, a Sobolev function behaves nicely when restricted to an $(N-1)$-dimensional manifold and I will present two different approaches to it. Approach 1. According to Theorem 2 p. 164 in [1] (I am referring to the first edition) any function $f\in W^{1,p}$ has a representative that is absolutely continuous on almost all lines. Here by a representative I mean a Borel function defined everywhere and equal to $f$ almost everywhere. If $M\subset\Omega$ is an $(N-1)$-dimensional manifold, then almost all lines pass through almost all points on $M$ so we can define restriction of $f$ to $M$ by looking at values of $f$ at the points where the lines intersect with $M$. Such a restriction is called a trace. Therefore $f$ may be discontinuous on $M$, but still it behaves nicely on $M$. Approach 2. Any Function in $W^{1,p}$, $p>N$, has a continuous representative by the Sobolev embedding theorem so there is no issue here.If $f\in W^{1,N}$, then $f\in W^{1,p}_{\rm loc}$ for ant $1\leq p0$, there is an open set $V\subset\Omega$ with $\operatorname{Cap}_p(V)<\epsilon$ such that $f|_{\Omega\setminus V}$ is continuous. Here $\operatorname{Cap}_p$ stands for the $p$-capacity. Capacity is a certain outer measure. While I will not recall its definition I will explain how it is related to the Hausdorff measure. The next result is Theorem 4 p. 156 and and Theorem 3 p. 193 in [1]. Theorem 2. If $1\leq pn-p$. Moreover if $A$ is compact, then $\operatorname{Cap}_1(A)=0$ if and only if $\mathcal{H}^{N-1}(A)=0$. Now Theorem 1 says that away of a set of arbitrarily small capacity, $f$ is continuous so the exceptional set has capacity zero and Theorem 2 says that this exceptional set has vanishing $(N-1)$-dimensional measure. Therefore is we consider an $(N-1)$-dimensional manifold $M$ in $\Omega$ this exceptional set has measure zero. [1] L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions. Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992. [2] P. Hajłasz, Non-linear elliptic partial differential equations .<|endoftext|> TITLE: Galois theory of modular functions QUESTION [6 upvotes]: Let $\mathcal M_m$ be the set of $2$-by-$2$ primitive (relatively prime entries) matrices with determinant $m$. Let $\alpha \in \mathcal M_m$ and let $\Gamma\subset \operatorname{SL}_2(\mathbb Z)$. Define $$R_{\alpha,\Gamma}=\lbrace M\in \Gamma\backslash\mathcal M_m : M \equiv \alpha \text{ in } \Gamma\backslash\mathcal M_m/\Gamma\rbrace.$$ Thus $R_{\alpha,\Gamma}$ is a set of representatives $M$ for the left action of $\Gamma$ on $\mathcal M_m$ such that there exist matrices $A,B\in \Gamma$ with the property $M=A\alpha B$. Let $F_{N,\mathbb C}$ be the field of modular functions of level $N.$ Suppose that $f\in F_{N,\mathbb C}$, $\mathbb C(j)\subset \mathbb C(f)$, and $\alpha \in \mathcal M_m$. The Galois group of $F_{N,\mathbb C}$ over $\mathbb C(j)$ is isomorphic to $\operatorname{SL}_2(\mathbb Z/N\mathbb Z)/\{\pm 1\}$. The subgroup corresponding to the subfield $\mathbb C(f)$ is $$\Gamma(f)=\lbrace A\in\operatorname{SL}_2(\mathbb Z/N\mathbb Z): f\circ A = f\rbrace.$$ It is easy to see that $f\circ \alpha \in F_{mN,\mathbb C}$, and that the minimal polynomial of the function $f\circ \alpha$ over $\mathbb C(f)$ is $$\prod_{M\in R_{\alpha, \Gamma(f)}}X-f\circ M.\label{m}\tag{1}$$ Suppose that in addition $f$ has rational Fourier coefficients. Is the polynomial above the also the minimal polynomial of $f\circ\alpha$ over the field $\mathbb Q(f)$? Update The result is true when $\alpha = \begin{pmatrix}m & 0 \\0& 1\end{pmatrix}$. In this case the function $f\circ \alpha$ has rational Fourier coefficients. It is therefore invariant under the action of $(\mathbb Z/mN\mathbb Z)^\times$ on $F_{N}$. To show that the polynomial $\eqref{m}$ in question is also the minimal polynomial of $f\circ\alpha$ over $\mathbb Q(f)$, we must show that the set of roots of $\eqref{m}$ is stable under all automorphisms fixing $\mathbb Q(f)$. Each such automorphism can be represented as $$\begin{pmatrix}d & 0 \\0& 1\end{pmatrix}\gamma,$$ where $\gamma \in \Gamma(f)$ and $d\in (\mathbb Z/mN\mathbb Z)^\times$. This is because $\operatorname{Gal}(F_{mN}/\mathbb Q(j)) \cong \operatorname{PGL}_2(\mathbb Z/ mN\mathbb Z)$. Since $f\circ\alpha(z)=f(mz)=\sum_ka_kq^{mk/N}$ has rational coefficients, the matrix $ \begin{pmatrix}d & 0 \\0& 1\end{pmatrix}$ acts trivially on $f\circ\alpha$. The set $R_{\alpha,\Gamma(f)}$ is stable under the action of $\gamma$ by definition. Update II We prove that the result is true when both $f$ and $f\circ\alpha$ have rational Fourier coefficients. Let $k=mN$. First note that if $d\in(\mathbb Z/k\mathbb Z)^\times$ and $\gamma \in \Gamma(f)$ then any lift to $\operatorname{SL}_2(\mathbb Z)$ of the matrix $$\begin{pmatrix}1 & 0\\0& d\end{pmatrix}\gamma\begin{pmatrix}1 & 0 \\0& d\end{pmatrix}^{-1}$$ also lies in $\Gamma(f)$ because it has determinant equal to $1$ and the automorphism induced by it fixes $f$ (here we use that $f$ has rational coefficients). We know that $F_k=\mathbb Q(j,h^{(r,s)}:(r,s)\in \mathbb Z^2, \not \in k\mathbb Z^2)$, where $$h^{(r,s)}(\tau)=\frac{g_2(\tau)}{g_3(\tau)}\wp_\tau\left(\frac{r\tau+s}{k}\right)$$ are the Fricke functions. They satisfy $h^{(r,s)\gamma}=h^{(r,s)}\circ\gamma$ for $\gamma\in \operatorname{SL}_2(\mathbb Z)$, and $(h^{(r,s)})^{\sigma_d}=h^{(r,sd)}$ where $\sigma_d$ is the automorphism induced by $d\in(\mathbb Z/k\mathbb Z)^\times$. Let $M\in R_{\alpha,\Gamma(f)}$. We must show that for each $d\in(\mathbb Z/k\mathbb Z)^\times$ we have $(f\circ M)^{\sigma_d}=f\circ M'$ for some $M'\in R_{\alpha,\Gamma(f)}$. Write $f\circ \alpha = Q(h^{(r,s)})$ where $Q$ is a rational function with rational coefficients. Here $(r,s)$ ranges over some finite set of representatives. We have $A\alpha B=M$ for some $A,B\in \Gamma(f)$; thus $$f\circ M=f\circ \alpha \circ B = Q(h^{(r,s)B}).$$ Let $D = \begin{pmatrix}1 & 0\\0& d\end{pmatrix}$. Then $$(f\circ M)^{\sigma_d}= Q(h^{(r,s)BD})=Q(h^{(r,s)DB'}),$$ where $B'=D^{-1}BD$ lies in $\Gamma(f)$ by above discussion. Therefore $$(f\circ M)^{\sigma_d}=Q(h^{(r,s)D}\circ B').$$ On the other hand, $(f\circ \alpha)^{\sigma_d}= Q(h^{(r,s)D})$, so $(f\circ M)^{\sigma_d}=(f\circ \alpha)^{\sigma_d}\circ B'$. Now we use the fact that $f\circ\alpha$ has rational Fourier coefficients to deduce that $(f\circ M)^{\sigma_d}=f\circ \alpha\circ B'$. Thus we can take $M' = \alpha B'$. For the Fricke functions see Shimura: An introduction to the theory of automorphic functions, Section 6.2. REPLY [2 votes]: EDIT. The answer is yes as soon as $f$ and $f \circ \alpha$ both have rational Fourier coefficients. To see this, we recall Shimura's theorem: for any modular form $f$ of level $N$, any $g \in \mathrm{SL}_2(\mathbb{Z})$ and any $\sigma \in \mathrm{Aut}(\mathbb{C})$, we have $(f | g)^\sigma = f^\sigma | g_\lambda$ where $g_\lambda \in \mathrm{SL}_2(\mathbb{Z})$ is any lift of the matrix $\begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix}^{-1} g \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$, and $\lambda$ is defined by $\sigma(e^{2\pi i/N})=e^{2\pi i\lambda/N}$. See the comments below for the precise reference. Note that $g \to g_\lambda$ defines an action of $(\mathbb{Z}/N\mathbb{Z})^\times$ on $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$. Shimura's theorem implies immediately the following. Lemma. Let $f$ be a modular function (or modular form) of level $\Gamma(N)$ with rational Fourier coefficients. Then the stabilizer $\Gamma(f)$ of $f$ in $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$ is stable under the action of $(\mathbb{Z}/N\mathbb{Z})^\times$ defined above. Now, let $\alpha \in M_2(\mathbb{Z})$ be any matrix with positive determinant, and assume that $f \circ \alpha$ has rational Fourier coefficients. The function $f \circ \alpha$ is modular of some level $N'$ divisible by $N$. We want to show that the polynomial $P_{f,\alpha}$ defined by (1) has coefficients in $\mathbb{Q}(f)$. For this, it is enough to show that the set of roots of $P_{f,\alpha}$ is stable by $\mathrm{Aut}(\mathbb{C})$. So let $M$ be any matrix in $\Gamma(f) \alpha \Gamma(f)$, which we write $M=\gamma' \alpha \gamma$. Then \begin{equation*} (f|M)^\sigma = (f | \alpha \gamma)^\sigma = (f \circ \alpha)^\sigma | \gamma_{\lambda'} = f | \alpha \gamma_{\lambda'} \end{equation*} for some $\lambda' \in (\mathbb{Z}/N'\mathbb{Z})^\times$. By the lemma above applied in level $N'$, the matrix $\gamma_{\lambda'}$ belongs to $\Gamma(f)$, which finishes the proof. In general however, this is not always true. For example, take a modular function $f$ of level $N$ with rational Fourier coefficients and $\Gamma(f)=\{\pm I\}$. Such an $f$ exists: the modular curve $Y(N)(\mathbb{C}) = \Gamma(N) \backslash \mathcal{H}$ can be defined over $\mathbb{Q}$, so we have an extension of function fields $\mathbb{Q}(Y(N))/\mathbb{Q}(j)$ with Galois group $\mathrm{PSL}_2(\mathbb{Z}/N\mathbb{Z})$, then we can take any $f$ generating this extension. Take $m=1$ and let $\alpha$ be any matrix in $\mathrm{PSL}_2(\mathbb{Z}/N\mathbb{Z})$. Note that $R_{\alpha,\Gamma_f} = \Gamma(N) \alpha$, so we simply have $P_{f,\alpha}(X)=X-f \circ \alpha$. In general $f \circ \alpha$ will not have rational Fourier coefficients. In fact, a theorem of Shimura tells us that for any $\sigma \in \mathrm{Aut}(\mathbb{C})$, we have $(f \circ \alpha)^\sigma = f^\sigma \circ \alpha_\sigma$ where $\alpha_\sigma = \begin{pmatrix} 1 & 0 \\ 0 & \chi(\sigma) \end{pmatrix}^{-1} \alpha \begin{pmatrix} 1 & 0 \\ 0 & \chi(\sigma) \end{pmatrix}$ and $\chi(\sigma) \in (\mathbb{Z}/N\mathbb{Z})^\times$ is the cyclotomic character, defined by $\sigma(e^{2\pi i/N})=e^{2\pi i \chi(\sigma)/N}$. See for example this MO answer. In general however, it seems that you are asking whether the Hecke operator $T_\alpha =\Gamma(f) \alpha \Gamma(f)$ is defined over $\mathbb{Q}$. Maybe there will still be a relation of the form $T_\alpha \circ \sigma = \sigma \circ T_{\alpha'}$ at least when $m$ is prime to $N$ (this is just a guess, I have to check). In any case, this kind of property is more naturally stated using the adelic language: you take a compact open subgroup $K$ of $\mathrm{GL}_2(\hat{\mathbb{Z}})$ and consider the Hecke correspondence $T_\alpha = K\alpha K$ where $\alpha$ is any matrix in $\mathrm{GL}_2(\mathbb{A}_f)$, here $\mathbb{A}_f = \hat{\mathbb{Z}} \otimes \mathbb{Q}$ are the finite adeles of $\mathbb{Q}$. Then the Galois action will essentially correspond to the determinant of $\alpha$ via abelian class field theory.<|endoftext|> TITLE: Ends of finitely generated torsion groups QUESTION [7 upvotes]: It is known that the number of ends of a finitely generated group is 0,1, 2 or $\infty$. Problem 1. What is known about the number of ends of infinite finitely generated torsion groups? In particular, Problem 2. Is there any description of finitely generated torsion groups with 1 end? REPLY [11 votes]: A direct proof. Following Yves' comments above, it is possible to give an easy proof of the fact that an infinitely-ended group must contain an infinite-order element, which implies that it cannot be a torsion group. (It is standard that two-ended groups are virtually infinite cyclic, so I focus on the infinitely-ended case.) Let $B$ be a ball inside your group $G$ such that $G \backslash B$ has $n \geq 3$ unbounded connected components. Let $C_1$ be one of them. Now define $a_1,a_2, \ldots \in G$ and $C_2,C_3,\ldots \subset G$ by induction in the following way: Fix an element $a_i \in G$ such that $a_iB$ disconnects $C_i$ into $n$ unbounded components. Let $C_{i+1}$ be such a component which is disjoint from $C_i$. If for every $i$ you number the components of $G \backslash a_iB$ from $1$ to $n$, then each $a_i$ naturally defines a bijection of $\{1, \ldots, n\}$ as it sends the components of $G \backslash B$ to the components of $G \backslash a_iB$. Up to extracting a subsequence, we may suppose that all the $a_i$'s define the same permutation. We get the following picture: If $a$ sends $3$ to $3$ or $2$, then you find easily a subset $X$ of the space of ends such that $aX \subsetneq X$. In this case, $a$ must have infinite order. Similarly, if $b$ sends $3$ to $2$ or $3$, there exists some $X$ inside the space of ends such that $b X \subsetneq X$. In this case, $b$ must have infinite order. Now, if $a$ and $b$ both send $3$ to $1$, then $ab^{-1}$ sends $1$ to $1$. Next you find easily a subset $X$ of the space of ends such that $ab^{-1}X \subsetneq X$, so that $ab^{-1}$ must have infinite order. More about infinitely-ended groups. As already mentioned, Stallings' theorem implies that an infinitely-ended group $G$ acts on a tree $T$ without inversions, fixed-point freely and with finite edge-stabilisers. As a consequence, finitely-generated torsion subgroups of $G$ must be included into vertex-stabilisers. So you know that $G$ cannot be itself a torsion group, but it also follows some control on torsion subgroups. An even stronger consequence of the action of $G$ on $T$ is that $G$ must be hyperbolic relatively to vertex-stabilisers. It implies that $G$ is very far from being a torsion group. For instance: $G$ is SQ-universal, ie., every countable group embeds into a quotient of $G$. The center $Z$ of $G$ is finite, and $G/Z$ satifies Property $P_\text{naive}$, ie., for every finite collection of elements $g_1, \ldots, g_n \in G/Z$, there exists some $g \in G/Z$ such that $\langle g,g_i \rangle = \langle g \rangle \ast \langle g_i \rangle$ for every $i$. A random element of $G$ (chosen thanks to a random walk) has infinite order almost surely. In fact, a random subgroup of $G$ (chosen thanks to a fixed number $n \geq 2$ of independent random walks) is almost surely free (of rank $n$).<|endoftext|> TITLE: Asymptotic of integral $\int_{1}^{e^n}(1-\frac{\ln(x)}{n})^n\,dx$ QUESTION [5 upvotes]: How could we find the large-$n$ asymptotic of $$\int_{1}^{e^n}\left(1-\frac{\ln x}{n}\right)^n\,dx.$$ I have a suspicion that this is $\sqrt{n}$. REPLY [11 votes]: Denote $t=\ln(x)/n$, then $t$ varies from 0 to 1 and the the integral reads as $$n\int_0^1 ((1-t) e^{t})^ndt.$$ We have $(1-t)e^t=(1-t)(1+t+t^2/2+\ldots)=1-t^2/2+O(t^3)=\exp(-t^2/2+O(t^3))$ for small $t$ and $(1-t)e^t<1$ for $0 TITLE: Fano Schemes of Intersections of Quadrics QUESTION [10 upvotes]: Let $g\geqslant 2$, and denote by $\mathrm{X}=\mathrm{Q}_1\cap\mathrm{Q}_2\subset\mathbf{P}^{2g+1}$ a smooth intersection of quadrics. By considering the pencil generated by $\mathrm{Q}_1,\mathrm{Q}_2$ we obtain an associated hyperelliptic curve $\mathrm{C}$ of genus $g$. The following results on the Fano scheme $\mathrm{F}_n(\mathrm{X})$ of $n$-planes on $\mathrm{X}$ are well-known: (i) $\mathrm{F}_n(\mathrm{X})=\emptyset$ for $n>g-1$; (ii) $\mathrm{F}_{g-1}(\mathrm{X})\simeq\mathrm{J}(\mathrm{C})$, the Jacobian of $\mathrm{C}$ (theorem of A. Weil); (iii) $\mathrm{F}_{g-2}(\mathrm{X})\simeq\mathrm{U}_{\mathrm{C}}(2,\mathscr{L})$, the moduli space of rank $2$ bundles on $\mathrm{C}$ with fixed determinant $\mathscr{L}$ of odd degree (theorem of Desale-Ramanan). Is there an interpretation of the $\mathrm{F}_n(\mathrm{X})$ for $n TITLE: Milnor immersion of circle, disks, and a ball QUESTION [6 upvotes]: Milnor surprisingly found an immersion of a circle in the plane which bounds two "incompatible" immersed disks (see [1] for example, picture included). I think I can glue these two disks together to define an immersion of a sphere $f:S^2\to\mathbb{R}^3$ where Milnor's immersed circle (the equator) sits in the $xy$-plane and contracts to a point (in the $\pm z$ directions) by running along either immersed disk. Does $f(S^2)$ bound an immersed ball, i.e. does $f$ extend to an immersion of a 3-ball? This question was sparked by glancing at Gromov's book [1] and a related paper of Eliashberg-Mishachev which argues that the projection $S^2\to\mathbb{R}^2$ (composing $f$ above with projection onto $xy$-plane) is not homotopic through folded maps to the "standard" folded map $(x,y,z)\mapsto (x,y)$ on the unit 2-sphere (the fold being the equator). REPLY [3 votes]: If one assumes a couple of simple conditions on $f$, namely that the restriction of $z$ on $f(S^2)$ is a Morse function with two critical points and that projections of two halves of $f(S^2)$ ($z>0$, $z<0$) onto the plane $z=0$ realize two original different immersions then $f$ can not be extended to an immersion of the ball. Indeed, assume by contradiction that such an immersion $F:B^3\to \mathbb R^3$ exists. Let $m<0$ be the minimum of $z$ on $f(S^2)$ and $M>0$ be the max. In this case the preimages of planes $z=c$ under $F$ will be $2$-disks in $B^3$ for all $c\in (m,M)$. And by continuity, (changing $c$ from $M$ to 0, or from $m$ to $0$) we will get that the map $F$ applied to the disk $F^{-1}(z=0)$ realizes both immersions, which is absurd. One can relax the two conditions by requiring that there is a regular function $z'$ on $\mathbb R^3$ such that $z'=0$ is the plane $z=0$ and $z'$ restricts to $f(S^2)$ as a Morse function with two critical points. If I understand correctly this condition is implicit in your construction of $f$. (One should express what it means that the two halves of $S^2$ realize two different immersions of a disk.) PS. To clarify the above reasoning I'll give a proof in the most restrictive case. This case already contains the main idea. So the setting will be the following Setting. Let $S^2$ be the unit $2$-sphere, $S^1\subset S^2$ be the equator, $O_+$ be the north pole and $O_-$ the south. Let $S^2_+$ be the open upper half-sphere and $S^2_-$ the lower one. Let us assume that the immersion $f: S^2\mathbb\to \mathbb R^3$ has the following properties. 1) The restriction of $f$ to the equator $S^1\subset S^2$ realizes the Milnor circle. 2) Function $z$ restricts to $f(S^2)$ as a Morse function with two critical points so that $\max_z f(S^2)=1$, $\min_z f(S^2)=-1$ and moreover the poles are sent to critical points $z(f(O_+))=1$, $z(f(O_-))=-1$. 3) Let $\pi: (x,y,z)\to (x,y)$ be the projection. Then the maps $\pi\circ f: S^2_+\to \mathbb R^2$ and $\pi\circ f: S^2_-\to \mathbb R^2$ realize the two Milnor immersions. So, suppose all these conditions hold. Then each set $F^{-1}(z=c)$ is a disk $D_c\subset B^3$ for $c\in (-1,1)$. Consider the family of immersed disks $\pi(F(D_c))\subset \mathbb R^2$ for $c$ varying from $1$ to $0$. I claim that this family of immersed disks is the same as the following second family of immersed disks: Second family. Let $D_c^+\subset S_+^2$ be the disk composed of points with $z\ge c$. Then we can consider the family of immersed disks $\pi(F(D_c^+))$. So the claim is that the two families $\pi(F(D_c))$ and $\pi(F(D_c^+))$ coincide. But then, the immersion $\pi(F(D_0))$ is the same as $\pi(F(D_0^+))=\pi(F(S_+^2))$. Just in the same way we prove that $\pi(F(D_0))$ is the same as $\pi(F(S_-^2))$. This contradicts 3).<|endoftext|> TITLE: Non-trivial factor splitting from vacuum in TQFTs. F-move not unity? QUESTION [6 upvotes]: Consider a unitary modular TQFT, defined by the F and R moves. More specifically, a braided tensor category relevant for anyon models in 2D topologically ordered phases of matter. I am interested in the value of the F-move, $[F^{abc}_1]_{\bar{a}\bar{c}}$, diagrammatically defined by where $1$ is the vacuum and $\bar{a}$ is the inverse of $a$, i.e. $1 \in \{a\times \bar{a}\}$ I would naively expect this to be the identity. However, in https://thesis.library.caltech.edu/2447/2/thesis.pdf the author states that this need not be the case. I think this is related to bending, and/or the $\mathbb{Z}_2$, or possibly the $\mathbb{Z}_3$, Frobenius Schur indicator. Any clarification (or references) on the value of this F-move would be much appreciated! REPLY [4 votes]: Consider the case $a=b=c$. Then the fusion space $V_{aaa}$ affords a representation of $\mathbb Z/3$ via a $2\pi/3$ rotation. The F-move in your question is essentially this $2\pi/3$ rotation, and the rotation need not act trivially (its eigenvalues might be non-trivial 3rd roots of unity). See for example eqn 470 of https://arxiv.org/pdf/1709.01941.pdf , which cites https://arxiv.org/pdf/0704.0208.pdf . [edit:] In response to the questions in the comments: (1) Yes, this is sometimes called the $\mathbb Z/3$ Frobenius-Schur indicator. (2) Away from the case $a=b=c$, it does not (strictly speaking) make sense to ask whether the F-moves in question are the identity, because they are linear maps between distinct vector spaces $V_{abc}$, $V_{bca}$ and $V_{cab}$. [Another edit: I'm using here the fact that $V_{1a\overline a}$ and $V_{1\overline cc}$ (from the figure in the question) are 1-dimensional and have canonical generators, so that in this particular case the F-move is equivalent to a map from $V_{abc}$ to $V_{cab}$.] However, if we chose bases of those vector spaces appropriately, we can arrange for the matrix representations of the F-moves to be the "identity" matrix (1s down the diagonal). Note that in general these matrices are only well-defined up to unitary change of bases in the various $V_{xyz}$, so with a different choice of basis we can arrange for the F-matrix in the question to be any unitary matrix that we like.<|endoftext|> TITLE: Has the geometry of the variety of nilpotent matrices over $\mathbb{C}$ been studied? QUESTION [8 upvotes]: Consider the complex projective variety given by $X^n = 0$, where $X\in \mathrm{M}_n(\mathbb{C})$ and, say, $n\geq 3$. Some basic properties of it are already mentioned in this question: https://math.stackexchange.com/questions/405291/variety-of-nilpotent-matrices I would like to know if its geometry has been studied in more detail in the sense of complex geometry (algebraic, differentiable, analytic). References are appreciated since the question above mentions only Jantzen's "Lie Theory". REPLY [9 votes]: Consider $PGL_n$ acting by conjugation on the space of $n\times n$ matrices $M_n$, and let the GIT quotient map be $\pi:M_n\to M_n//PGL_n$. I think you are asking about the geometry of the fibre of zero. Regardless, I believe this is a special case of the nullcone for a reductive group action on a vector space $V$. Two papers that come to mind about the geometry of the nullcone are: A Stratification of the Null Cone Via the Moment Map by Linda Ness (appendix by David Mumford), American Journal of Mathematics, Vol. 106, No. 6 (Dec., 1984), pp. 1281-1329 Irreducible components of the nullcone by Richardson, R. W., Invariant theory, 409–434, Contemp. Math., 88, Amer. Math. Soc., Providence, RI, 1989. The work of these two authors is worth reading if you are interested in these kinds of problems (and the references therein).<|endoftext|> TITLE: Relation between Frobenius, spectral norm and sum of maxima QUESTION [6 upvotes]: Let $A$ be a $n \times n$ matrix so that the Frobenius norm squared $\|A\|_F^2$ is $\Theta(n)$, the spectral norm squared $\|A\|_2^2=1$. Is it true that $\sum_{i=1}^n\max_{1\leq j\leq n} |A_{ij}|^2$ is $\Omega(n)$? Assume that $n$ is sufficiently large. I cannot find a relation between matrix norms that can show this. The idea behind this question is that there are many singular values of $A$ that are $\Theta(1)$. Thanks! REPLY [6 votes]: This is false in general, but true for matrices with non-negative entries. For a counterexample, suppose that $n=p$ is prime, and consider the matrix $$ A=\left\|p^{-1/2}\left(\frac{i-j}p\right)\right\|_{i,j=0,\dotsc,p-1} $$ where $(\cdot/p)$ is the Legendre symbol. This is a circulant matrix; its non-zero eigenvalues are normalized Gaussian sums, equal $1$ in absolute value; hence, $\|A\|_2\le 1$. Also, we have $\|A\|_F^2=p-1$. On the other hand, $$ \sum_i \max_j |A_{ij}|^2 = 1. $$ Suppose now that all elements of $A$ are non-negative. Let $u_i\in{\mathbb R}^n$ be the row vectors of $A$, and denote by $\|\cdot\|_p$ the $\ell^p$-norm over ${\mathbb R}^n$; when $p=2$, this is the standard Euclidean norm. The Frobenius norm of $A$ is $\|A\|_F^2=\sum_i\|u_i\|_2^2$. Assuming that $\|A\|_F^2\ge cn$ and $\|A\|_2^2\le C$, we show that $\sum_i\|u_i\|_\infty^2\ge C^{-1}c^2n$. Denoting by $\vec 1$ the all-$1$ vector, we have $$ C \ge \|A\|_2^2 = \max_x \frac{\|Ax\|_2^2}{\|x\|_2^2} \ge \frac{\|A\vec 1\|_2^2}{\|\vec 1\|_2^2} = \frac1n\sum_i \|u_i\|_1^2. $$ (It is this computation that uses the non-negativeness assumption.) This implies $$ \sum_i \|u_i\|_1^2 \le Cn $$ and, consequently, by Cauchy-Schwartz, $$ cn \le \|A\|_F^2 = \sum_i \|u_i\|_2^2 \le \sum_i \|u_i\|_\infty \|u_i\|_1 \le \left( \sum_i \|u_i\|_\infty^2\right)^{1/2} \left( \sum_i \|u_i\|_1^2\right)^{1/2} \le \left( Cn\sum_i \|u_i\|_\infty^2\right)^{1/2}, $$ which yields the desired estimate $$ \sum_i \|u_i\|_\infty^2 \ge C^{-1}c^2n. $$<|endoftext|> TITLE: Set-theoretical foundations of Mathematics with only bounded quantifiers QUESTION [14 upvotes]: It seems that outside of researchers in Mathematical Logic, mathematicians use almost exclusively bounded quantifiers instead of unbounded quantifiers. In fact, I haven't observed any other practice from the very first day on when I was a student. For example, a logician would write $\forall a : ( a \in \mathbb R ) \rightarrow ( a^2 \geq 0 )$ whereas most working analysists and algebraists write $\forall a \in \mathbb R : a^2 \geq 0$ On the other hand, most mathematicians I know accept the idea that all of mathematics can be built up from set-theoretical foundations alone (starting the natural numbers). So there seems to be a set of assumptions, almost universally agreed upon, which most working mathematicians assume implicitly for their practice. These assumptions start with set theory but apparently exclude unbounded quantifiers. In fact, unless you attend a class in formal logic you might never encounter unbounded quantifiers. It seems that most mathematicians use a subset of human language enhanced with a subset of mathematical language (avoiding universal quantifiers) as their working language. Question: Have there been attempts at precisely identifying this mathematical sublanguage and the rules that it governs? REPLY [12 votes]: Most mathematics can be done in logical systems which are far weaker than Zermelo-Fraenkel set theory. For example, something like structural set theory will suffice for a great deal of ordinary mathematics. It's not exactly true that mathematicians never use unbounded quantification. For example, in category theory universal properties quantify over all objects of a category. When the category in question is large this amounts to unbounded quantification. In a sense, such quantification is "harmless" because it is "on the outside", i.e., it is of the form $\forall X . \phi(X)$ where $\phi$ itself contains no further unbounded quantifiers. In many cases we can replace such a statement with a schema $\phi(X)$ where $X$ is a schematic symbol (that is, instead of having a single formula $\forall X . \phi(X)$ we have many separate formulas $\phi(X)$, one for each $X$). Occasionally one sees mathematical statements which do contain inner unbounded quantifiers, but those are not common. One example I can think of is the following. The notion of epimorphism in a category requires quantification over all objects: a morphism $f : A \to B$ is epi when for all all $C$ (unbounded quantifier!) and $g, h : B \to C$, if $g \circ f = h \circ f$ then $g = h$. Often in concrete example we can characterize epis equivalently with some statement that only contains bounded quantifiers (e.g., in the category of sets a map is epi if, and only if it is surjective), but if we make general statements about epis in large categories, large quantification will be required.<|endoftext|> TITLE: When does the dual to the space $K(X)$ of compact operators consist of nuclear functionals? QUESTION [8 upvotes]: Let $X$ be a Banach space and $B(X)$ be its space of all (bounded) operators. A nuclear functional on $B(X)$ is a linear functional $u:B(X)\to{\mathbb C}$ that can be represented in the form $$ u(A)=\sum_{n=1}^\infty \lambda_n\cdot f_n(Ax_n),\qquad A\in B(X), $$ where $\lambda_n\in{\mathbb C}$, $x_n\in X$, $f_n\in X^*$ are such that $$ \sum_{n=1}^\infty |\lambda_n|<\infty,\quad \sup_{n}||x_n||\le 1,\quad \sup_{n}||f_n||\le 1. $$ Let us denote by $N(X)$ the space of all nuclear functionals on $B(X)$. If $X$ is a Hilbert space, then it is well known (see G.J.Murphy, C*-Algebras and Operator Theory, Theorem 4.2.1) that the dual space $K(X)^*$ to the space of all compact operators $K(X)$ coinsides with the space of all nuclear functionals: $$ K(X)^*=N(X) $$ (this is an isomorphism of Banach spaces, but for me it is important that this is an equality of sets). Is the same true for all Banach spaces $X$? Or at least for all Banach spaces with the (classical) approximation property? I am mostly interested in the case when $X=C(T)$, the space of continuous functions on a compact topological space $T$. REPLY [7 votes]: This is the question of duality of injective and projective tensor products of Banach spaces, and the natural question would be whether the dual of $K(X)$ can be represented by the functionals in $N(X^*)$. A quick answer is: If $X^*$ or $X^{**}$ has the approximation property and if $X^*$ or $X^{**}$ has the Radon-Nikodym property, then this is so; see Chapter VIII in Diestel and Uhl's "Vector Measures'' or Ryan's "Introduction to Tensor Products of Banach Spaces'' or Defant and Floret's "Tensor Norms and Operator Ideals''. So if $X$ is reflexive with the approximation property things are ok.<|endoftext|> TITLE: What makes Graph invariants so useful/important? QUESTION [8 upvotes]: What makes Graph invariants so useful/important? If I were trying to create a useful graph invariant, what principles should I follow? My understanding is that they allow one to isolate and study specific properties of graphs algebraically or to classify graphs up to isomorphism (although, it seems to me that canonical labellings are the right tool for this). However, important graph invariants are constructed from counting proper colorings of a graph, for an appropriate definition of proper. A priori, why do we know that those graph invariants isolate and study specific properties or is there some other key motivation for graph invariants? REPLY [5 votes]: Some graph invariants are useful to separate NP-complete problems from tractable problems. Bounded treewidth is such a parameter. For example, the weighted max independent set problem is NP-complete but can be solved in linear time for graphs of bounded treewidth. Similarly, the NP-complete 3-coloring problem can be solved in linear time for graphs of bounded treewidth. The time complexities for these problems are $O(2^w n)$ and $O(3^w n)$ respectively, for graphs of tree-width $w$. Two classes of graphs with bounded treewidth are the Halin graphs and pseudoforests:                     Images from Wikipedia: Halin graph, pseudoforest.<|endoftext|> TITLE: Least quadratic residue under GRH: an explicit bound QUESTION [13 upvotes]: Let $m$ be a positive integer and $\chi$ a primitive character mod $m$. Let $x$ be such that $\chi(p)\ne 1$ for all primes $p TITLE: Explicit version of the Burgess theorem QUESTION [18 upvotes]: Does there exist a totally explicit version of the Burgess theorem? Precisely, let $m$ be a positive integer, and let $\chi$ be a primitive character mod $m$. A special case (sufficient for my purposes) of the Burgess theorem asserts that $\left| \sum_{a\le n\le a+x}\chi(n) \right|\ll_\varepsilon x^{1/2}m^{3/16+\varepsilon}$ I wonder if this was ever made totally explicit. In the case $m=p$ prime Iwaniec and Kowalski prove in their book the inequality with right-hand side $cx^{1/2}p^{3/16}(\log p)^{1/2}$ and claim (without going into details) that $c=30$ is nice. I need, however, the case of composite modulus as well. Was anything like this ever done? REPLY [18 votes]: There are now at least two instances of such an explicit result. Theorem 1.1 of Bordignon: https://arxiv.org/abs/2001.05114. Theorem 1.1 (and Corollary 1.2) of Jain-Sharma, Khale, and Liu: https://arxiv.org/abs/2010.09530v2, or https://www.worldscientific.com/doi/10.1142/S1793042121500834. Bordignon's result applies to convoluted Dirichlet characters; however, applying the result with $\psi$ identically $1$ (and $k=1$) gives the following \begin{equation*} \left\rvert\sum_{M TITLE: Divisibility of sum of multinomials QUESTION [7 upvotes]: Let $n, m$ and $t$ be positive integers. Define the multi-family of sequences $$S(n,m,t)=\sum_{k_1+\cdots+k_n=m}\binom{m}{k_1,\dots,k_n}^t$$ where the sum runs over non-negative integers $k_1,\dots,k_n$. These numbers are related to average distances (from the origin) of uniform unit-step random walks on the plane. QUESTION. Is it always true that $n$ divides $S(n,m,t)$? Observe that $S(n,m,1)=n^m$. REPLY [9 votes]: We count the number of $t$-tuples $(\xi_1,\ldots,\xi_t)$ of the colorings of $\{1,\ldots,m\}$ with $n$ given colors, for which any two colorings use the same multisets of colors. If $M$ is the number of such tuples in which 1 is colored red in the coloring $\xi_1$ (red is one of our $n$ colors), the total number of $t$-tuples equals $n\cdot M$.<|endoftext|> TITLE: Does the average primeness of natural numbers tend to zero? QUESTION [23 upvotes]: This question was posted in MSE. It got many upvotes but no answer hence posting it in MO. A number is either prime or composite, hence primality is a binary concept. Instead I wanted to put a value of primality to every number using some function $f$ such that $f(n) = 1$ iff $n$ is a prime otherwise, $0 < f(n) < 1$ and as the number divisors of $n$ increases, $f(n)$ decreases on average. Thus $f(n)$ is a measure of the degree of primeness of $n$ where 1 is a perfect prime and 0 is a hypothetical perfect composite. Hence $\frac{1}{N}\sum_{r \le N} f(r)$ can be interpreted as a the average primeness of the first $N$ integers. After trying several definitions and going through the ones in literature, I came up with: Define $f(n) = \dfrac{2s_n}{n-1}$ for $n \ge 2$, where $s_n$ is the standard deviation of the divisors of $n$. One reason for using standard deviation was that I was already studying the distribution of the divisors of a number. Question 1: Does the average primeness tend to zero? i.e. does the following hold? $$ \lim_{N \to \infty} \frac{1}{N}\sum_{r = 2}^N f(r) = 0 $$ Question 2: Is $f(n)$ injective over composites? i.e., do there exist composites $3 < m < n$ such that $f(m) = f(n)$? My progress $f(4.35\times 10^8) \approx 0.5919$ and decreasing so the limit if it exists must be between 0 and 0.5919. For $2 \le i \le n$, computed data shows that the minimum value of $f(i)$ occurs at the largest highly composite number $\le n$. Note: Here standard deviation of $x_1, x_2, \ldots , x_n$ is defined as $\sqrt \frac{\sum_{i=1}^{n} (x-x_i)^2}{n}$. Also notice that even if we define standard deviation as $\sqrt \frac{\sum_{i=1}^{n} (x-x_i)^2}{n-1}$ our questions remain unaffected because in this case in the definition of $f$, we will be multiplying with $\sqrt 2$ instead of $2$ to normalize $f$ in the interval $(0,1)$. REPLY [23 votes]: The answer to Question 1 is "yes". To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e. $$ s_n\leq \sqrt{\frac{\sum_{d\mid n}d^2}{\sum_{d\mid n} 1}}=\sqrt{\frac{\sigma_2(n)}{\sigma_0(n)}}, $$ where $\sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now, $$ \sigma_2(n)=n^2\sigma_{-2}(n), $$ so $$ \sigma_2(n)<\frac{\pi^2}{6}n^2 $$ for all $n$. Therefore we have $$ f(n)\leq \frac{2}{n-1} \sqrt{\frac{\pi^2}{6}n^2/\sigma_0(n)}\leq \frac{5.14}{\sqrt{\sigma_0(n)}} $$ for all $n$. Now, almost all $n\leq N$ have at least $0.5\ln\ln N$ distinct prime factors. In particular, for almost all $n\leq N$ we have $\sigma_0(n)\geq 0.5\ln\ln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0\leq f(n)\leq 1$ gives $$ \sum_{n\leq N} f(n)\leq \sum_{n\leq N, \sigma_0(n)\geq 0.5\ln\ln N} \frac{5.14}{\sqrt{\sigma_0(n)}}+\sum_{n\leq N, \sigma_0(n)<0.5\ln\ln N} 1= o(N), $$ as needed. Using contour integration method one can even prove something like $$ \sum_{n\leq N} f(n)=O(N(\ln N)^{1/\sqrt{2}-1}) $$<|endoftext|> TITLE: Geometric interpretations of nil-Hecke ring and affine Hecke algebra QUESTION [7 upvotes]: I am interested in two related constructions which give us either the cohomology or the $T \times \mathbb{C}^*$-equivariant $K$-theory of flag varieties. Let $G$ be a semisimple, simply connected algebraic group, with $T \subset B \subset G$ a chosen maximal torus and Borel subgroup. In order to gain geometric information about the flag variety $G/B$, we make use of a collection of $\mathbb{P}^1$ bundles. To be a little more explicit: In what I would (perhaps erroneously) call the Bernstein-Gelfand-Gelfand approach, to find the cohomology of $G/B$, we would use minimal parabolic subgroups $P_i$ and maps $G/B \rightarrow G/P_i$. This gives $G/B$ the structure of a $\mathbb{P}^1$-bundle over $G/P_i$. In this situation we can use the Leray-Serre spectral sequence to obtain $H^*(G/B)$ in terms of $H^*(\mathbb{P}^1)$ and $H^*(G/P_i)$, and in particular we can get our hands on classes $[\overline{X}_{s_i}]$ where $X_{s_i}$ is the Schubert cell associated to a simple reflection $s_i$. In the construction of $K^{T \times \mathbb{C}^*}(G/B)$, following Ginzburg (and possibly originally due to Kazhdan and Lusztig?), we construct a different $\mathbb{P}^1$-bundle. Namely, the $G$-diagonal orbits in $G/B \times G/B$ are parametrized by $w \in W$. Let $Y_{s_i}$ denote the orbit associated to $(B/B, s_iB/B)$, and $\overline{Y_{s_i}}$ its orbit closure. Then via projection onto the first factor, $\pi_1 :\overline{Y_{s_i}} \rightarrow G/B$ is a $\mathbb{P}^1$-bundle. Ginzburg goes on to construct all sorts of sheaves in this setup, and constructs the affine Hecke algebra geometrically. The two constructions should be related in the following way: "The $T_{s_i}$ action on $K^{T \times \mathbb{C}^*}(T^*G/B)$ is given by $e^{\lambda} \mapsto \frac{e^{\lambda}-e^{s_{i} (\lambda)}}{e^{\alpha_i}-1}-q \frac{e^{\lambda}-e^{s_{i}(\lambda)+\alpha_i}}{e^{\alpha_i}-1}.$ The fraction on the left is the Demazure operator associated to $s_i$, which is used to find the $K$-theory of $G/B$" (here I am paraphrasing from Chriss-Ginzburg Thm 7.2.16). I know I am mixing cohomology and all sorts of $K$ theory here, but it seems that there should be a more accessible topological relationship. My question is: on the most simple (purely topological) level, how do these two distinct $\mathbb{P}^1$-bundles give us similar information about the cohomology (or $K$-theory) of $G/B$? I feel like I've simultaneously included too many details and left out too many details, and I'd be happy to edit for clarification. REPLY [4 votes]: The subvariety $\overline{Y}_{s_i}\subset G/B \times G/B$ is the fiber product $G/B\times_{G/P_i}G/B$. The set $\overline{Y}_{s_i}$ is the saturation for the diagonal $G$-action of $\{B/B\}\times P_i/B$, by definition, and of course, that also lies in the fiber product; since they are smooth irreducible varieties of the same dimension, inclusion shows they are equal.<|endoftext|> TITLE: Measure of rational hyperplanes of $\mathbb{R}$ QUESTION [6 upvotes]: Let's view $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and pick some basis $(v_\alpha)_{0 \leq \alpha < \mathfrak{c}}$ of it. We can then consider the subspace $L$ spanned by $(v_\alpha)_{0 < \alpha < \mathfrak{c}}$ (ie leaving out one vector from the basis). Given the horrible way we have built $L$, I don't suppose there is much a priory reason for $L$ to be measurable. However, I am wondering whether we can say something about the outer measure of $L$. REPLY [8 votes]: This is the (a version of the) Vitali set, which is not Lebesgue measurable. A quick reason is: $\mathbb{R}= (v_0\mathbb{Q})\oplus_\mathbb{Q} V$ shows that $\mathbb{R}$ is a countable union of translates of $V$, so $V$ cannot be a Lebesgue set of measure zero, but must have positive outer measure. To show that the outer measure of $V$ is actually $+\infty$, note that, being $V$ a $\mathbb{Q}$-linear subspace, $2V=V$ so that its Lebesgue outer measure is $\lambda^*(V)=2\lambda^*(V)$, which has to be $+\infty$ because it is not $0$. On the other hand, for any Lebesgue measurable set of positive measure $S$, according to Steinhaus property, $S-S$ is a nbd of $0$. Therefore $V=V-V$, which is nowhere dense, contains no measurable set of positive measure.<|endoftext|> TITLE: If the Tate construction vanishes for all trivial $G$-actions, then does it vanish for all $G$-actions? QUESTION [5 upvotes]: Let $\mathcal{C}$ be a semiadditive $\infty$-category, complete and cocomplete, and let $G$ be a finite group. Then for any $X \in Fun(BG,\mathcal{C})$, there is a norm map $N_X: X_G \to X^G$. For each $X \in \mathcal{C}$, we have the constant $X^{triv} \in Fun(BG,\mathcal{C})$. Suppose that $N_{X^{triv}}$ is an equivalence for each $X \in \mathcal C$. Then does it follow that $N_X$ is an equivalence for each $X \in Fun(BG,\mathcal{C})$? Evidence that this might be true comes from thinking in terms of ambidexterity: the question is whether n-ambidexterity can be checked on trivial objects when $n=1$. Note that this is the case when $n=(-2),(-1)$ (vacuously), or $n=0$ (If a pointed category $\mathcal C$ has finite coproducts and finite products, and if $X \vee X \to X \times X$ is an equivalence for each $X \in \mathcal C$, then $X \vee Y \to X \times Y$ is an equivalence for each $X,Y \in \mathcal C$). Really, I'm interested in the question for all $n \in \mathbb N$: Question: Let $\mathcal C$ be an $(n-1)$-ambidextrous $\infty$-category, complete and cocomplete. Let $B$ be an $n$-truncated space with finite homotopy groups. Then for all $X \in Fun(B,\mathcal C)$, there is a norm map $N_X: \varinjlim X \to \varprojlim X$. Suppose that this map is an equivalence when $X$ is constant. Then is $N_X$ an equivalence for all $X$? Perhaps this doesn't hold "locally" (i.e. for a fixed $B$). Then can we at least say that if $N_X$ is an equivalence for all constant $X$ and all $n$-trucated, $\pi$-finite $B$, then $\mathcal C$ is $n$-ambidextrous? REPLY [5 votes]: There is a reference in my comment above (Lemma 2.1.5 in arxiv.org/pdf/1811.02057.pdf). Let me repeat the argument from there in the answer here. Let $X$ be a $\pi$-finite space and $q:X\to pt$ be the projection to the point. We assume that the norm map $Nm_q : q_! \to q_*$ is an isomorphism when evaluated at "trivial modules", which are just functors of the form $q^*A$. We want to prove that it is an isomorphism at every object. For an object $A$, the zigzag indentity gives us a retract diagram $q_*A \to q_*q^*q_*A \to q_*A$. Now, the norm map $Nm: q_! q^*q_*A \to q_*q^*q_*A$ is an isomorphism by the assumption, and the naturality of the norm map gives us that the composition $q_*A\to q_*q^*q_*A \stackrel{Nm^{-1}}{\to} q_!q^*q_*A \to q_!A \stackrel{Nm}{\to}q_*A$ is the identity (where the unlabeled arrows are the usual units and counits). Hence, $Nm: q_! A \to q_*A$ is invertible from the left. A similar dual argument with the Zigzag identity for $q_!$ and $q^*$ gives us that it is right invertible, hence an isomorphism. The intuition here is that taking homotopy orbits/fixed points twice just enlarge the object so this way we reduce from general module to a trivial one.<|endoftext|> TITLE: Number of Reflections in a Circle between Two Points QUESTION [8 upvotes]: For my research I am interested in the transmission characteristics between a transmitter (Tx) and a receiver (Rx) situated in a circular room. In particular, it is important for me to know the number of paths a ray can take such that it reflects exactly once off the walls of the room. Since reflections occur such that the incident ray has the same angle relative to the normal as the reflected ray, I tried to use vectors to attack the problem but the math became very unwieldy. Empirically, I have found that depending on the situation of the transmitter and receiver, there could be 2, 3, or 4 paths—no more, no less. There is an exceptional case where the transmitter and receiver are co-located at the centre, in which case there are infinitely many paths. Can my experimental result be validated (or denied) analytically? REPLY [13 votes]: We use complex numbers to prove that there are at most $4$ such points unless both transmitter and receiver are at the center. Identify the circular room with the unit circle $|z|=1$ in the complex plane, and let $r$ and $t$ be the complex numbers corresponding to the receiver and transmitter, with $|r|<1$ and $|t|<1$. If $z$ is a point of reflection then the condition $\theta_r = \theta_t$ comes down to $(z-r)(z-t)$ being a real multiple of $z^2$; that is, to the ratio $(z-r)(z-t)/z^2$ being a real number. Write $$ (z-r)(z-t) / z^2 = 1 - (r+t) z^{-1} + rt z^{-2}, $$ and note that a complex number $w$ is real if and only if it equals its own complex conjugate $\overline w$. Since $z$ is on the unit circle, $\overline z = z^{-1}$, so our condition is $$ \overline{rt} z^2 - (\overline r + \overline t) z + (r+t) z^{-1} - rt z^{-2} = 0. $$ Multiplying by $z^2$ yields a polynomial of degree $4$ in $z$. Thus there are at most $4$ solutions, even without the condition $|z|=1$, unless the polynomial vanishes identically, in which case every $z$ is a solution. But the polynomial vanishes identically if and only if $r+t = rt = 0$, which is to say $r=t=0$, so we recover the degenerate case where receiver and transmitter are both in the center of the circular room.<|endoftext|> TITLE: Does homotopy equivalence to a wedge of spheres imply shellability? QUESTION [7 upvotes]: It's pretty clear that for a simplicial complex $\Delta$, shellability of the complex implies that it is homotopy equivalent to a wedge of spheres. However, does the converse hold? That is, does $\Delta$ being homotopy equivalent to a wedge of spheres imply shellability? Thanks! REPLY [7 votes]: As @j.c. says, the answer is "no", and it is good to know that there are non-shellable triangulations of spheres. However, it is even easier to find counterexamples, that are not shellable because of topological obstructions. I'll give two examples: The annulus is homotopy equivalent to $S^1$, but no triangulation is shellable (since a shellable 2-dimensional complex is topologically a bouquet of 2-spheres, not 1-spheres). More deeply, the wedge of 2 or more spheres of dimension at least 2 is never shellable, nor even Cohen-Macaulay. (Note the lack of the word "homotopy type" in the previous sentence!) This is because if you take two $d$-spheres and glue them together at the point $v$, then the link of $v$ is disconnected of dimension $d-1$, hence not shellable. Since the link of every vertex in a shellable complex is shellable, this gives the desired non-shellability. The fragility of shellability in this respect can be seen as a good thing: there are lots of tools to compute homotopy type of a complex, but shellability (when carefully examined) tells you a fair bit about the "homeomorphism type" of the complex. By the way, the smallest dimension of a non-shellable sphere is 3. The 1978 papers of Danaraj and Klee are one reference for this; the first paper I'm aware of that uses the word shelling, "Isotopy in 3-manifolds- Isotopic deformations of 2-cells and 3-cells" by Donald Sanderson, is another.<|endoftext|> TITLE: Covering the primes with pairs of consecutive integers QUESTION [8 upvotes]: Is it true that for every sufficiently large positive integer $n$, one can always find at most $k=\lfloor\pi(n)/2\rfloor$ integers, $a_1,a_2,a_3,a_3,\dots a_k$, between $1$ and $n$, such that each of the $\pi(n)$ primes not greater that $n$ divides at least one of the integers $a_1,a_1+1,a_2,a_2+1,a_3,a_3+1,\dots,a_k,a_k+1$? REPLY [2 votes]: Even with relaxing the conditions , the answer is no for $n=44$ and most $94 \leq n \leq 200.$ I didn't check any further but I doubt it is true ever again. (I relaxed the condition even further than when I first wrote this and haven't rechecked.) We will say that $b$ covers the prime $p$ if $p \mid b.$ To review, we want a set of $k$ integers,$A=\{a_1,\cdots a_k\}$ where $k=\lfloor \frac{\pi(n)}2 \rfloor$ so that every prime less than $n$ is covered by at least one of the $2k$ members of $B=\{a_1,a_1+1,\cdots,a_k,a_k+1\}.$ The task is the same for $n=2m-1$ and $n=2m$ so there is no loss in assuming that $n$ is even. Actually, we could also restrict to the case that $2n-1$ is prime. Otherwise it is the same set of $\pi(n-2)=\pi(n)$ primes and the task is the same. As noted, if $m \lt p \lt 2m$ is prime then one of the chosen $a_i$ must be either $p$ or $p-1.$ The issue will be that there are only a relatively few more primes in $\{2 \cdots m\}$ compared to $\{m+1,\cdots 2m-1\}$ and this doesn't allow us enough flexibility to cover all the smaller primes. So if $\pi(2m)-\pi(m)=j$ we have $2^j$ starts to $A$ to consider. To simplify that we relax the problem: We will set $k=\lceil \frac{\pi(n)}2 \rceil.$ We will build the sets $A$ and $B$ as follows: Start with a list $L$ of the primes in question in decreasing order and $A=B=\emptyset$ . Take, p, the first so far uncovered prime in the list, choose a multiple $a=cp\leq n$ to add to $A.$ Add $a-1,a,a+1$ to $B$ and then move down $L$ to the largest still uncovered prime. Repeat until either all the primes are covered or $A$ is too big. Clearly this relaxed condition gives bigger sets $B$ and lets us cover anything we could cover in the original problem. In the case $m=22,n=44$ there are $8$ primes up to $22$ and $6$ more primes after that, $$23, 29, 31, 37, 41, 43.$$ These are all but one of of the members of $A.$ So, with the relaxed conditions, we so far have $B=\{22,23,24,28,29,20,31,32,36,37,38,40,41,42,43,44\}$ and can pick one more member of $A.$ Next down is $19$ which is already covered. For $p=17$ we need to pick an $a \in\{17,34\}$ But then we have all $7$ members of $A$ and $13$ is still uncovered. For $n=43,45,46$ we have the same primes to consider so those are also a failure. In the case above we had all but one prime covered. As $n$ increases, so does the discrepancy.<|endoftext|> TITLE: Carnot-Carathéodory metric QUESTION [21 upvotes]: The metric in sub-Riemannian geometry is often called the Carnot-Carathéodory metric. Question 1. What is the origin of this name? Who was the first to introduce it? I believe that the "Carathéodory" part of the name could be related to his work in theoretical thermodynamics [1], but I do not really know how it is related to his work. Question 2. How is the notion of Carnot-Carathéodory metric related to the work of Carathéodory? I know that Carnot groups are special examples of sub-Riemannian manifolds, but is it the reason for "Carnot" part in the name of the metric? Question 3. What does the "Carnot" part of the name of the metric stand for? [1] C. Carathéodory, Untersuchungen uber die Grundlagen der Thermodynamik. Math. Ann. 67 (1909), 355–386. REPLY [21 votes]: Pierre Pansu tells us that the terminology of the Carnot-Carathéodory metric is due to Mikhail Gromov [1]. Gromov himself explains the choice of the name: The metric is called the Carnot-Carathéodory metric because it appears (in a more general form) in the 1909 paper by Carathéodory on formalization of the classical thermodynamics where horizontal curves roughly correspond to adiabatic processes. The proof of this statement may be performed in the language of Carnot cycles and for this reason the metric was christened Carnot-Carathéodory. Pansu adds While the reference to Carathéodory is fundamental, the reference to Carnot must be seen as a place holder for the many authors who rediscovered accessibility criteria from the middle of the twentieth century back to a much earlier date. [1] M. Gromov – Structures métriques pour les variétés Riemanniennes, Textes Mathématiques, vol. 1, Paris, 1981, Edited by J. Lafontaine and P. Pansu.<|endoftext|> TITLE: Most general conditions for (weak or classical) solutions to Poisson's equation QUESTION [5 upvotes]: I thought I knew this but have found it surprisingly difficult to find good references. I am interested in solving $$ \left\{ \begin{align} & \Delta \psi = - \rho & & \mbox{in } \mathbb{R}^3, &(1) \\ & \psi(\infty) = 0. & & &(2) \end{align} \right. $$ where $\rho$ is a compactly supported function. We know the answer should be given (up to constant factors) by \begin{equation} \psi(x) = \int_{\mathbb{R}^3} \frac{\rho(y)dy}{|x-y|}. \qquad (3) \end{equation} My first question is what are the mildest conditions that can be imposed on $\rho$ for (1) and (2) to hold pointwise, and is the solution indeed given by (3)? I think Gilbarg and Trudinger give it to hold for $\rho$ Holder continuous with Holder exponent $\alpha \in (0,1]$. My second question is what if I relax the condition that (1) hold pointwise, and instead seek weak solutions i.e. $\psi$ satisfying $$ \int_{\mathbb{R}^3}\left[ \nabla\psi\cdot\nabla\varphi - \rho\varphi\right]dx = 0, \qquad \psi(\infty)=0 \qquad (4) $$ should hold for all test functions (i.e. $C_c^\infty$) $\varphi$? Then how does the generality improve? Can it be broadened to allow for measures $\rho$ if we replace $\int\varphi\rho dx$ by $\int \varphi d\rho$ in (4)? FYI the books I have been consulting include Evan's PDE, Gilbarg and Trudinger's Elliptic PDE, Landkof's Potential Theory, Helms' Potential Theory, Jackson's Electrodynamics. Thank you all in advance. REPLY [4 votes]: The name of the subject is Potential Theory (your integral (3) is called the Newtonian potential). Good references are: N. Landkof, Foundation of modern potential theory, Springer 1972, M. Brelot, Éléments de la théorie classique du potentiel. 3e édition. Les cours de Sorbonne. 3e cycle. Centre de Documentation Universitaire, Paris 1965, (these titles are slightly misleading: Brelot's book is more modern and more abstract than Landkof's book. But Landkof contains much more.) L. Hormander, Notions of convexity, Birkhauser, 1994. and also many books under the title Subharmonic functions, for example, W. Hayman and P. Kennedy (in 2 vols, first volume is about $R^n$, second about $n=2$.) In the most general setting $\Delta u$ is considered in the sense of Schwartz distributions, and $\rho$ is a charge (signed measure, difference of two Radon measures, difference of two positive distributions). Distributional solutions of this equation are given by the potential which you wrote, and in the case of positive charge $\Delta u$ (a measure) they exist as honest functions $R^n\to R\cup\{-\infty\}$ (defined pointwise everywhere, not just almost everywhere). These functions are upper semicontinuous and they are called subharmonic. For a signed charge $\rho$, function $u$ cannot be reasonably defined at every point, but can be defined a. e. and belongs to $L^p_{loc}$ for all $p<\infty$. In fact they are defined quasieverywhere, that is the exceptional set has zero Newton capacity (logarithmic capacity when $n=2$). They are called delta-subharmonic functions. Pointwise second derivative in the classical sense has little use in this theory and it exists under a suitable conditions on $\rho$ stated in the answer of user111. Potentials which are 2-ce differentiable in the classic sense do not have good properties, expecially with respect to taking limits. Also the important property that pointwise maximum of subharmonic functions is subharmonic is lost if we restrict ourselves to 2-ce differentiable functions. When $n=1$, subharmonic functions are nothing but convex functions, and it is not reasonable to restrict oneself to 2-ce differentiable convex functions. There is a different setting of "potentials of finite energy" which are elements of the completion of smooth functions with respect to energy norm. This has an advantage that they form a Hilbert space (See Landkof).<|endoftext|> TITLE: Density near at $0$ for the integral of the positive part of the Brownian motion QUESTION [5 upvotes]: This question was asked recently on MO and then deleted by the owner, user Aalon. I think the question deserves to be answered, which is what I will try to do here. Aalon was reading this paper, where it is claimed (on page 32, in line 5) that \begin{equation} P\Big(\int_0^1(W_t)_-\,dt\le\varepsilon\Big)\le\varepsilon \tag{1} \end{equation} for $\varepsilon>0$, where $W_\cdot$ is the standard Wiener process. The "reason" for this claim given in the mentioned paper (on page 32, in line 4) was that \begin{equation} \int_0^1(W_t)_-\,dt\ge\Big|\int_0^1 W_t\,dt\Big|. \tag{2} \end{equation} Aalon had doubts about both (1) and (2). REPLY [5 votes]: Thank you for your work on this question! I am one of the authors of the cited paper and, yes, we stated a wrong claim (1) in the appendix of this paper (let me remark that the results in the main part remain unaffected as they do not rely on the exact bound). I am sorry for the confusion. I want to add here how a true upper bound for the probability can be obtained. Denote $(W_t)_{t\ge 0}$ a standard Brownian motion and $(W_t)^+$ its positive part and $(W_t)^-$ its negative part, respectively. Observe that $\int_0^1(W_t)^-\,dt\stackrel{d}{=}\int_0^1(W_t)^+\,dt$, such that \begin{align*}P\Big(\int_0^1(W_t)^-\,dt\le x\Big)=P\Big(\int_0^1(W_t)^+\,dt\le x\Big)\,,x>0\,.\end{align*} Let us bound the second probability. For any $\epsilon>0$, the inequality \begin{align*}\int_0^1(W_t)^+\,dt\ge \int_0^1W_t \cdot\mathbb{1}(W_t>\epsilon)\,dt\ge \epsilon \int_0^1\mathbb{1}(W_t>\epsilon)\,dt\end{align*} leads us to \begin{align*}P\Big(\int_0^1(W_t)^+\,dt\le x\Big)&\le P\Big(\epsilon\int_0^1\mathbb{1}(W_t>\epsilon)\,dt\le x\Big)\\ &=P\Big(1-\int_0^1\mathbb{1}(W_t\le \epsilon)\,dt\le x/\epsilon\Big)\\ &=P\Big(\int_0^1\mathbb{1}(W_t\le \epsilon)\,dt\ge 1-x/\epsilon\Big)\,. \end{align*} For the last expression, we can use a nice generalization by Takács of Lévy's arc-sine law, see https://projecteuclid.org/download/pdf_1/euclid.aoap/1034968240. Using (15) and (16) from this work, we obtain that \begin{align*} P\Big(\int_0^1\mathbb{1}(W_t\le \epsilon)\,dt\ge 1-x/\epsilon\Big)&=\frac{1}{\pi}\int_{1-x/\epsilon}^{1}\frac{\exp(-\epsilon^2/(2u))}{\sqrt{u(1-u)}} du +2\Phi(\epsilon)-1\,, \end{align*} with $\Phi$ the cdf of the standard normal distribution. Thereby, we obtain that \begin{align*}P\Big(\int_0^1(W_t)^+\,dt\le x\Big)&\le \frac{1}{\pi}\int_{1-x/\epsilon}^{1}\frac{\exp(-\epsilon^2/(2u))}{\sqrt{u(1-u)}} du +2\int_0^{\epsilon}\frac{\exp(-u^2/2)}{\sqrt{2\pi}}du\,, \end{align*} and elementary bounds give the upper bound \begin{align*}P\Big(\int_0^1(W_t)^+\,dt\le x\Big)&\le \frac{2}{\pi}\sqrt{\frac{x}{\epsilon}}\frac{1}{\sqrt{1-x/\epsilon}} +\frac{2\epsilon}{\sqrt{2\pi}}\,. \end{align*} Since we are especially interested in the probability for $x\to 0$, the bound is not too bad for this case (for larger values of $x$ it can be crude). Choosing $\epsilon=x^{1/3}$, we obtain the upper bound \begin{align*}P\big(\int_0^1(W_t)^+\,dt\le x\big)&\le \frac{2}{\pi}x^{1/3}\frac{1}{\sqrt{1-x^{2/3}}} +\frac{2x^{1/3}}{\sqrt{2\pi}}\,. \end{align*} Observe that $P\big(\int_0^1(W_t)^+\,dt\le x\big)=\mathcal{O}(x^{1/3})$, $x\to 0$, is in line with the conjecture in (261) of Janson (https://projecteuclid.org/download/pdfview_1/euclid.ps/1178804352), where one can find an expansion with a precise constant of the leading term.<|endoftext|> TITLE: Finite etale covers of products of curves QUESTION [5 upvotes]: Probably this question can be phrased in a much greater generality, but I will just state it in the generality I require. I work over $\mathbb{C}$. Let $C_1, C_2 \subset \mathbb{P}^1$ be non-empty open subsets and $f: X \to C_1 \times C_2$ a non-trivial finite etale cover. Does there exist $i\in \{1,2\}$ such that the composition $X \to C_1 \times C_2 \to C_i$ has non-connected fibres? REPLY [6 votes]: The question already has a beautiful answer, but here's a different point of view which you may find helpful. Let $F_i = \pi_1(C_i, x_i)$, which is a free group on $\#(\mathbf{P}^1\setminus C_i) - 1$ generators (the etale fundamental group will be the profinite completion of this). Then $F = \pi_1(C_1\times C_2, x_1\times x_2) = F_1\times F_2$. A finite etale cover of $C_i$ or $C_1\times C_2$ corresponds to a finite set (its fiber at the basepoint $x_i$ or $x_1\times x_2$) with an action of $F_i$ or $F$. The cover is connected if and only if the action on that set is transitive. Let $\sigma_i \colon C_i\to C_1\times C_2$ be the section $\sigma_1(x) = (x, x_2)$, $\sigma_2(x) = (x_1, x)$. Then for a finite etale cover $X\to C_1\times C_2$, the composition $X\to C_1\times C_2\to C_i$ has connected fibres if and only if the pull-back of $X$ along $\sigma_{2-i}$ is connected. So now the question is equivalent to: suppose that $S$ is a finite set with more than one element with an action of $F=F_1\times F_2$. Is it possible that $F_1$ and $F_2$ both act transitively on $S$? It is very easy to construct such examples. The easiest one could be $S$ with two elements, with every generator of each $F_i$ acting by a nontrivial involution. If there are only two punctures on each curve, this coincides with Francesco Polizzi's construction, and we see that his example is in some sense minimal.<|endoftext|> TITLE: "Dimension" of discrete subgroups of infinite covolume in Lie groups QUESTION [5 upvotes]: Let $G$ be a semisimple Lie group with finite center, $K$ a maximal compact subgroup, and $d=\dim(G/K)$. Let $\Gamma$ be a non-cocompact discrete subgroup of $G$. [Edit: assume that $\Gamma$ is virtually torsion-free (which is automatic if $G$ is linear and $\Gamma$ is finitely generated).] Is it true that the virtual cohomological dimension (vcd) of $\Gamma$ is $ TITLE: Rigid versus log-rigid cohomology for semistable varieties QUESTION [7 upvotes]: If $K$ is a p-adic field, with maximal unramified subfield $K_0$, and $X$ is a proper semi-stable $O_K$-scheme, then there's a canonical way to make the special fibre $X_k$ into a log-scheme; and there's a theory of log-rigid cohomology adapted to such things, which gives $K_0$-vector spaces $H^i_{log-rig}(X_k / K_0)$ equipped with a Frobenius $\varphi$ and a monodromy operator $N$. However, if we forget the log-structure, we can also make sense of the rigid cohomology of $X_k$: rigid cohomology is defined for any $k$-variety (not necessarily smooth), and this gives spaces $H^i_{rig}(X_k / K_0)$ which have Frobenius actions (but not monodromy). How are the groups $H^i_{rig}(X_k / K_0)$ and $H^i_{log-rig}(X_k / K_0)$ related? I tried writing everything down for a semistable elliptic curve and convinced myself that in this case, the rigid cohomology was isomorphic to the kernel of the monodromy operator on the log-rigid cohomology. Is this true more generally? REPLY [6 votes]: $\require{AMScd}$I'll expand a little on my comment to give an answer to David's follow up question: Firstly, the general relationship is described in Chiarellotto's Duke 1999 paper "Rigid cohomology and invariant cycles for a semistable log scheme". He shows that the weight-monodromy conjecture implies that there is an exact sequence \begin{equation*} H_{rig}^{n}(X_{k}/K_{0})\rightarrow H_{log-cris}^{n}(X_{k}/K_{0})\xrightarrow{N}H_{log-cris}^{n}(X_{k}/K_{0}) \end{equation*} This is unconditional if, for example, $X$ is a semistable family of curves or surfaces. In these cases the first map is an injection for $n=1$. For searching purposes, the elements of $\ker N$ are often called "$p$-adic invariant cycles". As for your second question, Chiarellotto states in the same paper that one cannot expect the first arrow to be an injection for $n\geq 2$ in general. Here is my guess as to why: For simplicity, let us suppose that there exists a proper scheme $P$ over $W(k)$ such that $X\hookrightarrow P$ and $P$ is smooth around $X_{k}$. Let $X_{k}=\bigcup_{i\in I}X_{k,i}$ be the decomposition of $X_{k}$ into irreducible components. Then $\left\{]X_{k,i}[_{X_{K_{0}}^{rig}}\right\}_{i\in I}$ is an admissible covering of the rigid analytic space $X_{K_{0}}^{rig}$ associated to $X_{K_{0}}$, and similarly $\left\{]X_{k,i}[_{P_{K_{0}}^{rig}}\right\}_{i\in I}$ is an admissible cover of $]X_{k}[_{P_{K_{0}}^{rig}}$. The Čech cohomology of the first cover computes $H_{log-cris}^{\ast}(X_{k}/K_{0})$ and the Čech cohomology of the second cover computes $H_{rig}^{\ast}(X_{k}/K_{0})$. Let us call the first map in above sequence $\psi^{\ast}$; it is the map on cohomology induced by $\psi:X_{K_{0}}^{rig}\rightarrow ]X_{k}[_{P_{K_{0}}^{rig}}$. It induces the following commutative diagram where the rows are the Mayer-Vietoris sequences for the two coverings: \begin{CD} \displaystyle\bigoplus_{i\in I}H_{dR}^{1}(]X_{k,i}[_{X_{K_{0}}^{rig}}) @>\alpha>> \displaystyle\bigoplus_{i>> H_{log-cris}^{2}(X_{k}/K_{0}) @>>> \displaystyle\bigoplus_{i\in I}H_{dR}^{2}(]X_{k,i}[_{X_{K_{0}}^{rig}}) @>\beta>> \displaystyle\bigoplus_{i\alpha'>> \displaystyle\bigoplus_{i>> H_{rig}^{2}(X_{k}/K_{0}) @>>> \displaystyle\bigoplus_{i\in I}H_{dR}^{1}(]X_{k,i}[_{P_{K_{0}}^{rig}}) @>\beta'>> \displaystyle\bigoplus_{i\alpha>> \displaystyle\bigoplus_{i>> H_{log-cris}^{2}(X_{k}/K_{0}) @>>> \displaystyle\bigoplus_{i\in I}H_{rig}^{2}(U_{i}/K_{0}) @>\beta>> \displaystyle\bigoplus_{i\alpha'>> \displaystyle\bigoplus_{i>> H_{rig}^{2}(X_{k}/K_{0}) @>>> \displaystyle\bigoplus_{i\in I}H_{cris}^{2}(X_{k,i}/K_{0}) @>\beta'>> \displaystyle\bigoplus_{i>> \mathrm{coker}(\alpha) @>>> H_{log-cris}^{2}(X_{k}/K_{0}) @>>> \ker(\beta) @>>> 0\\ @. @AAA @A\psi^{\ast}AA @AAA @.\\ 0 @>>> \mathrm{coker}(\alpha') @>>> H_{rig}^{2}(X_{k}/K_{0}) @>>> \ker(\beta') @>>> 0 \\ \end{CD} and from the Gysin sequences we can see that the left vertical arrow is an injection, but the kernel of the right vertical arrow will be \begin{equation*} \ker\left(\bigoplus_{i\in I}\rho_{i}(H_{rig}^{0}(D_{i}/K_{0})\rightarrow\bigoplus_{i TITLE: prime ideals minimal over a zerodivisor QUESTION [6 upvotes]: Let $R$ be a commutative ring with identity. If $P$ is a prime ideal of $R$ that is minimal over some zerodivisor of $R$, then must $P$ consist only of zerodivisors? I suspect not but I can't figure out how to construct a counterexample. (The full context of the situation I am in is this. Suppose that $P$ is a prime ideal of $R$ such that $PP^{-1} \neq P$. Then there is an element $a$ of $P$ and an element $b$ of $R-P$ such that $P = (aR :_R bR)$. In this case, $P$ is necessarily minimal over $a$. My question is, if $a$ is a zerodivisor, can I conclude that $P$ consists only of zerodivisors?) REPLY [4 votes]: For the first question, consider $R = \mathbb{Z}[X]/(X^3 - 1)$, $P = (x+ 1)$ and $a = (x + 1)(1 + x + x^2)$ where $x$ denotes the image of $X$ in $R$. In order to see that it provides us with a reduced one-dimensional Noetherian counter-example, note that $R/P \simeq \mathbb{Z}/2\mathbb{Z}$ and that $(x - 1)a = 0$. Hence $P = (2, x -1)$ is a maximal ideal of $R$ and $a$ is a zero-divisor. Any other prime ideal of $R$ containing $a$ has to contain $Q = (1 + x + x^2)$. As $x + 1$ doesn't divide $1 + x + x^2$ (since otherwise $1 + x + x^2$ would map to an even integer modulo $(x - 1)$), $Q$ is not a subset of $P$. Therefore $P$ is a minimal prime over $a$ which contains regular elements, namely $2$ and $x + 1$. As the ideal $P$ is generated by the regular element $x + 1$, it is invertible, i.e., it satisfies $PP^{-1} = R \neq P$. It is easy to check that $P = ((a) :_R (1 + x + x^2))$, so that this counter-example satisfies also your extra requirements. On the positive side, it is well-known that any minimal prime of a commutative ring with identity consists only of zero-divisors. Thus the answer is yes if $a = 0$. Side note. In [1], D. Anderson and J. Pascual refers to the definition of Property (A): A commutative ring $R$ with identity enjoys property (A) if every faithful finitely generated ideal $I$ (i.e., $I$ is finitely generated and $(0:_R I) = 0$) is regular (i.e., $I$ contains at least one regular element). The authors remark that a ring in which $0$ has a primary decomposition satisfies Property $(A)$. I am not sure if this can be helpful in your context, but if your ring $R$ has Property $(A)$, then $P$ consists solely of zero-divisors if and only if $P$ is not faithful, which may be easier to check. [1] D. Anderson and J. Pascual, "Regular Ideals in Commutative Rings, Sublattices of Regular Ideals, and Prüfer Rings", 1987.<|endoftext|> TITLE: Earliest use of the term "Galois extension"? QUESTION [8 upvotes]: Does anyone know the earliest use of the term "Galois extension"? I thought it might be in Emil Artin's Notre Dame lectures but I couldn't find it there. (He does use the terms "normal" and "separable.") REPLY [10 votes]: Not an answer, but an extended comment. Probably the terminology is not due to Artin. The Notre Dame lectures were in 1942 and published in 1944. In Emil Artin's 1947 lecture notes (notes taken by Albert A Blank, and seems to have been printed by Courant Institute; I'm not sure as I only have a photocopy) what normally is now called a Galois extension he calls a "normal extension" (he used normal differently from how it is used today). This is also corroborated by a footnote in Hungerford's Algebra. In my edition the definition of a Galois extension (Definition 2.4 in Chapter V) has a footnote that reads (emphasis mine): A Galois extension is frequently required to be finite dimensional or at least algebraic ... equivalent to the usual one. Our definition is essentially due to Artin, except that he calls such an extension "normal." Since this use of "normal" conflicts (in case char $F\neq 0$) with the definition of "normal" used by many other authors, we have chosen to follow Artin's basic approach, but to retain the (more or less) conventional terminology.<|endoftext|> TITLE: Beck-Chevalley condition on pushouts QUESTION [6 upvotes]: Let $C$ be a regular category with pushouts and $S(X)$ is the lattice of subjects of $X$. For every arrow $f\colon X\to A$, pulling back along $f$ gives a map $f^*\colon S(A)\to S(X)$ which has a left adjoint $f_*$. These maps satisfy the Beck-Chevalley condition on a square $fg=hk$ if $h^*f_*=k_*g^*$. The BC condition holds on all pullbacks. Does it hold also on pushouts? If no, do you know a class of regular categories where this is true (exact, abelian, pretoposes...)? REPLY [5 votes]: No, the Beck-Chevalley condition does not hold for all pushout squares in a regular category, and not even if the category is exact, or a pretopos, or even a topos. In fact, here is a counterexample in $\rm Set$. Let $B = \{a,b\}$ and $C = \{\alpha,\beta\}$ and $A=\{0,1,2\}$, define $f:A\to B$ by $f(0)=f(1)=a$ and $f(2)=b$, and $g:A\to C$ by $g(0)=\alpha$ and $g(1)=g(2)=\beta$. Let $P$ be the pushout of $f$ and $g$ with coprojections $p:B\to P$ and $q:C\to P$. Then $$q(\alpha) = q(g(0)) = p(f(0)) = p(a) = p(f(1)) = q(g(1)) = q(\beta) = q(g(2)) = p(f(2)) = p(b)$$ so $P$ is a one-element set. This, if we let $U=\{b\} \in S(B)$, then $p_!(U)$ is all of $P$, hence $q^* p_!(U)$ is all of $C$. (The notation $p_!$ for left adjoints of $p^*$ is, I think, more common than $p_*$ in this field, the latter being used more often for right adjoints.) But $f^*(U) = \{2\}$, so $g_! f^*(U) = \{\beta\} \neq q^* p_!(U)$. There is something positive that can be said, however. Suppose $P = B \amalg_A C$ is a pushout of $f:A\to B$ and $g:A\to C$ such that the induced map $\Delta : A \to B\times_P C$ is a regular epimorphism. Then the pullback square defining $B\times_P C$, consisting of $h:B\times_P C\to B$ and $k:B\times_P C\to C$ with $p:B\to P$ and $q:C\to P$, does satisfy the Beck-Chevalley condition. Moreover, since $\Delta$ is a regular epimorphism, we have $\Delta_! \Delta^* = \rm Id$, and therefore $$ g_! f^* = k_! \Delta_! \Delta^* h^* = k_! h^* = q^* p_! $$ so the original pushout square does satisfy the Beck-Chevalley condition. The above "zig-zag" counterexample is simply the "minimal" situation in which $\Delta$ fails to be surjective. (One might call it "the minimal way to violate the hypotheses of the baby Blakers-Massey theorem".) Thus, it seems that $\Delta$ being a regular epi is probably the best possible hypothesis.<|endoftext|> TITLE: Adjoint of norm map QUESTION [5 upvotes]: Let $G$ be a finite group acting trivially on a spectrum $X$. Then the norm map $X_{hG} \to X^{hG}$ may be thought of as a map $\Sigma^\infty_+ BG \wedge X \to F(\Sigma^\infty_+ BG , X)$. By adjointness, this corresponds to a map $\Sigma^\infty_+BG\wedge \Sigma^\infty_+ BG \wedge X \to X$. By naturality, it really suffices to consider the case where $X=S$ is the sphere spectrum, so we have a pairing $\Sigma^\infty_+ BG \wedge \Sigma^\infty_+ BG \to S$. Question: What is this pairing? All I can tell is that it restricts to the multiplication-by-$|G|$ map on $S$. I'd be happy to understand the case $G = C_p$, $p$-locally. REPLY [4 votes]: To elaborate the information in the comments: Inside $G\times G$ you have the diagonal subgroup $\Delta$. This gives a map $$ p\colon E(G\times G)/\Delta\to E(G\times G)/(G\times G) $$ This is a covering map, so it induces a transfer map $p^!$ of suspension spectra in the opposite direction. We can identify $E(G\times G)/(G\times G)$ with $B(G\times G)=BG\times BG$, and $E(G\times G)/\Delta$ with $B\Delta\simeq BG$, so we have $p^!\colon BG_+\wedge BG_+\to BG_+$. There is also a collapse map $BG_+\to S^0$, and by composing these, we get a stable map $BG_+\wedge BG_+\to S^0$. This is the map that you are asking about. A proof is contained in Theorem 8.3 of my paper "K(n)-local duality for finite groups and groupoids". As you mention, that paper works in the $K(n)$-local context, but that does not really play any role in the proof of Theorem 8.3; it is easy to extract a proof of the unlocalised statement. Nonetheless, the proof is more complicated than I feel it ought to be; it would be interesting to revisit the question using more modern foundations.<|endoftext|> TITLE: Reference request: Oldest number theory books with (unsolved) exercises? QUESTION [10 upvotes]: Per the title, what are some of the oldest number theory books out there with (unsolved) exercises? Maybe there are some hidden gems from before the 20th century out there. I am already aware of the books of Dickson and Hardy. Motivation for this question. Some person came up to me before class and asked, are you the person asking the ridiculous "oldest book with exercises" questions on MO? I said yes, and he asked I could do one on number theory. So here we are. REPLY [7 votes]: The book "Théorie des nombres, Tome premier" by Edouard Lucas was published 1891. Many of the "Exemples" are actually exercises left to the reader. A scan is freely available in the archive.<|endoftext|> TITLE: The (co)tangent sheaf of a topological space QUESTION [9 upvotes]: Let $X$ be a topological space (assume additional assumptions if needed) and denote by $\mathcal O _X$ its sheaf of $\Bbbk$-valued continuous functions where $\Bbbk$ is $\mathbb{R}$ or $\mathbb{C}$ with standard topology. Then, as it is done in the differentiable setting or in algebraic geometry, one can define the following objects $$T_X:=\mathscr{Der}_\Bbbk (\mathcal O_X,\mathcal O_X)$$ the tangent sheaf, i.e. the sheaf of $\Bbbk$-linear derivations of $\mathcal O_X$ with values in $\mathcal O_X$ (on local sections, $\Bbbk$-linear maps $D:\mathcal O_X(U)\to\mathcal O_X(U)$ satisfying Leibniz: $D(f\cdot g)=f\cdot Dg + g\cdot Df$), and $$\Omega_X^1:=\mathcal I/\mathcal I^2$$ the sheaf of differentials, where $\mathcal I$ is the ideal sheaf of $X$ embedded diagonally $\Delta:X\hookrightarrow X\times X$ into $X\times X$ (i.e. $\mathcal I(U)=$ functions in $\mathcal O_{X\times X}(U)$ that are zero on every point of $\Delta(X)\subset X\times X$). Well, what can be said about these two sheaves? Anything interesting at all? Also, is there any relationship between $T_X$ and the "tangent microbundle" $\tau_X$ in case $X$ is a topological manifold? REPLY [11 votes]: Your $$ is always $0$. If $$ is a derivation and $$ is a function, then for every point $$ $$ vanishes at $$; it suffices to prove this when $()=0$, and in that case $=ℎ$ where $()=0=ℎ()$, so $=ℎ+ℎ$ vanishes at $$.<|endoftext|> TITLE: Mathematics of imaging the black hole QUESTION [52 upvotes]: The first ever black hole was "pictured" recently, per an announcement made on 10th April, 2019. See for example: https://www.bbc.com/news/science-environment-47873592 . It has been claimed that state-of-the-art imaging algorithms were an enabler for this historic success. Does anybody care to describe the difficulties, and (quite certainly non-trivial) mathematics that went into this effort ? REPLY [45 votes]: Essential elements$^*$ of the reconstruction algorithm were developed at MIT under the name CHIRP = Continuous High-resolution Image Reconstruction using Patch priors, as described in Computational Imaging for VLBI Image Reconstruction (2015). The difficulty of VLBI (Very Long Baseline Interferometry Image) reconstruction is that the inversion problem is highly-ill posed, there are many images that explain the data. The challenge is to find an explanation that respects our prior assumptions about the “visual” universe while still satisfying the observed data. Bayesian approaches are generally employed for that purpose, in CHIRP machine learning is used to automatically identify visual patterns --- obviating the need for hand training of the algorithm. A key technical innovation is a way to correct for the delays in the signal received from the various telescopes. The delays are difficult to predict, since they depend the local variations in the speed of the radio waves through the noisy atmosphere. CHIRP adopts an algebraic solution known as phase closure to this problem: If the measurements from three telescopes are multiplied, the extra delays caused by atmospheric noise cancel each other out. One test case that shows the resolving power of CHIRP, compared to a competing algorithm (BU) is shown below (taken from the MIT paper). Notice how CHIRP is able to resolve 2 separate, previously unresolved, bright emissions in the blazar OJ287. $^*$ UPDATE: This statement must be qualified, as stated by the lead author, Katie Bouman: “No one person or algorithm made the image. It was actually made by combining the images produced by 3 separate imaging pipelines, leading to a more powerful result than we could ever achieve with any single method.”<|endoftext|> TITLE: Reference request: uniformization theorem QUESTION [6 upvotes]: I would appreciate if someone could point me to some introductory literature/resources where I can learn about Poincaré's uniformization theorem at a basic level. Any good powerpoint notes, short papers or video lectures would be nice. I want to learn about the result in general, the proof, how it relates to other important theorem's in geometry and possible real-life applications. Thank you. REPLY [2 votes]: Let me second Alex Eremenko's suggestion for Donald Marshall, Complex Analysis, Cambridge 2019. The proof is based on the new notion of dipole Green's function, and is especially interesting in view of the following reasons: It does not require second countability in the definition of a Riemann surface (see Alex's answer), but rather obtain it as a corollary of the uniformization theorem. It gives existence of meromorphic functions that separate points for arbitrary Riemann surfaces.<|endoftext|> TITLE: Lower Bound of KL-Divergence Between Two Gibbs Measures QUESTION [8 upvotes]: Suppose we have two Gibbs measures with densities $$ p_f(x) \propto \exp(f(x)),\quad q_g(x)\propto \exp(g(x)). $$ Consider the KL-divergence between $p_f$ and $q_g$, as a functional of $f$ and $g$, that is, $$ D(f, g) := \text{KL}(p_f \| q_g). $$ Question: Do we have the following lower bound: $$ D(f, g) \geq \|f - g\|^2, $$ where we are interested in, for example, the $L_2$-norm of $f-g$. REPLY [5 votes]: The answer is no. Indeed, your question can be restated as follows: Is it true that for some constant $c>0$ we have $$\int P\ln\frac PQ\,d\mu\ge c\int(\ln P-\ln Q)^2\,d\mu, \tag{1} $$ where $P$ and $Q$ are probability densities with respect to a measure $\mu$? Let $\mu$ be the counting measure on the set $\{0,1\}$, and let $P(0)=p\in(0,1)$ and $Q(0)=q\in(0,1)$. Then (1) will become $$p\ln\frac pq+(1-p)\ln\frac{1-p}{1-q}\ge c\Big[\Big(\ln\frac pq\Big)^2+\Big(\ln\frac{1-p}{1-q}\Big)^2\Big]. \tag{2} $$ Letting now $p\to0$, we get a contradiction, because the left-hand side of (2) will go to $\ln\frac1{1-q}<\infty$, but the right-hand side of (2) will go to $\infty$.<|endoftext|> TITLE: Obstructions for finding a volume form for a given flow QUESTION [5 upvotes]: Let $X$ be a non-singular $C^\infty$ vector field on a three manifold $M$. There are some obvious obstructions for finding a volume form that is preserved under the flow given by $X$: If $X$ is singular, or If there's a sphere to which $X$ is transverse (which implies a singularity), or if $X$ is transverse to a torus. Could it be that these are all the obstructions? Or is there a theorem characterizing all topological obstructions? I have a Riemannian metric on the manifold to begin with, and I want to change the metric so that the vector field (that avoids the above obstructions) becomes divergence free, but is still smooth or at least $C^1$. Thanks for any hints/answers/references! REPLY [2 votes]: The existence of invariant volume form for a vector field $X$ is cohomological problem, i.e. one has to solve a cohomological equation to guarantee the existence of such a form. In fact, if $\Omega_0$ denotes an arbitrary volume form, one defines the $\Omega_0$-divergence of $X$ as the function $\mathrm{div} X\colon M\to\mathbb{R}$ such that $\mathcal{L}_X\Omega_0 = (\mathrm{div} X)\Omega_0$, where $\mathcal{L}_X$ denotes the Lie derivative. Then, if one considers a function $u\colon M\to\mathrm{R}$ and the volume form $\Omega := e^{-u}\Omega_0$, one can show that $\Omega$ is $X$-invariant if and only if $$Xu = \mathrm{div}X.$$ So, in order the existence of an invariant volume form is equivalent to prove the existence of a solution $u$ for this PDE, and there is no general theory to study such equations. As you mention above, there is some clear obstruction to solve such an equation, but in general we don't know how to characterize all of them. In some particular cases, for instance when the vector field induces an Anosov flow, we know that periodic orbits are all the obstructions to solve any (sufficiently smooth) cohomological equation, and so, such a vector field admits an invariant volume form if an only if the integral of $\mathrm{div} X$ on every periodic orbit is equal to zero. This is the so called Livsic theorem. A very good survey about cohomological equations is the following one by A. Katok and E. Robinson: http://www.personal.psu.edu/axk29/pub/KR.pdf<|endoftext|> TITLE: Lower bound for a combinatorial problem ($N$ students taking $n$ exams) QUESTION [10 upvotes]: We have $N$ students and $n$ exams. We need to select $n$ out of the students using the grade of those exams. The procedure is as follows: 1- We set some ordering on the exams. 2- Going through this order, In every exam, the best student will be selected and we forget about his/her grade in the next exams. Now we have all grades and we know that before setting the ordering on the exams, all the student are hopeful to be selected. How large can $N$ be in terms of $n$? (As far as I remember, we proved a $2n+c$ lower bound and an $O(n\log n)$ upper bound for $N$.) REPLY [3 votes]: My initial argument was erroneous. I'm rewriting everything completely. Denote by $f(n)$ the maximal value of $N$ for that value of $n$. I claim that $$ f(k+\ell)\geq f(k)+f(\ell)+k \qquad\text{for all $k\leq \ell$.} \qquad\qquad(*) $$ In a standard way, this shows that $f(n)\geq 1+\frac12n\log_2 n$. To prove $(*)$, assume that we have $k+\ell$ exams. Take a set $X$ of $f(k)$ students hopeful for the first $k$ exams and hopeless for the last $\ell$, and take a set $Y$ of $f(\ell)$ students performing in the opposite way. Finally, take a set $Z$ of $k$ students, each of which wins his own exam in the first group and his own exam in the last group and not attenfing the others. Now, if the first $k$ exams are taken into account first, then $Z$ win their exams and become out of competition, so $Y$ are still hopeful. Similarly, $X$ stay hopeful if the last $\ell$ exams are taken into account first.<|endoftext|> TITLE: Maps from mod-$p$ Eilenberg-MacLane spectrum to connective $K$-theory spectrum QUESTION [10 upvotes]: Let $ku$ be the connective cover of the complex $K$-theory spectrum $KU$. Consider the mod-$p$ Eilenberg-MacLane spectrum $H\mathbb{Z}/p$. I want to see that $[H\mathbb{Z}/p,ku]=0$. Since $H\mathbb{Z}/p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories") the result will follow once we know that $ku$ is harmonic (Corollary 4.11 in loc. cit.) Is it true that the spectrum $ku$ is harmonic? If not, how can I prove the claim? Certainly, an idea is to use directly the theorem of Margolis (“Eilenberg-MacLane Spectra”): For any spectrum $Y$ of finite type, there is an isomorphism $$[H\mathbb{Z}/p,Y]\to \text{Hom}_{\mathcal{A}}(H^{*}(Y,\mathbb{Z}/p),\mathcal{A})$$ where $\mathcal{A}$ is the Steenrod algebra. Is it true that $\text{Hom}_{\mathcal{A}}(H^{*}(ku,\mathbb{Z}/p),\mathcal{A})=0$? If the answer is yes, why is that? REPLY [15 votes]: You can also cheat: since HZ/p is connective,$$ [HZ/p,ku] = [HZ/p,KU] = [HZ/p \wedge KU,KU]_{KU-modules} = 0$$ since $HZ/p \wedge KU = 0$. Edit: Let me stress that the other answers give more information, namely the calculation of maps from HZ/p to ku in all degrees.<|endoftext|> TITLE: What is the official definition of $\mathcal{M}_g$ as an orbifold, and how much can I ignore it? QUESTION [9 upvotes]: There is a well-known description of $\mathcal{M}_g$ as $\mathcal{T}_g/\Gamma$ where $\mathcal{T}_g$ is the Teichmuller space and $\Gamma$ is the mapping class group. Teichmuller space is homeomorphic to a ball. But the quotient here is not exactly a manifold because $\Gamma$ has fixed points. Let's just consider the case $g = 1$, where $\mathcal{T}_g = \mathbf{H}$ and $\Gamma = \text{PSL}(2,\mathbf{Z})$. The quotient $\mathbf{H}/\text{PSL}(2,\mathbf{Z})$ has two orbifold points. Most talks I've seen about this topic ignore the orbifold behavior and pretend that $\mathcal{M}_g$ is a manifold. It seems the general sentiment is that this does not hurt us for most things we are interested in. But I would like to actually understand what the space really is at these points. So my general question is: What is the official definition of an orbifold? What is, for instance, a vector bundle (such as the Hodge bundle) on an orbifold? Is there a general reference somewhere that tells us what kinds of things about manifolds can be imported freely over to orbifolds, and which things require more delicate analysis? Explicit descriptions in the case of the modular curve above would be very welcome. REPLY [3 votes]: Not being a moduli space expert, I will give an answer to the more general question. To me an orbifold is a way to recover the action of a group $G$ on a space $X$ in terms of the quotient object $G\backslash X$. In the nicest possible case, $G$ acts freely, properly, and cocompactly on $X$. From the quotient $G\backslash X$ we can recover $G = \pi_1(G\backslash X)$ and recover the action of $G$ on $X$ via the deck transformations of $\pi_1(G\backslash X)$ on $X$. When the the action is no longer free, but the action is still nice, e.g. the action of the modular group on Teichmuller space, then information is lost in the quotient $G\backslash X$. An orbifold structure on $G\backslash X$ is the extra structure required to recover the action of $G$ on $X$. Every point of $G\backslash X$ is either regular if its lifts in $X$ have trivial stabilizers, or singular if its lifts have non-trivial stabilizers. The orbifold structure on $G\backslash X$ is the following extra information: at each singular point $x$ we record the action of the stabilizer of a lift $G_{\tilde x}$ on a small neighborhood $\tilde x \in \tilde U \subset X$. This group $G_{\tilde x}$ is called the isotropy group of $x$. This extra local information will be enough to patch together all the local actions in order to recover the orbifold universal cover $X \twoheadrightarrow G\backslash X$. You asked about what can be ignored. Here is a list of increasing detail: Simply the quotient $G\backslash X$. $G\backslash X$ where we record the order of the isotropy groups of each singular point. $G\backslash X$ where we record the isomorphism type of each isotropy group of each singular point. The full orbifold: where we record the how the isotropy groups act on small neighbourhoods in $X$. In my experience, sometimes just level 2 of detail is sufficient depending on what I need to prove. In the case of $\mathcal M_g$ the points represent shapes of the underlying surface (with fat and skinny bits.) By Nielsen realization all finite groups fix points, so in the full orbifold quotient the isotropy groups represent symmetries of a given shape. For example if your surface is very "lumpy", then it shouldn't have any symmetries so the shape should represent a regular point in $\mathcal M_g$. On the other hand a very symmetric shape would correspond to a singular point in $\mathcal M_g$ because you can flip it, reflect it, etc and the symmetry group will be represented by the isotropy group.<|endoftext|> TITLE: Relationship between Gromov-Witten and Taubes' Gromov invariant QUESTION [6 upvotes]: Fix a compact, symplectic four-manifold ($X$, $\omega$). Recall Taubes' Gromov invariant is a certain integer-valued function on $H^2(X; \mathbb{Z})$ defined by weighted counts of pseudoholomorphic curves in $X$. In particular, this count combines curves of any genus. On the other hand, the Gromov-Witten invariants for a fixed genus $g$ and homology class $A \in H_2(X; \mathbb{Z})$ are (very roughly) integers derived from the "fundamental class" of the moduli space $\mathcal{M}_g^A(X)$ of pseudoholomorphic maps from a surface of genus $g$ into $X$ representing the class $A$. Is there any sort of relationship, conjectural or otherwise, between these two invariants that is stronger than "they both count holomorphic curves"? Of course, this would require looking at Gromov-Witten invariants for any genus $g$. I personally don't understand the best way to define these for genus $g > 0$. I do understand that Zinger has a construction for $g = 1$ that is a bit more refined than looking at the entire moduli space. REPLY [6 votes]: Ionel and Parker worked out the relationship between Taubes' Gromov invariant and the usual Gromov--Witten invariants (which they refer to as "Ruan--Tian invariants") in this paper: https://arxiv.org/abs/alg-geom/9702008<|endoftext|> TITLE: open book decompositions of $\Sigma\times S^1$ QUESTION [6 upvotes]: Let $\Sigma$ be a closed orientable surface. Is there a standard open book decomposition on the $3$-manifold $M=\Sigma\times S^1$? If the binding is allowed to be empty in the definition of an open book decomposition, then this is obvious, since $M$ is the mapping torus of the identity of $\Sigma$. The literature is not clear on this, however. If the binding must be non-empty, then since it is contained in every page, it seems that the obvious fibering $M\to S^1$ is never the fibering of an open book. REPLY [5 votes]: I'm not sure about "preferred", but here is the simplest open book decomposition of $\Sigma_g \times S^1$ that I know: Take the open book $(\Sigma_{g,1},\rm{id})$ with page the genus-$g$ surface with one boundary component and trivial monodromy. The result of page-framed surgery (ie $0$-surgery) on the binding component will give us $\Sigma_g \times S^1$. There are several ways to implement that surgery on the open book, but the simplest (AFAIK) is to do a positive stabilisation (along a boundary-parallel arc) to get $(\Sigma_{g,2},T_{\partial_1})$ and then add a negative Dehn twist to $\partial_2$ (which is the same knot as we started with, but now with page framing $-1$). So $(\Sigma_{g,2},T_{\partial_1}T^{-1}_{\partial_2})$ will give you an open book for $\Sigma_g \times S^1$. Notice that when $g=0$, this recovers the standard open book $(S^1 \times [0,1],\rm{id})$ for $S^2 \times S^1$.<|endoftext|> TITLE: Unitary representations of finite groups over finite fields QUESTION [9 upvotes]: I would like to learn the basic theory of unitary representations of finite groups over finite fields. Here, the unitary group $\operatorname{GU}(n,\mathbb{F}_{q^2})$ consists of all invertible transformations of $\mathbb{F}_{q^2}^n$ that preserve the Hermitian form $\langle x, y \rangle = \sum_{i \in [n]} x_i y_i^q$, and "unitary representation" means a group homomorphism $\rho \colon G \to \operatorname{GU}(n,\mathbb{F}_{q^2})$. This is a special case of the usual notion of a representation $\rho \colon G \to \operatorname{GL}(n,\mathbb{F}_{q^2})$. Over the complex numbers, every representation $\rho \colon G \to \operatorname{GL}(n,\mathbb{C})$ of a finite group $G$ is similar to a unitary representation $\rho' \colon G \to \operatorname{GU}(n,\mathbb{C})$, in the sense that there is an invertible operator $M$ such that $\rho'(g) = M\rho(g) M^{-1}$ for every $g \in G$. In this sense and others, the theory of unitary representations over $\mathbb{C}$ is essentially the same as that of ordinary representations. However, over finite fields the notions are distinct. If $G$ is a finite group and $\rho \colon G \to \operatorname{GL}(n,\mathbb{F}_{q^2})$ is a representation, there might not be an invertible operator $M$ such that $M \rho(g) M^{-1} \in \operatorname{GU}(n,\mathbb{F}_{q^2})$ for every $g \in G$. For example, $\mathbb{Z}_5$ has a faithful 2-dimensional representation over $\mathbb{F}_{3^2}$ that is not similar to any unitary representation, since 5 divides $|\operatorname{GL}(2,\mathbb{F}_{3^2})|$ but not $|\operatorname{GU}(2,\mathbb{F}_{3^2})|$. Question: Have unitary representations of finite groups over finite fields been systematically studied, and if so where can I learn the basics? Here is one example of what I want to learn to do: Describe all the unitary representations of the dihedral group of order 8 when $q=11$. At the moment I do not even know how to: Describe all the unitary representations of $\mathbb{Z}_2 \times \mathbb{Z}_2$ when $q=3$. Some other things I want to learn include: Where Maschke's Theorem holds (i.e. $(|G|,q) = 1$ so that $\mathbb{F}_{q^2}[G]$ is semisimple), does every unitary representation decompose as an orthogonal direct sum of irreducible unitary subrepresentations? Again where Maschke's Theorem holds, is there any analogue of the Peter-Weyl Theorem to give the space $L^2(G)$ of functions $f \colon G \to \mathbb{F}_{q^2}$ an orthogonal basis consisting of matrix elements for irreducible unitary representations? What are some conditions for a modular representation to be similar to a unitary representation? (i.e. which subgroups of $\operatorname{GL}(n,\mathbb{F}_{q^2})$ are conjugate with subgroups of $\operatorname{GU}(n,\mathbb{F}_{q^2})$?) Bonus for answers understandable to a humble analyst. REPLY [8 votes]: We had to deal with this problem when classifying maximal subgroups of the finite classical groups, which is the aim of our book: The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, by John N. Bray, Derek F. Holt, Colva M. Roney-Dougal. The most difficult maximal subgroups to classify, are those in the so-called Aschbacher class ${\mathscr S}$, consisting of absolutely irreducible subgroups that are almost simple mod scalars. Many of these arise as reductions of complex representations over finite fields. Tables of complex representations of groups that are close to simple are available up to dimension about $250$, but we needed to know which classical group the reduction lies in, which means identifying the fixed form. We generally relied on Lemma 4.4.1 of the book, which says: For a given absolutely irreducible representation over ${\mathbb F}_{q^2}$ of a group $G$, with Frobenius-Schur indicator $\circ$, the image of $G$ under the representation consists of isometries of a unitary form if and only if the action of the field automorphism $\sigma :x \to x^q$ on the Brauer character is the same as complex conjugation. In many cases, such as when $q$ is coprime to the group order, the Brauer character is just the ordinary complex character. As an example, the reduction of the complex representation of degree $3$ of the $3$-fold cover $3.A_6$ of $A_6$ lies in ${\rm PSL}(3,p)$ for primes $p \equiv 1,4 \bmod 15$, in ${\rm PSU}(3,p)$ (as a subgroup of ${\rm PSL}(3,p^2)$) when $p \equiv 11,14 \bmod 15$ (or when $p=5$), and in ${\rm PSL}(3,p^2)$ without preserving a unitary form when $p \equiv 2,3 \bmod 5$.<|endoftext|> TITLE: Monodromy groups from Galois's viewpoint QUESTION [6 upvotes]: According to the Wikipedia article about monodromy, the monodromy group can be defined in terms of Galois theory in following way: Let $F(x)$ denote the field of the rational functions in the variable $x$ over the field $F$, which is the field of fractions of the polynomial ring $F[x]$. An element $y = f(x)$ of $F(x)$ determines a finite field extension $[F(x) : F(y)]$. This extension is generally not Galois but has Galois closure $L(f)$. The associated Galois group of the extension $[L(f) : F(y)]$ is called the monodromy group of $f$. In the case of $F = \mathbb{C} $, Riemann surface theory enters and allows for the geometric interpretation given above. In the case that the extension $[\mathbb{C}(x) : \mathbb{C}(y)]$ is already Galois, the associated monodromy group is sometimes called a group of deck transformations. This has connections with the Galois theory of covering spaces leading to the Riemann existence theorem. Furthermore, it is explicitly remarked that for the field $F = \mathbb{C} $, this definition of monodromy coincides with the classical one in light of complex analysis and cover theory. My question is: how, explicitly, do these two definitions of monodromy correspond to each other for $F = \mathbb{C} $? How do we obtain a cover map giving rise to the topological monodromy action from the field extensions above? REPLY [8 votes]: All of the following can be found in the third chapter of Szamuley's "Galois Groups and Fundamental Groups" but I will try to sum it up a bit: I think the starting point of making this precise is the fact that the function field $\mathbb{C}(t)$ is isomorphic to the field of meromorphic functions on the Riemann sphere $ \textbf{P}^1(\mathbb{C})$. In fact, mapping a Riemann suface to its field of meromorphic functions, gives a contravariant functor $M(-)$ from the category of Riemann surfaces to the category of commutative rings. Now it is shown in Theorem 3.3.7 of Szamuely that if we restrict this functor to the category of connected compact Riemann surfaces equipped with map to $ \textbf{P}^1(\mathbb{C})$, then $M(-)$ induces a contravariant equivalence between this category and the category of finite field extensions of $M(\textbf{P}^1(\mathbb{C})) \cong \mathbb{C}(t)$. Now it is shown before in Szamuely that if we have a compact Riemann surface and a map $\phi:X \rightarrow \textbf{P}^1(\mathbb{C})$, then there is a finite set of points $S \subseteq \textbf{P}^1(\mathbb{C})$ such that $\phi^\prime: \phi^{-1}( \textbf{P}^1(\mathbb{C}) \setminus S) \rightarrow \textbf{P}^1(\mathbb{C}) \setminus S$ is in fact a covering in the usual sense and this assertion also gives and equivalence of categories between compact Riemann surfaces mapping holomorphically onto $ \textbf{P}^1(\mathbb{C})$ and so called branched coverings of $ \textbf{P}^1(\mathbb{C})$ (i.e. continuous maps that are coverings outside of a finite set of points). So we have identified finite field extensions of $\mathbb{C}(t)$ with branched coverings of $ \textbf{P}^1(\mathbb{C})$. It can also be seen that under this correspondences a finite Galois extension $\mathbb{C}(t) \hookrightarrow L$ corresponds to a Galois (branched) cover $Y \rightarrow \textbf{P}^1(\mathbb{C})$. So we get an isomorphism $Gal(L,\mathbb{C}(t)) \cong Aut(Y \rightarrow X)$ between the galois group and the automorphism group of a Galois covering. Now it is a standard fact in covering theory that a covering is already completely determined by the monodromy action. In particular we get an isomorphism between our Galois group and the automorphism group of the monodromy action on the fibre of our covering.<|endoftext|> TITLE: How many cones with angle theta can I pack into the unit sphere? QUESTION [8 upvotes]: Given a unit sphere (radius 1), I would like to know how many cones I can pack into this unit sphere. Restrictions: The top of the cone needs to be in the center of origin. The bottom of the cone needs to form a circle on the unit sphere. I have found a related question, but with a cube: Packing space by cones: Translates best? I have also tried to find an upper bound myself by performing the following calculation: Surface of the projection of the base of the cone on the unit sphere: $$2 \pi r^2(1 + \sin(\theta) \pm \cos(\theta))$$ Surface of the unit sphere: $4 \pi r^2$ Now, a (very high) upper bound would be: $$\frac{2 \pi r^2(1 + \sin(\theta) \pm \cos(\theta))}{4 \pi r^2}$$ This however does not take into account the restrictions of the shapes, so the actual number will likely be much lower. Question 1: What would be a closer upper bound Question 2: If an example value is easier, what would be a realistic number of cones given $\theta = 5^{\circ}$ REPLY [4 votes]: A good reference for volumetric arguments for the maximum number of 'cones' or spherical 'caps' that one can fit, is a series of papers by Jon Hamkins. The density of a packing of these caps can be at most $\frac{\pi}{2\sqrt{3}}$, (this being known as the Fejes Tóth bound) and in the minimal distance between centers of a packing, $d$, the density is bounded above by $\frac{\pi}{2\sqrt{3}}-O(d^2)$. This follows from a bound on the maximum size of a code on the sphere with pairwise distances at least $d$, $$M(3,d)\leq 2\left(1-\frac{\pi}{6\cot^{-1}\sqrt{3-d^2}}\right)^{-1}$$ taken from chapter three of Hamkin's thesis here (being another form of Tóth's bound). For specific parameters, the problem can be pretty intricate. One of the best references has already been mentioned in Bullet's answer and is Sloane's webpage. Codes on Euclidean Spheres, by Ericson and Zinoviev is worth checking out as well as a reference on these specific packings (and is slightly outdated), along with Sphere Packings, Lattices and Groups by Conway and Sloane.<|endoftext|> TITLE: Free operad over a monoid object QUESTION [6 upvotes]: Let $\mathcal{O}$ be an operad in the monoidal category $M$. Then $\mathcal{O}(1)$ together with the morphisms $$\mathcal{O}(1)\otimes \mathcal{O}(1)\to \mathcal{O}(1)$$ and the unit $\eta:1\to \mathcal{O}(1)$ is a monoid object. Moreover, a morphism $\varphi:\mathcal{O}\to \mathcal{O}'$ of operads induces a morphism $\mathcal{O}(1)\to \mathcal{O}(1)$ of monoid objects. Therefore, we get a forgetful functor $$\mathrm{Operads}(M)\to \mathrm{Monoids}(M).$$ If we work with coloured operads, we get a functor from coloured $M$-operads to $M$-enriched categories. Conversely, if $T$ is a monoid object, we can build an operad by $$\mathcal{O}_T(r) := T^{\otimes r}.$$ and the structure maps $$T^{\otimes r}\otimes \bigotimes_{i=1}^r T^{\otimes k_i}\to T^{\otimes (k_1+\dotsb+k_r)}$$ as follows: Let $\Delta:T\to T^{\otimes k}$ be the diagonal (existence is clear if $\otimes$ is the categorical product). Then $$T\otimes T^{\otimes k}\stackrel{\Delta\otimes \mathrm{id}}{\to}T^{\otimes k}\otimes T^{\otimes k} \cong (T^{\otimes 2})^{\otimes k} \to T^{\otimes k}.$$ In $\mathbf{Set}$, this just means $t(t_1,\dotsc,t_k)=(tt_1,\dotsc,tt_k)$. It should be clear that this construction gives us an operad. Now two problems/questions: Does the morphism $\Delta$ always exist in the general setting? It is obviously not the same as $$T\cong T\otimes 1^{\otimes (k-1)}\stackrel{\mathrm{id}\otimes \eta^{\otimes (k-1)}}{\to} T^{\otimes k}.$$ Obviously $\mathcal{O}_T(1)\cong T$, but it seems to be not true that $\mathcal{O}_T$ is the free operad over the monoid object $T$. Is there another construction for the “free” operad over $T$? REPLY [9 votes]: Let me mention that this is related to this earlier question of mine (which is unanswered :-( ) and more generally to semi-direct products of operads by bialgebras. You construction is the semi-direct product $\mathtt{Com} \rtimes T$ of the commutative operad $\mathtt{Com}$ by $T$. No, there is no diagonal in general. You need a cocommutative bimonoid object, i.e. an object equipped with a multiplication $\mu : T \otimes T \to T$ and a comultiplication $\Delta : T \to T \otimes T$ such that $\mu$ is associative and unital, $\Delta$ is coassociative, cocommutative and counital, and they satisfy a compatibility relation $\Delta \circ \mu = \mu \circ (\Delta \otimes \Delta)$. I'll assume that by "free operad" you mean a left adjoint to the forgetful functor. Let $\varnothing$ be the initial object of your category and suppose that $\varnothing \otimes X = \varnothing = X \otimes \varnothing$ for all $X$. Then (as Simon Henry mentions in the comments), the free operad on $T$ is simply given by $\mathtt{O}(1) = T$ and $\mathtt{O}(n) = \varnothing$ for $n \neq 1$.<|endoftext|> TITLE: End-extending cardinals QUESTION [13 upvotes]: Let us say a cardinal $\kappa$ end-extending if there is a function $F : V_\kappa^{<\omega} \to V_\kappa$ such that: (a) If $M \subseteq V_\kappa$ is closed under $F$, then $M \prec V_\kappa$. (b) If $M$ is closed under $F$ and of size $<\kappa$, then there is $N \supseteq M$ closed under $F$ such that $N \cap \sup(M \cap \kappa) = M \cap \kappa$ and $N \cap \kappa \not= M \cap \kappa$. It is a standard argument to show that measurable cardinals are end-extending, and this reflects below them. It is fairly easy to see that end-extending cardinals must be regular and cannot be successor cardinals (except for the trivial case $\kappa = \omega_1$). (1) Is being an end-extending limit cardinal equivalent to a well-known large cardinal notion? (2) Must end-extending limit cardinals be strongly inaccessible? REPLY [6 votes]: Suppose $\kappa$ carries an $\omega_1$-saturated $\kappa$-complete ideal $I$, given $M\prec (V_{\kappa+2},\in , <)$ ($<$ well orders $V_{\kappa+2}$) of size $<\kappa$ containing $I$, we show how to find an end-extension $N\prec V_{\kappa+2}$ of $M$ (i.e. $N\cap \sup(M\cap \kappa)=M\cap \kappa$). Let $G\subset P(\kappa)/I$, and let $j: V\to N$ be the generic embedding in $V[G]$. Look at $M'=\{j(f)(\kappa): f\in M\}$. In $N$, $M'\prec j((V_{\kappa+2}, \in , <))$. Clearly $\kappa\in M'\cap j(\kappa)- j(M)$. Note $j(M)\cap j(\kappa)=j''M \cap j(\kappa)=j(M\cap \kappa)=j'' M\cap \kappa=M\cap \kappa$. If $\xi\in \kappa$ such that $\xi\in M'$, $\xi\in M$. The reason is that: fix $f$ such that $j(f)(\kappa)=\xi$. In $V$, $M$ contains a maximal antichain $\{A_i: i<\omega\}$ such that for each $i$, there exists $\delta_i<\kappa$, $A_i\Vdash j(f)(\kappa)=\delta_i$. Clearly $\{A_i: i<\omega\}, \{\delta_i: i<\omega\}\subset M$. Hence in $V[G]$, there exists $i<\omega$, $\xi=\delta_i \in M$. This shows $M'\cap \sup (j(M)\cap j(\kappa))=M'\cap \sup(M\cap \kappa)=M\cap \kappa=j(M)\cap j(\kappa)$. By elementarity, there exists $N\prec V_{\kappa+2}$ containing $M$ such that $N\cap \sup (M\cap \kappa)=M\cap \kappa$ and $N\cap \kappa \neq M\cap \kappa$. So $\kappa$ can be $2^\omega$.<|endoftext|> TITLE: Is Sp(1).Sp(1).Sp(1) the homotopy-fixed locus of Triality? QUESTION [7 upvotes]: The group $$ Sp(2)\cdot Sp(1) := \big( Sp(2) \times Sp(1)\big)/\big\{(1,1), (-1,1)\big\} $$ is canonically a subgroup of $Spin(8)$ in three different ways, these representing three distinct conjugacy classes of subgroups, as indicated here: Under the action of the triality group $$ Out(Spin(8)) \;\simeq\; S_3 $$ these three conjugacy classes are permuted into each other. The relevant definitions and references for these statements are collected on the nLab here. This means that there are automorphisms making the outer two circles in the following diagram commute with respect to the given subgroup inclusions: My question is about the inner circle: Each of the above quaternionic subgroups canonically contains the further subgroup $$ Sp(1) \cdot Sp(1) \cdot Sp(1) \;:=\; \big( Sp(1) \times Sp(1) \times Sp(1) \big)/ \big\{ (1,1,1), (-1,-1,-1) \big\} $$ Moreover, this subgroup has an evident automorphism action of the symmetric group $S_3$ on it, given by permuting its dot-factors. This suggests that the middle automorphism circle above also commutes with the inner one, the relevant automorphisms of the inner circle are these dot-factor permutations, and maybe that $Sp(1)\cdot Sp(1) \cdot Sp(1)$ is something like the homotopy-fixed locus of triality. I can see that a) and b) are true at least for the left sector of the above diagram, and it seems it must be at least close to being true also in the other two sectors, but I haven't tried yet to make that a full proof. Is this known? REPLY [5 votes]: This isn't quite an answer to your question so much as an elaboration of my comments. Each Dynkin diagram, or equivalently Cartan matrix $a_{ij}$, determines a presentation of the corresponding complex Lie algebra $\mathfrak{g}$ (which is much the same data as the compact Lie group). I have the name ``Chevalley'' associated to this presentation, but perhaps Cartan and Serre are also important. In any case, the presentation is the following. For each node $i$ of the Dynkin diagram, you introduce three generators $h_i,x^+_i,x^-_i$. You declare that the $h_i$s commute among themselves, that $[h_i, x^{\pm}_j] = \pm a_{ij}x^\pm_j$, and that $[x_i^+,x_j^-] = \delta_{ij}h_i$, so that for fixed $i$, $\{h_i,x_i^+,x_i^-\}$ forms an $\mathfrak{sl}(2)$-triple. So far this presents an infinite-dimensional Lie algebra with a unique finite-dimensional quotient. To cut down to the quotient, you need also to impose the Serre relations which say that, for $\epsilon \in \{+,-\}$, $(\operatorname{ad} x^\epsilon_i)^{1-a_{ij}} x^\epsilon_j = 0$, where of course $(\operatorname{ad} x)(y) = [x,y]$. This presentation of $\mathfrak{g}$ is detailed in various textbooks, for instance Section 5.6 of my lecture notes. This presentation has various advantages and disadvantages. It makes manifest the sub Lie groups corresponding to sub diagrams, and obscures most other subgroups. In particular, it makes manifest an action on $\mathfrak{g}$ by the diagram automorphisms — it is a standard, although nontrivial, result that the group of diagram automorphisms is isomorphic via this action to the outer automorphism group $\operatorname{Out}(\mathfrak{g}) = \operatorname{Aut}(\mathfrak{g})/\operatorname{Inn}(\mathfrak{g})$ of $\mathfrak{g}$ — and obscures other possible splittings of the map $\operatorname{Aut}(\mathfrak{g}) \to \operatorname{Out}(\mathfrak{g})$. In particular, taking the $D_4$ diagram and its subdiagram of the three outer nodes, we find a manifest triality-fixed (in the setwise sense) subgroup $\mathrm{Sp}(1)^3 \subset \mathrm{Spin}(8)$. Well, at the Lie algebra level it is $\mathfrak{sl}(2)^3 = \mathfrak{sp}(1)^3 \subset \mathfrak{spin}(8) = \mathfrak{so}(8)$; at the Lie group level one should take some central quotient. Clearly triality acts to permute the three $\mathrm{Sp}(1)$s, and so its fixed locus is a single copy of $\mathrm{Sp}(1)$. Note that the $\mathfrak{sl}(2)$-triple corresponding to the central node in the $D_4$ diagram is also fixed (pointwise!) by triality. Together, this fixed $\mathfrak{sl}(2)$ and the diagonal $\mathfrak{sl}(2) \subset \mathfrak{sl}(2)^3$ generate the full fixed subgroup $G_2$ — indeed, they correspond to the two nodes in the $G_2$ dynkin diagram. (The "diagonal" $\mathfrak{sl}(2)\subset \mathfrak{sl}(2)^3$ is "three times longer" than any coordinate, explaining why $G_2$ has roots of different lengths.) Actually, there is a copy of $\mathfrak{sl}(2) \subset \mathfrak{g}$ associated to each root, not just the simple ones. (This is a slightly subtle point. Given a root $\alpha$, you can take $x^\pm$ to be generators of the $\pm\alpha$ root spaces, and $h = [x^+,x^-]$. But the normalization of $x^\pm$ is ambiguous, so the root $\alpha$ does not uniquely determine a homomorphism $\mathfrak{sl}(2) \to \mathfrak{g}$, but rather determines a subalgebra of $\mathfrak{g}$ abstractly-isomorphic to $\mathfrak{sl}(2)$. Actually, using the theory of vertex algebras, you can almost specify the normalization of $x^\pm$, but there is a fundamental sign ambiguity. This sign ambiguity is reflected by the fact that the the Weyl group is not typically a subgroup of the Lie group, but that there is a typically-nonsplit extension of shape $2^{\text{rank}}.\text{Weyl}$, called the "Tits lift of the Weyl group", which is a subgroup of the Lie group.) In particular, you can take the three outer nodes of the $D_4$ Dynkin diagram together with the highest root, and these will together produce a subgroup $\mathrm{Sp}(1)^4 \subset \mathrm{Spin}(8)$ (I continue to ignore central quotients), which turns out to be maximal and to be a centralizer in $\mathrm{Spin}(8)$. You actually know this $\mathrm{Sp}(1)^4$. Identify $\mathrm{Spin}(4) = \mathrm{Sp}(1)^2$; but of course $\mathrm{Spin}(4)^2 \subset \mathrm{Spin}(8)$, up to central factors. This lets you see exactly how the vector and spin representations of $\mathrm{Spin}(8)$ decompose. Indeed, the vector $V^8$ decomposes over $\mathrm{Spin}(4)^2$ as $V^4 \otimes 1 \oplus 1 \otimes V^4$, where $1$ denotes the trivial module, and the isomorphism $\mathrm{Spin}(4) = \mathrm{Sp}(1)^2$ identifies $V^4$ with the tensor product of the two 2-dimensional modules, so all together $$ V^8|_{\mathrm{Sp}(1)^4} = 2 \otimes 2 \otimes 1 \otimes 1 + 1 \otimes 1 \otimes 2 \otimes 2.$$ On the other hand, the full spin module for $\mathrm{Spin}(m+n)$ decomposes over $\mathrm{Spin}(m)\times \mathrm{Spin}(n)$ as the product of the two full spin modules. In even dimensions, the full spin module breaks as a sum of half-spin modules. The result is that the half-spin module $S^8_+$ decomposes over $\mathrm{Spin}(4)^2$ as $S^4_+ \otimes S^4_+ \oplus S^4_- \otimes S^4_-$, where $S^4_\pm$ are the two 2-dimensional half-spin modules for $\mathrm{Spin}(4)$, whereas $S^8_-$ decomposes as $S^4_+ \otimes S^4_- \oplus S^4_+ \otimes S^4_-$. Under the identification $\mathrm{Spin}(4) = \mathrm{Sp}(1)^2$, the half-spin modules are identified with $2\otimes 1$ and $1\otimes 2$, respectively, and so $$ S^8_+ |_{\mathrm{Sp}(1)^4} = 2 \otimes 1 \otimes 2 \otimes 1 + 1 \otimes 2 \otimes 1 \otimes 2,$$ $$ S^8_- |_{\mathrm{Sp}(1)^4} = 2 \otimes 1 \otimes 1 \otimes 2 + 1 \otimes 2 \otimes 2 \otimes 1.$$ Triality permutes the modules $V^8 \mapsto S^8_+ \mapsto S^8_-$, and so permutes the three ways of pairing off these four $\mathrm{Sp}(1)$s. The four $\mathrm{Sp}(1)$s in $\mathrm{Sp}(1)^4 \subset \mathrm{Spin}(8)$ all look the same from this perspective. What we did when we fixed the presentation (aka Dynkin diagram) was to choose one of them to correspond to the highest root, and the other three to be the outer nodes of the dynkin diagram. The specific triality automorphism depended on this choice. (Different choices are related by inner automorphisms. In this case, if you chose a different $\mathrm{Sp}(1)$ from $\mathrm{Sp}(1)^4$ as your highest root, your choice would be related to mine by an element of the Weyl group. This is because implicitly we chose the Cartan of $\mathrm{Spin}(8)$ to be the Cartan of $\mathrm{Sp}(1)^4$.) I mention this to emphasize the subtleties when talking about "fixed points of triality". Maybe I'll end with one more comment, which is that I mentioned $G_2$ contains an $\mathrm{Sp}(1)^2$ (up to central blah blah). We can recognize is at spanned by the (fixed) "highest" $\mathrm{Sp}(1)$, which might as well be the first of the $\mathrm{Sp}(1)^4$, and a diagonal copy inside the other three. Then over this, how do the above 8-dimensional modules break? The answer is clear: $2\otimes 1\otimes 1$ restricts over the diagonal to $2$, whereas $1 \otimes 2 \otimes 2$ restricts as $3+1$. So we get $$ 8 = 2 \otimes 2 + 1 \otimes 3 + 1\otimes 1.$$ The first two of these are together the 7-dimensional representation of $G_2$, and the third is the trivial rep.<|endoftext|> TITLE: Name for abelian category in which every short exact sequence splits QUESTION [17 upvotes]: What is the name of the class of abelian categories defined by the following property: every short exact sequence splits? REPLY [9 votes]: For an abelian category $A$, the following are equivalent: Every short exact sequence splits. Every object is projective. Every object is injective. Every additive functor from $A$ to an abelian category is exact. (To show 4 $\Rightarrow$ 2, use the Hom functor.) If $A$ has these properties, any full pseudo-abelian subcategory of $A$ is a Serre subcategory (in particular, abelian) with the same properties. They are also preserved by Serre localisation. $A$ is semi-simple (in the sense that every left $A$-module is a direct sum of simple objects, or equivalently that every object of $A$ is a finite direct sum of simple objects) provided every object of $A$ is Noetherian (or, equivalently, Artinian), or if it is a category of modules over an appropriate additive category. A fun counterexample is the category of infinite-dimensional vector spaces over a field, localised by the Serre subcategory of finite-dimensional vector spaces. I am not sure that it is Grothendieck (does it have infinite direct sums?) Another example, still in the spirit of Leonid's answer, is the category of finitely presented modules over a von Neumann regular ring $R$. (Reduce to the case of one generator to show projectivity. We get a quotient of $R$ by a finitely generated ideal. This ideal is generated by an idempotent, hence a direct summand, hence the quotient is indeed projective.)<|endoftext|> TITLE: Are "most" spaces aspherical? QUESTION [25 upvotes]: There's a heuristic idea that "most" closed manifolds $M$ are aspherical (i.e. $\pi_{\geq 2}(M) = 0$). Does this heuristic extend usefully to all spaces -- or at least to all finite CW complexes? To make this question more precise, I should say something about in what sense "most" manifolds are aspherical. I don't know a lot about this heuristic, but here's where I'm coming from: It's true in low dimensions: trivially in 0 or 1 dimensions, and by classification of surfaces in 2 dimensions. In 3 dimensions, I've heard it said that part of the upshot of Thurston's Geometrization Conjecture is that "most" 3-manifolds are hyperbolic, and in particular aspherical. There's some discussion of this heuristic in this survey article of Luck (at the end). How do things look if we think about CW complexes? Well, every 0 or 1-dimensional CW complex is aspherical. And the Kan-Thurston theorem tells us that every space is homology-equivalent to an aspherical space. But it's really not clear to me whether I should think of "most" spaces as being aspherical. REPLY [7 votes]: One way to think about whether "most" spaces are aspherical is measure-theoretically. Here a few examples and non-examples of random topological spaces being aspherical. Examples Presentation complexes of density random groups are aspherical for every density $d < 1/2$, and for density $d> 1/2$ these groups collapse, so this is essentially the entire interesting range of parameter. Random 3-manifolds. N. Dunfield and W. Thurston introduced a model for random 3-manifold using a random walk on the mapping class group to generate a random Heegaard splitting. Joseph Maher showed that these random 3-manifolds are hyperbolic with high probability, so in particular their universal cover hyperbolic space $\mathbb{H}^3$ is contractible. Let $Y(n,p)$ denote the Linial-Meshulam random 2-dimensional simplicial complex. This complex has vertex set $[n]$, complete $1$-skeleton, and each $2$-face appears independently with probability $p=p(n)$. Costa and Farber showed that if $p \ll n^{-1/2 - \epsilon}$, $Y(n,p)$ is nearly aspherical, in the following sense: if you delete one 2-face from every sufficiently small sphere, pinched sphere (along a vertex, or an edge), or projective plane, the resulting complex is aspherical. It is easy to check that the expected number of these local obstructions is much smaller than the expected total number of 2-faces. So you can delete one face from each one to result in an aspherical complex and have almost all the faces remaining. In a similar spirit, Andrew Newman and I recently showed that random 2-dimensional hypertrees (random Q-acyclic complexes) according to a certain "determinantal measure" are are aspherical, in Topology and geometry of random 2-dimensional hypertrees. Non-examples If one considers the random 2-complex $Y(n,p)$ with $p \ge (\gamma n)^{-1/2}$ and $\gamma = 4^4 / 3^3$, Luria and Peled showed that $Y(n,p)$ is simply connected, so at that this point is homotopy equivalent to a bouquet of $2$-spheres, and is not aspherical. It is not "nearly aspherical" in the sense of Costa and Farber either, so there is a phase transition near $p = n^{-1/2}$ from nearly aspherical to not. What if we just count homotopy types of simplicial complexes on $n$ vertices? Andrew Newman showed that there are doubly exponentially many homotopy types, at least $2^{2^{0.02n}}.$ On the other hand, there are most $2^{n \choose 3}$ different fundamental groups, a much smaller number, so somehow "most" homotopy types of simplicial complexes can not be aspherical.<|endoftext|> TITLE: Have the Quantum Group Theorists taught the Group Theorists Anything? QUESTION [26 upvotes]: I will start with the general before moving to the specific. Consider for a moment the two (very) soft definitions. An abstraction of an object $X$ is a category $\mathcal{C}_0$ such that $X$ is an object in the class $\operatorname{ob}(\mathcal{C}_0)$. A generalisation of an abstraction $\mathcal{C}_0$ is a category $\mathcal{C}$ such that $\mathcal{C}_0$ is a proper subcategory of $\mathcal{C}$ (so that in this soft definition regime, a generalisation is also an abstraction). It is a storied theme of mathematics that by abstracting an object $X$ to $\mathcal{C}_0$, that we can prove theorems for a whole class of objects, rather than just for the single object $X$. Moreover, often when interested just in the object $X$, it can be easier to work in the abstraction $\mathcal{C}_0$, as this sometimes allows us to disregard the irrelevant idiosyncrasies of $X$. Mathematical history --- with all its humanity --- is littered with examples of theorems proved in an abstraction $\mathcal{C}_0$ before they were known or considered in the specific context of $X$. This is of a subjectively different flavour to just putting together a slicker proof or proving a general result. Of course, when you move from $X$ to $\mathcal{C}_0$ some theorems are no longer true. The same is true when we look at a generalisation $\mathcal{C}$ of $\mathcal{C}_0$. However, of course, theorems true in $\mathcal{C}$ will be true for $X$ but moreover $\mathcal{C}_0$. Moving towards the specific, the Peter-Weyl Theorem in the category of compact groups is also true (with suitable definitions) in the generalisation to compact matrix quantum groups. There are many definitions/categories of quantum groups. In those categories which are (in the sense above) generalisations of categories of classical groups (classical in the sense of "has a set of points $G$" --- I believe all such definitions of quantum groups include at the very least the category of finite groups), have the quantum group theorists ever 'discovered' something that group theorists either were interested in, or would plausibly be interested in? When a generalisation $\mathcal{C}$ of an abstraction $\mathcal{C}_0$ is developed to help study objects in $\mathcal{C}_0$, you can imagine that this happens, but as quantum groups are, arguably, studied for their non-commutative aspects, rather than as an attempt to understand classical groups better, this may not have happened. To bookend; my question: Have quantum group theorists discovered something new about groups that is interesting to group theorists? REPLY [5 votes]: The term "Quantum groups" itself, implies that the development of the hopf algebra theory generalizes -in some categorical sense- usual group theory. There are various points that might support this view (although i am not sure if this is what you are really looking for): If $H$ is a cocommutative, finite dimensional hopf algebra over an algebraically closed field $k$, of characteristic zero then $H\cong kG$, for some finite group $G$. Similarly, the commutative hopf algebras over $k$ are isomorphic to the duals of group hopf algebras of finite groups. The above results can be viewed categorically: there is an equivalence (but not necessarily an isomorphism) of categories, between the category $\mathcal{H}$, of commutative, cocommutative, f.d. hopf algebras over $k$ and the category $\mathcal{Ab}_f$ of finite abelian groups. This implies that in finite dimensions and under the constraints imposed by commutativity and cocommutativity, the hopf algebra theory is "essentially" the theory of finite abelian groups. If we drop commutativity and keep only cocommutativity we have the finite group theory. I do not know if these are new discoveries, in the sense that they are classical results of the hopf algebra theory; cocommutativity after all is an obvious property in the "tensoring" of group representations. (and of the lie algebra representations as well). However, -as mentioned in the OP- it is the noncommutative (and the non-cocommutative i would add) aspects of quantum group theory or hopf algebra theory that are really interesting. The notions of quasitriangularity (QT) and coquasitriangularity (CQT), generalize cocommutativity and commutativity respectively. However they still keep close touch to group theory: CQT group hopf algebras are abelian and equipped with a bicharacter $\langle . | . \rangle:G\times G\rightarrow k$. The set of bicharacters on $G$ is in bijection with the set of the homomorphisms of $G$ to its character group $\hat{G}$. In the f.d. case and for $k=\mathbb{C}$ the complex numbers, the bicharacters of the finite, abelian group $G$ are in bijection with the QT and the CQT structures of the group hopf algebra $\mathbb{C}G$ (that is, its universal $R$-matrices) and in bijection with the braidings of the monoidal category ${}_{\mathbb{C}G}\mathcal{M}$ of the group algebra representations. In this sense, the non-trivial (co)quasitriangular structures of the group hopf algebra (if the group is finite, abelian these are non-trivial $R$-matrices), correspond to non-trivial bicharacters of the group or to non-trivial braidings of its category of representations. These notions contribute to the expansion of the definition of quantum groups. Braided groups, are hopf algebras in the braided monoidal categories of representations of (co)quasitriangular group hopf algebras. (i.e. group hopf algebras equipped with non-trivial $R$-matrices or non-trivial bicharacters of the corresponding group). Edit: Since the OP cites generalizations of group theoretic results to the quantum groups/hopf algebra setting (like the Peter-Weyl theorem), maybe it would be interesting to mention results on the generalizations of Frobenius-Schur indicator for compact groups: In arXiv:math/0004097 [math.RT], the Frobenius-Schur theorem for finite groups, is generalized for semisimple hopf algebras over algebraically closed fields of zero char and to semisimple/cosemisimple hopf algebras if the characteristic is greater than zero. Some more recent results are presented in FSZ groups and Frobenius-Schur indicators for quantum doubles. There, the authors study the problem of when higher indicators of the reps of the quantum double of a finite group are all integers? They characterize this as an interesting group-theoretic question and proceed in finding groups which have this property and counterexamples as well. Concluding, i am not claiming that quantum group theory has answered unsolved problems of group theory but it may have contributed some ideas, or at least some descriptions, or even has posed some questions, of interest to a group theorist.<|endoftext|> TITLE: Weyl law for (non-semiclassical) Schrodinger operator QUESTION [9 upvotes]: The Weyl law for a semiclassical Schrodinger operator $$ A_h\ := \ -h^2\Delta+V(x) $$ on an $d$-dimensional complete Riemannian manifold $M$ says that the number $N(A_h,1)$ of eigenvalues of $A_h$ which are smaller than 1 has asymptotic behavior $$ N(A_h,1)\ \sim \ \frac1{(2\pi h)^d}\ \text{Vol}\ {\Large(}(x,\xi)\in T^*M:\ |\xi|^2+V(x)\le 1{\Large)}, \quad h\to 0.\ \ \ (\ast) $$ I am interested in a non-semiclassical Schrodinger operator $$ A\ := \ -\Delta+V(x).$$ I believe that the number $N(A,\lambda)$ of eigenvalues of $A$ smaller than $\lambda$ has a similar asymptotic $$ N(A,\lambda) \sim \ \frac1{(2\pi)^d} \text{Vol}\ {\Large(}(x,\xi)\in T^*M:\ |\xi|^2+V(x)\le \lambda{\Large)}, \qquad \lambda\to \infty. \quad(\ast\ast) $$ This is, of course, true on a compact manifolds, due to the classical Weyl's law. It is also not difficult to verify (**) for operator $-\Delta+|x|^n$ on $\mathbb{R}^d$ (where $A_h$ and $A$ can be related by rescaling of $x$). So my question: is (**) true and, if it is true, where can I find it? PS. Victor Ivrii in his review article https://arxiv.org/abs/1608.03963 mentions (page 5) that, using Birman-Schwinger principle, one can obtain a Weyl law for $N(A,\lambda)$ from (*). But I don't see how it can be done. REPLY [4 votes]: We are talking about a non-compact Riemannian manifold, right? Then Weyl's law may be incorrect, at least out of the box. Let us consider $H = -\Delta$ (so $V(x)=0$) in the domain $X \subset \mathbb{R}^d$ with the Dirichlet or Neumann boundary condition (the case of the closed manifold is very similar to the Neumann case). Let non-compact part of $X$ be $\{x\colon x_1 > c, \ x'\in \rho(x_1) \Omega\}$ with $x'=(x_2,\ldots, x_d)$, $\Omega$ a compact domain with a smooth boundary in $\mathbb{R}^{d-1}$ and $\rho \to 0$ as $x_1\to \infty$. Observe that depending on the rate of decay of $\rho$ the volume of $X$ could be finite or infinite (and the area of $\partial X$ also could be finite or infinite). The literal Weyl's law says: if $X$ has a finite volume then the spectrum is discrete and Weyl's law holds and if $X$ has an infinite volume then there is an essential spectrum. However the reality is more nuanced: Dirichlet Laplacian In this case spectrum is discrete for sure; simply in the case of the infinite volume the correct asymptotic expansion is different. Let $\rho (t)= t^{-\mu}$, $\mu >0$. Volume is finite for $\mu (d-1)> 1$ and infinite otherwise. Literal Weyl's law holds for $\mu (d-1)> 1$. For $\mu (d-1)=1$ (logarithmic divergence) $N(\lambda)\asymp \lambda ^{(d-1)/2}\ln (\lambda)$ and is given by the same formula but with integration over $x_1\colon \rho(x_1)>\lambda^{-1/2}$. For $\mu (d-1)<1$ (power divergence) $$ N(\lambda)\sim \sum_{j} n_j(\lambda) \tag{D} $$ where $n_j(\lambda)$ is a Weyl's expression for $1$-dimensional Schrödinger operator $\mathsf{h}_j = -\partial_1^2 +\nu_j \rho(x_1)^{-2}$ and $\nu_j >0$ are eigenvalues of $(d-1)$-dimensional Dirichlet Laplacian in $\Omega$. Neumann Laplacian Then things change drastically. There is an essential spectrum unless $\rho\to 0$ really fast. To understand why look at (D) but instead of a Dirichlet Laplacian in $\Omega$ one should consider a Neumann Laplacian, and $\nu_1=0$. Does it mean essential spectrum? If $\rho =x_1^{-\mu}$ or even $\rho =\exp (-c x_1)$ then yes. But if $\rho = \exp(-x_1^{k+1})$ with $k>0$ then no, and $$ N(\lambda) \sim N^W(\lambda) + N^c(\lambda) \tag{N} $$ where $N^W(\lambda)$ is a Weyl expression for an original operator, and $N^c(\lambda)$ is a Weyl expression for $1$-dimensional Schrödinger $\mathsf{h}_0= -\partial_1^2 + W(x_1)$ with a potential $W =\frac{1}{4} (\partial_{x_1} \log \rho (x_1))^2$ and depending on $k$ either the first or the second term in (N) may prevail. For Dirichlet/Neumann case in more general setting see subsections 3.2.2/3.2.3 of 100 years of Weyl's law either in arXiv or in Bull Math Sci<|endoftext|> TITLE: A continuum which is both Suslinean and non-Suslinean? QUESTION [15 upvotes]: Continuum means compact connected metrizable with more than one point. A continuum is Suslinean if every collection of pairwise disjoint subcontinua is countable. There is an apparent contradiction in the literature I would like to resolve... In Example 3 at the end of Paper A, there is constructed a continuum $Y:=X/\sim$ which is the quotient of another continuum $X$. It is clear from the construction that $X$ is Suslinean, and since the quotient map is monotone $Y$ is also Suslinean. Actually, the author claims $X$ and $Y$ are hereditarily locally connected, and such continua are automatically known to be Suslinean. On the other hand, $Y$ is the closure of a ray (a one-to-one continuous image of $[0,1)$) such that the ray is first category in $Y$. In other words, the ray and its complement are each dense in $Y$. It follows that there is a sequence of pairwise disjoint arcs in $Y$ which converges to $Y$ in the Hausdorff distance. By Theorem 30 in Paper B, $Y$ is non-Suslinean. Question 1: Am I correct in finding a contradiction? Question 2: Is Example 3 correct? Specifically, why is it Hausdorff (and therefore metrizable)? I found a couple of typos, e.g. $A_n$ should be $C_n\cup \bigcup ...$ and $\overline{z_1 z_1}$ should be $\overline{z_1 z_2}$, but otherwise it seems okay. If there is a problem with Theorem 30, then I think the error must be in the proof of Lemma 29. Specifically there is a claim that $H_W$ is non-degenerate since it is an inverse limit of non-degenerate continua. This is a false statement in general because we can take the continua $[0,1/n]$ with bonding maps the inclusions $[0,1/(n+1)] \hookrightarrow[0,1/n]$, and the inverse limit is just the single point $\langle 0,0,..\rangle$. However, since each factor of the inverse limit for $H_W$ projects into $\partial U$ and $\partial V$ in the first factor, the inverse limits on preimages of these boundary sets should be non-empty and disjoint subsets of $H_W$... Upon closer inspection, the proof of Lemma 29 (Paper B) is flawed in more ways than one. I'm not sure it can be saved. At this point I would lean toward the example being correct. I really just need to see that $Y=X/\sim$ is Hausdorff. Paper A: Tymchatyn, E. D., Some rational continua, Rocky Mt. J. Math. 13, 309-319 (1983). ZBL0514.54022. Paper B: Mouron, Christopher, The topology of continua that are approximated by disjoint subcontinua, Topology Appl. 156, No. 3, 558-576 (2009). ZBL1165.54012. REPLY [5 votes]: Example 3 in Paper A is indeed a counterexample to Theorem 30 in Paper B. Lemma 29 must also be false because it implies Theorem 30. The only thing left to verify is that $Y=X/\sim$ (from Example 3) is Hausdorff, so that $Y$ is actually a metrizable continuum. This can be proved in a few steps: For every compact $K\subseteq [0,1]$ the set $$\widehat{K}:=K\cup \bigcup \big\{\overline{z_1 z_2}:z_1\text{ and }z_2\text{ are consecutive endpoints of some }C_n \text{ and }(z_1\in K\text{ or }z_2\in K)\big\}$$ is a closed (compact) subset of $X$. If $U$ is any open subset of $X$ containing an element $y\in Y$, then $$V=X\setminus\widehat{[0,1]\setminus U}$$ is an open subset of $X$ which is a union of members of $Y$, and $y\subseteq V\subseteq U$. $Y$ is Hausdorff by 2 and normality of $X$. We can argue 1 as follows. Let $x$ be any point in the closure of $\hat K$. Suppose for a contradiction that $x\notin \hat K$. Then $x\in [0,1]\setminus K$ and there is a sequence of semi-circular arcs $\overline{z^n_1 z^n_2}\in Y$ with $z^n_1\in K$ and $z^n_2\to x$. (I assume without loss of generality that the subscripts on the $z$'s are arranged in this way). Since $d(x,K)>0$ this means infinitely many semicircles have radius greater than some positive constant. But this is impossible by the construction of $X$. Therefore $x\in \hat K$ and $\hat K$ is compact. Now 1 $\Rightarrow$ 2 $\Rightarrow$ 3 and the proof is complete.<|endoftext|> TITLE: The underlying space of a scheme remembers its affineness? QUESTION [6 upvotes]: Let $f:X\rightarrow Y$ be a morphism of schemes. We know that if $Y$ is affine and $f$ induces homeomorphism on the underlying spaces then $X$ is affine. Is it true that if $X$ is affine and $f$ induces a homeomorphism on the underlying spaces then $Y$ is affine? More generally, is it true that a scheme whose underlying space is homeomorphic (possibly via a homeomorphism that is not induced by a morphism of schemes) to the underlying space of an affine scheme is affine? EDIT: actually, the answer to the last question is a very strong "NO" as the underlying space of any scheme is sober and Noetherian sober spaces are spectral (i.e. homeomorphic to the underlying space of an affine scheme). REPLY [5 votes]: As pointed out by abx here, the following example fits the bill. Take a nodal cubic in $\mathbb{P}^2$, this is an irreducible scheme proper over $\mathbb{C}$, take its normalization and throw out a point from the inverse image of the node. We get a $\mathbb{C}$-morphism from the affine line to our cubic, which is bijective on the underlying spaces. Since a non-empty proper closed subspace of either the affine line or the cubic is a finite union of closed points, this morphism is also closed, thus it induces a homeomorphism on the underlying spaces. I find it somewhat amusing that this example was not pointed out earlier (indeed, it seems to be even simpler than Julian Rosen's much upvoted example).<|endoftext|> TITLE: Semisimplicity of the category of coherent sheaves? QUESTION [8 upvotes]: The category of coherent sheaves on a locally Noetherian scheme is abelian. Are there some geometric conditions on the scheme that imply that the category of coherent sheaves is semisimple? Edited in response to posic's comments. REPLY [12 votes]: Let $X$ be a locally Noetherian scheme. Then the abelian category of coherent sheaves on $X$ is semisimple if and only if $X$ is the disjoint union of finitely many reduced points. The if direction is clear: the category of coherent sheaves on a finite union of reduced points is a direct sum of categories of finite dimensional vector spaces (over fields), so semisimple. Only if direction. If the category of coherent sheaves is semisimple, then all $Ext^1$ vanish, in particular, for every closed point $x$ of $X$, we have $Ext^1(k_x,k_x)=0$, where $k_x$ is the skyscraper sheaf at $x$. But $Ext^1(k_x,k_x)$ is the Zariski tangent space at $X$ (e.g. see https://math.stackexchange.com/questions/75673/tangent-space-in-a-point-and-first-ext-group ). As $X$ is locally Noetherian, the local ring at $x$ is Noetherian and the vanishing of the Zariski tangent space at $x$ implies by Nakayama lemma that the local ring at $x$ is a field. Using the fact that in a locally Noetherian scheme, every point specializes to a closed point (e.g. see https://stacks.math.columbia.edu/tag/01OU), it follows that $X$ is a disjoint union of reduced points. If this union is infinite, then the category of coherent sheaves is not semisimple (the structure sheaf is not a finite direct sum of simple objects). So $X$ has to be a finite disjoint union of reduced points.<|endoftext|> TITLE: Reversing the CRT: Is $5$ tough? QUESTION [5 upvotes]: Given odd primes $p\ne q$, by the CRT we can find an integer $x$ such that $x\equiv 2^{p-1}\pmod q$ and $x\equiv 2^{q-1}\pmod p$. Can this procedure be reversed? For which integers $x$ there exist odd primes $p\ne q$ satisfying $$ \begin{cases} 2^{p-1}\equiv x\pmod q \\ 2^{q-1}\equiv x\pmod p \end{cases}\ ? $$ If such $p$ and $q$ exist, then I call $x$ nice; otherwise, $x$ is tough. Say, $1$ is nice (as witnessed by the prime pair $(11,31)$), while $0$ is tough. Computations show that in the range $-20\le x\le 20$, the numbers $$ -19, -18, -17, -16, -13, -11, -8, -6, -5, -4, 1, 2, 3, 4, 6, 8, 9, 10, 13, 14, 16, 17, 18, 19 $$ are nice, while $$ -20, -15, -14, -12, -10, -9, -7, -3, -2, -1, 0, 5, 7, 11, 12, 15, 20 $$ are good candidates to be tough, in the sense that there are no primes $3\le p\nu_2(p-1)$ where $\nu_2$ is the $2$-adic valuation function. On the other hand, from $d\mid(q-1)$ we get $\nu_2(d)\le \nu_2(q-1)$. Hence, $\nu_2(p-1)<\nu_2(q-1)$. Switching the roles of $p$ and $q$, in the very same way we obtain $\nu_2(q-1)<\nu_2(p-1)$, a contradiction. My motivation came originally from this problem (consider $n=pq$), but it will be fair to say that I am driven mainly by curiosity. REPLY [8 votes]: The system of congruences in question is equivalent to $2^{pq}\equiv 2x\pmod{pq}$, i.e., the question is whether there exists an odd semiprime solution $n$ to $2^n\equiv c\pmod{n}$ (where $c=2x$). You can exclude many candidates by looking at 2^n mod n page at OEIS Wiki (and the corresponding sequences). For example, the sequence for $c=10$ contains an odd semiprime 24430928839, which disproves that 5 is tough.<|endoftext|> TITLE: Categories with every indecomposable object being uniserial QUESTION [6 upvotes]: Let $A$ be an abelian category. A Jordan-Hölder series for an object $X$ is a filtration $0 TITLE: What does the torsion-free condition for a connection mean in terms of its horizontal bundle? QUESTION [17 upvotes]: I must have read and re-read introductory differential geometry texts ten times over the past few years, but the "torsion free" condition remains completely unintuitive to me. The aim of this question is to try to finally put this uncomfortable condition to rest. Ehresmann Connections Ehresmann connections are a very intuitive way to define a connection on any fiber bundle. Namely, an Ehressmann connection on a fiber bundle $E\rightarrow M$ is just a choice of a complementary subbundle to $ker(TE \rightarrow TM)$ inside of $TE$. This choice is also called a horizontal bundle. If we are dealing with a linear connection, then $E=TM$, and the Ehresmann connection is a subbundle of $TTM$. This makes intuitive sense -- basically it's saying that for each point in $TM$ it tells you how to move it to different vectors at the tangent spaces of different points. ($ker(TTM \rightarrow TM)$ will mean moving to different vectors at the same tangent space; so that is precluded.) I like this definition -- it makes more intuitive sense to me than the definition of a Koszul connectionan $\mathbb{R}$-linear map $\Gamma(E)\rightarrow\Gamma(E\otimes T^*M)$ satisfies some condition. Unlike that definition it puts parallel transport front and center. Torsion-Freeness A Levi-Civita connection is a connection that: 1. It preserves with the Riemannian metric. (Basically, parallel transporting preserves inner products.) 2. It is torsion-free. Torsion free means $\nabla_XY - \nabla_YX = [X,Y]$. This definition very heavily uses the less intuitive notion of connection. So: Questions How can you rephrase the torsion-free condition in terms of the horizontal bundle of the connection? (Phrased differently: how can it be phrased in terms of parallel transports?) I realized that I don't actually have handy an example of a connection on $\mathbb{R}^2$ that preserves the canonical Riemannian metric on $\mathbb{R}^2$ but that does have torsion. I bet that would help elucidate the answer to my first question. REPLY [2 votes]: We have some good threads on this topic on MO, but they're usually phrased in a different language. Here is how I like to think of torsion from the perspective of an Ehresmann connection. I will use the formalism that an Ehresmann connection on $TM$ is fibrewise linear projection $c : T^2 M \to TM$ whose kernel is complementary to the kernel of $D\pi : T^2 M \to TM$ where $\pi : TM \to M$ is the tangent bundle projection. From this perspective, take a function of two variables $$ f : \mathbb R^2 \to M $$ and consider its partial derivatives to be maps $$\frac{\partial f}{\partial x_i} : \mathbb R^2 \to TM$$ i.e. $\frac{\partial f}{\partial x_i}(p) = Df(p, e_i)$. Then from this perspective, the torsion of the connection $c$, $\tau_c$ is given by: $$\tau_c(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) = c\left(\frac{\partial^2 f}{\partial x \partial y}\right) - c\left(\frac{\partial^2 f}{\partial y\partial x}\right)$$ A more elaborate (and physical) way of phrasing this would be to describe $\tau_c(v,w)$ as the result of this process: take the geodesic along $v$. Parallel transport $w$ along this geodesic, and then take the geodesic out of this parallel transport of $w$. This gives you a function of two variables (the time parameter for the two geodesics). You differentiate it with respect to the first geodesic, and ask if this vector field is parallel along the 2nd parameter. It gives you the same formula.<|endoftext|> TITLE: Any holomorphic vector bundle over a compact Riemann surface can be defined by only one transition function? QUESTION [5 upvotes]: It is known that any holomorphic bundle of any rank over a noncompact Riemann surface is trivial. A proof can be found in Forster's "Lectures on Riemann surfaces", section 30. Let $E$ be a holomorphic vector bundle over a compact Riemann surface $X$ with gauge group $G$. A consequence of the above theorem is the restriction $E|_{X-\{p\}}$ for any point $p\in X$ is a trivial bundle. Thus $E$ can be recovered by specifying the transition function $g: D\cap (X-\{p\}) \rightarrow G$ where $D$ is a small disk containing $p$. Is this correct? If not, could you give a counter-example? I am mainly interested in learning about the moduli space of holomorphic bundles over $X$ in a concrete way, e.g. using transition functions. If the argument is correct, then there is another issue. Consider a one-parameter family of transition functions $g_{\alpha\beta}(t)$. Imposing the cocycles condition on $g_{\alpha\beta}' := g_{\alpha\beta}+\epsilon\dot{g}_{\alpha\beta}$ where $\epsilon$ is infinitesimal, one finds $\dot{g}_{\alpha\beta}$ defines a class in $H^1(X,\mathfrak{g})$. Thus the tangent space at $[E]$ to the moduli space of bundles $Bun_G(X)$ is $H^1(X,\mathfrak{g})$; equivalently, $$T_{[E]}^\ast Bun_G(X) \cong H^0(X,\mathfrak{g}\otimes K_X)$$ by Serre's duality. This is the standard argument in constructing e.g. the Hitchin's system. But now as we minimally only have one transition function, we have no cocycle conditions to impose. How do I still see that $T_{[E]}^\ast Bun_G(X) \cong H^0(X,\mathfrak{g}\otimes K_X)$? REPLY [4 votes]: It is known that any holomorphic bundle of any rank over a noncompact Riemann surface is trivial. The idea of proof is the following: non-compact curve is actually affine manyfold. On affine manyfolds cohomologies of coherent sheaves vanishes (Serre). Extensions of one bundle by another is controlled by Ext^1(V,W) which is coherent and hence vanishes. So non-triviality of vector bundles reduces to linear bundles. Linear bundles are trivial by exponential sequence (Pay attention here is difference between holomorphic and algebraic situation - algebraic linear bundles can be non-trivial on affine curves). Let $E$ be a holomorphic vector bundle over a compact Riemann surface $X$ with gauge group $G$. A consequence of the above theorem is the restriction $E|_{X-\{p\}}$ for any point $p\in X$ is a trivial bundle. Thus $E$ can be recovered by specifying the transition function $g: > D\cap (X-\{p\}) \rightarrow G$ where $D$ is a small disk containing $p$. Is this correct? If not, could you give a counter-example? I am mainly interested in learning about the moduli space of holomorphic bundles over $X$ in a concrete way, e.g. using transition functions. Yes, that is correct. Such point of view on bundles became very popular since end 1980-ies. It is widely used in reseach related to confomal field theory (Verlinde formula), integrbale systems (Hitchin system) and Langlands correspondence over C. However it is not something which makes things "explicit". The goal of that technique is to translate geometric problems about vector bundles to Lie group/algebra questions about loop groups. The benefit is that you have "uniformization" - you can think of transition function function which are actually loop group G((t)) as a kind of "universal" moduli space of vector bundles - universal in very strong way since curves of all genuses "sit inside". In a sense that is in a spirit of global to local principle - geometric questions (global) reduced to questions on G((t)) - which are local. The Verlinde formula story, Knizhnik-Zamolodchikov equation, Hitchin system etc benefit much from it, see e.g. https://mathoverflow.net/a/316733/10446 But now as we minimally only have one transition function, we have no cocycle conditions to impose. How do I still see that $T_{[E]}^\ast > Bun_G(X) \cong H^0(X,\mathfrak{g}\otimes K_X)$? I am not sure I understand question correctly. But let me try to answer. Let us think in terms of group theory. So as we discussed above above G((t)) is a kind of universal moduli space. Tanget space at "e" is Lie algebra g((t)). But we are interested in cotanget space ! The point is that natural pairing is given by $ \int f d g $ , so g((t)) are functions, but $g^*((t))$ are 1-forms ! That is how "K_x" appears on the level of Lie algebra. (Going further for central extension of g((t)) dual space will be identified with space of connections).<|endoftext|> TITLE: Classification of bundles, Postnikov towers, obstruction theory, local coefficients QUESTION [11 upvotes]: RECAP on classification of bundles We want to classify $G$-principal bundles over $X$ (smooth manifold, G compact Lie). These are in 1-1 correspondence with homotopy classes of maps $[X,BG]$ (where $EG\to BG$ is the universal bundle as usual). If $BG \simeq K(\pi,n)$ then it's easy: $$[X, BG]\leftrightarrow H^n(X,\pi),$$ therefore there is a cohomology class that gives us the classification (e.g. the 1st Chern class for the frame bundles of complex line bundles). In general, $BG$ can be more complicated, in any case $BG$ has a Postnikov tower which induces a factorization of the classifying map $f\in [X, BG]$ in $(f_i)_i$ $\require{AMScd}$ \begin{CD} \vdots@. \vdots\\ @| @VVV \\ X@>>f_2> P_2(BG)\\ @| @VVp_2V \\ X @>>f_1> P_1(BG)@.\simeq K(\pi_1(BG),1) \end{CD} The homotopy type of $f$ is given by the homotopy type of the $f_i$s. Since $P_1(BG)\simeq K(\pi_1(BG),1)$ $f_1$ is given by a cohomology class in $H^1(X, \pi_1(BG))$. However, not any choice of $f_2$ works, because it must lift $f_1$. From this answer* I learned that there is a Cartesian diagram $\require{AMScd}$ \begin{CD} X@>>f_2> P_{2}(BG) @>>> K(\pi_0G,1)\\ @| @VVp_2V @VVV\\ X @>>f_1> P_1(BG) @>>> K(\pi_2 G,3)_{h\pi_0G} \end{CD} 1)Explanation/references for this? I was expecting the second column to be something like $K(\pi_2(BG),2)\to K(\pi_1(BG),1)$, it reminds me of principal fibrations. 2)How to see that these lifts are parametrized by $H^{2}(X,\pi_{2}(BG))$ cohomology with local coefficients twisted by $f_1\in H^1(X,\pi_1(BG))$? Obstruction theory tells the necessary conditions to lift $f_1$ but not how many lifts there are. In the end we get that the principal bundle is classified by $f_1\sim \alpha_1 \in H^1(X,\pi_1(BG))$ and a sequence of cohomology classes $\alpha_k \in H^{k}(X, \pi_k(BG)) $ in the cohomology with local coefficients twisted by $\alpha_1$. 3) How to compute them? Is there any example for say $G=O(2)$? Any link with invariant polynomials in $\mathfrak{g}$ or the Weyl algebra of $\mathfrak{g}$? This is essentially Denis Nardin's answer. In his comment Nardin, says another interesting thing if $G=O(n)$, then $\alpha_1 = 0 $ iff the bundle is orientable, $\alpha_1, \alpha_2 = 0$ iff the bundle is spin and so on climbing the Whitehead tower of $O(n)$ $$O(n)\leftarrow SO(n)\leftarrow Spin(n)\leftarrow String(n)\leftarrow ...$$ 4)Is this true for any Whitehead tower of groups? Does this implicitly say that the Postnikov tower of $SO(n)$ $(Spin(n))$ is the one of $O(n)$ without the first (second) term? BG as a twisted product If $\pi_1(BG)$acts on $\pi_{n+1}(BG)$ trivially then the Postnikov tower gives us an expression for $BG$ in terms of a twisted product of Eilenberg-MacLane spaces $BG \simeq K(\pi_1(BG),1)\times_{k_1} K(\pi_2(BG),2)\times_{k_2} \dots$ There is no need of local coefficients for the $\alpha_k$ above. In the same answer*, Mark Grant says that in the case of $G=O(2)$: there is a fibration $$ K(\mathbb{Z},2)\to E\mathbb{Z}/2\times_{\mathbb{Z}/2} K(\mathbb{Z},2)\to B\mathbb{Z}/2 $$ given by the twisting of $w_1$ on the universal $SO(2)$ bundle, and this fibration agrees up to homotopy with the fibration $$BSO(2)\to BO(2)\to BO(1).$$ Question: 5) Can you explain this in the more general setting of a fibration $F\to E\to B$? Also I do not understand $E\mathbb{Z}/2\times_{\mathbb{Z}/2} K(\mathbb{Z},2)$, what is the $\mathbb{Z}/2$ action on $K(\mathbb{Z},2)$? Grant says $w_1$ is involved but I cannot imagine how. (I know that in general $w_1 \in H^1(X, \pi_1(BO(2))$ gives me an action of $\pi_1(X)$ on $\pi_n(BO(2))$). How does this relate to the Whitehead tower above? *Classification of $O(2)$-bundles in terms of characteristic classes REPLY [10 votes]: I'll try to answer question 1. Unfortunately, I know of no especially convenient reference for the case of understanding general Postnikov towers; however, everything I say below is "well known". In a Postikov tower for $X$, the map $p=p_n\colon P_n\to P_{n-1}$ is, by definition, a fibration with fiber equivalent to $K(A,n)$, where $A=\pi_n X$. The idea is that there exists an (essentially unique) homotopy pullback square $\require{AMScd}$ \begin{CD} P_{n} @>>> E_F\\ @VpVV @VVqV\\ P_{n-1} @>>> B_F \end{CD} where $q$ is the universal example of a fibration with fiber equivalent to $F:=K(A,n)$. This universal fibration is rarely ever a principal bundle. Here is the formula for $q$: Let $\def\Aut{\mathrm{Aut}}\def\Map{\mathrm{Map}}$ $\Aut(F)\subseteq \Map(F,F)$ be the union of all components of the mapping space which contain homotopy equivalences. Then $\Aut(F)$ is a topological monoid which is "grouplike" (i.e., $\pi_0\Aut(F)$ is a group), and which acts on $F$. Then the map $$ F_{h\Aut(F)} \to (*)_{h\Aut(F)}=B\Aut(F)$$ is the universal $F$-fibration. For $F=K(A,n)$, it turns out you can identify $\Aut(F)$ very explicitly. (I'm going to assume $n\geq 2$ here.) The construction of Eilenberg-MacLane $A\mapsto K(A,n)$ spaces lifts to a functor $$(\text{abelian groups})\to (\text{topological abelian groups}).$$ Therefore we get a topological group $K(A,n)\rtimes \Aut(A)$ acting on $K(A,n)$ (the $K(A,n)$ acts on itself by left-translation, and $\Aut(A)$ is the discrete automorphism group of $A$.) It turns out (by a computation) that this group is equivalent (up to homotopy) to the topological monoid $\Aut(K(A,n))$, so $$B_{K(A,n)} \approx B\Aut(K(A,n)) \approx B\bigl( K(A,n)\rtimes \Aut(A)\bigr) \approx BK(A,n)_{h\Aut(A)}\approx K(A,n+1)_{h\Aut(A)},$$ and $$E_{K(A,n)} \approx K(A,n)_{h\bigl(K(A,n)\rtimes\Aut(A)\bigr)} \approx (*)_{h\Aut(A)} \approx B\Aut(A).$$ So the desired pullback square has the form \begin{CD} P_{n} @>>> B\Aut(A)\\ @VpVV @VVqV\\ P_{n-1} @>>> K(A,n+1)_{h\Aut(A)} \end{CD} In practice, when you have a Postnikov tower of a space like $BG$, the action of the fundamental group $\pi_1 BG=\pi_0G$ on $A=\pi_nBG$ determines a homomorphism $\pi_0G\to \Aut(A)$, and you can "restrict" along this homomorphism to get a pullback square \begin{CD} P_{n} @>>> B\pi_0G\\ @VpVV @VVqV\\ P_{n-1} @>>> K(A,n+1)_{h\pi_0G} \end{CD}<|endoftext|> TITLE: Interesting examples of non-locally compact topological groups QUESTION [11 upvotes]: Harmonic analysis is concentrated mostly on studying locally compact groups. I would like to ask people about examples of non-locally compact topological groups that are interesting in connection with some applications in mathematics (or physics). I must confess that I know only two examples: topological vector spaces are studied as examples of topological groups (with the additive group operation) to which Pontryagin duality is sometimes transferred from the class of locally compact abelian groups, the groups of diffeomorphisms of smooth manifolds are studied in the theory of infinite dimensional manifolds. Can people enlighten me about other similar subjects? (If possible, with motivations.) REPLY [6 votes]: Quantum mechanics: The unitary group $U$ of a Hilbert space. If you use the norm topology, then it is a Banach Lie group with bounded skew hermitian operators as Lie algebra. But unbounded self-adjoint operators $A$ (Schrödinger operators) lead to 1-parameter semigroups $e^{itA}$ in $U$, where now the strong operator topology plays a role. REPLY [3 votes]: The question mentions topological vector spaces, but aren't Banach spaces a special case of sufficient interest to merit explicit mention? Every infinite dimensional Banach space is a non-locally compact topological group under addition and the norm topology.<|endoftext|> TITLE: Endomorphism rings of ordinary elliptic curves QUESTION [6 upvotes]: Let's say $p$ is a prime and $t\neq 0$ is a trace of Frobenius that occurs over $\mathbb{F}_p$. The discriminant of the Frobenius polynomial is $\Delta:=t^2-4p.$ So we obtain $4p=t^2-\Delta.$ If $E$ is an elliptic curve over $\mathbb{F}_p$ with trace of Frobenius $t$, then the Frobenius, call it $\sigma$, generates an order in the endomorphism ring of $E$, $End(E)$. Symbollically, $\mathbb{Z}[\sigma]\subseteq End(E).$ Put another way, the discriminant of $End(E)$ divides $\Delta$. Now $\Delta$ may not be the discriminant of a maximal order. We write $\Delta=B\Delta_{\mathcal{O}_K}$ where $\Delta_{\mathcal{O}_K}$ is the discriminant of the ring of integers of some imaginary quadratic field $K$ and $B\geq 1$. Suppose there is an integer $b>1$ with $b^2\mid B$ and $b$ prime to $p$. Then $b$, the conductor, is the index of the order of discriminant $b\Delta_{\mathcal{O}_K}$ in $\mathcal{O}_K$. There could be an elliptic curve over $\mathbb{F}_p$ whose endomorphism ring is the order of discriminant $b\Delta_{\mathcal{O}_K}$. My question is, in general, does every possible order occur as an endomorphism ring of an elliptic curve over $\mathbb{F}_p$? Kohel seems to indicate this is the case in his thesis, but I can't find a proof. By work of Deuring and because the $j$-invariants in question are algebraic, we know that over some finite extension of $\mathbb{F}_p$ there is an elliptic curve with an endomorphism ring that is isomorphic to each possible order. For my applications though, I'd like to have this elliptic curve be defined over $\mathbb{F}_p.$ REPLY [2 votes]: The paper "Abelian varieties over finite fields" of Waterhouse (Annales scientifiques de l'École Normale Supérieure, Série 4, Volume 2 (1969) no. 4, p. 521-560), available at http://www.numdam.org/item/?id=ASENS_1969_4_2_4_521_0 has, I think, relevant information (see Section 4). There might be more recent works in this direction also.<|endoftext|> TITLE: Questions about "The best card trick" QUESTION [5 upvotes]: Kleber's Best card trick proceeds as follows: The mark (audience member) freely selects five playing cards from a standard deck of $52$ and passes these five to the magician's assistant. The assistant studies those cards, returns one mystery card to the mark, and places the remaining four exposed cards, face up, in a sequence on a table. The magician then enters, inspects the sequence of exposed cards, and then correctly announces the full identity of the mystery card. The trick works through the clever use of mathematics. The five selected cards always contain at least one suit that is represented by two (or more) cards. The assistant will choose one of these as the mystery card, and another of the same suit to be the first in the sequence of exposed cards. Thus the magician learns the suit of the mystery card. Playing cards can be placed in a canonical order ($\clubsuit A, 2, \ldots K, \diamondsuit A, 2, \ldots K, $ etc.), and thus the three remaining exposed cards can be placed in $3!$ possible order sequences. Thus the assistant can signal six candidate card values, counting from the value of the first card (modulo 13). However, that approach alone will not cover all 12 potential card values. This issue is circumvented by the assistant being careful about which card of a pair with the same suit is given back and which is used as the first card in the sequence: use as the first card the one whose value is fewer than seven steps before the other of the same suit (modulo 13), which is the mystery card; that way the $3!$ possible steps ensure that the mystery card can be reached from the first card in the exposed sequence. Question 1: How many four-card exposed sequences can arise in such tricks? Question 2: How many sets of five selected cards have more than one acceptable sequence of exposed cards? Background The mark is free to choose any set of five cards, and there are of course ${52 \choose 5}$ ways to do this. Each such set of five is guaranteed to have at least one sequence of four visible cards. Moreover, according to the algorithm, each exposed sequence leads to a unique mystery card. Even given these facts, not all $52!/48!$ conceivable exposed sequences of four cards will appear. For example, the sequence $2 \spadesuit\ 3\spadesuit\ 4 \spadesuit\ 5 \spadesuit$ will never appear, because the inference algorithm implies the hidden card is the $3 \spadesuit$, which is instead in the exposed sequence. As @IlyaBogdanov points out in a comment, any set of five selected cards that has just one suit represented by two cards will have a unique exposed sequence, and because of the uniqueness of the inference method, these must be distinct sequences. The number of such cases is $4 {13 \choose 2} 13^3$. The number of cases that have two suits, each represented by two cards is ${4 \choose 2}{13 \choose 2}{13 \choose 2} 13$, and each of these will have two distinct possible exposed sequences. The trickier cases are when there is a single suit with $3$ or $4$ or even $5$ cards of that suit, and the mixture of $3$ cards in one suit and $2$ in another. The challenge is that the number of exposed sequences will depend upon the particular cards in the represented suit. For instance, even if all five cards are spades, there will be a different number of solutions in these two cases: $\{2 \spadesuit\ 4 \spadesuit\ 6 \spadesuit\ 8 \spadesuit\ 10 \spadesuit \}$ and $\{2 \spadesuit\ 3 \spadesuit\ 4 \spadesuit\ 5 \spadesuit\ 6 \spadesuit \}$. REPLY [6 votes]: The answer to Question 1 is $52!/48! - 4 \binom{13}{2} \binom{50}{2}= 6115200$. That is, I claim that the number of $4$-tuples that cannot occur is $4 \binom{13}{2} \binom{50}{2}$. To see this, note that a $4$-tuple $(a,b,c,d)$ cannot occur if and only if the fifth card $e$ that $(a,b,c,d)$ signifies is among $\{b,c,d\}$. Note that there are $4 \binom{13}{2}$ choices for $(a,e)$. Finally, for each of the $\binom{50}{2}$ $2$-subsets $\{x,y\}$ of $[52] \setminus \{a,e\}$, there is a unique ordering $(b,c,d)$ of $\{ x,y,e \}$ such that $(a,b,c,d)$ signifies $e$. As Ilya Bogdanov points out in the comments, a $5$-set has a unique $4$-tuple if and only if it contains exactly one pair of cards of the same suit. The number of such $5$-sets is $4 \binom{13}{2}13^3$. Thus, the answer to Question 2 is $\binom{52}{5}-4 \binom{13}{2}13^3= 1913496$.<|endoftext|> TITLE: Do we know the consistency strength of the Singular Cardinal Hypothesis failing on an uncountable cofinality? QUESTION [13 upvotes]: Suppose that $\kappa$ is a strong limit cardinal. The singular cardinal hypothesis states $2^\kappa=\kappa^+$. We know that the failure of SCH requires large cardinals, and in fact is equiconsistent with a measurable cardinal $\kappa$ satisfying $o(\kappa)=\kappa^{++}$. But this failure is at $\aleph_\omega$. Suppose we wanted more. Suppose that we wanted the failure to happen on a couple isolated points. Well, it's not hard to redo the standard constructions and get just that. But what happens when we have limit points? Even more, by Silver's theorem if SCH fails at $\kappa>\operatorname{cf}(\kappa)>\omega$, then there is a stationary subset of $\kappa$ where SCH failed. What would be the consistency strength when $\kappa$ is a singular limit of singular cardinals, and SCH fails cofinally below $\kappa$? What if we require $\kappa$ to be of uncountable cofinality? As a side question, what if $\kappa$, with uncountable cofinality, does satisfy SCH, but an unbounded subset (which has to be non-stationary, of course) of it does not? REPLY [10 votes]: Suppose $\kappa$ is a singular cardinal and there are $cf(\kappa)$-many measurable cardinals $\lambda < \kappa$ with $o(\lambda)=\lambda^{++}$ cofinal in $\kappa.$ Then you can perform a Prikry type iteration and get the failure of $SCH$ at cofinally many singular cardinals below $\kappa.$ Now suppose we also want for $SCH$ to fail at $\kappa$ itself. First let us consider the countable cofinality. Assume $\kappa$ is a measurable cardinal with $o(\kappa)=\kappa^{++}+1.$ Then we can get an extension in which $cf(\kappa)=\omega, 2^\kappa=\kappa^{++}$ and for cofinally many singular cardinals $\lambda$ below $\kappa$, we have $2^{\lambda}=\lambda^{++}$. I don't know if this assumption is really needed or if it can be reduced. For uncountable cofinality, say $\theta$, a measurable cardinal $\kappa$ with $o(\kappa)=\kappa^{++}+\theta$ is sufficient. As then you can first find an extension in which $2^\kappa=\kappa^{++}$ and such that in the extension, $o(\kappa)=\theta$. Then if you perform Magidor forcing for changing cofinality of $\kappa$ to $\theta,$ you can get a club $C$ of singular cardinals below $\kappa$ such that for all $\lambda \in C, 2^\lambda=\lambda^{++}$. As far as I know, if we require $\theta=cf(\kappa)> \omega_1$, the large cardinal assumption is optimal, but for $\theta=\omega_1,$ I think it is open if this assumption is optimal.<|endoftext|> TITLE: Double integral with logarithms QUESTION [5 upvotes]: Faster method to calculate the exact solution of the following integral (based on the ideas of Fedor Petrov: https://mathoverflow.net/users/4312/fedor-petrov, Double integral with logarithms, URL (version: 2019-04-15): https://mathoverflow.net/q/328126): $$J\equiv \int_{0}^{1}{\int_{0}^{1}{\frac{\ln x-\ln y}{x-y}}}dxdy .$$ Since $$f\left( x,y \right)=\frac{\ln x-\ln y}{x-y}=f\left( y,x \right),$$ the surface $f\left( x,y \right) $ is symmetric with respect to the bisector plane $x = y$; so, $$\frac{J}{2}=\int_{0}^{1}{dx\int_{0}^{x}{\frac{\ln x-\ln y}{x-y}}}dy.$$ With the change of variable$$y\equiv tx,\ t\in \left( 0,\ 1 \right),$$ the integral $$\int_{0}^{x}{\frac{\ln x-\ln y}{x-y}}dy,$$ is transformed into the following one that does not depend on $x$, $$ I\equiv -\int_{0}^{1}{\frac{\ln t}{1-t}\,}dt.$$ The integration on the unit square $(0, 0), (1, 0), (1, 1), (0,1)$ is reduced to the integration on the triangle $(0, 0), (1, 0), (1, 1).$ To solve the integral $I$ we will carry out the new change of variable, $$s\equiv 1-t,$$ by which $I$ is transformed into the integral that defines the dilogarithm, whose value for $s = 1$ coincides with the Riemann zeta $\zeta \left( 2 \right)$,whose value is well known: $$I=\int_{1}^{0}{\frac{\ln \left( 1-s \right)}{s}\,}ds=\text{L}{{\text{i}}_{2}}\left( 1 \right)=\zeta \left( 2 \right)=\frac{{{\pi }^{2}}}{6}.$$ Therefore, the solution to the proposed integral is $$J=\frac{{{\pi }^{2}}}{3}.$$ Note. I've tried it by polylogarithmic transformations, but I couldn't get the result $\frac{{{\pi }^{2}}}{3}.$ REPLY [15 votes]: By symmetry we have $J/2=\int_0^1 dx \int_0^x f(x, y) dy$ where $f(x, y) $ is your integrand. Integrating against $y$ for fixed $x$ we denote $y=tx$, $t$ varies from 0 to 1 and the integral against $y$ reads as $-\int_0^1 \frac{\log t} {1-t}dt$. It does not depend on $x$ and is well known to be equal to $\pi^2/6$ (you may use the geometric progression expansion $\frac{1}{1 - t} =\sum_{n>0 } t^{n-1}$ and integrate term-wise to get $\sum 1/n^2$).<|endoftext|> TITLE: Non-faithful irreducible representations of simple Lie groups QUESTION [6 upvotes]: For a complex simple Lie algebra $\frak{g}$, which of its finite dimensional irreducible representations give non-faithful representations of the corresponding simply-connected compact Lie group. More specifically, can somebody point me to a table detailing, for each series, those dominant weights for faithfulness fails. Edit: To make my question clearer, I am asking in relation to the answer of this question, which asks when can you build up all representations from the fundamental and antifundamental ones? It is answered that, for an irreducible Lie algebra representation $V$, when the corresponding representation of $G$ is faithful, any other irreducible Lie algebra can be found in a tensor product $V^{\otimes k}$, for sufficiently high $k$. The answer is qualified with the following comment: One mild warning: there is an obvious representation of $\frak{}()$ which is not a faithful representation of the corresponding simply-connected compact Lie group when $n\geq 3$ Is the "obvious representation" of $\frak{so}(n)$ the only representation for which this happens, or are there others? REPLY [10 votes]: Let $G_{sc}$ be as in the answer by Victor Protsak and let $\varpi_1$, $\ldots$, $\varpi_l$ be the fundamental dominant weights. Let $\lambda$ be a dominant weight and write $\lambda = \sum_{i = 1}^l a_i \varpi_i$ for $a_i \in \mathbb{Z}_{\geq 0}$. Then the irreducible representation of $G_{sc}$ with highest weight $\lambda$ is faithful precisely in the following cases: Type $A_{l}$ ($l \geq 1$): $\gcd(l+1, a_1+2a_2+\cdots+la_l) = 1$. Type $B_l$ ($l \geq 2$): $a_l$ is odd. Type $C_l$ ($l \geq 2$): $a_1 + a_3 + a_5 + \cdots$ is odd. Type $D_l$ ($l \geq 4$): $l$ is odd and $a_{l-1} + a_l$ is odd Type $G_2$: always Type $F_4$: always Type $E_6$: $a_1 - a_3 + a_5 - a_6$ is not divisible by $3$. Type $E_7$: $a_2 + a_5 + a_7$ is odd Type $E_8$: always This can be determined by a direct computation. The kernel of any irreducible representation of $G_{sc}$ lies in $Z(G_{sc})$, which is finite. Furthermore, you can describe $Z(G_{sc})$ explicitly and compute the action of any $z \in Z(G_{sc})$ in an irreducible representation (it is of course always multiplication by some scalar). See for example Chapter 3, Lemma 28 in "Lectures on Chevalley Groups" by Steinberg. Note above that in types $G_2$, $F_4$ and $E_8$ the center of $Z(G_{sc})$ is trivial so every irreducible representation is faithful. Also, for type $D_{2l}$ the center is not cyclic so no irreducible representation is faithful.<|endoftext|> TITLE: Unbounded Component of the Fredholm Domain QUESTION [5 upvotes]: Let $X$ be a Banach space and $T \in \mathcal L(X)$. The authors Engel and Nagel introduce in their book "One-Parameter Semigroups for Linear Evolution Equations" on p. 248 the concept of the Fredholm domain of $T$ defined by $$\rho_F(T) := \{\lambda \in \mathbb C: \lambda - T \text{ is a Fredholm operator} \}.$$ On the next page the following is stated: "Here, we only recall that the poles of $R(\cdot, T)$ with finite algebraic multiplicity belong to $\rho_F(T)$. Conversely, an element of the unbounded connected component of $\rho_F(T)$ either belongs to $\rho(T)$ or is a pole of finite algebraic multiplicity." I can prove the first statement very elementary just by using properties of spectral projections and some very basic functional calculus. But the second statement seems to be quite difficult to prove. In the cited literature I found a proof of the stament (cf. the proof Corollary XI.8.5 in "Classes of Linear Operators Vol. I" by Gohberg, Goldberg and Kaashoek). But it seems to rely on quite some theorems about Fredholm operator valued functions. So my question is whether there is a more elementary way to see that the statement holds, maybe just by using some basic facts on spectral projections? I thought quite some time about it but couldn't prove it. So is there maybe a reference for the statement which uses more elementary arguments? Or does someone know another way how to prove it? I am looking forward to your answers. REPLY [3 votes]: A maybe different proof can be obtained using techniques related with the single-valued extension property (SVEP for short) of an operator. A good reference for this topic is the book [Aiena]: Pietro Aiena. Fredholm and local spectral theory, with applications to multipliers. Kluwer, 2004. Let us denote by $\rho_F^\infty(T)$ the unbounded connected component of $\rho_F(T)$. First you have to show that $\rho_F^\infty(T)\setminus\rho(T)$ consists of isolated points of the spectrum of $T$. This fact can be proved as follows: (1) The index $\textrm{ind}(zI-T):= \dim \ker(zI-T) - \dim X/R(zI-T)$ is constant for $z$ in each connected component of $\rho_F(T)$. Hence $\textrm{ind}(zI-T)=0$ for $z\in\rho_F^\infty(T)$. Proof: This is not trivial, but it is a standard result. (2) The set $\rho_F^\infty(T)\setminus\rho(T)$ does not have accumulation points in the spectrum of $T$. Proof: If $T$ is a Fredholm operator with $\textrm{ind}(T)=0$, then so is the conjugate operator $T^*:X^*\to X^*$. Therefore you can find closed subspaces $M$ of $X$ and $N$ of $X^*$ such that $X=\ker(T)\oplus M$, $X^*=\ker(T^*)\oplus M$ and there exists $c>0$ such that $\|Tm\|\geq c \|m\|$ for each $m\in M$ ($T$ is bounded below on $M$) and $\|T^*n\|\geq c \|n\|$ for each $n\in N$. As a consequence, if $0<|z| TITLE: Is there a "higher Segal conjecture"? QUESTION [12 upvotes]: The Segal conjecture describes the Spanier-Whitehead dual $D \Sigma^\infty_+ BG$ for certain $G$. Is there a similar description of $D\Sigma^\infty_+ K(G,n)$ when $n \geq 2$ when $G$ is finite (and abelian)? Notes: I'd be happy to understand the case of cyclic groups $G = C_p$. $K(G,n)$ can be modeled by an abelian topological group, but I'm not sure it falls under the umbrella of other known generalizations of the Segal conjecture, although when $G = \mathbb Z$ and $n=2$ there is a known decomposition (see Ravenel). For $G = \mathbb Z^n$ and $n=2$ there is also this. Let me recall that the Segal conjecture (proved by Carlsson) says that when $G$ is finite, the Spanier-Whitehead dual $D\Sigma^\infty_+ BG$ is a certain completion of $\vee_{(H) \subseteq G} \Sigma^\infty_+ BW_G(H)$ where $(H) \subseteq G$ ranges over conjugacy classes of subgroups and $W_G(H) = N_G(H) / H$ is the Weyl group of $H$ in $G$. In particular, when $G = C_p$ it says that $$D\Sigma^\infty_+ BC_p = \mathbb S \vee(\Sigma^\infty_+ BC_p )^{\wedge}_p$$ where $\mathbb S$ is the sphere spectrum (corresponding to the subgroup $C_p \subseteq C_p$; the other term corresponds to the trivial subgroup $0 \subseteq C_p$) and $(-)^\wedge_p$ is $p$-completion. Lin showed that $D H G = 0$ when $G$ is a finite abelian group, where $H$ indicates taking Eilenberg-MacLane spectra. Since $HG = \varinjlim_n \Sigma^{\infty-n} K(G,n)$, we have $0 = DHG = \varprojlim_n \Sigma^n D\Sigma^\infty K(G,n)$, and from the Milnor exact sequence we conclude that $\varprojlim_n \pi_{\ast-n} D\Sigma^\infty K(G,n) = \varprojlim^1_n \pi_{\ast-n} D \Sigma^\infty K(G,n) = 0$. But I'm not sure how much information that is, really. If we work in the $K(h)$-local or the $T(h)$-local category then by ambidexterity we have $F(\Sigma^\infty_+ K(G,n), L\mathbb S) = L \Sigma^\infty_+ K(G,n)$ where $L$ is the relevant localization. But it seems that the relevant limit does not commute with localization here. REPLY [16 votes]: In the 1980's, Chun Nip Lee showed that the Spanier Whitehead dual of (the suspension spectrum of) $K(\mathbb Z/p, n)$ is contractible for $n >1$. (The key case is $n=2$. The idea: view $K(A,n+1)$ as the bar construction on $K(A,n)$.) (No time right now to write more ... but maybe this is enough.)<|endoftext|> TITLE: Closed manifold with non-vanishing homotopy groups and vanishing homology groups QUESTION [24 upvotes]: Is there a closed connected $n$-dimensional topological manifold $M$ ($n\geq 2$) such that $\pi_i(M)\neq 0$ for all $i>0$ and $H_i(M, \mathbb{Z})=0$ for $i\neq 0$, $n$? The manifold $S^1\times S^2$ satisfies the first requirement but not the second (generally, the direct product of two positive-dimensional manifolds does not seem to satisfy the second requirement because of Künneth). REPLY [18 votes]: As suggested by Lennart Meier, the connected sum $M=P\#P$ of two copies of the Poincaré homology sphere gives an example. The homotopy groups $\pi_n(M)$ are nonzero for all $n>1$ because the universal cover $\widetilde{M}$ has a retraction (not a deformation retraction) onto $S^2$ and it is known that all the higher homotopy groups of $S^2$ are nontrivial. A retraction $r:\widetilde M \to S^2$ can be obtained in the following way. The connected sum $P\#P$ is constructed by removing an open ball from $P$ to obtain a manifold $P'$, then gluing two copies of $P'$ together by identifying their boundary spheres. The universal cover $\widetilde {P'}$ is $S^3$ with $|\pi_1(P)|=120$ disjoint balls removed. The universal cover $\widetilde M$ is obtained by gluing infinitely many copies of $\widetilde {P'}$ together in a tree-like pattern, identifying boundary spheres in pairs. Each copy of $\widetilde {P'}$ retracts onto any one of its boundary spheres since $\widetilde {P'}$ is $S^2\times I$ with 118 balls removed and $S^2\times I$ retracts onto one of its boundary spheres hence $\widetilde {P'}$ also retracts onto this boundary sphere by restriction. We can build $\widetilde M$ as an infinite sequence of attachments of one copy of $\widetilde {P'}$ at a time, starting with a single copy. Each stage of this construction retracts onto the previous stage. The infinite composition of these retractions is well-defined (and continuous) since any compact subset of $\widetilde M$ is contained in a finite stage. The infinite composition gives a retraction of $\widetilde M$ onto $\widetilde {P'}$ which in turn retracts onto $S^2$. The fact that all the higher homotopy groups of $S^2$ are nontrivial was shown in a paper by S. O. Ivanov, R. Mikhailov, and Jie Wu in Homology Homotopy Appl. 18 (2016), no. 2, 337--344. The corresponding result holds also for $S^3$ since $\pi_n(S^3)=\pi_n(S^2)$ for $n\geq 3$, and the result is known for $S^4$ and $S^5$ as well. It fails for other spheres since $\pi_{n+4}(S^n)=0$ for $n\geq 6$ since this is in the stable range and the stable 4-stem is trivial. However, it's not clear how to apply these results to obtain homology $n$-spheres with all homotopy groups nontrivial when $n>3$. The argument for $P\# P$ can probably be extended to connected sums of arbitrary nonsimply-connected homology 3-spheres with a little more work to cover the case that the summands of $M$ have infinite fundamental groups.<|endoftext|> TITLE: Tower of moduli spaces in Scholze's theory QUESTION [16 upvotes]: My question is related to another one I read here in Overflow. I am reading Scholze's papers about moduli spaces of $p$-divisible groups and elliptic curves, and I am very interested in the formal geometry involved there. Actually, I noticed that there is a paper by Andreatta, Iovita and Pilloni, titled Le halo spectral, which seems to deal with formal integral models of Scholze's towers. First, if I well understand Scholze, talking about elliptic curves, there is a perfectoid space $\mathcal{X}_{\infty}(\epsilon)$ which gives the "tilda limit" of modular curves $\mathcal{X}_{\Gamma(p^n)}(\epsilon)$ where each $\mathcal{X}_{\Gamma(p^n)}(\epsilon)$ describes open neighborhoods of the ordinary locus of the $\Gamma_1(N)$ modular curve, where the universal elliptic curve coming from pullback is not too supersingular. Actually, the construction of this object is performed by computing the adic generic fiber of the formal scheme $\mathfrak{X}_{\infty}(\epsilon)$ which is the real limit (in the category of formal schemes) of integral models of $\mathcal{X}_{\Gamma_(p^n)}(\epsilon)$ where the maps in the inverse system are given by a lifting of mod $p$-Frobenius. A very similar construction is performed in chapter $6$ of Andreatta, Iovita and Pilloni's paper, where they construct the integral anticanonical tower $\mathfrak{X}_{\infty}$ exactly in the same way, but working over a basis which is a suitable blowup of an integral model of Coleman's weight space. Now, I'm just wondering whether or not it is possible to interpret these "infinite" level spaces as moduli spaces of elliptic curves plus a new kind of level structure. Somewhere in Scholze's paper it is mentioned that a point of $\mathcal{X}_{\infty}$ over $\text{Spa}(C,\mathcal{O}_C)$, where $C$ is a complete algebraically closed extension of $\mathbb{Q}_p$ corresponds to an elliptic curve over $C$ with a trivialization of its Tate module. Now, why is this true? It's not mentioned in Scholze and I cannot prove it. Moreover, does a similar description hold for different kind of points, e.g. $\text{Spa}(R,R^+)$ with $R$ a perfectoid $\mathbb{Q}_p$-algebra? Moreover, does the same intepretation hold for its formal integral model? And what about the Andreatta, Iovita and Pilloni's tower? Is it true that it parametrizes elliptic curves with $p$-divisible groups playing the role of the canonical subgroup? The point essentially is, does this object gives by pullback a universal elliptic curve? Which kind of level does it have a similar elliptic curve? REPLY [16 votes]: Wow, that's eight questions, plus more in the comments -- I don't think I can answer all of them, but I'll try to answer at least a few! :) First of all, let's fix the setting: It seems to me that you are using three different kinds of level structures, $\Gamma_0(p^n)$, $\Gamma_1(p^n)$ and $\Gamma(p^n)$, and some questions seem to be referring to different ones. Also, there are various possible different meanings of $\mathcal X$ and $\mathfrak X$, since these mean different things in each paper respectively. So we first have to agree on some uniform setting for your questions: Since you are interest in perfectoid moduli spaces, I suggest we follow Scholze (III.2.2 in the torsion paper) and denote by $\mathfrak X$ the formal completion of the modular curve over $\mathbb Z_p^\mathrm{cyc}$ of some fixed tame level. Denote by $\mathfrak X^{\ast}$ the completion of the compactified modular curve, and by $\mathcal X^{\ast}$ the adic generic fibre. Since you are interested in moduli of elliptic curves, I suggest we now deviate from Scholze's notation and denote by $\mathcal X$ the analytification of the modular curve over $\mathbb Q^{\mathrm{cyc}}$ (rather than the generic fibre of $\mathfrak X$, which is the good reduction locus). moduli interpretations of the spaces $\mathcal X_{\Gamma_0(p^\infty)}(\epsilon)_a$, $\mathcal X_{\Gamma_1(p^\infty)}(\epsilon)_a$ and $\mathcal X_{\Gamma(p^\infty)}(\epsilon)_a$ (Most of what I'm going to say in regards to this question can be found in more detail in this related article: https://nms.kcl.ac.uk/ben.heuer/PGp-adMC.pdf.) Let's follow the torsion paper and start with level $\Gamma_0(p^n)$ and the anticanonical locus $\mathcal X_{\Gamma_0(p^n)}(\epsilon)_a$ of some tame level. This represents the functor which sends a (sheafy) adic space $\mathrm{Spa}(R,R^{+})$ to the set of isomorphism classes of triples $(E,\alpha,D)$ where $E|R$ is an elliptic curve with some condition on the Hasse invariant which ensures that $E$ has a canonical subgroup $C=C(E)\subseteq E[p]$, where $\alpha$ is a tame level structure, and where $D\subseteq E[p^n]$ is an anticanonical cyclic subgroup scheme of rank $p^n$. Here "anticanonical" means $D\cap C=0$. Scholze now proves that there is a perfectoid space $$\mathcal X_{\Gamma_0(p^\infty)}(\epsilon)_a\sim \varprojlim\mathcal X_{\Gamma_0(p^n)}(\epsilon)_a.$$ Since perfectoid tilde-limits satisfy the universal property of the limit for perfectoid test objects, this space represents the functor which sends $\mathrm{Spa}(R,R^{+})$ for any perfectoid $\mathbb Q_p^\mathrm{cyc}$-algebra $R$ to the set of isomorphism classes of $(E,\alpha,D_\infty)$ where $D_\infty = (D_n\subseteq E[p^n])_{n\in\mathbb N}$ is a collection of anticanonical cyclic subgroup schemes with $D_{n+1}[p^n]=D_n$ (See Corollary 3.2 of the above document). So in regards to your question of moduli of $p$-divisible groups, one could call this data an "anticanonical $p$-divisible subgroup of height 1". Similar results hold for the perfectoid tilde-limits $\mathcal X_{\Gamma_1(p^\infty)}(\epsilon)_a\sim \varprojlim\mathcal X_{\Gamma_1(p^n)}(\epsilon)_a$ and $\mathcal X_{\Gamma(p^\infty)}(\epsilon)_a\sim \varprojlim\mathcal X_{\Gamma(p^n)}(\epsilon)_a$ by the same reasoning: The first represents the functor which sends $\mathrm{Spa}(R,R^{+})$ for any perfectoid $\mathbb Q_p^\mathrm{cyc}$-algebra $R$ to the set of isomorphism classes of $(E,\alpha,\beta: \mathbb Z_p\xrightarrow{\sim} T_pD_\infty(R))$ where $D_\infty$ is an anticanonical $p$-divisible subgroup of height 1 and beta is a trivialisation of its Tate module. The second uses instead isomorphism classes of tuples $(E,\alpha,\gamma: \mathbb Z_p^2\xrightarrow{\sim} T_pE(R))$ where the image of $\gamma(1,0)$ in $E[p](R)$ generates an anticanonical subgroup. the formal model of the anticanonical tower As with the last question, we first need to agree on a base: The torsion paper considers an anticanonical tower over $\mathbb Z_p^\mathrm{cyc}$ (whose limit you denote by $\mathfrak X_\infty(\epsilon)$), whereas Le halo spectral basically works over $\mathbb Z_p$ (as you say, they really work relatively to some weight space, which is great because it allows them to construct integral families of modular forms. But I think in order to understand what's going on in terms of moduli, it might be easier if we specialise to a point -- the weight space doesn't change much in that respect). Let's follow Scholze and work over $\mathbb Z_p^\mathrm{cyc}$ if you don't mind, so we simply base-change the constructions of Andreatta--Iovita--Pilloni to $\mathbb Z_p^\mathrm{cyc}$ (their constructions actually require Noetherianess in several places in order to construct normalisations, but once you got the spaces, you may still simply base-change to $\mathbb Z_p^\mathrm{cyc}$. The resulting spaces agree with Scholze's $\mathfrak X^{\ast}(\epsilon)$ up to a normalisation issue). Now there are arguably two "anticanonical towers", which are isomorphic: The first one, which gives the tower its name, is the tower $$\dots\to\mathcal X^{\ast}_{\Gamma_0(p^2)}(\epsilon)_a\to\mathcal X^{\ast}_{\Gamma_0(p)}(\epsilon)_a\to \mathcal X^{\ast}(\epsilon).$$ The second tower is used in the torsion paper to prove that the above tower has a perfectoid tilde limit $\mathcal X^{\ast}_{\Gamma_0(p^\infty)}(\epsilon)_a$: Let's recall how this works. Let $\mathfrak X^{\ast}(\epsilon)$ be like in Scholze's Definition III.2.12. As explained there, (away from the cusps) this represents the functor sending $\mathrm{Spf}(R)$ for $p$-adically complete $\mathbb Z_p^\mathrm{cyc}$-algebras $R$ to the set of isomorphism classes $(E,\alpha,\eta)$ where $E|R$ is an elliptic curve, $\alpha$ is a tame level and $\eta\in \omega_E^{\otimes(1-p)}$ such that $\eta \mathrm{Ha} = p^{\epsilon} \in R/p$. Scholze constructs Frobenius lifts $F:\mathfrak X^{\ast}(p^{-1}\epsilon)\to \mathfrak X^{\ast}(\epsilon)$ which on the level of moduli (away from the cusps) are given by quotienting by the canonical subgroup, i.e. sending $E\mapsto E/C$. In the limit, this gives rise to the space $\mathfrak X^{\ast}(p^{-\infty}\epsilon)=\varprojlim_{{F}} \mathfrak X^{\ast}(p^{-n}\epsilon)$ which is integrally perfectoid. In particular, its adic generic fibre is a perfectoid space. The relation to the anticanonical tower is that on the level of adic spaces over $\mathbb Q_p^\mathrm{cyc}$, there is a natural "Atkin-Lehner" isomorphism $$\varphi_n:\mathcal X^{\ast}(p^{-n}\epsilon)\to \mathcal X^{\ast}_{\Gamma_0(p^n)}(\epsilon)_a, \quad E\mapsto (E/C_{n},E[p^n]/C_{n})$$ where $C_n\subseteq E[p^n]$ is the rank $p^n$ canonical subgroup. Its inverse is given by sending $(E,D)\mapsto E/D$. One can now check on the level of moduli that for different $n$, these give a comparison isomorphism between the anticanonical tower and the Frobenius tower: $\require{AMScd}$ \begin{CD} \dots @>>> \mathcal X^{\ast}_{\Gamma_0(p^2)}(\epsilon)_a @>>> \mathcal X^{\ast}_{\Gamma_0(p)}(\epsilon)_a @>>> \mathcal X^{\ast}(\epsilon)\\ @AAA @AA\varphi_2A @AA\varphi_1A @|\\ \dots @>>> \mathcal X^{\ast}(p^{-2}\epsilon) @>>> \mathcal X^{\ast}(p^{-1}\epsilon) @>>> \mathcal X^{\ast}(\epsilon) \end{CD} We may thus see the tower of morphisms $F:\mathfrak X^{\ast}(p^{-(n+1)}\epsilon)\to \mathfrak X^{\ast}(p^{-n}\epsilon)$ as a formal model of the anticanonical tower. In particular, we may see $\mathfrak X^{\ast}(p^{-\infty}\epsilon)$ as a canonical formal model for $\mathcal X^{\ast}_{\Gamma_0(p^\infty)}(\epsilon)_a$. Alternatively, I think this should imply that we can regard $\mathfrak X^{\ast}(p^{-n}\epsilon)$ as representing (away from the cusps) tuples $(E,\alpha,D_n)$ where $D_n\subseteq E[p^n]$ is a cyclic rank $p^n$ subgroup scheme which is generically anticanonical (its special fibre may well be canonical). So what is the moduli interpretation of a $\mathrm{Spf}(R)$-point of $\mathfrak X^{\ast}(p^{-\infty}\epsilon)$ (away from the cusps) where $R$ is a complete $\mathbb Z_p^\mathrm{cyc}$-algebra? One answer is that, by definition, it is the data of $(E_0,E_1,E_2,\dots,\alpha, (\eta_n)_{n\in\mathbb N})$ where $(E_0,\alpha,\eta_0)$ is like before, $E_{n+1}/C(E_{n+1})=E_n$ for all $n$, and the $\eta_n$ are compatible under $F$. Alternatively, by the above tower this should be equivalent to the data of $(E,\alpha, (\eta_n)_{n\in\mathbb N},D_\infty)$ where $E:=E_0$ and $D_\infty=(D_n)_{n\in\mathbb N}$ is a generically anticanonical $p$-divisible subgroup of $E[p^\infty]$ of height 1. Here $D_n$ is defined as the kernel of the dual isogeny to $E_n\to E_0$, so that $E_n=E_0/D_n$, and the $\eta_n\in \omega_{E_n}^{\otimes(1-p)}$ are as before. the integral model for $\mathcal X^{\ast}_{\Gamma_1(p^\infty)}(\epsilon)_a$ of Andreatta--Iovita--Pilloni Now to the spaces in Le halo spectral, I'll try to elaborate on Leeeeroy_Jennnnkins' answer and answer a question raised in the comments. (If you allow another plug, most of this can be found in more detail in \S 4 of https://arxiv.org/pdf/1902.03985.pdf). Andreatta--Iovita--Pilloni go further than what you denote by "$\mathfrak X_\infty(\epsilon)$": They also consider the Igusa schemes $\mathfrak I\mathfrak G_n(p^{n}\epsilon)\to \mathfrak X^{\ast}(p^{-n}\epsilon)$ which relatively represent the choice of a trivialisation $\mathbb Z/p^n\mathbb Z\to C_n^\vee$, namely morphisms which are an isomorphism over the ordinary locus. They show that the Frobenius isogeny lifts to a "Frobenius" morphism $F:\mathfrak I\mathfrak G_{n+1}\to \mathfrak I\mathfrak G_n$ and form the "Igusa curve at infinite level" which in order to be consistent with my notation I should probably denote by $\mathfrak I\mathfrak G_{\infty}(p^{-\infty}\epsilon)=\varprojlim_{F}\mathfrak I\mathfrak G_{n}(p^{-n}\epsilon)$. Now how does this compare to Scholze's spaces? The short exact sequence of group schemes $$0\to C_n\to E[p^n]\to E[p^n]/C_n\to 0$$ shows that the Weil pairing canonically identifies $C_n^{\vee}$ with $E[p^n]/C_n$. Thus the Igusa tower equivalently parametrises trivialisations $\mathbb Z/p^n\mathbb Z\to E[p^n]/C_n$. But under the above "Atkin-Lehner" isomorphism, $E[p^n]/C_n$ is the corresponding anticanonical subgroup of $E/C_n$. This means that on the adic generic fibre, this isomorphism lifts to a canonical isomorphism \begin{CD} \mathfrak I\mathfrak G_n(p^{-n}\epsilon)^{\mathrm{ad}}_{\eta} @>\sim>>\mathcal X^{\ast}_{\Gamma_1(p^n)}(\epsilon)_a \\ @AAA @AAA \\ \mathcal X^{\ast}(p^{-n}\epsilon) @>\sim>> \mathcal X^{\ast}_{\Gamma_0(p^n)}(\epsilon)_a. \end{CD} In particular, this means that $\mathfrak I\mathfrak G_n(p^{-n}\epsilon)$ is a formal model of $\mathcal X^{\ast}_{\Gamma_0(p^n)}(\epsilon)_a$. In the limit, it follows that $$\mathfrak I\mathfrak G_\infty(p^{-\infty}\epsilon)^{\mathrm{ad}}_{\eta}=\mathcal X^{\ast}_{\Gamma_1(p^\infty}(\epsilon)_a.$$ So this gives you a canonical formal model of $\mathcal X^{\ast}_{\Gamma_1(p^\infty)}(\epsilon)_a$. Its moduli interpretation (away from the cusps) may be given in terms of tuples $(E,\alpha,(\eta_n)_{n\in\mathbb N},\beta:\mathbb Z_p\to T_pD_\infty)$ where $(E,\alpha,(\eta_n)_{n\in\mathbb N},D_\infty)$ is like above, and beta is a morphism that becomes an isomorphism over the ordinary locus. Finally, if you are interested in integral models for the full level modular curve $\mathcal X^{\ast}_{\Gamma(p^\infty)}$, you may want to have a look at Lurie's preprint http://www.math.harvard.edu/~lurie/papers/LevelStructures1.pdf. universal elliptic curves There are different universal elliptic curves over $\mathfrak X^{\ast}(p^{-\infty}\epsilon)$, and the "right" one depends on your choice of moduli interpretation. Looking at the above comparison map of towers again, as you say, we get a different "universal elliptic curve" by pullback along any $\mathfrak X^{\ast}(p^{-\infty}\epsilon)\to \mathfrak X^{\ast}(p^{-n}\epsilon)$. This is the universal $E_n$ in the moduli description in terms of data $(\alpha,\eta,E_0,E_1,E_2,\dots)$. Alternatively, the moduli interpretation in terms of $(E,\alpha,\eta,D_\infty)$ suggest to look at the pullback $\mathfrak E_\infty^{\mathrm{univ}}$ of the universal elliptic curve $\mathfrak E^{\mathrm{univ}}$ along $\mathfrak X^{\ast}(p^{-\infty}\epsilon)\to \mathfrak X^{\ast}$. Can we make sense of the adic generic fibre of $\mathfrak E_\infty^{\mathrm{univ}}\to \mathfrak X^{\ast}(p^{-\infty}\epsilon)$? Yes: The adic generic fibre of $\mathfrak E_\infty^{\mathrm{univ}}$ can be described as the fibre product of the relatively smooth rigid space $(\mathfrak E^{\mathrm{univ}})^{\mathrm{ad}}_{\eta}\to \mathcal X^{\ast}(\epsilon)$ with the perfectoid space $\mathcal X^{\ast}_{\Gamma_0(p^\infty)}(\epsilon)\to \mathcal X^{\ast}(\epsilon)$. I think this should exist as a sousperfectoid (hence sheafy) adic space $\mathcal E_\infty^{\mathrm{univ}}$. Is it a perfectoid space over $\mathcal X^{\ast}_{\Gamma_0(p^\infty)}(\epsilon)_a$? No: Fibre products of perfectoid spaces are perfectoid, but if you take the fibre product with any point $\mathrm{Spa}(\mathbb Q_p^\mathrm{cyc})\to \mathcal X^{\ast}_{\Gamma_0(p^\infty)}(\epsilon)$, you will get the analytification of an elliptic curve $E^{an}\to \mathrm{Spa}(\mathbb Q_p^{\mathrm{cyc}})$ which is certainly not perfectoid. If you want something perfectoid, I think it would be reasonable to guess that $\varprojlim_{[p]} \mathcal E_\infty^{\mathrm{univ}}$ is perfectoid -- this is true over the good reduction locus, but as far as I know, it's not currently known whether it's true over the whole space.<|endoftext|> TITLE: Alberti rank one theorem and a blow-up argument QUESTION [5 upvotes]: In this paper, it is written that Alberti’s rank says that the singular part $D^s u$ with respect to $\mathcal L^d$ of the distributional derivative $Du$ of a function $u \in BV_{loc}(\mathbb R^d; \mathbb R^m )$ can be written, in polar decomposition, as $D^s u = \xi \otimes \eta|D^s u|$. Then "by a standard blow-up argument this implies that near to $|D^s u|$-a.e. point $x$ asymptotically $u(y)$ behaves like a function having a single non-zero component, parallel to $\eta(x) \in \mathbb{S}^{m-1}$, and depending on a single scalar variable, the component of $y$ along $\xi(x) \in \mathbb{S}^{d-1}$". What does the statement in quotes mean heuristically? How can it be proved rigorously (that is, could you outline the details of the "standard blow-up argument" mentioned above? Where can I find a picture to represent this situation? A more generic question was asked in Meaning of Alberti rank-one theorem. A related issue is in Alberti rank-one theorem and reduction of the study of BV function to the two-dimensional case. REPLY [3 votes]: The answer to your questions 1. & 2. can be found in Theorem 3.95 of the book Ambrosio, Fusco, Pallara Functions of Bounded Variation and Free Discontinuity Problems. Another very recent and excellent reference is the book by Rindler Calculus of Variations, paragraph 10.4 (where you also find an excellent picture). It is not easy to sketch the proof: the underlying idea is to introduce a suitable rescaling of $u$ (see eq. (3.92) in the book by Ambrosio, Fusco and Pallara) for which you get automatically compactness. The study of the structure of the limit (which is the core of the Theorem) is rather delicate and relies on theorems about tangent measures and, of course, on Alberti's Rank One Theorem. In this way you are able to conclude that the blow up is a vector valued function which is "directed" in only one direction (say $\eta$) and depends only on the other direction (say $\xi$). This explanation is very poor and the best I can do is to refer you to the aforementioned books. Concerning the picture to represent the situation, I suppose the best is if you try to work out the details of what is going on by yourself. A good toy-model to work with is a "piecewise defined" vector field $u=u(x,y)$ in $\mathbb R^2$ (e.g. something like $u_1(x,y)$ if $x<0$ and $u_2(x,y)$ otherwise: compute the jacobian matrix and find by yourself $\eta,\xi$). Of course this example shows the rank-one property for the jump part and not for the Cantor part (which is actually the crucial point of Alberti's Theorem but is also considerably more difficult to "visualize"). Hope this helps.<|endoftext|> TITLE: What is the connection between Frechet Lie groups and Lie algebras? QUESTION [5 upvotes]: An ordinary Lie group has a differentiable manifold structure, i.e. it is locally isomorphic to a finite-dimensional Euclidean space. A Frechet Lie group, on the other hand, has a Frechet manifold structure, i.e. it is locally isomorphic to an infinite-dimensional Frechet space. My question is, what is the connection between Frechet Lie groups and Lie algebras? Wikipedia says this: Some of the relations between Lie algebras and Lie groups remain valid in this setting. But which ones remain valid and which ones don’t? REPLY [2 votes]: Karl Hermann Neeb has lecture notes from a course on infinite-dimensional Lie theory given at a summer school in Monastir. They can be found for example here.<|endoftext|> TITLE: A binary hook-length formula? QUESTION [6 upvotes]: This is purely exploratory and inspired by curiosity. Setup: For an integer $k>0$, let $k=\sum_{j\geq0}k_j2^j$ be its binary expansion and denote the sum of its digits by $\eta(k):=\sum_jk_j$. Further, introduce a binary factorial $$[n]!_b:=\eta(1)\eta(2)\cdots\eta(n).$$ Given an integer partition $\lambda$, let $Y_{\lambda}$ be the corresponding Young diagram. If $\square$ a cell in $Y_{\lambda}$, construct its hook-length $h_{\square}$ in the usual manner but replace it by $\eta(h_{\square})$. For example, take $\lambda=(3,2,1)$ then its multiset of hooks is $\{h_{\square}:\square\in Y_{\lambda}\}=\{5,3,1,3,1,1\}$ which shall be replaced by $\{\eta(h_{\square}):\square\in Y_{\lambda}\}=\{2,2,1,2,1,1\}$. Naturally, we define the (new) product of hook-lengths and denote (with an abuse of notation) $$H_{\lambda}=\prod_{\square\in\lambda}\eta(h_{\square}).$$ If $\lambda\vdash n$, it is easy to verify that $\frac{[n]!_b}{H_{\lambda}}$ is an integer. QUESTION. What do these integer count? $$\sum_{\lambda\vdash n}\frac{[n]!_b}{H_{\lambda}}.$$ The first few values are: $1, 2, 3, 7, 10, 23, 52, 82, 117, 258, \dots$ but not listed on OEIS. REPLY [11 votes]: I am afraid they are not always integers. Take large $p$ and $n=2^{2p}-1$. Then $[n]!_b$ is divisible by $p^N$ for $N={2p\choose p}+1$. And $[2n+1]!_b$ by $p^K$ for $K={2p+1\choose p}+2p+1<2N$. Then for $2\times N$ diagram we get $p$ in the denominator.<|endoftext|> TITLE: Explicit surjection $\mathbb{A}^n\longrightarrow \mathbb{P}^n$ QUESTION [9 upvotes]: Let $k$ be an algebraically closed field. I have been told several times that for any $n\geq 0$, there exists a morphism of $k$-schemes $\mathbb{A}^n_k\rightarrow \mathbb{P}^n_k$ that is surjective on the underlying topological spaces. I was not able to find an explicit example of a such a morphism. Now, the talk about "a sufficiently general collection of polynomials" nerves me out. If I give you $\mathrm{char}\,k$ and $n$, can you give me an explicit example of a surjection $\mathbb{A}^n_k\rightarrow \mathbb{P}^n_k$? I believe I can do this for $n=0$. Pretty much by definition we have identifications $\mathbb{A}^0_k\approx \mathrm{Spec}\,k$ and $\mathbb{P}^0_k\approx \mathrm{Spec}\,k$, so we can take the identity morphism $\mathrm{Spec}\,k\rightarrow \mathrm{Spec}\,k$. EDIT: to state the obvious, in our case it is enough to verify surjectivity on closed points. The set-theoretic image of a morphism of finite presentation is constructible. A constructible set containing all closed points of a scheme of finite type over an algebraically closed field should be the whole space (since otherwise its complement would be a non-empty constructible set containing no closed points; a constructible set contains an open dense subset of its closure and in a scheme of finite type over an algebraically closed field, closed points are dense in any closed subset). REPLY [20 votes]: Let me give a solution for $n=2$, that can be easily generalized to all $n$. Let us consider the morphism $$f \colon \mathbb{P}^2 \to \mathbb{P}^2, \quad f([x:y:z]) =[x^2:y^2:z^2].$$ This is a Galois cover, with Galois group isomorphic to the Klein group $\mathbb{Z}_2 \times \mathbb{Z}_2$ and ramified on the union of the three lines $x=0$, $y=0$, $z=0$, with ramification index generically $2$. There are precisely three points with total ramification, namely $[1:0:0]$, $[0:1:0]$, $[0:0:1]$. If we take a general line $H$ disjoint from the total ramification locus, then the restriction of $f$ to $\mathbb{P}^2 - H$ remains surjective onto $\mathbb{P}^2$. But $\mathbb{P}^2 - H$ is clearly isomorphic to $\mathbb{A}^2$, so we are done.<|endoftext|> TITLE: All quadratic forms of given genus over $\mathbb{Z}$ QUESTION [7 upvotes]: Given a (ternary) quadratic form over $\mathbb{Z}$ how can I find all quadratic forms (up to equivalence over $\mathbb{Z}$) in the same genus? REPLY [4 votes]: Now that I think of it, people may not realize that the entirety of the tables by Brandt and Intrau are available, compiled by Alexander Schiemann TABLE 1 and TABLE 2. As far as the words "odd" and "even," Alexander followed usage from integral lattices, as in SPLAG by Conway and Sloane added: I do not know of any software that deals with indefinite ternaries. What follows is about positive forms as long as the power of two dividing the discriminant is not too large, you will get the genus correctly split into spinor genera with my Magma program. I guess I will put samples first. Note that it correctly says the genus of $x^2 + 24 y^2 + 576 z^2$ has four classes, but it prints out four spinor genera with repeat of the one spinor genus, which is nonsense. The actual genus has two spinor genera, --------------------------------------- g0 : 55296 : 1 24 576 0 0 0 auto 8 g1 : 55296 : 24 25 25 14 0 0 auto 8 --------------------------------------- g2 : 55296 : 4 24 145 0 4 0 auto 8 g3 : 55296 : 9 24 64 0 0 0 auto 8 --------------------------------------- jagy@phobeusjunior:~$ sage ┌────────────────────────────────────────────────────────────────────┐ │ SageMath Version 6.9, Release Date: 2015-10-10 │ │ Type "notebook()" for the browser-based notebook interface. │ │ Type "help()" for help. │ └────────────────────────────────────────────────────────────────────┘ sage: q1 = QuadraticForm(ZZ,3,[1,0,0, 24,0, 576] ) sage: q1 Quadratic form in 3 variables over Integer Ring with coefficients: [ 1 0 0 ] [ * 24 0 ] [ * * 576 ] sage: q1.det() 110592 sage: q1.number_of_automorphisms() 8 sage: q1.conway_mass() 1/2 sage: sage: quit Exiting Sage (CPU time 0m0.41s, Wall time 1m5.23s). jagy@phobeusjunior:~$ The trouble first came to my attention about 1996 when Manjul Bhargava was corresponding with Irving Kaplansky, Manjul asked Magma to find all forms alone in a genus, and it gave the wrong answer for $x^2 + 8 y^2 + 64 z^2 \; . \;$ I put a good deal of relevant material at TERNARY Oh, precisely because of these occasional errors, I now have a number of error correcting programs. The final say is the Mass Formula, which is given correctly in Sage. I am hoping that I can get someone competent to finish some fairly basic routines in Sage so that checking is less annoying. Gonzalo Tornaria wrote the handful of commands they have. It is especially important at this time to get such a thing: I searched for spinor regular forms from about 2004-2007. Just recently, Andrew Earnest and Anna Haensch (student of Wai Kiu Chan) completed a proof that my list SPINOR REGULAR is complete, but I suspect this is contingent on my list being complete up to the bound I used. I would like someone to do my search over; I had written some elaborate workarounds, that is why it took three years of my time to run, but it is always possible that I did not really correct every error. ============================================================================== // http://magma.maths.usyd.edu.au/calc Q:=RationalField(); Z:=Integers(); M3:=MatrixRing(IntegerRing(),3); tolettuce:=function(sixlist) temp := LatticeWithGram(M3![2 * sixlist[1],sixlist[6],sixlist[5], sixlist[6], 2 * sixlist[2],sixlist[4], sixlist[5],sixlist[4],2 * sixlist[3]]); return temp ; end function; tohex := function(lettuce) tripe := Basis(lettuce); return [ Norm( tripe[1] ) div 2 , Norm( tripe[2] ) div 2 , Norm( tripe[3] ) div 2 , InnerProduct( tripe[2],tripe[3] ) , InnerProduct( tripe[3],tripe[1] ) , InnerProduct( tripe[1],tripe[2] ) ] ; end function; temp2 := tolettuce([1, 24, 576, 0, 0, 0]); // CHANGE !!!!!!!!!! tempgenus := GenusRepresentatives(temp2); tempSG := SpinorGenera(Genus(temp2)); print "=====Discriminant " , " ==Genus Size== ", #tempgenus , "\n"; // print "This genus has " , #tempSG , " spinor genera ------\n"; reps := [ Representatives(S) : S in tempSG]; for i in [1..#reps] do tempspin := reps[i]; // print "-------------**---------------------- ", " s. g. size--- ", #tempspin , "\n"; for j in [1..#tempspin] do tohex(tempspin[j]); end for; print "\n---**----- end of spinor genus ", i, " --------\n"; // print "--------------------------------------------------"; end for; =====================================================================================<|endoftext|> TITLE: Is there a holomorphic function on open unit disc with this property? QUESTION [10 upvotes]: Let $D=\{z\in \mathbb{C}\mid |z|<1\}$. Is there a holomorphic function $f:D\to \mathbb{C}$ such that for every $n\in \mathbb{N} \cup \{0\},\;f^{(n)}$ has a continuous extension to $\bar D$ but $f$ has no a holomorphic extension to any open neighborhood of $\bar D$? REPLY [16 votes]: Yes, for example $$f(z)=\sum_{n=1}^\infty e^{-\log^2n}z^{n^2}.$$ The radius of convergence is $1$ by Hadamard's test. The function has gaps (the density of the sequence $n^2$ among the integers is zero) which are sufficient to apply Fabry's gap theorem (or Polya's gap theorem) to ensure that this $f$ does not have any analytic continuation to a larger region than the unit disk, and all derivatives are continuous in the closed disk because their power series are absolutely and uniformly convergent in the closed disk. Refs. It is difficult to find a good exposition of Fabry's work in English, one can read his original papers (in French) or the exposition of Bieberbach (in his book Analytische Fortsetzung, in German). Polya work is also in German. But for the Polya gap theorem one can consult Koosis, The Logarithmic integral, vol. II.<|endoftext|> TITLE: Where are Serre’s lectures at Collège de France to be found? QUESTION [12 upvotes]: Having run into several references, at various places and occasions, to "Serre’s Course at Collège de France, 19xy-19xy+1" for various values of xy, I would genuinely like to know where these lectures are written down and archived. Has there been an effort to preserve them in any form? There are overviews of them in Serre’s collected work, but they have no references and don’t give full proofs. REPLY [17 votes]: The lecture notes for many of Serre's courses have been published: Groupes algébriques et corps de classes (transl. Algebraic Groups and Class Fields) (Course at Collège de France 1956-1957) Algèbre Locale, Multiplicités (Course at Collège de France 1957-1958) Corps locaux (transl. Local Fields) (Course at Collège de France 1958-1959) Groupes proalgébriques (Course at Collège de France 1959-1960) Cohomologie galoisienne (transl. Galois cohomology) (Course at Collège de France 1962-1963) Arbres, amalgames (transl. Trees) (Course at Collège de France 1968-1969) Lectures on the Mordell-Weil theorem (Course at Collège de France 1980-1981) Représentations linéaires des groupes finis (Course at École Normale 1966) Cours d'arithmétique (transl. A course in arithmetic) (Course at École Normale 1962-1964) Groupes finis (transl. Finite groups) (Course at ENSJF, 1978-1979) Topics in Galois theory (Course at Harvard University 1988) REPLY [8 votes]: The detailed abstracts (résumés détaillés) that are referenced are supposedly available in the "Annuaire du Collège de France". The journal's website unfortunately only contains newer entries (2005–). However, your library might have paper versions. According to Mathdoc, some physical versions are available in a few French libraries (notably at the IHP, the ENS, at the CIRM – with a few gaps in each case, unfortunately). Some of the published versions that Carlo Beenakker mentions are available at Numdam. You can directly search for Jean-Pierre Serre's works. It might take some digging to know which lecture corresponds to which published paper, but the titles would probably help.<|endoftext|> TITLE: Intersection of spaces with Schauder basis QUESTION [5 upvotes]: Let $\{v_n\}_{n \in \mathbb{N}}$ be a Schauder basis of $V$ subspace of $\ell^2$ over $\mathbb{C}$ and $\forall m \in \mathbb{N}$ let $V_m = \overline{\operatorname{span}} \{v_n\}_{n \geq m}$ Let $\{u_n\}_{n \in \mathbb{N}}$ be a Schauder basis of $U$ subspace of $\ell^2$ over $\mathbb{C}$ and $\forall m \in \mathbb{N}$ let $U_m = \overline{\operatorname{span}} \{u_n\}_{n \geq m}$ Under the hypothesis that $\forall m \in \mathbb{N}: V_m + U_m$ is closed, is it true that: $$ \bigcap_{m=1}^\infty \left( V_m + U_m \right) = \{0\} $$ REPLY [6 votes]: In general it is not true: $V_m$ and $U_m$ could even be transverse for all $m$, giving $$ \bigcap_{m=1}^\infty \left( V_m + U_m \right) = \ell_2. $$ Let $\kappa:\mathbb{N}\to\mathbb{N}$ be a map such that every $p\in\mathbb{N}$ has fiber $\kappa^{-1}(p)$ of infinite cardinality. Let $\{v_n\}_{n\ge1}$ be the orthonormal basis of $\ell_2$ and let $K:\ell_2\to\ell_2$ be the bounded linear operator defined by $Kv_n= 2^{-n}v_{\kappa(n)}$ for all $n\ge1$. Then $$\|K\|\le\|K\|_{HS}=\Big(\sum_{n\ge1}\|Kv_n\|^2\Big)^{1/2}=\Big(\sum_{n\ge1}4^{-n}\Big)^{1/2}=3^{-1/2}<1$$ Therefore $I+K$ is invertible and $u_n:=(I+K)v_n$ for $n\ge1$ defines a Schauder basis. However, by the hypothesis on $\kappa$, for any $1\le p TITLE: Irreducible of finite Krull dimension implies quasi-compact? QUESTION [6 upvotes]: Let $X$ be the underlying space of a scheme. If $X$ is irreducible of finite Krull dimension, is it necessarily quasi-compact? Is it necessarily Noetherian? What if we assume not only that Krull dimension is finite but also that it is 1? REPLY [10 votes]: The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $\mathbb{A}^1$ along the open subsets $\mathbb{G}_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.<|endoftext|> TITLE: Wu formula for manifolds with boundary QUESTION [18 upvotes]: The classical Wu formula claims that if $M$ is a smooth closed $n$-manifold with fundamental class $z\in H_n(M;\mathbb{Z}_2)$, then the total Stiefel-Whitney class $w(M)$ is equal to $Sq(v)$, where $v=\sum v_i\in H^*(M;\mathbb{Z}_2)$ is the unique cohomology class such that $$\langle v\cup x,z\rangle=\langle Sq(x),z\rangle$$ for all $x\in H^*(M;\mathbb{Z}_2)$. Thus, for $k\ge0$, $v_k\cup x=Sq^k(x)$ for all $x\in H^{n-k}(M;\mathbb{Z}_2)$, and $$w_k(M)=\sum_{i+j=k}Sq^i(v_j).$$ Here the Poincare duality guarantees the existence and uniqueness of $v$. My question: if $M$ is a smooth compact $n$-manifold with boundary, is there a similar Wu formula? In this case, there is a fundamental class $z\in H_n(M,\partial M;\mathbb{Z}_2)$ and the relative Poincare duality claims that capping with $z$ yields duality isomorphisms $$D:H^p(M,\partial M;\mathbb{Z}_2)\to H_{n-p}(M;\mathbb{Z}_2)$$ and $$D:H^p(M;\mathbb{Z}_2)\to H_{n-p}(M,\partial M;\mathbb{Z}_2).$$ Thank you! REPLY [18 votes]: A relative Wu formula for manifolds with boundary is discussed in Section 7 of Kervaire, Michel A., Relative characteristic classes, Am. J. Math. 79, 517-558 (1957). ZBL0173.51201. In particular, there are relative Wu classes $U^q\in H^q(M;\mathbb{Z}/2)$ for $q=0,1,\ldots , n$ defined by the property that $$ Sq^q(x)=U^q\cup x \in H^n(M,\partial M;\mathbb{Z}/2) $$ for all $x\in H^{n-q}(M,\partial M;\mathbb{Z}/2)$. Kervaire deduces the relative Wu formula $w(M)=Sq(U)$ from the absolute Wu formula for the double $N=M\cup_{\partial M} M$ (the closed manifold obtained by gluing two copies of $M$ along the identity map of $\partial M$), using naturality with respect to the inclusion $i:M\hookrightarrow N$.<|endoftext|> TITLE: What can be said about the map $K(n)_\ast(X) \to K(n)_\ast(\tau_{\leq n}X)$ when $X$ is a finite complex? QUESTION [5 upvotes]: Ravenel and Wilson showed that $K(\mathbb Z / p^j,q)$ is $K(n)$-acyclic for any $q \geq n+1$, and that $K(\mathbb Z, q)$ is $K(n)$-acyclic for $q \geq n+2$. It follows that $K(A,q)$ is $K(n)$-acyclic for $q \geq n+2$ when $A$ is finitely-generated. From here, a Serre spectral sequence argument reveals that the map $\tau_{\leq m} X \to \tau_{\leq n+1} X$ (where $\tau$ is Postnikov truncation) is a $K(n)$-local equivalence for all $m \geq n+1$ when $X$ has finitely-generated homotopy groups. (For $\pi$-finite spaces, $\tau_{\leq m} X \to \tau_{\leq n} X$ is in fact a $K(n)$-local equivalence, as observed by Carmeli,Schlank, and Yanovski). It's tempting to conclude that $X \to \tau_{\leq n+1} X$ is a $K(n)$-local equivalence for any $X$ with finitely-generated homotopy groups, but this can't possibly be true. If it were true, then in particular $X \to \tau_{\leq n+1} X$ would be an equivalence for all simply-connected finite spaces $X$. Then we could conclude $K(n)_\ast(S^2) = K(n)_{\ast+n+1}(\Sigma^{n+1} S^2) = K(n)_{\ast+n+1}(pt)$, which is false. Indeed, according to Bauer, the convergence of the spectral sequence for the Postnikov tower of $X$ only holds when $X$ is $n$-truncated. This leads to my Question: If $X$ is a space with infinitely many nontrivial homotopy groups, is there any meaningful relationship between $K(n)_\ast(X)$ and $K(n)_\ast(\tau_{\leq n} X)$ or $K(n)_\ast(\tau_{\leq n+1} X)$? (Beyond the mere existence of a map -- for all I know, this map is zero!) How about if $X$ is finite? Or perhaps, what if $X$ is $(n-1)$-connected? REPLY [6 votes]: Let me make more concrete a comment I already wrote. In [Adv. Math. 201 (2006), 318-378], my example 2.22, illustrating a theorem just before it when $n=1$, says that, for any spectrum $Y$, there is a short exact sequence of Hopf algebras over $K(1)_*$ as follows: $$ K(1)_*(\mathbb PY) \rightarrow K(1)_*(\Omega^{\infty}Y) \rightarrow K(1)_*(\tau_{\leq 2}\Omega^{\infty}Y).$$ Here $\mathbb PY$ is the free $E_\infty$--algebra generated by $Y$: the wedge of all the extended powers of $Y$. So if $X = \Omega^{\infty}Y$, then $K(1)_*(X) \rightarrow K(1)_*(\tau_{\leq 2}X)$ is onto. One could wonder if one has an epimorphism for other $X$. Okay, lets try something else: Bousfield gave a cute short argument that if $X$ is $E_*$--acyclic, so is $K(\pi_j(X), j)$ for any $j$. So if $X$ is $K(1)_*$--acyclic, so is $\tau_{\leq 2}X$, and thus our map is still an epi. This is enough fun for one answer, so I'll leave things here.<|endoftext|> TITLE: Dimension-specific phenomena in algebraic geometry QUESTION [9 upvotes]: In differential topology, there are some funny phenomena that can only happen in dimension 4. For example, only in dimension 4 you can have a closed topological manifold admitting infinitely many distinct smooth structures. Is there something like this happening in algebraic geometry? Does there exist an integer $k>1$ and some interesting geometric statement that only holds for varieties of dimension$\neq k$? P.S. It should be noted that a professional topologist has said that he has doubts about one important paper on manifold topology. He did not give a lot of details, and not being a topologist, I can not judge the veracity of his claims. I do not know if the existence statement in the second sentence of this post can be established independently of that paper. If somebody does know, let us know. REPLY [3 votes]: Let $X,L$ be a smooth polarized projective variety of dimension $n$ with $K_X =\mathcal{O}_X$. Let $v \in H^{\bullet}(X,\mathbb{Q})$ a primitive vector and consider $M_{L,v}$ the moduli space of $L$-stable sheaves with Chern character equal to $v$. Assume that the actual dimension of $M_{L,v}$ is equal to its virtual dimension. Then $M_{L,v}$ has an open subset which is smooth, say $M_{L,v}^{sm}$. Miracle when $\dim X =2$ : $M_{L,v}^{sm}$ carries a holomorphic symplectic form! This is because only when $\dim X = 2$ is there a symplectic isomorphism $\mathrm{Ext}_X^1(F,F) \simeq \mathrm{Ext}_X^{1}(F,F)^*$ (induced by Serre duality). For the time being, there is no Theorem which implies that all moduli spaces of sheaves (or rather objects) on higer dimensional projective varieties that carry a holomorphic symplectic form are necessarily moduli spaces of objects on a $\mathrm{K}$-trivial surface. In fact, we know examples of such moduli spaces which original constructions go via higher dimensional manifolds (the Fano scheme of lines on a cubic fourfold for instance). But in every such example known, there is a (possibly non commutative) $\mathrm{K}$-trivial surface hidden in the story. For instance, the Fano scheme of lines on a cubic fourfold can be realized as the moduli space of objects in the derived category a non-commutative $\mathrm{K}$-trivial surface which sits inside the derived category of the cubic fourfold. So it seems that having moduli spaces carying holomorphic symplectic form is a miracle related to (possibly non-commutative) two-dimensional $\mathrm{K}$-trivial varieties of dimension 2. Note the miraculous numerical coincidence of my answer with your question : complex dimension $2$ corresponds to real dimension 4 ;)<|endoftext|> TITLE: Non-zero winding number on a space curve implies a linked curve in the zero set? QUESTION [5 upvotes]: The following statement has been largely improved from my original post thanks to discussions with @DmitriPanov ad well as the comment from @Wojowu. Let $f \colon \mathbb{S}^3 \to \mathbb{R}^2$ be real analytic. Let $C$ be a closed space curve in $\mathbb{S}^3$, which I might need to assume to be unknot. If $f(C) \subset \mathbb{R}^2$ has a non-zero winding number around $0$, then $f^{-1}(0)$ contains a space curve in $\mathbb{S}^3$ that is linked to $C$. I consider this as a generalization of Kronecker's existence theorem. I sketched a plausible argument in my comment to the answer of @DmitriPanov, which I did not check with great care. The statement and argument is obviously generalizable to other dimensions, using topological degree. I believe that this result is not new. So where do I find a reference for this and its higher dimensional versions? REPLY [2 votes]: Now the statement is indeed correct. What follows below is and answer to the previous version of the question which I'll keep for the moment. Note that the condition for $f(C)$ to have non-zero winding number around $0$ is more-less empty. Indeed, to fix the winding number of $f(C)$ around $0$ one needs to choose infinity. And for more-less any closed curve $\eta$ in $S^2$ that doesn't pass through $0$ one can choose $\infty \in S^2$ so that the winding number of $\eta$ in $S^2\setminus \infty$ around $0$ is non-zero. Now, for a concrete set of counter-examples suppose that $f: S^3\to S^2$ is ANY map with non-zero differential at a point $x\in S^3$ such that $f(x)\ne 0$. Then take a small ball $U$ containing $x$ such that $0\notin f(U)$ and a curve $\gamma\subset U$ that projects to a small circle in $S^2$. Choose $\infty\in S^2$ such that $f(\gamma)$ separates $0$ from $\infty$. Then clearly the winding number of $f(\gamma)$ around $0$ is $\pm 1$ but $f^{-1}(0)$ doesn't contain a component linked to $c$. I can see only one way to fix this. Ask $f$ not to be null-homotopic and ask $C$ to be a full premiage of a point $x\in S^2$ different from $0$ ...<|endoftext|> TITLE: Property of the trace on finitely generated projective modules QUESTION [8 upvotes]: Let $A$ be a commutative ring with unit and let $P$ be a projective $A$-module finitely generated. By definition, there exists an $A$-module $P'$ such that $P\oplus P'$ is free of finite rank $r$. If $f$ is an endomorphism of $P$, its trace $\operatorname{Tr}(f|P)$ is defined to be $\operatorname{Tr}(f\oplus 0|P\oplus P')$. We can show that this definition does not depend on the choice of $P'$. Let $Q$ be a submodule of $P$ stable by $f$ which is projective and finitely generated. Assume that there is a positive integer $n$ such that $f^n(P)\subset Q$. If $P/Q$ would have been projective and finitely generated, I would have conclude that $\operatorname{Tr}(f|P)=\operatorname{Tr}(f|Q)+s$, where $s$ is a nilpotent element of $A$, as $f$ acts nilpotently on $P/Q$. But, does this still holds in general? Any idea is welcome. Many thanks! (EDIT: As Darij Grinberg pointed me out in comments, the trace of a nilpotent endomorphism of a projective finitely generated module is a sum of nilpotents elements of $A$. As $A$ is commutative, the trace is again nilpotent and not necessarily zero) REPLY [6 votes]: This is true. It's more natural to prove a generalisation: Lemma. Let $A$ be a commutative ring, let $F_\bullet$ be a bounded complex of finite projective $A$-modules (say in degrees $[a,b]$), and let $f_\bullet \colon F_\bullet \to F_\bullet$ be an endomorphism such that $f_\bullet^n$ is homotopic to $0$ for some $n \in \mathbf Z_{>0}$. Then $\operatorname{tr}(f_\bullet) := \sum_i (-1)^i\operatorname{tr}(f_i)$ is nilpotent. The original question is the case $[a,b] = [0,1]$ and $d \colon F_1 \to F_0$ is injective. The advantage of the version above is that it's stable under base change (whereas injectivity of $d$ is not). We use homological indexing to avoid confusion when writing $f_\bullet^n$. I'm not sure if there is a fancy homological algebra explanation for this lemma, but here is a hands-on proof: Proof of Lemma. When $A = k$ is a field, this follows by additivity of trace and nilpotence of $H_i(f)$ for all $i$. In general, note that the question is stable under base change: for any ring map $\phi \colon A \to B$, we have $$\operatorname{tr}\left(f_\bullet \underset A\otimes \mathbf 1_B\right) = \phi\big(\operatorname{tr}(f_\bullet)\big).$$ Applying this to $A \to \kappa(\mathfrak p)$ for any prime $\mathfrak p \subseteq A$, we conclude from the field case that $\operatorname{tr}(f_\bullet) \in \mathfrak p$. Applying this to all $\mathfrak p$, we see that $\operatorname{tr}(f_\bullet)$ is nilpotent. $\square$<|endoftext|> TITLE: Classification of 2-types -- crossed modules vs. Postnikov data? QUESTION [8 upvotes]: A few questions about the equivalence between 2-types and crossed modules. For simplicity, assume everything is connected. What is the precise statement? Is there an equivalence of categories (or at least a bijection of isomorphism classes) of the form $$\{\text{2-truncated spaces}\}[\text{weak homotopy equivalence}^{-1}] \simeq \{\text{crossed modules}\}[\mathcal{W}^{-1}]$$ for some class of morphisms $\mathcal W$? If so, what is $\mathcal W$? Is it exactly the isomorphisms? What is a precise reference? Crossed modules $1 \to A \to H_2 \to H_1 \to G \to 1$ are classified by $H^3(G;A)$ up to zigzags of morphisms of extensions. Does this translate to a classification of connected homotopy 2-types $X$ with $\pi_1(X) = G$ and $\pi_2(X) = A$ up to homotopy equivalence, or is it only up to a coarser equivalence relation? If this is a classification up to homotopy equivalence, then how does one see that a non-invertible morphism of crossed modules induces a homotopy equivalence of classifying spaces? In any event, from every crossed module $1 \to A \to H_2 \to H_1 \to G \to 1$, I can extract an element of $H^3(G;A)$. Homotopically, this corresponds to a Postnikov invariant for a possibly non-principal Postnikov tower. Where is the theory of non-principal Postnikov invariants written, and in particular does this invariant (exist and) completely classify a 2-type? REPLY [2 votes]: The book Nonabelian Algebraic Topology discusses the classifying space $BC$ of a crossed complex $C$ and the homotopy classification $$[X,BC] \cong [\Pi X_* , C] $$ for a CW-complex $X$ with skeletal filtration $X_*$. (Theorem 11.4.19). The functor $\Pi$ from filtered spaces to crossed complexes goes back in essence to Blakers (1948) and to Whitehead (1949). The paper Modelling and Computing homotopy types:I explains something of the origin and methodology lying behind the above work, including how we should represent cohomology classes. Maybe helpful in relation to the question of dealing with finite crossed modules is the paper "Homotopy 2-Types of Low Order" G Ellis, LV Le - Experimental Mathematics, 2014 - Taylor & Francis April 23,2019: I mention also that p.428ff of the NAT book discusses the notion of crossed $n$-fold extension and the bijection $$\text{OpExt}^n(G,M) \cong H^{n+1}(G,M) .$$ An advantage of the approach is that we work in the realm of free crossed resolutions, and that we can use a standard free crossed resolution $SF(G)$ which is related to the crossed complex of the Nerve of the group $G$ and so to more classical cocycle conditions; thus conditions in several dimensions are replaced by the condition of giving a morphism of crossed complexes $SF(G) \to C$. See also the paper Bullejos, M., Faro, E., and García-Muñoz, M.A., Postnikov invariants of crossed complexes. J. Algebra 285 (1) (2005) 238-291 for a different approach.<|endoftext|> TITLE: Is $\operatorname{dim} H^1$ of an abelian variety the same for any Weil cohomology? QUESTION [7 upvotes]: Let $A$ be an abelian variety over a field $k$ of dimension $g$, and $H$ be a Weil cohomology theory for smooth projective varieties over $k$ with characteristic $0$ coefficient field $E$. Is it true that $\operatorname{dim} H^1(A)=2g$? I think this will follow from some standard conjectures, but do we know this unconditionally at present (at least for some low dimensional cases)? For example, one can prove $\operatorname{dim} H^1(A) \leq 2g$ by some power tricks, see Chapter 3. Theorem 8.1 in https://pages.uoregon.edu/ddugger/wbook.pdf (it may contain some typos, but the idea works). And if $k$ is algebraically closed and $A$ is superspecial, then $End_k(A)\otimes \mathbb Q \cong M_g(D_{p,\infty})$ is central simple algebra of dimension $4g^2$ over $\mathbb Q$, hence $End(A) \otimes E \hookrightarrow End(H^1(A))$, and $dim H^1(A)=2g$. REPLY [6 votes]: Yes, it's the same for all Weil cohomology theories --- see 2A.8 of Kleiman, S. L. Algebraic cycles and the Weil conjectures. Dix exposés sur la cohomologie des schémas, 359--386, Adv. Stud. Pure Math., 3, North-Holland, Amsterdam, 1968.<|endoftext|> TITLE: Subalgebra of a group algebra QUESTION [12 upvotes]: Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra. Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$. Question: Is there any criterion for when $A = k[H]$ for some subgroup $H$? Also, in that case, how do we read of the generating subgroup $H$? Will the situation become better/easier if I assume $A$ to be a sub-Hopf-algebra? REPLY [7 votes]: Yes, there is a criterium. Assuming $G$ finite, $A$ is of the form $k[H]$ for some subgroup of $G$ if and only if $A$ is a sub-bialgebra (and since $G$ is finite, if and only if is Hopf subalgebra). Clearly $k[H]$ is a Hopf subalgebra of $k[G]$, but if you take any subalgebra $A$, the fact that $A$ is also a subcoalgebra means that $A$ is a subcomodule of $k[G]$, hence $G$-graded, so, the $G$-homogeneous components are $1$-dimentionals and $A$ is generated by group-like elements. But a subcoalgebra of the form $k[X]$ with $X\subseteq G$ is subalgebra only when $X$ is a subgroup. (By the way, if $G$ is infinite everything works fine except that $H$ is maybe a submonoid and not a subgroup). An alternative proof avoiding the grading/comodule argument is the following: $A$ being subcoalgebra means $A^*$ is an algebra quotient of $k[G]^*=k^G$= the algebra of functions from G to k. But it is clear that any quotient of $k^G$ identifies with $k^X$ with $X\subseteq G$. The rest of the argument is the same.<|endoftext|> TITLE: Does anybody do $p$-adic Teichmüller theory? QUESTION [20 upvotes]: In "Foundations of $p$-adic Teichmüller theory", Mochizuki describes a theory one of whose goals (according to the author) is to generalize Fuchsian uniformization of Riemann surfaces to the $p$-adic context. A quick glance through this book reveals a lot of new concepts (with fancy names!) introduced by Mochizuki. The question is: what are some landmark papers in this theory (after this textbook)? Did anybody other than Mochizuki and his postdocs/students make contributions to this theory? A less mathematical question: are there any people in Western Europe/U.S. working on this topic in the present? For example, while I think that Jakob Stix is doing some anabelian geometry (and Mochizuki has made major contributions to it), I am not sure if any of his work is specifically building upon "Foundations of $p$-adic Teichmüller theory". P.S. Not being an inter-universalist, I do not know whether "Foundations of $p$-adic Teichmüller theory" have anything to do with the IUT. The question, however, is not about IUT. To avoid off-topic debate, let us pretend in this thread that IUT papers do not exist. REPLY [2 votes]: From MathSciNet: MR3905130 Lan, Guitang ; Sheng, Mao ; Yang, Yanhong ; Zuo, Kang . Uniformization of p-adic curves via Higgs–de Rham flows. J. Reine Angew. Math. 747 (2019), 63--108. R3700485 Wakabayashi, Yasuhiro . Duality for dormant opers. J. Math. Sci. Univ. Tokyo 24 (2017), no. 3, 271--320. MR3658210 Joshi, Kirti . The degree of the dormant operatic locus. Int. Math. Res. Not. IMRN 2017, no. 9, 2599--2613. MR3615581 Zhao, Yifei . Maximally Frobenius-destabilized vector bundles over smooth algebraic curves. Internat. J. Math. 28 (2017), no. 2, 1750003, 26 pp. MR3417530 Hoshi, Yuichiro . Nilpotent admissible indigenous bundles via Cartier operators in characteristic three. Kodai Math. J. 38 (2015), no. 3, 690--731. MR3318144 Joshi, Kirti ; Pauly, Christian . Hitchin-Mochizuki morphism, opers and Frobenius-destabilized vector bundles over curves. Adv. Math. 274 (2015), 39--75. MR3296806 Gabber, Ofer ; Gille, Philippe ; Moret-Bailly, Laurent . Fibrés principaux sur les corps valués henséliens. (French) [[Principal bundles over Henselian valued fields]] Algebr. Geom. 1 (2014), no. 5, 573--612. MR3262443 Wakabayashi, Yasuhiro . An explicit formula for the generic number of dormant indigenous bundles. Publ. Res. Inst. Math. Sci. 50 (2014), no. 3, 383--409. MR3103899 Bouw, Irene I. ; Zapponi, Leonardo . Existence of covers with fixed ramification in positive characteristic. Int. J. Number Theory 9 (2013), no. 6, 1475--1489. MR2680418 Bouw, Irene I. ; Möller, Martin . Teichmüller curves, triangle groups, and Lyapunov exponents. Ann. of Math. (2) 172 (2010), no. 1, 139--185. MR2566970 Ducrohet, Laurent . The Frobenius action on rank 2 vector bundles over curves in small genus and small characteristic. Ann. Inst. Fourier (Grenoble) 59 (2009), no. 4, 1641--1669. MR2518167 Osserman, Brian . Logarithmic connections with vanishing p-curvature. J. Pure Appl. Algebra 213 (2009), no. 9, 1651--1664. MR2384903 Joshi, Kirti . Two remarks on subvarieties of moduli spaces. Internat. J. Math. 19 (2008), no. 2, 237--243. MR2365412 Lange, Herbert ; Pauly, Christian . On Frobenius-destabilized rank-2 vector bundles over curves. Comment. Math. Helv. 83 (2008), no. 1, 179--209. MR2317114 Osserman, Brian . Mochizuki's crys-stable bundles: a lexicon and applications. Publ. Res. Inst. Math. Sci. 43 (2007), no. 1, 95--119. MR2285248 Benedetto, Robert L. Wandering domains in non-Archimedean polynomial dynamics. Bull. London Math. Soc. 38 (2006), no. 6, 937--950. MR2266885 Conrad, Brian . Relative ampleness in rigid geometry. Ann. Inst. Fourier (Grenoble) 56 (2006), no. 4, 1049--1126. MR2255181 Osserman, Brian . The generalized Verschiebung map for curves of genus 2. Math. Ann. 336 (2006), no. 4, 963--986. MR2231194 Joshi, Kirti ; Ramanan, S. ; Xia, Eugene Z. ; Yu, Jiu-Kang . On vector bundles destabilized by Frobenius pull-back. Compos. Math. 142 (2006), no. 3, 616--630. MR2223683 Liu, Fu ; Osserman, Brian . Mochizuki's indigenous bundles and Ehrhart polynomials. J. Algebraic Combin. 23 (2006), no. 2, 125--136. MR2219211 Bouw, Irene I. ; Wewers, Stefan . Indigenous bundles with nilpotent p-curvature. Int. Math. Res. Not. 2006, Art. ID 89254, 37 pp. MR2118045 Mochizuki, Shinichi . Categories of log schemes with Archimedean structures. J. Math. Kyoto Univ. 44 (2004), no. 4, 891--909. MR2095769 Moriwaki, Atsushi . Diophantine geometry viewed from Arakelov geometry [translation of Sūgaku 54 (2002), no. 2, 113–129; MR1911908]. Sugaku Expositions. Sugaku Expositions 17 (2004), no. 2, 219--234. MR1859246 Ogus, Arthur . Elliptic crystals and modular motives. Adv. Math. 162 (2001), no. 2, 173--216. MR1834911 Nakamura, Hiroaki ; Tamagawa, Akio ; Mochizuki, Shinichi . The Grothendieck conjecture on the fundamental groups of algebraic curves [translation of Sūgaku 50 (1998), no. 2, 113–129; MR1648427 (2000e:14038)]. Sugaku Expositions. Sugaku Expositions 14 (2001), no. 1, 31--53. MR1812812 Edixhoven, S. J. ; Moonen, B. J. J. ; Oort, F. Open problems in algebraic geometry. Bull. Sci. Math. 125 (2001), no. 1, 1--22.<|endoftext|> TITLE: Notational question about quadratic differentials in Strebel's book "Quadratic differentials" QUESTION [6 upvotes]: In Kurt Strebel's book "Quadratic Differentials", in Chapter 2, $\S4$, he begins by saying: "Every analytic function $\varphi$ is a domain $G$ of the $z$-plane defines, in a natural way, a field of line elements $dz$, namely by the requirement that $\varphi(z)dz^2$ is real and positive. This means of course that $\text{arg }dz = -\frac{1}{2}\text{arg }\varphi(z)\mod \pi$, and thus $dz$ is determined, up to sign, for every $z$, where $\varphi(z)\ne 0,\infty$. One may then ask for the integral curves of this field of line elements." I am having some trouble with the language used here. Note that I am not a differential geometer by training. My background in differential geometry mostly comes from Voisin's first book on Hodge theory, Bott-Tu's "Differential forms in Algebraic Topology", and a bit of Kobayashi-Nomizu and a few snippets from elsewhere. This book began its first chapter on background material on Riemann surfaces, and the point of the book seems to be to study the differential geometry of Riemann surfaces. Thus, I'm sure $G$ must be a domain in $\mathbb{C}$, $z$ is a holomorphic coordinate, and "analytic" probably means "complex analytic". Now, normally, for me, a line element should be a differential 1-form. Though, for him, since he says "field of line elements", I'm assuming he uses "line element" to refer to a cotangent vector at a point, and thus his "field of line elements" should be taken to be a differential 1-form. Okay, fine, but what would it mean for $\varphi(z)dz^2$ to be real and positive? In all analogous texts, $dz^2$ is really short for $dz\otimes dz$, ie a holomorphic section of the tensor square of the complex-valued cotangent bundle, but presumably for him he really means $\varphi(z_0)(dz|_{z_0}\otimes dz|_{z_0})$ as an element of the tensor square of the complex-valued cotangent space at $z_0\in G$? In this case, if we view $dz = dx + idy$ as a complex-valued differential 1-form, one might interpret his requirement as saying that $$\varphi(z_0)(dz|_{z_0}\otimes dz|_{z_0})(X\otimes X) := \varphi(z_0)\cdot dz(X)\cdot dz(X)\in\mathbb{R}_{>0}\qquad\text{for all $X\in T_{G,z_0}$}$$ where $T_{G,z_0}$ is the (real) tangent space at $z_0$, but then this condition will never be satisfied since if it holds for $X$, then it will fail for $iX$, where $i$ is the complex involution on $T_{G,z_0}$. But maybe there is hope, as he explains that this means $\text{arg }dz = -\frac{1}{2}\text{arg }\varphi(z)\mod \pi$. However, this only confuses me more. For a complex number $z_0 = r\cdot e^{i\theta}$ with $r\in\mathbb{R}_{>0},\theta\in\mathbb{R}$, $\text{arg }(z_0) := \theta\mod 2\pi$. But what do they mean by the argument of a differential/cotangent vector? The best I can think of is: Identify $T_{G,z_0}$ with $\mathbb{C}$ via $\frac{\partial }{\partial x}\mapsto 1$ and $\frac{\partial}{\partial y}\mapsto i$, and then $\text{arg }dz$ is how much the $\mathbb{C}$-linear map $dz|_{z_0} : T_{G,z_0} = \mathbb{C}\rightarrow\mathbb{C}$ "rotates" the tangent vector (viewed as an element in $\mathbb{C}$). However, this still does not resolve the previous issue with $\varphi(z)dz^2$ being "real and positive". Lastly, what are "the integral curves of this field of line elements"? Usually one takes integral curves of a vector field. Are the $dz$'s really tangent vectors? REPLY [6 votes]: Here is a translation of Strebel's definition. First of all, a quadratic differential on a complex manifold $M$ (holomorphic or not) is a smooth section of the symmetric square $S^2(T^{*(1,0)}M)$ of the holomorphic cotangent bundle; if $M$ is a complex curve (which I will assume from now on), we can identify such a section with a section of the tensor square of $T^{*(1,0)}M$. Every quadratic differential $\omega$ defines the notion of positive tangent vectors $v\in T_pM$, namely, vectors satisfying $\omega(v,v)\ge 0$. If $\omega$ does not vanish at $p$ then the set of positive tangent vectors in $T_pM$ is a real line $L_p\subset T_pM$. Therefore, one obtains a rank 1 smooth distribution (a line field) $L$ on the complement $M'$ to the set of zeroes of $\omega$ in $M$. (In other words, this is a real line subbundle in $TM'$.) This distribution $L$ is the field of "line elements" that Strebel is defining. The integrability condition of the Frobenius (integrability) theorem is trivially satisfied in this situation and, hence, $L$ is tangent to a foliation ${\mathcal H}$ on $M'$, called the horizontal foliation of $\omega$. Leaves of this foliation are the "integral curves" in Strebel's book. Similarly, given an angle $\theta\in S^1$ one defines the distribution $L^{\theta}$ on $M'$ by the condition that a nonzero vector $v\in T_pM$ is in $L^{\theta}_p$ iff $$ arg(\omega(v,v))=\theta. $$ The line field $L^{\theta}$ is again tangent to a foliation on $M'$. Besides the horizontal foliation ($\theta=0$), one frequently uses the vertical foliation ($\theta=-1$). The horizontal and vertical foliations are orthogonal to each other with respect to the flat metric on $M'$ defined by $\omega$.<|endoftext|> TITLE: Integral surgery on $S^2 \times S^1$ QUESTION [5 upvotes]: It is a well-known fact that $S^2 \times S^1$ can be obtained by $0$-surgery on unknot. What about the $(-1)$-surgery on $S^2 \times S^1$? It seems the resulting manifold, say $W$, bounds contractible manifold. But I cannot prove it yet or refutes my argument. Any help will be appreciated. REPLY [8 votes]: This is true, with some points to clarify. First, you are presumably talking about surgery along a knot that generates the first homology (and hence fundamental group) of $S^1\times S^2$. Then the result of adding the corresponding 2-handle to $S^1\times B^3$ is contractible. The construction (called a `Mazur manifold') goes back to Mazur's paper, A Note on Some Contractible 4-Manifolds, Annals 1961. The other point is that framing as an integer is not a priori defined for a knot that represents a homology class of infinite order. But fortunately the statement is true for any framing (as in choice of trivialization of the normal bundle). I'd suggest some basic reading about 4-dimensional handle calculus, as in the book of Gompf-Stipsicz.<|endoftext|> TITLE: Are there two deformations of a wire loop such that neither can pass through the other? QUESTION [16 upvotes]: My 10-year old son asked me this simple question, and I've been unable to answer it. Suppose we start with two (unlinked) circular wire loops (maybe different sizes), and allow both to be continuously deformed in three-dimensional space, the only constraint being that different parts of the loop mustn't touch. The two loops can be deformed in different ways. Is it possible to do so in such a way that neither of the loops (now considered rigid) can be passed through the other by rigid motion? Consider the wire to be one-dimensional; the definition of 'passing through' permits the two loops to touch (as is obviously necessary in the case of two identical circles), but not to cross each other. For a formal definition of 'passing through', suppose the two loops are A (fixed) and B (moving). If there is a point on B that traces a path C through space as we move B, such that C is a closed loop that is linked with A, then we say B has passed through A. Edit Joseph O'Rourke has given an interesting example where the wires are 'interlocked'. So I'd now like to add a condition to exclude this situation, and say that, prior to the 'passing through', the wires must be separated (occupying different sides of some plane). Edit Experimenting with garden wire, inspired by some of the comments below, I came up with a configuration that may work: But even with this I am not certain. Ideally I'd like a simple provable example. In particular, could there be an example where both loops are planar? REPLY [8 votes]: I think there are quite a few examples. Here is one that is the tip of a family of ideas. Let's call the two wire loops $L_1$ and $L_2$. I'll start by describing a slightly more pliable definition of what it means for one loop to pass through another. This can be done without too much technical detail, as we are moving our loops with rigid motions. Fix an embedded spanning disc $D_1$ for $L_1$ and $D_2$ for $L_2$. As you move $L_1$ and $L_2$ by rigid motions, imagine doing the same rigid motion to $D_1$ and $D_2$. The loop $L_1$ passes through $L_2$ (once) if, in the family of loops, any point on the spanning disc $D_1$ intersects the spanning disc $D_2$ transversely exactly once (through the motion). To make this definition fully general you'd need to bring in some transversality theory, but let's ignore that detail for now. Example. $L_1$ is a circle of radius 1, and $L_2$ is a "double wound" helix about another circle of radius 1, as in the picture. Basically this is a gimmick to make the loop $L_2$ behave as if it were a "thick" circle. If you think about the rigid motions that put the loop $L_1$ so that its centre is on the spanning disc $D_2$, they either force $L_1$ and $L_2$ to intersect, or they link. I hope that makes some sense. This example makes me think you should also be able to come up with examples of loops $L_1$ and $L_2$ where $L_1$ can't pass through $L_2$, but $L_2$ can pass through $L_1$ $n$ times for some $n>1$, but not with $n=1$. i.e. to have one loop pass through another, it might have to do it multiple times, but it won't be able to do it just a single time.<|endoftext|> TITLE: Lagrange four-squares theorem --- deterministic complexity QUESTION [18 upvotes]: Lagrange's four-squares theorem states that every natural number can be represented as the sum of four integer squares. Rabin and Shallit gave a randomised algorithm that finds one of these solutions in quadratic time. My question is if anything is known about the deterministic time complexity of finding one of the solutions? Any pointers would be appreciated. (It seems that enumerating all the solutions is hard as factoring in certain cases (via Jacobi's four-square theorem), but correct me if I am wrong.) REPLY [19 votes]: As far as I know, this is still an open problem. This is discussed in Section $5$ of the paper Finding the four squares in Lagrange's theorem by Pollack and Treviño. They mention that there is a deterministic polynomial-time algorithm when $n$ is a prime via quaterion multiplication, due to Bumby. Assuming a conjecture of Heath-Brown, there is a deterministic polynomial-time algorithm that works for all $n$. Finally, they mention that a positive proportion of all numbers can be written as the sum of four squares in deterministic polynomial time. Under the Extended Riemann Hypothesis, almost all numbers can be written as the sum of four squares in deterministic polynomial time.<|endoftext|> TITLE: Preparation for GIT (Geometric Invariant Theory) QUESTION [5 upvotes]: I am trying to read Mumford's Geometric Invariant Theory, however, I find my knowledge in algebraic geometry is inadequate. My knowledge is at the level of Hartshorne's Algebraic Geometry. Mumford cites a lot of results from EGA and SGA, but I cannot read French. Therefore, I want to ask: What should I read if I want to read GIT? I should mention that most of my knowledge is in differential geometry. I want to read Mumford's work to understand stable bundles and moment maps from another perspective; different from differential geometry. Any suggestions for alternatives to Mumford's work will also be acceptable. REPLY [8 votes]: Before I started reading/referencing Geometric Invariant Theory by Mumford, Fogarty, Kirwan I read Dolgachev's book Lectures on Invariant Theory. I also like Peter Newstead's book Introduction to Moduli Problems and Orbit Spaces. Given your stated interest, these treatments might be sufficient. Even if they are not, you should be able to read them after going through Hartshorne and then I would guess GIT will be "easier" to read (the technicality will still be the same, but your intuition will be better). Addendum (given OP's comment): As far as a prerequisite for GIT itself, I don't know (beyond Hartshorne). I think when you come across a concept or term you are unfamiliar with you need to just "bite the bullet" and find a reference for that definition/concept and think about the concept. So you can use EGA/SGA as referenced directly in GIT or you can perhaps find what you need in the StacksProject (see this MO post about the pros/cons of this). But if you have first read, say Dolgachev's notes, then when you are trying to put together these new ideas into a coherent whole it will be easier since you will have some sense of what the bigger picture looks like.<|endoftext|> TITLE: What was the first elementary proof that $\pi(x)=o(x)$? QUESTION [5 upvotes]: Denote by $\pi(x)$ the number of primes $\leq x$. I'm interested in knowing who came up with the first elementary proof that $\pi(x)=o(x)$. I know that Chebyshev demonstrated elementarily before Hadamard and de la Vallee-Poussin the slightly stronger result that $\pi(x)=O(x/\log x)$. REPLY [4 votes]: Leonhard Euler knew that the infinite product: $$ \prod_{p \textrm{ prime}} \left(1 - \frac{1}{p} \right)^{-1} = \sum_{n=1}^{\infty} \frac{1}{n} $$ is divergent (and used this to prove the infinitude of primes), so would have also known that the product: $$ \prod_{p \textrm{ prime}} \left(1 - \frac{1}{p} \right) $$ tends to zero. In other words: $$ \lim_{n \rightarrow \infty} \prod_{p < n\textrm{ prime}} \left(1 - \frac{1}{p} \right) = 0 $$ Observe that the product in the last expression is the density of integers which are coprime to the set $\{ p < n \textrm{ such that } p \textrm{ is prime} \}$; this trivially implies that primes have zero upper density.<|endoftext|> TITLE: How many morphisms from 1 to 1+1 can there be? QUESTION [36 upvotes]: Here is an interesting question raised by Alice Rhyl. Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 \to 1 + 1$ can there be? There are always two obvious morphisms $f : 1 \to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one. Can there be three different morphisms $f : 1 \to 1 + 1$? I don't know. There can be four. Take $C = \mathrm{Set}^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) \to (1,1) + (1,1)$. Indeed, any power of two is possible; just take $C = \mathrm{Set}^n$. What other numbers are possible? (I find finite cardinals more interesting here.) So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 \to 1 + 1$ can there be in this case? (All the examples I've given above are categories of this sort.) REPLY [13 votes]: If all we want is a category with a terminal object and finite coproducts, the following works to show that any $n$ can occur. Let $R$ be a commutative ring of cardinality $n$ (for $n$ finite, it could be $\mathbb{Z}/(n \mathbb{Z})$). Let our category consist of pairs $(M, \mu)$ where $M$ is an $R$-module and $\mu$ is an $R$-module map $M \to R$, and where $\mathrm{Hom}((M,\mu),\ (N, \nu))$ is the set of $R$-module maps $\phi: M \to N$ such that $\phi \circ \nu = \mu$. Then $(R, \mathrm{Id})$ is the terminal object, and the coproduct of $(M, \mu)$ and $(N, \nu)$ is $(M \oplus N,\ \mu + \nu)$. So $1+1$ is $(R^{\oplus 2},\ [1\ 1])$ and $\mathrm{Hom}(1, 1+1)$ is $\left\{ \left[ \begin{smallmatrix} x \\ y \end{smallmatrix} \right] : x+y=1 \right\}$. There are $n$ solutions to $x+y=1$ in the ring $R$. Furthermore, this category has all small limits and co-limits. Namely, let $D$ be a directed graph with vertex set $D_0$ and edge set $D_1$, and suppose we have elements $(M_v, \mu_v)$ for $v \in D_0$ and maps $\phi_e$ for edges $e \in D_1$. I claim that the co-limit is $(M, \mu)$ where $M$ is the colimit of the diagram of $M_v$ in the category of $R$-modules and the map $\mu: M \to R$ comes from the universal property of co-limits and the maps $\mu_v$. To compute the limit, build a new diagram in the category of $R$-modules by adding one more vertex $\infty$ to $D_0$, one edge from every other vertex to $\infty$, with $M_{\infty} = R$ and the map $M_v \to M_{\infty}$ given by $\mu_v$ for each $v$. Then I claim that the limit of our original diagram in our funny category is $(M, \mu)$ where $M$ is the limit of the new diagram in the category of $R$-modules and $\mu$ is the projection $M \to M_{\infty}=R$. There is a very similar category which is a variety of algebras, thus fitting into Emil's answer. Let $R$ be a commutative ring of cardinality $n$. Define an $R$-affine space to be a set $A$ which, for every $k$-tuple $(r_1, \ldots, r_k)$ of elements of $R$ obeying $\sum r_j=1$, has a $k$-ary operation $\phi_{r_1, \ldots, r_k} : A^k \to A$ obeying certain conditions. The idea is that $\phi_{r_1, \ldots, r_k}(a_1, \ldots, a_k) = \sum r_j a_j$. The conditions can be found at nlab, but you have to ignore all places where they assume $R$ is a field. Basically, the axioms say that $\phi$ is invariant under permuting the $r_j$ and $a_j$ by the same permutation; that, if $a_j = a_j$, then we can replace $r_j$ and $r_k$ by $r_j+r_k$; and that we can expand nested $\phi$'s in the obvious way. The free $R$-affine space on one element is a single point, and the free $R$-affine space on $2$ points has cardinality $|R|$. To see the relation between this part of the answer and the part above the line, note that this is the "unbiased definition" of an affine space at nlab (except that we allow the empty set) and the part above the line is the "slice of Vect" definition (except that we don't require the map to $R$ to be surjective). Useless generalizations: $R$ doesn't have to be commutative, and could be a rig (ring without negation), in which case the free $R$-affine space on two elements has cardinality $\# \{ (x,y) \in R^2 : x+y=1 \}$.<|endoftext|> TITLE: Eulerian number identity QUESTION [16 upvotes]: The Eulerian number $A(n,m)$ is defined as the number of permutations $\sigma \in S_n$ having precisely $m$ descents, i.e. indices $i$ such that $\sigma(i)>\sigma(i+1)$. The wikipedia entry on these numbers mentions one identity for which I cannot locate a proof, and I would be grateful for a reference. Namely $$\sum_{m=0}^{n-1} (-1)^m \frac{A(n,m)}{\left(n\atop m\right)}=(n+1) B_n\ \ \forall n\ge 2.$$ Here $\{B_n\}$ are the Bernoulli numbers. (I do understand the two related identities immediately preceding this one. The first of these can be found in many places, and the second is easy.) REPLY [3 votes]: A scent that leads to interesting creatures and vistas (and several different derivations of this Eulerian identity) is a blend of $h(t) = e^{t}-1 \; \; \; $ and $\; \; \; h^{(-1)}(t) = \ln(1+t)$. Follow this and you encounter paths blazed by Bernoulli, Stirling, Euler, and even the ancient wizard Archimedes with the help of the bold Blissard. First denote the Bernoulli numbers $b_n$ and polynomials $Ber_n(x) = (b.+x)^n = \sum_{k=0}^n \binom{n}{k} \; b_k \; x^{n-k} ,$ then $$ \; t\;D_t \ln(e^{-t}-1) = \frac{t}{e^t-1} = e^{b.t} .$$ (Notice this relates the zeros of $e^{-t}-1$ to the Riemann zeta fct. and the Bernoullis, as Riemann noted in his famous paper.) Transform this with the inverse $t = -\ln(1+u)$ to get $$\frac{\ln(1+u)}{u} (1+u) = e^{b.(-\ln(1+u))}= (1+u)^{-b.},$$ or $$\frac{\ln(1+u)}{u} = e^{a.u} = (1+u)^{-b.-1},$$ where $a_n= \frac{(-1)^nn!}{n+1} .$ Then since the Bernoulli polynomials, as do all Appells, have the shift property $(Ber.(x)+y)^n= (b.+x+y)^n = Ber_n(x+y)$, the binomial expansion gives $$\frac{a_n}{n!} = \binom{-b.-1}{n} = (-1)^n \binom{b.+n}{n}$$ $$ =\binom{-Ber.(1)}{n}=(-1)^n\binom{Ber.(n)}{n}$$ $$ = \binom{Ber.(0)}{n} = \binom{b.}{n},$$ or, equivalently, $$ a_n = ST1_n(-Ber_.(1)) = (-1)^n ST1_n(Ber.(n)) = ST1_n(b.). $$ where $ST1_n(x)$ are the Stirling polynomials of the first kind. (The relation $(-1)^n Ber_n(1) = Ber_n(0)=b_n$ follows from the e.g.f. for the Bernoulli polynomials that follow from their natural operational definition $f(Ber.(x+1)) - f(Ber.(x)) = \frac{d}{dx} f(x)$. The sign only makes a difference for $n=1$ since all other odd order $b_n$ equal zero.) Then, using the umbral compositional inverse relation between the Stirling polynomials of the first and second kinds $$ST1_n(ST2.(x))=x^n=ST2_n(ST1.(x)),$$ we have $$ST2_n(a.) = ST2_n(ST1.(b.)) = (b.)^n = b_n,$$ and the less elegant, trickier relation $$ST2_n(-a.) = \sum_{k=0}^n ST2_{n,k} ST1_k(B.(k)) = \sum_{k=0}^n ST2_{n,k} \sum_{k=0}^m ST1_{k,m}B_m(k)).$$ Now take one of the Archimedean solids--the truncated octahedron--generalize it to all dimensions, and call the resulting simple convex polytopes the permutahedra (or permutohedra). Then we can relate the Stirling numbers of the second kind, $ST2_{n,k}$, and, consequently, the first kind, $ST1_{n.k}$, to the Eulerian numbers, for the h-polynomials of the permutahedra are the Eulerian polynomials $E_n(x) = \sum_{k=0}^n E_{n.k} \ x^k$, and the corresponding face polynomials $F_n(x)$ are related via $F_n(x) = E_n(1+x)$ (see footnote for examples). Now the reverse face polynomials are $$RF_n(x) = x^n F_n(1/x) = \sum_{k=-0}^n ST2_{n,k}\;k!\; x^k= x^n E_n(1+1/x) ,$$ or, changing signs, $$\sum_{k=0}^n ST2_{n,k}(-1)^k \;k!\; x^k = \sum _{k=0}^n E_{n,k} \sum_{m=0}^k (-1)^{n+m} \binom{k}{m} x^{n-m}.$$ Integrating from $0$ to $1$, $$ST2_n(a.) = \sum _{k=0}^n E_{n,k} \sum_{m=0}^k \binom{k}{m} (-1)^{n+m}\frac{1}{n-m+1}.$$ Focusing on the last summation, $$ \frac{1}{(n+1)!} = D_{x=1}^{j-n-1} \frac{x^{j}}{j!} = \int_{0}^1 \frac{u^{n-j}}{(n-j)!}\frac{(1-u)^j}{j!} du$$ $$ = \frac{(-1)^{j+n}}{(n-j)!j!} \int_{0}^1 (-u)^n \sum_{m=0}^{j} (-u)^{-m} \binom{j}{m}du$$ $$ = \frac{(-1)^{j+n}}{(n-j)!j!} \sum_{m=0}^j \frac{(-1)^{n-m}}{n-m+1} \binom{j}{m},$$ giving $$ \sum_{m=0}^j \frac{(-1)^{n-m}}{n-m+1} \binom{j}{m} = (-1)^{j+n}\frac{(n-j)!j!}{(n+1)!} \; .$$ So equating the two expressions for $ST2_n(a.)$, we arrive at, for $n \geq 0$, $$ b_n = \sum_{j=0}^{n} E_{n,j} \sum_{m=0}^j \frac{(-1)^{n-m}}{n-m+1} \binom{j}{m}= \sum_{j=0}^{n} (-1)^{j+n} E_{n,j} \frac{(n-j)!j!}{(n+1)!}.$$ These are the infants mimicking the adults-- $$Ber_n(x) = ST2_n(q.(ST1.(x))),$$ with $$q_n(x) = (q.(0)+x)^n = (a. + x)^n.$$ If we go deeper into the forest, we encounter Worpitzky, Dobinski, Kummer, Laguerre, Lah, the positroids and their consorts the stellahedra, among others, but that's another episode. (Only the Shadow knows what magic dwells in the heart of forests.) (But like John Mayer in his song) one more thing. Returning to the relation between the f-polynomials and the Eulerians, we see that $$Bell_n(x) = ST2_n(x) = \sum_{j=0}^n ST2_{n,j} \; x^j = \sum_{j=0}^n \frac{x^j}{j!}\sum_{k=0}^n E_{n,k} \binom{k}{n-j}$$ $$ = \sum_{k=0}^n E_{n,k} \sum_{m=0}^n \binom{m}{n} \frac{x^{j-m}}{(j-m)!} = \sum_{k=0}^n E_{n,k} L_n^{k-n}(-x),$$ introducing the associated Laguerre polynomials $$L_n^{\alpha}(-x) = \sum_{k=0}^n\; \binom{n+\alpha}{k+\alpha}\;\frac{x^k}{k!}.$$ (Befitting the entire reciprocal Euler gamma function, I use the convention $1/(-n)! =0 $ for $n=1,2,3,...$ .) Invoking the Chu-Vandermonde identities (alternatively see this MO-Q), $$ L_n^{k-n}(-St1.(x)) = \sum_{j=0}^n \binom{k}{n-j} \frac{ST1_{j}(x)}{j!} $$ $$= \sum_{j=0}^n \binom{k}{n-j} \binom{x}{j} = \binom{x+k}{n},$$ implying after composing with $ST2.(x) = Bell.(x)$ that $$ L_n^{k-n}(-x)= \binom{Bell.(x)+k}{n},$$ giving the Bell morphed Worpitzky identity, $$ Bell_n(x)= \sum_{k=0}^n E_{n,k} \binom{Bell.(x)+k}{n}.$$ By umbrally composing this with $ST1.(x)$, we morph back to the canonical WI, $$ x^n= \sum_{k=0}^n E_{n,k} \binom{x+k}{n},$$ and, umbrally morphing again, $$ Ber_n(x)= \sum_{k=0}^n E_{n,k} \binom{Ber.(x)+k}{n} = \sum_{k=0}^n E_{n,k} \binom{Ber.(x+k)}{n}.$$ Evaluating at $x=0$ and comparing with the previous identities, $$ \binom{Ber.(m)}{n} = \binom{b.+m}{n} = (-1)^{m+n} \frac{(n-m)!m!}{(n+1)!}.$$ (Edit, Dec 12, 2020: This identity is valid only for $m \leq n = 0,1,2,...$. For quick checks, $ D_x \binom{x}{n+1} = \binom{Ber.(x)}{n}$. This and the WI imply $\sum_{k=0}^{n} E_{n,k} \binom{Ber.(k)}{n-1} = 0$, consistent with palindromic symmetry of the Eulerians and skew (signed) palindromic symmetry of the umbral binomials of the Bernoulli numbers. Actually, this last identity allows the extension of the validity of the chief identity to $m = n$.) (Again, there is a deeper story here based on a dance between the $h(t)= e^t-1$ and its consort $h^{(-1)}(t)= \ln(1+t)$ and similar couples.) For nailing down indices, definitions, and spot checking these identities and any new ones someone may discover: $Ber_n(x) = (b. +x)^n = \sum_{k=0}^n \binom{n}{k} \; b_n \;x^{n-k}$ $$e^{B.(x)t} = \frac{t}{e^t-1} e^{x t} = e^{b.t} e^{xt}=e^{(b.+x)t}$$ $b_0 = 1\;\;\;\;\;\;\;\;\;\;$ $B_0(x)=1$ $b_1 = -1/2\;\;\;\;\;\;\;\;\;\;$ $B_1(x)= \frac{-1+2x}{2}$ $b_2 = 1/6\;\;\;\;\;\;\;\;\;\;$ $B_2(x)= \frac{1-6x+6x^2}{6}$ $b_3 = 0\;\;\;\;\;\;\;\;\;\;$ $B_3(x)= \frac{x-3x^2+2x^3}{2}$ $ST1_n(x) = \sum_{k=0}^n ST1_{n,k}\; x^k \;\;\;\;\;$ (a.k.a. the falling factorials) $$e^{ST1.(x) t} = e^{x\ln(1+t)} = (1+t)^x$$ $ST1_0(x) =1 $ $ST1_1(x) = x$ $ST1_2(x) = -x+x^2$ $ST1_3(x) = 2x-3x^2+x^3$ $ST2_n(x) = \sum_{k=0}^n ST2_{n,k}\; x^k \;\;\;\;\; $ (a.k.a. the Touchard / Bell / Exponential polynomials) $$e^{ST2.(x) t} = e^{x(e^t-1)}$$ $ST2_0(x) =1 $ $ST2_1(x) = x$ $ST2_2(x) = x+x^2$ $ST2_3(x) = x+3x^2+x^3$ $RF_n(x) = \sum_{k=0}^n RF_{n,k}\; x^k$ $$e^{RF.(x)t} = \frac{1}{1+x-xe^{t}}$$ $RF_0(x) =1 $ $RF_1(x) = x$ $RF_2(x) = x+2x^2$ $RF_3(x) = x+6x^2+6x^3$ $F_n(x) = \sum_{k=0}^n F_{n,k}\; x^k$ $$e^{F.(x)t} = \frac{x}{1+x-e^{tx}}$$ $F_0(x) =1 $ $F_1(x) = 1$ $F_2(x) = 2+x$ $F_3(x) = 6+6x+x^2$ $E_n(x) = \sum_{k=0}^n E_{n,k}\; x^k$ $$e^{E.(x)t} = \frac{x-1}{x-e^{(x-1)t}}$$ $E_0(x) =1 $ $E_1(x) = 1$ $E_2(x) = 1+x$ $E_3(x) = 1+4x+x^2$<|endoftext|> TITLE: Complexity of the set of closed subsets of an analytic set QUESTION [8 upvotes]: Let $X$ be a compact Polish space and $K(X)$ the hyperspace of closed subspaces of $X$ with the Vietoris/Hausdorff metric topology. Question: If $A$ is an analytic subset of $X$, what is the complexity of the set $\{C\in K(X): C\subseteq A\}$? This set is clearly $\mathbf{\Pi}^1_2$, but can this be reduced to $\mathbf{\Delta}^1_2$ or even to (co-)analytic? REPLY [3 votes]: As far as I see, the set can be $\mathbf{\Pi}^1_2$-complete. Take a $\mathbf{\Pi}^1_1$ set $C \subset 2^\omega \times 2^\omega$, such that $\pi_1(C)$ is a $\mathbf{\Sigma}^1_2$-complete set. Then consider the mapping $x \mapsto \{x\} \times 2^\omega$. This is a continuous mapping from $2^\omega$ to $\mathcal{K}(2^\omega \times 2^\omega)$ and a reduction of $2^\omega \setminus \pi_1(C)$ (which is a $\mathbf{\Pi}^1_2$-complete set) to $\{K \in \mathcal{K}(2^\omega \times 2^\omega): K \subset 2^\omega \times 2^\omega \setminus C\}$, that is, the collection of the compact subsets of $C$'s complement.<|endoftext|> TITLE: Cohomology of simple finite groups remembers the group? QUESTION [9 upvotes]: Let $G$ and $H$ be finite simple groups. I expect that if $G$ and $H$ are not isomorphic, then their cohomology groups with integral coefficients are not all isomorphic, that is, $H^*(G,\mathbb{Z})$ and $H^*(H,\mathbb{Z})$ are not isomorphic graded abelian groups. Is there any proof of this? Notice, I do not want to compute those cohomology groups (and as far as I know it hasn't been done yet completely), just to show that they are not allisomorphic. REPLY [8 votes]: This is a remark rather than an answer to your question. If you remove the word `simple' it is easy to find such pairs of finite groups. The first examples I learned were (I think) constructed by Atiyah. Each of the two finite groups has a normal subgroup that is cyclic of order three and quotient dihedral of order 8. In each case the dihedral group acts non-trivially on the $C_3$, but in one case the kernel of the action is $C_2\times C_2$ whereas in the other case the kernel of the action is $C_4$. The Lyndon-Hochschild-Serre spectral sequence gives a complete calculation of the integral cohomology ring, and it is just $H^*(C_3)^{C_2}\otimes H^*(D_8)$ in each case. Somehow cohomology cannot see that the centralizers of the $C_3$ subgroups are different (which is how one can see that the groups are not isomorphic). I computed the integral cohomology rings of a family of $3$-groups of nilpotence class two in my thesis, and I observed that for each $n\geq 5$ there are a pair of groups of order $3^n$ with isomorphic integral cohomology rings. I also gave a more conceptual proof that there are pairs of groups of order $p^n$ for each odd $p$ and $n\geq 5$ that are not isomorphic but have isomorphic integral cohomology groups. My articles are '$3$-groups are not determined by their integral cohomology rings' JPAA Vol 103 (1995) 61-79 and '$p$-groups are not determined by their integral cohomology groups' Bull London Math Soc Vol 27 (1995) 585-589. None of these arguments are any help for finite simple groups though.<|endoftext|> TITLE: Generalized Hodge Decomposition on Manifolds with Boundary QUESTION [9 upvotes]: This question is motivated by the problem of finding heat kernels to use for the renormalization of quantum field theories on manifolds with boundary. If $(\mathscr{E}, Q)$ is an elliptic complex on a closed manifold $M$, then if one chooses a metric $(\cdot,\cdot)$ for $\mathscr{E}$, one can define the formal adjoint $Q^*$ to $Q$ and the cohomological degree-zero operator $QQ^*+Q^*Q$ is elliptic. Its kernel is precisely the cohomology of $\mathscr{E}$. Moreover, there is a decomposition $\mathscr{E}= \ker(Q)\cap\ker(Q^*)\oplus \text{Im}Q\oplus \text{Im}Q^*.$ On a compact manifold with boundary, there is a similar decomposition (see Günter Schwarz: Hodge Decomposition - A Method for Solving Boundary Value Problems) for differential forms: a $k$ form can be written uniquely as the sum of a harmonic $k$-form (one which is both $d$- and $\delta$-closed), the boundary of a form with vanishing tangential component on $\partial M$, and the coboundary of a form with vanishing normal component on the boundary. Is there a similar statement which applies for a general elliptic complex on $M$? I would be happy enough to know this for an elliptic complex which is isomorphic to one of the form $(\mathscr{E’}\otimes \Omega^\bullet_{[0,\epsilon)},Q’\otimes 1+1\otimes d)$ near $\partial M$ (after a choice of collar neighborhood for $\partial M$), and the metric on $\mathscr{E}$ is a product metric under this identification. Here, $(\mathscr{E}’, Q’)$ is an elliptic complex on $\partial M$. In this situation, it still makes sense to define the tangential and normal parts of an element $e\in \mathscr{E}$ near $\partial M$. So, more precisely, my question is: in the situation just described, can one write $\mathscr{E} = \ker Q\cap \ker Q^*\oplus Q\mathscr{T}^{k-1}\oplus Q^* \mathscr{N}^{k+1},$ where $\mathscr{T}^{k-1}$ is the space of elements of degree $k-1$ in $\mathscr{E}$ with vanishing tangential component on the boundary and $\mathscr{N}^{k+1}$ is the space of elements of degree $k+1$ in $\mathscr{E}$ with vanishing normal component on the boundary? REPLY [3 votes]: The answer to this question as asked is no. However, you generally obtain something similar. Consider $D = Q + Q^*$. By standard arguments, $$\ker(D) = \ker(Q)\cap \ker(Q^*).$$ (Of course $D\Phi = 0$ means that $D^2\Phi=0$. But $D^2\Phi = 0$ implies $$0 = \langle \Phi, (Q^*Q+QQ^*) \Phi \rangle = \|Q\Phi\|^2 + \|Q^*\Phi\|^2,$$ hence $Q\Phi = Q^*\Phi = 0$. In other words, $$\ker(D) \subseteq \ker (D^2) \subseteq \ker(Q)\cap \ker(Q^*).$$ The other direction is trivial.) Now we have the integration by parts rule $$\int_X\Bigl( \langle D\Phi, \Psi\rangle - \langle \Phi, D \Psi \rangle \Bigr) = \int_{\partial X} \langle \Phi|_{\partial X}, \sigma(\nu)\Psi|_{\partial X}\rangle,$$ where $\sigma$ is the principal symbol of $D$, $\nu$ is the normal vector to the boundary (choose the one that makes the sign correct :P). This shows that a smooth $D\Psi$ is orthogonal to $\ker(D)$ if and only if $$ \sigma(\nu)\Psi|_{\partial X} \perp \{\Phi|_{\partial X} \mid D\Phi = 0\}. \qquad (*)$$ Hence we obtain the splitting $$\mathscr{E} = \ker(D) \oplus D\mathscr{R},$$ where $\mathscr{R}$ is the space of $\Psi$ satisfying the orthogonality requirement $(*)$. If we further take into account the grading and figure out what the integration by parts formula gives us, we get $$\mathscr{E}^k = \mathscr{E}^k \cap \ker(D) \oplus Q \mathscr{T}^{k-1} \oplus Q^* \mathscr{N}^{k+1},$$ where $$ \begin{aligned} \mathscr{T}^{k-1} &= \{ \Psi \in \mathscr{E}^{k-1} \mid \rho(\nu)\Psi|_{\partial X}, \perp L^{k}\},& \quad L^{k} &= \{\Phi|_{\partial X} \mid Q^*\Phi = 0, \Phi \in \mathscr{E}^k\}\\ \mathscr{N}^{k+1} &= \{ \Psi \in \mathscr{E}^{k+1} \mid \rho^*(\nu)\Psi|_{\partial X} \perp M^{k}\},& \quad M^{k} &= \{\Phi|_{\partial X} \mid Q\Phi = 0, \Phi \in \mathscr{E}^k\}.\end{aligned}$$ Here $\rho$, $\rho^*$ are the principal symbols of $Q$ and $Q^*$, respectively, so that $\sigma = \rho + \rho^*$. In your specific example, it so happens that $L^k = M^k = \Gamma(X, \Lambda^k T^*X)$, while $\rho(\nu)$ and $\rho^*(\nu)$ are wedging with, respectively insertion of the normal vector. Therefore $\mathscr{T}$ and $\mathscr{N}$ have the description you gave. As an example where this is not the case, take $$0 \longrightarrow \mathscr{S}^+ \stackrel{Q}{\longrightarrow} \mathscr{S}^- \longrightarrow 0,$$ where $\mathscr{S} = \mathscr{S}^+ \oplus \mathscr{S}^-$ are the smooth sections of the spinor bundle over an even-dimensional spin manifold, and $Q= D^+$ is (half of) the Dirac operator. In this case, $Q$ and $Q^* = D^-$ are elliptic, and $L^-$, $M^+$ are not everything. In fact, it is a theorem that $$\mathscr{S}^+|_{\partial X} = \rho^*(\nu) L^- \oplus M^+, \qquad \text{and} \qquad \mathscr{S}^-|_{\partial X} = L^- \oplus \rho(\nu) M^+,$$ where each of the sums is direct. (Here I wrote $\pm$ instead of $0$, $1$ for the grading.)<|endoftext|> TITLE: Tannaka duality for semisimple groups QUESTION [9 upvotes]: Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $\mathcal{C}$ equipped with a fiber functor $F: \mathcal{C} \to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G \cong Aut(F)$ such that $\mathcal{C} \cong Rep(G)$. This means that any such category is associated with a root datum. Is there a version of this reconstruction theorem that will tell us when a category $\mathcal{C}$ is the category of finite dimensional representations of a semisimple group? I would like to be able to associate with a Tannakian category a root system, and not just a root datum. REPLY [2 votes]: I've decided to turn my comments into an answer. (a) The conditions characterizing the Tannakian categories attached to connected reductive groups can be found in Chapter 2 (2.20, 2.22, 2.23) of the notes by Deligne and Milne on Tannakian categories. (b) The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $D\to Z(G)$ is the same as giving an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne. (c) To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any lattice $X$ in $P$ containing $Q$. This gives a complete description of the Tannakian categories corresponding to root systems (better, diagrams) without using algebraic groups. See arXiv:0705.1348 –<|endoftext|> TITLE: Big list of comonads QUESTION [27 upvotes]: The concept of a monad is very well established, and there are very many examples of monads pertaining almost all areas of mathematics. The dual concept, a comonad, is less popular. What are examples of comonads, in different categories, and different fields of math? REPLY [3 votes]: The so-called game comonads have been recently studied in the context of finite model theory. The main references are the following: Samson Abramsky, Anuj Dawar, and Pengming Wang. "The pebbling comonad in finite model theory." 2017 32nd Annual ACM/IEEE Symposium on Logic in Computer Science (LICS). IEEE, 2017. (see also arXiv:1704.05124) Samson Abramsky and Nihil Shah. "Relating Structure and Power: Comonadic semantics for computational resources." International Workshop on Coalgebraic Methods in Computer Science. Springer, Cham, 2018. (see also arXiv:1806.09031) Game comonads are an active area of research. In nutshell, some of the well-known model comparison can be turned into a comonad. This is done by representing the states of the game semantically. For example, the Ehrenfeucht-Fraïssé comonad $\mathbb E_k$ is a comonad on the category of relational structures. The universe of $\mathbb E_k(A)$ consists of non-empty words of length ${\leq}k$, where the alphabet is taken to be the universe of $A$. A word $[a_1,a_2,\dots,a_n]$ represents spoiler's moves in $A$ in the one-way Ehrenfeucht-Fraïssé game.<|endoftext|> TITLE: What is the geometrical meaning of higher Chern forms and classes? QUESTION [7 upvotes]: Let $M$ be a complex manifold, $R^{\nabla}$ be the curvature operator for connections $\nabla$. Consider a polynomial function $f:\operatorname M_n(\mathbb{C})\to\mathbb{C}$. For the gauge group $\operatorname{GL}_n(\mathbb{C})$, if $f(A)=f(gAg^{-1})$ where $A \in \operatorname M_n(\mathbb{C})$ and $g \in \operatorname{GL}_n(\mathbb{C})$, then $f$ is said to be an invariant polynomial function. Let $I^k(\operatorname M_n(\mathbb{C}))$ be the set of all such polynomials of degree $k$. Also, $\bigoplus_{k\geq0}I^{k}(\operatorname M_n(\mathbb{C}))=I(\operatorname M_n(\mathbb{C}))$. We shall need $\phi_n(A)=det(A)$, $n>1$, and $\phi_1(A)=Tr(A)$. Define a global differential form ($2k$-form) $f(R^{\nabla}) \in \mathbb{A}^{2k}(M)$. If we have the de Rham cohomology group $H^{2k}(M)$, then the Weil homomorphism is defined as the map $\omega:I(\operatorname M_n(\mathbb{C}))\to \bigoplus_{k\geq 0}H^{2k}(M)$. The Chern forms $c_{i}(R^{\nabla})=\phi_{i}(\frac{\sqrt{-1}}{2\pi}R^{\nabla})$. For the complex vector bundle $(\mathbb{E},\pi,M)$, where $\mathbb{E}$ is the total space, the Chern classes are defined as $c_{i}(\mathbb{E}) \in H^{2k}(M)$. Therefore $c_{i}(\mathbb{E})\mathrel{:=}\omega(c_{i}(R^{\nabla}))=[c_{i}(R^{\nabla})]$ (de Rham cohomology class). The Chern forms $c_{i}(R^{\nabla}) \in \mathbb{A}^{2i}(\mathbb{E})$. $\mathbb{A}^{2i}(\mathbb{E})$ is sheaf of smooth $\mathbb{E}$-valued $2i$ forms on $M$. Cohomology groups are very important in geometry for understanding the invariants that can be defined on manifolds. That is, the transformations that keep some special properties of the manifold and which analogous to the gauge transformations in physics. Chern classes are special type of cohomology classes. If the first Chern class vanishes for a particular manifold, then it must be a Ricci-flat manifold. For example the Calabi–Yau manifolds (they have lots of other special properties, e.g., trivial canonical bundle, etc.). But what do the higher Chern classes mean? What uses are those higher cohomology classes corresponding to the higher Chern classes of? REPLY [7 votes]: This is a big topic, which should be covered in the union of many standard texts (Chern, Griffiths-Harris, Milnor-Stasheff...). I'll list a few answers off the top of my head. Suppose that $L$ is the tautological line bundle on the complex projective space plane, then it's clearly not trivial, and neither is $V= L\oplus L^{-1}$. But $c_1(V)=0$, because the trace of curvature is zero for an induced connection. Of course, $V$ is certainly not trivial. So it's natural to look for higher cohomological obstructions (as Arun suggested), e.g. $c_2(V) = -c_1(L)^2\not=0$ would work. For universal bundles on Grassmanians, Chern classes have natural geometric interpretations involving Schubert cycles. Chern classes come up in formulas expressing answers to natural geometric questions: Gauss-Bonnet, Riemann-Roch, or more general index theorems.<|endoftext|> TITLE: Comparison of Rademacher processes QUESTION [5 upvotes]: Suppose that $T$ is a bounded set in $\mathbb{R}^n$ and $f,g$ are two nonnegative functions such that $0\leq f(x)\leq g(x)$ for all $x\geq 0$. Let $\epsilon_1,\epsilon_2,\dots,$ be a Rademacher sequence. Does it hold that $$ \mathbb{E} \sup_{t\in T} \left| \sum_i \epsilon_i f(|t_i|) \right| \leq C\cdot \mathbb{E} \sup_{t\in T} \left| \sum_i \epsilon_i g(|t_i|) \right| \qquad (*) $$ for some absolute constant $C$? It looks intuitively true to me but I cannot find it anywhere in the literature. To compare it with the contraction principle that $$ \mathbb{E} \sup_{t\in T} \left| \sum_i \epsilon_i \alpha_i g(|t_i|) \right| \leq \mathbb{E} \sup_{t\in T} \left| \sum_i \epsilon_i g(|t_i|) \right|, \quad |\alpha_i|\leq 1 $$ it seems that the $\alpha_i$ in my case would depend on $t$ and the proof of the contraction principle above does not seem to go through. Question: Does (*) hold? Is there a reference in the literature or is there a particularly simple proof? REPLY [3 votes]: $\newcommand{\ep}{\varepsilon}$ Letting $a_{it}:=f(|t_i|)$ and $b_{it}:=g(|t_i|)$, rewrite your inequality ($\ast$) as \begin{equation} E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i a_{it}\Big|\le C\,E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i b_{it}\Big| \end{equation} for any natural $N$ and any set $T$, with the condition that $0\le a_{it}\le b_{it}$ for all $i,t$. Let now $T:=2^{[N]}$, the power set of the set $[N]:=\{1,\dots,N\}$. Let $b_{it}:=1$ and \begin{equation} a_{it}:=1_{i\in t} \end{equation} for all $i\in[N]$ and $t\in T=2^{[N]}$. Then \begin{equation} E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i a_{it}\Big| \ge E\sum_{i=1}^N\max(0,\ep_i)=N/2, \end{equation} whereas \begin{equation} E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i b_{it}\Big| =E\Big|\sum_{i=1}^N\ep_i\Big|\le\sqrt{E\Big(\sum_{i=1}^N\ep_i\Big)^2}=\sqrt N. \end{equation} So, your inequality ($\ast$) will fail to hold for any $C>0$ if $N>4C^2$.<|endoftext|> TITLE: To what extent is homological localization determined by its values on $K(G,n)$'s? QUESTION [5 upvotes]: Consider a homological localization $L$ of $p$-local spaces or (I think equivalently) a localization of $p$-local spectra. If a space $X$ is $L$-acyclic, then so is $K(\pi_n(X),n)$ for each $n$. To what extent does the converse hold? That is, if $X$ is a space and $K(\pi_n(X),n)$ is $L$-acyclic for each $n$, then is $X$ $L$-acyclic? If the converse always holds, then it's not known, because this would provide an affirmative answer to the telescope conjecture. So I suppose I should ask: Question: What is an example of a pair of homological localizations $L,L'$ of $p$-local spaces which agree on Eilenberg-MacLane spaces but not on all spaces? If $L,L'$ are not $H\mathbb Z_{(p)}$-localization, then they kill some $K(\mathbb Z/p,n+1)$ and thus (if $B\mathbb Q$ is acyclic) all $K(A,m)$'s for $m \geq n+2$. In this case, I believe that $L,L'$ agree on $n+1$-truncated spaces if and only if they agree on Eilenberg-MacLane spaces. On the other hand, if $X$ is $n+1$-connected, then the value of $LX$ or $L'X$ depends only on the $m$-connective cover $\tau_{\geq m} X$ for any $m \geq n+2$. So any difference between $L,L'$ is kind of hard to think about in terms of homotopy groups. REPLY [5 votes]: You ask two different questions. Regarding the first, there are easy counterexamples: e.g. $\widetilde K(1)_*(S^3) \neq 0$, but $\widetilde K(1)_*(K(\pi_n(S^3),n)) = 0$ for all $n$.<|endoftext|> TITLE: Examples of problems where considering "discrete analogues" has provided insight or led to a solution of the original problem QUESTION [16 upvotes]: The Kakeya conjecture posits that any Kakeya set in $\mathbb{R}^n$ has dimension $n$. A discrete (finitized?) version of this problem is the Finite Field Kakeya conjecture, which was proved by Dvir in 2008. My understanding is that the Finite Field Kakeya Conjecture was proposed at the end of the twentieth century with the hope that it would lead to methods that could be applied to the original Kakeya conjecture. However, it seems that in this case the approach used for resolving the discrete analogue is not easily applied to the original problem, so that the Kakeya conjecture still remains open. My question is asking for examples of problems where discretizing "succeeded." Question: Are there examples of math problems where looking at a finite or discrete variant of the initial statement did lead to a solution (or if not a complete solution, at least significant progress) of the original problem? If so, what are they? Edit: As commenters pointed out, this question is similar to the one here which requests examples where a discrete version of a theory was developed before the continuous version of the same theory. My question is different from that previous question, in that I am not interested in cases where discrete problems predated continuous problems. Instead, I'd like to learn about instances where a continuous problem was already proposed, and studying a discrete version of that problem helped inform a solution to the continuous variant. REPLY [10 votes]: In my paper Tao, Terence, Norm convergence of multiple ergodic averages for commuting transformations, Ergodic Theory Dyn. Syst. 28, No. 2, 657-688 (2008). ZBL1181.37004. I was able to settle a question in ergodic theory (namely, the norm convergence of averages $\frac{1}{N} \sum_{n=1}^N \int_X T_1^n f_1 \dots T_k^n f_k\ d\mu$ for $k$ commuting measure-preserving transformations $T_1,\dots,T_k$ on a probability space $(X,\mu)$ and bounded functions $f_1,\dots,f_k$), by abandoning all the usual ergodic theory machinery (e.g., characteristic factors), and translating the problem to a purely finitary one without explicit use of limits. This could then be attacked by methods related to graph and hypergraph regularity. The argument was then significantly generalised in Walsh, Miguel N., Norm convergence of nilpotent ergodic averages, Ann. Math. (2) 175, No. 3, 1667-1688 (2012). ZBL1248.37008. It should however be pointed out that an alternate, ergodic-theoretic proof of these results was also subsequently given in Austin, Tim, On the norm convergence of non-conventional ergodic averages, Ergodic Theory Dyn. Syst. 30, No. 2, 321-338 (2010). ZBL1206.37003. Austin, Tim, A proof of Walsh’s convergence theorem using couplings, Int. Math. Res. Not. 2015, No. 15, 6661-6674 (2015). ZBL1372.37012. I should also mention that there are several papers of Bourgain in which he establishes various ergodic theorems by converting them to quantitative questions in harmonic analysis which are strictly speaking not discrete or finitary, but are amenable to many of the same techniques (in particular, a focus on "hard analysis" estimates) as such discrete problems. A typical such paper is Bourgain, J., On the pointwise ergodic theorem on $L^p$ for arithmetic sets, Isr. J. Math. 61, No. 1, 73-84 (1988). ZBL0642.28011. REPLY [5 votes]: As a non-expert, I will tell a story (since I am not qualified to do more): Once upon a time (1859) there was conjecture about the zeros of a complex function known as the Riemann Hypothesis (RH). Years later Weil formulated a "discrete version" (in terms of finite fields $\mathbb{F}_q$) of this conjecture; the third of his Weil Conjectures (1949). In 1974, Deligne proved this discrete version to much acclaim. Here is a well regarded exposition by Milne. Years later still, Connes is motivated to explore a potential proof of the original RH along the lines of Deligne's proof by taking a "limit as $q\to 1$". These attempts are probably uncontroversially considered mathematical advances. Some, presumably Connes himself, consider this work an advance in the problem at hand (and so making this hopefully an appropriate response to this MO question). However, there are other opinions, and so this MO post might prove controversial too.<|endoftext|> TITLE: Why shouldn't this prove the Prime Number Theorem? QUESTION [13 upvotes]: Denote by $\mu$ the Mobius function. It is known that for every integer $k>1$, the number $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free. Letting $k\rightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form $$\sum_{n=1}^{\infty} \frac{\mu(n)}{n}=0,$$ since the probability that an integer is ``$1$-free'' is zero ? REPLY [20 votes]: You ask: Denote by $\mu$ the Mobius function. It is known that for every integer $k>1$, the number $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free. Letting $k\rightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form $$\sum_{n=1}^{\infty} \frac{\mu(n)}{n}=0,$$ since the probability that an integer is ``$1$-free'' is zero ? As pointed out by the users @wojowu and @PeterHumphries, it is true that the PNT is equivalent to $$\lim_{x \to \infty} \sum_{n\leq x} \frac{\mu(n)}{n}=0,$$ and it is relatively easy to prove that $$\lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}=0.$$ The real difficulty lies in proving that $$\lim_{x\rightarrow \infty} \sum_{n\leq x} \frac{\mu(n)}{n}= \lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s},$$ which is highly nontrivial and requires intricate arguments. In particular, as pointed out by @TerryTao in the comments: if $t\neq 0$ is real, then $$ \lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{n^{it}}{n^s},$$ can be shown to converge to a finite value, whereas $$\lim_{x\rightarrow \infty} \sum_{n\leq x} \frac{n^{it}}{n}$$ is undefined. So at a bare minimum one has to somehow stop $\mu(n)$ from "pretending" to be like $n^{it}$. This turns out to be basically equivalent to preventing $\zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the PNT.<|endoftext|> TITLE: Simplicial Objects in Additive Categories QUESTION [5 upvotes]: I am looking for a reference, preferably as elementary as possible, for the following statement. Let $X_{m,n}$ be a bi-simplicial object in an additive category $\mathcal{A}$. Then the complex $|X_{m,n}|$ obtained from iterative realization (i.e. the total complex of the associated bi-complex) is homotopy equivalent to the complex obtained by realizing the diagonal $|X_{n,n}|$. I know that it should follow from the cofinality of the map $\Delta^{op} \to \Delta^{op} \times \Delta^{op}$ together with the identification of the realization as the colimit of a simplicial object in the $\infty$-category of bounded above complexes in $\mathcal{A}$. However, all this seem to me like a real overkill for this problem, and I would expect a much more elementary explanation can be given. I also remember seeing a proof along this lines in the case where $\mathcal{A}$ is Abelian, but I really need a more general result, since the example I have in mind is not abelian. REPLY [2 votes]: For abelian categories this is known as the Eilenberg–Zilber theorem, see, for instance, Theorem 8.5.1 in Weibel's book. One can write down explicit comparison maps in both directions (namely, the Alexander–Whitney and Eilenberg–Zilber maps) and also write down an explicit chain homotopy that shows these maps to be chain homotopy equivalences. The constructions are explicit and concrete, and perhaps the proof goes through for additive categories, but I have not checked this.<|endoftext|> TITLE: Traveling Salesman Problem on finite group QUESTION [11 upvotes]: Given a finite group $H$, define a norm on $H$ to be a function $f : H \rightarrow \mathbb{R}_{\geq 0}$ satisfying: $f(x) = 0 \iff x = e$ is the identity; $\forall x \in H$, we have $f(x) = f(x^{-1})$; $\forall x, y \in H$, we have $f(xy) \leq f(x) + f(y)$. This induces a metric $d : H \times H \rightarrow \mathbb{R}_{\geq 0}$ as follows: $$ d(x, y) := f(x y^{-1}) $$ Is there an efficient algorithm for solving the Traveling Salesman Problem on such a finite metric space? (This problem arises when you have some group $G$ endowed with a Cayley graph, and want to find an optimal tour that visits every element of a subgroup $H \leq G$. Specifically, we can calculate the norm $f$ on $G$ by a single application of Dijkstra's algorithm, then restrict to $H$.) REPLY [3 votes]: Here is an observation which suggests that it might not be that easy: There are a number of versions of the longstanding Lovasz Conjecture among them Every finite connected vertex transitive graph contains a Hamilton path. Every finite connected vertex transitive graph (except for five known exceptions) contains a Hamilton cycle. None of the five exceptions is a Cayley Graph, although, of course, Cayley Graphs are vertex transitive. There is also an over $20$ years old conjecture of Babai which denies this. Namely that there is some $c \lt 1$ such that there are infinitely many Cayley Graphs with no cycle of length as great as $cn.$ Consider a finite group $H$ with $n$ elements and a distinguished set of generators and let $\Gamma$ be the associated Cayley Graph. Then we can define a norm on $H$ (essentially an edge weighting of the complete graph $K_n$) by distance in $\Gamma.$ Then the Lovasz conjecture implies that the traveling salesman has a route of weightii $n$ while Babai's conjecture implies that this does not always happen. It is not know if the Lovasz conjecture holds for all (Cayley Graphs arising from generating sets of) dihedral groups (or at least it wasn't when this article was published.) So this argues against an efficient algorithm. To balance that, here is a question which asks how hard it could be at least in a certain case: What have you found by looking at small examples? One might suspect that, if there is an $k \in H$ such that only $k$ and $k^{-1}$ have the minimum positive norm, then an optimal route uses $k$ and/or $k^{-1}$ for $n-j$ steps where $m=\frac{n}j$ is the order of $k.$ Specifically, one can start at $e$ and use $k$ the first $m-1$ times to go through $K=\langle k \rangle$ then use some other element landing in a coset of $K$ and then either $k$ or $k^{-1}$ the next $m-1$ times to go through this coset and continue on in this manner until all the cosets have been traversed. Perhaps there are easy counter-examples. But do you know?<|endoftext|> TITLE: Does a random sequence of vectors span a Hilbert space? QUESTION [8 upvotes]: Let $\mathcal{H}$ be a separable Hilbert space. Let $v$ be a random variable taking values in $\mathcal{H}$ such that $P(v \perp h) < 1$ for all $h \in \mathcal{H}.$ Suppose we sample an infinite sequence $v_1, v_2, \ldots.$ Is it the case that, almost surely, the closed span of $v_1, v_2, \ldots$ is all of $\mathcal{H}?$ REPLY [6 votes]: (This may turn out to be a simplified version of J. E. Pascoe's answer). The support of (the distribution of) $v$, that we denote by $\operatorname{supp} v$, is the set of vectors $h \in \mathcal{H}$ such that $P(v \in B(h, \varepsilon)) > 0$ for every $\varepsilon > 0$. We list some properties of this set. The set $\operatorname{supp} v$ is the complement of the union of all open sets $B$ such that $P(v \in B) = 0$. Thus, the support is a closed set. Since $\mathcal{H}$ is a separable metric space, it has a countable topological base $\mathcal{B}$, and $\operatorname{supp} v$ is the complement of the union of all $B \in \mathcal{B}$ such that $P(v \in B) = 0$. By countable additivity, it follows that $P(v \in \operatorname{supp} v) = 1$ (the support is a set of full measure). With probability one, the closure of the random set $V = \{v_1, v_2, \ldots\}$ contains $\operatorname{supp} v$. Indeed, let $\{h_1, h_2, \ldots\}$ be a countable, dense subset of $\operatorname{supp} v$. For every $i, n = 1, 2, \ldots$ we have $P(v \in B(h_i, \tfrac{1}{n})) > 0$, and thus, by Borel–Cantelli, $P(V \cap B(h_i, \tfrac{1}{n}) = \varnothing) = 0$. It follows that $h_i \in \overline{V}$ for every $i = 1, 2, \ldots$, and consequently $\operatorname{supp} v \subseteq \overline{V}$. For every $h \in \mathcal{H}$, we have $P(h \perp v) < 1$, and therefore $h$ is not orthogonal to $\operatorname{supp} v$. It follows that the closed span of $\operatorname{supp} v$ is $\mathcal{H}$. It remains to note that the closed span of $V$ is the same as the closed span of the closure of $V$, which with probability one contains the closed span of $\operatorname{supp} v$, which we have shown to be equal to $\mathcal{H}$. (Item 1 is valid for any topological space; items 2 and 3 work in an arbitrary separable metric space.)<|endoftext|> TITLE: An isoperimetric-type inequality inside a cube QUESTION [8 upvotes]: I am looking for a reference for the following inequality: if $\Omega \subset [0,1]^d$ satisfies $\mbox{vol}(\Omega) \leq 1/2$, then $$ \mathcal{H}^{d-1}\left( \partial\Omega \cap (0,1)^d\right) \geq c_d \mbox{vol}(\Omega)^{\frac{d-1}{d}},$$ where $\mathcal{H}^{d-1}$ is the $(d-1)-$dimensional Hausdorff measure and $c_d > 0$ is a universal constant depending only on $d$. This is a variation of the classical isoperimetric inequality with the interesting addition that surface 'on the boundary of the cube' does not count. This seems like it should be known. A discrete version of this inequality (for subsets of the grid graph) was proven by Bollobas and Leader (Edge-isoperimetric inequalities in the grid, Combinatorica 1991) and it seems there is a wealth of information for the discrete case. Has anybody seen the continuous case stated somewhere? REPLY [7 votes]: This result is known as the relative isoperimetric inequality, see e.g. Functions of Bounded Variation by L. Ambrosio, N. Fusco and D. Pallara (2000), Eq. (3.43). It follows from Poincare inequality (see e.g. Eq. (3.41) in the cited book) applied to $\chi_\Omega$ (the indicator of the set $\Omega$). Indeed, by Poincare inequality it holds $\|\chi_\Omega - \mbox{vol}(\Omega)\|_{L^p((0,1)^d)} \le C |D\chi_\Omega|((0,1)^d)$, where $p=\frac{d}{d-1}$. Here $|D\chi_\Omega|((0,1)^d)=\mathcal{H}^{d-1}(\partial \Omega \cap (0,1^d))$ if $\partial \Omega$ is sufficiently smooth. And $$ \|\chi_\Omega - \mbox{vol}(\Omega)\|_p = \bigl((1 - \mbox{vol}(\Omega))^p \mbox{vol}(\Omega) + \mbox{vol}(\Omega)^p (1 - \mbox{vol}(\Omega))\bigr)^{1/p} \ge \frac{1}{2} \mbox{vol}(\Omega)^{1/p} $$ since $\mbox{vol}(\Omega) \le \frac{1}{2}$.<|endoftext|> TITLE: Research topics in Microlocal Analysis QUESTION [7 upvotes]: Before asking this question here I did some research on web but I would like to get the opinion of those directly interested if there are any , (as I did in this thread Research topics in distribution theory since a good connection exists). I have seen that the applications and/or research topics are varied. I also tried to see more about mathematical analysis, but every time I am more passionate about these topics. I also mention this thread for applications: Applications of Microlocal Analysis? For example, even if the question I'm doing is quite generic, these theories that are mostly applied to the study of linear PDEs, as can be applied in non-linear PDE contexts (I read that they are very important for this case too very difficult). The non-linear case techniques on what are based? Clearly I refer for example to the existence and uniqueness of the solution or regularity. Another thing to add, I have read of applications in various classification of PDEs, ellipticals obviously (fairly well known), but also hyperbolic or parabolic. I would be grateful for every answer. Clearly any reference is welcome REPLY [5 votes]: I can give a brief description of a (perhaps, currently the main) topic pertaining the application of microlocal analysis to the study of nonlinear PDEs, which seem the argument you are most interested to, since it is also the one to which I am more accustomed, having been interested in it many years ago: to my knowledge, the main (if not the only) method used to extend this analysis to nonlinear PDEs, is the use of paradifferential operators and paraproducts (introduced by Jean-Michel Bony), so let's sketch how this approach works. Notation I am taking the monograph [2] as a reference, and assume as known, at least formally, the basic linear theory of pseudodifferential operators calculus. As it is customary when dealing with $\psi$DOs, we have $$ \begin{split} D_k&\overset{\text{Def}}{=}\frac{1}{i}\frac{\partial}{\partial x_k}=\frac{1}{i}\partial_{x_k}\quad k=1,\ldots,n\\ D&\overset{\text{Def}}{=}\frac{1}{i}\left(\frac{\partial}{\partial x_1},\cdots,\frac{\partial}{\partial x_n}\right)\\ D^\beta&\overset{\text{Def}}{=}\left(\frac{1}{i}\frac{\partial}{\partial x_1}\right)^{\beta_1}\!\!\!\!\cdot\ldots\cdot\left(\frac{1}{i}\frac{\partial}{\partial x_n}\right)^{\beta_n}\quad\beta=(\beta_1,\ldots,\beta_n)\in\Bbb N^n\\ \end{split} $$ Introduction Microlocal analysis is, roughly speaking, the analysis of the wavefront set ${W\!F}(u)$ of a distribution $u\in \mathscr{D}^\prime(\Omega)$, $\Omega\subseteq\Bbb R^n$ which solves, in the standard case, a given linear PDE (or $\psi$DE). This is generically achieved with the use of the instruments of pseudodifferental calculus (parametrices, composition operators and commutator formulas) and a priori estimates. Applying this machinery to nonlinear problems implies being able to represent nonlinear operators in terms of standard pseudodifferential operators: this is the approach taken by Bony, Meyer and their pupils. Paradifferential operators and paraproducts Consider a $C^\infty$ nonlinear function $F:\Bbb R\to\Bbb R$ (just to focus on the basic ideas: the method is well suited also for vector functions), a function $u\in C^r(\Bbb R^n)$ and a dyadic partition of unity of $\Bbb R^n$, $\{\psi_j(\xi)\}_{j\in \Bbb N}$ (see Taylor [2], §1.3 pp. 41-42, for the details): then we can define a countable family constant coefficients pseudodifferential operators $\{\psi_j(D)\}_{j\in \Bbb N}$ and the associated family $\{\Psi_j(D)\}_{j\in \Bbb N}$ $$ \Psi_j(D)=\sum_{i\le j}\psi_i(D)\quad j\in \Bbb N $$ Applying it to $u$ we obtain $$ u_j=\Psi_j(D)u\implies u(x)=\sum_{j=0}^\infty \psi_j(D)u(x)=\lim_{j\to\infty}u_j(x) $$ and $$ F(u)=F(u_0)+\big[F(u_1)-F(u_0)\big]+\big[F(u_2)-F(u_1)\big]+\cdots+\big[F(u_{k+1})-F(u_k)\big]+\cdots $$ Now we have that $$ \begin{split} F(u_{k+1})-F(u_k)&= F\big(u_k+\psi_{k+1}(D)u\big)-F(u_k)\\ &=m_k(x)\psi_k(D)u \end{split}\quad\forall k\in\Bbb N $$ where $$ m_k(x)=\int_0^1\!\!\! F^\prime\big(u_k+t\psi_{k+1}(D)u\big)\mathrm{d}t = \int_0^1\!\!\! F^\prime\big(\Psi_j(D)u+t\psi_{k+1}(D)u\big)\mathrm{d}t\quad\forall k\in\Bbb N $$ Thus we have obtained the following decomposition $$ F(u) = F(u_0)+\sum_{k=1}^\infty m_k(x)\psi_k(D)u = M_F(x,D)u + R(u)\label{1}\tag{1} $$ where $M_F(x,D)u\overset{\text{Def}}{=} \sum_{k=1}^\infty m_k(x)\psi_k(D)u$ is a paradifferential operator, which is a particular kind of $\psi$DO by construction. $R(u)=F(u_0)$ is a remainder term, which is $C^\infty$-smooth by construction, since each $\psi_j(\xi)$, $j\in \Bbb N$, is a compactly supported $C^\infty$-function. We have thus obtained a general nonlinear function composition as the sum of a (infinitely) smooth remainder term $R$ and of a paradifferential operator: now, considering a nonlinear partial differential equation $$ F(x,u,\ldots,\partial^\beta u,\ldots)_{|\beta|\le m}=f\quad m\in\Bbb N\label{nlpde}\tag{NLPDE} $$ we can proceed in a similar way as for \eqref{1} and obtain the following formula $$ \begin{split} F(x,u,\ldots,\partial^\beta u,\ldots)_{|\beta|\le m}&=\sum_{|\beta|\le m} M_{\partial_{\partial^\beta u}F}(x,D)\partial_\beta u + F(x,u_0,\ldots,\partial^\beta u_0,\ldots)\\ &=\sum_{|\beta|\le m} M_{\partial_{\partial^\beta u}F}(x,D)\partial_\beta u + R(x,u_0,\ldots,\partial^\beta u_0,\ldots) \end{split}\label{2}\tag{2} $$ Thus again we have obtained an expansion expressing a nonlinear PDE as the sum of a finite number of paradifferential operators and a $C^\infty$-smooth remainder term: then you can apply to formula \eqref{2} the machinery of microlocal analysis developed for the linear theory and get a priori estimates and information on the structure of the wavefront set $W\!F(u)$. The paraproduct is instead a linear operator related to the operator $M_F(x,D)$ in formula \eqref{1}. There are several versions of it: one is the following $$ \pi(a,u)=\sum_{j\ge 1}^\infty \big(\Psi_{j-1}(D)a\big)\big(\psi_{j+1}(D)u\big) \quad a, u\in C^r\label{3}\tag{3} $$ Apart from its elementary properties, \eqref{3} can be used in approximating $M_F(x,D)u$ and obtain a formula analogous to \eqref{1} (see [2], p. 76, proposition 3.2.C) $$ F(u)=\pi\big(F^\prime(u),u\big)+R(u)\label{4}\tag{4} $$ where $R$ belongs now to an appropriate Sobolev space. The analogies between this formula and \eqref{1} can be used in the study of \eqref{nlpde} to obtain from \eqref{4} a representation similar to \eqref{2} and study it by the techniques of microlocal analysis. Not being an expert in microlocal analysis nor having been involved in studying the subject recently, I address to the monographs [1] and [2] for the details of this sketchy answer. Notes The book [1] is a nice and well organized illustration of known results in the microlocal analysis of nonlinear PDEs, and exposes many results on the Navier-Stokes system including the ones obtained by Marco Cannone, who proved an important global existence theorem for the solution of this system of PDEs, assuming the Cauchy data belongs to an appropriate Besov space. See also the MathOverflow question:-"Why is paraproduct or paradifferential calculus important in PDE theory? In particular, the original paper of J-M. Bony linked in the answer to that question is very clear and informative. Bibliography [1] Hajer Bahouri, Jean-Yves Chemin, Raphaël Danchin (2011), Fourier Analysis and Nonlinear Partial Differential Equations, Grundlehren der mathematischen Wissenschaften, vol. 343, Berlin-Heidelberg-Dordrecht-London-New York: Springer-Verlag, pp. XVI+524, DOI 10.1007/978-3-642-16830-7, ISBN 978-3-642-16829-1, e-ISBN 978-3-642-16830-7, MR2768550, Zbl 1227.35004. [2] Michael Eugene Taylor (1991), Pseudodifferential operators and nonlinear PDE, Progress in Mathematics, vol. 100. Boston, MA-Basel-Berlin: Birkhäuser Verlag, pp. 213, ISBN: 0-8176-3595-5, MR1121019, Zbl 0746.35062.<|endoftext|> TITLE: Reference requests: Integral cohomology of $G_2$-homogeneous spaces QUESTION [11 upvotes]: Do you know a place where the integral cohomology of $G_2$-homogeneous spaces is computed? Great computational efforts using representation theory in order to determine the characteristic classes of homogeneous spaces were done by Borel and Hirzebruch in a series of papers: Characteristic classes and homogeneous spaces. I. Amer. J. Math. 80 1958, 491–504. Characteristic classes and homogeneous spaces. II. Amer. J. Math. 81 1959 315–382. Characteristic classes and homogeneous spaces. III. Amer. J. Math. 82 1960 458–538. I was unable to find there a systematic answer to the question. REPLY [4 votes]: I don't know how "systematic" the answer you are looking has to be, but for the quotients of the form $G_2/T$ and $G_2/P$ (P: Parabolic subgroup) you can find the results in Schubert presentation of the integral cohomology ring of the flag manifolds G/T, by Haibao Duan and Xuezhi Zhao<|endoftext|> TITLE: Non-equivalent eulerian trails in $K_{2n+1}$ QUESTION [6 upvotes]: Two eulerian trails of $K_{2n+1}$ are defined to be equivalent if the orientations obtained by orienting the edges as traversed by the trails are isomorphic as digraph. How many non-equivalent trails are there? REPLY [4 votes]: If I understand your question, you are asking for the number of isomorphism classes of regular tournaments. There are no exact formulas. See http://oeis.org/A096368 for counts up to 15 vertices. The asymptotic number is known. It is $RT(2n+1)/(2n+1)!$, where $RT(2n+1)$ is given in the abstract of this paper.<|endoftext|> TITLE: Diophantine equation $3^a+1=3^b+5^c$ QUESTION [11 upvotes]: This is not a research problem, but challenging enough that I've decided to post it in here: Determine all triples $(a,b,c)$ of non-negative integers, satisfying $$ 1+3^a = 3^b+5^c. $$ REPLY [29 votes]: I can't resist this: The young Chris Skinner showed that if $a$, $b$, $c$, $d$ are fixed positive integers, and $p$ and $q$ are positive coprime integers then the equation $$ ap^x + bq^y = c+ dp^z q^w $$ has a bounded number of solutions in $(x,y,z,w)$ and that a bound on these could be computed (and the equation solved in practice). This solves (in principle) the more general equation $1+3^a 5^d = 3^b+ 5^c$. Anyway, there is a large literature around such exponential diophantine equations, and Skinner's paper will give some references. Following a reference from Skinner's paper, Theorem 4.01 of Brenner and Foster gives an explicit treatment of the equation $$ 3^a + 7^b=3^c+5^d, $$ which completely resolves the problem in this question (take $b=0$). Their proof is elementary, and the only non-trivial solution to the equation in the question is $3^3+1 = 3 + 5^2$.<|endoftext|> TITLE: Maximum rotation made by a symmetric positive definite matrix? QUESTION [10 upvotes]: Say we have some symmetric positive definite $n\times n$ matrix $M$ with $n$ distinct eigenvalues $\{\lambda_1,...,\lambda_n\}$. Is there a general formula for the maximum angle $\theta$ for which $M$ can rotate some vector, in terms of matrix invariants? I worked out the $2\times 2$ case and the answer is $$\theta=\text{arccos}\Big(2\sqrt{\frac{\text{det}M}{(\text{tr}M)^2}}\Big)$$ In general, the answer is $$\theta=\text{arccos}\Bigg(\min_{v\neq0}\frac{v^TMv}{||v||\cdot||Mv||}\Bigg)$$ However I would like to find an answer analogous to the $2\times 2$ case for the general $n\times n$ case. In trying to work out the $3\times 3$ case, the minimization procedure became extremely complicated. The best I could do was the case where two of the eigenvalues are equal $\lambda_1,\lambda_2,\lambda_2$. In that case, the angle turns out to be the same angle you would calculate for a $2\times 2$ matrix with the same eigenvalues $\lambda_1,\lambda_2$. In fact, upon further analysis it should be true that for an $n\times n$ matrix with $k$ distinct eigenvalues, the formula for $\theta$ is equal to that which you would get from a $k\times k$ matrix with the corresponding eigenvalues. Edit: Using the above fact, my guess for the $3\times 3$ case with $3$ distinct eigenvalues is $$\theta=\text{arccos}\Bigg(2\sqrt{\frac{6\text{det}M}{(\text{tr}M)^3-\text{tr}M^3}}\Bigg)$$ And my guess for the $4\times4$ case is $$\theta=\text{arccos}\Bigg(2\sqrt{\frac{48\text{det}M}{(\text{tr}M)^4-3(\text{tr}M^2)^2-4\text{tr}M\text{tr}M^3+6\text{tr}M^4}}\Bigg)$$ Technically there are $6$ possible solutions to the $3\times3$ case, however this is the only solution with rational coefficients on the traces. The $4\times4$ solution shown is also the only solution whose coefficients are rational. In general, for an $n\times n$ matrix there are at most $\prod_{k=3}^{n}2\times p(k)$ possible solutions based on the statement just above the edit. Where $p(k)$ is the partition function. These guesses are based on the assumption that you can write the solution as a ratio of linear combinations of power traces, for which each terms power sums to $n$. Based on Carlo Beenakker's answer this is an incorrect assumption. REPLY [20 votes]: As explained in these notes, the maximum rotation angle $\theta$ of a symmetric positive definite matrix $M$ is related to the condition number $K=\mu_{\rm max}/\mu_{\rm min}$ of the matrix (the ratio of largest and smallest eigenvalue) by $$K=\frac{1+\sin\theta}{1-\sin\theta}\Leftrightarrow\cos\theta=\frac{2\sqrt{K}}{{1+K}}.$$ For $n=2$ this reduces to the first equation in the OP. The formulas for $\cos\theta$ in the OP for $n=3,4$ do not agree with the above. I just noticed a similar answer at MSE.<|endoftext|> TITLE: pair of injective morphisms of simplicial groups QUESTION [5 upvotes]: Let $X$ and $Y$ be two simplicial groups such that there exists $f:X\rightarrow Y$ and $g:Y\rightarrow X $ two injective morphisms of simplicial groups. Suppose that $X$ is contractible, does it follow that $Y$ is contractible ? REPLY [2 votes]: Pick pointed topological spaces $A$ and $B$ which admit pointed injective continuous maps $A \to B$ and $B \to A$ for which $A$ is contractible but $B$ has nonvanishing reduced homology. For example, we could take $A = [0,1]$ and $B = [0,1] \cup \{2\}$ pointed at $0$. Now apply the functor $\mathbb{Z}[\mathrm{Sing}(-)]/\mathbb{Z}[\mathrm{Sing}(*)]$ to obtain simplicial (abelian) groups $X$ and $Y$ and maps $X \to Y$ and $Y \to X$, which are still injective. We have $\pi_*(X) = \tilde H_*(A)$ and $\pi_*(Y) = \tilde H_*(B)$, so $X$ is contractible but $Y$ is not.<|endoftext|> TITLE: Volume comparison on Grassmannian QUESTION [5 upvotes]: Let $G(r,n-r)$ be the Grassmannian, which can be identified with the space of all rank $r$ projection matrices in $\mathbb{R}^n$. Let $\mu$ be the uniform measure on $G(r,n-r)$. For any $\lambda_1,...\lambda_r>0$ and $\sum_{i=1}^r\lambda_i=1$, we can define a Pseudo distance metric $d_\Lambda$ on $G(r,n-r)$: $$d_\Lambda(P,Q)=\sqrt{trace\left\{(P-Q)\Lambda(P-Q)\right\}}$$ where $$\Lambda=diag\{\lambda_1,...,\lambda_r,0,0,...,0\}$$ Define the Balls in $G(r,n-r)$ centered at $I_{r,n-r}$ with respect to $d_\Lambda$ with radius $a$ as: $$B_a(\Lambda)=\{P\in G(r,n-r): d_\Lambda(P,I_{r,n-r})\leq a\}$$ where $I_{r,n-r}$ is the diagonal matrix putting $r$ 1s on the upper left corner and $n-r$ 0s on the bottom right corner. Note that $d_\Lambda(P,I_{r,n-r})=0$ implies $P=I_{r,n-r}$. I'm trying to prove the following conjecture: for any $1>a>b>0, a/b>100$, there exists a constant $C>0$, independent of $\Lambda$, such that $$\frac{\mu(B_a(\Lambda))}{\mu(B_b(\Lambda))}\leq\left(\frac{Ca}{b}\right)^{r(n-r)}$$ I'm trying to find a way to use Bishop-Gromov inequality. However, $d_\Lambda$ is not the geodesic distance metric. This is the tricky part. If we upper bound the nominator and lower bound the denominator separately by geodesic balls on the Grassmannian, the constant $C$ will depend on $\Lambda$, which is not desirable. This is like bounding the volume ratio of ellipsoids, the ratio should not depend on the length of axis. Update: I have a very naive idea and do not know if it will lead to anything (with high probability nothing, I'm really new to this field). Construct n dimensional diagonal matrix $\tilde{\Lambda}=diag\{\lambda_1,...,\lambda_r,b^2/r,...,b^2/r\}$. This defines a new metric on the space of all symmetric matrices: $=trace(A\tilde{\Lambda}B)$. Let $\tilde{\mu}$ be the new volume measure on the Grassmannian with metric induced from the new metric defined above. Then: $$\frac{\mu(B_a(\Lambda))}{\mu(B_b(\Lambda))}\leq\frac{\mu(B_{a+b}(\tilde{\Lambda}))}{\mu(B_b(\tilde{\Lambda}))}\leq\frac{\mu(B_{1.01a}(\tilde{\Lambda}))}{\mu(B_b(\tilde{\Lambda}))}$$ Now the questions is to relate $\mu$ with $\tilde{\mu}$ and relate $B_{1.01a}(\tilde{\Lambda}),B_b(\tilde{\Lambda})$ with the geodesic ball on the Grassmannian with the newly defined metric. Then we can apply Bishop-Gromov. However, I do not know if this is doable... REPLY [4 votes]: $\DeclareMathOperator{\Gr}{Gr}$ It is convenient to identify $G(r,n-r)$ with the space $\newcommand{\bR}{\mathbb{R}}$ $\Gr_r(\bR^n)$ of $r$-dimensional subspaces of $\bR^n$ via the map $$ \Gr_r(\bR^n)\ni L\mapsto P_L \in G(r,n-r), $$ where $P_L$ is the orthogonal projection on $L$. The operator $I_{r,n-r}$ corresponds to the subspace $L_0=\bR^r\oplus 0\subset \bR^n$. Denote by $\newcommand{\eO}{\mathscr{O}}$ $\eO$ the open subset of $\Gr_r(\bR^n)$ consisting of subspaces that intersect $L_0^\perp=0\oplus \bR^{n-r}$ transversally. $\DeclareMathOperator{\Hom}{Hom}$ The open set $\eO$ can be identified with $\Hom(\bR^r,\bR^{n-r})$ via the graph map $$ \Hom(\bR^r, \bR^{n-r})\ni S\mapsto \Gamma_S\in \eO, $$ $$ \Gamma_S=\big\{(x,Sx)\in\bR^r\times \bR^{n-r}:\;\;x\in\bR^r\big\}. $$ We think of matrices $S\in\Hom(\bR^r,\bR^{n-r})$ as defining coordinates on $\eO$ Using the decomposition $\bR^n=\bR^r\times\bR^{n-r}$ we can describe an $n\times n$ matrix $X$in blocks $$ X=\left[ \begin{array}{cc} A & B\\ C & D \end{array} \right], $$ where $A$ is $r\times r$, $D$ is $(n-r)\times (n-r)$ etc. For $S\in\Hom(\bR^r,\bR^{n-r})$ (or equivalently a $(n-r)\times r$ matrix) the orthogonal projection onto $\Gamma_S$ has the block decomposition (see Eq. (1.2.5) p.17 of this book) $$ P_{\Gamma_S}=\left[ \begin{array}{cc} (1+S^*S)^{-1} & (1+S^*S)^{-1}S^*\\ S (1+S^*S)^{-1} & S (1+S^*S)^{-1} S^* \end{array} \right]. $$ For simplicity we set $W(S):=(1+S^*S)^{-1}\in\Hom(\bR^r,\bR^r)$. The plane $L_0=\bR^r\oplus 0$ has coordinate $S_0=0$ and $P_{L_0}=I_{r, n-r}$. For $L\in \eO$ with coordinates $S=S(L)$ we have $$ P_L-P_{L_0}= . \left[ \begin{array}{cc} W(S)-1 & W(S)S^*\\ S W(S) & S W(S) S^* \end{array} \right] $$ If we denote by $d_\Lambda$ the metric you defined. Then, setting $W=W(S)$, we get$\DeclareMathOperator{\tr}{tr}$ $$ d_\Lambda(L,L_0)^2=\tr (W-1)\Lambda(W-1) +\tr (SW\Lambda WS^*) $$ $$ =\tr(\Lambda(W-1)^2)+\tr(\Lambda WS^*SW) $$ ($S^*S$ commutes with $W(S)$) $$ =\tr(\Lambda(W-1)^2)+\tr(\Lambda W^2S^*S) $$ $$ =\tr\Lambda \big(\; (W-1)^2+W^2 S^*S\;\big) $$ For simplicity we set $R=R(S)=S^*S$ Thus $$ d_\Lambda(L,L_0)^2=\tr\Lambda \big(\; (W-1)^2+W^2 R\;\big). $$ Note that $$ W-1=R(1+R)^{-1}=WR,\;\;(W-1)^2=W^2R^2, $$ $$(W-1)^2+W^2R=W^2(1+R)R=WR. $$ Hence $$ d_\Lambda(L,L_0)^2=\tr\Lambda R W=\tr \Lambda R \big(1+R\big)^{-1},\;\;R=R\big(S(L)). $$ Let us see how this looks, when $r=m$, $n=n+1$. Then $S$ is a $1\times m$ matrix $$ S=(s_1,\dotsc,s_m) $$ Then $$ R(1+R)^{-1}= R- R^2+ O(\Vert R\Vert^3) $$ Denote by $E_a(\Lambda)$ the ellipsoid $$ \big\{S\in\bR^m;\;\;\tr \Lambda R(S)\leq a^2\big\}=\big\{S\in\bR^m:\;\;\lambda_1s_1^2+\cdots+\lambda_ms_m^2\leq a^2\big\}. $$ Then $$ \lim_{a\searrow 0}\frac{\mu\big( B_a(\Lambda)\big)}{ \mu\big( E_a(\Lambda)\big)}=1,\;\;\forall \Lambda,\;\;\prod_i\lambda_i>0. \tag{1} $$ The proof uses the change in coordinates $$ S=T\Lambda^{-\frac{1}{2}} ,\;\;T=(t_1,\dotsc, t_m), $$ so $$ R(S)= \Lambda^{-\frac{1}{2}}R(T)\Lambda^{-\frac{1}{2}}. $$ If the convergence in (1) were uniform in $\Lambda$, it would confirm your claim in the case $r=n-1$ and supports your initial intuitive comparison with ellipsoids. Perhaps it would be good to look at the case $m=2$ first. In this case $$ R(S)=\left[ \begin{array}{cc}s_1^2 & s_1s_2\\ s_1s_2 & s_2^2 \end{array} \right],\;\; W(S)=\frac{1}{1+s_1^2+s^2} \left[\begin{array}{cc}1+s_2^2 & -s_1s_2\\ -s_1s_2 & 1+s_1^2 \end{array} \right]. $$ $$ RW= \frac{1}{1+s_1^2+s^2} \left[\begin{array}{cc} s_1^2 &\ast\\ \ast & s_2^2 \end{array} \right] $$ $$ \tr(\Lambda RW)=\frac{\lambda_1s_1^2+\lambda_2s_2^2}{1+s_1^2+s^2},\;\;\lambda_1+\lambda_2=1. $$ As $\lambda_1\searrow 0$ the ball $B_a(\Lambda)$ becomes unbounded in the $S$ coordinates and $\mu$ can no longer be approximated by the Euclidean volume. The precise form of the invariant volume on $\Gr_r(\bR^m)$ is described in Sec. 9.1.2 of the same notes linked above. The computations in the case $r=2$, $n=3$ are eminently doable and you can test your hypothesis in this case. Remark Suppose that $m=2$ and $a^2=\lambda_1$. Here $a$ is meant to be small. Then $$ B_a(\Lambda)=\big\{ (a_1,s_2)\in\bR^2;\;\;(1-a^2)s_2^2\leq a^2\big\} $$ so that in the $(s_1,s_2)$ coordinates $B_a(\Lambda)$ is, up to a $\mu$-negligible set, the strip $$ |s_2|\leq \frac{a}{\sqrt{1-a^2}}. $$ For $b^2=1-a^2$ the region $B_b(\Lambda)\cap \eO$, $\lambda_1=a^2$ has the description $$ a^2s_1^2+(1-a^2)s_2^2)\leq (1-a^2)(1+s_1^2+s_2^2) $$ i.e., $$ (1-a^2)+(1-2a^2)s_1^2\geq 0. $$ Thus $B_b(\Lambda\cap \eO=\eO$ so $$ \mu(B_b(\Lambda))= \mu(B_b(\Lambda)\cap \eO)= \mu(\eO)=\mu(\Gr_m(\bR^{m+1}). $$ If your estimate were to hold uniformly we would have $$ \mu(B_{\sqrt{\lambda_1}}(\Lambda))=O(\lambda_1). $$ Given that $B_{\sqrt{\lambda_1}}(\Lambda)$ is the strip $|s_1|\leq \sqrt{\lambda_1}$ its volume is $\geq const.\sqrt{\lambda_1})$. To see this consider the thin rectangle $$ D_{\lambda_1}=\{|s_1|\leq 1,\;\;|s_2|\leq \sqrt{\lambda_1}\}. $$ We write $\mu$ ast $$ \mu=\rho(s_1,s_2) $$ and we observe that there exists $C_\mu>0$, independent of $\Lambda$ such that $$ \rho(s_1,s_2)\geq C_\mu,\;\;\forall |s_1|,|s_2|\leq 1. $$ Then $$ \mu(B_{\sqrt{\lambda_1}}(\Lambda))\geq \mu(D_{\lambda_1})\geq C_\mu\sqrt{\lambda_1}. $$<|endoftext|> TITLE: Why these surprising proportionalities of integrals involving odd zeta values? QUESTION [22 upvotes]: Inspired by the well known $$\int_0^1\frac{\ln(1-x)\ln x}x\mathrm dx=\zeta(3)$$ and the integral given here (writing $\zeta_r:=\zeta(r)$ for easier reading)$$\int_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx=12\zeta(5)-\pi^2\zeta(3)=6(2\zeta_5-\zeta_3\zeta_2),$$ I have looked at the integrals $$I_{n,m}=:\int_0^1\frac{\ln^n(1-x)\ln^m x}x\mathrm dx$$ for $m,n\in\mathbb N$. From the power series of $\ln(1-x)$ and recursion, it is not hard to see that $I_{m,n}$ can be written as a rational combination of products of integer zeta values, with the arguments ($\ge 2$) for each product summing up to $m+n+1$, e.g. $$I_{n,1}=\begin{cases} -n!\Bigl(\dfrac{n-1}4\zeta_{n+2}-\dfrac12\sum\limits_{j=1}^{k-1}\zeta_{2j+1}\zeta_{n+1-2j} \Bigr)\ \ &\text{ for }n=2k\\ -n!\Bigl(\dfrac{n+1}2\zeta_{n+2}- \sum\limits_{j=1}^{k-1}\zeta_{2j+1}\zeta_{n+1-2j}\Bigr)\ \ &\text{ for }n=2k-1 \end{cases}.$$ So far, so good. Now the intriguing (numerical) discovery is that there are several families of pairs of those integrals that have rational ratios. For instance, $$\frac{I_{n+2,n-2}}{I_{n-1,n+1}}=\frac{n+2}{n-1}\quad \text{or}\quad\frac{I_{n+1,n-1}}{I_{n,n}}=\frac{n+1}{n}\quad \text{or}\quad\frac{I_{n,2}}{I_{3,n-1}}=\frac{n}{3},$$ e.g. $$I_{5,2}=\color{red}{10}\color{blue}{(61\zeta_8-72\zeta_5\zeta_3+12\zeta_3^2\zeta_2)} $$ and $$I_{3,4}=\color{red}6\color{blue}{(61\zeta_8-72\zeta_5\zeta_3+12\zeta_3^2\zeta_2)}. $$ Note that each such pair has the same proportion as the respective first arguments. I don't see how that could be possibly proven by recursion, so there must be some deeper connection between those integrals. How to explain that? REPLY [37 votes]: For $n\geq 1$ and $m\geq 0$, an application of integration by parts ($u=\log^n(1-x)$, $dv=\log^m(x)\,dx/x$) followed by the substitution $x\mapsto 1-x$ shows that $$ \frac{I_{n,m}}{I_{m+1,n-1}}=\frac{n}{m+1}. $$ All of your examples are special cases of this identity.<|endoftext|> TITLE: Universal homotheties for elliptic curves QUESTION [24 upvotes]: Let $K$ be a number field and $E_1, \cdots, E_n$ elliptic curves over $K$. Let $\ell$ be a prime. Then there exists an element $\sigma \in \text{Gal}(\overline{K}/K)$ such that $\sigma$ acts on $T_\ell(E_i)$ via a non-root-of-unity scalar for all $i$. (Here $T_\ell(E_i)$ denotes the $\ell$-adic Tate module of $E_i$.) This is a result of Bogomolov, who shows that for any Abelian variety $A/K$, there exists an element of $\text{Gal}(\overline{K}/K)$ which acts on $T_\ell(A)$ via a non-root-of-unity scalar; applying this result to $\prod_i E_i$ gives the claim. I would like to know if an analogous statement is true for infinite sets of elliptic curves. In particular: Fix a prime $\ell$. Does there exist an element $\sigma\in \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ such that for all elliptic curves $E/\mathbb{Q}$, $\sigma$ acts on $T_\ell(E)$ via a non-root-of-unity scalar? I am also interested in the following stronger variant: Fix a prime $\ell$. Does there exist an element $\sigma\in \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ such that for all number fields $K\subset \overline{\mathbb{Q}}$ and all elliptic curves $E/K$, there exists an integer $N$ such that $\sigma^N$ acts via a non-root-of-unity scalar on $T_\ell(E)$? Note that the above makes sense because for $N$ sufficiently divisible, $\sigma^N\in \text{Gal}(\overline{\mathbb{Q}}/K)$. Of course this question is related to uniform boundedness conjectures, but my hope is that it can be answered independently from them. REPLY [3 votes]: Fix the prime $\ell$. I hope: There exists a constant $C$ such that for any elliptic curve $E/\mathbb{Q}$ the image of the Galois representation $\rho\colon \operatorname{Gal} ( \bar{\mathbb{Q}}/\mathbb{Q}) \to \operatorname{GL}_2(\mathbb{Z}_{\ell})$ contains $\bigl\{x^C \bigm\vert x\in \mathbb{Z}_{\ell}^{\times}\bigr\}$. Carola Eckstein in her thesis http://cds.cern.ch/record/897530/files/cer-002567978.pdf answers this partially. If $E$ has complex multiplication then $C=54$ will do. If $\ell>163$ then $C=12$ is ok. Agnès David https://arxiv.org/abs/1007.4725 makes further progress and shows that for $\ell>23$ and different from $37$, $43$, $67$ and $163$ then $C=2$ works. I don't know more about the question for smaller $\ell$, but as ulrich hints in his comments above, one should be able to prove the existence of $C$. Suppose there is such a $C$. Given an elliptic curve $E/\mathbb{Q}$, let $K_E$ be the field fixed by the image of $\operatorname{im}(\rho)\to \operatorname{PGL}_2(\mathbb{Z}_\ell)$ (as in ulrich's comment). Set $K_\infty = \mathbb{Q}(\mu_{\ell^{\infty}})$. The image of the homotheties under $\det:\operatorname{im}(\rho) \to \operatorname{Gal}(K_\infty/\mathbb{Q})\cong \mathbb{Z}_{\ell}^{\times}$ will contain all $2C$-th powers. Therefore there is a finite extension $F/\mathbb{Q}$ inside $K_\infty$ such that $K_E\cap K_\infty\subset F$ for all elliptic curves $E$. In particular the compositum $\mathcal{K}$ of any (or all) $K_E$ intersects $K_\infty$ in $F$ and hence any element of $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathcal{K})$ that maps to an element $\sigma$ of infinite order in $\operatorname{Gal}(K_\infty/F)$ will work. The reference above will also show that this works for many $\ell$ over any fixed number field instead of $\mathbb{Q}$. I have no answer for the second question, yet.<|endoftext|> TITLE: Do "seemingly impossible functional programs" work with arrow types interpreted as Turing machines? QUESTION [5 upvotes]: The title is a reference to this article by Martin Escardo, referring to work by originally by Ulrich Berger. It occurred that the programs described in this article can interpreted in the Turing machine model for higher order functions, where an element of type a -> b is encoded by a Turing machine takes encodings of elements of a and outputs encodings of elements of b. I'm wonder whether such programs are correct in this model. More precisely, we say a natural number encodes a Cantor space element if it encodes a Turing machine which for any natural number input halts and returns a binary output (either 0 or 1). There's an equivalence relation $x \sim_C y$ iff $x$ and $y$ both encode Cantor space elements and the corresponding Turing machines return the same outputs for the same inputs. Next, a natural number encodes a predicate on Cantor space if it encodes a Turing machine M such that For any $x$ encoding a Turing machine element $M (x)$ halts and outputs a binary value. For any $x, y$ with $x \sim_C y$ the outputs for $M (x), M (y)$ are the same. Question: Consider the Haskell function forsome implemented in the link above, which depends on find_i (I'm not so interested in the more complicated versions of find described later in the article). Can this program be translated in a Turing maching which takes as input encodings of predicates $p$ on the Cantor, always halts for such inputs and outputs a binary value, and outputs $1$ iff there exists an encoding for a Cantor space element $x$ such that $p (x) = 1$? I believe that translating the program into a Turing machine is (in principle) straightforward, and the question is whether this Turing machine has the desired behavior. Side Question: Suppose the answer to the above question is yes. Does that imply that in the internal logic of the effective topos, the question of whether a decidable subset of the Cantor space is empty is itself decidable? I believe it does, but I'm not confident in my understanding of the semantics of the effective topos. REPLY [9 votes]: $\newcommand{\E}[1]{\mathtt{E}_{#1}}$ You are indeed using the model of computability that corresponds to the effective topos. It is also called hereditarily effective operators (HEO), Type I computability, and Russian constructivism. These things got invented several times. Let us first recall a couple of definitions, and generally set things up in a reasonable way so that we do not have to fiddle with Turing machines. We work constructively, so that everything we say works in the effective topos, as well as in many other toposes (such as classical set theory). For the purposes of this answer, say that a set $X$ is searchable if there exists a map $\E{X} : 2^X \to 2$, where $2 = \{0,1\}$ is the discrete two-point set, such that for all predicates $p : X \to 2$ $$\E{X}(p) = 1 \iff \exists x \in X \,.\, p(x) = 1.$$ We would like to know whether the Cantor space $\mathsf{C} = 2^\mathbb{N}$ is searchable in the effective topos. We do not know how to directly construct the map $\E{\mathsf{C}}$, but we do know how to write it down as a relation, i.e., define $F \subseteq 2^\mathsf{C} \times 2$ by $$F(p, b) \iff ((\exists \alpha \in \mathsf{C} \,.\, p(\alpha) = 1) \iff b = 1).$$ This is a single-valued relation: for every $p \in 2^\mathsf{C}$ there is at most one $b \in 2$ such that $F(p, b)$. Therefore, a priori $F$ yields a partial map $E : 2^\mathsf{C} \rightharpoonup 2$, defined by, $$E(p) = b \iff F(p, b),$$ with the domain of definition $$D = \{ p \in 2^\mathsf{C} \mid \exists b \in 2 \,.\, F(p, b) \}.$$ The map $E$ is our candidate for $\E{\mathsf{C}}$, but we do not know whether its domain of definition $D$ is all of $2^\mathsf{C}$. Recall that the Cantor space is a complete separable metric space, with the (ultra)metric defined by $$d(\alpha, \beta) = \lim_{k \to \infty} 2^{- \min \{i \leq k \mid i = k \lor \alpha_i \neq \beta_i\}}$$ The formula is written this way so that it works constructively (and therefore computably). We have $d(\alpha, \beta) = 2^{-i}$ if $i$ is the least index for which $\alpha_i \neq \beta_i$. The domain $D$ is not small: Proposition 1: $D$ contains the uniformly continuous predicates. Proof. A uniformly continuous predicate $p : \mathsf{C} \to 2$ has a nice tree representation from which it is easy to calculate whether $p$ attains $1$. $\Box$ Proposition 1 explains that the Cantor space is searchable in toposes that validate the statement "every map $\mathsf{C} \to 2$ is uniformly continuous" (because then $2^\mathsf{C} \subseteq D \subseteq 2^\mathsf{C}$). An example is the Kleene-Vesley topos, also known as Type II computability and Brouwerian intuitionism. However, in the effective topos things are strange. Recall that the Baire space $\mathsf{B} = \mathbb{N}^\mathbb{N}$ is also a complete separable metric space, with the metric defined by the same formula above as for the Cantor space. Many questions about Cantor space in the effective topos may be answered using the following theorem. Theorem 2 (effective topos): The Cantor space and the Baire space are homeomorphic as complete metric spaces. The proof relies on the existence of a Kleene tree. This is a strange theorem indeed, since classically the Cantor space is compact and the Baire space is locally non-compact (every open ball contains a sequence without an accumulation point). Proposition 3: If $X$ is searchable and there is a surjection $s : X \to Y$ then $Y$ is searchable. Proof. $\E{Y}(p) = \E{X}(p \circ s)$. $\Box$ Proposition 4: If $\mathbb{N}$ is searchable then LPO holds. Proof. Exercise $\Box$. Theorem 5 (effective topos): The Cantor space is not searchable. Proof. Suppose we had $D = 2^\mathsf{C}$, so that $\mathsf{C}$ is searchable with $\E{\mathsf{C}} = E$. By Theorem 2 there is a surjection $\mathsf{C} \to \mathsf{B}$, therefore $\mathsf{B}$ is searchable. There is also a surjection $\mathsf{B} \to \mathbb{N}$, namely $\alpha \mapsto \alpha(0)$, hence $\mathbb{N}$ is searchable. By Proposition 4 LPO holds. But LPO is invalid in the effective topos, because it implies the existence of a Halting oracle. $\Box$ And lastly, let us explain why it is so easy to get confused about searchability of Cantor space. While we proved that in the effective topos $D$ is a proper subset of $2^\mathsf{C}$, the complement $2^\mathsf{C} \setminus D$ is always empty! Proposition 6: $2^\mathsf{C} \setminus D = \emptyset$. Proof. Notice that $$D = \{ p \in 2^\mathsf{C} \mid (\exists \alpha \in \mathsf{C} \,.\, p(\alpha) = 1) \lor (\forall \alpha \in \mathsf{C}\,.\, p(\alpha) = 0)\}.$$ Suppose we had $q \in 2^\mathsf{C} \setminus D$. Then it follows that $$\lnot ((\exists \alpha \in \mathsf{C} \,.\, q(\alpha) = 1) \lor (\forall \alpha \in \mathsf{C}\,.\, q(\alpha) = 0))$$ which is equivalent to $$(\forall \alpha \in \mathsf{C} \,.\, q(\alpha) = 0) \land \lnot (\forall \alpha \in \mathsf{C}\,.\, q(\alpha) = 0)$$ which is false. $\Box$ The conclusion is strange but not contradictory: we can never produce a concrete predicate $p \in 2^\mathsf{C}$ for which $E$ does not work, while on the other hand in the effective topos $E$ does not work correctly on all of $2^\mathsf{C}$. Translated to classical computability, the same conclusion is perhaps a little bit easier to swallow: every computable predicate on the computable Cantor space either attains $1$ or it does not, but there is no computable procedure for determining whether computable predicates on Cantor space attain $1$.<|endoftext|> TITLE: BV functions and wave equation QUESTION [7 upvotes]: What is the role played by BV functions in the study of (possibly nonlinear) wave equations? I think that one would need assume small initial data in $L^1$ or $H^1$ to get a well-posedness result (is that correct?). Has the case of initial data in BV been studied? REPLY [7 votes]: The answer to this question depends a lot on the space dimension $n$. It is true that if $n=1$, the Cauchy problem has been studied with data in either $L^\infty(R)$ or $BV(R)$. For superlinear wave equation, every $L^\infty$-data yields at least one bounded global-in-time "entropy" solution. This is done by Compensated Compactness (DiPerna 1983). However, nobody knows whether this solution is unique. The $BV$ space is better suited in some sense, because Bressan was able to prove uniqueness of the entropy solution ; however, existence is obtained only if the initial data $u_0$ is not too large, in the sense that $$\|u_0\|_\infty TV(u_0)<\delta$$ for some absolute finite constant $\delta>0$. In some sense, the Cauchy problem is well-posed in $BV$, in a neighbourhood of constant data. In several space dimensions, Rauch remarked that you should forget the $BV$ space. The Cauchy problem cannot be well-posed in this topology. The reason is that the Cauchy problem for the linear wave equation itself is ill-posed in $BV$. This is a consequence of a theorem by Brenner in the 60's, which tells that this Cauchy problem is ill-posed in every $L^p$-space, except for the Hilbert case $p=2$. Brenner's theorem is a bit more complete. It says that for a first-order hyperbolic system $$\partial_tf+\sum_{j=1}^nA_j\partial_jf=0,$$ the Cauchy problem is well-posed in some $L^p$ with $p\ne2$ if, and only if the matrices $A_j$ commutte to each other. This very strong condition amounts to saying that the system can be rewritten as a list of decoupled transport equations. It is interesting to notice that the opposite of Brenner's condition is that the characteristic cone of the differential operator has maximal curvature, hence the Cauchy problem admits a Strichartz-like estimate. You can read the discussion in the 1st chapter of my book co-authored with S. Benzoni-Gavage.<|endoftext|> TITLE: Proving inequality for positive definite matrix QUESTION [6 upvotes]: For a positive definite diagonal matrix $A$, I want to prove that for any $x$: $$\frac{x^T \sqrt{A} x}{\|\sqrt{A}x\|_2} \geq \frac{x^T A x}{\|Ax\|_2}$$ So far I cannot find any counterexamples, and it intuitively makes sense since the $\sqrt{\cdot}$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this. EDIT: changed $>$ to $\geq$ REPLY [14 votes]: Your inequality says $$\frac{\sum\sqrt{\lambda_j}x_j^2}{\left(\sum\lambda_j x_j^2\right)^{1/2}}\geq \frac{\sum\lambda_jx_j^2}{\left(\sum\lambda_j^2x_j^2\right)^{1/2}},$$ or after a simple transformation $$\sum\lambda_j x_j^2\leq\left(\sum\sqrt{\lambda_j}x_j^2\right)^{2/3} \left(\sum\lambda_j^2x_j^2\right)^{1/3}$$ And this is Holder's inequality with $p=3/2$ and $q=3$. The strict inequality does not always hold.<|endoftext|> TITLE: Obstructions to $E_2$-algebra structure on $E_1$-algebra QUESTION [9 upvotes]: Let $A$ be an $E_1$-algebra in chain complexes over $\mathbb Q$. Is there an easy way to check if $A$ admits the structure of an $E_2$-algebra (or $E_\infty$-algebra)? REPLY [5 votes]: This isn't a complete answer, as it would fill a whole paper; and I haven't checked everything. But I think you can use a strategy similar to what is done in the paper The Intrinsic Formality of $E_n$-operads by Fresse and Willwacher. I'm not saying it's easy, but it's systematic. The situation is the following. You have the diagram: $$\require{AMScd} \begin{CD} E_1 @>>> \mathsf{End}_A \\ @V{\iota}VV \\ E_2 \end{CD}$$ and you want to know whether there exists a lift. (Of course there is nothing special about $\mathsf{End}_A$ here, it could be any operad.) This is a two-pronged question: Does there exist a map $E_2 \to \mathsf{End}_A$ at all? Can you find one that makes the diagram commute? (The problem I see with this approach is that the second part depends on the first, but the first could have many different answers...) You can use Bousfield's obstruction theory to answer these question. You want to find a path-component in the mapping space $\operatorname{Map}(E_2, \mathsf{End}_A)$, i.e. you're looking at $\pi_0$. Of course, the very first obstruction is to find a Poisson $2$-algebra structure on $H_*(A)$. If you deal with the resolutions correctly and everything, I believe that such a structure would correspond to a class in the first page of the Bousfield spectral sequence. You want to know whether it survives until the end. Bousfield's obstruction theory should tell you that obstructions live in the cohomotopy groups $\pi^{s+1} \pi_s X$ where $X$ is a certain bicosimplicial space. You want all these obstructions to vanish. Based on the work of Fresse–Willwacher, my (perhaps wild) guess is that you get obstructions indexed by classes in Kontsevich's graph complex $\mathrm{GC}_2$. Now you have your morphism and you want to know if the diagram commutes, at least up to homotopy I guess. In other words, you want to know if the two maps $E_1 \to \mathsf{End}_A$ (the first one is given by the $E_1$-structure, the second by restricting the $E_2$-structure you found through $\iota$) are in the same path component, i.e. they are joined by a path. Relative Bousfield obstruction theory should tell you that obstructions to finding such a path live in $\pi^s \pi_s X$. If you check that they all vanish, this means that your two morphism are equal up to homotopy. Again, my guess is that obstructions should be indexed by the hairy graph complex $\mathrm{HGC}_{12}$.<|endoftext|> TITLE: What is nullification with respect to an Eilenberg-MacLane space? QUESTION [8 upvotes]: By nullification with respect to $K(A,n)$, I mean the Bousfield localization $L_{A,n}$ where a spectrum $E$ is $L_{A,n}$-local if and only if $\tilde E^\ast(K(A,n)) = 0$. Question: Is there a good description of which spectra are $L_{A,n}$-local? Is there a good description of the localization functor $L_{A,n}$? Notes: I'd be happy to understand the case $A = \mathbb Z/p$ when $E$ is also $p$-local. For example, $K(n)$-local spectra are $L_{\mathbb Z/p,m}$-local for $m \geq n+1$. If $E$ is $L_{A,n}$-local, then $E$ is also $L_{A,n+1}$-local. This follows from the Zabrodsky lemma. As a consequence, $HA$ is $L_{A,n}$-acyclic for all $n$. It's a fact (due to Bousfield I think) that $K(\mathbb Z/p, n)$ is always either $E$-local or $E$-acyclic (as a space -- note that $E$-acyclicicity the same whether we consider $K(\mathbb Z/p,n)$ as a space or suspension spectruem, but $E$-locality may be different I think). And I believe any nontrivial homological localization of $p$-local spectra [EDIT: assuming it kills some suspension spectrum -- e.g. not the harmonic localization!] must kill some $K(\mathbb Z/p,n)$. So I'm asking about the "minimally nontrivial" localizations of $p$-local spectra. This follows from the theorem of Bousfield that if a space $X$ is $E$-acyclic, then $K(\pi_n(X), n)$ is $E$-acyclic, and then you can show that if some $K(A,n)$ is $E$-acyclic, so is $K(\mathbb Z/p, n)$ for some $p$. One form of the Sullivan conjecture (proved by Miller says that in the unstable case, every finite-dimensional complex is $L_{A,n}^{unst}$-local for every finite $A$ and $n \geq 1$. A related theorem of Lee says that in the stable case, every finite spectrum is $L_{A,n}$-local for every finite $A$ and $n \geq 2$. Unlike the unstable case, I don't think this extends to finite-dimensional spectra -- e.g. I think it already fails for any infinite wedge of equidimensional spheres. Certainly it fails even for the finite case at $n=1$ by the Segal conjecture (proved by Carlsson). Nick Kuhn points out in his answer below that in the unstable case, Neisendorfer showed the following: Let $X$ be the $p$-completion of a simply-connected finite space with $\pi_2(X)$ finite. Then for all $m \in \mathbb N$, we have $L_{\mathbb Z/p, 1}^{unst} \tau_{\geq m} X = X$. I believe this extends to say that if $X$ is the $p$-completion of an $n$-connected finite space with $\pi_{n+1}(X)$ finite, then for all $m$ we have $L_{\mathbb Z/p, n}^{unst} \tau_{\geq m} X = X$. The stable analog of Neisendorfer's theorem then says that if $L$ is a localization of $p$-complete spectra which kills $H\mathbb Z/p$, then for any spectrum $E$, $\tau_\geq n E \to E$ is an $L$-equivalence for all $n \in \mathbb N$. REPLY [3 votes]: This question is more fun at the space level: see [Neisendorfer, Joseph A. Localization and connected covers of finite complexes. The Čech centennial (Boston, MA, 1993), 385–390, Contemp. Math., 181, Amer. Math. Soc., Providence, RI, 1995.]<|endoftext|> TITLE: Example of a central simple algebra QUESTION [6 upvotes]: Let $A$ be a finite dimensional central simple algebra over a field $F$ of characteristic $0$. So by Weddernburn's theorem, $A\cong M_n(D)$ for some division algebra $D$ over $F$. Let $\dim_F(D)=m^2$. Then $m$ is called the index of $A$. Assume that $A$ is crossed product: there is a finite Galois extension $E$ of $F$, say of degree $n$, which is (isomorphically sitting) inside $A$ and $n$ invertible elements $u_1,\ldots, u_n\in A$ such that $Eu_1\oplus \cdots \oplus Eu_n=A$ (direct sum as $F$-vector space). The product of elements of $A$ from this decomposition are determined by an equivalence class of $2$-cocycle $f$ in $H^2({\rm Gal}(E/F), E^*)$ (together with action of Galois group on $E^*$). Q. In general, it is known that the order of $[f]$ in $H^2({\rm Gal}(E/F), E^*)$ divides the index $m$ of $A$. Further, if $F$ is a number field, then equality holds. So, what is an example where order of $[f]$ properly divides the index $m$ of $A$? The results mentioned in question are taken from Albert's book Structure of Algebras. REPLY [7 votes]: You can take: $F=\mathbb{C}(X_1,Y_1,\ldots,X_n,Y_n)$ and take the tensor product of quaternion algebras $$A=(X_1,Y_1)_F\otimes_F\cdots\otimes (X_n,Y_n)_F.$$ Here $A$ contains a subfield $E$ isomorphic to $F(\sqrt{Y_1},\ldots \sqrt{Y_n})$ for example. Now $A$ is a division algebra, so it has index $2^n$, but it has order $2$ in $\mathrm{Br}(E/F)\simeq H^2(\mathrm{Gal}(E/F),E^\times)$. PS. If you need details, I can post something this evening. Edit. Here is a proof that $A$ is division. I will assume some knowledge on quadratic forms. Note that I replace $\mathbb{Q}$ by $\mathbb{C}$ to get rid of some ennoying coefficients (see below), but the result is the same. If $A$ is a central simple algebra over a field $F$, let $T_A:A\to F$ the quadratic form defined by $T_A(a)=Trd_A(a^2)$. Lemma. If $T_A$ is anisotropic, then $A$ is division. Proof. Assume $A$ is not division, so $A\simeq M_r(D)$, where $D$ is division. The matrix $a_0$ filled with zero, except on the top right corner where the entry is $1_D$, satisfies $a_0^2=0$, hence $T_A(a_0)=0$, while $a_0\neq 0$. Hence, $T_A$ is isotropic. Prop. Let $F$ be a field of characteristic different from $2$. Assume that $q_1,q_2$ are anistropic quadratic forms over $F$. Then the form $q_1+Xq_2$ is anistropic over $F(X)$. Proof. Omitted. This is a standard result. The rest of the proof is devoted to show that the trace form of our example is anisotropic. Easy fact. We have $T_{A\otimes B}\simeq T_A\otimes T_B$ (by standard properties of the reduced trace) Now if $Q=(a,b)_F$, the standard basis $1,i,j,ij$ is orthogonal wrt $T_Q$, and $$T_Q\simeq \langle 2,2a,2b,-2ab\rangle\simeq \langle 1,a,b,ab\rangle\simeq \langle 1,a\rangle\otimes \langle 1,b\rangle.$$ The second isomorphism comes from the fact that $2$ and $-2$ are squares in $\mathbb{C}$; this is why I changed. Hence $T_A\simeq \langle 1,X_1\rangle\otimes \langle 1,Y_1\rangle\otimes\cdots\otimes \langle 1,X_n\rangle\otimes \langle 1,Y_n\rangle$. Using the previous proposition and induction (starting form the fact that $\langle 1\rangle$ is anisotropic), we get the desired result. Of course, you can do all of this for $n=2$...<|endoftext|> TITLE: Discernible Objects in a Topos QUESTION [5 upvotes]: Perhaps an overly elementary question: let $\mathcal{E}$ be a topos and let $X, Y$ be non-isomorphic objects in $\mathcal{E}$. Is it always true that there exists a formula $\phi$ of $\mathcal{E}$'s Mitchell-Benabou language with one free variable which is true of one but not both of $X$ and $Y$ (on the Kripke-Joyal semantics)? If so, what about when $X$ and $Y$ are isomorphic? REPLY [10 votes]: Summary: The answer may depends on the choice of a precise interpretation of the question, which is too vague. But for all the interpretations I can think of the Kripke-Joyal semantics do not distinguishes between objects that are "locally isomorphic". Conversely objects that satisfies the exact same formulas are locally isomorphic simply because the existence of a local isomorphism can be expressed in the Mitchell-Bénabou language. So what you can distinguishes is exactly the equivalence class of object up to the "locally isomorphic" equivalence relation. Let's start with some observations: The Mitchell-Bénabou language of a large topos has a proper class formula. For example for any object $X$ of the topos, "$\exists x \in X$" is a formula in the language of the topos, so there is no set theoretic obstruction for formula to distinguish all objects of a large topos. The question as asked do not quite make sense. Variables of of formula in the Mitchell-Bénabou language are not object of the topos, and object of the topos cannot be substitued to these variables. Objects of the topos corresponds to the type of the language. Examples of formula of the language are things like: $$ \forall x,y \in X, (x= y) \quad\text{ (holds if and only if $X$ is subterminal)}$$ $$ \forall y \in Y, \exists x \in X, f(x)=y \quad \text{ (holds if and only if $f$ is an epimorphism)}$$ $$ \forall x,y \in Y, f(x)=f(y) \Rightarrow x = y \quad \text{ (holds if and only if $f$ is monomorphism)}$$ where $X$ and $Y$ are fixed object of the topos and $f: X \rightarrow Y$ a morphism. As you can see, I can try to substitute $X$ by another object in the first formula and it somehow makes sense, but if I try to do it in the other two, "f(x)" no longer means anything. So it is not quite clear how you want to define this in general. I'll propose a precise meaning of the question below, and answer this precise version of the question. If this is not the meaning you intented, then try to make your question more precise. In the meantime, here are a few general remarks one can make that should be true for any interpretation of the question and that already almost answers it: The Kripke-Joyal semantics is invariant under isomorphisms. Meaning that if I replace every objects appearing in the topos by an isomorphic one, and translate appropriately all the terms and functions in the formula using these isomorphisms this do not modifiy the validity of the formula. Hence isomorphic objects should be not be discernable by any formula whatever that means. The Kripke-Joyal semantics only see "local" properties. I.e. if $\Phi$ is some formula in the language of $\mathcal{T}$. And if $p:Y \twoheadrightarrow 1 $ is a cover of the terminal object of $\mathcal{T}$, then the formula $\Phi$ is valid in the Kripke-Joyal semantics in $\mathcal{T}$, if and only if $p^* \Phi$ is valid in the Kripke-Joyal semantics in $\mathcal{T}_{/Y}$. This suggest that objects that are "locally isomorphic" should satisfies the same formulas as well, though we are already in the area where one needs to make the question precise as it is not clear what it means that they "satisfies the same formulas", but I can't think of a way to make sense of the question that will be able to distinguish locally isomorphic objects. Of course there are plenty of example of objects of a topos that are locally isomorphic without being isomorphic: any locally trivial structure on a space (like a vector bundle) is locally isomorphic to a trivial one, in the topos of $G$-sets, any $G$-set $X$ is locally isomorphic to $X$ endowed with the trivial $G$-action. The interpretation of the question I will give below will have this property, and it will distinguishes exactly objects that are not locally isomorphic. Here is my proposed interpretation. The goal is to make sense of formulas where one can substitute an abstract type symbol $\mathbb{O}$ by an arbitrary object $X$ of $\mathcal{T}$. For this, I consider the "Free topos $\mathcal{T} \{ \mathbb{O}\}$ over $\mathcal{T}$". Here I am not talking of the object classier of $\mathcal{T}$, but freeness is in the sense of logical morphisms (so that the Mitchell-Bénabou language will be preserved). More precisely I want $\mathcal{T} \{ \mathbb{O}\}$ to have the following universal property: For any other elementary topos $\mathcal{E}$, a logical morphism $\mathcal{T} \{ \mathbb{O}\} \rightarrow \mathcal{E}$ is given by, a logical morphism $\mathcal{T} \rightarrow \mathcal{E}$ together with the choice if an object $X \in \mathcal{E}$ (the image of $\mathbb{O}$). The existence of $\mathcal{T} \{ \mathbb{O}\}$ is clear if $\mathcal{T}$ is small (because 'elementary topos' is a quasi-algebrique theory). If your topos is not small... justs take a larger universe. I'm interpreting a formula in "one free object parameters" as a formula in the Mitchell-Bénabou language of $\mathcal{T} \{ \mathbb{O}\}$. Given $\Phi$ such a formula, and $X$ an object of $\mathcal{T}$, I have a unique logical morphism $ev_X: \mathcal{T} \{ \mathbb{O}\} \rightarrow \mathcal{T}$, I'm calling $\Phi(X)$ the image of the formula $\Phi$ by this logical morphism $ev_X$. The two observation above applies immediately to this interpreation: If $X \simeq X'$ then I have a natural isomorphism $ev_X \simeq ev_{X'}$ and the validity of the formula $\Phi(X)$ and $\Phi(X')$ will be equivalent. Let $X$ and $Y$ be two objects that are locally isomorphic, i.e. become isomorphic in some étale cover $\mathcal{T}_{/Z} \rightarrow \mathcal{T}$ (for $Z \twoheadrightarrow 1$ a cover). It means that their image by the logical morphism $\mathcal{T} \rightarrow \mathcal{T}_{/Z}$ are isomorphic. It means that the two composites $$ \mathcal{T}\{\mathbb{O}\} \rightrightarrows \mathcal{T} \rightarrow \mathcal{T}_{/Z} $$ are isomorphic, hence the validity of the image of $\Phi$ by these is equivalent. As the second morphism is conservatif (and preserves interpretation of formula), the validity of $\Phi(X)$ in the Kripke-Joyal semantics in $\mathcal{T}$ is equivalent to the validty of $\Phi(Y)$. So one cannot distinguish between locally isomorphic objects. Now, given any fixed object $Y \in \mathcal{T}$ consider the following formula the Mitchell-Bénabou Language of $\mathcal{T}\{\mathbb{O}\}$ : $$\Phi_Y = \exists f: Y \rightarrow \mathbb{O}, \exists g: \mathbb{O} \rightarrow Y,( \forall x : \mathbb{O}, f(g(x)) =x \text{ and } \forall y : Y, g(f(y))=y) $$ Then $\Phi_Y (X)$ is exaclty the formula " there exists an isomorphism between $X$ and $Y$", which holds in the Kripke-Joyal semantics if and only if $X$ and $Y$ are locally isomorphic. Hence if $X$ and $Y$ satisifes the same formula, one has that $X$ satisfies $\Phi_Y(X)$ and hence $X$ and $Y$ are locally isomorphic. An alternative to make sense of it would be to consider the extention of Mitchell-Bénabou language and the Kripke-Joyal Sémantics introduced by M.Shulman which allows to have object as variable and quantifying on them. I believe it provides a larger class of formula than what I've done here, but it has the same properties of only seeing local properties, so the final answer will be the same.<|endoftext|> TITLE: Two definitions of horofunction for Gromov hyperbolic spaces QUESTION [8 upvotes]: Let $X$ be a proper, geodesic, $\delta$-hyperbolic metric space (e.g. a hyperbolic group), and let $x_0$ be a basepoint for $X$. There seem to be two different definitions of "horofunction" for $X$, and I'd like to understand the relationship between them. First Definition Definition 1. For each $p\in X$ let $f_p\colon X\to\mathbb{R}$ be the function $$ f_p(x) = d(x,p)-d(x_0,p). $$ A function $f\colon X\to \mathbb{R}$ is called a horofunction if there exists an unbounded sequence $\{p_n\}$ in $X$ such that $f_{p_n}$ converges uniformly to $f$ on compact sets. This definition is due to Gromov, and the set of all horofunctions on $X$ is known as the horofunction boundary. Note that this definition works for any metric space. Second Definition The following definition seems to come out of the work of Coornaert and Papadopoulos on the symbolic dynamics of the visual boundary of a hyperbolic group, though it is similar to the "local" description of horofunctions using cocycles given by Gromov in his essay on hyperbolic groups. Definition 2. A function $f\colon X\to \mathbb{R}$ with $f(x_0)=0$ is called a horofunction if it satisfies the following conditions: There exists an $\epsilon>0$ so that $f$ is $\epsilon$-convex, in the sense that $$ f(\gamma_t)\leq (1-t)f(\gamma_0) + t f(\gamma_1) + \epsilon $$ for every constant-speed geodesic $\gamma\colon [0,1]\to X$. The function $f$ is distance-like, in the sense that $$ f(x) = \lambda + d\bigl(x, f^{-1}(\lambda)\bigr) $$ for every $x\in X$ and every $\lambda\in (-\infty,f(x)]$. My Question What, exactly, is the relationship between these two definitions? Are they equivalent? Is the second a generalization of the first? I'd particularly appreciate a reference to a paper that discusses both definitions. REPLY [3 votes]: I don't know whether this was known before, but Collin Bleak, Francesco Matucci, and I have settled this question in the course of our work on our recent paper [1]. The answer is that any horofunction satisfying Definition 1 satisfies Definition 2, but there exist hyperbolic groups with horofunctions satisfying Definition 2 and not Definition 1. Definition 1 implies Definition 2 This is based on the following claim. Claim. Each of the functions $f_p$ is $2\delta$-convex. Proof: Let $\gamma\colon [0,1]\to X$ be a constant-speed geodesic of length $L$, and let $a=\gamma(0)$ and $b=\gamma(1)$. Let $t\in [0,1]$, and choose geodesics $[p,a]$ and $[p,b]$ from $p$ to $a$ and $b$, respectively. Since $X$ is $\delta$-hyperbolic, there exists a point $q$ on $[p,a]\cup [p,b]$ so that $d(q,\gamma_t)\leq \delta$. If $q\in [p,a]$ then \begin{multline*} d(\gamma_t,p) \leq d(q,p)+\delta = d(a,p) - d(a,q) + \delta \\ \leq d(a,p) - d(a,\gamma_t) + 2\delta = d(a,p)-tL+2\delta \end{multline*} so $f_p(\gamma_t)\leq f_p(a) - tL+2\delta$, and a similar statement holds if $q\in[p,b]$. Thus $$ f_p(\gamma_t) \leq \max\bigl(f_p(a)-tL,f_p(b)-(1-t)L\bigr)+2\delta $$ for all $t\in[0,1]$, and it follows that $$ f_p(\gamma_t)\leq (1-t)f_p(a)+tf_p(b)+2\delta $$ for all $t\in[0,1]$. $\square$ Taking a limit, we deduce that any horofunction satisfying Definition (1) is $2\delta$-convex. The following claim finishes the proof. Claim. Any horofunction satisfying Definition (1) is distance-like. Proof: Let $\{p_n\}$ be an unbounded sequence of points in $X$ so that $f_{p_n}$ converges to a function $f\colon X\to \mathbb{R}$ uniformly on compact sets. Let $x\in X$ and let $\lambda\in (-\infty,f(x))$, so $f_{p_n}(x) > \lambda$ for large enough $n$. Since the sequence $\{p_n\}$ is unbounded, we also know that $f_{p_n}(p_n)<\lambda$ for large enough $n$. For each such $n$, choose a geodesic $[p_n,x]$. By the Intermediate Value Theorem, there is a point $y_n$ on this geodesic so that $f_{p_n}(y_n) = \lambda$. Note then that $d(x,y_n) = f_{p_n}(x) - \lambda$ for each $n$, so $d(x,y_n) \to f(x)-\lambda$ as $n\to\infty$. Since $X$ is proper, the sequence $\{y_n\}$ must have a limit point $y$, which satisfies $f(y) = \lambda$ and $d(x,y) = f(x)-\lambda$. Thus $d\bigl(x,f^{-1}(\lambda)\bigr) \leq f(x)-\lambda$, and the opposite inequality follows from the fact that $f$ is $1$-Lipschitz (since each $f_p$ is). $\square$ Definition 2 does not imply Definition 1 As Anton Petrunin suggests, it is sometimes possible to take something like a minimum of functions that satisfy Definition (1) to get a function that satisfies Definition (2) but not Definition (1). For example, let $X$ be the Cayley graph of the group $$ G = \langle a,b \mid ab=ba,b^3=1\rangle \cong \mathbb{Z}\times\mathbb{Z}_3 $$ where $x_0$ is the identity vertex, and note that $G$ acts on $X$ in a natural way. Let $T$ be the triangle of edges connecting $1$, $b$, and $b^2$. Using Definition (1), each point $p\in T$ has two associated horofunctions, namely those associated to the sequences $\{a^n p\}$ and $\{a^{-n}p\}$, and it is not hard to check that these are all of the horofunctions on $X$ determined by Definition (1). In particular, the horofunction boundary of $X$ by Definition (1) is homeomorphic to the disjoint union of two copies of $T$. However, there are horofunctions for $X$ satisfying Definition (2) but not Definition (1). For example, let $\{g_n\}$ be the sequence of functions $$ g_n(x) = \min\bigl(d(x,a^n),d(x,a^nb),d(x,a^nb^2)\bigr) - n $$ where $a^n,a^nb,a^nb^2$ denote the corresponding vertices in $X$. Then $\{g_n\}$ converges uniformly on compact sets to a function $g\colon X\to\mathbb{R}$ which satisfies Definition (2) but not Definition (1). In particular, $g(1)=g(b)=g(b^2)=0$, but this is not true for any horofunction satisfying Definition (1). More generally, for any values $u,v\in [-1,1]$ satisfying $|u-v|\leq 1$, there are exactly two Definition (2) horofunctions on $X$ that satisfy $g(1)=0$, $g(b)=u$, and $g(b^2)=v$. The set of all such $u$ and $v$ is a closed hexagonal region in $\mathbb{R}^2$, and the set $\Phi$ of horofunctions satisfying Definition (2) is the union of two disjoint copies of this region. Note that $G\times \mathbb{Z}_3$ will have similar behavior for any hyperbolic group $G$. In particular, there exist non-elementary hyperbolic groups for which Definition (1) and Definition (2) are not equivalent. Note also that the function $g$ defined above takes integer values on vertices, so it corresponds to a point in the space $\Phi_0$ of integral coycles defined by Coornaert and Papadopoulos.<|endoftext|> TITLE: The 'gros' functor from schemes into (strictly) locally ringed topoi QUESTION [10 upvotes]: Consider the category of finitely presented commutative rings, and equip its opposite category with either the Étale or Zariski topology. These give rise to topoi $\operatorname{Sh}(\mathbf{Et})$ and $\operatorname{Sh}(\mathbf{Zar})$ by taking sheaves. By the descent theorem for schemes, the functors of points of schemes are all sheaves in either of these topoi. In what follows, we will consider the Étale case: Given a scheme $S$ viewed as an object of $\operatorname{Sh}(\mathbf{Et})$, we can form the slice topos $\operatorname{Sh}(\mathbf{Et})_{/S},$ and by the Yoneda lemma, we see that the functor taking a scheme to this slice topos is fully faithful into the category of categories over $\operatorname{Sh}(\mathbf{Et})$. However, given a map $f:S\to T$ of schemes, the induced functor $$f_!:\operatorname{Sh}(\mathbf{Et})_{/S}\to \operatorname{Sh}(\mathbf{Et})_{/T},$$ is not a geometric morphism. It is however a morphism in $\mathbf{Cat}_{/\operatorname{Sh}(\mathbf{Et})}$ with respect to the projection functors. Since all of the functors in this triangle admit right adjoints that themselves admit right adjoints, the calculus of mates tells us that we have a triangle of essential (in fact Étale) geometric morphisms: $$\begin{matrix} \operatorname{Sh}(\mathbf{Et})_{/S} & \xrightarrow{f_*} & \operatorname{Sh}(\mathbf{Et})_{/T}\\ S_* \searrow&\overset{\alpha}{\Leftarrow}&\swarrow T_*\\ &\operatorname{Sh}(\mathbf{Et}) \end{matrix},$$ where the natural transformation $\alpha$ is the mate of the mate of the commuting triangle arising from the Yoneda embedding: $$\begin{matrix} \operatorname{Sh}(\mathbf{Et})_{/S} & \xrightarrow{f_!} & \operatorname{Sh}(\mathbf{Et})_{/T}\\ S_!\searrow&=&\swarrow T_!\\ &\operatorname{Sh}(\mathbf{Et}) \end{matrix},$$ In particular, since $\operatorname{Sh}(\mathbf{Et})$ is the classifying topos of strictly local rings (also called strictly henselian rings), the geometric morphism $f_*$ together with the 2-cell $\alpha$ determines a map of strictly locally ringed topoi. Since everything is natural and functorial, this defines a functor from schemes (in fact from $\operatorname{Sh}(\mathbf{Et}$)) to the category of strictly locally ringed topoi. Question: Is this assignment fully faithful in the category of strictly locally ringed topoi? REPLY [9 votes]: So the final answer, is 'no', but there is still something interesting to say: Given any topos $\mathcal{T}$, the construction you are describing produces an equivalence of category between $\mathcal{T}$ and the full subcategory of $Top_{/\mathcal{T}}$ (where $Top$ is the 2-category of toposes) of toposes $\mathcal{E} \rightarrow \mathcal{T}$ that are "étale over $\mathcal{T}$", i.e. of the form $\mathcal{T}_{/X}$. (This is the topos theoretic version of the representations of sheaves by their etale spaces) As the category of scheme indentifies in the way described in the question with a full subcategory $Sh(Et)$ this shows that morphisms of scheme $X \rightarrow Y$ corresponds exactly to morphisms between their (gros) étale topos, "over the étale topos of Spec $\mathbb{Z}$, i.e. geometric morphisms $f : Et_X \rightarrow Et_Y$ endowed with an isomorphism $f^* \mathcal{O}_Y \simeq \mathcal{O}_X$ between their structural sheaves (as described in the question) as sheaves of rings, so they corresponds to a special kind of morphisms of locally ringed toposes. One might wonder whether all morphisms of locally ringed toposes between étale toposes of scheme are actually of this form, but unfortunately, the answer is no. I'm quite far from algebraic geometry so I could be completely off, but I believe the following gives a counter example: Consider $X = Spec (\mathbb{Z}/p \mathbb{Z} )$ and $\mathcal{T}$ its gros étale topos. Its structural sheaf is a sheaf of strict local $\mathbb{Z}/p\mathbb{Z}$ algebra, in fact the universal one. I claim the Frobenius endomorphism of this sheaf (which is always a local morphism) provides a non-trivial non-invertible endormorphism of this sheaf: if it were trivial or invertible, then by universality, it would mean that for any sheaf of strictly local $\mathbb{Z}/p\mathbb{Z}$-algebra on any topos the Forbenius endomorphism would be trivial/invertible. Hence the identity of the topos together with the Frobenius endomorphism provides a morphism of locally ringed topos that is not of the form above and hence do not corresponds to a morphism of scheme.<|endoftext|> TITLE: Are the polynomials in $\{1/t\}$ dense in $L^2(0,1)$? QUESTION [7 upvotes]: Added. My question in the title was solved (in the negative) by Nik Weaver (in the answer below) and Mateusz Kwaśnicki (in the comments). In both solutions, the reason is that the $L^2$ density fails even on the smaller interval $[1/3,1]$, where the function $\{1/t\}$ is piecewise glued by $1/t - 2$ (from $1/3$ to $1/2$) and $1/t - 1$ (from $1/2$ to $1$). In view of this feature that I had neglected (existence of an obstruction already on some compact subsegment $[c,1] \subset (0,1]$), and also in view of my own example below the line, I still wanted to raise the follow-up in this linked question. The original question. (Which turned out easy.) The question of the title: Is $\{1/t\}^k$, $k = 0, 1, \ldots$, an $L^2$ spanning set on the interval $(0,1)$? Background. For amusement, let me include some background on what leads me to raise this kind of question (besides idle curiosity). The following has no strictly mathematical bearing on the actual question above, which could for all I know turn out to be easy and unrelated to RH. I had an observation some time ago that the Riemann Hypothesis implies (and is reversely implied by; which is the trivial part from the viewpoint of the Mellin transform) the $L^2$ density of the linear span of the countable sequence of functions $$ \mathcal{B}: \quad t^{d}\big(B_{d+1}(\{1/t\}) - B_{d+1}(1/t)\big) - t^{d-1}\big(B_d(\{1/t\}) - B_d(1/t)\big), \quad d = 1, 2, \ldots. $$ Here $B_m(x)$ are the Bernoulli polynomials; so this (RH conditional) $L^2$ spanning set consists of certain polynomials in $\{1/t\}$ and $t$. Observe that the Mellin transform of this linear space is equal to $$ \{ p(s) \zeta(s) \big/ s(s+1) \cdots (s+N) \quad \mid \quad p(s) \in \mathbb{R}[s], \, \deg{p} \leq N; \quad N = 0, 1, \ldots \}. $$ I was somehow led to think that this should probably be straightforward to prove for anyone who cared enough about such an implication. (I could be wrong to think that. Anyhow it may hopefully be appropriate to leave this statement here as an exercise "for the interested reader.") I did however find this to be quite different in at least one regard than the classical Nyman-Beurling result, which expressed RH as the question of the $L^2$ density of linear span of the continuum of functions $$ \mathcal{N}: \quad \Big\{ \alpha \{1/t\} - \{ \alpha / t\} \mid \alpha \in [0,1] \Big\}. $$ In either question, the full problem reduces to expressing the constant function $1$ as an $L^2$-limit of linear combinations from the designated set of functions. But while in Nyman-Beurling one knows that $1$ is in the $L^2$ closure of span of $\mathcal{N}$ if and only if it is already in the $L^2$ closure of the linear subspace spanned by the countable sequence $\alpha = 1, 1/2, 1/3, \ldots$, the latter countable sequence is known to not be a topological spanning set for the full $L^2(0,1)$; indeed it has a rather small closure. In contrast, under RH, the Bernoulli sequence $\mathcal{B}$ is a countable spanning set for all of $L^2(0,1)$. REPLY [9 votes]: They are not. The function $g(t) = \begin{cases}\frac{-1}{(1-t)^2}& \frac{1}{3} < t < \frac{1}{2}\cr 1& \frac{1}{2} < t < 1\end{cases}$ is orthogonal to all of them. That is because $\{\frac{1}{t}\} = \frac{1}{t} - 2$ on the first interval and $\frac{1}{t} - 1$ on the second. So integrating $g$ against $\{\frac{1}{t}\}^k$ on $[\frac{1}{3}, \frac{1}{2}]$ yields $-\int_{1/3}^{1/2} (\frac{1}{t} - 2)^k \frac{dt}{(1-t)^2}$, and integrating the product on $[\frac{1}{2}, 1]$ yields $\int_{1/2}^1 (\frac{1}{t} - 1)^k dt$. A simple change of variable with $s = \frac{t}{1-t}$ (so $\frac{1}{s} = \frac{1}{t} - 1$) turns the first integral into minus the second, so their sum is zero.<|endoftext|> TITLE: Pushforward of invertible sheaf on $P^1$ by finite flat map QUESTION [5 upvotes]: Let $k$ be an algebraically closed field. Any finite $k$-morphism $P^1_k\rightarrow P^1_k$ is flat (miracle flatness) and surjective on the underlying spaces. Therefore, the pushforward of a coherent locally free sheaf is coherent locally free (on $P^1_k$, such sheaves can be described by a finite sequence of integers using the fact that Picard rank is 1 and there is Birkhoff--Grothendieck splitting). Assume we have a finite $k$-morphism $P^1_k\rightarrow P^1_k$ such that the inverse image of the generic point has cardinality $n\geq 2$. Which sheaves can we get as the pushforward of a locally free sheaf of rank 1? REPLY [3 votes]: Here is another way to view this. Let's look at local sections of $\mathbb{P}^1_w$. Cover $\mathbb{P}^1_w$ by affine charts $U_0$ and $U_1$. Then $\mathcal{O}_{\mathbb{P}^1_w}$ and $\mathcal{O}_{\mathbb{P}^1_w}(-1)$ can be viewed as following pictures: $$ \rlap{\underbrace{\phantom{\cdots w^{-3}, w^{-2}, w^{-1}}}_{U_0}} w^{-3}, w^{-2}, w^{-1}, \overbrace{1, w , w^2, w^3, \cdots}^{U_1} $$ $$ \underbrace{\cdots w^{-3}, w^{-2}, w^{-1}, 1\,}_{U_0} , \, \overbrace{ w , w^2, w^3, \cdots}^{U_1} $$ For example, the local sections of $\mathcal{O}_{\mathbb{P}^1_w}$ over Spec$k[w]$ are $1, w, w^2,\cdots$ and over Spec$k[w^{-1}]$ are $1, w^{-1}, w^{-2},\cdots$. So the global section is only $k$. Now suppose we have the morphism given by $z=w^3$. Then we can decompose $\mathcal{O}_{\mathbb{P}^1_w}$ as following three: $$ \rlap{\underbrace{\phantom{\cdots w^{-9}, w^{-6}, w^{-3}}}_{U_0}} w^{-9}, w^{-6}, w^{-3}, \overbrace{1, w^3 , w^6, w^9, \cdots}^{U_1} $$ $$ \underbrace{\cdots w^{-8}, w^{-5}, w^{-2}}_{U_0}, \, \overbrace{ w^{1}, w^4 , w^7, \cdots}^{U_1} $$ $$ \underbrace{\cdots w^{-7}, w^{-4}, w^{-1}}_{U_0}, \, \overbrace{ w^{2}, w^5 , w^8, \cdots}^{U_1} $$ Hence $$f_* \mathcal{O}_{\mathbb{P}^1_w} = \mathcal{O}_{\mathbb{P}^1_z} \oplus \mathcal{O}_{\mathbb{P}^1_z}(-1)^{2}.$$<|endoftext|> TITLE: Subspaces of $L^2(0,1)$ dense on every truncation $L^2(c,1)$ QUESTION [9 upvotes]: It may be better to move this to a separate question. Let me call a linear subspace $V \subset L^2(0,1)$ to be tame if, for every linear subspace $W \subset V$, either $W$ is dense in $L^2(0,1)$, or else there exists a $c > 0$ such that already the restriction $W|_{[c,1)}$ is not dense in $L^2(c,1)$. (We think of $0$ as a ''point at infinity,'' cf. the example in (ii*) below; it is a matter of convention whether to formulate it this way or for $L^2(0,\infty)$ and its finite truncations $L^2(0,T)$.) It is not a priori clear to me whether there is any tame and dense subspace of $L^2(0,1)$, nor indeed that this is a reasonable notion. In relation to this question and its solutions, I wonder: Question. (i) Is the space of polynomials tame in $L^2(0,1)$? Or is there another explicit example of a tame and dense subspace of $L^2(0,1)$? (ii*) Is the space $$ V := \big\{ P(t,\{ 1/t \}) \mid P \in \mathbb{R}[x,y] \big\} \subset L^2(0,1) $$ tame? (Here, $\{\cdot\}$ denotes the fractional part function.) A motivation for asking this is that, as described in the linked question, the Riemann Hypothesis can be expressed as an $L^2(0,1)$ density of the particular linear subspace $$ \mathcal{B} := \big\{ \sum_d c_d \cdot t^d (B_{d+1}(\{1/t\}) - B_{d+1}(1/t)) \quad \sum_d c_d = 0 \big\} $$ of the space $V$ in (ii*). (To give an indication of why the $L^2$ density of this latter space is equivalent to the RH, observe that the Mellin transform of the space is equal to the multiples of the Riemann zeta function by the rational functions of the form $p(s) / s(s+1) \cdots (s+N)$ with $\deg{p} \leq N$; and that $t^{\rho-1} \in L^2(0,1)$ if and only if $\mathrm{Re}(\rho) > 1/2$ strictly.) Regarding (i), observe at least that by the Müntz–Szász theorem, the condition to be checked is fulfilled for the particular case that $W$ is the span of a subset of the monomials. Regarding (ii*), it can be shown that $\mathcal{B}|_{[c,1)}$ is $L^2(c,1)$-dense for every $c > 0$ and hence a positive answer would imply RH. That made me to wonder about constructing any subspace $W \subset V$ of the space $V$ in (ii*) such that $W|_{(c,1)}$ is $L^2$-dense on every truncation $(c,1)$ but $W$ is not $L^2$-dense on the full segment $(0,1)$. Lastly, the example of $\mathcal{B}$ (resp. its restriction to $d \geq 3$) shows easily that the answer in (ii*) certainly becomes negative if the Hilbert space $L^2(0,1)$ gets replaced by any of $p > 2$ Banach spaces $L^p(0,1)$, or by $C[0,1]$. The same questions may be asked also for $L^1(0,1)$, a variant which is perhaps more plausible but without interesting implications. REPLY [8 votes]: It is not a priori clear to me whether there is any tame and dense subspace of $L^2(0,1)$. Indeed, no such space exists. To see it, choose any sequence of functions $f_k\in L^2([0,1])$ such that $f_k|_{[c,1]}$ are dense in $L^2([c,1])$ for all $c>0$ but $\int_0^1 f_k=0$ for all $k$. Now if $V$ is dense in $L^2([0,1])$, then we can find $g_k\in V$ such that $\|g_k-f_k\|_{L^2([0,1])}<\frac 1k$ and $\int_0^1 g_k=0$ (just approximate $f_k\pm \frac 1{3k}$ with precision $\frac 1{6k}$ and take an appropriate convex combination of these 2 approximations). Then the linear span of $g_k$ is dense in any $L^2([c,1])$ but not in $L^2([0,1])$. Unfortunately, I cannot offer an equally trivial proof of RH :-)<|endoftext|> TITLE: Map of coherent sheaves inducing isomorphism on the stalks at the generic point QUESTION [5 upvotes]: Let $f:X\rightarrow Y$ be a finite morphism between Noetherian integral schemes that is surjective on the underlying topological spaces. Does there exist an integer $n>0$ and a coherent $O_X$-module $F$ such that there is a morphism of $O_Y$-modules $O_Y^n\rightarrow f_*F$ inducing an isomorphism on the stalks at the generic point of $Y$? It is pretty easy to see that there must be a quasi-coherent $O_X$-module like this. Pick an affine open $V\subset Y$, then its inverse image is an affine open $U\subset X$. It is easy to see that $O_U$ answers the question for the morphism $f|_{U}$. Therefore, if $i:U\rightarrow X$ denotes the inclusion, $i_* O_U$ is a quasi-coherent $O_X$-module (because $i$ is qcqs) answering the question (because the generic stalk can be computed on any non-empty open). It appears, however, that this way we will rarely get a coherent sheaf. EGA IV, partie quadrieme, Cor. 21.12.7 says that if $X$ is separated, the complement of $U$ has codimension 1 so $i_*$ has little chance to preserve coherence. P.S. This is Hartshorne Ch. III Ex. 4.2 (a). I have consulted some solutions online and could not convince myself they really produce coherent (rather than quasi-coherent) sheaves. The same question has been asked on Math Stackexchange but we have different perspective on the direction of the attack. REPLY [2 votes]: I think this can also be solved by taking the "coherent subsheaf generated by finitely many global sections of $i_* \mathcal{O}_U$". I'm not sure this is fundamentally different from Aknazar's solution, but it avoids the direct mention of (co)limits, and might be what Hartshorne had in mind. As Aknazar, suppose we can solve the problem for affine schemes, and choose an open affine $j: V \hookrightarrow Y$ and let $i: U \hookrightarrow X$ be its preimage. Suppose we have a morphism $$ \alpha_V: \mathcal{O}_V^n \to (f|_U)_*\mathcal{O}_U$$ which is an isomorphism at the generic point $\eta$ of $Y$. The problem is that $i_* \mathcal{O}$ is not in general coherent on $X$. But $\alpha_V$ chooses $n$ global sections $s_1,\dotsc,s_n \in \Gamma(X, i_* \mathcal{O}_U) = \Gamma(U, \mathcal{O}_U)$, which can be used to define a morphism $\alpha_X: \mathcal{O}_X^n \to i_* \mathcal{O}_U$. Let $\mathscr{G}$ be the image of this morphism. Then $\mathscr{G}$ is coherent, because for every open affine $\text{Spec }A = W \subset X$, $\mathscr{G}(W) \subset (i_*\mathcal{O}_U)(W)$ is the $A$-submodule generated by $s_1|_W,\dotsc,s_n|_W$. This allows us to define the morphism $$ \alpha_Y: \mathcal{O}_Y^n \to f_* \mathscr{G} \subset (f i)_* \mathcal{O}_U$$ by takting the same global sections $s_1,\dots,s_n \in \Gamma(Y, f_* \mathscr{G})$. At the generic point this yields $$ \mathcal{O}^n_{Y, \eta} \xrightarrow{\alpha_{Y, \eta}} (f_*\mathscr{G})_\eta \hookrightarrow ((fi)_* \mathcal{O}_U)_\eta, $$ and the composition is $\alpha_{U, \eta}$ which is an isomorphism. Hence $\alpha_{Y, \eta}$ is an isomorphism as well.<|endoftext|> TITLE: Integral of top forms in terms of Čech representative QUESTION [7 upvotes]: Let $X$ be a compact connected Riemann surface and let $\omega$ be a two-form on $X$. We can view the cohomology class $[\omega]$ as an element of the Čech cohomology group $\check{H}^2(X,\mathbb{R})$, represented by some real numbers $\omega_{\alpha\beta\gamma}\in\mathbb{R}$ on triple intersections of a cover $\{U_\alpha\}$. Is there an explicit formula for the integral $\int_X\omega$ in terms of the real numbers $\omega_{\alpha\beta\gamma}$? REPLY [6 votes]: For any covering $\mathcal U = \{U_\alpha\}_{\alpha\in A}$ of $X$, we can form its Cech nerve $C(\mathcal U)$, an abstract simplicial complex with vertex set $A$ whose simplices are exactly the finite subsets $I\subset A$ such that $\bigcap_{\alpha\in I} U_\alpha\neq\emptyset$. We can then glue standard simplices together as specified by this abstract simplicial complex to get a CW complex $|C(\mathcal U)|$. If $\mathcal U$ is a good cover, i.e. all finite intersections are empty or contractible, and $X$ is sufficiently nice, e.g. a manifold, there is a homotopy equivalence $|C(\mathcal U)|\to M$ constructed by mapping the $0$-simplex corresponding to $\alpha$ to some point of $U_\alpha$ and then iteratively extending it over simplices (which is always possible and unique up to homotopy by contractibility). The cellular complex of $|C(\mathcal U)|$ is precisely the Cech complex of $U$. Thus the Cech cocycle $\omega_{\alpha\beta\gamma}$ is just a representative of the pullback of $\omega$ in the cellular complex. Integration is pairing against the fundamental class, which is an element of $H_2(X)\cong H_2(|C(\mathcal U)|)$, so it can also be expressed as a sum of formal elements corresponding to nonempty intersections of $U_\alpha$'s. In fact, for special coverings one can identify this class explicitly: Choose a triangulation of $X$, i.e. an abstract simplicial complex $K$ and a homeomorphism $|K|\cong X$. To every vertex $x$ of $K$ corresponds a contractible open subset of $|K|$, the star of $x$, which is the union of the interiors of all simplices which have $x$ as a vertex (together with $x$ itself). Together, these form a good cover of $|K|$, and the corresponding Cech nerve is just $K$ itself. The fundamental class is given as the sum of all top-dimensional (in this case $2$-dimensional) simplices, with signs corresponding to the orientations. Thus pairing with the fundamental class sends a Cech cocycle $\{\omega_{xyz}\}$ to the sum over all triangles $t$ of $K$, with vertices $x_t,y_t,z_t$ in this order, of $\omega_{x_t,y_t,z_t}$.<|endoftext|> TITLE: Proper scheme such that every vector bundle is trivial QUESTION [6 upvotes]: It is claimed here that there exist proper schemes (probably over a field but not explicitly stated) with trivial Picard group. This means that every locally free $O_X$-module of rank 1 is trivial. Do there exist proper schemes over a field such that every locally free $O_X$-module of finite rank is trivial? Maybe it is easier to give some examples with algebraic spaces, but the accepted answer should give a scheme. EDIT: examples should be positive-dimensional. REPLY [5 votes]: According to this paper it is not known if every proper algebraic scheme admits nontrivial vector bundles. Partial results can be found here.<|endoftext|> TITLE: Do codimension 1 subsets of a scheme cover it? QUESTION [6 upvotes]: Let $X$ be an irreducible scheme. A point $p\in X$ is said to have codimension $n\in\mathbb{Z}_{\geq 0}\cup \{\infty\}$ if $\overline{\{p\}}$ has codimension $n$. Is it true that any point of positive codimension lies in the closure of a point of codimension 1? REPLY [3 votes]: This is false, already for spectra of valuation rings. Recall that ideals in a valuation ring $R$ correspond to ideals in the value group $\Gamma$, which can be any totally ordered abelian group. Moreover, since the ideals are totally ordered, so are the prime ideals, so it suffices to construct a totally ordered abelian group that does not have a height $1$ prime. This is easy: Example. Let $\Gamma$ be the group $\mathbf Z[x]$, with the ordering defined by $f \geq g$ if and only if $f(n) \geq g(n)$ for $n \gg 0$. This is equivalent to the lexicographic ordering with $1 < x < x^2 < \ldots$. Suppose $I \subseteq \Gamma$ is a height $1$ prime ideal, i.e. $I+\Gamma_{\geq 0} \subseteq I$, if $i + j \in I$ then $i \in I$ or $j \in I$, and if $J \subsetneq I$ is another prime ideal, then $J = 0$. Let $f \in I$ be a nonzero element, and assume $\deg f = n$. Then $x^{n+1} > f$, so $(x^{n+1}) = x^{n+1} + \Gamma_{\geq 0} \subsetneq I$ is a strictly smaller nonzero prime ideal inside $I$, contradicting the height $1$ hypothesis. $\square$ Remark. An example of a valuation ring with this value group can be constructed as follows: let $K = k(x_0, x_1, \ldots)$ be a rational function field in (countably) infinitely many variables. It has a valuation $K \to \mathbf Z[x]$ defined on monomials in $k[x_0,\ldots]$ by $$\prod_i x_i^{n_i} \mapsto \sum_i n_ix^i,$$ then on polynomials by mapping to the leading term (with respect to the lexicographic ordering with $x_0 < x_1 < \ldots$), and finally on $K$ by extending multiplicatively. Then one may take $R$ to be the value ring for this valuation. Remark. Obviously the result is true if every point has finite height; for example if $X$ is Noetherian.<|endoftext|> TITLE: Perfect Runs of Consecutive Integers QUESTION [11 upvotes]: A run of 2 or more consecutive integers is said to be perfect if the sum of its terms equals the sum of all the their proper divisors, adding common divisors as often as they occur. For example, 672, 673 is such a run (672 is a triperfect number whose proper divisors add up to 1344, while 673 is a prime, 1 being its only proper divisor). Are there arbitrarily long perfect runs of consecutive numbers? Infinitely many of any length greater than 1? REPLY [6 votes]: 523776, 523777 is another example of a 2-run. Again, 523776 is a triperfect number. An example of a 3-run is 5829840, 5829841, 5829842. No apparent pattern in these three numbers: Factorization Sum of proper divisors 5829840 2^4 * 3^3 * 5 * 2699 14258160 5829841 29 * 53 * 3793 316439 5829842 2 * 2914921 2914924 Here are two more perfect 3-runs of consecutive numbers: Factorization Sum of proper divisors 3414097920 2^16 * 3 * 5 * 23 * 151 8061430272 3414097921 13 * 31 * 41 * 206627 473814527 3414097922 2 * 1707048961 1707048964 39339578248 2^3 * 4917447281 34422130982 39339578249 29 * 4561 * 297421 1365596671 39339578250 2 * 3^2 * 5^3 * 7 * 2497751 82231007094 The first of these two, like the earlier example, were found by computer search with PARI/GP. I noticed that their middle odd terms turn up in A072188: Numbers $n$ such that $n$ divides $\sigma(n-1)+\sigma(n)+\sigma(n+1)$, where $\sigma(n)$ is the sum of integer divisors of $n$. The final of the above examples of perfect 3-runs comes from that sequence.<|endoftext|> TITLE: Additive group of local rings QUESTION [7 upvotes]: Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity? REPLY [9 votes]: Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring. Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ \mathbb{Z}/p^k \mathbb{Z} \times M $ where $ p^k = \exp (G) $. Then $ M $ naturally has the structure of a $ \mathbb{Z}/p^k \mathbb{Z} $ - module. Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity. Now let's prove it's local with maximal ideal $ \{ (x,m) \mid p $ divides $x \} $. Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local. This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.<|endoftext|> TITLE: Global to local principle for f.g. $\mathbb{Z}[x]$ modules QUESTION [8 upvotes]: In graduate school, while I was working on the maximal subgroup growth of certain metabelian groups, I discovered and proved a lemma which gave me the impression that it was already known. Do you know of any reference for this result? Does it follow easily from a known result? Here is the result: Let $N$ be a finitely generated $\mathbb{Z}[x]$-module. Suppose (by using the fundamental theorem of f.g. modules over PIDs) we have the following isomorphism of $\mathbb{Q}[x]$-modules: $$\mathbb{Q} \otimes_\mathbb{Z} N \cong \left( \bigoplus_{j=1}^{s_1} \mathbb{Q}[x]/(a_j) \right) \oplus \mathbb{Q}[x]^{s_2} $$ for some $a_j \in \mathbb{Q}[x]$ that are not units and such that $a_1 \mid a_2 \mid \ldots \mid a_{s_1}$. Then for all large primes $p$, we have the following isomorphism of $\mathbb{F}_p[x]$-modules: $$N/pN \cong \left( \bigoplus_{j=1}^{s_1} \mathbb{F}_p[x]/(\overline{a_j}) \right) \oplus \mathbb{F}_p[x]^{s_2}.$$ While my proof of the above result is about three pages long, I do have a short heuristic argument: When doing the computation required in finding the decomposition of $\mathbb{Q} \otimes N$ into a direct sum of cyclic modules, the only thing keeping us from doing this computation to $N$ itself (as a $\mathbb{Z}[x]$-module) is that we may need to divide by finitely many integers. So if $p$ is large enough, then in $\mathbb{F}_p$ we can divide by all those integers (i.e. their residues mod $p$). For such $p$, the steps of the algorithm would be the same for $N/pN$ as for $\mathbb{Q} \otimes N$. In case you are interested, this is Lemma 41 in my arXiv paper Maximal subgroup growth of some metabelian groups. I actually proved a slight generalization, concerning finitely generated $D^{-1}\mathbb{Z}[x]$-modules, where $D$ is the multiplicative closure of a finite set of primes. REPLY [7 votes]: Your lemma easily follows from the Smith Normal Form Theorem, a result you already referred to. The short heuristic argument that you gave can indeed be turned into a short proof. Nothing in the sequel should be new to you. But it is shorter and it also makes clear that your lemma generalizes immediately when replacing $\mathbb{Z}$ by any integral domain with infinitely many maximal ideals ("for all prime $p$ large enough" becomes "for all but finitely many maximal ideals"). Given a rational prime number $p$, let $\mathbb{Z}_{(p)} = \left\{ \frac{r}{s} \, \vert \, r, s \in \mathbb{Z},\, \mathbb{Z}s + \mathbb{Z}p = \mathbb{Z} \right\}$ and let $\phi_p: \mathbb{Z}_{(p)} \rightarrow \mathbb{Z}_{(p)} / p\mathbb{Z}_{(p)} \simeq \mathbb{F}_p$ be the natural epimorphism. Restricted to $\mathbb{Z}$, the map $\phi_p$ is the reduction modulo $p$. Given a commutative and unital ring $R$, we denote by $\text{M}_{m,n}(R)$ the $R$-module of the $m$-by-$n$ matrices over $R$. Abusing notation, we denote also by $\phi_p$ the epimorphisms $ \mathbb{Z}_{(p)}[x] \rightarrow \mathbb{F}_p[x]$ and $\text{M}_{m,n}(\mathbb{Z}_{(p)}[x]) \rightarrow \text{M}_{m,n}(\mathbb{F}_p[x]) $ induced by the reduction of coefficients. Proof of OP's Lemma. Let $N$ be a finitely generated module over $\mathbb{Z}[x]$ and let \begin{equation} \label{EqSeq} \mathbb{Z}[x]^n \xrightarrow[]{f} \mathbb{Z}[x]^m \rightarrow N \rightarrow 0 \quad\quad (1) \end{equation} be an exact sequence that provides us with a presentation of $N$ where $m$ is the minimal number of generators of $N$. Let $M(f) \in \text{M}_{m, n}(\mathbb{Z}[x])$ be the matrix of $f$ with respect to the canonical bases. Applying the tensor functors $-\otimes_{\mathbb{Z}[x]} \mathbb{Q}[x]$ and $-\otimes_{\mathbb{Z}[x]} \mathbb{F}_p[x]$ to $(1)$, we obtain two other exact sequences. The exactness of the first is immediate as $-\otimes_{\mathbb{Z}[x]} \mathbb{Q}[x]$ is exact since we just inverted non-zero rational integers. For the second, some easy diagram chasing involving also $(1)$ settles the claim. Thus we get a presentation of $N \otimes_{\mathbb{Z}[x]}\mathbb{Q}[x]$ as the cokernel of the matrix $M(f\otimes_{\mathbb{Z}[x]}\mathbb{Q}[x]) = M(f)$ and a presentation of $N \otimes_{\mathbb{Z}[x]}\mathbb{F}_p[x] \simeq N/pN$ as the cokernel of $M(f\otimes_{\mathbb{Z}[x]}\mathbb{F}_p[x]) = \phi_p(M(f)) \in \text{M}_{m,n}(\mathbb{F}_p[x]))$. Since $\mathbb{Q}[x]$ is a principal ideal ring, the Smith Normal Form Theorem applies. Therefore we can find $A \in \text{GL}_m(\mathbb{Q}[x]), B \in \text{GL}_n(\mathbb{Q}[x])$ such that $$AM(f)B = \text{diag}(a_1, \dots, a_{s_1}, \underbrace{0, \dots, 0}_{s_2 \text{ zeroes}}) \quad (2)$$ with $a_j \in \mathbb{Q}[x]$ for all $j$, $a_j \vert a_{j + 1}$ and $a_1 \notin \mathbb{Q} \setminus \{0\}$. For all $p$ sufficiently large, we have $A \in \text{GL}_m(\mathbb{Z}_{(p)}[x]), B \in \text{GL}_n(\mathbb{Z}_{(p)}[x])$ and hence $a_j \in \mathbb{Z}_{(p)}[x]$ for all $j$. Applying $\phi_p$ to both sides of $(2)$, we get $$\phi_p(A) \phi_p(M(f)) \phi_p(B) = \text{diag}(\phi_p(a_1), \dots, \phi_p(a_{s_1}), 0, \dots, 0).$$ The proof is then complete.<|endoftext|> TITLE: Are these two notions of unstable localization suitably equivalent? QUESTION [7 upvotes]: It seems to me that although homological localization (i.e. formally inverting $E$-homology equivalences for some $E$) is a reasonable thing to do to a spectrum, it's a pretty brutal thing to do to a space, particularly if that space is not simply connected. To put a finer point on it: shouldn't there be a better notion of localization for non-simply-connected spaces? Shouldn't it at least take into account local coefficients, for example? As soon as one says "local coefficients" and has general cohomology theories in mind, one should start thinking in terms of parameterized homotopy theory. Recall that for $X$ a space, the $\infty$-category of $X$-parameterized spectra is just the functor category $Fun(X,Sp)$ where $Sp$ is spectra. For $E \in Sp$, let $Sp_E$ be the category of $E$-local spectra, and define the category of $E$-local $X$-parameterized spectra to be $Fun(X,Sp_E)$. Definition: A map $f: Y \to X$ of spaces is a strong $E$-local equivalence if the induced functor $f_\ast: Fun(Y,Sp_E) \to Fun(X,Sp_E)$, between categories of parameterized $E$-local spectra, is an equivalence. (Note that $f_\ast$ has a left adjoint $f^\ast$ which has a further left adjoint $f_!$; it's equivalent to ask for any one of these to be an equivalence). I'm curious to what extent this notion agrees with just Bousfield localizing at $E$-homology equivalence. So, some test cases: Questions: Let $E = H \mathbb Q$. Is a rational equivalence between simply-connected spaces a strong $H\mathbb Q$-equivalence? Let $E = H\mathbb Z_{(p)}$. Is a $p$-local equivalence between (sufficiently connected?) spaces a strong $H\mathbb Z_{(p)}$-equivalence? Let $E = M(p)$. Is a $p$-complete equivalence between sufficiently connected spaces a strong $M(p)$-equivalence? Let $E = K(n)$. Is a $K(n)_\ast$-equivalence between $(n+1)$-connected spaces a strong $K(n)$-equivalence? Guiding analogy: My perspective is that a space is like a scheme which is etale over a point. A parameterized spectrum is like a quasicoherent sheaf over this scheme. So we're asking for an equivalence of localized categories of quasicoherent sheaves. From this perspective, an $E_\ast$-homology equivalence $f: Y \to X$ looks like pretty a crude thing -- it just means that $t_! f_! \mathbb S_Y \to t_! \mathbb S_X$ is an $E$-local equivalence, where $t: X \to pt$ is the unique map. In the algebraic geometry analogy, this is like saying that the map induces an $E$-local isomorphism of the cohomology-with-compact-support of the structure sheaves, which seems like a pretty weak condition. So perhaps strong $E$-equivalences are truly a stronger notion than $E$-equivalences... REPLY [2 votes]: Let $E \neq 0$ be a spectrum. Here is a classification of the strong $E$-local equivalences, a proof that the localization with respect to them exists, and an analysis of some special cases. Classification of strong $E$-local equivalences: Claim 1: A map of spaces $f: X \to Y$ is a strong $E$-local equivalence if and only if it is a $\pi_0$-equivalence and each map of connected components is a strong $E$-local equivalence. The proof is straightfoward. Claim 2: If $X,Y$ are connected spaces, then $f: X \to Y$ is a strong $E$-local equivalence if and only if $E_\ast(\Omega X) \to E_\ast(\Omega Y)$ is an isomorphism. Proof: Note that $f^\ast: Sp_E^Y \to Sp_E^X$ is conservative and has a left adjoint $f_!$, so is an equivalence if and only if $M \to f^\ast f_! M$ is an equivalence for all $M \in Sp_E^Y$. Note that $Sp_E^X$ is equivalent to the category of modules for $L_E \Sigma^\infty_+ \Omega X \in Alg^{E_1}(Sp_E)$, and likewise $Sp_E^Y$ is the category of modules for $L_E \Sigma^\infty_+ \Omega Y$. So $f_!(M) = L_E \Sigma^\infty_+ \Omega Y \wedge_{L_E \Sigma^\infty_+ \Omega X} M$. So if $L_E \Sigma^\infty_+ \Omega X \to L_E \Sigma^\infty_+ \Omega Y$ is an equivalence, then $f^\ast$ is an equivalence, and taking $M = L_E \Sigma^\infty_+ \Omega X$, we see that the converse also holds. And this can be tested by taking $E$-homology, yielding the claim. The local objects with respect to strong $E$-local equivalences form an accessible localization of all spaces: Claim 3: Strong $E$-local equivalences are closed under 2-out-of-3 and under filtered colimits and coproducts in the arrow category. The proof is straightforward. Claim 4: Strong $E$-local equivalences are closed under cobase change (i.e. pushout along an arbitrary map). For reference, consider a pushout square and its image under $\Omega$: $$\require{AMScd} (\ast) \begin{CD} X @>>> Y\\ @VVV @VVV\\ W @>>> Z \end{CD} \qquad (\ast\ast) \begin{CD} \Omega X @>>> \Omega Y\\ @VVV @VVV\\ \Omega W @>>> \Omega Z \end{CD}$$ Observation 5: The monad for $E_1$-spaces commutes with $\pi_0$, and preserves $E_\ast$-equivalences. As a result, Lemma 6: If $(\ast)$ is a pushout, all spaces are connected, and $X \to Y$ is a $\pi_1$-isomorphism, then $(\ast\ast)$ is also a pushout of $E_1$-spaces. Proof: By the equivalence of pointed connected spaces and grouplike $E_1$-spaces, it suffices to check that the pushout of the relevant $E_1$ spaces is grouplike. This can be checked on $\pi_0$, and so follows from the observation. Lemma 7: $E_\ast$-equivalences of $E_1$-spaces are stable under pushout. Proof: Because the $E_1$ monad commutes with $E_\ast$-equivalences, it descends to the Bousfield localization at $E_\ast$-equivalences, and the result follows. Proof of Claim 4: Suppose that $X \to Y$ is a strong $E$-local equivalence in $(\ast)$; we want to show that $W \to Z$ is as well. It suffices to consider the case where $W$ is connected. We may take the pushout in two stages: first we take the pushout along $\amalg_i X_i \to \vee_i X_i$ where the $X_i,Y_i$ are the connected components of $X,Y$. Second, we take the pushout along $\vee_i X_i \to W$, i.e. we consider the case where $X$ is connected. The first case reduces to showing that if $X \to Y$ is a strong $E$-local equivalence of connected spaces, then so is $X \vee U \to Y \vee U$ where $U$ is connected, which reduces to the second case. Thus we may assume that $X$ is connected. But then by the two lemmas, because $\Omega X \to \Omega Y$ is an $E_\ast$-equivalence, so is $\Omega W \to \Omega Z$ as desired. Claim 8: Every strong $E$-local equivalence is a highly filtered colimit of a small set of strong $E$-local equivalences. Proof: This is clear in the category of pointed connected spaces, since $E_\ast$ and $\Omega$ are accessible functors. Then an arbitrary strong $E$-local equivalence is a coproduct strong $E$-local equivalences of pointed connected spaces, and because filtered colimits are computes the same in both categories the result follows. Thus the strong $E$-local equivalences are an accessible subcategory of the arrow category of spaces closed under 2-out-of-3, colimits, and cobase change. It follows that the objects local with respect to them form an accessible localization of the category of spaces. Some special cases: Case A: Suppose that $f: X \to Y$ is a map of simply-connected spaces and an $H_\ast(-;R)$-equivalence, where $R$ is a PID. Then $\Omega f: \Omega X \to \Omega Y$ is also an $H_\ast(-;R)$-equivalence, i.e. $f$ is a strong $HR$-equivalence. Proof: We have a map of Serre spectral sequences: $$ H_p(X;R) \otimes_R H_q(\Omega X; R) \oplus Tor_1^R(H_{p-1}(X;R),H_q(\Omega X; R)) \Rightarrow R \\ \qquad \qquad \qquad \qquad \qquad\qquad \downarrow \qquad \qquad \qquad \qquad \qquad\qquad \qquad \quad = \\ H_p(Y;R) \otimes_R H_q(\Omega Y; R) \oplus Tor_1^R(H_{p-1}(Y;R),H_q(\Omega Y; R)) \Rightarrow R$$ Let $q$ be minimal such that $(\Omega f)_q: H_q(\Omega X;R) \to H_q(\Omega Y; R)$ is not an isomorphism. Then $q \geq 1$ because $\Omega X, \Omega Y$ are connected. If $(\Omega f)_q(\xi) = 0$ with $\xi\in H_q(\Omega X; R)$, then there are no differentials that can kill $1 \otimes \xi$, a contradiction. If $\zeta \in H_q(\Omega Y;R)$ is not in the image of $(\Omega f)_q$, then there are no differentials which can kill $1 \otimes \zeta$, another contradiction. So $(\Omega f)_\ast$ is an isomorphism as desired. Case B: Let $\Phi$ be the Bousfield-Kuhn functor (for a fixed prime $p$ and $n\geq 1$). Suppose that $f: X \to Y$ is a map of connected spaces and $\Phi f$ is an equivalence. Then $f$ is a strong $T(n)$-equivalence (a similar statement holds for $L_{K(n)} \Phi$ and strong $K(n)$-equivalences). Proof: $\Phi$ commutes with fiber sequences, so $\Phi(\Omega f)$ is an equivalence, so $T(n)_\ast \Omega f$ is an equivalence as desired.<|endoftext|> TITLE: Indecomposable, non-simple, modules of quantum groups at roots of unity QUESTION [5 upvotes]: Let us consider the quantum group $U_q(\mathfrak{sl}_2)$ (as defined in Kassel's book on quantum groups), for $q$ being a root of unity of order $d$ (i.e., $d$ is the smallest positive integer for which $q^d=1$). If $n$ is the dimension of an irreducible, finite-dimensional representation of $U_q(\mathfrak{sl}_2)$ (over a complex vector space), then it is known that $n$ is bounded above by $$ e=\begin{cases} d, & \text{$d$: odd} \\ d/2, & \text{$d$: even.} \end{cases} $$ As far as I know, there are indecomposable, non-simple modules of dimension higher than $e$. I have made some small search on the structure of such modules, but I have not found anything substantial apart from Chari and Premet - Indecomposable restricted representations of quantum $sl_2$ (pdf abstract MSN), which however refers to the restricted case. So my questions are: Is there some reference on the structure of indecomposable, non-simple modules of quantum groups at roots of unity? Are there infinite dimensional, indecomposable, non-irreducibles? How can the limits of such representations (either fin or inf dimensional) at $q\to 1$, be computed? I would be interested either on references or on some short—if possible-—description of such modules, mainly for the case of $U_q(\mathfrak{sl}_2)$ and more generally for $U_q(\mathfrak{g})$, where $\mathfrak{g}$ is a finite-dimensional, simple, complex Lie algebra. Related: Indecomposable modules for the big quantum group What are the indecomposable $U_q\mathfrak{sl}(2)$-modules? REPLY [3 votes]: I apologize for self-citation, however in an old paper of mine, together with R. Giachetti The two-dimensional Euclidean quantum algebra at roots of unity, in the process of describing decomposition of tensor products of irreps at roots of unity we listed some explicit indecomposable modules for $E_q(2)$. Let me mention that this is a way in which irreps at roots of unity behave like infinite-dimensional ones: in that he tensor products of irreps are not completely reducible any more (and this happens each time the dimension exceeds the roots of unity degree, which then, in a way, plays the role of $\infty$). I think you may find something of this kind also in papers referring to tensor product decompositions for $SL_q(2)$ at roots of unity.<|endoftext|> TITLE: Transitive embedding of the projective plane $\Bbb R P^2$ into the $4$-sphere QUESTION [7 upvotes]: Is there an embedding (i.e. injective continuous map) $$\phi:\Bbb R \Bbb P^2\hookrightarrow S^4\subseteq\Bbb R^5$$ of the projective plane $\Bbb R\Bbb P^2$ into the $4$-sphere, that is transitive, i.e. for any two $x,y\in \Bbb R\Bbb P^2$ there is an orthogonal transformation $T\in\mathrm{O}(\Bbb R^5)$ that fixed the image $\mathrm{im}(\phi)$ set-wise, and has $Tx=y$? Is $\Bbb R^5$ the lowest dimensional space in which such an embedding is possible, or do we need even more dimensions? I may ask the same question for $\Bbb R\Bbb P^n$: what is the lowest dimensional Euclidean space needed for such an embedding. REPLY [8 votes]: Since you may ask about $\mathbb{RP}^n$, I may point out that you can embed $\mathbb{RP}^n$ into $\mathbb{R}^{(n+2)(n+1)/2-1}$ with a transitive action of $O(n+1)$. From e.g. Exercise 5-C of Milnor-Stasheff, $\mathbb{RP}^n$ is the space of symmetric idempotent $(n+1)\times (n+1)$ matrices with trace $1$. This lives inside $\mathbb{R}^{(n+2)(n+1)/2-1}$ as the space of symmetric $(n+1)\times (n+1)$ matrices with trace $1$. $O(n+1)$ acts by conjugation on this space as isometries, and preserving the subspace of idempotent matrices. An exceptional case is $\mathbb{RP}^3 \cong SO(3)$. The group $SO(3) \subset \mathbb{R}^9$ as $3\times 3$ matrices. But we can do a bit better: we may think of $SO(3)$ as pairs of orthogonal unit vectors $(v_1,v_2)\in (\mathbb{R}^3)^2$. This gives an embedding of $SO(3)\subset S^5\subset \mathbb{R}^6$ with a transitive group action. I believe that this special embedding exists since $so(4)=so(3)\oplus so(3)$. One might be able to construct similar smaller dimensional embeddings using fibrations $S^1\to \mathbb{RP}^{2n+1}\to \mathbb{CP}^n$ and $\mathbb{RP}^3\to \mathbb{RP}^{4n+3}\to \mathbb{HP}^n$, but I haven't checked if they give smaller embeddings with isometric actions. However, $\mathbb{RP}^{2n}$ is not a fibration (for the same reason that $S^{2n}$ is not a fibration). Hence, if we have an embedding $\mathbb{RP}^{2n}\subset \mathbb{R}^k$ and a transitive action by isometries $G\leq O(k)$, then the representation of the compact group $G$ must be irreducible. If not, then there is a splitting $\mathbb{R}^k=\mathbb{R}^{k_1}\times \mathbb{R}^{k_2}$ which is invariant under $G$. In this case, we get $v=(v_1,v_2)\in \mathbb{RP}^n\subset \mathbb{R}^k$, $v_i\in \mathbb{R}^{k_i}$, and $\mathbb{RP}^n = G\cdot v \to G\cdot v_1$. Hence we have a fibration $\mathbb{RP}^{2n} \to G\cdot v_1$, a contradiction unless $G\cdot v_1=\mathbb{RP}^{2n}$, in which case $k$ was not minimal.<|endoftext|> TITLE: Isotropy subgroupoid of a regular Lie groupoid QUESTION [5 upvotes]: Let $(G\rightrightarrows M)$ be a Lie groupoid (i.e. a groupoid with source map $s$ and target map $t$ such that $G,M$ are smooth manifolds and the structural maps are all smooth (and $s$,$t$ are submersions). Defining $$IG := \{g \in G \mid s(g) = t(g) \}$$ we obtain the so called isotropy subgroupoid $IG \rightrightarrows M$ which is in general only a subgroupoid but not a sub LIE groupoid, in that $IG$ need not be a submanifold of $G$ (for an example consider the action groupoid of the canonical action $\mathbb{S}^1 \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$ of the circle group via rotation. In this case $G=\mathbb{S}^1 \times \mathbb{R}^2$ and thus $IG = \mathbb{S}^1 \sqcup \bigsqcup_{x\in \mathbb{R}^2} \{1\}$, whence the isotropy subgroupoid of the action groupoid can not be a Lie groupoid with respect to the subspace topologies. Now I have read in several places (cf. e.g. 1) that the isotropy groupoid is an embedded Lie subgroupoid if the Lie groupoid is regular. Recall that a Lie groupoid is regular if its anchor $$(s,t)\colon G \rightarrow M\times M, g \mapsto (s(g),t(g))$$ is a mapping of constant rank (there are some other equivalent formulations, using e.g. the orbit foliation of the Lie groupoid, see e.g. 2). Unfortunately, I was not able to track down a proof of this folklore fact in the literature. The old book by Mackenzie (Lie Groupoids and Lie Algebroids in Differential Geometry) proves a similar statement using foliation charts for the orbit foliation. Note that the statement in the book claims that it is true for all Lie groupoids (there called differentiable groupoids), but the proof holds only for locally trivial Lie groupoids (and seems to be intended to be only for this class). For this class however, the anchor $(s,t)$ is a submersion (and $IG = (s,t)^{-1} (\Delta M)$, where $\Delta M$ is the diagonal) whence it is easy to see why $IG$ is an (embedded) submanifold of $G$. I wonder now how one tackles the general case for regular Lie groupoids. Surely there must be a proof somewhere in the literature? Or alternatively an easy argument which eludes me? REPLY [3 votes]: This is not exactly an answer to your question, but I hope it helps. If it would be fine to first pass to the connected components of the identity of each isotropy group, I believe you find the proof of the analogous statement (that you then obtain an embedded subgroupoid) in Proposition 2.5 of I. Moerdijk, On the Classification of Regular Groupoids P.S. - See update 1 for a counterexample, and update 2 for an even easier counterexample to the general case. Update 1 After thinking a bit more about this, I believe that in general one really has to first pass to the connected component of the identity in the isotropy Lie groups. One problem that might arise in general is seen in the following example of an action groupoid (which comes from family of actions of the reals on a torus). Consider the action of the Lie group $(\mathbb{R},+)$ on $T \times (-1,1)$, where $T=\mathbb{R}^2/ \mathbb{Z}^2$ is a 2-torus, given by $$\lambda \cdot ([x,y],\epsilon)= ([x+\lambda, y+\lambda\epsilon],\epsilon), $$ for $\lambda\in \mathbb{R}$, $[x,y]\in T$ and $\epsilon\in (-1,1)$. The associated action groupoid $G=\mathbb{R}\times T \times (-1,1) \rightrightarrows T \times (-1,1)$ is a regular Lie groupoid, because all the orbits are diffeomorphic to either circles or lines. The isotropy of a point $([x,y],\epsilon)$ is: a copy of $\mathbb{Z}$ inside of $\mathbb{R}\times \{([x,y],\epsilon)\}$ if $\epsilon$ is rational; the trivial group $\{ 0\} \times \{([x,y],\epsilon)\}$ if $\epsilon$ is irrational. Moreover, all points along a same orbit have ''the same'' isotropy in the sense that if $([x,y],\epsilon)$ and $([x',y'],\epsilon)$ are in the same orbit, and $(\lambda , [x,y],\epsilon)$ is in the isotropy of $([x,y],\epsilon)$, then $(\lambda , [x',y'],\epsilon)$ is in the isotropy of $([x',y'],\epsilon)$. This means in particular that the isotropy subgroupoid, if it were an embedded submanifold should have dimension at least one. So take for example $\epsilon=0$, $\lambda = 1$. Then $(1,[x,y],0)$ is a point in $IG$, but there is no neighborhood of it inside of $G$, which when intersected with $IG$ is diffeomorphic to some euclidean space, because of all the points $([x,y],\epsilon)$ with $\epsilon$ small that have trivial isotropy. So $IG$ is not an embedded submanifold. I get the impression this would work one dimension lower, taking just a circle instead of a torus, but for me it is easier to visualize the situation when the orbits are one-dimensional. Update 2 (May 7, 2019) Thanks to a question of Alexander Schmending, I later realised that a comment I previously made (and now deleted to avoid confusion) was not true. In the comment I suggested that if $G$ is regular and proper, $IG$ should be a Lie subgroupoid. But again, this is not necessarily the case, and the example is way easier than the one above. This can be seen by considering the action groupoid $G= \{-1,1\}\times \mathbb{R^2}$ associated with the action of $\{-1,1\}$ on the plane $\mathbb{R^2}$ by reflection through the origin. This Lie groupoid is proper and regular. All points except the origin have trivial isotropy, while the origin has $\{-1,1\}$ as isotropy. The source map $s$ restricted to $IG$ cannot be a submersion, as $s_{|IG}$ has rank 0 at $(-1,0)\in G=\{-1,1\}\times \mathbb{R^2}$. So things go wrong in a different way than that of the example of update 1: $IG$ is Hausdorff, it is even an embedded submanifold of $G$ (with components of different dimensions), just not a Lie groupoid.<|endoftext|> TITLE: Etale sheaves on algebraic spaces vs. Etale sheaves on affines QUESTION [6 upvotes]: Let's fix a field $k$. First, consider $Aff_k$ to be the category of affine finite type $k$ schemes. On this category, one can define the etale topology and thus consider the site $Aff_k^{et}$, then define the category of sheaves: $Sh(Aff^{et}_k)$ Similarly, it seems to me that one could also take the category $ \text{AlgSp}_k$ of finite type algebraic spaces and define the etale topology and get a site $\text{AlgSp}_k^{et}$, and then also consider the category of sheaves: $Sh(\text{AlgSp}_k^{et})$ My question: can I expect an equivalence of topoi between $Sh(Aff^{et}_k)$ and $Sh(\text{AlgSp}_k^{et})$? Secretly, when I write "sheaves", I mean sheaves of spaces in the $\infty$-categorical sense, which are not necessarily hyper-complete. But I really don't have so much intuition for this, so even if we take sheaves of sets I'd be interested in the answer. Thanks! REPLY [3 votes]: Yes, the two $\infty$-topoi are equivalent. Let $u: \mathrm{Aff} \to \mathrm{AlgSp}$ be the inclusion. Then $u$ preserves étale covering families and pullbacks, hence commutes with the formation of the Čech nerve of such a covering. This implies that $u^*$ preserves sheaves. The functor $u$ is also cocontinuous: if $X\in\mathrm{Aff}$ and $R\subset \mathrm{AlgSp}_{/X}$ is a covering sieve, then $u^*(R)\subset \mathrm{Aff}_{/X}$ is still a covering sieve. This implies that the right Kan extension functor $u_*$ preserves sheaves also. We therefore have an adjunction $(u^*,u_*)$ between the $\infty$-categories of sheaves, where $u_*$ is fully faithful. By looking at the triangle identities, it remains to show that $u^*$ is conservative. To see this we can look at the two inclusions $\mathrm{Aff} \subset \mathrm{SepSch} \subset \mathrm{AlgSp}$. Each inclusion $\mathcal{C}\subset\mathcal{D}$ has the property that every object of $\mathcal D$ admits a covering by objects of $\mathcal C$ such that all the fiber products occurring in the Čech nerve are still in $\mathcal C$ (because the diagonal of a separated scheme is affine and the diagonal of an algebraic space is representable by separated schemes); this immediately implies that restriction from $\mathcal D$ to $\mathcal C$ detects equivalences between sheaves. Remark. We also have an equivalence between the $\infty$-topoi of Nisnevich sheaves. The cocontinuity of $u$ and the conservativity of $u^*$ in that case are less obvious; they follow from a result of Gruson and Raynaud (see Prop. 3.7.5.3 in Lurie's Spectral Algebraic Geometry).<|endoftext|> TITLE: Finite-dimensional algebras isomorphic as commutative unital rings QUESTION [9 upvotes]: For which fields $k$ do there exist two finite-dimensional $k$-algebras of different dimension which are isomorphic as commutative unital rings? Some thoughts: for a finite field this can not happen, because the cardinality of the algebra determines its dimension. consider any field $K$ and the function field $k=K(x^2)$, then the one-dimensional algebra over $k$ and the two-dimensional algebra $K(x^2)[x]$ are isomorphic as commutative unital rings. REPLY [3 votes]: To find examples of such fields, it is sufficient to look for fields $k$ having isomorphic finite extensions of differing degrees, just like the example given by schematic_boi. Claim: For a field $k$, the following are equivalent: Finite dimensional commutative $k$-algebras that are isomorphic as rings always have the same $k$-dimension. Finite field extensions of $k$ that are isomorphic as fields always have the same degree over $k$. Proof: Clearly 1 implies 2. Suppose 2 holds; to deduce 1, it is sufficient to let $A$ be a finite-dimensional commutative $k$-algebra and to show that the $k$-dimension of $A$ can be recovered from the ring-theoretic structure of $A$. To this end, consider the filtration given by the Jacobson radical $J = J(A)$: $$ A \supseteq J \supseteq J^2 \supseteq \cdots \supseteq J^n = 0. $$ First note that $A/J \cong F_1 \times \cdots \times F_r$ is isomorphic to a product of finite field extensions of $k$. By hypothesis 2, the dimension $[F_i : k]$ of each of these is uniquely determined by the isomorphism type of the field $F_i$. Each filtration factor $J^i/J^{i+1}$ is a module over the semisimple algebra $A/J$ and therefore is a direct sum of simple modules. These simple modules, including their multiplicities, are uniquely determined (Jordan-Hölder). Because every simple $A$-module is isomorphic to a 1-dimensional vector space over some $F_i$, its $k$-dimension is determined by that $F_i$. Thus the dimension of each $J^i/J^{i+1}$ is determined by its decomposition into simple modules. It follows that $\operatorname{dim}_k(A) = \sum_{i=0}^{n-1} \operatorname{dim}_k(J^i/J^{i+1})$ is determined entirely by the ring-theoretic struture of $A$, from which 1 follows. QED Remark 1: This begs the question of which fields have property 2, and I certainly do not know the answer. It is true for algebraically closed fields (since there are no nontrivial finite extensions), for finite fields (as noted in the OP), and for algebraic number fields (it holds for $\mathbb{Q}$ since any field isomorphism fixes the prime subfield, and property 2 also passes to finite extensions). But Hagen Knaf's answer to this MSE question provides examples of a geometric nature. Remark 2: If desired, one should be able to remove the assumption of commutativity from property 1 in the claim above. One would need to analyze the Artin-Wedderburn structure of $A/J$ more carefully, but this reduces to the case of a simple finite-dimensional algebra $S$. One would then apply property 2 to the center $F$ of $S$ (which is a field), and then note that the dimension of $S$ over $F$ is determined by the $F$-dimension of its unique simple module and the length $S$ as an $S$-module, all of which is purely ring-theoretic information about the structure of $S$.<|endoftext|> TITLE: Ádem relations for the Steenrod and the Dyer–Lashof algebra QUESTION [5 upvotes]: In this paper by Nondas Kechagias, the Steenrod algebra and the Dyer–Lashof algebra are compared. The rough difference ist: The Steenrod algebra arises by dividing out the “cohomological” Ádem relations and $Q^0=1$. The Dyer–Lashof algebra arises by dividing out the “homological” Ádem relations and $Q^I=0$ where $I$ has negative excess. The cohomological Ádem relations are of the form $$Q^iQ^j =\sum_{k=0}^{[i/2]}\binom{j-k-1}{i-2k}Q^{i+j-k}Q^k~~~\text{for }i<2j$$ The homological Ádem relations are of the form $$Q^iQ^j-\sum_{k>0}\binom{k-j-1}{2k-i}Q^{i+j-k}Q^k~~~\text{for }i>2j.$$ In what way are they dual to each other? If I consider the dual Steenrod sqares $\mathrm{Sq}_i:H_*(X)\to H_{*-i}(X)$ for $X$ of finite type, the should satisfy other Adem relations than the “homological ones”, namely the cohomological ones with the only difference that the composition has to be read contravariantly. Why do we call both relations Ádem relations? REPLY [12 votes]: See Peter May's "A General Algebraic approach to Steenrod Operations". Both arise from considering the inclusion of a Sylow $p$-subgroup into $\Sigma_{p^2}$. There is a universal $\mathbb{Z}$-indexed set of operations with the Steenrod algebra and Dyer-Lashof algebra arising as the quotients acting on the homology of appropriate non-negative (resp., non-positive) complexes. Here's a nice example: consider the Steenrod algebra acting on the cohomology of graded cocommutative Hopf algebras. (In this, $Sq^0 \neq 1$.) One can grade the operations homologically, $Q^i : Ext^{s,t} \to Ext^{t+s-i,2t}$ or cohomologically, $Sq^i : Ext^{s,t} \to Ext^{s+i,2t}$. The Dyer-Lashof operation $Q^i$ raises the `total degree' $t-s$ by $i$, and the Steenrod operation does what you expect from the cohomology of groups, for example, if everything is in internal degree $t=0$. It is the same set of operations on $Ext$, but in the first grading you get the homological form of the Adem relations and in the second grading you get the cohomological form. One reason to consider the homological grading is that if $R$ is an $E_\infty$-ring spectrum, then $Q^i$ in the $E_2$-term of the Adams spectral sequence $Ext_{A}(H^*R, F_2) \Rightarrow \pi_* R$ agrees with the $Q^i$ in $H_* R$ under the edge homomorphism and the Hurewicz map.<|endoftext|> TITLE: Ideals of commutative Frobenius algebras QUESTION [5 upvotes]: Given a finite dimensional commutative (connected=local) Frobenius algebra $A$ over a field $K$. Question 1: Does $A$ have only finitely many ideals? (the answer should be no in the non-commutative case, but Im not sure about the commutative case) Question 2: In case the poset of ideals of two such algebras $A_1$ and $A_2$ are isomorphic, are $A_1$ and $A_2$ isomorphic in case they have the same vector space dimension? Question 3:Is there a way to obtain the ideal lattice of a finite dimensional algebra via QPA? (Im not sure whether there is a good method to check whether one module is a submodule of another one (not up to isomorphism)) REPLY [2 votes]: Question 1: Let $A=k[x,y]/\langle x^3, y^3\rangle$. A $k$-basis for $A$ is $\{1,x,y,x^2,xy,y^2,x^2y,xy^2, x^2y^2\}$. Let $I_q = \langle x^2y + qxy^2\rangle$, and a $k$-basis for $I_q$ is $\{ x^2y + qxy^2, x^2y^2\}$. Hence all the ideal $I_q$ are different, and if the field $k$ is infinite, there exists infinite many ideals. Question 2: Don't know. Question 3: For an infinite field the above example shows that there are infinitely many ideals, so I assume a finite field. Define the following algebra in QPA: Q := Quiver( 1, [ [ 1, 1, "a" ], [ 1, 1, "b" ] ] ); KQ := PathAlgebra( GF(5), Q ); AssignGeneratorVariables( KQ ); relations := [ a^3, a * b - b * a, b^3 ]; A:= KQ / relations; M := AlgebraAsModuleOverEnvelopingAlgebra( A ); Finding all the submodules of $A$ as a bimodules, that is, all ideals in $A$. test := AllSubmodulesOfModule( M ); Say we want to check which $3$ dimensional modules the first $2$ dimensional module is contained in. Then we can do the following: # # Creating a vectorspace of the image of a homomorphism # f := test[ 3 ][ 1 ]; Imf := List( BasisVectors( Basis( Source( f ) ) ), m -> ImageElm( f, m )![ 1 ]![ 1 ] ); Imf := List( Imf, b -> Flat( b ) ); # # Creating all the vectorspaces for the 3 dimensional submodules # dim3submodules := [ ]; for g in test[ 4 ] do Img := List( BasisVectors( Basis( Source( g ) ) ), m -> ImageElm( g, m )![ 1 ]![ 1 ] ); Img := List( Img, b -> Flat( b ) ); Add( dim3submodules, Subspace( GF(5)^9, Img ) ); od; # # Checking which ones contains Imf # List( dim3submodules, V -> ForAll( Imf, m -> m in V ) ); The output looks like this from the last line: gap> List( dim3submodules, V -> ForAll( Imf, m -> m in V ) ); [ true, true, true, true, true, true, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false, false ] I hope that these comments are helpful. The QPA-team.<|endoftext|> TITLE: Classification of commutative Frobenius algebras QUESTION [8 upvotes]: Are there attempts to classify commutative finite dimensional Frobenius algebras? They appear often in mathematics, such as in algebraic geometry and the famous category equivalence between commutative Frobenius algebras and 2-dimension topolocial quantum field theories. However, I have not yet seen attempts to classify this class of algebras (up to isomorphism of k-algebras). Recall that a commutative Frobenius algebra is a finite dimensional algebra $A$ with $A \cong D(A)$ or equivalently simple socle in case it is local. Here are two questions related to such a classification (we can assume that commutative Frobenius algebra are connected): Question 1: Is a commutative Frobenius algebra "field-independent"? This means that in its presentation $KQ/I$ by quiver and relations, there exists such $I$ which only contains the field element 1 or -1 so that a given commutative Frobenius algebra is defined over all fields. In case question 1 has a positive answer, this would mean that a classification is independet of the field (maybe excluding characteristic 2). Question 2: For a given integer $d$, are the only finitely many $d$-dimensional commutative Frobenius algebras of vector space dimension $d$? (here we say that two algebras are isomorphic in case they are isomorphic as $K$-algebras) A positive answer to question 2 would be surprising, but I think it should be true for $d \leq 5$ at least. REPLY [3 votes]: I believe the answer to question 1 is no. But the question is slightly ambiguous. I can give a quiver presentation with only $\pm 1$ appearing as coefficients in the relations and where being Frobenius depends on the characteristic of the field. Maybe you are asking if you can always find some quiver presentation of a commutative Frobenius algebra with $\pm 1$ coefficients that remains Frobenius over all fields. I suspect that the examples I give would fail this somewhat different question but I will stick with my interpretation. I'll give a construction via a quiver presentation but the quiver just has loops at one vertex so this is basically a monoid presentation. Let $A$ be an $n\times n$ symmetric matrix with 0/1 entries with $n\geq 2$. We have a quiver $Q$ with $n$-loops at a single vertex labeled $x_1,\ldots,x_n$. We take as relations that all paths of length $3$ are zero, $x_ix_j =0$ if $A_{ij}=0$ and make all $x_ix_j$ with $A_{ij}=1$ equal to each other. This gives an admissible ideal $I$ and all the generators of $I$ are either monomial or a difference of monomials. The algebra is commutative since $A$ is symmetric. It’s essentially an algebera of a finite $3$-nilpotent commutative semigroup with adjoined identity. Notice all paths of length two which are not zero in $KQ/I$ are equal. Also $KQ/I$ is local since there is one vertex in the quiver. I claim that if $K$ is a field, $KQ/I$ is Frobenius iff $\det A\neq 0$ in the field $K$, i.e., the characteristic of $K$ does not divide $\det A$. The radical of $KQ/I$ is spanned by $x_1,\ldots, x_n$ (or rather their cosets but I ignore the distinction). If $A=0$, then the socle is spanned by $x_1,\ldots, x_n$ and hence is not simple. So assume $A\neq 0$. So there is at least one nonzero product of two loops. All these nonzero products are the same element $z$ of $KQ/I$. Clearly $z$ is in the socle. So we need that nothing else is in the socle. First note that the $1,x_1,\ldots, x_n$ and the element $z$ form a basis for $KQ/I$ and I shall identify the former elements with their coset. To be in the socle is to be annihilated by $x_1,\ldots, x_n$. Such an element must not be a unit, so it belongs to the radical. Therefore, it must be of the form $a=c_1x_1+\cdots+c_nx_n+cz$. But then it is straightforward from the defining relations to see $x_ia=\sum_{j=1}^nA_{ij}c_jz$ and so $x_ia=0$ if and only if row $i$ of $A$ dot product with $(c_1,\ldots, c_n)=0$. Thus $a$ is in the socle iff $(c_1,\ldots, c_n)$ is in the nullspace of $A$ and so the socle is just $Kz$ iff $\det A\neq 0$. For a concrete example let $J_n$ be the $n\times n$ all ones matrix and $B_n=J_n-I_n$. So all diagonal entries are zero and all other entries are $1$. So our defining relations are $x_i^2=0$ for all $i$ and $x_ix_j=x_kx_\ell$ whenever $i\neq j$ and $k\neq \ell$. It is well known that $\det B_n =\pm (n-1)$. This is because the characteristic polynomial of $J_n$ is $p(\lambda)=(\lambda -n)\lambda^{n-1}$ and so $\det B_n = (-1)^np(1)=\pm (n-1)$. If $p$ is a prime, taking $n=p+1$ you get an algebra which is Frobenius in all characteristics except $p$. More generally, for any finite set of primes, by taking $n$ one larger than the product of those primes you get an algebra which is Frobenius in all characteristics except those in your finite set.<|endoftext|> TITLE: Inductive limit commutes with topological tensor product QUESTION [6 upvotes]: Consider $H \left(U \right); U \subset \mathbb{C}$ - space of holomorphic functions with compact-open topology. In this topology, this space is Montel, nuclear and Frechet. I want to take the inductive limit of tensor product of 2 copy of this space and I am wondering: $$ \varinjlim\limits_{U \supset K} \left( H \left(U \right) \hat{\bigotimes} H \left(U \right) \right) \stackrel{?}{=} H \left(K \right) \hat{\bigotimes} H \left(K \right) = H \left(K \times K\right), $$ where $H \left( K \right)-$ space of germs of holomorphic functions on compact set $K \subset \mathbb{C}$. It doesn't matter which type of tensor product we take because space is nuclear. REPLY [4 votes]: Only some ideas, but too long for a comment. I think there are at least to approaches which might help here: Since the spaces here are Frechet spaces and therefore every partially continuous bilinear map is continuous, you can work with the "inductive tensor product" (which is different from the injective tensor product) introduced by A. Grothendieck which works well with inductive limits - here the question reduces to the question of completeness of the inductive limit on the left hand side, see Proposition 14 on p. 76 of Grothendieck's thesis. The completeness of inductive limits of course is also quite nontrivial. You could try to use the commutativity results of R. Hollstein (Inductive limits and ε-tensor products.) to show the desired results. Since every nuclear space is an $\varepsilon$-space in his sense the problem here is again to show that your spectrum satisfies one of his conditions.<|endoftext|> TITLE: Extracting Dirichlet series coefficients QUESTION [5 upvotes]: Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$\sum_n a_n n^{-s},$$ is there any known method to compute a desired coefficient $a_n$? REPLY [7 votes]: Even for more general Dirichlet series $$f(z)=\sum_{0}^\infty a_n e^{-\lambda_nz}$$ there is the formula $$a_ne^{-\lambda_n\sigma}=\lim_{T\to\infty}\frac{1}{T}\int_{t_0}^Tf(\sigma+it)e^{\lambda_n it}dt,$$ where $t_0$ is arbitrary (real) and $\sigma>\sigma_u$, the abscissa of uniform convergence. This formula determines both $\lambda_n$ and $a_n$: the RHS=0 when we integrate against $e^{i\lambda t}$ with $\lambda\neq \lambda_n$. The class of functions which can be represented by such a series is called (analytic) almost periodic functions (on a vertical line $\{s=\sigma+it:t\in R\}$). The "number-theoretic case" corresponds to $\lambda_n=n$. Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969. Dirichlet series with complex $\lambda_n$ have been also studied (by A. F. Leont'ev and his school).<|endoftext|> TITLE: Injection into a proper class and choice without regularity QUESTION [14 upvotes]: In $\sf ZF$, we have that the axiom of choice is equivalent to: For all sets $X$, and for all proper classes $Y$, $X$ inject into $Y$ and For all sets $X$, and for all proper classes $Y$, $Y$ surject onto $X$ To see that those are indeed equivalent to choice we have for one direction to inject a set $X$ into $Ord$ and this will give well ordering for $X$(and because $Ord$ well ordered, we can easily construct an injective from $X$ to $Ord$ using a surjective from $Ord$ to $X$) To see that the other direction is true, take a set $α$ and a class $Y$, because we are assuming $\sf AC$ we may assume WLOG that $α∈Ord$. Then we may use induction to create a sequence $(x_β)$ of ordinals such that for $β<γ$ we have $Y∩V_{x_β}\subsetneq Y∩V_{x_γ}$, then we look at $V_{x_α}$, and by well ordering it find an injective $α→Y$(and surjective $Y→α$). In the proof use relied heavily on the axiom of foundation, so we can ask are those 3 equivalent in $\sf ZF-\mbox{regularity}$? When talking with @Wojowu he told me that his intuition told him that $\sf AC$ is not equivalent to the other 2, saying that he thinks that there is a model of $\sf ZFC-\mbox{regularity}+\mbox{a proper class of atoms}+\mbox{only finite sets of atoms}$, in which case no infinite set inject into the class of atoms, but after searching I couldn't find any reference to such model. My questions: If such model exists, can someone direct me to a reference, or explain it's construction? If not, how those 2 behave in $\sf ZF-\mbox{regularity}$? What about the other 2? Does the surjective version implies the injective version in $\sf ZF-\mbox{regularity}$? REPLY [8 votes]: The results appear in Jech's "The Axiom of Choice" in the problem section of Chapter 9 (Problems 2,3, and 4). Indeed, it is easy to see that the injections into classes imply the surjections from classes which imply choice. Exactly by means of the class of ordinals. So the point is to separate the others. And if we have a proper class of atoms whose subsets are all finite, then it is a class which does not map onto $\omega$, but every set has only finitely many in its transitive closure, so it can be well-ordered. The last model is described well in Jech, this is Problem 4 in the aforementioned reference, and the key point is that the atoms are indexed by countable sequences of ordinals, so that there are always surjections onto every set, but there is no $\omega$ sequence of atoms, which form a proper class, so there is no injection from any infinite set into the class of atoms. (And indeed, that implies all sets of atoms are finite.)<|endoftext|> TITLE: Ricci Curvature on Grassmannian QUESTION [8 upvotes]: Suppose $G_r(n)$ is the Grassmannian, which is the collection of all $r$ dimensional subspace in $\mathbb{R}^{n}$ equipped with the usual invariant metric. Let $Ricc(G_r(n))$ be the Ricci curvature tensor. What are the best known constants $0 TITLE: Find a function $F$ on $[0,1]$ with moments decaying as $(\ln n)^{-n}$ QUESTION [5 upvotes]: Let $F:[0,1]\to\mathbb{R}$ be a measurable function such that $$ \mu_n(F)=\int_0^1F(t)t^ndt\sim\frac1{(\ln n)^n}\quad\mbox{as}\quad n\to\infty. $$ More precisely, $$ 00$ and all complex $z$. So, $f(z)$ is an entire function of exponential type $a$ for any real $a>0$. Hence, by a Paley--Wiener theorem (more specifically, see e.g. Theorem 19.3 on page 375), for each real $a>0$ there is an $L^2$ function $F_a$ such that for all complex $z$ \begin{equation*} f(z)=\int_{-a}^a F_a(t)e^{itz}\,dt \end{equation*} Taking the inverse Fourier transform, we see that for each $a\in(0,1)$ and all complex $z$ \begin{equation*} f(z)=\int_0^a F(t)e^{itz}\,dt. \end{equation*} Thus, for all complex $z$ \begin{equation*} 0=f(z)=\int_0^1 F(t)e^{itz}\,dt. \end{equation*} So, indeed $F=0$ a.e. on $[0,1]$. Consider now the more general setting when $F(t)\,dt$ is replaced by $\rho(dt)$, where $\rho$ is a signed measure over $\R$ with support $\supp\rho\subseteq[0,1]$. Then condition (1) is replaced by \begin{equation*} |\mu_n(\si)|(\ln(n+2))^n0$. For real $b>0$ and $t\in\R$, let \begin{equation*} G_b(t):=(g_b*d\si)(t):=\int_\R g_b(t-s)\,\si(ds) =\frac{\si([t-b,t+b])}{2b}, \end{equation*} where $g_b:=\frac1{2b}\,1_{[-b,b]}$. Then $G_b\in L^2(\R)$, since the signed measure $\si$ is finite with $\supp\si\subseteq[0,1]$. Also, \begin{equation*} \hat g_b(z):=\int_\R e^{itz}g_b(t)\,dt=\frac{\sin bz}{bz} \end{equation*} for $z\ne0$, and hence \begin{equation*} f_b(z):=\int_\R e^{itz}\,G_b(t)\,dt =\int_\R e^{itz}\,(g_b*d\si)(t)\,dt=\hat g_b(z)f(z) \end{equation*} is of exponential type $b+a$ for all real $a>0$. Thus, by the cited Paley--Wiener theorem, $\supp(g_b*d\si)\subseteq[-b-a,b+a]$ for all real $a>0$ and hence \begin{equation*} \supp(g_b*d\si)\subseteq[-b,b]. \end{equation*} On the other hand, because $\supp g_b$ and $\supp\si$ are both compact, by Theorem 4.3.3 on page 117, $\cch\supp(g_b*d\si)=\cch\supp g_b+\cch\supp\si$, where $\cch$ denotes the closed convex hull. Since $\cch\supp g_b\ni0$, we conclude that $\cch\supp\si\subseteq\cch\supp(g_b*d\si)\subseteq[-b,b]$, whence $\supp\rho\subseteq\supp\si\cup\{0\}\subseteq[-b,b]$, for any real $b>0$. Thus, condition (1a) implies \begin{equation*} \supp\rho\subseteq\{0\}. \tag{2} \end{equation*} Vice versa, trivially (2) implies (1a). Thus, (1a) holds iff $\supp\rho\subseteq\{0\}$. Response to the second comment by the OP: Let us try to sort all this out. Your condition \begin{equation} 0 TITLE: Foliation with trivial leaf holonomy QUESTION [6 upvotes]: In 1960, R. Hermann showed the following: Theorem Let $M$ be a manifold with a foliation $F$ and a bundle-like metric, if all leaves are compact and the holonomy group of each leaf is trivial, then $M/F$ is a smooth manifold. (It is the partially result of the main theorem on Hermann, R., On the differential geometry of foliations, Ann. Math. (2) 72, 445-457 (1960). ZBL0196.54204.) Q If we drop the condition on bundle-like and admit the trivial holonomy group, can we get the same result? That is to say: Let $M$ be a manifold with a foliation $F$, if all leaves are compact and diffeomorphism to each other, and the holonomy group of each leaf is trivial, is it true that $M/F$ is a smooth manifold? Any reference is welcome. REPLY [4 votes]: This follows from Theorem 2 in Thurston's 1974 paper "A generalization of the Reeb stability theorem", at least if $H^1(L,R)=0$. https://core.ac.uk/download/pdf/82172971.pdf<|endoftext|> TITLE: Ergodic without atoms implies completely conservative? QUESTION [5 upvotes]: I am currently trying to learn Patterson-Sullivan theory, but I am getting stuck on basic questions about ergodic theory. Here is one of them, given as an exercise in one of the texts I am trying to read. If you need wider context, the text (in French) is here, with the relevant question towards the bottom of page 17 (next-to-last sentence of the long paragraph). Let ($\Omega$, $\Gamma$, $\mu$) be a dynamical system satisfying the following assumptions: $\Omega$ is a Hausdorff, locally compact and $\sigma$-compact topological space; $\Gamma$ is a countable group acting on $\Omega$; $\mu$ is a Radon measure on $\Omega$ that is quasi-invariant by $\Gamma$ (for every $g \in \Gamma$, $g_* \mu$ is absolutely continuous with respect to $\mu$). Let us recall some definitions: a subset $W \subset \Omega$ is called wandering if, for $\mu$-almost all $w \in W$, the intersection of the orbit $\Gamma w$ with $W$ is finite; we say that the system is completely conservative if $\Omega$ has no wandering subsets of positive measure; we say that the system is completely dissipative if $\Omega$ has some wandering subset $W$ such that $\Omega = \bigcup_{g \in \Gamma} g W$. Statement to prove: If the system is ergodic and the measure $\mu$ does not have any atoms, then the system is completely conservative. Even though it is supposed to be an easy exercise ("on vérifie aisément que..."), I am at a loss. I have found a partial proof, that works only if you can assume that the set $W$ introduced in the proof is closed. Unfortunately, I have no idea whether it is true, in general, that every completely dissipative dynamical system is generated by a closed wandering set. Can someone give me a proof of this last statement, or find some other way to fix my proof presented below? In fact, I have remarkably little intuition about what "completely dissipative" actually means. I am tempted to think that a completely dissipative action should in particular be properly discontinuous, but apparently this is false (though I have never seen an actual counterexample). I would be very happy if I could find, somewhere, an arsenal of examples and counterexamples of completely dissipative and completely conservative systems, that is sufficiently well-stocked to forge an intuition about these properties and to be able to test whether some reasonable-sounding statement is actually true. The partial proof: Let $\Omega = \Omega_C \sqcup \Omega_D$ be the Hopf decomposition of $\Omega$, i.e. a partition of $\Omega$ into two $\Gamma$-invariant subsets such that $\Omega_C$ is completely conservative and $\Omega_D$ is completely dissipative. Since the system is ergodic, they cannot both have positive measure: this means that $\Omega$ is either completely conservative or completely dissipative. By contradiction, assume the latter: let $W \subset \Omega$ be a wandering subset such that $$\Omega = \bigcup_{g \in \Gamma} g W.$$ Note that the set $$\{x \in W \mid \Gamma x \cap W \text{ is finite}\} \cap \operatorname{Supp}(\mu)$$ has the same (positive) measure as $W$, hence is nonempty. Let $x_0$ be any point in this set. Let $F := \{ g \in \Gamma \mid g x_0 \in W \}$; this is a finite set by construction. Since $x_0 \in \operatorname{Supp}(\mu)$, every neighborhood of $x_0$ has positive measure; on the other hand, since the measure is Radon and without atoms, any point has neighborhoods of arbitrarily small measure. Let $U$ be some neighborhood of $x_0$ such that for every $g \in F$, we have $\mu(g U) < \frac{\mu(W)}{\# F}$. We can also assume (provided that $W$ is closed!) that for all $g \not\in F$, the image $g U$ is disjoint from $W$. Then consider the set $X := \bigcup_{g \in \Gamma} g U$: this set is $\Gamma$-invariant by construction; it has positive measure by construction; its complement $\Omega \setminus X$ contains in particular $W \setminus X$, the complement of $W \cap X$ in $W$. But by construction, the intersection $W \cap X$ is contained in $\bigcup_{g \in F} g U$, so that we have $$\mu(W \cap X) \leq \sum_{g \in F} \mu(g U) < \mu(W).$$ Hence the complement $W \setminus X$, and hence also the complement $\Omega \setminus X$, has positive measure. But these three properties of the set $X$ contradict ergodicity, QED. REPLY [2 votes]: First a general comment. Unfortunately, the categorical approach is completely missing in the way the measure theory is still taught nowadays, and the fact that there actually is only one "reasonable" purely non-atomic probability measure space (so called Lebesgue, Lebesgue--Rokhlin, or standard probability space) isomorphic to the unit interval endowed with the classical Lebesgue measure is not very widely known. Any Borel probability measure on a Polish space makes it a Lebesgue space, so that in particular all measure spaces one has to deal with in the Patterson theory are from this class. The definition of a wandering set you are using is a non-standard one. According to the usual definition a measurable set is wandering if all its translates are pairwise disjoint mod 0 (i.e., the intersection with almost any orbit consists of at most one point). However, it is very easy to see that the existence of a wandering set $W$ in your sense implies the existence of a "usual" wandering set $W'$. Namely, realize your measure space as the unit interval, and then for each orbit take the minimal point from its intersection with $W$. This operation is clearly measurable and provides you with a measurable set $W'\subset W$, which is wandering in the usual sense. You are asking why ergodicity implies complete conservativity. The reason (with the right definition of wandering sets) is very simple: the union of the translates of any non-trivial subset of a wandering set is obviously a non-trivial invariant set. Actually, more generally, the union of the free discrete ergodic components of the action is precisely the dissipative part of the Hopf decomposition. See Kaimanovich for more details and Grigorchuk - Kaimanovich - Nagnibeda for various examples concerning the relationship between dissipativity and minimality of boundary actions.<|endoftext|> TITLE: Combinatorial system with parity QUESTION [8 upvotes]: Let $S = \{1,...,n\}$ be a set of $n$ elements and let $Y$ be a set of non-empty subsets of $S$ with the following requirements: $|Y|$ is odd. Every element in $S$ is contained in an even number of sets in $Y$. Every pair of elements of $S$ is contained in an even number of sets in $Y$. Question: Does there exist $y_1 \neq y_2 \in Y$ such that $(\exists_{a \in y_1} \forall_{b \in y_2} : a < b) \wedge (\exists_{a \in y_1} \forall_{b \in y_2} : a > b)$, that is, such that $y_1$ contains an element that is strictly smaller than all elements in $y_2$ and an element that is strictly larger than all elements in $y_2$? Below you find two examples of systems that satisfy the requirements. Set containment of the elements are displayed using lines. In both examples the answer is YES no matter the ordering of the elements, is this always the case? REPLY [6 votes]: After numerous failed attempts to prove this, I found a counterexample with $n=6$ and $|Y|=13$: $$\{1\}, \{1,2\}, \{3,4\}, \{5,6\}, \{1,2,3\}, \{2,3,4\}, \{3,4,5\}, \{4,5,6\}, \{1,3\}, \{2,4\}, \{3,5\}, \{3,4,6\}, \{3,6\}$$ or in the form of $13\times 6$ incidence matrix: $$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 1\\ 1 & 1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 1 & 0\\ 0 & 0 & 1 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 & 0 & 1\\ \end{bmatrix} $$<|endoftext|> TITLE: Extension of 2-adic valuation to the real numbers QUESTION [12 upvotes]: I just want to know what properties of valuations extend to $\mathbb R$... Denote an extension of the 2-adic valuation from $\mathbb Q$ to $\mathbb R$ by $\nu$. Suppose $\nu(x)=\nu(y)=0$. Is it true that $\nu(x+y)\ne 0$? What about $\nu(x^2+y^2)\le 1$? I'm interested in knowing both whether these are true for every extension, as well as knowing whether there is some extension for which they are true (for every $x$ and $y$). REPLY [16 votes]: No. The important thing to know is that, if $K \subseteq L$ is a field extension and $v: K \to \mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by working in some easy to handle subfield of $\mathbb{R}$. I'll work in $K = \mathbb{Q}(\sqrt{5})$ for the first question and in $K = \mathbb{Q}(\sqrt{3})$ for the second. The ring of integers in $\mathbb{Q}[\sqrt{5}]$ is $\mathbb{Z}[\tau]$ where $\tau = \tfrac{1+\sqrt{5}}{2}$, with minimal polynomial $\tau^2=\tau+1$. Note that $\mathcal{O}_K/(2 \mathcal{O}_K)$ is the field $\mathbb{F}_4$ with four elements. Your first statement is true in $\mathbb{Q}$ only because $\mathbb{Z}/(2 \mathbb{Z})$ has two elements. Specifically, both $1$ and $\tau$ are in $\mathcal{O}_K$ but not $2 \mathcal{O}_K$, so $v(1) = v(\tau) = 0$, but $1+\tau$ is also not in $2 \mathcal{O}_K$ so $v(1+\tau)=0$ as well. Similarly, the ring of integers in $\mathbb{Q}(\sqrt{3})$ is $\mathbb{Z}[\sqrt{3}]$ and the prime $2$ is ramified, with $2 = (1+\sqrt{3})^2 (2-\sqrt{3})$ (note that $2-\sqrt{3}$ is a unit). We have $v(1) = v(\sqrt{3}) = 0$, but $v(1+\sqrt{3}^2) = 2$. In this case, the result is true in $\mathbb{Q}$ because $2$ is unramified.<|endoftext|> TITLE: Box stacking problem QUESTION [5 upvotes]: Real world problem alert: I am moving from my house to another one, and the problem below arised when I tried to fit some little boxes of various shapes into a large box: We are given a positive integer $n$ and $3$-tuples consisting of positive rationals $(w_i, l_i, h_i)$ for $i \in \{1,\ldots,n\}$. For each $i$, the tuple $(w_i, l_i, h_i)$ represents a box of width $w_i$, length $l_i$, and height $h_i$. We want to put these $n$ boxes into one big box of rational dimensions $(W, L, H)$ such that the volume $W\cdot L\cdot H$ of the big box is minimal. Is the problem of finding (one possible choice of) $(W,L,H)$, such that $WLH$ is minimized, computable? EDIT. Thanks to Reid Barton and Tony Huynh for their comments below. The boxes can be rotated arbitrarily. REPLY [4 votes]: Yes, this is computable. And the infimum is attained. To see this, observe you can express a placement of your boxes by giving the coordinates of each box corner together with the angles. Then this placement is a valid packing if a (big) collection of simple inequalities are satisfied. It's clear that there are only finitely many combinatorially different placements (as defined, say, by extending all the bounding planes and looking at the combinatorial structure of the corresponding arrangement of planes) and hence one can simply minimise (continuously, hence attaining the minimum) over each structure. This is (a) possible, actually fairly easily, but Tarski's theorem is easier to quote, and (b) not in any way an efficient algorithm..!<|endoftext|> TITLE: Relationship between AC, WO, and Zorn's lemma in ZF-Powerset QUESTION [8 upvotes]: In regular ZF, AC, WO, and Zorn's Lemma are equivalent, but every proof I know (of the implication AC -> WO and AC -> Zorn) uses the axiom of choice on the powerset of X (where X is the Set which is to be well-ordered). My question is, whether or not there is a proof of this equivalence that doesn't use the axiom of choice or whether there is a Model of ZF-Powerset in which AC holds but the well-ordering principle (or Zorn's Lemma) fails. REPLY [13 votes]: This is a classic theorem of Zarach, that it is consistent that ${\sf ZF}^-$ holds with the Axiom of Choice, but not every set can be well-ordered. Zarach, Andrzej, Unions of ${\sf ZF}^-$models which are themselves ${\sf ZF}^-$ models, Logic colloquium ’80, Eur. Summer Meet., Prague 1980, Stud. Logic Found. Math. 108, 315-342 (1982). ZBL0524.03039.<|endoftext|> TITLE: Complete dual of bornological space QUESTION [6 upvotes]: A bornologigal topological vector space is such that any bounded linear function on it is continuous. It is a standard result [Jarchow, Locally convex spaces, 1981] that if the dual $E'$ of a Mackey space $E$ is complete for the topology $\mathcal{T}_{\mathcal{B}_0}$ of uniform convergence on bipolars of null sequences, then the Mackey space is bornological. In a report thesis [Gach, Topological versus Bornological Concepts in Infinite Dimensions, 2004, Thm 6.1.16], this result even in stated for the strong topology on $E'$. Does this result appears anywhere else ? It seems quite strong to me, and I am not sure I understand the proof. REPLY [3 votes]: Jochen is quite right. I have another example, just using any irreflexive Banach space $A$. The space $E = (A^*,\mu(A^*,A))$ is Mackey, by definition. The bounded sets in $E$ are the same as the norm-bounded sets, because $A$ is Banach, and therefore barrelled, so all dual topologies on $A^*$ have the same bounded sets (because $\sigma(A^*,A)$-bounded $\Leftrightarrow$ equicontinuous). So the canonical embedding $i(A)$ of $A$ in the strong dual of $E$ is an isomorphism, and so $E^* \cong A$ is complete. But $E$ is not bornological, because by the same characterization of the bounded sets of $E$, the bounded linear functionals on $E$ are exactly $A^{**}$, which contains $E^* = i(A)$ as a proper subspace by the assumption that $A$ be irreflexive. I think this also shows that Gach's proof of (4) $\Rightarrow$ (1) is at fault when he says "apply (4.1.5)", because if we apply his proof to $E$, the $F$ obtained in the proof will be $A^*$ with its norm topology, and $A^*$ and $E$ don't have the same set of continuous linear functionals.<|endoftext|> TITLE: Origin of $j$-invariant QUESTION [7 upvotes]: It is often asserted that the $j$-invariant was first introduced by Felix Klein. Is there any evidence for this claim? What works of Felix Klein do deal with it? What is the origin of the symbol $j$ used for it? Where it had first appeared? REPLY [5 votes]: Felix Klein, Über die Transformation der elliptischen Funktionen und die Auflösung der Gleichungen fünften Grades, Math. Ann. 14,‎ 111-172 (1878-1879). There is also a slightly earlier brief note in Italian, Sull' equazioni dell' Icosaedro nella risoluzione delle equazioni del quinto grado, Rendiconti Reale Istituto Lombardo, Serie 2, vol. 10, p. 253-255‎ (1877). Dedekind may have gotten there first, see Dedekind or Klein (p. 67). Notation: Klein used capital $J$, there is a factor 1728 difference with the $j$-invariant. (Read more about the origin of this factor.)<|endoftext|> TITLE: Does symplectic morphism after geometric quantization induce Hilbert spaces morphism? QUESTION [5 upvotes]: Let us have two symplectic manifolds $(M, \,\omega)$ and $(N, \,\omega')$ and morphism between them: $$ \varphi \ :\ M \to N.$$ Then we geometrically quantize these systems: we add a prequantum line bundle, $ L_M \to M$ and $L_N \to N$ correspondingly; choose some real polarizations $P_M$ and $P_N$ such that $(\varphi_*P_M)(x) \neq P_N(x), \, \text{for }x \in \varphi(M)$; let the quotient maps $\pi_{M}\ :\ M \to M/P_M$ and $\pi_N\ :\ N \to N/P_N$ be fibration. Then the space of compactly supported sections of $L_M/P_M \otimes |T(M/P_M)|^{1/2} $ after completion realise Hilbert space $H_M$. The same procedure for $N$ gives $H_N$. I don't understand how to build morphism between $H_M$ and $H_N$ from $\varphi$. Is there any natural way to do it? I'm not sure whether it's a trivial question or not. I don't have deep understanding of geometric quantization, I've only read this review by Lerman. So if some comprehensible text about the question exists, I'll gladly accept a reference. P.S. I'm reading this paper by Nekrasov which describes morphism between classical Calogero-Moser and Calogero-Sutherland models, and derives from it morphism between their quantum versions. Nekrasov briefly describes how he gets quantum morphism from classical one in the beginning of 4th section. But I don't understand it. So I'm interested whether there is a general way to do it. If it is possible only in this particular case, I'd like to get a more thoroughly explanation of Nekrasov's procedure. REPLY [2 votes]: The slogan to keep in mind is that "quantization is not a functor". For details on this slogan, see the (decade old!) Mathoverflow question What does “quantization is not a functor” really mean?. Basically, "is not a functor" means that there is no canonical way to build morphisms of Hilbert spaces from morphisms of symplectic manifolds. Of course, in many practical cases there may be some "best" morphism of Hilbert spaces for a given "nice" morphism of symplectic manifolds, or some family of "best" morphisms. For instance, perhaps in your practical case you have some control over how your $\varphi$ interacts with the prequantum and polarization data.<|endoftext|> TITLE: Tight error terms for partial sums $\sum_{n\leq x} 1/n^s$ QUESTION [6 upvotes]: (a) Let $s>1$, $x>0$ be real. Then it is not hard to see that $$\sum_{n\leq x} \frac{1}{n^s} \leq \zeta(s) - \frac{1}{(s-1) x^{s-1}} + \frac{1}{2 x^s},$$ basically because $x\mapsto 1/x^s$ is convex. (The naive bound would have $1/x^s$ instead of $1/2 x^2$.) By Euler-Maclaurin, this bound is tight, in the sense that the inequality would not be valid for large $x$ if $1/2$ were replaced by a smaller constant. This bound looks as if it should be completely standard (in fact, known since the umpteenth century). Is there an easy reference? Also, what happens for real $0 TITLE: A group theoretic interpretation of Lagarias inequality QUESTION [11 upvotes]: Let $G$ be a finite group, $S \subset G$ a generating set. Set $\sigma(G):=\sum_{U \subset G} |U| $, where the sum runs over all subgroups $U$ of $G$. Set $H_G := \sum_{g \in G} \frac{1}{|g|+1}$, where $|g|:= $ word length (with respect to $S$). For $G:=\mathbb{Z}/(n)$ we get $\sigma(G) = \sigma(n)=$ sum of divisors of $n$. and $H_{\mathbb{Z}/(n)} = H_n=n$-th harmonic number, where $S=\{+1\}$. My naive conjecture inspired by Lagarias inequality is $$ \sigma(G) \le H_G + \exp(H_G) \log(H_G)$$ For $G:=\mathbb{Z}/(n)$ and $S:=\{+1\}$ this is the Lagarias inequality. I have checked in Sagemath for the symmetric group up to $n=6$: def sigmaGr(G): return sum([len(U.list()) for U in (G.subgroups())]) def wordLen(g): return g.length() def HG(G): return sum([1/(wordLen(g)+1) for g in G.list()]) def LG(G): H = HG(G) return (H+exp(H)*log(H)).N() for n in range(1,6): G = SymmetricGroup(n) print sigmaGr(G),LG(G) My question is, if this inequality can be proved for the generating set $S:=G$ or if there are finite groups and generating sets for which this inequality is false? For $S=G$ it is $H_G=(|G|+1)/2$ and for $G$ the cyclic group, the inequality reduces to $$\sigma(n)\le (n+1)/2+\exp((n+1)/2)\log((n+1)/2)$$ so the question is if one can prove this inequality? Related: https://math.stackexchange.com/questions/3204237/a-group-theoretic-interpretation-of-lagarias-inequality Edit 24.05.2019: It seems that it is better to define $\sigma(G)$ as $= \sum_{H \le G} [G:H]$ which in the case of cyclic groups is equal to the first definiton $=\sigma(n)$. Also this notion of $\sigma(G)$ is related to the zeta function of the finite group $G$ as we have: $$\zeta_G(-1) = \sigma(G)$$ where $$\zeta_G(s) = \sum_{H \le G} \frac{1}{[G:H]^s}$$ REPLY [10 votes]: The general inequality is interesting, but the special case you want is easy to prove -- at least if $n$ is large, but the proof can easily be quantified. Note that every finite group $G$ of size $n$ can be generated by at most $\lfloor \log n/\log 2 \rfloor$ elements. You can see this greedily. Suppose a set $S$ generates a subgroup of $G$ of size $k$. Add an element not in this subgroup to $S$. The group generated by $S$ plus this element has size at least $2k$. Therefore the number of subgroups of $G$ is bounded by the number of all possible subsets of $G$ with size at most $\lfloor \log n/\log 2\rfloor$, which is $$ \sum_{k=0}^{\lfloor \log n/\log 2\rfloor} \binom{n}{k}. $$ (See also General bound for the number of subgroups of a finite group ) Thus $$ \sigma(G) \le n \sum_{k=0}^{\lfloor \log n/\log 2\rfloor} \binom{n}{k} =\exp(O((\log n)^2), $$ which is certainly smaller than $\exp(n/2)$. It should be easy enough to compute a reasonable value of $n$ from which the inequality holds.<|endoftext|> TITLE: Is every universally catenary ring a going-between ring? QUESTION [6 upvotes]: This question asks for the necessity of a noetherian hypothesis in a certain relation between properties of rings concerning chains of prime ideals. We use the following definitions. A ring $R$ is called universally catenary if every $R$-algebra of finite type is catenary. (Note that $R$ need not be noetherian.) A ring $R$ is called a going-between ring if for every integral ring extension $R\subseteq S$, every saturated chain of primes in $S$ contracts to a saturated chain of primes in $R$. From results by Ratliff we get the following. Theorem: Every noetherian universally catenary ring is a going-between ring. The proof is rather complicated, as it relies on several non-trivial results concerning relations between different chain conditions. In particular, it is not clear to me whether one can get through without noetherianness. Is it known whether or not we can omit the noetherian hypothesis in the above result? REPLY [4 votes]: OK, let $R \subset S$ be an integral ring extension with $R$ universally catenary. Let $\mathfrak q \subset \mathfrak q'$ be primes in $S$ such that there is no prime strictly in between them. We have to show that the same is true for the corresponding primes $\mathfrak p \subset \mathfrak p'$ of $R$. Reduction to the finite case. Suppose to the contrary that $\mathfrak p''$ is strictly in between. Write $S$ as the filtered colimit $S = \text{colim} S_i$ of its finite sub $R$-algebras. Denote $E_i$ the set of primes of $S_i$ which lie over $\mathfrak p''$ and are strictly in between $\mathfrak q \cap S_i$ and $\mathfrak q' \cap S_i$. If we know the result in the finite case then $E_i$ is nonempty. The inclusions between the rings $S_i$ define transition maps between the sets $E_i$ (this is a standard fact about finite ring maps). Now $E = \text{lim} E_i$ is nonempty as a cofiltered limit of finite nonempty sets. An element of $E$ is the same thing as a prime of $S$ strictly between $\mathfrak q$ and $\mathfrak q'$. Finite case. Choose a surjection $P = R[x_1, \ldots, x_n] \to S$. The primes $\mathfrak q \subset \mathfrak q'$ correspond to primes $\mathfrak r \subset \mathfrak r'$ of $P$ with no prime strictly in between. Note that $\mathfrak r$ defines a closed point of the fibre of $\text{Spec}(P) \to \text{Spec}(R)$ over $\mathfrak p$. Hence there is a length $n$ saturated chain of primes starting with $\mathfrak p P$ and ending with $\mathfrak r$. Thus we have a saturated chain of length $n + 1$ between $\mathfrak p P$ and $\mathfrak r'$. But if we have $\mathfrak p''$ as above, then we can make a chain longer than this (and arrive at a contradiction). Namely, we can first consider $\mathfrak p P \subset \mathfrak p'' P \subset \mathfrak p' P$ and then make a chain between $\mathfrak p' P$ and $\mathfrak r'$ in the fibre as before.<|endoftext|> TITLE: Analogues of Hecke relations for Maass forms QUESTION [6 upvotes]: If a (suitably normalised) holomorphic cusp newform has q-expansion $$f(z) = \sum_n \lambda_f(n) e(nz),$$ then we know the Hecke relations for $(mn,q)=1$, $$(\star) \qquad \lambda_f(m)\lambda_f(n) = \sum_{d | (m, n)} \lambda_f\left( \frac{mn}{d^2} \right).$$ In the case of a Maass cusp form, we can write the expansion $$u(z) = \sum_{n \geqslant 1} \rho_u(n) W_s(nz),$$ where $s$ is related in an explicit way to the associated eigenvalue and $W_s$ is a Whittaker function. Similarly, the continuous part of the spectrum made of Eisenstein series at a cusp $a$ admits expansion of the form $$E_a(s,z) = \phi_a y^s + \phi_a(s) y^{1-s} + \sum_{n \geqslant 1} \phi_a(n,s)W_s(nz)$$ I am interested in the following question : Are there analogous "Hecke relations" for the coefficients of Maass forms, $\rho_u(n)$ and $\phi_a(n,s)$? I suppose so, but I do not have any good reference for these matters. REPLY [4 votes]: Two remarks supplementing the excellent comments below your post. The relation $(\star)$ holds for every $m$ and $n$ if your restrict the sum to $(d,q)=1$. This is because $f$ is a newform (and we assume that $f$ has trivial nebentypus). The same is true for Maass cuspidal newforms. You can get a Hecke eigenbasis of Eisenstein series if you induce the basis vectors from Dirichlet characters modulo $q$. See Section 2.1.1 of this article for a quick introduction. I also recommend you to study Chapter 7 of Miyake's book, which discusses these induced Eisenstein series in detail and in classical language.<|endoftext|> TITLE: Packing rectangles: Does rotation ever help? QUESTION [10 upvotes]: Dominic van der Zypen posed an interesting Box stacking problem. This is a spin-off question. Let a collection of rectangles $r_1,\ldots,r_n$ be given by their side lengths in $\mathbb{R}$. Let $R$ be a rectangle of minimum area enclosing the rectangles arranged in the plane without overlap (i.e., with disjoint interiors). Q. Is there an example where not all the rectangles have sides aligned with the sides of $R$? In other words, where at least one rectangle's sides are not parallel to the sides of $R$? Is it ever advantageous to "tilt" one or more rectangles to achieve a minimal area? REPLY [12 votes]: The classic answer to this is a paper of Erdos and Graham 'On packing squares with equal squares'. Given a square of side $n+\varepsilon$, where $0<\varepsilon<1$, we can obviously fit in $n^2$ unit squares, and it's fairly trivial to check that if the squares are axis aligned this is best possible (count the squares intersecting vertical lines on the integers). Obviously, this means about $2\varepsilon n$ area is going to waste. But Erdos and Graham show one can cover asymptotically all but $n^{7/11}$ area, using skew angles - this is maybe more surprising than Yosemite Stan's example (which also works perfectly well).<|endoftext|> TITLE: What is the relationship between spectral sequences and obstruction theory? QUESTION [11 upvotes]: Let $X,Y$ be objects of some category $\mathcal C$, and suppose I want to study homotopy classes of maps from $X$ to $Y$ (almost everything one does in algebraic topology can be viewed this way). It seems as though there are two classes of methods available to me: I can approach this via some kind of obstruction theory, or else I can use some kind of spectral sequence. But what makes a method of computation "obstruction-theoretic" versus "spectral sequence-theoretic"? It seems as though both types of method typically take as their starting point some kind of filtration of one of the objects involved (or else a filtration of whatever data defines an object of $\mathcal C$), and proceed by analyzing the problem in terms of the layers of the filtration -- they yield the obstruction groups / the starting page of the spectral sequence. Perhaps the methodologies diverge in an essential way after that? For instance, spectral sequences seem like a much more "structured" methodology -- are they available only in a more restricted class of situations? Questions: What is the relationship between obstruction theory and spectral sequences in general? Is there a systematic way to pass back and forth between the two viewpoints? For example, does every spectral sequence have an associated obstruction theory or vice versa? For a very concrete example, suppose that by "obstruction theory", I mean the most classical possible thing -- computing the space of lifts in a commutative square of spaces via a Postnikov tower. Is there an associated spectral sequence (perhaps under certain conditions)? How does one pass back and forth? Conversely, take your favorite spectral sequence. Is there an associated obstruction theory? REPLY [14 votes]: This is a partial answer, but every obstruction theory (in some precise sense) provides you with a spectral sequence (in fact several). Let me clarify what do I mean with obstruction theory. All this material is a modern reinterpretation of Bousfield's amazing paper Homotopy Spectral Sequence and Obstructions, although I am describing a slightly different thing than he is because I want the Goerss-Hopkins-Miller obstruction theory to be a special case of what I'm writing here. Suppose you have a moduli space $\mathcal{M}$ of some kind of objects you wish to study, which has a forgetful map to some moduli space of objects you understand $\mathcal{M}_1$. Typically $\mathcal{M}$ is going to be a moduli space of "topological" objects, while $\mathcal{M}_1$ is a moduli space of "algebraic" objects, but of course this is not required. For example $\mathcal{M}$ could be the space of sections of some fibrations, and $\mathcal{M}_1$ the groupoid of the possible lifts at the level of fundamental groupoids. Then an obstruction theory is a factorization of $\mathcal{M}\to \mathcal{M}_1$ as a tower $$\mathcal{M}\to \cdots \to \mathcal{M}_2\to \mathcal{M}_1$$ where $\mathcal{M}\cong \lim \mathcal{M}_n$ and for every $i\ge 0$ there's a cartesian square $$\require{AMScd} \begin{CD} \mathcal{M}_{i+1} @>>> BG_i\\ @VVV @VVV \\ \mathcal{M}_i @>>> (A_i)_{hG_i} \end{CD}$$ where $G_i$ is some discrete group and $A_i$ is some group-like $E_1$-space equipped with a $G_i$-action, such that the vertical map is the one induced by the zero map $*\to A_i$. For example, if $E\to B$ is a map, $\mathcal{M}=\mathrm{Map}_B(B,E)$ is the space of sections, we can let $E\to E_n\to B$ be the $n$-th relative Postnikov truncation and $\mathcal{M}_n=\mathrm{Map}_B(B,E_n)$, so that $\mathcal{M}_1$ is the space of lifts at the level of fundamental groupoids. Why am I calling this an obstruction theory? Well, the reason is simple: suppose we have a point $x_1\in \mathcal{M}_1$ and we want to lift it to $\mathcal{M}$. First we need to lift it to $\mathcal{M}_2$. This is the same as lifting its image in $(A_1)_{hG_1}$ to $BG_1$, wich is possible iff its image in $\pi_0(A_1)_{hG_1}=(\pi_0A_1)/G_1$ is zero. Moreover the space of possible lifts is parametrized by the fiber, which is $\Omega A_1$, and in particular the equivalence classes of lifts are parametrized by $\pi_1A_1$. Iterating this procedure we obtain an infinite sequence of obstruction classes. If all these obstruction classes vanish, we have reached a point of $\mathcal{M}$. This in the above example recovers the classical obstruction theory for the sections of a fibration. Ok, we know what we mean with an "obstruction theory" now, but where do spectral sequences come in? The point is that it is all well and good to be able to construct points of $\mathcal{M}$, but it would be nice to have a better description of its homotopy types (i.e. how many equivalence classes are in $\mathcal{M}$ and what are their automorphism groups). Let us fix a basepoint $x\in\mathcal{M}$, constructed with the above procedure. Then, Bousfield and Kan taught us that to every tower of fibrations there is an associated fringed spectral sequence. Note that the pullback square forces the fiber of $\mathcal{M}_{i+1}\to\mathcal{M}$ to be equivalent to $\Omega A_i$, so we can write the spectral sequence as $$\pi_{t+1}A_s\Rightarrow \pi_{s-t}\mathcal{M}$$ where for convenience of notation I write $A_0=\mathcal{M}_1$ (i.e. I'm fixing $\mathcal{M}_0=*$). Note that this spectral sequence is "fringed" (i.e. the low degree terms are not abelian groups), and so it needs to be handled with care, but it still very useful. An example is the Goerss-Hopkins-Miller obstruction theory, where they compute the moduli space of $E_∞$-structures on Morava E-theory and show it is essentially algebraic. Note Bousfield actually works with a slightly different tower, the Tot-tower of a cosimplicial resolution of $\mathcal{M}$. In practice the towers that appear tend to be of the above form, and the above description covers obstruction theories that do not come from cosimplicial resolutions, but the difference is probably something worth being aware of.<|endoftext|> TITLE: Pressure to defend the relevance of one's area of mathematics QUESTION [118 upvotes]: I am a set theorist. Since I began to study this subject, I became increasingly aware of negative attitudes about it. These were expressed both from an internal and an external perspective. By the “internal perspective,” I mean a constant expression of worry from set theorists and logicians about the relevance of their work to the broader community / “real world”, with these worries sometimes leading to career-defining decisions on the direction of research. For me, this situation is unwanted. I studied set theory because I thought it was interesting, not because I wanted to be a soldier in some kind of movement. Furthermore, I don’t see why an area needs defending when it produces a lot of deep theorems. That part is hard enough. To what degree does there exist, in the various areas of mathematics, a widespread feeling of pressure to defend the relevance of the whole subject? Are there some areas in which it is enough to pursue the research that is considered interesting, useful, or important by experts in the field? Of course there will always be a demand to explain “broader impacts” to funding agencies, but I am talking about situations where the pressure comes from one’s own colleagues or even one’s own internalized sense of what is proper research. REPLY [3 votes]: this notion that there is X and then there is rest of mathematics is a very very strange view of mathematics. mathematics is a fluid subject, ideas float around yielding results in many different places. partitioning mathematics into groups and putting names on them certainly has a value and purpose, but neither its value nor its purpose is to classify some of them as "normal" and some as "not normal". for me personally, it is incredibly difficult to say what set theory is. often i have difficulty saying whether a paper is set theory or not. perhaps it is possible to make a distinction between computational math and more abstract math, like calculus vs abstract algebra, but it doesn't work. the boundaries are blurred. there is an example above where a user says that you cannot claim that the notion of being countable is set theoretic. very well! exactly the point! mathematical concepts are fluid concepts that appear in many forms and many places. in the case of the notion of "being countable", if you are not giving this to set theorists then you are admitting that the very action of splitting math into such boxes is silly, it is much better to think that "being countable" is a mathematical notion rather than set theoretic. of course, there are concepts and notions that fall into groups. it will be hard to argue that triangles are not geometric. but the point is that the boundaries of these subjects are so confused and meshed that you will ultimately run into problems. geometry has been connected to algebra more times than you can count. is a group a geometric concept or an algebraic concept? we can debate this for the sake of debating mathematical points of views, but not for the sake of defining subjects, like drawing a boundary around geometry and algebra with no intersections, such an endeavor is ultimately divisive, meaningless and above all pointless. there is no set theory or number theory, there is mathematical thought process, and people respond to that thought process differently when they are subjected to it, some pursue very concretely defined objects such as numbers, some pursue less concretely defined geometric shapes and some pursue the mysteries of infinity. more people pursue number theory, why? i don't know, i suspect it is because it is easier to grow in the subject, but if you are like me then you would leave number theory because number theory never made me feel satisfied with myself, i was never happy with myself while growing up in number theory, it was like i was living somebody else's life, set theoretic ideas and its mysteries did make me feel fulfilled and happy. i believe what we study has a lot to do with our personalities. i do not like the concrete, it bores me, i like what i don't understand, i like that there is independence and vagueness and my job is to clear it. i have noticed this exact characteristics in myself in many aspects of life, in my choice of music, films, books and etc. there are different personalities. some mathematicians are problem solvers, and they want problems you can state in a few minutes and that they spend a month thinking about it. set theory of course has many of them of various difficulty. some were mentioned earlier by others. some people like theories and stories. there are people in between. i cannot say that i am doing math so that i can then tell it to others, i believe this is more like a personal quest, at least that is how i got into it. however, this is a profession, and we need to be professionals. while hearing things like "X is disconnected from the rest of math" is stressful and frustrating, there are many situations where i enjoy defending myself and explaining the meaning of what i do. i enjoy giving general audience talks or write introductions to my papers. i think this is part of the profession and is important, given that the society ultimately pays our salary. i do agree, though, that there is too much out there claiming that set theory is somehow irrelevant. but as i hinted above, most cases of this that i have encountered were expressed by people who didn't really know what they were claiming. already the meaning of the word "relevance" is problematic for me.<|endoftext|> TITLE: Problem based representation theory book QUESTION [5 upvotes]: I am trying to find books similar in the spirit of Ram Murty's Problems in Analytic Number theory or Murty Esmonde's Problems in Algebraic number theory in the field of Representation Theory (of groups, algebras, Lie algebras, possibly even goes in the direction of quivers). I want a book which doesn't provide details but directs me in the right direction via lots of exercises. One other book that kind of fits this category maybe Farb and Dennis's Noncommutative Algebra. But it is not purely focused on representation theory although some of the topics are. Any other suggestions? REPLY [3 votes]: There is also the book by William Fulton and Joe Harris : Representation Theory (Springer, corrected edition,551 pages). It contains many exercises, with hints, answers, and references (pp. 516-535).<|endoftext|> TITLE: Are there universal homological functors? QUESTION [5 upvotes]: There is a bifunctor $H: Stab^{op} \times Ab \to Top$ where $H(C,A)$ is the space of homological functors $C \to A$. Is this bifunctor left or right representable? That is, for each small abelian category $A$ does there exist a small stable $\infty$-category $\underline A$ and a homological functor $\underline A \to A$ such that every homological functor $C \to A$ (where $C$ is a small stable $\infty$-category) factors uniquely through $\underline A \to A$ via an exact $\infty$-functor $C \to \underline A$? On the other hand, for each small stable $\infty$ category $C$, does there exist an abelian category $\overline C$ and a homological functor $C \to \overline C$ such that every homological functor $C \to A$ (where $A$ is a small abelian category) factors uniquely through $C \to \overline C$ via an exact functor $\overline C \to A$? REPLY [8 votes]: Here is half an answer (i.e. an answer to the last question). Given a small stable $\infty$-category $C$ and a small abelian category $A$, any functor $C\to A$ factors through the homotopy category $ho(C)$ of $C$. Hence the category of homological functors from $C$ to $A$ is equivalent to the category of homological functors from the triangulated category $ho(C)$ to $A$. There is a universal homological functor from $ho(C)\to A(C)$: this is documented in Verdier's thesis (see Théorème 3.2.1 in Astérisque 239, an electronic copy being available here), although Verdier thanks Freyd for sharing the construction with him. A modern reference in English is Theorem 5.1.18 in Neeman's book.<|endoftext|> TITLE: How to study the global stability for this 3D system? QUESTION [7 upvotes]: I am studying a biological system (HIV) and arrived at this simplified dynamical system: \begin{align} x_1' &= a_1 + a_2x_2 - a_1x_2 - a_4x_1 - a_5\frac{1+a_6x_3}{1+a_7x_3}x_1\\ x_2' &= a_5\frac{1+a_6x_3}{1+a_7x_3}x_1 - a_2x_2\\ x_3' &= a_8x_1 - a_9x_3 \end{align} Where as the context and notation dictate: all coefficients $a_i$ are strictly positive. I also showed all $x_i$ to be strictly positive and bounded. I have obtained locally-asymptotically stability for the unique positive fixed point. I also tried various lyapunov functions, but without much luck. I would truly appreciate any idea to approach this problem, perhaps a form of Lyapunov function, a direct method, etc that I should try. Additionally, I have done extensive numerical simulations, which suggests the positive steady state is globally asymptotically stable. Edit: for those interested in the original system - especially to back up the various claims that I made in my question. Please note that the parameters do not match to the system in the question. It's just a convenient way to write the parameters. \begin{align} x_1' &= a_1x_2 -a_2x_1\\ x_2' &= -(f(y_2) + a_1) x_2 + a_3x_3 + a_2x_1\\ x_3' &= f(y_2)x_2 - a_3x_3\\ y_1' &= b_1x_2 - (b_2+b_3)y_1\\ y_2' &= b_2y_1 - b_4y_2\\ f(y_2) &= \frac{c_1}{c_2} \frac{1 + c_2(y_2/c_3)^n}{1+(y_2/c_3)^n}. \end{align} The only conditions on the parameters are that they are positive and $c_2>1$. Note that the first 3 equations are conservative, hence it can be reduced. And at any time, $x_1 + x_2 + x_3 = 1$. This system is part of a larger system, but is decoupled from the rest. REPLY [5 votes]: This is more a long comment than an answer, but I know that a similar problem, also originated from (macromolecular) biology, was studied and solved by Gaetano Fichera, Maria Adelaide Sneider and Jeffreys Wyman. The system of ODEs they studied was the following one $$ \left\{ \begin{split} x_1' &= Lc_1^2 - (\rho_1 + 2Lc_1 + Qc_2)x_1 + (P - 2Lc_1)x_2 — Lc_1x_3 + Lx_1^2 \\ &\quad+ (2L + Q)x_1 x_2 + (L + Q)x_1x_3 + Lx_2^2 + Lx_2x_3,\\ \\ x_2' &=Qc_2x_1- \rho_2 x_2 + Nc_1x_3- Qx_1x_2- (N+Q)x_1x_3- Qx_2x_3,\\ \\ x_3' &=Mc_1c_2- Mc_2x_1+\big[\rho_2 - P- M(c_1+c_2)\big]x_2- \big[\rho_3+Nc_1+M(c_1+c_2)\big]x_3\\ &\quad+ Mx_1x_2+ (M+N)x_1x_3+Mx^2_2+(2M+N)x_2x_3 + Mx_2^2. \end{split} \right. $$ where $\rho_1, \rho_2, \rho_3, c_1, c_2 L, M, P, Q$ are positive constants satisfying only the condition $\rho_2 > P$. This system of ODEs is related to the behavior of the pigment rhodopsin in the presence of light and the problem studied in [1] (of which [2] is a summary of results) is equivalent to prove the existence of a unique critical point $\xi=(\xi_1,\xi_2,\xi_3)$ and its asymptotic stability. I read about this problem from the obituary of Maria Adelaide Sneider and from the "Last lesson" of Gaetano Fichera: the initial condition $x(0)$ was required to belong to a given polyhedron $K\in\Bbb R^3$, and in [1] the existence and uniqueness of a critical point was proven without restriction on the coefficients, while the asymptotic stability was proved only locally (and globally only under restrictions on the coefficients). The problem was fully solved only ten years after by Maria Adelaide Sneider in the paper [3], which have been recently digitalized by the Italian Mathematical Union. According to Fichera, she was able to find a continuous Lyapunov function having only piecewise continuous first derivatives: then, by decomposing $\Bbb R^3$ in 16 "pyramidal" regions, she was able to prove that, in the intersection of each of them with the polyhedron $K$, conditions for the global asymptotic stability are attained. [1] Gaetano Fichera, Maria Adelaide Sneider, Jeffreys Wyman (1977), "On the existence of a steady state in a biological system", Atti della Accademia Nazionale dei Lincei. Memorie. Classe di Scienze Fisiche, Matematiche e Naturali, Serie VII, Sezione III, XIV (1): 1–26, Zbl 0414.92004. [2] Gaetano Fichera, Maria Adelaide Sneider, Jeffreys Wyman (1977a), "On the existence of a steady state in a biological system", PNAS, 74 (10): 4182–4184. [3] Sneider, Maria Adelaide (1988), "Steady state in a biological system: global asymptotic stability", Atti della Accademia Nazionale dei Lincei. Rendiconti. Classe di Scienze Fisiche, Matematiche e Naturali, Serie VIII (in Italian), LXXXII (2): 345–352, MR 1152650, Zbl 0964.92502.<|endoftext|> TITLE: Do $p$-adic topological modular forms exist? QUESTION [8 upvotes]: Are there $p$-adic topological modular forms? What is the analogue of finite slope and overconvergent? REPLY [12 votes]: There are very natural analogues of $p$-adic modular forms in the study of topological modular forms ($tmf$)--they are $K(1)$-local topological modular forms. One can obtain the classical ring of p-adic modular forms by first inverting the Eisenstein series $E_{p-1}$ and then $p$-adically completing (at least for large primes; at 2 and 3 we have to be a little more careful). The results are modular forms that are defined on ordinary elliptic curves over p-complete rings with no supersingular contributions; i.e. on a moduli of elliptic curves whose formal group has exact height $1$. The keyword for the analogue in topology is called $K(1)$-localization, and it is carried out in roughly the same way--we would like to narrow in on the portion that carries exact height $1$, by inverting a class (topologists call this "$v_1$") and $p$-adically completing. The technical details are a little trickier because it applies to a wider class of objects than $tmf$, and every prime behaves like the prime 2 or 3 does in the classical case. The problem can be solved in a similar way: for every $k$, some power $v_1^{p^d}$ of $v_1$ exists mod $p^k$, and so it still makes sense to invert $v_1$ mod $p^k$. We can then assemble these into an inverse system and take the limit. As a sub-comment: the weight of a $p$-adic modular form is no longer necessarily an integer. On the topological side, weight corresponds to grading: the coefficient group $\pi_n tmf$ is a set of homotopy classes of maps $S^n \to tmf$. Generally, there are not spheres with non-integer dimensions, but in the $K(1)$-local category there are: the full set of gradings is called the $K(1)$-local Picard group. At odd primes it is $Hom(\Bbb Z_p^\times, \Bbb Z_p^\times) \cong \Bbb Z_p \times \Bbb Z/(p-1)$, and so we can define groups $\pi_n tmf_{K(1)}$, a group of weight-$n$ topological modular forms where $n$ is now drawn from this larger group that expresses something like $p$-adic interpolation. At $p=2$ there is an additional "exotic" part of the grading with no algebraic analogue. I don't have a good answer to the second half of your question. As far as I know, the analytic techniques you would need to give a definition of overconvergence are something that don't yet exist on the topological side. I think that this analytic point of view leads to really interesting questions, but while I've thought some about trying to answer them I don't have anything of substance I can offer right now.<|endoftext|> TITLE: Dirac operator on a Morita equivalent algebra QUESTION [5 upvotes]: Let $(A,H,D)$ be a spectral triple and let $B$ be an algebra which is Morita equivalent to $A$. Then there exists a finitely generated, projective $A$ module $E$ such that $B=End_A(E)$. Endow $E$ with $A$-valued inner product $(\cdot,\cdot)$ and form a Hilbert space $H':=E \otimes_A H$ with the inner product given by $$\langle x \otimes \xi, y \otimes \eta \rangle_{H'}:=\langle (x,y)\xi,\eta \rangle_H$$ One would like impose the structure of a spectral triple on $B$: one naive choice is to put $D'(x \otimes \xi)=x \otimes D\xi$ however this map is not well defined since $H'$ is a tensor product over $A$ In order to fix this one uses a connection $\nabla$ i.e. a map $\nabla:E \to E \otimes_A \Omega_D^1$ where $\Omega^1_d=\{\sum_{j=1}^N a_jdb_j: a_j,b_j \in A, N \in \mathbb{N}\}$ and $db:=[D,b]$. This map has to satisfy $\nabla(xa)=\nabla(x)a+x\otimes da$ (note that due to Jacobi idenity for commutators the space $\Omega_D^1$ is in fact $A-A$-bimodule thus $E \otimes_A \Omega^1_D$ is a right $A$-module). Having fixed a connection one defines $$D'(x \otimes \xi)=x\otimes D\xi + \nabla(x)\xi$$ (where $\nabla(x)\xi$ is understood as follows: if $\nabla(x)=\sum_j x_j \otimes \omega_j$ then $\nabla(x)\xi=\sum_j x_j \otimes \omega_j(\xi)$ which is fine since each $\omega_j \in \Omega^1_D$ still acts on $H$). The natural action of $B$ on $H'$ is as follows $b(x \otimes \xi)bx \otimes \xi$ which is fine since $b \in B=End_A(E)$. How to show that for each $b \in B$ the commutator $[D',b]$ (extends to a/) is a bounded operator on $H'$? Simple computation shows that this commutator evaluated on $x \otimes \xi$ is equal to $\nabla(bx)\xi-b(\nabla x(\xi))$ and I don't see how to obtain that this operator is bounded-do we have to assume something on $\nabla$ (some sort of compatibility with the $B$-action)? REPLY [5 votes]: Your question is entirely covered by Section 2 of Brain–Mesland–Van Suijlekom, but the fgp case is simple enough to ultimately boil down to folklore proved by Chakraborty–Mathai. Let me summarise what happens, while incorporating some technical simplifications from Blecher–Kaad–Mesland. For convenience, let me take $A$ and $B$ to be unital $C^\ast$-algebras and let $\Omega_D$ be the unital $C^\ast$-subalgebra of $B(H)$ generated by $A$ and $[D,\mathcal{A}]$. Let $\mathcal{A} := \{a \in A \mid a \operatorname{Dom}(D) \subset \operatorname{Dom}(D),\, [D,a] \in B(H)\}$ be given the Lipschitz norm $\|a\| := \|a\| + \|[D,a]\|$. Then $\mathcal{A}$ defines an involutive operator algebra and the inclusion $\mathcal{A} \hookrightarrow A$ is completely bounded with dense range that is closed under the holomorphic functional calculus; in particular, it follows that $M_N(\mathcal{A})$ is dense in $M_N(A)$ and closed under the holomorphic functional calculus for any $N \in \mathbb{N}$. You should think of $\mathcal{A}$ as defining a Lipschitz structure on the NC space $A$. Let $\mathcal{E}$ be a dense subspace of $E$ satisfying $\mathcal{E} \cdot \mathcal{A} \subset \mathcal{E}$ and $(\mathcal{E},\mathcal{E})_A \subset \mathcal{A}$. By part 1 and the properties of $\mathcal{E}$, we can find $A$-module generators $\{\xi_1,\dotsc,\xi_N\} \subset \mathcal{E}$ for $E$, such that $$ \forall e \in E, \quad e = \sum_{i=1}^N \xi_i \cdot (\xi_i,e)_A. $$ Thus, if $p := \left((\xi_i,\xi_j)_A\right)_{i,j=1}^N \in M_N(\mathcal{A})$, then $e \mapsto \left((\xi_i,e)_A\right)_{i=1}^N$ defines an isomorphism $E \cong pA^N$ of Hilbert $A$-modules that restricts to an isomorphism $\mathcal{E} \cong p\mathcal{A}^N$ of pre-Hilbert $\mathcal{A}$-modules; this now makes $\mathcal{E}$ into a (finitely generated) projective operator module over $\mathcal{A}$ in a manner that depends on the choice of $\{\xi_1,\dotsc,\xi_N\}$ only up to completely bounded isomorphism. You should think of $\mathcal{E}$ as defining a Lipschitz structure on the NC vector bundle $E$; since $E$ is fgp, this Lipschitz structure is canonically induced by the choice of Lipschitz structure on $A$. Note that such $\mathcal{E}$ exists: given an isomorphism $E \cong p_0 A^N$ of Hilbert $A$-modules for $p_0 \in M_N(A)$ an orthogonal projection, you can use the properties of $\mathcal{A}$ to find an orthogonal projection $p \in M_N(\mathcal{A})$, such that $E \cong p A^N$, in which case, you can take $\mathcal{E}$ to be the pre-image of $p\mathcal{A}^N$ and $\{\xi_1,\dotsc,\xi_N\}$ to be given by the pre-images in $E$ of the columns of $p$. Let $\mathcal{B} := \{b \in B \mid b \cdot \mathcal{E} \subset \mathcal{E}\}.$ Since $$ \mathcal{B} = \left\{b \in B \mid \left((\xi_i,b\xi_j)_A\right)_{i,j=1}^N \in M_N(\mathcal{A})\right\}, $$ it follows that $\mathcal{B}$ is $\ast$-closed and that the $\ast$-isomorphism $B \cong p M_N(A) p$ induced by the Hilbert $A$-module isomorphism $E \cong p A^N$ restricts to a $\ast$-isomorphism $\mathcal{B} \cong p M_N(\mathcal{A}) p$. Thus, $\mathcal{B}$ can be topologised as a closed $\ast$-subalgebra of the involutive operator algebra $M_N(\mathcal{A})$, thereby making $\mathcal{E}$ into a Lipschitz $(\mathcal{B},\mathcal{A})$-correspondence. In particular, you can view $\mathcal{B}$ as defining a Lipschitz structure on the NC space $B$ compatible with the Lipschitz structure $\mathcal{A}$ on the NC space $A$. Let $\nabla : \mathcal{E} \to E \hat\otimes_A \Omega_D$ be a Hermitian connection, i.e., a $\mathbb{C}$-linear map satisfying $$ \forall e \in \mathcal{E},\, \forall a \in \mathcal{A}, \quad \nabla(ea) = \nabla(e)a + e \hat\otimes [D,a],\\ \forall e_1,e_2 \in \mathcal{E}, \quad (\nabla(e_1),e_2 \hat\otimes 1)_{\Omega_D} +(e_1 \hat\otimes 1,\nabla(e_2))_{\Omega_D} = [D,(e_1,e_2)_A]; $$ for example, the Graßmann connection $\nabla_0$ induced by the frame $\{\xi_1,\dotsc,\xi_N\}$ is defined by $$ \forall e \in \mathcal{E}, \quad \nabla_0(e) := \sum_{i=1}^N \xi_i \hat\otimes [D,(\xi_i,e)_A], $$ and indeed, if $\nabla$ is any other Hermitian connection, then $\nabla = \nabla_0 + \omega$ for $\omega \in B(E,E \hat\otimes_A \Omega_D)$ defined by $$ \forall e \in \mathcal{E}, \quad \omega(e) := \sum_{i=1}^N \nabla(\xi_i)(\xi_i,e)_A. $$ Given a Hermitian connection $\nabla$, define $1 \hat\otimes_\nabla D : \mathcal{E} \otimes^{\mathrm{alg}}_{\mathcal{A}} \operatorname{Dom}(D) \to E \hat\otimes_A H$ by $$ \forall e \in \mathcal{E}, \, \forall h \in H, \quad 1 \hat\otimes_\nabla D(e \hat\otimes h) := \nabla(e)h + e \hat\otimes Dh = \omega(e)h + 1 \hat\otimes_{\nabla_0}D(e \hat\otimes h). $$ Then, by Section 2 of Chakraborty–Mathai, the operator $1 \hat\otimes_\nabla D$ is essentially self-adjoint and defines a spectral triple $(B,E \hat\otimes_A H,1 \hat\otimes_\nabla D)$ with Lipschitz algebra $\mathcal{B}$. In particular, by a direct computation, $$ \forall b \in \mathcal{B}, \, \forall e \in E, \, \forall h \in H, \quad [1 \hat\otimes_{\nabla_0} D,b](e \hat\otimes h) = \sum_{i,j=1}^N \xi_i \hat\otimes [D,(\xi_i,b\xi_j)_A](\xi_j,e)_A h; $$ so that $1 \hat\otimes_\nabla D$, which is a bounded perturbation of $1 \hat\otimes_{\nabla_0} D$, also has bounded commutators with $\mathcal{B}$. To conclude, the missing ingredient was a Lipschitz structure on the fgp $A$-module $E$, i.e., a dense subspace $\mathcal{E}$ of $E$, such that $\mathcal{E} \cdot \mathcal{A} \subset \mathcal{E}$ and $(\mathcal{E},\mathcal{E})_A \subset \mathcal{A}$. Because $E$ is fgp, this suffices to pick out a dense $\ast$-subalgebra $\mathcal{B}$ of Lipschitz elements of $B$ and to pin down any Hermitian connection $\nabla$ on $E$ with domain $\mathcal{E}$ up to bounded perturbation, so that $\mathcal{B}$ will necessarily have bounded commutators with $1 \hat\otimes_\nabla D$; indeed, by using a normalised tight frame for $E$ consisting of vectors in $\mathcal{E}$, you can boil everything down to the compatibility of $\mathcal{A}$ with $D$.<|endoftext|> TITLE: Relations between two tower numbers QUESTION [8 upvotes]: A tower is a subset $T\subset [\omega]^\omega$ of the family $[\omega]^\omega$ of all infinite subsets of $\omega$ such that $T$ is well-ordered by the relation $\supset^*$ of almost inclusion and has no infinite pseudointersections. A tower is regular if its cardinality is a regular cardinal. Consider two small uncountable cardinals: $\mathfrak t=\min\{|T|:T\subset[\omega]^\omega$ is a tower$\}$; $\hat{\mathfrak t}=\sup\{|T|:T\subset[\omega]^\omega$ is a regular tower$\}$. It is clear that $$\mathfrak t\le\hat{\mathfrak t}\le\mathfrak c.$$ Martin's Axiom implies $\mathfrak t=\hat{\mathfrak t}=\mathfrak c$. On the other hand the strict inequality $\mathfrak t<\hat{\mathfrak t}$ is known to be consistent (but I do not know the reference). The cardinal $\mathfrak t$, called the tower number, is well-studied in Set Theory. I am interested in the cardinal $\hat{\mathfrak t}$, more precisely in its relation to other known cardinal characteristics of the continuum. Problem 1. Is there any non-trivial upper or lower bound for the cardinal $\hat{\mathfrak t}$? In particular, is $\mathfrak b\le\hat{\mathfrak t}$? Problem 2. In which known models of ZFC does the strict inequality $\mathfrak t<\hat{\mathfrak t}$ hold? Maybe there is some fixed standard notation for $\hat{\mathfrak t}$? REPLY [2 votes]: Assuming I understand the definitions correctly, I can give you a couple of references. (1) Dordal (see below) gives a model in which $\mathfrak{b}=\mathfrak{c}=\aleph_2$ and all towers have cardinality $\aleph_1$. Thus, in his model $\mathfrak{t}=\hat{\mathfrak{t}}=\aleph_1<\aleph_2=\mathfrak{b}=\mathfrak{c}$, and so the last question in your first problem has a negative answer. Edit: In fact, $\mathfrak{h}=\aleph_2$ in the model since he is using Mathias forcing, so I'm not sure of any good candidate for a non-trivial lower bound for $\hat{\mathfrak{t}}.$ (2) On the other hand, the model of Blass and Shelah discussed here gives a model in which $\mathfrak{t}=\mathfrak{b}=\aleph_1$ and there is a tower of length $\aleph_2$ (namely, the generating tower for the simple $P_{\aleph_2}$-point). Thus, this gives a model where $\mathfrak{t}<\hat{\mathfrak{t}}$. Edit 3/17/20 I tracked down another reference (due to Dordal [2]) with some more information on this question. Theorem 1.3 in the paper gives a connection between towers in the structures $^\omega\omega$ and $[\omega]^\omega$: Theorem (Dordal, Theorem 1.3 of [2]) Let $\kappa$ be an uncountable regular cardinal, and suppose there is a $\kappa$-tower in $^\omega\omega$ but not $[\omega]^\omega$. Then there is a $\kappa$-scale in $^\omega\omega$ (that is, $\mathfrak{b}=\mathfrak{d}=\kappa$). With regard to Problem 1, we may say that if $\mathfrak{b}<\mathfrak{d}$ then $\mathfrak{b}\leq\hat{\mathfrak{t}}$ (apply the above theorem, and use the fact that there is a tower of length $\mathfrak{b}$ in the structure $^\omega\omega$. On the other hand, it is consistent that $\mathfrak{b}=\mathfrak{d}=\aleph_2$ and $\hat{\mathfrak{t}}=\aleph_1$ (this occurs in the model from [1]). The paper has many other forcing constructions showing that not much more can be said. [1] Dordal, Peter Lars, A model in which the base-matrix tree cannot have cofinal branches, J. Symb. Log. 52, 651-664 (1987). ZBL0637.03049. [2] Dordal, Peter Lars, Towers in $[\omega]^{\omega}$ and $^{\omega}\omega$, Ann. Pure Appl. Logic 45, No. 3, 247-276 (1989). ZBL0686.03024.<|endoftext|> TITLE: Injectivity radius on complete manifolds with positive and bounded curvature QUESTION [7 upvotes]: I have two question: 1) Are there any examples of complete manifold with strictly positive and bounded section curvature which has zero injectivity radius? 2) Is there a sequence of non-compact complete manifolds with strictly positive and bounded section curvature with injectivity radius approach to zero? I think one may construct these examples from Beger's spheres, but I cannot do it rigorously. REPLY [5 votes]: If by strictly positive, you mean that there is a $\epsilon>0$ so that the $Sect >\epsilon$, then there are no such examples. The reason for this is that this curvature assumption implies that a manifold is compact. Since the injectivity radius of a point is a positive and continuous function on a manifold, it has a non-zero minimum. Igor Belegradek pointed out (thanks for this) that strictly positive also has a another meaning, which is that the sectional curvature is everywhere positive, without a positive lower bound. If we allow this as the definition of "strictly positive", there is a simple counterexample, which is just to take a disjoint union of countably many collapsing Berger spheres. Some of the sectional curvature of the Berger spheres converge to zero as the metric collapses (See here), but the curvature is always positive throughout the collapse. This also addresses your second question in terms of this definition of positive curvature. However, it seems worth asking whether we can create a connected manifold with positive curvature and zero injectivity radius. I was unable to find a simple argument to rule out such examples, but I strongly suspect that none exist. As I mentioned before, the only spaces we need to worry about are non-compact. In this case, one can either use the Soul Theorem or results of Gromoll and Meyer to see that the manifold must be diffeomorphic to $\mathbb{R}^n$. Such a manifold also has no closed geodesics and there is a lower bound on the distance between conjugate points, which is strong evidence that there should be control on the injectivity radius. However, I'm not sure how to derive such an estimate. Maybe someone more familiar with this can help.<|endoftext|> TITLE: How strong is the requirement of being a Gelbart-Jacquet lift? QUESTION [8 upvotes]: Let $\pi$ be a cuspidal automorphic representation of $\mathrm{GL}(3)$ over a number field $F$. I am wondering how general are Gelbart-Jacquet lifts of automorphic representations of $\mathrm{GL}(2)$ among these. I have a very limited understanding of Gelbart-Jacquet lifts: they are the cuspidal automorphic representations of $\mathrm{GL}(3)$ such that $L(s, \mathrm{sym}^2 \pi)$ has a pole at $s=1$. I would like to precisely understand what they are as automorphic objects, in order to understand how strong is this hypothesis and how it can be used in practice to appeal to the $\mathrm{GL}(2)$ setting. Any precise reference for it would be welcome. REPLY [11 votes]: The Gelbart-Jacquet lift of a $\mathrm{GL}_2$ cuspidal automorphic representation $\pi$ is an automorphic representation $\Pi$ of $\mathrm{GL}_3$. More generally, the Gelbart-Jacquet lift $\Pi$ is a $\mathrm{GL}_3$ automorphic representation associated to a $\mathrm{GL}_2$ automorphic representation $\pi$ and a Hecke character $\chi$ such that $L(s,\pi \otimes \widetilde{\pi} \otimes \chi) = L(s,\Pi) L(s, \chi)$. one can use the $\mathrm{GL}_3$ converse theorem to show that $L(s,\Pi)$ is an $L$-function of an automorphic representation of $\mathrm{GL}_3$. When $\chi = 1$, $\Pi$ is equal to $\mathrm{ad} \pi$, the adjoint. When $\chi = \omega_{\pi}^{-1}$, the inverse of the central character of $\pi$, so that $\widetilde{\pi} \otimes \chi = \pi$, $\Pi$ is equal to $\mathrm{sym}^2 \pi$. (In particular, these two are equal up to a twist.) As an aside, note that the Gelbart-Jacquet lift need not be cuspidal; see Poles of $L$-functions associated to Maass forms. Gelbart-Jacquet lifts are rare among all $\mathrm{GL}_3$ automorphic representations. The first way to quantify this is a result of Ramakrishnan; which states that there is (essentially) a bijection (after twisting, if necessary) between Gelbart-Jacquet lifts and self-dual automorphic representations of $\mathrm{GL}_3$. Unlike for $\mathrm{GL}_2$, being self-dual is a very restrictive statement for $\mathrm{GL}_n$ with $n \geq 3$. The second way to quantify this is via the Weyl law. A result of Guerreiro gives a Weyl law for Gelbart-Jacquet lifts with main term $\asymp T^4/\sqrt{\log T}$, whereas the Weyl law for all $\mathrm{GL}_3$ forms has main term $\asymp T^5$. A third way of viewing this is to look locally. Let $\Pi$ be an automorphic representation of $\mathrm{GL}_3$. At almost all places $v$, the local component $\Pi_v$ is an unramified principal series representation $|\cdot|_v^{t_{v,1}} \boxplus |\cdot|_v^{t_{v,2}} \boxplus |\cdot|_v^{t_{v,3}}$, where the spectral parameters $t_{v,1},t_{v,2},t_{v,3}$ satisfy $t_{v,1} + t_{v,2} + t_{v,3} = 0$ and $|\Re(t_{v,j})| < 1/2$ (the best known bound towards the Ramanujan conjecture in this case is $5/14$ due to Blomer-Brumley). Self-dual representations satisfy the additional (quite stringent!) restriction that $t_{v,j} = 0$ for some $j \in \{1,2,3\}$.<|endoftext|> TITLE: What is the simplest known finite presentation of a free nilpotent group? QUESTION [9 upvotes]: Finitely generated nilpotent groups are always finitely presented. This is true for abelian groups, and can be shown by induction for nilpotent ones, using the classical lift of a presentation of $N$ and a presentation of $G/N$ to a presentation of $G$. As a consequence, the free $c$-nilpotent group of rank $n$, $F_n/\Gamma_{c+1}(F_n)$, is finitely presented. In fact, I do know a finite presentation : kill all iterated commutators of the generators of length between $c+1$ and $2c$. This works because all iterated commutators of length greater than $c$ can be written as a product of iterated commutators of the generators of length greater than $c$ (using basic commutator calculus), and every such commutator has a sub-commutator of length between $c+1$ and $2c$ (it can help to think of commutators as rooted planar binary trees). But I am unable to find a simpler presentation (say, with less relations). In fact, I can think of many presentations making the lower central series to stop, and giving the right Lie algebra (using linear trees, or Lyndon trees), but I do not know how to show that the result is nilpotent, and it may well not be. And I have not been able to find a simpler presentation from the induction process described above, either. Whence my question. REPLY [2 votes]: Taking "simplest" to mean "having fewest generators", the question is how many generators you need to normally generate $\Gamma_n$ in a free group. As mentioned already in the comments, $\Gamma_n / \Gamma_{n+1}$ is free abelian with rank given by Witt's formula, so certainly you need at least this many generators. It seems to be unknown (!) whether the basic commutators of weight $n$ normally generate $\Gamma_n$. A couple of references I found: Sims, Charles C. Verifying nilpotence. J. Symbolic Comput. 3 (1987), no. 3, 231--247. https://doi.org/10.1016/S0747-7171(87)80002-0 Jackson, David A. Basic commutators in weights six and seven as relators. Comm. Algebra 36 (2008), no. 8, 2905--2909. https://doi.org/10.1080/00927870802108148<|endoftext|> TITLE: Complex manifold with boundary QUESTION [10 upvotes]: My question is of local nature. Let $$f:\mathbb C^n\to\mathbb R$$ be a $C^\infty$ function that vanishes at $0\in \mathbb C^n$, with non-zero derivative. Then, around $0\in \mathbb C^n$, $$M:=f^{-1}(0)$$ is a CR manifold. Let me assume that $M$ is the simplest possible kind of CR manifold, namely that it is foliated by real-codimension-one complex submanifolds. [Equivalently, for those who don't know what CR manifolds are, consider the hyperplane distribution $L:=TM\cap i\cdot TM\subset TM$. I require the distribution $L$ to be integrable, i.e., to come from a (real codimension $1$) foliation of $M$.] Under the above assumptions, is $f^{-1}\big([0,\infty)\big)$ locally isomorphic to $$\big\{(z_1,...z_n)\in\mathbb C^n\,:\,\mathrm{im}(z_1)\ge 0\big\}?$$ I.e., does there exist a neighbourhood $U\subset f^{-1}([0,\infty))$ of zero and an isomorphism $\varphi:U\to \big\{z\in\mathbb C^n\,:\,\sum|z_i|^2<1,\,\mathrm{im}(z_1)\ge 0\big\}$ which is holomorphic in the interior and smooth all the way to the boundary. REPLY [2 votes]: Perhaps Giuseppe Della Sala's paper might be useful here: https://www.ams.org/journals/proc/2011-139-07/S0002-9939-2010-10746-3/home.html It precisely deals with the equivalence of smooth Levi-flats. There are examples in the paper<|endoftext|> TITLE: Homotopy fibers of infinity functors QUESTION [9 upvotes]: Let $F: C \to D$ be an infinity functor. Is it true that the homotopy fiber at $y$ can be described as $C \times_D D^{\simeq}_{/y}$? If not, is there a simple formula resembling this one? Beside the infinity structure, its points are pairs $(x \in C, s: F(x) \to y \text{ equivalence})$. Models I am using are the following: Infinity categories are thought as quasi-categories, i.e. simplicial sets in which every inner horn can be filled. If also outer horns can be filled, it is an infinity groupoid. The homotopy category of $C$ has vertex of $C$ as objects, and $Hom_{Ho(c)}(x,y) := \pi_0( \{x\} \times_{\Delta^{\{0\}} } Map( \Delta^1, C) \times_{\Delta^{\{1\}}} \{y\}) $. In other words, we take the simplicial set of "maps from x to y" and we take the connected components. This gives a definition of what an equivalence is, namely something that is iso in $Ho(C)$. Given an infinity category $C$, we denote by $C^{\simeq}$ the subsimplicial set spanned by invertible arrows. One can verifies that this is the maximal subgroupoid of $C$. An infinity functor is just a map of simplicial sets; the set of infinity functors $Fun(C,D)$ has a quasi category structure given by $Map(C,D)^{\simeq}$ We have a cofibrantly generated model structure on simplicial sets, the Joyal model (see here for a definition https://ncatlab.org/nlab/show/model+structure+on+simplicial+sets). Generally, if one has a finite diagram D, we have an induced model structure on $C^D$ such that $\lim : C^D \to C$ is right Quillen adjoint to $\Delta: C \to C^D$. The structure is such that cofibrations and weak equivalences are levelwise, and fibrations are obviously a shift. Thus we have an induced functor $\mathrm{Holim}: Ho(C^D) \to Ho(C)$. Recall that this means holim is computed by taking a fibrant replacement of the diagram in the above model structure, and then taking the ordinary limit. Homotopy fibers are taken in the context of 6. Namely consider a map $f:C \to D$ and an inclusion $i: \{y\} \to D$. The homotopy fiber at $y$ is defined as the homotopy pullback of the diagram given by $i,f$. Thanks, Andrea REPLY [9 votes]: In general the homotopy pullback of the diagram given by $i:\{y\} \to \mathcal{D}$ and $f:\mathcal{C} \to \mathcal{D}$ is given by first replacing $i$ and $f$ by fibrations between fibrant objects (so that the diagram formed by $f$ and $i$ is fibrant in the injective model structure), and then taking the actual pullback. It turns out that specifically for pullback diagram a slightly weaker condition suffices: it is enough to have one of $i,f$ represented by a fibration between fibrant objects and the other just by any map between fibrant objects (this can be deduced, for example, from Proposition 13.1.2 of Hirschhorn's book on model categories and localizations). In the case at hand $\mathcal{C}$ and $\mathcal{D}$ are already fibrant in the Joyal model structure, and so it will suffice to replace $i$ by a fibration. The map $\tilde{i}:\mathcal{D}^{\simeq}_{/y} \to \mathcal{D}$ (obtained by composing the projection $\mathcal{D}^{\simeq}_{/y} \to \mathcal{D}^{\simeq}$ with the inclusion $\mathcal{D}^{\simeq} \subseteq \mathcal{D}$) is a fibration in the Joyal model structure by the following criterion: a map between $\infty$-categories is a categorical fibration if and only if it is an inner fibration and has a certain lifting property for equivalences (see HTT, Corollary 2.4.6.5). On the other hand, the inclusion $\{Id_y\} \to \mathcal{D}^{\simeq}_{/y}$ is an equivalence of $\infty$-categories, and so the map $\tilde{i}$ can be considered as a fibration replacement for $i$. We may now conclude that $\mathcal{C} \times_{\mathcal{D}} \mathcal{D}^{\simeq}_{/y}$ is a model for the homotopy pullback of $i$ and $f$.<|endoftext|> TITLE: Decompose $MT(E(d)\times_{\mathbb Z_2} SU(2))$ as the wedge sum or smash product of spectra QUESTION [5 upvotes]: Consider the extension $$1\to SU(2)\to X\to O\to1,$$ there are 4 possibilities for $X$: $X=O\times SU(2)$ or $E\times_{\mathbb{Z}_2}SU(2)$ or $Pin^+\times_{\mathbb{Z}_2}SU(2)$ or $Pin^-\times_{\mathbb{Z}_2}SU(2)$ where $E$ is defined in Freed-Hopkins's work1 as the colimit of the group $E(d)$, the group $E(d)$ is defined to be the subgroup of $O(d)\times\mathbb{Z}_4$ consisting of the pairs $(A,j)$ such that $\det A=j^2$, where $\mathbb{Z}_4=\{\pm1,\pm\sqrt{-1}\}$ is the multiplicative group of order 4. Here the notation $G_1\times_{\mathbb{Z}_2} G_2 :=\frac{G_1\times G_2}{\mathbb{Z}_2} $ is defined as mod out the common $\mathbb{Z}_2$ of $G_1\times G_2$. The question is about computing $MT(E(d)\times_{\mathbb Z_2} SU(2))$ and the bordism group $\Omega_d^{E \times_{\mathbb Z_2}SU(2)}$. (1) There is a short exact sequence of groups: $1\to SO(d)\to E(d)\to\mathbb{Z}_4\to 1$. So naively, people may suspect that $$MT(E(d)\times_{\mathbb Z_2} SU(2))=MT E(d)\wedge\Sigma^{-3}M SO(3)=MSO(d)\wedge\Sigma^{-2}M\mathbb Z_4\wedge\Sigma^{-3}M SO(3).$$ However, this is likely to be incorrect. (2) The space $B(E \times_{\mathbb Z_2}SU(2))$ sits in a homotopy pullback square: a map $M \to B(E \times_{\mathbb Z_2}SU(2))$ is determined by two maps $M \to BO$ and $M\to BSO(3)$ which correspond to bundles $TM$ and $V_{SO(3)}$ such that $w_1(TM)^2=w_2(V_{SO(3)})$. To compute the bordism group $\Omega_d^{E \times_{\mathbb Z_2}SU(2)}$, we need to know the Madsen-Tillmann spectrum $MT(E \times_{\mathbb Z_2}SU(2))$ and decompose it as the wedge sum or smash product of some familiar spectra. The figure attachment here is my own attempt, but the map $f$ is not a homotopy equivalence. I actually obtain an identification $$ \text{Thom$(B(E \times_{\mathbb Z_2}SU(2)),-2V)=MT(Pin^+ \times_{\mathbb Z_2}SU(2))$} $$ which is already known in 1604.06527 paper, but we need to know $$ \text{Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)=MT(E \times_{\mathbb Z_2}SU(2))$,}$$ where $V$ is the induced virtual bundle of dimension 0 by $B(E \times_{\mathbb Z_2}SU(2)) \to BO$. Is Thom$(B(E \times_{\mathbb Z_2}SU(2)),-2V)$=smash product of Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)$ and Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)$? If so, how to obtain Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)$ as the "square root" of Thom$(B(E \times_{\mathbb Z_2}SU(2)),-2V)$? 1 Reflection positivity and invertible topological phases Daniel S. Freed, Michael J. Hopkins, arXiv:1604.06527 REPLY [5 votes]: Let $Y$ be a space, $V$ be a virtual bundle of dimension $0$ over $Y$ (this $V$ is your $-V$). Then $Thom(Y,2V)$ is almost never (except when $Y$ is contractible, or something like that) a smash product of the form $Thom(Y,V)$ with itself. You can see this by looking at the homology: the homology of the former is more or less isomorphic to that of $Y$, whereas the latter is more or less isomorphic to its square.<|endoftext|> TITLE: Is every dg-coalgebra the colimit of its finite dimensional dg-subcoalgebras? QUESTION [5 upvotes]: I saw this result in A Model Category Structure for Differential Graded Coalgebras by Getzler-Goerss, but when the coalgebra is non-negatively graded, is this property also satisfied when the dg coalgebra is $\mathbb{Z}$-graded?. Thanks. REPLY [5 votes]: For coassociative dg-coalgebras over any field $k$ the answer is positive, because: Let $C$ be a $\mathbb Z$-graded coalgebra and $D\subset C$ a finite-dimensional ungraded subcoalgebra (of the underlying ungraded coalgebra) of $C$. Let $D^{gr}\subset C$ denote the graded vector subspace spanned by all the grading components of the elements of $D$. Then $D\subset D^{gr}$ and $D^{gr}$ is a finite-dimensional graded subcoalgebra of $C$. Let $(C,d)$ be a dg-coalgebra and $D\subset C$ be a finite-dimensional graded subcoalgebra of $C$. Set $D^{dg}=D+d(D)\subset C$. Then $D\subset D^{dg}$ and $D^{dg}$ is a finite-dimensional dg-subcoalgebra of $C$. Using the observations 1. and 2. and the fact that any ungraded coassociative coalgebra is the union of its finite-dimensional subcoalgebras, one deduces the assertion that any $\mathbb Z$-graded dg-coalgebra is the union of its finite-dimensional dg-subcoalgebras. Possible generalizations: One can replace a field $k$ by a Noetherian commutative ring $k$ and speak about subcoalgebras that are finitely generated as $k$-modules (instead of "finite-dimensional"). All the assertions remain true. EDIT: I've realized that the preceding paragraph is problematic for the following reason: given a $k$-submodule $D$ is a $k$-module $C$, the tensor product $D\otimes_k D$ is not a submodule of the tensor product $C\otimes_k C$, generally speaking. So the very notion of a $k$-subcoalgebra for a commutative ring $k$ is problematic, or at least requires extra care with nonexact tensor products. So, I retract the preceding paragraph. One cannot drop the coassociativity condition. Indeed, even for ungraded coalgebras over a field of characteristic $0$, there is an example of infinite-dimensional Lie coalgebra $L$ having no nonzero finite-dimensional subcoalgebras. The Lie coalgebra $L$ is simplest described in terms of its dual topological Lie algebra structure (on a pro-finite-dimensional topological vector space): $L^*=\mathfrak g=k[[z]]\,d/dz$, the Lie algebra of vector fields on the formal disk.<|endoftext|> TITLE: Non-Gorenstein Curves QUESTION [11 upvotes]: I am interested in (as explicit as possible) descriptions of non-Gorenstein integral projective curves. Most of the literature on singular curves appears to be focused around the Gorenstein case, with a notable exception being the paper of Kleiman, The Canonical Model of a Singular Curve, and a handful of others. Any other references would be appreciated. Specifically I am interested in knowing what has been understood about their derived categories of coherent sheaves (or the subcategory of perfect complexes), so I would very interested in any developments along those lines. REPLY [7 votes]: I don't think there is a reasonably short explicit description of non-Gorenstein curves. They could be described as lacking some of those conditions that define Gorenstein curves. In any case, here is a simple example: Let $C$ be the union of three coordinate axis in $\mathbb A^3$. Then $C$ is not Gorenstein. It is probably a good exercise for you to check that this is true. What I find fascinating about this is that the union of three lines contained in $\mathbb A^2$ is a Gorenstein curve, so this shows that being Gorenstein or not could depend on very subtle details. Of course, $C$ can be projected to a plane and if the projection is general, then it will be one-to-one on $C$ and an isomorphism on each component to their respective images, however, it is not an isomorphism on $C$. This follows from the fact that the image of $C$ is Gorenstein, or more directly from the fact that the Zariski tangent space of $C$ at its singular point is $3$-dimensional while that of its image is $2$-dimensional, so they can't be isomorphic. (In case you are worried that this example is reducible, then think about how to get this singularity on an irreducible curve).<|endoftext|> TITLE: Constructing a model of $\mathrm{DCF}_0$ via forcing QUESTION [5 upvotes]: As is mentioned in the introduction of this paper of Spodzieja there is a lack of 'natural' examples of differentially closed fields. The immediate naive guesses, namely the field of germs of meromorphic functions at some point, only partially work in that these structures contain differentially closed fields as substructures by Seidenberg's embedding theorem but are not differentially closed themselves. It seems that the while the sheaf of meromorphic functions is fairly rich, it's always possible to find differential equations that are badly behaving at a given point. As someone who has been in the same room as people talking about set theoretic forcing, this made me wonder whether or not a 'direct' construction of a differentially closed field could be achieved with forcing. Essentially I want to take the field of germs of meromorphic functions at a generic point. Does there exist a forcing extension $V[G]$ such that for some $z\in\mathbb{C}^{V[G]}\setminus \mathbb{C}^V$ the field of germs at $z$ of meromorphic functions in $V$ is differentially closed? Of course this is equivalent to considering the ring of germs of holomorphic functions in $V$ at $z$, since no such function can have a zero or pole at $z$. A potentially more general question could be asked in terms of direct limits along directed sets of open subsets of $\mathbb{C}$, but I wanted to focus on this simpler form first. REPLY [4 votes]: I had a conversation about this with James Freitag at a conference recently and he pointed me to a fundamental result of Seidenberg that seems to resolves this problem in the positive. The relevant result is from Seidenberg's paper 'Abstract Differential Algebra and the Analytic Case II.' Here it is restated slightly: Let $U$ be a non-empty connected open subset of $\mathbb{C}$ and let $K$ be a finitely generated differential field of meromorphic functions on $U$. If $L \supseteq K$ is a finitely generated abstract differential field extension of $K$, then there exists a non-empty connected open set $W\subseteq U$, a finitely generated differential field of meromorphic functions $L^\prime$, and a differential field isomorphism $f:L\rightarrow L^\prime$ extending the restriction map $\mathrm{res}_{W,U}$ on $K$. For anyone familiar with forcing, this result has clearly done most of the work for us. Define a forcing poset $\mathbb{P}$ whose conditions are non-empty connected open subsets of $\mathbb{C}$, with $\mathbf{1}=\mathbb{C}$, ordered by inclusion, i.e. $W \leq U$ if and only if $W \subseteq U$. (This should be equivalent to Cohen forcing building a single new real, or should I say a single new complex? It is certainly $ccc$.) Now in the extension $V[G]$, we have that $\bigcap G$ contains a single complex number, $z$. I want to argue that the field of germs of meromorphic functions in $V$ at $z$ is differentially closed. Call this field $K$. (Note that since these are meromorphic functions it is automatically a field and not just a ring, but since no non-zero element of this collection can vanish at $z$ it's actually the same as the ring of germs of holomorphic functions at $z$.) Let $L \supseteq K$ be any abstract differentially closed field extension of $K$ (in $V[G]$). Let $b\in L$ be any element and $\overline{a} \in K$ be any finite tuple of elements. Let $L_0$ be the differential field generated by $\overline{a}b$. Since $\mathrm{DCF}_0$ is $\omega$-stable, it doesn't gain any new types over $\varnothing$ in $V[G]$ (since there are only countably many of them), so there exists a differential field $K_0 \in V$ isomorphic to $L_0$. Let $U\in G$ be a common domain of the elements of the tuple $\overline{a}$. Now consider the set $D\subseteq \mathbb{P}$ of opens $W \subseteq U$ such that there exists a meromorphic function $c$ on $W$ such that the field generated by $\overline{a}c$ is isomorphic to $K_0$. By Seidenberg's result this is clearly dense below $U$ (as a condition), so by genericity some $W\in D$ is also in $G$. So the germ of $c$ at $z$ is also in $K$. This says that every quantifier free type over a finite tuple $\overline{a} \in K$ is realized in $K$, implying that it is existentially closed and therefore differentially closed. Moreover by quantifier elimination we actually get that $K$ is $\omega$-saturated. Since $\mathbb{P}$ is $ccc$, $K$ has size continuum, since it contains $\mathbb{C}^V$. I'd be really curious to know more about the structure $K$ model theoretically, since it seems fairly natural to me. In particular, is it $\omega_1$-saturated or more? Can it consistently have a Vaughtian pair (by which I mean a definable infinite subset of size strictly less than continuum)? Does its isomorphism type depend on the choice of the generic point $z$? What if we chose some other $w\in \mathbb{C}^{V[G]} \setminus \mathbb{C}^V$? What are its non-trivial automorphisms?<|endoftext|> TITLE: Finite posets for which all intervals are atomic QUESTION [8 upvotes]: Let $P$ be a finite poset which is a lattice with $0,1 \in P$. An atom in $P$ is an upper cover of $0$ and a coatom is a lower cover of $1$. $P$ is atomic if every element is a join of atoms and coatomic if every element is a meet of coatoms. Say that $P$ is locally branched if all intervals of length $2$ contain at least four elements. This is if, for a saturated chain $a \prec b \prec c$ in $P$, there is an element $d \neq b$ with $a < d < c$. In a paper I am currently writing, I make use of the following equivalence: $P$ is locally branched Every interval in $P$ is atomic Every interval in $P$ is coatomic This property is not hard to prove and is very handy. For example, face lattices of polytopes have the diamond property (every interval of length $2$ contains exactly four elements), so this equivalence shows that they are atomic and coatomic. Has the property of being locally branched been observed or even used before in the literature? Is this equivalence already known in the literature? REPLY [3 votes]: This struck me as a sort of opposite to thin, so googled for "thick" and found arXiv:math/0101075 by Bayer and Hetye. In this paper a poset is called $r$-thick if every nonempty open interval contains at least $r$ elements. So $2$-thick is locally branched. I gave the paper a quick glance and do not see your equivalence. Google Scholar says the paper has 14 citations. I did not try to follow these citations, but this paper gives a start into the literature.<|endoftext|> TITLE: "Noetherian" and "finitely generated" for polynomial algebras QUESTION [11 upvotes]: Let $k$ be a field. Does there exist a positive integer $n$ such that there is $k$-subalgebra of $k[x_1, \dots, x_n]$ which is Noetherian but not finitely generated? REPLY [15 votes]: The following example of Eakin [Eak72] says that $n = 2$ already suffices. I have tried to fill in some details to (hopefully) make the example independent from the rest of Eakin's paper. Example [Eak72, Example on p. 79]. Let $k$ be a field, and consider the formal power series ring $k[[t]]$ in one variable over $k$. By an argument of Schmidt (see [Sch33, Hilfssatz 5] or [MLS39, Second Proof of Lemma 1]), there exist two algebraically independent elements $x,y \in k[[t]]$ that have the same positive valuation with respect to the valuation on $k[[t]]$. For a specific example, Schmidt's result implies that one can choose $$x = t \qquad\text{and}\qquad y = \sum_{n=1}^\infty t^{n!}.$$ Now let $V = k[[t]] \cap k(x,y)$ and $R = k[x,1/y] \cap V$. Setting $X = x$ and $Y = 1/y$, we have $$k \subseteq k[X,XY] \subseteq R \subseteq k[X,Y].$$ We then show the following: Claim. $R$ is noetherian but not finitely generated as a $k$-algebra. To show that $R$ is noetherian, we note that $R$ is a Krull ring by [Mat89, Theorem 12.4(ii)] since both $k[x,1/y]$ and $V$ are Krull. A theorem of Heinzer [Hei69, Theorem 9] (see also [Eak72, footnote 1 on p. 78]) then implies that $R$ is noetherian. To show that $R$ is not finitely generated as a $k$-algebra, we first note that the prime ideal $\mathfrak{p} = \mathfrak{m}_V \cap R$ has height one in $R$ and that $V = R_{\mathfrak{p}}$, since $V$ is a DVR that must appear when $R$ is written as the intersection of DVR's in $k(X,Y)$ [Mat89, Theorem 12.3]. The residue field of $V$ is $k$, since it must simultaneously contain $k$ and also be contained in the residue field of $k[[t]]$, which is $k$. Thus, $\mathfrak{p}$ is a prime ideal of height one in $R$ whose residue field is not transcendental over $k$, which cannot be the case if $R$ is finitely generated over $k$ [ZS75, Corollary on p. 92]. References [Eak72] Paul Eakin. "A note on finite dimensional subrings of polynomial rings." Proc. Amer. Math. Soc. 31 (1972), 75–80. DOI: 10.2307/2038515. MR: 289498. [Hei69] William Heinzer. "On Krull overrings of a Noetherian domain." Proc. Amer. Math. Soc. 22 (1969), 217–222. DOI: 10.2307/2036956. MR: 254022. [Mat89] Hideyuki Matsumura. Commutative ring theory. Second ed. Translated from the Japanese by M. Reid. Cambridge Stud. Adv. Math. 8. Cambridge Univ. Press, Cambridge, 1989. DOI: 10.1017/CBO9781139171762. MR: 1011461. [MLS39] Saunders Mac Lane and O. F. G. Schilling. "Zero-dimensional branches of rank one on algebraic varieties." Ann. of Math. (2) 40, (1939), 507–520. DOI: 10.2307/1968935. MR: 158. [Sch33] Friedrich Karl Schmidt. "Mehrfach perfekte Körper." Math. Ann. 108 (1933), no. 1, 1–25. DOI: 10.1007/BF01452819. MR: 1512831. [ZS75] Oscar Zariski and Pierre Samuel. Commutative algebra. Vol. II. Reprint of the 1960 edition. Grad. Texts in Math. 29. Springer-Verlag, New York-Heidelberg, 1975. DOI: 10.1007/978-3-662-29244-0. MR: 389876.<|endoftext|> TITLE: Classifying Space of "Valuation Ringed Spaces over a Topos" QUESTION [6 upvotes]: The classifying topos for local rings is the big Zariski topos of $\text{Spec}(\mathbb{Z})$. Call this topos $T$. Geometric maps of topoi from a topos $T'$ to $T$ are in correspondence with sheaves of rings on $T'$ which are locally local rings. This is related to how $\text{Spec} : \text{Ring} \rightarrow \text{Loc}$ from rings to locally ringed spaces is adjoint, while if we replace $\text{Loc}$ with ringed spaces, it is not an adjunction. I wonder if there is a similar situation for $\text{Spa}$ and some classifying topos involving valuation rings. That is, I want some topos $T_{val}$ such that maps of a topos $T'$ into $T_{val}$ are in correspondence with sheaves of rings on $T'$ which are locally certain valuation rings. REPLY [7 votes]: Since the axioms describing what a valuation ring can be put as what's called geometric sequents [*], by the fundamental theorem on classifying toposes, there is a topos $T_{val}$ with precisely the universal property you're asking for. (See for instance Section 2.1.2 and more specifically Theorem 2.1.8 in Olivia Caramello's book Theories, Sites, Toposes.) But then there is the additional question whether we can recognize this topos as one of the toposes commonly used in algebraic geometry, just as we can recognize the classifying topos of local rings as the big Zariski topos of $\operatorname{Spec}(\mathbb{Z})$. As far as I know, this question is open. The answer linked to by Robert in the comments gives some indication that $T_{val}$ might coincide with the big rh topos of $\operatorname{Spec}(\mathbb{Z})$, but just that $T_{val}$ and the big rh topos have the same topos-theoretic points, as set out in the linked answer, doesn't yet mean that the toposes are equivalent. You could also ask the analogous question for algebraically closed valuation rings. In this case again a classifying topos exists, by the same abstract nonsense, and conditional on some representability conjecture it coincides with the big ph topos of $\operatorname{Spec}(\mathbb{Z})$ (see Proposition 21.18 in these notes of mine). [*] A geometric sequent is a logical formula of the form $\forall \cdots \forall. (\cdots) \Rightarrow (\cdots)$ where in the two bracketed subformulas the connectives $\Rightarrow$ $\neg$ $\forall$ may not occur. For instance the field axiom "$\forall x : R. x = 0 \vee (\exists y : R. xy = 1)$" can be put as a geometric sequent (just imagine an invisible "$\top \Rightarrow$"), but the (classically but not intuitionistically) equivalent axiom "$\forall x : R. \neg(\exists y : R. xy = 1) \Rightarrow x = 0$" cannot.<|endoftext|> TITLE: Are triangulations of compact manifolds PL homeomorphic? QUESTION [6 upvotes]: I have frequently come across the statement "Any two triangulations of a compact n-manifold are related by bistellar moves" attributed to Pachner via Lickorish's paper 'Simplicial moves on complexes and manifolds'. The theorem this refer to is the following: Closed combinatorial n-manifolds are PL homeomorphic if and only if they are bistellar equivalent. My question is: Considering that Hauptvermutung is not true for manifolds of dimension more than 3, how can we justify this statement? REPLY [11 votes]: Pachner‘s Theorem is about PL-homeomorphic manifolds, while the Hauptvermutung is asking for a PL-homeomorphism between manifolds which a priori are only homeomorphic.<|endoftext|> TITLE: Compactness theorem for minimal surfaces QUESTION [6 upvotes]: I am a bit confused about the statement of Theorem 1.1 in this paper by Brian White. For convenience, I will restate it here. Theorem: Let $\Omega$ be an open subset of a Riemannian $3$-manifold. Let $g_i$ be a sequence of smooth Riemannian metrics on $\Omega$ converging smoothly to a Riemannian metric $g$. Let $M_i \subseteq \Omega$ be a sequence of properly embedded surfaces such that $M_i$ is minimal with respect to $g_i$. Suppose also that the area and the genus of $M_i$ are uniformly bounded on compact subsets of $\Omega$. Then (after passing to a subsequence) the $M_i$ converge to a smooth, properly embedded $g$-minimal surface $M$. For each connected component $\Sigma$ of $M$, either the convergence to $\Sigma$ is smooth with multiplicity one, or the convergence is smooth (with some multiplicity $> 1$) away from a discrete set $S$. In the second case, if $\Sigma$ is two-sided, then it must be stable. My question concerns the last sentence of the statement above. It seems very surprising to me the conclusion about the stability. What if $g$ is a metric with strictly positive Ricci curvature? I was thinking about the following 2-dimensional situation. Imagine to have a sequence of metrics $g_i$ on the 2-sphere such that the limit metric $g$ has positive curvature. Imagine that each $g_i$ carries two closed geodesics which are converging to an unstable geodesic with multiplicity 2. See this poor quality picture: The compactness theorem above says that one cannot construct such an example on the $3$ sphere and using minimal surfaces instead of geodesics. Can anyone give me an intuition of what is going on here? Thank you! REPLY [5 votes]: What's going on with the picture you drew is that because the $g_i$ have regions of negative curvature (in the "valleys" where the $M_i$ sit) that become arbitrarily close to the "hilltop" where $M$ sits), one must have that the curvature of $g$ along $M$ is identically zero. This means that $M$ is actually weakly stable as the constant function $1$ is an eigenfunction of the stability operator (and is the lowest as it doesn't change sign) with eigenvalue $0$.<|endoftext|> TITLE: When are valuative criteria useful? QUESTION [5 upvotes]: We have valuative criteria for properness and universal closedness of morphisms of schemes. I would agree that these criteria shed some light on the geometric nature of these morphisms. However, are there any situations in which these criteria are more convenient to use than simply starting from the basic definitions? Unfortunately, all the applications I have seen so far felt like "Surely I could argue from the definitions themselves but the word "valuative" is so fancy and makes me sound smart so I am going to use it anyway." I think that once you become experienced enough, the difference between different approaches might seem negligible; when giving examples assume you are a person who just finished studying Hartshorne (so your intuition is more-or-less at that level). REPLY [5 votes]: I'm not sure if you're looking only for examples where checking properness/separatedness/etc. is easier using valuative criteria; this is an example where proving something is made easier. The example is liftability of arcs under proper birational morphisms. A bit of setup: Say $X$ is a variety over a field $k$; an arc $\gamma$ on $X$ is a morphism $\gamma: \mathrm{Spec}\, k[[t]] \to X$. Write $X_\infty$ for the set of arcs on $X$ (in fact, $X_\infty$ carries a natural scheme structure, but we ignore this). The set of arcs on $X$ carries a great deal of information about the singularities of $X$, so it's of interest in birational geometry to understand how sets of arcs behave under proper birational maps; in particular, given $f:Y\to X$ a proper birational map, we want to know what the relation between $Y_\infty$ and $X_\infty$ is. Say that $f$ is an isomorphism over $X-Z$ for some closed subset $Z$ of $X$. "Most" arcs on $X$ don't have scheme-theoretic image contained in $Z_\infty$ (i.e., the morphism $\mathrm{Spec}\, k[[t]] \to X$ doesn't factor through $Z$). Write $Z_\infty$ for the set of arcs on $X$ that do factor through $Z$, and likewise $(f^{-1}(Z))_\infty$ for the arcs on $Y$ factoring through the (set-theoretic) preimage $f^{-1}(Z)$. We claim that the proper birational morphism $f:Y\to X$ induces a bijection between $Y_\infty-(f^{-1}(Z))_\infty$ and $X_\infty-Z_\infty$, which should be thought of as a bijection away from a "measure zero" set. Using the valuative criterion for properness though, this is super easy to see: take $\gamma $ in $X_\infty-Z_\infty$. Since $\gamma$ is not in $Z_\infty$ the image of the generic point $ \mathrm{Spec}\, k((t)) \to \mathrm{Spec}\, k[[t]] \to X$ lies in the locus where $f$ is an isomorphism, so we can lift $\mathrm{Spec}\, k((t)) \to X$ to a map $ \mathrm{Spec}\, k((t)) \to Y$, giving a diagram $\require{AMScd}$ \begin{CD} \mathrm{Spec}\, k((t)) @>>> Y\\ @V V V @VV f V\\ \mathrm{Spec}\, k[[t]] @>>> X \end{CD} Now, the valuative criterion for properness of $f$ says exactly that there is a unique lift of $\gamma$ to an arc $\mathrm{Spec}\, k[[t]] \to Y$, giving the desired bijection. Without using the valuative criterion, the other way I can think to prove this is to recognize $f:Y\to X$ as the blowup of $X$ at some ideal sheaf, and then work locally with the natural charts on the blowup; this isn't hard, but involves involves using the nontrivial fact that all proper birational maps are blowups of some ideal.<|endoftext|> TITLE: What are these 3-manifolds from surgery? QUESTION [6 upvotes]: I know that surgery on the unlink with +0 slope gives $S^2 \times S^1$ (where all the links above are embedded in $S^3$). I figured (I think) that surgery on the hopf link (with +0 on both) returns $S^3$ by performing the surgery on one of the unknots and pulling the second unknot into the new torus (the 'slam dunk' trick) - but here is my first question: In "On Kirby's Calculus of Links", Fenn and Rourke prove that you can get from any one integral surgery presentation of a 3-manifold to any other presentation through 'Kirby Moves' - where these are performing surgery on +1 unknots (or adding +1 unknots into the link) changing the surgery coefficients and linking of the components correctly. How do I perform these kirby moves to get that the surgery on the hopf link shown above equals $S^3$? (Actually, I think this picture solves my first question:) I'm still not sure what 3-manifold the final surgery corresponds to - I couldn't simplify it. REPLY [2 votes]: The manifold obtained from the figure has homology $Z/2Z \times Z/2Z$ so I don't know that you will be able to simplify it more. At the very least, you will need a two component link in $S^3$ to define it and arguably by most measures of complexity, the picture above will be the simplest possible description. As for what the manifold is it is the Seifert fibered space with base orbifold $S^2(2,2,2)$. Using Hatcher's description this should be the Seifert fibered space over $S^2$ with (2,1), (2,1) and (2,-1) as its exceptional fibers. The fundamental group of this orbifold is the $Q_8=\langle a, b | aba^{-1}b, aba^{-1}b\rangle.$ In particular, it is finite and the manifold is covered by $S^3$.<|endoftext|> TITLE: Automorphism group of the special unitary group $SU(N)$ QUESTION [6 upvotes]: Let us consider the automorphism group of the special unitary group $G=SU(N)$. We know there is an exact sequence: $$ 0 \to \text{Inn}(G) \to \text{Aut}(G) \to \text{Out}(G) \to 0. $$ For $G=SU(2)$, we have: $\text{Z}(SU(2)) =\mathbb Z_2$, $\text{Inn}(SU(2)) = SO(3)$, $\text{Out}(SU(2)) = 0$, And so $\text{Aut}(SU(2))=SO(3)$. For $N > 2$, we have: $\text{Z}(SU(N)) =\mathbb Z_{N}$, $\text{Inn}(SU(N)) = PSU(N)$, $\text{Out}(SU(N)) = \mathbb Z_2$. My question is: Does $\text{Aut}(SU(N))=PSU(N) \times \mathbb Z_2$? If not, does this answer depend on whether $N$ is odd or even? It looks to me that there is a nontrivial fibration depending on something like $H^2(B\mathbb Z_2,PSU(N))$ due to $$ B\text{Inn}(G) \to B\text{Aut}(G) \to B\text{Out}(G) \to B^2\text{Inn}(G) \to$$ and thus $$ BPSU(N) \to B\text{Aut}(G) \to B\mathbb Z_2 \to B^2PSU(N) \to$$ But I do not know how to define $H^2(B\mathbb Z_2,PSU(N))$, if this is a correct thing to ponder. REPLY [13 votes]: Let $G$ be a compact simple simply connected Lie group. Then any automorphism of $G$ determines an automorphism of its Lie algebra $\mathfrak{g}$ and visa versa. So $\mathrm{Aut}(G)$ is naturally isomorphic to the linear group $\mathrm{Aut}(\mathfrak{g})$. The sequence $1\to \mathrm{Inn}(\mathfrak{g})\to \mathrm{Aut}(\mathfrak{g})\to \mathrm{Out}(\mathfrak{g})\to 1$ is split. Moreover, $\mathrm{Out}(G)\cong\mathrm{Out}(\mathfrak{g})\cong \mathrm{Aut}(D_\mathfrak{g})$ where $D_\mathfrak{g}$ is the Dynkin diagram of $\mathfrak{g}$. The upshot is: $$\mathrm{Aut}(G)\cong \mathrm{Aut}(\mathfrak{g})\cong \mathrm{Inn}(\mathfrak{g})\rtimes \mathrm{Out}(G)\cong \mathrm{Inn}(\mathfrak{g})\rtimes \mathrm{Aut}(D_\mathfrak{g}).$$ For types $A_1, B_n, C_n, G_2, F_4, E_7, E_8$ there are no symmetries of the Dynkin diagram. For $A_n$ ($n>1$), $D_n$ ($n\not=4$), and $E_6$, we have $\mathrm{Aut}(D_\mathfrak{g})\cong \mathbb{Z}/2\mathbb{Z}$. And in the final case of $D_4$, the symmetry group is the symmetric group on three letters. In particular, as stated in the comments: $$\mathrm{Aut}(\mathrm{SU}(n))\cong\left\{\begin{array}{ll}\mathrm{PSU}(n)\rtimes \mathbb{Z}/2\mathbb{Z},&\text{ if } n\geq 3\\ \mathrm{PU}(2),&\text{ if }n=2. \end{array}\right.$$<|endoftext|> TITLE: Coefficients of linear dependency QUESTION [7 upvotes]: Let $L_1, \ldots, L_m \in \mathbb{Z}[x_1, \ldots, x_n]$ be polynomials of the form $L_i = l_{i1} \cdot l_{i2} \ldots \cdot l_{ik}$, where every $l_{ij}$ is an integer linear form. Assume that magnitudes of all coefficients of all $l_{ij}$ are bounded by some integer $H$. (So, every $l_{ij}$ has form $A_1 \cdot x_1 +\ldots + A_n \cdot x_n$, where every $A_i$ is an integer such that $|A_i| \le H$.) Suppose that polynomials $L_1, \ldots, L_m$ are linearly dependent over $\mathbb{Z}$, i.e. there exists $B_1, \ldots, B_m \in \mathbb{Z}$ (not all equal to zero) such that $B_1\cdot L_1 + \ldots + B_m\cdot L_m \equiv 0$. Is it true that these coefficients $B_1, \ldots, B_m$ can be chosen in such a way so that every $|B_i| \le H^{\text{poly}(n,m,k)}$ for some polynomial $\text{poly}(n,m,k)$? This question is motivated by studing depth-$3$ arithmetic circuits. REPLY [9 votes]: Yes. The coefficients of the $L_j$ are $O(k! H^k)$ (the $k!$ is a bound for how many different products can contribute to the same term, a multinomial coefficient with $k$ on top). The $B_i$ are given as the kernel of a matrix whose entries are the coefficients of $L_j$. That kernel can be computed by Cramer's rule, giving $m \times m$ determinants, so the $B$'s are $O((m!) (k!)^m H^{mk}) = O(H^{m k} e^{m k \log k + m \log m})$.<|endoftext|> TITLE: Algebras such that the tensor product with any Noetherian algebra is Noetherian QUESTION [6 upvotes]: Let $R$ be a Noetherian commutative unital ring. It is generally speaking not true that the tensor product of two Noetherian $R$-algebras is Noetherian (e.g. take $R$ to be a field, and consider the tensor square of some crazy field extension). What is true is that a tensor product of a finite type $R$-algebra and a Noetherian $R$-algebra is Noetherian. It is not true, however, that an $R$-algebra whose tensor product with any Noetherian $R$-algebra is Noetherian has to be of finite type (consider the power series ring, for example). What is the name for the class of $R$-algebras that have this property? REPLY [3 votes]: A (not necessarily commutative) algebra $A$ over a commutative noetherian ring $R$ is called strongly noetherian if for every noetherian $R$-algebra $R'$ the extension $A \otimes_R R'$ is noetherian. See this paper of Artin, Small, and Zhang for a reference. This property has also been discussed in this MO question. This is an issue of importance in noncommutative algebra, where there exist finitely generated noetherian algebras that do not remain noetherian after extending the base field. (There is a well-known example of this phenomenon due to Resco and Small.) But as your question indicates, it is already an interesting property to consider for commutative algebras.<|endoftext|> TITLE: Bounded weak derivative QUESTION [5 upvotes]: Let $f \in L^{\infty}$ be a function such that $f$ and the weak derivatives $D^{\alpha}f\in L^{\infty}$ exist for all $\vert \alpha\vert\ge 2$. Does this imply that also $D^{\alpha}f$ with $\vert \alpha\vert=1$ exists in $L^{\infty}?$ It sounds somehow natural and I could not find a counterexample, but I don't quite know. REPLY [4 votes]: It depends on the geometry of the domain. Consider e.g. the function $f(x,\,y) = xy$ on $\{|xy| < 1\}$, which is bounded with constant Hessian on this domain but has linearly growing gradient. Or take $f$ defined on a union of disjoint intervals in $\mathbb{R}$ of length $2^{-k}$ to be linear with slope $2^k$ on each and vanishing at the midpoints. The result will hold locally by mollification and interpolation, and globally on domains where interpolation estimates between $C^0$ and $C^2$ hold, e.g. convex domains (like $\mathbb{R}^n$, as Iosif shows) or bounded $C^2$ domains.<|endoftext|> TITLE: What Morrey and Campanato space characterize QUESTION [5 upvotes]: Morrey and Campanato space is some subspace of $L^p$. We know that for a bounded domain $\Omega$, $L^p$ space characterize how the function blow up at some point. I want to know what Morrey and Campanato space characterize? For bounded domain $\Omega$ and fixed $p$, if $\lambda<\mu$, we have $$L^{p,\mu}\subset L^{p,\lambda}$$ for Morrey space ,then since $L^{p,n}\cong L^{\infty}$, so it seems that Morrey space is another way to characterize blow-up. It is a bit different for Campanato space when $\lambda\geq n$, for $\mathcal{L}^{p,\mu}\cong C^{0,\alpha}$. Thus it seems it characterize kind of oscillation. REPLY [4 votes]: The lecture notes by Melanie Rupflin answer the question "What is a Morrey Space? What is a Campanato Space?" The Morrey space $L^{p,\lambda}$ is a subset of $L^p$ containing functions $f$ on a domain $\Omega\in\mathbb{R}^m$ such that the integral of $|f|^p$ over a ball $B$ of radius $r$ centered at $x_0$ goes to zero at least as fast as $r^\lambda$ uniformly in $x_0$. The purpose of the restriction of $L^p$ to the subspace $L^{p,\lambda}$ is that it allows for an alternative to the Sobolev-Embedding Theorem which can be used with exponent $p$ adapted to the equation (and not, for example, fixed by dimensionality). Choosing a smaller $p$ results in having to show a Morrey-Property with larger $\lambda$. The Campanato space ${\cal L}^{p,\lambda}$ is defined similarly as the corresponding Morrey space, but the integral of $|f|^p$ is replaced by the integral of $|f-\bar{f}|^p$, where $\bar{f}$ is the average of $f$ over the ball. This is less restrictive, so $L^{p,\lambda}\subset {\cal L}^{p,\lambda}$. The less restrictive definition of Campanato spaces allows for an integral characterization of Hölder continuity: $C^{0,\alpha}={\cal L}^{p,m+p\alpha}$. Of particular interest is the case $\lambda=m$ where $\lambda$ equals the dimensionality of the domain $\Omega$. Then ${\cal L}^{p,m}={\cal L}^{1,m}$ characterizes functions of bounded mean oscillations, meaning bounded $|B|^{-1}\int_B|f(x)-\bar{f}|dx$. It lies in between $\cap_{p>1}L^p$ and $L^\infty$. A simple example of a function in ${\cal L}^{1,1}$ is $\log x$ for $0 TITLE: How to spot a moduli space? QUESTION [7 upvotes]: This is admittedly a vague question. I wonder if there is a way - a criterion - to tell if an algebraic scheme/stack $M$ is a (fine) moduli space, i.e., if parametrizes all families of isomorphism classes of algebro-geometric objects of a certain kind. By definition it must allow a map from the "universal family" of such objects, and entertain maps from families of the same over arbitrary bases such that pullback of the universal family agrees with the family over the base. This is a strong constraint; however, if all one has to begin with is $M$, can there be any way at all to tell that this collection of compatible maps exists? Even if we give ourselves a morphism $U \rightarrow M$ at the outset, can there be a criterion whereby $M$ turns out to be a moduli space with universal family $U$? In this case, we have the fibres to work with, so it's conceivable. REPLY [5 votes]: You cannot, even for some very natural moduli spaces. Take a look for instance here: M. Kapovich, J. Millson, On representation varieties of Artin groups, projective arrangements and the fundamental groups of smooth complex algebraic varieties, Math. Publications of IHES, Vol. 88 (1999) p. 5-95. M. Kapovich, J. Millson, Universality theorem for configuration spaces of planar linkages, Topology, Vol. 41 (2002), no. 6, p. 1051--1107. R. Vakil, Murphy's Law in algebraic geometry: Badly-behaved deformation spaces, Invent. Math. Vol. 164 (2006), 569-590. M. Reineke, Every projective variety is a quiver Grassmannian, Algebras and Representation Theory, Vol. 16 (2013) 1313–1314.<|endoftext|> TITLE: Basis for free modules over an affine domain QUESTION [8 upvotes]: Let $A=k[x_1,\cdots,x_r]/I$ for some prime ideal $I$ and some field $k$. Consider the free $A$-module $A^n$. Question 1. Given an element $e\in A^n$, is there a method to tell whether $e$ can be admitted as an element of some $A$-basis for $A^n$? Obviously, the components of $e$ need to generate $A$, but I doubt this is enough. What about if instead of one element $e$ we have a set of $A$-linearly independent elements $e_1, \dots, e_k$ for $k < n$. Question 2. Is there a method to tell whether these can be admitted as basis elements? I'm looking for more practical methods than theoretical ones. For example, you can consider the determinant polynomial with one column equal to $e$ and equate it to $1$ and say whenever this has a solution in $A$, then $e$ is an element of some basis. But this method is not practical. These were my main questions. Let $S$ be the set of elements in $A^n$ that can be part of some basis. Now define an equivalence relation on $S$ by setting two elements to be equivalent iff there is some basis containing both of them as basis elements. Question 3. After taking the quotient under this equivalence, is $S$ connected? If not, what are its connected components? REPLY [4 votes]: The first question is already answered very nicely by Steven Landsburg. My answer adds references and further results while addressing, to some extent, your second and third questions (see the last two sections). Some of the remarks below are very general, but each paragraph concludes with statements that are specific to affine algebras or affine domains, the setting of your post. Most of the content of my answer has been extracted from three textbooks: [2, VIII.2] (Projective Modules over Affine Algebras) , the end of [1, 11.5] (Suslin's theorems on affine algebras) and some items of [3]. I elaborate on each excerpt, trying to address also some practical aspects. Your first question asks about practical means to decide whether an element $e$ of the free module $A^n$, with $A$ an affine domain, can be completed in a basis of $A^n$. I will tackle this question first, and then consider the case of multiple rows. Completing a single unimodular row into an invertible square matrix (1st question). The more general question where $A$ is replaced by an arbitrary unital ring $R$ (I will assume throughout that $R$ is commutative for simplicity), has been investigated extensively in the context of the study of stably free modules, particularly in the resolution of Serre's problem on projective modules and its generalizations [9]. Let us make this connection explicit. An $R$-module $P$ is said to be stably free of type $m$ ($0 \le m < \infty$) if $P \oplus R^m$ is free. The module $P$ is stably free if it is stably free of type $m$ for some $m$. We say that $e \in R^n$ is a unimodular row if the components of $e$ generate $R$ as an $R$-module. In other words, the row $e$ is a right-invertible matrix. We have the following Proposition A [9, Corollary I.4.5]. The following are equivalent: Any finitely generated stably free $R$-module is free; Any finitely generated stably free $R$-module of type $1$ is free; Any unimodular row over $R$-module can be completed to a square invertible matrix over $R$ (by adding a suitable number of rows). Let us denote by $\text{Um}_n(R) \subset R^n$ the set of unimodular rows over $R$. The following proposition highlights the link between completable unimodular rows and the action of $\text{GL}_n(R)$ on $\text{Um}_n(R)$ by matrix right-multiplication. Proposition B [9, Proposition I.4.8]. The orbits of $\text{Um}_n(R)$ under the $\text{GL}_n(R)$-action are in $1$-$1$ correspondence with the isomorphism classes of $R$-modules $P$ for which $P \oplus R \simeq R^n$. Under this correspondence, the orbit of $(1, 0, \dots, 0)$ corresponds to the isomorphism classes of the free module $R^{n - 1}$. In particular $e \in \text{Um}_n(R)$ is completable if and only if $e \sim_{\text{GL}_n(R)} (1, 0, \dots, 0)$. A ring which satisfies the equivalent properties of Proposition A is called a Hermit ring (T. Lam's terminology). So, if $A$ is a Hermit ring, then it suffices for $e$ to be unimodular in order to be part of a basis of $A^n$. Commutative semilocal rings, Dedekind rings and Bézout rings are Hermit rings [9, Examples I.4.7]. By the Quillen-Suslin Theorem [9, Theorem V.2.9] [6, Corollary 11.5.5], the ring $k[x_1, \dots, x_r]$ is a Hermit ring if $k$ is a field or a principal ideal domain. This theorem admits several generalizations which can be useful in the context of this post, see, e.g., [9, Theorems V.2.11, V.3.3] and [6, Theorem 11.5.6] = [3]. For instance, it follows from the latter theorem that $A[x_1 x_1^{-1},x_2, x_2^{-1} \dots, x_t, x_t^{-1}, x_{t + 1}, \dots, x_r]$ is a Hermit ring if $A$ is any commutative Noetherian ring of Krull dimension at most $1$. In addition, it follows from a theorem of Murthy and Swan [2], that an affine algebra $A$ over an algebraically closed ground field $k$ is a Hermit ring if its Krull dimension is $2$ ($A$ is an affine surface). With Hermit rings, we are of course asking for too much: we want unimodular rows of any size to be completable. Let us consider individual sizes. First, there is a trivial, but effective observation: for any commutative unital ring $R$, any unimodular row of size $2$ is completable in a matrix of $\text{SL}_2(R)$. Focusing now on size-specific statements, we are led to the following definitions. Definition [9, I.$(4.5)_d$]. Let $d$ be a non-negative integer. A commutative unital ring $R$ is a $d$-Hermit ring if it satisfies any of the following equivalent properties. Any finitely generated stably free $R$-module of rank $> d$ is free. Any unimodular row over $R$-module of size $\ge d + 2$ can be completed to a (square) invertible matrix over $R$. For $n \ge d + 2$, $\text{GL}_n(R)$ acts transitively on $\text{Um}_n(R)$. Definition [6, 11.1.14]. The general linear rank $\text{glr}(R)$ of $R$ is the least integer $n \ge 1$ such that $\text{GL}_{n + 1}(R)$ acts transitively on $\text{Um}_{n + 1}(R)$. A $d$-Hermit ring $R$ in the sense of T. Lam is thus a ring $R$ which satisfies $\text{glr}(R) \le d + 1$. As already observed by Steven Landsburg, a Noetherian ring $R$ of Krull dimension $d$ is a $d$-Hermit ring. This follows from Bass's Stable Range Theorem [9, Theorem II.7.3] [6, Theorem 11.3.7] [11, Section 1.1]. This result can be strengthen for affine algebras over algebraically closed fields and also for few other classes of ground fields, see [9, Section VIII.2]. More specifically, an affine algebra $A$ over an algebraically closed field is a $(d - 1)$-Hermit ring if its Krull dimension is $d$ (this follows from [3]). By the main result of [5], such an $A$ is even a $(d - 2)$-Hermit ring if we assume moreover that $A$ is of the form $R[x]$ with $R$ regular. We have just seen that the $d$-Hermit property for a ring $R$ ensures that every unimodular row of $R$ with size $\ge d + 2$ is completable in a matrix of $\text{GL}_{d +2}(R)$. There are also criteria that apply to an individual row of a given size. Suslin's $n!$ Theorem [9, Theorem III.4.1] has already been mentioned in Steven Landburg's answer. Section III of [9] contains several other interesting criteria of the same flavor, e.g., Corollaries III.5.5 and III.5.8. Completing a matrix with multiple rows into an invertible square matrix (2nd question). The $d$-Hermit property, or equivalently the property $\text{glr}(R) \le d + 1$ is also relevant in the case of multiple rows over $R$ of the same size. Indeed, it follows from [9, Proposition I.4.3 and I.$(4.5)_d$] that an $m$-by-$d$ matrix $M$ over $R$ $(m \le d + 2)$ is completable in a matrix of $\text{GL}_{d +2}(R)$ if and only if $M$ is right-invertible. Thus, if $R$ is $d$-Hermit, then there is a practical means to determine whether such a matrix $M$ can be completed: it is necessary and sufficient that the $m$-by-$m$ minors of $M$ are unimodular, i.e., the ideal generated by the $m$-by-$m$ minors of $M$ is $R$ [8, Theorem 3.7]. Identifying rows of invertible matrices (3rd question). This section is dedicated to the third question. Let $R$ be a commutative and unital ring. Let $\Gamma_n(R)$ be the graph whose vertex set is $\text{Um}_n(R)$ and for which an edge connects two distinct vertices if the corresponding unimodular rows are the rows of a same matrix in $\text{GL}_n(R)$. Note that no unimodular row is isolated in $\Gamma_n(R)$ if $n = 2$ or if $R$ is a $(n - 2)$-Hermit ring. Third question. Is $\Gamma_n(R)$ connected? If not, what are the connected components? Let $H_n(R)$ denote the Hilson graph of $R$ [7], that is, the graph whose vertex set is $\text{Um}_n(R)$ and for which an edge connects two vertices (possibly the same) if the corresponding unimodular row $\alpha$ and $\beta$ satisfy $\alpha \beta^t = 1$ (i.e., the dot product is the identity element of $R$). Let $\text{E}_n(R)$ be the subgroup of $\text{GL}_n(R)$ generated by the elementary matrices, i.e., the matrices which differ from the identity by a single off-diagonal element. We will prove the following Claim. The following are equivalent. $(i)$ $\Gamma_2(R)$ is connected. $(ii)$ $H_2(R)$ is connected. $(iii)$ $R$ is a $GE_2$-ring in the sense of P. Cohn [1], i.e., $\text{SL}_2(R) = \text{E}_2(R)$. Combining the above claim with [1, Proposition 7.3], we obtain for instance that $\Gamma_2(k[x_1, \dots, x_r])$, with $k$ a field, is connected if and only if $r \le 1$. In general, Lemma 2 below shows that $\Gamma_n(R)$ is connected only if $H_n(R)$ is. In addition, the following theorem of Hinson shows that the number of connected components of $\Gamma_n(R)$ is not less that the cardinality of the orbit space $W_n(R) \Doteq \text{Um}_n(R)/ \text{E}_n(R)$. Theorem [9, Theorem III.6.4]. Let $n \ge 3$. Two vertices $\alpha$ and $\beta$ of $H_n(R)$ are connected by an edge path of even length if and only if $\alpha = \beta E$ for some $E \in \text{E}_n(R)$. The orbit space $W_n(R)$ turns to be a group under some conditions on the spectrum of $R$ and its Krull dimension. The orbit space $W_n(R)$ is the subject of a series of papers by L. Vaserstein, A. Suslin, W. van der Kallen, M. Roitman, R. Rao and J. Fasel. A thorough account on this work is given in [9, VIII.5]. Two recent contributions to this topic are [10] and [12]; both contain results specific to affine algebras. The proof of the claim relies on the next two lemmas. Lemma 1 [9, Proposition III.6.6]. Let $n \ge 2$. Let $e_1$ be the first the vector of the canonical basis of $R^n$. Then $e_1\text{E}_n(R)$ coincides with the connected component of $e_1$ in $H_n(R)$. Lemma 2. The two following hold. $(1)$ If two vertices belong to the same connected component of $\Gamma_n(R)$, then they are connected by an edge path of even length in $H_n(R)$. $(2)$ If two vertices are connected by an edge path of even length in $H_2(R)$, then they belong to the same connected component of $\Gamma_2(R)$. Proof of Lemma 2. We shall prove that if two unimodular rows $\alpha, \beta \in \text{Um}_n(R)$ are connected by an edge in $\Gamma_n(R)$ then they are connected by an edge path of length two in $H_n(R)$, which establishes $(1)$. Indeed, consider a matrix $M \in \text{GL}_n(R)$ whose first two rows are $\alpha$ and $\beta$. Set $\gamma \Doteq (M^{-1}(e_1 + e_2))^t$, then we have $\alpha \gamma^t = \beta \gamma^t = 1$. Assume now that $n = 2$ and let $\alpha, \beta, \gamma \in \text{Um}_n(R)$ be such that $\alpha \gamma^t = \beta \gamma^t = 1$. Write $\gamma = (\gamma_1, \gamma_2)$ and set $\gamma' = (-\gamma_2, \gamma_1)$. As the $2$-by-$2$ matrices $\begin{pmatrix} \alpha \\ \gamma' \end{pmatrix}$ and $\begin{pmatrix} \beta \\ \gamma' \end{pmatrix}$ have determinant $1$, the rows $\alpha$ and $\beta$ are connected by an edge path of length two in $\Gamma_2(R)$. Assertion $(2)$ follows immediately. We are now in position to prove the claim. Proof of the claim. The equivalence $(i) \Leftrightarrow (ii)$ is essentially given by Lemma 2. For the implication $(i) \Leftarrow (ii)$, we use Lemma 1 and observes that, since there is an edge looping on $e_1$ ($e_1 e_1^t = 1$), any two vertices can be connected by an edge path of even length in $H_2(R)$. The equivalence $(ii) \Leftrightarrow (iii)$ is given by Lemma 1. [1] P. Cohn, "On the structure of the $\text{GL}_2$ of a ring", 1966. [2] M. Murthy and R. Swan, "Vector bundles over affine surfaces", 1976. [3] A. Suslin, "On the structure of the special linear group over polynomial rings", 1977. [4] A. Suslin, "A cancellation theorem for projective modules over algebras", 1977. [5] H. Lindel, "On the Bass-Quillen conjecture concerning projective modules over polynomial rings", 1981. [6] J. McConnell, J. Robson, "Noncommutative Noetherian Rings", 1987. [7] E. Hinson, "Path of unimodular vectors", 1989. [8] K. Rao, "Theory of Generalized Inverses Over Commutative Rings", 2002. [9] T. Lam, "Serre's Problem on Projective Modules", 2006. [10] J. Fasel, "Some remarks on orbit sets of unimodular rows", 2011. [11] C. Weibel, "The K-book: An introduction to algebraic K-theory", 2013. [12] A. Gupta, A. Garge, R. Rao, "A nice group structure on the orbit space of unimodular rows - II", 2013.<|endoftext|> TITLE: Representation over matrices $A_i^3=I$, $A_0A_1^\dagger+A_1A_2^\dagger+A_2A_0^\dagger=0$, $A_0^\dagger A_1+A_1^\dagger A_2+A_2^\dagger A_0=0$ QUESTION [6 upvotes]: I would like to know what all the possible finite-dimensional representations of the following relations are. $$A_0^3 = A_1^3 = A_2^3 = I \tag{1}$$ $$A_0 A_1^\dagger + A_1 A_2^\dagger + A_2 A_0^\dagger = 0 \tag{2}$$ $$A_0^\dagger A_1 + A_1^\dagger A_2 + A_2^\dagger A_0 = 0 \tag{3}$$ where $I$ is the identity matrix. In other words, what are the matrices (in any dimension) satisfying $(1)$, $(2)$, $(3)$? A first step is to characterize what the $3 \times 3$ matrices satisfying it are. Is there any numerical way to tackle this problem? Behind this should be the clock/shift algebra (Weyl theorem). Considering the clock and shift operators (of size 3, see https://en.wikipedia.org/wiki/Generalizations_of_Pauli_matrices) $X$, $Z$ such that $ZX=ωXZ$ with $ω=e^{2iπ/3}$. Then $A_k=ω^{k(k+1)}XZ^k$, is a solution. A second solution is obtained with $\omega$ changed into its conjugate. I expect that all solutions are block diagonal matrices of those two examples. REPLY [2 votes]: $\DeclareMathOperator\deg{deg}\DeclareMathOperator\dim{dim}\DeclareMathOperator\span{span}$ I decided to split my answer into a more direct "answer" post and a "tidbits" post. This is the "tidbits" post, where I point out some things I determined about $S$ that may help with other analysis. $S$ inherits a grading $S = \bigoplus_{k = 0}^\infty S^k$ from $TV$, where $\deg(x) = \deg(y) = \deg(z) = 1$, as the quotient of a graded ring by a homogeneous ideal. Claim: $S^k = q(\sum_{\ell = 0}^k \left( T^\ell\span(x, y) z^{k - \ell}\right))$, where $\span(x, y) \subseteq V$ is the vector space spanned by $x$, $y$; $T \span(x, y)$ is its tensor algebra; and $T^k$ denotes the $k$th graded component of the tensor algebra. Conjecture: Further, $q$ is an isomorphism. Equivalently, $\langle xy + yz + zx, yx + zy + xz\rangle \cap \sum_{\ell = 0}^k \left( T^\ell\span(x, y) z^{k - \ell}\right) = \{0\}$. Less formally, this is saying that every element of $S^k$ has a "normal representative" in $TV$ such that each monomial only has $z$ at the ends, with no $x$ or $y$ after a $z$. The conjecture is that this representative is unique. Proof: We work by induction. If $k = 0$, this is trivial. Otherwise, let $s \in S^k$, with some representative $t \in T^kV$. By the definition of $TV$, we have that $t = x C_x + y C_y + z C_z$ for some $C_x, C_y, C_z \in T^{k - 1}V$. Then $s = p(x) p(C_x) + p(y) p(C_y) + p(z) p(C_z)$. By the induction step, $p(C_x)$, $p(C_y)$, $p(C_z)$ have such "normal representatives" $C'_x$, $C'_y$, $C'_z$. Write $C'_z = x D_x + y D_y + z D_z$. By the definition of "normal representative", we have that $D_z$ consists only of linear combinations of strings of $z$s. Let $E_x$ be the "normal representative" of $z D_x$ and $E_y$ be the "normal representative" of $z D_y$. Then: \begin{align*} s &= p(x) p(C_x) + p(y) p(C_y) + p(z) p(C_z) \\ &= p(x) p(C_x) + p(y) p(C_y) + p(z) p(x) p(D_x) + p(z) p(y) p(D_y) + p(z) p(z) p(D_z) \\ &= p(x) p(C_x) + p(y) p(C_y) - p(x) p(y) p(D_x) - p(y) p(z) p(D_x) - p(x) p(z) p(D_y) - p(y) p(x) p(D_y) + p(z) p(z) p(D_z) \\ & = p(x) p(C_x) + p(y) p(C_y) - p(x) p(y) p(D_x) - p(y) p(E_x) - p(x) p(E_y) - p(y) p(x) p(D_y) + p(z) p(z) p(D_z) \end{align*} Clearly, $x C_x + y C_y - x y D_x - y E_x - x E_y - y x D_y + z z D_z$ is a normal representation of $x$. $\square$ Uniqueness shouldn't be too hard to prove by induction, but I haven't worked out the exact "trick". The idea is that if $t, t'$ are distinct "normal representatives" of $s$, then $t - t'$ is a nonzero "normal representative" of $0$, so we can assume WLOG that $t'$ is the obvious representation of $0$. Then $t = x C_x + y C_y + z C_z$ for some $C_x, C_y, C_z \in T^{k - 1}V$, where $C_x$, $C_y$ are "normal representatives" and $C_z = a z^{k - 1}$ for some scalar $a$. But if $C_z \neq 0$, then $t$ can't be in the ideal (as all of the monomials in the ideal contain some non-$z$ letter), so $C_z = 0$. There should be a relatively simple demonstration then that $C_x = C_y = 0$, which would finish the proof. Corollary of conjecture: $\dim(S^k) = 2^{k + 1} - 1$. I've done some minor independent checking (using the ideal and inclusion-exclusion), and this seems to hold at least up to $k = 5$, if my calculations are correct. So we have bounded the growth of $S^k$ (and found its dimension exactly if the conjecture is correct). This should help if there is an analogue of the Peter-Weyl theorem. Final tidbit: it may be useful to look at $S$ as being "noncommutatively graded" over $S_3$, with $\text{"deg"}(x) = (12), \text{"deg"}(y) = (23), \text{"deg"}(z) = (13)$. This can give us some idea of likely "useful elements" to consider: $xyxyxy + yzyzyz + zxzxzx$ should be interesting.<|endoftext|> TITLE: Fourier coefficients of a periodic distribution? QUESTION [6 upvotes]: Let $\tau>0$, and let $T\in \mathcal{D}'(\mathbb{R})$ be a $\tau$-periodic distribution (that is, $ \langle T, \varphi(\cdot+\tau)\rangle= \langle T,\varphi\rangle $ for all $\varphi \in \mathcal{D}(\mathbb{R})$). Then $$ T=\sum_{n\in \mathbb{Z}} c_n e^{i 2\pi t/\tau}, $$ for some $c_n\in \mathbb{C}$, and where the equality means that the symmetric partial sums of the series on the right hand side converge in $\mathcal{D}'(\mathbb{R})$ to $T$. What are the $c_n$s in terms of $T$? One would think that they are given by $c_n=\langle T|_{(0,2\pi)}, e^{-in2\pi /\tau}\rangle/\tau$, but $e^{-in2\pi/\tau}$ is not a test function in $\mathcal{D}((0,2\pi))$. REPLY [3 votes]: Just a quick complement to what Paul said, in order to explain more concretely what "descends" means. Take $\tau=1$ for simplicity. Let $\rho$ be a function in $\mathscr{D}(\mathbb{R})$ ($\mathscr{S}(\mathbb{R}$) would work too) which gives a partition of unity of the form $$ \sum_{n\in\mathbb{Z}}\rho(t+n)=1 $$ for all $t\in\mathbb{R}$. then $c_n$ can be extracted as $$ c_n=\langle T,\rho(t)e^{-2i\pi n t}\rangle\ . $$ Indeed if $n\in\mathbb{Z}$, $$ \int_{\mathbb{R}}\rho(t)e^{2i\pi nt}\ dt =\sum_{m\in\mathbb{Z}}\int_{m}^{m+1} \rho(t)e^{2i\pi nt}\ dt $$ $$ =\sum_{m\in\mathbb{Z}}\int_{0}^{1} \rho(t+m)e^{2i\pi n(t+m)}\ dt $$ $$ =\int_{0}^{1}\left(\sum_{m\in\mathbb{Z}}\rho(t+m)\right)e^{2i\pi nt}\ dt $$ $$ =\int_{0}^{1}e^{2i\pi nt}\ dt\ = \delta_{n,0}\ . $$<|endoftext|> TITLE: Is there a proof of the Hawking bound for the efficiency of a black holes merger? QUESTION [10 upvotes]: Consider two black holes with masses $m_1,m_2$ and zero angular momenta merging to form a single one with the mass $m$ and the rotation parameter $a=J/m$. Hawking, in "Black Holes in General Relativity" Commun. math. Phys. 25 (1972), 152—166 proposed an inequality $$m^2+m\sqrt{m^2-a^2}>2(m_1^2+m_2^2)$$ for this process (in fact, for a more general one, see p. 14 of the paper). I learned about this bound ages ago from the Lightman-Press-Price-Teukolsky relativity problem book and had no doubt about it. But now I think that the proof given in this paper is total rubbish despite being published in a supposedly mathematical journal. The inequality is derived from what is now called an area theorem which sates that the area of the event horizon never decreases. There is nothing wrong with the theorem itself except the way it is formulated makes it completely useless for obtaining an inequality of this sort. (And probably for any other meaningful conclusion.) The fishy point here is the assumption that the area of a black hole event horizon is given by the formula (in geometric units $c=G=1$) $$A=8\pi m(m+\sqrt{m^2-a^2}).$$ No doubt, this assumption is true for a Kerr black hole but there is a big problem. The event horizon as it is defined in the formulation of the area theorem depends on the (arbitrarily distant) future evolution of a black hole, so even it the thing looks exactly as a standard Kerr black hole now its event horizon may still well be very different from what one of a Kerr hole is supposed to be, with very different area. There is no formula for the actual area of this event horizon in terms of the mass and the angular momentum. To see where the problem really lies it is convenient to consider a scattering of two black holes instead of their merger. This process has an inverse which is also perfectly physical even if not likely to ever happen in reality. (Because general relativity dynamics is, of course, time-symmetric.) Then exactly the same argument as in the paper when applied to both processes gives two inequalities which contradict each other. Admittedly, from reading more recent physical literature I have the impression that the problem is more or less known. However, it is never mentioned explicitly. Apparently, physicists believe that the inequality is true anyway and do not care much about gaps in its proof. A mathematician like myself would rather like to see an actual proof though. Is such a proof already known or, at the very least, was the problem ever considered seriously? This is my question. REPLY [5 votes]: A mathematical proof of Hawking's area theorem has been given by Chruściel, Delay, Galloway, and Howard, in Regularity of Horizons and The Area Theorem (2001). The proof identifies the conditions under which the area of sections of future event horizons in space–times satisfying the null energy condition is non–decreasing towards the future. The monotonicity is shown to hold without making requirements on the differentiability of event horizons. The specific relation between the area of the event horizon and the angular momentum in the OP is model dependent, it does not have general validity.<|endoftext|> TITLE: Zero divisors in compact quantum groups QUESTION [5 upvotes]: Let $\mathcal{G}$ be compact quantum group in the sense of S. L. Woronowicz. As is well-known, every compact quantum group contains a dense Hopf algebra, called the polynomial Hopf algebra Pol$(\mathcal{G})$. For example, consider the famous $C(SU_q(2))$, the $q$-deformation of $SU(2)$, with generators $\alpha$ and $\beta$ : https://en.wikipedia.org/wiki/Compact_quantum_group The dense Hopf algebra is now the polynomial $*$-algebra generated by $\alpha$ and $\beta$. A well-known fact about Pol($SU_q(2)$) is that it has no zero-divisors, that is, it is a domain. What is a good example of a compact quantum group $\mathcal{G}$ such that Pol$(\mathcal{G})$ has zero divisors? On the other hand, is there any abstract characterization of the compact quantum groups such that the polynomial Hopf algebra is a domain? REPLY [5 votes]: Concerning the first question, consider the quantum permutation group $S_N^+$. Its polynomial algebra is generated by an $N\times N$ matrix $u=(u_{ij})$ of self-adjoint projections which sum up to $1$ on each row and column. As a consequence, it has lots of zero divisors. As for a general criterion, note that for any discrete group $\Gamma$, there is a "dual" compact quantum group with polynomial algebra $\mathbb{C}[\Gamma]$. Characterizing when this is a domain is an open problem. For instance, Kaplansky's zero divisor conjecture asserts that if $\Gamma$ is torsion-free then $k[\Gamma]$ is a domain for any field $k$ and even for $k = \mathbb{C}$ this is, as far as I know, still open in full generality.<|endoftext|> TITLE: Floor of Riemann zeta function QUESTION [14 upvotes]: How to show that $$\left\lfloor\zeta\left(1+\frac{1}{n}\right)\right\rfloor=n$$ for every positive integer $n$? REPLY [23 votes]: It is known that (see Corollary 1.14 in Montgomery-Vaughan: Multiplicative number theory I) $$\frac{1}{\sigma-1}<\zeta(\sigma)<\frac{\sigma}{\sigma-1},\qquad \sigma\in(0,1)\cup(1,\infty).$$ In particular, taking $\sigma=1+\frac{1}{n}$, we get $$n<\zeta\left(1+\frac{1}{n}\right) TITLE: Cardinality of certain subsets in vector spaces over finite fields QUESTION [5 upvotes]: Assume that you have an $n$-dimensional vector space over a finite field (therefore the number of elements in the vector space is finite) and $F$ is a subset of this vector space which contains $m$ elements. Let $A$ be a subset of this vector space such that the intersection of $A+A$ and $F$ is empty. The question is: What is a non trivial lower bound for the maximal possible cardinality of such an $A$? REPLY [8 votes]: Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|$, it has an independent set of size at least $2^n/(|F|+1)$. In general, this bound is best possible: it is attained when $F$ is a subgroup of $\mathbb Z_2^n$ with the zero element removed (so that $|F|=2^k-1$, where $k$ is the rank of $F$), and $A$ contains a unique element from each $F$-coset. Better bounds can be given if some information about the set $F$ is available. Say, if $F=\{f_1,\dotsc, f_m\}$ is an independent set, then one can find a subgroup $H<\mathbb Z_2^n$ such that $\mathbb Z_2^n=\langle F\rangle\oplus H$, and take $$ A := \{ c_1f_1+\dotsb+c_mf_m+h\colon c_1+\dotsb+c_m=0,\ h\in H \} $$ to have $|A|=2^{n-1}$.<|endoftext|> TITLE: Want to prove an inequality QUESTION [5 upvotes]: I want to show that $9*\left[\frac{xy}{x+y}-q(1-q)\right]-12*[xy-q(1-q)]+(1-q-x)^{3}+(x+y)^{3}+(q-y)^{3}-1\geq0$ where $0 TITLE: Is an HNN extension of a virtually torsion-free group virtually torsion-free? QUESTION [15 upvotes]: This is a cross post from Math.StackExchange after 2 weeks without an answer and a bounty being placed on the question. Let $G=\langle X\ |\ R\rangle$ be a (finitely presented) virtually torsion-free group. Let $H,K TITLE: History of the kernel of a homomorphism? QUESTION [8 upvotes]: This previous question traces the notion of group homomorphism to Jordan (1870) and the term "homomorphic" to Fricke and Klein (1897) and to earlier lectures of Klein: Whence “homomorphism” and “homomorphic”? What about the concept of the kernel of a homomorphism? This webpage traces it to Fredholm's (1903) use of the french "noyau" for the nullspace of a system of integer linear equations. Then Hilbert put this into German as "Kern" (1904). http://jeff560.tripod.com/k.html But the concept of a kernel is structural, equivalent to the concept of normal subgroup. I guess that this insight is due to Emmy Noether but I can't find a reference. Probably what I am looking for is the first enunciation of the theorem $$G/\ker\varphi\cong\mathrm{im}\,\varphi$$. REPLY [12 votes]: The word at least, seems to originate with Pontryagin (1931, p. 186): 28) Wenn eine Gruppe $A$ auf eine Gruppe $B$ homomorph abgebildet ist, so heißt die Untergruppe von $A$, die aus allen Elementen besteht, welche auf das Einheits- (Null-) element von $B$ abgebildet werden, der Kern der homomorphen Abbildung. E.g. van der Waerden (1930, p. 35) still states the isomorphism $G/\ker\varphi\cong\mathrm{im}\,\varphi$ without $“\ker”$, as $\mathfrak{G/e\cong\overline{G}}$ with $\mathfrak e$: derjenige Normalteiler von $\mathfrak G$, dessen Elementen das Einselement in $\overline{\mathfrak G}$ entspricht. Added: Noether wrote things like $G/\varphi^{-1}(H')\cong G'/H'$, but I’m not sure she ever spelled out the case $H'=\{e\}$. Anyway, the theorems whose “wording and proof” van der Waerden (1975, p. 34) attributes to Noether are those for “groups with operators” (1930, p. 136). For groups, he says he followed Speiser, who has indeed (1923, p. 19): Satz 14: Ist $\mathfrak G'$ mit $\mathfrak G$ isomorph, so entspricht dem Einheitselement von $\mathfrak G'$ ein Normalteiler $\mathfrak N$ von $\mathfrak G$, und $\mathfrak G'$ ist homomorph mit der Faktorgruppe $\mathfrak{G/N}$. and Burnside, who has (1897, pp. 36, 38): Theorem VII. If a group $G$ is multiply isomorphic with a group $G'$, then (i) the operations of $G$, which correspond to the identical operation of $G'$, form a self-conjugate sub-group of $G$; (...) a group $G'$ with which a group $G$ is multiply isomorphic, in such a way that to the identical operation of $G'$ there corresponds a given self-conjugate sub-group $\Gamma$ of $G$, is completely defined (as an abstract group) when $G$ and $\Gamma$ are given. (...) Herr Hölder (1889, p. 31) has introduced the symbol $ \smash{\frac G\Gamma} $ to represent this group; he calls it the quotient of $G$ by $\Gamma$, and a factor-group of $G$. I think it’s fair to say that Hölder (pp. 32–33) already has all of the above, except the word kernel. (E. g. he writes that normal subgroups of $G|\mathsf H$ make normal subgroups of $G$, with the identity making $\mathsf H$ itself, and that one could start from a morphism rather than a normal $\mathsf H$.) But it goes further: kernels are already in Dyck (1882, p. 12, cited by Hölder): Operationen der Gruppe $G$, welche sonach der Identität in $\overline G$ entsprechen, bilden eine Gruppe $H$ und diese ist (...) in $G$ ausgezeichnet enthalten. in Capelli (1878, p. 36), who calls them primi periodi (denoted $\mathrm O_0$) and shows: Affinchè un gruppo $\mathrm O$ contenuto in $\mathrm G$ possa esser preso come primo periodo, à necessario che esso si permutabile a tutte le sostituzioni di $\mathrm G$. Vedremo più tardi (III, 3) che questa condizione à anche sufficiente, vale a dire, che si può sempre costruire un gruppo $\Gamma$ isomorfo a $\mathrm G$, il quale ad $\mathrm O$ faccia corrispondere l’unità. in Jordan (1870, §67, cited by Capelli): Le groupe $\Gamma$ contient la substitution $ı$. Soient $h_1,\dots,h_m$ les substitutions correspondantes de $\mathrm G$ : elles forment un groupe auquel toutes les substitutions de $\mathrm G$ sont permutables. and finally(?), as per Noether’s “Es steht alles schon bei Dedekind”, in Dedekind (1855–58, p. 440), for a morphism $M\to M_1$: Der Komplex aller der $n$ in $M$ enthaltenen Objekte $\varphi$, denen das Objekt $1$ entspricht, bildet eine Gruppe, und zwar einen eigentlichen Divisor von $M$; dann ist $m = m_1n$. REPLY [8 votes]: Two original references are: E. Noether, Hyperkomplexe Größen und Darstellungstheorien (1929): see page 648. E. Noether, Abstrakter Aufbau der Idealtheorie in algebraischen Zahl- und Funktionenkörpern (1927): see page 40. Noether phrases the isomorphism theorems in terms of "Normalteiler" (normal subgroups) rather than "Kerne" (kernels). Van der Waerden recalls that the 1929 paper was based on a course with the same title that Noether taught at Göttingen in 1926/1927.<|endoftext|> TITLE: Reference for Schur multiplier identity QUESTION [9 upvotes]: Let $G$ be a finite group and $H$ a normal subgroup of $G$. I recently stumbled upon the following identity for the Schur multiplier of $G/H$: $$\operatorname{H}_2(G/H,\mathbb{Z}) \cong \frac{\overline{H} \cap [\overline{G},\overline{G}]}{[\overline{H},\overline{G}]}$$ Here $\overline{G}$ is a Schur covering group of $G$ and if $$1 \to K \to \overline{G} \xrightarrow[]{\lambda} G \to 1$$ is a universal cover of $G$, then $\overline{H} = \lambda^{-1}(H)$. Note in particular that for $H=1$ one recovers the fact that $H_2(G,\mathbb{Z}) \cong K$. I've found this formula as a special case of an algebro-geometric identity (namely, via an explicit formula for the unramified Brauer group of a smooth compactification of a certain family of tori). My question is the following: Is there purely group-theoretic proof of the above identity? The identity quite resembles Hopf's formula for the Schur multiplier, but I was unable to derive it from that result. I also could not find it in some of the standard references for this topic, e.g. Karpilovski's book on the Schur multiplier. REPLY [6 votes]: Let $G = F/R$ with $F$ free and $H = S/R$. Since everything is happening modulo $[F,R]$, I am just going to work modulo $[F,R]$. Then, by the Hopf formula, $M(G)$ (the Schur Multiplier) is isomorphic to $[F,F] \cap R$, which has free abelian complements $C$ in $R$ with $R = ([F,F] \cap R) \times C$, and $\bar{G} = F/C$. (Note that different complements can give non-isomorphic covering groups $\bar{G}$.) Now, since $[F,F] \cap C = 1$, we have $$M(G/H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} = \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$<|endoftext|> TITLE: The correct determinant exponent of the weight $k$-operator for defining Hecke operators/adelizing modular forms QUESTION [8 upvotes]: For $g \in \operatorname{SL}_2(\mathbb R)$, and $\mathbb H$ the upper half plane, and $k\geq 1$ an integer, the weight $k$-operator on functions $f: \mathbb H \rightarrow \mathbb C$ is defined by $$f[g](z) = f(g.z) j(g,z)^{-k}$$ where $j(g,z) = (cz+d)^{-1}$, $g = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$. In order to define Hecke operators, or to adelize modular forms, or to identify modular forms as functions on $\operatorname{GL}_2(\mathbb R)^+$, it is necessary to extend this definition to $g \in \operatorname{GL}_2^+(\mathbb R)$. In A First Course in Modular Forms, in Chapter 5.1 Diamond and Shurman set $$f[g](z) = f(g.z)j(g,z)^{-k} \det(g)^{k-1}$$ In Automorphic Forms and Representations, in Chapter 1.4 Bump sets $$f[g](z) = f(g.z)j(g,z)^{-k} \det(g)^{k/2}$$ Which exponent of the determinant is better to use, and why? If we adelize a Hecke eigenform for $\operatorname{SL}_2(\mathbb Z)$ and look at the corresponding automorphic representation $\pi = \otimes_p \pi_p$, which normalization is better to define Hecke operators with, if we want the classical Hecke operator $T_p$ to coincide naturally with an action of the spherical Hecke algebra $\mathscr H(\operatorname{GL}_2(\mathbb Q), \operatorname{GL}_2(\mathbb Z_p))$ on the local component $\pi_p$? Recall that to adelize a modular form $f$ of $\operatorname{SL}_2(\mathbb Z)$ of some given weight, we would first identify $f$ with a function $\phi$ on $\operatorname{GL}_2^+(\mathbb R)$ by setting $$\phi(g) = f[g](i)$$ and then we would define an automorphic form $\varphi$ on $\operatorname{GL}_2(\mathbb Q) \backslash \operatorname{GL}_2(\mathbb A)$ by using the decomposition $\operatorname{GL}_2(\mathbb A) = \operatorname{GL}_2(\mathbb Q) \operatorname{GL}_2^+(\mathbb R)K$ for $K$ a suitable compact subgroup, writing $g = \alpha g_{\infty}k$, and setting $\varphi(g) = \phi(g_{\infty})$. REPLY [14 votes]: This is a question which has no "right" answer. A posh interpretation of the choice of exponent is that a Hecke eigenform $f$ determines an equivalence class of irreducible representations $\Pi = \bigotimes'_v \Pi_v$ of $GL_2(\mathbb{A}_\mathbb{Q})$, differing by twists by powers of the character $g \mapsto \|\det(g) \|$, and the power of $\det$ that you put in the action of $GL_2^+(\mathbb{Q})$ determines which twist you get. The normalisation that Diamond and Shurman use is the one that makes the eigenvalue of the double coset $\begin{pmatrix} p & 0 \\ 0 & 1 \end{pmatrix}$ on $\Pi_p$ correspond to the Fourier coefficient $a_p(f)$; while Bump's normalisation makes it correspond to $a_p(f) / p^{(k/2-1)}$. From the perspective of the analytic theory of automorphic forms, Bump's choice is the "obviously right" one, since it makes $\Pi$ be unitary. Then you can find $\Pi$ as a subrepresentation of $L^2(GL_2(\mathbb{Q}) \backslash GL_2(\mathbb{A}))$ and the analytic theory works as it should. Since Bump's text emphasises the analytic theory of automorphic forms, this is the convention he chooses (and Paul Garrett's comment seems to be coming from the same viewpoint). On the other hand, from the viewpoint of the algebraic theory (Galois representations, special values of L-functions, etc), the factor $k/2$ is extremely inconvenient, particularly when $k$ is odd. The $a_p$'s all lie in some common finite extension of $\mathbb{Q}$, but the eigenvalue of $\Pi_p$ has been multiplied by $p$ to a half-integer power; so the extension of $\mathbb{Q}$ generated by the eigenvalues is not finite, and correspondingly the critical values of the $L$-series are at half-integers rather than integers, meaning that the $L$-series of $\Pi$ cannot correspond to a motive. With Diamond and Shurman's normalisation, the finite part of $\Pi$ is definable over a number field, and its $L$-series is motivic. (This would work with any integer power of $\det$, but $\det^{k-1}$ is the minimal one which makes the Hecke eigenvalues algebraic integers.) So the Diamond-Shurman normalisation is better for the algebraic theory, and the Bump normalisation for the analytic one. There is a great quote on this (attributed to Deligne, but possibly apocryphal): "Langlands is very convinced he knows what the square root of $p$ is. I have never been so sure."<|endoftext|> TITLE: Dividing a cake between $n-1$, $n$, or $n+1$ guests QUESTION [34 upvotes]: A housewife is waiting for guests and has prepared a cake. She doesn't know how many guests will come, but it will be $n-1$, $n$, or $n+1$. What is the minimal number $f(n)$ of pieces the cake should be cut to make it possible to divide between guests equally? For $n=2$, $f(n)=f(2)=4$: The problem was posed 16.10.2018 by Oleksandr Maksymets on page 76 of Volume 2 of the Lviv Scottish Book. The prize: Cooked duck or lunch + beer! REPLY [4 votes]: $f(6) = 13$ with a cake of size $210$ and piece sizes: $$\{3, 5, 8, 10, 12, 13, 17, 18, 20, 22, 25, 27, 30\},$$ where $$17+25 = 5+10+27 = 20 + 22 = 3+8+13+18 = 12+30,$$ $$10+25 = 5+30 = 3+12+20 = 17+18 = 13 + 22 = 8+27,$$ $$5+25 = 3+27 = 10+20 = 12+18 = 13+17 = 30 = 8+22.$$ It was computed with via solving MILP as explained in my other answer.<|endoftext|> TITLE: What's the minimal weight of a maximal space? QUESTION [7 upvotes]: A non-empty topological space without isolated points is called maximal if every finer topology on that space has at least an isolated point. The existence of a (Hausdorff) maximal space is a simple consequence of Zorn's Lemma. Note that in a maximal space $(X, \tau)$, nowhere dense sets are closed (and discrete). Indeed, if there were a nowhere dense set $N$ that is not closed then $\tau \cup \{X \setminus N \}$ would be a subbase for a finer topology without isolated points on $X$. In particular, a maximal space can't contain any non-trivial convergent sequences, because a discrete set is nowhere dense in a space without isolated points. Let $\mathfrak{max}$ be the minimal weight of a Hausdorff maximal space. Clearly $\aleph_1 \leq \mathfrak{max} \leq \mathfrak{c}$, but we can prove a better lower bound, namely: $\mathfrak{d} \leq \mathfrak{max}$. If $X$ is a countable Hausdorff maximal space then $w(X) \geq \mathfrak{d}$. Proof: Fix a point $x \in X$ and let $\{U_n: n < \omega \}$ be a maximal pairwise disjoint family of non-empty open sets with the property that $x \notin \overline{U_n}$, for every $n< \omega$: Clearly $x \in \overline{\bigcup \{U_n: n < \omega \}}$. Let $\{x^n_k: k < \omega \}$ be an enumeration of $U_n$. Suppose by contradiction that $w(X)=\kappa < \mathfrak{d}$ and let $\{B_\alpha: \alpha < \kappa \}$ enumerate a local base at $x$. For every $\alpha < \kappa$, the set $B_\alpha$ intersects infinitely many $U_n$'s, so we can find an integer-valued function $f_\alpha$ with infinite domain $\subseteq \omega$ such that $x^n_{f_\alpha(n)} \in B_\alpha \cap U_n$, for every $n \in dom(f_\alpha)$. Since $\kappa < \mathfrak{d}$, we can find a function $f: \omega \to \omega$ such that, for every $\alpha < \kappa$, there is $n \in dom(f_\alpha)$ with $f_\alpha(n) < f(n)$. Let $D=\{x^n_k: k \leq f(n), n < \omega \}$. Then $D$ is discrete and $x \in \overline{D} \setminus D$, so $D$ is a nowhere dense set in $X$ which is not closed and that contradicts maximality. QUESTION: Is $\mathfrak{max}=\mathfrak{d}$ in ZFC? EDIT(10/05/2019): Will Brian answered the above question in the negative, but the question of the title is still open. What is $\mathfrak{max}$? Is it equal to some product of known cardinal invariants of the continuum? REPLY [5 votes]: No, these cardinals are not provably equal. This follows from a result of El'kin 1: Let $X$ be a set and let $x \in X$. If $\tau$ is a maximal topology on $X$, then $\{U \setminus \{x\} \,:\, x \in U \in \tau \}$ is a base for a non-principal ultrafilter on $X$. If $X$ is countable (as in your question), it follows that any local basis at $x$ has size at least as big as the ultrafilter number $\mathfrak u$, defined as the smallest possible cardinality of a base for a non-principal ultrafilter on a countable set. Hence we have found another lower bound for your cardinal: $\mathfrak u \leq \mathfrak{max}$. The reason this answers your question is that it is consistent to have $\mathfrak d < \mathfrak u$. (This happens for example in the random real model.) By the previous paragraph, any model in which $\mathfrak d < \mathfrak u$ is also a model in which $\mathfrak d < \mathfrak{max}$. 1 A. G. El'kin, "Ultrafilters and irresolvable spaces," Vestnik Moskov. Univ. Ser. I Mat. Mekh. 24 (1969), no. 5, pp. 51-56. $\ $ I was unable to find an online version of this paper, and in fact I don't even know whether it's been translated from the Russian I presume it was written in. But you can find the result mentioned above quoted in this book (page 54), or mentioned (a little vaguely) in this paper (page 2).<|endoftext|> TITLE: The (fiber of the) cofiber of the fiber of a map of spaces QUESTION [11 upvotes]: Consider a fiber sequence of spaces $$F \overset{i}{\to} E \to B$$ The cofiber $C(i)$ of the inclusion of the fiber comes with a canonical map $C(i) \to B$. Its possible to show (using some point set topology) that the fiber of that map is homotopy equivalent to $\Omega B \ast F$ hence we obtain a new fiber sequence $$\Omega B \ast F \to C(i) \to B$$ To demonstrate the power of this statement consider the case of the path space fibration where $B= X , E = PX \cong \ast , F = \Omega X$. The new fiber sequence we obtain would be $$\Sigma (\Omega X \wedge \Omega X) \cong \Omega X \ast \Omega X \to \Sigma \Omega X \to X$$ This implies immediately that whenever $X$ is $n$-connective for $n\ge 1$ then the canonical map $\Sigma \Omega X \to X$ is ($2n-1$)-connective (which is sort of the Eckman Hilton dual of Freudenthal suspension theorem). Question: Is there an Eckman-Hilton dual to this statement? To be precise, let $X \to Y \overset{\pi}{\to} Z$ be a cofiber sequence and let $F := fib(\pi)$ be the fiber $\pi$. Again there's a canonical (upto homotopy) map $f: X \to F$ and we may take its cofiber to obtain a new cofiber sequence $$X \to F \to C(f)$$ Is there a "closed formula" for $C(f)$ (and for the maps involved) in terms of $X,Y$ and $Z$ (and the maps) similar to the one from above? REPLY [12 votes]: This sort of question was studied a lot by Ganea (see here and here). The first piece of bad news is that duality is not perfect: there is no formula for that cofiber that depends only on $\Sigma X$ and $Z$. A counterexample was supplied by Barratt (as Remark 3.5 in the first linked paper above). Even worse, the homotopy type of that cofiber does not have a formula just in terms of $X$ and $Z$! A counterexample is supplied in the second linked paper as 1.4. The best we can do is that the suspension of the cofiber is given by $\Omega Z \ast X$. This observation leads to a nice proof of the James splitting, incidentally- with no calculation.<|endoftext|> TITLE: Dirichlet series with a single zero QUESTION [16 upvotes]: I need to find a Dirichlet series f that has the following property. f is zero in only one point s such that Re(s) > $\sigma_c $. REPLY [23 votes]: That such a Dirichlet series exists was a conjecture of Balazard, which was recently resolved by Hilberdink and Saias. If the Riemann Hypothesis is true, then $1/\zeta(s)$ would provide such an example (with the abscissa of conditional convergence being $1/2$), and the goal was to find an unconditional example.<|endoftext|> TITLE: Which definition of "proper" is better? QUESTION [14 upvotes]: It is well known that topology and algebraic geometry assign different meanings to the word "proper". Let us recall the relevant definitions from topology (and we work in the context of topological spaces): A map $f:X\to Y$ is called separated iff the diagonal $X\hookrightarrow X\times_YX$ is a closed inclusion. A map $f:X\to Y$ is called proper iff for every net $(x_\alpha)_\alpha$ in $X$ such that $(f(x_\alpha))_\alpha$ converges to $y\in Y$, there exists a subnet converging to $x\in X$ with $f(x)=y$. Equivalently, for every compact subspace $K\subseteq Y$ its inverse image $f^{-1}(K)\subseteq X$ is compact (compact means every open cover admits a finite subcover) EDIT: The condition of compact sets having compact inverse images is strictly weaker than properness. Here is a counterexample: take the identity map $\{0,1\}\to\{0,1\}$ where the domain has the discrete topology and the target has the topology whose open sets are $\{\},\{0\},\{0,1\}$. The inverse image of a compact set under this map is compact (every subspace of $\{0,1\}$ with the discrete topology is compact), however this map is not proper in the sense of the definition above. The above are the definitions from topology. In algebraic geometry, a map is called "proper" iff it is proper and separated in the sense defined above. Is there a good reason to prefer one definition of the word "proper" over the other? Does the answer depend on whether one is doing topology or algebraic geometry? Note: I am not asking for which meaning was first historically, nor for whether someone should be blamed for changing the meaning of an existing mathematical term, nor for people's personal preferences between the two meanings. Rather I am looking for a mathematical reason why one definition might be preferred over the other. For example, both "separated" and "proper" are preserved under pullback, thus so is "separated and proper". But: Is there another important formal property of "separated and proper" which doesn't immediately follow from corresponding formal properties of "separated" and "proper" individually? Or, in the other direction: Is there some important property of "proper" maps which cannot be formally derived from results about "separated" and "separated and proper" maps? Is there any argument using the notion of a "proper" map which would become inconvenient if one only had access to the notions of "separated" and "separated and proper"? For clarity, let me suggest that answers should specify whether they are using the word "proper" as I have defined it above or in the algebro-geometric sense (called "separated and proper" above). REPLY [2 votes]: In topology, separated and proper maps satisfy a certain lifting property which allows to show that separated and proper maps are contained in a certain orthogonal of several closed maps of finite topological spaces. See https://ncatlab.org/nlab/show/Tychonoff+theorem#ProofViaTaimanovTheoremAndLiftingProperties , Remark 2.4 and Question 2.5 for details. In view of the remark in the previous answer one wants a stronger closure than just closure under pullbacks: one also wants closure under passing to diagonals. This stronger closure is in essence used to build up a six functor formalism, surely among other things. the class of separated and proper maps can probably be defined as the orthogonal wrt unique lifting property, and such orthogonals (wrt unique lifting property) are necessarily closed under passing to diagonals. Well, I have not checked that being separated is enough, but it shouldn't be hard. When the separated and proper maps is a map to a single point, this lifting property is known as Taimanov theorem and says that under some obviously necessary conditions a map to a compact Hausdorff space can be extended from a dense subspace. https://ncatlab.org/nlab/show/Taimanov+theorem<|endoftext|> TITLE: Sum of random variables are equal in distribution QUESTION [5 upvotes]: Suppose that $X,Y$ are scalar random variables supported on some standard Lebesgue probability space $(\Omega, \mathrm{P})$, such that $X \overset{\mathrm{d}}{=} Y$ in the sense that their pushforward measures are equal, $X_*(\mathrm{P}) = Y_*(\mathrm{P})$. Does there exist a nondegenerate random variable $Z$ on $(\Omega, \mathrm{P})$ satisfying $X + Z \overset{\mathrm{d}}{=} Y + Z$? In the case that $Y = X\circ T$ for some measure preserving automorphism $T: \Omega \rightarrow \Omega$ (this appears to be often the case by: Random variables with same distribution), and we have in addition the factorization $T = S^2$ for some automorphism $S: \Omega \rightarrow \Omega$, then clearly we can set $Z = X \circ S$, whence $X + Z = X + X\circ S \overset{\mathrm{d}}{=} X \circ S + X \circ S^2 = Z + Y$. Are there more general conditions than this, or perhaps conditions under which the answer to the question is negative? Edit: it also turns out that if $Y = X \circ T$ where $T$ is measure preserving and non-ergodic then $Z$ exists; just take $Z$ such that $Z \circ T = Z$ almost surely and $Z$ non-constant. Then a computation like the one above shows equality in distribution. REPLY [6 votes]: There is$^*$ a counterexample in the atomic case, see below, so we will assume that $(\Omega, \mathrm{P})$ is a non-atomic standard Lebesgue probability space (so it is Isomorphic to the unit interval with Lebesgue measure, see https://en.wikipedia.org/wiki/Standard_probability_space). In that setting we claim: If $X \overset{\mathrm{d}}{=} Y$ are two random variables on $(\Omega, \mathrm{P})$ , then there always exists a nondegenerate random variable $Z$ on $(\Omega, \mathrm{P})$ satisfying $X + Z \overset{\mathrm{d}}{=} Y + Z$. Proof: We consider three cases. Case 1 : $X+Y$ is not an almost sure constant. Then defining $Z:=-X-Y$ proves the claim. Case 2: $X+Y=c$ almost surely but $X$ takes more than 2 values a.s. (That is, for every pair of reals $a,b$ we have $ \mathrm{P}[X \in\{a,b\}]<1.$ ) Then $Z:=X(c-X)$ is nondegenerate. Denoting $f(x)= x+x(c-x)$, we have $f(X)=X+Z$ and $f(Y)=Y+Z$ almost surely, so $X + Z \overset{\mathrm{d}}{=} Y + Z$. Case 3: $X+Y=c$ almost surely, and there exist $a,b$ such that $ \mathrm{P}[X \in\{a,b\}]=1.$ In this case let $D_a$ be a subset of $\{\omega: X(\omega)=a\}$ with $\mathrm{P}(D_a)=\mathrm{P}(\{\omega: X(\omega)=a\})/2$. (To see that $D_a$ exists we may assume that $(\Omega, \mathrm{P})$ is the unit interval with Lebesgue measure, and observe that the mapping $t \mapsto \mathrm{P}([0,t] \cap \{\omega: X(\omega)=a\})$ is continuous on $[0,1]$ so its image includes $\mathrm{P}(\{\omega: X(\omega)=a\})/2$.) Similarly, let $D_b$ be a subset of $\{\omega: X(\omega)=b\}$ with $\mathrm{P}(D_b)=\mathrm{P}(\{\omega: X(\omega)=b\})/2$. Let $Z(\omega)=1$ if $\omega \in D_a \cup D_b$ and $Z(\omega)=0$ otherwise. Then $Z$ is independent of $X,Y$ so $X + Z \overset{\mathrm{d}}{=} Y + Z$. QED (*) Counterexample in the atomic case: If $\Omega=\{0,1\}$ with the uniform measure and maximal $\sigma$-algebra, and $X(\omega)=\omega=1-Y(\omega)$, then for every nonconstant random variable $Z$ on $\Omega$, the laws of $X+Z$ and $Y+Z$ are different; e.g., if $Z(1)>Z(0)$ then the maximum of $X+Z$ exceeds the maximum of $Y+Z$.<|endoftext|> TITLE: A question regarding Bourgain's paper on $\Lambda(p)$-subsets QUESTION [11 upvotes]: I'm trying to understand Bourgain's proof of Proposition 1.10 on page 304-307 in On $\Lambda(p)$-subsets of squares which states Given $p>4$, we have the estimate \begin{align} \left\|\sum_{n=1}^N a_ne^{in^2t} \right\|_{L^p(\mathbb{T})} \lesssim N^{\frac{1}{2}-\frac{2}{p}}\left( \sum^N_{n=1}|a_n|^2\right)^{\frac{1}{2}}. \end{align} More precisely, I am trying to understand the following on page 305 Let $t_1, \ldots, t_R$ be $1/N^2$-separated points in $[0, 1]$ satisfying \begin{align} \left|\sum^N_1 a_n e^{2\pi i n^2t_r} \right|>\delta N^{\frac{1}{2}} \ \ (1\leq r \leq R). \ \ \ \ \ \ (4.11) \end{align} Here we already assumed $\sum |a_n|^2 \leq 1$ and $0<\delta<1$. Then the paper continues Our purpose is to estimate $R$. By linearization, (4.11) yields \begin{align} \sum_{1\leq r, r'\leq R}\left|\sum^N_1 \exp[2\pi i n^2(t_r-t_{r'})] \right|>\delta^2NR^2. \ \ \ \ \ (4.12) \end{align} Here I have a potentially very elementary question. What does he mean by linearization in this context, that is, how did he deduced (4.12) from (4.11)? REPLY [3 votes]: The method given by Mayank Pandey is pretty much it, but here's an alternative presentation: There exists some choice of signs $c_r\in \mathbb{C}$ with $\lvert c_r\rvert =1$ such that $$ \delta N^{1/2}R < \sum_{1\leq r\leq R}\lvert \sum_{1\leq n\leq N} a_n e(n^2t_r) \rvert = \sum_{1\leq n\leq N}a_n \sum_{1\leq r\leq R}c_r e(n^2t_r). $$ By the Cauchy-Schwarz inequality $$ \delta^2NR^2 < \left(\sum \lvert a_n\rvert^2\right)\left(\sum_{1\leq r,r'\leq R}c_rc_{r'}\sum_{1\leq n\leq N}e(n^2(t_r-t_{r'}))\right). $$ The result follows by the triangle inequality and fact that $\sum \lvert a_n\rvert^2\leq 1$. Note that it was not needed in this argument that the points $t_r$ are $1/N^2$-separated. The 'trick' of changing the order of summation 'through an absolute value sign' (by introducing the sign factors $c_r$ which are then discarded in the final conclusion), following by Cauchy-Schwarz and/or Holder's inequality, is extremely powerful and appears often throughout Bourgain's work (and elsewhere - it also appears under a slightly different guise in the large sieve-type techniques). It is usually what Bourgain refers to when he invokes linearisation. I'm not sure if the use of this term for this technique appears outside of Bourgain's work -- presumably it refers to being able to create information about linear relations between the $t_r$ from just a pointwise bound.<|endoftext|> TITLE: Peter–Weyl theory for vector fields QUESTION [7 upvotes]: Let $G$ be a compact Lie group. The classical Peter-Weyl theorem shows that $L^2(G)$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$. This is a powerful statement as it allows to answer questions about functions on $G$ in terms of matrix coefficients of irreducible representations. I was wondering if there exist a decomposition of the space $\mathfrak{X}(G)$ of vector fields on $G$ that has a similar spirit than the Peter-Weyl theorem. In particular, I was hoping that the gradient map $C^\infty(G) \to \mathfrak{X}(G)$ (with respect to the Riemannian metric induced by the Killing form of $G$) has a nice behavior with respect to the decompositions of the spaces on both sides. I'm a bit vague here because I don't know what I can hope for. In the best case, the gradient map is diagonalized (similar to how the Laplacian is diagonalized by the classical Peter-Weyl theorem). REPLY [6 votes]: Let $\mathfrak g \to \mathfrak{X}(G)$ be the inclusion of right-invariant vector fields on $G$ into vector fields on $G$. Then we have an isomorphism $C^\infty(G) \otimes_{\mathbb R} \mathfrak g \to \mathfrak{X}(G)$ of left $C^\infty(G)$-modules defined by $f \otimes_{\mathbb R} x \mapsto fx$. It is $G \times G$-equivariant. Applying Peter-Weyl to $C^\infty(G)$ gives the desired decomposition. The gradient $C^\infty(G) \to \mathfrak{X}(G)$ is $G\times G$-equivariant if the metric is biinvariant.<|endoftext|> TITLE: For which tempered distributions is the fractional derivative well-defined? QUESTION [5 upvotes]: Let $\gamma \geq 0$ and consider the fractional derivative operator defined in Fourier domain by $$\mathcal{F} \{\mathrm{D}^{\gamma} \varphi \} (\omega) = (\mathrm{i} \omega)^{\gamma} \mathcal{F}\{\varphi\} (\omega),$$ where $\varphi \in \mathcal{S}(\mathbb{R})$ is a smooth and rapidly decaying function. Of course, the definition can be extended to much more functions than $\varphi \in \mathcal{S}(\mathbb{R})$, including some, but not all, tempered distributions. It is for instance possible to extend $\mathrm{D}^{\gamma}$ to any compactly supported distribution (as for any convolution operator from $\mathcal{S}(\mathbb{R})$ to $\mathcal{S}'(\mathbb{R})$). My question is the following: Is there a good notion of the "domain of definition" of the operator $\mathrm{D}^{\gamma}$, understood as the largest topological vector space $\mathcal{S}(\mathbb{R}) \subseteq \mathcal{X} \subseteq \mathcal{S}'(\mathbb{R})$ such that $\mathrm{D}^{\gamma} : \mathcal{X} \rightarrow \mathcal{S}'(\mathbb{R})$ is well-defined and continuous? Or at least, if the question is somehow meaningless, any natural construction that will include many tempered distributions in a satisfactory* manner? *To give a bit of context, I am especially interested by the fractional case where $\gamma \notin \mathbb{N}$. The question is obvious for $\gamma = n \in \mathbb{N}$, since one can select $\mathcal{X} = \mathcal{S}'(\mathbb{R})$. However. when $\gamma$ is purely fractional, there is no hope to define the product $(\mathrm{i} \omega)^{\gamma} \mathcal{F}\{u\} (\omega)$ when $u \in \mathcal{S}'(\mathbb{R})$ is too irregular around the origin, which means morally that $u$ growth too fast at infinity. "In a satisfactory manner" would be a way of specifying properly a good "growth property" of $u \in \mathcal{X}$. REPLY [3 votes]: A preliminary remark. The operator $(d/dx)^\gamma$ is never continuous on the Schwartz space (and thus on tempered distributions) except if $\gamma$ is a non-negative integer, since you introduce a singularity at 0 by multiplication by $(i\omega)^\gamma=\exp(\gamma \log(i\omega))$ (you have to choose a determination of the logarithm on the imaginary axis and you get a singularity). Now you can also define what is called an homogeneous distribution with degree $\lambda$, where $\lambda$ is a given complex number: a distribution $T$ on $\mathbb R$ is said to be homogeneous with degree $\lambda$ whenever $$ x\frac{d T}{dx}=\lambda T. $$ It is an exercise to prove that an homogeneous distribution is actually tempered. Examples are $$ \chi_{+,\lambda}=(x_+)^{\lambda}/\Gamma(\lambda +1),\quad \chi_{-,\lambda}=(x_-)^{\lambda}/\Gamma(\lambda +1), $$ and it is possible to prove that homogenous distributions of degree $\lambda\notin \mathbb Z_-$ are $$ c_{+}\chi_{+,\lambda}+c_{-}\chi_{-,\lambda} \quad\text{where $c_\pm$ are constants.} \tag{$\ast$} $$ Note that for $\lambda=-1$, homogeneous distributions of degree $-1$ on the real line are linear combinations of $ \text{pv}\frac{1}{x},\ \delta_0. $ With $\mathscr S'_\lambda$ standing for homogeneous distributions with degree $\lambda$, we get that for $\lambda, \lambda-\gamma\notin \mathbb Z_-$, $$ D^\gamma:\mathscr S'_\lambda\longrightarrow \mathscr S'_{\lambda-\gamma}. $$ To prove this you check that $ (d/dx)^\gamma\chi_{+,\lambda}=\chi_{+,\lambda-\gamma}. $ N.B. Multi-dimensional versions are available, more information in Lars Hörmander's ALPDO 256, Section 3.2. REPLY [2 votes]: Perhaps another way to approach a characterization of such tempered distributions is indeed to look at their Fourier transforms. First, the easy case is that distributions with support not containing $0$ admit fractional differentiation. Second, distributions $u$ with $\widehat{u}=\varphi\cdot v$ for a tempered distribution $v$ and $\varphi$ a smooth function vanishing to infinite order at $0$ (and perhaps identically $1$ outside an $\varepsilon$-ball around $0$). Then Fourier inversion gives something...<|endoftext|> TITLE: Mapping a loop space to quaternionic projective space QUESTION [16 upvotes]: Let $\mathbf{H}P^\infty$ denote the infinite-dimensional quaternionic projective space. The inclusion of its bottom cell defines a map $S^4 \to \mathbf{H}P^\infty$. Does this extend to a map $\Omega S^5 \to \mathbf{H}P^\infty = BSU(2)$? Since $\Omega S^5$ is the James construction on $S^4$, this question would be very easy to answer (in the positive) if $\mathbf{H}P^\infty$ was a homotopy associative H-space --- but it's known that this is not true. (If $Y$ is a homotopy associative H-space, then any map $X\to Y$ from a path-connected space $X$ admits a unique extension to a H-space map $\Omega \Sigma X \to Y$.) However, the composite $S^4\to BSU(2) \to BSU$ does extend to a map $f_\xi:\Omega S^5\to BSU$ classifying a bundle $\xi$ over $\Omega S^5$; it is easy to see that the Chern classes $c_i(\xi)$ vanish for $i\geq 3$, so the map $f_\xi$ factors, at least on cohomology, through $BSU(2)$. One natural expectation for the desired map is that it gives a map of fiber sequences from the EHP fiber sequence $S^2 \to \Omega S^3 \to \Omega S^5$ to the Hopf invariant fiber sequence $$S^2 \to \mathbf{C}P^\infty = BS^1 \to \mathbf{H}P^\infty = BS^3$$ via the map $\Omega S^3 \to \mathbf{C}P^\infty$ extending the inclusion of the bottom cell of the target. In fact, thinking along these lines shows that we'd get the desired map if $S^2$ was a loop space, which it isn't. An approach to constructing the desired map comes from equivariant considerations. Namely, the bottom $C_2$-equivariant cell of $\mathbf{C}P^\infty$ under the complex conjugation action is the one-point compactification $S^\rho$ of the regular representation $\rho$ of $C_2$. This gives a map $\Omega S^{\rho+1} \to \mathbf{C}P^\infty$, and hence a map $(\Omega S^{\rho+1})_{hC_2} \to (\mathbf{C}P^\infty)_{hC_2} = \mathbf{H}P^\infty$. To get the desired map, it therefore suffices to construct a nonequivariant map $\Omega S^5 \to (\Omega S^{\rho+1})_{hC_2}$, but it's not clear to me how/whether such a map exists. I'd like to remark that looping the map $\Omega S^5\to \mathbf{H}P^\infty$ defines a map $\Omega^2 S^5\to S^3$. Such a map is known to exist if we require that it be degree $2$ on the bottom cell of $\Omega^2 S^5$ (it is the map appearing in work of Cohen-Moore-Neisendorfer). REPLY [2 votes]: [I should've posted this answer a long time ago, given that it was answered in the comments.] In order to extend the map $S^4\to \mathbf{H}P^\infty$ to a map from $J_2(S^4)$, the composite of $[\iota_4,\iota_4]:S^7\to S^4$ with the inclusion of $S^4$ into $\mathbf{H}P^\infty$ must be null. This is not true: the Whitehead product $[\iota_4, \iota_4]\in \pi_7(S^4)$ would have to be $2\nu$ to get the desired extension, but Equation 5.8 of Toda's composition methods book says that $[\iota_4, \iota_4] = \pm (2\nu - \Sigma \nu')$, where $\nu'\in \pi_6(S^3)$ is the Blakers-Massey element. As Gustavo Granja points out in the comments, there is a $p$-local analogue of this (with $p>2$): the map $S^4\to \mathbf{H}P^\infty$ extends to a map $J_{(p-1)/2}(S^4)\to \mathbf{H}P^\infty$, and the composite with the $(p+1)/2$-fold Whitehead product $[\iota_4, \cdots, \iota_4]: S^{2p+1}\to J_{(p-1)/2}(S^4)$ produces $\alpha_1\in \pi_{2p+1}(\mathbf{H}P^\infty)$. (See also this answer: Is $\mathbb{H}P^\infty_{(p)}$ an H-space?.) This implies that there is no extension to a map $J_{(p+1)/2}(S^4)\to \mathbf{H}P^\infty$.<|endoftext|> TITLE: Multiplication in Deligne cohomology: explicit formula for $p=q=1$ QUESTION [6 upvotes]: [This is a double of my question of math.stackexchange https://math.stackexchange.com/questions/3214962/multiplication-in-deligne-cohomology-explicit-formula-for-p-q-1] In the very beginning of [1] the geometric meaning of Deligne cohomology $H^q(X, \mathbb{Z}(p))_D$ and multiplicative structure on it is being discussed. In particular, it is not hard to see that $H^q(X, \mathbb{Z}(1))$ can be canonically identified with $H^{q-1}(X, \mathcal{O}^{\times}_X)$. The group $H^2(X, \mathbb{Z}(2))_D$ is identified with the group of holomorphic rank $1$ bundles with holomorphic connection (group structure is given by tensor product) The $\cup$-multiplication gives us a map $$ H^1(X, \mathbb{Z}(1))_D \otimes H^1(X, \mathbb{Z}(1))_D = H^0(X, \mathcal{O}^{\times}_X) \otimes H^0(X, \mathcal{O}^{\times}_X) \to H^2(X, \mathbb{Z}(2))_D $$ In other words, given two nowhere vanishing holomorphic functions $f$ and $g$ on $X$ we obtain a holomorphic line bundle with holomorphic connection on $X$. Though in [1] the explicit formula for this in terms of Čhech cocycles is given, I am looking for another description of the same operation. First of all, observe that each pair of functions $f, g \in H^0(X, \mathcal{O}^{\times}_X)$ define a holomorphic map $F_{f,g} \colon X \to (\mathbb{C}^{\times})^2$. Following Esnault and Viehweg, denote the resulting line bundle with holomorphic connection $f \cup g$ by $r(f, g)$. Then it seems clear from functoriality reasons that $$r(f, g) = F_{f,g}^*r(z,w),$$ where $z$ and $w$ are coordinate functions on $\mathbb{C}^{\times}\times \mathbb{C}^{\times}$. Thus, I'd be happy to understand, what $r(z, w)$ is. Since $(\mathbb{C}^{\times})^2$ is a product of two Stein manifolds, there are no non-trivial holomorphic line bundles. Therefore, the only ''interesting'' part of $r(z,w)$ is the holomorphic connection. Any holomorphic connection on trivial bundle is given by $\nabla = d + \eta$, where $\eta$ is a holomorphic $1$-form. So my questions are: What is this $1$-form $\eta$ on $\mathbb{C}^{\times} \times \mathbb{C}^{\times}$? It seems to me, that $\frac{dz}{z} - \frac{dw}{w}$ would be nice (at least, if this is the case, it satisfies the properties of $r(f, g)$ given in [1]), however I'm not able do deduce this explicitly form Esnault-Viehweg formulae. From my speculations it follows that the underlying line bundle for any $r(f,g)$ is trivial. Is this at least true? If not, than where is my mistake? Thank you for any comments! [1] -- H. Esnault, E. Viehweg. Deligne-Beilinson cohomology. in: Beilinson's Conjectures on Special Values of L-Functions ( Ed.: Rapoport, Schappacher, Schneider ). Perspectives in Math. 4, Academic Press (1988) 43 - 91 (http://page.mi.fu-berlin.de/esnault/preprints/ec/deligne_beilinson.pdf) REPLY [7 votes]: There is indeed a "universal" holomorphic bundle with connection on $\mathbb{C}^\times \times \mathbb{C}^\times$ which induces the bundles $r(f,g)$ defined in Esnault-Viehweg. This universal bundle has been constructed by D. Ramakrishnan using the Heisenberg group (Bulletin AMS vol. 5 n. 2, 1981, https://doi.org/10.1090/S0273-0979-1981-14942-9 ). It is nicely explained in R. Hain, Classical polylogarithms. Put \begin{equation*} H_{\mathbb{C}} = \begin{pmatrix} 1 & \mathbb{C} & \mathbb{C} \\ 0 & 1 & \mathbb{C} \\ 0 & 0 & 1 \end{pmatrix} \end{equation*} and \begin{equation*} H_{\mathbb{Z}} = \begin{pmatrix} 1 & \mathbb{Z}(1) & \mathbb{Z}(2) \\ 0 & 1 & \mathbb{Z}(1) \\ 0 & 0 & 1 \end{pmatrix}. \end{equation*} The exponential map gives a canonical projection $H_{\mathbb{Z}}\backslash H_{\mathbb{C}} \to \mathbb{C}^\times \times \mathbb{C}^\times$ with fiber $\mathbb{C}/\mathbb{Z}(2) \cong \mathbb{C}^\times$, which is the bundle you want. This bundle is not trivial but becomes trivial after pulling-back to $\mathbb{C} \times \mathbb{C}$ using this exponential map. Denoting by $u,v$ the coordinates on $\mathbb{C} \times \mathbb{C}$ (they are just the matrix coefficients using the Heisenberg description), the pull-back of the connection is given by \begin{equation*} \nabla s = ds - s \cdot u dv/2\pi i \end{equation*} This defines a connection on $\mathbb{Z}(2)\backslash H_{\mathbb{C}}$ which descends to $H_{\mathbb{Z}} \backslash H_{\mathbb{C}}$.<|endoftext|> TITLE: Involutions in $\mathbb{F}_p[[x]]$ QUESTION [9 upvotes]: Question: For a prime $p$, is every involution in $\mathbb{F}_p[[x]]$ with a zero constant term a reduction modulo $p$ of some involution in $\mathbb{Z}[[x]]$? Here involution in $A[[x]]$ means $f\in A[[x]]$ such that $f\circ f=x$. Obviously, a reduction mod $p$ of an involution in $\mathbb{Z}[[x]]$ yields an involution in $\mathbb{F}_p[[x]]$. Moreover, at least one involution in at least one $\mathbb{F}_p[[x]]$ is not obtainable by reducing mod $p$, namely, $1+x\in\mathbb{F}_2[[x]]$, exactly because its constant term is nonzero. (Update: The second sentence in this paragraph confuses the rings of polynomials and power series, see comments and answers for corrections.) But do we get all involutions in $\mathbb{F}_p[[x]]$ with the zero constant term this way? If there is a reference where this is discussed, that would be very helpful, too. REPLY [11 votes]: This is algebraic dynamics, a big field that I’m outsider to, though I have made some peripheral contributions to it. First, I’m assuming that you’re asking about ring automorphisms of $\Bbb Z[[x]]$ and of $\Bbb F_p[[x]]$, and the involutions from among these. Under this assumption, $x\mapsto1+x$ is not an involution of $\Bbb F_2[[x]]$, since it wants to take the nonunit $x$ to the unit $1+x$. Second, and more generally, in view of the fact that all automorphisms of $\Bbb F_q[[x]]$ are necessarily continuous, we don’t need to specify that our involution $x\mapsto u(x)$ have no constant term in $u$. Third, though Will has described an infinite set of nonconjugate involutions (in the group of all invertible series over $\Bbb F_2$), as I recall, according to a theorem of Klopsch, these are all that there are out there, in this $\Bbb F_2$-case. Finally, all the involutions of $\Bbb Z[[x]]$ that I know about are (naturally) the $[-1]$-automorphisms of $\Bbb Z$-formal group laws, and these look like $-x+cx^q\pmod{x^{q+1}}$ for a prime power $q$. Surely there are other involutions in this situation, but I don’t know them. In particular, if you try $n=5$ here, you get a series $x+x^6+\cdots\in\Bbb F_2[[x]]$ that doesn’t seem to come from $[-1]_F(x)\in\Bbb Z[[x]]$ for any $\Bbb Z$-formal group law $F$. But globally-defined formal group laws I’m rather ignorant about. Nobody’s saying, either, that a $\Bbb Z$-involution has to belong to a formal group over $\Bbb Z$. I think your question has to remain unanswered, at least with the expertise exhibited here so far. EDIT — Addendum: All thanks to Will Sawin for his persistence in the face of my repeated careless computational errors. The principle involved is that for an integral domain $R$, the map on the semigroup of formal series under substitution $u(x)\in R[[x]]$ with no constant term and $u'(0)\ne0$, namely $$ u(x)=xw(x)\mapsto \bigl(u(x^m))^{1/m}=x\bigl(w(x^m)\bigl)^{1/m}\,, $$ is a homomorphism, as long as it’s defined. For definedness, you need $u'(0)=w(0)$ to have an $m$-th root in the base, and more broadly, you need some sort of convergence theorem to tell you that you can take the $m$-th root of the series $w(x)$. In the comments, I pointed out that $(1+p^2x)^{1/p}\in\Bbb Z[[x]]$, and this means that $(1+m^2x)^{1/m}\in\Bbb Z[[x]]$ as well, for any positive integer $m$. Will says, start with the fundamental involution $u(x)=-x/(1+x)=\sum_1^\infty(-x)^n$, first apply $u(x)\mapsto u(m^2x)/m^2=U(x)$, and apply the “stretching-out” operation above, which will be defined as long as $U'(0)=-1$ has an $m$-th root in $\Bbb Z$, namely as long as $m$ is odd. Since $U(X^m)=-x^m+m^2x^{2m}-m^4x^{3m}+\cdots=-x^m[1-m^2x^m+m^4x^{2m}-\cdots]$, the conditions for the existence of the $m$-th root are satisfied. Thus $\bigl(U(x^m)\bigr)^{1/m}$ is a good lifting to $\Bbb Z[[x]]$ of the characteristic-two involution of the form $-x(1+g(x^m))$ for $m$ odd.<|endoftext|> TITLE: Does a conservativity conjecture imply the standard conjectures? QUESTION [9 upvotes]: Does a conservativity conjecture (e.g. Conjecture 2.1 of http://user.math.uzh.ch/ayoub/PDF-Files/Article-for-Steven.pdf) imply the standard conjectures? Specifically I am confused with Beilinson's article arXiV:1006.1116 "Remarks on Grothendieck’s standard conjectures", where it seems to show that the conservativity of a realization functor implies the standard conjectures on algebraic cycles. REPLY [9 votes]: Here is a partial answer: In characteristic 0 it is known that conservativity + algebraicity (edit: modulo rational equivalence) of the Künneth projectors implies the remaining standard conjectures. Indeed, if the Künneth projectors are algebraic, then one may use conservativity to show that the inverse of the Lefschetz operator is algebraic (i.e. Lefschetz standard conjecture). Once that is known, the hom=num standard conjecture also follows. For example, André's category of “motivated” motives would be equivalent to the categories of homological/numerical motives.<|endoftext|> TITLE: Hopf algebra in derived category vector spaces QUESTION [6 upvotes]: Let $H$ be a complex of vector spaces over some field $k$ which is endowed with the structure of a Hopf algebra object. I have heard several times that if $H$ is concentrated in positive or negative degrees, then the cohomology in degree zero of $H$ is a Hopf algebra in the usual sense. How does one prove such a statement and why is the condition on the degrees important? REPLY [2 votes]: If $C$ is a graded coalgebra (e.g. $C$= the homology of a d.g. Hopf algebra), then $C_0$ is not necesarily a subcoalgebra, because $$\Delta(C_0)\subset (C\otimes C)_0=\oplus_{n\in\mathbb Z}C_n\otimes C_{-n}$$ For example, $H=k\{x,y\}/(x^2=y^2=xy+yx)$ is a graded Hopf algebra with $|x|=1$ and $|y|=-1$, both $x$ and $y$ primitives (in particular a d.g. Hopf algebra with $d=0$ and agree with its homology). $H_0=k\oplus kx\wedge y$, and $$ \Delta(x\wedge y)= (x\otimes 1+1\otimes x)(y\otimes 1+1\otimes y)$$ $$= x\wedge y\otimes 1+1\otimes x\wedge y+ x\otimes y-y\otimes x \notin H_0\otimes H_0$$ Of course if $C=\oplus_{n\geq 0} C_n$, then $C_n=0$ for $n<0$ and $C_{-n}=0$ for $n>0$, so, the only nonzero summand in $\oplus_{n\in\mathbb Z}C_n\otimes C_{-n}$ is $C_0\otimes C_0$.<|endoftext|> TITLE: Optimization problem on trigonometric polynomials QUESTION [5 upvotes]: I would like to maximize $$ \int_0^{2\pi} \frac{(f'(x))^2}{f(x)}dx $$ subject to $f(x)\leq 1$ for all $x$ over the space of nonnegative trigonometric polynomials of degree smaller or equal to $n$. The application would be to design a (theoretically) optimal 1d-image for position tracking via template matching. Here is short explanation why the maximum of this interval gives the (theoretical) optimal image. The values of $f$ correspond to the intensity of the image, this is why it must be positive. The image is assumed to be periodic for simplification and because of practical reasons. The optics acts as a low-pass filter cutting of completely all frequencies above some threshold. Hence the image will be a trigonometric polynomial. With one measurement at $x$ we can determine the position with an std of $\sigma /f'(x)$. If the noise is dominated by shot noise we will have $\sigma=\sqrt{f(x)}$. The variance therefore is $f(x)/(f'(x))^2$. Combining a series of independent measurement at positions $x_1, \dots, x_n$ with corresponding weights $w_1, \dots, w_n$ with $\sum_{i=1}^n w_i=1$ the variance of the estimated position becomes $$ \sum_i w^2_i \frac{f(x_i)}{(f'(x_i))^2}\geq \frac{\left(\sum_i w_i \right)^2}{\left(\sum_i \frac{(f'(x_i))^2}{f(x_i)},\right)}=\left(\sum_i \frac{(f'(x_i))^2}{f(x_i)}\right)^{-1} $$ where we used Cauchy-Schwarz. Hence in order to minimize the variance/std we need to maximize $\sum_i \frac{(f'(x_i))^2}{f(x_i)}$. In the limit this becomes the integral. As an extension it would also be interesting to solve a similar problem in two-dimensions where we need to track position (2dof's) and rotation (1 dof). I will think about a problem formulation in this case. REPLY [5 votes]: $f(x)=\frac 12(1+\sin nx)$ is, indeed, the optimal choice. To see it, let's normalize a bit differently by $0\le f\le 2$. Then $f=1+g$ where $g$ is a real trigonometric polynomial of degree $n$ bounded by $1$. We need the following classical Lemma. $g^2+n^{-2}(g')^2\le 1$ (I learned it from Alexandre Eremenko when we were discussing another MO question). Assume the lemma. Then $g$ has at most $2n$ intervals of monotonicity. Let $I$ be one such interval on which $g$ changes between $a$ and $b$, say. Then $$ \int_I \frac{g'(x)^2}{1+g(x)}\,dx=\int_a^b \frac{|g'\circ g^{-1}(y)|}{1+y}\,dy\le n\int_a^b\frac{\sqrt{1-y^2}}{1+y}\le n\int_{-1}^1\frac{\sqrt{1-y^2}}{1+y} $$ and for $g=\sin nx$ we have equalities everywhere. There are many proofs of the lemma, which is actually just Bernstein's inequality in disguise. I'll present one that derives it directly from the classical Bernstein inequality. Let $N>n$ be any real number. Consider the function $G=g^2+N^{-2}(g')^2$. At the point of maximum, we must have $(g+N^{-2}g'')g'=0$. If $g'=0$, we are done. Otherwise $g=-N^{-2}g''$ and, thereby, $G_1=N^{-2}((g')^2+N^{-2}(g'')^2)$ has a not smaller maximum. Continuing this way and recalling that, by the classical Bernstein inequality, $\max|g^{(m)}|\le n^m\max|g|=n^m$, we conclude that either we stop at some point and get $\max G_m\le (n^2N^{-2})^m\le 1$ because $g^{(m+1)}=0$ at the point of maximum of $G_m$, or go so far that $\max G_m\le (n^2N^{-2})^m+(n^2N^{-2})^{m+1}<1$ by the Bernstein bound and the observation that a decreasing geometric progression tends to $0$.<|endoftext|> TITLE: Motives of complex-analytic spaces QUESTION [8 upvotes]: In any setting where we have a notion of space and a notion of cohomology theory, we in principle could ask "what are motives in this setting?". In some settings the question can be interesting (i.e. algebraic varieties and Weil cohomology theories), in some settings not so much (i.e. topological spaces and cohomology theories satisfying all of the Eilbenberg--Steenrod axioms). For complex-analytic spaces, I am not completely sure how should we define a good cohomology theory. First question: what are some ways to define this notion? Presumably rational Betti cohomology and de Rham cohomology should be examples of good cohomology theories. Second: can we study motives in this setting? I would guess that the answer turns out to be boring, but it would be nice if there is a 3-page paper where this is proved. REPLY [8 votes]: See theorem 1.8 of (Joseph Ayoub. Note sur les opérations de Grothendieck et la réalisation de Betti. J. Inst. Math. Jussieu.) You can set up a theory of analytic motives, and the resulting category is equivalent to the derived category of vector spaces (over your field of coefficients, typically $\mathbb{Q}$). In https://arxiv.org/pdf/1810.04968v1.pdf the analogue for p-adic analytic geometry is proven by Bambozzi and Vezzani. (Maybe also relevant: https://arxiv.org/pdf/1708.04284.pdf)<|endoftext|> TITLE: Motives and homotopy theories of algebraic varieties QUESTION [6 upvotes]: The theory of motives is an attempt to cope with the fact that there are many reasonable cohomology theories of algebraic varieties. Now, sometimes your cohomology theory does not just give you a bunch of groups/vector spaces; it gives you a full-fledged (pro-)homotopy type (though based on limited responses to this question, I think there is no formal way to functorially produce homotopy types given a cohomology theory). The question is: what should be the extension of motives to homotopy types? What are some notable works in this direction? I have heard there is something called motivic homotopy theory, is it really relevant here or it just happens to have a similar name? More specific questions: First off, we need to show that usual motives are not good enough. This is probably obvious to anyone working in the field, but not everybody on this site is an expert on motives so an explicit example would be great. I think there is a rigorously constructed triangulated category of motives which is supposed but not known to be the derived category of some abelian category of motives. There is also a somewhat more pedestrian Grothendieck ring of varieties. I think the class in the Grothendieck ring does not determine the motive nor does the motive determine the class in the Grothendieck ring. Is there an example of two varieties with the same class in the Grothendieck ring which have different etale fundamental groups? An example of two varieties with the same motive in the triangulated category which have different etale fundamental groups? Before you can say "every Weil cohomology theory factors through motives", you have to define a Weil cohomology theory. What should the definition be for homotopy types? Has somebody written down a manageable list of axioms definining exactly what we are interested in? Before we can seriously talk about motivic homotopy types, we should understand on the categorical level what we expect the relevant category to be. In the case of motives, the relevant piece of category theory is Tannakian formalism, I believe (then we can say smart words like "a Weil cohomology theory is just a fiber functor blah blah"). What should the category theory look like for motivic homotopy types? A kind of non-abelian Tannakian formalism? P.S.: yeah, I know the question is super naive, you are free to call me an idiot in the comments. EDIT: I did not think about the coefficients when asking this. I believe Voevodsky's category makes sense with $\mathbb{Z}$-coefficients, so take that in the first question. REPLY [2 votes]: The projective plane and fake projective planes should have isomorphic motives in the triangulated category, at least with rational coefficients, but different etale fundamental groups. This is because they have isomorphic cohomology.<|endoftext|> TITLE: $l$-adic periods? QUESTION [6 upvotes]: For an algebraic variety $X$ over $\mathbb{Q}$ the comparison isomorphism between Betti and de Rham cohomologies provides the theory of periods with a motivic context whose reformulation as motivic periods with the associated Tannakian symmetry group has shown great success and promise recently - at least in genus 0 (mixed Tate motives) and genus 1 (mixed elliptic or modular motives) - with applications ranging from structure of the absolute Galois group (Deligne-Ihara conjecture) to transcendental number theory (multi-zeta values) to perturbative quantum field theory (Feynman amplitudes). Betti and de Rham are only two of the realizations of motives. I wonder if there is a similar exploration of another comparison isomorphism, between de Rham and $l$-adic cohomologies. Is there a notion of "$l$-adic periods", concretely thought of as matrix coefficients of the isomorphism from de Rham to $l$-adic cohomology after some choice of bases? If so, what arithmetic significance and applications are they known to have? Of course, there are other comparison isomorphisms as well, for which the question may well be trivial or uninteresting, just as it may be for de Rham/$l$-adic. REPLY [6 votes]: Yes. There are multiple isomorphisms between de Rham and $\ell$-adic cohomology. You can get one by combining the de Rham - Betti and Betti - $\ell$-adic isomorphisms. But this is silly as the periods you will get, in $\mathbb C\otimes_{\mathbb Q} \mathbb Q_\ell$, will just be the usual Betti periods. A better approach is to use $\ell$-adic Hodge theory, which gives an isomorphism between de Rham cohomology and Betti cohomology when both are tensored to the field $B_{dR}$. However, in this case, there is no need to use the motivic Tannakian group to study the periods. This is because of the identity $$H^*_{dR} (X, \mathbb Q) \otimes_{\mathbb Q} \mathbb Q_\ell = (H^*(X, \mathbb Q_\ell) \otimes B_{dR})^{ \operatorname{Gal} (\overline{\mathbb Q_\ell} | \mathbb Q_\ell )} $$ which implies that the periods in $B_{dR}$ are determined by the $\ell$-adic Galois representation restricted to $\operatorname{Gal} (\overline{\mathbb Q_\ell} | \mathbb Q_\ell )$ and thus are related to the Tannakian group of the category of $\operatorname{Gal} (\overline{\mathbb Q_\ell} | \mathbb Q_\ell )$-representations. This Tannakian group can be much smaller than the motivic Galois group.<|endoftext|> TITLE: Unoriented bordism with twisted orientation QUESTION [12 upvotes]: The computation of the unoriented bordism group of the point $N_*=\Omega_*^O$ is a classic result. I would like to know a related bordism group, where we specify the twisted fundamental class $[M]\in H_d(M,\mathbb{Z}^w)$ as part of the data. More precisely, I would like to consider the pairs $$ (M, [M]\in H_d(M,\mathbb{Z}^w)) $$ where $\mathbb{Z}^w$ is the coefficient system twisted by $w_1(M)$; I then call two $(M,[M])$ and $(M',[M'])$ bordant when there is $(N,[N]')$ such that $\partial N=M \sqcup M'$ where the twisted orientation of $N$ induces that of $M$ and $M'$, as in the case of oriented bordism. Most probably this is well known to the experts... REPLY [11 votes]: I think this theory is the same as unoriented bordism, when one tries to make sense of it. As it stands I don't think it makes sense, because I think that the expression "$\mathbb{Z}^w$" does not describe a local system on $M$, but rather an isomorphism class of local systems, and hence $H_d(M ; \mathbb{Z}^w)$ is an isomorphism class of abelian group, and hence it is not meaningful to talk about an element of it. A manifold $M$ has an "orientation" local system(=locally constant sheaf). This is the locally constant sheaf with values $\mathcal{O}_M(U) = H_d(M, M \setminus U; \mathbb{Z})$ on all balls $U \cong \mathbb{R}^d$ in $M$. I am sure that $\mathcal{O}_M$ is isomorphic to whatever might be meant by $\mathbb{Z}^w$, but $\mathcal{O}_M$ is an actual local system. Any $d$-dimensional closed manifold $M$ has a unique fundamental class $[M] \in H_d(M ; \mathcal{O}_M)$, where $\mathcal{O}_M$ is the "orientation" local system of $M$, and fundamental class means that it restricts to the canonical generator of $$H_d(\mathbb{R}^d, \mathbb{R}^d\setminus \{0\} ; \mathcal{O}_{\mathbb{R}^d}) \cong_{UCT} H_d(\mathbb{R}^d, \mathbb{R}^d\setminus \{0\} ; \mathbb{Z}) \otimes H_d(\mathbb{R}^d, \mathbb{R}^d\setminus \{0\} ; \mathbb{Z})$$ for every ball $\mathbb{R}^d \cong U \subset M$. Note that this rank 1 $\mathbb{Z}$-module indeed has a preferred generator as it is the tensor square of a rank 1 $\mathbb{Z}$-module. That such a fundamental class is unique if it exists, and that it exists, is by the usual proof of existence of fundamental classes. Made precise in this way the question is vacuous, because $[M]$ is seen to not be a choice of data.<|endoftext|> TITLE: Convergence of random measure QUESTION [6 upvotes]: Suppose that $S$ is a separable metric space or Polish. Let $μ_{n},n∈N $ be a random probability measures and let μ be a deterministic probability measure on $S$. That is to say, that the $ μ_{n}$ are measurable maps from a probability space $(Ω,T,P)$ to the space of $M_1(S) $ equipped with the Borel-σ-algebra generated by the topology of weak convergence. Assume that the expected measures $ν_{n}:=Eμ_{n}$, defined via duality as $∫f dν_{n}:=E∫f dμ_{n}$ for all $f∈C_{b}(S)$, converge weakly to μ, i.e. that for all $f∈C_{b}(S)$ we have the convergence of $E∫fdμ_{n}$ to $∫fdμ$. For all ϵ>0 the sequence $P(d_{BL}(μ_{n},μ)>ϵ)$ converges to zero, where $d_{BL}$ is the bounded Lipschitz metric $$ d_{BL}(μ,ν)=\sup\Big\{\Big\vert∫f dμ−∫f dν\Big|: f:S\to\mathbb{R} \; \mbox{is 1-Lipschitz and $\Vert f\Vert_\infty\leq 1$}\Big\} $$ which completely metrizes the topology of weak convergence) For all $f∈C_{b}(S)$ it holds that $∫f dμ_{n}$ converges in probability to $∫f dμ$. please i want to know if 1) implies 2) or if 2) implies 1) or if 1) is equivalent to 2) and in the case where $μ_{n}$ is not necessary a probability measure but just a random measure, do the same results remain true? REPLY [5 votes]: In your setting, the two are equivalent. Let's first show that your assumptions imply that the $\mu_n$, viewed as random variables in $M_1(S)$, are tight. Since $S$ is Polish and the $\nu_n$ converge weakly to a limit, we know from Prokhorov that the sequence $\nu_n$ is tight so, for every $\varepsilon > 0$ there exists a compact set $K_\varepsilon \subset S$ such that $\nu_n(K_\varepsilon) > 1-\varepsilon $, uniformly in $n$. A simple application of Markov's inequality then shows that $P(\mu_n(K_\varepsilon) > 1-\sqrt \varepsilon) > 1-\sqrt \varepsilon$. Define now $$\mathcal{K}_\delta = \{\mu \in M_1(S)\,:\, \mu(K_{2^{-2m}\delta^2}) > 1- 2^{-m} \delta \quad \forall m \in \mathbb{N}\}.$$ The set $\mathcal{K}_\delta$ is tight and therefore precompact in $M_1(S)$. Furthermore, we know from above that $P(\mu_n(K_{2^{-2m}\delta^2}) > 1-2^{-m} \delta) > 1-2^{-m} \delta$ so that, summing all of these probabilities, one has $P(\mu_n \in \mathcal{K}_\delta) > 1-\delta$. Now that we know that the $\mu_n$ are tight, we know that there are (possibly random) accumulation points $\hat \mu$. The claim now is that both of your conditions are equivalent to the statement that $\hat \mu = \mu$ almost surely. It is clear that this statement implies your second claim by continuity of the maps $\eta \mapsto \int f\,d\eta$. It also implies the first claim since $d_{BL}$ is a metric for $M_1(S)$ and convergence in law to a deterministic element is the same as convergence in probability. The fact that your claims both imply that $\hat \mu$ cannot be anything other than $\mu$ is pretty much immediate. I haven't checked the details, but I guess that if the $\mu_n$ are allowed to be random positive measures, then the claim still holds. (Testing with $f=1$ gives an a priori bound on the mass of the measure and identifies the mass of the limit.)<|endoftext|> TITLE: Existence of a weight of a representation in the fundamental Weyl chamber QUESTION [6 upvotes]: Let $\mathfrak g$ be a complex simple Lie algebra. Fix a Cartan subalgebra $\mathfrak h$ of $\mathfrak g$, let $\Delta$ denote the corresponding root system. Pick a partial order on $\mathfrak h$, which induces a positive root system $\Delta^+$, fundamental Weyl chamber, etc. I wish a reference for the following result, which must be well known. My proof is case-by-case and a bit long. For all but finitely many finite dimensional irreducible representations $(\pi, V)$ of $\mathfrak g$, there is a weight $\mu$ of $\pi$ lying in the fundamental Weyl chamber. Recall that $\mu\in\mathfrak h^*$ is called a weight of $\pi$ if $V(\mu):=\{v\in V: \pi(X)\cdot v=\mu(X)\, v\quad\forall \, X\in\mathfrak h\}$ is non-zero. Furthermore, the fundamental Weyl chamber is given by elements $\mu\in\mathfrak h^*$ satisfying $\langle \mu,\alpha\rangle >0$ for all $\alpha\in\Delta^+$. The proof is immediately reduced to consider irreducible representations of $\mathfrak g$ with highest weight lying in a face of the fundamental Weyl chamber. Furthermore, it reduces to check irreducible representations with highest weight a multiple of a fundamental weight. It is easy to see that the result is no longer valid when $\mathfrak g$ is assumed semisimple in place of simple. REPLY [7 votes]: Here's maybe another (more conceptual?) way to think about it. First of all, if $\mu_1$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_1}$, and $\mu_2$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_2}$, then $\mu_1+\mu_2$ is a dominant weight which appears with nonzero multiplicity in $V^{\lambda_1+\lambda_2}$. (This is just saying that if $\lambda_1-\mu_1$ is a nonnegative sum of simple roots, and ditto for $\lambda_2-\mu_2$, then definitely $(\lambda_1+\lambda_2)-(\mu_1-\mu_2)$ is as well.) The previous paragraph reduces the problem to showing that if $\lambda=\omega_i$ is a fundamental weight, then for all $m\gg0$ sufficiently large, $V^{m\omega_i}$ contains a weight strictly in the fundamental chamber. Let's show this. Recall the Weyl vector $\rho:=\frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha$, which is well known to satisfy $\rho= \sum_{j=1}^{n}\omega_j$. Let $\Phi_i$ denote the maximal parabolic root system generated by the simple roots $\alpha_j$ for $j\neq i$, and let's use the notation $\rho_i :=\frac{1}{2}\sum_{\alpha\in\Phi_i^+}\alpha$ for its "parabolic Weyl vector." Note that $\langle \rho_i,\alpha^\vee_j\rangle = 1$ for all $j\neq i$ precisely because this is the Weyl vector of the parabolic root system, while $\langle \rho_i, \alpha^\vee_i \rangle < 0$ since no factor of $\alpha_i$ appears at all in $\rho_i$ but some $\alpha_j$ for $j$ a node adjacent to $i$ in the Dynkin diagram does appear. Therefore $2\rho - 2\rho_i = \sum_{\alpha\in \Phi^+\setminus \Phi_i^+}\alpha= c\omega_i$ for some integer $c > 0$. But one of the $\alpha \in \Phi^+\setminus \Phi_i^+$ is the highest root $\theta$ of $\Phi$, which when written $\theta=\sum_{j=1}^{n}a_j\alpha_j$ satisfies $a_j \geq 1$. So there is some multiple $k$ of $2\rho - 2\rho_i$ for which $k(2\rho - 2\rho_i) - 2\rho$ is a nonnegative sum of simple roots. This is to say the strictly dominant weight $2\rho$ appears in $V^{kc\omega_i}$. And then for any $m\geq kc$, the strictly dominant weight $2\rho + (m-kc)\omega_i$ appears in $V^{m\omega_i}$, completing the proof. REPLY [4 votes]: I don't know about a reference, but here is a short uniform proof. I assume that all of your representations are finite dimensional. Let $\alpha_1$, ..., $\alpha_n$ be the simple roots and let $\omega_1$, ..., $\omega_n$ be the dual weights, so $\langle \alpha_i, \omega_j \rangle = \delta_{ij}$. So, if $\mu$ is an integral positive weight, then $\mu = \sum c_j \omega_j$ for $c_j \in \mathbb{Z}_{\geq 0}$. Let $A_{ij}$ be the Cartan matrix. Let $\Gamma$ be the Dynkin diagram, so this is a graph with vertices $1$, $2$, ..., $n$ and an edge $(i,j)$ if $A_{ij} \neq 0$. For a positive root $\mu$, let $a_r(\mu) = \{ j : c_j \leq r \}$. We partially order the integral positive weights as follows: We put $\mu \leq \nu$ if there is some index $q$ for which $a_q(\mu) > a_q(\nu)$ and $a_r(\mu) = a_r(\nu)$ for $0 \leq r < q$. Fix your representation $V$ and let $\mu = \sum c_j \omega_j$ be minimal among the integral positive weights of $V$, with respect to the above order. Key lemma: For every edge $(i,j)$ of $\Gamma$, we have $c_i - c_j \leq 2$. Proof: Suppose for the sake of contradiction that $c_i \geq c_j + 3$. Since $c_i = \langle \alpha_i, \mu \rangle > 0$, the weight $\mu' := \mu - \alpha_i$ is also a weight of $V$. We will verify that $\mu'>\mu$. We have $\mu' = \sum c'_k \omega_k$ where $c'_i = c_i-2$ and $c'_k = c_k + (-A_{ik})$ for $k \neq i$. Put $q = c_j$. I claim that $a_q(\mu') > a_q(\mu)$ and $a_r(\mu') \geq a_r(\mu)$ for $r < q$. Since $c'_i = c_i-2 \geq c_j+1=r+1$, the change between $c'_i$ and $c_i$ won't change the value of $a_q$ for $q \leq r$. Each coefficient other than the $i$-th gets larger going from $\mu$ to $\mu'$, so all the $a_q$ get weakly larger. Moreover, $c_j=r$ and $c'_j = r + (-A_{ij}) > r$, so $a_r$ gets strictly larger. $\square$ Let $\delta$ be the diameter of the graph $\Gamma$. (Here is where we use that the Lie algebra is simple, so $\Gamma$ is connected.) So, if $\mu = \sum c_j \omega_j$ is the above minimal weight, and one of the $c_j$ are $0$, then all of the $c_j$ are bounded by $2 \delta$. In particular, there are only finitely many choices for $\mu$. Remark Experimentally, every representation seems to have a weight with $c_i - c_j \leq 1$, but it isn't always the minimum in the order I've defined, and I don't want to work hard enough to get this better bound.<|endoftext|> TITLE: Generators for the semigroup $\mathrm{SL}(n,\mathbb{N})$ QUESTION [7 upvotes]: For $2\times 2$ matrices we have the following result. Any matrix in $\mathrm{SL}(2,\mathbb{Z})$ with nonnegative entries can be obtained from $\mathrm{Id}_2$ by repeatedly adding one column to another. Proof: It is enough to prove that if $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in\mathrm{SL}(2,\mathbb{Z})\setminus \{\mathrm{Id}_n\}$$ has nonnegative entries, then either $$\begin{pmatrix} a-b & b \\ c-d & d \end{pmatrix} \text{ or }\begin{pmatrix} a & b-a \\ c & d-a \end{pmatrix}$$ has nonnegative entries as well. After this you can finish by induction. Now to prove that, suppose $a$ is the biggest entry of the matrix, if $a=1$ then we obtain the matrices $$\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \tag{$\star$} $$ and we are done. Otherwise $a>1$, hence $$d-c\leq d-bc/a=(ad-bc)/a=1/a$$ from which $d-c\leq 0$ and we arrive in the first case. The cases in which the maximal entry is different from $a$ are done similarly $\blacksquare$ This can be restated as saying that the elementary matrices in ($\star$) generate the semigroup $\mathrm{SL}(2,\mathbb{N})$. Here $\mathbb{N}$ denotes the non-negative integers (including $0$). My question is: Is it true that $\mathrm{SL}(n,\mathbb{N})$ can be generated by elementary matrices, similarly as in the case $n=2$? I would guess this has already been discussed in the literature so a good reference would be enough. Edit: Probably the question above is stated better in terms of $\mathrm{SL}^{\pm}(n,\mathbb{N})$, the set of square $n\times n$ matrices with nonnegative integral entries and determinant $1$ or $-1$. This set has the following properties: $\mathrm{Id}_n\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ If $A\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ and we change a column of $A$ by the addition of it with other column, the result is still on $\mathrm{SL}^{\pm}(n,\mathbb{N})$. If $A\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ and we switch two columns of $A$, the result is still in $A$. The problem is to show that the set $\mathrm{SL}^{\pm}(n,\mathbb{N})$ is the minimum set of matrices with this three properties. Notice that property $2$ is equivalent to If $A\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ and $L_{i,j}(1)$ is the elementary matrix that acts by changing column $i$ by the addition of column $i$ and $j$ (see for example Wikipedia) then $$A\cdot L_{i,j}(1)\in \mathrm{SL}^{\pm}(n,\mathbb{N}).$$ and property 3 is equivalent to If $A\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ and $T_{i,j}$ is the elementary matrix that acts by switching column $i$ and $j$ then $$A\cdot T_{i,j}\in \mathrm{SL}^{\pm}(n,\mathbb{N}).$$ As $L_{i,j}(1), T_{i,j}\in \mathrm{SL}^{\pm}(n,\mathbb{N})$, the problem above is equivalent to Every matrix in $\mathrm{SL}^{\pm}(n,\mathbb{N})$ can be written as a multiplication of matrices of the form $L_{i,j}(1)$ and $T_{i,j}$. As operation 2 and 3 commute with each other. We can put all permutation matrices at the end, then by multiplying them we can use only one permutation matrix. If the final matrix has determinant 1 so will have this permutation matrix. In this way we see that this question is equivalent to the original one stated in terms of $\mathrm{SL}(n,\mathbb{N})$. REPLY [3 votes]: I believe that the answer is negative. A positive answer would mean that for any $A\in\mathrm{SL}^\pm(n,\mathbb{N})$, $A\neq1$ there exists $i,j$ and some other element $B\in\mathrm{SL}^\pm(n,\mathbb{N})$ such that $A=L_{ij}(1)\cdot B$ or $A=B\cdot L_{ij}(1)$, possibly after some permutation of rows and/or columns of $A$. If, however, one can find an $A$ such that $L_{ij}(-1)\cdot A', A'\cdot L_{ij}(-1) \notin \mathrm{SL}^\pm(n,\mathbb{N})$ for any $i,j$ and any permutation $A'$ of columns/rows of $A$, then $A$ can not be decomposed into the product of $L_{ij}(1)$'s and $T_{ij}$'s. A quick computer search reveals that $$ A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 3 & 1 \\ 2 & 0 & 1 \end{pmatrix} $$ is one such matrix.<|endoftext|> TITLE: Two (probably) equal real numbers which are not proved to be equal? QUESTION [53 upvotes]: Can someone give me a nice example of two computable real numbers which are believed but not proved to be equal? I never really understood the assertion that "the reals do not have decidable equality" until I saw concrete examples such as this gem. It's clear that both sides are well-defined real numbers, however it's not at all clear how one might go about proving their equality. However, perhaps a little disappointingly, on page 10 of this article by Bailey, Borwein, Broadhust and Zudilin, the equality is proved (and that article appeared on ArXiv before Stanley's post). It's very easy to give "silly" examples of real numbers which are probably equal but not provably equal, e.g. the real number which is 0 if the Birch and Swinnerton-Dyer conjecture is true and 1 if not, is probably equal to zero, but we can't prove it. So for the pedants, can anyone give me an example of two computable real numbers whose equality is believed, but not known? More informally, I am looking for an example like the one Richard Stanley mentions in the link above where computer scientists can compute both sides to a gazillion decimal places but mathematicians can't prove equality rigorously? Stanley calls such things rare. Several potential examples are mentioned on p13 of the BBBZ paper I cited above, however that paper was written ten years ago and things seem to move fast in this area. Are any of these still open? It looks to me like the authors are developing techniques which might solve them. For example equation (10): if $$Cl_2(\theta):=-\int_0^\theta\log|2\sin(\sigma)|d\sigma$$ and $\theta_2:=2\tan^{-1}(\sqrt{2})$ then is $$27Cl_2(\theta_2)−9Cl_2(2\theta_2) + Cl_2(3\theta_2)= 8Cl_2(\pi/4)+ 8Cl_2(3\pi/4)$$ still an open problem? REPLY [2 votes]: In December of the year 2017 I discovered the following formula (but I could not prove it), see the addendum of this: $$ \sum_{n=1}^{\infty} \frac{(4n)!(2n)!(n!)^2}{(8n)!} \frac{120-1273n+2210n^2}{7^4(-2n+1)n^3} (-2401)^n \, {\overset{?}=} \, \sum_{n=1}^{\infty} \frac{\chi_{-7}(n)}{n^2}= L_{-7}(2), $$ a Dirichlet $L$ value (see here).<|endoftext|> TITLE: Nonnegativity of an integral over the unitary group QUESTION [10 upvotes]: For an $n$-by-$n$ unitary matrix $U$ and a permutation $\sigma\in S_n$, let $$w_\sigma=(-1)^\sigma\det(U^*)\prod_{i=1}^n U_{i,\sigma(i)}.$$ Is $\int_{U(n)}\mathrm{Re}(w_{\sigma_1})\mathrm{Re}(w_{\sigma_2})dU\ge 0$ for all $\sigma_1,\sigma_2\in S_n$? The integral can be expanded to $$\frac{1}{4}\int_{U(n)} (w_{\sigma_1}+\bar w_{\sigma_1})(w_{\sigma_2}+\bar w_{\sigma_2})dU=\frac{1}{2}\int_{U(n)} w_{\sigma_1}w_{\sigma_2}+w_{\sigma_1}\bar w_{\sigma_2}dU.$$ Once the $\det(U^*)$ factors have been expanded as sums over permutations, this can be evaluated using Weingarten functions (see Collins 2003). The latter term is \begin{align}\int_{U(n)} w_{\sigma_1}\bar w_{\sigma_2}dU&=(-1)^{\sigma_1\sigma_2}\int_{U(n)}U_{1\sigma_1(1)}\cdots U_{n\sigma_1(n)}U^*_{\sigma_2(1)1}\cdots U^*_{\sigma_2(n)n}dU\\ &=(-1)^{\sigma_1\sigma_2}Wg(\sigma_2^{-1}\sigma_1,n).\end{align} By Novak 2010, the element $\sum_\sigma Wg(\sigma,n)\sigma$ of the group algebra $\mathbb{C}[S_n]$ can be written as a product of elements $(n+J_k)^{-1}$ of the form $\sum_\sigma (-1)^\sigma|a_\sigma|\sigma$, hence $(-1)^\sigma Wg(\sigma,n)\ge 0$. Computer experiments suggest the former term is nonnegative as well, but I haven't been able to prove it. Note that $\int_{U(n)}\mathrm{Re}(w_\sigma)dU=1/n!$ by a symmetry argument (see here). I can show that $$I_{\sigma_1,\sigma_2}=\frac{1}{n!(n+1)!}\sum_{\pi\in C}\sum_{\tau_1,\tau_2\in R}(-1)^{\sigma\pi} [\tau_1\pi\tau_2=\sigma],$$ where $C$ and $R$ are the Young subgroups $\mathrm{Sym}(1,\dots,n)\times\mathrm{Sym}(n+1,\dots,2n)$ and $\mathrm{Sym}(1,n+1)\times\cdots\times\mathrm{Sym}(n,2n)$ respectively and $\sigma=\sigma_1\oplus\sigma_2\in C$. Is there a way to link this with Carlo's formula or otherwise show it's nonnegative? Proof: By expanding the $\det U^*$ factors and applying Weingarten functions, \begin{align*} I_{\sigma_1,\sigma_2}&=(-1)^{\sigma_1\sigma_2}\sum_{\pi_1,\pi_2}(-1)^{\pi_1\pi_2}\int_{U(n)}\prod_{i=1}^n U_{i\sigma_1(i)}U_{i\sigma_2(i)}U^*_{\pi_1(i)i}U^*_{\pi_2(i)i}dU\\ &=(-1)^{\sigma}\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}Wg(\sigma^{-1}\tau_1\pi\tau_2,n)\\ &=(-1)^{\sigma}\sum_{\lambda\vdash 2n}\frac{\chi^\lambda(1)^2}{(2n)!^2s_{\lambda,n}(1)}\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}\chi^\lambda(\sigma^{-1}\tau_1\pi\tau_2). \end{align*} For each $\lambda$, $\chi^\lambda$ is being summed over cosets of $R$ and the dual character $\chi^{\lambda'}=\mathrm{sgn}\cdot\chi^\lambda$ is being summed over cosets of $C$. The sum of a character $\chi$ over a coset of a subgroup $H$ is $0$ unless $\mathrm{res}_H\chi$ contains the trivial character with multiplicity at least $1$. By the Frobenius-Young correspondence, the only irreducible contained in both $\mathrm{ind}^{S_{2n}}_R 1$ and $\mathrm{ind}^{S_{2n}}_C \mathrm{sgn}$ is $\chi^\mu$ with multiplicity $1$, where $\mu=(2^n)$. Therefore, $$c_\lambda:=\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}\chi^\lambda(\sigma^{-1}\tau_1\pi\tau_2)=\begin{cases}0 & \text{if }\lambda\neq\mu\\ c_\mu&\text{if }\lambda=\mu.\end{cases}$$ Evaluating the above formula for the character of the regular representation yields $$\chi^\mu(1)c_\mu=\sum_{\lambda\vdash 2n}\chi^\lambda(1)c_\lambda=c_{reg}=\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}(2n)![\sigma^{-1}\tau_1\pi\tau_2=1].$$ It's easy to show with the hook-length formula that $\chi^\mu(1)=\frac{(2n)!}{n!(n+1)!}$ and even easier to show that there is only one semistandard Young tableau of shape $\mu$ with entries in $[n]$, so $s_{\mu,n}(1)=1$. REPLY [4 votes]: I don't understand the matrix integrals, but the sum in the last paragraph is easy to deal with. Here is a story to visualize the permutation groups. Imagine $n$ couples lined up in two lines in a hall, perhaps to dance. A permutation in $C$ permutes each line within itself, breaking up partnerships. A permutation in $R$ makes some couple switch places with each other. Note that every permutation in $R$ is an involution. Let's first prove positivity. Let $\tau_s$ switch $k_s$ pairs of couples. If we are to have $\tau_1 \sigma \tau_2 \in C$, then each person who switches under $\tau_1$ must switch again until $\tau_2$ to get back to their original line. So $k_1 = k_2$ and $(-1)^{\tau_1} = (-1)^{\tau_2}$. This implies that $(-1)^{\pi} = (-1)^{\tau_1 \sigma \tau_2}$ and thus $(-1)^{\sigma \pi}=1$. Now, let's compute the sum. Let $T_s$ be the set of indices $k$ for which $\tau_s(k) = k+n$. In order to have $\tau_1 \sigma \tau_2 \in C$, each person must cross the hall under $\tau_1$ if and only if they do under $\tau_2$. Looking at the people who start on the $\{1,2,\ldots, n \}$ side of the hall, we must have $\sigma_2(T_1) = T_2$. Looking at the people who start on the other side, we must also have $\sigma_1(T_1) = T_2$. Thus $\sigma_1^{-1} \sigma_2 (T_1) = T_1$. So $T_1$ is a union of orbits of $\sigma_1^{-1} \sigma_2$. Each such union of orbits has a unique compatible $T_2$, by the equation $\sigma_1(T_1) = T_2$, and each such $(\tau_1, \tau_2)$ determines a unique $\pi$ by $\pi = \tau_1 \sigma \tau_2$. So the number of nonzero terms is the number of subsets of orbits of $\sigma_1^{-1} \sigma_2$. The number of orbits is $n-\ell_T(\sigma_1^{-1} \sigma_2)$ where $\ell_T$ is the length with respect to the set of all reflections, so the sum is $2^{n-\ell_T(\sigma_1^{-1} \sigma_2)}$ as requested.<|endoftext|> TITLE: Examples where existence is harder than evaluation QUESTION [36 upvotes]: In expressions involving an infinite process (infinite sum, infinite sequence of nested radicals), sometimes the hardest part is proving the existence of a well-defined value. Consider, for example, Ramanujan's infinite nested radical: $$ \sqrt{1+2\sqrt{1+3\sqrt{1+\ldots}}}. \qquad(*) $$ Assuming the above is well-defined, there is a slick trick showing that it evaluates to $3$. But such careless assumptions can lead to trouble, as in the example of the expression: $$ -5 + 2(-6 + 2(-7 + 2(-8 + \ldots))). \qquad(**) $$ Applying the identity $n = -(n + 2) + 2(n + 1)$ repeatedly for $n=3,4,5,\ldots$, we get \begin{align} 3 &= -5 + 2(4) \\ &= -5 + 2(-6 + 2(5))\\ &= -5 + 2(-6 + 2(-7 + 2(6))\\ &= -5 + 2(-6 + 2(-7 + 2(-8 + 2(7)))\\ &=\ldots, \end{align} which would falsely suggest that $(**)$ evaluates to $3$. What are some interesting examples where evaluating an expression assuming its existence is much easier than proving existence? Edit: clarifying in light of some of the discussion in the comments. I can see how $(**)$ can also invite examples of false conclusions from an assumption of existence. That was not the intent of the question; the sole point of $(**)$ was to show that the solution technique to $(*)$ provided at the link can in general yield false conclusions if existence is assumed without additional proof. The spirit of the question is to exhibit cases where the limit exists, but its value given the existence is much easier to establish than the existence itsef. REPLY [3 votes]: Minimal surfaces would provide a lot of examples. Construction of minimal surfaces often reduces to finding a meromorphic and a holomorphic functions (Weierstrass representation). Often we have a general idea about these functions up to some parameters, which are determined by "closing the periods", meaning that some path integrals should vanish. In many cases the periods are easily closed numerically, leading to beautiful pictures, and the surface is naturally named after the discoverer (they deserve). But existence of period-closing parameters could be very hard prove, sometimes never done. The Horgan (non)surface is a famous example that computer closes periods which is later proved impossible. Then there is the embeddedness to prove, which could be even harder.<|endoftext|> TITLE: Equivalent forms of the P vs. NP problem QUESTION [14 upvotes]: Many things in math can be formulated quite differently; see the list of statements equivalent to RH here, for example, with RH formulated as a bound on lcm of consecutive integers, as an integral equality, etc. I wonder about equivalent formulations of the N vs. NP problem. Formulations that are very much different from the questions such "Is TSP in P?", formulation that may seem unrelated to complexity theory. REPLY [2 votes]: The P vs NP problem can be formulated in terms of incomplete sets in NP. Ladner theorem can be stated as: $P \ne NP$ if and only if there is an incomplete set in NP. Incomplete set is a set that is not complete for $NP$ under many-one polynomial time reductions (Karp reductions). Another formulation in terms of sparse sets is Mahaney's Theorem: There is no sparse NP-complete set if and only if $P \ne NP$ (under Karp reduction). Complexity Theory and Cryptology: An Introduction to Cryptocomplexity By Jörg Rothe, page 106<|endoftext|> TITLE: How come mathematicians published in Annals of Eugenics? QUESTION [26 upvotes]: I was surprised to see that On the construction of balanced incomplete block designs by Raj Chandra Bose was published (in 1939) in a journal named Annals of Eugenics (see here) (published between 1925 and 1954). I double checked and it seems that the journal was indeed focused on eugenics and the article does not relate to eugenics or genetics in any way. So why would a mathematician decide to publish in such a journal and why was the article accepted? It appears that there are also other mathematics articles published in the same journal. REPLY [3 votes]: The zbMATH journal profile https://zbmath.org/serials/?q=se%3A00009697 has 137 reviews for documents from this journal indexed in old Jahrbuch and Zentralblatt issues, of overall 49 mathematicians. Apart from Fisher's own contributions, Bose's paper on block designs Bose, R. C., On the construction of balanced incomplete block designs, Ann. Eugenics 9, 353-399 (1939). ZBL0023.00102. mentioned in the question might have been the most influential (at least, it is the one most frequently cited in subsequent reviews).<|endoftext|> TITLE: Method to draw 3-connected planar graph on a sphere QUESTION [8 upvotes]: The Tutte embedding is a way to create a "nice" drawing of a 3-connected planar graph in the plane, after having chosen an outer face. Is there a similar method to draw such a graph on a sphere? Steinitz's theorem suggests that such a drawing is possible. Of course, we could just map the planar coordinates to a sphere with some arbitrary projection, but this does not produce a "nice" drawing, e.g. it will not reflect the natural symmetries of the graph, and will not produce a dodecahedron for a dodecahedral skeleton. While "nice" is not a very well defined term, I assume there must be some existing work on this topic. REPLY [6 votes]: If the graph $G=(V,E)$ has a lot of symmetries, then using spectral realizations might give you nice drawings that reflects these symmetries. The success of this method (e.g. whether the drawing is planar, whether all vertices are on a sphere) depends on a lot of factors, some of which are not completely clear to me. However, what I can tell you is that it works for the graphs of all uniform 3-polytopes (so, e.g., the dodecahedron). I explain the most straight forward way to do it, some tweaks might be neccessary for the general case: Costruction. Let $\theta$ be an eigenvalue of (the adjacency matrix of) $G$, and $v_1,v_2,v_3\in\Bbb R^n$ three ortho-normal eigenvectors to $\theta$. Construct the matrix $M:=(v_1,v_2,v_3)\in\Bbb R^{n\times 3}$ with the $v_i$ as columns. The rows of that matrix are a 3-dimensional embedding of the vertices of $G$. Usually, you should take $\theta_2$, i.e., the second-largest eigenvalue of the adjacency matrix of $G$. Surprisingly, this eigenvalue has multiplicity three for most symmetric graphs that come from 3-polytopes (exceptions are, as far as I know, only prisms). This means, you cannot do anything wrong by choosing just any orthonormal basis of eigenvectors. Here is code for Mathematica to automatically find a nice drawing of the dodecahedral graph: G = GraphData["DodecahedralGraph"]; A = AdjacencyMatrix[G]; n = VertexCount[G]; eval = Eigenvalues[A // N]; th2 = RankedMax[eval, 2]; evec = NullSpace[A - th2*IdentityMatrix[n]]; GraphPlot3D[G, VertexCoordinateRules -> Table[i -> evec[[{1,2,3}, i]], {i, 1, n}] ] Output: If $\theta_2$ has multiplicity $<3$, you can add eigenvectors of other eigenvalues until you have three, preferably from the next largest eigenvalues. Just do not use the largest eigenvalue. For your example in the comments (the capped cube), we have the problem that $\theta_2$ has multiplicity one. However, as explained, we could use eigenvectors of $\theta_3$ (which has multiplicity two) to complete to a set of three vectors. Use this: th2 = RankedMax[eval, 2]; th3 = RankedMax[eval, 3]; evec = NullSpace[A - th2*IdentityMatrix[n]] ~Join~ NullSpace[A - th3*IdentityMatrix[n]]; Output: For general graphs one should certainly follow a more dynamic approach, e.g. sorting the eigenvectors by their eigenvalues and taking the largest three (but not the largest one). Something like this: vec = SortBy[Transpose[Eigensystem[A // N]], First][[{-2, -3, -4}, 2]]; GraphPlot3D[G, VertexCoordinateRules -> Table[i -> vec[[;; , i]], {i, 1, n}] ]<|endoftext|> TITLE: Third differential in the homology AHSS QUESTION [8 upvotes]: I need some guidance in identifying the third differential in the homology AHSS for $\Omega_{\ast}^{\text{Spin}^c}(X)$ in degrees $\leq 4$. Remember that $\pi_0(M\text{Spin}^c)=\Bbb Z$, $\pi_2(M\text{Spin}^c)=\Bbb Z$, $\pi_4(M\text{Spin}^c)=\Bbb Z\oplus \Bbb Z$ and zero in degree $1,3$. So if we consider the (homologically graded) AHSS $$ E_{p,q}^2 = H_p(X;\Omega_{q}^{\text{Spin}^c} ) \Rightarrow \Omega_{p+q}^{\text{Spin}^c}(X)$$ we immediately see that there are no second differentials and the third page looks like this (in the portion I'm interests in) Can I conclude something akin to the cohomology AHSS (See here) where the first non-trivial differential is a stable cohomology operation? what bothers me is that having to deal with integer coefficients homology I don't have (in general) a perfect pairing with cohomology where the linked question would apply. Is there an isomorphism in low degree homotopy group for $M\text{Spin}^c$ (I'm thinking of something like $M\text{Spin}$ is weakly homotopy to $ku$ up to degree $7$.) We could leverage this to compute the AHSS, but I can't find much online. Any comments/observations are really appreciated! Thanks a lot! REPLY [8 votes]: The realification map $BU \to BSO$ lifts through $BSpin^c$, because the first Chern class provides an integral lift of the second Stiefel-Whitney class. Calculating with the Serre spectral sequence for the homotopy fiber sequence $K(\mathbb{Z}, 2) \to BSpin^c \to BSO$ I get that $H^*(BSpin^c; \mathbb{F}_2)$ agrees with $\mathbb{F}_2[w_2, w_4, w_6, w_7]$ in degrees $* \le 7$, so that $BU \to BSpin^c$ is $6$-connected after $2$-completion. (I have not checked this against other sources, so caveat emptor.) It follows that $MU \to MSpin^c$ is $6$-connected, also after $2$-completion. Hence $Spin^c$-bordism agrees with complex bordism in the range of degrees you say you are interested in, and the Atiyah-Hirzebruch spectral sequence for $MSpin^c_*(X)$ agrees with that for $MU_*(X)$, after $2$-completion, in vertical degrees $* \le 5$. (Do you care about odd primes?) Since $MU$ splits $2$-locally as $BP \vee \Sigma^4 BP \vee \dots$, you may as well work out the spectral sequence for $BP_*(X)$. The differential $d^3_{n,0} : H_n(X; BP_0) \to H_{n-3}(X; BP_2)$ is the integral homology operation induced by the $k$-invariant $\tilde Q_1 : H\mathbb{Z} \to \Sigma^3 H\mathbb{Z}$ with homotopy fiber the second Postnikov section of $BP$. (This is the same as the second Postnikov section of $MU$, and of $ku$.) The differential $d^3_{n,2} : H_n(X; BP_2) \to H_{n-3}(X; BP_4)$ is induced by the $k$-invariant connecting $BP_2$ and $BP_4$, which is also $\tilde Q_1$. Here $\tilde Q_1$ is an integral lift of the Milnor primitive operation $Q_1 = [Sq^2, Sq^1]$. Hence, apart from one of the two $\mathbb{Z}$-summands in bidegree $(0,4)$ in your picture, $MSpin^c_*(X)$ will agree after $2$-completion with $ku_*(X)$ for $* \le 4$. If you are interested in $X = BG$, the book by R.R. Bruner and J.P.C. Greenlees on $ku_*(BG)$ should be helpful.<|endoftext|> TITLE: $\int_{-\infty}^\infty \frac{e^{-y^2/2}}{((y+y_0)^2+x_0^2)^r} dy$ QUESTION [7 upvotes]: I have to estimate the integral $$\int_{-\infty}^\infty \frac{e^{-y^2/2}}{((y+y_0)^2+x_0^2)^r} dy,$$ for $r\in \mathbb{R}^+$. I am a little amazed that Sage and Wolfram Alpha have nothing to say about it, and that Gradshteyn-Ryzhik doesn't seem to have anything on it either; it feels like a rather natural integral, the denominator being a distance function. Of course I realize that, if I just expand $1/((y+y_0)^2+x_0^2)^r$ into a Taylor series around $y=-y_0$ and integrate term-by-term, I am not going to get something convergent; what I get is an asymptotic formula. But what is the right order of magnitude of the error term when the formula gets cut off at the $k$th term? For instance: let $f(y)=1/((y+y_0)^2+(x_0^2))^r$ and write $$f(y) = f(y_0) + \frac{(y-y_0)^2}{2} O^*(\max_t f''(t)).$$ Then we get an error term of size $O(1/x_0^{2 r + 3})$. If we go up to a higher-order approximation, we obtain an error term of the form $O(1/x_0^{2(r+k)+1})$ for higher $k$. But can one also give an error term that depends on $y_0$ and not just on $x_0$, and thus is better when $x_0$ is large and $y_0$ is much larger still? Note: also asked on MSE. Note 2: just to be clear - the easy estimate (obtained by a cheap version of Laplace) is $$\frac{\sqrt{2 \pi}}{(x_0^2+y_0^2)^r} + O^*\left(\frac{2 r \cdot \sqrt{2 \pi}}{e^{3/2} x_0^{2 r + 2}}\right).$$ That is fine if $x_0 \asymp y_0$, but we would like to do better than that in general, particularly in the tricky case $y_0\ggg x_0\ggg 1$. Here $O^*(\text{X})$ means "something of absolute value at most $X$". REPLY [4 votes]: I would have liked to say this in comments, but the tex integrals make it impractical. This is not an answer but a reformulation. We can change the integrand slightly: $$I(r)=I(r,x_0,y_0)= \int _{-\infty} ^{+\infty} e^{-y^2} \frac{dy}{((y+y_0)^2+x_0^2)^r}.$$ Multiplying by $\Gamma (r)$ we see that this becomes (after a change of variables) $$ \Gamma (r)I(r)= \sqrt{2\pi}\int _0^{\infty} \frac{du}{u} u^r \frac{1}{\sqrt{(1+u)}} e^{-(ux_0^2+y_0^2\frac{u}{1+u})}.$$ Hopefully, the latter integral can be estimated more easily.<|endoftext|> TITLE: Computability-theoretic results relevant to realizability QUESTION [10 upvotes]: This may be a very naive question which only reflects my failure at literature search, but: Although realizability (in its original form at least) is grounded in computability, the details of computability theory itself don't seem too relevant for realizability as far as I can tell. For example, I don't know of any result in realizability which relies on (say) the Low Basis Theorem or on the solution to Post's problem. Basic computability of course plays a role - e.g. whipping up realizability semantics appropriate for computability notions which lack universal machines, such as the primitive recursive functions, is difficult and interesting - but (to my practically-nonexistent knowledge) the deeper results don't seem to play a role. My question is whether this is accurate. In particular, I'd be extremely interested in any significant role played by priority arguments in realizability. The most relevant thing I've found is Charles McCarty's paper "Realizability and recursive set theory", which established a connection between isols/recursive equivalence types and realizability. There are many results about isols, of course, which are proved via complicated priority arguments. However, unless I'm missing something this seems to be a situation where realizability sheds light on the isols, rather than results about isols being relevant to realizability. REPLY [9 votes]: Many results of computability theory can be stated synthetically in the effective topos. I have dubbed this idea "synthetic computability" and have tried to develop as much computability this way as I can. Some examples of how classical theorems in computability theory are stated synthetically: "There is no computable enumeration of total recursive functions" becomes "The set of functions $\mathbb{N} \to \mathbb{N}$ is uncountable" in the internal language of the effective topos. "There is a computable enumertion of computably enumerable sets" becomes "There are countably many countable subsets of $\mathbb{N}$". "There is a c.e. set whose complement is not c.e." becomes "there is a countable subset of $\mathbb{N}$ whose complement is not countable". "There is an immune set" becomes "there is a subset of $\mathbb{N}$ which is neither finite nor infinite". One can keep going this way. So far I have been stuck on obtaining a suitable synthetic definition of Turing reducibility. I have one, but I am not completely happy with it. Intermediate degrees are also out of my reach, although it looks like they should translate to results about measure theory, or perhaps descriptive set theory, I am not sure. Essentially, a priority argument feels a bit like showing that a certain intersection of open sets is non-empty, but it's subtle. The only other place where I would look for the kinds of results that you are asking about is perhaps the work of Jaap van Oosten, in particular his study of the subtoposes of the effective topos (I am not sure how much is published). It seems exceedingly hard to get a grip on the Lawvere-Tierney $j$-operators in the effective topos, and this may be related to some complicated structures in computability. There is a general principle: in order for a result $A$ in computability theory to have significance in realizability, you need to find an intuitionistic statement $B$ such that validity of its realizability interpretation is related to $A$. I started to study synthetic computability more seriously when I discovered that the Recursion theorem corresponds to a variant of Lawvere's fixed-point theorem (see this paper1). Until we find out what priority arguments correspond to under realizability interpretation, we pretty much have no clue how to use them. 1Bauer, Andrej, On fixed-point theorems in synthetic computability, Tbil. Math. J. 10, No. 3, 167-181 (2017). Author's website, ZBL1403.03076, MR3725758.<|endoftext|> TITLE: Reciprocal sum of binomials and divisibility by $3$ QUESTION [8 upvotes]: We all know that $\sum_{k=0}^n\binom{n}k$ is not divisible by $3$. QUESTION. Is it true that the numerator of $a_n$ (in reduced form) is never divisible by $3$? $$a_n=\sum_{k=0}^n\frac1{\binom{n}k}.$$ POSTSCRIPT. This the most detailed and pedagogical answer to any question that I asked here on MO. Thank you, Darij Grinberg. REPLY [12 votes]: Here is the answer I hinted at in a comment, in real detail. Took me a while, but I had no idea how tiresome such arguments are to expose... Yes, it is true: see Corollary 6 (b) below. The proof relies on the concept of a $p$-adic valuation of a rational number. It is defined as follows: Definition. Let $p$ be a prime. For each rational number $r$, we shall now define an element $v_{p}\left( r\right) $ of the set $\mathbb{Z}\cup\left\{ \infty\right\} $ (where $\infty$ is understood to be a symbol that satisfies the rules $\infty\geq n$ and $\infty+n=\infty$ for all $n\in\mathbb{Z} \cup\left\{ \infty\right\} $). We define it in three steps: If $r=0$, then we set $v_{p}\left( r\right) =\infty$. If $r\in\mathbb{Z}$, then we let $v_{p}\left( r\right) $ be the largest $k\in\mathbb{N}$ satisfying $p^{k}\mid r$. (Here and in the following, $\mathbb{N}$ means the set $\left\{ 0,1,2,\ldots\right\} $.) If $r$ is any nonzero rational number, then we define $v_{p}\left( r\right) $ by $v_{p}\left( r\right) =v_{p}\left( a\right) -v_{p}\left( b\right) $, where $a$ and $b$ are two nonzero integers satisfying $r=a/b$. (This is well-defined, since any nonzero rational number $r$ can be written as $r=a/b$ for two nonzero integers $a$ and $b$, and since $v_{p}\left( a\right) -v_{p}\left( b\right) $ does not depend on the specific choice of $a$ and $b$. Also, this new definition of $v_{p}\left( r\right) $ for rational numbers $r$ does not conflict with the previous definition of $v_{p}\left( r\right) $ for integers $r$. This is all easy to check.) Thus, an element $v_{p}\left( r\right) \in\mathbb{Z}\cup\left\{ \infty\right\} $ has been defined for each $r\in\mathbb{Q}$. This element $v_{p}\left( r\right) $ will be called the $p$-adic valuation of $r$. We will need the following rules for $p$-adic valuations: Proposition 1. Let $p$ be a prime. (a) We have $v_{p}\left( ab\right) =v_{p}\left( a\right) +v_{p}\left( b\right) $ for any $a,b\in\mathbb{Q}$. (b) We have $v_{p}\left( a+b\right) \geq\min\left\{ v_{p}\left( a\right) ,v_{p}\left( b\right) \right\} $ for any $a,b\in\mathbb{Q}$. (c) We have $v_{p}\left( a^{k}\right) =kv_{p}\left( a\right) $ for any $a\in\mathbb{Q}$ and $k\in\mathbb{N}$. (d) For any $i\in\mathbb{N}$ and $n\in\mathbb{Z}$, we have the equivalence $\left( p^{i}\mid n\right) \ \Longleftrightarrow\ \left( v_{p}\left( n\right) \geq i\right) $. (e) If $a,b\in\mathbb{Q}$ satisfy $v_{p}\left( a\right) >v_{p}\left( b\right) $, then $v_{p}\left( a+b\right) =v_{p}\left( b\right) $. (f) Let $s$ and $t$ be two coprime integers such that $t\neq0$ and $v_{p}\left( \dfrac{s}{t}\right) \leq0$. Then, $p\nmid s$. Proof of Proposition 1. This is all well-known. Just in case, here are a few pointers: Parts (a) and (b) of Proposition 1 are obvious in the case when one of $a$ and $b$ is $0$ (because $v_{p}\left( 0\right) =\infty\geq g$ for all $g\in\mathbb{Z}\cup\left\{ \infty\right\} $). Thus, they only need to be proven in the case when both $a$ and $b$ are nonzero. But in this case, they are exactly the parts (c) and (d) of Exercise 3.4.1 in my Introduction to Modern Algebra notes (version of 31 May 2019) (the numbering might change in the future, but you can always find the version of 31 May 2019 frozen on github). (Be warned that, in the latter notes, I use two different notations for what I am here calling $v_{p}\left( r\right) $: The first notation is "$v_{p}\left( r\right) $", which I use only in the case when $r$ is an integer; the second notation is "$w_{p}\left( r\right) $", which I use in the general case of rational $r$. The reason why I do this is to avoid a conflict of notations, even a theoretical one that doesn't actually happen; the notes are written for undergraduates.) Proposition 1 (c) follows by induction on $k$. (The induction base uses $v_{p}\left( 1\right) =0$, while the induction step uses Proposition 1 (a).) Proposition 1 (d) is Lemma 2.13.25 in my Introduction to Modern Algebra notes (version of 31 May 2019) (the numbering might change in the future, but you can always find the version of 31 May 2019 frozen on github). It is essentially a direct consequence of the definition of $v_{p}\left( n\right) $. (e) Let $a,b\in\mathbb{Q}$ satisfy $v_{p}\left( a\right) >v_{p}\left( b\right) $. We must prove that $v_{p}\left( a+b\right) =v_{p}\left( b\right) $. Assume the contrary. Thus, $v_{p}\left( a+b\right) \neq v_{p}\left( b\right) $. But Proposition 1 (b) yields $v_{p}\left( a+b\right) \geq\min\left\{ v_{p}\left( a\right) ,v_{p}\left( b\right) \right\} =v_{p}\left( b\right) $ (since $v_{p}\left( a\right) >v_{p}\left( b\right) $). Combined with $v_{p}\left( a+b\right) \neq v_{p}\left( b\right) $, this yields $v_{p}\left( a+b\right) >v_{p}\left( b\right) $. On the other hand, $v_{p}\left( \underbrace{-a}_{=a\left( -1\right) }\right) =v_{p}\left( a\left( -1\right) \right) =v_{p}\left( a\right) +v_{p}\left( -1\right) $ (by Proposition 1 (a), applied to $b=-1$). Since $-1$ is an integer, we have $v_{p}\left( -1\right) \geq0$. (Actually, $v_{p}\left( -1\right) =0$, but we don't need this.) Now, $v_{p}\left( -a\right) =v_{p}\left( a\right) +\underbrace{v_{p}\left( -1\right) }_{\geq0}\geq v_{p}\left( a\right) >v_{p}\left( b\right) $. Hence, both $v_{p}\left( a+b\right) $ and $v_{p}\left( -a\right) $ are $>v_{p}\left( b\right) $ (since we know that $v_{p}\left( a+b\right) >v_{p}\left( b\right) $). Thus, $\min\left\{ v_{p}\left( a+b\right) ,v_{p}\left( -a\right) \right\} >v_{p}\left( b\right) $ (since $\min\left\{ v_{p}\left( a+b\right) ,v_{p}\left( -a\right) \right\} $ must be one of the numbers $v_{p}\left( a+b\right) $ and $v_{p}\left( -a\right) $). But $b=\left( a+b\right) +\left( -a\right) $. Hence, \begin{align*} v_{p}\left( b\right) =v_{p}\left( \left( a+b\right) +\left( -a\right) \right) \geq\min\left\{ v_{p}\left( a+b\right) ,v_{p}\left( -a\right) \right\} \end{align*} (by Proposition 1 (b), applied to $a+b$ and $-a$ instead of $a$ and $b$). This contradicts $\min\left\{ v_{p}\left( a+b\right) ,v_{p}\left( -a\right) \right\} >v_{p}\left( b\right) $. This contradiction shows that our assumption was wrong. Hence, Proposition 1 (e) is proven. (f) Assume the contrary. Thus, $p\mid s$. Hence, $p^{1}=p\mid s$. But Proposition 1 (d) (applied to $i=1$ and $n=s$) shows that we have the equivalence $\left( p^{1}\mid s\right) \ \Longleftrightarrow\ \left( v_{p}\left( s\right) \geq1\right) $. Hence, we have $v_{p}\left( s\right) \geq1$ (since $p^{1}\mid s$). But Proposition 1 (a) (applied to $a=\dfrac{s}{t}$ and $b=t$) yields $v_{p}\left( \dfrac{s}{t}\cdot t\right) =\underbrace{v_{p}\left( \dfrac {s}{t}\right) }_{\leq0}+v_{p}\left( t\right) \leq v_{p}\left( t\right) $. Hence, $v_{p}\left( t\right) \geq v_{p}\left( \underbrace{\dfrac{s}{t}\cdot t}_{=s}\right) =v_{p}\left( s\right) \geq1$. But Proposition 1 (d) (applied to $i=1$ and $n=t$) shows that we have the equivalence $\left( p^{1}\mid t\right) \ \Longleftrightarrow\ \left( v_{p}\left( t\right) \geq1\right) $. Hence, we have $p^{1}\mid t$ (since $v_{p}\left( t\right) \geq1$). Thus, $p=p^{1}\mid t$. Now, $p$ is a common divisor of $s$ and $t$ (since $p\mid s$ and $p\mid t$). This shows that $s$ and $t$ have a common divisor larger than $1$ (since $p>1$); but this contradicts the fact that $s$ and $t$ are coprime. This contradiction shows that our assumption was false. Thus, Proposition 1 (f) is proven. $\blacksquare$ Next, we need a binomial identity: Proposition 2. Let $n\in\mathbb{N}$. Then, \begin{align*} \sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}=\dfrac{n+1}{2^{n+1}}\sum\limits_{k=1} ^{n+1}\dfrac{2^{k}}{k}. \end{align*} Proposition 2 is the equality (7) in https://math.stackexchange.com/a/481686/ ; it is also Exercise 3.20 (b) in my Notes on the combinatorial fundamentals of algebra, version of 2019-01-10. $\blacksquare$ Next, we need two specific $3$-adic valuations: Lemma 3. Let $k$ be a positive integer. (a) We have $v_{3}\left( \dfrac{2^{k}}{k}\right) =-v_{3}\left( k\right) $. (b) If $k$ is odd, then $v_{3}\left( \dfrac{2^{k}}{k}+\dfrac{2^{2k}} {2k}\right) =-v_{3}\left( k\right) $. Proof of Lemma 3. Clearly, $2^{k}$ is an integer; thus, $v_{3}\left( 2^{k}\right) \in\mathbb{N}$. Similarly, $v_{3}\left( 1+2^{k-1}\right) \in\mathbb{N}$. We don't have $3^{1}\mid2^{k}$ (since the only prime divisor of $2^{k}$ is $2$). Proposition 1 (d) (applied to $p=3$, $i=1$ and $n=2^{k}$) yields the equivalence $\left( 3^{1}\mid2^{k}\right) \ \Longleftrightarrow\ \left( v_{3}\left( 2^{k}\right) \geq1\right) $. Hence, we don't have $v_{3}\left( 2^{k}\right) \geq1$ (since we don't have $3^{1}\mid2^{k}$). In other words, we have $v_{3}\left( 2^{k}\right) <1$; thus, $v_{3}\left( 2^{k}\right) =0$ (since $v_{3}\left( 2^{k}\right) \in\mathbb{N}$). (a) Proposition 1 (a) (applied to $p=3$, $a=\dfrac{2^{k}}{k}$ and $b=k$) yields $v_{3}\left( \dfrac{2^{k}}{k}\cdot k\right) =v_{3}\left( \dfrac{2^{k}}{k}\right) +v_{3}\left( k\right) $. Comparing this with $v_{3}\left( \underbrace{\dfrac{2^{k}}{k}\cdot k}_{=2^{k}}\right) =v_{3}\left( 2^{k}\right) =0$, we obtain $v_{3}\left( \dfrac{2^{k}} {k}\right) +v_{3}\left( k\right) =0$. Hence, $v_{3}\left( \dfrac{2^{k}} {k}\right) =-v_{3}\left( k\right) $. This proves Lemma 3 (a). (b) Assume that $k$ is odd. Thus, $k-1$ is even. Now, $2\equiv -1\operatorname{mod}3$ and thus $2^{k-1}\equiv\left( -1\right) ^{k-1}=1\operatorname{mod}3$ (since $k-1$ is even). Hence, $1+\underbrace{2^{k-1}}_{\equiv1\operatorname{mod}3}\equiv1+1=2\not \equiv 0\operatorname{mod}3$, so that $3\nmid1+2^{k-1}$. But Proposition 1 (d) (applied to $p=3$, $i=1$ and $n=1+2^{k-1}$) yields the equivalence $\left( 3^{1}\mid1+2^{k-1}\right) \ \Longleftrightarrow\ \left( v_{3}\left( 1+2^{k-1}\right) \geq1\right) $. Hence, we don't have $v_{3}\left( 1+2^{k-1}\right) \geq1$ (since we don't have $3^{1}\mid1+2^{k-1}$ (because $3^{1}=3\nmid1+2^{k-1}$)). In other words, we have $v_{3}\left( 1+2^{k-1}\right) <1$; thus, $v_{3}\left( 1+2^{k-1}\right) =0$ (since $v_{3}\left( 1+2^{k-1}\right) \in\mathbb{N}$). Now, Proposition 1 (a) (applied to $p=3$, $a=\dfrac{2^{k}}{k}$ and $b=1+2^{k-1}$) yields \begin{align*} v_{3}\left( \dfrac{2^{k}}{k}\left( 1+2^{k-1}\right) \right) =v_{3}\left( \dfrac{2^{k}}{k}\right) +\underbrace{v_{3}\left( 1+2^{k-1}\right) } _{=0}=v_{3}\left( \dfrac{2^{k}}{k}\right) =-v_{3}\left( k\right) \end{align*} (by Lemma 3 (a)). In view of \begin{align*} \dfrac{2^{k}}{k}\left( 1+2^{k-1}\right) =\dfrac{2^{k}}{k}+\underbrace{\dfrac {2^{k}}{k}\cdot2^{k-1}}_{=\dfrac{2^{2k-1}}{k}=\dfrac{2^{2k}}{2k}}=\dfrac {2^{k}}{k}+\dfrac{2^{2k}}{2k}, \end{align*} this rewrites as $v_{3}\left( \dfrac{2^{k}}{k}+\dfrac{2^{2k}}{2k}\right) =-v_{3}\left( k\right) $. This proves Lemma 3 (b). $\blacksquare$ Lemma 4. Let $n$ be a positive integer. Let $m$ be the largest nonnegative integer such that $n\geq3^{m}$. Set \begin{align*} a=\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k} }\dfrac{2^{k}}{k}\qquad\text{and}\qquad b=\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\mid k}}\dfrac{2^{k}}{k}. \end{align*} Then: (a) We have $v_{3}\left( a\right) >-m$. (b) We have $v_{3}\left( b\right) =-m$. Proof of Lemma 4. (a) Define a subset $G$ of $\mathbb{Q}$ by \begin{align*} G=\left\{ r\in\mathbb{Q}\ \mid\ v_{3}\left( r\right) >-m\right\} . \end{align*} Then, $0\in G$ (since $v_{3}\left( 0\right) =\infty>-m$). Next, we shall show that the set $G$ is closed under addition. Indeed, let $s,t\in G$. Then, $s\in G=\left\{ r\in\mathbb{Q}\ \mid\ v_{3}\left( r\right) >-m\right\} $; in other words, $s\in\mathbb{Q}$ and $v_{3}\left( s\right) >-m$. Likewise, $t\in\mathbb{Q}$ and $v_{3}\left( t\right) >-m$. Both numbers $v_{3}\left( s\right) $ and $v_{3}\left( t\right) $ are $>-m$ (since $v_{3}\left( s\right) >-m$ and $v_{3}\left( t\right) >-m$). Thus, $\min\left\{ v_{3}\left( s\right) ,v_{3}\left( t\right) \right\} >-m$ (since $\min\left\{ v_{3}\left( s\right) ,v_{3}\left( t\right) \right\} $ is one of these two numbers $v_{3}\left( s\right) $ and $v_{3}\left( t\right) $). Now, Proposition 1 (b) (applied to $3$, $s$ and $t$ instead of $p$, $a$ and $b$) yields $v_{3}\left( s+t\right) \geq\min\left\{ v_{3}\left( s\right) ,v_{3}\left( t\right) \right\} >-m$. In other words, $s+t\in G$ (by the definition of $G$). Now, forget that we fixed $s,t$. We thus have proven that $s+t\in G$ for all $s,t\in G$. In other words, the set $G$ is closed under addition. Hence, $G$ is a submonoid of the additive monoid $\left( \mathbb{Q},+\right) $ (since $0\in G$). Thus, any finite sum of elements of $G$ is an element of $G$. Now, let $k\in\left\{ 1,2,\ldots,n\right\} $ be such that $3^{m}\nmid k$. Proposition 1 (d) (applied to $3$, $m$ and $k$ instead of $p$, $i$ and $n$) shows that we have the equivalence $\left( 3^{m}\mid k\right) \ \Longleftrightarrow\ \left( v_{3}\left( k\right) \geq m\right) $. Thus, we do not have $v_{3}\left( k\right) \geq m$ (since we do not have $3^{m}\mid k$ (because $3^{m}\nmid k$)). In other words, we have $v_{3}\left( k\right) -m$. In other words, $\dfrac{2^{k}}{k}\in G$ (by the definition of $G$). Forget that we fixed $k$. We thus have shown that $\dfrac{2^{k}}{k}\in G$ for each $k\in\left\{ 1,2,\ldots,n\right\} $ satisfying $3^{m}\nmid k$. In other words, all addends in the sum $\sum\limits_{\substack{k\in\left\{ 1,2,\ldots ,n\right\} ;\\3^{m}\nmid k}}\dfrac{2^{k}}{k}$ belong to $G$. Hence, $\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k} }\dfrac{2^{k}}{k}$ is a finite sum of elements of $G$, and thus must be an element of $G$ itself (since any finite sum of elements of $G$ is an element of $G$). In other words, $\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k}}\dfrac{2^{k}}{k}\in G$. Hence, $a=\sum\limits_{\substack{k\in \left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k}}\dfrac{2^{k}}{k}\in G$. In other words, $v_{3}\left( a\right) >-m$ (by the definition of $G$). This proves Lemma 4 (a). (b) We are in one of the following two cases: Case 1: We have $2\cdot3^{m}\leq n$. Case 2: We have $2\cdot3^{m}>n$. Let us first consider Case 1. In this case, we have $2\cdot3^{m}\leq n$. We have defined $m$ to be the largest nonnegative integer such that $n\geq3^{m}$. Thus, $n\geq3^{m}$ but $n<3^{m+1}$. Hence, $3^{m}\leq n$ and $3^{m+1}>n$. Thus, $3^{m}\in\left\{ 1,2,\ldots,n\right\} $ (since $3^{m}\leq n$) and $2\cdot3^{m}\in\left\{ 1,2,\ldots,n\right\} $ (since $2\cdot3^{m}\leq n$), but $3\cdot3^{m}\notin\left\{ 1,2,\ldots,n\right\} $ (since $3\cdot 3^{m}=3^{m+1}>n$). Hence, the only multiples of $3^{m}$ that belong to the set $\left\{ 1,2,\ldots,n\right\} $ are $3^{m}$ and $2\cdot3^{m}$. In other words, the only $k\in\left\{ 1,2,\ldots,n\right\} $ that satisfy $3^{m}\mid k$ are $3^{m}$ and $2\cdot3^{m}$. But \begin{align*} b=\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\mid k} }\dfrac{2^{k}}{k}=\dfrac{2^{3^{m}}}{3^{m}}+\dfrac{2^{2\cdot3^{m}}}{2\cdot 3^{m}} \end{align*} (since the only $k\in\left\{ 1,2,\ldots,n\right\} $ that satisfy $3^{m}\mid k$ are $3^{m}$ and $2\cdot3^{m}$). Hence, \begin{align*} v_{3}\left( b\right) =v_{3}\left( \dfrac{2^{3^{m}}}{3^{m}}+\dfrac {2^{2\cdot3^{m}}}{2\cdot3^{m}}\right) =-v_{3}\left( 3^{m}\right) \end{align*} (by Lemma 3 (b), applied to $k=3^{m}$), since $3^{m}$ is odd. Thus, \begin{align*} v_{3}\left( b\right) =-\underbrace{v_{3}\left( 3^{m}\right) }_{=m}=-m. \end{align*} Hence, Lemma 4 (b) is proven in Case 1. Let us next consider Case 2. In this case, we have $2\cdot3^{m}>n$. Thus, $3^{m}\in\left\{ 1,2,\ldots,n\right\} $ (since $3^{m}\leq n$), but $2\cdot3^{m}\notin\left\{ 1,2,\ldots,n\right\} $ (since $2\cdot3^{m}>n$). Hence, the only multiple of $3^{m}$ that belongs to the set $\left\{ 1,2,\ldots,n\right\} $ is $3^{m}$. In other words, the only $k\in\left\{ 1,2,\ldots,n\right\} $ that satisfies $3^{m}\mid k$ is $3^{m}$. But \begin{align*} b=\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\mid k} }\dfrac{2^{k}}{k}=\dfrac{2^{3^{m}}}{3^{m}} \end{align*} (since the only $k\in\left\{ 1,2,\ldots,n\right\} $ that satisfies $3^{m}\mid k$ is $3^{m}$). Hence, \begin{align*} v_{3}\left( b\right) =v_{3}\left( \dfrac{2^{3^{m}}}{3^{m}}\right) =-v_{3}\left( 3^{m}\right) \end{align*} (by Lemma 3 (a), applied to $k=3^{m}$). Thus, \begin{align*} v_{3}\left( b\right) =-\underbrace{v_{3}\left( 3^{m}\right) }_{=m}=-m. \end{align*} Hence, Lemma 4 (b) is proven in Case 2. We have now proven Lemma 4 (b) in both Cases 1 and 2. Hence, Lemma 4 (b) always holds. $\blacksquare$ Theorem 5. Let $n$ be a positive integer. Let $m$ be the largest nonnegative integer such that $n\geq3^{m}$. Then, \begin{align*} v_{3}\left( \sum\limits_{k=1}^{n}\dfrac{2^{k}}{k}\right) =-m. \end{align*} Proof of Theorem 5. Define $a$ and $b$ as in Lemma 4. Then, Lemma 4 (a) yields $v_{3}\left( a\right) >-m$, but Lemma 4 (b) yields $v_{3}\left( b\right) =-m$. Thus, $v_{3}\left( a\right) >-m=v_{3}\left( b\right) $. Hence, Proposition 1 (e) (applied to $p=3$) yields $v_{3}\left( a+b\right) =v_{3}\left( b\right) =-m$. But each $k\in\left\{ 1,2,\ldots,n\right\} $ satisfies either $3^{m}\nmid k$ or $3^{m}\mid k$ (but not both at the same time). Hence, we can split the sum $\sum\limits_{k=1}^{n}\dfrac{2^{k}}{k}$ as follows: \begin{align*} \sum\limits_{k=1}^{n}\dfrac{2^{k}}{k}=\underbrace{\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k}}\dfrac{2^{k}}{k}} _{\substack{=a\\\text{(by the definition of }a\text{)}}}+\underbrace{\sum\limits _{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\mid k}}\dfrac{2^{k} }{k}}_{\substack{=b\\\text{(by the definition of }b\text{)}}}=a+b. \end{align*} Therefore, \begin{align*} v_{3}\left( \sum\limits_{k=1}^{n}\dfrac{2^{k}}{k}\right) =v_{3}\left( a+b\right) =-m. \end{align*} This proves Theorem 5. $\blacksquare$ Corollary 6. Let $n\in\mathbb{N}$. Then: (a) We have \begin{align*} v_{3}\left( \sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}\right) \leq0. \end{align*} (b) Assume that $\sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}=\dfrac{s}{t}$ for two coprime integers $s$ and $t$ satisfying $t\neq0$. Then, $3\nmid s$. Proof of Corollary 6. (a) Let $m$ be the largest nonnegative integer such that $n+1\geq3^{m}$. Then, $n+1\geq3^{m}$ but $n+1<3^{m+1}$. If we had $3^{m+1}\mid n+1$, then we would have $3^{m+1}\leq n+1$ (since $3^{m+1}$ and $n+1$ are positive integers), which would contradict $n+1<3^{m+1}$. Hence, we do not have $3^{m+1}\mid n+1$. Theorem 1 (d) (applied to $3$, $m+1$ and $n+1$ instead of $p$, $i$ and $n$) yields the equivalence $\left( 3^{m+1}\mid n+1\right) \ \Longleftrightarrow\ \left( v_{3}\left( n+1\right) \geq m+1\right) $. Hence, we do not have $v_{3}\left( n+1\right) \geq m+1$ (since we do not have $3^{m+1}\mid n+1$). In other words, we have $v_{3}\left( n+1\right) TITLE: Cryptography and elliptic curves QUESTION [20 upvotes]: Cryptography sometimes uses elliptic curves over finite fields. Does cryptography also use elliptic curves over $\mathbb{Q}$ or rational points on them? REPLY [31 votes]: Not directly, as far as I know, since explicitly computing large multiples of points in $E(\mathbb Q)$ is infeasible. However, people have considered lifting points from $E(\mathbb F_p)$ to $E(\mathbb Q)$ or to the $p$-adics $E(\mathbb Q_p)$ in order to devise algorithms to solve the discrete log problem in $E(\mathbb F_p)$ (although, unsuccessfully so far). Here are a few papers to get you started: Elliptic curve discrete logarithms and the index calculus. Advances in cryptology—ASIACRYPT'98 (Beijing), 110–125, Lecture Notes in Comput. Sci., 1514, Springer, Berlin, 1998 The xedni calculus and the elliptic curve discrete logarithm problem. Des. Codes Cryptogr. 20 (2000), no. 1, 5–40; Analysis of the xedni calculus attack. Des. Codes Cryptogr. 20 (2000), no. 1, 41–64. Lifting and elliptic curve discrete logarithms, Selected Areas of Cryptography (SAC 2008), Lecture Notes in Computer Science 5381, Springer-Verlag, Berlin, 2009, 82--102.<|endoftext|> TITLE: Extending Kan fibrations, without using minimal fibrations QUESTION [14 upvotes]: $\require{AMScd}$One thing that needs to be checked to give an interpretation of type theory in simplicial sets (as in Kapulkin-Lumsdaine) is that "the base of the universal fibration is fibrant". Expanding the definitions reduces this to the following statement. $(*)$ Given a Kan fibration $p : X \to A$ and an acyclic cofibration $i : A \to B$, there exists a square $$ \begin{CD} X @>j>> Y \\ @V p VV @VV q V \\ A @>>i> B \end{CD} $$ in which $q$ is a Kan fibration and the square is a pullback square. It suffices to handle the case when $i : A \to B$ is one of the generating acyclic cofibrations $\Lambda^n_k \to \Delta^n$. I have seen only one proof of $(*)$, which relies on the theory of minimal fibrations. I'm wondering about alternative approaches, so my main question is Are there proofs of $(*)$ which do not use minimal fibrations? In particular, a probably related question is whether, for fixed $i : A \to B$, the construction in $(*)$ can be carried out functorially in $X$ in an appropriate sense. Observe that in a square such as in $(*)$, the map $j$ is automatically an acyclic cofibration (it is a monomorphism, and a weak equivalence because $\mathrm{SSet}$ is right proper). So $X \xrightarrow{j} Y \xrightarrow{q} B$ is an acyclic cofibration-fibration factorization of $ip : X \to B$, but with the added constraint that $j : X \to Y$ can only attach simplices which live above $B \setminus A$. Perhaps it is possible to modify a known method for constructing factorizations so that it adheres to this additional constraint? REPLY [2 votes]: Despite having left academia, I still work on this problem from time to time, and I (hope I) recently worked out a combinatorial proof for the base case of extension along a horn inclusion: https://github.com/woupiestek/really-hot/blob/master/fibrant-universes.tex I want these constructions to work for realizability toposes and related categories--in particular for the category of simplicial assemblies. In that case, monics aren't necessarily cofibrations as in Karol's paper and horns aren't necessarily $\kappa$-representable for any classical cardinality as in Mike's paper, hence the complicated workaround. The category of assemblies has a full internal subcategory--the modest sets--and I believe that the bundle of 'modest Kan complexes' is an interesting model for homotopy type theory--one that could serve as a counterexample to many Grothendieck-topos based intuitions. Four years ago, there was too little work in constructive simplicial homotopy for me to build on. It is good to see that much progress has been made in this area.<|endoftext|> TITLE: Peculiarities in low dimensions or low order or etc QUESTION [5 upvotes]: I have been pondering about certain conjectures and theorems viewed as either low vs high dimensions, or smaller vs larger primes, or anything of the sort "low vs high order". Let me mention a couple of such mathematical phenomena that might be more familiar. Poincaré's conjecture (now a theorem) in dimension $3$ persisted much longer than in higher dimensions. Congruence modulo primes for the partition function $p(n)$ lingers for primes $p=2, 3$ while a recent work on Maass forms settles such for higher primes. Hoping that these citations shed light, I like to ask: QUESTION. Do you know of conjectures (problems) which manifested to be either notoriously harder or unsolved for "lower dimensions/orders/primes" compared to their "higher dimensional/order/prime" cousins? REPLY [3 votes]: All irreducible spherical buildings (i.e. with finite Weyl group) of rank greater than 2 are associated to simple algebraic or classical groups. This is not the case for rank $\leq$ 2. See https://en.wikipedia.org/wiki/Building_(mathematics)<|endoftext|> TITLE: E[X|Y] and E[Y|X] QUESTION [7 upvotes]: Suppose $x, y$ are random variables jointly distributed on $[0,1]^2$. The marginal distribution of $x$ is uniform. It is also known that $E[y]=E[x]=\frac12$ and $E[x|y]=y$, so $y$ second-order stochastically dominates $x$. We also know that $E[y|x]$ is non-decreasing in $x$ (not sure whether this is helpful.) I am trying to further characterize $E[y|x]$. It seems that the following might be true for any $c\in[0,1]$ but I have no idea how to prove it: $\int_0^c E[y|x]dx\ge \int_0^c xdx$. Imagine if $x$ and $y$ are independent, then $LHS=\frac12c\ge\frac12 c^2=RHS$. Imagine if $y$ reveals $x$ completely, then $E[y|x]=E[E[x|y]|x]=x$. Moreover, the inequality seems to be true if $y$ is a function of $x$. REPLY [5 votes]: Your conjecture is true. Indeed, take any $c\in[0,1]$. Then $$\int_0^c E(y|x)\,dx=Ey1_{x\le c}\quad\text{and}\quad Ex1_{x\le c}=\int_0^c x\,dx=c^2/2, $$ because $x$ is uniform on $[0,1]$. Next, $$E(x-y)1_{y\le c}=0, $$ because $E(x|y)=y$. Also (which is the key point), $$(y-x)(1_{x\le c}-1_{y\le c})\ge0. $$ So, \begin{align} &\int_0^c E(y|x)\,dx-c^2/2+0 \\ &=Ey1_{x\le c}-Ex1_{x\le c}+E(y-x)1_{y\le c} \\ &=E[(y-x)(1_{x\le c}-1_{y\le c})]\ge0, \end{align} whence $$\int_0^c E(y|x)\,dx\ge c^2/2, $$ as claimed.<|endoftext|> TITLE: Teaching cohomology via everyday examples QUESTION [15 upvotes]: This question is a "sequel" to my similar questions about the fundamental group and homology. All of these questions were inspired by seeing a talk, by Tadashi Tokieda, about the interesting physics that appears in toys. Short version: What stories, puzzles, games, paradoxes, toys, etc from everyday life are better understood after learning about cohomology? Long version: I am teaching a short course on cohomology, from chapter three of Hatcher's book. I would like to present a collection of real-life phenomena that are greatly illuminated by actually knowing about cohomology. Ideally, I would refer back to these examples as the course progressed and explain them with the new tools the students learn. There are some interesting examples, best explained via cohomology, given as answers to my previous questions. These include: Impossible objects such as the Penrose tribar that exist locally, but not globally. (I would be very very happy to get a much longer, more in-depth reference.) Currency arbitrage. Addition with carrying. Here is a non-example: The belt trick; this relies on the fundamental group, not on cohomology. And here is one example firmly on the border: Kirchhoff's laws for electrical circuits. Now, these are cohomology made flesh, but they are not quite an "everyday" example... REPLY [5 votes]: A very similar question was already asked on this site: Teaching homology via everyday examples. One of the answers mentioned the book Robert Ghrist, Elementary applied topology which has a lot of everyday examples of cohomology (and many other everyday examples).<|endoftext|> TITLE: Polynomials (or analytic functions) vanishing on a real algebraic set QUESTION [5 upvotes]: I have seen the following result stated several times in the literature, without proof: Let $\mathbb{K}$ be $\mathbb{R}$ or $\mathbb{C}$, and assume $P\in\mathbb{K}[X_{1},\ldots,X_{n}]$ is an irreducible polynomial of $n$ variables, and let $$ V = \{z\in\mathbb{C}^{n},~P(z)=0\},\quad I(V)=\{Q\in\mathbb{C}[X_{1},\ldots,X_{n}],~Q=0\text{ on }V\}.$$ Assume that $V$ contains a real point $a\in\mathbb{R}^{n}$ which is regular, that is $$\text{dim }V=n-\text{rank }J_{a}(V),$$ where $J_{a}(V)$ denotes the Jacobian of a family of generators of $I(V)$, evaluated at $a$. Then the set $V_{\mathbb{R}}=V\cap\mathbb{R}^{n}$ of real points of $V$ is Zariski dense in $V$, or equivalently any polynomial that vanishes on $V_{\mathbb{R}}$ must vanish on $V$. I am also interested by the analytic version of this result (if true) where $V$ is still the zero set of a polynomial $P$ but $I(V)$ is the ideal of analytic functions vanishing on $V$, and the vanishing of an analytic function $f$ on $V_{\mathbb{R}}$ implies the vanishing of $f$ on $V$. Could someone provide a proof for the algebraic or analytic case? Many thanks in advance. REPLY [6 votes]: The more general statement that is true is the following: Let $X \subset \mathbb{C}^d$ be an irreducible affine variety defined by real polynomials. If $X$ has a smooth real point, then $X(\mathbb{R})$ is Zariski dense in $X.$ A sketch of the proof and some good examples are given here: Real Algebraic Geometry for Geometric Constraints by Frank Sottile (page 8). The main fact used is, under the assumption there is a real smooth point, the smooth points of the $\mathbb{R}$-locus is a real manifold of the same dimension. The main reference is: Real Algebraic Geometry by Bochnak, Coste, Roy. Fun facts: If you ask instead for the density of $\overline{\mathbb{Q}}$-points then that is true more generally for any $d$–dimensional affine variety $V$ defined over $\mathbb{Q}$ by Noether Normalization. If you remove the assumption that there is a smooth point, then the statement is false as $V(x^2+y^2)$ has only one $\mathbb{R}$-point which is not smooth, namely $(0,0)$, and $x^2+y^2$ is irreducible over $\mathbb{R}$ and the $\mathbb{C}$-locus is dimension 1 (so the $\mathbb{R}$-locus not Zariski dense).<|endoftext|> TITLE: Examples of smooth Hurewicz fibrations which are not smooth fiber bundles QUESTION [7 upvotes]: In the category of smooth manifolds (without corners), what are some examples of Hurewicz fibrations which are not fiber bundles? The minimal topological example I know is to project the standard 2-simplex onto the $x$-axis. The right-most fiber degenerates into a point while the others are homeomorphic to a closed interval. I don't understand how to produce such a "dimensional degeneration" phenomenon in the smooth world. In fact this seems sort of impossible to me: a Hurewicz fibration is a submersion and the vertical bundle of a submersion has locally constant rank, so the fibers are homotopy equivalent equidimensional embedded submanifolds which foliate the source (assume the base is connected). I don't see what other kind of degeneration (other than dimensional) might preclude a fibration from being a fiber bundle. Out of helplessness, since fibrations are submersions, I was tempted to examine submersions which are not fiber bundles. The classical example $\mathbb R^2\to \mathbb R,\; (x,y)\mapsto(x^2-1)e^y$ is not a Hurewicz fibration because fibers of negative numbers are connected while those of non-negative numbers are disconnected. I don't know any other examples. REPLY [5 votes]: Some poking around led to an example in G. Meigniez, Submersions, fibrations, and bundles, Trans. Amer. Math. Soc. 354 (2002), 3771-3787. It's Example 21 in that paper and described briefly as follows: Let $W\subset\mathbb R^3$ be the Whitehead manifold, an open, contractible subset not homeomorphic to $\mathbb R^3$. Let $E$ be the set of $(x,y,z,t)\in\mathbb R^4$ with $(x,y,z)\in W$ or $t \neq 0$. Then the projection $\pi\colon E\to \mathbb R$, $\pi(x,y,z,t) = t$, is a smooth submersion, a fibration (with contractible fibers), but not locally trivial because the fiber over $0$ is not homeomorphic to the others. That paper also refers to other counterexamples given in S. Ferry, Alexander duality and Hurewicz fibrations, Trans. Amer. Math Soc. 327 (1991), 201-219.<|endoftext|> TITLE: What are the possible linking matrices of a quasi-positive link? QUESTION [10 upvotes]: I was surprised recently to come across a 3-component link where the linking number of two of the components was negative. For a while I thought I had made a mistake, then I thought a little more and realized that there was no particular reason that the linking numbers had to be all positive. The linking matrix I found was $$\begin{pmatrix} * & 3 & 3 \\ 3 & * & -1 \\ 3 & -1 & * \end{pmatrix}. $$ Here is the link. (I haven't attempted to find a quasi-positive representative yet, the proof it's quasi-positive was pretty indirect.) REPLY [2 votes]: Any linking matrix appears as linking matrix of a strongly quasipositive link. This follows from a result of Rudolph ("Constructions of Quasipositive Knots and Links, I") that any Seifert matrix appears as Seifert matrix of a strongly quasipositive link.<|endoftext|> TITLE: For a fixed dominant weight $\lambda$, are almost all dominant weights in the same coset above it? QUESTION [5 upvotes]: First some notation as in e.g. the book by Humphreys on Lie Algebras. Let $E$ be an Euclidean space with inner product $(-,-)$, and denote $\langle v,w \rangle = \frac{2(v,w)}{(w,w)}$. Let $\Phi$ be an irreducible root system on $E$, so $\langle \beta, \alpha \rangle \in \mathbb{Z}$ and $\alpha - \langle \beta, \alpha \rangle \beta \in \Phi$ for all $\alpha, \beta \in \Phi$. Fix a set of simple roots $\alpha_1$, $\ldots$, $\alpha_l$. Let $\Lambda$ be the set of weights, i.e. the set of $\lambda \in E$ such that $\langle \lambda, \alpha \rangle \in \mathbb{Z}$ for all $\alpha \in \Phi$. Let $\Lambda^+$ be the set of dominant weights, that is, the set of $\lambda \in \Lambda$ such that $\langle \lambda, \alpha_i \rangle \geq 0$ for all $i$. We have the usual partial order on $\Lambda$, by defining $\mu \preceq \lambda$ iff $\lambda - \mu = \sum_{i = 1}^k k_i \alpha_i$ for some integers $k_i \in \mathbb{Z}_{\geq 0}$. It is well known that for a fixed $\lambda \in \Lambda^+$, there are only finitely many $\mu \in \Lambda^+$ such that $\mu \preceq \lambda$ (See for example 13.2 in Humphreys). But is the following true? Fix $\lambda \in \Lambda^+$. Then for all but finitely many $\mu \in \Lambda^+$ with $\lambda - \mu \in \mathbb{Z}\Phi$, we have $\lambda \preceq \mu$. If the answer is yes, this could be used to give a different solution to a previous question asked here: link. REPLY [3 votes]: Here's basically the same answer as Mikko Korhonen, but written in a way that's slightly easier for me to understand. Let me use $\leq$ to denote the partial order on all of the vector space $E$ with $u \leq v$ for $u,v \in E$ if and only if $v-u = \sum_{i=1}^{n}a_i \alpha_i$ with all $a_i \geq 0$ (but not necessarily integral). Considering $\leq$ instead of $\preceq$ means I don't have to deal with the different classes of the root lattice mod the weight lattice. Fix any $v \in E$. We claim there are only finitely many dominant weights $\mu \in \Lambda^{+}$ for which we don't have $v \leq \mu$, from which the desired claim obviously follows. Indeed, let $\omega$ be a fundamental weight. As Mikko explains (and as is also fundamental to the answer I gave to the linked question), we have $\omega = \sum_{i=1}^{n} a_i \alpha_i$ with all $a_i > 0$. Hence clearly there is some $m \geq 0$ so that $v \leq m\omega$. Now let $M$ be the maximum over all fundamental weights of such $m$. Then writing $\mu = \sum_{i=1}^{n} c_i \omega_i$, the only way we could fail to have $v \leq \mu$ is if $c_i < M$ for all $i$. (Here we are using the fact that if $v \leq u$ and $\nu \in \Lambda^{+}$, then $v \leq u+\nu$, which can be seen again from the fact that writing $\omega = \sum_{i=1}^{n} a_i \alpha_i$ for any fundamental weight $\omega$, we have $a_i > 0$.) There are clearly only finitely many $\mu=\sum_{i=1}^{n} c_i \omega_i \in \Lambda^{+}$ with $c_i < M$ for all $i$. EDIT: Here is an even more general statement/context. Let $V$ be an n-dimensional Euclidean vector space with inner product $\langle \cdot, \cdot \rangle$ and let $v_1,\ldots,v_n \in V$ be a collection of vectors such that: $v_1,\ldots,v_n$ form a basis of $V$; $\langle v_i , v_j \rangle \leq 0$ for all $i \neq j$ (in other words, the vectors are pairwise non-acute); there is no nontrivial decomposition of $V = V_1 \oplus V_2$ into orthogonal subspaces $V_1$ and $V_2$ such that $v_i \in V_1 \cup V_2$ for all $i$ (this is an irreducibility condition- equivalently it says that if we draw the graph on the $v_i$ with $v_i$ adjacent to $v_j$ if $\langle v_i, v_j\rangle < 0$ then that graph will be connected). Then let $Q_{\geq 0} := \{ v=\sum_{i=1}^{n}a_iv_i, a_i \geq 0\}$ be the cone generated by the $v_i$. And let $P_{\geq 0} := \{v \in V\colon \langle v,w\rangle \geq 0 \textrm{ for all $w\in Q_{\geq 0}$}\}$ be the dual cone to $Q_{\geq 0}$. Then the claim is that for any $v\in V$ we have that $P_{\geq 0} \setminus (v+Q_{\geq 0})$ is a bounded subset of $V$. To prove this, observe that any nonzero $w \in P_{\geq 0}$ (in particular, any generator of this cone) has all $a_i > 0$ when we write $w=\sum_{i=1}^{n} a_i v_i$. The reason for this is that the matrix $M=(\langle v_i, v_j \rangle)$ is a nonsingular, irreducible $M$-matrix, and hence $M^{-1}$ (which expresses the coordinates of the generators of $P_{\geq 0}$) is a matrix with all entries strictly positive (see e.g. Theorem A of https://core.ac.uk/download/pdf/82640451.pdf). Then we can apply the same argument as above to the generators of the cone $P_{\geq 0}$. (The particular situation above corresponds to the $v_i$ being the simple roots and the cone $P_{\geq 0}$ being the dominant cone, i.e., cone spanned by the fundamental weights.)<|endoftext|> TITLE: Serre's formula for $\Delta^{1/3}$ QUESTION [8 upvotes]: Let $$E\quad\colon \quad y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ be an elliptic curve, and let $b_4 = a_1a_3+2a_4$. Serre in Propriétés galoisiennes des points d'ordre fini des courbes elliptiques (Section 5.3, b), p. 305) gives the following formula $$\Delta^{1/3}=b_4-3(x_i x_j+x_kx_l),\label{1}\tag{1}$$ where the $x_i$ are the $x$-coordinates of points of $E$ of order $3$, and $\{1,2,3,4\}=\{i,j\}\cup\{k,l\}$. This formula can be rewritten as \begin{equation*} j^{-1/3}=\frac{\frac{1}{2}\frac{27j}{j-12^3}+3(x_ix_j+x_kx_l)}{\frac{27j}{j-12^3}}.\label{2}\tag{2} \end{equation*} The relation \eqref{2} is a relation between modular forms, therefore its proof comes for free. Questions: Is there an algebraic reason why \eqref{1} should be true? Is there an algebraic proof of \eqref{1}? Are there similar formulas for $\Delta^{1/d}$ where $d$ is a divisor of $24$? Are there similar formulas for $\Delta^{1/d}$ if $d$ is not a divisor of $24$? What is the algebraic significance of the number $24$ in this context ($\Delta$ is the $24$-th power of the Dedekind eta function, but what does this mean algebraically)? REPLY [5 votes]: Regarding Q5, the modular form $\Delta$ has weight 12, so it is a section of the line bundle $\omega^{\otimes 12}$ on the modular curve $Y$. Here $\omega$ is the Hodge bundle, which can be defined as $\omega = \pi_* \Omega^1_{E/Y}$ where $\Omega^1_{E/Y}$ is the sheaf of relative Kähler differentials on the universal elliptic curve $\pi : E \to Y$ (note that here $Y=Y(1)$, so there is no universal elliptic curve, to get around this problem we must treat $Y(1)$ as an algebraic stack). If $d$ is a divisor of $12$ then $\Delta^{1/d} = \eta^{24/d}$ will be a cusp form of weight $12/d$ (I don't recall the precise level), so everything is algebraic. If you want to get to $\eta$ (in other words $d=24$) then you need half-integral weight modular forms. The interpretation of such modular forms uses the metaplectic cover of $\mathrm{SL}_2(\mathbb{R})$, which is not an algebraic group anymore, only a Lie group (see David Loeffler's answer for more details). The point is that the Hodge bundle $\omega$ is not the square of a line bundle, but it becomes a square once you pull back to the metaplectic group.<|endoftext|> TITLE: Where does the $\hat A$ class get its name? QUESTION [11 upvotes]: In K-theory we have the Todd class and the $\hat A$ class. The Todd class is named after the Cambridge geometer John Arthur Todd. Where does the name $\hat A$ come from? Does the A stand for Atiyah? REPLY [15 votes]: It seems that the $\hat{A}$ genus was first introduced in Section 23 of the paper "Characteristic Classes and Homogeneous Spaces, II", by Borel and Hirzebruch (1959). It is presented as a small modification of some $A$-genus previously introduced by Hirzebruch in his book "Neue Topologische Methoden in der Algebraischen Geometrie" (1956) (the $A$-genus is the genus associated to the power series $2z^{1/2}/sh(2z^{1/2})$, whereas the $\hat{A}$-genus is associated to the power series $(z^{1/2}/2)/sh(z^{1/2}/2)$). It is a resonable explanation of the hat on $\hat{A}$: first some $A$-genus was introduced, then some small modification $\hat{A}$, and it was only gradually realized that the $\hat{A}$-genus was more important that the $A$-genus (to the point that today I don't know if anyone still uses the terminology $A$-genus). It seems unclear if there is a good explanation for the $A$ in $A$-genus except that $A$ is a quite common letter. When Hirzebruch used the letter $A$ in his book published in 1956, I don't think that Atiyah was already thinking about these questions and so it seems very unlikely that $A$ stand for Atiyah.<|endoftext|> TITLE: Abstract proof that $\lvert H^2(G,A)\rvert$ counts group extensions QUESTION [8 upvotes]: (This question is originally from Math.SE, where it didn't receive any answers.) $\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\ext}{Ext} \newcommand{\Z}{\mathbb{Z}}$ Let $G$ be a group, let $A$ be a $G$-module, and let $P_3\to P_2\to P_1\to P_0\to\Z\to0$ be the start of a projective resolution of the $G$-module $\mathbb{Z}$. Consider the cohomology group $$H^2(G,A)=\frac{\ker(\Hom_{\Z G}(P_2,A)\to\Hom_{\Z G}(P_3,A))}{\im(\Hom_{\Z G}(P_1,A)\to\Hom_{\Z G}(P_2,A))}.$$ It can be shown that $\lvert H^2(G,A)\rvert$ counts the number of equivalence classes of group extensions $0\to A\to E\to G\to0$. The only proof that I know of this result involves choosing a specific projective resolution (namely, the bar resolution). Is there a proof of this result that does not require choosing a specific projective resolution? For context, $\lvert\ext_R^n(M,N)\rvert$ counts the number of equivalence classes of extensions $0\to N\to X_n\to\ldots\to X_1\to M\to0$. The proof of this result is fairly abstract and does not require picking a specific projective resolution of $M$ or a specific injective resolution of $N$. Also, I am aware that we actually have isomorphisms in both of these results but I am more interested in the existence of an explicit bijection. Here is one approach for constructing an element of $H^2(G,A)$ from an extension $0\to A\to E\to G\to0$: Treat $A$ as an $E$-module and consider the transgression map $H^1(A,A)^{E/A}\to H^2(E/A,A^A)$. Rewriting this gives a homomorphism $\Hom(A,A)^G\to H^2(G,A)$. The image of $\id_A$ under this map will be an element of $H^2(G,A)$. To make this work, this map would need to be a bijection from equivalence classes of group extensions and elements of $H^2(G,A)$. Another approach that I considered was to work directly with the arbitrary projective resolution (similar to the proof of the Yoneda Ext result). Suppose that we are given a group extension $0\to A\to E\to G\to0$. We want to construct an element of $\ker(\Hom_{\Z G}(P_2,A)\to\Hom_{\Z G}(P_3,A))$. Equivalently, we want to construct a $\Z G$-module homomorphism $P_2/\im(P_3\to P_2)\to A$. However, $\im(P_3\to P_2)=\ker(P_2\to P_1)$ and $P_2/\ker(P_2\to P_1)\cong\im(P_2\to P_1)=\ker(P_1\to P_0)$. Thus, we want to construct a $\Z G$-module homomorphism $f\colon\ker(P_1\to P_0)\to A$. Furthermore, if we unwind some more definitions, we see that we only need to construct $f$ up to the restriction of a $\Z G$-module homomorphism $P_1\to A$. Unfortunately, the only information we have about $A$ is the short exact sequence $0\to A\to E\to G\to0$ which makes it hard to define a $\Z G$-module homomorphism to $A$. REPLY [15 votes]: Here is a simple way. The extension $A \to E \to G$ induces a map of classifying spaces $BA\to BE \to BG$, which is a principal fibration, so classified by (homotopy class of) a map $BG \to BBA=K(A,2)$, i.e., an element of $H^2(BG,A)$.<|endoftext|> TITLE: Why are Thompson's groups called $F$, $T$ and $V$? QUESTION [17 upvotes]: Why are Thompson's groups called $F$, $T$ and $V$? I never saw Thompson's unpublished notes, in which he introduces these groups; maybe an explanation can be found there? REPLY [5 votes]: Matt Zaremsky's justification of "$F$" for "free" can also be found in Ross Geoghegan's review MR1239554 on Mathscinet. I also took a look at McKenzie and Thompson's article An elementary construction of unsolvable word problems in group theory (MSN) (which seems to be the first publication mentioning the groups constructed by Thompson), and there is some history in §8. Below is a screenshot: The groups $\mathfrak{P}$ and $\mathfrak{P}'$ correspond to the group now called $F$. From the context, $\mathfrak{P}$ probably refers to the "p" of "permutation". The group $\mathfrak{C}'$ corresponds to the group now called $V$. I would guess that $\mathfrak{C}$ refers to the "C" of "Cantor space".<|endoftext|> TITLE: On $\{P(x)+Q(y):\ x,y=0,\ldots,p-1\}$ with $p$ prime QUESTION [7 upvotes]: QUESTION: Is my following conjecture (formulated in 2016) true? How to solve it? Conjecture. For any non-constant polynomials $P(x)，Q(x)\in\mathbb Z[x]$, there is a positive integer $N(P,Q)$ depending on $P(x)$ and $Q(x)$ such that for any prime $p>N(P,Q)$ the set $$\{P(x)+Q(y):\ x,y=0,1,\ldots,p-1\}$$ contains a complete system of residues modulo $p$. For example, it seems that we may take $$N(x^4,x^2)=5,\ N(x^5,x^2)=11,\ N(x^3,x^3)=7,\ N(x^6,x^3)=31.$$ REPLY [6 votes]: The conjecture is true when the degrees of $P(x)$ and $Q(y)$ are relatively prime. For general polyomials $F(x,y)$ in two variables, a variant of the conjecture is true. Theorem 1. Let $P(x)\in\mathbb{Z}[x]$ and $Q(y)\in\mathbb{Z}[y]$ be two polynomials of relatively prime degrees. For any prime $p$, and for all residues $r$ modulo $p$, the congruence $P(x)+Q(y)\equiv r\pmod p$ has $p+O(p^{1/2})$ solutions. The implied constant depends only on $P(x)$ and $Q(y)$. Sketch of proof. Without loss of generality, the prime $p$ exceeds the leading coefficients of $P(x)$ and $Q(y)$. Then, for all residues $r$ modulo $p$, the polynomial $P(x)+Q(y)-r$ is absolutely irreducible over $\mathbb{F}_p$, hence the congruence $P(x)+Q(y)\equiv r\pmod p$ has $p+O(p^{1/2})$ solutions by the celebrated theorem of Weil (1940). For more details on the quoted results, see Theorems 1A and 1B in Section III.1 of Schmidt: Equations over finite fields (Springer, 1976). Theorem 2. Let $F(x,y)\in\mathbb{Q}[x,y]$ be an absolutely irreducible polynomial. There exists a positive integer $N(F)$ depending on $F(x,y)$ such that for any prime $p>N(F)$, and for all but $O(1)$ residues $r$ modulo $p$, the congruence $F(x,y)\equiv r\pmod p$ has $p+O(p^{1/2})$ solutions. The implied constants depend only on the degree of $F(x,y)$. Sketch of proof. It follows from a theorem of Noether (1922) that there exists a positive integer $N(F)$ depending on $F(x,y)$ such that for any prime $p>N(F)$, and for all but $O(1)$ residues $r$ modulo $p$, the polinomial $F(x,y)-r$ is absolutely irreducible over $\mathbb{F}_p$. From here we finish as in the previous proof. For more details on Noether's theorem, see Theorem 2A in Section V.2 of Schmidt: Equations over finite fields (Springer, 1976).<|endoftext|> TITLE: Vanishing of L-function of elliptic curve over $\mathbb{Q}$ QUESTION [5 upvotes]: For an elliptic curve $E$ over $\mathbb{Q}$, it is not very difficult to show $L(E,1)\not=0$ (when the analytic rank$=0$) by computing the several Fourier coefficients but seem to be difficult to determine whether one has $L(E,1)=0$ (when the analytic rank$\not=0$). Is there a small constant $c$ such that, if we have $\mid L(E,1)\mid TITLE: Taylor-like expansion for a holomorphic function in non-simply-connected domain QUESTION [5 upvotes]: Suppose $f$ is a holomorphic function in a simply connected open set $U$, and we know it's Taylor expansion at a point $p\in U$. We can then find a holomorphic map $g$ of $U$ to the unit disc which sends $p$ to 0, and obtain Taylor series of $f(g^{-1}(z))$ near $z=0$. Mapping back to $U$ this then yields an expansion of $f$ in a series of functions, which is convergent in all of $U$. It is essential that if we knew only first $N$ terms if the Taylor expansion, we can still obtain first $N$ terms of the final expansion. Can we find a similar expansion for $f$ if $U$ is not simply-connected? I.e. write $f(z)=\sum_n a_n h_n(z)$, convergent in all of $U$, with $h_n$ single valued in $U$, and $a_n$ being a linear combination of first $n$ terms of Taylor expansion of $f$ at $p$? Of course, the functions $h_n$ should not depend on choice of $f$. REPLY [3 votes]: OK, here goes the "annulus". Let's say that a domain $\Omega$ has a good approximation property at a point $w\in\Omega$ if for every $\rho>1$ there is a compact set $K=K(\rho)\subset \Omega$ and a constant $C=C(\rho)>0$ such that for every function $f$ analytic in $\Omega$ and every $m\ge 0$ there is a function $g$ analytic in $\Omega$ such that $g(z)-f(z)=O(|z-w|^{m+1})$ as $z\to w$ and $\sup_\Omega|g|\le C(\rho)\rho^m\sup_K|f|$. Claim 1: The unit disk has a good approximation property at $0$. Proof: Just let $K$ be the circle of radius $r\in(\rho^{-1},1)$ and take for $g$ the Taylor polynomial of $f$ of degree $m$. Claim 2: Good approximation property is a conformal invariant on the Riemann sphere (with the understanding that if $w=\infty$, then the condition $g(z)-f(z)=O(|z-w|^{m+1})$ is replaced with $g(z)-f(z)=O(|z|^{-m-1})$, of course). Proof: Obvious (map everything). Claim 3: The Riemann sphere $\widehat{\mathbb C}$ with finitely many (reasonable) cuts $\Gamma_j\subset \mathbb C$ has good approximation property at $\infty$ (I assume $\Gamma_j$ are reasonable so that they are compact, pairwise disjoint, and $\widehat{\mathbb C}\setminus\Gamma_j$ is conformally equivalent to the unit disk for every $j$) Proof: This is just a version of the simplest lemma on the separation of singularities. Fix $\rho>1$. First, each $\widehat{\mathbb C}\setminus\Gamma_j$ has good approximation property at $\infty$, so we can surround $\Gamma_j$ by some very close to it smooth contour $C_j$ that has the property of the compact $K$ in the definition of the good approximation property for $\widehat{\mathbb C}\setminus\Gamma_j$. Next, the function $f$ can be written as the sum $f(\infty)+\sum_j f_j$ where $f_j$ is the (clockwise) Cauchy integral of $f$ on any contour surrounding $\Gamma_j$ and having the evaluation point and other $\Gamma_k$ outside it. We can now readily estimate $f_j$ on $C_j$ by $\max_{C_j}|f|+|f(\infty)|+\sum_{k\ne j}\frac{\ell(C_k)}{2\pi\operatorname{dist}(C_j,C_k)}\max_{C_k}|f|$ by the triangle inequality, which yields $\max_{C_j}|f_j|\le C\max_K|f|$ where $K=\cup_j C_j$. Now, given $m$, just find an appropriate $g_j$ for each $f_j$ separately and put $g=\sum_j g_j$. Assume now that $\Omega$ is a bounded domain with good approximation property at $0$ (we can always map the Riemann sphere with cuts and a fixed point conformally to this configuration). Let $H$ be the Hilbert space of functions analytic in $\Omega$ and square integrable with respect to the area measure (you can also introduce some reasonable weight, if you feel like it). Consider the subspaces $H_m=z^mH$ (those are just subspaces of functions in $H$ vanishing to order $m$ at the origin). Clearly, we have $H=H_0\supset H_1\supset H_2\supset\dots$ and each $H_{m+1}$ is a closed subspace of $H_m$ of codimension $1$. Let $h_m$ be the function in $H_m$ orthogonal to $H_{m+1}$ of unit norm. Then, since $\cap_m H_m$ consists of analytic functions vanishing at $0$ with all derivatives, i.e., of just $0$, the functions $h_m$ form an orthonormal basis in $H$, so every function in $H$ can be written uniquely as its Fourier series in $h_m$, which converges in $H$ and, thereby, uniformly on compact subsets of $\Omega$ with all derivatives. Notice now that the coefficient at $h_0$ is uniquely determined by $f(0)$ (all other terms vanish at $0$), the coefficient at $h_1$ is thus determined by $f(0)$ and $f'(0)$, and so on. This allows to write a formal Fourier decomposition into $h_m$ for any formal Taylor series at $0$. The only task is to show that the resulting series converges uniformly on compact sets in $\Omega$ if that Taylor series represents a function analytic in $\Omega$. Note that due to the high order of $0$ of $h_m$ at the origin and the uniform bound for the $L^2$ norm with respect to the area measure, for every compact set $Q\subset\Omega$, there exists $B=B(Q)>0$ and $r=r(Q)<1$ such that $\max_Q|h_m|\le Br^m$. Now choose $\rho>1$ such that $\rho r<1$ and use the good approximation property. It yields a compact $K\subset\Omega$ and, for every $m$, a function $g$ that is bounded by $C\rho^m\max_K|f|$ and has the same first $m$ derivatives as $f$ at $0$. Then the formal Fourier coefficient of $f$ at $h_m$ is the same as of $g$, but $g\in H$ with norm bounded by the square root of the area of $\Omega$ times the uniform norm, so its Fourier coefficient is bounded by $C\rho^m$. This immediately implies the geometric convergence of the formal Fourier series of $f$ on $Q$. Now it remains to note that $Q$ was arbitrary and that the sum of the formal Fourier series of $f$ (which we now know to converge) has the same derivatives at $0$ as $f$, so it must converge to $f$ itself. Formally this doesn't work for general domains, but I still wonder if one can find a counterexample to this approach. Of course, one needs to replace the area measure by some weight very fast decreasing near the boundary, but it is not at all clear to me what the convergence properties of the corresponding formal Fourier series will be. The simplest case to consider is the punctured unit disk. There is no hope to get the good approximation property there (no matter what reasonable norm of $g$ you want to bound geometrically), so the last trick that yields a good bound for Fourier coefficients out of nothing won't work anymore, but can we still do something? Or can somebody come up with a completely different idea?<|endoftext|> TITLE: A paper seems to have completely disappeared from editor's hands QUESTION [14 upvotes]: I have submitted a paper to a journal on June 2017. The corresponding author is my coauthor. The first response of the editor was after a year, June 2018: Your paper has been sent, consecutively, to four referees. Of the preceding three, one declined to review it and two never responded to the invitation or to multiple reminders. The fourth one accepted; a report is expected in September [2018]. Since then, the editor disappeared. No notice at all, neither in the positive, nor in the negative. Obviously we tried to contact him again multiple times (approximately every two-three months) through the journal's platform. No response at all. We sent countless emails to the editor, either via the platform or through his personal email (my coauthor knows him in person). No response at all. We tried to contact the chief editor. No response at all. In a few days, it will be exactly two years since the submission: this is an incredible amount of time, especially since we absolutely don't know what is the motivation for this delay. What shall we do? [I'm not posting this on academia, since this is a paper in Mathematics; but feel free to migrate the discussion elsewhere if you feel like so] Unfortunately I am unable to comment as I lack sufficient reputation. Yes, "consecutively" means exactly that 3 people in a row refused to referee the paper; the fourth accepted and then disappeared since June 2018. The editor-in-chief should be aware of what's going on, more so because we wrote him an email two weeks ago or so. No answer. The journal is a pretty good and reputable one. At least until I spread this voice. My coauthor has nothing against me if I reveal the name of the journal: it's JoA https://www.journals.elsevier.com/journal-of-algebra REPLY [9 votes]: Here are some additional things to consider: Sometimes email addresses are blocked, for one reason or another. One of my coauthors had his institution block all hotmail addresses, which led to some trouble in our communications. In your post you mention "countless" emails; this can cause certain servers to start blocking emails. Conversely, their emails to you might be blocked. In your correspondence you can mention that you haven't heard anything, and perhaps give them your phone number. Sometimes people are away from computers for multiple weeks. Then it takes a while to contact the referee. My personal rule of thumb is to contact an editor and if I receive no reply then I follow up a week later. I repeat this one more time. After 3 emails (and 3 weeks of waiting), I believe it is then time to contact the editor-in-chief. During this process you might try to use alternate email addresses which haven't been potentially blocked. You do not want to pester an editor with "countless" emails. Some of the issues above can be avoided if the journal in question has its own way to contact editors, which it appears you already tried to use. Nobody should wait multiple months between trying to contact an editor, if they are receiving no responses. That is unreasonably long. This noncommunication issue should have been dealt with near the date the editor gave for the report to be done. Given the time-frame, if you can't get a response from the editor-in-chief through the methods people gave in the comments, I would send one final (kindly worded!) email saying that if you don't hear from them within two weeks, then your paper should be considered as withdrawn.<|endoftext|> TITLE: How does Otto theory work in this example of Wasserstein a.c. curve of probabilities? QUESTION [6 upvotes]: I'm trying to read chapter 8 of the book on gradient flows by Ambrosio-Gigli-Savaré. In this context, I would like to better understand how the theory works for the following specific example. Take the family of probability measures on the real line $$\mu_t=(1-t)\delta_0+t\delta_1, \quad t\in(0,1),$$ where $\delta_x$ denotes the Dirac delta at $x$. It seems that this probability-valued curve is absolutely continuous with respect to the Wasserstein metric, but it does not seem to satisfy the conclusions of Theorem 8.3.1 in the book. In other words, there does not seem to be a vector field satisfying the continuity equation for this family of probabilities. As far as I can see the proof would already fail at equation (8.3.10), since it is possible to produce a test function with $\partial_t\phi\neq 0$ everywhere, yet $\partial_x\phi=0$ on the support of $\mu_t$, namely, $\{0,1\}$. So my question is, what am I missing here? Thanks a lot in advance. REPLY [10 votes]: This is the typical example of a curve of measures which is as nice as it gets for any "linear structure" (e.g. TV norm or whatever) but is somehow the worst case scenario for quadratic Wasserstein optimal transport. In fact your curve is NOT absolutely continuous in the Wasserstein metric, contrarily to what you claim: indeed you can compute explicitly $$ W_2(\mu_t,\mu_s)=\sqrt{|t-s|}. $$ You can check this easily by moving a mass $|t-s|$ from $x=0$ to $x=1$, hence the cost $|t-s|$ for the squared distance $W_2^2$. More precisely, it is easy to check that the $W_2$ geodesic from $\mu_t$ to $\mu_s$ is of the form $$ (\tilde\mu_{\tau})_{\tau\in[0,1]}=s\delta_0+(t-s)\delta_{\tau}+(1-t)\delta_1, $$ say for $t>s$ (proving that this ansatz is optimal is a good exercise) As a consequence $$ \lim\limits_{h\to 0}\frac{W_2(\mu_{t+h},\mu_t)}{h}=+\infty \qquad\mbox{ for all }t\in(0,1), $$ and therefore your curve is clearly not absolutely continuous. Heuristically this is due to the fact that, since your supports are fixed at distance $|1-0|>0$ from each other, you need to send mass infinitely fast at a positive distance thus with infinite cost (infinitesimally in time).<|endoftext|> TITLE: Topological category of topological monoids / operads QUESTION [5 upvotes]: The category of topological monoids can be made into a topological category in a naive way. Namely, the space of all continuous homomorphisms between two topological monoids is a closed subspace of the space of all continuous maps between the underlying topological spaces. My question is that, regarding homotopy theory, is this a "good" construction? More precisely, if we take the homotopy category of this topological category, which is a category enriched over the category of homotopy types, does it give the "correct" homotopy theory? More generally, consider all the topological operads (with a single color). There is also a naive way of making the category of all topological operads into a topological category, similar to the above one. Does the homotopy category of this topological category behave as we would expect? REPLY [4 votes]: I don't think that there's necessarily a right answer to this question. Any category with spaces of maps like you describe has a homotopy category, as well as a lot of other attendant structure. You have to decide what you are interested in. Let's step back from topological monoids and just talk about topological spaces. There are two different homotopy categories in common usage: the ordinary homotopy category of topological spaces, and the weak homotopy category where you invert maps that are isomorphisms on all homotopy groups. In theory, the first category is a much stronger invariant of the category of topological spaces. In practice, we can say an awful lot more about the second category: it's more amenable to descriptions from algebra; it's equivalent to the homotopy category of CW-complexes (which encode most of the spaces that we are interested in on a day-to-day basis); it's also equivalent to the homotopy category of simplicial sets. Whether you're interested in the ordinary homotopy category or the weak homotopy category depends a lot on whether you're interested in pathological spaces. There is an exactly analogous question for topological monoids (or topological groups). We could either construct the ordinary homotopy category that you describe, which is a strong invariant, or we could build the weak homotopy category where we invert maps $M \to N$ of topological monoids that are weak equivalences on the level of spaces. These are very different homotopy categories. As an example of this, if you restrict your attention to topological groups then the homotopy theory of topological groups is equivalent to the homotopy theory of pointed connected spaces, by the correspondence $G \leftrightarrow BG$. The space $BG$ doesn't remember a lot about the strict group-level structure, such as identities that are satisfied by the multiplication in $G$. When we looked at this question for topological spaces, it was a little easier to make the decision because most topological spaces of interest are equivalent to CW-complexes, which are built up freely from basic building blocks. The analogues of CW-complexes are topological monoids that don't really arise in nature very often: they include free groups, but not Lie groups. This means that the weak homotopy category of topological monoids ignores a lot of the features that one might be interested in. However, it is a lot easier to build things there: homotopy colimits, homotopy limits, Postnikov towers, and so on. As a result, what you pick probably depends on the applications you have in mind. (Similar remarks apply to operads.)<|endoftext|> TITLE: How many permutations are there at a given Cayley distance from the identity? QUESTION [5 upvotes]: Permutations $\sigma$ in the symmetric group $S_n$ can be characterized by their Cayley distance $C_\sigma$, being the minimal number of transpositions needed to convert $\{1,2,3,\ldots n\}$ into $\sigma$. The sign of the permutation is $(-1)^{C_\sigma}$. For example, when $\sigma=\{2, 3, 4, 5, 1\}$, one has $C_\sigma=4$ and for $\sigma=\{1, 2, 3, 5, 4\}$ one has $C_\sigma=1$. Of the $5!$ permutations in $S_5$ there are, respectively, $1,10,35,50,24 $ with Cayley distance $C_\sigma=0,1,2,3,4$. Question: What is the general formula that counts the number of permutations at a given Cayley distance? This question was motivated by my attempt to check an integral formula in the unitary group. REPLY [7 votes]: The Cayley distance of a permutation is also known as its absolute length, as can be found out by supplying a few values at https://findstat.org/StatisticFinder/Permutations, which yields https://findstat.org/St000216. There, we also find that for a permutation in $\mathfrak S_n$ with $k$ cycles it is simply $n-k$. This fact is, for example, Problem 5.6 in [1]. [1] Petersen, T. Kyle, Eulerian numbers, Birkhäuser Advanced Texts. Basler Lehrbücher. New York, NY: Birkhäuser/Springer (ISBN 978-1-4939-3090-6/hbk; 978-1-4939-3091-3/ebook). xviii, 456 p. (2015). ZBL1337.05001.<|endoftext|> TITLE: Degree-3 curves on the Calabi–Yau quintic QUESTION [5 upvotes]: Robbert Dijkgraaf said,1 concerning the simplest Calabi–Yau space, the quintic: "A classical result from the 19th century states that the number of lines — degree-one curves — is equal to 2,875. The number of degree-two curves was only computed around 1980 and turns out to be much larger: 609,250. But the number of curves of degree three required the help of string theorists." I gather from OEIS sequence A076912 that the number is 317,206,375. Q. Can anyone point me to (or describe) how the degree-3 count was settled with "the help of string theorists"? 1Robbert Dijkgraaf. "Quantum Questions Inspire New Math." Quanta. The Best Writing in Mathematics 2018, Ed. M. Pitici. p.80. Princeton. Publisher's link. REPLY [6 votes]: The physicists Candelas, De La Ossa, Green, Parkes predicted the virtual number $n_d$ of rational curves of degree $d$ on a quintic threefold for any $d\geqslant 1$. The numbers $n_d$ are defined by the 'multiple cover' formula $$\sum_{d=1}^\infty N_d q^d=\sum_{d=1}^\infty \sum_{k=1}^\infty \frac{n_d}{k^3}q^{kd}$$ in terms of other numbers $N_d$, the genus $0$ Gromov-Witten invariants (which, roughly speaking, count maps $f$ – not just embedded curves – from $\mathbf{P}^1$ to the quintic threefold with $f_\ast[\mathbf{P}^1]=d$). The $N_d$ need not be integers, and are in general only rational numbers – e.g. one has $N_2=4876875/8$. How the physicists used a change of variables, the 'mirror map', to equate a generating series of the $n_d$ with an explicitly calculable series associated to the 'mirror' of the quintic is best described elsewhere (e.g. in this book, in various levels of details, or the Perutz article linked in the comments). In any case, I would like to point out that it is a subtle question if the numbers $n_d$ are enumerative, that is, $n_d$ is the actual number of rational curves of degree $d$ in the quintic*. For example, the $n_d$ are deformation-invariant (because the $N_d$ are) and so independent of the quintic $X$, but the actual number is certainly not deformation-invariant. Think of the classical case $d=1$ (lines), $n_1=2875$ computed by Schubert. If you take the equation of the quintic $X$ to be $x_0^5+\cdots+x_3^5=0$, then the number of lines on $X$ is not even finite, in fact the moduli space of lines on $X$ has dimension $1$. However, if $X$ is 'generic', then a fairly classical and elementary argument shows that the moduli space of lines is $0$-dimensional and reduced, and the number of points can easily be computed by using the intersection theory of the Grassmannian $\mathrm{Gr}(2,5)$. For $d>1$ the analysis of the moduli space of rational curves of degree $d$ on $X$ and the intersection theory calculations become much more complicated. For $d=2$ this was carried out by Katz, while the $d=3$ case was handled by Ellingsrud and Strømme. Curiously, the number first computed by Ellingsrud and Strømme in $1991$ was $2682549425$, in disagreement with the number $n_3=317206375$ computed by Candelas, De La Ossa, Green, Parkes. Ellingsrud and Strømme then found an error in their computer program – in a sense, string theory served as a debugger. (The history of the $d=3$ case is described in great detail in this article by P. Galison.) *Indeed, the number $n_{10}$ is not enumerative.<|endoftext|> TITLE: Anabelian geometry ~ higher category theory QUESTION [9 upvotes]: Note: I'm worried this question might be taken as controversial, because it relates to Shinichi Mochizuki’s work on the abc conjecture. However, my question has nothing to do with the correctness of the proof; I'm specifically trying to get a handle on the methods being used. Please assume I'm asking in the most naive sense possible. In this article on inference review, it's explained that Mochizuki's approach differs from standard category theory in refusing to identify isomorphic objects. Since complete ordered fields are rigid, there is really only one way to do this. But for other categories, there are many choices to be made, and the choices must be made in a compatible way. Sometimes it is best to avoid making such choices, but it is possible to do if desired. After all, a pair of equivalent categories cannot distinguish between themselves using only categorical properties. It can thus be a deep theorem to establish such an equivalence, and highly nonobvious. This way of thinking is becoming more and more entrenched in certain disciplines of mathematics, especially those where category theory has been used extensively. Algebraic geometry is one such discipline. One can, with care, sometimes work as if isomorphic objects are identical. When Mochizuki insists that the isomorphic objects he describes must be distinguished at all costs, and so labelled to keep them distinct, it feels like prohibiting a boxer the use of his fists. On the other hand, at the beginning of the Wikipedia page on higher category theory it says: higher category theory is the part of category theory at a higher order, which means that some equalities are replaced by explicit arrows in order to be able to explicitly study the structure behind those equalities. So isn't higher category theory precisely studying differences between isomorphic objects? Is Mochizuki talking about higher category theory? REPLY [18 votes]: I am the author of that article in Inference. Mochizuki has explicitly said he is working with the truncation of the natural 2-categories of objects he wants to work with, for instance categories and isomorphism classes of functors, rather than categories, functors and natural transformations. This is a loss of information, even when two functors might be uniquely isomorphic. As far as I can tell, this leads to a complication when one wants to treat diagrams in categories as being made up of specific objects, rather than isomorphism classes of objects: a diagram is a functor, after all. This leads to the 'solution' of considering only small subcategories $D \hookrightarrow C$ as diagrams in $C$. Up to cofinality (replacing $D$ by a (co)final subcategory), equivalence (one might need to replace $C$ by an equivalent category) and natural isomorphism (and finally the functor by a naturally isomorphic one), this is perfectly fine. But then if someone comes along who wasn't privy to this private fan dance, and who is ok with diagrams as functors, and in particular non-injective-on-objects functors, they will disagree that every node of the diagram must be unequal to every other node of the diagram, and you are going to disagree that one can have all nodes of the diagram equal, with no ill-effects. Demanding that nodes of a diagram are unequal isn't a statement compatible with the principle of equivalence, since it is perfectly consistent with structural mathematics that the objects of a large category don't even have an equality predicate, or more prosaically, one cannot tell the difference between naturally isomorphic diagrams. Category theory is agnostic on whether objects are isomorphic or not, as opposed to replacing equal things with unequal but isomorphic things. Mochizuki is very much using the language of category theory, but he is not doing category theory, nor is he working in the spirit of it, and certainly is not using higher category theory, even if that would in fact tighten up some of his argument (though not the exposition). Just because a computer scientist uses natural numbers, it doesn't mean they are doing number theory.<|endoftext|> TITLE: Disjunction number of a graph QUESTION [5 upvotes]: Let $S\neq \emptyset$ be a set. We make its powerset ${\cal P}(S)$ into a simple, undirected graph by saying that $A, B\in{\cal P}(S)$ form an edge if and only if $A\cap B=\emptyset$. The disjunction number of a graph $G=(V,E)$ is the smallest cardinal $\kappa$ such that $G$ is isomorphic to an induced subgraph of ${\cal P}(\kappa)$, and we denote the disjunction number of $G$ by $\delta(G)$. Is $\delta(G) \leq |V|$ for any (finite or infinite) graph $G=(V,E)$? (As an aside remark, it would also be interesting to know whether what I call "disjunction number" has a proper name.) REPLY [2 votes]: The mathematical questions have been answered by Fedor Petrov and Gerhard Paseman. My purpose is to offer a partial answer to the historical/bibliographical question suggested in parentheses at the the end; namely, that the same notion, or something closely related to it (being vague because I'm too lazy to read the paper in detail), can be found in: K. Čulík, Applications of graph theory to mathematical logic and linguistics, in: Theory of Graphs and its Applications, Proceedings of the Symposium held in Smolenice in June 1963; Publishing House of the Czechoslovak Academy of Sciences, Prague; Academic Press, New York and London; pp. 13–20. Čulík defines the number of completeness of a graph $G$, denoted by $\omega(G)$, as the smallest cardinal number of a collection of complete subgraphs covering all the edges and vertices of $G$. As a slight modification, let me define $\varepsilon(G)$ as the smallest number of complete subgraphs covering all the edges of $G$. (I'm sorry if $\varepsilon$ is a bad choice of notation; I don't know if there is any Greek letter that does not already have a reserved meaning in graph theory.) If $\overline G$ denotes the complement of $G$, it is easy to see that $$\delta(G)=\varepsilon(\overline G).$$ The answers to the mathematical questions for infinite and finite graphs follow from the fact that $\varepsilon(G)\le|E(G)|$ in all cases, while $\varepsilon(G)=|E(G)|$ if $G$ is triangle-free. Thus, if $G$ is an infinite graph, then $\delta(G)=\varepsilon(\overline G)\le|E(\overline G)|\le|V(G)|$; if $G$ is a finite graph with $n$ vertices and $\overline G$ is bipartite with $\left\lfloor\frac{n^2}4\right\rfloor$ edges, then $\delta(G)=\varepsilon(\overline G)=|E(\overline G)|=\left\lfloor\frac{n^2}4\right\rfloor$. The papers citing Čulík's paper may also be relevant, but I don't have time to look into them.<|endoftext|> TITLE: Applications of $h$-topology and $h$-descent QUESTION [7 upvotes]: This is a technical problem about applications of Grothendieck topologies. In some recent works, the technique of $h$-topology and $h$-descent is very useful, for an introduction see https://stacks.math.columbia.edu/tag/0ETQ. However, it seems that such technique is not well-known (for example, there are very few discussions on MathOverflow). What are some good applications of such technique? Some examples are Original work by Voevodsky to study the homology of schemes. Beilinson's approach to p-adic Hodge theory. Bhatt and Scholze's work on projectivity of Witt affine Grassmanian. And why is $h$-topology and $h$-descent technique useful ? In philosophy, it captures mainly the topological information of schemes (although it's not subcanonical). But I hope to understand more about technical advantages. REPLY [10 votes]: I guess a relevant part of the philosophy is that schemes are smooth locally in the $h$-topology. So, if we are in characteristic zero, we can use resolution of singularities to produce an $h$-hypercovering $U_\bullet\to X$ of a scheme $X$. Then, if we have something like a cohomology theory, which is defined on smooth schemes and satisfies $h$-descent, then a very natural way of extending it to arbitrary schemes $X$ is to evaluate it on $h$-hypercoverings $U_\bullet\to X$. This way, we can extend things from to schemes without changing anything on smooth schemes. The other relevant point is that a lot of things we know and love on smooth schemes have $h$-descent. (Maybe this is a combination that we know many things that have étale descent and where we can control what happens in blowups.) For example, differential forms and the de Rham complex. The above strategy to take something on smooth schemes and extend via $h$-hypercoverings was used implicitly in Deligne's Theorie de Hodge III to define the mixed Hodge structure for a proper variety, way before the explicit definition of the $h$-topology. More recent examples of the use of the $h$-topology to deal with differential forms on non-smooth schemes can be found in the following papers: A. Huber and C. Jörder. Differential forms in the h-topology, Alg. Geom. 1(4), 449-478, (2014). https://arxiv.org/abs/1305.7361 A. Huber, S. Kebekus and S. Kelly. Differential forms in positive characteristic avoiding resolution of singularities. Bull. Soc. Math. France 145 (2017), 305-343. https://arxiv.org/abs/1407.5786 A. Huber. Differential forms in algebraic geometry - a new perspective in the singular case, Port. Math. 73 (2016), no. 4, 337–367. Modifications of this are possible. Motivic cohomology or K-theory for smooth schemes don't have etale descent and therefore can't have h-descent. However, they satisfy cdh-descent, so for instance the definition of motivic cohomology for non-smooth schemes is as cdh-hypercohomology of a suitable cycle complex. In characteristic $p$, we don't have resolution of singularities, and there are modifications of $h$-related topologies where we can use alterations instead.<|endoftext|> TITLE: Construct a non-connected graph with a given degree sequence QUESTION [7 upvotes]: Is there a known (efficient) algorithm to construct a non-connected graph with a given degree sequence (if it exists)? Examples The sequence $\{3, 2, 2, 2, 2, 2, 1\}$ has both connected and non-connected realizations as simple graphs: All non-connected realizations of this sequence are isomorphic to the graph shown above. The algorithm should construct such a graph. All realizations of the sequence $\{3, 3, 1, 1, 1, 1\}$ are connected (it's a forcibly connected sequence) and isomorphic to: The algorithm should either fail on this sequence, or construct a connected graph like the one above. To help with experimentation, the following is an exhaustive list of degree sequences of size $\le 7$ that have both connected and non-connected realizations: {2, 2, 2, 1, 1} {{3, 2, 2, 1, 1, 1}, {2, 2, 2, 2, 1, 1}, {3, 3, 2, 2, 1, 1}, {2, 2, 2, 2, 2, 2}, {3, 3, 3, 3, 1, 1}} {{4, 2, 2, 1, 1, 1, 1}, {3, 3, 2, 1, 1, 1, 1}, {4, 3, 2, 2, 1, 1, 1}, {4, 4, 2, 2, 2, 1, 1}, {3, 2, 2, 2, 1, 1, 1}, {4, 2, 2, 2, 2, 1, 1}, {2, 2, 2, 2, 2, 1, 1}, {3, 2, 2, 2, 2, 2, 1}, {3, 3, 2, 2, 2, 1, 1}, {3, 3, 3, 2, 1, 1, 1}, {4, 3, 3, 2, 2, 1, 1}, {4, 3, 3, 3, 1, 1, 1}, {4, 4, 3, 3, 2, 1, 1}, {2, 2, 2, 2, 2, 2, 2}, {3, 3, 2, 2, 2, 2, 2}, {3, 3, 3, 3, 2, 1, 1}, {4, 3, 3, 3, 3, 1, 1}, {4, 4, 4, 3, 3, 1, 1}, {3, 3, 3, 3, 2, 2, 2}, {4, 4, 4, 4, 4, 1, 1}} Other interesting sequences: {4, 4, 4, 3, 3, 3, 3, 2, 2} and {4, 4, 4, 3, 3, 3, 2, 2, 1}. Both of these have a single non-connected realization (ignoring isomorphic duplicates) and none of the components of these realizations are cliques. REPLY [6 votes]: A graphic degree sequence is called forcibly connected if all realizations are connected graphs. So, you want to know a given degree sequence is not forcibly connected and then to find a disconnected graph with the degree sequence. Not forcibly connected is also known as potentially disconnected. More generally, there exists literature of forcibly P and potentially P for some property P. These keywords may be helpful in finding results. One recent paper I found that may be of interest to you is An efficient algorithm to test forcibly-connectedness of graphical degree sequences by K. Wang. The complexity of the algorithm given in the paper is exponential, but the author performs some experiments showing the algorithm can be used in some cases. The problem of testing forcibly connectedness is co-NP, the author believes co-NP hardness is open. The bibliography of the paper points to some sufficient conditions for forcibly connectedness. This literature deals with the decision problem rather than the construction you ask for. However, the hardness of the decision problem appears to be open. The algorithm in the paper works by partitioning the degree sequence into two parts and testing if each in graphic. The paper suggests some ways to speed up the purely naive approach of testing of splits, but the algorithm is still exponential. If you can split into to graphical sequences you can then construct the graph with the Havel–Hakimi algorithm on each smaller degree sequence.<|endoftext|> TITLE: How much Replacement does this axiom provide? QUESTION [19 upvotes]: (There have been many questions on MathOverflow about the axiom scheme of replacement, including a few with a similar flavour to mine. Some have very informative answers and link to excellent papers and blog posts. I've spent a while reading the older questions etc., and as far as I know, my question isn't answered in any of them — but it's entirely possible that it is and I missed it. In that case, I'll be grateful if someone points out where.) Consider the following statement about sets and functions, which I'll call "axiom A": A. For all well-ordered sets $(B, \leq)$, there exist a set $X$ and a function $p: X \to B$ with the following property: for all $b \in B$, the fibre $p^{-1}(b)$ is an infinite set of smallest cardinality greater than that of $p^{-1}(b')$ for each $b' < b$. In the traditional notation of set theory, if $(B, \leq)$ is an ordinal $\beta$, then $p^{-1}(\alpha) \cong \aleph_\alpha$ for each $\alpha \in \beta$. So, $X$ is the disjoint union $\coprod_{\alpha \in \beta} \aleph_\alpha$ and $p: X \to \beta$ is the obvious projection. My question is about ETCS (Lawvere's Elementary Theory of the Category of Sets) together with axiom A. In what follows, I'm going to take ETCS as the background theory. Thus, when I say "this is weaker than that", I mean weaker in the presence of ETCS. On the one hand, A isn't a theorem of ETCS (unless, of course, ETCS is inconsistent). That's because ETCS+A proves the existence of $\aleph_\omega$ but ETCS alone doesn't. On the other, if I'm not mistaken, A is weaker than replacement. That's because ETCS+replacement is bi-interpretable with ZFC, and unless I'm misremembering, the fact that ETCS+A is a finite list of axioms (not involving axiom schemes) somehow implies that it can't be as strong as ZFC. So, it seems that axiom A is a weaker form of replacement. My question: To what fragment of replacement is axiom A equivalent (in the presence of the axioms of ETCS)? That question is a little vague, so let me focus it more: What's the simplest statement you can think of that's true in ETCS+replacement (or equivalently ZFC) but not provable in ETCS+A? I don't mind whether the statement is purely set-theoretic or from another part of mathematics. Added later Incidentally, "axiom A" is equivalent (under ETCS) to the following simpler statement: for every set $B$, there exists a map into $B$ whose fibres all have different cardinalities. Formally: for all sets $B$, there exist a set $X$ and a map $p: X \to B$ such that for all $b, b' \in B$, if $p^{-1}(b) \cong p^{-1}(b')$ then $b = b'$. REPLY [6 votes]: The principle is essentially asserting that $\aleph_\alpha$ exists for every ordinal $\alpha$. More precisely, it asserts that for every well-order type $\alpha$, there is a set of cardinality $\aleph_\alpha$; the difference is that without the replacement axiom, one doesn't necessarily have the usual von Neumann ordinals. This axiom is not provable in Zermelo set theory, since $V_{\omega+\omega}$ models the Zermelo theory, but it has only countably many infinite cardinals, and so $\aleph_{\omega+\omega}$ does not exist in this model, even though there are orders of type $\omega+\omega$ there. The axiom is provable in ZFC and is strictly weaker than ZFC, as has been pointed out by B2C, since $V_\kappa$ is a model of the principle if and only if $\kappa$ is a beth-fixed point $\kappa=\beth_\kappa$. What I would like to mention is that the principle is a natural instance of the principle of recursion: Principle of transfinite recursion. (See the account on my blog) If $A$ is any set with well-ordering $<$ and $F:V\to V$ is any class function, then there is a function $s:A\to V$ such that $s(b)=F(s\upharpoonright b)$ for all $b\in A$, where $s\upharpoonright b$ denotes the function $\langle s(a)\mid a TITLE: A finite generating set for the group $SL_{2} (\mathbb{F}_{2} [t , t^{-1} ] )$ QUESTION [8 upvotes]: $SL_{2} ( \mathbb{F}_{2} [t , t^{-1} ])$ is defined to be the group of $2 \times 2$ matrices with determinant one with entries polynomials in $t$ and $t^{-1}$ with coefficients in the field of two elements, $\mathbb{F}_{2}$. By the work of Stuhler (Stuhler, U., Zur Frage der endlichen Prasentierbarkeit gewisser arithmetischer Gruppen im Funktiönenkorperfall, Math. Ann. 224 (1976), 217–232.), it is known that $SL_{2} (\mathbb{F}_{2} [t , t^{-1} ])$ is finitely generated. Is any finite generating set known? REPLY [10 votes]: I guess it was known earlier (the paper you mention deals with finite presentability, a harder issue). Denoting $(E_{ij})$ the basis of the space of matrices, $e_{ij}(t)=I+tE_{ij}$, $d_{ij}(a)=aE_{ii}+a^{-1}e_{jj}$, the group $\mathrm{SL}_2(\mathbf{F}_p[t,t^{-1}])$ for $p$ prime is generated by $\{e_{12}(1),e_{12}(t),e_{21}(1),e_{21}(t),d_{12}(t)\}$. Indeed, $\mathbf{F}_p[t,t^{-1}]$ being a Euclidean ring, $\mathrm{SL}_2(\mathbf{F}_p[t,t^{-1}])$ is generated by elementary matrices. And using conjugation by $d_{12}(t)$ and the four given elementary generators, one obtains all other basis elements among elementary matrices. (It's not hard to check that $e_{21}(t)$ is redundant in this generating subset, and also that no two of these 5 elements generate the group.)<|endoftext|> TITLE: Kronheimer's results on ALE spaces as hyperkahler quotients QUESTION [8 upvotes]: Background: In his two papers from late 80s Kronheimer proved that any 4-dimensional ALE space is given by a hyperkahler quotient, say $X_{{\zeta_\mathbb{R}},{\zeta_\mathbb{C}}}(Q)$ where Q is a Dynkin graph of type ADE, and ${\zeta_\mathbb{R}},{\zeta_\mathbb{C}}$ are parameters of hyperkahler moment map that satisfy a certain genericity conditions. He also proves that, fixing a graph Q all these spaces are diffeomorphic to a space, call it $X(Q),$ which is given as the minimal resolution of $$\pi: X(Q)\rightarrow \mathbb{C}^2/G,$$ (where G is a certain subroup of $SU(2)$). From earlier work of Du Val we know the topology of $X_Q$ - its deformational rectract is the exceptional divisor $$\pi^{-1}(0)=\cup_{i\in Q^0} \mathbb{C}P^1_i$$ which is a Dynkin Q tree of spheres that intersect transversely according to the graph, and whose self-intersections are -2. In particular, $$H_2(X(Q))=\oplus_{i\in Q^0} [\mathbb{C}P^1_i].$$ Now the question: It seemed to me that Kronheimer also proves the following: In the ALE space $X_{{\xi_\mathbb{R}},0}(Q)$ the $\omega_I$-volumes of those exeptional spheres are given \begin{equation} \tag{1} \langle \omega_I, [\mathbb{C}P^1_i] \rangle= \zeta_\mathbb{R}^i \end{equation} exactly by the components of the real moment map parameters $\zeta_\mathbb{R}=(\zeta_{\mathbb{R}}^i)_{i\in{Q^0}}.$ Now this seems to be false, as those exceptional spheres are $\omega_I$-symplectic, hence their $\omega_I$-volumes are positive, whereas moment parameters $\zeta_{\mathbb{R}}^i$ can be negative. So, does anyone knows ''the cure'' to the formula (1) to make it true? REPLY [6 votes]: $\mathbb C P^1_i$ does not make sense universally on $X_\zeta$ for all $\zeta$. When a parameter $\zeta$ cross a wall, the homology class $[\mathbb CP^1_i]$ is changed by the Weyl group reflection. Therefore (1) is not correct.<|endoftext|> TITLE: Integral $p$-adic Hodge theory and the space of comparisons of cohomology theories QUESTION [6 upvotes]: Weil cohomology theories can be considered as fibre functors from the category of motives. Given two such functors, we have an affine scheme of invertible natural transformations between them, and rational Hodge theory can be considered as providing $R$-valued points of this scheme (where $R$ is the coefficient ring). See this post, for example. Can integral Hodge theory be viewed in a similar way? We are not over a field, so there might be Tannakian issues. Maybe it is pretty trivial, but then it would be nice if somebody provided full details. REPLY [2 votes]: See the works of Bhatt--Morrow--Scholze. Possibly they provide what you want.<|endoftext|> TITLE: Do Grothendieck universes matter for an algebraic geometer? QUESTION [36 upvotes]: I recently learned that some parts of SGA require axioms beyond ZFC. I am just a simple algebraic geometer so I am trying to understand how can this fact impact my life (you may have engaged in a similar game with the axiom of choice at some point of your mathematical career). I think this was only needed in the volume developing cohomology of topoi, correct me if I am wrong. The question: is there some statement about schemes not involving the word "topoi" that you know how prove using the additional axioms, but do not know how to prove without them? I believe this question has nothing to do with the notion of completeness in mathematical logic (because I am not asking if something can be proved in principle, just if there is an obvious argument). If there is some logical argument showing that the answer is negative, I would like to learn about that too. Bonus points if the proof of the statement in your answer relies on a Weil cohomology theory, rather than something random (I do not know if this is even possible). To be more precise, a Weil cohomology theory is only defined for smooth projective schemes over a field, for which all of this should probably be irrelevant, so I mean that the cohomology theory you use restricts to a Weil cohomology theory for smooth projective schemes over a field. P.S. There was similar discussion in the context of Fermat's theorem, and if I understand correctly, the user BCnrd insists that universes are useless for etale cohomology without giving a reference. I believe there are few people on this planet who know etale cohomology better than BCnrd so that's some useful information. The question is not, however, limited to etale cohomology. REPLY [15 votes]: Tim Chow drew my attention to this thread, and asked if I cared to comment. Actually I wrote up a detailed version of my thoughts several years ago, On doing category theory within set theoretic foundations, and to my surprise, Solomon Feferman thought enough of it for that to appear in the book "What is Category Theory". Their website has disappeared, so here is the version I found in my back-ups. As a homotopy theorist, I have no idea of what is need for Wiles' proof of FLT, or what is used in EGA or SGA. For homotopy theory, the crux of the matter comes down to two things. First, we construct towers iteratively (potentially of transfinite length) and then take its limit or colimit. This assumes that the whole tower is small, while all we know is that the length is small. This is an implicit use of replacement. [Granted, if the iterative step is predicative, ZFC is enough. But see the next paragraph] The earliest use of this I know of is the proof of Brown Representability Theorem dating back 1960. I recall the same kind of argument somewhere in Neeman's book on triangulated categories. So may be it also needed for derived categories. Another similar situation: I have a functor F and a small subcategory D of the domain of F. I want to know that F(D) is small so that I can talk about its limit/colimit. This already happens with the Milnor short exact sequence for arbitrary generalized cohomology theories [and can enter into play when studying generalized cohomology of infinite dimensional spaces such as classifying spaces of groups.] If F is itself constructed as the (co)limit of a tower constructed by transfinite induction, trying to stay within ZFC proper is not trivial. People seem to get hung up on indexed categories (just use fibered categories all the time) and functor categories. But there is another use of universes that is more troublesome: Prove a theorem about small categories (or quasicategories or Segal spaces or ...) and then apply to "the category of all spaces" (or ...). If the theorem about small categories was proved using only ZFC, then ZFC + global reflection does the trick. This was a suggestion of Feferman from the 60's, and my write-up was about getting around the second issue above. But published proofs ignore the subtleties: They implicitly assume Morse-Kelly. Then we need to invoke Grothendieck universes. [If you don't understand the fuss in the third paragraph, think about this: $W$ is a set that satisfies all the ZFC axioms. For each natural number n, I construct an element $x_n$ of $W$. The details of the construction are hidden in a back box. Is the set $\{x_n\}$ an element of $W$? Or do you need to know the details of the construction?]<|endoftext|> TITLE: Does the statement 'there exists a first-order theory $T$ with no saturated models' have any set theoretic strength? QUESTION [11 upvotes]: Exercises 6, 7, and 8 in section 10.4 of Hodges' big model theory textbook contain an outline of a proof of the consistency of the following statement There exists a countable first-order theory $T$ such that $T$ has no saturated models. Call this statement $\mathrm{NSat}_\omega$ and call the analogous statement with no cardinality restriction $\mathrm{NSat}$. The proof outlined by Hodges uses a result of Woodin's that it is consistent with $\mathrm{ZFC}$ that for every infinite cardinal $\kappa$, $2^\kappa=\kappa^{++}$, but he mentions that this result assumes some large cardinals and in particular 'a supercompact is more than enough'. Are large cardinals necessary for this result? There are different meanings of a question like this but a typical formulation would be something like: Does $\mathrm{ZFC}$ + 'there exists a transitive model of $\mathrm{ZFC}+\mathrm{NSat}_\omega$' prove the existence of transitive models of any typical large cardinal axiom? What about $\mathrm{ZFC}+\mathrm{NSat}$? REPLY [15 votes]: Unless I'm missing something, if $\vert T\vert+\aleph_0<\kappa$ and $\kappa^+=2^\kappa$, then we can build a saturated model of $T$ of cardinality $2^\kappa$. So: If we want a countable theory with no saturated model, we need GCH to fail everywhere. If we want some theory with no saturated model, we need GCH to hold at most boundedly often. And these principles have rather high consistency strength. A failure of GCH at a singular limit cardinal is in particular a failure of the Singular Cardinal Hypothesis, the failure of which is known to be equiconsistent with the existence of a $\kappa$ such that $o(\kappa)=\kappa^{++}$. So this gives a lower bound to both principles. A better analysis of the strength involved, at least for NSat (although I suspect they'll be identical), is this wonderful MSE answer by Andres Caicedo (in particular, my previous paragraph is just a rephrasing of his second-to-last paragraph). However, that's from a while ago, so more might be known now.<|endoftext|> TITLE: Kernel of the determinant morphism from the first algebraic K-theory QUESTION [5 upvotes]: If $A$ is the coordinate ring of a smooth variety over a finite field is it known whether the kernel of the determinant map $K_1(A)\rightarrow A^{\times}$ is torsion or not? REPLY [3 votes]: I think the following affine deleted quadric is an example where the kernel of the determinant fails to be torsion. Assume we are working in odd characteristic. Then we have the smooth affine quadric $Q_2$ defined by the equation $XY+Z^2=1$. We have $K_0(Q_2)=\mathbb{Z}\oplus\mathbb{Z}$ (either by direct computation or using that $Q_2$ is $\mathbb{A}^1$-equivalent to $\mathbb{P}^1$). Consider the localization sequence involving $Q_2\subseteq \mathbb{A}^3$ with open complement $U=\mathbb{A}^3\setminus Q_2$: $$ K_1(\mathbb{A}^3)\to K_1(U)\to K_0(Q_2)\to K_0(\mathbb{A}^3)\to K_0(\mathbb{A}^3\setminus Q_2). $$ The first group is torsion and the last map is an isomorphism, so that $K_1(U)$ has rank 2. On the other hand, the variety $U$ has coordinate ring $\mathbb{F}_q[X,Y,Z][(XY+Z^2-1)^{-1}]$. The element $XY+Z^2-1$ in the polynomial ring (which is a UFD) is irreducible, hence the multiplicative set $\{(XY+Z^2-1)^n\mid n\in \mathbb{N}\}$ is saturated and therefore the group of units of the coordinate ring has rank 1. So the determinant map on $K_1(U)$ must have non-torsion kernel. (This should work more generally whenever we consider the complement of an irreducible smooth hypersurface $Z$ in affine space where the hypersurface $Z$ has non-torsion reduced K-theory.) You probably know that the Parshin conjecture would imply that the determinant map $K_1(X)\to \mathbb{F}_q^\times$ has torsion kernel, but that's for smooth projective $X$. I think the above example shows that it's not quite so clear what the right formulation of consequences of Parshin's conjecture for K-theory of smooth affine varieties should be. In any case, the Parshin conjecture isn't known at this point, except of course for curves. To emphasize, the simplest case where we don't know what's happening is related to Bloch's conjecture for curves over global function fields. If $X/K$ is a curve over a finite extension $K/\mathbb{F}_q(T)$, then there is a smooth projective curve $C/\mathbb{F}_q$ with function field $K$ and a smooth projective surface $\mathfrak{X}$ with a proper flat map $\mathfrak{X}\to C$ whose generic fiber is $X$. Asking if the determinant map $K_1(\mathfrak{X})\to \mathbb{F}_q^\times$ has torsion kernel is equivalent to asking if the kernel $V(X)$ of the pushforward map $CH^2(X,1)=H^1_{\rm Zar}(X,\mathbf{K}_2)\to K_1(K)$ is torsion. This equivalence is established in R. Akhtar. Cycles on curves over global fields of positive characteristic. Trans. Amer. Math. Soc. 357(7), 2557-2569, 2005. I guess the best we know about the group $V(X)$ at this point is that the maximal divisible subgroup is uniquely divisible, and the reduced quotient (mod max divisible) is finite. This is proved in S. Kondo and S. Yasuda. First and second K-groups of an elliptic curve over a global field of positive characteristic. Ann. Inst. Fourier (Grenoble) 68 (2018), no. 5, 2005–2067. So even in the smooth projective case, with appropriate reformulations, the question seems fairly open and it seems we don't know the simplest cases beyond curves.<|endoftext|> TITLE: Telling right from left QUESTION [15 upvotes]: I know a lot of people, some of them mathematicians, who have trouble telling right from left. This can lead to problems when you are composing functions, for example. When did this seemingly innocuous confusion lead to wrong results being published? Here is an example from arithmetic geometry. REPLY [7 votes]: I don’t know about wrong results, but compare Godement’s Algèbre (1966, and still 1980): on dit que l’ensemble $x\mathrm H$ est une classe à droite modulo $\mathrm H$ (translation: “we say that the set $x \mathrm{H}$ is a right class modulo $\mathrm{H}$”) with everyone else, e.g. his associate Bourbaki (1970): les ensembles $x\mathrm H$, qu'on appelle classes à gauche suivant $\,\mathrm H$ (ou modulo $\,\mathrm H$) (translation: “the sets $x \mathrm{H}$, which are called left classes following $\mathrm{H}$ (or modulo $\mathrm{H}$)”)<|endoftext|> TITLE: Restricted independent set of the cycle graph $C_{3n}$ QUESTION [6 upvotes]: Let $V$ be the vertices of the cycle graph $C_{3n}$. Suppose there is a partition of $V$ into sets of $3$, i.e. $V=\cup_{k=1}^{n}{V_k}$ where $|V_k|=3$ for $k$ in $1..n$. QUESTION: Is it possible to find an independent set of $V$ with exactly one vertex from each $V_k$? By the Lovasz Local Lemma, it's possible if the $3$ is replaced with some larger number, say, $11$. REPLY [5 votes]: The answer is yes, via a stronger proposition: for each part in the partition, add edges between the vertices the part contains; the question now is, can you vertex-3-color the resulting graph? This is the well-known cycle plus triangles problem, originally posed by Erdos and solved by Fleischner and Stiebitz using the Alon-Tarsi theorem, and later again by Sachs (inductively), affirmatively. The proofs are far from trivial or easy (I happen to like the inductive proof). So to answer your question, just pick a color class ...<|endoftext|> TITLE: Is $\operatorname{Hom}(F,G)$ finite if $F$ and $G$ are endofunctors of the category of finite sets? QUESTION [9 upvotes]: I asked this question on Mathematics Stackexchange but got no answer. Are there endofunctors $F$ and $G$ of the category of finite sets such that there are infinitely many natural transformations from $F$ to $G$? Same question with $F$ and $G$ contravariant. REPLY [10 votes]: Thank you to Achim Krause for pointing out that the first version was broken. Let's try again. Let $k$ be a finite field. For a set $X$, let $kX$ be the free vector space on $X$. Let $\bigwedge^{\bullet} kX$ be the exterior algebra on $kX$. Then $X \mapsto \bigwedge^{\bullet} kX$ becomes a functor in an obvious way. Choose any scalars $a_0$, $a_1$, $a_2$, ... in $k$ and define the natural transformation of $\bigwedge^{\bullet} kX$ by multiplying $\bigwedge^{j} kX$ by $a_j$. This gives infinitely (even uncountably) many natural transformations from $X \mapsto \bigwedge^{\bullet} kX$ to itself as a functor from finite sets to finite sets (or even from finite sets to vector spaces). When studying functors $F$ from the category of finite sets or related categories, one usually wants to impose some sort of finite generation condition, saying roughly that there is some integer $N$ such that any subfunctor of $F$ which agrees with $F$ on sets of size $\leq N$ is the same as $F$. One does this precisely to avoid this sort of trickery with the functor $X \mapsto 2^X$. For example, Eric Ramos, Graham White and I classify functors from FI to FinSet with a finite generation hypothesis and my student John Wiltshire-Gordon classified functors from FinSet to $\mathbb{Q}$-Vect under a similar hypothesis. These are the two papers I know which come closest to studying functors from FinSet to FinSet.<|endoftext|> TITLE: History of the Frobenius Endomorphism? QUESTION [11 upvotes]: The existence of the Frobenius endomorphism probably goes back to Euler's proof of Fermat's little theorem. But why is it named after Frobenius? Who gave it this name? When was it first stated in full generality? How did people refer to this concept before the language of ring homomorphisms? REPLY [17 votes]: Since you reach back to Euler, who proved Fermat's little theorem in the form $a^p \equiv a \bmod p$ by using induction on $a$ and the binomial theorem, I think your "Frobenius endomorphism" is the $p$th-power map in characteristic $p$ (or $p^k$-th power map if the base field has order $p^k$). Frobenius has his name associated to this rather elementary operation, used long before him, because he proved the existence of lifts of the $p$-th power map to finite Galois groups over $\mathbf Q$. Those automorphisms, which Francois mentions in his answer as Frobenius substitutions, are more intricate than the $p$-th power map in characteristic $p$ and it is not surprising that the person who first published a paper about them got them named after him. (Dedekind wrote to Frobenius that he had gotten the existence of such lifts to Galois groups over $\mathbf Q$ earlier, but this was just a private communication.) Once the name Frobenius substitution was used, it is not surprising that the map in characteristic $p$ got named after Frobenius too. However, I don't know who first used his name for the characteristic $p$ operation.<|endoftext|> TITLE: What is the most direct proof of the Riemann hypothesis for Dirichlet L-functions over function fields? QUESTION [33 upvotes]: Let $\mathbb{F}$ be a finite field of order $q$, let $m$ be an irreducible polynomial in the ring $\mathbb{F}[T]$, and let $\chi$ be a Dirichlet character modulo $m$. Define the function field Dirichlet $L$-function $$ L(s,\chi) := \sum'_f \frac{\chi(f)}{|f|^s}$$ where the sum is over monic polynomials in $\mathbb{F}[T]$, and $|f| = q^{\mathrm{deg}(f)}$ is the usual valuation. The Riemann hypothesis for this $L$-function asserts that the non-trivial zeroes of this $L$-function all lie on the line $\mathrm{Re}(s) = \frac{1}{2}$. An equivalent form of this result is that the error term in the prime number theorem for arithmetic progressions is of square root type in the function field setting. The only way I know how to prove this is to show that such a Dirichlet $L$-function can be multiplied with other Dirichlet $L$-functions or zeta functions to create (up to some local factors) a Dedekind zeta function over some finite extension of $\mathbb{F}[T]$, which is essentially the local zeta function of some curve over $\mathbb{F}$, and at this point one can use any of the usual proofs of RH for such curves (Weil, Bombieri-Stepanov, etc.). But to get the finite extension I either need to appeal to some general theorem in class field theory (existence of ray class fields, which I understand to be a difficult result) or to explicitly construct the extension using Carlitz modules or something equivalent to such modules (the latter is discussed for instance in the answer to this other MathOverflow post). My question is whether there is a more direct way to establish RH for Dirichlet L-functions over function fields without having to locate a suitable field extension (or whether there is some "soft" way to abstractly demonstrate the existence of such an extension without a huge amount of effort). For instance, is it possible to interpret the Dirichlet $L$-function directly as the zeta function of some $\ell$-adic sheaf? Or can the elementary methods of Stepanov type be adapted directly to the Dirichlet $L$-function (or perhaps the product of all $L$-functions of the given modulus $m$? REPLY [16 votes]: Switching from comment to answer because the comment thread is getting too long. Let $\# \mathbb{F}=q, t=q^{-s}$ and consider $L(t,\chi)$. Then (by taking the logarithmic derivative of the Euler product) $$L(t,\chi) = \exp (\sum_{n=1}^{\infty} S_n t^n/n )$$ where $S_n = \sum_{\deg P | n} \chi(P)\deg P$ and $P$ runs through irreducible polynomials of $\mathbb{F}[T]$. Then for any integer $d>0$, $$\prod_{\zeta^d =1} L(\zeta t,\chi) = \exp (\sum_{n=1}^{\infty} S_{dn} t^{dn}/n )$$ Now, if $Q$ is an irreducible polynomial of degree $n$ over the field of $q^d$ elements, then $Q$ has $m$ (some $m|d$) conjugates $Q_i$ over $\mathbb{F}$ and the product of these conjugates is an irreducible polynomial $P$ in $\mathbb{F}[T]$ so (edit: fixed error pointed out in comments) $$\sum_i \chi(Q_i)\deg Q_i = (\sum \chi(Q_i))\deg Q = \chi(P)m\deg(Q) = \chi(P)\deg(P).$$ Using this, one checks that the equivalent of $S_n$ over the field extension equals $S_{nd}$ and this gives $\exp (\sum_{n=1}^{\infty} S_{dn} t^n/n )$ is the $L$-function in the extension field, say $L_d(t,\chi)$. Another way of stating this is $\prod_{\zeta^d =1} L(\zeta t,\chi)= L_d(t^d,\chi)$.The relation with the zeros follows. (This is e.g. in Weil, Basic Number Theory, Appendix 5, lemma 4 in much more generality and fancier language). REPLY [13 votes]: Is it possible to interpret the Dirichlet $L$-function directly as the zeta function of some $\ell$-adic sheaf? Yes and no. Yes in that it's possible to express the Dirichlet $L$-function directly as the zeta function of some $\ell$-adic sheaf. No in that doing so is essentially the same as constructing the relevant Galois extension, and so you're forced to again rely on an explicit construction or an abstract existence result. Indeed the relevant $\ell$-adic sheaf has rank one, so you get a homomorphism $\pi_1^{et}( \operatorname{Spec} \mathbb F [T , m^{-1} ] ) \to \overline{\mathbb Q}_\ell^\times$ but $\pi_1^{et}( \operatorname{Spec} \mathbb F [T , m^{-1} ] )$ is just a quotient of $\operatorname{Gal} (\mathbb F (T))$ so you can derive the abelian Galois extension from the $\ell$-adic sheaf. whether there is some "soft" way to abstractly demonstrate the existence of such an extension without a huge amount of effort. I don't think the explicit existence proof of the field extension is really that bad. I might regret saying this, but I don't think it's that much worse than the proof of power reciprocity. An (I think) essentially self-contained proof using mainly elementary algebraic number theory is below: Lemma: Fix $m$ a polynomial of degree $d$ over $\mathbb F_q$. Let $k$ be an algebraically closed field containing $\mathbb F_q$ and let $c_0,\dots, c_{d-1}$ be elements of $k$. If $\operatorname{Res} ( c_0 + \dots + c_{d-1} X^{d-1} , m(X) ) \neq 0$, then the solutions $a_0,\dots, a_{d-1} \in k$ with $$(a_0^q + \dots + a_{d-1}^{q} X^{d-1} ) = ( c_0 + \dots + c_{d-1} X^{d-1} ) ( a_0 + \dots + a_{d-1} X^{d-1} ) \mod m(X)$$ and $$\operatorname{Res} ( a_0 + \dots + a_{d-1} X^{d-1} , m(X) ) \neq 0$$ form exactly one orbit under the action of $(\mathbb F_q[X]/m(X))^\times$. Proof: Multiplying by an element of $(\mathbb F_q[X]/m(X))^\times$ gives another solution, so the solutions form a union of orbits. The ratio between any two solutions is an element of $(k[X]/m(x))^\times$ with each coordinate equal to its own $q$th power, hence an element of $(\mathbb F_q[X]/m(X))^\times$, so there is at most one orbit. To check that there is at least one orbit, first note that if $a_0,\dots, a_{d-1}$ are independent transcendentals, then $c_0,\dots, c_{d-1}$ would be independent transcendentals (or otherwise there would be infinitely many solutions), so there exist solutions when $c_0,\dots, c_{d-1}$ are independent transcendentals. Given any tuple $c_0,\dots, c_{d-1}$ in $k$, take $c'_0,\dots, c'_{d-1}$ independent transcendentals in $k' = k(c_0',\dots, c_{d-1}'$, and then the coefficients of $$( c_0 + \dots + c_{d-1} X^{d-1})( c_0' + \dots + c_{d-1}' X^{d-1})\mod m$$ are themselves independent transcendentals, so dividing the solutions for these two, there are solutions in $k'$, and then because $k$ is algebraically closed, solutions in $k$. QED Now adjoin to $\mathbb F_q(T)$ a root in $\overline{\mathbb F_q(T)}$ of the system of equations $a_0,\dots, a_{d-1} \in k$ with $$(a_0^q + \dots + a_{d-1}^{q} X^{d-1} ) = ( X-T ) ( a_0 + \dots + a_{d-1} X^{d-1} ) \mod m(X)$$ and $$\operatorname{Res} ( a_0 + \dots + a_{d-1} X^{d-1} , m(X) ) \neq 0$$ Because the set of roots forms an orbit under $(\mathbb F_q[X]/m(X) )^\times$, the Galois group is a subgroup of $(\mathbb F_q[X]/m(X) )^\times$. We want to check that the Frobenius element associated to a prime $\pi(T)$ not dividing $m(T)$ inside this Galois group is equal to the reduction of $m(X)$ mod $(X)$. It follows immediately from the definition of the zeta function of a global field that the zeta function of the function field generated by this root is a product of Dirichlet $L$-functions for characters of $(\mathbb F_q[X]/m(X) )^\times$. The Frobenius element in the Galois group is the same as the Frobenius element in the local field $K_\pi$ for $K = \mathbb F_q(T)$. By Hensel's lemma, each solution in the algebraic closure of the residue field $\overline{k}_\pi$ lifts to a solution in an unramified extension of $K_\pi$ and thus to an element of the algebraic closure $\overline{K}_\pi$. Because $\overline{k}_\pi$ and $\overline{K}_\pi$ have the same number of solutions, all solutions in $\overline{K}_\pi$ arise this way, so the action of Frobenius on solutions over the local field $\overline{K}_\pi$ is equal to its action on solutions over the residue field $k_\pi = \mathbb F_q[T]/\pi(T)$. But the Frobenius in $\overline{ \mathbb F_q[T]/\pi(T)}$ sends $a_i$ to $a_i^{q^{\deg \pi}}$, so it acts on $(a_0 + \dots + a_{d-1} X^d)$ by multiplication by $$(X-T) (X-T^q) \dots (X - T^{ q^{ \deg \pi -1 } } ) .$$ Because $T \in \mathbb F_q[T]/\pi(T)$ is a root of $\pi(T)$, $T^q,T^{q^2}, \dots$ must be the remaining roots, and so Frobenius acts by multiplication by $\pi(X)$, as desired.<|endoftext|> TITLE: Plucker relations in orthogonal Grassmannian QUESTION [5 upvotes]: Let $G=SO_7$ and let $P$ be the maximal parabolic corresponding to the fundamental weight $\varpi_3$. Since $\varpi_3$ is minuscule, $G/P$ may be fairly easy to study. Is the structure of $G/P$ known in literature ? What are the Plücker type relations (quadratic) in this case ? Are the relations same as they are in the Grassmannian $G(3,7)$ ? I have a feeling that the quotient is $\mathbb P^2$. REPLY [12 votes]: If $G = SO(7)$ and $P$ corresponds to $\varpi_3$, then $$ G/P \cong Q^6 $$ is a 6-dimensional quadric. In particular, there is a unique Plücker relation in this case (the equation of the quadric). The isomorphism can be sen as a combination of a general isomorphism $$ SO(2n-1)/P_{\varpi_{n-1}} \cong SO(2n)/P_{\varpi_{n}} $$ and the triality isomorphism $$ SO(8)/P_{\varpi_{4}} \cong SO(8)/P_{\varpi_{1}} \cong Q^6. $$<|endoftext|> TITLE: Coincidences between average Catalan tableaux QUESTION [13 upvotes]: There are Catalan number $C_n$ of standard Young tableaux of shape $(n,n)$, which we view as $2\times n$ matrices. Denote by $P_n$ the average of these matrices: $$ P_n \, := \, \frac{1}{C_n} \, \sum_{A \, \in \, \text{SYT}(n,n)} A $$ Note that $P_n=(p_{ij})$ is a rational matrix with entries monotone increasing in rows and columns. Also, by the 180$^\circ$ rotational symmetry, $p_{ij} + p_{3-i,n+1-j} = 2n+1$. For example, $p_{11}=1$, $p_{2n}=2n$. The asymptotic density of $P_n$ as $n\to \infty$ is easy to obtain by a direct calculation or via the Brownian excursion (see e.g. here or there), but my question is different. Question. Let $\beta(n):= \min_{(ij)\ne (kl)} |p_{ij}-p_{kl}|$. Is it true that $\beta(n) = o(1)$? It would be cool if there was an easy way to see this. I really want a generalization of this result to all large partitions, but at the moment even this is confounding. UPDATE (May 18, 2019): Let me explain the motivation behind the question. Recall the 1/3-2/3 conjecture that every poset $\mathcal P=(X,\prec)$ that is not a linear order contains two elements $x,y\in X$ such that $$\frac13 \le P(x\prec y) \le \frac23 $$ For width 2 posets this was shown by Linial in this paper, but I thought that for shapes $(n,n)$ one can improve $1/3$ to perhaps $(1/2-\varepsilon)$, since we know so much about Catalan numbers (including the average of Catalan objects). Linial's proof cannot be easily improved, unfortunately. Now, a beautiful Kahn-Linial proof of the weaker $1/2e$ bound starts with the average LE of $\mathcal P$. If $\beta(n)=o(1)$, their argument plus the (earlier) Grünbaum Theorem implies the $(1/e-\varepsilon)$ bound, already a nice result. Now, Richard's calculaitons give $p_{17} \to 9949/1024 \approx 9.7158$, $p_{23} \to 9.75$. This means that taking $x=(1,7)$ and $y=(2,3)$ in $(n,n)$ gives $$0.3553 < \frac1e \left(1-\frac{35}{1024}\right) < P(x\prec y) < \frac2e \left(1+\frac{35}{1024}\right) < 0.6447 $$ for $n$ large enough (unless I miscalculated). This improvement over $1/3$ bound is good to know, but surely one can do better. UPDATE (May 27, 2020): In our most recent paper we found an upper bound $O\bigl(n^{-5/4}\bigr)$ for the sorting probability for these Catalan posets. This problem was stated in the previous update. Note that we obtained only the upper, but not the lower bounds! Swee Hong Chan, Igor Pak, Greta Panova, Sorting probability of Catalan posets, preprint, 2020. REPLY [7 votes]: This is not a solution, but rather a long comment. Let $f^{a,b}$ denote the number of standard Young tableaux (SYT) of shape $(a,b)$. The number of SYT $T$ of shape $(n,n)$ with $T_{1d}=k$ is $f^{d-1,k-d}f^{n-k+d,n-d}$. Hence $$ p_{1d} = \frac{1}{C_n}\sum_{k=d}^{2d-1} kf^{d-1,k-d}f^{n-k+d,n-d}. $$ There is a similar formula for $p_{2,d}$, though the number of terms in the sum increases as $n\to\infty$. In particular, \begin{eqnarray*} p_{12} & = & \frac{1}{C_n}\left( 2f^{n,n-2}+3f^{n-1,n-2}\right)\\ & = & \frac{1}{C_n}\left( \frac{2\cdot 3(2n-2)!}{(n+1)!(n-2)!} +\frac{3\cdot (2n-2)!}{n!(n-1)!}\right)\\ & = & \frac{3(3n-1)}{2(2n-1)}. \end{eqnarray*} Write $\bar{p}_{ij}=\lim_{n\to\infty}p_{ij}$ (assuming this limit exists, which I believe is always the case). Thus $\bar{p}_{12}=\frac 94$. Similarly, $$ p_{21} = \frac{1}{C_n}\left( \sum_{k=2}^{n+1} kf^{n-1,n-k+1}\right). $$ Now \begin{eqnarray*} \frac{f^{n-1,n-k+1}}{C_n} & = & \frac{(2n-k)!(k-1)n!(n+1)!}{n!(n-k+1)!(2n)!}\\ & \to & \frac{k-1}{2^k}. \end{eqnarray*} Thus (assuming we can interchange a limit and an infinite sum) $$ \bar{p}_{21} = \sum_{k\geq 2}\frac{k(k-1)}{2^{k-1}} = 4 $$ (modulo computational error). In general, $\bar{p}_{1d}$ will be given by a finite sum, and $\bar{p}_{2d}$ by an infinite series. Addendum. I worked out $\bar{p}_{1d}$ in general, namely, \begin{eqnarray*} \bar{p}_{1d} & = & \sum_{k=d}^{2d-1} k(2d-k+1)f^{d-1,k-d}2^{-k}\\ & = & 2^{-2d+1}(d+1)\left(4^d-{2d+1\choose d}\right). \end{eqnarray*} Beginning with $d=2$, the numbers are $$ \frac 94,\ \frac{29}{8},\ \frac{325}{64},\ \frac{843}{128},\ \frac{4165}{512},\ \frac{9949}{1024},\ \frac{185517}{16384},\dots. $$ We can also write $$ \bar{p}_{1,d-1} =2d-\frac{d{2d\choose d}}{4^{d-1}}. $$ Addendum #2. I worked out $\bar{p}_{2d}$. If my computation is correct, then $$ \bar{p}_{2d} = 2d+\frac{d{2d\choose d}}{4^{d-1}}. $$ Compare with the formula for $\bar{p}_{1,d-1}$ above. Is there a less computational reason for such simple formulas? Do they extend to shapes other than $n(1,1)$, e.g., $n(1,1,1)$ or $n(2,1)$?<|endoftext|> TITLE: Does a non-singular matrix have a large minor with disjoint rows and columns and full rank? QUESTION [7 upvotes]: Say I have an $n$-by-$n$ non-singular matrix $A$ all of whose diagonal entries are $0$. We call an $m$-by-$m$ minor of $A$ good if its set $I$ of row indices and its set $J$ of column indices ($I,J\subset \{1,2,\dotsc,n\}$) are disjoint. Can one give a good lower bound on the size $m$ of the largest non-singular good minor of $A$? (Perhaps $m = \lfloor n/2\rfloor$?) EDIT: All right, so obviously there aren't enough conditions - the answer is too easy. What if $A$ is antisymmetric? REPLY [17 votes]: I am assuming the question is for antisymmetric matrix. Then $n$ is even. The claim follows from the properties of Pfaffian (see wikipedia): If $M$ is $2n$ by $2n$ anti-symmetric matrix, then $\det(M)=Pf(M)^2$, where $Pf(M) = 2^{-n} \sum_{I\sqcup J=[1,2n]} \pm \det(M_{I,J})$, where $I, J$ specify partition of the set $\{1,\dots,2n\}$ into two subsets of size $n$. For each such partition we take the corresponding minor. The sign is the sign of the permutation $(i_1,j_1,i_2,j_2,\dots,i_n,j_n)$ where $I=\{i_1,\dots,i_n\}$ and $J=\{j_1,\dots,j_n\}$ so that $i_1<\ldots TITLE: Independence of Duistermaat-Heckman measure QUESTION [7 upvotes]: Suppose that a compact Kähler manifold $(X,\omega)$ has a real torus acting on it by symplectomorphisms in a Hamiltonian way (the torus is not necessarily of maximal rank). Then for any smooth function from $PSH_{tor}(X,\omega) := \{\phi \in PSH(X,\omega)\;|\; \phi\text{ - invariant}\}$ the form $\omega_\phi := \omega + i\partial\bar{\partial}\phi$ is another invariant symplectic form in the same cohomology class, thus it has the same moment polytope. Now the question is: if $J$ is the moment map for $\omega$ and $J_\phi$ is the moment map for $\omega_\phi$, do they produce the same Duistermaat-Heckman measure on the polytope? I suppose they do, but why? REPLY [4 votes]: We just need to show that $(X,\omega)$ and $(X,\omega_\phi)$ are $T$-equivariant symplectomorphic, then there is a commutative diagram relating the moment maps $J$ and $J_\phi$ over the same polytope through the equivariant symplectomorphism, and this directly implies their induced Duistermaat-Heckman measure agree. We apply Moser's argument. Let $\omega_s=(1-s)\omega+s\omega_\phi$. Then $\omega_s-\omega=sd(\Re i\overline{\partial}\phi)$. Let $Y_s$ be the vector field satisfying $\iota_{Y_s}\omega_s=-\Re i\overline{\partial}\phi$, and let $\varphi_Y^s$ be the time $s$ flow for $Y_s$. Then we have $$\frac{d}{ds}(\varphi_Y^s)^*\omega_s=(\varphi_Y^s)^*(\mathcal{L}_{Y_S}\omega_s+\frac{d}{ds}\omega_s)=(\varphi_Y^s)^*(d\iota_{Y_s}\omega_s+d\Re i\overline{\partial}\phi)=0.$$ Hence $(\varphi_Y^1)^*\omega_\phi=\omega$. Since $\varphi_Y^1$ commutes with the $T$-action (because by definition $Y_s$ is $T$-invariant), it gives a $T$-equivariant symplectomorphism between $(X,\omega)$ and $(X,\omega_\phi)$.<|endoftext|> TITLE: Hopf algebra kernels vs. algebra kernels QUESTION [9 upvotes]: Let $f: H_1 \rightarrow H_2$ be a map of graded connected cocommutative Hopf algebras over a perfect field. Let $H \subset H_1$ be the Hopf algebra kernel of $f$, and let $I \subset H_1$ be the kernel of $f$, viewed as an algebra map. Let $\bar H$ be the positive dimensional part of $H$, and let $(\bar H) \subset H_1$ be the algebra ideal generated by $\bar H$. Clearly $(\bar H) \subseteq I$. Questions: Does $(\bar H) = I$? It seems likely that this is a standard fact. If so, where is this in the literature? (The hypotheses that the Hopf algebras are cocommutative, and having the field be perfect, just happen to hold in the situation I am considering, and perhaps are irrelevant.) REPLY [5 votes]: The paper "A correspondence between bi-ideals and sub-Hopf algebras in cocommutative Hopf algebras" by K. Newman (J. Algebra, Volume 36, Issue 1, July 1975, Pages 1-15) may answer your question. See also Susan Montgomery, "Hopf algebras and their actions on rings," in particular the question at the bottom of p. 36, Theorem 3.4.6, and ensuing discussion. (This is what led me to Newman's paper.) And as I said in a comment, Proposition 1.3 in Wilkerson's paper "The Cohomology Algebras of Finite Dimensional Hopf Algebras" may be relevant (and see also Theorem 4.9 in Milnor-Moore), although it is only stated for surjections of Hopf algebras. Wilkerson is the only one of these working in the graded connected setting, so if you have graded connected Hopf algebras which are not actually Hopf algebras if you forget the grading, you should take care with the other results. REPLY [3 votes]: -too long for a comment- I am a little confused about the way terminology is used in the OP. Maybe i'm missing the point; in case i do not, the closest result i know of -quite general and does not refer specifically to graded or connected or cocommutative case- is Lemma 16.0.2, p. 306, of Sweedler's book. Copying verbatim: Let $K$ and $L$ be hopf algebras and $\pi:K\to L$ a surjective hopf algebra map. Let $A=\{ g\in K / (Id\otimes\pi)\Delta(g)=g\otimes 1$. $A$ is a subalgebra of $K$ and $\varepsilon_K|A$ is an augmentation. The augmentation ideal $(\ker\varepsilon)\cap A$ is denoted $A^+$. If $A^+ K$ denotes the right ideal in $K$, generated by $A^+$, we have: $$\ker\pi=A^+K$$<|endoftext|> TITLE: What are the modularity properties of Weierstrass sigma function? QUESTION [7 upvotes]: I'm a little confused at the sigma orientation of tmf, see e.g. Witten genus and its references. The Weierstrass sigma function can be written as $$\sigma_L(z)(q)=\frac{z}{\exp\left(\sum_{k\ge 2} G_k(q) \frac{z^k}{k!}\right)}$$ and is the exponential for the sigma orientation of Tate K-theory over $\mathbb{Q}[[q]]$. Doesn't the sigma orientation correspond to the universal elliptic formal group law in the neighborhood of that cusp? And as such, shouldn't $\sigma_L^{-1}(z)(q)$ be a logarithm for the universal elliptic formal group law? If so, its inverse (in $z$) should be some kind of weight-1 modular form in $q$, no? A logarithm for the universal formal group law has the same coefficients as an invariant differential fiberwise, so it should transform like a section of the pushforward of the relative differentials $\pi_* \Omega_{E/S}$, whose global sections down below are weight-1 modular forms. However, the expression I wrote above doesn't really seem to transform in the correct way; I've tried computing the action of $\Gamma$ on $\sigma_L^{-1}$ with the usual substitution $\tau\mapsto \frac{a\tau + b}{c\tau + d}$ and $z\mapsto \frac{z}{c\tau + d}$, but all I end up getting is a mess. Is this coordinate $z$ not the usual coordinate coming from $(\tau, z)\in \mathbb{H} \times \mathbb{C}$? Where am I going wrong? REPLY [5 votes]: The classical Weierstrass sigma function is not exactly $\sigma_L$ but rather, with your notations, $\sigma(z,\tau)=e^{az^2} \sigma_L(2\pi iz)(q)$ for some constant $a$, see Remark 5.3 in Ando, Basterra, The Witten genus and equivariant elliptic cohomology. (It is more standard to put $u=e^{2\pi iz}$ instead of $u=e^z$, this is for example the convention in Silverman's books.) The modularity property cannot be found using the infinite $q$-product of $\sigma$. Rather we can use the definition of $\sigma$, namely a product over the lattice $\mathbb{Z}+\tau\mathbb{Z}$. The link between the two representations is worked out in details in Silverman, Advanced topics in the arithmetic of elliptic curves (Chapter 1, Sections 5 and 6). You're right that $\sigma$ basically transforms like a modular form of weight -1. Using the product over the lattice it is immediate that \begin{equation*} \sigma\bigl(\frac{z}{c\tau+d},\frac{a\tau+b}{c\tau+d}\bigr) = (c\tau+d)^{-1} \sigma(z,\tau) \end{equation*} for any $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ in $\mathrm{SL}_2(\mathbb{Z})$. The way to produce modular forms out of that is to specialize $z=\alpha+\beta\tau$ where $\alpha,\beta$ are fixed rational numbers (the level of the modular form is the denominator of $(\alpha,\beta)$). But here it won't work as easily because $\sigma$ is not periodic with respect to $z$, you have to take suitable products or quotients of specializations to get true modular forms. EDIT. Actually we can also use the formula you mention to show the modularity property. The only nontrivial step is the transformation rule for the Eisenstein series $G_2$, which is only quasimodular. This is classical and can be found for example in the book of Cohen and Strömberg on modular forms. The quasi-modularity will then correspond to this additional factor $e^{az^2}$.<|endoftext|> TITLE: Yau's problem: Construct a triangle given a side, an angle, and an angle bisector QUESTION [11 upvotes]: In Shing-Tung Yau's autobiography The Shape of a Life, he mentions a problem that he came up with as a teenager. Suppose you know the length of one side of a triangle, one angle, and the length of one angle bisector. Can you construct the corresponding triangle, using just a compass and ruler? I worked on this problem for the better part of a year and made little headway. … One day, I found a book that discussed [this problem]. I learned that it could not be solved, which came as quite a relief. The book cited a recent argument that proved you could not construct one, and only one, triangle that satisfied three of these conditions. I was excited to see that "my problem" had stumped other people and was only recently shown to be insoluble. I further realized that this same problem was similar to one that dated back many centuries: Could you trisect an angle if you had only a ruler and compass? No, you could not. Nor could you solve another long-standing problem, "squaring a circle." … I was proud to find out that my problem was in the same category as these two classic problems. I'm curious to learn more about the history and literature of this problem. I tried doing some searching but when I use the obvious keywords, I get too many hits on unrelated problems. Possibly this question belongs on some other stackexchange site; I'm willing to migrate it if people think it should be. EDIT: I tried contacting Yau directly. He confirmed that there is indeed a non-constructibility theorem here, but he was not able to come up with a literature reference on the spot. In particular he couldn't remember much about the book where he first saw the result in print, other than that it was in Japanese. REPLY [6 votes]: This isn‘t really an answer but gives information that might be of interest to you. The problem you quote is an example of a class which goes under the heading „recovering a triangle from three parts“. There are two types—-recovering from special points and recovering from quantities. The titles are self-explanatory and the one you mention belongs to the second type. There are three questions of interest—-the existence of a solution, its uniqueness, and its constructibility (in the sense of a ruler and compass construction). These questions have had a prominent position at varios epochs——in the nineteenth century during the resurgence of triangle geometry; and recently (applications of computers for finding and proving results). Their rebirth in modern times goes back to Euler, who considered some examples in a paper with the title „A simple proof of some difficult geometrical problems“ (my translation from the latin). There are 27 possible choices of the quantities you mention for a given triangle and so the question has several variants (not 27, of course). There is a systematic method to reduce such questions to ones about equations in two or three variables—-again about existence, uniqueness and constructibility. I have done this explicitly for one of your cases and here it is, for what it is worth: the equations are $b^2-p^2=p^2 B$ and $$A(a+b)^2 =((b-ap-bp)^2+(a+b)^2(b^2-p^2). $$ Here, these are equations in the variables $a$ and $b$, where $A>0$ and $B>0$ are given and $p=\frac 12(b^2-a^2+1)$.<|endoftext|> TITLE: Amenable groups with special presentations QUESTION [5 upvotes]: Is there a group with a presentation $\left< X \mid r_i, i \in \mathbb{N} \right>$ (where $X$ is finite) with $\left< X \mid r_i, i \in A \right>$ is amenable if and only if $A\subset \mathbb{N}$ is infinite. REPLY [5 votes]: An almost yes. For $n,m\ge 2$ consider the group $$H=H(n,m)=\langle t,x,y\mid txt^{-1}=x^n,\;t^{-1}yt=y^m\rangle$$ Remark: $H$ is a semidirect product $\mathbf{Z}\ltimes(\mathbf{Z}[1/n]\ast\mathbf{Z}[1/m])$, where the positive generator of $\mathbf{Z}$ acts by multiplication by $n$ on the first factor, and by $1/m$ on the second factor. We have an automorphism $\psi$ of $H$, mapping $(t,x,y)\mapsto (t,x^n,y)$. (It is not inner.) Define $r_i=[t^{-i}xt^i,y]$; note that $r_i=\psi^{-i}(r_0)$. Then in $H(n,m)$, the relator $r_i$ is a consequence of $r_{i+1}$ for each $i$ (i.e., is in the normal subgroup generated by $r_{i+1}$). Define $G_i$ as quotient of $H$ by the $r_j$ for $j\le i$. Then $G_i$ is isomorphic to $G_0$ for all $i$. Note that this is a non-ascending HNN extension, namely of $\mathbf{Z}^2$, using the isomorphism between $\mathbf{Z}\times m\mathbf{Z}$ and $n\mathbf{Z}\times \mathbf{Z}$ given by the diagonal matrix $(n,1/m)$. In particular it has a non-abelian free subgroup. On the other hand, $$G_\infty=\langle t,x,y\mid txt^{-1}x^{-n},\;t^{-1}yty^{-m},\;\;r_i:i\ge 0\;\;\rangle=H/\langle\!\langle r_i:i\ge 0\rangle\!\rangle$$ is the metabelian semidirect product $\mathbf{Z}\ltimes(\mathbf{Z}[1/n]\times\mathbf{Z}[1/m])$, where the positive generator of $\mathbf{Z}$ acts by the diagonal matrix $(n,1/m)$. So, $H$ modulo any finite subfamily of the family $(r_i)$ is non-amenable, and modulo any infinite subfamily is metabelian. This almost answers the question, except that we have these two additional initial relators. I hope this is enough for your purposes.<|endoftext|> TITLE: Quotient of Three Dimensional Torus by Permutation on Coordinates QUESTION [14 upvotes]: The Mobius Strip can be realized as a quotient of $T = (S^1)^2$ via the identifications $(x,y) \sim (y,x)$. I tried to generalized this concept to a higher dimension, and consider the quotient of $(S^1)^3$ by the action of the symmetric group $S_3$ on the coordinates. I was able to compute the homology of this space: $H_n = \mathbb{Z}$ for $n = 0,1$, and 0 otherwise (with reduced homology being 0 at $n=0$ as well). Even with this information I wasn't able to identify said space in any other way. Is it well known, or, can it be described in any other fashion? What can be said about higher dimensions? REPLY [12 votes]: What you are describing are the symmetric powers of $S^1$. The symmetric power $SP^n(S^1)$ is a fibre bundle over $S^1$ with fibre an $(n-1)$-simplex. (In particular your calculation of the homology for $n=3$ checks out.) This bundle is orientable if $n$ is odd, and non-orientable if $n$ is even. This result is due to H. R. Morton, and is nicely written up at the nLab page https://ncatlab.org/nlab/show/symmetric+product+of+circles<|endoftext|> TITLE: Intuition for the expected number of returns of a Dyck path QUESTION [5 upvotes]: It is known that the expected number of returns of a Dyck path of semilength $n$ to the $x$-axis is $3n/(n+2)$, so it tends to 3 as $n\to\infty$. (This was proved in the Dyck path context by Deutsch in Equation (6.23) of his 1999 paper "Dyck path enumeration", but in the equivalent context of ordered trees, the result dates back to at least Dershowitz and Zaks' 1980 paper "Enumerations of ordered trees", where it is Corollary 4.1.) The computations necessary to compute the exact value of this expected value are not difficult, and there is a fairly intuitive explanation of why the answer should tend to 3, given that Dyck paths are counted by the Catalan numbers, which are approximately $4^n$. Instead I ask is there a simple and intuitive explanation for why the expected number of returns of a Dyck path should be finite in the first place? (In the language of Chapter V.2 of Flajolet and Sedgewick's Analytic Combinatorics, I am essentially asking for an intuitive explanation of why the Catalan numbers do not form a supercritical sequence.) REPLY [4 votes]: I am not sure that the following qualifies for a simple and intuitive explanation, but it may shed some light from a probabilistic perspective in a more general context of random trees: as in the Dershowitz and Zaks' paper, the number returns of a random uniform Dyck path of semilength $$ to the $$-axis has the same distribution as the outdegree of a uniform random plane tree with $2n$ edges. it is well known that a uniform random plane tree with $2n$ edges has the same distribution as a Bienaymé-Galton-Watson random tree with geometric $1/2$ offspring distribution, conditioned to have $2n$ edges. Now consider a more general critical offspring distribution $\mu$ (a probability distribution on the non-negative integers with mean $1$), and assume that: $\mu(0)+\mu(1)<1$ (to avoid trivial cases), $\mathbb{P}(T \text{ has } n \text{ vertices})>0$ for every $n$ sufficiently large (to avoid periodicity) and that the variance of $\mu$ is finite (these assumptions are satisfied for the geometric $1/2$ offspring distribution). Let $T$ be a non-conditioned Bienaymé-Galton-Watson tree with offspring distribution $\mu$ (roughly speaking, one starts with an ancestor which has a random number of children distributed according to $\mu$, then each one of its children has an independent random number of children distributed according to $\mu$, and so on) and let $T_n$ be a Bienaymé-Galton-Watson tree with offspring distribution $\mu$ conditioned on having $n$ vertices (which is well defined for $n$ suffficiently large). By using the coding of BGW trees by the so-called Lukasiewicz path (see e.g. Section 1.1 here), if $(S_n)_{n \geq 0}$ is a random walk on the integers starting from $0$ and with jump distribution given by ${S}_1=\mu(i+1)$ for $i \geq -1$, and using the cyclic lemma (see e.g. Section 5 here), one gets that \begin{eqnarray*} \mathbb{P}(\text{degree root}(T_n) =k) & = \mu(k) \frac{\mathbb{P}(\text{a random forest with } k \text { trees has } n-1 \text{ vertices})}{\mathbb{P}(T \text{ has n vertices})} \\ & = \mu(k) \frac{ \frac{k}{n-1} \mathbb{P}(S_{n-1}=-k)}{\frac{1}{n}\mathbb{P}(S_n=-1)}. \end{eqnarray*} By the local limit theorem, there exists a constant $C>0$ such that $\frac{\mathbb{P}(S_{n-1}=-k)}{\mathbb{P}(S_n=-1)} \leq C$ for every $n$ sufficiently large and $k \geq 1$, so that for every $n$ sufficiently large and $k \geq 1$ $$ \mathbb{P}(\text{degree root}(T_n) =k) \leq C k \mu(k).$$ It follow that $\mathbb{E}[\text{degree root}(T_n)]$ is bounded since $\mu$ has finite variance. Also, another reason that the expected number of returns of a Dyck path of semilength $$ to the $$ tends to $3$ is that $3$ is the expectation of a size-biased geometric $1/2$ random variable. Indeed (see this survey for details): Large random Bienaymé-Galton-Watson trees with critical offspring distribution converge in distribution for the local topology to the so-called Kesten's infinite random tree. In this infinite random tree, the outdegree of the root follows a size biased distribution. As a consequence, the number of returns of a Dyck path of semilength $$ to the $$-axis converges in distribution to a size-biased geometric $1/2$ random variable, and one can see that this convergence also holds in expectation.<|endoftext|> TITLE: Large dominating sets in tournaments QUESTION [6 upvotes]: It is known that in any tournament with $n$ vertices, there is a dominating set of size no more than $\lceil \log_2 n\rceil$. (See Fact 2.5 here.) What are tournaments such that any dominating set is of size $\Omega(\log n)$? No example is given in the link above. A tournament that does not work is one where the vertices are on a cycle and each vertex has an edge to $(n-1)/2$ following vertices clockwise -- in this case taking two opposite vertices already gives a dominating set. REPLY [10 votes]: All logarithms are base-$2$ here. With high probability, a random tournament has no dominating sets of size $k\approx\log n-2\log\log n$, as shown by Erdős [1]. Paley tournaments: if $q\equiv3\pmod4$ is a prime power, define a tournament whose vertex set is the finite field $\mathbb F_q$ by $$x\to y\iff y-x\text{ is a square in }\mathbb F_q.$$ Graham and Spencer [2] proved that the Paley tournament has no dominating sets of size $k\approx\frac12\log q-\log\log q$, as a consequence of Weil’s bound on character sums. Blass and Rossman [3] gave a somewhat complicated, but explicit construction of a tournament with an elementary proof that it has no dominating sets of size $k\approx(\log n)^{1/4}$. In all these examples, the tournaments actually have stronger properties: they satisfy the extension axioms up to size $k$ (i.e., whenever you select a set of $k$ vertices, and prescribe for each of them individually whether it wins or loses, there exists a vertex that fits these constraints). References: [1] Paul Erdős: On a problem in graph theory, Mathematical Gazette 47 (1963), no. 361, pp. 220-223. [2] Ronald L. Graham, Joel H. Spencer: A constructive solution to a tournament problem, Canadian Mathematical Bulletin 14 (1971), no. 1, pp. 45-48. [3] Andreas Blass, Benjamin Rossman: Explicit graphs with extension properties, Bulletin of the EATCS 86 (2005), pp. 166–175.<|endoftext|> TITLE: Number of numbers in $n$th difference sequence QUESTION [7 upvotes]: Suppose that $r$ is an irrational number with fractional part between $1/3$ and $2/3$. Let $D_n$ be the number of distinct $n$th differences of the sequence $(\lfloor{kr}\rfloor)$. It appears that $$D_n=(2,3,3,5,4,7,5,9,6,11,7,13,8,\ldots),$$ essentially A029579. Can someone verify that what appears here is actually true? Example: for $r=(1+\sqrt{5})/2$, we find $$(\lfloor{kr}\rfloor)=(1,3,4,6,8,9,11,12,14,16,17,\ldots) = A000201 = \text{ lower Wythoff sequence,}$$ \begin{align*} \Delta^1 =&(2,1,2,2,1,2,1,2,2,1,2,2,1,\ldots), & D_1=2, \\ \Delta^2 =&(1,-1,1,0,-1,1,-1,1,0,-1,1,\ldots), & D_2=3, \\ \Delta^3 =&(-2,2,-1,-1,2,-2,2,-1,-1,2,\ldots), & D_2=3. \end{align*} REPLY [3 votes]: It's easy to see that $D_n\leq \texttt{A029579}(n)$. Indeed, $\Delta^1$ is a Sturmian word, which is known to have exactly $n+1$ factors of length $n$. Now, $\Delta^n$ is formed by values of the $(n-1)$-th difference operator on the factors of $\Delta^1$ of length $n$, i.e., $$\Delta^n = \left(\sum_{i=0}^{n-1} \binom{n-1}{i} (-1)^{n-1-i} \Delta^1_{k+i}\quad \big|\quad k=1,2,\dots\right).$$ For even $n$, we immediately have $D_n\leq n+1 = \texttt{A029579}(n)$. For odd $n$, we additionally notice (I did not verify this carefully) that (i) the reverse of a Sturmian factor is a factor itself, (ii) values of the operator on a factor and its reverse are the same, and (iii) there are exactly two symmetric factors. Hence, here we have $$D_n\leq \frac{n+1-2}2 + 2 = \frac{n+3}2 = \texttt{A029579}(n).$$ It remains to prove that, besides the aforementioned cases, the operator values on factors of length $n$ are distinct.<|endoftext|> TITLE: How do we explain the use of a software on a math paper? QUESTION [24 upvotes]: Suppose one has written a math/computer science paper that is more focused in the math part of it. I had a very complicated function and needed to find its maximum, so I used Mathematica (Wolfram) to do it. How do I explain that? "Using wolfram we find the maximum of $f$ to be $1.0328...$ therefore...". It looks very sloppy. REPLY [5 votes]: How do I explain that? "Using wolfram we find the maximum of $f$ to be $1.0328...$ therefore...". It looks very sloppy. Well, it looks sloppy because it kind of is sloppy by your description. Maybe the right framework for thinking about your question is to imagine that the maximization you are claiming is a component in a claimed proof of, say, the Riemann hypothesis. Would you pay serious attention to a paper containing such a claimed proof that purported to rely on an opaque software-based maximization without providing any code, let alone any guarantees that the code and the underlying software platform it runs on do what they say they do? I wouldn’t, nor do I expect any serious person to allow such a paper to be published. Now, I‘m guessing the problem your paper claims to solve isn’t as important as the Riemann hypothesis. The question of what authors of math papers can get away with and still get their paper published is distinct from the question of what authors should do as an ideal to aspire to that would make their paper’s claims truly convincing and watertight. People do get away with all kinds of minor (or even major) sloppiness, of the kinds you mention and of other kinds having nothing to do with the use of computers, all the time. I know I have. Nonetheless, since you seem to be asking about what is the “right” way to explain your result, the answer (assuming you are publishing what is meant to be a rigorous theorem with a proof that’s up to the standards of a good pure math journal) is: provide as many details about your claimed maximization as are needed to convince almost everyone in the research community of the validity of your claims. For myself, seeing your Mathematica code would be an absolute minimum to satisfy this condition. Depending on the precise nature of your calculations, I may also want to see that they satisfy some combination of the following conditions: They are based on algorithms that have been around for a long time and everyone is sure are correct. They are based on algorithms that have themselves been published in peer-reviewed journals. They can be replicated with relative ease in software packages other than Mathematica. Hope this helps.<|endoftext|> TITLE: Long time existence for heat flow in Corlette-Donaldson Theorem QUESTION [5 upvotes]: I'm having some minor confusion about the proof of the Corlette-Donaldson Theorem found here (Theorem 3.14) https://arxiv.org/pdf/1402.4203.pdf For completeness, the statement is as follows. Theorem: Let $M$ be a closed manifold and $\rho: \pi_1(M)\to SL_n(\mathbb{C})$ be a semisimple representation. Then there is a $\rho$-equivariant harmonic map from $\tilde{M}\to SL_n(\mathbb{C})/SU_n(\mathbb{C})$. (Look at the survey starting from page 27, for the notations and conventions). The proof uses the well-known heat flow method, pioneered by Eells and Sampson. Namely, one considers the equation $$\frac{\partial}{\partial t} u_t = \tau(u_t)$$ where $\tau(\cdot)$ denotes the tension fields, $t\in [0,\infty)$, and $u_0$ is some fixed $\rho$-equivariant map. Note that if $u_t$ is connected to $u_0$ along the heat flow then $u_t$ is necessarily $\rho$-equivariant as well. One notes you can solve this equation for all times and there is a limiting map that is necessarily harmonic. Here is the source of my confusion. Eells-Sampson prove that given a map $f$ from a compact manifold to some other manifold, there is a short-time solution to the heat equation. They then give some sufficient conditions for long time existence. In particular, if this solution remains bounded and the target manifold has negative curvature, then we get long time existence and convergence to a harmonic map. Hamilton proves a similar statement but for maps between compact manifolds with boundary. However, Wentworth is working on some cover of $M$, not necessarily compact. To obtain long time existence, namely a solution, Wentworth quotes the papers of Eells-Sampson, and Hamilton. He then shows the solution remains bounded and seems to apply some results from Eells-Sampson. I am wondering, why is he able to use these results? There must be something about the $\rho$-equivariance that is making this possible, but I'm not seeing it. (A similar thing is done in, say, Donaldson's original paper, and the same issue arises). Thanks. REPLY [2 votes]: I won't answer your question "why can he use these results" directly, but let me indicate the structure of the proof and hopefully this will alleviate some of your confusion. For the long time existence argument, I don't think you need look at Eells-Sampson or Hamilton, or any other paper. The limiting convergence to a harmonic map is in the end an application of Arzela-Ascoli to a new sequence of maps $g_{n}\cdot f_{n}$ where $g_{n}\in {SL}_{n}(\mathbb{C});$ this sequence is cooked up to ensure for some $x\in \widetilde{X},$ we have $g_{n}\cdot f_{n}(x)=p$ is fixed, which will allow us to apply Arzela-Ascoli. Here, the $f_{n}: \widetilde{X}\rightarrow SL(n,\mathbb{C})/SU(n)$ is some sequence of maps along the flow. The non-compactness of the universal cover $\widetilde{X}$ is not an issue, because you only need to prove convergence on a compact fundamental domain for the action of the fundamental group, since the map is equivariant. Using Arzela-Ascoli and the previously obtained estimates in the paper, you always obtain a limiting harmonic map $f_{\infty}: \widetilde{X}\rightarrow SL(n,\mathbb{C})/SU(n).$ But, the key point is that these maps $g_{n}\cdot f_{n}$ are now $\rho_{n}:=g_{n}\circ \rho \circ g_{n}^{-1}$-equivariant. Since $SL(n, \mathbb{C})$ is non-compact, it's not clear that the $\rho_{n}$ sub-converge to some $\rho_{\infty}$ for which $f_{\infty}$ is $\rho_{\infty}$-equivariant. Moreover, even if you achieve this, it's not clear that $\rho_{\infty}$ is conjugate to $\rho,$ which is what needs to be proved. What he proves, using the semi-simple hypothesis, is that the $g_{n}$ sub-converge to some $g\in SL_{n}(\mathbb{C})$ and therefore $\rho_{\infty}=g\circ \rho \circ g^{-1}$ and $f_{\infty}$ is $\rho_{\infty}$-equivariant. If $\rho$ is not reductive, what is happening? Well, it's a key fact that the orbit closure of $\rho$ under the conjugation action contains a unique reductive orbit. The limiting harmonic map you get $f_{\infty}$ will be equivariant for a new reductive representation $\rho_{\infty},$ which lies in the orbit closure of $\rho,$ but is not conjugate to $\rho.$ A helpful alternative viewpoint getting rid of the non-compactness is to see your maps $g_{n}\cdot f_{n}$ as sections of the bundles $E_{n}:=\widetilde{X}\times_{\rho_{n}}SL(n,\mathbb{C})/SU(n)$ which are now bundles over the compact Riemann surface $X.$ But of course now to apply Arzela-Ascoli, you'll need to understand the sense in which the $E_{n}$ converge to $E_{\infty}$ and use some version of Arzela-Ascoli which takes into account varying targets. This is perhaps a bit more conceptual than the proof I describe above, but it turns out to be technically harder to work out all the details, at least I assume, since I haven't seen a proof along these lines in any of the sources I know.<|endoftext|> TITLE: Word length zeta function QUESTION [7 upvotes]: Let $G$ be a group with a finite symmetric set $S$ of generators. Let $\ell_S(x)$ denote the word-length of a given $x\in G$. For $s\in\mathbb C$ set $$ Z(s)=\sum_{x\in G^*}\ell_S(x)^{-s}, $$ where $G^*=G\smallsetminus\{1\}$. It may happen that this sum converges for some $s$. By Gromov's Theorem on groups of polynomial growth, this is the case if and only if the group has a nilpotent subgroup of finite index. In the very few cases I have computed, this function turned out to be a linear combination of Riemann zetas with shifted arguments. My question is a reference request: Has this kind of Dirichlet series been investigated? If so, I would like to have references. Thank you. REPLY [4 votes]: Pierre de la Harpe's Topics in Geometric Group Theory has a chapter (VI) on series using the length function on words, but these may be traditional power series. Zeta functions are usually introduced when there is some sort of product formula, as in Euler's product formula for the Riemann zeta-function. So does this happen here, at least for some reasonable class of groups?<|endoftext|> TITLE: A Besicovitch-type Covering Theorem QUESTION [9 upvotes]: In the book The Geometry of Domains in Spaces by Krantz and Parks, the authors proved the weak $(1,1)$-type estimate of the maximal function $M_\mu f$, where $\mu$ is a Radon measure, using their version of the Besicovitch covering theorem. Let $d$ be a positive integer. Then there exists a constant $C=C(d)$ such that for any finite collection of balls $\mathcal B = \{B_i\}_{i=1}^m$ in $\Bbb R^d$ with the property that no ball contains the center of any other ball, we can partition the family $\mathcal B$ into $$ \mathcal B = \mathcal B_1 \cup\mathcal B_2 \cup \dots \cup \mathcal B_C, $$ where each subfamily $\mathcal B_j$ consists of disjoint balls. This version of the covering theorem seems pretty restrictive, especially the part about no ball contains the center of any other balls. Indeed, in theor proof of the weak $(1,1)$-type estimate, they relied on a certain claim that they did not prove. Edit As Skeeve mentioned in the comment, this claim is not explicitly stated in the book but more of a paraphrasing of the part the authors left out in a proof. Claim: Let $K\subset \Bbb R^d$ be a compact set such that each $x\in K$ is associated with a real number $r_x>0$. Then $K$ can be covered by a family of balls $$ \mathcal B = \{ B(x_i,r_i) : i=1,\dots,k\ \}, $$ where $r_i := r_{x_i}$, such that for any distinct $i,j \le k$, we have $$ x_i\notin B(x_j,r_j) \quad\text{and}\quad x_j\notin B(x_i,r_i). $$ I don't find this claim to be trivial at all. In fact, I tried many different methods but failed to prove it. Note that the mapping $x\mapsto r_x$ doesn't enjoy any nice property like continuity of any kind. While the usual version of Besicovitch covering theorem circumvents this problem, I still would like to know how to prove the above claim (or a counter example if it is actually false). REPLY [11 votes]: This trivial counterexample in $\mathbb R^2$ should have taken me five minutes. Instead, I spent almost two days. The moral is the usual one: after 50 you'd better give up on mathematics. Let $y,z$ be 2 points at distance $1$ from each other. We shall construct by induction a sequence of points $x_j$ and radii $r_j>\max(d(x_j,y),d(x_j,z))$ such that $x_j\to y$ when $j$ is odd, $x_j\to z$ when $j$ is even, $x_j$ do not lie on the line $yz$, $\max(d(x_j,y),d(x_j,z))<1$, the disk $D(x_j,\max(d(x_j,y),d(x_j,z)))$ contains $x_1,\dots, x_j$ but $x_{j+1}\notin D(x_j,r_j)$ . If you choose the radius $\rho$ for $y$ and $z$ small enough so that the corresponding disks do not contain $x_1$, you'll get a bad configuration. Indeed, an attempt to choose $y$ or $z$ as one of the centers results in the exclusion of all the centers $x_j$, after which covering $x_1$ gets impossible. Out of $x_i$, we can choose only one (if $i\max_{i\le j}d(z,x_i)+d(z,x_{j+1})\ge \max_{i\le j}d(x_i,x_{j+1})$ and choose $r_j$ anywhere between $d(x_j,z)$ and $d(x_j,x_{j+1})$. Clearly, we can keep $x_j$ with odd indices at the distance $<1/3$ to $y$ and converging to $y$ and similarly for even indices and $z$.<|endoftext|> TITLE: An inner product approach to Hopf algebras QUESTION [8 upvotes]: We fix the standard inner products on $\mathbb{C}^n$ and $\mathbb{C}^n\otimes \mathbb{C}^n$. Is there an algebra structure on $\mathbb{C}^n$ with multiplication $m$ such that the adjoint operator $m^*: \mathbb{C}^n \to \mathbb{C}^n \otimes \mathbb{C}^n$ is a coproduct (coalgebraic operation)? Is there a bialgebra structure whose product and coproduct are adjoints of each other? Is there a Hopf algebra with the latter property, and the additional condition that the antipode map is an isometry? Note: The adjoint operators are taken with respect to the corresponding inner products. REPLY [9 votes]: This doesn't directly answer your question concerning Hopf structures on $\mathbb{C}^n$, but a particularly well-studied class of Hopf algebras for which the product is the adjoint of the coproduct are known as positive self-adjoint Hopf (PSH) algebras. These were originally introduced by Zelevinsky to study representations of finite classical groups. A concise, modern introduction can be found in notes from Grinberg and Reiner which have a focus on the application of PSH algebras to combinatorics. To say a little more, a PSH algebra is a graded, connected Hopf algebra having a $\mathbb{Z}$ basis of homogeneous elements so that the structure constants for the algebra and coalgebra structures agree and are positive integers. Note that in any Hopf algebra with chosen basis the agreement of the structure constants for the algebra and coalgebra structures is equivalent to the condition that they are adjoint to one another. The structure theorem of PSH algebras says that every PSH algebra is isomorphic to a tensor product of degree-shifts of the ring of symmetric functions, with one copy for each primitive basis element.<|endoftext|> TITLE: Differentiability of the distance function from a (variable) point to a (fixed) set QUESTION [7 upvotes]: The distance of from a point $x$ to a set $A$ is defined by $$ d(x,S) = \inf\{d(x,a)\mid a\in A\}, $$ where you may choose the setting to be $\mathbb R^n$, a Banach space or a complete metric space. This is of interest in constructive analysis and my own theory called abstract Stone duality, as explained below, but I believe that my question is meaningful to classical real analysts, whose intuition I am seeking. Here $A$ is fixed and I am considering the real valued function $d(x)=d(x,A)$. My picture is that $A$ is the sea and we are on an island that is foggy (so you can't see the sea) but you can measure the altitude, which is equal to the (nearest) distance to the sea. So the geography of the island consists of conical hills. Although the altitude is a scalar (it doesn't tell you which way to get to the sea), so long as you're not at the summit of a hill, there is a unique path that goes straight down to the sea on a $45^\circ$ slope. My conjecture is that $D\cup A$ is dense, where $D$ is the subspace on which $d(x)$ is differentiable. For any point $x$, even a summit, there is a straight line $[x,a]$ of length $d(x)$, with $a\in A$. Let's call any open segment of such a line a downward path and let $D_0$ be the union of all such segments. Then $D_0\cup A$ is dense. By simple Euclidean geometry (if we're working in $\mathbb R^n$), any two downward paths through a point of $D_0$ are co-linear. If $D_0$ is open then it coincides with $D$ and my conjecture follows. Is this necessarily true? Here is the context of my question: In Bishop-style constructive analysis, a subset $S$ of a metric space is said to be located if $f(x,S)$ is a real number and not just a lower real. Many classical theorems become constructively valid once locatedness is added as a hypothesis. In various forms of constructive topology (including mine, called abstract Stone duality), there is a concept called overtness that is lattice-dual to compactness. For example, familiar results using with words closed, compact and Hausdorff have duals with the words open, overt and discrete. However, overtness is invisible (vacuously true) in the classical setting, so I am trying to show how it is related to more familiar ideas, by concentrating on the metric case. What is important is not the overt subspace $A$ qua subspace, but the predicate $\lozenge U$ on open subspaces of whether $U\cap A$ is inhabited. This satisfies $$ \lozenge \bigcup\{U_i\mid i\in I\} \Longrightarrow \exists i\in I. \lozenge U_i. $$ Since $\lozenge$ preserves unions in this sense, we need only consider its values at \emph{basic} open subspaces, which are the balls $B_r(x)$ in the case of a metric space. So we write $$ d(x) \lt r \iff \lozenge B_r(x). $$ This function has the following properties: $$ d(x) < r \iff \exists r'. d(x) < r' < r $$ $$ d(x) < r \land d(x,y)\lt s \Longrightarrow d(y)< r+s $$ $$ d(x) < r \land \epsilon\gt 0 \Longrightarrow \exists y. d(x,y)< r\land d(y)<\varepsilon $$ In fact $d(x)$ is in general only an upper real number, so the function $d$ is upper semi-continuous. They are real-valued and continuous iff the subspace is also closed; for simplicity we assume this here. The subspace in question is $A = \{a\mid d(a)= 0\}$, where $d(a)=0$ means $\forall\varepsilon\gt 0.d(a)<\varepsilon$. It is not difficult to show that this function $d$ is exactly the point--set distance $d(x,A)$ above. However, where this may be of interest to classical real and numerical analysts is this analogy with the Newton--Raphson algorithm: Newton's idea is that, given an approximant $x_0$ to a solution to $f(x)=0$ where $f$ is differentiable, we get a better approximant $$ x_1=x_0-\delta(x_0) \quad\mbox{where}\quad \delta(x) = f(x)/f'(x), $$ at least in certain good circumstances. My observation is that $\delta(x)$ is like $d(x)$ above, except that $\delta$ is a vector, but $d$ is a scalar; and $d$ satisfies a triangle law, but $\delta$ doesn't. Adding the triangle law is easy in logic but not of course in numerical analysis. The reason for asking about differentiability is to turn a scalar into a vector. I would like then to find some weaker notion than the join-preserving operator that would be an abstraction of the Newton--Raphson algorithm. The connection between locatedness and overtness above was first considered by Bas Spitters. My draft paper about these topics is Overt Subspaces of ${\mathbb R}^n$ NB This question is not my mathematical territory, so I do not know what are the appopriate tags, so others are welcome to change these. REPLY [6 votes]: For finite dimensional spaces, as distance functions are Lipschitz, they are differentiable almost everywhere. So a stronger form of your conjecture is true, namely the complement of $D\cup A$ has measure zero. In infinite dimensions this fails in general as remarked in the comments. I'm not sure what happens for infinite dimensional Hilbert spaces though. Also, the set of points where the distance function to $A$ is non differentiable is known as the medial axis of the complement of $A$, and a great deal is known about its properties. Note: http://annals.math.princeton.edu/wp-content/uploads/annals-v157-n1-p05.pdf seems to be a relevant reference for positive results in the infinite dimensional case<|endoftext|> TITLE: Taylor's polynomials and loss of real roots QUESTION [7 upvotes]: Real-rootedness, log-concavity, and unimodality are intertwined properties. It's in this light that I was prompted to ask the question below. Suppose the roots of a polynomial $p(x)$ are all real and $p(0)>0$. Fix an integer $k\geq0$ and consider the function $f=\frac1{p^{2k+1}}$. I like to consider the partial sums (polynomial) $$\sum_{j=0}^{2k}\frac{f^{(j)}(0)}{j!}x^j; \tag1$$ where $f^{(j)}$ means the $j$-th derivative. QUESTION. Is it true that the polynomial in (1) has no real roots? REPLY [7 votes]: It immediately follows from the observations that every even order Taylor polynomial of $e^{ax}$ is strictly positive for any $a\in\mathbb R$ and that $\frac 1{b-x}=\int_0^\infty e^{ax}e^{-ab}\,da$ and $\frac 1{b+x}=\int_0^\infty e^{-ax}e^{-ab}\,da$ for $b>0$ and $|x| TITLE: Matrix of cosecants appearing in equivariant index computations QUESTION [9 upvotes]: In a computation of characters of certain representations of finite cyclic groups which appear as equivariant indices of Dirac operators (using the Atiyah-Bott fixed point formula, cf. [1, Theorem 8.35]), the following elementary problem emerged. Question 1 Let $d > 2$ be an integer and set $n = \lceil\frac{d}{2}\rceil -1$. For $k \in \{1, \dotsc, n\}$ with $\gcd(k,d)=1$, we set $$ v_k = \begin{pmatrix} \csc^2(\frac{k \pi}{d})& \csc^2(\frac{2k \pi}{d}) &\cdots&\csc^2(\frac{n k\pi}{d}) \end{pmatrix} \in \mathbb{R}^n, $$ where $\csc x = \frac{1}{\sin x}$ denotes the cosecant. If $\gcd(k,d) \neq 1$, we let $v_k = (0\ \cdots\ 1\ \cdots 0)$ be the standard basis vector with entry $1$ at the $k$-th position. Question: Do the vectors $v_1, \dots, v_n$ generate the vector space $\mathbb{R}^n$? As this appears somewhat intimidating at first glance, let us simplify it to the case where $d$ is an odd prime number. Then it reads as follows: Question 2 Let $p$ be an odd prime number. Question: Is the following $\frac{p-1}{2} \times \frac{p-1}{2}$-matrix invertible? $$M_p = \begin{pmatrix} \csc^2(\frac{\pi}{p}) & \csc^2(\frac{2\pi}{p}) &\cdots & \csc^2(\frac{(p-1)\pi}{2p}) \\ \csc^2(\frac{2\pi}{p}) & \csc^2(\frac{4\pi}{p}) &\cdots & \csc^2(\frac{2(p-1)\pi}{2p}) \\ \vdots & \vdots & \ddots & \vdots\\ \csc^2(\frac{(p-1)\pi}{2p}) & \csc^2(\frac{2(p-1)\pi}{2p}) &\cdots & \csc^2(\frac{(p-1)^2\pi}{4p}) \end{pmatrix} $$ Discussion The questions appears to be of a number-theoretic nature. Potentially relevant formulas involving the cosecant appear for instance in Cauchy's elementary solution to the classical Basel problem. Moreover, numerical experiments suggest that Question 1 has an affirmative answer at least for $d \leq 200$. The determinant of the matrix in Question 2 is an integer times $\frac{1}{\sqrt{p}}$ if $p \equiv 1 \mod 4$, and an integer otherwise. This suggests a relation to quadratic reciprocity. Spotting a concrete formula for the determinants from numerical computations is difficult, however, because the numbers grow very rapidly. But in [2, Lemma 3.1], the formula $$ \prod_{j=1}^{\lfloor\frac{d}{2}\rfloor} \sin(\frac{j \pi}{d}) = \frac{\sqrt{d}}{2^{\frac{d-1}{2}}} $$ is provided which is reminiscent of our numerical observations. A more complicated formula involving quadratic reciprocity is provided in [2, Lemma 3.2]. All of this makes it plausible that it should be possible to derive a concrete formula for the determinants of the relevant matrices appearing in Question 1 and 2, but it remained elusive to me so far. References [1] Atiyah, Michael F.; Bott, Raoul, A Lefschetz fixed point formula for elliptic complexes. II: Applications, Ann. Math. (2) 88, 451-491 (1968). ZBL0167.21703. [2] Miatello, Roberto J.; Podestá, Ricardo A., Eta invariants and class numbers, Pure Appl. Math. Q. 5, No. 2, 729-753 (2009). ZBL1183.58021. REPLY [4 votes]: As suggested in the comment of JP McCarthy, the Gershgorin circle theorem indeed leads to a solution: Each row of the matrix $M_p$ is just a permutation of the first row and the entry $\csc^2(\frac{\pi}{p})$ occurs precisely once in each column. So we can permute the rows of $M_p$ to ensure that the entry $\csc^2(\frac{\pi}{p})$ appears on each diagonal position of the matrix. Denote the resulting matrix by $\tilde{M}_p$. Next, we apply Gershgorin to $\tilde{M}_p$ to show that it is invertible. For this, it suffices to show that in each row the sum of the absolute values of the non-diagonal entries is strictly smaller than the absolute value of the diagonal entry. But since each row is just a permutation of the same vector of positive numbers and we have arranged the diagonal position, this just amounts to $$\sum_{l=2}^{n} \csc^2 \left( \frac{l \pi}{p} \right) < \csc^2\left(\frac{\pi}{p} \right),$$ where $n = \frac{p-1}{2}$. To prove the latter, we use the formula $$\sum_{l=1}^{d-1} \csc^2\left(\frac{l \pi}{d}\right) = \frac{d^2-1}{3},$$ which holds for any $d \in \mathbb{N}$, see the previously mentioned Wikipedia entry on the Basel problem or this Math Stack Exchange question. From this, we conclude \begin{align}\sum_{l=2}^n \csc^2\left(\frac{l \pi}{p} \right) &= \frac{p^2-1}{6} - \csc^2\left(\frac{\pi}{p}\right) \\ &<\frac{p^2}{6} - \csc^2\left(\frac{\pi}{p}\right)\\ &< 2 \frac{p^2}{\pi^2} - \csc^2\left(\frac{\pi}{p}\right)\\ &\leq 2 \csc^2\left(\frac{\pi}{p}\right)- \csc^2\left(\frac{\pi}{p}\right) = \csc^2\left(\frac{\pi}{p}\right), \end{align} where we used $ \frac{\pi^2}{2} < 6 $ for the second inequality and $\frac{1}{x^2} \leq \csc^2(x)$ for the third inequality. This is what we needed to prove via Gershgorin that $\tilde{M}_p$ is invertible. Hence also $M_p$ is invertible. This answers Question 2 affirmatively. The same argument also applies to Question 1. Just put the $v_k$ as row vectors into a matrix and reorder it in the same way as before putting $\csc^2(\pi/d)$ in the diagonal position whenever the row index $k$ has $\gcd(k, d)=1$. The additional unit vectors can also be appropriately reordered so that they do not disturb the Gershgorin argument. We have also written up the detailed argument for this on p.12-13 of arXiv:1712.03722v2.<|endoftext|> TITLE: Asking SnapPy for core curves after surgery QUESTION [5 upvotes]: Suppose I give SnapPy a cusped hyperbolic 3-manifold (using, say, the link editor) and specify some filling. SnapPy can then provide a presentation of the fundamental group of the filled manifold. Can it tell me what the core curve of the added solid torus is, as a word in the fundamental group? REPLY [3 votes]: I think you want to use Snappy's 'fillings_may_affect_generators=False' flag for your purposes. Although you can also extract the information directly if for some reason this becomes inefficient. In[1]: M = Manifold('m004') not needed but M.fundamental_group? will give all possible flags In[2]: G = M.fundamental_group? Docstring: Manifold.fundamental_group(self, simplify_presentation=True, fillings_may_affect_generators=True, minimize_number_of_generators=True, try_hard_to_shorten_relators=True) there is more but it is redacted here In[3]: G = M.fundamental_group() In[4]: G Out[4]: Generators: a,b Relators: aaabABBAb In[5]: m=G.meridian(); G.meridian() Out[5]: 'ab' In[6]: l=G.longitude(); G.longitude() Out[6]: 'aBAbABab' In[7]: M.dehn_fill((5,1),0) In[8]: H = M.fundamental_group(fillings_may_affect_generators=False) In[9]: H Out[9]: Generators: a,b Relators: aaabABBAb ababababababaBAbAB Here, the second relation is m^5*l. The core curve will be isotopic to any curve p*[m]+q*[l] (in boundary M) such that |5q-1p|=1. More generally for filling along r,s, we want |rq-sp|=1. There are a number of simple python scripts to do that for example the extended gcd script taken from https://www.kkhaydarov.com/greatest-common-divisor-python/ def egcd(r, s): if r == 0: return (s, 0, 1) else: g, x, y = egcd(s % r, r) return (g, y - (b // a) * x, x) Here the g is the gcd of r and s, p=y - (s // r) * x and q=x. To complete the example where we will fill along (5,1), (p,q)=(1,0) so m is sufficient.<|endoftext|> TITLE: Generalization of Cauchy's eigenvalue interlacing theorem? QUESTION [26 upvotes]: Cauchy's Interlacing Theorem says that given an $n \times n$ symmetric matrix $A$, let $B$ be an $(n-1) \times (n-1)$ principal submatrix of it, then the eigenvalues of $A$ and those of $B$ interlace. Using this property, one can obtain a lower bound on the $k$-th largest eigenvalue of a $t \times t$ principal submatrix of $A$, using the $(k+n-t)$-th largest eigenvalue of $A$. This lower bound is best possible, for example when $A$ is diagonal. But for many interesting (fixed) matrices, such bound is usually far from being optimal. For example, let $$A=\begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1\\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}$$ The eigenvalues of $A$ are $2, 0, 0, -2$. So if we would like to bound from below the largest eigenvalue of its $3 \times 3$ principal submatrix using Cauchy's Theorem, we only get a lower bound of $0$. However it is straightforward to check that it is always at least $\sqrt{2}$. I am wondering if there is a more "quantitative" Interlacing Theorem, say if your matrix satisfies some additional properties (non-negative, binary, etc.), then one can obtain a better lower bound on the $k$-th largest eigenvalue of a $t \times t$ principal submatrix of $A$? REPLY [2 votes]: This is only a partial answer to the case when the symmetric matrix is nonnegative and only applies for bounding the so-called Perron root (i.e., the spectral radius). If $$r_i = r_i(A) := \sum_{k=1}^n a_{ik}$$ and $\rho = \rho(A) := \max_{\lambda \in \sigma(A)}\{|\lambda|\}$, then $$\min_{1 \le i \le n} r_i \le \rho \le \max_{1 \le i \le n} r_i.$$ This result is due to Frobenius [MR0235974]. For example, for the $3$-by-$3$ leading principal sub-matrix $$ B:= \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}, $$ we obtain $1 \le \rho(B) \le 2$. One way to try and improve these bounds is by applying the same bounds to $D^{-1} A D$, where $D$ is a diagonal matrix with positive diagonal entries. There are many other works in the literature that are dedicated to improving the Frobenius bounds above (too many to list here, but I would be happy to email you a list if you'd like), but many are quite complicated (it seems to me that you want to get $\sqrt{2}$ as a lower bound).<|endoftext|> TITLE: Almost complex manifold of dimension 2... locally isomorphic to ℂ? QUESTION [6 upvotes]: I know that this is supposed to be standard, but I don't know how to search for it... hence the question: Let $J$ be an almost complex structure on $M:=\mathbb R^2$, i.e., a $C^\infty$ section of $\mathrm{End}(TM)$ which squares to $-\mathrm{id}_{TM}$. How does one prove that $(M,J)$ is integrable, i.e. how does one prove that it is locally isomorphic to $\mathbb C$? REPLY [9 votes]: Eckmann-Frölicher say so right after they introduce the (“Nijenhuis”) torsion tensor in (1951, p. 2284), bottom: “dans $\mathrm E_2$, $t^j_{kl}$ est toujours $=0$.” And this was known to imply integrability in the plane, as Newlander-Nirenberg start by recalling with proof in (1957, p. 394): On p. 392 they also write For $n=1$ the compatibility conditions are vacuous and the problem becomes that of introducing isothermal coordinates with respect to the Riemannian metric $ds^2=|dz + ad\bar z|^2$ — a problem whose solution Lichtenstein (1919, §24e, p. 264) traces back to Gauss (1825); I would add Lagrange’s Sur la construction des Cartes Géographiques (1781, pp. 167-169).<|endoftext|> TITLE: A metric has positive sectional curvature if and only if ${\rm Ric}_{ij} < \frac{r}{2}g_{ij}$ QUESTION [6 upvotes]: This is a cross-post from my question on MSE. It is well known that In dimension three a metric has positive sectional curvature if and only if ${\rm Ric}_{ij} < \frac{r}{2}g_{ij}$. where $r$ denotes the scalar curvature. Is there a higher dimensional analogues of this theorem? Any reference or counterexample? REPLY [6 votes]: This depends on how strictly one interprets the word 'analogues'. For example, there cannot be any inequality on the Ricci tensor alone that is equivalent to positive sectional curvature in dimensions higher than $3$. The reason is that such a condition would then imply that every Einstein metric of positive scalar curvature in dimension greater than $3$ would have positive sectional curvature (since the criterion would have to hold for the round spheres), and this is known to be false. In fact, in higher dimensions, the irreducible symmetric spaces of compact type are Einstein with positive scalar curvature, but most of them have some sectional curvatures vanishing. Moreover, it would not be hard to construct Einstein metrics in dimension $4$ with positive scalar curvature that have some negative sectional curvatures. For example, many Kähler-Einstein metrics in dimension $4$ with positive scalar curvature do have some negative sectional curvatures. On the other hand, in dimension $n$, the space of Riemann curvature tensors with positive sectional curvature is an open cone on the space of Riemann curvature tensors and hence it can be described by some set of (strict) inequalities that are invariant under the natural action of $\mathrm{O}(n)$. When $n>3$, these inequalities will have to involve not only the Ricci curvature but the Weyl curvature as well. If you are willing to consider these as analogues of the inequality that works in dimension $3$ (where the Weyl curvature vanishes identically), then the answer would be 'yes, there is an analogue in each dimension'.<|endoftext|> TITLE: Hodge structure not coming from the cohomology of a manifold QUESTION [13 upvotes]: What is an explicit example a pure polarizable finite-dimensional $\mathbb{Q}$-Hodge structure that is not a subquotient of the cohomology of a scheme smooth proper over $\mathbb{C}$? What if replace "a scheme smooth proper over $\mathbb{C}$" with "closed complex manifold bimeromorphic to a closed Kaehler manifold"? Note that we are asking for an explicit example, not "Hilbert scheme blah blah countable blah blah". If it helps, assume any famous conjecture you want to assume. REPLY [9 votes]: The existence of non-motivic (not coming from algebraic geometry) polarized Hodge structures is currently non-constructive. For period domains which are not Hermitian symmetric, motivic Hodge structures are contained in countably infinite proper submanifolds, so in a sense most Hodge structures in the domain are not motivic. However the explicit construction of a single example is still an open problem.<|endoftext|> TITLE: Eigenvalues of the Laplace-Beltrami operator on a compact Riemannnian manifold QUESTION [16 upvotes]: Let $(M,g)$ be a compact Riemannian manifold, and let $\Delta_g$ be its Laplace-Beltrami operator. A "well-known fact" is that the eigenvalues of $\Delta_g$ have finite multiplicity and tend to infinity. What is the easiest/simplest way to estblish this fact? By simplest, I mean an abstract, formal, approach to the problem, not too heavily based on technical calculation. In the same line of thought, is there a philosophical/intuitive reason to see why the spectrum should behave like this? REPLY [11 votes]: A modern "simple philosophical" explanation is that this problem can be restated as an eigenvalue problem for a compact operator in an appropriate Hilbert space, whose eigenvalues are reciprocal to those of the Laplacian. Then two properties that you stated follow from the general properties of compact operators. This approach is due to Hilbert. He wrote a long series of papers on the subject in 1904-1912. (see also his book with Courant, Methods of mathematical physics). Modern expositions are usually based on Hilbert's ideas. The notions of Hilbert space and compact operator were essentially distilled from these works. An earlier philosophy of Poincaré interprets these eigenvalues as poles of certain meromorphic function in the plane (in modern language it is essentially the resolvent), and the poles of a meromorphic function are isolated and tend to infinity. Poincaré was the first to prove under general conditions the existence of an infinite sequence of eigenvalues tending to infinity. (Sur les équations de la physique mathématique, Rend. Circ. mat. Palermo, 1894 8, 57-155.) Three remarks should be made: a) At the time of Poincaré and Hilbert, the modern formal notion of compact Riemannian manifold did not exist. (The notion of compact was introduced by Aleksandrov and Urysohn in 1924, and the notion of manifold by Weyl 1913, and only for dimension 2). Even in the classical book on the subject by Titchmarsh, Eigenfunction expansions..., 1958, the words "manifold" and "compact" are not mentioned! b) There was a very large number of problems about vibrations which were solved "explicitly" in 18th and 19th century. So Hilbert and Poincaré had a lot of "empirical material" to generalize. Fourier should be mentioned: his work inspired Hilbert and Poincare. He solved many concrete eigenvalue problems but had no tools to attack the general problem. c) As a physical fact, existence of infinite discrete spectrum was first discovered (for the case of a string) by a music theorist Marin Mersenne in 1637. This started a long story of research about these eigenvalues.<|endoftext|> TITLE: Growing a chain of unit-area triangles: Fills the plane? QUESTION [11 upvotes]: Define a process to start with a unit-area equilateral triangle, and at each step glue on another unit-area triangle.                     $50$ triangles. Above, the red dot marks the starting triangle $T_0$. Then outside one edge $e$ of $T_0$, a point $p$ is chosen to form a unit-area triangle $T_1$. Now there are two edges of $T_1$ for possible growth. One edge is chosen randomly, and another unit-area triangle $T_2$ is erected on that edge. Then $T_3$ is built on an edge of $T_2$. And so on. The point $p$ at each step is chosen according to a normal distribution $\cal N(0,1)$ centered on the midpoint normal vector to $e$, as illustrated below:                     The location of $p$ follows a normal distribution. My question is: Q. Does this process fill the plane in the limit? I think Yes but am not certain. Here are two more examples (using different random seeds):                     $500$ triangles.                     $1,000$ triangles.                     $50,000$ triangles. REPLY [9 votes]: Here is a sketch of the "cheapest" way I know how to prove something like that. Filling in the details may still be a bit lengthy but should be essentially routine. To set things up, let's write $X$ for the space of oriented non-degenerate area $1$ triangles, so $\tau \in X$ is a pair $(v_1,v_2)$ with $v_i \in \mathbf{R}^2$ and $|v_1 \wedge v_2| = 2$. I then view your process as a Markov chain $(\tau_n,p_n)$ on $X \times \textbf{R}^2$ with the second coordinate $p_n$ being the $n$th value $p$ chosen in your description and $\tau_n$ the $n$th triangle obtained in your construction, translated and oriented in such a way that the edge $\overline{p_{n} p_{n+1}}$ of that triangle corresponds to $(v_1)_n$, say. We are then in the general setting that we have an autonomous Markov chain $\tau_n$ on $X$, a function $f\colon X \to \mathbf{R}^2$, and a random walk on $\mathbf{R}^2$ given by $p_{n+1} = p_n + f(\tau_n)$. In this case $f(v_1,v_2) = v_1$, but the argument doesn't care about $f$ and the Markov chain $\tau$ as long as they are "nice" enough as detailed below. The space $X$ comes with a natural 'size' $D(v_1,v_2) = \max\{|v_1|,|v_2-v_1|\}$ which has compact sublevel sets. It is also not hard to check that, if we denote by $P$ our Markov transition operator on $X$, the function $D$ is such that for every $q > 1$ there exist constants $c \in (0,1)$ and $C > 0$ such that $P(D^q) \le c D^q + C$, i.e. $D^q$ is a Lyapunov function. Even better, for $\kappa >0$ small enough $\exp(\kappa D)$, one even has $P \exp(\kappa D) \le \exp(\kappa' D) + K$ for some $\kappa' < \kappa$. It is also the case that for every compact set $K \subset X$ one can find $n > 0$ and a strictly positive measure $\nu$ on $X$ such that $P^n(x,\cdot) \ge \nu$, uniformly over $x \in K$. Together with the existence of a Lyapunov function, this tells us that $P$ admits a unique invariant measure $\mu$, as well as a spectral gap in spaces of bounded functions weighted by $D^p$. (See Harris's theorem.) By uniqueness and rotation invariance of the problem, $\mu$ must be rotation invariant so $\int f\,d\mu = 0$. We need two more ingredients inspired from stochastic homogenisation: a corrector $\phi\colon X \to \mathbf{R}^2$ solving $P\phi = \phi - f$ (which exists and is bounded by $D$ at infinity by the spectral gap mentioned above) a second-order corrector $\psi \colon X \to \mathbf{R}^2 \otimes \mathbf{R}^2$ solving $P \psi = f \otimes f + f \otimes P\phi + P\phi \otimes f - A$, where $A \in \mathbf{R}^2 \otimes \mathbf{R}^2$ is the symmetric matrix such that the average of the right hand side against $\mu$ vanishes and therefore the equation for $\psi$ is solvable. The matrix $A$ is strictly positive definite, by some algebraic manipulation combined with Cauchy-Schwartz and the non-degeneracy of the Markov chain on $X$; then rotation invariance guarantees that it is a constant multiple of the identity. The constant $A$ turns out to be the diffusion coefficient of the limiting Brownian motion in the heuristic picture. Let's write $Q$ for the Markov operator of the full Markov chain on $Y = X \times \mathbf{R}^2$, so that $(Qg)(\tau,p) = (Pg)(\tau,p+f(\tau))$ (with the obvious abuse of notation). To show that the process eventually covers the whole plane, note that we can find a countable sequence of open sets $U_n \subset Y$ such that if the chain visits each $U_n$ at least once, then the corresponding triangle process will have covered the plane. (Simply take a dense enough grid $y_n \in \mathbf{R^2}$ and, for each $n$ take $U_n$ to be a small enough ball around $(\Delta, y_n)$ with $\Delta$ the standard equilateral triangle.) In other words, it suffices to show that the process is recurrent. By standard criteria (see for example Section 8.4.2 in the book by Meyn and Tweedie), it is enough to find a function $W \colon Y \to \mathbf R$ with compact sublevel sets such that $Q W < W$ outside of a compact set. For the $2D$ Brownian motion, a possible choice of such a function is given by $V(p) = (\log(1+p^2))^\alpha$ for any choice $\alpha \in (0,1)$. The idea now is to use the correctors to build a function $W$ that explicits the intuitive fact that our random walk behaves like a Brownian motion with diffusion coefficient $A$. I claim (modulo silly mistakes) that a possible choice is $$ W(\tau,p) = V(p) + \langle \nabla V(p),\phi(\tau)\rangle + {1\over 2}\langle \nabla^2 V(p), \psi(\tau)\rangle + \exp(\kappa D(\tau))\hat V(p)\;, $$ where $\hat V(p) = 1/(1+|p|^{5/2})$ for example. This certainly has compact sublevel sets since the middle two terms are dominated by the first and last term outside of a compact set. We also have \begin{align} (QW-W)(\tau,p) &= V(p+f(\tau)) + \langle \nabla V(p+f(\tau)),(P\phi)(\tau)\rangle + {1\over 2}\langle \nabla^2 V(p+f(\tau)), (P\psi)(\tau)\rangle + (PD^4)(\tau)\hat V(p+f(\tau)) - W(\tau,p)\\ &\approx {1\over 2} A\Delta V(p) + (P\exp(\kappa D(\tau)))(\tau)\hat V(p+f(\tau))-\exp(\kappa D(\tau))\hat V(p)\\ &\le {1\over 2} A\Delta V(p) \end{align} outside of a compact set, which is precisely what we wanted. Here, to get from the first to the second line, we set $$V(p+f(\tau)) \approx V(p) + \langle \nabla V(p),f(\tau)\rangle + {1\over 2}\langle \nabla^2 V(p),(f\otimes f)(\tau)\rangle$$ and similarly for $\nabla V(p+f(\tau))$, and we exploit the definitions of the correctors $\phi$ and $\psi$ to get a bunch of cancellations. This approximation can be justified by absorbing the error terms into the terms $\Delta V$ and $\exp(\kappa D(\tau))\hat V(p)$.<|endoftext|> TITLE: What are quadric bundles? QUESTION [8 upvotes]: In Mori program in dimension $3$ there is a class of Mori contractions $\phi: X\to C$ called quadric bundles, where $X$ is a three-dimensional manifold and $C$ is a curve. As far as I understand, such contractions have the property that the morphism $\phi$ is flat and $\phi^{-1}(C)$ is a smooth quadric surface for a generic point $x\in C$ . Let's assume that we work over $\mathbb C$. Questions. 1) What is the definition of quadric bundles? 2) Suppose that $X$ and $C$ are smooth and complex. Is it true that there is a projective $\mathbb CP^3$ bundle over $C$ into which $X$ naturally embeds - so that in each $\mathbb CP^3$ a quadric surface sits? 3) If there a classification of quadric bundles over $\mathbb CP^1$ with at most two singular fibres? 4) Is there some local model for neighborhoods of singular fibres? 5) Is there some relatively pedagogical place where I can read about this staff? I would be grateful for answers to any of these questions. REPLY [3 votes]: have you consulted chapter one of: Variétés de Prym et jacobiennes intermédiaires Beauville, Arnaud Annales scientifiques de l'École Normale Supérieure, Serie 4, Volume 10 (1977) no. 3, p. 309-391<|endoftext|> TITLE: Bass' conjecture implies the Parshin's conjecture QUESTION [9 upvotes]: In the appendix of this paper. It is proved that Bass' conjecture for $K_n$ implies the rational Beilinson-Soulé conjecture for $K_n$. Then at the end the author claims that the same method can be applied to prove that the Bass' conjecture implies the Parshin's conjecture but I can't figure it out. The general idea behind the Beilinson-Soulé conjecture is that he proves the Theorem A.1 for fields then uses the Quillen spectral sequence to prove it for the general regular scheme $X$ which I can't see how this idea can be used for the Parshin conjecture. I wonder whether this proof is written anywhere with more details. I'd appreciate if anyone of the experts in the field can explain the sketch of the proof for the Parshin's conjecture or answer this question: Is the finite generation of $K_0$ required to imply the Parshin's conjecture or the finite generation of $K_n$ for $n\geq 1$ is enough? Edit: Some thoughts: I believe the proof should be some sort of induction and using the Quillen spectral sequence we should be able to reduce the problem to the function field of projective varieties. The inductive step should be something like showing the algebraic $K$-groups of the function field $F$ of a smooth projective variety over a finite field (of char $p$) is uniquely $p$-divisible after multiplying it with some finite number $N$. Since the Milnor $K$-theory of fields of finite char coincides with the algebraic $K$-theory up to uniquely divisible groups this reduces to showing that $K^M_i(F)$ is uniquely $p$-divisible after multiplying it with some finite number $N$. Is this something known? REPLY [2 votes]: I think the claim in the original paper linked above is not correct. It is either totally wrong or it is meant to be something else (like etale motivic version of Bass conjecture) which again I'm not sure whether the same argument given for the Beilison-Soule works or not (I cannot see how it works.). As a reference you can take a look at here page 15 the paragraph after Theorem 8.1.<|endoftext|> TITLE: Existence of a model of ZFC in which the natural numbers are really the natural numbers QUESTION [14 upvotes]: I know that, from compactness theorem, one can prove that there are models of first order arithmetic in which there is some "number" which is not a successor of zero, in the sense that it is strictly bigger than any successor of zero (i.e. any element of the model obtained by applying the successor function to zero finitely many times). From the axiom of infinity, it follows that every model of ZFC must contain an element which one can think of as "the natural numbers", in the sense that it is a model of Peano Axioms. Peano Axioms are a second-order theory, since the principle of induction is a second order axiom, and from the principle of induction it follows that the Peano Axioms have a unique model in ZFC, so that we can call this model among the others of first-order arithmetic, the "standard" model of arithmetic. But picking one model of ZFC, how do we know what's really inside its standard model of arithmetic? How do we know there is nothing else than the successors of zero? After all, every model of ZFC thinks that his natural numbers are the standard ones, so one can use compactness to produce a model of ZFC which has non-standard natural numbers from an external point of view, and in which there will be a standard natural numbers object containing elements bigger than any successor of zero. So, if one wants to do mathematics inside a model of a first order theory of sets, how can one know that he is able to pick a model in which the natural numbers are not non-standard? REPLY [29 votes]: This, in fact, cannot be proven, even in $ZFC+Con(ZFC)$. This is because $ZFC$ proves the following statement: If we have a model $M$ of $ZFC$ whose natural numbers are standard, then $M$ satisfies $ZFC+Con(ZFC)$. Indeed, $Con(ZFC)$ is an arithmetic statement. Since we are assuming that $ZFC$ has a model $M$, and hence that $ZFC$ is consistent, $\mathbb N\vDash Con(ZFC)$, and since $\mathbb N^M\cong\mathbb N$, $\mathbb N^M\vDash Con(ZFC)$, and hence $M\vDash Con(ZFC)$. Now if $ZFC$ could prove "if there is a model of $ZFC$, then there is a model of $ZFC$ with standard $\mathbb N$", then we would get that $ZFC+Con(ZFC)$ proves that $ZFC+Con(ZFC)$ has a model, contradicting Godel's second incompleteness theorem. Therefore, we cannot conclude, from existence of a model, existence of a model with standard $\mathbb N$. Thought it might be worth mentioning that this reasoning is under the assumption that $ZFC+Con(ZFC)$ is consistent. In the other case, $ZFC$ proves that $ZFC$ has no models, so the implication I discuss holds vacuously. Will Sawin asks whether existence of a model of $ZFC$ with standard $\mathbb N$ (apparently called $\omega$-models, as Gro-Tsen's comment to the question notes) is equivalent to ZFC being arithmetically sound. The answer is negative (of course, again, under blanket consistency assumptions). The idea is very similar. Suppose $ZFC$ is arithmetically sound, and that there is an $\omega$-model $M$ of $ZFC$. We claim $M$ satisfies "$ZFC$ is arithmetically sound". If we show that then we're done, since "$ZFC$ is arithmetically sound" cannot prove $Con$("$ZFC$ is arithmetically sound"). Arithmetic soundness is equivalent to "for all $n$, $ZFC$ is $\Sigma^0_n$-sound, and $\Sigma^0_n$-soundness is an arithmetic statement for each $n$. Since $M$ has standard $\mathbb N$, $\Sigma^0_n$-soundness holds in $M$ as well. Now we use standardness of $M$'s $\mathbb N$ again, to observe that $\Sigma^0_n$-soundness in $M$ for all $n$ in (external) $\mathbb N$ is equivalent to arithmetic soundness internally in $M$. As you can see, existence of an $\omega$-model is a fairly strong property, much stronger than any consistency or soundness assumption.<|endoftext|> TITLE: Results relying on higher derived algebraic geometry QUESTION [10 upvotes]: Are there any results in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures but whose most natural (known) proof uses derived $n$-Artin stacks for $n>1$? We are using Toen--Vezzosi terminology. EDIT: the $n$-Artin stack in question should not to be obtained as the quotient of the constant groupoid associated to a $(n-1)$-Artin stack. I did not specify this initially, my fault, but I think the requirement is pretty natural. Neither of the two answers I can see at the time of the edit address this point. REPLY [7 votes]: The derived moduli stack of perfect complexes $RPerf$ is a derived Artin stack which admits a filtration by open sub stacks $RPerf^{[a,b]}$. The latter is a derived $(b-a+1)$-Artin stack. See: Moduli of objects in dg-categories. Annales scientifiques de l'École Normale Supérieure, Serie 4, Volume 40 (2007) no. 3, pp. 387-444. doi : 10.1016/j.ansens.2007.05.001. http://www.numdam.org/item/ASENS_2007_4_40_3_387_0/ This derived stack plays a critical role in the recently hot topic of shifted symplectic structures and shifted deformation quantization, see: Pantev, T., Toën, B., Vaquié, M. et al. Publ.math.IHES (2013) 117: 271. https://doi.org/10.1007/s10240-013-0054-1<|endoftext|> TITLE: What is a conic bundle and why is it called so? QUESTION [8 upvotes]: I am desperately trying to understand what is a conic bundle. It seems like this is a completely standard term in algebraic geometry, there is even a page on wiki about it, but this doesn't really help. For example, I have the following test question. Test question. Let $S$ be a smooth complex ruled surface $S\to C$ over a curve $C$. Suppose we blow up $S$ twice in the same fibre, so that the preimage of one point in $C$ is a union of three lines. Is this a conic bundle or not? My problem is the following. According to some definitions that I saw, in a conic bundle the preimage of a point should be a conic. And a union of three lines is not a conic. However, I tried to trace back the definition of conic bundles, and one of the earliest versions that I found is an article of Sarkisov 1980 (in Russian): http://www.mathnet.ru/links/b10a1373601dacdd9b7debba2b3e1c8f/im1862.pdf Sarkisov is just asking that the preimage of a generic point be a rational curve. I fear that my main question (what is a conic bundle) is a complicated one, for example judging by the fact that the answer to the following question was not given by the mathoverflow community: References about conic bundles Question 2. Why Sarkisov calls his bunldes conic bundles? Is there some relatively pedagogical place where on can read about this? (I fear that I can't understand this because I misinterpret the expression "generic fiber") REPLY [8 votes]: I think the reasonable definition is to ask for a flat morphism $f$ whose generic fiber is a rational curve. Then you may put more conditions according to your needs (for instance, there is a more strict notion of standard conic bundle). But the answer to Question 2 is, I think, quite simple. A rational curve $C$ over a field $k$ is not necessarily isomorphic to $\mathbb{P}^1_{k}$, because there will usually exist no line bundle on $C$ of degree $1$. However, there is always a line bundle of degree $2$, namely the tangent bundle $T_C$. The global sections of $T_C$ define an embedding $C\hookrightarrow \mathbb{P}^2_{k}$, whose image is a conic. Thus the generic fiber (and the general fibers as well) of $f$ are conics, hence the name.<|endoftext|> TITLE: Do we have $G(\mathbb A_S) G(k) = G(\mathbb A)$ for sufficiently large $S$? QUESTION [5 upvotes]: Let $G$ be a linear algebraic group over a number field $k$. If necessary, assume $G$ is connected and reductive. Let $\mathbb A$ be the ring of adeles of $k$, and $\mathbb A_S = \prod\limits_{v \in S} k_v \prod\limits_{v \not\in S} \mathcal O_v$ for any (large) finite set of places $S$ containing the archimedean ones. Is it the case that $$G(\mathbb A_S) G(k) = G(\mathbb A)$$ for sufficiently large $S$? This is claimed in Moeglin and Waldspurger's book on Spectral Decomposition and Eisenstein Series, in the proof that $Z(\mathbb A)G(k)$ is closed in $G(\mathbb A)$ when $G$ is connected reductive. This is easy to see in the case $G = \operatorname{GL}_1$. We have a copy of $H = (0,\infty)$ in $G(\mathbb A) = \mathbb A^{\ast}$ by sending $\rho$ to $(\rho^{1/n}, ... , \rho^{1/n}, 1, 1, ...)$ in $\prod\limits_{v \mid \infty} k_v$, where $n = [k : \mathbb Q]$. The quotient $\mathbb A^{\ast}/H k^{\ast}$ is compact, and is covered by the images of the open sets $\mathbb A_S^{\ast}$. REPLY [7 votes]: Yes, $G(\mathbb A_S) G(k) = G(\mathbb A)$ holds when $S$ is sufficiently large and contains the set of archimedean places $\infty$. This is because the double coset space $G(A_\infty)\backslash G(\mathbb A)/G(k)$ is finite (its cardinality is called the class number), and a set of representatives can be chosen from $G(\mathbb A_S)$ when $S\supset\infty$ is sufficiently large. For more details, see Theorem 5.1 in Platonov-Rapinchuk: Algebraic groups and number theory (Academic Press, 1994).<|endoftext|> TITLE: Illustrating that universal optimality is stronger than sphere packing QUESTION [12 upvotes]: I'm a physicist interested in the conformal bootstrap, one version of which was recently shown to have many similarities to the problem of sphere packing. Sphere packing in $\mathbf{R}^d$ has been solved in $d=8$ and $24$, and recently those solutions were shown to be universally optimal among point configurations, i.e., "they minimize energy for every potential function that is a completely monotonic function of squared distance (for example, inverse power laws or Gaussians)." I'm trying to appreciate the difference between sphere packing and universal optimality-- to this end, does anyone know of a simple example for which it is clear that the densest sphere packing is not universally optimal? My initial intuition was that sphere packings would be necessarily universally optimal since you could imagine a spherical equipotential surface surrounding each particle, and that minimizing the energy of the configuration would amount to finding the optimal packing of these equipotential surfaces. Evidently this is false. The paper with the aforementioned proof mentions that it was found in 3 dimensions that the conjectured optimal lattice solutions for potential functions of the form $r \mapsto \mathrm{e}^{- \pi r^2}$ are not optimal when nonlattice configurations are considered, but I am hoping for a more obvious illustration. REPLY [18 votes]: In three dimensions you don’t need to go beyond lattices to see the failure of universal optimality. When the potential function is sufficiently steep (e.g., a narrow Gaussian), the face-centered cubic lattice is optimal, but for wide Gaussians the body-centered cubic beats it. You can see this using Poisson summation (the face-centered and body-centered lattices are duals, so if one wins for narrow Gaussians its dual must win for wide Gaussians) or just by direct calculation. Your intuition seems reasonable for very steep potential functions, where the energy is dominated by nearby particles, but when longer-range interactions need to be taken into account there’s no reason dense sphere packing should be optimal.<|endoftext|> TITLE: Are schemes obtained from reduced schemes by gluing infinitesimal stuff along a reduced closed subscheme? QUESTION [8 upvotes]: Let $X$ be a scheme of finite type (say over the complex numbers). The set of points for which the local ring is reduced is then an open subset $U\subseteq X$. Is it true that there is a closed subscheme $Z\hookrightarrow X$ such that $Z$ is supported on $X\smallsetminus U$ and $X$ is the coproduct $X\simeq X_{\mathrm{red}}\cup_{Z_{\mathrm{red}}}Z$ in the category of schemes? An affine example: $$X=\mathrm{Spec}\frac{k[x,y]}{(xy,y^2)}, \quad Z=\mathrm{Spec}\frac{k[y]}{(y^2)},\quad X_{\mathrm{red}}=\mathrm{Spec}(k[x])=\mathbb{A}^1_k, \quad Z_{\mathrm{red}}=\mathrm{Spec}(k),$$ $Z\to X$ given by $y\mapsto y$, $Z_{\mathrm{red}}\to Z$ given by $y\mapsto 0$, $Z_{\mathrm{red}}\to X$ given by $x\mapsto 0, y\mapsto 0$. so $X$ is a line with an embedded point sticking out from the origin, $U=\mathbb{A}^1\smallsetminus\{0\}$, and $Z$ is the infinitesimal first-order "segment". Now, if I'm not mistaken, $$\frac{k[x,y]}{(xy,y^2)}\simeq k[x]\times_k \frac{k[y]}{(y^2)}:=\{ (a(x),b(y))\in k[x]\times \frac{k[y]}{(y^2)} \mid a(0)=b(0)\}$$ by $x\leftrightarrow(x,0), y\leftrightarrow (0,y)$, so the algebra of $X$ is the fibered product of those of $X_{\mathrm{red}}$ and $Z$ over the evaluations to $k$, hence $X$ is the corresponding coproduct of schemes. REPLY [10 votes]: If this happens then the cotangent bundle $\mathcal I_{Z_{red}} / \mathcal I_{Z_{red}}^2$ of $Z_{red}$ in $X$ is equal to the sum of the contangent bundle of $Z_{red}$ in $Z$ and the cotangent bundle of $Z_{red}$ in $X$. This is because locally a coproduct of schemes gives a fibered product of rings which therefore gives a product of ideals. So if we chose $X$, letting $Y$ be the induced reduced subscheme structure on $X \setminus U$, where the natural map from the cotangent bundle of $Y$ in $X$ to the cotangent bundle of $Y$ in $X^{red}$ does not split, such $Z$ will not exist. For instance we can choose $X = \operatorname{Spec} k[x,y]/(x^2y)$. Or we can pick a non-split exact sequence of vector bundles $0 \to V_1 \to W \to V_2 \to 0$ on some other scheme and take the relative spectrum of the symmetric algebra of $W$ modulo the ideal generated by $W V_1$, so that the cotangent space is $W$ and the cotangent space of the reduced subscheme of the non-reduced locus is $V_2$.<|endoftext|> TITLE: Generic saturation of inner models QUESTION [6 upvotes]: Say that an inner model $M$ of $V$ is generically saturated if for every forcing notion $\Bbb P\in M$, either there is an $M$-generic for $\Bbb P$ in $V$, or forcing with $\Bbb P$ over $V$ collapses cardinals. What is the consistency strength of "$L$ is generically saturated"? If the answer is $0^\#$ exists, is this sort of a general answer for relative constructibility (i.e. $L[A]$ is generically saturated if and only if $A^\#$ exists)? If the answer is negative, what can we conclude from this principle? Note, Mohammad Golshani remarks (also this), that assuming $0^\#$ exists for every $\kappa$, $\operatorname{Add}(\kappa,1)^L$ collapses $\kappa$ to $\omega$ (in particular, if $\kappa$ is countable in $V$, it is just a Cohen forcing). So in the presence of $0^\#$ at least we know that a lot of the forcings in $L$ do collapse cardinals, even if they do not admit generics in $V$ (e.g., $\operatorname{Col}(\omega,\omega_1^V)$ cannot admit a generic, although it does collapse cardinals). (The idea here is to marry Foreman's maximality principle that states that every forcing adds a real or collapses cardinals, with inner model hypothesis-like ideas.) REPLY [10 votes]: The concept is inconsistent. Consider $\mathbb{P}=Add(\omega, \kappa)_L=Add(\omega, \kappa),$ where $\kappa$ is $(2^{\aleph_0})^+$ of $V$. Forcing with $\mathbb{P}$ over $V$ doesn't collapse cardinals (it is ccc.c. in $V$) and there is no $M$-generic filter for $\mathbb{P}$ in $V$.<|endoftext|> TITLE: Derived algebraic geometry and virtual fundamental cycles: cotangent complexes QUESTION [5 upvotes]: I have been thinking of a way to apply the derived algebraic geometry of Toen-Vezzosi to construct virtual fundamental cycles on moduli spaces of pseudo-holomorphic curves. This seems to be the natural setting for which to define symplectic topological invariants when the relevant moduli spaces aren't necessarily cut out transversally. Indeed, Joyce adopts this philosophy in his "D-orbifolds" project. There, Joyce defines a d-orbifold to be a kind of Deligne-Mumford $C^{\infty}$- stack equipped with a sequence of quasi-coherent sheaves on this stack that should be thought of a cotangent complex. Should one expect that this "cotangent complex" is a literal cotangent complex for some derived geometric $C^{\infty}$ - stack? More precisely - should there correspond, to a D-orbifold, a derived stack $\mathcal{X}^{der}$ which is a derived extension of some geometric stack $\mathcal{X}$, and where this derived structure is the analog of a perfect obstruction theory? REPLY [6 votes]: I don't know how to define a VFC for the moduli of pseudoholomorphic curves other than using Pardon's paper. Nevertheless here's the general picture (couldn't tell from your question whether you're familiar with this story). Let $X$ be a Deligne-Mumford stack. Say we want to define the virtual fundamental class $[X]^{vir}$. The optimal way to proceed is to construct a quasi-smooth derived stack $X^{der}$ whose classical truncation is $X$. For historical and cultural reasons there are various shadows of the notion of quasi-smooth derived stack people came up with that are "good enough", meaning that you can still define a VFC out of that data. Perfect obstruction theories and d-orbifolds are examples of such approximations. The relevant point here is that a QS derived DM stack gives you a perfect obstruction theory via the cotangent complex. Quasi-smooth means that the cotangent complex of $L_{X^{der}}$ is a perfect complex concentrated in degrees $-1$ and $0$ (Tor-amplitude $[-1,0]$). Therefore the canonical thickening map $i : X \to X^{der}$ induces a morphism of complexes $$ i^*(L_{X^{der}}) \to L_X $$ which is a perfect obstruction theory for $X$. The d-orbifold construction is similar (presumably). Anyway then you can follow any of the usual paths to get your VFC. The question of how to define $X^{der}$ deserves to be addressed. The stack $X$ defines some moduli problem, say $X$ is the moduli of thingamajiggies. So for every affine scheme $Spec(R)$, you have a groupoid $X(R)$ of thingamajiggies over $Spec(R)$. The way to define $X^{der}$ is to figure out how to define the space $X^{der}(R)$ of thingamajiggies over $Spec(R)$ when $R$ is a derived commutative ring. If you do this correctly (it's a "you'll know it when you see it" kinda thing) then you'll end up with something quasi-smooth, hopefully actually with the cotangent complex you expected, and you'll be good to go. A relevant example of this general picture, for the moduli of stable maps (GW-theory), is worked out in the paper of Toen, Vezzosi, Schuerg in Crelle. They define a derived moduli stack classifying derived stable maps and "surprise surprise", it's quasi-smooth and gives the expected perfect obstruction theory and VFC. To answer your question, it is reasonable to expect that every perfect obstruction theory arising in practice actually comes from a qs derived structure (there's no reason it shouldn't). At the same time there's also no reason to believe that an arbitrary/abstract perfect obstruction theory necessarily comes from a qs derived structure (this is highly unlikely IMO).<|endoftext|> TITLE: Geometric dissection theory QUESTION [6 upvotes]: A few days ago, i realized that one way to prove the Pythagorean Theorem is to dissect the given right-angled triangle into 2 triangles similar to it, and apply well-known properties of ratios of areas. So this fundamental theorem can be viewed as a consequence of the possibility to dissect a given right-triangle into 2 similar sub-triangles. My question is: Which branch of geometry systematically studies dissections of general polygons (or other shapes) into other sub-polygons (sub-shapes) of a given type, and the metrical properties that emerge as consequences of these dissections? My search is on-going, i have seen some relevant results & books, for example the problem of dissecting a random triangle into n similar triangles, and books on discrete & combinatorial geometry that border on the subject (but not studied them yet), so i'm asking for some further guidance towards a systematic exposition of the relevant theory, in 2D, 3D or more dimensions.. REPLY [3 votes]: The following paper proves (1) that every polygon may be partitioned (dissected) into (a finite number of) non-obtuse triangles, and (2) into acute triangles: Maehara, Hiroshi. "Acute triangulations of polygons." European Journal of Combinatorics 23, no. 1 (2002): 45-55.           Dissections specifically into similar triangles: Jones, C. A., P. Jones, and A. B. Bolt. "Dissections of triangles into five similar triangles." The Mathematical Gazette 82, no. 494 (1998): 225-234.           See also: MathWorld on Triangle Dissection. The textbook Discrete and Computational Geometry has a chapter on polygon dissections. MO question: 3-piece dissection of square to equilateral triangle? MO question: Inside-out polygonal dissections.<|endoftext|> TITLE: What is classified by generalised Eilenberg MacLane spaces? QUESTION [18 upvotes]: Given an abelian group $A$, the Eilenberg MacLane spaces $K(A,n)$ represent the the nth cohomology group in $A$. In a similar vein, given an arbitrary group $G$ and a space $X$, maps to the classifying space $X\to BG$ classify principal $G$-bundles on $X$. In the literature I have encountered spaces $K(G,V,n)$, where $G$ is a group and $V$ is a finite dimensional $G$-representation, called generalised (or sometimes twisted) Eilenberg MacLane spaces. These spaces are determined up to homotopy by the property that $\pi_{i}(K(G,V,n)) = \begin{cases} G,\ \ i=1,\\ V,\ \ i=n,\\ 0,\ \ \text{else}. \end{cases} $ My first question is, what do generalised Eilenberg MacLane spaces classify? Am I correct in thinking that they represent cohomology in the local system determined by $G$ and $V$? My second question is, what does it mean to localise with respect to generalised Eilenberg MacLane spaces? By this I am thinking of a Bousfield localisation on the model category of spaces in which a local equivalence is declared to be a map $f:X\to Y$ which induces a weak equivalence $$ f^{*}:\text{Map}(Y, K(G,V,n))\to \text{Map}(X,K(G,V,n)), $$ and a space $Z$ is local if any local equivalence $f:X\to Y$ induces a weak equivalence $$ f^{*}:\text{Map}(Y, Z)\to \text{Map}(X,Z). $$ Specifically, I would like to know what information is isolated by performing these localisations for a given group $G$ and representation $V$. REPLY [20 votes]: To answer your first question, take a look at the reference Gitler, Samuel, Cohomology operations with local coefficients, Am. J. Math. 85, 156-188 (1963). ZBL0131.38006. In particular, Theorem 7.18 in Chapter III does what you want. To paraphrase, let $\mathcal{V}$ be the local system of groups on your Eilenberg--Mac Lane space $K(G,V,n)$ determined by the representation $G\to \operatorname{Aut}(V)$. Given pointed connected CW-complexes $X$ and $Y$ and a homomorphism $\alpha:\pi_1(X)\to \pi_1(Y)$ between their fundamental groups, let $[X,Y]_\alpha$ denote the set of pointed homotopy classes of pointed maps $f:X\to Y$ such that $f_*=\alpha:\pi_1(X)\to \pi_1(Y)$. Then there is a (natural in an appropriate sense) isomorphism $$ [X,K(G,V,n)]_\alpha \cong H^n(X;\alpha^*\mathcal{V}). $$ The isomorphism is given as in the untwisted case by pulling back an appropriately defined fundamental class $\iota\in H^n(K(G,V,n);\mathcal{V})$.<|endoftext|> TITLE: Stronger exact pairs QUESTION [6 upvotes]: Question: Suppose $\{a_n : n < \omega \}$ is a $<_T$-ascending sequence in $2^{\omega}$. Can we find $x, y \in 2^{\omega}$ such that for every $z \in 2^{\omega}$, the set of reals computable from each one of $x \oplus z$ and $y \oplus z$ is the Turing ideal generated by $\{a_n : n < \omega\} \cup \{z\}$? So taking $z$ to be computable, this implies that $x, y$ form an exact pair for the Turing ideal generated by $\{a_n : n < \omega\}$. My thoughts: The usual construction of exact pair has only countably many splitting requirements ($\Phi^{x} = \Psi^{y} = w \implies (\exists n)(w \leq_T a_n)$) but this construction would require continuum many. I suspect such pairs don't exist but I don't have a proof. REPLY [5 votes]: You could look at Theorem 3.1 of this paper: On the Σ2-Theory of the Upper Semilattice of Turing Degrees Author(s): Carl G. Jockusch, Jr. and Theodore A. Slaman Source: The Journal of Symbolic Logic, Vol. 58, No. 1 (Mar., 1993), pp. 193-204 Published by: Association for Symbolic Logic 3.1 characterizes the possible upper-semi-lattice end-extensions of a countable ideal that can be generated by the ideal and a single new element. Basically, anything that is consistent as an upper-semi-lattice can be realized. In particular, for any $x$ and $y$ as above, there is a $z$ such that $z\oplus x$ and $z\oplus y$ are Turing equivalent, but for all $a_n$, $z\oplus a_n$ computes neither $x$ nor $y$. Your instincts were correct: there is no pair of the described type.<|endoftext|> TITLE: Why do Maynard-Tao weights succeed? QUESTION [13 upvotes]: I'm attempting to understand why the Maynard-Tao weights are successful in proving bounded gaps between primes, but the GPY weights are not. These two posts do an excellent job in giving an overview of how one arrives at the Maynard-Tao weights and the role of the weights themselves: What is the crucial difference the Maynard/Tao approach and Goldston-Pintz-Yildirim that extends to prime k-tuples with $k>2$ Why do the Maynard-Tao weights work so well? The additional flexibility in allowing the $\lambda_{d_{1},\dots,d_{k}}$ to depend on the divisors $d$ individually allows the method to be succeed. If possible I'd like a deeper understanding why the additional flexibility allows it to succeed. Soundararajan's expository paper: https://arxiv.org/pdf/math/0605696.pdf (pages 9 - 14) explain the core ideas of the GPY method and that the ratio of $$\dfrac{\Big(\sum_{\substack{x \leq n \leq 2x} \\ n + h_{j} \text{prime}}a(n)\Big)}{\Big(\sum_{x \leq n \leq 2x}a(n)\Big)}$$ cannot be made greater than $\frac{1}{k}$ and we therefore fail to just prove bounded gaps between primes. Therefore the GPY method finds a probability distribution such that Probability$(n+h_{i} \text{is prime}) \asymp \frac{1}{k}$ and this differs from the optimal value by a factor of about $k.$ The Maynard-Tao weights clearly give a better weighting so this issue is overcome and find a different probability distribution to enable the inequality to hold but I'm a bit confused. Why is it that because the $\lambda_{d_{1},\dots,d_{k}}$ to depend on the divisors $d$ individually, this issue is overcome? What is the probability distribution that the Maynard-Tao weights find such that the inequality holds? Apologies if this isn't the appropriate place to ask this question. I'm an Undergraduate interested in analytic number theory... REPLY [4 votes]: TLDR: I have two questions thanks in advance: Can we assume that the $w_{n}$ are smooth approximations and if so why? I've always struggled in knowing when it is permissible / how to find smooth approximations Is it correct that in the Maynard case the probability distribution that is found is $\mathbb{P}(n+h_{i}\text{ is prime}) \asymp \frac{\log k}{k}$? I came to this by looking at equation 8.19 of the Maynard paper (last page) and corollary 6.4 (page 44) of the following paper: https://arxiv.org/pdf/1407.4897.pdf Further detail: Having looked at this more I've found the following: The overall goal is we want to find weights $w_{n}$ to be larger than but as close as we can get to $\mathbb{1}_{\text{all $n + h_{i}$ prime}}$ It is best to have deviations between our chosen weights $w_{n}$ and $\mathbb{1}_{\text{all $n + h_{i}$ prime}}$ when we have many of the $n+h_{i}$ being prime because then we will have $\mathbb{P}(n + h_{i} \text{is prime})$ being large for each $i$. In the GPY case we are considering Selberg type weights of the form $w_{n} = \Big(\sum_{d|n}\lambda_{d}\Big)^{2}$ where $\lambda_{d} = \mathbb{1}_{d \leq R}\mu(d)F(\frac{\log R}{d})$ where $F$ is a polynomial. In other words, the GPY weights are like the square of $$\sum_{d|\prod(n+h_{i})}\mu(d)\log\Big(\dfrac{\prod_{i=1}^{k}(n+h_{i})}{d}\Big)^{k}$$ In the Maynard case we are considering weights that are like the square of $$\Big(\sum_{d_{1}|n+h_{1}}\mu(d_{1})\dfrac{\log(n+h_{1})}{d_{1}}\Big)\dots\Big(\sum_{d_{k}|n+h_{k}}\mu(d_{k})\dfrac{\log(n+h_{k})}{d_{k}}\Big)$$ for $k$ sufficiently large. The above GPY weights vanish when not all of the translates, $n+h_{i}$ are prime. The above Maynard weights vanish when not all of the translates, $n+h_{i}$ are prime powers. So if we view the $w_{n}$ as a smooth approximation to the square of the expression above in the Maynard case. Then if the first factor was not a prime power (but the remaining factors were) it would vanish. If we were to approximate it using $w_{n}$ it would be a small factor, but would be compensated for by larger values at the remaining translates. We also have that $w_{n}$ is larger when more and more of the $n+h_{i}$ are prime. However in the GPY case even if we again view the $w_{n}$ as a smooth approximation we don't get this compensation of by larger values for the remaining translates... Furthermore since GPY found $\mathbb{P}(n+h_{i} \text{is prime}) \asymp \frac{1}{k}$ the smooth weights $W_{n}$ will be getting worse as $k$ gets larger. Whereas in the Maynard case it looks like we can save a factor of $\log k$ and we have $\mathbb{P}(n+h_{i}\text{ is prime}) \asymp \frac{\log k}{k}$ and thus this tends to infinity as $k$ gets larger which is what we want.<|endoftext|> TITLE: Estimation of a sum of algebraic numbers QUESTION [10 upvotes]: Let $\alpha_1, \ldots, \alpha_n$ be algebraic numbers and let $p_1, \ldots, p_n$ be the corresponding minimal polynomials with integer coefficients. Denote by $H$ the maximal magnitude among all coefficients of all $p_i$. Assume that $\alpha_1 + \ldots + \alpha_n \not= 0$. Is it true that $$|\alpha_1 + \ldots + \alpha_n | \ge {(CH)}^{-q(\text{deg}(p_1) +\ldots \text{deg}(p_n) )}$$ for some fixed polynomial $q$ and constant $C$? Warning: $C$ and $q$ should not depend on $\alpha_1, \ldots, \alpha_n$. Updates, since the question appears difficult: Does the statement hold when all $\alpha_i$ are quadratic irrationals? Does the statement hold under reasonable hypotheses like the effective $abc$ conjecture or Vojta conjecture? Is there a $C$ such that $|\alpha_1 + \alpha_2| \ge H^{-C(\deg(p_1) + \deg(p_2))}$ whenever $\alpha_1 + \alpha_2 \not=0$? REPLY [6 votes]: In the answer below I assume that $q$ may depend on $n$. It seems plausible that the assertion is true for $q(x)=x^2$ or something similar, but I don't know how to prove it. It may be hard. The answer below is based on Liouville's inequality, that is on estimates for $P(\alpha_1,\dots,\alpha_n)$ which are essentially the best possible for arbitrary polynomials and algebraic numbers. But for a fixed $P$ much better estimates can be available as illustrated by Roth's theorem. Unfortunately, Roth's theorem and Schmidt's Subspace theorem are ineffective, so can't be used here. Old answer: Yes, that's true. I will write the estimates in terms of height. The usual height $H(p)$ of a polynomial $p$ with integer coefficients is defined as the maximum of absolute values of its coefficients. Then the usual height $H(\alpha)$ of an algebraic number $\alpha$ is the usual height of its minimal polynomial. The absolute logarithmic height of $\alpha$ is $$h(\alpha)=\frac{1}{d}\log H(\alpha)\,,$$ where $d$ is the degree of $\alpha$. The basic version of Liouville's inequality says that if $\alpha\neq0$ then $$|\alpha|\geq\frac{1}{H(\alpha)}\,.$$ Now, combine that with $$h(\alpha_1+\dots+\alpha_n)\leq\log n+h(\alpha_1)+\dots+h(\alpha_n)$$ and $$\deg(\alpha_1+\dots+\alpha_n)\leq \deg(\alpha_1)\cdot\dots\cdot\deg(\alpha_n)\,.$$ All these claims about height can be found in many places. I like Chapter 3 of Waldschmidt "Diophantine approximation on linear algebraic groups".<|endoftext|> TITLE: Status of proof by contradiction and excluded middle throughout the history of mathematics? QUESTION [21 upvotes]: Occasionally I see the claim, that mathematics was constructive before the rise of formal logic and set theory. I'd like to understand the history better. When did proofs by contradiction or by excluded middle become accepted/standard? Can one find them for instance in classical works (Archimedes, Euclid, Euler, Gauss, etc.)? Was there ever a debate about their validity before Brouwer? My interest was sparked after reading the following "proof" from Newton's Principia that seems to use contradiction: LEMMA I. Quantities, and the ratios of quantities, which in any finite time converge continually to equality, and before the end of that time approach nearer the one to the other than by any given difference, become ultimately equal. If you deny it, suppose them to be ultimately unequal, and let $D$ be their ultimate difference. Therefore they cannot approach nearer to equality than by that given difference $D$; which is against the supposition. REPLY [16 votes]: I want to echo the other answer, that Brouwer presents the first robust challenge to the Law of Excluded Middle (LEM), but I do want to add some history and background, and maybe recommend a related paper. The main paper by Brouwer is De onbetrouwbaarheid der logische principes or in English The unreliability of the logical principles published in 1908. In some sense, this is a continuation of his dissertation in 1907, but I think it's better to just speak of 1908 as the starting point for Brouwer's views on LEM. In a paper [1] offering a new translation of Brouwer's paper, the authors have a section devoted to "precursors" (this is only four pages long, too long to quote in full, but worth reading). At first glance, the references given do not mention LEM, and the relationship to Brouwer seems slightly strained, but I think the authors adequately defend the relevance of the section. It starts with a quote from Kronecker in 1882 about constructability issues. It seems this was extended by Kronecker's student Molk, who said in 1885 The definitions should be algebraic and not only logical. It does not suffice to say: ‘A thing exists or it does not exist’. One has to show what being and not being mean, in the particular domain in which we are moving. Only thus do we make a step forward. The authors then devote an entire page of quotes from Molk (in 1904), noting the similarities "even down to some of the finer detail" to Brouwer's line of thought, though Molk doesn't continue as far as Brouwer. For instance, here's one sentence from the page of quotes: In order to give a mathematical demonstration of a proposition, it does not suffice, for example, to establish that the contrary proposition implies a contradiction. I think it should be pointed out that the authors tried but failed to establish any record that Brouwer was aware of Molk's writings. A footnote points out that Brouwer himself seems to have become aware of the similarities some years later. Following the discussion on Molk, the authors give a correspondence by Lebesgue in 1905 which contains an afterthought very similar to LEM: Although I strongly doubt that one will ever name a set that is neither finite nor infinite, the impossibility of such a set seems to me not to have been demonstrated. but this seems to be more a curious coincidence than serious mathematical reasoning. I think the final sentence summarizes the section well: However, in spite of the early efforts by Kronecker, only with Brouwer do we get a comprehensive development of mathematics excluding any ‘unreliable’ use of the principle of excluded middle. Edit: Adding a little bit more history, just some more sources that say the same as above. In [4], Kleene devotes the very first sentence to a brief background on intuitionism (he goes on to talk about Brouwer in the next paragraph, not included here): The constructive tendency in mathematics has been represented, prior to or apart from intuitionism, in the criticism of "classical" analysis by Kronecker around 1880-1890, and in the work of Poincaré 1902, 1905-6, Borel 1898, 1922 and Lusin 1927, 1930 (cf. Heyting 1934 or 1935). Above, Kleene references his Introduction to Meta-Mathematics [5], which says the following at the opening of section 13 In the 1880's, when the methods of Weierstrass, Dedekind and Cantor were flourishing, Kronecker argued vigorously that their fundamental definitions were only words, since they do not enable one in general to decide whether a given object satisfies the definition. Poincaré, when he defends mathematical induction as an irreducible tool of intuitive mathematical reasoning (1902, 1905-6), is also a fore-runner of the modern intuitionistic school. In 1908 Brouwer, in a paper entitled "The untrustworthiness of the principles of logic", challenged the belief that the rules of the classical logic, which have come down to us essentially from Aristotle (384-322 B.C.), have an absolute validity, independent of the subject matter to which they are applied. ... I extracted the references for Poincaré and Borel, they are listed at the end of the references section below. I could not find a copy of Brouwer’s Intuitionism. Studies in the History and Philosophy of Science, vol. 2 by W. van Stigt (1990) to read, so that may or may not have other useful information. However, I did find the following paragraph in a book review [2], I wanted to include this just to give some context for LEM and when it gained prominence. From his philosophy of mathematics Brouwer draws the remarkable conclusion that the Principle of the Excluded Middle of logic is not reliable. His paper The Unreliability of the Logical Principles of 1908, written in Dutch, did not attract the attention it deserved. Van Stigt notes that even Brouwer himself originally did not appreciate its revolutionary character. Even in 1912 his attack on the Law of Excluded Middle is only an added footnote to the English translation of his inaugural address. Only in 1923 does Brouwer return to logic, criticizing the principle of double negation elimination $ \lnot \lnot \phi \rightarrow \phi $. By November he discovers $\lnot \lnot \lnot \phi \leftrightarrow \lnot \phi $. The attacks on the Principle of the Excluded Middle became a propagandistic rallying point. The development of intuitionistic logic, however, is left to Kolmogorov, Glivenko, and, finally, Heyting in 1928, just when Brouwer retires into silence. Finally, there is a related paper [3] that might be relevant to the original question (though I think the "crisis" talk is a bit sensationalized from a modern perspective, and could have been moderated a bit more in the paper). It looks into the history of science and math about Intuitionism in particular, discusses Kuhnian paradigm shifts, and some difficulties in why Intuitionism never "caught on." There's a lot of background, probably the only section of interest is section 4 which begins In the early years of the twentieth century, classical mathematics entered a period of crisis. Paradoxes had sprung up in and around Cantor’s set theory, and ultimately these ‘crisis-provoking anomalies’ (to transplant a Kuhnian phrase) led concerned classical mathematicians to investigate the foundations of their subject, searching for a way to bring certainty back to mathematics. At the height of this crisis, in his 1907 dissertation and subsequent papers, the Dutch mathematician L. E. J. Brouwer offered a new paradigm for mathematics. Intuitionism promised to secure the foundations of mathematics and explain away the anomalies. Yet it failed to convert the mathematical community. [1] Mark van Atten, Göran Sundholm. "L.E.J. Brouwer's `Unreliability of the logical principles'. A new translation, with an introduction" https://arxiv.org/abs/1511.01113 [2] Wim Ruitenburg. "Review of W. P. van Stigt, Brouwer's Intuitionism". Modern Logic 2 (1992), no. 4, 424--430. https://projecteuclid.org/euclid.rml/1204834908 [3] Bruce Pourciau. "Intuitionism as a (failed) Kuhnian revolution in mathematics" Studies in History and Philosophy of Science Part A. Volume 31, Issue 2, June 2000, Pages 297-329. https://doi.org/10.1016/S0039-3681(00)00010-8 [4] S. C. Kleene, R. E. Vesley. "The Foundations of Intuitionistic Mathematics" North-Holland, 1965, page 1. [5] S. C. Kleene. "Introduction to Meta-Mathematics" Ishi Press, New York 2009. Page 46. Kleene's references (apologies for bad French) Borel 1898 Leçons sur la théorie des fonctions. Paris (Gauthier-Villars), 8+136 pp. 4th ed. (Leçons sur la théorie des fonctions; principes de la théorie des ensembles en vue des applications à la théorie des fonctions). Paris (Gauthier-Villars), 1950, xii+295 pp. Available on archive.org, e.g. https://archive.org/details/leconstheoriefon00borerich Poincaré 1902. La Science et l'hypothèse. Paris, 284 pp. Translated by G. Bruce Halstead as pp. 27-197 of The foundations of science by H. Poincaré, New York (The Science Press) 1913; reprinted 1929. Poincaré 1905-6 Les mathématiques et la logique. Revue de métaphysique et de morale, vol. 13 (1905), pp 815-835, vol. 14 (1906), pp. 17-34, 294-317. Reprinted in 1908 with substantial alterations and additions.<|endoftext|> TITLE: Alexandrov's generalization of Cauchy's rigidity theorem QUESTION [5 upvotes]: Wikipedia states that A. D. Alexandrov generalized Cauchy's rigidity theorem for polyhedra to higher dimensions. The relevant statement in the article is not linked to any source. The sources at the end of the Wikipedia page seem to be only about $3$-dimensional polyhedra as well, in particular Alexandrov's book "Convex polyhedra". Where can I find a reference for that statement? REPLY [5 votes]: Wikipedia is correct. This is discussed in Alexandrov's book "Convex polyhedra" in Section 3.6.5.<|endoftext|> TITLE: Comparison of several topologies for probability measures QUESTION [6 upvotes]: Let $X$ be a compact metric space and denote $\mathcal M(X)$ the set of probability measures on $X$. For $\mu\in\mathcal M(X)$ we write $\operatorname{supp} \mu$ for the support of $\mu$. As is well known, if a sequence of measures $(\mu_n)_{n=1}^\infty$ weak$^*$ converges to some $\mu\in\mathcal M(X)$, then $$ \operatorname{supp}\mu\subset \lim_{n\to\infty}(\operatorname{supp}\mu_n). $$ and the inverse is not necessarily true. We can define the following metric on $\mathcal{M}(X)$: $$ d^{PH}(\mu,\mu_n)=\text{max}\{d^P(\mu,\mu_n),d^H(\mu,\mu_n)\}$$ where $d^P(\mu,\mu_n)$ is the Prokhorov metric on $\mathcal{M}(X)$ and $d^H(\mu,\mu_n)$ is a metric between $\mu$ and $\mu_n$ defined by the Hausfdorff distance between $\operatorname{supp}\mu$ and $\operatorname{supp}\mu_n$. Finally denote as $d^{TV}(\mu, \mu_n)$ the total variation metric on $\mathcal{M}(X)$, given by $$d^{TV}(\mu, \mu_n)= \sup_{A \in \mathcal F}|\mu_n (A) - \mu(A)| $$ where $\mathcal F$ is the underlying $\sigma$-algebra on $X$. I have the following question: Is it true that the topology on $\mathcal{M}(X)$ generated by $d^{TV}(\cdot)$ is strictly finer than the topology generated by $d^{PH}(\cdot)$ and how could one prove this? Is the topology generated by $d^{PH}(\cdot)$ separable? REPLY [5 votes]: The topology generated by $d^{PH}$ is, in general, neither coarser nor finer than $d^{TV}$. The Hausdorff distance between the supports of two measures, even when combined with the Prokhorov distance, has very little to do with the total variation distance between them. To see one direction, let $X$ have at least two points $x,y$ and let $\mu_n = \frac{n-1}{n} \delta_x + \frac{1}{n} \delta_y$, $\mu = \delta_x$. Then the support of $\mu_n$ is $\{x,y\}$ while the support of $\mu$ is $\{x\}$, so the Hausdorff distance between their supports is $d(x,y)$ for every $n$. Hence $\mu_n$ does not converge to $\mu$ in $d^{PH}$. On the other hand, $d^{TV}(\mu_n, \mu) = 1/n$ so $\mu_n$ does converge to $\mu$ in $d^{TV}$. For the other direction, take $X = [0,1]$ and let $\mu_n = \frac{1}{n} \sum_{k=1}^n \delta_{k/n}$, with $\mu$ being Lebesgue measure. Then $\mu_n \to \mu$ weakly, and the Hausdorff distance between $\{1/n, 2/n, \dots, 1\}$ and $[0,1]$ is $1/n$. So $\mu_n \to \mu$ in $d^{PH}$. But $\mu_n$ is singular to $\mu$ for every $n$, so $d^{TV}(\mu_n, \mu) = 1$, and $\mu_n$ does not converge to $\mu$ in $d^{TV}$. I think that $d^{PH}$ is separable. Fix a countable dense set $\{x_1, x_2, \dots\}$ in $X$ and consider the set $M_0$ of all probability measures $\nu$ of the form $\nu = \sum_{i=1}^m q_i \delta_{x_i}$ where the values $q_i$ are nonnegative rationals that sum to $1$. This is dense in the weak topology, and given a sequence $\nu_n$ in $M_0$ converging weakly to some $\mu$, you ought to be able to modify the measures $\nu_n$ so as to ensure that all their supports (which are finite sets) are Hausdorff close to that of $\mu$. I leave the details to you, if you wish to work it out.<|endoftext|> TITLE: History of the notion of $(G,X)$-structure QUESTION [13 upvotes]: I'm currently searching for sources and historical basis on the notion of $(G,X)$-structure as it appears in Thurston's work. So far, it appears that he was the first to set it. Many mathematicans cite Ehresmann's work, also I'm not sure to what precisely. It also seems that Kuiper (On compact conformally Euclidean spaces of dimension $> 2$, 1950) also had some ideas close to what we call a developing map, and related to the idea to put a structure on a manifold. Maybe someone here already have made such researches, or could testify ? Thanks edit : As requested, I specify a sense of $(G,X)$-structure. I mean by that a pair of $G$ a group (or pseudo-group if wanted) of analytic diffeomorphisms of $X$, a smooth analytic connected manifold. The point is to have $(G,X)$-structure notion clean enough to get the pair of a developing and holonomy maps. REPLY [3 votes]: how do i guess n. kuiper learned about the developing mapping of a G structure.? kuiper and everyone before the war learned and used projective geometry. also in holland brouwers tradition was felt and he and other differential geometers thought geometrically in terms of developable surfaces, and the developing idea of curvature. I think ehreshmann was formalizing this intuition. kuipers papers 49-50 were also formalizing this in a more geometric way....ereshmanns lectures were very formal and abstract ( reported by tony phillips who attended them in early 60's as a fulbright scholar in france.) covering spaces and these ideas are taught everywhere but they are non trivial. one sees that in the work of thurstion who found important new phenomena of holonomy of developments in very classical examples like gluing two ideal noneuclidean triangle together. dennis sullivan<|endoftext|> TITLE: Example of a ring where every module of finite projective dimension is free? QUESTION [7 upvotes]: I'm interested in seeing an example of a ring which is not self-injective where every module admitting a finite projective resolution is free, or at least projective. Note that self-injectivity says that every monomorphism $R \to M$ splits (where $M$ is an arbitrary module), whereas the property I'm looking for looks a bit weaker, requiring this only when $M$ is free. I'm happy to look at noncommutative rings. REPLY [10 votes]: The classes of ring you look at are precisly the rings of finitistic dimension zero for the question with projective instead of free. I will give a large class of example for finite dimensional algebras. Let $A$ be a local finite dimensional (over a field $K$) non-selfinjective algebra. Then every non-projective module has infinite projective dimension and thus the only modules of finite projective dimension are the projective modules, which are all free. Concrete example: Take a quiver $Q$ with one point and $n \geq 2$ loops and let $KQ/I$ the quiver algebra with the relations $I$ such that $I=J^2$ consists of all paths of length 2. Alternatively, this is $K[x_1,...,x_n]/I$ with $I$ the ideal generated by all monomials of length 2. So the "smallest" example is probably the 3-dimensional algebra $K[x,y]/(x^2,y^2,xy)$. For finite dimensional algebras, the local finite dimensional non-selfinjective algebras are exactly those with the property that every module of finite projective dimension is free. When you look instead at the property "every module of finite projective dimension is projective", then you arrive at the finite dimensional algebras with finitistic dimension zero (the finitistic dimension is defined as the supremum of all projective dimensions of modules having finite projective dimension), which is a very large class that contains for example all selfinjective and all local algebras. You need two results about finite dimensional algebras: 1.For a local algebra, every projective module is free. Local algebras have finitistic dimension zero. The first result is well known and 2. is easy to prove, but I will search for a reference now. (https://link.springer.com/chapter/10.1007/978-3-0348-8658-1_8 proposition 2.1. states that the finitistic dimension of local algebras is 0, this is not a good reference but I leave it here until a better one is found).<|endoftext|> TITLE: Why is the Eisenstein ideal paper so great? QUESTION [64 upvotes]: I am currently trying to decipher Mazur's Eisenstein ideal paper (not a comment about his clarity, rather about my current abilities). One of the reasons I am doing that is that many people told me that the paper was somehow revolutionary and introduced a new method into number theory. Could somebody informed about these matters explain exactly what subsequent developments did the paper bring, what ideas in the paper were considered more-or-less original (at the time it was published), and exactly what difficulties did these ideas resolve that people failed to resolve before the paper was published (if any)? REPLY [97 votes]: First, Mazur's paper is arguably the first paper where the new ideas (and language) of the Grothendieck revolution in algebraic geometry were fully embraced and crucially used in pure number theory. Here are several notable examples: Mazur makes crucial use of the theory of finite flat group schemes to understand the behavior of the $p$-adic Tate modules of Jacobians at the prime $p$. He studies modular forms of level one over finite rings (which need not lift to characteristic zero when the residue characteristic is $2$ or $3$). He proves theorems about mod-$p$ modular forms using what are essentially comparison theorems between etale cohomology and de Rham cohomology, and many more examples. The proof of the main theorem ($\S5$, starting at page 156) is itself a very modern proof which fundamentally uses the viewpoint of $X_0(N)$ as a scheme. Second, there are many beautiful ideas which have their original in this paper: it contains many of the first innovative ideas for studying $2$-dimensional (and beyond) Galois representations, including the link between geometric properties (multiplicity one) and arithmetic properties, geometric conceptions for studying congruences between Galois representations, understanding the importance of the finite-flat property of group schemes, and the identification of the Gorenstein property. There is a theoretical $p$-descent on the Eisenstein quotient when previously descents were almost all explicit $2$-descents with specific equations. It introduces the winding quotient, and so on. Third, while it is a dense paper, it is dense in the best possible way: many of the small diversions could have made interesting papers on their own. Indeed, even close readers of the paper today can find connections between Mazur's asides and cutting edge mathematics. When Mazur raises a question in the text, it is almost invariably very interesting. One particular (great) habit that Mazur has is thinking about various isomorphisms and by pinning down various canonical choices identifies refined invariants. To take a random example, consider his exploration of the Shimura subgroup at the end of section 11. He finishes with a question which to a casual reader may as well be a throw-away remark. But this question was first solved by Merel, and more recently generalized in some very nice work of Emmanuel Lecouturier. Lecouturier's ideas then played an important role in the work of Michael Harris and Akshay Venkatesh. Again, one could give many more such examples of this. Very few papers have the richness of footnotes and asides that this paper does. Never forget that one of the hardest things in mathematics is coming up with interesting questions and observations, and this paper contains many great ones - it is bursting with the ideas of a truly creative mathematician. Finally, the result itself is amazing, and (pretty much) remains the only method available for proving the main theorem (the second proof due to Mazur is very related to this one). To give a sense of how great the theorem is, note that if $E$ is a semistable elliptic curve, then either $E$ is isogenous to a curve with a $p$-torsion point, or $E[p]$ is absolutely irreducible. This result (added for clarity: explicitly, Mazur's Theorem that $E/\mathbf{Q}$ doesn't have a $p$-torsion point for $p > 7$) is crucially used in Wiles' proof of Fermat. One could certainly argue that without this paper (and how it transformed algebraic number theory) we would not have had Wiles' proof of Fermat, but it's even literally true that Mazur's theorem was (and remains so today, over 40 years later) an essential step in any proof of Fermat.<|endoftext|> TITLE: Quantum corrections to geometry QUESTION [17 upvotes]: In this video Alain Connes made a comment about the ,,quantum corrections'' of the geometry. I would like to understand this notion in some details since I haven't found anything about this in the literature. I'm guessing that this notion is different to the notion of so called ,,inner fluctuations'' of the metric (see for example this discussion) and unlike the inner fluctuations, quantum corrections is less understood. So making the long story short my question is What are quantum corrections of a geometry? EDIT: Let me be a little bit more precise. One of the highlights of noncommutative geometry is that it can reproduce the complicated Lagrangian of the Standard Model from the so called spectral action and can be seen as pure gravity on some slightly noncommutative space of the form $M \times F$. Here $F$ is not correctly defined as a set but manifests itself as a noncommutative algebra: the idea is that $F$ represents some noncommutative space. Since this noncommutativity is already present on the electroweak scale (i.e. distances approx. $10^{-16}\text{cm}$) I guess that the quantum corrections of geometry should be: already visible on this (much lower than Planck) energy scale understood within the context of noncommutative geometry i.e. Dirac operators and spectral triples (if we take the lesson from the Standard Model seriously). So let me state the problem once again, this time more precisely: How to understand quantum corrections of geometry within the context of noncommutative geometry? REPLY [19 votes]: Physicist chiming in; "quantum corrections of a geometry" represents the vague idea that gravity should be quantum. Classically, gravity is a geometric theory: "reality" is modelled as a (Lorentzian) manifold, whose metric can be found -- in principle -- by solving a system of PDEs. To be more specific, the metric $g$ is such that the classical action $S[g]$ is stationary, $$ S'[g]\equiv0 $$ where $S$ is, say, the Hilbert action $S=\int_X R$. (This is the canonical choice, which agrees with experiments to a great level of accuracy, but there are other choices that may work better in extreme regimes; a popular alternative is $S=\int_X f(R)$ for some appropriate function $f$, cf. f(R) gravity). Quantumly, this does not longer hold. Gravity should still be a geometric theory, but "quantum corrected", meaning that the paradigm changes but should reduce to the standard (classical) framework in the classical limit (schematically written as $\hbar\to0$). This is the problem of quantum gravity. The intuitive idea is that the metric is no longer fixed by a system of differential equations, but "fluctuates": there should be a "measure" $\mu[g]$ in the "space of metrics" such that, when we perform a given experiment, a single metric $g$ is measured, drawn at random from this space of metrics, and with probability $\mu[g]$. The "expectation value" for the geometry of spacetime is $$ \langle g\rangle\sim\int \mu[g] $$ where the integral is over "the space of all metrics", whatever this means. A popular choice for the measure is $\mu[g]=e^{-S[g]/\hbar}$, which has the nice property that, in the $\hbar\to0$ limit, is peaked around the stationary point $S'[g]=0$, and so this measure appears to single out the classical configuration as a special point in this space. Obviously, this is not a well-defined procedure, but it represents what we physicists would like to have. We leave it to the mathematicians to develop a theory of "integration over function spaces" strong enough to accommodate this our intuition. But anyway, the take home message is that, classically, nature is supposed to be described as a manifold with a fixed (Lorentzian) metric. But if you go to very high energies (or very short distances), the classical picture breaks down and you need to take into account "quantum corrections": the metric is no longer fixed but becomes a random variable, peaked around the classical value but "fluctuating" according to some measure. Making this precise is the open problem in physics. REPLY [11 votes]: I presume this refers to quantum corrections to the metric tensor. These are expected to be important in general relativity when the curvature of space-time approaches the Planck scale $\sqrt{G\hbar/c^3}\simeq 10^{-33}$ cm. Some pointers to the literature: Einstein Equation with Quantum Corrections Quantum Corrections to the Schwarzschild and Kerr Metrics Quantum Corrections to the Reissner-Nordstrom and Kerr-Newman Metrics The edit also asks "How to understand quantum corrections of geometry within the context of noncommutative geometry (NCG)?".$^*$ I do not think this understanding exists at present. NCG has given a geometrical description of the Standard Model of elementary particles, and there might be ways to unify this with gravity, but that development is still very much in its infancy. $^*$ The name noncommutative geometry is misleading, in the sense that it suggests a quantum theory (since we all know that non-commuting operators play such an essential role in the quantum world), but in fact the homological theory that constitutes NCG does not itself contain quantum effects.<|endoftext|> TITLE: Is completion of isolated singularity isolated? QUESTION [5 upvotes]: Let $K$ be an algebraically closed field and let $A=K[x_1,\dots,x_n]/I$ be a $K$-algebra of finite type which has only an isolated singularity at the origin. Let $\mathfrak{m}=(x_1,\dots,x_n)$ and consider the localization $A_{\mathfrak{m}}$ and its $\mathfrak{m}$-adic completion $B$. Is $B$ also an isolated singularity? The localization at $\mathfrak{m}$ should not be a problem, but completion might. In general, the completion of an isolated singularity need not to be isolated. However, I was wondering whether in this situation the answer might be positive. I am willing to assume some stronger hypothesis. Can anybody point out any results in this direction? Thank you in advance. REPLY [4 votes]: I believe that in your situation, $B$ indeed has an isolated singularity at the maximal ideal $\mathfrak{n} \subseteq B$. Let me first give two possible definitions for “isolated singularity”; please let me know if there are standard definitions for these notions, and I will edit this answer accordingly! Definition. Let $X$ be a locally noetherian scheme. We say that $X$ has an isolated singularity at a closed point $x \in X$ if there is an open neighborhood $U \ni x$ such that the scheme $U \smallsetminus \{x\}$ is regular. Suppose that $X$ is a scheme over a field $k$. We say that $X$ has a geometric isolated singularity at a closed point $x \in X$ if there is an open neighborhood $U \ni x$ such that the scheme $U \smallsetminus \{x\}$ is geometrically regular over $k$. Adopting either definition, the ring $B$ in your situation has a isolated singularity (resp. geometric isolated singularity) at the maximal ideal $\mathfrak{n} \subseteq B$. We can in fact prove the following more general result: Proposition. Let $X$ be a locally noetherian scheme (resp. locally noetherian scheme over a field $k$) with an isolated singularity (resp. geometric isolated singularity) at a closed point $x \in X$, such that $\mathcal{O}_{X,x}$ is a $G$-ring. Then, $\operatorname{Spec}\widehat{\mathcal{O}}_{X,x}$ has an isolated singularity (resp. geometric isolated singularity) at the unique closed point $\widehat{x}$. Note that this proposition applies to your situation, since $A$ is of fintie type over a field, hence a $G$-ring in the sense of [Mat89, p. 256] by [Mat89, Cor. to Thm. 32.6] ($A$ is also excellent, but we only need the $G$-ring part of the definition of excellence). Proof. Let $U \subseteq \operatorname{Spec} \mathcal{O}_{X,x}$ be the intersection of $\operatorname{Spec} \mathcal{O}_{X,x}$ and a neighborhood of $x$ in $X$ satisfying the condition in the definition above. Let $f\colon \operatorname{Spec} \widehat{\mathcal{O}}_{X,x} \to \operatorname{Spec} \mathcal{O}_{X,x}$ be the morphism induced by the completion homomorphism. The morphism $$f^{-1}(U) \smallsetminus \{\widehat{x}\} \longrightarrow U \smallsetminus \{x\}\tag{1}\label{eq:completion}$$ is regular in the sense of [Mat89, p. 255] since the morphism $f$ is regular by the $G$-ring condition, and regular morphisms are preserved under base change [EGAIV$_2$, Prop. 6.8.3(iii)]. For isolated singularities, we apply [Mat89, Thm. 23.7] to the (localizations of the) morphism \eqref{eq:completion} to conclude that $f^{-1}(U) \smallsetminus \{\widehat{x}\}$ is regular. For geometric isolated singularities, the composition $$f^{-1}(U) \smallsetminus \{\widehat{x}\} \longrightarrow U \smallsetminus \{x\} \longrightarrow \operatorname{Spec} k$$ is regular by [Mat89, Thm. 32.1], i.e., $f^{-1}(U) \smallsetminus \{\widehat{x}\}$ is geometrically regular over $k$. $\blacksquare$ References [EGAIV$_2$] Alexander Grothendieck and Jean Dieudonné. “Éléments de géométrie algébrique. IV. Étude locale des schémas et des morphismes de schémas. II.” Inst. Hautes Études Sci. Publ. Math. (1965), no. 24, 1–231. DOI: 10.1007/BF02684322. MR: 199181. [Mat89] Hideyuki Matsumura. Commutative ring theory. Second ed. Translated from the Japanese by M. Reid. Cambridge Stud. Adv. Math. 8. Cambridge Univ. Press, Cambridge, 1989. DOI: 10.1017/CBO9781139171762. MR: 1011461.<|endoftext|> TITLE: Exterior powers and choice QUESTION [7 upvotes]: Under the assumption that any vector space has a basis (so under the assumption of the axiom of choice), we can prove the following algebraic statements : 1) If $\varphi:V\to W$ is an injective linear map between vector spaces, then the exterior powers $\Lambda^k \varphi : \Lambda^k V\to \Lambda^k W$ are injective. (See Corollary 5.9 here) 2) If $v_1,\cdots,v_k$ are linearly independent vectors in a vector space $V$ then $v_1\wedge \cdots\wedge v_k\neq 0$. (See Theorem 7.1 here) My question is (similar to this question about tensor products) : Are those statements (1 and 2) still true without the help of the axiom of choice? Or do they imply a form of choice in some sense ? REPLY [5 votes]: As YCor notes in comments, (2) is a special case of (1), so I'll only address (1). Suppose $\Lambda^k\varphi:\Lambda^kV\to\Lambda^kW$ is not injective, and let $x\neq0$ be in the kernel. Then $x$ can be written in the form $$x=\sum_{i=1}^mv_{i1}\wedge\dots\wedge v_{ik}.$$ Let $V'$ be the finite dimensional subspace of $V$ spanned by $\{v_{ij}\mid1\leq i\leq m,1\leq j\leq k\}$, and $\varphi':V'\to W$ the restriction of $\varphi$ to $V'$. Let $$x'=\sum_{i=1}^mv_{i1}\wedge\dots\wedge v_{ik},$$ considered as an element of $\Lambda^kV'$. Then $x'\neq0$, since it is sent to $x$ by the map induced by the inclusion of $V’$ into $V$. Also $$\Lambda^k\varphi'(x')=\Lambda^k\varphi(x)=0,$$ so $x'$ is a nonzero element of the kernel of $\Lambda^k\varphi'$. Hence we may as well assume that $V$ is finite dimensional. The fact that $\Lambda^k\varphi(x)=0$ follows from a finite number of the relations defining the exterior power $\Lambda^kW$, involving only finitely many elements of $W$. If we replace $W$ by the finite dimensional subspace $W''$ spanned by the image of $\varphi$ and these finitely many elements, then $\varphi$ induces a map $\varphi'':V\to W''$, and $\Lambda^k\varphi''(x)=0$, since the same relations that implied $\Lambda^k\varphi(x)=0$ in $\Lambda^kW$ also imply that $\Lambda^k\varphi''(x)=0$ in $\Lambda^kW''$. Hence we can also assume that $W$ is finite dimensional, and what remains is a problem about finite dimensional vector spaces that can easily be answered without choice by choosing bases.<|endoftext|> TITLE: Example of two exotic closed 4-manifolds s.t. SW(X)=0 QUESTION [13 upvotes]: I am interested in seeing examples of two closed 4-manifolds $X_1,X_2$ such that $SW(X_i)=0$ and they are homeomorphic but not diffeomorphic. So far in the literature I've only found examples which somehow used symplectic geometry and operations on the Seiberg-Witten invariant (knot surgery, rational blow down, star surgery, etc.) to distinguish one from the other. So I am interested in knowing some other (if known) techniques in this direction. The reason I've mentioned $SW(X_i)=0$ is to ensure that none of them admits symplectic structures (by work of Taubes). Trisection theory is potentially a way but I am not sure if anything is known in that direction or not. Thanks in advance. REPLY [2 votes]: Here is an example as Kyle suggested in his comment. Claim is that $K3 \#K3 \# \bar{\mathbb {CP^2}}$ and $\mathbb{ \#_6 CP^2}\#_{39}\mathbb{ \bar{CP^2}}$ are homeomorphic but not diffeomorphic. Notice that they have same intersection form and since they are simply-connected, by Freedman, they are hoeomorphic. Observe that they have trivial Seiberg Witten invariant follows from Taubes result of vanishing Seiberg Witten invariant under connected sum. In order to distinguish their diffeomorphism type, we are going to use stable cohomotopy Seiberg Witten invariat (ScSW). Here are two important theorem in this context, one is corresponds to vanishing and the other one for non-vanishing Theorem 1(Bauer) If a closed smooth connected 4 manifold satisfies one of the following 1) or 2) then they have non vanishing ScSW 1) If $X = X_1 \#X_2$ and $X_1$ has non vanishing ScSW and $b_2^+(X_2)=0$ 2) If $X= \#_n X_i$ such that $b_2^+(X_i)=3 (mod 4)$, $b_1(X_i)=0$. Each $X_i$ has a compatible spinc structure $s_i$ such that $SW_{s_i}(X_i)=1(mod 2)$ and $11$. Suppose $Y$ is a codim 1 embedded manifold which admists a positive scalar curvature. And the inclusion map induced non trivial map $H^2(X,\mathbb Q)\to H^2(Y,\mathbb Q)$, then ScSW vanishes. Observe that $\mathbb {CP^2}$ sits in $\mathbb{ \#_6 CP^2}\#_{39}\mathbb{ \bar{CP^2}}$. There is a self intersection 1 sphere in $\mathbb CP^2$. If we blown up once, that will give us a self intersection 0 sphere in $\mathbb{ \#_6 CP^2}\#_{39}\mathbb{ \bar{CP^2}}$. Inface the boundary of such a neighbourhood is $S^1\times S^2$. And the copy of $S^2$ gives a non-torsion element in $H^2$. So by Theorem 2, ScSW=0. And thus they are EXOTIC.<|endoftext|> TITLE: A spectral description of Fredholm operators QUESTION [6 upvotes]: Let $L:H \to H$ be a bounded operator on a Hilbert space $H$, with finite dimensional kernel, and whose adjoint also has finite dimensional kernel. Is it true that $L$ is Fredholm if and only if its spectrum is norm bounded below by a non-zero constant? REPLY [6 votes]: I assume when you say 'the spectrum is bounded below' you mean that there exists $c>0$ so that no $\lambda$ with $0 < |\lambda| \leq c$ is in the spectrum. In fact, for any bounded operator $L$ on a Hilbert space, this condition is equivalent to the demand that $L$ has closed range. Each of these conditions for $L$ are equivalent to the same condition for $L^*L$, so we may as well assume $L$ is self-adjoint. Second, both conditions are preserved by the addition of a map $\epsilon p: H \to H$ which factors as an orthogonal projection $p: H \to \text{ker}(L) \hookrightarrow H$ (the first condition is preserved so long as $\epsilon$ is taken sufficiently small, though of course the constant $c$ will change). So we may as well assume that $L$ is injective. From here on $L$ is an injective self-adjoint operator. If $L$ has closed range, then the identification $\text{coker}(L) \cong \text{ker}(L^*) \cong \text{ker}(L) = 0$ and the closed graph theorem imply $L$ is an isomorphism; because $\|L^{-1}v\| \leq C\|v\|$, we see that $\|Lv\| \geq \frac 1C \|v\|,$ and in particular $|\text{Spec}(L)| \geq 1/C$. Conversely, suppose $L$ has no eigenvalues near zero, and suppose $Lv_n \to w \in \overline{\text{Im}(L)}$. Then because $\frac 1C \|v_n\| \leq \|Lv_n\|$, we see that $$\limsup \frac 1C \|v_n\| \leq \|w\|,$$ and in particular $\|v_n\|$ is bounded, so $v_n$ converges in the weak* topology to some $v'$; correspondingly, we see that $Lv_n \to Lv'$ in the weak* topology, and hence $w = Lv'$. Thus the range of $L$ is closed, as desired. Adding on the conditions about the kernel of $L$ and its adjoint you get your result. I'm sure you can get the same answer for Banach spaces, but the argument would need to be different.<|endoftext|> TITLE: Does every large $\mathbb{R}^4$ embed in $\mathbb{R}^5$? QUESTION [25 upvotes]: This question was prompted by my answer to this question. An exotic $\mathbb{R}^4$ is a smooth manifold homeomorphic to $\mathbb{R}^4$ which is not diffeomorphic to $\mathbb{R}^4$ with its standard smooth structure. An exotic $\mathbb{R}^4$ is said to be small if it can be embedded in the standard $\mathbb{R}^4$ as an open subset. An exotic $\mathbb{R}^4$ which is not small is called large. Does every large $\mathbb{R}^4$ embed in $\mathbb{R}^5$? Freedman and Taylor showed there is a maximal exotic $\mathbb{R}^4$, into which all other $\mathbb{R}^4$'s can be smoothly embedded as open subsets. So it would be enough to show that this one embeds in $\mathbb{R}^5$. By the Whitney Embedding Theorem, every large $\mathbb{R}^4$ embeds in $\mathbb{R}^8$, but I suspect one can do better. REPLY [21 votes]: The product of any two smooth open contractible manifolds is diffeomorphic to a Euclidean space, see e.g. remark 5.3 in my paper; the result is of course not due to me but I don't know any other place where all references are collected. In particular, this applies to the product of $\mathbb R$ and any exotic $\mathbb R^4$, so the latter embeds into $\mathbb R^5$. In this case the point is that the product is simply-connected at infinity, and hence by a result of Stallings it is PL homeomorphic to $\mathbb R^5$, but any PL structure on $\mathbb R^5$ is induced by a unique smooth structure (as proved by Munkres).<|endoftext|> TITLE: What is a tamely-ramified Weil-Deligne representation? QUESTION [28 upvotes]: Let $W_F$ denote the Weil group of a finite extension of $\mathbb{Q}_p$. Let $I$ denote the inertia subgroup and $I^{>0}$ the (pro-$p$) subgroup of wild inertia. (I hope I've got my notation right...apologies if I haven't.) We have $W_F/ I = \mathbb{Z}$ (canonically), and $I/I^{>0} \cong \prod_{\ell \ne p} \mathbb{Z}_\ell$ (non-canonically). Consider a Weil-Deligne representation $(V, \rho, N)$, where $V$ is a finite-dimensional complex vector space, $\rho : W_K \to GL(V)$ is a rep of $W_K$, and $N$ is a nilpotent operator satisfying a well-known condition. Naively, I would have thought that a Weil-Deligne representation $(V,\rho,N)$ is tamely ramified if $\rho$ vanishes on $I^{>0}$, no conditions on $N$ However, when I read literature on the Deligne-Langlands conjecture (e.g. Kazhdan-Lusztig, Chriss-Ginzburg) it seems that tamely ramified actually means $\rho$ vanishes on $I$, no conditions on $N$ [Under this assumption an $F$-semi-simple, tamely ramified, Weil-Deligne representation becomes a (conjugacy class of a) pair $(s,N)$ consisting of a semi-simple element $s$, a nilpotent element $N$, such that $sNs^{-1} = qN$, which is what one expects from affine Hecke algebras, and what we are told the Deligne-Langlands conjecture predicts that the answer should be. [EDIT (following Ben Zvi's comments below): Actually we also need an irreducible representation of the component group of the centraliser of this pair. This extra piece of data is absent for $GL_n$ as this group is always connected.] Passing back through Grothedieck's equivalence, the above seems to give that a tamely-ramified and continuous representation of $W_K$ in a $\mathbb{Q_\ell}$-vector space, should have each $\mathbb{Z}_{\ell'}$ factor (for $\ell' \ne \ell$) acting trivially, which seems a little strange. (I.e. the finite monodromy which could come from the other factors should magically be trivial.) Finally, an admittedly lazy question. I can read in several places that, under the local Langlands correspondence, tamely ramified should correspond to representations admitting an Iwahori fixed vector. Is this explained somewhere? Why should tamely ramified (whatever that actually means, c.f. the above question) translate into having an Iwahori fixed vector? I'm sure this is easy for the pros! Thanks in advance. REPLY [4 votes]: This is more of an extended comment. If I'm understanding the question correctly, I don't believe it's correct to say that tamely ramified Galois representations with unipotent monodromy correspond precisely to representations with Iwahori fixed vector, outside of say $GL_n$ -- it seems to me one can't characterize reps with an Iwahori-fixed vector in terms of only their Langlands parameter. [Edit: More precisely, there are strictly more unipotent tamely ramified representations than reps with an Iwahori-fixed vector. This is precisely analogous to the situation with Chevalley groups: there are more representations which are unipotent in the Deligne-Lusztig sense, i.e. are attached the trivial semisimple conjugacy class in the dual group, than there are unipotent principal series, i.e., reps with a Borel-fixed vector. This is accounted for by Lusztig's generalized Springer correspondence -- geometrically, it's the statement that not all equivariant perverse sheaves on the nilpotent cone are generated by the Springer sheaf, outside of $GL_n$.] Kazhdan-Lusztig's result identifies representations with Iwahori-fixed vector as having unipotent Langlands parameter - i.e. q-commuting pairs of a semisimple and nilpotent in the dual group. However representations aren't classified by their Langlands parameter - rather one should think of them as suitable sheaves on a stack of Langlands parameters, and one has to identify the action of the stabilizer (i.e. the L-packet). (This I think is a modern form of local Langlands correspondence, one seeks to identify categories of reps of groups over local fields as or inside suitable categories of sheaves on stacks of Langlands parameters). Kazhdan-Lusztig say we only get SOME of the objects in the corresponding L-packets: the representations of stabilizers that appear are only those that take part in a form of the Springer correspondence. This has to do with the existence of cuspidal local systems on Levis (as in the generalized Springer correspondence) and hence is invisible for $GL_n$. Lusztig then proved [Classification of unipotent representations of simple p-adic groups. Internat. Math. Res. Notices 1995, no. 11, 517–589] an "affine generalized Springer correspondence". To paraphrase: unipotent Langlands parameters exactly correspond to unipotent representations -- a generalization of reps with an Iwahori fixed vector, in which the notion of Iwahori fixed vector (given by the trivial representation of the finite torus, hence of the finite Borel, hence of Iwahori) is replaced by a unipotent cuspidal representation of a Levi of the corresponding Chevalley group. This distinction (unipotent vs Iwahori-fixed) does not arise in geometric Langlands -- it appears that the categorical analogs of unipotent representations have Iwahori-fixed vectors. This is e.g. well understood for reductive groups (rather than loop groups), where all unipotent representations come from decategorifying the finite Hecke category. Likewise, if you formally take the categorical trace of Frobenius on a (suitably mixed!!) version of Bezrukavnikov's Langlands duality for the Iwahori-Hecke category (cited by Dr. Evil), you get (up to singular support issues) the full category of coherent sheaves on the stack of unipotent Langlands parameters -- i.e. the home (thanks to Lusztig's theorem) of all unipotent reps of our original group.<|endoftext|> TITLE: Notion of linking between two general $p$ and $q$ manifolds embedded in a higher dimensional manifold QUESTION [6 upvotes]: Let $M$, $N$ be compact, connected, oriented manifolds without boundary embedded in $\mathbb{R}^m$ of dimensions $p$ and $q$ respectively. I know that when $m=p+q+1$ we can define the linking between $M$ and $N$ in the sense given here, but what if $M$ and $N$ do not satisfy this constraint? Example: Can we 'link' two surfaces in $\mathbb{R}^3$? What is the definition of linking in this case ? REPLY [7 votes]: Sure, there are ways to make sense of linking beyond the constraints you mention. I suppose the most basic would be: two disjoint submanifolds $A$ and $B$ of a manifold $M$ are unlinked if you can find disjoint embedded $m$-dimensional discs $D_1, D_2 \subset M$ such that $A \subset D_1$ and $B \subset D_2$. Here $m=dim(M)$. The manifolds would be linked if they are not unlinked. A nice feature of this definition is you can talk about linking of three or more objects, so you can similarly talk about Brunnian properties, i.e. $k$ objects can be linked but any collection of $k-1$ sub-objects are unlinked. For example, there is a classical theorem of Schubert that says if you have two (or more) disjoint non-trivial knot exteriors in $S^3$, then they are unlinked. By knot exterior I mean a compact 3-manifold diffeomorphic to the complement of an open tubular neighbourhood of a knot in $S^3$. As you likely know, if the knot exterior is trivial it would be diffeomorphic to $S^1 \times D^2$, and such objects can be linked in $S^3$. I believe this is a commonly-used notion of linking that generalizes the definition you mention. Using this definition, linking of surfaces in $\mathbb R^3$ or $S^3$ is a fairly easy definition to work with, once you are comfortable with the tools of 3-manifold theory. In your question you refer to "the linking between $M$ and $N$" but really you are talking about linking numbers. Interesting numerical linking invariants can take a fair bit of work to define, and John Klein's answer is directed towards that kind of question. But if all you care about is what the term "linking" means, that is much more elementary, and what my answer is directed towards. edit: here are more details to support my comment below. This argument is typical of Hassler Whitney's work. I'm not certain if he ever made this specific argument, but the general mode of argument very much is due to him. For convenience, let's say $M$ and $N$ are compact submanifolds of $\mathbb R^k$. We can get by without this assumption, but it makes the arguments less technical. By compactness, there is some vector $v \in \mathbb R^k$ so that $M+v$ is disjoint from $N$. Consider the function $$F: [0,1] \times M \to [0,1]\times \mathbb R^k$$ given by $F(t,p) = (t, tv+p)$. Thus $F(0,M) = \{0\} \times M$ and $F(1,M) = \{1\} \times (v+M)$. So $F([0,1] \times M)$ is disjoint from $[0,1] \times N$ at the start and the end of the interval, but maybe not in the interior. We can perturb $F$ a little bit (not affecting its values at the endpoints) to make it transverse to $[0,1]\times N$. Provided $k+1 > (m+1)+(n+1)$ the transverse map would be disjoint from $[0,1]\times N$, i.e. $k \geq m+n+2$. Since the transversal perturbation can be made arbitrarily small in the $C^\infty$-topology, we can assume it represents an isotopy of $M$. Thus $M$ can be isotoped to be separable from $N$ by disjoint embedded discs. So by isotopy-extension, $M$ and $N$ can be separated by disjoint embedded discs.<|endoftext|> TITLE: How weird can a ring spectrum be if all of its modules are free? QUESTION [13 upvotes]: Let $R$ be a ring spectrum. If $\pi_\ast(R)$ is a graded field, then all module spectra over $R$ are free. But I don't believe the converse holds. How badly can it fail? I'm assuming that $R$ is at least an $A_3$ ring spectrum, and probably $A_\infty$ (I'm also interested in more highly-structured cases); if the answer depends on just how highly structured $R$ is, I'd find that interesting. Note that if $\alpha \in \pi_\ast(R)$ is non-nilpotent, then the freeness of the localization $\alpha^{-1} R$ entails that $\alpha$ is a unit. So if all module spectra over $R$ are free, then $\pi_\ast(R)$ is a local ring of dimension 1. But $\pi_\ast(R)$ might have nonzero nilpotents as far as I can see. Question: What is an example of a ring spectrum $R$ such that $\pi_\ast(R)$ has nonzero nilpotents and yet every module spectrum over $R$ is free? REPLY [5 votes]: I think what you are looking for is the notion of a semisimple ring spectrum, as studied by Hovey and Lockridge. For such a ring spectrum, $E$, every module spectrum is projective, i.e. is a retract of a coproduct of suspensions of $E$. The homotopy groups $E_*$ of such spectra are characterized in Theorem 1.2 of this paper: $E_* \cong R_1 \times \dots \times R_n$ where $R_i$ is either a graded field $k$ or an exterior algebra $k[x]/(x^2)$ over a graded field with a unit in degree $3|x|+1$. Such rings $E_*$ are called graded commutative $\Delta^1$-rings, and clearly have plenty of non-zero nilpotent elements. If $E$ is commutative, then $E$ is semisimple if and only if $E_*$ is a graded commutative $\Delta^1$-ring and for every factor ring of the form $k[x]/(x^2)$, we have $x\cdot \pi_*(C) \neq 0$ where $C$ is the cofiber of $x\cdot E$. I'll finish by mentioning classic work of Hopkins and Smith, that the OP is surely aware of but future readers might not be, that defines the fields of stable homotopy theory as ring spectra $E$ such that $E_*X$ is a free $E_*$-module for all spectra $X$, and that characterizes fields as spectra having the homotopy type of a wedge of suspensions of Morava $K$-theory $K(n)$ for some fixed $n$ and prime $p$.<|endoftext|> TITLE: Grade-school elementary algebra presented in an abstract-algebra style? QUESTION [12 upvotes]: I remember once hearing a (probably apocryphal) story about a university math professor that tried to teach a gradeschool class about algebra by telling them a few simple axioms and definitions and then making deductions. As the story goes, he thought he would be making things as easy as possible by minimizing the number of things the students would have to learn, but that turned out to be a bad idea because the way you get kids to do good on the SAT is by minimizing how much they have to think. I am curious about what that lecture series would actually look like. I would like to find a presentation of elementary algebra that treats it from an abstract standpoint, but that requires no prior knowledge. Essentially, I am looking for algebra explained in the "professional" style that the above story depicts. I have no idea where to find it. Textbooks that present the abstract algebra view of elementary algebra tend to assume you already know elementary algebra. University textbooks about elementary algebra written in the 1700s (when elementary algebra was a dominant research topic) come close, but abstract algebra was not around back then. Does such a book or series of notes exist? Can it exist? REPLY [3 votes]: Perhaps you mean the series of school books by Georges Papy titled Mathématique moderne. (Image from images.slideplayer.fr)<|endoftext|> TITLE: Characterization of operator ranges QUESTION [5 upvotes]: My question is motivated by the following little proposition: Proposition. For a vector subspace $V$ of a Banach space $(X, \|\cdot\|_X)$ the following assertions are equivalent: (i) There exists a Banach space $Z$ and a bounded linear operator $T: Z \to X$ with range $V$. (ii) There exists a complete norm $\|\cdot\|_V$ on $V$ such that the canonical embedding of $(V, \|\cdot\|_V)$ into $(X,\|\cdot\|_X)$ is continuous. (See below for a proof.) $\,$ Question. Are (i) and (ii) also equivalent to the following assertion (iii)? (iii) There exists a bounded linear operator $S: X \to X$ with range $V$. $\,$ Proof of the Proposition. Obviously (ii) implies (i), so assume that (i) holds. Let $\tilde T: Z / \ker T \to X$ denote the injective operator induced by $T$; then $\tilde T$ also has range $V$. The inverse ${\tilde T}^{-1}$ is a closed linear operator $X \supseteq V \to Z / \ker T$, so $V$ becomes a Banach space with respect to the graph norm given by $\|x\|_V := \|x\|_X + \|{\tilde T}^{-1}x\|_{Z / \ker T}$ for all $x \in V$. Remark. For Hilbert spaces results of this type can, for instance, be found in the paper "Fillmore and Williams: On Operator Ranges (1971)". In fact, the above proof is an adaptation of an argument that appears in the proof of Theorem 1.1 of this paper. REPLY [6 votes]: The answer to your question is "No". It can be seen in the following way: If there exists an operator $S$ mentioned in the Question, then, using the standard techniques, one can show that $V$ has to be isomorphic to a quotient space of $X$. So it remains to show that there exists $X$ and an operator range in $X$ for which this condition fails. This can be done by using injective nuclear operators with non-closed range from any separable Banach space $V$ into $X$ and by picking $V$ and $X$ in such a way that $V$ is not a quotient of $X$. For example, let $X$ be a separable Hilbert space and $V$ be a separable Banach space which is not isomorphic to a Hilbert space. Example 4.12 in Cross, R. W.; Ostrovskiĭ, M. I.; Shevchik, V. V. [Operator ranges in Banach spaces. I. Math. Nachr. 173 (1995), 91–114] is a much stronger example. In the same paper you can find more details on the proof sketched above.<|endoftext|> TITLE: What are the applications of topological quantum field theory to continuous-time dynamical systems? QUESTION [10 upvotes]: From wikipedia: In dynamics, all continuous time dynamical systems, with and without noise, are Witten-type TQFTs and the phenomenon of the spontaneous breakdown of the corresponding topological supersymmetry encompasses such well-established concepts as chaos, turbulence, 1/f and crackling noises, self-organized criticality etc. These are all very "deep" phenomena which are studied in continuous-time dynamics, and its somewhat surprising that TQFT has such a wide range of applications. Unfortunately, that wikipedia article doesn't include inline citations, so it isn't clear which source that information comes from. Could someone point to a textbook or review on the applications of TQFT to continuous-time dynamical systems? EDIT: Is the "topological supersymmetry" bascially the structural stability of the system? That would make sense, except structural stability doesn't imply topological invariance, just invariance under a certain restricted set of diffeomorphisms (i.e. perturbations). REPLY [5 votes]: The Wikipedia entry cited in the OP is paraphrased from Topological field theory of dynamical systems by Igor Ovchinnikov (2012): It is shown that the path-integral representation of any stochastic or deterministic continuous-time dynamical model is a cohomological or Witten-type topological field theory, i.e., a model with global topological supersymmetry (Q-symmetry). As many other supersymmetries, Q-symmetry must be perturbatively stable due to what is generically known as non-renormalization theorems. Ovchinnikov discusses the notion of topological supersymmetry in a dynamical system, and the breaking of that symmetry at the transition to chaotic dynamics, in Introduction to Supersymmetric Theory of Stochastics (2016). A more colloquial description is given in Chaos or Order (2017). I should add that (judging from the citation trail) this line of research does not seem to have been taken up by other groups, and it remains to be seen how productive the connection identified by Ovchinnikov will turn out to be.<|endoftext|> TITLE: If a (distance) metric on a connected Riemannian manifold locally agrees with the Riemannian metric, is it equal to the induced metric? QUESTION [7 upvotes]: Let $(M,g)$ be a connected Riemannian manifold. Let $d_g$ be the induced distance metric of $g$. Now let $d$ be some other metric on $M$. Suppose that for each $x \in M$, there is a neighborhood $U$ of $x$ so that $d = d_g$ on $U \times U$. Question: Does this imply that $d = d_g$ on $M \times M$? I am not assuming that $d$ is the metric of some Riemannian metric; I know the answer is yes in that case. Remarks: It's clear to me that $d = d_g$ on a neighborhood of the diagonal in $M \times M$; this essentially is just a rephrasing of the hypothesis. Connectedness is necessary to avoid stupid counter examples. For example, we could take two disjoint points $\{x,y\}$. Then $d_g$ between them is infinity, but we can take $d(x,y) = 1$. They both agree in a neighborhood of $x$, namely $x$. (We can build similar examples by taking disjoint unions of non-zero dimensional manifolds.) Maybe the 'right' question is actually: let X be a (path) connected topological space and consider two metrics $g$ and $h$ on X, compatible with the topology on X. If these metrics are locally equal in the sense of the question (i.e. agree on a neighborhood of the diagonal), are they equal? (I feel like this either has an obvious proof or some terrible counter example. It appears to be true on graphs metrized by assigning edge lengths - oops, this is wrong by the cut off metric example.) I (think) I can prove a special case. See motivation section below. Motivation: Let $(N,h)$ be a connected Riemannian manifold, with geodesic distance function $d_N$. Let $G$ be the group of a covering space action by isometries on $N$, with quotient map $\pi$, with $G$ finite. Then $N / G$ inherits a Riemannian metric $g$ by using local trivializations. $N / G$ also inherits a metric, defined by $d([x], [y]) = \inf_{g \in G} d_N(gx, y)$. Locally $d = d_g$ because of local trivializations. I want to know if $d = d_g$ on all of $N / G$. I think this is true if the geodesic distance between any two points in $N$ and $N / G$ is always realized by some shortest path, by using the covering space path lifting + being a geodesic is a local condition + isometry of covering actions in order to relate shortest paths in $N$ to those in $N / G$. That is, if $g$ minimizes $d(gx, y)$ let $\gamma : [0,1] \to N$ be a shortest path in $N$ between $gx$ and $y$. Then $\pi( \gamma)$ is a (potentially self intersecting) geodesic, of the same length as $\gamma$, connecting $[x]$ and $[y]$. Hence $d_g \leq d$. On the other hand, if we have a geodesic path $\gamma$ from $[x]$ to $[y]$, we can lift it to a geodesic from $x$ to $hy$ of the same length, for some $h \in G$, and hence $d_g \geq d$. I would like to be able to drop this condition about there always being shortest path witnesses to shortest distances (I mean that the inf in geodesic distance is achieved). I think that maybe in this covering space case one can achieve this by taking a sequence of paths approximating the geodesic distance, but the argument starts to get a lot fuzzier, and I am already outside my comfort zone and procrastinating on my actual work as it is... I would appreciate a reference for either the main question or the motivation. REPLY [13 votes]: Take $d(x,y) = {\rm min}(|x-y|, 1)$ on $\mathbb{R}$,<|endoftext|> TITLE: Is categoricity retained when reducing the language? QUESTION [13 upvotes]: Suppose $\mathcal L \subseteq \mathcal L’$ are first-order languages, $\kappa$ is a cardinal, and $T’$ is a theory in $\mathcal L’$ that is $\kappa$-categorical. Let $T = T’ \restriction \mathcal L$. Is $T$ $\kappa$-categorical? If $|\mathcal L’| = \kappa = \aleph_0$, then I can show the answer is yes using the Ryll-Nardzewski Theorem, which says that a countable theory is $\aleph_0$-categorical iff for each $n$, the number of $n$-types is finite. REPLY [5 votes]: One can also break countable categoricity when $|\mathcal{L}'|>{\aleph_0}$. This example comes from an undergraduate course I took with Malliaris. Let $\mathbb{P}$ be the collection of primes and let $\mathcal{P}(\mathbb{P})$ be the powerset of $\mathbb{P}$. Let $\mathcal{L'} = (+,\times,0,1;(D_{\alpha}(x))_{\alpha \in \mathcal{P}(\mathbb{P})})$ and let $T \models Th_{\mathcal{L}'}(\mathbb{N})$ with the usual interpretation of the symbols, and for each $\alpha \in \mathcal{P}(\mathbb{P})$, $\models D_{\alpha}(n)$ if and only if $n \in \alpha$. One can show that $T$ is $\aleph_0$-categorical and that the only countable model is the standard model. This follows from the fact that if one adds a single non-standard element, one must necessarily add $2^{\aleph_0}$ many elements. However, if we let $\mathcal{L} = \{+\}$, then by Ryll-Nardzewski, we no longer have countable categoricity.<|endoftext|> TITLE: Non-containing subsets of two sizes QUESTION [6 upvotes]: Let $T=\{1,2,\dots,2n+1\}$. What is the largest $k$ such that we can choose $k$ subsets of size $n$ and $k$ subsets of size $n+1$ of $T$ so that no chosen subset contains another? $k=\binom{2n}{n-1}$ is possible: Choose all sets of size $n$ containing $1$, and all sets of size $n+1$ not containing $1$. Does it follow from Sperner's theorem that this is the best possible? REPLY [2 votes]: Here's another approach: Given families $\mathcal{A}$ and $\mathcal{B}$ of $n$-sets and $(n+1)$-sets (respectively) that obey the condition, replace $\mathcal{B}$ with the family $\mathcal{B}'$ of complements of sets in $\mathcal{B}$. Now the condition that no set in $\mathcal{A}$ is contained in one in $\mathcal{B}$ becomes that every set in $\mathcal{A}$ intersects non-trivially with every set in $\mathcal{B}'$. Such a system is called cross-intersecting. Now you can apply combinatorial shifting to $\mathcal{A}$ and $\mathcal{B}'$, and show that it preserves the intersection property. This reduces showing the proof of your inequality to shifted families, where it is likely to be much easier. (I haven't worked out all the details here.)<|endoftext|> TITLE: Strict unfriendly partitions QUESTION [8 upvotes]: Given a graph $G$, an unfriendly partition of $G$ is a partition of $V(G)$ into two classes, such that for every class, every vertex has at least as many neighbors in the other class as in its own class. It is known that every finite graph has an unfriendly partition (just take a partition maximizing the number of edges between two classes) and not every infinite graph has an unfriendly partition (Milner and Shelah). However, it's open whether every countable graph has an unfriendly partition. Define a strict unfriendly partition of a graph $G$ to be a partition of $V(G)$ into two classes, such that for each class, each vertex has strictly more neighbours in the other class than in its own class. Not all finite graphs have strict unfriendly partitions. Indeed, it's easy to see that $K_n$ or $C_n$ has a strict unfriendly partition if and only if $n$ is even. Is there a characterization of all finite graphs $G$ such that $G$ has a strict unfriendly partition? REPLY [5 votes]: This is more an extended comment than an actual answer. If I'm not mistaken, then deciding whether a given graph has a strict unfriendly partition is NP hard, i.e. there probably is no easy-to-decide characterisation. Here is a reduction from 3-SAT to this decision problem. Given a 3-SAT formula $F$, define a graph $G$ as follows: Take a path on 5 vertices $(a,u,b,v,c)$. For every variable $x_i$ of $F$ take a path on 4 vertices $(t_i,u_i,v_i,f_i)$, and attach to every $t_i$ and every $f_i$ as many leaves as there are clauses in $F$. for every clause $C_j$ of $F$ take a vertex $y_j$ and connect this vertex to $t_i$ if $x_i$ appears in $C_j$, and to $f_i$ if $\lnot x_i$ appears in $C_j$. Furthermore, add edges from $a$, $b$, and $c$ to every $y_j$. We claim that $G$ has a strict unfriendly partition if and only if $F$ is satisfiable. Assume that $G$ has a strict unfriendly partition. The vertices $u$,$v$,$u_i$, and $v_i$ make sure that $a,b,c$ end up in the same part whereas $t_i$ and $f_i$ end up in different parts. Denote the part containing $a,b,c$ by $A$. Since 3 of the 6 neighbours of any $y_j$ are contained in $A$, we know that $y_j \notin A$. Furthermore, at least one of the remaining neighbours must be in $A$, otherwise we wouldn't have a strict unfriendly partition. By construction of $G$, this shows that every clause is satisfied if we set $x_i = \mathrm{true}$ for $t_i \in A$ and $x_i = \mathrm{false}$ for $f_i \in A$, and thus $F$ is satisfiable. Conversely, assume that $F$ is satisfiable and let $x_i = \hat x_i$ be an assignment of values that satisfies $F$. Define a partition of $V(G)$ by $$A = \{a,b,c\} \cup \{t_i, v_i, \text{leaves attached to }f_i \mid \hat x_i = \mathrm true\} \cup \{f_i, u_i, \text{leaves attached to }t_i \mid \hat x_i = \mathrm false\},$$ and $B = V(G) \setminus A$. This is strictly unfriendly in $a,b,c,u,v,u_i,v_i$ because all their neighbours are in in the other part, respectively. It is strictly unfriendly in $t_i$ and $f_i$ because the leaves attached to them outweigh the neighbours of the form $y_j$. It is strictly unfriendly in $y_j$ because $a,b,c \in A$ and $x_i = \hat x_i$ was assumed to satisfy $F$ (and thus at least one more neighbour of $y_j$ is contained in $A$).<|endoftext|> TITLE: q-difference equations and quantum mechanics QUESTION [9 upvotes]: I have been trying to understand why the term quantum is so easily accepted for calculus based on q-numbers $[n]_q=\frac{q^n-1}{q-1}$ and q-analogs of classical operators (derivatives, integrals,...). But the best hints I could find is this question Why are quantum groups so called? this one Intuition behind the definition of quantum groups, and this answer of Pavel Etingof on mathoverflow What is the relation between quantum symmetry and quantum groups?. But I could not find any attempt to connect the q-calculations arising from the mathematical idea of deformation (in algebra or combinatorics) to precise concepts in quantum mechanics. Have there ever been actual uses of q-calculus and quantum groups to computing or understanding solutions of Schrodinger equations, or functions actually arising from (physical) quantum mechanics, like states of the harmonic oscillator or some simple atomic hamiltonian like that of the hydrogen atom? Pavel Etingof says that "the main mechanisms through which quantum groups appear in physics is the same as for usual Lie groups" but I have seen hundreds of physics books with applications of classical Lie groups (classical groups usually, $SO,U,SU,...$) to quantum mechanical problems but none for quantum groups. Also when did the term q-calculus, quantum calculus, q-hypergeometric series first appear -in particular in what order did they appear? Thank you. REPLY [2 votes]: Regarding the first part of the question: Have there ever been actual uses of q-calculus and quantum groups to computing or understanding solutions of Schrodinger equations, or functions actually arising from (physical) quantum mechanics, like states of the harmonic oscillator or some simple atomic hamiltonian like that of the hydrogen atom? In my understanding, the answer is yes, there have been lots of applications in various different settings coming from quantum mechanical problems. As far as i know, the first applications of this kind were the algebraic descriptions of the solutions of the Schrodinger equations for (deformed) $q$-oscillators and $q$-rotators together with the computations of the corresponding energy spectrums, transition rates etc. Among the pioneering papers: The quantum group $SU_q(2)$ and a q-analogue of the boson operators, L C Biedenharn, J. of Phys. A: Math. and Gen., vol 22, 18, L873, 1989 On q-analogues of the quantum harmonic oscillator and the quantum group $SU(2)_q$, A J Macfarlane, J. of Phys. A: Math. and Gen., vol 22, 21, 4581, 1989 Since then, there has been quite a lot of literature on similar topics. If we confine ourselves to low and intermediate energy QM (that is: mainly models of atomic and nuclear physics), an early overview can be found at: Quantum groups and their applications in nuclear physics, D.Bonatsos and C.Daskaloyannis, Prog. in Part. and Nucl. Phys., Vol 43, 1999, Pages 537-618 For a more complete list and further discussion on "actual uses of $q$-calculus and quantum groups ... ", maybe you will be interested in the answers (and the references included) to the following question: Is there any published physics article where $q$-mathematics is applied?<|endoftext|> TITLE: Why are Stein manifolds/spaces the analog of affine varieties/schemes in algebraic geometry? QUESTION [15 upvotes]: I presume this is a GAGA-style result, but I cannot find a reference. REPLY [24 votes]: Also like affine varieties, we have: Theorem. A complex manifold is Stein if and only if it embeds into some $\mathbb{C}^N$ as a closed complex submanifold. For the "only if" direction, see Hörmander, An Introduction to Complex Analysis in Several variables, Theorem 5.3.9. For the converse, an argument is contained on pp 109-110 of Hörmander, immediately after the definition of Stein manifold.<|endoftext|> TITLE: Under what conditions is a symmetric tensor category equivalent to $\operatorname{\mathsf{Rep}}G$ for some group $G$? QUESTION [5 upvotes]: Deligne's theorem on tensor categories states that for any symmetric tensor category $\mathcal{C}$ satisfying the subexponential growth condition, there is a fiber functor to $\mathsf{sVec}$ and that $\mathcal{C}$ is equivalent to the representations of a supergroup. Under what conditions can we draw the stronger conclusion that $\mathcal{C}$ is equivalent to $\operatorname{\mathsf{Rep}}G$ for some group $G$? Does it suffice for the simple objects in $\mathcal{C}$ to all have positive dimension? REPLY [8 votes]: If you are in characteristic zero and the dimension of every object is a positive integer then it admits a fiber functor to $Vec$ and is equivalent to $Rep(G)$ for some (pro-algebraic) group $G$. This is theorem 7.1 in Deligne's "Categories Tannakiennes".<|endoftext|> TITLE: RSK and crystal operators QUESTION [6 upvotes]: Is there a good reference on how RSK (and the 3 other variants) interact with crystal operators on the semi-standard tableaux $(P,Q)$ in the image? That is, we have biwords, $W$ which are in bijection with pairs of semi-standard tableaux $(P,Q)$ under RSK. Now, we act on $P$ or $Q$ with crystal raising/lowering operators $e_i$ and $f_i$. These actions are defined on SSYTs. What happens with $W$? Theorem 2.2 here almost answers my question (although it just references a paper by Lascoux that is a bit hard to parse), but I would like to know if there is a survey/book that goes into depth on this - in particular when considering the other variants of RSK. I am in particular interested in variant III in Christian's survey. For a quick reference, I describe this variant on my page here, including an example. REPLY [2 votes]: I figured out the details, and wrote it up here. I did not manage to find a good reference. There are a few nice surveys on RSK and on crystals, but a survey covering how different tableau operators interact would be nice to see someone type up. For the interested, these properties above were needed for this project, where we look at a type of skew q-Whittaker functions. We manage to give a Schur-expansion for a class of LLT polynomials where the expansion was previously unknown.<|endoftext|> TITLE: Is $\mathrm{SL}_n(\mathbb{Q}_p)$ virtually torsion-free? QUESTION [5 upvotes]: Recall that a group is virtually torsion-free if it admits a finite index subgroup which is torsion-free. Question. Is $\mathrm{SL}_n(\mathbb{Q}_p)$ virtually torsion-free for $n > 1$? Comments. Note that $\mathrm{GL}_1(\mathbb{Q}_p) = \mathbb{Q}_p^*$ is virtually torsion-free. We know by a theorem of Selberg that for a field $K$ of characteristic 0, any finitely generated subgroup of $\mathrm{GL}_n(K)$ is virtually torsion-free. However, this does not apply to $\mathrm{SL}_n(\mathbb{Q}_p)$ as it is not finitely generated; the diagonal matrices give a copy of $\mathbb{Q}_p^*$, which is uncountably infinite. A related question can be found here where it is shown that $\mathrm{SL}_n(\mathbb{Z}_p)$ is virtually torsion-free as it is a compact $p$-adic analytic group. Thanks in advance for the help! REPLY [2 votes]: No (this was already answered in comments). $\mathrm{SL}_n(\mathbf{Q}_p)$ is generated by its unipotent 1-parameter subgroups isomorphic to $\mathbf{Q}_p$, hence it has no proper subgroup of finite index. Hence, if it were virtually torsion-free, it would be torsion-free, which is not the case (for $n\ge 2$) as it has an element of order $2$.<|endoftext|> TITLE: A mistake in "Chtoucas de Drinfeld et correspondance de Langlands" QUESTION [8 upvotes]: I have been told by some people that there is a mathematical mistake in the last section of "Chtoucas de Drinfeld et correspondance de Langlands" (L. Lafforgue) concerning the hyperplane section arguments which does not invalidate the main results. I was not able to identify the mistake myself. Could somebody equipped to answer this question tell me what the mistake was and how it could be fixed? REPLY [6 votes]: I think that you refer to a rather small point, which is discussed by Deligne in Sections 0.7 and 1.5-1.9 of https://www.math.ias.edu/files/deligne/FrobTraces.pdf The main results of the Lafforgue's paper are about curves over finite fields. In the last section, he discusses applications to higher dimensional varieties, reducing to the case of curves by hyperplane sections arguments. In one of such arguments (proof of Proposition VII.7), he cites some version of Bertini theorem stated in Hartshorne's book with projectivity assumption, whereas he applies it to some non-projective variety (the locus of smoothness of some l-adic sheaf). In his paper, Deligne writes down some clarification of the argument, with reference to some version of Bertini theorem due to Jouanolou.<|endoftext|> TITLE: Area of $n$-sphere contained outside $\ell_1$ ball QUESTION [6 upvotes]: For a given $r>1$, what is the surface area of $\mathbb S^{n-1}$ (the sphere of radius 1 in $\mathbb R^n$) which is contained outside of the $\ell_1$ ball of radius $r$? Or equivalently, if $X\sim U(\mathbb S^{n-1})$, a point sampled uniformly from the sphere, what is the probability that $\Vert X\Vert_1\geq r$? This is easy to compute for $r\geq \sqrt{n-1}$, as the area is exactly $2^n$ spherical caps, and this has a clean, closed-form formula. For smaller values of $r$, however, these caps intersect, and the algebra gets worse. The exact value of this probability matters less than approximate asymptotic bounds for $n$ large, with $r$ growing in $n$ (like $n^c$ for $c>0$) REPLY [2 votes]: We use the gaussian construction of $U(\mathbb{S}^{n-1})$ to write $$\|X\|_{\ell^1} = \frac{\sum_i |N_i|}{\sqrt{\sum_i |N_i|^2}}$$ with $N_i$ iid random gaussian variables with variance 1. We have $$\mathbb{P}(\|X\|_{\ell^1} \leq \alpha\sqrt{n})\leq \mathbb{P}(\sum_i |N_i| \leq \alpha_1 n) + \mathbb{P}(\sum_i |N_i|^2 \geq \alpha_2 n)$$ for all $\alpha_1,\alpha_2 $ such that $\frac{\alpha_1}{\sqrt{\alpha_2}}\geq \alpha$. Then if $\alpha<\mathbb{E}(|N_1|)=\sqrt{\frac{2}{\pi}}$ you can choose $\alpha_2 >1$ and $\alpha_1 <\mathbb{E}(|N_1|)$ to write a large deviation principle (Cramer theorem). The upper bound is $$\mathbb{P}(\sum_i |N_i| \leq \alpha_1 n)\leq \Big(\frac{\mathbb{E}(e^{\lambda|N_1|})}{e^{\alpha_1 \lambda}}\Big)^n $$for any $\lambda$. Because $\alpha_1<\mathbb{E}(|N|_1)$ you can find $\lambda$ such that $\frac{\mathbb{E}(e^{\lambda|N_1|})}{e^{\alpha_1 \lambda}}=\kappa_{\alpha_1}<1$. We do the same for $|N_i|^2$ and get $$\mathbb{P}(\sum_i |N_i|^2 \geq \alpha_2 n)\leq (\kappa_{\alpha_2})^n $$ with $\kappa_{\alpha_2}<1$. Conclusion for $\alpha<\mathbb{E}(|N_1|)$ we have that $$\mathbb{P}(\|X\|_{\ell^1} \leq \alpha\sqrt{n})\leq (\kappa_{\alpha_1})^n+(\kappa_{\alpha_2})^n$$ decays exponentially.<|endoftext|> TITLE: Riemann Hypothesis, Primes and Groups QUESTION [8 upvotes]: Let $G$ be a finite group $S\subset G$ a generating set $|g|=$ word length with respect to $S$. Set $$ \sigma(G) = \sum_{H \le G} [G:H]$$ Let $\rho$ be the regular representation and set $A_G := \sum_{g \in G} \frac{1}{1+|g|} \rho(g)$. Then $|A_G| = H_G = \sum_{g \in G} \frac{1}{1+|g|}$, $A_G$ is a normal matrix (and possibly not singular), and $A_G/H_G$ is a doubly stochastic matrix. Let $L(x) = x+\exp(x)\log(x)$ be the Lagarias operator. The "group-theoretic" Lagarias inequality might then be stated as: $$\sigma(G) \le L(H_G) = L(|A|) (=^? |L(A)|)$$ where $|B| = $ spectral norm. (For $G=C_n$ the cyclic group and $S = \{a\}$ this is equivalent as shown by Lagarias to Riemann Hypothesis.) I have put an $?$ on the equality, since numerics suggest this is true, but I have no proof yet. For a possibly infinte group $G$, Marcus du Sautoy and Fritz Grunewald define: $$\zeta_G(s) = \sum_{n=1}^\infty \frac{a(G,n)}{n^s} $$ where $a(G,n) = |\{H \le G : [G:H] = n\}|$. For $G = \mathbb{Z}$ this is the Riemann Zeta function. For $G= C_n$ the cyclic group we get: $$\zeta_{C_n}(s) = \sum_{d|n} \frac{1}{d^s} = \sigma_{-s}(n)$$ which was studied by Ramanujan to give (under the RH) an upper bound for $\sigma(n)$. Now, what does that have to do with primes: Intuitively if $$\hat{\pi}(n):= \{ p | p \text{ prime }, p \le n \}$$ then $\hat{\pi}($|G|$)$ will determine how much $\sigma(G)$ can grow at most, since by Lagranges theorem for each $H \le G$ we have $[G:H]||G|$, hence $[G:H]$ must be divisible by some primes $p \in \hat{\pi}(|G|)$. For instance by Sylows theorems, we get a tight ($G=C_{p^k}=$ cyclic p-group) lower bound for $\sigma(G)$: $$\sigma(G) \ge |G| + |G| \sum_{1\le i \le r} \sum_{1 \le n_i \le \alpha_i} \frac{1}{p_i^{n_i}}$$ where $|G| = \prod_{1 \le i \le r} p_i^{\alpha_i}$ is the prime factorization. So there seems to be a conjectured relationship between the Riemann Hypothesis, primes and finite groups. My question is, what is the relationship between: $$\zeta_{\mathbb{Z}}(s), \zeta_G(s) \text{ and } \zeta_{\mathbb{Z}\times G}(s)$$ where $G$ is a finite group and the group $\mathbb{Z}\times G := \mathbb{Z}\times_S G$ is defined in my previous question and is not the direct product. (It turns out, it is the direct product!) Thanks for your help! Edit 25.05.2019: All these upper bounds have something in common: Let $G = C_n$ be the cyclic group. 1) Lagarias inequality: zeta_(Cn) ( -1) = sigma(n) <= L(H_n) 2) Ramanujans upper bound under RH for zeta_(Cn)(s) = sigma_{-s}(n) 3) There is also an upper bound equivalent to RH for the number of divisors: zeta_(Cn)(0) = tau(n) <= .... some upper bound 4) The "group theoretic" conjectured Lagarias upper bound, can be interpreted as: zeta_G(-1) = sigma(G) <= L(H_G) . These are all upper bounds for the zeta functions of some finite group and have directly or indirectly something to do with RH. Edit 04.06.2019: I found some (unpublished) upper bounds to $\zeta_{C_n}(s) = \sigma_{-s}(n)$ which are equivalent to Lagarias inequality and hence to RH. REPLY [8 votes]: Unless I'm mistaken, the group that you're constructing here is an extension of $\mathbb{Z}$ by $G$. It would cause less confusion if you reserved the notation $\mathbb{Z} \times G$ for the direct product. A fairly common notation for this extension is $\mathbb{Z} : G$. As far as your question on the relationship between zeta functions, a relationship of this type is worked out for any normal subgroup $N$ of a finite group $K$ in a paper of Ken Brown: Brown, Kenneth S., The coset poset and probabilistic zeta function of a finite group, J. Algebra 225, No. 2, 989-1012 (2000). ZBL0973.20016. Note that Brown's $P(G,s)$ is (at least in the case where $G = \mathbb{Z}$, haven't completely carefully checked elsewhere) the reciprocal of the $\zeta$ function. Brown also uses a theorem of Gaschütz which is stated only for finite groups. It seems likely to me, however, that you might be able to extend some of what Brown does to your situation.<|endoftext|> TITLE: Poisson summation formula and its implication for the spectrum of the flat torus QUESTION [5 upvotes]: I usually ask questions on math.stackexchange but I figure this one is more suited to being asked here. I should preface that I am a complete novice undergraduate, and unlikely to understand answers which refer to complicated differential geometry or number theory. The Poisson summation formula (also called Jacobi Inversion Formula?) for a lattice in $\mathbb{R}^n$ is given for $t\in(0,+\infty)$ by \begin{align} \sum_{\gamma^*\in \Gamma^*} e^{-4\pi^2||\gamma^*||^2t} = (4\pi t)^{-n/2}\text{Vol}(\Gamma)\sum_{\gamma\in \Gamma} e^{-\frac{||\gamma||^2}{4t}}. \end{align} I am interested in the connection to the Laplace-eigenvalues of the flat torus $\mathbb{R}^n/\Gamma$ which we know are given in the spectrum as $\left\{ 4\pi^2||y||^2 : y\in \Gamma^* \right\}$. In particular I do not understand the following statement from Gordon: "Next observe that the geodesic length spectrum of a torus $\Gamma\backslash \mathbb R^n$ coincides precisely with the length spectrum of the lattice $\Gamma$, if we define the multiplicity of a length in the geodesic length spectrum to be the number of free homotopy classes of loops containing a geodesic of the given length. The Jacobi inversion formula implies that two lattices $\Gamma_1$ and $\Gamma_2$ have the same length spectrum if and only if the dual lattices $\Gamma_1^*$ and $\Gamma_2^*$ have the same length spectrum. We conclude that two flat tori are isospectral if and only if they have the same geodesic length spectrum." This is put in another way in the book Old and New Aspects in Spectral Geometry as simply saying that due to the Poisson summation formula, two flat tori $\mathbb{R}^n/\Gamma$ and $\mathbb{R}^n/\Gamma'$ are isospectral if and only if the share the same length spectrum $\{||y||:y\in\Gamma\}=\{||y'||:y'\in\Gamma'\}$. I have tried playing with the formula by changing $t\mapsto 1/t$ and subsequently trying to show that the RHS determines the eigenvalues by taking away the zeroth eigenvalue from the series and multiplying with an exponential $e^{rs}$ ($r\in\mathbb{R}$) similarly to how one shows it for the LHS. But I can't get this approach to work due to the presence of $(4\pi)^{-n/2}s^{n/2}$. I have also searched the web (and textbooks) to the best of my ability, but have come out empty. Any help would thus be greatly appreciated, thank you. REPLY [5 votes]: The theta function (the left-hand side of the Jacobi identity) uniquely determines the values of ${||\gamma^*||:\gamma^*\in\Gamma^*}$ counted with multiplicities, just by looking inductively at its asymptotic expansion as $t\to +\infty$: the leading term will be $1$ since the multiplicity of $0$ is 1, the next term will be $Ne^{-4\pi^2\alpha t}$, where $\alpha$ is the smallest norm and $N$ its multiplicity, and so on. Alternatively, you can use the inverse Laplace transform formula. Similarly, the RHS uniquely determines the values of ${||\gamma||:\gamma\in\Gamma}$ by looking at the asymptotic expansion as $t\to 0+$. So the two tori are isospectral iff they have the same theta function iff they have the same length spectrum.<|endoftext|> TITLE: Is the field of q-series 'dead'? QUESTION [24 upvotes]: I had a discussion with my advisor about what am I interested as my future research direction and I said it is special functions and q-series. He laughed and said that the topic is essentially dead and the people who study it are dinosaurs. I'm really confused by this statement and don't know what to think. Is this area really dead and not worth pursuing a research in it? REPLY [13 votes]: The reports of the death of the field of $q$-series and special functions are greatly exaggerated. George Andrews wrote a book Q-Series on the subject published in 1986. He and Bruce Berndt last year completed the publication of a five volume edition about the results in Ramanujan's Lost Notebook. An important subject for Ramanujan was $q$-series and special functions. George Andrews is a former President of the AMS with many awards. Ken Ono, the Vice President of the AMS, has done research on Rogers-Ramanujan identities, Mock theta functions, and recently he and his coworkers have proved the Umbral Moonshine Conjecture. In joint work with Jan Bruinier, he discovered a finite algebraic formula for computing partition numbers. The partition numbers are the coefficients of the reciprocal of the fundamental $q$-series $(q;q)_\infty$. There are many other examples that I could cite, but this should be enough. REPLY [9 votes]: I think that (properly understood) theory of $q-$series is one of the most fashionable areas of mathematics right now. One reason is the appearance of $q-$series in topology. Namely, Turaev-Viro TQFT associates to each $3-$manifold with boundary a $q-$special function, defined for $q$ being a root of unity. These functions are expected to be $q-$holonomic, so should satisfy interesting difference equations. The building block for these invariants is so-called quantum $6j-$symbol, which coincides with ${}_4F_3-$ basic hypergeometric series with general parameters at $z=1$.<|endoftext|> TITLE: Are there more than two rational solutions to $a^b= b^a$? QUESTION [11 upvotes]: I have wondered for a while if there are any interesting rational solutions to $a^b = b^a$. I have tried but cannot find any solutions other than $a=2$ and $b=4$, or vice versa. Thank you in advance. REPLY [27 votes]: There is an infinite number of rational solutions $$a=\left(\frac{n+1}{n}\right)^n,\;\;b=\left(\frac{n+1}{n}\right)^{n+1},\;\;n\in\mathbb{Z},\;\;0\neq n\neq -1.$$ For a proof that these are all the rational solutions of $a^b=b^a$ with $a\neq b$, see Marta Sved's article (1990). As she describes, this question has a long history, it was first answered by Euler in 1748 and has been generalized in various ways. I show a screen shot from Euler's proof that there is an infinite number of rational solutions (Euler uses the word "innumerabilia" -- uncountable, obviously not in the technical sense of the word). REPLY [11 votes]: There are infinitely many solutions. For example, take any positive integer $n$ and set $c=(n+1)/n, a=c^n, b=c^{n+1}$. Then $$ a^b=c^{nb}=c^{nc^{n+1}} $$ and $$ b^a=c^{(n+1)a}=c^{(n+1)c^n}. $$ These two expressions are equal because $$ nc^{n+1}=(nc)c^n=(n+1)c^n. $$<|endoftext|> TITLE: Is every finite quantum group a quantum symmetry group? QUESTION [13 upvotes]: This post is basically a quantum extension of Is every finite group a group of “symmetries”? Here finite quantum group means finite dimensional Hopf ${\rm C}^{\star}$-algebra. Frucht's theorem states that every finite group is the automorphism group of a finite graph. Wang defined here a notion of quantum automorphism group. The application to a finite space of $n$ elements is called "the quantum permutation group of $n$ symbols", and its quantum subgroups are called quantum permutation groups of degree $n$. Bichon introduced here the quantum automorphism groups of finite graphs, these are quantum permutation groups. See also this survey of Banica-Bichon-Collins. Kojima proved here that every finite group is realized as the full isometry group of some compact hyperbolic $3$-manifold. This book of Goswami-Bhowmick introduces the notion of quantum isometry group. General question: Is every finite quantum group a quantum symmetry group? Sub-question 1: Is a finite quantum permutation group a twisted finite group? Sub-question 2: Is every finite quantum group a quantum permutation group? Answer: no, the smallest counter-example has dim. $24$ (see this paper of Banica-Bichon-Natale). Sub-question 3: Is every finite quantum permutation group of dimension $n$ a quantum permutation group of degree $n$? Sub-question 4: Is every finite quantum permutation group a quantum automorphism group of a finite graph? Sub-question 5: Is every finite quantum group $\mathbb{G}$ a quantum subgroup of the quantum automorphism group of a finite dimensional ${\rm C}^{\star}$-algebra $\mathcal{A}$? Ok for $\mathcal{A} = C(\mathbb{G})$? Answer (Bhowmick): yes, it follows trivially from the definition, using Haar state. Sub-question 6: Is every finite quantum group a quantum isometry group? REPLY [4 votes]: The answer to sub-question 4 is no. See here, The Frucht property in the quantum group setting. To expand slightly, this paper gives four explicit finite quantum permutation groups which are not the quantum automorphism groups of a finite graph: the duals of $S_3$, $A_4$, and $A_5$, and the Kac--Paljutkin quantum group of order eight, denoted $G_0$. If the dual of $S_3$ is the quantum automorphism group of a graph, then, taking abelianisations, the (classical) automorphism group of the graph must be $\mathbb{Z}_2$. The paper above shows that if the dual of $S_3$ acts on a graph then a (classical) abelian group larger than $\mathbb{Z}_2$ acts on the graph too. Therefore the dual of $S_3$ is not the quantum automorphism group of a graph. In fact it is shown that if the dual of $S_3$ acts on a graph then the dual of a(n infinite) free product acts on the graph too. A similar story holds for the duals of $A_4$ and $A_5$. If $G_0$ is the quantum automorphism group of a graph, then the graph has classical automorphism group $\mathbb{Z}_2\times\mathbb{Z}_2$. The above shows that if $G_0$ acts on a graph then so does the dihedral group of order eight. Therefore $G_0$ is not the quantum automorphism group of a graph.<|endoftext|> TITLE: A closed formula for $\det(\partial/\partial U)^p\prod_{i=1}^n\prod_{j=1}^p U_{i\sigma_j(i)}$ QUESTION [5 upvotes]: This is a continuation of this question. Is there a simple formula for $$I(\sigma_1,\cdots,\sigma_p)=(-1)^\sigma\left(\left(\det\left(\frac{\partial}{\partial U_{ij}}\right)_{i,j=1}^n\right)^p \prod_{i=1}^n\prod_{j=1}^p U_{i\sigma_j(i)}\right)|_{U=0},$$ where all $\sigma_i\in S_n$? Equivalently $$I(\sigma_1,\cdots,\sigma_p)=\sum_{\pi\in C}\sum_{\tau_1,\tau_2\in R}(-1)^{\sigma\pi} [\tau_1\pi\tau_2=\sigma],$$ where $C$ and $R$ are the Young subgroups of $S_{pn}$ $$C=\mathrm{Sym}(1,\dots,n)\times\cdots\times\mathrm{Sym}((p-1)n+1,\dots,pn)\cong S_n^p$$ and $$R=\mathrm{Sym}(1,\cdots,(p-1)n+1)\times\cdots\times\mathrm{Sym}(n,\cdots,pn)\cong S_p^n$$ and $\sigma=\sigma_1\oplus\cdots\oplus\sigma_p\in C$. In the linked question, Carlo Beenakker guessed and Abdelmalek Abdesselam and David Speyer both proved that $I(\sigma_1,\sigma_2)=2^{\#\mathrm{cyc}(\sigma_1\sigma_2^{-1})}$ when $p=2$. This can also be expressed in the group algebra $\mathbb{C}[S_n]$ as $$\sum_{\sigma\in S_n} I(1,\sigma)\sigma=(2+J_1)\cdots(2+J_n),$$ where the $J_k$ are the Jucys-Murphy elements. Note that $I$ is symmetric and $I(\sigma_1,\cdots,\sigma_p)=I(\rho\sigma_1,\cdots,\rho\sigma_p)$ for all $\rho$. When $(p,n)\ge(4,3)$ or $(3,5)$, $I$ can be negative. It appears that the prime factors of $I$ are at most $p$. Here are the values of $I(\sigma_1,\sigma_2,1)/12$ when $p=3$ and $n=2$. Here are the values of $I(\sigma_1,\sigma_2,1)/24$ when $p=3$ and $n=3$. Here are the values of $I(\sigma_1,\sigma_2,1)/24$ when $p=3$ and $n=4$. When $p=3$ and $n=5$, $I/24$ takes the values $-1,0,1,2,4,6,12,36,108,324$. REPLY [6 votes]: Perhaps for small values of $p$ one can find reasonable formulas but not for general $p$, I think. Take the cycle $\sigma_1=(12\cdots n)$ and then let $\sigma_j=(\sigma_1)^j$. Then, $I$ is, up to scale, $$ \int_{SU(n)} \prod_{i=1}^{n}\prod_{j=1}^{n} U_{ij}\ dU $$ which counts the difference between even and odd latin squares. It is an open problem, called the Alon-Tarsi conjecture to show that it is nonzero for $n$ even. See this article by Kumar and Landsberg.<|endoftext|> TITLE: When is the conductor of an elliptic modular curve equal to its level? QUESTION [5 upvotes]: Suppose the usual modular curve $E=X_0(N)$ over $\mathbb{Q}$ has genus 1 (e.g. $N=15$). Define the conductor of $E/\mathbb{Q}$ as the ideal/integer: $$M=\prod_{p}p^{f(E/\mathbb{Q}_p)},$$ where $$f(E/\mathbb{Q}_p)=\begin{cases}0 & E\text{ has good reduction mod }p\\1 & E\text{ has multiplicative reduction mod }p\\ 2 & E\text{ has additive reduction mod }p\end{cases}$$ is the "exponent of the conductor of $E/\mathbb{Q}_p$" (see for example Silverman's book Advanced Topics in the Arithmtic of Elliptic Curves, chapter IV, section 10). In the case of $p=2,3$ the exponent $f(E/\mathbb{Q}_p)$ might have extra terms depending on its wild ramification. How do you prove that $N=M$? If $E$ had a Weierstrass equation, the proof would be straight forward since $f(E,\mathbb{Q}_p)$ is easy to calculate. You could also use the Ogg-Saito formula if you can calculate the Neron model of $E/\mathbb{Q}_p$, but this also requires a Weierstrass equation. I know that the function field of $E/\mathbb{C}$ is equal to $\mathbb{C}(j,j_N)$ and that an algebraic relation between $j$ and $j_N$ gives an equation for $E$, but this polynomial is extremely inconvenient for calculations. Is there a way to calculate a simple equation for $E/\mathbb{Q}$? REPLY [5 votes]: Already in the beginning of the 20th century, Fricke had determined explicit equations of the modular curves $X_0(N)$ which are of genus 1. To do this he constructed, in each case, two explicit functions $\sigma, \tau$ on $X_0(N)$ such that $j$ and $j_N$ are rational functions of $\sigma,\tau$ with coefficients in $\mathbb{Q}$. In this way he gets an equation of the form $\sigma^2 = P(\tau)$ where $P$ is a certain polynomial of degree 3 or 4 with integer coefficients. From this, it is not difficult to compute a minimal Weierstrass equation and the conductor (which were not defined at the time of Fricke), using Tate's algorithm and Ogg's formula. As you have correctly guessed, one finds that the conductor is equal to $N$. The very question of determining the conductor of $X_0(N)$ in the case of genus 1 has in fact been solved a long time ago by Ligozat in his PhD thesis (1974), where he proved BSD for such curves (assuming the Tate-Shafarevich group is trivial). His article can be found here: Courbes modulaires de genre 1. The book by Fricke is Die elliptischen Funktionen und ihre Anwendungen (II), which I believe is available on archive.org (Ligozat gives all the precise references). Since this proof is essentially case-by-case, you may ask for a more abstract-oriented proof. This is indeed possible but uses deep results. It is a theorem of Carayol that the conductor of a modular elliptic curve is equal to the level of the associated modular form. One way to explain this is to consider the associated $L$-functions. The conductor $M$ of an elliptic curve $E$ is determined by the $L$-function $L(E,s)$, in the sense that it is the only integer such that $M^{s/2} (2\pi)^{-s} \Gamma(s) L(E,s)$ has a functional equation of the standard type. On the modular side, it is easy to show that the $L$-function of a modular form of level $N$ has conductor $N$ (in the previous sense). So if you accept to use Carayol's theorem, you get $N=M$ (Wiles's theorem is not needed in your case since $X_0(N)=E$). Using this more advanced method, you can in fact determine the conductor of the abelian variety $J_0(N)$, the Jacobian variety of $X_0(N)$, for any integer $N$. The precise formula depends on the dimensions of the space of cusp forms of weight 2 and level $N'$ dividing $N$ (see Ligozat's article, at that time it was only conjectured, and proved only in the case $N$ is squarefree using Deligne's results).<|endoftext|> TITLE: Casson invariant and Euler characteristic QUESTION [5 upvotes]: A slogan I frequently hear is: "the Casson invariant is the Euler characteristic of the Floer homology of flat SU(2)-connections on the integral homology sphere". Is there a single paper/reference that essentially states this as a result? It has been difficult to locate precise definitions. In addition, a sketch of how this works would be very helpful. REPLY [6 votes]: Just to finalize comments since people are upvoting the question: The canonical reference is Taubes' "Casson's invariant and gauge theory" which makes the statement rigorous and has all the definitions (based on the relevant Chern-Simons functional). Floer's paper "An instanton-invariant for 3-manifolds" does build the Floer homology using this functional, and asserts that its Euler characteristic is (twice) the Casson invariant, but he only quotes Taubes' paper (that's what the reference [3] in Floer's paper really is). Although I think Taubes' paper is beautiful enough that a sketch isn't needed, you can find a background sketch/summary in Saveliev's book ("Invariants of homology 3-spheres"), Chapter 5: "Casson Invariant and Gauge Theory".<|endoftext|> TITLE: Symmetric and anti-symmetric parts of the covariant derivative of a connection QUESTION [9 upvotes]: The following is an excerpt from Sharpe's Differential Geometry - Cartan's Generalization of Klein's Erlangen Program. Now we come to the question of higher derivatives. As usual in modern differential geometry, we shall be concerned only with the skew-symmetric part of the higher derivatives. In essence, what we shall be doing is taking the partial derivatives with respect to the base (i.e., manifold) variables and skew-symmetrizing the result, thus forgetting about the part of the higher derivatives that vanish under this procedure. However, this will not be made explicit in our treatment. The part of the higher derivative that disappears has not been studied much in differential geometry since Elie Cartan showed how useful it is to consider only the skew-symmetric part, that is, the exterior derivative. The old masters did use the symmetric part... "Partial derivatives with respect to the base" must be the covariant derivative of the connection. If I correctly understand what's written in this answer, then we have for any torsion free connection on a manifold the equality $ \mathrm dw=\operatorname{Alt}(\nabla w)$. This already seems rather remarkable since the exterior derivative is intrinsic. Question 1. What is the geometric meaning of the above fact for torsion-free connections? How does taking anti-symmetrization "forget" the structure of a horizontal bundle (connection)? Question 2. Still for a torsion-free connection, what is the geometric meaning of "the part of the higher derivative that vanish under this procedure" of anti-symmetrization (i.e the symmetric part)? Question 3. What's the conceptual picture for an arbitrary connection on a manifold (possibly with torsion)? Is the anti-symmetrization still the exterior derivative? What is the symmetric part? Remark on Q3. I vaguely recall being told that torsion measures the failure of the fundamental theorem of calculus, and also that in torsion-free connections the parallel transport commutator is given by the Lie bracket (again, the latter is intrinsic). REPLY [2 votes]: Although higher derivatives are best thought through with jet bundles where we actually differentiate a bundle and not just a manifold, there is a useful description of second derivatives using secondary tangent bundles, this is just iterating the tangent bundle, ie $TTM$. This is described in one of the early chapters in Kolár, Michor & Slováks book on natural bundles. (They also describe a generalisation of this in one of the later chapters which they call section forms - but they don't do anything with it). Question 3. What's the conceptual picture for an arbitrary connection on a manifold (possibly with torsion)? Is the anti-symmetrization still the exterior derivative? What is the symmetric part? On the last part of this question you may be interested in this paper, A Description of the Derivations of the Algebra of Symmetric Tensors, by A. Heydari, N. Boroojerdian, E. Peyghan and published in Archivum Mathematicum. They introduce symmetric forms, bracket, Lie derivative and differential. Whilst the symmetric differential is not nilpotent, ie $(d^s)^2$ does not vanish, they show proposition 5: Let $(M, g)$ be a Riemmanian manifold with Levi-Civita connection $∇$. The 1-form $ω$ is Killing if and only if $d^sω = 0$. and proposition 7: Every derivation of degree 1 on symmetric forms, whose value for a function is the differential of the function, is of the form of a symmetric differential of a connection which is also unique.<|endoftext|> TITLE: Generalised Chinese Reminder Theorem - How to compute the cokernel? QUESTION [6 upvotes]: Let $R$ be a commutative ring of dimension one with minimal prime ideals $P_1,\ldots,P_n$. We have the canonical injective map $$\phi_n: R/(P_1 \cap \ldots \cap P_n) \to \prod_{i=1}^n R/P_i.$$ My question is: Is there a formula for the cokernel of $\phi_n$? For instance as a product of rings again. The case $n=2$ is well known and here $\operatorname{coker}(\phi_2) \cong R/(P_1+P_2)$. That this does not generalize to $\operatorname{coker}(\phi_n) \cong \prod_{i < j}^n R/(P_i+P_j)$ can be seen by a counter-example, see this answer. I am grateful for any kind of insights, references and proofs! REPLY [3 votes]: Let us first observe that $M \Doteq \text{coker}(\phi_n)$ is naturally an $R/\text{rad}(R)$-module with $$\text{rad}(R) \Doteq P_1 \cap \cdots \cap P_n.$$ So, there is no actual loss in generality if we suppose that $R$ is reduced, i.e., $\text{rad}(R) = \{0\}$: if there is a formula for $M$ as an $R/\text{rad}(R)$-module, we should be able to derive a formula for $M$ taken as an $R$-module by re-injecting $\text{rad}(R)$. Let us assume that $R$ is reduced and let us identify $R$ with its image by $\phi_n$. There is of course a formula that expresses $M \Doteq \text{coker}(\phi_n) = \left(\prod_{i = 1}^nR/P_i\right)/R$ as a function of the ideals $P_i$. This is for instance the following presentation of $M$ over $R$ $$ M = \left\langle e_1, \dots, e_n \, \vert \, \sum_{j = 1}^n e_j = P_ie_i = 0,\quad i = 1, \dots, n \right\rangle $$ where $e_i$ corresponds to the identity element of $R/P_i$. You may object that it doesn't to tell much about the structure of $M$ and I would agree. Still, note that it makes clear that $M$ can be generated by $n - 1$ elements, so that if $n = 2$, we see almost immediately that $M$ is the cyclic $R$-module $R/(P_1 + P_2)$. Now comes the bad news. In general, the $R$-module $M$ does not split as a direct sum of factor rings of $R$, or equivalently, cyclic submodules of $M$. Here is a counter-example: Claim 1. Let $R$ be the integral group ring of $C_4$, the cyclic group with $4$ elements, that is, $R = \mathbb{Z}[C_4] = \mathbb{Z}[X]/(X^4 - 1)$. Let $\tilde{R}$ be the integral closure of $R$, that is, $\tilde{R} = \mathbb{Z}[X]/(X - 1) \times \mathbb{Z}[X]/(X + 1) \times \mathbb{Z}[X]/(X^2 + 1) \simeq \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}[i]$. Then $M \Doteq \text{coker}(\phi_3) = \tilde{R}/R$ is a non-cyclic two-generated $R$-module which is indecomposable. According to [1, pages 19--20], the integral group ring $R = \mathbb{Z}[G]$ of a finite Abelian group $G$ is always Gorenstein, but it is a Bass ring if and only if the cardinality $\vert G \vert$ of $G$ is square-free. By a Bass ring, we mean a Noetherian reduced commutative unital ring $R$ of Krull dimension one and such that $\tilde{R}/R$ is a cyclic $R$-module where $\tilde{R}$ is the integral closure of $R$. Thus the fact that the module $M$ of Claim 1 is not cyclic is already predicted by Bass's theorem, see [2, Theorem 2.1] and the rings $\mathbb{Z}[G]$ may yield further counter-examples. Proof of Claim 1. Let us denote by $x$ the image of $X$ in $R$ and set $P_1 = R(x - 1), P_2 = R(x + 1)$ and $P_3 = R(x^2 + 1)$. We denote by $\epsilon_i$ the identity element of $R/P_i$ and by $e_i$ its image in $M$ for $i = 1, 2, 3$. It is easily checked that $M \simeq \left(R/P_1 \times R/P_2 \right) /P_3(\epsilon_1 + \epsilon_2)$ and that $P_3(\epsilon_1 + \epsilon_2) = \mathbb{Z} \cdot 2(\epsilon_1 + \epsilon_2) + \mathbb{Z} \cdot 2(\epsilon_1 - \epsilon_2)$. As a result, $M$ has the following $R$-module presentation: $$M = \langle e_1, e_2 \, \vert \, 4e_1 = 4e_2 = 2(e_1 + e_2) = 0 \rangle.$$ In particular $M$ is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$ as an Abelian group and we can take $e \Doteq e_1 + e_2$ as the canonical generator of $\mathbb{Z}/2\mathbb{Z}$ and any of $e_1$ or $e_2$ as the canonical generator of $\mathbb{Z}/4\mathbb{Z}$. Now we claim that: $M$ is not a cyclic $R$-module. $Re_1, Re$ and $Re_2$ are the only $R$-submodules of $M$ with $4$ elements. $R \cdot 2e_1 = R \cdot 2e_2$ is the only $R$-submodule of $M$ with $2$ elements. The fact that $R \cdot 2e_1$ is the unique minimal non-zero $R$-submodule of $M$ proves instantly that $M$ is indecomposable. For the first assertion, let us reason by contradiction, assuming that $M = Rf$ with $f = ae + be_1, a, b \in \mathbb{Z}$. Then $M$ is generated as a $\mathbb{Z}$-module by $f$ and $xf = -ae + (2a + b)e_1$, so that the $\mathbb{Z}$-module $M/(R \cdot 2e_1) \simeq_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ is generated by the image of $f$, a contradiction. The last two assertions are routine. $\square$ Remark. In the answer to your MSE post, the following example was considered. Let $R = \mathbb{C}[X, Y]/XY(X - Y)$, with minimal primes $P_1 = Rx, P_2 = Ry$ and $P_3 = R(x - y)$ where $x, y$ are the images of $X$ and $Y$ in $R$. Then the integral closure of $R$ is $\tilde{R} = R/P_1 \times R/P_2 \times R/P_3$ and it is not difficult to show that $$\tilde{R}/R \simeq (R/(P_1 + P_2))^2.$$ Indeed, we have $\tilde{R}/R \simeq \left(R/P_1 \times R/P_2 \right) /P_3(\epsilon_1 + \epsilon_2)$ and $P_3(\epsilon_1 + \epsilon_2) = P_2\epsilon_1 + P_1\epsilon_2$. Hence this $M \Doteq \tilde{R}/R $ does split as a direct sum of two indecomposable cyclic $R$-modules. On the positive side, we have: Claim 2. Assume that $M \Doteq \text{coker}(\phi_n)$ can be generated by $d \le n - 1$ over $R$. Then we have $\text{Fitt}_0(M) \subseteq \text{ann}(M)$ and $\text{ann}(M)^{d} \subseteq \text{Fitt}_0(M)$ where the Fitting ideal $\text{Fitt}_0(M)$ is given by $$ \text{Fitt}_0(M) = \sum_{1 \le i_1 < \cdots < i_{n - 1} \le n} P_{i_1} \cdots P_{i_{n - 1}}. $$ In particular, $\text{ann}(M) = \text{Fitt}_0(M)$ if $M$ can be generated by one element over $R$. Proof. This is [3, Proposition 20.7]. [1] H. Bass, "On the ubiquity of Gorenstein rings", 1963. [2] L. Levy, R. Wiegand, "Dedekind-like behaviors of rings with $2$-generated ideals", 1985. [3] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 1995.<|endoftext|> TITLE: Can a non-trivial algebraic variety carry a vector bundle whose total space is affine space? QUESTION [10 upvotes]: Suppose $X$ is an algebraic variety over $\mathbb{C}$, and let $Y\to X$ be an algebraic vector bundle. Suppose $Y$ is algebraically isomorphic to $\mathbb{C}^n$ for some $n$. Does it follow that $X$ is algebraically isomorphic to $\mathbb{C}^m$ for some $m$? REPLY [8 votes]: Summing up the discussion in the comments: As user ulrich observed, the vector bundle has to be trivial. First, since $X$ is a closed subscheme of $Y\simeq \mathbf{A}^n$ via the zero section, it has to be affine. It is also smooth. Finally, it is $\mathbf{A}^1$-contractible since $Y$ is, since $Y\to X$ induces an equivalence in $\mathbf{A}^1$-homotopy. We conclude by a theorem of Morel (see Chapter 7 in $\mathbf{A}^!$-algebraic topology over a field, here), saying that vector bundles on a smooth affine $\mathbf{A}^1$-contractible variety are trivial. A simpler argument using Quillen-Suslin was given by Anton: $X$ is a retract of $Y \simeq \mathbf{A}^n$ via the zero section, and since every vector bundle on $Y$ is trivial, the same is true for $X$. This turns the question into an important special case of the Cancellation Problem. Cancellation Problem. If $X\times \mathbf{A}^m \simeq Z \times \mathbf{A}^m$, can we conclude that $X\simeq Z$? The answer to this general problem is no (there are famous counterexamples already in dimension two, known as Danielewski surfaces). However, in the special case where $Z$ is an affine space ($\mathbf{A}^{r}$, $r+m=n$ in our case) the answer is known to be yes for $r\leq 2$, and open for $r>2$. (In positive characteristic, the answer is no for all $r>2$: arxiv.org/abs/1208.0483 ) This answer is "Community Wiki".<|endoftext|> TITLE: What is a really good book for complex variables? QUESTION [5 upvotes]: I'm an engineering student but I self-study pure mathematics. I am looking for a Complex Variables Introduction book (to study before complex analysis). I have the Brown and Churchill book but I was told that's for engineers and physicist mostly, not for mathematicians. I also looked for Fisher and Flanigan, but they don't seem to have as many topics as Brown. I wonder which book is best for the subject or if one of the two previously mentioned will do to master most of the topics of complex variables as a mathematician. Thanks. REPLY [2 votes]: The Complex Analysis Project is a rich-in-content textbook on this topic which realises modern approach with Maple. Also see William T. Shaw, Complex Analysis with MATHEMATICA®<|endoftext|> TITLE: Roots of $x^n-x^{n-1}-\cdots-x-1$ QUESTION [12 upvotes]: It is easy to see that $f(x)=x^n-x^{n-1}-\cdots-x-1$ has only one positive root $\alpha$ which lies in the interval $(1,2)$. But it is claimed that this root is a Pisot number (a.k.a. PV number), i.e., the other roots are in the open disk $\{z\in \mathbb{C}: |z|<1\}$. I have tried the following, but I failed. I considered $$P(x)=(1-x)f(x)=x^{n+1}-2x^n+1=x^n(x-2)+1$$ and then I tried to use Rouché's theorem, by picking $g(x)=-x^n(x-2)$ and trying to show that $1<|g(z)|+|P(z)|$ for $|z|=1$. Proving this implies that $P(x)$ has $n$ roots in $\{z\in \mathbb{C}: |z|<1\}$ which demonstrates the claim. But this inequality fails at $z=1$. Can you give me an idea how one can prove this claim? REPLY [4 votes]: You may want to take a look at pages 155-156 of the problems and solutions section of the February 1989 issue of the American Mathematical Monthly. You are going to find there two proofs of the irreducibility over $\mathbb{Q}$ of the polynomial $$p(x)=x^{n}-x^{n-1}-\cdots-1.$$ The first proof depends crucially on the fact that you wished to establish (which is settled therein via Rouché's theorem along the lines of the above proof by GH from MO). The editorial comment you are to find on page 156 is noteworthy, too: in his paper "On algebraic equations with all but one root in the interior of the unit circle", Alfred T. Brauer proved that if $a_{1}, a_{2}, \ldots, a_{n}$ are integers with $a_{1} \geq a_{2} \geq \cdots \geq a_{n} > 0$, then the polynomial $$x^{n}-a_{1}x^{n-1}-a_{2}x^{n-2}-\cdots-a_{n}$$ has one of its roots in the exterior of the unit circle and all the others in its interior. Let me conclude this intervention by echoing, once again, Leo Sauvé's famed remark on the problems and solutions department of the Monthly: "it seems like all problems have once been published in the American Mathematical Monthly".<|endoftext|> TITLE: Maxima of Brownian motion QUESTION [6 upvotes]: It is well-known that Brownian motion attains infinitely many maxima in each time interval $[0,T]$ a.s.. From a physics perspective it seems reasonable that when the disorder of the path of a particle decreases and the motion becomes more deterministic, then the number of maxima should decrease. But I could not find anything on that. Now, there were two natural things to look at: Is there a way to quantify that a Brownian motion with large variance (large disorder) has more maxima than one with little disorder? Or is there a way to say that a diffusion process $dX_t = \mu (X_t) \ dt + \alpha dB_t $ has "less" maxima when $\alpha $ is small compared to $\alpha$ large? I guess it is hard to make this question more precise, since this is not a question of cardinality of maxima but more about finding a suitably chosen measure that could capture such an effect. REPLY [9 votes]: A good way to measure the set of maxima is the Hausdorff dimension of the set of records, which for BM is a.s. 1/2. Because of time/scale invariance, the dimension is the same for $\alpha B_\cdot$, and if the drift is nice enough to have absolute continuity, the same holds for your drifted diffusion. This however changes when you move from BM to fractional BM, which I would suggest is the right frame for your question. There, the change in regularity is reflected in the Hausdorff dimension of the set of zeros. See https://projecteuclid.org/download/pdfview_1/euclid.ecp/1522375381 for details.<|endoftext|> TITLE: Quantum Field Theory: completing the "A Bridge between Mathematicians and Physicists" series QUESTION [6 upvotes]: I decided to read the series "A Bridge between Mathematicians and Physicists" written by Eberhard Zeidler. But when I read the preface of the first book I realized that at first this series should be composed of six volumes, namely: Quantum Field Theory I: Basics in Mathematics and Physics Quantum Field Theory II: Quantum Electrodynamics Quantum Field Theory III: Gauge Theory Quantum Field Theory IV: Quantum Mathematics Quantum Field Theory V: The Physics of the Standard Model Quantum Field Theory VI: Quantum Gravity and String Theory I have only the first three books. Searching for the last three I discovered that Eberhard Zeidler died and therefore the last three volumes will never be published. For this reason I ask: could someone please indicate me books that cover the subject of the last three books and that focus a lot on the mathematical part? Thank you for your attention. REPLY [4 votes]: The volume number 4 has a vague title. 2 volumes of "Quantum fields and strings: a course for mathematicians" probably cover whatever that term is supposed to mean (there is also a partial QFT-mathematics dictionary in the books, very nice). The books are outdated by this point, but they should be enough to get you started (then you should be able to read physical mathematics papers without too much pain).<|endoftext|> TITLE: A comprehensive list of random walk inequalities? QUESTION [13 upvotes]: I am interested in finding a comprehensive list of all noticeable random walk inequalities. ie. $S_n = \sum_{k\leq n} X_i$ for i.i.d symmetric $X_i$ I can only seem to find books/papers that list the already well known ones like Kolmogorov's maximal inequality. Does anyone know of such a paper/book? REPLY [26 votes]: There I will list the inequalities and asymptotic theorems on random walks, that are currently known to me: Notation that will be used in the list: $\{X_n\}_{n = 1}^\infty$ are i.i.d. random variables. $\{S_n\}$ is the random walk ($S_n = \Sigma_{k = 1}^n X_k$) $\nu(t) = \max\{n \in \mathbb{N}_0 | S_n < t \}$ - the corresponding renewal process (well defined if $P(X_1 > 0) = 1$) $U(t) = 1 + E(\nu(t))$ -the corresponding renewal function (well defined if $P(X_1 > 0) = 1$) The types of convergence will be denoted in the following way: $\to_D$ is convergence by distribution $\to_P$ is convergence by probability $\to_{a.s.}$ is convergence almost surely $\Rightarrow$ is convergence of stochastic processes in Skorokhod space. THE LIST: Law of Large Numbers If $|E(X_1)| < \infty$, then $$\frac{S_n}{n} \to_{a.s.} E(X_1)$$ Laws of Iterated Logarithm 1.Suppose $E(X_1) = 0$ and $Var(X_1) = 1$, then $$P(\overline{\lim_{n \to \infty}} \frac{S_n}{\sqrt{2n\log\log n}} = 1) = 1$$ 2.Suppose $E(X_1) = 0$ and $Var(X_1) = 1$, then $$P(\underline{\lim_{n \to \infty}} \frac{S_n}{\sqrt{2n\log\log n}} = -1) = 1$$ Central Limit Theorem If $|E(X_1)| < \infty$ and $Var(X_1) < \infty$, then $$\frac{S_n - nE(X_1)}{\sqrt{n}} \to_{D} Z \sim {N}(0, Var(X_1))$$ Berry-Esseen Inequality If $E(X_1) = 0$, $0 < Var(X_1) < +\infty$ and $E(|X_1|^3) < +\infty$, then $$|P(S_n \leq x \sqrt{n Var(X_1)}) - \frac{e^{\frac{x^2}{2}}}{\sqrt{2\pi}}| < \frac{0.4748 E(|X_1|^3)}{(Var(X_1))^{\frac{3}{2}} \sqrt{n}}$$ Shevtsova inequalities 1.If $E(X_1)= 0$, $0 < Var(X_1) < +\infty$ and $E(|X_1|^3) < +\infty$, then $$|P(S_n \leq x \sqrt{n Var(X_1)}) - \frac{e^{\frac{x^2}{2}}}{\sqrt{2\pi}}| < \frac{0.33554 E(|X_1|^3) + 0.415 (Var(X_1))^{\frac{3}{2}}}{(Var(X_1))^{\frac{3}{2}} \sqrt{n}}$$ 2.If $E(X_1) = 0$, $0 < Var(X_1) < +\infty$ and $1.286(Var(X_1))^{\frac{3}{2}} < E(|X_1|^3)< +\infty$, then $$|P(S_n \leq x \sqrt{n Var(X_1)}) - \frac{e^{\frac{x^2}{2}}}{\sqrt{2\pi}}| < \frac{0.3328 E(|X_1|^3) + 0.429 (Var(X_1))^{\frac{3}{2}}}{(Var(X_1))^{\frac{3}{2}} \sqrt{n}}$$ Hoeffding Inequalities 1.If $P(X_1 \in [0;1])=1$ and $t > 0$, then $$P(S_n - nE(X_1) \geq t) \leq e^{\frac{2t^2}{n}}$$ 2.If $P(X_1 \in [0;1])=1$ and $t > 0$, then $$P(|S_n - nE(X_1)| \geq t) \leq 2e^{\frac{2t^2}{n}}$$ Bennet Inequality Suppose $E(X_1) = 0$, $0 < Var(X_1) < +\infty$, $P(X_1 < a) = 1$, for some $a < +\infty$ and $t > 0$ then $$P(S_n > t) \leq {(\frac{Var(X_1) + at}{Var(X_1)})}^{-\frac{Var(X_1) + at}{a^2}} e^{\frac{t}{a}}$$ Bernstein Inequalities 1.If $E(X_1) = 0$ and $P(|X_1| \leq M) = 1$, then $$P(S_n > t) \leq e^{- \frac{3t^2}{6n Var(X_1) + 2Mt}}$$ 2.If $\exists L >0$ $\forall k > 1$ $2E(|X_1^k|) \leq k!L E(X_1^2)$ and $0 < t < \frac{n E(X_1^2)}{L}$ then $$P(S_n > t) < e^{-\frac{t^2}{4n E(X_1^2)}}$$ 3.If $\exists L >0$ $\forall k > 3$ $4!5^{k - 4}E(|X_1^k|) \leq k!L^{k - 4}$, and $0 < t < \frac{5}{4L}$, then $$P(|S_n - \frac{2}{3}nE(X_1^3)t^2| \geq 2nE(X_1^2)t(1 + \frac{E(X_1^4)t^2}{3E(X_1^2)}))) < 2e^{-nE(X_1^2)t^2}$$ Kolmogorov Inequality If $E(X_1) = 0$, $Var(X_1) < +\infty$ and $t > 0$, then $$P(\max_{1 \leq k \leq n} S_k \geq t) \leq \frac{n Var(X_1)}{t^2}$$ Law of Large Numbers for Renewal Process If $P(X_1 > 0) = 1$, $E(X_1) < +\infty$ , $t \to +\infty$ then $$\frac{\nu(t)}{t} \to_{P} \frac{1}{E(X_1)}$$ Central Limit Theorem for Renewal Process If $P(X_1 > 0) = 1$, $E(X_1) < +\infty$ , $Var(X_1) < +\infty$, $t \to +\infty$ then $$\frac{(E(X_1))^\frac{3}{2} \nu(t) - t (E(X_1))^{\frac{1}{2}} }{t^{\frac{1}{2}}(Var(X_1))} \to_{D} Z \cong Z \sim {N}(0, 1)$$ Wald Equality If $P(X_1 > 0) = 1$, $E(X_1) < +\infty$ and $t > 0$ then $$E(S_{\nu(t) + 1})=U(t)E(X_1)$$ Fundamental Renewal Theorem If $P(X_1 > 0) = 1$, $h > 0$ and $E(X_1) < +\infty$ then $$\lim_{t \to \infty} (U(t + h) - U(t))= \frac{h}{E(X_1)}$$ Integral Renewal Theorem If $P(X_1 > 0) = 1$ and $E(X_1) < +\infty$ then $$\lim_{t \to \infty} \frac{U(t)}{t} = \frac{1}{E(X_1)}$$ Donsker invariance principle If $E(X_1) = 0$, $0 < Var(X_1) < +\infty$ and $t \in [0; 1]$ then $$\frac{S_{\lfloor nt \rfloor}}{\sqrt{n Var(X_1)}} \Rightarrow W(t)$$ where, $W(t)$ stands for Brownian motion. Liggett invariance principle If $E(X_1) = 0$, $0 < Var(X_1) < +\infty$, $t \in [0; 1]$ and $a > 0$ then $$\frac{S_{\lfloor nt \rfloor}}{\sqrt{n Var(X_1)}} | S_n \in (-a; a] \Rightarrow B(t)$$ where, $B(t)$ stands for Brownian bridge. Eagleheart invariance principle If $E(X_1) = 0$, $0 < Var(X_1) < +\infty$ and $t \in [0; 1]$ then $$\frac{S_{\lfloor nt \rfloor}}{\sqrt{n Var(X_1)}} | \min\{n > 0| S_n \leq 0\} > n \Rightarrow W^+(t)$$ where, $W^+(t)$ stands for Brownian meander. Hopf lemma If $E(X_1) < +\infty$, $p \in \mathbb{R}$ and $n \in \mathbb{N}$ then $$E(X_1 ; \{max_{k \leq n} \frac{S_k}{k} > t\}) \geq tP(\{max_{k \leq n} \frac{S_k}{k} > t\})$$ He-Zhang-Zhang inequality If $P(X_1 > 0) = 1$ and $EX_1 = 1$, then $$P( \frac{S_n}{n} - 1 \geq \frac{1}{n}) \leq \frac{7}{8}$$ Van Zuijlen bounds If $P(X_1 = 1) = P(X_1 = -1) = 0.5$, then $$P(|S_n| \leq \sqrt{n}) \geq 0.5$$ If $X_1 \sim N(0, 1)$, then $$P(|S_n| \leq \sqrt{n}) \geq 0.31$$ Elementary Renewal-Reward Theorem Suppose $P(X_1 > 0) = 1$, $E(X_1) < +\infty$, $\{Y_n\}_{n = 1}^{\infty}$ is a sequence of i.i.d. random variables with finite expectation, then $$\lim_{t \to \infty} \frac{E(\Sigma_{i = 1}^{\nu(t)}Y_i)}{t} = \frac{E(Y_1)}{E(X_1)}$$ Law of Large Numbers for Renewal-Reward processes Suppose $P(X_1 > 0) = 1$, $E(X_1) < +\infty$, $\{Y_n\}_{n = 1}^{\infty}$ is a sequence of i.i.d. random variables with finite expectation, and $t \to +\infty$ then $$\frac{\Sigma_{i = 1}^{\nu(t)}Y_i}{t} \to_{a.s.} \frac{E(Y_1)}{E(X_1)}$$ Renewal Equation Suppose $P(X_1 > 0) = 1$, $E(X_1) < +\infty$, $t > 0$ and $P(X_1 \leq x) \in C^1[0; 1]$ then $$E(\nu(t)) = P(X_1 \leq t) + \int_0^t E(\nu(t - s))\frac{\partial P(X_1 \leq s)}{\partial s}ds$$ Inspection Paradox Suppose $P(X_1 > 0) = 1$, $x > 0$ and $t > 0$, then $$P(X_{\nu(t) + 1} > x) \geq P(X_1 > x)$$ Local limit theorem Suppose $A \subset \mathbb{Z}$ is finite and $P(X_1 \in A) = 1$. Then $\exists 0 < C_1 < C_2 < +\infty$, such that $$\frac{C_1}{\sqrt{n}} \leq sup_{k \in \mathbb{Z}} P(S_n = k) \leq \frac{C_2}{\sqrt{n}}$$ Kurtosis Equality Suppose $E(X_1^4)$ is finite. Then $$\frac{E((S_n - nE(X_1))^4)}{(nVar(X_1))^2} - 3 = \frac{1}{n}(\frac{E((X_1 - E(X_1))^4)}{(Var(X_1))^2} - 3)$$ Erdos-Renyi counting inequality Suppose $P(X_1 \geq 0) = 1$ and $P(X_1 > 0) > 0$, then $$P(S_n > 0) \geq \frac{nP(X_1 = 0)}{1 + (n-1)P(X_1 = 0)}$$ Durrett Finite Moment Theorem Suppose $E(X_1) = 0$ and $$\frac{S_n}{n^{\frac{1}{p}}} \to_{a.s.} 0$$ then $E(|X_1|^p)<+\infty$ Cramer Theorem If $\forall t \in \mathbb{R}$ $E[e^{tX_1}]<+\infty$, then $$\lim_{n \to \infty} \frac{1}{n}\ln(P(S_n \geq nx)) = \inf_{t \in \mathbb{R}}(\ln(E[e^{tX_1}])-tx)$$ If you already know all these facts and want something more exotic, then sorry (however, if I find anything else, I will expand this list)<|endoftext|> TITLE: Integers with a Hamiltonian Square Path QUESTION [5 upvotes]: Let $n>1$ be an integer and set $[n]=\{1,\ldots,n\}$. We say that $n$ has a "Hamiltonian Square Path" if there is a bijection $\varphi:[n]\to[n]$ such that for all $k\in [n-1]$ we have that $\varphi(k)+\varphi(k+1)$ is a square number. For instance $15$ and $16$ have this property. Question. Is there an integer $N>1$ such that every integer $n\geq N$ has a Hamiltonian Square Path? Note. This problem can be formulated in the language of graph theory and Hamiltonian paths. We say that $a\neq b\in [n]$ form an edge if their sum is square, and the above question is about integers such that the resulting graph has a Hamiltonian path. REPLY [7 votes]: Post #22 at https://mersenneforum.org/showthread.php?p=477787 by R. Gerbicz claims a proof that $N=25$ is the answer, and that for $N\ge32$ there is a Hamiltonian cycle. See also the tabulation and discussion at the Online Encyclopedia of Integer Sequences.<|endoftext|> TITLE: Which complete orthocomplemented lattices arise as the lattice of 'regular opens' in a closure space? QUESTION [5 upvotes]: Every complete Boolean algebra arises as the lattice of regular open sets in some topological space, namely given a complete Boolean algebra $B$, the corresponding Stone space $S(B)$ will be extremally disconnected, so in particular its regular open sets are precisely its clopen sets. A semi-common generalization of topological spaces is the class of spaces $(X,\tau)$ such that $\varnothing,X\in\tau$ and such that $\tau$ is closed under arbitrary unions (no assumption about intersections). This is equivalent to a weakening of the Kuratowski closure axioms to the axioms of a general closure operator. (Wikipedia says that these are sometimes called 'closure spaces,' but I've also seen them called 'generalized topological spaces' in some papers. If anyone knows another name for these things I would really like to know it because searching for relevant literature has been difficult.) A closure operator is equivalently specified by an interior operator, a function $\mathrm{int}:\mathcal{P}(X)\rightarrow \mathcal{P}(X)$ satisfying: $\mathrm{int}(A)\subseteq A$ $A\subseteq B \Rightarrow \mathrm{int}(A)\subseteq \mathrm{int}(B)$ $\mathrm{int}(\mathrm{int}(A))=\mathrm{int}(A)$ (This definition makes sense in an arbitrary poset.) Under such an operator, the collection of open sets (sets such that $\mathrm{int}(U)=U$) forms a complete lattice with arbitrary meets given by set theoretic union. Unlike in a topological space finite joins are given by $U \wedge V = \mathrm{int}(U \wedge V)$. (This lattice is dual to the lattice of closed sets, but I'm phrasing this question in terms of open sets since I find it a little more familiar and because there's no standard name for the closure of complement operation.) An interior operator gives rise to a natural exterior operator: $\mathrm{ext}(A)=\mathrm{int}(X \setminus A)$. Just like in a topological space one can show that for an open set $U$, $\mathrm{ext}(\mathrm{ext}(\mathrm{ext}(U)))=\mathrm{ext}(U)$, so call an open set $U$ 'regular' if it satisfies $\mathrm{ext}(\mathrm{ext}(U))=U$. (Note that unlike in topology it seems that the exterior operator is not definable from the structure of the lattice of opens alone.) The map $U\mapsto \mathrm{ext}(\mathrm{ext}(U))$ gives a closure operator on the lattice of open sets, so we get that the collection of regular opens also forms a complete lattice whose join is the join of open sets (although again unlike in topology this is not set theoretic intersection) but whose meet is different, so in particular neither operation is set theoretic. One can show that the collection of regular opens together with their meet, join, and the exterior operator form an orthocomplemented lattice, with the exterior playing the role of orthocomplementation, i.e. $ext$ is an order-reversing involution such that $U$ and $\mathrm{ext}(U)$ are always complements. This is a purely algebraic fact about Boolean algebras with interior operators satisfying the above axioms. With a finite counterexample search I found that in general the lattice does not need to be orthomodular. I convinced myself that every complete lattice arises as the lattice of opens of a closure space, although it's easier to phrase in terms of closed sets and the dual lattice. Given a lattice $(L,0,1,\wedge,\vee)$ we can consider the collection of subsets of $L\setminus \{0\}$ and let a set be closed if it is of the form $(0,a]$ for some $a\in L\setminus \{0\}$ (so in particular the closure operator is $\mathrm{cl}(A)=(0,\bigvee A]$). By completeness of the lattice any intersection of sets of this form is of this form or empty and in particular the intersection's top is the greatest lower bound of the tops of the original sets, so the lattices agree (if we identify $\varnothing$ with $0$, this is why we need to remove $0$, otherwise we get an extra bottom element). To get a lattice of opens we can just start with the dual lattice. That construction always produces trivial lattices of regular opens (no non-trivial open sets are disjoint, so the exterior of any non-empty set is empty), which shows that the exterior operator really does depend on the underlying set and not just the lattice of opens (since there are closure spaces with non-trivial lattices of regular opens). So finally we can come to the question: Which complete orthocomplemented lattices arise as the lattice of regular opens in a closure space? REPLY [3 votes]: All orthocomplemented lattices may be so represented. A complete ortholattice L is the system of regular closed sets in a closure space generated on the family F of all maximal increasing orthocomplement-free sets of L. The embedding of L in this space associates with each x in L a clopen set of F, whose clopen complement is the set associated with x', the orthocomplement of x. L is separated by F, so the lattice order corresponds isomorphically to the order on the clopen sets. Since these sets are clopen, they also form the system of regular open sets of L. The work toward this result began in investigations into reconciliations of classical (Boolean) and quantum logics, and culminated with the independent and nearly simultaneous papers by Iturrioz, Katrnoska and Mayet in 1982. The relevant references are below. L. Iturrioz, A simple proof of a characterization of complete orthocomplemented lattices, Bull. London Math. Soc. 14, no. 6 (1982), 542–544. F. Katrnoˇska, On the representation of orthocomplemented posets, Comment. Math. Univ. Carolin. 23, no. 3 (1982), 489–498. A.R. MARLOW: Quantum theory and Hilbert Space, Journal of Math. Phya. 19(1978), 1-15. R. Mayet, Une dualit´e pour les ensembles ordonn´es orthocompl´ement´es, C. R. Acad. Sci. Paris S´er. I Math. 294, no. 2 (1982), 63–65. Luigi Santocanale, Friedrich Wehrung. Lattices of regular closed subsets of closure spaces. International Journal of Algebra and Computation (IJAC), 2014, 24 (7), pp.969–1030. M. Sekanina, On a characterisation of the system of all regularly closed sets in general closure spaces, Math. Nachr. 38 (1968), 61–66. N. ZIERLER - M. SCHLESINGER: Boolean Embeddings of orthomodular sets and Quantum logics, Duke Math. Journal 32(1965), 251.<|endoftext|> TITLE: Do i.i.d. sums concentrate any faster than martingales? QUESTION [7 upvotes]: Suppose $X_1,X_2, \ldots, X_N \in \mathbb R^d$ are random variables with each $\|X_n\|_2 \le 1/2$ (this choice of the constant simplifies later formulae). The simplest concentration inequality I know only applies in the case $d=1$ and only when $X_1,X_2, \ldots, X_N$ are i.i.d. The Hoeffding Lemma gives for each $\epsilon >0$ the bound $$P(|X_1 + \ldots + X_N| \ge \epsilon) \le \exp\left (-\frac{2\epsilon^2}{N} \right).\tag{1}$$ On the other end of the spectrum are results that work under the weaker assumption that $X_1,X_2, \ldots, X_N$ is a martingale, and work for any $d \in \mathbb N$, or indeed for infinite dimensional Banach spaces provided some variant of the parallelogram is satisfied. For example Theorem 3.5 of this paper of Pinelis leads to the following variant of the Azuma-Hoeffding inequality. $$P(\|X_1 + \ldots + X_n\|_2 \ge \epsilon \text{ for some }n\le N) \le \exp\left (-\frac{\epsilon^2}{2N} \right).\tag{2}$$ The exponent is the same as the scalar Azuma Hoeffding. Notice the $2$ is now downstairs rather than upstairs like before. If we are only dealing with i.i.d scalars and only interested in the final element, we should use $(1)$ because it gives a better bound. If we are dealing with either vectors, martingales, or want a uniform inequality we better use $(2)$ instead. My problem is between the two extremes. I am dealing with a sequence of i.i.d vectors in $\mathbb R^d$ and I am interested in a uniform bound. I wonder does there exist an appropriate middle-ground between these two results? Perhaps combining the $-2\epsilon^2/N$ of the first with the uniform nature of the second, at the expense of only applying to i.i.d sequences as opposed to martingales. REPLY [4 votes]: Other than the presence of an extra factor $D$ or $D^2$, depending on the context, the coefficients in the bounds for martingales in $(2,D)$-smooth spaces in the paper you cite are exactly the same as the ones for sums of independent real-valued random variables. There is no gap in this sense. Also, you are citing Hoeffding's inequality incorrectly: the restriction on the $X_n$'s is of course in terms of the $\infty$-norm, rather than in terms of the $2$-norm.<|endoftext|> TITLE: Is the set of separable quantum states closed? QUESTION [12 upvotes]: Let $\mathcal H,\mathcal H'$ be Hilbert spaces (not necessarily separable). A "separable state" is a trace-class operator of the form $\sum_i \rho_i\otimes\rho_i'$ where $\rho_i,\rho_i'$ are positive trace-class operators over $\mathcal H,\mathcal H'$, respectively. (Convergence of the sum is with respect to the trace norm. $\otimes$ represents the tensor product.) Is the set of separable states closed with respect to the trace norm? REPLY [8 votes]: The answer is negative if we believe [1] (which comes without proofs, unfortunately): [1] defines the set of separable states as the convex closure of $\{\sum_{i=1}^n\sigma_i\otimes\tau_i\}$. I will call that set $S$. The set as defined in the original question I call $T$. Then $T\subseteq S$, and $T=S$ iff the answer is yes (i.e., iff $T$ is closed). [1] shows that there is a $\rho\in S$ such that $\rho$ cannot be represented as a Bochner integral $\int\psi\psi^*\otimes\phi\phi^* \pi(d(\psi,\phi))$ for an atomic measure $\pi$. ($\rho$ is not "countably decomposable" in their language.) Any $\rho\in T$ can be represented as such an integral with a discrete and hence atomic measure $\pi$. Thus $T\subsetneq S$. On the positive side, if we define $T':=\{\int\psi\psi^*\otimes\phi\phi^* \pi(d(\psi,\phi))\}$ for probability measures $\pi$, then [1] shows $S=T'$. In particular, $T'$ is closed. So, in the spirit of the original question, the set of infinite convex combinations of $\sigma_i\otimes\tau_i$ is closed, only the notion of "infinite convex combination" must be changed: Not infinite sums, but integrals. [1] Werner, R. F.; Kholevo, A. S.; Shirokov, M. E., On the concept of entanglement in Hilbert spaces., Russ. Math. Surv. 60, No. 2, 359-360 (2005); translation from Usp. Mat. Nauk 60, No. 2, 153-154 (2005). ZBL1098.47019.<|endoftext|> TITLE: Is the composition of two nowhere differentiable functions still nowhere differentiable? QUESTION [13 upvotes]: Let $f,g:\mathbb R\to\mathbb R$ be two continuous but nowhere differentiable functions. By the Denjoy–Young–Saks theorem for almost every point $x_0\in\mathbb R$ one has $$ \limsup\limits_{x\to x_0}\frac{|f(x)-f(x_0)|}{|x-x_0|}=+\infty. $$ Is $f\circ g$ still nowhere differentiable? It seems possible that the high frequency parts of $g$ lie in the low frequency parts of $f$, which might lead to a cancellation of the oscillation. I am trying to prove non-differentiability through the following: Fix $x_0$ and consider $$ I_{f,x_0}^k=\{x\in\mathbb R:|f(x)-f(x_0)|\ge k|x-x_0|\}, $$ the set of points outside a double-sided cone with slope $k$. Is it true that $$ \lim\limits_{\delta\to0}|I_{f,x_0}^k\cap[x_0-\delta,x_0+\delta]|/(2\delta)=1? $$ If yes, then we know that “most” of the points near $x_0$ have values relatively far away from $f(x_0)$, and so there will not be much frequency cancellation. REPLY [10 votes]: The composition may have points of differentiability. Let $f_0(x)=x$ for $x\geq 0$ and $f_0(x)=2x$ for $x<0$. Let $g_0(x)=2x$ for $x\geq 0$ and $g_0(x)=x$ for $x<0$. Then none of them is differentiable at $0$, but $f_0\circ g_0=2x$ is differentiable everywhere. Let $h$ be a function that is nowhere differenitable, except at $0$ where it is very rapidly decaying. E. g. I can take $h(x)=e^{-\frac{1}{x^2}}B_{|x|}$, where $B_t$ is a sample of Brownian motion. Now just take $f=f_0+h$ and $g=g_0+h$.<|endoftext|> TITLE: Positive proportion of logarithmic gaps between consecutive primes QUESTION [5 upvotes]: For $x, \lambda > 0$, define $$S_\lambda(x) := \#\{p_{n+1} \leq x : p_{n+1} - p_n \geq \lambda \log x\} ,$$ where $p_n$ is the $n$th prime number. It is known [1] that an uniform version of the Hardy-Littlewood prime k-tuples conjecture implies that for fixed $\lambda > 0$, $$S_\lambda(x) \sim e^{-\lambda} \frac{x}{\log x}$$ as $x \to +\infty$. My question is: If we want only a lower bound of the form $$S_\lambda(x) \gg_\lambda \frac{x}{\log x}, \quad x > 2,$$ for every fixed $\lambda > 0$, has this been proved unconditionally? Thank you for any reference or suggestion. [1] Funkhouser,Goldston, Ledoan, Distribution of Large Gaps Between Primes, https://doi.org/10.1007/978-3-319-92777-0_3 REPLY [11 votes]: I think I can establish the claim for all $\lambda < 1/4$. In principle one should be able to get up to $\lambda < 1.145\dots$ but as noted in other comments this would require upper bounds on the density of large intervals free of primes that do not appear to be known unconditionally. For any $\lambda > 0$ and any $X \geq 1$, let $f_X(\lambda)$ be the proportion of $1 \leq x \leq X$ such that the interval $[x, x+\lambda \log X]$ is free of primes (one can replace $\log X$ here by $\log x$ if desired, it doesn't really change things). This is clearly a monotone decreasing function of $\lambda$ taking values in $[0,1]$. The problem is essentially equivalent to establishing a noticeable dip in the value of $f_X$. More precisely: Proposition 1 Let $\lambda_0>0$. Then the following are equivalent: For all $\lambda < \lambda_0$, one has $S_\lambda(X) \gg_\lambda \frac{X}{\log X}$ for sufficiently large $X$. For all $\lambda < \lambda_0$, there exists $\Lambda>\lambda$ such that $f_X(\lambda) - f_X(\Lambda) \gg_\lambda 1$ for sufficiently large $X$. Proof Suppose 1 holds and $\lambda < \lambda_0$. From the PNT and Markov's inequality one has $S_\Lambda(X) \ll \frac{1}{\Lambda} \frac{X}{\log X}$, thus for $\Lambda = O_\lambda(1)$ large enough one has $S_\lambda(X) - S_\Lambda(X) \gg_\lambda \frac{X}{\log X}$. Thus there are $\gg_\lambda \frac{X}{\log X}$ prime gaps in $[1,x]$ of size between $(\lambda-o(1)) \log X$ and $(\Lambda+o(1)) \log X$ (since $\log p_n = (1+o(1)) \log X$ for almost all primes $p_n \leq x$). This easily implies that $f_X(\lambda) - f_X(\Lambda) \gg_\lambda 1$ for sufficiently large $X$ (possibly after tweaking $\lambda$ and $\Lambda$ by $o(1)$ first). Conversely, if 2 holds, then for a proportion $\gg_\lambda 1$ of $x \in [1,X]$, the interval $[x,x+\lambda \log X]$ is free of primes while the interval $[x,x+\Lambda \log X]$ has a prime. Let $A$ be large. The proportion of $x$ for which $[x,x+\Lambda \log X]$ has a prime but $[x-A\Lambda \log X, x]$ does not is $O(1/A)$, so by taking $A = O_\lambda(1)$ large enough, we see that for a proportion $\gg_\lambda 1$ of $x \in [1,X]$, the intervals $[x-A \Lambda \log X, x]$ and $[x+\lambda \log X,x+\Lambda \log X]$ contain primes but $[x,x+\lambda \log X]$ does not, thus $x$ lies in a prime gap of size between $\lambda \log X$ and $(A+1)\Lambda \log X$. From this one can establish $S_\lambda(X) \gg_\lambda \frac{X}{\log X}$ without difficulty. $\Box$ The claim now follows from this proposition and Proposition 2 For any $\lambda > 0$, one has $$ 1 - \lambda+o(1) \leq f_X(\lambda) \leq 1 - \frac{\lambda}{1+4\lambda}+o(1)$$. Proof Let $N_\lambda$ be the number of primes in $[x,x+\lambda \log X]$ where $x$ is drawn uniformly at random from $[1,X]$, so $f_X(\lambda)$ is the probability that $N_\lambda=0$. From the prime number theorem and the first moment method one has $$ {\bf E} N_\lambda = \lambda + o(1)$$ which by Markov's inequality gives that ${\bf P}(N_\lambda \geq 1) \leq \lambda + o(1)$, which gives the lower bound on $f_X(\lambda)$. For the upper bound we need to control the second moment $$ {\bf E} \binom{N_\lambda}{2}.$$ Gallagher's calculation assuming Hardy-Littlewood predicts this quantity to be $\lambda^2/2 + o(1)$ (and more generally $N_\lambda$ should be Poisson distributed with parameter $\lambda$). This is not known unconditionally, but if one uses the standard Selberg sieve estimate coming from Bombieri-Vinogradov (e.g. Theorem 7.16 of Opera del Cribro), which is an upper bound that is worse than Hardy-Littlewood by a factor of $4$, one gets an upper bound of $2\lambda^2 + o(1)$. In particular $$ {\bf E} N_\lambda^2 \leq \lambda + 4 \lambda^2 + o(1).$$ But from Cauchy-Schwarz one has $$ {\bf E} N_\lambda^2 \geq ({\bf E} N_\lambda)^2 / (1-f_X(\lambda))$$ and if one puts together all these inequalities one obtains the claim. $\Box$ If one could prove an upper bound on $f_X(\Lambda)$ that went to zero as $\Lambda \to \infty$ then we could use the above arguments to show that $S_\lambda(x) \gg_\lambda \frac{x}{\log x}$ for all $\lambda < 1$. The paper https://www.ams.org/journals/tran/2010-362-05/S0002-9947-09-05009-0/ linked to in the previous answer would extend this to $\lambda < 1.145...$ due to slightly improved lower bounds on $f_X(\lambda)$ for $\lambda$ near $1$. One can probably improve the upper bound on $f_X(\lambda)$ a little bit by using the inequality $f_X(\lambda) \binom{-k}{2} \leq {\bf E} \binom{N_\lambda - k}{2}$ for any natural number $k$, then optimising in $k$, but I don't think this actually improves the $1/4$ threshold. The Elliott-Halberstam conjecture would let one replace $1/4$ with $1/2$ though. Meanwhile, the Maynard sieve should be able to slightly improve upon the Markov inequality lower bound for $f_X(\lambda)$ for any value of $\lambda>0$ and thus lead to modest improvement in the $1/4$ threshold (this may possibly already follow from the paper linked to above). On the other hand getting the upper bound for $f_X(\Lambda)$ below $1/2$ seems to require getting around the parity problem (in particular eliminating the scenario in which the integers split into long intervals, with the Liouville function on almost primes biased to be +1 on half of these long intervals (as measured by density), and biased to be -1 on the other half).<|endoftext|> TITLE: Does this degree 12 genus 1 curve have only one point over infinitely many finite fields? QUESTION [5 upvotes]: Let $F(x,y,z)$ be the degree 12 homogeneous polynomial: $$x^{12} - x^9 y^3 + x^6 y^6 - x^3 y^9 + y^{12} - 4 x^9 z^3 + 3 x^6 y^3 z^3 - 2 x^3 y^6 z^3 + y^9 z^3 + 6 x^6 z^6 - 3 x^3 y^3 z^6 + y^6 z^6 - 4 x^3 z^9 + y^3 z^9 + z^{12}$$ Over the rationals it is irreducible and $F=0$ is genus 1 curve. Numerical evidence in Sagemath and Magma suggests that for infinitely many primes $p$, the curve $F=0$ is irreducible over $\mathbb{F}_p$ and $F=0$ has only one point over $\mathbb{F}_p$, the singular point $(1 : 0 : 1)$. Q1 Is this true? Set $p=50033$. Then we have only one point over the finite field and the curve is irreducible of genus 1. This appears to violate the bound on number of rational points over finite fields given in the paper "The number of points on an algebraic curve over a finite field", J.W.P. Hirschfeld, G. Korchmáros and F. Torres ,p. 23. Q2 What hypothesis am I missing for this violation? Sagemath code: def tesgfppoints2(): L1=5*10^4 L2=2*L1 for p in primes(L1,L2): K.=GF(p)[] F=x^12 - x^9*y^3 + x^6*y^6 - x^3*y^9 + y^12 - 4*x^9*z^3 + 3*x^6*y^3*z^3 - 2*x^3*y^6*z^3 + y^9*z^3 + 6*x^6*z^6 - 3*x^3*y^3*z^6 + y^6*z^6 - 4*x^3*z^9 + y^3*z^9 + z^12 C=Curve(F) ire=C.is_irreducible() if not ire: continue rp=len(C.rational_points()) print 'p=',p,';rp=',rp,'ir=',ire,'g=',C.genus() REPLY [14 votes]: The polynomial you wrote is the product of the four polynomials $x^3 - r y^3 - z^3$, where $r$ is a root of the polynomial $t^4 - t^3 + t^2 - t + 1$. I did not read your reference, but likely they assume that the curves are geometrically irreducible.<|endoftext|> TITLE: Relative Dickson (trace) criterion for Jacobson radical? QUESTION [6 upvotes]: In the following, all algebras are associative and unital. Let $J\left(A\right)$ denote the Jacobson radical of an arbitrary algebra $A$. Recall that this is defined as the set of all $a \in A$ such that for each $s \in A$, the element $1 - as \in A$ is invertible. The following is Dickson's trace criterion for the Jacobson radical (slightly generalized): Theorem 1. Let $\mathbb{K}$ be a field, and let $n \in \mathbb{N}$. Assume that the integers $1, 2, \ldots, n$ are invertible in $\mathbb{K}$. Let $A$ be a $\mathbb{K}$-algebra that is $n$-dimensional as a $\mathbb{K}$-vector space. Let $a \in A$. Then, $a \in J\left(A\right)$ if and only if every $s \in A$ satisfies $\operatorname{Tr}\left(as\right) = 0$. Here, the trace $\operatorname{Tr}\left(b\right)$ of an element $b \in A$ is defined as the trace of the endomorphism $A \to A, \ u \mapsto bu$ of the $\mathbb{K}$-vector space $A$. Note that the "$\Longrightarrow$" direction of Theorem 1 does not require the assumption that $1, 2, \ldots, n$ be invertible; only the other direction requires it. The definition of $\operatorname{Tr}\left(b\right)$ does not require $\mathbb{K}$ to be a field; it suffices that $A$ is a free $\mathbb{K}$-module. (It even suffices that $A$ is a projective $\mathbb{K}$-module, but I don't want to go that far afield.) Of course, we cannot straightforwardly generalize Theorem 1 to arbitrary commutative rings $\mathbb{K}$, since it would fail even for $A = \mathbb{K}$. But here is an attempt at a generalization that would work: Conjecture 2. Let $\mathbb{K}$ be a commutative ring, and let $n \in \mathbb{N}$. Assume that the integers $1, 2, \ldots, n$ are invertible in $\mathbb{K}$. Let $A$ be a $\mathbb{K}$-algebra that is a free $\mathbb{K}$-module of rank $n$. Let $a \in A$. Then, $a \in J\left(A\right)$ if and only if every $s \in A$ satisfies $\operatorname{Tr}\left(as\right) \in J\left(\mathbb{K}\right)$. Question. Is this conjecture correct? If not, is at least one of its two directions correct? Almost nothing in rschwieb's proof of Theorem 1 generalizes easily to this setting; Artinianity breaks down, and elements of Jacobson radicals don't have to be nilpotent. I'm also not exactly swamped by good examples of Jacobson radicals, so there might be a counterexample much shorter than this post. The only thing I was able to do is make some questionable headway into the "$\Longleftarrow$" direction of Conjecture 2. Namely, assume that every $s \in A$ satisfies $\operatorname{Tr}\left(as\right) \in J\left(\mathbb{K}\right)$. Consider the commutative ring $\overline{\mathbb{K}} := \mathbb{K} / J\left(\mathbb{K}\right)$, which is well-known to satisfy $J\left(\overline{\mathbb{K}}\right) = 0$. Let $\overline{A}$ be the $\overline{\mathbb{K}}$-algebra $\overline{\mathbb{K}} \otimes_{\mathbb{K}} A$; this is a free $\overline{\mathbb{K}}$-module of rank $n$. Now, consider the projection $\overline{a}$ of $a$ onto $\overline{A}$. Then, our assumption yields that every $s \in \overline{A}$ satisfies $\operatorname{Tr}\left(\overline{a}s\right) \in J\left(\overline{\mathbb{K}}\right) = 0$ (here we are using the fact that surjective ring homomorphisms induce homomorphisms between the Jacobson radicals). Now, a well-known fact from linear algebra (e.g., Corollary 4.1 (b) in my note The trace Cayley-Hamilton theorem) yields that for each $s \in \overline{A}$, the element $\overline{a}s$ is nilpotent, whence the element $1 - \overline{a}s$ is invertible; hence, $\overline{a} \in J\left(\overline{A}\right)$. It sounds reasonable to expect that this entails $a \in J\left(A\right)$, although the exact mechanics of how this should follow eludes me. REPLY [4 votes]: $\newcommand{\m}{\mathfrak{m}} \newcommand{\K}{\mathbb{K}} \DeclareMathOperator{\Tr}{Tr}$Let me prove the $\Longrightarrow$ implication. Let $a\in J(A)$. Let $\m$ be a maximal ideal of $\K$. Then $A/\m$ is a finite-dimension algebra over the field $\K/\m$, and is therefore Artinian, and moreover $a\bmod \m \in J(A/\m)$, so we have $\Tr(a \bmod \m)=0$, i.e. $\Tr(a)\in \m$. So we get $\Tr(a) \in \bigcap_{\m \in \mathrm{maxspec}(\K)}\m = J(\K)$. The full implication follows from the fact that $J(A)$ is a two-sided ideal.<|endoftext|> TITLE: Bounding higher derivatives of $f(x) = 1/(1+x^2)^r$ QUESTION [5 upvotes]: Let $r\in \lbrack 0,\infty)$. Define $f(x) = 1/(1+x^2)^r$. It would seem to be the case that $$|f^{(k)}(x)|\leq \frac{2r \cdot (2r+1) \dotsb (2r + k-1)}{(1+x^2)^{r + k/2}}$$ for all even $k\geq 0$. What would be a clean, short proof? Note that $$\left(\frac{1}{(1+x^2)^r}\right)^{(k)} = \frac{P_k(x)}{(1+x^2)^{r+k}},$$ where $P_k\in \mathbb{Z}\lbrack x\rbrack$ is given by the recurrence formula $$P_k(x)=1,\;\;\;\;\;\;P_{k+1}(x) = P_k'(x) (1+ x^2) + 2 (r+k)\cdot x P_k(x).$$ It should not be too hard to work out all the coefficents of $P_k(x)$; the leading terms are $$(-1)^{k+1} \cdot 2r \cdot (2r+1) \dotsb (2r + k-1)\cdot \left(x^k - \binom{k}{2} \frac{x^{k-2}}{2 r + 1} + \dotsc\right).$$ The desired bound should then follow after some work -- but I am hoping for a better proof. REPLY [13 votes]: Well, $$ (a^2+x^2)^{-r}=(x+ai)^{-r}\cdot (x-ai)^{-r}. $$ Differentiate this $k$ times, we get $$ \left((a^2+x^2)^{-r}\right)^{(k)}=k!\sum_{s=0}^k {-r\choose s}{-r\choose k-s} (x+ai)^{-r-s}(x-ai)^{-r-(k-s)}. $$ Estimate this by the triangle inequality, you get some constant times $(x^2+a^2)^{-r-k/2}$. But for $a=0$ we have equality (because the signs of all summands are equal to $(-1)^k$), thus the constant is $$ (-1)^kx^{2r+k}\left(x^{-2r}\right)^{(k)}=2r(2r+1)\ldots(2r+k-1). $$<|endoftext|> TITLE: What is the difference between total integral closure and integral closure? QUESTION [6 upvotes]: I was advised here to make this a new question: What is the difference between total integral closure and integral closure (geometrically, in the context of rigid analytic geometry)? I have read in multiple places that the difference is basically rank one valuations vs higher rank valuations but how does one make this precise? It is surprisingly hard to even find a definition of total integral closure... REPLY [4 votes]: A basic example: Let $R$ be the ring of Laurent series $k[x^{\pm}, y^{\pm}]$ with coefficients in some field $k$. Let $A$ be the subring generated by $x$ and by all monomials of the form $x^{-j} y$. The element $x^{-1}$ is not integral over $A$; equivalently, the $A$-submodule $M$ of $R$ generated by all powers of $x^{-1}$ is not finitely generated as an $A$-module. However, $M \subset y^{-1} A$, so $M$ is contained in a finitely generated $A$ module, and $x^{-1}$ is in the total integral closure of $A$. One can see the connection to valuations easily in this example: Let $\Lambda$ be $\mathbb{Z}^2$ ordered lexicographically (with the second coordinate more important than the first) and define $v : R_{\neq 0} \to \Lambda$ by $$v\left( \sum_{(i,j)} c_{ij} x^i y^j \right) = \min\left\{ \vphantom{\sum_{(i,j)} c_{ij} x^i y^j} (i,j) : c_{ij} \neq 0 \right\}.$$ Then $v(x^{-1}) = (-1, 0)$, detecting that $v$ is not in the integral closure. However, no rank $1$ valuation takes negative value on $x^{-1}$.<|endoftext|> TITLE: Is $π:\mathcal{C}^∞(M,N)→\mathcal{C}^∞(S,N)$, $π(f)=f|_S$ a quotient map in the $\mathcal{C}^1$ topology? QUESTION [6 upvotes]: This question was previously posted on MSE. Let $M, N$ be smooth connected manifolds (without boundary), where $M$ is a compact manifold, so we can put a topology in the space $\mathcal C^\infty(M, N)$ using $\mathcal{C}^1$ Whitney Topology. Now, consider $S\subset M$ a compact submanifold of $M$ with boundary such that $\text{dim}S=\text{dim}M$, using the same process we can put a topology in $\mathcal C^\infty(S,N)$ using the $\mathcal{C}^1$ Whitney Topology. There is a natural continous projection of $\mathcal C^\infty(M, N)$ on $\mathcal C^\infty(S,N)$, definided by \begin{align*} \pi: \mathcal C^\infty(M, N) &\to \mathcal C^\infty(S,N)\\ f&\mapsto \left.f\right|_{S}. \end{align*} My Question: Is $\pi$ an open map or at least a quotient map? Some comments $\mathcal{C}^1$-Whitney Topology is also called $\mathcal{C}^1$-strong topology. As noticed for the user Adam Chalumeau, on the book "Morris W. Hirsh Differential Topology" there is the following exercise [Exercise 16, page 41]: Let $M, N$ be $\mathcal{C}^r$ manifolds. Let $V⊂M$ be an open set then The restriction map $$δ:\mathcal{C}^r(M,N)→\mathcal{C}^r(V,N)$$ $$δ(f)=f|_V$$ is continuous for the weak topology, but not always for the strong. $δ$ is open for the strong topologies, but not always for the weak". Since our $M$ is compact weak topology = strong topology. However, I don't know how to solve this exercise let alone adapt such proof to the case that I want. Does anyone know anything about this problem? REPLY [6 votes]: Let me try to give at least some partial answers (and apologies that I am leaving the realm of topological spaces to give you these answers, but I am a differential geometer, so you have been warned): In general the restriction map will fail to be a quotient map, as it will fail to be surjective (and quotient maps are by definition surjective mappings). Extending a $C^\infty$-mapping from a compact submanifold with boundary to a larger manifold is not always possible due to topological reasons. As a simple example take a smooth map $f \colon \mathbb{S}^2\supseteq C \rightarrow \mathbb{S}^1$, where the $\mathbb{S}^i$ are the unit spheres in $\mathbb{R}^{i+1}$ and $C$ is an 'equatorial belt' (which is a compact submanifold with smooth boundary). Then $f$ cannot extend to $\mathbb{S}^2$ if has non-zero winding number (cf. Lee, Introduction to smooth manifolds, Corollary 6.27). It might be helpful to note that the $C^k$-Whitney topologies on all your spaces coincide with the compact open $C^k$-topology (not sure whether you meant this topology with 'weak topology', just wanted to be sure we are on the same page). If you had endowed your spaces of smooth functions with the $C^\infty$-topology (instead of the $C^1$-topology), then the answer to your question is positive: The continuous projection $\pi$ is open. This follows from a recent paper of mine (and D. Roberts): Extending Whitney’s extension theorem: nonlinear function spaces Theorem B establishes that $\pi \colon C^\infty (M,N) \rightarrow C^\infty (S,N)$ is a submersion (but in general NOT surjective, see 1.) between infinite-dimensional manifolds (the meaning of this is that it is in local charts a projection, thus an open map by the usual proof). Note that in the cited paper, the compact submanifold $S$ is allowed to have a much more irregular boundary then the smooth boundary you want to consider. To adapt what was said in 2. to your case with the much weaker $C^1$-topology, I guess there would be two (relatively straight forward) ways to achieve this: (a)Try to copy the argument in 2.: The function space techniques which allow one to drag the Whitney extension theorem from chart domains to a compact manifold are (apart from the Whitney extension theorem) all amenable to change from $C^\infty$ to $C^k$ ($k$ finite), i.e. if you change the topology on the spaces, you obtain smooth maps with respect to the weaker topologies (to see this, compare with the description of the manifold charts etc. for $C^k (M,N)$ in Appendix A of Lie groupoids of mappings taking values in a Lie groupoid). Hence you can just copy the proof steps and either see that the Whitney map (from the extension theorem) is also continuous in the $C^1$-topology (Check this!) or you use a corresponding extension theorem for $C^1$-maps due to Fefferman (and check that it yields $C^\infty$-mappings if it gets a $C^\infty$-map as input [I have no idea if this will work]). The arguments then yield a local continuous inverse for $\pi$ and this should imply that $\pi$ is open. (b) Ignore all the infinite-dimensional manifold stuff and try to construct something in the topology. As a subbase ($\{f\in C^\infty (M,N) \mid Tf (L) \subseteq O \subseteq TN\}, L \subseteq TM$ compact and $O$ open) of the compact open $C^1$-topology is well known you can check openness of $\pi$ on the BASE generated by the subbase. Since $S$ is compact we can identify $TS$ with a closed subset of $TM$, whence $L \cap TS$ is compact in the subspace topology. Now you probably want to distinguish cases of whether $L \cap TS =\emptyset$ or not. Note however, that it is not quite clear to me how to identify the image of the (sub)basic neighborhoods under $\pi$. Since we are dealing here with a mapping which is in general non surjective, one probably needs at this stage an extension argument to see that the image of such a neighborhood is an open neighborhood of each of its elements.<|endoftext|> TITLE: Good upper-bound for $\mathbb E[|X-np|^r]$ where $X \sim \text{Binomial}(n,p)$ and $r \ge 1$ QUESTION [5 upvotes]: Disclaimer. Question moved from SE. Setup Let $X \sim \text{Binomial}(p, n)$, and $r \ge 1$. Question What is a good upper-bound for $\mathbb E[|X-np|^r]$ ? Solution for small $r$ If $r=2$, then $\mathbb E[|X-np|^2] = np(1-p)$. If $1 \le r < 2$, then Jensen's inequality gives $$ \mathbb E[|X-np|^r] =(\mathbb E[|X-np|^r])^{2/r})^{(r/2)} \le (\mathbb E[|X-np|^2])^{r/2} = (np(1-p))^{(r/2)}. $$ Notes I'm ultimately interested in bounding (via McDiamid's inequality) the $\ell_r$-distance between a distribution $P$ and it's empirical version $\hat{P}_n$. REPLY [4 votes]: By the main result of the paper Exact Rosenthal-type bounds, we have $$E|X-np|^r\le c^r E|\Pi_\lambda-\lambda|^r $$ for real $r\in(2,\infty)\setminus(3,4)\setminus(4,5)$, where real $c>0$ and $\lambda>0$ are defined by the conditions $$c^r\lambda=n(q^rp+p^rq)\quad \text{and}\quad c^2\lambda=npq; $$ $q:=1-p$; and $\Pi_\lambda$ is a Poisson random variable with parameter $\lambda$. Other results on Rosenthal-type bounds can be found e.g. in this paper or its arXiv version, and in references therein. Added: In a comment, the OP stated that it may be assumed that $1\le r\le 2$. This simplifies the matter a great deal. Indeed, in this case we have \begin{equation} E|X-np|^r\le\min((npq)^{r/2},2npq(q^{r-1}+p^{r-1}))\ll s^r\wedge s^2,\tag{1} \end{equation} where $(npq)^{r/2}$ is the bound the OP obtained by using Jensen's inequality; and $2npq(q^{r-1}+p^{r-1})$ is a bound immediately obtained by using the von Bahr--Esseen inequality -- see e.g. this paper or its arXiv version; $s:=\sqrt{npq}$. For positive expressions $e_1$ and $e_2$, we write $e_1\ll e_2$ or, equivalently, $e_2\gg e_1$ if $e_1\le C e_2$ for some universal positive real constant $C$. The upper bound on $E|X-np|^r$ in (1) is optimal up to a universal constant factor: for $r\in[1,2)$ \begin{equation} E|X-np|^r\gg s^r\wedge s^2.\tag{2} \end{equation} This lower bound on $E|X-np|^r$ is obtained by using the log-convexity of $m_t:=E|X-np|^t$ in $t>0$ and an upper Rosenthal-type bound -- as follows: By (say) Theorem 1.5 in the already cited paper Exact Rosenthal-type bounds, we have $m_3\ll s^3\wedge s^2$. Now using the log-convexity of $m_t:=E|X-np|^t$ in $t>0$ or, equivalently, the Hölder inequality, we have \begin{equation*} m_2\le m_r^{1/a}m_3^{1-1/a}, \end{equation*} where $a:=3-r$. Hence, \begin{equation*} E|X-np|^r=m_r\ge m_2^{3-r}m_3^{r-2}\gg s^{6-2r}(s^{3r-6}\wedge s^{2r-4}) =s^r\wedge s^2. \end{equation*} Thus, (2) indeed holds.<|endoftext|> TITLE: Terminology about G- simplicial complexes QUESTION [8 upvotes]: For a simplicial complex $X$ with an action of a discrete group $G$, we can impose the following condition, namely that if $g\in G$ stabilizes a given simplex $\sigma\subseteq X$, then $g:\sigma\to\sigma$ is in fact the identity map. This condition is nice because it implies that the quotient $X/G$ is glued out of simplices (rather than simplices modulo a finite group of permutations of their vertices). Is there a standard term for this condition (or any related one) on the action $G\curvearrowright X$? REPLY [7 votes]: I've never stumbled across a standard one in all of the literature, though I've definitely appreciated a remark by Ken Brown in his group cohomology bible: "The hypothesis that $G_\sigma$ fixes $\sigma$ pointwise is not very restrictive in practice. In the case of a simplicial action, for example, it can always be achieved by passage to the barycentric subdivision." Some references have the pointwise-fixed assumption baked into the definition of a $G$-CW-complex, but not all. Ken Brown's book doesn't, and he calls a $G$-CW-complex "admissible" if we include the pointwise-fixed constraint. This doesn't seem to be standard though. (Now as for definite nonstandard terminology, Farb-Margalit's book "A Primer on Mapping Class Groups" calls it a group action "without rotations".)<|endoftext|> TITLE: When the image of a convex set in $\mathbb{R}^n$ is still a convex set? QUESTION [6 upvotes]: Here is tricky problem I came across when writing a paper. But I can't figure it out, so I ask for help here. Let $M$ be a $n$-dimensional smooth manifold which is also a complete geodesic space. We can map a small ball around a point in $M$ to $\mathbb{R}^n$ endowed with the Euclidean metric such that the map is a homeomorphism. The small ball is a convex set in $M$ (i.e the geodesic lines which connect any two points in the small ball are still in it). Can we find a homeomorphism such that the image of the small ball in $\mathbb{R}^n$ is also convex set? REPLY [5 votes]: Sometimes it is not possible. Let me describe a metric on the 2-disk $\mathbb{D}$ that is Riemannian everywhere except the center of $\mathbb{D}$. Make a sequence of holes in $\mathbb{D}$ that converge to the center. Attach a finger to each hole such that the length of fingers converge to zero, but it still much larger than the distance to the center. The construction could be made in such a way that in the obtained metric any small ball around the center has fingers that sticks out of it. In particular any such ball is not homeomorphic to the disk.<|endoftext|> TITLE: What is the generator of $\pi_9(S^2)$? QUESTION [18 upvotes]: This is more or less the same question as [ What is the generator of $\pi_9(S^3)$? ], except what I would like to know is if it is possible to describe this map in a way not only topologists can make sense of. [EDIT] (Following the advice of Ryan Budney.) A purely geometric construction like the famous Hopf fibration for $\pi_3(S^2)$ would be perfect. (Something like a map $S^9\to S^2$ or $S^9\to S^3$ which may be written down in equations.) But I understand that there is little hope for that. A less explicit but probably more reasonable approach is to try and represent this homotopy class by a framed 7-manifold in ${\mathbb R}^9$ following Pontryagin. In fact, any information about such a manifold may be of help. Is it really complicated? I am not really familiar with the work of Jie Wu, but what I have read this far makes sense to me. So, the answer can also be along this lines, but if so I would like to see more then hints. (This computation looks horrendous, and I probably cannot handle it by myself.) REPLY [8 votes]: I posted an answer to the previous question describing one way to view a generator of $\pi_9 S^3$. From the Hopf fibration $$S^1\to S^3 \overset{\eta}{\to} S^2$$ and fibration long exact sequence, we know that $$\pi_9 S^1 = 0 \to \pi_9 S^3 \overset{\eta}{\to} \pi_9 S^2\to \pi_8 S^1=0 $$ is exact, so $\pi_9 S^3 \cong \pi_9 S^2$ via the Hopf map $\eta$ (comment made by @skd above). I'm not sure if these answers are only understandable by topologists, but all the information is available in Hatcher's book Algebraic Topology (however, some of the information is not proved there, e.g. Bott periodicity).<|endoftext|> TITLE: Ample vector bundles and embeddings QUESTION [5 upvotes]: If $X$ is a complete variety over a field, a line bundle $L$ is said to be very ample if there is a closed immersion from $X$ into a projective space, such that the pullback of $\mathcal{O}(1)$ is isomorphic to $L$. Consequently, an ample line bundle can be defined as a line bundle such that some tensor power of it is very ample. There is also a notion of amplitude for vector bundles, which seems to have mostly been developed by Hartshorne. One definition that can be taken is that if $E$ is a vector bundle, then $E$ is ample if $\mathcal{O}(1)$ on $\mathbb{P}(E)$ is ample. This is the definition taken in Positivity in Algebraic Geometry II, by Robert Lazarsfeld, and in Example 6.1.6, he states that the tautological bundle on some Grassmanians is nef but not ample, and hence any provisional definition of ample vector bundles in terms of embeddings into Grassmanians is destined to fail. My questions are the following: What might be some conditions on $E$ so that it does give some embedding into a Grassmanian? Can this be geometrically realized as in the case of ample vector bundles, i.e., is there a notion of very ample vector bundle (which hopefully would include ample vector bundles) which do give embeddings? Given any coherent sheaf $F$ over $X$, we can form its symmetric algebra $S(F)$, and then form $\tilde X = \underline{\operatorname{Proj}}(S(E))$. Could then one define an ample coherent sheaf as above, where $\mathcal{O}(1)$ on $\tilde X$ is ample? Could one also generalize this to a notion of embeddings, i.e., very ample coherent sheaves? If the answer to the above is no; what about the special case of divisorial sheaves (reflexive, generically rank one)? Specifically I am interested in some notion which generalizes the canonical model of a smooth variety to that of Cohen-Macaulay varieties by somehow defining amplitude (in terms of embeddings) of the canonical (generalized) divisor $\omega_X=h^{-n}(\omega_{\overline X}^\bullet)|_X=\mathscr Ext^{N-n}_{\mathbb P^N}(\mathscr O_{\overline X},\omega_{\mathbb P^N})|_X$, where $X \subset \mathbb{P}^N$ is of dimension $n$. REPLY [4 votes]: The answer to your first question is yes. Gieseker (https://projecteuclid.org/download/pdf_1/euclid.nmj/1118798367) calls such bundles as strongly ample. ( see p 92 of above paper). To see that such bundles give embeddings into grassmanians (the universal subbundle on the grassmanian (or maybe its dual, depending on your convention) pulls back to it), you might have to look at the proof of Prop 2.1, p101 of the above paper. Strongly amples bundles are ample , Theorem 2.1, p102. Also look at p 172 of Fulton's https://link.springer.com/content/pdf/10.1007/BF01389960.pdf Its seems the answer to your second question is also yes. See Kubota, K., Ample sheaves, J. Fac. Sci. Univ. Tokyo Sect. I A Math<|endoftext|> TITLE: Calculate number of vertices adjacent to a clique, but not each other QUESTION [5 upvotes]: I have a network of N sensors and a test that tells me whether two sensor outputs are definitely not causally related. This allows me to construct a causality graph where each sensor is a vertex and two vertices share an edge if they have passed the test. Naturally, I'm looking for large cliques to see if the sensor network got anything of note. Causality graphs for one of the event types that I observe have an interesting property: they have one big clique and a set of vertices that are adjacent to the clique, but not each other. Like nodes 4 and 5 here: Is there an optimal way to count these vertices? Right now I simply look for one-element subtractions of large cliques, but there is probably a body of work on the subject. More generally, what kind of graph property am I dealing with and what terminology should I look up? REPLY [2 votes]: You want to find independent sets (or the maximal size of an independent set) in the subgraph induced by the common neighbors of the vertices of a clique. More specifically, if $K$ is a clique in the graph $G$ and $S$ is the set of vertices which are adjacent to each vertex of $K$, then any independent set of the induced subgraph $G[S]$ will give a set of vertices which are adjacent to vertices of the clique but not adjacent to each other. In the special case when the clique contains a single vertex $v$, the problem is to find the maximum size of an independent set in the induced subgraph $G[\Gamma(v)]$, where $\Gamma(v)$ is the set of neighbors in $G$ of vertex $v$. This quantity seems NP-hard to compute because given a graph $H$, the maximum size of an independent set of $H$ can be obtained by adding a new vertex $v$ and joining $v$ to each vertex of $H$, and then computing in this new graph the maximum number of vertices adjacent to $v$ which are not adjacent to each other. This invariant (for the $|K|=1$ case) reminds me of the induced star number or interference degree of a graph, and has appeared in recent literature in wireless adhoc networks - the worst-case performance of certain distributed algorithms in wireless sensor networks is a factor of exactly this quantity away from optimal; see https://arxiv.org/abs/1810.04109 and the references therein.<|endoftext|> TITLE: Lie Algebra of Automorphism Group of $\mathbb{P}_k^1$ QUESTION [7 upvotes]: Let $X$ be a scheme over an algebraically closed field $k$ and let $\operatorname{Aut}(X)$ denote the functor sending a $k$-scheme $T$ to the group $\operatorname{Aut}_T(X \times_k T)$ of automorphisms of $X \times_k T$ over $T$. My goal is to have a better grasp of the equality $\operatorname{Lie}(\operatorname{Aut}(X))= H^0(X, \mathcal{T} X)$. Therefore I am trying to work through the example where $X = \mathbb{P}_k^1$ so that $\operatorname{Aut}(X)= PGL(2,k)$. The global sections of $\mathcal{T} X$ are are of the form $a_0 \partial_z + a_1 z \partial_z+ a_2 z^2 \partial_z$ where $z=v/u$ is a choice of homogeneous coordinates on $X$. On the other hand, I know every $\phi \in \operatorname{Aut}_k(X)$ is given by the following map of $k$-algebras. $$ z \mapsto \frac{az+b}{cz + d}.$$ What is the identification between the global sections of $\mathcal{T} X$ and the $k$-algebra maps $z \mapsto \frac{az+b}{cz + d}$? Solving for the integral curve I end up with the equation $z'(t) = a_0 + a_1 z(t) + a_2z^2(t)$. If $a_0=0$, this would be a Bernoulli differential equation and I can solve it to find $z(t)= \frac{a_1 z_0 e^{a_1 t}}{a_1 -a_2 z_0 e^{a_1 t}}$. I think that this corresponds to the $k$-algebra map $z \mapsto (a_1 a^*) z /(a_1 - a_2 a^* z )$ where $a^* \in k^*$. This is close but not exactly right. However, the affine subset itself has automorphisms given by $z \mapsto \alpha + \beta z$. If I compose these maps with the maps $z \mapsto (a_1 a^*) z /(a_1 - a_2 a^* z )$ I get from integrating the tangent space I do get the Mobius tranformation. Is this the correct approach? REPLY [8 votes]: Elements of $\operatorname{Lie}(\operatorname{Aut}(X))$ are not $k$-algebra maps, but rather maps over the ring $k[\epsilon]/(\epsilon^2)$ that reduce to the identity $k$-algebra map under $\epsilon\mapsto 0$. For $X=\mathbb{P}^1$, such maps can be identified with elements of the kernel of $PGL_2(k[\epsilon]/(\epsilon^2))\to PGL_2(k)$, or in other words maps $$ z\mapsto \frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)}, $$ with $a$, $b$, $c$, $d\in k$. We can compute $$ \frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)}=\big((1+a\epsilon)z+b\epsilon\big)\big(1-d\epsilon-c\epsilon z\big)=z+\epsilon\big( b+(a-d)z-cz^2 \big), $$ and the vector field corresponding to the map above is $b\partial_z+(a-d)z\partial_z-cz^2\partial_z$.<|endoftext|> TITLE: Zero tensor product over a complex algebra? QUESTION [5 upvotes]: Let $A$ be an algebra over $\mathbb{C}$. Let $M$ be a left $A$-module, let $N$ be a right $A$-module and consider the tensor product $N \otimes_A M$, which is a complex vector space. Q1: Can this tensor product be the trivial vector spacefor non-zero $M$, $N$? If yes, what are examples where this happens? Q2: Are there criteria on $M$ and $N$ which ensure that the tensor product is non-zero? Edit: Thanks for the answers below. It seems that Q2 is too broad to answer in general. However, I am mostly interested in the case of an infinite-dimensional Clifford algebra; in other words, $A$ is central simple. REPLY [4 votes]: I give an answer in case the modules are finite dimensional (not 100% sure whether the algebra also has to be finite dimensional but I think it is not needed). The base field can be arbitrary and does not need to be $\mathbb{C}$. Let the algebra $A$ be over a field $K$ and $D=Hom_K(-,K)$. Then we have $N \otimes_A M \cong D Hom_A(N,D(M))$. This makes it very easy to at least calculate the vector space dimension of $N \otimes_A M$ and give alot of examples for Q1: Just take $M$ and $N$ simple such that $N$ is not isomorphic to $D(M)$ and by Schur's lemma $N \otimes_A M \cong D Hom_A(N,D(M))=0$. Example: Let $A$ be a finite dimensional algebra with $n$ fixed idempotent $e_1,...,e_n$ and simple right modules $S_1,...,S_n$ and simple left modules $G_1,...,G_n$ (corresponding to the idempotents ). Then $D(G_i) \cong S_i$ and thus $S_j \otimes_A G_i \neq 0$ if and only if $i=j$. At least when the algeba is a finite dimensional quiver algebra deciding when $D Hom_A(N,D(M))=0$ can be reduced to linear algebra and answers Q2.<|endoftext|> TITLE: Cube root of the $j$-invariant QUESTION [5 upvotes]: Let $$\Gamma=\bigg \lbrace \begin{pmatrix} a&b\\c&d\end{pmatrix}\in\Gamma(1):b\equiv c~(\text{mod }3)\text{ or } a\equiv d\equiv 0~(\text{mod }3)\bigg \rbrace.$$ Then $\Gamma$ has exactly one cusp and its hauptmodul is the modular function $g=j^{1/3}$ usually denoted by $\gamma_2$. Now pretend that we do not know that $ g^3=j $ and let $h$ be a Hauptmodul for $\Gamma$ which is holomorphic on the upper half-plane $\mathfrak H$. Now let's try and find a relation between $j$ and $h$ using the description of $\Gamma$. Since $j$ is holomorphic on $\mathfrak H$, there is a polynomial $P$ such that $j=P(h)$. Because the only zero of $j$ is at $\rho=e^{2\pi i /3}$ (modulo the action of $\Gamma(1)$),the coset representatives for $\Gamma$ in $\Gamma(1)$ are $I,T,T^2$, and $h$ has a simple pole at infinity (that is, its Fourier expansion begins with $q^{-1/3}$), we can write $$j=(g-g(\rho))(g-g(T\rho))(g-g(T^2\rho)).$$ On the other hand, $ST\rho= \rho$. Therefore, as $S\in \Gamma $, we have $g(T\rho)=g(\rho)$. However, I am at loss trying to prove that $g(\rho)=g(T^2\rho)$. What am I missing? REPLY [5 votes]: Because $S \in \Gamma$ and $\Gamma$ is a normal subgroup of $\Gamma(1)$, we may as well take the coset representatives to be $1, ST , (ST)^2$. Or alternately we have $\rho = ST ST \rho = S (TS T^{-1}) T^2 \rho$ and $S (TST^{-1}) \in \Gamma$ because $\Gamma$ is a normal subgroup (or by explicit calculation).<|endoftext|> TITLE: How to calculate the volume of a section of a convex body? QUESTION [6 upvotes]: The following is essentially a partial case for my previous question. Let $B\subset\mathbb{R}^m$ be the unit ball with respect to a concrete norm on $\mathbb{R}^m$, say $l^p$-norm, $p\in (1,\infty)$. Let $v_1,...,v_n\in \mathbb{R}^m$ be linearly independent. How to calculate the $n$-dimensional volume of $B\cap span\{v_1,...,v_n\}$? I need to express this volume through the coordinates of $v_1,...,v_n$, or perhaps through some distances between certain combinations of them. I know that there is extensive literature on related matters, but I hope that this specific question has a specific answer.. REPLY [5 votes]: See http://matwbn.icm.edu.pl/ksiazki/sm/sm88/sm8817.pdf and references therein to other papers by same author to see how to calculate these sections.<|endoftext|> TITLE: Isomorphism between the hyperbolic space and the manifold of SPD matrices with constant determinant QUESTION [5 upvotes]: I'm studying properties of the manifold of symmetric positive-definite (SPD) matrices and I've learnt about the following connection to the hyperbolic space [1, Section 2.2], $$\mathcal{P}(n) = \mathcal{SP}(n) \times \mathbb{R}^+,\qquad \mathcal{SP}(n) \cong \mathbb{H}^p,$$ where $\mathcal{P}(n)$ is the space of SPD matrices, $\mathcal{SP}(n) = \{ A \in \mathcal{P}(n) : \lvert A \rvert = 1 \}$, and $p = n (n + 1) / 2 - 1$. In words, $\mathcal{P}(n)$ is a foliated manifold whose codimension-one leaves are isomorphic to the hyperbolic space $\mathbb{H}^p$. I wanted to see a proof of this isomorphism (and what it means more precisely) and I found [2] and its follow up [3] which focus on $2 \times 2$ SPD matrices. They mention certain connections between the standard metrics in these spaces, e.g. [3, p4], \begin{align}\tag{1}\label{eq:1} d_{\mathcal{P}}(X_1,X_2) = \sqrt{\frac{1}{2} \big(\log \lvert X_1 \rvert - \log \lvert X_2 \rvert \big)^2 + d_{\mathbb{D}}^2(y_1,y_2)}, \end{align} where $d_{\mathcal{P}}$ is the canonical metric on $\mathcal{P}(n)$ (see [2, eq. (6)]), $d_{\mathbb{D}}(y_1,y_2)$ is the distance in the Poincaré disk between two points that "correspond" to the SPD matrices $X_1, X_2$ (more precisely, their scaled versions $\tilde{X}_1,\tilde{X}_2 \in \mathcal{SP}(n)$; see [2, p4-6]). That being said, I couldn't find a proof for the general case of $n \times n$ SPD matrices. Is anyone aware of other relevant resources? Or is it really obvious? Does a generalized form of \eqref{eq:1} still hold? Thank you. [1]: Moakher, M. (2005). A differential geometric approach to the geometric mean of symmetric positive-definite matrices. SIAM Journal on Matrix Analysis and Applications, 26(3), 735-747. [2]: Chossat, P., & Faugeras, O. (2009). Hyperbolic planforms in relation to visual edges and textures perception. PLoS Computational Biology, 5(12), e1000625. [3]: Faye, G., Chossat, P., & Faugeras, O. (2011). Analysis of a hyperbolic geometric model for visual texture perception. The Journal of Mathematical Neuroscience, 1(1), 4. REPLY [7 votes]: This 'isomorphism' does not hold for $n>2$, in the sense that the 'natural' $\mathrm{SL}(n,\mathbb{R})$-invariant metric on what you are calling $\mathcal{SP}(n)$ and the constant sectional curvature hyperbolic metric on $\mathbb{H}^p$ for $p = n(n{+}1)/2-1$ are only isometric when $n=1$ (the trivial case) and $n=2$. The reason is that $\mathcal{SP}(n) = \mathrm{SL}(n,\mathbb{R})/\mathrm{SO}(n)$ as symmetric spaces while $\mathbb{H}^p = \mathrm{SO}(p,1)/\mathrm{SO}(p)$ as symmetric spaces. In fact, for $n>2$, the symmetric space $\mathrm{SL}(n,\mathbb{R})/\mathrm{SO}(n)$ does not have constant sectional curvature, while hyperbolic space $\mathrm{SO}(p,1)/\mathrm{SO}(p)$ has constant sectional curvature for all $p$. When $n=2$, it's one of the 'accidental' isomorphisms that $\mathrm{SO}(2,1)$ is double-covered by $\mathrm{SL}(2,\mathbb{R})$. For $p>2$, the group $\mathrm{SO}(p,1)$ is not evenly-covered by any $\mathrm{SL}(n,\mathbb{R})$ for any $n$.<|endoftext|> TITLE: Iwasawa theory and perfectoid spaces QUESTION [10 upvotes]: Have there been any applications of perfectoid theory to Iwasawa theory? At a first glance, this seems like a natural choice. For instance, the field $\mathbb Q_p(\mu_p^{1/p^\infty})$ is studied in both theories (the class group in Iwasawa theory and as a natural example of a perfectoid field (possibly after completion)). Can we use the tilting correspondence to get information about class groups as in Iwasawa theory? REPLY [4 votes]: A flippant response is that people had the idea of using perfectoid theory in Iwasawa theory long before perfectoid theory even existed. What I'm referring to here is the work of Fontaine--Wintenberger, who studied the "field of norms" of a tower of p-adic fields, or the "tilt" as youngsters like you would call it, way back in the 1970's. Scholze's tilting equivalence generalises this to a much wider class of rings, of course, but for fields it's all there in Fontaine--Wintenberger. Fontaine--Wintenberger's construction of the field of norms is the starting point for the theory of $(\varphi, \Gamma)$-modules, and these are essential tools for handling Iwasawa theory for p-adic Galois representations; see e.g. Cherbonnier and Colmez's JAMS paper "Theorie d'Iwasawa des representations p-adiques d'un corps local".<|endoftext|> TITLE: Universal locally countable partial order QUESTION [16 upvotes]: Call a poset locally countable if the set of predecessors of every member of the poset is countable. Is the following consistent? There is no locally countable poset $P$ of size continuum such that every locally countable poset of size continuum embeds into $P$? Remark: under ZFC+CH, there is such a universal locally countable poset: the Turing degrees. This is due to Sacks. REPLY [7 votes]: The answer to this question, at least in ZFC, is "no." It is provable in ZFC that there is a universal locally countable partial order of size continuum. In other words, it is provable in ZFC that there is a locally countable partial order of size continuum into which every other such partial order embeds. This is true even if the embedding is required to be an initial segment embedding. I'm not sure what happens in ZF though. I will describe a locally countable partial order of size continuum and show that every locally countable partial order of size continuum has an initial segment embedding into it. (Edited to add: This example has been known for a while but I'm not sure if it's written up anywhere. As pointed out in a comment below, it is mentioned in the paper "Independence Results from ZFC in Computability Theory: Some Open Problems" by Marcia Groszek.) Definition of the partial order: The partial order (one of several that works) is basically the "hereditarily a column of" partial order on reals. For convenience (and to make sure we can actually get an initial segment embedding), I will actually use a slight variation on this partial order. To define the partial order, first pick your favorite bijection $f \colon \omega\times \omega \to \omega$. This lets us think of subsets of $\omega$ as being subsets of $\omega \times \omega$ and justifies us when we talk about the "columns" of a real $x \in 2^\omega$: the $n^\text{th}$ column of $x$ is the set $\{m \mid f(n, m) \in x\} \subseteq \omega$. It will be convenient for us to assume that $f$ has the property that for all $(n, m)$, $m < f(n, m)$ (which is easy to arrange). We will also assume that $f(0, 0) = 0$. Next, define the "is a column of" relation, $<_{col}$ on $2^\omega$. This is the relation such that $x <_{col} y$ holds if there is some $n$ such that $x$ is the $n^\text{th}$ column of $y$. We will modify this in two ways: first, we'll require that $n > 0$. So we basically ignore column $0$ of $y$ when deciding if $x$ is a column of $y$. Second, we will not call $x$ a column of $y$ if $x$ is the empty set (i.e. the all zeros binary sequence). This basically gives us a way to ignore columns, so we can have reals which have no columns (or rather, reals which we pretend have no columns). Now define the "hereditarily a column of" partial order, $<_{hcol}$, to be the transitive closure of $<_{col}$. In other words $x <_{hcol} y$ if there is a sequence $x = x_0 <_{col} x_1 <_{col} \ldots <_{col} x_k = y$. There is one problem: this relation is not necessarily irreflexive. We can fix this by simply deleting all elements $x$ such that $x <_{hcol} x$ (note that this means we automatically will ignore the all zeros sequence, so we didn't really need to specify that in the previous step). Let's call the resulting partial order $(P, <_{hcol})$. This partial order is locally countable because it is a subset of the transitive closure of a binary relation with countable sections (namely, $<_{col}$). Hopefully it is obvious that it has size continuum. Initial segment embedding: Let $(Q, <_Q)$ be a locally countable partial order of size continuum. We will show how to embed $Q$ as an initial segment of $P$. First, let $g \colon Q \to 2^\omega$ be an injection (which we can find because $Q$ has size continuum). Also, for each element $x$ of $Q$, pick an enumeration of its predecessors in $Q$, $x_1,x_2,\ldots$ (this is the only part where we need to use choice). Now define the embedding $\psi : Q \to P$ as follows. For each $x$, let $\psi(x)$ be the real whose $0^\text{th}$ column is $g(x)$ and whose $n^\text{th}$ column is $\psi(x_n)$ (recall that $x_n$ is the $n^\text{th}$ predecessor of $x$ in $Q$). If $x$ has only finitely many predecessors then fill the remaining columns with all zeros. This definition of $\psi$ probably looks circular. It's actually not! The point is that to decide the $f(n, m)^\text{th}$ bit of $\psi(x)$ we need to know the $m^\text{th}$ bit of $\psi(x_n)$. Since $m < f(n, m)$, this process is actually well-defined. Here's another way to think about the definition of $\psi$. Imagine that every $\psi(x)$ is initially a sequence of blanks (rather than a binary sequence). We will fill in these blanks in a series of steps. In the first step, we fill in the bits in column $0$ of each $\psi(x)$, as well as the bits of columns that are supposed to be all zeros. In each subsequent step, we look at each pair $x <_Q y$ and fill in bits of $y$ corresponding to bits of $x$ that were filled in on the previous step. One can show by induction (using the special properties of $f$ that we assumed) that by the end of step $n$, the $n^\text{th}$ bit of every $\psi(x)$ has been filled in. Note first that $\psi$ is an injective map from $Q$ into $2^\omega$ (because $g$ is injective and column $0$ of $\psi(x)$ is $g(x)$). Next, the columns of $\psi(x)$ are exactly the images of the predecessors of $x$ (except for the special column $0$ and the all zeros columns, both of which are ignored in the definition of $<_{col}$). This also shows that the "hereditary columns" of $\psi(x)$ are exactly the images of the predecessors of $x$. Since $\psi$ is injective, this shows that we never have $\psi(x) <_{hcol} \psi(x)$ and thus the image of $\psi$ actually lands in the partial order $P$. It also shows that $\psi$ is truly an initial segment embedding of $Q$ into $P$. Note: I believe the arithmetic degrees are also a universal locally countable partial order of size continuum but I don't think it's written up anywhere. The proof is an amalgamation of the proof I gave above and the counterexample to Martin's conjecture on the arithmetic degrees proved by Slaman and Steel and written up in this paper. I'm not sure if you can always get an initial segment embedding into the arithmetic degrees though.<|endoftext|> TITLE: Trace of inverse of random positive-definite matrix in high dimension? QUESTION [7 upvotes]: Consider a random matrix $A \in \mathbb{R}^{n\times n}$ with i.i.d. entries, with symmetric law and finite variance. I am curious about the behavior of $$\mathrm{Tr}( (A^T A + \lambda \mathrm{Id})^{-1})$$ when the size of the matrix $n$ goes to infinity. Here, $\lambda > 0$ is fixed and ensures that $A^T A + \lambda \mathrm{Id}$ is invertible as a positive definite matrix. Typically, I am wondering if this quantity behaves asymptotically like $n^{\gamma}$ for some $\gamma$. REPLY [7 votes]: If the elements of the $n\times n$ matrix $A$ are independent identically distributed random variables with mean 0 and variance $\sigma^2$, the $n$ eigenvalues $x_i$ of $A^{\rm T}A$ have for large $n$ the Marchenko-Pastur distribution $$P(x)=\frac{ \sqrt{4 n\sigma^2-x}}{2 \pi n \sigma^2 \sqrt{x}},\;\;0 TITLE: How to prove this inequality of Karamata type? QUESTION [9 upvotes]: Question 1: Let $x_{i}>0$, ($i=1,2,\cdots,n$) and such that $$x_{1}+x_{2}+\cdots+x_{n}=\pi.$$ Show that $$ \dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\le\left(\dfrac{\sin{\frac{\pi}{n}}}{\sin{\frac{2\pi}{n}}}\right)^n $$ Question 2 (may not hold): if $f''(x)\le 0,x\in I$, can we prove the following inequality? $$ \begin{split} f(x_{1}+x_{2})+&f(x_{2}+x_{3})+\ldots+f(x_{n}+x_{1})+nf\left(\dfrac{x_{1}+x_{2}+\ldots+x_{n}}{n}\right)\\ &\ge f(x_{1})+f(x_{2})+\ldots+f(x_{n})+nf\left(\dfrac{2(x_{1}+x_{2}+\ldots+x_{n})}{n}\right), \end{split} $$ where $x_{i}\in I$, $i=1,2,3\ldots,n$. I tried everything, but failed. As an example of Question 2, consider $f(x)=\ln{\sin{x}}$, $00$ and $\alpha_k\in(0,\frac 12)$ are some appropriate numbers. The rest is trivial. If you want more analytic flavor, just iterate this inequality like crazy until everything except $\prod_1$ wears out on the RHS, but you can also do it by completely elementary means (you have some fancy kind of log-concavity here). The upshot is that you get an inequality $\prod_k\ge C_k\prod_1$ in which the regular $n$-gon produces an identity. Now just take $k=2$. I'm answering here on MO because the MSE thread is cluttered enough already, but your best bet for asking such stuff is, probably, AoPS.<|endoftext|> TITLE: Arranging all permutations on $\{1,\ldots,n\}$ such that there are no common points QUESTION [8 upvotes]: If $n>0$ is an integer, let $[n]=\{1,\ldots,n\}$. Let $S_n$ denote the set of all permutations (bijections) $\pi:[n]\to[n]$. For which positive integers $n$ is there a bijection $\Phi:[n!]\to S_n$ such that for all $x\in[n!-1]$ we have $ (\Phi(x))(k) \neq (\Phi(x+1))(k) \text{ for all }k\in [n]$? Note. This can be formulated in the language of Hamiltonian paths: Put a graph structure on $S_n$ saying that $\pi_1,\pi_2\in S_n$ form an edge if $\pi_1(k)\neq \pi_2(k)$ for all $k\in[n]$, and find a Hamiltonian path in this graph. REPLY [11 votes]: it is still of interest to me, I would be glad to hear about it! OK, sorry for the delay. This elementary argument is, probably, well-known but I was too lazy to make a thorough search, so if somebody has a reference to something similar, it will be appreciated. Consider all cyclic shifts of a permutation $\pi$. Note that any two of them can be placed next to each other, so we can try to make them single blocks. Let's denote the corresponding blocks by $C$ (there are $(n-1)!$ of them). Now two blocks $C'$ and $C''$ are possible to place next to each other if the corresponding permutations $\pi'$ and $\pi''$ have the property that some cyclic shift of $\pi''$ has no common elements with $\pi'$. When we try to consider all cyclic shifts of $\pi''$, each element of $\pi'$ has only one chance to coincide with the element of the shift of $\pi''$ in the same position, so we have total of $n$ coincidences. If some shift of $\pi''$ has 2 or more common elements with $\pi'$, then there is another shift with no common elements. Now, shift both $\pi'$ and $\pi''$ so that $1$ is in the same position. If no other elements coincide, we have a derangement of length $n-1$. Thus, only about $(n-1)!/e$ blocks are incompatible with a given block, so we can make a Hamiltonian cycle of blocks such that every 2 adjacent blocks are compatible (with accurate count, it works for $n\ge 4$). It remains to place permutations without common elements at the junction points. Note that once we have $\pi'\in C'$ and $\pi''\in C''$ without common elements, any simultaneous cyclic shift of those will give you another pair without common elements. Thus we have $\ge n$ pairs that we can place at any particular junction. Just place them one by one in any order and notice that when you do any junction, you can have at most 2 conflicts with 2 adjacent junctions, so if $n\ge 3$, the Hamiltonian cycle of blocks can be upgraded to a Hamiltonian cycle of permutations.<|endoftext|> TITLE: Italian-style algebraic geometry in homotopy theory? QUESTION [8 upvotes]: In homotopy theory, stacks can be occasionally useful (i.e. in the chromatic story). I come from a differential geometry background, so when people say that algebraic geometry is useful in homotopy theory, I have mixed feelings (like the stack classifying formal groups of height $n$ is basically a quotient of a single point, not a particularly geometric object from some perspectives). Has Italian-style algebraic geometry ever shed light on homotopy theory? A caricaturistic (and probably wrong) example of how an an answer should look like: "27 lines on a cubic surface are actually in bijection with this stable homotopy group of spheres!" P.S. "Italian-style" can be understood in many ways; one possible interpretation is "the study of non-trivial facts about separated schemes of finite type over $\mathbb{C}$." REPLY [2 votes]: This is just a long comment. Homotopy theory is a rather broad field, so the answer to your question depends on what part of homotopy theory you'd like to see having interactions with "Italian-style" algebraic geometry. I assume that there are more interactions of the sort you desired in motivic homotopy theory (particularly given your definition of "Italian-style" algebraic geometry), e.g., results relating $\mathbf{A}^1$-homotopy theory to birational geometry. I know very little about this area, though, so hopefully somebody with more expertise could elaborate on this. If I recall correctly, Wickelgren gave some lectures about this topic at the Arizona Winter School this year, but their website is down at the moment, so I can't link to the notes. In chromatic homotopy theory (which you brought up with the example of the moduli of formal groups), however, the situation is different. (Denis brought up the example of elliptic curves, but you said that didn't count.) The reasoning behind this is partly due to the fact that I haven't seen such interactions in the literature, and partly philosophical: most interesting algebro-geometric phenomena in chromatic homotopy theory stem from stacky phenomena. For instance: the $E_2$-page of the homotopy fixed points spectral sequence calculating $\pi_\ast KO$ is exactly the cohomology of the classifying stack $B\mathbf{Z}/2$; the $E_2$-page of the Adams spectral sequence is the cohomology of the classifying stack $B\mathrm{Aut}(\widehat{\mathbf{G}}_a)$. Already at the level of $E_2$-pages (the "purely algebraic" input into homotopy theory), one might say that interesting homotopy theoretic phenomena wouldn't be visible without the introduction of stacks (over $\mathbf{Z}$). (This is obviously an overblown statement, because topologists did calculations way before stacks were introduced into homotopy theory.)<|endoftext|> TITLE: Knight's tour problem QUESTION [8 upvotes]: It is known that on an infinite board, if all squares of the form $(ki,kj)$ are removed, $k$ even, $i,j\in\mathbf{Z}$, then there is no knight's tour due to unbalanced black and white squares. My questions are the following: If $k$ is odd, does there exist a knight's tour? For example, $k=3$. If finitely many black squares are to be removed, can there possibly exist a knight's tour? Any results on these problems? Thanks. REPLY [18 votes]: $k=3$ and $k=5$ are certainly possible, and I believe the procedure for $k=5$ can be extended to other odd $k$ as well. Here is a picture of $k=3$: (click to enlarge; Java source code to generate this SVG file is here on GitHub) (click to enlarge; Java source code to generate this SVG file is here on GitHub) The procedure for $k=3$ is as follows: divide the board into 3-by-3 boxes, centered around the removed squares. There are eight ways for the knight to enter a box, and once it's in, it will make a complete tour in the box and end on one of the two other squares a knight's move away, so there are 16 possibilities labelled 0 through F: (green = starting square, red = finishing square) From one box, it can then jump to another box. The idea is to make a spiral of these boxes, centered at the origin. Therefore, we need to figure out for each box which other boxes can be next, depending on which direction we want to go. For example, from box 0 (and box D), we can't go left (we would end up in the box itself), we can go up to box A or B, down to box 2 or 3 and right to boxes 2, 3, A and B. The table is as follows: Note that it's possible to go right by alternating between E and C, go down by F and 1, go left by 5 and 3 and go up by 9 and 7. This is the key to fill the spiral, which will look like this: (For the first image, I ended up with C and F instead of 0 and 2 in the center.) Since the length of the spiral's horizontal sides are always even, and odd for the vertical sides, this pattern can be extended indefinitely. For $k \ge 5$, it makes more sense to leave the removed square in the corner. I had a computer program (linked above) brute-force all possible knights tours inside a $k$ by $k$ box with the top left square removed; then, it's a matter of finding the right 'connections' between the boxes like I did for the $k=3$ case.<|endoftext|> TITLE: Rationally connected Kähler manifolds are projective QUESTION [7 upvotes]: I would like to find a proof for Remark 0.5 in the following article of Claire Voisin: https://webusers.imj-prg.fr/~claire.voisin/Articlesweb/fanosymp.pdf She writes in this remark the following: Remark 0.5 A compact Kähler manifold $X$ which is rationally connected satisfies $H^2(X, {\cal O}_X) = 0$, hence is projective. I understand that a Kähler manifold with $H^2(X, {\cal O}_X) = 0$ is projective. However, I don't understand why a Kähler manifold that is rationally connected has $H^2(X, {\cal O}_X) = 0$. Indeed, the definition for rational connectedness that Voisin is using is the following: Definition 0.3 A compact Kähler manifold $X$ is rationally connected if for any two points $x, y\in X$, there exists a (maybe singular) rational curve $C\subset X$ with the property that $x\in C$, $y\in C$. So my question is the following: How to prove that $H^2(X, {\cal O}_X)$ for a compact Kähler manifold $X$ that satisfies Definition 0.3? Is this easy/hard/well-known? PS. As Donu Arapura correctly says below the vanishing of $H^2(X, {\cal O}_X)$ for rationally connected projective manifolds is a classical fact. However I want a proof of such a vanishing for Kähler manifolds (to show that they are projective). So I want to know if this vanishing is a well known fact or a couple of pages are needed to prove it? REPLY [4 votes]: This result follows from Corollaire (p.212) of this paper by F. Campana, Coréduction algébrique d'un espace analytique faiblement kählérien compact, Invent. Math. 63 (1981), no. 2, 187–223. I had the same problem of finiding a citable reference for this result some time ago, and this one did the job.<|endoftext|> TITLE: Orientable with respect to complex cobordism? QUESTION [11 upvotes]: I have learned that an orientation of a manifold $M$ with respect to ordinary cohomology is an ordinary orientation, that an orientation with respect to complex K-theory is a Spin$^c$ structure, and that an orientation with respect to real K-theory is a spin structure. I think this is a very beautiful picture and I am wondering if orientations with respect to other theories like elliptic cohomology, G-equivariant cohomology, quaternionic K-theory, or spin cobordism correspond to interesting and well-studied differential-geometric structures. Complex manifolds should be oriented with respect to any complex-oriented cohomology theory. Indeed, if $E$ is a complex-oriented cohomology theory then all complex vector bundles carry $E$-orientations. In particular, if $X$ is a complex manifold then its tangent bundle $TX$ has a complex structure making $X$ an $E$-oriented manifold. Given that complex cobordism is the universal complex-oriented cohomology theory, I would guess that an orientation with respect to complex cobordism is a complex structure. I have been unable to find any literature on this and I am unsure how to approach the problem rigorously on my own. Maybe someone knows? REPLY [14 votes]: Let me expand a bit on my comments. If $E$ is a nice enough ring spectrum (e.g. an $\mathbb{E}_2$-ring spectrum; there is also a slightly modified version that works for an $\mathbb{E}_1$-ring) then the story of orientations work like this: If you have a vector bundle, or, more generally, a stable spherical fibration (of rank 0, say) on a space $X$, this will be classified by a map $X \to \mathrm{BGL}_1(S^0)$ where $\mathrm{GL}_1(S^0)$ is the space of self-equivalences of the sphere spectrum. Let $\mathrm{GL}_1(E)$ denote the union of those components of $\Omega^{\infty}E$ corresponding to units in $\pi_0\Omega^{\infty}E = \pi_0E$. An $E$-orientation is a nullhomotopy of the composite $X \to \mathrm{BGL}_1(S^0) \to \mathrm{BGL}_1(E)$. (Again, there are some slight modifications if $E$ is less nice; there are also definitions one can make without anything more than a homotopy ring structure on $E$, but that is a slightly less intuitive picture I think.) To summarize in a more informal way: if you have a vector bundle $V$ of rank $n$ on $X$, you can form the corresponding stable spherical fibration which, intuitively, means you associate to each point of $x$ the spectrum $\Sigma^{-n}S^{V_x}$; to each path $x \to y$ in $X$ you get an equivalence $\Sigma^{-n}S^{V_{x}} \to \Sigma^{-n}S^{V_y}$; a homotopy of paths gives a homotopy between equivalences, etc. etc. This spells out a map $X \to \mathrm{BGL}_1(S^0)$. This is a local system of spectra which all look like $S^0$ up to equivalence. You can fiberwise smash with $E$ to get a local system of spectra that all look like $E$ up to equivalence, and that gives you the map $X\to \mathrm{BGL}_1(E)$. An $E$-orientation is a trivialization of this local system. It's saying that, through the eyes of $E$, the bundle looks like the trivial bundle (whence the Thom isomorphism: the Thom spectrum looks like a suspension (i.e. the Thom spectrum of a trivial bundle) through the eyes of $E$). Now, if you want to $E$-orient all spin bundles, or spin-c bundles, etc., then you just have to do the above in the universal case, when $X$ is the classifying space for such bundles. So, for example, to show that every oriented vector bundle is $\mathrm{H}\mathbb{Z}$-oriented, we consider the map $\mathrm{BSO} \to \mathrm{BGL}_1(\mathrm{H}\mathbb{Z})$. This factors as $\mathrm{BSO} \to \mathrm{BO} \to \mathrm{BGL}_1(\mathrm{H}\mathbb{Z})$. But $\mathrm{GL}_1(\mathrm{H}\mathbb{Z}) = \mathbb{Z}/2=O(1)$, the discrete group, so its classifying space is $\mathrm{BO}(1)$ and the sequence $\mathrm{BSO} \to \mathrm{BO} \to \mathrm{BO}(1)$ is the defining sequence for $\mathrm{BSO}$. In other words: not only is it the case that every oriented bundle is $\mathrm{H}\mathbb{Z}$-oriented, but the converse also holds because a nullhomotopy of the composite $X \to \mathrm{BO} \to \mathrm{BO}(1)$ is exactly the data of an orientation. But this is a happy accident. For example, it is not the case that we have fiber sequences $\mathrm{BSpin} \to \mathrm{BO} \to \mathrm{BGL}_1(\mathrm{KO})$, nor do we have fiber sequences $\mathrm{BU} \to \mathrm{BO} \to \mathrm{BGL}_1(\mathrm{MU})$. Instead, in each instance the composite has a nullhomotopy (which is, in the first case, a difficult theorem of Atiyah-Bott-Shapiro, and, in the latter case, sort of tautological) but the first term is not the fiber of the second map. Some added stuff in response to the OP and Mark: Suppose you've got some classifying space for vector bundles with extra structure, $\mathrm{BG}$, and you provide a nullhomotopy for $\mathrm{BG} \to \mathrm{BO} \to \mathrm{BGL}_1(E)$. This buys you a map $\mathrm{BG} \to \mathrm{GL}_1(E)/\mathrm{O}$ (where you should interpret this symbol carefully- really it's just notation for the fiber of $\mathrm{BO} \to \mathrm{BGL}_1(E)$ but you can realize it as a sort of homotopy quotient if you want). The failure of this map to be an equivalence will be the failure of "$E$-orientation" to be the same as "$G$-structure". As an explicit example, let's consider the difference between $U$-structures and $\mathrm{MU}$-orientations. The nontrivial map $S^9 \to \mathrm{BO}$ certainly doesn't lift to $\mathrm{BU}$ (since $\pi_9\mathrm{BU} = 0$), but it does become nullhomotopic in $\pi_9\mathrm{BGL}_1(\mathrm{MU})$. Indeed, every bundle on every sphere $S^n$ for $n>1$ is $\mathrm{MU}$-orientable because the map $\pi_n\mathrm{BGL}_1(S^0) \to \pi_n\mathrm{BGL}_1(\mathrm{MU})$ is trivial for $n>1$ (since the source is torsion and the target is torsion-free when $n>1$).<|endoftext|> TITLE: Reference request: When is the variance in the central limit theorem for Markov chains positive? QUESTION [5 upvotes]: I'm looking for a reference which gives sufficient conditions for the variance to be positive in the central limit theorem for Markov chains (cf https://en.wikipedia.org/wiki/Markov_chain_central_limit_theorem). In other words, using the notation on the wikipedia page, I'm looking for a sufficient condition for $\sqrt{n}(\hat\mu_n-\mu)$ to legitimately be a random variable in the limit. In my situation, the Markov chain is simply a random walk on a finite, aperiodic, strongly-connected graph, so a result relating to this simpler situation would be good enough if nothing more general exists. Also, in my situation, actually attempting to estimate $\sigma^2$ via the formula is pretty much out of the question. I beleive that I can argue that the variance will be positive if there are two directed cycles $v_0\to v_2\to\dots\to v_n=v_0$ and $u_0\to u_2\to\dots\to u_m=u_0$ with ${1\over n}\sum_{t=1}^n g(v_t)\neq{1\over m}\sum_{t=1}^m g(u_t)$. Intuitively, this is because the chain will always have a chance to move around either of these cycles and these cycles contribute different values to the sum. I'm not looking for anyone to write down a proof since, if necessary, I can likely prove it myself. However, I would really like to avoid having to spend the necessary pages to do so in a paper I'm writing. It seems very likely that such a result must have appeared in either a paper or book somewhere, but I haven't had any luck in hunting down a reference so far. REPLY [4 votes]: Fluctuations in Markov processes by Komorowski, Landim and Olla has more than you could ever hope for in this direction. Also, the Courant lecture notes of Varadhan have, if I recall correctly, a chapter in this direction.<|endoftext|> TITLE: Intuition behind the growth condition in the result of Griffin, Ono, Rolen and Zagier on Jensen polynomials QUESTION [5 upvotes]: With great pleasure I read the recent paper of Griffin, Ono, Rolen and Zagier proving the surprising result that the Jensen polynomials $J^{d, n}_\alpha$ for a sequence $\alpha = \{\alpha(0), \alpha(1), \ldots \}$ of real numbers whose growth (?) is controlled in a certain way converges for fixed $d$ to a limiting polynomial of the same degree uniformly on compact subsets of $\mathbb{R}$. Their main application is to some sequence of real numbers coming from the Riemann Xi function (see also this other MO question) but I already had lots of fun trying to see how this works out for much simpler sequences such as $\alpha(n) = 1$ or $\alpha(n) = 2^n$. My question is however with the conditions in their main non-RH-related result: theorem 8 which reads: Suppose that $\{E(n)\}$ and $\{\delta(n)\}$ are positive real sequences with $\delta(n)$ tending to $0$, and that $F(t) = \sum_{i =1}^\infty c_i t^i$ is a formal power series with complex coefficients. For a fixed $d \geq 1$, suppose that there are real sequences $\{C_0(n)\},\ldots,\{C_d(n)\}$, with $\lim_{n \to \infty} C_i(n) = c_i$ for $0 \leq i \leq d$, such that for $0 \leq j \leq d$, we have $$\frac{\alpha(n+j)}{\alpha(n)} E(n)^{-j} = \sum_{i =0}^d C_i(n) \delta(n)^i j^i + o(\delta(n)^d) \qquad (*)$$ as $n \to \infty$. Then we have: $$\lim_{n \to \infty} \frac{\delta(n)^{-d}}{\alpha(n)} J^{d, n}_\alpha \left(\frac{\delta(n)X - 1}{E(n)}\right) = H_{F, d}$$ uniformly on compact subsets of $\mathbb{R}$ where $H_{F, d}$ is defined by the generating function $F(−t) e^{Xt}=\sum_{m=0}^\infty H_{F,m}(X) \frac{t^m}{m!}$. My question is about (*). Hopefully it is clear why I wrote above that I already had fun seeing what this theorem means even for really simple sequences $\alpha$: it is a priori not at all clear what $\delta, E$ or $C_i$ to take and one surprising thing I found is that (unlike their limits $c_i$) the sequences $C_i$ may depend non-trivially on the choice of the fixed value of $d$ even in cases we know a priori that the the limits exist for all $d$. But managing to find sequence $E, \delta, C_i$ that 'work' is something quite different from understanding what is going on. My question is: what is, intuitively speaking, the set of conditions (*) trying to convey? Is it saying that the sequence $\alpha$ cannot grow too fast? Something else? Is it reasonable to think of finite sum on the right hand side as 'roughly a constant' so that the condition says that $\alpha$ grows more or less as $E^j$ where $E$ is the 'typical' value of $E(n)$. Ugh, as soon as I type it it stops making sense. Any enlightenment is welcome here. REPLY [5 votes]: The condition does restrict the rate of growth of the functions considered, and $E$, $C_i$, and $\delta$ do have a certain amount of freedom, however they are meant to encode specific information. The $E$ term is meant to account for the exponential part of the growth of $\alpha(n)$. Once $E$ has been fixed, the right hand side of (*) still has a certain amount of freedom due to the little-oh term, but notice that as $d$ becomes large, the $C_i(n)\delta(n)^i$ terms are forced to approximate the coefficients of a series expansion in $j$ of the left hand side. The $\delta$ term should account for the residual decay of the $C_i(n)$, so that at least some of the $C_i(n)$ have non-zero limits as $n\to \infty$. In fact in the example in the paper, $\delta(n)^2$ essentially measures the log concavity of the LHS. Once $\delta(n)$ and $E(n)$ have been fixed, there is still some freedom in choosing the $C_i(n)$ functions, however they are always approximations of the series expansion of the LHS, with more freedom allowed for small $d$ and less for large $d$. The theorem itself is not very instructive in how to find these numbers, but the proof demonstrates a nice method in the case that the $\alpha(n)$ are values of a sufficiently smooth function. The first thing we do is expand $\log(\frac{\alpha(n+j)}{\alpha(n)})$ as a power series in $j$. I'll walk through a couple examples in a moment, but for now, let's assume we have such an expansion, so that $$\log\left(\frac{\alpha(n+j)}{\alpha(n)}\right)=g_1(n)\cdot j+g_2(n)\cdot j^2+\dots,$$ with $g_i(n)\to 0$ as $n\to \infty$ for $i\geq 2$. We'll take $E(n)=\exp(g_1(n)),$ which is the primary exponential contribution. The numbers $\delta(n)$ should be chosen to be positive numbers which decay like the slowest of $\sqrt[i]{|g_i(n)|}$. In the cases considered in the paper, we took $\delta(n)=\sqrt{-g_2(n)}$, for $n\geq 6$. The fact that $g_2(n)<0$ for large $n$ is connected to the log-concavity of these sequences. The two examples you gave, $\alpha(n)=1$ and $\alpha(n)=2^n$ will be the same except for the $E(n)$ term. In the first case we take $E(n)=1$ and in the second we take $E(n)=2^n.$ At this point it doesn't matter what we take $\delta(n)$ to be. As long as they are non-zero, the $C_i(n)$ must all be $0$. The function $F(t)=1$, and the renormalized polynomial will be (no limit needed) $X^d$, which are generated as desired by $$F(-t)e^{Xt}=e^{Xt}.$$ We get a more interesting example when we consider $\alpha(n)=\frac{k^n}{n}$ for non-zero $k$. Different choices of $k$ only result in different $E$ terms, so for simplicity, lets assume $k=1$, so $\alpha(n)=1/n$. As before, in order to find $E(n)$, we consider $$\log\left(\frac{\alpha(n+j)}{\alpha(j)}\right)=\log\left(\frac{1}{1+j/n}\right)=-\frac{j}{n}+\frac{j^2}{2n^2}-\frac{j^3}{3n^3}\dots,$$ so we take $E(n)=e^{-1/n}.$ For $\delta(n)$, we could take $\delta(n)=\sqrt{|g_2(n)|}=\frac{1}{\sqrt{2}n},$ but it works just as well (and we get simpler expressions) if we take $\delta(n)=\frac{1}{n}.$ Then we have $$F(t)=\frac{1}{1+t}e^{t}.$$ This gives us $$F(-t)\exp(tX)=1+\frac{X}{1!}t+\frac{X^2+1}{2!}t^2+\frac{X^3+3X+2}{3!}t^3+\dots.$$ If we calculate the degree $d=3$ re-normalized polynomial for $n=1000$ using these choices of $E(n)$ and $\delta(n)$, we get $$\sim 1.003X^3-.007X^2+2.997X+1.992,$$ which matches with our expected limit of $X^3+3X+2$.<|endoftext|> TITLE: Computing the differentials in the Adams spectral sequence QUESTION [12 upvotes]: Assume you are given an explicit presentation of the $E_2$-terms of the Adams spectral sequence. Are the differentials on $E_2$ and further algorithmically computable? I do not care how practical it is, just if it can be done. I have heard that the homotopy groups of finite CW complexes are algorithmically computable. REPLY [12 votes]: In principle, everything is algorithmically computable, but the proof does not lead to practical algorithms. In practice, you can find some differentials by ad hoc means and then deduce many more differentials. You can get some of these by just using the ring structure and the Leibniz property, but then you can get much further by using algebraic Steenrod operations. Bob Bruner has lots of relevant code at http://www.rrb.wayne.edu/papers/#code. REPLY [8 votes]: The work of Baues and collaborators on the secondary Steenrod algebra gives a category in which Ext is the $E_3$-term of the Adams spectral sequence. This must in some sense decide the $d_2$ differential, especially if there is a clear relation between Baues' E_3 = secondary-Ext and the usual $E_2$-term. I am sorry that I can't be more explicit off the top of my head. As for Steenrod operation differentials, they only apply in $Ext(H^*R,F_p) \Longrightarrow \pi_*R$ for an $H_\infty$-ring spectrum $R$. Fortunately, the sphere spectrum is $H_\infty$ and perhaps this is what you care about. For such spectra, at $p=2$, the 'generic' differential is $d_2(\cup_i(x)) = h_0\cup_{i-1}(x)$ if $i \equiv n \pmod{2}$, and similarly at odd $p$. Unfortunately, 'most' elements are not in the image of a $\cup_i$ operation. The computability result you have heard about is probably that of Ed Brown: MR0083733 Brown, Edgar H., Jr. Finite computability of Postnikov complexes. Ann. of Math. (2) 65 (1957), 1--20. As Neil says, it shows that the homotopy groups of a finite simplicial complex are effectively computable, but the result is not a practical algorithm. It also does not directly say anything about the Adams spectral sequence, which is a rather different approach to those homotopy groups.<|endoftext|> TITLE: Is differential topology a dying field? QUESTION [16 upvotes]: I recently had a post doc in differential topology advise against me going into the field, since it seems to be dying in his words. Is this true? I do see very little activity on differential topology here on MO, and it has been hard for me to find recent references in the field. I do not mean to offend anyone who works in the field with this, I do love what I’ve seen of the field a lot in fact. But I am a little concerned about this. Any feedback would be appreciated, thanks! REPLY [40 votes]: I don't think differential topology is a dying field. I'll interpret this as the classification of smooth manifolds and, more broadly, maps between them (immersions, embeddings, diffeomorphism groups). Also, I'll restrict to the finite-dimensional case. There are related topics which are very active, usually studying smooth manifolds with extra structure, e.g. exterior differential systems, foliations and contact structures, symplectic and Riemannian geometry. I won't comment much on these areas. The classification of smooth manifolds was quite successful in the 60s with the h- and s-cobordism theorems framing many classification problems in terms of surgery problems. The classification of exotic spheres was more-or-less reduced to problems in homotopy theory, the stable homotopy groups and Kervaire invariant problems. The study of these invariants is still active, but the techniques are more algebraic. Moreover, there is still an industry of studying Riemannian metrics on exotic spheres. Maybe one of the biggest open problems now in differential topology is the cobordism hypothesis, originally formulated by Baez-Dolan, but reformulated by Lurie. This is formulated as a classification of "fully extended topological field theories" in terms of $(\infty,n)$-categories. His sketch of proof is regarded as incomplete, and a few groups are trying to fill in the details. From discussions I've had with experts, a big issue here is foundational results in differential topology. Lurie's outline relies on results about manifolds with corners, and I think that Schommer-Pries has filled in some details, but I think that the proof of the cobordism hypothesis is still incomplete. Another (very special) problem that has received some attention is the Hirzebruch Prize Question: Does there exist a 24-dimensional compact, orientable, differentiable manifold $X$ (admitting the action of the Monster group) with $p_1(X) = 0$, $w_2(X) = 0, \hat{A}(X) = 1$, and $\hat{A}(X, T_C) = 0$? Here $\hat{A}$ is the A-hat genus. The twisted Witten genus is supposed to be related to certain modular functions (McKay-Thompson series) associated with Monstrous Moonshine. I believe that Hopkins proved that a manifold with the right properties exists, but only in the topological category, and without the action of the Monster group. Daniel Allcock is working on constructing this manifold. Shmuel Weinberger has championed the study of decidability questions in differential topology. The Novikov conjectures would imply that $\mathcal{L}$-classes (certain combinations of Pontryagin classes) are invariant under homotopy equivalence of smooth aspherical manifolds. See a recent survey. There is still active study of diffeomorphism groups. An active topic here is the study of homological stability for diffeomorphism groups, which is an understanding of the homology of the classifying spaces for such groups.<|endoftext|> TITLE: Are semisimplicial hypercoverings in a hypercomplete $\infty$-topos effective? QUESTION [5 upvotes]: A hypercovering in an $\infty$-topos $E$ is a simplicial object $U \in {\rm Fun}(\Delta^{\rm op},E)$ such that each map $U_n \to ({\rm cosk}_{n-1} U)_n$ is an effective epimorphism. Theorem 6.5.3.12 of Higher Topos Theory shows that if $E$ is hypercomplete, then all hypercoverings are effective, i.e. ${\rm colim}(U)$ is terminal. The notion of hypercovering makes sense also for semisimplicial objects $U \in {\rm Fun}(\Delta_s^{\rm op},E)$, with $\Delta_s$ the subcategory of injective maps in $\Delta$. I believe the argument of Lemma 6.5.3.7 in HTT shows that the underlying semisimplicial object of a simplicial object $U$ is a semisimplicial hypercovering if and only if $U$ is a simplicial hypercovering, and since the inclusion $\Delta_s \subseteq \Delta$ is homotopy cofinal (Lemma 6.5.3.7 in HTT) their colimits also coincide. However, not every semisimplicial object underlies any simplicial object. On the other hand, we can left Kan extend a semisimplicial object from $\Delta_s$ to $\Delta$ to obtain a simplicial object with the same colimit, but this operation doesn't preserve the property of being a hypercovering. This exhausts my ideas for reducing the semisimplicial case to the simplicial one; I've also inspected the proof of 6.5.3.12, but it's not clear to me how to generalize it to the semisimplicial case, since it seems to use the degeneracies (even though, intriguingly, the way it uses them is to reduce one of the main questions to the semisimplicial case, in 6.5.3.9!). So is a semisimplicial hypercovering in a hypercomplete $\infty$-topos necessarily effective? REPLY [3 votes]: You should check out appendix A of SAG, which has a much-expanded section on semisimplicial hypercoverings. In particular, take a look at A.5.3.3. http://www.math.harvard.edu/~lurie/papers/SAG-rootfile.pdf<|endoftext|> TITLE: Original reference for Adams Riemann-Roch theorem QUESTION [8 upvotes]: Let $f\colon Y\to X$ be a proper morphism between smooth quasiprojective $k$-algebraic varieties. Denote by $\psi^j$ the $j$-th Adams operation on the Grothendieck group of vector bundles and $\theta^j(T_f)$ the $j$-th cannibalistic class of the relative tangent bundle $T_f$. The Adams Riemann-Roch theorem states that for any $j$ the diagram $\require{AMScd}$ $$ \begin{CD} K_0(Y)@>f_*>>K_0(X)\\ @V\theta^j(T_f)\cdot \psi^jVV@VV\psi^jV\\ K_0(Y)\otimes\mathbb{Z}[\tfrac{1}{j}]@>f_*>>K_0(X) \otimes \mathbb{Z}[\tfrac{1}{j}] \end{CD} $$ commutes. The oldest reference I know is Theorem 7.6 of Chapter V in W. Fulton; S. Lang: Riemann-Roch algebra. Grundlehren der Mathematischen Wissenschaften , 277. Springer-Verlag, New York, 1985. x+203 pp.. However, that reference is already very general (it does not even require schemes to be over a field) and 1985 is very "late" for such a Riemann-Roch type statement. Therefore my question: $ \phantom{aaaaaaaa}$What is the original reference for the Adams Riemann-Roch theorem? I am looking for something as Borel-Serre's paper is for Grothendieck-Riemann-Roch. REPLY [7 votes]: This seems both interesting and hard to pinpoint. Köck (1991; 1998, 4.6b) credits Fulton-Lang (1985) and Soulé (1985, cf. Thm 7), but your suggestion that the theorem must have been “known” earlier is also well-supported. For one thing, Rössler (1999, §1) finds it in Manin (1969, Thm 16.6). For another, Dyer (1962) (cited by Eckmann at ICM (1963), Adams (1965, (iii) p. 152), Fuchs (1973, pp. 349–350), and reprinted in Adams (1972)) starts: This lecture is principally an exposition of a folk theorem of a Riemann-Roch type for general cohomology theories known to Adams, Atiyah, Hirzebruch... . While these authors don’t spell out how this “folk” theorem includes yours, Panin (2004, §0.1; 2004, p. 823) says at least that his Theorem 2.5.3 “inspired by a Riemann-Roch theorem from [Dy]”, does. See also Smirnov (2006, 2.5.3). So my impression is that specialists understood the Adams Riemann-Roch theorem as an instance of the more general Dyer-Riemann-Roch theorem, long before any of them bothered to name it.<|endoftext|> TITLE: Finite dimensional commutative algebras containing infinitely many nilpotents whose $d$-way products are nonzero QUESTION [7 upvotes]: I'm interested in the following strange question: for some $d > 1$, what is the minimum dimension of a commutative $\mathbb{C}$-algebra containing infinitely many elements that square to zero, but where the product of any $d$ of these elements is nonzero? The best upper bound I know is $d\cdot \binom{\lceil 3d/2 \rceil}{\lfloor d/2 \rfloor} < d\cdot 2.6^d$, and can be obtained as follows. Let $f = \sum_{i_1 < i_2 < \cdots < i_d} \prod_{1\le j < k \le d}(i_j - i_k)^2 x_{i_1} \cdots x_{i_d}$ be an element of $S = \mathbb{C}[[x_1, x_2, \ldots]]$, the ring of formal power series in infinitely many variables. Let $T = \mathbb{C}[[\partial_{x_1} , \partial_{x_2}, \ldots]]$ be the ring of partial differential operators/dual power series in infinitely many variables. $T$ acts on $S$ via differentiation, which I denote by $\circ$. Now let $f^\perp = \{g \in T : g \circ f = 0\}$ be the "apolar ideal" to $f$. I claim that the ring $R = T/f^\perp$ has the desired properties. First, it can be shown that that $\dim \text{span}\{g \circ f : g \in T\} \le d\cdot \binom{\lceil 3d/2 \rceil}{\lfloor d/2 \rfloor}$, and hence the stated bound holds for $\dim R$. Now since each variable appearing in $f$ has degree at most $1$, we have that $\partial^2_{x_i} \circ f = 0$ for all $i$. Furthermore, for all $i_1<\cdots < i_d$ we have $\partial_{x_{i_1}} \cdots \partial_{x_{i_d}} \circ f = \prod_{j,k}(i_j - i_k)^2 \neq 0$. This shows that the images of $\partial_{x_1}, \partial_{x_2}, \ldots$ under the quotient map have the desired properties in $T$. I would appreciate any pointers to concepts or related work that could be useful here. EDIT As YCor points out, there is a lower bound of $2^d$: if $x_1, \ldots, x_d$ are elements of such an algebra with the desired properties, then the products $\{x_S : S \subseteq [d]\}$ must be linearly independent (if $\sum_{S \subseteq [d]} \alpha_S x_S = 0$ is a nontrivial relation, then letting $U \subseteq [d]$ be such that $\alpha_U \neq 0$ and $U$ is minimal with respect to set inclusion among the sets in the support of this relation, multiplying by $x_{[d]-U}$ we find that $x_{[d]}= 0$, a contradiction.) Also, the stated upper bound actually holds for the following more general family of algebras: let $(a_i)_{i \in \mathbb{N}}$ be a sequence of distinct elements in $\mathbb{C}$, let $f_a = \sum_{i_1 < i_2 < \cdots < i_d} \prod_{1\le j < k \le d}(a_{i_j} - a_{i_k})^2 x_{i_1} \cdots x_{i_d}$, and take $R =T/f_a^\perp$. (Even more generally, one can take a matrix $A \in \mathbb{C}^{d \times \mathbb{N}}$ with nonvanishing $d \times d$ minors, and take $f = \sum_{S \subset \mathbb{N}, |S| = d} \det(A_S)^2 x_S$ and $R = T/f^\perp$, although the dimension bound, while finite, ends up being worse.) REPLY [5 votes]: $\def\CC{\mathbb{C}}\def\fm{\mathfrak{m}}\def\PP{\mathbb{P}}$Here are some ideas, but not a solution. First, I will reduce the problem to the sort of examples the OP is already considering. I will then make some connections to secant varieties. I will prove the optimum value for $d=3$ to be $12$, given by the OP's construction. Finally, I will report on some numeric investigations of the OP's construction. Calling the OP's ring $R^d$, my data suggests that $$\dim R^d_k = \begin{cases} \binom{2d-k}{k} & 0 \leq k \leq d/2 \\ \binom{d+k}{d-k} & d/2 \leq k \leq d \end{cases}.$$ We assume throughout $d \geq 2$. In this section, our goal is to reduce to the case that $R$ is a graded Gorenstein ring, with the $x_i \in R_1$ and with socle in degree $d$. Let $V$ denote the vector space spanned by the $x_i$ and let $\fm$ be the ideal they generate. Let $X$ be the Zariski closure of $\{ x_i \}$ in $\PP(V)$ and let $CX$ be the cone on $X$ in $V$. Reduction 1 We can assume that the $x_i$ generate $R$ as a $\CC$-algebra. Proof Passing to the subalgebra they generate reduces the dimension of $R$ and preserves the hypotheses. $\square$ Having made this reduction, I claim that $\fm$ is a maximal ideal and that $R$ is a local ring. Proof: Since $\fm$ is nilpotent and $R \neq 0$, we have $R \neq \fm$ and $R/\fm$ is nonzero. But, since the $x_i$ generate $R$, the dimension of $R/\fm$ is at most $1$. We conclude that $R/\fm \cong \CC$, so $\fm$ is maximal. Also, since this maximal ideal is nilpotent, we deduce that $R$ is local. Reduction 2 We may assume that $X$ is irreducible. Proof If $X$ has multiple irreducible components, one of them must contain infinitely many $x_i$. Reduction 3 We may assume that $\fm^{d+1}=0$. Proof Note that the condition on $d$-fold products ensures that $\fm^d \neq 0$ so, by Nakayama, we have $\fm^d/\fm^{d+1} \neq 0$. Replace $R$ by the ring $R/\fm^{d+1}$. We will now construct a new infinite subset of $CX$ with the given conditions. We note that, for any $x \in CX$, we have $x^2=0$. Since $CX$ spans $V$, which generates $\fm$, the set of $d$-fold products of elements of $CX$ spans $\fm^d/\fm^{d+1}$. In particular, there are some $x_1$, $x_2$, \dots, $x_d$ in $CX$ with $x_1 x_2 \cdots x_d \not \in \fm^{d+1}$. Now, suppose inductively that we have constructed $x_1$, $x_2$, ..., $x_N$ in $CX$, for $N \geq d$, so that every $d$-fold product of $x_i$ is nonzero in $\fm^d/\fm^{d+1}$. For every $x_{i_1}$, ..., $x_{i_{d-1}}$, there exists an $y \in CX$ such that $x_{i_1} \cdots x_{i_{d-1}} y \neq 0$ modulo $\fm^{d+1}$ (namely, any $x_j$ distinct from $x_{i_1}$, ..., $x_{i_{d-1}}$). The condition that $x_{i_1} \cdots x_{i_{d-1}} y \neq 0$ is an open condition on $y$, and we already reduced to the case that $X$ is irreducible. The intersection of finitely many nonempty opens on an irreducible variety is nonempty, so we can find some $x_{N+1}$ in $CX$ so that all the $x_{i_1} \cdots x_{i_{d-1}} x_{N+1}$ are simultaneously nonzero modulo $\fm^{d+1}$. This concludes the inductive construction. $\square$. Reduction 4 We may assume that $R$ is graded. Proof Let $\hat{R}$ be the associated graded of $R$ with respect to the filtration by powers of $\fm$. So we have $\hat{R}_1 \cong \fm/\fm^2 \cong V$, and we can thus think of $CX$ as a subset of $\hat{R}_1$. The condition that $x^2=0$ for $x \in CX$ implies the same claim in $\hat{R}$. Since $\fm^{d+1}=0$, we have $\hat{R}_d \cong \fm^d$, so our condition on $d$-fold products in $R$ implies the same in $\hat{R}$. $\square$. Reduction 5 We may assume that $R_d$ has dimension $1$. Proof If $\dim R_d>1$, then quotient $R$ by a generic element of $R_d$. If this element is chosen generically enough, it will not conicide with any of the countably many products $x_{i_1} \cdots x_{i_d}$. $\square$. Reduction 6 We may assume that $R$ is Gorenstein. Proof We have already assumed that $\dim R_d=1$ and $R_k=0$ for $k>d$. If $R$ has socle in degree $k TITLE: Greatest prime factor of n and n+1 QUESTION [15 upvotes]: For a positive integer $n$ we denote its largest prime factor by $\operatorname{gpf}(n)$. Let's call a pair of distinct primes $(p,q)$ $\textbf{nice}$ if there are no natural numbers $n$ such that $\operatorname{gpf}(n)=p, \operatorname{gpf}(n+1)=q$ or $\operatorname{gpf}(n)=q, \operatorname{gpf}(n+1)=p$. For example, $(2,23)$ is nice. Are there nice pairs $(p,q)$ with $p,q>100$? REPLY [8 votes]: So running a simple sieve algorithm allows for recording pairs which are not nice, and there are many of them below 9 million. I get that the complement includes (2,q) for q=23,29,37,47, and more, (3,q) for q=89,103,113,131,137 and more, (5,q) for q=307,503,509,613,619 and some more, (7,q) for q=967,971,1031,1039,1049 and some more, (11,q) for q=2381,2543,2551,2591,2801 and a few more, and (13,q) for q=2531,2689,2797. For larger values of 13 $\lt p \lt q \lt$ 3000, there are no nice pairs. I am willing to believe there is a q less than 2^2^101 for which (101,q) is nice. Based on preliminary data, I doubt q would be less than 2^101. Gerhard "These Are Rather Big Numbers" Paseman, 2019.05.31.<|endoftext|> TITLE: Sub-Gaussian decay of convolution of $L^1$ function with Gaussian kernel QUESTION [6 upvotes]: I think it might be helpful to put the new statement at the beginning and put the original post at the end. This new statement is more mathematically elegant. Let $f\geq0$ be in $L^1(\mathbb{R}^d)$ and $g(x)=\exp(-\|x\|^2)$. Let $1_{B_1}$ be the indicator function of the unit ball centered at origin. Let $*$ be the convolution operation. Does the condition $$(f*g)(x)\leq C_1\exp(-C_2\|x\|^2)$$ for some $C_1,C_2>0$ imply $$\lim_{n\to+\infty}\frac{(f*1_{B_1})(\mu_n)}{(f*g)(\mu_n)}=0$$ for some sequence $\mu_n\in\mathbb{R}^d$? If this is not true, what additional regularity conditions on $f$ do we need? Any idea or possibly useful reference would be appreciated! The result can be verified easily when $f$ is another Gaussian function as well as some linear combination of Gaussian functions. ----------------Original post--------------------- Let $X$ be a random vector in $\mathbb{R}^d$ satisfying the following property: there exists $C_1,C_2>0$ such that $$\int_0^{+\infty}\mathbb{P}(\|X-\mu_0\|\leq\sqrt{t})\exp(-t)dt\leq C_1\exp(-C_2\|\mu_0\|^2)$$ for any $\mu_0\in\mathbb{R}^d$. Here $\|\|$ is the Euclidean norm in $\mathbb{R}^d$. If the above property holds, is the following statement true: there exists a sequence of vectors $\mu_n$ in $\mathbb{R}^d$ and a sequence of real numbers $t_n\to+\infty$ ($t_n$ may depend on $\mu_n$ for example $t_n=\|\mu_n\|^2/4$) such that: $$\lim_{n\to+\infty}\frac{\mathbb{P}(\|X-\mu_n\|\leq1)}{\mathbb{P}(\|X-\mu_n\|\leq \sqrt{t_n})\exp(-t_n)}=0$$ If this is not true, is there a counter example? Or is the the following result true? $$\lim_{n\to+\infty}\frac{\mathbb{P}(\|X-\mu_n\|\leq1)}{\int_0^{+\infty}\mathbb{P}(\|X-\mu_n\|\leq\sqrt{t})\exp(-t)dt}=0$$ REPLY [5 votes]: We assume that $f$ is not identically zero, whence $f*g(x)>0$ for all $x \in \mathbb{R}^d$. The answer is positive, and this is also true when the density f is replaced by an arbitrary positive finite measure (or equivalently, a probability measure, as in the original formulation). Moreover, for any fixed nonzero vector $v \in \mathbb{R}^d$, the sequence $\{\mu_n\}$ can be taken as a subsequence of the positive integer multiples $\{k v\}_{k \ge 0}$ of $v$. Indeed, if this fails, then there exists $\alpha>0$ and an integer $k_0>0$ such that $$\frac{(f*1_{B_1})(kv)}{(f*g)(kv)} \ge \alpha $$ for all $k \ge k_0$. Let $$\beta=\beta(v):=\inf \{g(x) \, : \, x \in v+B_1\}>0. $$ Then $$(f*g)((k+1)v) \ge \beta (f*1_{B_1})(kv),$$ so the first display gives $$(f*g)((k+1)v) \ge \alpha \beta (f*g)(kv) \,. $$ We infer inductively that $$(f*g)(kv) \ge (\alpha\beta)^{k-k_0} (f*g)(k_0 v) \,. $$ This exponentially decaying lower bound contradicts the hypothesis $$(f*g)(x)\leq C_1\exp(-C_2\|x\|^2).$$<|endoftext|> TITLE: Tricks for getting a creative idea QUESTION [5 upvotes]: Caveat: I fear that people will criticize me for asking this potentially inappropriate question here, but I guess that the community here is quite unique in the ability of potentially answering my question (if you don't have an answer, then probably there is no), and that there is some (little) chance of getting some good answers to the question - and not asking this question here or banning it right away would reduce the chance of getting a good answer to 0. The question is: If one tries to prove something, are there some tricks for getting a creative idea? Ok, I now what you think, yes, there is no algorithm to finding a creative idea, otherwise the idea wouldn't be creative. However, there are some general "tricks": if for minutes one stares at ones sheet of papers with no new ideas, just moving in the same thought cycles, it certainly helps to go and talk to a colleague, because somehow talking awakes the creative ability of the brain (and, additionally, together with a colleague one can mutually pick up an idea of the other and think it a bit further). Also, forgetting the problem for a moment and go and attend talks (even if they are about another topic) or even just rest helps. Do you have any other general "tricks" for getting creative ideas for solving mathematical problems? To make the question a bit more concrete, do you know of any tricks for finding or looking for a good lemma (or several lemmas)? I have the feeling that often the most creativity in proving a theorem lies in finding the right lemma (not even the proof of it, but just the statement). I noticed that whenever I see a proof about which I afterwards say "wow, that's genius, I don't even rudimentally see how one could have come up with it", the crucial point was a lemma (or several lemmas). This also seems to me to be one difference between doing research and doing like homework problems: in homework assignments the proofs usually require only one or two main ideas, and if it requires a lemma, this lemma often is stated in the task as a subtask - while in research, one doesn't even know how much one has to "go down", how many levels of lemmas one has to show. REPLY [4 votes]: I am not addressing your request for "tricks." One can hardly do better than Poincaré's description of creativity in mathematics: Poincaré, Henri, translated by Francis Maitland. Science and method. Courier Corporation, 2003. Preface by Bertrand Russell: "The writing of professional philosophers on such subjects has too often the deadness of external description. Poincaré's writing, on the contrary, ..., has the freshness of actual experience, of vivid, intimate contact with what he is describing." More recently, one can find literature on what constitutes mathematical creativity, often in the math-education literature. Two examples: Ervynck, Gontran. "Mathematical creativity." In Advanced mathematical thinking, pp. 42-53. Springer, Dordrecht, 2002. Springer link. "We hypothesize that the context for creativity is set by a preparatory stage in which mathematical procedures become interiorized through action before they can be the objects of mathematical thought."   Sriraman, Bharath. "The characteristics of mathematical creativity." Mathematics Educator 14, no. 1 (2004): 19-34. Journal link. "The results indicate that, in general, the mathematicians’ creative processes followed the four-stage Gestalt model of preparation-incubation-illumination-verification." REPLY [4 votes]: Here's a simple but extremely effective technique to generate creativity: 1) Work hard on a problem until you reach an impasse. 2) Take an extended break from the problem (it could be a week or up to a few months, it's your call). 3) Repeat For creativty, Take Breaks - Article And a piece of historical evidence to consider: When Isaac Newton was forced to return home from school due to the Black Plague, he began to generate vast swaths of groundbreaking ideas including laying the foundation for classical mechanics and calculus. His productivity lasted for decades.<|endoftext|> TITLE: Non-standard tensor products of inner product spaces QUESTION [5 upvotes]: For two inner product spaces $(\mathcal{V}, (\cdot,\cdot)_V)$ and $(\mathcal{W}, (\cdot,\cdot)_W)$, we can put an inner product on their tensor product in the obvious way: $$ (1) ~~~~ \langle v \otimes w, v' \otimes w'\rangle := \langle v,v'\rangle_V \langle w,w'\rangle_W. $$ This then implies that $$ (2) ~~~~ \|v \otimes w\| = \|v\|_V \|w\|_W. $$ Is there an example of an inner product on $\mathcal{V} \otimes \mathcal{W}$ such that (2) holds, but the inner product is not of the form (1)? Are such things of interest? Do they have a name? REPLY [6 votes]: Assuming complex scalars, no, any inner product which satisfies (2) for all $v \in \mathcal{V}$ and $w \in \mathcal{W}$ has the given form. To see this, let $[\cdot,\cdot]$ be any inner product which satisfies (2). So right away we know that $[v\otimes w,v\otimes w] = \langle v\otimes w, v\otimes w\rangle$ for all $v$ and $w$. Next, for $v,v' \in \mathcal{V}$ and $w \in \mathcal{W}$, we know that $$[(v + v')\otimes w, (v + v')\otimes w] = \langle (v + v')\otimes w, (v + v')\otimes w\rangle.$$ Expanding this out and applying $[v\otimes w,v\otimes w] = \langle v\otimes w,v\otimes w\rangle$ plus the same for $v'\otimes w$ yields $2{\rm Re}[v\otimes w, v'\otimes w] = 2{\rm Re}\langle v\otimes w, v'\otimes w\rangle$. Replacing $v$ with $iv$ yields equality of the imaginary parts too. So now we know that $[v\otimes w,v'\otimes w] = \langle v\otimes w,v'\otimes w\rangle$ for all $v$, $v'$, and $w$. Finally, for any $v,v' \in \mathcal{V}$ and $w,w' \in \mathcal{W}$ we have $$[(v + v')\otimes (w + w'), (v + v')\otimes (w + w')] = \langle (v + v')\otimes (w + w'), (v + v')\otimes (w + w')\rangle.$$ Expanding this out and cancelling all the equalities we already have then leaves only $$2{\rm Re}([v\otimes w, v'\otimes w'] + [v\otimes w', v'\otimes w]) = 2{\rm Re}(\langle v\otimes w, v'\otimes w'\rangle + \langle v\otimes w', v'\otimes w\rangle).$$ Replacing $v$ with $iv$ yields equality of the imaginary parts too, so that $$[v\otimes w, v'\otimes w'] + [v\otimes w', v'\otimes w] = \langle v\otimes w, v'\otimes w'\rangle + \langle v\otimes w', v'\otimes w\rangle.$$ Then replacing $w'$ with $iw'$ yields $$[v\otimes w, v'\otimes w'] - [v\otimes w', v'\otimes w] = \langle v\otimes w, v'\otimes w'\rangle - \langle v\otimes w', v'\otimes w\rangle,$$ so that $[v\otimes w, v'\otimes w'] = \langle v\otimes w,v'\otimes w'\rangle$. As every element of the algebraic tensor product $\mathcal{V}\otimes\mathcal{W}$ is a linear combination of elementary tensors, this shows that $[\cdot,\cdot] = \langle\cdot,\cdot\rangle$. I feel there ought to be a one-line proof of this, but I don't quite see it. REPLY [3 votes]: Let $V, W$ be 2-dimensional with orthonormal bases $\{v_0, v_1\}$ and $\{w_0, w_1\}$. If you expand out your condition (2), you obtain the following conditions: $$ \langle v_i \otimes w_j, v_i \otimes w_j \rangle = 1 \\ \langle v_i \otimes w_j, v_i \otimes w_k \rangle = \langle v_i \otimes w_j, v_k \otimes w_j \rangle = 0 \\ \langle v_0 \otimes w_0, v_1 \otimes w_1 \rangle + \langle v_1 \otimes w_0, v_0 \otimes w_1 \rangle = 0 $$ Clearly you can pick any value for $\langle v_0 \otimes w_0, v_1 \otimes w_1 \rangle$. Any nonzero choice will not satisfy your condition (1), since the RHS will be 0. So it is certainly possible. I'm not sure if they're of interest. EDIT: Since there's disagreement in the comments, I'm posting my calculation. Consider general vectors $a_0 v_0 + a_1 v_1 \in V$, $b_0 w_0 + b_1 w_1 \in W$. Let's mangle Einstein notation by putting $\langle v_i \otimes w_j, v_k \otimes w_l\rangle = g_{ij}^{kl}$. (2) says that $ \langle a_0 v_0 + a_1 v_1, a_0 v_0 + a_1 v_1 \rangle = (a_0^2 + a_1^2)(b_0^2 + b_1^2)$. Expanding out the LHS gives $$ a_0^2 b_0^2 g_{00}^{00} + a_1^2 b_0^2 g_{10}^{10} + a_0^2 b_1^2 g_{01}^{01} + a_1^2 b_1^2 g_{11}^{11} + \\ 2\left(a_1 a_0 b_0^2 g_{10}^{00} + a_0^2 b_1 b_0 g_{01}^{00} + a_1^2 b_1 b_0 g^{10}_{11} + a_1 a_0 b_1^2 g^{01}_{11}\right) + \\ 4a_1 a_0 b_1 b_0 \left(g^{00}_{11} + g^{10}_{01} \right).$$ By choosing unit vectors, it is clear that $g^{ij}_{ij} = 1$. So the first line cancels with the RHS. By choosing a sum of two unit vectors, it is clear that $g^{ij}_{ik} = g^{ij}_{kj} = 0$, so the second line vanishes. We are left with $4a_1 a_0 b_1 b_0 (g^{00}_{11} + g^{10}_{01})$ = 0. If we choose our $g$ so that $g^{00}_{11} + g^{10}_{01} = 0$, this quantity will always vanish, and so (2) will hold for any pair of vectors.<|endoftext|> TITLE: Compact objects in the $\infty$-category presented by a simplicial model category QUESTION [5 upvotes]: Let $\mathsf{M}$ be a simplicial model category presenting an $\infty$-category $\mathcal{M}$. I'm interested in a general statement relating compact objects in $\mathcal{M}$ (in the $\infty$-categorical sense) with the compact objects in $\mathsf{M}$. Here's roughly what I expect to be true but if i'm missing some assumptions or if some are redundant feel free to phrase the correct statement as an answer. Suppose further that $\mathsf{M}$ satisfies that weak equivalences between fibrant objects are stable under filtered colimits. Then is the following true Let $X$ be a compact cofibrant object in $\mathsf{M}$. Is $X$ compact as an object in $\mathcal{M}$? Is every compact object in $\mathcal{M}$ a retract of (the image in $\mathcal{M}$) of some compact object in $\mathsf{M}$? REPLY [3 votes]: If $X$ is such that $X \times \Delta^n$ is compact for every $n$ then yes. This happens, for example, if the cotensor functor $(-)^{\Delta^n}$ preserves filtered colimits, a condition which is quite common. Sufficient conditions of a similar nature are described in Proposition 5.3.1 of this paper. That said, you should expect the answer to your question to be negative in general (though I don't have a counter-example of the top of my head, at least not with $X$ cofibrant).<|endoftext|> TITLE: The free loop space of spheres QUESTION [21 upvotes]: Let $n>1$. The homology of the free loop space $\Lambda S^n$ of the sphere $S^n$ contains two torsion if $n$ is even. Thus the fibration $$ \Omega S^n\rightarrow \Lambda S^n\rightarrow S^n $$ is not trivial if $n$ is even (Here $\Omega S^n$ denotes the based loop space.). The odd dimensional spheres do not have torsion in the homology of the free loop space and one can compute that $H_*(\Lambda S^{2k+1};\mathbb Z)\cong H_*(\Omega S^{2k+1}\times S^{2k+1};\mathbb Z)$. Hence the homology does not obstruct the existence of a trivialization of the free loop fibration. Indeed $\Lambda S^3$ and $\Omega S^3\times S^3$ are homeomorphic, which can be proven using the group structure on $S^3$. Is the free loop space fibration always trivial for odd dimensional spheres ? My guess would be that this is not the case, with a possible exception of $S^7$. REPLY [16 votes]: I am grateful to Tobias Barthel, who sent me the following paper of J. Aguadé: "On the space of free loops of an odd sphere". Pub. Mat. UAB No 25, June 1981. Aguadé proves the following theorem Theorem: The following are equivalent: $\Lambda S^{2n+1}\simeq S^{2n+1}\times \Omega S^{2n+1}$; The free loop space fibration is homotopically trivial $n=0,1,3$. That 3 implies 2 is due to the fact that these spheres are $H$-spaces, which was already noted. 2 implies 1 is trivial. Hence the only thing Aguadé shows is that 1 implies 3. For this it is assumed that there is a map $f$ inducing a homotopy equivalence. From this there is an induced map $h:S^1\times S^{2n+1}\times \Omega S^{2n+1}\rightarrow S^{2n+1}$. Aguadé applies the Hopf construction to this map to get a map $$ \tilde g:S^{2n+1}*(S^1\times S^{2n})\rightarrow S^{2n+2}. $$ The space on the left is a wedge of spheres $S^{2n+3}\vee S^{4n+2}\vee S^{4n+3}$, thus there is an induced map $g:S^{4n+3}\rightarrow S^{2n+2}$. Aguadé shows that this map has Hopf invariant one, hence the result follows.<|endoftext|> TITLE: How non-projective can a complete variety be? QUESTION [12 upvotes]: Given a positive integer $n$, can you give an example of a connected smooth proper $\mathbb{C}$-scheme such that any $n$ closed points are contained in a common affine open, but there is $n+1$ closed points that are not contained in a common affine open? What is the minimum dimension of such an example (probably larger than 2, because every smooth proper surface is projective)? REPLY [10 votes]: For every positive integer $n$, Farnik has constructed a smooth complete variety over an algebraically closed field with $n=\mathrm{sup}\{$$m$: every $m$ points of in $X$ are contained in a common affine open subset of $X\}$. The construction suggests that the varieties are threefolds so the minimum dimension appears to be $3$. M. Farnik, On strengthening of the Kleiman-Chevalley criterion, Proceedings of the AMS 141 no 11 (2013) pp 4005-4013, doi:10.1090/S0002-9939-2013-11695-3. Link at AMS site.<|endoftext|> TITLE: A sequence of points in the spectrum of a subhomogeneous C$^{*}$-algebra can converge to at most finitely many points QUESTION [5 upvotes]: Let $A$ be a subhomogeneous C$^{*}$-algebra (i.e., there is a finite upper bound on the size of the irreducible representations of $A$). Let $\hat{A}$ denote its spectrum. I heard of a result that states that If $(\pi_{n})$ is a sequence in $\hat{A}$, then $(\pi_{n})$ can converge to at most finitely many points. I'd like to know why this statement is true. For instance, if we look at the algebra $$ B:=\left\{f\in C([0,1],M_{2}(\mathbb{C})):\exists \lambda,\mu,\nu\in\mathbb{C}\text{ with }f(0)=\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix}\text{ and }f(1)=\begin{pmatrix}\mu & 0\\ 0 & \nu\end{pmatrix}\right\} $$ it is clear that there is a sequence of $2\times 2$ irreducible representations converging to two $1\times 1$ irreducible representations (the point evaluations converging to either the upper-left or bottom-right evaluation at $1$). But there is not sequence of $2\times 2$ irreducible representations converging to $\lambda,\mu$, and $\nu$, which makes sense to me because we cannot stick all three of them down the diagonal, as we are in $M_{2}(\mathbb{C})$. If $A$ is homogeneous, then $\hat{A}$ is Hausdorff, so the sequence can converge to at most one point in this case. I guess, in the subhomogeneous case the result has something to do with the fact that there are only finitely many choices for the dimensions of the irreducible representations. I also know that each subspace $\hat{A}_{n}:=\{\pi\in\hat{A}:\dim\pi=n\}$ is Hausdorff, but I'm not sure how to put this together. Any help is appreciated. REPLY [7 votes]: See JMG Fell, The Dual Spaces of C$^*$-Algebras, Trans. Amer. Math. Soc. 94 (1960), 365-403, Corollary 1 on p. 388: Let $A$ be a C$^*$-algebra with dual space $\hat{A}$. Let $T^i$ be a net of elements of $\hat{A}$, all of dimension equal to or less than the integer $n$; and let $S^1, \ldots S^r$ be distinct elements of $\hat{A}$ such that $T^i\to_i S^k$ for each $k$. Then $$\sum_{k=1}^r \dim S^k\le n.$$<|endoftext|> TITLE: Irreducibility of root-height generating polynomial QUESTION [17 upvotes]: The height $ht(\alpha)$ of a positive root $\alpha$ in a (finite, crystallographic) root system $\Phi$ is $\sum_{i=1}^n c_i$ where $\alpha = \sum_{i=1}^n c_i \alpha_i$ is its decomposition as a sum of simple roots. I am interested in the polynomial $$R_{\Phi}(x):=\sum_{\alpha \in \Phi^+} x^{ht(\alpha)-1}.$$ For example (using the usual Cartan-Killing nomenclature), we have: $R_{A_n}(x)=n+(n-1)x+\cdots +2x^{n-2}+x^{n-1}$ $R_{B_n}(x)=R_{A_n}(x^2)+xR_{A_{n-1}}(x^2)$ $R_{D_n}(x)=R_{B_n}(x)-\sum_{i=n-1}^{2n-2} x^i$ Questions: When $\Phi$ is an irreducible root system, is $R_{\Phi}(x)$ an irreducible polynomial over $\mathbb{Q}$? This has been checked for the exceptional types ($G_2, F_4, E_6, E_7,$ and $E_8$) and for the infinite families listed above for $n \leq 500$. If the answer to (1) is "yes", is there some uniform Lie-theoretic reason that this is true (that is, a proof not relying on the classification)? REPLY [8 votes]: (Turning Gjergji Zaimi's comment into a community wiki answer.) A problem equivalent to the case of $R_{A_n}$ is discussed in Classes of polynomials having only one non-cyclotomic irreducible factor, by Borisov, Filaseta, Lam, and Trifonov (Acta Arithmetica 90 (1999), 121–153). They prove irreducibility in many special cases but the problem remains open (or at least, none of the papers that Google Scholar lists as citing this paper solves the problem).<|endoftext|> TITLE: Orbit of an irreducible representation of a surface group under that action of the mapping class group QUESTION [9 upvotes]: Let $F$ be a closed oriented surface of negative Euler characteristic. Let $X^i(F)$ be the subset of the $SL_2\mathbb{C}$-character variety of the fundamental group of $F$ corresponding to irreducible representations. The mapping class group of $F$, $\mathcal{M}(F)$ acts on $X^i(F)$. Let $[\rho]\in X^i(F)$, is the orbit of the $[\rho]$ under the action of the mapping class group, $$\mathcal{M}(F).[\rho]$$ Zariski dense in $X(F)$? REPLY [3 votes]: As some further evidence for a positive answer to your question, there is a paper of Cantat and Loray that proves a relative version of this for mapping class group the 4-holed sphere (rel boundary). In this case, the variety of representations (with fixed holonomy around each boundary curve) is a cubic surface of the form $$x^2+y^2+z^2+xyz= Ax+By+Cz+D,$$ where $A,B,C,D$ depend on the holonomy of the boundary components, and $x,y,z$ are the traces of simple loops meeting pairwise in two points. In Theorem D of the paper, they prove that there is no invariant curve. Hence any orbit must be either finite or Zariski dense. I believe one could leverage this theorem into a proof in the closed case. The idea would be to find for an irreducible (Zariski dense) representation $\rho$ a 4-holed sphere subsurface whose orbit under the relative mapping class group of this subsurface is Zariski dense. Then the Zariski closure would be at least two dimensional. Repeating this for other subsurfaces by induction, I believe one could show that the mapping class group orbit is Zariski dense.<|endoftext|> TITLE: Definability of the ring of integer in algebraic extensions of $\mathbb Q$ QUESTION [10 upvotes]: J. Robinson has proved that exists a formula $\psi(x)$ in the language of rings which,applied to the rational numbers, defines the the ring integers (making the theory of $\mathbb{Q}$ undecidable, due to Godel's theorem). However, due to Tarski's results, the theory of the real field is complete and therefore the ring of integers $\mathbb{Z}$ is not definable in $\mathbb{R}$. Let $K/\mathbb{Q}$ be an algebraic extension of the rational numbers. I will ask two related questions: Without any further assumptions on the extension, is $\mathbb{Z}$ necessarily a definable subset in $\langle K,+,\times,0,1\rangle$? What purely algebraic properties of the extension might yield a more conclusive answer to the above questions? Specifically: separability, dimension, normality, finiteness, simplicity (maybe even Galois group structure). Moreover, is there a deeper side to this? Do there exist two extensions $K_{1}/\mathbb{Q},K_{2}/\mathbb{Q}$ that satisfy exactly the same algebraic properties denoted above, yet in one of them $\mathbb Z$ is a definable subset, and in the other it is not? These are not questions on which I have pondered for long, but I am curious whether there exist any interesting examples or logical tools needed in order to tackle these questions. REPLY [4 votes]: I hope this is not seen as self-promoting, but I just stumbled upon this question when browsing things related to that recent question. As Tom Scanlon said, it is very difficult if not impossible to get a complete picture of the algebraic extensions of $\mathbb{Q}$ in which $\mathbb{Z}$ is definable. However, one can say that in a certain sense, $\mathbb{Z}$ is not definable in $K$ for most algebraic extensions $K/\mathbb{Q}$: The set of fields $K\subseteq\overline{\mathbb{Q}}$ in which $\mathbb{Z}$ is definable is meager in the set of all such $K$, in the topology induced from $2^{\overline{\mathbb{Q}}}$. This is part of what is shown in this very short note of Philip Dittmann and myself. The reason for the non-definability used there is a purely algebraic one, as asked for in the question: Namely $\mathbb{Z}$ is not definable in any so-called pseudo-algebraically closed field $K$ (meaning every geometrically integral $K$-variety has a $K$-rational point), and the set of $K\subseteq\overline{\mathbb{Q}}$ that are pseudo-algebraically closed is non-meager.<|endoftext|> TITLE: The square modulus of coordinates of a uniformly chosen point in complex projective space is uniform in the simplex QUESTION [5 upvotes]: I can't recall where I learned this (beautiful) fact, and I would like a reference (if possible, in a textbook): Let $(z_0:\cdots:z_n) \in \mathbb{P}^n(\mathbb{C})$ be chosen uniformly at random w.r.t. the Fubini-Study metric, and normalized so that $|z_0|^2 + \cdots |z_n|^2 = 1$. Then the point $(|z_0|^2, \ldots, |z_n|^2)$ is uniformly distributed on the $n$-simplex (the set of $(p_0,\ldots,p_n)\in\mathbb{R}^{n+1}$ such that all $p_i\geq 0$ and $p_0+\cdots+p_n = 1$) w.r.t. its Euclidean metric. This affords the quantum-mechanical interpretation that if we draw a random quantum entanglement of $n+1$ pure states, and we observe the pure state it is in, we get a uniform probability measure on the pure states. However, I'm not asking for a reference in relation to quantum mechanics. This is, for example, proposition 1 in this paper but the only reference the authors give after calling the fact “known” is a 600-page book without any specific page number (shame!). It is also mentioned in this blog post (and attributed to Bill Wootters); and the particular case $n=1$ is mentioned in this other blog post (in relation to the Box-Muller transformation); but I would like something more tangible than a blog post. As a bonus question, if we take a uniformly random $(n+1)\times(n+1)$ unitary matrix (uniformly w.r.t. the Haar measure) and we look at the square norms of its columns, we get $n+1$ points on the $n$-simplex, each uniformly distributed by the above fact: does this distribution of $n+1$ points on the $n$-simplex have a standard name, and where might I learn more about it? REPLY [4 votes]: Long ago, I learned this from a MO answer by Greg Kuperberg. A more general fact is that the moment map of a projective toric variety is measure-preserving. As explained in the linked email by Yael Karshon, this is a special case of the Duistermaat-Heckman formula (see Corollary 3.3 in Duistermaat and Heckman's 1982 paper and note that $n=l$, where $l$ is the dimension of the torus acting on the symplectic manifold of dimension $2n$. Aside: I recommend the paper of Atiyah and Bott on this topic for a nice discussion of Duistermaat-Heckman in the context of equivariant cohomology and Witten's work on supersymmetry and Morse theory. There ought to be some nice textbook treatments on this aspect (maybe Guillemin and Sternberg's "Supersymmetry and Equivariant de Rham Theory"?), but I haven't looked to see whether your formula is mentioned explicitly.<|endoftext|> TITLE: What is an example of a quasicategory with an outer 4-horn which has no filler? QUESTION [8 upvotes]: A quasicategory has fillers for all inner horns $\Lambda^i[n]$ where $n\geq 2$ and $0 TITLE: Why to believe the Fargues geometrization conjecture? QUESTION [24 upvotes]: In the study of the arithmetic local Langlands correspondence, there is a conjecture that was recently (in this decade) formulated by Fargues. I can't even concisely state the conjecture so I will defer this to the answerers who will presumably do a much better job of it. I can, however, give a link to a paper by Fargues from which, with sufficient effort, you should be able to extract understanding of the conjecture. Can you show us "a mathematical path" one can walk along to arrive at the conjecture, starting from the facts which you might expect an average third-year graduate student in specializing in arithmetic geometry or automorphic forms to know? A sort of a one-page (or so) explanation of why one should believe the geometrization conjecture. It would be also nice if you explain what is the relation to the geometric Langlands theory as developed by Gaitsgory and others (my guess is that they are only similar in the name but that is just a guess). The paper of Fargues already provides a 46-page explanation but it would much easier to follow if one has a specific picture in mind to begin with. There is probably more than one answer to this question (all roads lead to Rome!) but there are not that many people in the world who understand the geometrization conjecture so if we get at least one answer that would be fortunate. REPLY [26 votes]: Why believe in it ? Because I said so. There is absolutely no doubt it is true. This is in some sense evident to me now, even more than in 2014 after my talk at the MSRI, in particular after the release of our joint paper with Peter. As said by Peter. There are numerous evidences coming from everywhere. At some point this is not a coincidence anymore and there is something as I could begin to see in 2014. Here are different steps from my point of view: Already in my PHD work, trying to prove some particular cases of this Kottwitz conjecture describing the discrete part of the cohomology of Rapoport-Zink spaces in terms of local Langlands, I could see that Kottwitz sets are very important (see for example my conjecture that says that for each element of $B(G,\mu)$ the associated Newton stratum is non-empty (now proven in a lot of cases)) When working on those twin-towers thing I tried further to geometrize Jacquet-Langlands. I mean in some sense the isomorphism (with modern notations) $[\Omega^\diamond / \underline{\mathrm{GL}_n (\mathbb{Q}_p)]}\cong [\mathbb{P}^{n-1,\diamond} / \underline{\mathrm{D}^\times}]$ is a geometric form of Jacquet-Langlands. I did not know what to do with this, tried a few things but this was hopeless (forget about this now). Btw I really understood at this point that those Hodge-Tate periods are a big thing, see further the link with modifications of vector bundles on the curve. Then there there was this curve thing. It was clear during my work with Fontaine that modifications of vector bundles on the curve are a key thing in the theory. I perfectly remember being super excited when I understood that the Hodge filtration of a filtered $\varphi$-module allows you to modify a vector bundle on the curve (see the talk I gave at Fontaine's conference where I say that we apply a Hecke correspondence (which gave rise to protestations by Fontaine, of course)). It was more and more clear that modifications of vector bundles were a key thing. In particular I really tried to understand this in my article at Laumon's conference: certain modification of vector bundles are the same as local Shtukas !!!! (aka BKF modules). This was later confirmed by the work of Scholze and Weinstein that says that Rapoport-Zink spaces are moduli of modifications of vector bundles. But just pronouncing the word Shtuka is a great indication that we are in a good direction : some Hecke property (aka modifications of vector bundles) is linked to moduli of Shtukas ;) There is this property that the curve is geometrically connected and thus its $\pi_1$ is $\mathrm{Gal} ( \overline{\mathbb{Q}}_p | \mathbb{Q}_p)$...this is the starting point I took at the MSRI in 2014 : let's say there is a geometric Langlands, let's writte the Hecke property...you find Kottwitz conjecture describing the discrete part of the cohomology of local Shtuka moduli spaces... The work of Kottwitz and Kaletha was fundamental at some point to try to link this to the classical local Langlands as formulated by Kottwitz and Kaletha. The work of Caraiani and Scholze is fundamental too to me, I really wanted to understand were does this perverse sheaf come from ? Coupled with the understanding of the cohomology of Igusa varieties (Harris-Taylor, Mantovan, Shin, Caraiani-Scholze) One great thing was the classification of G-bundles on the curve and realizing this is Kottwitz set B(G). This is super striking, in particular when you study reduction of G-bundles. There are plenty of other things that come to my mind. But since the release of my joint paper with Peter I am definitely convinced. In particular the categorical conjecture is more or less evident. The functor \begin{align*} \mathrm{Perf} ( \mathrm{LocSys}_{\hat{G}_{\overline{\mathbb{Q}}_\ell}}) & \longrightarrow D_{lis} ( \mathrm{Bun}_G,\overline{\mathbb{Q}}_\ell ) \\ M & \longmapsto M \ast \mathcal{W}_{\psi} \end{align*} defined by the spectral action can really be understood as some kind of non-abelian Fourier transform. In some sense the Whittaker sheaf is some kind of kernel. Then the geometric conjecture states that this extends (you define the Fourier transform on some subdomain and hope this "extends by continuity") to an equivalence $$ D^b_{coh} ( \mathrm{LocSys}_{\hat{G}_{\overline{\mathbb{Q}}_\ell}}) \xrightarrow{\sim} D_{lis} ( \mathrm{Bun}_G,\overline{\mathbb{Q}}_\ell )^\omega $$ This is really a Fourier type equivalence equivalence like Fourier-Mukai or Fourier-Deligne and so on, outside of the fact that here one side is coherent and the other one is étale (like in the Fourier-Mellin transform for a torus for example). If you put for $M$ a skycraper sheaf on a discrete parameter for $\mathrm{GL}_n$ (an irreducible $Frob$-semi-simple $\ell$-adic representation of $W_E$) this non-abelian Fourier transform is computed locally on a Whittaker stack by Drinfeld ($n=2$) and Laumon more generally. The fact that this is non-zero and gives the good thing should be an analog of Frenkel-Gaitsgory-Vilonen and Gaitsgory in the function field case over a finite field. Typicaly this geometric conjecture implies that the two possible natural definitions of the stable Bernstein center are the same, see Example X.1.6. This implies this existence of a nice kernel of functoriality too, see Example X.1.7.. Anyway, this conjecture is true, no doubt. I think that one reason why peoples have difficulties to understand why this has to be true is that they are too focused on classical geometric Langlands that is weird thing to me in some sense. But I think at the end it may be possible that the Langlands program has to be retaken from the beginning, typically the good object locally at a place $p$ are not smooth representations $\pi$ of $G(E)$ but complexes $A\in D_{lis} (\mathrm{Bun}_G,\overline{\mathbb{Q}}_\ell)$. As you can see the kernel of functoriality is defined at the level of $D_{lis}$ naturally. It may be possible that in the next years the Langlands program has to be re-though from the beginning, maybe the good objects that can be transferred "naturally" are not automorphic representations but some object in some $D_{lis}(whatever)$. Stay tuned for this.<|endoftext|> TITLE: Motivation for Suslin’s Rigidity Conjecture QUESTION [7 upvotes]: Suslin Rigidity conjecture states that motivic cohomology $$ H_{\mathcal{M}}^1(\operatorname{Spec}(F),\mathbb{Q}(n)) $$ of the field $F$ coincides with motivic cohomology for the subfield of constants $F_0$. The fact that first motivic cohomology don’t change under pure transcendental extensions gives some evidence this conjecture. Question: Does there exist a more conceptual reason for validity of this conjecture? Does it tell us something new about algebraic cycles (under assumption that Standard Conjectures hold)? REPLY [4 votes]: First off, I'm not sure that the assertion that ${\rm H}^1(-,\mathbb{Q}(n))$ doesn't change under purely transcendental extensions is known. The Gersten complex for the affine line provides the Milnor exact sequence $$ 0\to {\rm H}^1(F,\mathbb{Q}(n))\to {\rm H}^1(F(T),\mathbb{Q}(n))\to \bigoplus_{x\in (\mathbb{A}^1)^{(1)}} {\rm H}^0(\kappa(x),\mathbb{Q}(n-1))\to 0 $$ The vanishing of the groups ${\rm H}^0(\kappa(x),\mathbb{Q}(n-1))$ is only known for $n=2$ because we understand $\mathbb{Q}(1)$ in terms of units. For $n>2$, the birational invariance would assume the Beilinson-Soule vanishing conjecture which isn't generally known (but ok for finite and global fields). A significant part of the motivation seems to come from the rigidity of regulators. This is discussed for example in Section 7 of D. Ramakrishnan. Regulators, algebraic cycles, and values of L-functions. In: Algebraic K-theory and algebraic number theory, Contemp. Math. 83, Amer. Math. Soc. 1989, pp. 183-310. The conjecture is formulated in 7.1.8 (but the remark concerning indecomposable $K_3$ seems incorrect). The main point is that for any algebraically closed field $F$ with $\overline{\mathbb{Q}}\subseteq F\subseteq \mathbb{C}$ the image of the regulator map on ${\rm H}^1(F,\mathbb{Q}(n))$ in Deligne cohomology coincides with the image of ${\rm H}^1(\overline{\mathbb{Q}},\mathbb{Q}(n))$ (which we know from Borel's computations). This follows from the argument for Assertion 2.3.4 and a rigidity statement in Lemma 1.6.6.2 in A.A. Beilinson. Higher regulators and values of L-functions. (english translation) Journal of Soviet Math. 30 no.2 (1985), 2036-2070. In the specific case of ${\rm H}^1(F,\mathbb{Q}(2))$ (indecomposable $K_3$) we can express the regulator in terms of the dilogarithm which has a rigidity property formulated in Corollary 6.2.2. of Bloch's Irvine notes: S. Bloch. Higher regulators, algebraic K-theory and zeta functions of elliptic curves. CRM Monograph Series 11, Amer. Math. Soc. 2000. Similar rigidity properties are also true for the polylogarithms, cf. Theorem 50 in chapter 12 (Function theory of polylogarithms) of L. Levin's "Structural properties of polylogarithms". So this ties in with the picture that the regulators should be expressed in terms of polylogarithms, together with the expectation that the regulator should be injective on ${\rm H}^1(F,\mathbb{Q}(m))$. Formulated differently, the rigidity conjecture (say for characteristic zero fields) would be a consequence of the Beilinson conjecture: any element of ${\rm H}^1(F,\mathbb{Q}(n))$ would always come from some finitely generated arithmetic scheme over $\mathbb{Z}[1/N]$ and then rigidity for motivic cohomology would follow from a similar rigidity property of Deligne cohomology.<|endoftext|> TITLE: On norming weakly$^*$ sequences in the dual of the Banach space $c_0$ QUESTION [7 upvotes]: A bounded subset $B$ of the dual $X^*$ of a Banach space $X$ is called norming if the formula $\|x\|:=\sup\{|x^*(x)|:x^*\in B\}$ determines an equivalent norm on $X$. Observe that the sequence $(e_n^*)_{n\in\omega}\subset c_0^*=\ell_1$ of coordinate functionals in the dual of $c_0$ is norming and weakly$^*$ null. Question 1. Is it true that every absolutely convex bounded norming set $B\subset c_0^*$ in the dual of the Banach space $c_0$ contains a norming weakly$^*$ null sequence? The same question can be asked for subspaces of $c_0$. Question 2. Let $X$ be a closed subspace of the Banach space $c_0$ and $B\subset X^*$ is a norming bounded absolutely convex set. Is it true that $B$ contains a norming weakly$^*$ null sequence? Remark. If a Banach space $X$ admits a norming weakly$^*$ null sequence of functionals $\{f_n\}_{n\in\omega}\subset X^*$, then the operator $T:X\to c_0$, $T:x\mapsto (f_n(x))_{n\in\omega}$, is an isomorphic embedding of $X$ into $c_0$. Therefore, $X$ is isomorphic to a subspace of the Banach space $c_0$. REPLY [4 votes]: The answer to your Question 1 (and thus Question 2, for general subspace) is "No". Example. Let $B=B_{\ell_1}\cap \ker {\mathbf{1}}$, where $B_{\ell_1}$ is the unit ball of $\ell_1$ and ${\mathbf{1}}=(1,\dots,1,\dots)\in\ell_\infty$ (it is well-known and easy to check that $B$ is norming). Suppose that there is a weak$^*$ null norming sequence $\{x_i^*\}_{i=1}^\infty$ inside $B$. We may assume, slightly decreasing the norming constant, that (1) $x_i^*$ are finitely supported. (2) eventually they get more and more zeros at the beginning. Let $c$ be the norming constant of this sequence, that is $\sup_i|x_i^*(x)|\ge c||x||$ for every $x\in c_0$. Let $\alpha\in(0,1)$ be such that $1-\alpha\le c$. We consider the following sequence $u$ in $c_0$: $$u=(1,\underbrace{\alpha,\dots,\alpha}_{n_1}, \underbrace{\alpha^2,\dots,\alpha^2}_{n_2},\alpha^3,\dots),$$ where $n_1$ is such that all $x_i^*$ whose support contains $1$, are supported in $I_1:=\{1,\dots,n_1+1\}$; $n_2$ is such that all $x_i^*$ whose support intersects $I_1$ are supported in $I_2:=\{1,\dots,n_1+n_2+1\}$, so on. We claim that $|x_i^*(u)|\le\frac{c}2$ for each $i$, thus getting a contradiction. In fact, the vector $u$ is such that the support of each $x_i^*$ is contained in the set, where the coordinates of $u$ are either $\alpha^k$ or $\alpha^{k+1}$ for some $k\in\{0,1,2,3,\dots\}$, also $||x^*_i||_{\ell_1}\le 1$ and ${\mathbf{1}}(x_i^*)=0$. Thus $|x_i^*(u)|$ does not exceed the value which we get if the positive support of $x_i^*$ corresponds to $\alpha^k$ and the negative support of $x_i^*$ corresponds to $\alpha^{k+1}$, or the other way around. Therefore $|x^*_i(u)|\le \frac12(\alpha^k-\alpha^{k+1})\le\frac12(1-\alpha)\le \frac c2$.<|endoftext|> TITLE: Finiteness of $H_1 \backslash G / H_2$ and the geometry of the orbits QUESTION [11 upvotes]: Let $G$ be a connected reductive group over an algebraically closed field $k$. By the Bruhat decomposition, $P \backslash G/P \cong W_P \backslash W / W_P$ is a finite set for any parabolic subgroup $P$ of $G$. So there is some connection between the combinatorics of the Weyl group and the geometry of the orbits. Now consider two closed subgroups $H_i$ of $G$. When is $ H_1 \backslash G / H_2$ finite? In such cases, can we describe the $H_1$-orbits in $G/H_2$ using some group theoretic data or combinatoric data in general? If $H_1$ is a Borel subgroup, the finiteness is equivalent to $G/H_2$ being a spherical variety. And things become trivial if $H_1=G$. If $H_1=H_2$, is the finiteness equivalent to $H_1=H_2$ being a parabolic subgroup? For low rank $G$, can we classify all such pair $(H_1,H_2)$? REPLY [2 votes]: This doesn't answer all your questions, but it seems relevant enough to post. In The orbits of affine symmetric spaces under the action of minimal parabolic subgroups by T. Matsuki(1977), the double coset spaces $H\backslash G/P$ are described and shown to be finite where $G$ is a real semisimple Lie group, $P$ is a minimal parabolic, and $G^{\sigma}_0\subset H\subset G^\sigma$ where $G^{\sigma}$ denotes the closed subgroup of $G$ consisting of all the elements fixed by an involution $\sigma$. When $H$ is the maximal compact, the quotient is trivial because of the Iwasawa decomposition. When $H$ is a real form of a complex semisimple Lie group $G$, this is a result of Aomoto from 1965 (in that case $P=B$, the Borel). And in the case where $G$ is a product of a semisimple Lie group with itself and $H$ is the diagonal, then this reduces to the Bruhat decomposition you mention in your question.<|endoftext|> TITLE: Origin of Hecke operators QUESTION [5 upvotes]: What is the original paper in which Erich Hecke had first introduced the Hecke operators? REPLY [16 votes]: Über Modulfunktionen und die Dirichletschen Reihen mit Eulerscher Produktentwicklung. I, Math. Ann. 114 (1937), 1-28; II, ibid., 316-351. These two papers are available here and here.<|endoftext|> TITLE: Errata for Bott and Tu's book "Differential Forms in Algebraic Topology" QUESTION [14 upvotes]: My book is Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott of which An Introduction to Manifolds by Tu is a prequel. Is there a good list of errata for Bott and Tu available? A cursory Google search reveals not much except this: Some possible mistakes in Bott and Tu, and possibly more here though uncompiled. Is there any source available online which lists inaccuracies and gaps? Copying this: Errata for Atiyah-Macdonald Closed here: Errata for Bott Tu Differential Forms Even though I was suggested here: An old “list question” edited to include other points My recent question: How does one handle books without errata? I'm hoping something off-topic on stackexchange would be on topic on overflow. REPLY [4 votes]: Here is the comment to this book in author's web page: Differential Forms in Algebraic Topology (with Raoul Bott), third corrected printing, Graduate Text in Mathematics, Springer, New York, 1995. The third printing published in 1995 corrects misprints in earlier printings; after that, the book has remained stable. Any printing dated 1995 or later should be fine. Earlier printings should be discarded. Also you can check my web page for all possible math books errata (not completed yet).<|endoftext|> TITLE: On $C(K)$ spaces embeddable into the Banach space $c_0$ QUESTION [8 upvotes]: Problem 1. Characterize compact Hausdorff spaces $K$ for which the Banach space $C(K)$ of continuous real-valued functions embeds into the Banach space $c_0$. Since $c_0$ has separable dual, such $K$ must me countable. So, we can make Problem 1 more precise: Problem 2. Is it true that for every compact countable space $K$ the Banach space $C(K)$ is isomorphic to a subspace of $c_0$? Another possible option: Problem 3. Let $K$ be a compact Hausdorff space. Is it true that the Banach space $C(K)$ is isomorphic to $c_0$ if $C(K)$ is isomorphic to a subspace of $c_0$? REPLY [13 votes]: The Szlenk index is the answer. A space $C(K)$, where $K$ is infinite compact Hausdorff space, is embeddable into $c_0$ if and only if $K$ is homeomorphic to an ordinal below $\omega^\omega$ and if this is the case (and $K$ is infinite) the space itself is isomorphic to $c_0$. So the answer to problem 2 is no however the answer to problem 3 is yes. For details see Rosenthal's chapter in the Handbook of Banach spaces.<|endoftext|> TITLE: Useful invariants of etale topoi not coming from the shape QUESTION [5 upvotes]: Given a "nice" scheme over a number field or a finite field, etale cohomology (in part because of its functoriality) provides some powerful invariants of the underlying scheme. Sometimes non-abelian invariants like etale fundamental group are also useful. All of these IIUC factor (in the sense of functors) through the shape of the etale topos. Are there any useful invariants of the etale topos of a scheme that are not completely determined by the shape of the topos? A possibly useful link. REPLY [10 votes]: As Harry Gindy as said in the comments, there is a refinement of the notion of shape due to Barwick, Glasman and Haine that contains much more information that just the shape. This is not a pro-space, but a pro-(stratified space), and it essentially remembers the Galois groups of every points of your scheme plus how they are glued together along the specialization poset. The paper is available on the ArXiv as C. Barwick, S. Glasman, P. Haine, Exodromy, arXiv:1807.03281. This is maybe a bit unfair to call an "invariant" because it remembers the whole étale topos of a qcqs scheme, thanks to what they call "stratified Hochster duality". Recall that classical Hochster duality tells you that taking the specialization order gives an equivalence between spectral spaces and profinite posets. Then in the exodromyy paper they prove Theorem 10.10 The functor sending a profinite stratified space $\Pi$ to the ∞-topos of constructible sheaves over $\Pi$ is fully faithful. Moreover the essential image consists exactly of the spectral ∞-topoi. The paper gives a concrete description of spectral ∞-topoi and even better it proves that the étale ∞-topos of every quasi-compact and quasi-separated scheme is spectral. Thus the profinite stratified shape of a qcqs scheme describes completely the étale ∞-topos. Even better, the stratified shape of a scheme is always 1-truncated, so you can think of it as a category object in Stone spaces. This is a surprisingly concrete object to manipulate, even if it is not necessarily easy to compute (you need, for example, to know the Galois group of every residue field, and the decomposition group for every specialization).<|endoftext|> TITLE: Arriving at the same result with the opposite hypotheses QUESTION [55 upvotes]: I am pretty distant from anything analytic, including analytic number theory but I decided to read the Wikipedia page on the Riemann hypothesis (current revision) and there is some pretty interesting stuff there: Some consequences of the RH are also consequences of its negation, and are thus theorems. In their discussion of the Hecke, Deuring, Mordell, Heilbronn theorem, (Ireland & Rosen 1990, p. 359) say The method of proof here is truly amazing. If the generalized Riemann hypothesis is true, then the theorem is true. If the generalized Riemann hypothesis is false, then the theorem is true. Thus, the theorem is true!! What is surprising is that both a statement and its negation are useful for proving the same theorem. Do similar situations arise with other major, notorious conjectures in mathematics? I only care about algebraic geometry and algebraic number theory for the most part but I guess it will make little sense to have such questions devoted to each area of mathematics so post whatever you've got. To give an initial direction: are there any interesting statements one can prove assuming both some hard conjecture about motives (e.g. motivic $t$-structure, the standard conjectures, Hodge/Tate conjectures) and its negation? REPLY [27 votes]: Here is a baby answer. Theorem: There exist two irrational numbers $p,q$ such that $p^q$ is rational. Proof: In case $\sqrt 2^{\sqrt 2}$ is rational we can take $p=q=\sqrt 2$. Otherwise take $p= \sqrt 2^{\sqrt 2}$ and $q=\sqrt 2$. We have $p^q = \big(\sqrt 2^{\sqrt 2}\big)^{\sqrt 2} = \sqrt 2^{\sqrt 2 \cdot \sqrt 2} = \sqrt 2^2=2$ is rational.<|endoftext|> TITLE: Conditions on matrices imply that 3 divides $n$ QUESTION [6 upvotes]: A quite popular exercise in linear algebra is the following (or very related exercises, see for example https://math.stackexchange.com/questions/299651/square-matrices-satisfying-certain-relations-must-have-dimension-divisible-by-3 and https://math.stackexchange.com/questions/3109173/ab-ba-invertible-and-a2b2-ab-then-3-divides-n): Let $K$ be a field of characteristic different from 3 and $X$ and $Y$ two $n\times n$-matrices with $X^2+Y^2+XY=0$ and $XY-YX$ invertible. Then 3 divides $n$. A (representation-theoretic) proof can be given as in the answer of Mariano Suárez-Álvarez in https://math.stackexchange.com/questions/299651/square-matrices-satisfying-certain-relations-must-have-dimension-divisible-by-3 . Question: Is this also true for fields of characteristic 3? edit: So it turned out that the result holds for any field. A bonus question might be to find a proof that works independent of the characteristic of the field. REPLY [5 votes]: Let me start from $M^2=[M,Y]$ with $M\in {\bf GL}_n(k)$, as suggested by Andrea. Wlog, I assume that the characteristic polynomial of $M$ splits over $k$, and I decompose $k^n$ as the direct sum of characteristic subspaces $E_\mu=\ker(M-\mu I_n)^n$. It is enough to prove that the dimension of each $E_\mu$ is a multiple of $3$. To this end, observe that $Y$ acts over $E_\mu$. In details, let $x$ be an eigenvector, $Mx=\mu x$. Then $(M-\mu)Yx=\mu^2x$. Because $M$ is invertible, $\mu\ne0$ and therefore $$Yx\in\ker(M-\mu)^2\setminus\ker(M-\mu).$$ Likewise $$(M-\mu)Y^2x=2\mu^2Yx+2\mu^3x,$$ hence $Y^2x\in\ker(M-\mu)^3\setminus\ker(M-\mu)^2$. Eventually, using again ${\rm char}(k)=3$, we have $$(M-\mu)Y^3x=0.$$ We deduce that a basis $E_\mu$ is obtained by taking a basis $\cal B$ of $\ker(M-\mu)$, and adjoining the vectors of $Y\cal B$ and $Y^2\cal B$ ; all of them are linearly independent, as seen above. Thus $\dim E_\mu=3\dim\ker(M-\mu)$.<|endoftext|> TITLE: Order types of models of theories of ordinals QUESTION [8 upvotes]: For $C$ a set of ordinals, let $\mathcal{L}(C)$ be the language with identity, a relation symbol for less than, function symbols for successor, addition, multiplication, and exponentiation, and a constant for every $\alpha \in C$. If $\beta$ is an epsilon number with $\sup (C) \leq \beta$, let $\mathsf{T}_\beta(C)$ be $\{ \phi \in \mathsf{Sent}(\mathcal{L}(C)) : \mathcal{N}_\beta \vDash \phi \}$, where $\mathcal{N}_\beta$ is the model with domain $\beta$ and all symbols in $\mathcal{L}(C)$ interpreted standardly. (We require $\beta$ to be an epsilon number in order to ensure that the required functions are total.) Thus $\mathcal{N}_\omega = \mathcal{N}$, and $\mathsf{T}_\omega(\{ 0 \})$ is true arithmetic. Let us say that a model of $\mathsf{T}_\beta(C)$ is standard if its domain is order-isomorphic to $\beta$ and nonstandard otherwise. We use $\vert A \vert $ to denote the order type of $A$. As is well known, every nonstandard model of true arithmetic has order type $\omega + \vert A \vert \centerdot (\omega^* + \omega)$ for $A$ a dense linear order without endpoints. What can we say about the order types of nonstandard models of $\mathsf{T}_\beta(C)$ for various choices of $\beta$ and $C$? REPLY [4 votes]: First, I'll discuss the case of empty $C$. Observe that the structures $(\varepsilon_{\alpha};+,\cdot,\mathsf{exp})$ are definitionally equivalent with the structures $HF(\alpha;<)$ (the structure of hereditarily finite sets with ordinals $<\alpha$ as urelements). Namely we consider the bijection $f\colon HF(\alpha)\to \varepsilon_{\alpha}$ define by recursion on $HF(\alpha)$: $f\colon \beta\longmapsto \varepsilon_{\beta}$; $f\colon \{a_1,\ldots,a_n\}\longmapsto 2^{f(a_1)}+\ldots+2^{f(a_n)}$, where $f(a_1)>\ldots>f(a_n)$. Using the high expressive power of $HF(\alpha;<)$ it is easy to show that we could define the predicates $f(x)+f(y)=f(z)$, $f(x)f(y)=f(z)$, and $f(x)^{f(y)}=f(z)$. And in $(\varepsilon_{\alpha};+,\cdot,\mathsf{exp})$ we could easily define the predicates $\mathsf{Ur}(f^{-1}(x))$ as "$x$ is an $\varepsilon$-number"; $f^{-1}(x)\in f^{-1}(y)$ as "$y$ isn't an $\varepsilon$-number and $2^x$ is in $2$-base Cantor normal form for $y$" or equivalently as "$y$ isn't an $\varepsilon$-number and $2^{x+1}z+2^{x}\le y< 2^{x+1}z+2^{x+1}$, for some $z$"; $f^{-1}(x) TITLE: Multiplicativity of the homology Atiyah-Hirzebruch spectral sequence for a ring spectrum QUESTION [11 upvotes]: Let $E$ be a ring spectrum and $F$ a connective spectrum. Then we have a convergent Atiyah-Hirzebruch spectral sequence $H_s(F,E_t) \Rightarrow E_{s+t}(F)$. Suppose now that $F$ is also a ring spectrum. Then does the multiplication on the $E^2$ page induce multiplications on all subsequent pages, and do they agree with the multiplication on $E_\ast(F)$? Bizarrely, the only reference I can find for the multiplicative properties of the Atiyah-Hirzebruch spectral sequence is another MO question, and there is only treated the case of the cohomological AHSS where $F$ is a space, rather than the homological spectral sequence where $F$ is a ring spectrum. REPLY [5 votes]: Just a brief response to Tyler and John's answers- but not brief enough to fit into a comment. Here is one way to make the filtered object $\{\tau_{\ge n}E\}$ as structured as you'd like (just spelling out exactly what Tyler indicated). Consider the $\infty$-category $\mathsf{Fun}(\mathbb{Z}, \mathsf{Sp})$ of filtered spectra, where $\mathbb{Z}$ is regarded as a poset. This has a symmetric monoidal structure coming from Day convolution. We have a colimit functor $\mathsf{Fun}(\mathbb{Z}, \mathsf{Sp}) \to \mathsf{Sp}$ and its right adjoint the 'constant tower' functor. The constant tower functor can be promoted to a symmetric monoidal functor $\mathsf{Sp} \to \mathsf{Fun}(\mathbb{Z}, \mathsf{Sp})$ (this is true in general for Day convolution stuff, but here it's extra true since there's only one colimit preserving symmetric monoidal functor from $\mathsf{Sp}$ to any other stable, presentably symmetric monoidal $\infty$-category anyway...). Consider the full subcategory $\mathcal{C} \subseteq \mathsf{Fun}(\mathbb{Z}, \mathsf{Sp})$ spanned by the objects $\{E_n\}$ such that $E_n$ is $n$-connective, i.e. $E_n = \tau_{\ge n}E_n$. Staring at the formula for Day convolution, we learn that this subcategory is closed under the symmetric monoidal structure because smashing $n$-connective and $m$-connective thing gets you an $(n+m)$-connective thing (and $k$-connective things are closed under hocolims). Now general nonsense says we get a colocalization $\mathsf{Fun}(\mathbb{Z}, \mathsf{Sp}) \to \mathcal{C}$ which is canonically lax symmetric monoidal. (HA.2.2.1.1). The composite $\mathsf{Sp} \to \mathsf{Fun}(\mathbb{Z}, \mathsf{Sp}) \to \mathcal{C} \to \mathsf{Fun}(\mathbb{Z}, \mathcal{C})$ is then canonically lax symmetric monoidal and is given on objects by $E \mapsto \{\tau_{\ge n}E\}$. From here you can get whatever you want! For example, this induces a lax symmetric monoidal functor on homotopy categories, which gives you the pairings you needed in Tyler's answer. But it does much more: it also tells you that if you start with $E$, and algebra over any operad $\mathcal{O}$, then the Whitehead tower is canonically a filtered algebra over that operad. (This is not the end of the story, I think. I always get confused about this but, if I remember correctly, people often like to ask for some filtered version of the operad to act on this filtered gadget, in order to get the story of power operations in the spectral sequence? I might be confusing this with something else though... again- I never learned that story properly).<|endoftext|> TITLE: Introduction to Finsler manifolds from the metric geometry point of view (possibly from the Busemann's approach) QUESTION [7 upvotes]: This question is a cross post from Math.SE. I have requested the migration of the question, but unfortunately it is not possible after two months of posting. I also have found this related question, but in my opinion it is not a duplicate from mine. I was reading about geometry in metric spaces from different books, two of them are: (1) A course in metric geometry by Y. Burago, D. Burago and S. Ivanov; and (2) Metric spaces of non-positive curvature by M. Bridson and A. Häfliger. Both develop the Alexandrov's approach to curvature, which uses comparison triangles with the constant curvature model spaces. For a normed space $X$, the following statements are equivalent: $X$ has curvature $\leq\kappa$ in Alexandrov's sense, for some real number $\kappa$. $X$ has curvature $\leq 0$ in Alexandrov's sense. The norm on $X$ is induced by an inner product. So it seems to me that Alexandrov's approach is not very informative in the normed case. On the other hand, a geodesic space has non-positive curvature in the Busemann's sense if its metric is convex, in general this is a weaker notion than Alexandrov's, and in the normed case the following statements are equivalent: $X$ has non-positive curvature in the Busemann's sense. $X$ is uniquely geodesic, that is, every pair of points is joined by a unique geodesic (the linear segment between them). $X$ is strinctly convex, that is, the ball in $X$ is strictly convex which means that for every pair of different vectors $v$ and $w$ of norm equal to $1$ we have that $tv+(1-t)w$ has norm strictly less than $1$ for every $t$ in $(0,1)$. So it seems to me that this weaker notion is the appropriate notion for non-positive curvature in the normed case and I think also for finsler manifolds. I have never studied finsler geometry, but I am very interested in studying metric geometry from this approach. And I do not know where I should start. My question is: What is a good introductory book about finsler manifolds from the metric geometry point of view? What is a good introductory book for the Busemann's approach? If there was not an introductory book available, a reference to an advanced one along with references that cover the necessary background would be very welcome. In Math.SE, user @HK Lee has suggested the paper On intrinsic geometry of surfaces in normed spaces by D. Burago and S. Ivanov. And I have found the following references, although I need the advice of the experts: An introductory textbook by A. Papadopoulos about the Busemann's approach: Metric Spaces, convexity and non-positive curvature. A textbook by H. Busemann: The geometry of geodesics Two interesting papers by H. Busemann: The geometry of finsler spaces and Spaces with non-positive curvature. Thanks in advance! REPLY [3 votes]: Busemann's 1955 book The Geometry of Geodesics is a great presentation of his approach. It includes almost all of the content of his other two papers mentioned in the post. By contrast, Papadopoulos's book is only partly about Busemann's approach to metric geometry. Busemann also wrote a 1970 book called "Recent Synthetic Differential Geometry" with more advanced follow-ups from the intervening years. If you think "the weaker notion is the appropriate notion", you're likely to enjoy Busemann a lot.<|endoftext|> TITLE: Describing fiber products in stable $\infty$-categories QUESTION [14 upvotes]: Let $f\colon X \rightarrow Z$ and $g\colon Y \rightarrow Z$ be two morphisms in a stable infinity category $\mathcal{C}$. How does one show that the $\infty$-categorical fiber product $X \times_Z Y$ can be canonically identified with the fiber of the map $f - g \colon X \oplus Y \rightarrow Z$? I am sure that the question is quite basic (for instance, something quite similar is used implicitly in the proof of Proposition 1.1.3.4 of Lurie's "Higher algebra"), but I would nevertheless appreciate an explanation. The type of answer I am not looking for is "this is trivial consequence of the definitions" because I believe that there is something to check: $X \times_Z Y$ makes sense in any $\infty$-category that has fiber products, whereas the other description uses the stability assumption. While we are at it, let me ask a related bonus question: with $f$ and $g$ as above, suppose that we also have maps $a\colon F \rightarrow X$ and $b\colon F \rightarrow Y$ that make the evident diagram commute and such that $\mathrm{fib}(a) \simeq \mathrm{fib}(g)$ via the induced map (a five lemma type of setting where one of the vertical maps is an isomorphism). Do these conditions imply that the map $F \rightarrow X \times_Z Y$ is an isomorphism? REPLY [3 votes]: A proof of this in the context of stable derivators can be found in my paper Mayer-Vietoris sequences in stable derivators with Moritz Groth and Kate Ponto. It is rather more verbose than Denis's, but one reason for that is that it's intended to be fully precise. In an $\infty$-category, of course, you can't actually just write down some objects and arrows and have a "diagram"; there are coherences that need to be produced too, and kept track of all the way through. Derivators are a way of forcing oneself to remember that. (With that said, it's quite possible that if Denis's proof were made precise it would still be shorter than ours.)<|endoftext|> TITLE: Practical example of Hamiltonian reduction QUESTION [7 upvotes]: I know what is the Liouville integrability: given a Hamiltonian with $n$ degrees of freedom, with $n$ independent constants of motion in involution, the Hamiltonian can be brought to the form $H(p_1, \dots, p_n)$ (i.e. independent on the $q$s) by a canonical transformation. Many persons told me that something similar can be done in the case of one constant of motion, i.e., given a Hamiltonian, if we know that it has one constant of motion (independent on $H$ itself), then we bring the Hamiltonian to the form $H(p_1, q_1, \dots, p_{n-1}, q_{n-1}, p_n)$ (i.e. we remove the dependence on one of the $q$s, $q_n$) by means of a canonical transformation. It should be called "Hamiltonian reduction", however, no one was able to provide me a reference. Someone says that this is called "Marsden-Weinstein-Meyer reduction". I tried to read the corresponding theorem but it looks something different, moreover, it is written in a very technical way, too hard for a simple physicist like me. I found this, saying "reduction by stages", but it refers to a completely different idea of "stages", not the reduction of the degrees one by one. So, I'm looking for one of the following: a reference where I can find the theorem (clearly stated in terms of Hamiltonians and constants of motion), or an example of how to carry out the reduction (also in this case, giving an explicit Hamiltonian and an explicity constant of motion). REPLY [2 votes]: The reduction you require is a (very) special case of Marsden-Weinstein (1974). Your one constant of the motion, say $\psi$, is the moment map of an action of the additive group $G=\mathbf R$ on the symplectic manifold with coordinates say $(x_1,y_1,\dots,x_n,y_n)$ — viz. $\psi$’s hamiltonian flow, obtained by solving $$ \frac{d}{dt}\begin{pmatrix}x_i\\y_i\end{pmatrix}= \begin{pmatrix}-\partial\psi/\partial y_i\\\phantom{-}\partial\psi/\partial x_i\end{pmatrix}. \tag1 $$ Their Theorem 1 says that each level $\psi^{-1}(\mu)$ is a coisotropic submanifold whose null leaves are the $G$-orbits, so the symplectic form descends to the leaf space $\psi^{-1}(\mu)\,/\,G$. This is the reduced space, of dimension 2n – 2. Their Theorem 2 adds that $H$, being constant on leaves since $\{H,\psi\}=0$, descends to a function $H_\mu$ on the reduced space. This is the reduced system. (This is all subject to technical conditions: (1) complete, $\mu$ “weakly regular” value of $\psi$, $G$-action on levels free and proper — which I don’t think matter much in your coordinate formulation. E.g. taking $\psi$ as $p_n$, the subquotient means “fix $p_n$ and ignore $q_n$”, and Darboux charts on the symplectic manifold $\psi^{-1}(\mu)\,/\,G$ give your desired new coordinates $(p_1,q_1,\dots,p_{n-1},q_{n-1})$.) As their introduction points out, this special case $G=\mathbf R$ had long been known as the theory of “ignorable coordinates”, exposed with plenty of examples in e.g. Whittaker (1904, §38 sq). Other nice example from Souriau, who had the theory for abelian $G$ (1970, Chap. III, 12.153 sq): at a negative level of a hydrogen atom’s energy $\psi$, the reduced space is $\smash{\mathrm S^2\times\mathrm S^2}$, and any component $H$ of angular momentum or the eccentricity (a.k.a. Lenz) vector descends there (together they generate an $\mathrm{SO}(4)$ action on the subquotient).<|endoftext|> TITLE: A curious prime counting approximation or just data overfitting? QUESTION [7 upvotes]: I am not sure, if this is a research problem. If not I will move this question to ME: Let $\Omega(n) = \sum_{p|n} v_p(n)$, which we might view as a random variable. Let $E_n = \frac{1}{n} \sum_{k=1}^n\Omega(k)$ be the expected value and $V_n=\frac{1}{n} \sum_{k=1}^n(E_n-\Omega(k))^2$ be the variance. Then $$\pi(n) \approx \frac{n\gamma(\frac{V_n}{E_n},1.4854177\cdot \frac{V_n}{E_n^2})}{\Gamma(\frac{V_n}{E_n})}$$ where $\Gamma=$ Gamma function, $\gamma=$ lower incomplete gamma function. Background: I was trying to fit the gamma distribution to the random variable $\Omega(k)$ ,$1 \le k \le n$. The value $1.4854177$ is fitted to some data. My question is, if there is any heuristic why this approximation should be good, if at all, or if this is just an overfitting problem? Below you can find some sage code which implements this: def Omega(n): return sum([valuation(n,p) for p in prime_divisors(n)]) means = [] variances = [] xxs = [] omegas = [Omega(k) for k in range(1,10^4)] for nn in range(10^4,10^4+3*10^3+1): n = nn omegas.append(Omega(n)) print "---" m = mean(omegas[1:-1]) v = variance(omegas[1:-1]) shape,scale = m^2/v,v/m xx = find_root(lambda xx : n*(lower_gamma(shape,xx*1/scale)/gamma(shape) ).N()-prime_pi(n),1,2) xx = 1.4854177706344873 approxPrimePi2 = n*(lower_gamma(shape,xx*1/scale)/gamma(shape) ).N() primepi = prime_pi(n) print primepi, approxPrimePi2,shape.N(),scale.N(),xx print "---" print "err2 = %s" % (abs(primepi-approxPrimePi2)/primepi) xxs.append(xx) means.append(m.N()) variances.append(v.N()) REPLY [21 votes]: Your heuristic approximation is not correct. It was proved by Turán (1934) that $E_n$ and $V_n$ are both asymptotically $\log\log n$. As a result, the RHS of your display is $$n\frac{\gamma\left(1+o(1),\frac{1.4854177+o(1)}{\log\log n}\right)}{\Gamma\bigl(1+o(1)\bigr)}=n\frac{1.4854177+o(1)}{\log\log n}.$$ On the other hand, $\pi(n)$ is asymptotically $n/\log n$ by the prime number theorem.<|endoftext|> TITLE: Quasi-compact quasi-separated induction? QUESTION [5 upvotes]: I believe I've encountered the statement below, but I've lost my reference and am unable to find another one. So, I'm posting this question to see if someone can give a reference, or at least confirm the statement is true (or make a correction). Proposition: Let $\mathcal{C}$ be a class of schemes with the following properties: $\mathcal{C}$ contains all affine schemes If $X$ is a scheme, $\{ U, V \}$ is an open cover of $X$, and $\mathcal{C}$ contains the three schemes $U$, $V$, and $U \cap V$, then $\mathcal{C}$ contains $X$. Then $\mathcal{C}$ contains every quasi-compact quasi-separated scheme. Conversely, the class of quasi-compact quasi-separated schemes has the properties above. REPLY [4 votes]: This is proposition 3.3.1 and remark 3.3.2 of Generators and representability of functors in commutative and noncommutative geometry by A. Bondal, M. van den Bergh. I originally found this reference via the MO question The biggest class of schemes which the reduction principle holds, and I have now found it again by browsing the the Related links to my question, so hurrah for the MO system.<|endoftext|> TITLE: Product of $q$-analogues QUESTION [8 upvotes]: Background Recall that the $q$-analogue $[n]_q\in\mathbb Z[q]$ of a natural number $n\in\mathbb N$ is defined as $$ [n]_q := \frac{q^n -1}{q-1}$$ the idea being that formulas involving $q$ will specialize along $\mathbb Z[q]/(q-p^n)$ to counting formulas about finite fields, while specializing along $\mathbb Z[q]/(q-1)$ will yield counting formulas about finite sets. For example, defining the $q$-factorial by $$[n]_q! := [1]_q [2]_q \dotsm [n]_q,$$ we have that the number of $k$-dimensional subspaces of $\mathbb F_q^n$ is $$\#\mathrm{Gr}_{n,k}(\mathbb F_q) = \binom nk_q :=\frac{[n]_q!}{[n-k]_q![k]_q!}$$ and $\binom nk_q$ reduces to $\binom nk$ when $q\to 1$. Question Now let $p$ be a fixed prime. In my work, I've come across the expressions $$ [p]_q^{k_1} [p^2]_q^{k_2} \dotsm [p^n]_q^{k_n} $$ with $k_i\ge0$, $i=1,\dotsc, n$, as well as funny things like $$ [p^r]_{q^{p^s}} = \frac{q^{p^{r+s}}-1}{q^{p^s}-1}$$ Do these have any known combinatorial interpretation? I will usually want to consider the above product when the values of $n$ and $\sum k_i$ are fixed, so we can consider it ranging over all partitions of $\sum k_i$ into $n$ non-negative integers. Motivation Ultimately I'm interested in the specialization $\mathbb Z[q] \to \mathbb Z_p[\![q-1]\!] \to \mathbf A_{\mathrm{inf}}(R)$, where $R$ is a perfectoid ring containing a compatible choice of roots of unity $\zeta_{p^\infty}$. Letting $$\epsilon = (1,\zeta_p,\dots) \in R^\flat,$$ the structure map $\mathbb Z[\![q-1]\!] \to \mathbf A_{\mathrm{inf}}$ is given by $q\mapsto[\epsilon]$. Then the Fontaine map $$\tilde\theta_r\colon \mathbf A_{\mathrm{inf}}(R) \to W_r(R)$$ has kernel generated by $[p^r]_q$ (Example 3.16), while the map $\theta_r = \tilde\theta_r \varphi^r$ has kernel generated by $[p^r]_{q^{1/p^r}}$. You can get the product of these ideals, above, out of some $RO(S^1)$-graded THH calculations. REPLY [2 votes]: Recall Legendre's formula $$ v_p(n!) = \sum_{s=1}^\infty\left\lfloor\frac n{p^s}\right\rfloor = \sum_{r=0}^\infty a_r[r]_p $$ where $n = \sum a_r p^r$ is the base-$p$ expansion of $n$. A $q$-analogue of this formula is provided by Lemma 4.8 of The $p$-completed cyclotomic trace in degree 2, which states that \begin{equation}\label{AClB}\tag{$\ast$} \large [n]_q! = u\prod_{r=1}^\infty \varphi^{r-1}([p]_q)^{\lfloor n/p^r\rfloor} \end{equation} for a unit $u\in\mathbb Z_p[\![q-1]\!]^\times$. While my computations yielded products $$ [p]_q^{k_1} [p^2]_q^{k_2} \dotsm [p^n]_q^{k_n}, $$ for arbitrary $k_i\ge0$, the ones most relevant to my eventual application (the regular slice filtration on THH) were the principal ideal generated by the RHS of \eqref{AClB} (when $p$ divides $n$).<|endoftext|> TITLE: Properness of moment map QUESTION [5 upvotes]: Suppose that a torus $T$ acts on a non-compact symplectic manifold $M$. Assume that this action is Hamiltonian and that the fixed point set of $T$ is compact. Let $\mu:M\to\mathfrak{t}^{*}$ denote the moment map of the action, where $\mathfrak{t}$ denotes the Lie algebra of $T$. If there exists an $X\in\mathfrak{t}$ such that $\mu(X):M\to\mathbb{R}$ is a proper function that is bounded below, why is $\mu:M\to\mathfrak{t}^{*}$ necessarily proper? REPLY [3 votes]: This follows from a point set topology argument. The assumptions that the fixed point set is compact, or that $\mu(X)$ is bounded below, are not necessary. Consider the linear map $F: \mathfrak{t}^{*} \rightarrow \mathbb{R}$ given by $L \mapsto L(X)$. Note that $\mu(X) = F \circ \mu$. Suppose $C \subset \mathfrak{t}^{*}$ is compact, then $F(C)$ is compact since $F$ is continuous. Now, $\mu^{-1}(C) \subset \mu(X)^{-1}(F(C))$, which is compact since $\mu(X)$ is proper by assumption. $\mathfrak{t}^{*}$ is homeomorphic to a Euclidean space, so $C$ is closed and bounded in $\mathfrak{t}^{*}$. Since $\mu$ is continuous, $\mu^{-1}(C)$ is closed. On the other hand, we saw in the previous paragraph that it was contained in the compact subset $\mu(X)^{-1}(F(C))$, so it is itself compact. Hence, $\mu$ is proper.<|endoftext|> TITLE: Covering manifolds with some other manifolds QUESTION [5 upvotes]: Let $M$ ,$N$ be $n$-dimensional manifolds. Let $D_{1},D_{2},\dots D_{k} $ be $n$-dimensional manifolds embedded in $M$ and $\cup_{i=1}^kD_{i}=M$ and each $D_i$ is homeomorphic to $N$. Question is following. Problem Consider all embedding of $\cup_{i=1}^kD_{i}$. Determine the minimum of the value $k$. Example When we consider $M$ is 2-dimensional torus and each $D_i$ is a 2-dimensional disk, the minimum value is 3. I thought $M=L(5,2)$ and each $D_i$ is a 3-dimensional ball, and I couldn't determine the minimum. Can you answer? REPLY [10 votes]: A first observation is that such a $k$ may not exist, for example if $N$ does not embed in $M$. When $N$ is a disk, then $k$ equals the ball category of $M$, denoted $\operatorname{ballcat}(M)$ or $\operatorname{bcat}(M)$ in the literature. See for instance Gavrila, Caius, The Lusternik-Schnirelmann theorem for the ball category, Cornea, O. (ed.) et al., Lusternik-Schnirelmann category and related topics. Proceedings of the 2001 AMS-IMS-SIAM joint summer research conference on Lusternik-Schnirelmann category in the new millennium, South Hadley, MA, USA, July 29-August 2, 2001. Providence, RI: American Mathematical Society (AMS). Contemp. Math. 316, 113-119 (2002). ZBL1029.58007 or Chapter 3 of the book Lusternik-Schnirelmann category by Cornea, Lupton, Oprea and Tanré. In particular if $M$ is a smooth manifold then there is a chain of inequalities $$ \operatorname{Crit}(M)\geq \operatorname{ballcat}(M)\geq \operatorname{cat}(M), $$ where $\operatorname{cat}(M)$ is the Lusternik-Schnirelmann category of $M$ (the minimal $k$ such that $M$ has a cover by $k$ open sets which are contractible in $M$) and $\operatorname{Crit}(M)$ denotes the minimal number of critical points for any smooth function on $M$. In your example of $M=L(5,2)$ a $3$-dimensional lens space, $\operatorname{cat}(M)=4$ by a result of Fadell and Husseini and $\operatorname{Crit}(M)\le 4$ by a result of Takens (in general for a connected $n$ manifold $\operatorname{Crit}(M)\le n+1$). Hence $\operatorname{ballcat}(M)=4$.<|endoftext|> TITLE: Fibre of GIT morphism QUESTION [5 upvotes]: Let $ V $ be an affine variety (over $ \mathbb C$) with an action of a reductive group $ G$. I would like to consider the morphism $$ \pi : V \rightarrow V // G = Spec \, \mathbb C[V]^G $$ Let $ v \in V $. Assume that the orbit $ Gv $ is closed in $ V $. Assume also that the stabilizer of $ v $ in $ G $ is finite. Question: Is the following true? What additional hypothesis should I place on $ v $ in order to ensure that the following is true? The scheme-theoretic fibre $ \pi^{-1}(\pi(v)) $ equals $ G v $. I looked in Mumford's book, but I could not find this. Example: Here is an example where this does work. Suppose that $ G = SL_k $ and $ V = Hom(\mathbb C^k, \mathbb C^n) $ and let $ v $ be an injective map. Then $ V // G $ is the variety of pure $k$-tensors (the cone on the Grassmannian). REPLY [2 votes]: I am just recording what was said in the comments so this question does not appear completely unanswered. Let $\pi_X:X\to X//G$ be the GIT quotient of an affine variety over $\mathbb{C}$ by a reductive group $G$. WLOG assume the action is effective. First, a point is properly stable if its orbit is closed and it has finite stabilizer. The locus of properly stable points is Zariski open. Then since each fibre $\pi_X^{-1}(\pi_X(x))$ is a union of orbits, this union contains a unique closed orbit, two such orbits are in the same fibre if and only if their closures intersect, and such intersections can be detected by 1-parameter subgroups, we can conclude that if $x$ is properly stable then $\pi_X^{-1}(\pi_X(x))$ is set-theoretically the orbit $Gx$. Please see Stability of Affine G-varieties and Irreducibility in Reductive Groups by Casimiro and Florentino as a reference. As pointed out by Jason Starr in the comments, the fibre is not generally scheme-theoretically the orbit however. A counter-example is the action of $\mathbb{G}_m$ on $\mathbb{A}^2\times \mathbb{G}_m$ by $s\cdot((x,y),t)=(sx,sy,s^{-n}t)$ for $n>1.$ As noted by the OP, this is apparent even for $n=2$. We now refer to the Luna Slice Theorem; see Luna’s slice theorem and applications by Drézet. Let $V$ be a slice at a properly stable point $x$, and let $\pi_V:V\to V//S$ be the corresponding quotient where $S$ is the stabilizer of $x$ (necessarily a reductive subgroup). Then there is an isomorphism: $$G\times_S \pi^{-1}_V(\pi_V(x))\cong \pi_X^{-1}(\pi_X(x)).$$ So, the fibre is the scheme-theoretic orbit if it is smooth which, by the Chevalley-Shephard-Todd Theorem, occurs if and only if the stabilizer is generated by pseudoreflections.<|endoftext|> TITLE: Reference request: Extensions of Wiener's Tauberian Theorem QUESTION [10 upvotes]: Wiener's Tauberian Theorem says that linear combinations of translations of a function $f$ are dense in $L^1(\mathbb{R})$ if and only if the zero set of the Fourier transform of $f$ is empty. This is an old theorem, from 1932(ish). I am interested in generalizations such as: When can a nonnegative $L^1$ function approximated by nonnegative linear combinations of translations of $f\ge 0$? What about convex combinations for approximating a density function? Other spaces: What about $L^1(X)$ or $L^p(X)$ for $X\ne\mathbb{R}$, $p\ne1,2$? (This question is also related.) I am looking for any modern reference covering these types of extensions, or other interesting extensions of this theorem. Edit: I eventually found this tome, which covers Tauberian theory in great detail, but not necessarily the approximation problems above. At 500 pages, it will take me some time to go through it. Korevaar, Jacob, Tauberian theory. A century of developments, Grundlehren der Mathematischen Wissenschaften 329. Berlin: Springer (ISBN 3-540-21058-X/hbk). xvi, 483 p. (2004). ZBL1056.40002. REPLY [9 votes]: Let me record a couple of simple observations regarding question 1. First, it is clear that the answer is "very rarely", since there are very simple obstructions. For example, the step function $\mathbb{1}_{(0,1)}$ cannot be approximated by positive linear combinations of translates of $\mathbb{1}_{(0;3)}$. Indeed, if such a translate equals 1 at some point of $(0,1)$, then it is identically $1$ either on $(-1;0)$ or on $(1,2)$. Hence, any positive linear combination that is $\geq3/4$ at some point of $(0,1)$ will be $\geq 3/8$ either at all points of $(-1,0)$, or at all points of $(1,2)$. Admittedly, the Fourier transform of $\mathbb{1}_{(0;3)}$ does have zeros, but this is not relevant - the same argument can be adapted e. g. to Gaussians. Second, assuming without loss of generality that $\int f=\int g=1$, I can try to find a probability measure $\mu$ on $\mathbb{R}$ such that $$ f\star \mu = g. $$Passing to Fourier transforms, this gives $\hat{\mu}=\hat{g}/\hat{f}$. We are asking how to determine whether a given function is a Fourier transform of a probability measure, which is classical - the characterization is given by Bochner's theorem and a useful sufficient condition - by Polya's criterion, see e. g. Durrett, Thm 3.3.22. Note that, in particular, we must have $|\hat{\mu}(t)|\leq \hat{\mu}(0)=1$, so that if $|\hat{f}(t)|<|\hat{g}(t)|$ for some $t$, they we have no solutions. I guess if I have a sequence of probability measures such that $f\star \mu_n \to g$ in $L_1$, then $\mu_n$ must be tight and I could pass to a weakly convergent subsequence, and the limit must satisfy the equation above. So, Bochner's theorem really gives an "if and only if" characterisation.<|endoftext|> TITLE: Cotangent Complex in Analytic Category QUESTION [13 upvotes]: I am looking for a reference which develops the theory of the cotangent complex for complex analytic spaces. I need this to justify some computations I did assuming some formal properties which hold in the algebraic category. Here is a list of properties I want (some might be deducible from others, I haven't checked this carefully): (1) To each analytic space $X$, assign a well-defined object $\mathbb{L}^\bullet_X$ in the derived category of complexes of $\mathcal O_X$-modules with coherent cohomology sheaves, supported in degrees $\le 0$. (2) For smooth $X$, we must have a (quasi-)isomorphism $\mathbb L^\bullet_X = \Omega^1_X[0]$, where the right hand side is the holomorphic cotangent sheaf in degree $0$. For each morphism $f:X\to Y$, we have a morphism in the derived category $f^\dagger:Lf^*\mathbb L^\bullet_Y\to\mathbb L^\bullet_X$ (which is the obvious one if $X,Y$ are smooth, i.e., the adjoint of $df$ between the cotangent spaces) and whose cone defines the relative cotangent complex $\mathbb L^\bullet_{X/Y}$ (or $\mathbb L^\bullet_f$), which is an object living on $X$. The assignment $f\mapsto f^\dagger$ respects composition of maps. (3) For a closed immersion $X\subset Y$ with $Y$ smooth and $\mathscr I$ the ideal sheaf of $X$ in $Y$, we have $h^0(\mathbb L^\bullet_{X/Y})=0$ and $h^{-1}(\mathbb L_{X/Y}^\bullet) = \mathscr I/\mathscr I^2$. Further, the truncation $\tau_{\ge-1}\mathbb L^\bullet_X$ can be identified with $d:\mathscr I/\mathscr I^2\to\Omega^1_Y|_X$ (living in degrees $-1$ and $0$) in a way that is compatible with the exact triangle $Lf^*\mathbb L^\bullet_Y\to\mathbb L^\bullet_X\to\mathbb L^\bullet_{X/Y}\to Lf^*\mathbb L^\bullet_Y[1]$ coming from part (2). (4) More generally, for any sequence of maps $X\to Y\to Z$, there's an exact sequence $Lf^*\mathbb L^\bullet_{Y/Z}\to\mathbb L^\bullet_{X/Z}\to\mathbb L^\bullet_{X/Y}\to Lf^*\mathbb L^\bullet_{Y/Z}[1]$ where $f:X\to Y$ is the first map in the above. (5) Crucially, for a morphism $X\to Y$ of algebraic schemes, $\mathbb L^\bullet_{X/Y}$ should be functorially (quasi-)isomorphic to the analytification of the cotangent complex defined in the Stacks Project (tag 08T1). I'm actually happy if I just have a reference which treats the corresponding statements for just $\tau_{\ge-1}\mathbb L^\bullet_X$ (called the "Naive cotangent complex" in the Stacks Project). Also, if you can't find any reference which treats this, I'd also find useful a sketch of how to go about proving this in the analytic category. Apologies in advance if this is too basic a question, I'm rather new to these concepts. REPLY [6 votes]: I will attack the problem three ways, in increasing level of elaboration. 1. Here is a recent paper that proves all this. Compared to older sources, this paper uses more machinery. It uses stabilization / Goodwillie calculus, which is more directly about deformations and has to be proved to be about tangents, rather than vice versa. It uses derived algebraic geometry, which is hardly more work because, say, Illusie, was secretly building DAG. But it puts that in the foreground, especially the language of ∞-categories. Whereas, you might prefer black box statements about ordinary categories that bury the ∞-categories in proofs. 2. Lots of sources (eg, the Stacks project and Illusie) define the cotangent complex for ringed spaces (and ringed topoi). So that gives you a definition. They probably prove (4) in that generality. Illusie II 6.6.2.3 explicitly mentions the analytic case, but maybe he just says that it’s easy and doesn’t prove anything. Someone must have elaborated on this, but I don’t know where. 2a. As I understand it, Illusie takes the cotangent complex for rings and (homotopy) sheafifies it. This is difficult, but once he has created a coherent global object, we can study it locally and reduce to statements about rings. I believe that I have a slick proof of the comparison (5). Consider the map from the analytic space to the algebraic space. This has a relative cotangent complex. Since the map of rings is formally étale this cotangent complex is locally zero (ie, acyclic) to the sheaf is globally zero. Now apply (4) and the acyclic cone implies that the comparison map (5) is a quasi-isomorphism. Moreover, (5) implies (2)+(3) because every analytic morphism is locally isomorphic to an algebraic morphism. 2b. To build the analytic theory on its own and calculate the cotangent complex (2) and (3) without recourse to comparison (5), we would like to reduce to the case of rings. Rings of analytic rings are very similar to algebraic rings and (2) and (3) follow mutatis mutandis. We start with rings and sheafify, but we have to worry that sheafification is destructive. What we really want for a nice theory is that the presheaf that we are sheafifying was already a (homotopy) sheaf (ie, satisfied Mayer-Vietoris); we want sheafification to extend it to a global object, but not to change it on nice open sets (ie, Stein or affine ones). This is not true in general, as in the example of big topoi. But I believe that it is true for the Zariski and analytic topologies because the restriction map of the structure sheaf is formally smooth, so the restriction of the cotangent complex is just given by tensoring. Thus we can reduce to the cotangent complex of rings. (Actually, maybe you don’t need the general statement, but could do (2) and (3) by hand because (2) is a sheaf in a single degree, which is a homotopy sheaf and (3) is more subtle, but you only care about a single cohomology sheaf, which is a sheaf; you just need to check that the lower cohomology doesn’t mess it up.) Luc Illusie. Complexe cotangent et déformations. I, 1971, LNM. 239; II, 1972, LNM 283.<|endoftext|> TITLE: Textbook recommendation: Metric Geometry QUESTION [7 upvotes]: I’m currently reading Burago, Burago, Ivanov’s book A Course in Metric Geometry. I’m really enjoying it so far - what would be a good continuation to the book once I’m done? REPLY [10 votes]: As you have been reading Loh's book, I recommend you take a look at Metric spaces of non-positive curvature by M. Bridson and A. Häfliger. See also An invitation to Alexandrov geometry: CAT(0) spaces by S. Alexander, V. Kapovitch and A. Petrunin, or its older version.<|endoftext|> TITLE: Concrete example to illustrate the theory about blocks of groups with cyclic defect groups QUESTION [6 upvotes]: I'd like to to have a concrete example to illustrate the theory about blocks of groups with cyclic defect groups. Thus, I am looking for a finite group $G$ and a prime $p$ dividing $|G|$ satisfying all of the following properites: The defect group $D$ of a $p$-block $B$ of $G$ is cyclic and of order $p^r$, where $r\geq 2$ $G$ is neither $p$-solvable nor solvable (or at least not solvable) The shape of the graph of the Brauer tree associated to $B$ is not only a (part of a) star, but more complicated (and, if it doesn't make things too complicated, the Brauer tree associated to $B$ has an exceptional vertex) All inclusions are proper in the chain $1 \leq D_1 \leq D\cong C_{p^r} \leq N_G(D)\leq N_G(D_1)\leq G $, where $D_1$ is the unique subgroup of $D$ of order $p$ All inclusions are proper in the chain $C_G(D_1) \leq T(c) \leq N_G(D_1)$ It is still possible to express these groups relatively nicely, such that one does not have to say that some group is an extension of a semidirect product of the double cover of the first non-split extension of... with ...of... Remark for the penultimate point: Let $b$ be the block of $N_G(D_1)$ which is Brauer-correspondent to $B$, let $c$ be a block of $C_G(D_1)$ covered by $b$ and let $T(c)$ be the inertial goup. Unfortunately, I'm not aware of such an example. Neither was I able to construct one. If it is not possible to find an example with all inclusions proper, then maybe with not all but many of them? Thanks for the help. REPLY [6 votes]: You probably know this, but your conditions can't be met by any principal $p$-block of any finite group $G$ which is not $p$-solvable, although I think the Classification of Finite Simple Groups (CFSG) is necessary for that. For suppose that $G$ is a finite group with cyclic Sylow $p$-subgroup $D$ and with $|D| >p,$ but that $G$ is not $p$-solvable. Then the principal $p$-block $B$ of $G$ has defect group $D$, and this does not change on passage to $G/O_{p^{\prime}}(G),$ so we might as well suppose that $O_{p^{\prime}}(G) = 1.$ Now $O_{p}(G) = 1,$ for otherwise we have $D_{1} \lhd G,$ and then $C_{G}(D_{1}) \lhd G.$ But $C_{G}(D_{1})$ has a (characteristic) normal $p$-complement ( which must be trivial) as $O_{p^{\prime}}(G) = 1.$ Then $C_{G}(D_{1}) = D \lhd G,$ and $G$ is certainly $p$-solvable, a contradiction. Let $H$ be a Hall $p^{\prime}$-subgroup of $N_{G}(D).$ Then $D = [D,H] \times C_{D}(H)$ since $D$ is Abelian of order coprime to $|H|$. Since $D$ is cyclic, we either have $C_{D}(H) = D$ or $C_{D}(H) = 1.$ In the former case, $N_{G}(D)$ has a normal $p$-complement, and then so does $G,$ contrary to the fact that $G$ is not $p$-solvable. Hence $D = [D,H] \leq G^{\prime}.$ Now $G^{\prime}$ is not $p$-solvable, as $G$ is not, so we then obtain $D \leq G^{\prime \prime}$ by the same argument, and ultimately $D \leq G^{(\infty)},$ the terminal member of the derived series for $G$, by repeating the argument. Let $M$ be a minimal normal subgroup of $G.$ Then $M$ has order divisible by $p$, so that $D_{1} \leq M.$ Also $M$ is not a $p$-group, so $M$ is non-Abelian simple (using the fact that $M$ has cyclic Sylow $p$-subgroup). Then $G = MN_{G}(D_{1})$ by a Frattini (type) argument.Now $MC_{G}(D_{1}) \lhd G$ and $G/MC_{G}(D_{1})$ is a homomorphic image of the Abelian $p^{\prime}$-group $N_{G}(D_{1})/C_{G}(D_{1}),$ so is itself Abelian of order prime to $p$. Now we have $D \leq MC_{G}(D_{1})$ and $MC_{G}(D_{1})/M$ (being a homomorphic image of $C_{G}(D_{1})$ ) has a normal $p$-complement. But $MC_{G}(D_{1})$ is certainly not $p$-solvable, so arguing as before, we have $D \leq [MC_{G}(D_{1})]^{\prime}.$ Hence $MC_{G}(D_{1})$ has no factor group of order $p$, so that $MC_{G}(D_{1})/M$ is a $p^{\prime}$-group. Now $G/M$ is a $p^{\prime}$-group, so that $D \leq M.$ Now $G = MN_{G}(D)$ by a Frattini argument. I claim that we now have $N_{G}(D) = N_{G}(D_{1}).$ This is the (first and only) time we require CFSG, though the previous argument could have been shortened considerably using CFSG. Note that $N_{G}(D_{1}) = N_{M}(D_{1})N_{G}(D),$ so to prove the claim, it suffices to prove that $N_{M}(D_{1}) = N_{M}(D).$ But it is a Theorem of H. Blau that whenever a finite non-Abelian simple group $X$ has a cyclic Sylow $p$-subgroup $P \neq 1$, then $P$ is TI in $X$, that is, $P \cap P^{x} = 1$ for all $x \in X \backslash N_{X}(P)$. In our situation, this immediately yields that $N_{M}(D_{1}) = N_{M}(D),$ as required. Hence your fourth bulletpoint can't be satisfied (for principal blocks of non-$p$-solvable groups) as remarked in comments.<|endoftext|> TITLE: The distribution of the area of a region cut out by chordal SLE? QUESTION [6 upvotes]: Let $\mathbb{D}$ be the unit disc. Let $a,b \in \partial \mathbb{D}$. Let $\gamma$ be a chordal $SLE_{k}$ from $a$ to $b$. For $k \leq 4$, $\gamma$ is a simple curve, and so $\mathbb{D} \setminus \gamma$ has two components. Say that $A$ is the left-component of $\mathbb{D} \setminus \gamma$ when we traverse $\gamma$ from $a$ to $b$. Is there anything known about the distribution of the (Euclidean) area, $|A|$, of the random set $A$? Of course this depends on $(a,b)$. Here are some more concrete questions: If $a,b$ are antipodal, then by symmetry the expected area of $A$ is $|\mathbb{D}|/2$. What is the expected area when $a, b$ are not antipodal? I know that one can apply a conformal transformation to get back the antipodal case, but it feels unlikely that such a transformation will distort equal areas in the same way. In particular, if $a$ and $b$ were close, I would be surprised if the distribution was symmetric. What is $Var |A|$, as a function of $(a,b)$? REPLY [3 votes]: The expected area of $A$ is easy to compute, in principle explicitly: $$ \mathbb{E}(\text{Area}(A))=\mathbb{E}\left(\int_\mathbb{D}\mathbb{1}_{z\in A}\right)=\int_\mathbb{D}\mathbb{P}(z\in A). $$ The probability inside the integral is given by so-called Schramm's formula and can be computed by standard SLE techniques: since in the half-plane geometry, the probability that $z$ is to the left of the curve is scale invariant, it is given by $F(\arg(z))$ for some function $F$. On the other hand, conditionally on the curve up to time $t$, it is $F(\arg(g_t(z)-\sqrt{\kappa}B_t)).$ Hence the latter quantity is a martingale. Applying Ito's formula, this yields after some computations $$ \frac{1}{2}F''(\theta)+\left(1-\frac{4}{\kappa}\right)\tan(\theta)F'(\theta)=0, $$ which, together with boundary conditions $F(0)=1$, $F(\pi)=0$, gives an explicit formula for $F$ (for $\kappa=4$ we actually have $F(\theta)=1-\theta/\pi$, and for other $\kappa$ we have a hypergeometric function). Since the quantity in question is Mobius invariant, in the disc geometry it is given by $F(\text{hm}_z(ba))$, where $\text{hm}$ is the harmonic measure. In order to compute variance in a similar way, one would need to compute two-point function. This function is in general not known: although by an argument as above one can derive some PDEs for it, they are hard to solve in a closed form.<|endoftext|> TITLE: A map of non-pathological topology? QUESTION [38 upvotes]: I think of topological spaces as coming in several "islands of interestingness" (the CW island, the Zariski archipelago,...) dotting a vast "pathological sea" (the long line ocean, the gulf of the lower limit...). That is, I only know how to think about a topological space if it happens to live on one of these islands, the methods appropriate to one island may be completely unrelated to those of another, and a "random" topological space is probably unrelated to anything I know how to think about and is thus "pathological". I'd like to get a better perspective on how many of these islands there are -- and perhaps whether some which I think are distinct are actually connected by some isthmus. Here's what my current map looks like: CW complexes (and spaces homotopy equivalent to such) Zariski spectra of commutative rings (and schemes) Stone spaces (totally disconnected compact Hausdorff spaces) Infinite-dimensional topological vector spaces (and spaces locally modeled on them) One of the characteristic features of this map is that there is little overlap between the islands although there is overlap between these islands in a literal sense, what really sets them apart is that the tools used in exploring one island bear little resemblance to those used for another. For example, when studying spaces using CW complex tools, non-Hausdorffness is regarded as pathological, clopen sets as uninteresting, and infinite-dimensionality as an annoyance whereas such features are respectively embraced when studying Zariski spectra, Stone spaces, and functional-analytic spaces. Questions: Are there other classes of topological spaces which are interesting to study (and not just as a source of pathologies)? Are these islands less isolated than I'm making them out to be? E.g. are there interesting topological considerations to be made which apply simultaneously to, say, Banach spaces and Stone spaces? Is it correct to think that the ocean is vast, i.e. that "most" topological spaces are "pathological"? REPLY [16 votes]: I'll go ahead and say that Polish spaces are an interesting and almost sui generis class. There is a rich literature of applications to and from descriptive set theory, with layers of "pathology" hierarchically organized along lines closely related to the arithmetical and analytic hierarchies. They are also widely used, as Nate mentioned, in abstract measure and probability theory. I would say there are isthmuses connecting this class to the class of continua (mentioned by D.S. Lipham; consider for example the theory of pseudo-arcs) as compact connected metric spaces, and to some extent to locally compact Hausdorff spaces (e.g., a locally compact Hausdorff space is Polish iff it is second countable). REPLY [10 votes]: o-minimal structures, as generalisation of “tame” topology of semi-algebraic and semi-analytic sets, are quite interesting, in applications in particular. See e.g. the book https://books.google.co.uk/books/about/Tame_Topology_and_O_minimal_Structures.html<|endoftext|> TITLE: What is a name for co-Sobczyk Banach spaces? QUESTION [5 upvotes]: Definition. Let us define a Banach space $X$ to be co-Sobczyk if every linear bounded operator $T:Z\to c_0$ defined on a separable subspace $Z$ of $X$ extends to a bounded operator $\bar T:X\to c_0$. By the classical Sobczyk Theorem, each separable Banach space is co-Sobczyk. But the class of co-Sobczyk spaces includes many non-separable Banach spaces. In particular, a Banach space $X$ is co-Sobczyk if each separable subspace of $X$ is contained in a complemented separable subspace. So, all classical Banach spaces $c_0(\Gamma)$ and $\ell_p(\Gamma)$ for $1\le p<\infty$, are co-Sobczyk for any set $\Gamma$. I have a strong feeling that co-Sobczyk spaces have been studied in the theory of non-separable Banach spaces, so asking the MO commubnity for a proper reference and an existing terminology (I suspect that co-Sobczyk spaces are called differently). REPLY [5 votes]: Nigel Kalton studied a similar but stronger notion: for $\lambda \geqslant 1$, he termed a Banach space $X$ to have the $(\lambda, \mathcal{C})$-extension property, when for any compact space $K$ you may find extensions of operators $T$ from subspaces of $X$ into $C(K)$ to operators from $X$ to $C(K)$ with norm at most $\lambda \|T\|$. It is thus natural to term your spaces as having the $(\lambda, c_0)$-separable extension property if you care about the extension constant. Update: Correa and Tausk call this separable $c_0$-extension property.<|endoftext|> TITLE: Subspaces of metric spaces having prescribed dimension QUESTION [6 upvotes]: Let $(X,d)$ be a metric space having Hausdorff dimension $\alpha>0$ and let $0<\beta<\alpha$. Is there a metric subspace of $X$ having Hausdorff dimension $\beta$? REPLY [2 votes]: By Corollary 7 in [How95], every analytic subset of a complete separable metric space which has positive (or infinite) Hausdorff measure of dimension s contains a compact set which has finite and positive Hausdorff measure of dimension s. [How95] J.D. Howroyd. On dimension and on the existence of sets of finite positive Hausdorff measure. Proc. London Math. Soc. (3), 70:581–604, 1995.<|endoftext|> TITLE: Projective subvarieties of a quasiprojective variety QUESTION [5 upvotes]: Let $X$ be a quasiprojective variety over $\mathbf C$. Take the union of all projective subvarieties $W \subseteq X$ that have dimension at least $1$. Is the result Zariski closed? (I was wondering this in the particular setting $X = \mathcal M_g$, where the projective subvarieties have been the subject of some study. But the general question seems natural as well.) REPLY [10 votes]: No. For example take $X = \mathbb{A}^1 \times \mathbb{P}^1$ minus one point, say $(x,y)$. Then $W = (\mathbb{A}^1 \setminus x) \times \mathbb{P}^1$ is not closed.<|endoftext|> TITLE: Quantifier elimination in uncountable elementary "Fraïssé classes" QUESTION [7 upvotes]: Let $\mathcal{L}$ be an infinite relation language (this question is trivial in a finite relational language). Suppose that $\mathcal{K}$ is the class of finite models of some $\mathcal{L}$-theory (since we're only concerned with finite models this is equivalent to saying that for each $n$ the class of structures in $\mathcal{K}$ of size $n$ is closed under ultraproducts). Furthermore suppose that $\mathcal{K}$ satisfies all of the properties of Fraïssé classes, except for the countability requirement, i.e. it satisfies the hereditary property, the joint embedding property, and the amalgamation property. Call such a class an uncountable elementary Fraïssé class. Despite the fact that $\mathcal{K}$ does not in general have a unique countable generic structure associated to it, there is still a unique generic theory $T_\mathcal{K}$ associated to it which is formally analogous to the theory of a Fraïssé limit. If $\mathcal{L}$ is countable this theory can be described indirectly in terms of a game: Let $X$ be a set of complete $\mathcal{L}$ theories. Player I picks a structure $\mathfrak{A}_0 \in \mathcal{K}$. Then the players alternate picking $\mathfrak{A}_{i+1}\supseteq \mathfrak{A}_i$ with $\mathfrak{A}_{i+1} \in \mathcal{K}$. Player II wins if $\mathrm{Th}\left(\bigcup_{i<\omega} \mathfrak{A}_i\right) \in X$ and player I wins otherwise. $T_\mathcal{K}$ is the unique theory such that player II has a winning strategy when $X=\{T_\mathcal{K}\}$. Equivalently, $T_\mathcal{K}$ is the unique theory such that player II has a winning strategy if and only if $T_\mathcal{K} \in X$ and player I has a winning strategy if and only if $T_\mathcal{K} \notin X$. For an uncountable language this describes $T_\mathcal{K}$ 'locally' in each countable reduct of $\mathcal{L}$. (There might be a subtlety here in that the game needs to be played with structures in the full language and only victory is checked in the countable reduct.) A more set theoretical way of constructing $T_\mathcal{K}$ is to force to make $\mathcal{K}$ countable. Then in the forcing extension $\mathcal{K}^V$ literally is a Fraïssé class and has a unique Fraïssé limit. You can show that the theory of this structure is independent of the generic, so in particular it's definable in the ground model. I think this should be basically the same as treating the normal construction of a Fraïssé limit as a forcing poset and building a generic model in some extension. A less set theoretical but essentially equivalent way to get $T_\mathcal{K}$ is to construct an appropriate Kripke frame out of $\mathcal{K}$. Namely nodes are indexed by finite chains of members of $\mathcal{K}$ ordered by end extension. Each index $p = \{\mathfrak{A}_i\}_{i\leq n}$ has as an associated structure $\mathfrak{A}_n$. Then you argue that for any index $p = \{\mathfrak{A}_i\}_{i\leq n}$, any $\overline{a} \in \mathfrak{A}_n$, and any formula $\varphi(\overline{x})$, $p \Vdash \varphi(\overline{a})$ only depends on the isomorphism type of $\mathfrak{A}_n$, so that we can sensibly write $\mathfrak{A} \Vdash \varphi(\overline{a})$. Then you argue that for any sentence $\varphi$, and any $\mathfrak{A},\mathfrak{B} \in \mathcal{K}$, if $\mathfrak{A} \Vdash \neg \varphi$ then $\mathfrak{B} \Vdash \neg \varphi$ (this follows from JEP alone). Furthermore for any sentence $\varphi$, by definition every $\mathfrak{A} \in \mathcal{K}$ either has $\mathfrak{A} \Vdash \neg \varphi$ or has an extension $\mathfrak{B} \supseteq \mathfrak{A}$ such that $\mathfrak{B} \Vdash \varphi$ and so a fortiori $\mathfrak{B} \Vdash \neg \neg \varphi$. Together these imply that for any sentence $\varphi$, either for every $\mathfrak{A} \in \mathcal{K}$, $\mathfrak{A}\Vdash \neg \varphi$ or for every $\mathfrak{A} \in \mathcal{K}$, $\mathfrak{A}\Vdash \neg \neg \varphi$. This implies that there is a unique consistent complete classical theory $T_\mathcal{K}$ whose double negation translation is forced by every node of the frame. (As an aside this last construction of $T_\mathcal{K}$ only depends on the fact that $\mathcal{K}$ has JEP, so in particular the notion of $T_\mathcal{K}$ makes sense for any class of structures with the joint embedding property. The set theoretical argument gives you a generic model of $T_\mathcal{K}$ but in general this model is not going to be unique in the forcing extension, so you need to do more work to see that the theory is unique. This whole question was partially inspired by conversations with Noah Schweber relating to this fact and this question of his, although that question is concerned with structures rather than theories. I'm broadly curious about what is known about the theory $T_\mathcal{K}$ for arbitrary classes with JEP and would appreciate any references.) Questions I'm interested in the question of how 'nice' $T_\mathcal{K}$ is and more generally what's known about these theories. In particular I'm wondering If $\mathcal{K}$ is an uncountable elementary Fraïssé class, does $T_\mathcal{K}$ admit quantifier elimination? Recall that Fraïssé limits in finite relational languages always admit quantifier elimination. I have an idea for a (bad) proof of this but it has a couple holes: We pass to a forcing extension $V[G]$ in which $\mathcal{K}$ and $\mathcal{L}$ are countable. There we construct $\mathfrak{M}$, the Fraïssé limit of $\mathcal{K}^V$. I want to say that because $\mathcal{K}$ is an elementary class, $\mathfrak{M}$ is computably saturated (hole #1). This structure is homogeneous for quantifier free types, i.e. for any two tuples $\overline{a}$ and $\overline{b}$, if $\overline{a}$ and $\overline{b}$ have the same quantifier free type, then there is an automorphism taking $\overline{a}$ to $\overline{b}$. In general this property for a single model is not enough to witness that the theory has quantifier elimination but I want to say that if the model is computably saturated then this implies the theory has quantifier elimination (hole #2). Then quantifier elimination is absolute so it holds for $T_\mathcal{K}$ in the ground model as well. Part of the problem is that all of the examples I can think of look something like this: Let $\mathcal{L}=\{E_i\}_{i<\omega}$ where each $E_i$ is a binary relation. Consider the class $\mathcal{K}$ of finite $\mathcal{L}$-structures in which each $E_i$ is a graph, but no other relations are imposed. The 'Fraïssé limit' of this is the 'random $2^\omega$-colored graph.' The theory of this structures does admit quantifier elimination but this is because it can be 'approximated' by finite reducts which are themselves bona fide Fraïssé limits. So every finite reduct of the theory admits quantifier elimination and therefore the whole theory does as well. So a related question is whether or not this always happens (and I seriously doubt it does but I don't know a counterexample): If $\mathcal{K}$ is an uncountable elementary Fraïssé class, does there always exist a directed set $R \subseteq \mathcal{P}\mathrm(\mathcal{L})$ of sub-languages such that $\bigcup R = \mathcal{L}$ and for every $\mathcal{L}_0 \in R$, $T_\mathcal{K} \upharpoonright \mathcal{L}_0$ is the theory of a Fraïssé limit? Can the languages be required to be finite? I suspect that by picking a carefully chosen subclass of the above $\mathcal{K}$ (or something similar) you can find an example showing that if $\mathcal{K}^\prime$ is not elementary then $T_{\mathcal{K}^\prime}$ may not have QE, but I haven't worked out an example yet. EDIT: Here's a relatively simple counterexample when the class in question is not elementary. Let $\mathfrak{N}$ be the structure $(\mathbb{N},+,\cdot)$ where $+$ and $\cdot$ are taken as ternary relations. Also add predicates $P_n(x)$ such that $\mathfrak{N} \models P_n(a)$ if and only if $a=n$. Note that $\mathrm{Th}(\mathfrak{N})$ does not admit quantifier elimination. Let $\mathcal{K}_N$ be the class of finite substructures of $\mathfrak{N}$. Clearly the generic theory is just going to be $\mathrm{Th}(\mathfrak{N})$, which doesn't admit quantifier elimination. Strictly speaking this is a countable class, but we can make it 'uncountable Fraïssé class' by tacking on $2^\omega$-colored graphs: $\mathcal{K}_N^\prime$ is a class of structures in a language containing $+$, $\cdot$, $P_n$, $U$, and $E_n$. If $\mathfrak{A} \in \mathcal{K}_N^\prime$ then $U(\mathfrak{A})$ is a structure in the class $\mathcal{K}_N$ and $\neg U(\mathfrak{A})$ is a $2^\omega$-colored graph with edge relations among $E_n$. Now this is an uncountable Fraïssé class whose limiting theory does not admit quantifier elimination. REPLY [5 votes]: I'm thinking that a certain choice of "generalized metric space" should accomplish this, but it depends on if the "generic theory" ends up right. First I'll describe the general setup. Let $S$ be a fixed subset of $\mathbb{Q}^{\geq 0}$. Assume $0\in S$. Let $L$ be a language with a binary relation symbol $d_s(x,y)$ for all $s\in S$. Any metric space has a canonical interpretation as an $L$-structure where $d_s(x,y)$ is "$d(x,y)\leq s$". Let $\mathcal{K}$ be the class of finite metric spaces with distances in $S$. (This may not be the finite models of an elementary class. I'll get to this later.) Delhomme, LaFlamme, Pouzet, and Sauer showed that $\mathcal{K}$ is a Fraisse class if and only if for any $s_1,s_2,s_3,s_4\in S$, if there is some $t\in S$ such that $\{s_1,s_2,t\}$ and $\{s_3,s_4,t\}$ are "metric triangles" (i.e., any combination satisfies the triangle inequality), then there is some $u\in S$ such that $\{s_1,s_3,u\}$ and $\{s_2,s_4,u\}$ are metric triangles. In other words, this describes the amalgamation of two triangles $\{s_1,s_2,t\}$ and $\{s_3,s_4,t\}$ over the common distance $t$. In this case, we say that $S$ satisfies the four-values condition, and let $T$ be the theory of the Fraisse limit of $\mathcal{K}$. Now I claim that there is a choice of $S$ so that $T$ does not have QE. Set $S:=([2,3)\cup(3,\infty))\cap\mathbb{Q}$. It can be checked that $S$ satisfies the four-values condition (cutting $S$ off below at $2$ is needed for this). We show that $T$ does not have quantifier elimination. Let $\varphi(x_1,x_2;y)$ be the formula $d_2(x_1,y)\wedge\neg d_5(x_2,y)$. Consider the types: $$ p(x_1,x_2)=\{d_s(x_1,x_2):s>3\}\cup\{\neg d_s(x_1,x_2):s<3\} $$ $$ p_1(x_1,x_2)=p(x_1,x_2)\cup\{\exists y\varphi(x_1,x_2;y)\} $$ $$ p_2(x_1,x_2)=p(x_1,x_2)\cup\{\neg\exists y\varphi(x_1,x_2;y)\} $$ We show that $p_1$ and $p_2$ are both consistent. Suppose $\pi(x_1,x_2)$ is some finite subset of $p_0(x_1,x_2)$. Let $s_1=\max\{s:\neg d_s(x_1,x_2)\in \pi\}$ and $s_2=\min\{s:d_s(x_1,x_2)\in \pi\}$. Then the metric space $\{a_1,a_2,b\}$ such that $d(a_1,a_2)=s_2$, $d(a_1,b)=2$, and $d(a_2,b)=2+s_2$ witnesses the consistency of $\pi(x_1,x_2)\cup\{\exists y\varphi(x_1,x_2;y)\}$. On the other hand, the metric space $\{a'_1,a'_2\}$, where $d(a'_1,a'_2)$ is an element of $S$ strictly between $s_1$ and $3$, witnesses that $\pi(x_1,x_2)\cup\{\neg\varphi(x_1,x_2;y)\}$ is consistent. For most "reasonable" choices of $S$ (e.g., $S=\mathbb{Q}^{\geq 0}$), if $T$ exists then it has QE. But QE can be ruined by poking "holes" in $S$ like in the previous example. Now I will discuss how to get a class of finite models of a theory. Let $S$ be arbitrary again (not necessarily satisfying four-values). Let $T_0$ be the universal theory consisting of: $\forall x\forall y(d_0(x,y)\leftrightarrow x=y)$ for all $s\in S$, $$ \forall x\forall y(d_s(x,y)\leftrightarrow d_s(y,x)) $$ for all $r,s,t\in S$ such that there is no $x\in S$ satisfying $t TITLE: Evaluating a binary quadratic form at convergents QUESTION [5 upvotes]: We use the notation $$\displaystyle [a_0; a_1, \cdots, a_n] = a_0 + \cfrac{1}{a_1 + \cfrac{1}{\ddots + \cfrac{1}{a_n}}}$$ to denote a finite continued fraction, and for a given real number $\alpha$, we attach the (possibly infinite) sequence of partial quotients $[a_0; a_1, a_2, \cdots] = \alpha$. It is well-known by classic theorems of Euler and Lagrange that a real number has eventually periodic continued fraction expansion if and only if $\alpha$ is a quadratic irrational. Moreover, it is known that if $\alpha$ is an irrational number and $p/q$ is a reduced fraction with $q > 0$ such that $|\alpha - p/q| < 1/(2q^2)$, then $p/q$ is in fact a convergent of $\alpha$; i.e., there is a positive integer $k$ such that $p/q = [a_0; a_1, \cdots, a_k]$. Suppose that $\theta$ is a purely periodic quadratic irrational, so that its continued fraction is of the form $[\overline{a_0; a_1, \cdots, a_n}]$. Let $f(x) = ax^2 + bx + c$ be the minimal polynomial of $\theta$. What do we know about the evaluation of the quadratic form $F(x,y) = y^2 f(x/y)$ at the pairs $(p_0, q_0), \cdots, (p_n, q_n)$ corresponding to the convergents? In particular, what are the sizes of these values compared to the discriminant $\Delta(F)$ of $F$? REPLY [4 votes]: The item you want is the neighbor method, a version of continued fractions. I learned this from Buell, Binary Quadratic Forms. It is also in a 1929 Introduction by Dickson, and a book by Matthews I've never seen. For a positive but not square discriminant, all primitively represented numbers with absolute value below $\frac{1}{2} \sqrt \Delta$ appear as coefficients (first or third) in the chain. There are usually a few more coefficients with slightly larger absolute values. Each triple in a cycle is a "reduced" indefinite form. A form $ax^2 + b xy + c y^2$ is reduced if and only if $ac<0$ and $b > |a+c| \; .$ jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 1 3 -2 0 form 1 3 -2 1 0 0 1 To Return 1 0 0 1 0 form 1 3 -2 delta -1 ambiguous 1 form -2 1 2 delta 1 2 form 2 3 -1 delta -3 3 form -1 3 2 delta 1 ambiguous -1 composed with form zero 4 form 2 1 -2 delta -1 5 form -2 3 1 delta 3 6 form 1 3 -2 form 1 x^2 + 3 x y -2 y^2 minimum was 1rep x = 1 y = 0 disc 17 dSqrt 4 M_Ratio 16 Automorph, written on right of Gram matrix: -9 -32 -16 -57 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 1 13 -13 0 form 1 13 -13 1 0 0 1 To Return 1 0 0 1 0 form 1 13 -13 delta -1 ambiguous 1 form -13 13 1 delta 13 ambiguous 2 form 1 13 -13 form 1 x^2 + 13 x y -13 y^2 minimum was 1rep x = 1 y = 0 disc 221 dSqrt 14 M_Ratio 196 Automorph, written on right of Gram matrix: -1 -13 -1 -14 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 1 7 -3 0 form 1 7 -3 1 0 0 1 To Return 1 0 0 1 0 form 1 7 -3 delta -2 ambiguous 1 form -3 5 3 delta 2 2 form 3 7 -1 delta -7 3 form -1 7 3 delta 2 ambiguous -1 composed with form zero 4 form 3 5 -3 delta -2 5 form -3 7 1 delta 7 6 form 1 7 -3 form 1 x^2 + 7 x y -3 y^2 minimum was 1rep x = 1 y = 0 disc 61 dSqrt 7 M_Ratio 49 Automorph, written on right of Gram matrix: -79 -585 -195 -1444 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$<|endoftext|> TITLE: Is there a set of positive integers of density 1 which contains no infinite arithmetic progression? QUESTION [8 upvotes]: Let $V$ be a set of positive integers whose natural density is 1. Is it necessarily true that $V$ contains an infinite arithmetic progression?—i.e., that there are non-negative integers $a,b,\nu$ with $0\leq b\leq a-1$ so that: $$\left\{ an+b:n\geq\nu\right\} \subseteq V$$ In addition to an answer, any references on the matter would be most appreciated. REPLY [17 votes]: Yet another concrete counterexample: $$ \bigcup_{n=1}^\infty [n^3+n,(n+1)^3]. $$ More generally, any set containing arbitrarily long gaps is free of infinite arithmetic progressions, and has natural density $1$ if the gaps are properly spaced. Incidentally, an incomparably subtler question is whether any set of positive natural density contains arbitrarily long arithmetic progressions. This was conjecture by Erdős and Turán in 1936 and famously proved by Szemerédi almost 40 years later. REPLY [11 votes]: Another construction is to let $n \notin V$ if and only if $n$ begins with at least $\sqrt{\log{n}}$ consecutive '9's when written in decimal. This satisfies the stronger property that there is no non-constant polynomial $f : \mathbb{N} \rightarrow \mathbb{N}$ whose image is contained entirely in $V$.<|endoftext|> TITLE: Why is it difficult to solve the Monge problem directly? QUESTION [6 upvotes]: I'm trying to understand something about the Monge problem. The Monge problem is: Let $c(x,y): \mathbb{R}^d \times \mathbb{R}^d \rightarrow \mathbb{R}^d$ and $$\mathcal{T}(\mu_1,\mu_2) = \{ T: \mathbb{R}^d \rightarrow \mathbb{R}^d | \text{ Borel maps with condition} \, \, T\#\mu_1 = \mu_2 \}$$ where $\mu_1$ and $\mu_2$ are given, compactly supported measures which are absolutely continuous with respect to Lebesgue measure. The Monge problem is to find: $$\inf_{T\in\mathcal{T}(\mu_1,\mu_2)}C_M[T] = \int_{\mathbb{R}^d}c(x,Tx) \mu_1(dx)$$ A book I am reading provides the following discussion on why finding a minimizer directly using "usual" methods (ie taking a minimizing sequence) is tough: Usually, what one does is the following: take a minimizing sequence $T_n$, find a bound on it giving compactness in some topology (here, if the support of $\mu_2$ is compact, the maps $T_n$ take value in a common bounded set $\text{spt}\mu_2$, and so one can get compactness of $T_n$ in the weak-* $L^{\infty}$ convergence), take a limit $T_n \rightharpoonup T$, and prove that $T$ is a minimizer. This requires semicontinuity of the functional $C_M$ with respect to this convergence (which is true in many cases, for instance, if $c$ is convex in its second variable): we need $T_n \rightharpoonup T \implies \liminf_nC_M[T_n] \geq C_M[T]$, but we also need that the limit $T$ still satisfies the constraint. Yet, the nonlinearity of the pushforward condition prevents us from proving this stability when we only have weak convergence. Loosely, this all makes sense to me: but I'm getting turned around working the details. Explicitly: Suppose $\text{spt}\mu_2$ is compact, why does $T_n \rightarrow T$ weak-* in $L^{\infty}$. Explicitly: what are the sufficient conditions on $C_M[T]$ to ensure the implication in the above statement $$T_n \rightharpoonup T \implies \liminf_nC_M[T_n] \geq C_M[T]$$ Do I just need to say $C_M[T]$ is weakly lower semicontinuous and bounded from below? (i.e. I think this is the same as convexity) Suppose I show that a minimizer of $C_M[T]$ exists. How do I see that minimizing sequence isn't preserving the pushforward condition in the limit? Apprently "non-linearity" is preventing this -- but I don't see how. EDIT: For 1&2: Okay then combining all your comments: The statement: Let $C_M[T]: L^{\infty} \rightarrow \mathbb{R}$ be weak-* lower semicontinous and bounded from below. Then $C_M[T]$ has a minimizer. Pf: $C_M[T]$ is bounded from below so $\inf C_M[T]$ exists. Let $T_n \in L^{\infty}$ be a minimizing sequence and let $T$ denote the minimizer. Consider a closed ball in the weak-* topology of positive radius centered at $T$. This ball is compact in weak-* topology by banach-aloglu - so there exists a subsequence of $T_n$, call it $T_{n_k} \rightarrow T$. This gives us $$\liminf C_M[T_{n_k}] \leq C_M[T_i] \quad \forall T_i \in L^{\infty}$$ -- in particular $\liminf C_M[T_{n_k}] \leq C_M[T]$. Weak-* LSC gives the other inequality, so $C_M[T]$ achieves its min. for 3 It doesn't make sense to write: $$(aT_1+bT_2)_{\#}\mu = a(T{_1}_{\#}\mu) + b(T{_2}_{\#}\mu)$$ Let $a=b=1$, $T_1 = 2x$, $T_2 = x^2$, $\mu$ to be leb msr, and consider a set $A = [0,-1]$. Then $$(x^2 + 2x)_{\#}\mu\big([0,-1]\big) = \mu \bigg((x^2+2x)^{-1}\big([0,-1]\big)\bigg)$$ But $(x^2+2x)^{-1}[0,-1]$ can be $[0,1]$ or $[1,2]$ so LHS of such a statement is not well defined. So in what sense is the push forward non linear? REPLY [3 votes]: It is not true that $T_n\to T$ in any sense. However, since $L^\infty=(L^1)^*$, the unit ball in $L^\infty$ is weak* compact by Banach-Alaogu theorem. Hence, one can extract from $T_n$ a weak* convergent subsequence. Yes, this is a standard notion of lower semi-continuity from calculus of variations. For $d=1$, let $\mu_1$ be the restriction of Lebesgue measure to $[0,1]$, and $\mu_2$ the sum of $1/2$-point masses at $-1$ and $1$. Let $T_n(t):=\text{sgn}\sin(2\pi n t)$. Then, clearly, $T_n\sharp\mu_1=\mu_2$ for all $n$. But $T_n\to T\equiv 0$ in the weak* topology and $T$ maps the Lebesgue measure to the point mass at $0$.<|endoftext|> TITLE: Arithmetically random bitstreams QUESTION [5 upvotes]: Motivation (informal). When trying to generate a random bit-stream, we expect that "half of the" bits are $0$, and the "other half" are $1$. So, how about $010101\ldots$? Well, we would also expect that if we look at every second member of the sequence, then "half of" those bits are $0$ and the other half are $1$. So, let's make this precise. Formal version. Let $\mathbb{N}$ denote the set of non-negative integers. We can identify every bit-stream, that is function $f:\mathbb{N}\to \{0,1\}$ with some $A\in{\cal P}(\mathbb{N})$ (take $A = f^{-1}(\{1\})$). Given any $S\subseteq \mathbb{N}$ we define maps $\mu_S^+, \mu_S^-:{\cal P}(\mathbb{N})\to[0,1]$ by $$\mu^{+}_S(A)= \lim \sup_{n\to\infty}\frac{|A\cap S \cap\{1,\ldots,n\}|}{1+|S\cap \{1,\ldots,n\}|}, \text{ and } \mu^{-}_S(A)= \lim \inf_{n\to\infty}\frac{|A\cap S \cap\{1,\ldots,n\}|}{1+|S\cap \{1,\ldots,n\}|}.$$ We say that $A$ is well-balanced with respect to $S$ if $\mu^+_S(A) = \mu^-_S(A) = 1/2$. For $a,b\in \mathbb{N}$ with $a>0$ we set $S_{a,b} = \{an+b:n\in\mathbb{N}\}$ and we say that $A\in{\cal P}(\mathbb{N})$ is arithmetically random if $A$ is well-balanced with respect to $S_{a,b}$ for any $a,b\in\mathbb{N}$ with $a>0$. What is an example of an arithmetically random set $A\in{\cal P}(\mathbb N)$? REPLY [2 votes]: Mauduit and Sarkozy have studied essentially this and other related pseudorandomness measures for finite as well as infinite $\{\pm 1\}-$valued sequences, see here (not paywalled)and the references therein. Briefly, for a finite sequence $(e_1,\ldots, e_N)\in \{\pm 1\}^N$ of length $N,$ they define the well-distribution measure of the sequence by $$ W(e_1,\ldots,e_N)=\max_{a,b,t \in \mathbb{N}} \left| \sum_{j=0}^{t-1} e_{a+jb} \right| $$ where the maximum is taken over all AP's within $\{1,2,\ldots,N\}$. Another measure they define is the correlation measure of order $k$ $$ C_k(e_1,\ldots,e_N)=\max_{M,0\leq d_1 TITLE: Unstable Integers QUESTION [7 upvotes]: There is a question that has been bothering my mind for quite a while now. I will present it and my current thoughts and progress on it. Let the prime factorization of an integer $n$ be $$n = p_1^{e_1}\cdot p_2^{e_2}\cdot p_3^{e_3}...$$ And define $n$ to be "unstable" if there exists a prime triplet in its factorization $p_ie_i,e_k$ or $e_j TITLE: Tychonoff-ization and Urysohn (functionally Hausdorff) topological spaces QUESTION [6 upvotes]: Let me first make sure I have the correct definitions because my question will be about the difference about the two and there may be some massive confusion on my part. A topological space $X$ is said to be completely regular or Tychonoff when it is Hausdorff and satisfies the following equivalent conditions: For every $x \in X$ and closed $F \subseteq X$ such that $x\not\in F$, there exists a continuous $f\colon X\to\mathbb{R}$, which we can assume to have values in $[0,1]$, such that $f(x) = 0$ and $f|_F = 1$. The map $X \to [0,1]^{C(X,[0,1])}$ taking $x\in X$ to the family $(f(x))_{f\in C(X,[0,1])}$ of its images under every continuous $f\colon X\to[0,1]$ defines a homeomorphism of $X$ to its image. The Stone-Čech compactification map $X \to \beta X$ defines a homeomorphism of $X$ to its image. There exists a compact [Hausdorff] space $K$ such that $X$ is homeomorphic to a subspace of $K$. On the other hand, a (necessarily Hausdorff) topological space $X$ is said to be functionally Hausdorff (or Urysohn, but some people use this to mean something different, so it's probably best to avoid this terminology) when it satisfies the following equivalent conditions: For every $x,y \in X$ such that $x\neq y$, there exists a continuous $f\colon X\to\mathbb{R}$, which we can assume to have values in $[0,1]$, such that $f(x) = 0$ and $f(y) = 1$. The map $X \to [0,1]^{C(X,[0,1])}$ taking $x\in X$ to the family $(f(x))_{f\in C(X,[0,1])}$ of its images under every continuous $f\colon X\to[0,1]$ is injective. The Stone-Čech compactification map $X \to \beta X$ is injective. There exists a continuous injective map $X \to K$ with $K$ a compact [Hausdorff] space. I note that example 91 (the “deleted Tychonoff corkscrew”) in Steen & Seebach's Counterexamples in Topology gives an example of a functionally Hausdorff space which is not completely regular, showing that the two notions are not equivalent. Since until recently I thought these two notions were equivalent (I somehow thought that $X \to \beta X$ was automatically an embedding when it is injective), my goal is essentially to dispel the confusion I had; I first have to ask: Question 0a: Is the above account correct? (Are the properties I claim to be equivalent indeed equivalent, and equivalent to standard definitions for the terms they claim to define?) Every topological space $X$ has a complete regularization or Tychonoff-ization, namely a continuous map $X \to X'$ with $X'$ a completely regular space, such that every continuous map $X \to Y$ with $Y$ completely regular uniquely factors as $X \to X'\to Y$. (I.e., the functor $X \mapsto X'$ is left adjoint to the inclusion functor from the full subcategory of completely regular spaces to that of topological spaces.) This $X'$ can be defined as the image of the Stone-Čech compactification map $X \to \beta X$ with the subspace topology; in particular, $X \to X'$ is always surjective. Question 0b: Is this still correct? Edit (2019-06-08): Essentially the above description of complete regularization functor is given, under the name “Tychonoff functor” in the paper “The Tychonoff Functor and Related Topics” by T. Ishii, chapter 6 (p.203–243) in K. Morita & J. Nagata (eds.), Topics in General Topology (1989). So it would appear that it is correct. Now I thought $X'$ was a quotient space of $X$. This can't be the case because, if what I wrote above is correct, the equivalence relation (“having the same image in $X'$”) is simply “having the same image under every continuous function $X\to\mathbb{R}$ (or equivalently $X\to[0,1]$)”, which is trivial for a functionally Hausdorff space, yet the latter is not necessarily completely regular. But this is problematic because in section d-2 (“Higher Separation Axioms”, p.158–159) of the Encyclopedia of General Topology (Hart, Nagata & Vaughan eds.) one reads: “To every space $X$ one can associate a Tychonoff space $Y$ as follows. Two points $x$ and $y$ in $X$ are equivalent if $f(x) = f(y)$ for all continuous real-valued functions $f$ on $X$. The corresponding quotient space $Y$ is Tychonoff and the rings $C(X)$ and $C(Y)$ of real-valued continuous functions are isomorphic; the same holds for the rings $C^*(X)$ and $C^*(Y)$ of bounded real-valued continuous functions.” It would seem to me that this assertion is contradicted by the existence of the aforementioned counterexample in Steen & Seebach. Question 0c: Am I correct in believing that the above quote is in error? (Or did I miss some fine print or hidden assumption?) Now assuming all of the above is correct, there are two natural questions which are left open: Question 1a: Does every topological space $X$ have a “functional Hausdorffization” (or “Urysohnization”), namely, does the inclusion functor from the full subcategory of functionally Hausdorff spaces to that of topological spaces have a left adjoint? • Question 1b: If so, is it given by quotienting by the equivalence relation “$f(x) = f(y)$ for all continuous real-valued functions $f$ on $X$” or is there some subtlety? Edit (2019-06-08): The paper “A universal factorization theorem in topology” (Canad. Math. Bull. 9 (1966) 201–207), by R. Sharpe, M. Beattie & J. Marsen defines an equivalence relation $T_{3/2}$ (it is not clear to me whether this is a typo for $T_{3\frac{1}{2}}$, and whether they may be under the same confusion as mentioned above) on any topological space by $x\,\mathbin{T_{3/2}}\,y$ when for every continuous $f\colon X\to [0,1]$ we have $f(x)=f(y)$, i.e., the relation defined above. Calling just $R$ this equivalence relation, according to this paper's main theorem, the crucial point in answering questions 1ab positively is to check that $R = \lim R$ in the notation of the paper, where $\lim R$ is the equivalence relation defined by $x\mathbin{(\lim R)}y$ when for every continuous $f\colon X\to Z$ with $Z$ such that $R$ is just equality (i.e., in our case, every functionally Hausdorff $Z$) we have $f(x)=f(y)$; but this is trivial: so it appears that questions 1ab have a positive answer. Question 2a: Even if complete regularization is not given by a quotient, is there still a unique coarsest equivalence relation $R$ on any topological space such that $X/R$ is completely regular? • Question 2b: If so, can we describe $R$ concretely, and can we describe the natural continuous map $X' \to X/R$ (where $X'$ is complete regularization as defined above)? REPLY [5 votes]: 0a Correct, except for one point: is $X$ is not completely regular then there is no $\beta X$, so the third equivalence in `functionally Hausdorff' does not exist. 0b Not quite, the paper mentioned in your edit maps $X$ into a Tychonoff cube and lets $X'$ be the image, see 0a: there is no $\beta X$ available. 0c That is indeed erroneous; it is the quotient of the set $X$ endowed with the (weaker) topology generated by the continuous functions. 1a This is correct, as you point out below the question. 1b There is no subtlety, every continuous $f:X\to[0,1]$ will factor through that equivalence relation. 2a Assuming this definition of coarse: the coarsest equivalence relation on the underlying set (all points equivalent) gives a one-point quotient, which is certainly completely regular. In that case the answer to 2b is an obvious yes. 2b Let $X$ be the deleted corkscrew; its complete regularization $X'$ is just the set underlying set $X$ with a weaker topology: that generated by the continuous functions. It has many completely regular quotients: for every $\alpha<\omega_1$ and $n<\omega$ let $C(\alpha,n)$ be the set of points $(i,\beta,m)$ with $i\in\mathbb{Z}$, $\beta<\alpha$ and $m>n$. The set $C(\alpha,n)$ is clopen in the corkscrew-without-$\{a^+,a^-\}$, hence so is $\{a^+\}\cup C(\alpha,n)$ in $X$. Identifying that set to a point yields a completely regular quotient. There is no finest equivalence relation among these and their common refinement of all these yields the identity relation, which does not yield a completely regular quotient.<|endoftext|> TITLE: Is a smooth family of vector spaces always locally trivial? QUESTION [10 upvotes]: Let's define a smooth family of vector spaces to be the following data $(E,M,p,a,s,z)$, where (1) $p:E\to M$ is a smooth submersion of smooth manifolds (2) There is some fixed integer $n\ge 0$ such that the fibre $E_m = p^{-1}(m)$ has the structure of an $n$-dimensional vector space over $\mathbb R$ for each point $m\in M$. [The smooth structure of $E_m$ as a vector space coincides with its smooth structure as a submanifold of $E$.] (3) For this fibrewise vector space structure, the addition map $a:E\times_ME\to M$ is smooth, the scaling map $s:\mathbb R\times E\to E$ is smooth and the zero section $z:M\to E$ is a smooth embedding. Then, is $p:E\to M$ a locally trivial vector bundle (with $a,s,z$ being its addition, scaling and zero section maps)? Just by virtue of $p:E\to M$ being a smooth submersion, we have a locally trivial vector bundle $T_{E/M}$ on $E$ (the "vertical tangent bundle" of $p$), and we can pull it back to $M$ to get a locally trivial vector bundle $z^*T_{E/M}$. Now, define a map $\phi:E\to z^*T_{E/M}$ given by $e\mapsto\frac{\partial}{\partial t}|_{t=0}s(t,e)$. This is a smooth map (since $s,z$ are smooth) that commutes with the projections to $M$ and is a (fibrewise linear) bijection. Finally, if we look at the map $\phi$ along the image $z(M)$ of the zero section, it quite clearly has full rank. Thus, it has full rank on a neighborhood of $z(M)\subset E$. But now, since $\phi$ commutes with scaling, it follows that $\phi$ has full rank on all of $E$ and thus, it is a diffeomorphism by the inverse function theorem. Thus, $E$ being isomorphic (as a smooth family of vector spaces) to $z^*T_{E/M}$ means it is locally trivial. My question is: Is there any (possibly subtle) mistake in this proof (and if so, is local triviality true nevertheless)? Curiously, this argument does not seem to need even the smoothness of $a$ (it follows as a consequence of the isomorphism $\phi$, which itself depends only on the smoothness of $s$ and $z$). Also, the same argument seems to work if we replace all the objects involved by complex manifolds, holomorphic maps and complex vector spaces. The bracketed assumption in (2) seems to be used when we show that $\phi$ is a fibrewise linear bijection and that it has full rank along the zero section. I think it can be removed and that the proof should still go through - is this true? If you think the answer to the first question is NO, then I'd still like to hear about it in the comments. My actual motivation is to apply this in the setting of infinite dimensional Banach manifolds, and an answer to the second question covering this case would also be appreciated. REPLY [3 votes]: You are right. Also, it can be added that a submersion with every fibre merely diffeomorphic to R^n is always a locally trivial fibration (my paper "Submersions, fibrations and bundles", Trans. Amer. Math. Soc. 354 (2002), 3771-3787).<|endoftext|> TITLE: Ratio of integrals with increasing dimension over Euclidean balls QUESTION [5 upvotes]: Let $f_n(x)\geq0$ be any sequence of nonnegative $L^1(\mathbb{R}^n)$ functions such that $\int_{\mathbb{R}^{n}}f_n(x)dx=1$ where $dx$ is the Lebesgue measure on $\mathbb{R}^n$. For any $a>1,\epsilon>0$, does there exist a sequence $x_n\in\mathbb{R}^n$ such that $$\lim_{n\to+\infty}a^n\frac{\int_{\|x-x_n\|^2\leq n^{1-\epsilon}}f_n(x)dx}{\int_{\|x-x_n\|^2\leq n}f_n(x)dx}=0$$? If it does not hold, is there any counter example of the sequence $f_n$? What additional conditions do we need on the sequence $f_n$? A similar question is to show that $$\lim_{n\to+\infty}a^n\frac{\int_{\|x-x_n\|^2\leq n^{1-\epsilon}}\exp(-\|x_n-x\|^2)f_n(x)dx}{\int_{\|x-x_n\|^2\leq n}\exp(-\|x_n-x\|^2)f_n(x)dx}=0$$ ？This can be implied by the first display. Therefore, this is a weaker statement. REPLY [2 votes]: Such a sequence $(x_n)$ always exists. Indeed, the displayed ratio expression under the limit sign equals $R_n(x_n)$, where \begin{equation*} R_n(y):=\frac{g_{n,r_n}(y)}{g_{n,s_n}(y)},\quad r_n:=n^{(1-\epsilon)/2}, \quad s_n:=n^{1/2}, \end{equation*} \begin{equation*} g_{n,r}(y):=\int f_n(x)1_{|x-y|_n0$ \begin{equation*} \int g_{n,r}(y)\, dy=\int dx\, f_n(x)\int dy\,1_{|x-y|_n0$, so that $a^nR_n(x_n)\to0$, as desired. Details on the inequality in (1): Let \begin{equation*} c:=\inf_y \frac{g_{n,r_n}(y)}{g_{n,s_n}(y)}. \end{equation*} Then $g_{n,r_n}(y)\ge c g_{n,s_n}(y)$ for all $y$, whence $\int g_{n,r_n}(y)\, dy\ge c\int g_{n,s_n}(y)\,dy$ and \begin{equation*} \frac{\int g_{n,r_n}(y)\, dy}{\int g_{n,s_n}(y)\,dy}\ge c=\inf_y \frac{g_{n,r_n}(y)}{g_{n,s_n}(y)}. \end{equation*}<|endoftext|> TITLE: Conceptual (operadic?) reason for the generalized EHP fiber sequence $J_{q-1}(S^{2n}) \to J S^{2n} \to JS^{2nq}$? QUESTION [11 upvotes]: Let $q$ be a prime and $q=p^r$ a power. Then there is a $p$-local fiber sequence from the $q-1$st stage of the James construction on $S^{2n}$, to $J(S^{2n}) = \Omega \Sigma S^{2n}$, to $J(S^{2nq}) = \Omega \Sigma S^{2nq}$. Here the first map is the natural inclusion map for the James filtration, and the second map is the James-Hopf map, adjoint (under $\Sigma \dashv \Omega$) to the projection coming from the Snaith splitting $\Sigma J(X) = \Sigma \vee_{q\geq 1} X^{\wedge q}$ (with $X = S^{2n}$). The EHP sequence arises when $p=q=2$. What I'd like to understand is the fact that this is a fiber sequence. The proof I'm familiar with (from lectures 3-5 of notes by Akhil Mathew on a course by Mike Hopkins) can be seen by computing what it does on $\mathbb F_p$ homology and considering the Serre spectral sequence. The homology picture is rather suggestive: Recall that the James construction is the free $E_1$-space on an $E_0$-space (i.e. a pointed space), and the Snaith splitting tells us that correspondingly $H_\ast(J(X))$ is the free associative algebra on the augmented vector space $H_\ast(X)$: $H_\ast(J(X)) = T \tilde H_\ast(X)$. Moreover, the James filtration corresponds to the filtration of the tensor algebra by tensor rank: $H_\ast(J_k(X)) = T_{\leq k} \tilde H_\ast(X)$. The James-Hopf map is rather complicated, but one works out using the comultiplication (and computing some mod $p$ multinomials) that $p$-locally, and when $X = S^{2n}$, it looks additively like the obvious projection $H_\ast(J(S^{2n})) = \mathbb F_p[x_{2n}] \to \mathbb F_p[y_{2nq}] = H_\ast(J(S^{2nq}))$. So the fiber sequence boils down to the additive decomposition $\mathbb F_p[x_{2n}] \cong (\mathbb F_p[x_{2n}] / x_{2n}^q) \otimes \mathbb F_p[x_{2n}^q]$. This closely mirrors the universal properties of $J_{q-1} S^{2n}$ and $J(S^{2nq})$. So in some sense, The fiber sequence $J_{q-1} S^{2n} \to J S^{2n} \to J S^{2nq}$ says something about decomposing operations in the $E_1$ operad into operations of arity $ TITLE: Ordinal analysis and proofs of consistency QUESTION [6 upvotes]: $\epsilon_0$ is the proof-theoretic ordinal of PA. Gentzen proved the consistency of Peano's first-order axioms for arithmetic using primitive recursive arithmetic and induction up to $\epsilon_0$. For what other interesting theories $T$ can induction up to the proof-theoretic ordinal of $T$, together with some weak principles such as primitive recursive arithmetic, prove the consistency of $T$? For example: is there a recursive ordinal $\alpha$ such that induction up to $\alpha$, together with primitive recursive arithmetic, can prove the consistency of ZF? REPLY [11 votes]: In the field of ordinal analysis people are typically interested in finding "natural" computable ordinal notation systems corresponding to various theories; those are called proof-theoretic ordinals of the theories. Unfortunately we don't have any formal definition of "naturality" of ordinal notation systems, those is notion is vague. Nevertheless there is the following empirical fact about what a calculation of proof-theoretic ordinal $|T|$ for natural systems $T$ give: theory $\mathsf{PRA}+\text{transfinite induction up to }|T|$ proves consistency of $T$; if the language of $T$ could express the sentences $\mathsf{WO}(R)$ "a primitive-recursive binary relation $R$ is a well-order" then $\sup \{|R|\mid T\vdash \mathsf{WO}(R)\}=|T|$; theory $T$ proves totalitly of all computable functions $H_{\alpha}(x)$, for $\alpha<|T|$, but doesn't prove totality of the computable function $H_{|T|}(x)$, where Hardy hierarchy $H_{\alpha}(x)$ is defined by transfinite recurtion for $\alpha\le|T|$: $$H_0(x)=x+1,\;\;\;H_{\alpha+1}(x)=H_{\alpha}(x+1)\;\;\;H_{\lambda}(x)=H_{\lambda[x]}(x),$$ here we presume that for limit $\lambda\le |T|$ we have fixed fundamental sequences $\lambda[0]<\lambda[1]<\ldots$ such that $\sup_{i\in \omega}\lambda[i]=\lambda$. Some relatively well-known systems and their proof-theoretic ordinals in the order of strength increase $|\mathsf{PA}|=|\mathsf{ACA}_0|=\varepsilon_0$ $|\mathsf{ATR}_0|=\Gamma_0$ $|\mathsf{KP}\omega|=|\mathsf{ID}_1|=|\textsf{parameter free }\Pi^1_1\text{-}CA_0|=\psi_0(\varepsilon_{\Omega+1})$ $|\Pi^1_1\text{-}CA_0|=\psi_0(\Omega_{\omega})$ Note that the ordinals ordinals for the last two group are from ordinal notation systems defined by people doing ordinal analysis. And hence would be hardly familiar to the people that aren't really familiar with the field. The strength of the theories for which we do now know proof-theoretic ordinals are fairly limited. Namely the strongest systems for which a computation of proof-theoretic ordinal have been published (by M.Rathjen) is the (fairly strong) fragment of second-order arithmetic $\text{parameter free }\Pi^1_2\text{-}CA_0+\mathsf{BI}$. It is stronger than $\Pi^1_1\text{-}CA_0$, e.g. the strongest fragment in Simpson's big five of systems typically appearing in the context of reverse math. But the calculations of ordinals of $\mathsf{ZFC}$ and full second-order arithmetic arithmetic are beyond the reach of current methods. However if one isn't interested in "naturality" of computable ordinal notation systems. Then it is fairly trivial to craft computable ordinal notation system corresponding to any given $\Pi^1_1$-sound theory $T$. Namely the idea is to craft computable linear ordering with the order type $$\sum\limits_{T\vdash\mathsf{WO}(R_i)} R_i,$$ where $R_0,R_1,\ldots$ is an enumeration of primitive-recursive binary relations. For more details you could look at some surveys of the subject. For example [1]Rathjen, M. (2006, August). The art of ordinal analysis. In Proceedings of the International Congress of Mathematicians (Vol. 2, pp. 45-69). REPLY [8 votes]: This depends on how you define "proof-theoretic ordinal." I would say that to seriously address this question we need to restrict attention to a fully-objective definition (as opposed to one dependent on choice of notation; the answer then is that the proof-theoretic ordinal of an appropriate (= "does enough arithmetic") theory always exists, and all existing evidence points to it having the desired consistency-proving property (and so we expect the same to hold for ZF), but there is (to my knowledge) no metatheorem establishing that this in fact holds. In terms of how far ordinal analysis has been carried out in a way considered satisfying by the community (see the subtleties below), I believe the state-of-the-art is around $\Pi^1_2$-CA$_0$. My understanding is that ATR$_0$ represents the upper limit of the "relatively simple" analyses, and after that things get quite weird. But I'm not an expert. Related: This paper by Rathjen is a good survey of general ordinal analysis. I vaguely remember an excellent paper (by Avigad?) summarizing different definitions of proof-theoretic ordinal and analyzing their relationships; if I can find it, I'll add it. This paper by Arai (and its sequel) isn't directly relevant, but looks at a different way to analyze (set) theories in terms of ordinals - focusing not on establishing consistency but on determining how complicated constructions the theory can establish. This approach is sufficiently "coarse" that it is quite feasible even at the level of ZF. CAVEAT: Below I'm being a bit vague about what "proves induction along" means. In a theory in second-order arithmetic we can talk directly about well-orderedness; in a first-order theory we need to talk instead about fragments of induction (e.g. $\Sigma_1$-induction along the ordering in question). A snappy definition you'll often see is indeed "The smallest ordinal such that induction along that ordinal proves (over PRA) the consistency of $T$." Of course this isn't really precise. Rather, we need to talk about ordinal notations: ignoring the details, these are just primitive recursive well-orderings of $\omega$ (ignoring the finite ones for now). Note that the set of actual notations is $\Pi^1_1$-complete. The problem is that notations can be quite pathological. Consider for example the following way to build a linear order $\triangleleft$: we lay down points in order $$0\triangleleft 1\triangleleft 2\triangleleft ...$$ until we see a proof of $\perp$ in $T$, at which point we switch to building a copy of the "reversed" natural numbers. E.g. if the shortest proof of $\perp$ in $T$ has length $17$, we wind up building $$... 19\triangleleft 18\triangleleft 17\triangleleft 0\triangleleft 1\triangleleft 2\triangleleft 3\triangleleft ...\triangleleft 16.$$ There is a notation capturing this construction, and trivially induction along this notation proves (over PRA) the consistency of $T$. But its ordertype is just $\omega$ - did we just prove that every (appropriate) theory has proof-theoretic ordinal $\omega$? No - when we say "induction along $\alpha$ proves the consistency of $T$," what we really mean is that there is some natural system of unique notations for ordinals up to $\alpha+1$ such that induction along the notation for $\alpha$ in this system proves (over PRA) the consistency of $T$, but induction along any smaller notation doesn't. Of course this is a fundamentally vague task. In general though three other phenomena help motivate the specific choice: The systems so produced tend to "cohere" - e.g. there's a "simple" translation according to which the system for ATR$_0$ extends the system for PA. $T$ itself tends to (indeed, in every case I'm aware of does) prove induction along each smaller notation in that system. The system itself emerges naturally from an analysis of proofs from $T$ (e.g. by looking at cut elimination processes). By contrast, a fully notation-independent definition is the following: The computational proof-theoretic ordinal $\vert T\vert_{comp}$ of $T$ is the supremum of the computable ordinals $\alpha$ such that there is some notation $n$ for $\alpha$ which $T$ proves is well-founded. This makes sense for any (appropriate) theory $T$, and - so long as $T$ is sufficiently sound - is always striclty less than $\omega_1^{CK}$ (by $\Sigma^1_1$ bounding). It turns out that we have the following phenomenon: For every "natural" theory $T$, there is a "natural" notation for $\vert T\vert_{comp}$ such that induction along that notation proves (over PRA) the consistency of $T$. But as far as I know there is no meta-theorem which says that this will always be the case, and indeed I don't even know what such a theory would look like (you'd have to somehow pin down what a "reasonable" ordinal notation is). So while (as far as I know) this phenomenon is expected to hold for all natural theories, and in particular for ZF, that is open. Another reasonable attempt at a "notation-independent" definition would be the following: The inevitable proof-theoretic ordinal $\vert T\vert_{inev}$ is the least computable ordinal $\alpha$ such that for every notation $n$ for $\alpha$, induction along $n$ proves (over PRA) the consistency of $T$. But it's not clear to me that such a thing exists in general. Indeed, even for PA I'm not sure if such a thing exists, let alone that it is $\epsilon_0$: why shouldn't there be useless notations for long ordinals? EDIT: Fedor Pakhomov pointed out below that Beklemishev showed that this never exists.<|endoftext|> TITLE: Probability of words summing to $1$ in $S_n$ or $\mathrm{PGL}_2(n)$ QUESTION [11 upvotes]: $\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Conj{Conj}$Let $G$ be the symmetric group $S_n$ or the projective general linear group $\PGL_2(n)$. Let $X$ be a cyclically reduced word in the abstract variables $x_1, x_2, \ldots,x_k$, i.e. $X$ is a product containing $x_1, x_2, \ldots,x_k$ and their inverses, without any element appearing next to its own inverse in any cyclic permutation. (Only words with length $4$, $6$, $8$ are needed in my research.) Consider the probability $P$ that the word sums to $1$, with each $x_i$ chosen uniformly and independently from $G$. Question: What are the upper bounds of $\log_{|G|}P$? If $\log_{|G|}P$ converges when $n\to\infty$, what's the value? Answers are acceptable for either $G=S_n$ or $G=\PGL_2(n)$. Known: If there's a variable occurring only once in $X$, then $P$ is exactly $1/|G|$. If $X=x_1^k$, then the limit is $-1/k$ for symmetric groups by David E Speyer's argument. As Richard Stanley pointed out, if $X=x_1x_2x_1^{-1}x_2^{-1}$, then $P=|\Conj(G)|/|G|$. ($|\Conj(G)|$ is the number of conjugacy classes of $G$) The formula $P=|\Conj(G)|/|G|$ holds for the words $x_1x_1x_2x_2$ and $x_1x_2x_1x_2^{-1}$ if all the characters of $G$ are real, and that's exactly the case for $S_n$ and $\PGL_2(n)$. REPLY [4 votes]: $\DeclareMathOperator\PGL{PGL}$I believe the best result in this direction is due to M. Larsen & A. Shalev (2012); see this paper. I'll summarize their results here. This doesn't answer the questions whether the limit exists or what its precise value is, though, so this is not a full answer. Denote the length of the word $X$ by $\ell$ (while $k$ is the size of the alphabet). As for $S_n$, Proposition 2.3 gives $P < n!^{-\eta}$ for large enough $n$, if $\eta < \frac{2k-1}{4((2k)^{2\ell+1}-1)}$, so $\log_{|S_n|}P < -\eta. $ As for $\PGL_n(\mathbb{F}_q)$, Proposition 3.3 gives $\log_{|G|}P \le \eta$ for large enough $n$ (and every $q$), if $\eta < \frac{1}{1800\ell^2}$. (This holds for all classical groups of Lie type). Shalev and Larsen handle all finite simple groups in their paper. They also refer to this paper, which gives a bound for simple groups $G$ of Lie type of bounded rank $r$: $ P \le \frac{C}{q}$ where $C = C(r, X)$ is a constant, and $q \ge G^{\epsilon}$ for some $\epsilon = \epsilon(r)$, hence $\log_{|G|}P \le -\eta$ for large enough $G$ if $\eta < \epsilon$. EDITED: this last paragraph answers the $n=2, q\to\infty$ case.<|endoftext|> TITLE: Can you make an identity from this product? QUESTION [13 upvotes]: Start with the product $$(1+x+x^2) (1+x^2)(1+x^3)(1+x^4)\cdots$$ (The first polynomial is a trinomial..The others are binomials..) Is it possible by changing some of the signs to get a series all of whose coefficients are $ -1,0,$or $1$? A simple computer search should suffice to answer the question if the answer is "no." I haven't yet done such a search myself. This question is a takeoff on the well known partition identities like: $$\prod_{n=1}^{\infty} (1-x^n)= 1-x-x^2+x^5+x^7-\ldots$$ REPLY [12 votes]: Proof of Doriano Brogloli's answer: Call $a(n)$ the $n$th coefficient of $A(x)=(1-x)(1-x^2)\cdots$. By Euler's pentagonal theorem we have $a(n)=0$ unless $n=m(3m\pm1)/2$ for some $m$, in which case $a(m)=(-1)^m$. Call $b(n)$ the $n$th coefficient of $B(x)=(1-x^2)(1-x^3)...$. Since $A(x)=(1-x)B(x)$ we have $b(n)-b(n-1)=a(n)$, so $b(n)=\sum_{0\le j\le n}a(j)$. Now for any integer $n$ there exists a unique $m$ such that $(m-1)(3(m-1)+1)/2\le n TITLE: Notions in the literature capturing the "symmetric" or "homogeneous" flavour of $L_p$? QUESTION [11 upvotes]: This post/question is admittedly vague, but I hope that with some feedback in comments it could be made more precise. For $E$ a Banach space, $K(E)$ and $B(E)$ will denote the Banach algebras of compact and bounded linear operators on $E$, respectively. There is a tradition / industry in the world of Banach algebras that asks whether $K(E)$ or $B(E)$ can display various properties if we choose $E$ judiciously. As part of this we would like to understand these algebras when $E$ is one of the "classical" Banach spaces such as $\ell_p$ or $L_p$ (with the convention that the unadorned $L_p$ is an abbreviation for $L_p[0,1]$). On the other hand, we have made progress in establishing "bad" behaviour of $B(E)$ when $E$ is a space such as $\ell_p\oplus \ell_q$ when $p\neq q$, or various $\ell_p$-sums of $\ell_{k(n)}^n$ ($n\in {\bf N}$), not to mention "exotic" examples that have been constructed for the explicit purpose of giving $B(E)$ certain properties. Here are my questions. Q1. Is there any way to justify, with mathematical definitions or results, the feeling that $\ell_p$ and $L_p$ are more "symmetrical" examples than the various other examples mentioned above? Note that in various senses $\ell_p\oplus\ell_q$ is better behaved than $L_1$, e.g. the latter space does not have an unconditional basis, while the former space has a very nice unconditional basis. Q2. $\ell_p\oplus\ell_q$ and the $\ell_p$-sums of finite-dimensional $\ell_k$ have a flavour of "gluing together things which look different", and hence one might argue that it is not so striking that for such $E$ the algebra $B(E)$ has behaviour very unlike $B(\ell_2)$. Are there established notions in the Banach-space literature that capture the way $L_p$ and $\ell_p$ look much more "homogeneous"? (The phrase "homogeneous" is meant in a loose sense, not in the sense of https://arxiv.org/abs/math/9205207 ) Regarding Q2: I know that $\ell_p$ is a prime Banach space for each $p$ (every complemented subspace is isomorphic to the whole space) and that $L_p$ is primary (in any direct sum decomposition of the space, at least one of the summands is isomorphic to the whole space), but my understanding is that primeness is too restrictive while being primary allows many more examples. So candidates for answers to Q2 would be notions that are intermediate between "being prime" and "being primary", if such exist. REPLY [4 votes]: Fraïssé theory, which identifies structures that are homogenous and universal in some sense, has departed from its model-theoretic roots and is now available in the so-called metric model theory as well as it has a category-theoretic framework developed by W. Kubiś. In either case, one can show that $L_p[0,1]$ is the Fraïssé limit (in a suitable sense) of $\ell_p^n$s simply because they can be amalgamated together and stay in the class. In the metric case, this was observed by Ferenczi, Lopez-Abad, Mbombo, and Todorcević (see this recent preprint) and in the category-theoretic setting, this is an unpublished work of Kubiś and myself from 2016 or so. One has to be careful with the exact translation of Fraïssé's theorem as certainly the class of $\ell_p^n$-spaces is not closed under taking subspaces so perfect homogeneity is not feasible unless $p=2$. As for your second question, you mentioned primarity. A Banach space $X$ is primary whenever $X = E\oplus F$, to some closed subspaces $E,F$, then at least one of them is isomorphic to $X$. Bill Johnson and Detelin Dosev defined the following set: $$\mathscr{M}_X = \big\{T\in B(X)\colon I_X \neq ATB\; \big(A,B\in B(X)\big)\big\}.$$ This set need not be closed under addition but when it is, it is the unique maximal ideal of $B(X)$. For non-primary spaces (that are also isomorphic to their infinite sums, say), the set $\mathscr{M}_X$ cannot be closed under addition. The spaces $\ell_p$, $L_p$, $C[0,1]$ are primary, whereas $\ell_p\oplus\ell_q$ for $p\neq q$ are clearly not. So perhaps closedness of $\mathscr{M}_X$ under addition is the condition you seek?<|endoftext|> TITLE: Flat with smooth fibers implies formally smooth? QUESTION [7 upvotes]: Is every flat morphism of schemes $X\rightarrow S$ such that the fiber over any point is smooth necessarily formally smooth? There are formally smooth morphisms that are not flat so the converse fails. If we assume that the morphism is locally of finite presentation, then it is necessarily smooth so a fortiori formally smooth. REPLY [6 votes]: Welcome new contributor. This is too long for a comment. Let $R$ be a DVR with a uniformizing element $t$, e.g., $k[[t]]$ where $k$ is a field. Let $C$ be the $R$-algebra $$C=R[x_n:n\in \mathbb{Z}_{\geq 0}].$$ Let $J$ be the ideal in $C$, $$J=\langle x_n-tx_{n+1} :n\in \mathbb{Z}_{\geq 0} \rangle.$$ Let $A$ be the $R$-algebra $C/J$. Since $A$ has no $t$-torsion, the $R$-module $A$ is $R$-flat. The fiber ring $A\otimes_R (R/t R)$ equals $R/tR$, which is smooth over $R/tR$. The fiber ring $A\otimes_R R[1/t]$ equals $R[1/t][x_0]$, which is smooth over $R[1/t]$. Now consider the $R$-algebra $B=C/J^2$. Let $I$ be the image ideal in $B$, $$I=J/J^2.$$ This is a square-zero ideal whose quotient algebra equals $A$. Thus, if $A$ is formally smooth, there exists a retraction of $R$-algebras, $$s:A\to B.$$ In particular, $s(x_0)$ is a $t$-divsible element of $B$ that maps to $x_0$ in the quotient algebra $A$. Consider the quotient of $B$ by the ideal $$K = \langle x_mx_n :m,n\in \mathbb{Z}_{\geq 0} \rangle.$$ The quotient $B$-algebra equals $R\oplus M$ where $M$ is a square-zero ideal that is the free, countably-generated $R$-module with basis $$\{ x_n : n\in \mathbb{Z}_{\geq 0} \}.$$ In particular, the element $x_0$ in this quotient algebra is not $t$-divisible in $M$. Thus, there is no $t$-divisible element of $B$ that maps to $x_0$ in the quotient algebra $A$. Therefore, $A$ is not formally smooth over $R$.<|endoftext|> TITLE: Does a (nice) centerless group always have a centerless profinite completion? QUESTION [12 upvotes]: This is an extension of a question I asked here on Math.SE Assume that I have a finitely generated residually finite centerless group $G$. Is it true that the profinite completion $\hat{G}$ also has trivial center? In the linked question, user YCor was able to show that this fails in general if you do not assume either finite generation or residually finite. However, the result happens to be true if $G$ is a surface group. I’d like to know if this is a phenomenon specific to surface groups, or if this is a more general fact. REPLY [20 votes]: The answer is No in general. Let $n\geq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=\mathrm{SL}_n({\mathbb Z})$. There exists a subgroup $\Gamma \subset \mathrm{SL}_n({\mathbb Z})$ of finite index which is torsion-free and centreless (the centre can only be $\pm 1$ and because $n$ is odd the centre can only be trivial). However, $\mathrm{SL}_n({\mathbb Z})$ has the congruence subgroup property which means that we have the following inclusion of the profinite completions: $$\widehat {\mathrm{SL}_n({\mathbb Z})}= \prod _{q \; \mathrm{prime}} \mathrm{SL}_n({\mathbb Z}_q)\supset \widehat {\Gamma} \supset \prod _{p\in S} U_p \times \prod _{ \ell \notin S} \mathrm{SL}_n({\mathbb Z}_\ell),$$ where $S$ is a finite set of primes, $U_p$ is an open subgroup of finite index in $\mathrm{SL}_n({\mathbb Z}_p)$, and $\ell$ runs through primes in the complement of $S$. Since for infinitely many $\ell$ (for example, all $\ell$ with $\ell\equiv 1 \; (\mathrm{mod}\;n)$), the group $\mathrm{SL}_n({\mathbb Z}_\ell)$ has $n$-th roots of unity in the centre, it follows that the profinite completion of $\Gamma $ is not centreless.<|endoftext|> TITLE: Infinite noncyclic groups with abelian automorphism group QUESTION [5 upvotes]: Is there any infinite noncyclic group whose automorphism group is abelian..can we find a sufficient condition for infinite group to have an abelian automorphism group Thank you REPLY [7 votes]: The finite abelian groups that can be the automorphism group of an infinite abelian group have been classified by Fournelle in [Finite groups of automorphisms of infinite groups II, J. of Algebra 80, 1983, 106 - 112, Theorem 1.2]: There is an infinite abelian group $A$ with $Aut(A) = G$ for a finite abelian group $G$ iff $G$ is of even order and is a direct product of cyclic groups of orders 2, 3, and 4 with the property that if $G$ has an element of order 12 it also has an element of order 2 that is not a sixth power. Examples of torsion-free groups $A$ with $Aut(A)=G$ for $G$ as above are constructed in [Fuchs: Infinite abelian groups II, Chap. XVI, Sect. 116] in examples 1, 2 and Theorem 116.2.<|endoftext|> TITLE: Connections between linear representations and permutation representations QUESTION [9 upvotes]: A finite group $\Gamma$ might be represented by a linear transformation $$\rho : \Gamma\to\mathrm{GL}(\Bbb R^d),$$ or by permutations $$\phi :\Gamma\to\mathrm{Sym}(n).$$ Of course, latter ones can be interpreted as linear representations with permutations matrices. It appears to me that there are many interesting and non-trivial connections between these two types of representations. In particular, one seems to be able to derive properties for one of these by studying it from the perspective of the other. Question: Is there any literature that explores these connections in detail? What are the relevant search terms? I am especially interested in real representations, i.e., over $\Bbb R$. Here, I see many applications to geometry, e.g. symmetric polytopes, rigidity of symmetric frameworks, etc. Here are examples of what I would consider interesting connections: Can transitivity, primitivity, 2-closedness or any other property of a permutation group nicely characterized in terms of its decomposition into irreducible real linear representations? Can a geometric property of a symmetric point arrangement (an orbit polytope, if you want) nicely characterized by a property of the induced permutation group on the points? REPLY [5 votes]: The comments give a reasonable selection of textbooks. I'll add a few more recent papers here, concentrating on the first part of the question. These generalize the three results mentioned by Benjamin Steinberg in the second comment above. In Tsuzuku On multiple transitivity of permutation groups, Nagoya Math. J. 18 (1961), 93–109 there is a modern proof of a theorem of Frobenius that a permutation group $G \le S_n$ is $t$-transitive if and only if all irreducible characters $\chi^\lambda$ of $S_n$ with $n-\lambda_1 \le t/2$ restrict irreducibly to $G$. In particular, taking $\lambda = (n-r,r)$, this says that $G$ is $2r$-transitive only if $\chi^{(n-r,r)}$ restricts irreducibly to $G$. The main theorem of Saxl Characters of multiply transitive groups, J. Alg. 34 (1975), 528–539 is the following refinement: Let $G$ be a permutation group of degree $n > 4r$. If the character $\chi^{(n-r,r)}$ restricts irreducibly to $G$ then either $G$ is $2r$-transitive or $r = 2$, $n = 9$ and $G = \mathrm{P\Gamma L}_2(\mathbb{F}_8)$. Here $\mathrm{P\Gamma L}_2(\mathbb{F}_8)$ is the extension of $\mathrm{PGL}_2(\mathbb{F}_8)$ by the order $3$ Frobenius automorphism. This group is $3$-transitive but $4$-homogeneous. The permutation character of $S_9$ acting on the cosets of $\mathrm{P\Gamma L}_2(\mathbb{F}_8)$ is multiplicity-free. Such permutation characters of symmetric groups were partially classified by Saxl. The classification was completed by Godsil and Meagher, and independently, by me. Peter M. Neumann also has many relevant papers: for example Generosity and characters of multiply transitive groups, Proc. Lond. Math. Soc. 31 (1975), 457–481. Incidentally, all irreducible characters of symmetric groups are real (in fact rational) valued, and have the stronger property (see Geoff Robinson's comment) that they can be defined by homomorphisms into real general linear groups, as in the question.<|endoftext|> TITLE: Why do Bhargava-Skinner-Zhang consider the ordering by height? QUESTION [6 upvotes]: A result of Bhargava-Skinner-Zhang says that a majority of elliptic curves over $\mathbb{Q}$ satisfy the BSD rank conjecture. The are infinitely many isomorphism classes of $\mathbb{Q}$ so one naturally has to be more precise about what "a majority" means; in their case, the elliptic curves are done by height. The question is what are the arguments for the ordering by height being the correct ordering to consider (in the sense that Bhargava-style results actually constitute non-zero progress to the BSD rank conjecture)? The appearance of percentages in these theorems for some reason reminds me of traders on a bazaar advertising their products ("Ours can deal with 50% of the elliptic curves", "Forget him, mine can do 60%"). I just want to understand whether there is in fact a deep connection between the results in question and BSD or whether it is extraordinarily good sales pitch of the authors (this question is not meant to cast a shadow of doubt on anything other than the OP's knowledge of the subject). REPLY [18 votes]: Roughly speaking, the height of an arithmetic object (number, variety, ...) is a natural measure of its complexity, say in the sense of "how many bits does it take to describe the object." (This is not meant to be rigorous, but you seem to want to know why people use "heights".) One can then ask for heights (complexity measures) that have nice properties, for example, transform nicely (functorially) for maps, i.e., ht(f(object)) is related to some nice function of ht(object). Weil heights and morphisms constitute a nice example of this, and canonical heights on abelian varieties behave even more nicely. For [BSZ], counting elliptic curves by height is more-or-less counting by (1) # of bits to describe the $j$-invariant (i.e., the $\bar{\mathbb Q}$ isomorphism class of $E$) plus (2) # of bits to describe how twisted the curve is.<|endoftext|> TITLE: Coarse index of Dirac operator on $\mathbb{R}$ QUESTION [6 upvotes]: Let $D=i\frac{d}{dx}$ be the Dirac operator on $\mathbb{R}$, acting on the spinor bundle $\mathbb{R}\times\mathbb{C}$. The bounded operator $F=\frac{D}{\sqrt{D^2+1}}$ has a coarse index $$\text{Ind}(F)\in K_1(C^*(|\mathbb{R}|)),$$ where $C^*(|\mathbb{R}|)$ is the Roe algebra of the metric space (as opposed to the group) $\mathbb{R}$. It has been stated in various places (for example, Higson & Roe's book "Analytic $K$-Homology" chapter 12) that this coarse index in fact generates the group $K_1(C^*(|\mathbb{R}|))\cong\mathbb{Z}$. However, I haven't seen this explicitly computed anywhere, and it would be nice and instructive to see a detailed calculation showing. For example, the analogous fact that the Dirac operator on $\mathbb{R}^n$ has non-trivial coarse index can be used to show that there exists no Riemannian metric of positive scalar curvature on $\mathbb{T}^n$. To be specific, my question is just about the one-dimensional case, from which I would assume higher-dimensional computations follow: Question: Show that $\text{Ind}(F)$ generates $K_1(C^*(|\mathbb{R}|))\cong\mathbb{Z}$. So if there are experts in the area reading this who wouldn't mind sharing their thoughts, that would be very helpful. REPLY [6 votes]: There are a number of ways to do this calculation, but at risk of shamelessly plugging my own work there is a nice way to see it using a Mayer-Vietoris principle. Decompose $\mathbb{R}$ as the union of the rays $\mathbb{R}^+ = [0, \infty)$ and $\mathbb{R}^- = (-\infty, 0]$, intersecting at a point. This decomposition admits Mayer-Vietoris sequences in both K-homology and and coarse K-theory, and there is a coarse index map from the former sequence to the latter. This gives a commuting square: $$ \require{AMScd} \begin{CD} K_1(\mathbb{R}) @>Index>> K_1(C^*(\mathbb{R})) \\ @VVM{V}V @VVM{V}V \\ K_0(\text{point}) @>Index>> K_0(C^*(\text{point})) \end{CD} $$ Now, the Mayer-Vietoris boundary maps are isomorphisms because the rays $\mathbb{R}^\pm$ have trivial K-homology and trivial coarse K-theory - this follows from homotopy invariance of K-homology and an Eilenberg swindle argument for coarse K-theory. The group $K_1(\mathbb{R})$ is generated by the K-homology class of the Dirac operator, and a quick calculation shows that the Mayer-Vietoris boundary map in this case is just the ordinary boundary map induced by the inclusion of a point into $\mathbb{R}$. A difficult but standard K-homology calculation shows that this boundary map sends the Dirac class to a generator of $K_0(\text{point})$, i.e. a Fredholm operator of index 1. That completes the calculation, up to the construction of the Mayer-Vietoris sequences and and the index maps between them. I worked all that out in my PhD thesis, and you can see a preprint here. By inductively chopping $\mathbb{R}^n$ into half-spaces you get the Gromov-Lawson theorem about the torus for free, and the argument has a nice and geometric flavor to it. That said, it's hard to turn the argument into explicit formulas - calculating the boundary of the Dirac class is hard, and I doubt it's any cleaner to go the other way around the diagram.<|endoftext|> TITLE: What does the homotopy coherent nerve do to spaces of enriched functors? QUESTION [11 upvotes]: Suppose I have two fibrant simplicially enriched categories $C$ and $D$. Then I can consider the collection of simplicial functors $sFun(C,D)$ which, if I recall correctly, has the structure of a simplicial category (though this enrichment does not cooperate with the Bergner model structure, for instance). I can apply the coherent nerve functor and produce two quasicategories $N(C)$ and $N(D)$ which have a Kan complex of morphisms between them $Map(N(C),N(D))$. I'm interested in knowing if there's any way to control $Map(N(C),N(D))$ (even just up to homotopy) if I know what $sFun(C,D)$ is. I could control the homotopy type of $Map(N(C),N(D))$ if I knew, for instance, that $C$ was also cofibrant, but this seems to be an extremely strong condition to require. In general, is there just no useful way to get information about $Map(N(C),N(D))$ from $sFun(C,D)$? I should also add that in the example I'm most interested in, $C$ is just a discrete category (thought of as a simplicial category), so maybe that gives me extra control over its cofibrant replacement? REPLY [13 votes]: Have you looked in the paper by Jean-Marc Cordier and myself: Homotopy Coherent Category Theory, Trans. Amer. Math. Soc. 349 (1997) 1 - 54? We defined a simplicial set of coherent natural transformations between two simplicial functors $F,G:C\to D$. That may be useful as an intermediate setting. There are 'rectification' results if $D$ is complete/cocomplete and especially in the case in which $C$ is an ordinary category, one has an augmentation from its simplicial resolution $S(C)$ to $C$ itself which is well understood. (This may be how you can get at its cofibrant replacement as $S(C)$ is explicitly combinatorially specifiable and is better understood now than when we wrote that paper 22 years ago!)) In general I would not expect the simplicial category $sFun(C,D)$ to be that good a model for the other one. Some of our earlier papers perhaps : Vogt's Theorem on Categories of Homotopy Coherent Diagrams, Math. Proc. Camb. Phil. Soc. 100 (1986) pp. 65 - 90. might also contain useful ideas. The explicit combinatorial arguments we gave can be useful in specific cases although a model / infinity category approach may suit your aims as well. I will not attempt to give more recent references than our papers as there are other regular contributors here who know that approach better than I do.<|endoftext|> TITLE: On Euler's polynomial $x^2+x+41$ QUESTION [47 upvotes]: This is an elementary question about something way outside my area of expertise. A well-known observation due to Euler is that the polynomial $P(x)=x^2+x+41$ takes on only prime values for the first 40 integer values of $x$ starting with $x=0$, namely the values $41,43,47,53,61,71,83,\cdots,1601$. In particular this gives a rather long sequence of primes such that the differences between successive terms form an arithmetic progression, namely $2,4,6,\cdots$, which is a consequence of $P(x)$ being quadratic. All this is related to the fact that the discriminant of $x^2+x+41$ is $-163$ and the field ${\mathbb Q}(\sqrt{-163})$ has class number 1. Suppose one asks about the next 40 values of $P(x)$ after the value $P(40)=41^2$. We have $P(41)=1763=41\cdot 43$, also not a prime. After this the next two values $P(42)=1847$ and $P(43)=1933 $ are primes. Then comes $P(44)=2021=43\cdot 47$, then four primes, then $P(49)=2491=47\cdot 53$, then six primes, then $P(56)=3233=53\cdot 61$, then eight primes, then $P(65)=4331=61\cdot 71$, then ten primes, then $P(76)=5893=71\cdot 83$. The next four values are prime as well for $x=77,\ 78,\ 79,\ 80$, completing the second forty values. But then the pattern breaks down and one has $P(81)=6683=41\cdot 163$. Thus, before the breakdown, not only do we get sequences of $2,\ 4,\ 6,\ 8,\ 10$ primes but the non-prime values are the products of two successive terms in the original sequence of prime values $41,\ 43,\ 47,\ 53,\ 61,\ \cdots$. There is a simple explanation for this last fact, the easily-verified identity $P(40+n^2)=P(n-1)P(n)$, so when $n=1,2,3,\cdots$ we get $P(41)=P(0)P(1)=41\cdot 43$, $P(44)=P(1)P(2)=43\cdot47$, etc. However, this does not explain why the intervening values of $P(x)$ should be prime. I've done an online search to find where this might be discussed, without success, so my question is, what is a reference for this curious behavior of the second 40 values of $P(x)$ (or anything related)? I'm also a little puzzled by the identity for $P(40+x^2)$, though perhaps it comes from the norm function in ${\mathbb Q}(\sqrt{-163})$. REPLY [2 votes]: Please check my post. It indicates that the number of divisors of $x^2+x+41$ is equal to the number of lattice points of $X^2+163Y^2-2(2x+1)Y-1=0$. This formula is transformed in this way. $$163X^2+163^2Y^2-2\cdot163(2x+1)Y=163$$ $$163X^2+\{163Y-(2x+1)\}^2-(2x+1)^2=163$$ $$163X^2+(163Y-2x-1)^2=4x^2+4x+164$$ $X':=163Y-2x-1,\ Y':=X$ and we divide both sides by 4, $$\frac{{X'}^2+{163Y'}^2}{4}=x^2+x+41=P(x)$$ $$N\left(\frac{X'+Y'\sqrt{-163}}{2}\right)=P(x)$$ $\frac{X'+Y'\sqrt{-163}}{2}$ is an element of $\mathbb Q(\sqrt{-163})$. This formula indicates that the elements of $\mathbb Q(\sqrt{-163})$ with norm $P(x)$ is linked to the product pattern of two integers of $P(x)$. The phenomenon you are interested in is based on this fact. For example, let $x$ be $76$. \begin{eqnarray*} P(x)&=&76^2+76+41\\ &=&1\cdot5893\\ &=&71\cdot83 \end{eqnarray*} The number of product pattern of $P(x)$ is $2$. On the other hand, the elements of $\mathbb Q(\sqrt{-163})$ with norm $P(x)$ (ignore sign) are $$\frac{153+\sqrt{-163}}{2},\ 5+6\sqrt{-163}\ .$$ The number is $2$. The left one is a trivial element $\frac{(2x+1)+\sqrt{-163}}{2}$.The other one is a non-trivial element that indicates that $P(x)$ is a composite number.<|endoftext|> TITLE: Ordinal notations within non-standard models of arithmetic QUESTION [9 upvotes]: It is well-known that the order type of any countable non-standard model of arithmetic $\mathfrak{A}$ is $\omega+(\omega^*+\omega)\eta$. My question is what could be said about the order types of ordinal notation systems within non-standard models of arithmetic? The question is motivated by my answer to the question Order types of models of theories of ordinals about the order types of non-standard models of ordinal arithmetic: some group of this non-standard models corresponded to the ordinal notation systems within non-standard models of arithmetic, and I could say very little about their order types. Now I'll provide a bit more details. Let us consider some standard computable ordinal notation system, say Cantor ordinal notation for the ordinals $<\varepsilon_0$ (although the same question could be asked for different ordinal notation systems). The ordinal notation system essentially is an arithmetical formula $x\prec y$ such in the standard model it defines a well-ordering with the order type $\varepsilon_0$. For each $\alpha<\varepsilon_0$ let me denote as $\hat\alpha$ the Gödel number for the ordinal $\alpha$ within the ordinal notation system. And for $\alpha<\varepsilon_0$ and a model $\mathfrak{A}\models \mathsf{PA}$ let me denote as $(\alpha)^{\mathfrak{A}}$ the order type in $\mathfrak{A}$ of the arithmetical relation $\prec\upharpoonright_{\hat\alpha}$ that is the restriction of $\prec$ to the elements $\prec \hat\alpha$. It is easy to see that for each model of arithmetic $\mathfrak{A}\models \mathsf{PA}$ with the order type $A$ and an ordinal $\alpha<\omega^{\omega}$ with the Cantor normal form $\omega^{k_1}+\ldots +\omega^{k_n}$ the order type $(\alpha)^{\mathfrak{A}}$ is precisely $A^{k_1}+\ldots+ A^{k_n}$. However I have no clue of what happening even with $(\omega^{\omega})^{\mathfrak{A}}$. Does $(\omega^{\omega})^{\mathfrak{A}}$ depends only on the order type of $\mathfrak{A}$? Even if the latter isn't the case, is $(\omega^{\omega})^{\mathfrak{A}}$ the same for all countable non-standard models $\mathfrak{A}$? In the other side of the spectrum of possibilities, one could imagine that $(\omega^{\omega})^{\mathfrak{A}}$ reflects a lot of information about the model $\mathfrak{A}$. Could we recover model $\mathfrak{A}$ from the order type $(\omega^{\omega})^{\mathfrak{A}}$? Could we recover $\mathsf{SSy}(\mathfrak{A})$ from the order type $(\omega^{\omega})^{\mathfrak{A}}$? Of course I would be interested in the answers to the same kind of questions for other ordinals $\alpha$ and for more restricted classes of models $\mathfrak{A}$ (e.g. models of true arithmetic, or recursively saturated models, etc.). REPLY [4 votes]: I spent some time thinking about order-types of models of arithmetic, and of order-types of structures they interpret (my PhD). I think there are several formulations of your question: (1) about ordinals, (2) about ordinal notations (M-coded), (3) about "provably- (if we are talking of a theory) or satisfied- (if we are in a model of arithmetic) -well-orderededness of systems of ordinal notations", (4) well-orders if your model has a second-order structure (5) constructible or recursive ordinals (as in Turing, Feferman and Sacks's book), from the point of view of a nonstandard model. Also, Kleene's O in a nonstandard model. Each formulation leads to interesting answers (some known). I remember being very proud when, at Richard Kaye's hint, I found what a nonstandard model M (the interesting case is uncountable) thinks about the order-type of models of DLO, DIS,.....or PA itself for that matter (using arithmetized completeness inside M). The answers for DLO and PA are Q(M) and M+Q(M).(M*+M). It would be nice to see the spectrum of answers to (1)-(5) above.<|endoftext|> TITLE: Enumeration of field generators of a finite field over $\mathbb{F}_{q}$, which are $m^{th}$ powers in the same field QUESTION [6 upvotes]: Consider the finite field extension $\mathbb{F}_{{q}^{d}}$ over $\mathbb{F}_{q}$, where $q=p^a$ for some prime $p$. We assume $d\geq 2$. Let, $$ S=\{ \alpha \in \mathbb{F}_{q^d}\hspace{0.1 cm} | \hspace{0.1 cm} \mathbb{F}_{q}(\alpha)=\mathbb{F}_{q^d} \}$$ In other words, $S$ consists of all those elements in $\mathbb{F}_{q^d}$, whose minimal polynomial over $\mathbb{F}_{q}$ has degree $d$, or to say another way $S$ consists of all field generators of $\mathbb{F}_{q^d}$ Over $\mathbb{F}_{q}$. Let, $S^m= \{ s^m | s\in S\} $, where $m$ is a positive integer $\geq 2$. Then, $$ |S\cap S^m|=?$$ More, precisely, $S\cap S^m$ is the set of all field generators in $\mathbb{F}_{q^d}$ (over $\mathbb{F}_q$), which are $m^{th}$ powers in $\mathbb{F}_{q^d}$. I calculated, this for $m=2$. The answer depends on whether $d$ is odd or even. We have, $$ |S\cap S^2|= \begin{cases} \frac{|S|}{2} & if \hspace{0.2 cm} d \text{ is odd}\\ \frac{1}{2}[|S|-\frac{(q^{d/2}-1)}{d}] & if \hspace{0.2 cm} d \text{ is even} \end{cases} $$ And, $|S|=dM(d,q)$, where $M(d,q)$ denote the number of irreducible polyomials of degree $d$ over $\mathbb{F}_{q}$. I couldn't generalize my method to $m>2$. My, idea is that this problem seems to have been well studied in the literature of finite fields. So, I am hoping for some kind of help or suitable references in this case. Thank you! REPLY [6 votes]: $$| S \cap S^m | = \sum_{n|d} \mu(n) \frac{ \gcd ( m (q^{d/n}-1), q^{d}-1 )}{ \gcd (q^{d} - 1, m) } $$ First note that $|S \cap S^m | = |S \cap \mathbb F_{q^d}^m |$ because every $m$th power that generates is an $m$th power of a generator. We can count elements of $S$ by an inclusion-exclusion argument, subtracting and adding the number of elements in subfields. This gives the term $ \mu(n) q^{d/n}$, or $ \mu(n)( q^{d/n}-1)$ if we only count nonzero elements. To count elements of $S$ that are $m$th powers, we use inclusion-exclusion to count the number of $m$th powers in subfields. To count the number of elements of $\mathbb F_{q^{d/n}}^\times$ that are $m$th powers in $\mathbb F_{q^d}^\times$, we observe that their $m$th roots are both $ m (q^{d/n}-1)$st roots of unity and $(q^d-1)$st roots of unity, hence are $\gcd ( m (q^{d/n}-1),(q^d-1))$th roots of unity, and each of them has $\gcd(q^d-1, m)$ $m$th roots in $\mathbb F_{q^d}^\times$, so the total number of them is $\frac{ \gcd ( m (q^{d/n}-1), q^{d}-1 )}{ \gcd (q^{d} - 1, m) } $. Inclusion-exclusion gives our formula.<|endoftext|> TITLE: Best orthogonal approximation of rank 1 matrix QUESTION [5 upvotes]: Let $X=\lambda_0u_0v_0^T\in\mathbb{R}^{n\times n}$ be a rank 1 matrix where $\lambda_0\in\mathbb{R}$, $u_0,v_0$ are of unit Euclidean norm. What is the solution of the following problem? $$\hat{X}=\arg\min_{\substack{Y:Y=aU\\U\in O(n)\\a\in\mathbb{R}}}\|Y-X\|_F^2$$ where $U$ is orthogonal and $\|\|_F$ is the Frobenius norm. In other words, what is the best scaled orthogonal approximation of rank 1 matrix? REPLY [5 votes]: As suggested by Federico Poloni in a comment, it suffices via the SVD to consider the case when $X$ has $x_{1,1} > 0$ and $x_{i,j} = 0$ when $(i,j) \neq (1,1)$. Then for any orthogonal matrix $U$ with columns $\mathbf{u}_1,\ldots,\mathbf{u}_n$ and scalar $a \in \mathbb{R}$ we have \begin{align*} \|X - aU\|_F^2 & = \|x_{1,1}\mathbf{e}_1 - a\mathbf{u}_1\|^2 + \sum_{j=2}^n\|a\mathbf{u}_j\|^2 \\ & = \big(x_{1,1}^2 - 2ax_{1,1}u_{1,1} + a^2\big) + \big(\sum_{j=2}^na^2\big) \\ & = x_{1,1}^2 - 2ax_{1,1}u_{1,1} + na^2, \end{align*} where $\mathbf{e}_1 \in \mathbb{R}^n$ is the first standard basis vector. There is only one term in this expression that depends on $U$ at all, so (for any given value of $a$) it is minimized when we make $u_{1,1}$ as large as possible (i.e., $u_{1,1} = 1$). The remaining columns of $U$ are arbitrary (but of course must be orthogonal to the first column $\mathbf{e}_1$). Our goal is then to minimize \begin{align*} x_{1,1}^2 - 2ax_{1,1} + na^2, \end{align*} which is a quadratic in $a$ with vertex (i.e., minimum) at $a = x_{1,1}/n$.<|endoftext|> TITLE: Elementary proof of growth estimate for a polynomial via size from its zero set QUESTION [5 upvotes]: The paper Asymptotic properties of polynomials and algebraic functions of several variables by Gorin contains the following. Lemma 3.1. Let $f\in \mathbb R[x_1,\dots,x_n]$. Suppose $f$ has a root in the interior of the unit circle. Then there exists a positive constant $c$ such that $$\|f(x)\|\geq c \cdot d(x,f^{-1}(0))^{\deg f},$$ where $d$ denotes the standard metric on $\mathbb R^n$. The proof involves a previous theorem which is itself rather involved. Question. Is there a quick direct proof of the above lemma? REPLY [2 votes]: The Russian original text clearly states that the exponent is not $\deg f$ but just some $\alpha>0$. For $\alpha=\deg f$ is it not true: a polynomial $f(x,y)=x^{2n}+(x-y^n)^2$ has a unique zero $(0,0)$, but $f(\delta^n,\delta)=\delta^{2n^2}$. I remember obtaining a cubic upper bound for $\alpha$ in the case of a polynomial in two variables with a single zero (but the estimate holds only in a disk centred at the root, not on the whole plane). I may try to find it if you wish.<|endoftext|> TITLE: Categorical description of adjoint representation QUESTION [9 upvotes]: My apologies for a slightly vague question here. Let us fix a Lie algebra $\mathfrak{g}$ over a field $k$. Consider the symmetric monoidal category $\operatorname{Rep}_{k}^{\otimes}(\mathfrak{g})$ of representations of $\mathfrak{g}$. If you wish, you may consider variants of this with dg-categories, stable $(\infty, 1)$-categories etc. There is an element $\mathfrak{g}^{ad}$ of this category, moreover it is a Lie algebra object with respect to the monoidal product $\otimes$. Is there some way to characterize this representation categorically? Feel free to add additional structure so that the answer is positive, basically I'd like to know if there is a sensible axiomatization of "(symmetric monoidal) category with adjoint object". REPLY [2 votes]: For reference, here is the answer suggested by Adrien above. Let $(C, \otimes) $ be an abelian complete and co-complete Cartesian-closed symmetric monoidal category over a field $k$. We write $Map$ for internal mapping objects and $\mathbb{1}$ for the unit. An augmented algebra in $(C, \otimes) $ is an associative unital algebra object, $X$ together with a map of algebras $\eta:X\rightarrow\mathbb{1}$. We define the cotangent object of $(X, \eta)$ as $\Omega_{\eta}X:= ker(\eta) /ker(\eta) ^{2}$. We now define the adjoint object of $C$ as the cotangent object of the categorical end, $\int_{C} Map$, at its natural augmentation. (The augmentation is induced from the map $\int_{C} Map\rightarrow End(\mathbb{1})\cong\mathbb{1}$). In the case of lie algebras the end is $(U\mathfrak{g})^{ad}$ and so we have $$\mathfrak{g}^{ad} = \Omega_{\eta}\int_{Rep(\mathfrak{g})} Map$$. Edit. This is incorrect, as pointed out in the comments. Adrien's comments above still suffice to produce $U\mathfrak{g}^{ad}$ categorically, however the attempt in this post to leverage this to a construction of $\mathfrak{g}^{ad}$ is flawed.<|endoftext|> TITLE: When is compact induction cuspidal? QUESTION [6 upvotes]: Let $G=GL_2(\mathbb{Q}_p)$, and let $K$ be a compact-modulo-center subgroup of $G$, $\rho$ an irreducible smooth representation of $K$. Question 1: Is $\mathrm{ind}_K^G \rho$ cuspidal? Here cuspidal is meant in the sense that matrix coefficients are compactly supported. More precisely, I know already from Bushnell-Henniart, The Local Langlands Conjecture for $GL(2)$, Thm. 11.4, that $\mathrm{ind}_K^G \rho$ is irreducible cuspidal if the following condition holds: $g\in G$ intertwines $\rho$ if and only if $g\in K$. This condition is obviously necessary for irreducibility, but is it also necessary for cuspidality? Question 2: For which groups $G$ other than $GL(2)$ does this result hold? REPLY [4 votes]: Suppose (1) $K$ is open and contains a finite index subgroup of the center (2) For every parabolic subgroup $P$ of $G$ with unipotent radical $N$, $\rho^{ N \cap K}=0$. Then $\rho$ is cuspidal (i.e. a finite direct sum of super cuspidal representations). I think the argument is (or finishes on) page 28 of the book Harmonic analysis on reductive $p$-adic groups by Harish-Chandra, if you can understand the notation. But my understanding of the argument is that it is easy to check from (2) and the definition of the Jacquet module that the Jacquet module vanishes. If the Jacquet module vanishes, and the representation is admissible, then it is a finite direct sum of supercuspidals. To check that the induced representation is admissible, we must find its $K'$ invariants, which by Frobenius reciprocity reduces to finding the $g K' g^{-1} \cap K$ invariants for many all $g \in K' \backslash G/K$. Using the (affine) Bruhat decomposition you can check that, for all but finitely many $g$'s, $g K' g^{-1} \cap K$ contains $N \cap K $ for $N$ the unipotent radical of a parabolic subgroup. That these conditions are necessary is straightforward. A fun (?) exercise is to reprove that certain known constructions of irreducible supercuspidal representations are at least supercuspidal using this method instead of the $\rho$-intertwining condition.<|endoftext|> TITLE: Coarse moduli space versus Kuranishi family QUESTION [5 upvotes]: We will work over complex number field $\mathbb{C}$. Let $\mathscr{M}_h$ be the moduli functor for canonically polarized manifolds with $h$ the Hilbert polynomial. Let us denote by $M_h$ the coarse moduli space of $\mathcal{M}_h$. For any $X\in \mathscr{M}_h({\rm Spec}(\mathbb{C}))$, we know that its Kuranishi family $f:\mathscr{X}\to S$ exists, with $f^{-1}(s_0)\simeq X$. After shrinking $S$, we may assume that $f:\mathscr{X}\to S\in \mathscr{M}_h(S)$. Since $H^0(X,T_X)=0$, the Kuranishi family $f:\mathscr{X}\to S$ is universal. Moreover, the finite automorphism group $\mathrm{Aut}(X)$ acts on $S$ with $s_0$ the fixed point due to the definition of Kuranishi family. My question is: can $S/\mathrm{Aut}(X)$ be seen as a neighborhood of $[X]\in M_h$? In other words, $M_h$ can be obtained by glueing together the quotient of Kuranishi spaces. REPLY [4 votes]: The answer is yes: the germ of complex space $(M_h, \, [X])$ is analytically isomorphic to the quotient $S/\mathrm{Aut}(X)$. This is true not only for moduli spaces of hyperbolic curves, but also in the (much more difficult) context of Gieseker moduli space of (canonical models of) surfaces of general type, see Remark 3.7 in F. Catanese, A superficial working guide to deformations and moduli, Farkas, Gavril (ed.) et al., Handbook of moduli. Volume I. Somerville, MA: International Press; Beijing: Higher Education Press (ISBN 978-1-57146-257-2/pbk; 978-1-57146-265-7/set). Advanced Lectures in Mathematics (ALM) 24, 161-215 (2013). ZBL1322.14002.<|endoftext|> TITLE: About the paper by Buekenhout, Delandtsheer, Doyen, Kleidman, Liebeck and Saxl QUESTION [10 upvotes]: The paper by Buekenhout, Delandtsheer, Doyen, Kleidman, Liebeck and Saxl called Linear spaces with flag transitive automorphism groups (Geom. Dedicata) from 1990 annonces a very powerful classification result for the objects mentioned in the title. However, it does not contain any proofs. In a recent paper (Feng 2017) the paper is cited together with another one by Kantor, 2-transitive and flag transitive designs, but the latter is a survey and does not contain proofs, either. Surely in 2019 the details must be available somewhere! And I keep seeing the Buekenhout paper mentioned in different places, so the confidence in the result seems high. What is a good reference, with proofs, for this classification result? Thanks! REPLY [13 votes]: The proof for this appeared over a series of papers. The final one was Jan Saxl, `On Finite Linear Spaces with Almost Simple Flag-Transitive Automorphism Groups' Journal of Combinatorial Theory, Series A 100, 322–348 (2002). which includes references for all the papers.<|endoftext|> TITLE: (Stable) Auslander algebras in a specific example QUESTION [5 upvotes]: Let $k$ be a field and $Q$ be the quiver with two vertices 1 and 2 and three arrows: $a$ from 2 to 1, $b$ from 2 to 1 and $c$ from 2 to 2. Let $I_1=\langle ab-c^2,ba\rangle$ and $I_2=\langle ab-c^2,c^4,ba-bca\rangle$ and $A_1:=kQ/I_1$ and $A_2:=kQ/I_2$ be the corresponding quiver algebras. It is known that $A_1$ and $A_2$ are isomorphic if and only if the characteristic of the field is not equal to two. Let $M_1$ be the direct sum of all indecomposable $A_1$ modules and $M_2$ be the direct sum of all indecomposable $A_2$ modules. Let $B_1=\operatorname{End}_{A_1}(M_1)$ and $B_2=\operatorname{End}_{A_2}(M_2)$ be the Auslander algebras of $A_1$ and $A_2$. Let $C_1=\underline{\operatorname{End}_{A_1}}(M_1)$ and $C_2=\underline{\operatorname{End}_{A_2}}(M_2)$ be the stable Auslander algebra of $A_1$ and $A_2$. Question 1: Exercise 2 (e) in chapter VII. in the book on Artin algebras by Auslander, Reiten and Smalo asks for a prove that when the characteristic of the field $k$ is equal to two, $A_1$ and $A_2$ have the same Auslander–Reiten quiver (this follows because they are socle-equivalent), but $B_1$ and $B_2$ are not isomorphic. Is there an easy prove/argument for this using only the techniques developed in the book? A direct proof calculating quiver and relations of $B_1$ and $B_2$ would probably be very tedious (both algebras have 20 indecomposable modules) and longer than 10 pages. But maybe there is an elementary argument. Question 2: Are $B_1$ and $B_2$ derived or stable equivalent in characteristic 2? Question 3: Are $C_1$ and $C_2$ isomorphic? REPLY [2 votes]: Question 1: Yes, there is an elementary argument that $B_1$ and $B_2$ are not isomorphic. One uses Theorem VI.5.7, which states that there is a bijection between Morita equivalence classes of nonsemisimple artin algebras of finite type and Morita equivalence classes of nonsemisimple Auslander algebras. If $B_1 \cong B_2$, then they would be Morita equivalent Auslander algebras, and hence $A_1$ and $A_2$ would be Morita equivalent. Since $A_1$ and $A_2$ are both basic, they must be isomorphic, a contradiction. Question 3: It is known that the stable Auslander algebras $C_1$ and $C_2$ are not isomorphic (in characteristic $2$), but this is much harder to show. This is equivalent to the fact that $A_1$ and $A_2$ (as well as the other standard-nonstandard pairs of socle equivalent self-injective algebras of finite type) are not stably equivalent, which was shown by Asashiba (The derived equivalence classification of representation-finite selfinjective algebras. J. Algebra 214 (1999), no. 1, 182–221). Question 2: I would guess that $B_1$ and $B_2$ are neither derived nor stably equivalent (in characteristic $2$), but I have not checked the details. For stably equivalent, I expect that one could apply Martinez-Villa's result that a stable equivalence induces a stable equivalence between certain associated self-injective algebras, which should turn out to be $A_1$ and $A_2$ (Properties that are left invariant under stable equivalence. Comm. Algebra 18 (1990), no. 12, 4141–4169). However, we know these are not stably equivalent (see above). For derived equivalence, one might be able to apply ideas of Hu and Xi on almost $\nu$-stable derived equivalences, as in [Derived equivalences and stable equivalences of Morita type, I. Nagoya Math. J. 200 (2010), 107–152] and subsequent papers.<|endoftext|> TITLE: Question about parametric representations of solutions to $x^3+y^3+z^3=n \in \mathbb N$ QUESTION [7 upvotes]: There are such representations for $n=1,2$. However, by the Wikipedia article, it seems that there are no known parametric (polynomial) representations $P,Q,R$ such that $(P(m))^3+(Q(m))^3+(R(m))^3=3$. Is there any $n \in \mathbb N$ (other than those trivially excluded with conguence conditions) for which it is known that there do not exist three polynomials $A,B,C$ with integer coefficients such that $(A(d))^3+(B(d))^3+(C(d))^3=n$. And are there some general conjectures about non-representability of some natural numbers? Do some heuristics suggest that it is rare to have such a representation, as we have for $n=1,2$? REPLY [2 votes]: Above equation shown below: $x^3+y^3+z^3=n$ There is a parametric solution for rational values of $(x,y,z)$ and $(n= 3)$ given by Seji Tomita & is shown below. $x=w(27a^3+9ab^2-4b^3)$ $y=-w(27a^3+9ab^2+4b^3)$ $z=wb(27a^2+5b^2)$ $w=[(1)/((b(3a+b)(3a-b))]$ For, $(a,b)=(1,2)$ we have, $(x,y,z)=(31/10,-95/10, 94/10)$ The link to his site is given below. Click on "Computational number theory" & select article #104. http://www.maroon.dti.ne.jp/fermat<|endoftext|> TITLE: Rational functions on reduced complex varieties that extend to global holomorphic functions QUESTION [7 upvotes]: Suppose $A$ is an integral domain and a finite type $\mathbb{C}$-algebra. Let $X := \text{Spec}(A)$ and $K := \text{Frac}(A)$ be the fraction field. Suppose $h \in K$ is a rational function that extends to a global complex analytic function on $X(\mathbb{C}).$ Can we conclude that $h \in A$? If $A$ is integrally closed then, (it seems to me that) just the fact that $h$ extends continuously to $X(\mathbb{C})$ suffices to conclude that $h \in A$ and if $A$ is not necessarily normal, I realize that a continuous global extension is not sufficient to draw the required conclusion. Hence I'm interested in what happens in the absence of normality/regularity hypotheses and when we additionally require a holomorphic global extension? REPLY [2 votes]: Let $A$ be a noetherian integral domain, $K$ its field of fractions, and $f \in K$. Assume that for each maximal ideal $\frak m$ of $A$ the element $f \in K \subseteq K\otimes_{A}\hat{A}_{\frak m}$ is in $\hat{A}_{\frak m} \subseteq K\otimes_{A}\hat{A}_{\frak m}$ (here $\hat{A}_{\frak m}$ denotes the completion of $A$ at $\frak m$). Then $f \in A$. Here is the proof. It is enough to show that $f \in A_{\frak m}$ for all maximal ideals $\frak m$; hence we can assume that $A$ is local. Set $\hat K = K \otimes_A \hat A$; then $\hat K$ contains both $K$ and $\hat A$, and the statement is that $A = K \cap\hat A \in \hat K$. So, $f \in K \cap\hat A$. By the easy part of descent theory, it is enough to show that $f \otimes 1 = 1 \otimes f \in \hat A \otimes_A \hat A$. But $f \otimes 1 = 1 \otimes f \in \hat K \otimes_K \hat K$, because $f \in K$, and $\hat A \otimes_A \hat A$ injects into $K \otimes_A (\hat A \otimes_A \hat A) = \hat K \otimes_K \hat K$. By the way, I really don't like to interact with anonymous users, so I would appreciate it if you sent me a private email telling me who you are. You can find my email address on my homepage.<|endoftext|> TITLE: Prisms and Hodge-Tate comparisons QUESTION [5 upvotes]: A few weeks ago, Bhatt and Scholze uploaded a preprint of their paper 'Prisms and Prismatic Cohomology' to arxiv. In Theorem 6.3 they state their Hodge-Tate comparison. Recently, I started reading on Hodge decomposition and Hodge-Tate composition. However, all Hodge-Tate comparison theorems I encountered so far (see for example Theorem 1.3 here) are of a different form and neither could I relate the two to each other nor do I see a connection to Hodge-Tate composition. I would be happy if someone could give a (short) explanation for the name or point out references which I could consult. REPLY [11 votes]: tl;dr The Hodge-Tate comparison isomorphism relates the reduction mod $I$ of prismatic cohomology to something similar to the "Hodge-Tate cohomology" $\bigoplus_{i+j = k} H^i(X, \Omega^j_{X/K})$. Together with the étale comparison theorem relating prismatic cohomology away from $V(I)$ to étale cohomology, this gives the integral $p$-adic Hodge theory version of the usual Hodge-Tate comparison theorem. Let $X$ be a smooth and proper variety over a $p$-adic field $K$ (i.e. a finite extension of $\mathbf{Q}_p$). Let $C = \mathbf{C}_p$ be the $p$-adic completion of an algebraic closure of $\overline{K}$. The usual Hodge-Tate comparison theorem relates the $p$-adic étale cohomology of $X$ to its "Hodge-Tate cohomology". On one side, we consider the étale cohomology $\mathrm{H}^k_{\mathrm{et}}(X_{\overline{K}}, \underline{\mathbf{Z}_p})$: this is a $\mathbf{Z}_p$-module with an action of $G_K$, the absolute Galois group of $K$. On the other side, we consider the "Hodge-Tate cohomology" of $X$. This is: $$ \mathrm{H}^k_{\mathrm{HT}}(X/K) = \bigoplus_{i + j = k} \mathrm{H}^i(X, \Omega^j_{X/K}) $$ It is a graded $K$-vector space with $\mathrm{gr}^j \ \mathrm{H}^k_{\mathrm{HT}}(X/K) = \mathrm{H}^i(X, \Omega^j_{X/K})$. Now, there is a Hodge-Tate spectral sequence, compatible with the Galois actions on both sides. $$ \mathrm{E}_2^{i,j} = \mathrm{H}^i(X, \Omega^j_{X/K}) \otimes_K C(-j) \Rightarrow \mathrm{H}^k_{\mathrm{et}}(X_{\overline{K}}, \underline{\mathbf{Z}_p}) \otimes_{\mathbf{Z}_p} C $$ This spectral sequence is defined more generally when $X$ is any rigid-analytic space over $C$, and is known to always degenerate at $\mathrm{E}_2$ (Thm. 1.7 in "Integral $p$-adic Hodge theory" by Bhatt, Morrow, and Scholze). In the case that $X$ is actually defined over $K$, there is even a canonical splitting, giving a canonical $G_K$-equivariant isomorphism $$ \mathrm{H}^k_{\mathrm{et}}(X_{\overline{K}}, \underline{\mathbf{Z}_p}) \otimes_{\mathbf{Z}_p} C \simeq \bigoplus_j \mathrm{H}^i(X, \Omega^j_{X/K}) \otimes_K C(-j) $$ Now, prismatic cohomology is about integral $p$-adic Hodge theory, meaning that we want to get comparison theorems between cohomology theories without having to invert $p$. More precisely, assume that $X$ is the generic fiber of a smooth proper (formal) scheme $\mathscr{X}$ over $\mathscr{O}_K$. Then we can ask for a relationship between the $\mathbf{Z}_p$-module $\mathrm{H}^k_{\mathrm{et}}(X_{\overline{K}}, \underline{\mathbf{Z}_p})$ and the $\mathscr{O}_K$-module $\mathrm{H}^k_\mathrm{HT}(\mathscr{X}/\mathscr{O}_K) := \bigoplus_{i+j =k} \mathrm{H}^i(\mathscr{X}, \Omega^j_{\mathscr{X}/\mathscr{O}_K})$. Unfortunately, a comparison theorem of the above form does not hold in this context. Instead, prismatic cohomology constructs a "universal" cohomology object which "interpolates" between étale cohomology and Hodge-Tate cohomology (as well as de Rham cohomology, crystalline cohomology, etc). Assume for ease of notation that $K$ is unramified over $\mathbf{Q}_p$. Let $A$ be the ring $\mathscr{O}_K[[u]]$, equipped with a natural Frobenius endomorphism sending $u$ to $u^p$. There is an $\mathscr{O}_K$-linear surjection $A \rightarrow \mathscr{O}_K$ defined by $u \mapsto p$, with kernel $I = (u - p)$. Then $(A, I)$ is an example of a prism. Cohomology of the structure sheaf on the prismatic site of $\mathscr{X}$ relative to $A$ gives a complex $\mathrm{R} \Gamma_{\mathrm{prism}}(\mathscr{X}/A)$ in the derived category of $A$-modules. Loosely, $\mathrm{R} \Gamma_{\mathrm{prism}}(\mathscr{X}/A)$ recovers Hodge-Tate cohomology on the closed set $\mathrm{Spec }\ \mathscr{O}_K = V(u - p) \subseteq \mathrm{Spec }\ A$, and recovers $p$-adic étale cohomology of $\mathscr{X}$ on the open set $\mathrm{Spec }\ A[\frac{1}{u-p}] \subseteq \mathrm{Spec }\ A$. Bhatt and Scholze refer to the former as the Hodge-Tate comparison theorem, and the latter as the étale comparison theorem. Putting these two statements together describes the integral version of the relationship between Hodge-Tate and étale cohomologies. More precisely, there is an object $\Delta_{\mathscr{X}/A}$ in the derived category of Zariski (or étale) sheaves of $A$-modules on $\mathscr{X}$ such that $\mathrm{R} \Gamma(\Delta_{\mathscr{X}/A}) = \mathrm{R} \Gamma_{\mathrm{prism}}(\mathscr{X}/A)$. We consider its (derived) restriction to $\mathrm{Spec} \ A/I$, given by $\overline{\Delta}_{\mathscr{X}/A} := \Delta_{\mathscr{X}/A} \otimes^L_A (A/I)$, so we have $\mathrm{R} \Gamma_{\mathrm{prism}}(\mathscr{X}/A) \otimes^L_A (A/I) \simeq \mathrm{R} \Gamma(\overline{\Delta}_{\mathscr{X}/A})$. Now, the Hodge-Tate comparison theorem as stated in the Bhatt-Scholze paper gives a canonical isomorphism of $\mathscr{O}_K$-modules. Here, if $M$ is an $\mathscr{O}_K$-module, $M\{j\} = M \otimes_{\mathscr{O}_K} (I/I^2)^{\otimes j}$. $$ \mathrm{H}^j(\overline{\Delta}_{\mathscr{X}/A}) \simeq \Omega^j_{\mathscr{X}/\mathscr{O}_K}\{-j\} $$ This gives us a hypercohomology spectral sequence $$ \mathrm{E}_2^{i,j} = \mathrm{H}^i(\mathscr{X}, \Omega^j_{\mathscr{X}/\mathscr{O}_K})\{-j\} \Rightarrow \mathbf{H}^{i+j}(\mathscr{X}, \overline{\Delta}_{\mathscr{X}/A}) $$ The right side should be thought of as a "derived correction" of $\mathrm{H}^{i+j}_{\mathrm{prism}}(\mathscr{X}/A) \otimes_A (A/I)$. To relate this to étale cohomology, we must pass to a bigger prism. Let $A_{\mathrm{inf}} = \mathrm{W}(\mathscr{O}_{C^\flat})$. This has a natural Frobenius automorphism $\varphi$ lifting the one on $\mathscr{O}_{C^\flat}$ and a surjection $\widetilde{\theta} \colon A_{\mathrm{inf}} \rightarrow \mathscr{O}_C$ with kernel $J = (d)$ for a certain element $d \in A_{\mathrm{inf}}$. ($\widetilde{\theta} = \varphi^{-1} \circ \theta$, where $\theta$ is the usual map as defined by Fontaine). There is a map of prisms $(A, I) \rightarrow (A_{\mathrm{inf}}, J)$ which lifts the inclusion $\mathscr{O}_K \hookrightarrow \mathscr{O}_C$. Then we have $\mathrm{R} \Gamma_{\mathrm{prism}}(\mathscr{X}_{\mathscr{O}_C}/A_{\mathrm{inf}}) \simeq \mathrm{R} \Gamma_{\mathrm{prism}}(\mathscr{X}/A) \otimes_A^L A_{\mathrm{inf}}$. In particular, the above gives us a spectral sequence $$ \mathrm{E}_2^{i,j} = \mathrm{H}^i(\mathscr{X}, \Omega^j_{\mathscr{X}/\mathscr{O}_K}) \otimes_{\mathscr{O}_K} \mathscr{O}_C\{-j\} \Rightarrow \mathbf{H}^{i+j}(\mathscr{X}_{\mathscr{O}_C}, \overline{\Delta}_{\mathscr{X}_{\mathscr{O}_C}/A_\mathrm{inf}}) $$ where the right side should be thought of as a "derived correction" of $\mathrm{H}^{i+j}_{\mathrm{prism}}(\mathscr{X}_{\mathscr{O}_C}/A_{\mathrm{inf}}) \otimes_{A_{\mathrm{inf}}, \widetilde{\theta}} \mathscr{O}_C$ On the other hand, the étale comparison theorem gives an isomorphism, equivariant with respect to the natural $G_K$-actions on both sides: $$ \mathrm{R}\Gamma_{\mathrm{et}}(X_C, \underline{\mathbf{Z}_p}) \otimes_{\mathbf{Z}_p} A_{\mathrm{inf}}[1/d] \simeq \mathrm{R}\Gamma_{\mathrm{prism}}(\mathscr{X}_{\mathscr{O}_C}/A_{\mathrm{inf}})\otimes_{A_{\mathrm{inf}}}^L A_{\mathrm{inf}}[1/d] $$ Note that the right side is isomorphic to $\mathrm{R} \Gamma_{\mathrm{prism}}(\mathscr{X}/A) \otimes_A^L A_{\mathrm{inf}}[1/d]$. Thus, the étale and Hodge-Tate comparison theorems together show that $\mathrm{R} \Gamma_{\mathrm{prism}}(\mathscr{X}/A)$ determines a deformation from $\mathrm{H}^{i+j}_{\mathrm{et}}(X, \underline{\mathbf{Z}_p})$ to an object related to the Hodge-Tate cohomology $\bigoplus_j \mathrm{H}^i(\mathscr{X}, \Omega^j_{\mathscr{X}/\mathscr{O}_K})\{-j\}$.<|endoftext|> TITLE: Fully extended TQFT and lattice models QUESTION [19 upvotes]: I often read that fully extended TQFTs are supposed to classify topological phases of matter. So I would like to understand the formal nature of fully extended TQFTs on a more direct physical level (without having to read up on a huge amount of category theory language): 1) Many algebraic structures are given by a finite set of tensors (linear maps if you wish) which fulfil a finite set of tensor-network equations (equations between different compositions of the linear maps if you wish). For example, fusion categories are given by a 10-index $F$-tensor satisfying the pentagon equation, or (ordinary axiomatic) 2-dimensional TQFTs are given by a bunch of tensors associated to the pair of pants and a few other cobordisms, satisfying the axioms of (something like) a Frobenius algebra. Can $n$-dimensional fully extended TQFTs be formulated as a finite set of tensors obeying a finite set of tensor-network axioms? Is it known what these tensors and axioms are? 2) Is there any idea for a construction of a local partition function (similar to the Turaev-Viro construction) that takes data related to a fully extended TQFT as input? I guess for Turaev-Viro models (and any other topological state-sum construction) it's possible to find a corresponding extended TQFT. Are there any examples of extended TQFTs that are conjectured to correspond to phases without known state-sum constructions (such as chiral topological phases in 2+1D)? REPLY [20 votes]: It may take a bit of extraction, but positive answers to both of your questions follow from my results joint with Gaiotto in Condensations in higher categories (arXiv:1905.09566). In that paper we build a $\mathbb{C}$-linear $(d+1)$-category that we call $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$. ($d$ is arbitrary, and so is the ground field, but $\mathbb{C}$ is natural for bosonic physics.) One of our main theorems is that there is a natural equivalence of $(d+1)$-categories between $\Sigma^n\mathrm{Vect}_{\mathbb{C}}$ and the fully dualizable subcategory of the $(d+1)$-category $n\mathrm{Cat}_{\mathbb C}$ of all $\mathbb{C}$-linear $d$-categories [1]. Thus, given the cobordism hypothesis, $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$ is the universal target for $(d+1)$-dimensional TQFTs, and in particular every object in $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$ determines a (framed) TQFT [2, 3]. Second, our construction of $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$ is sufficiently explicit so that, for each object of $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$, we are able to build a (gapped topological) commuting projector Hamiltonian lattice system, which is, more or less, a very special type of the tensor-network models that you ask about. (Turaev–Viro is, essentially, the $d=2$ case of our construction.) Similarly, to each $k$-morphism, associate a (gapped topological) interface which is again commuting projector Hamiltonian. All together, this gives a realization of $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$ in terms of lattice models. Conversely, we argue that any $(d+1)$-dimensional gapped topological system which can be "condensed from the vacuum" is (in the same gapped phase as) an object of $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$. There are systems, e.g. Kitaev's honecomb model of $E_{8,1}$, which are gapped topological and have commuting projector Hamiltonian models, but which cannot be condensed from the vacuum, and are not realized in our $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$. Such systems correspond to TQFTs (e.g. TQFTs of Reshetikhin–Turaev type), but those TQFTs (probably) are not "fully extended" in any reasonable sense [4]. Indeed, the value a fully extended $(d+1)$-dimensional TQFT assigns to a point is (probably) the $d$-category of (local) gapped topological boundary conditions for the TQFT, which is (probably) empty unless the TQFT can be condensed from the vacuum. In any case, our theorems do provide an isomorphism between some large families of TQFTs and lattice models, which I think includes all the ones you are asking about. Footnotes. [1] I take arbitrary linear functors as my morphisms in $n\mathrm{Cat}_{\mathbb C}$. This choice picks out the "naive tensor product" as the symmetric monoidal structure. There are other natural $(d+1)$-categories of $n$-linear categories. One expects, given the appendix to Bartlett–Douglas–Schommer-Pries–Vicary, Modular categories as representations of the 3-dimensional bordism 2-category (arXiv:1509.06811), that all these different $(n+1)$-categories should have equivalent fully dualizable subcategories, but this is not known. Scheimbauer calls this the "bestiary hypothesis". [2] We conjecture that in fact each object of $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$ determines an oriented TQFT in an essentially-canonical way. If this conjecture is true, then it is special to the complex numbers $\mathbb{C}$ — I expect it to fail over other fields. Specifically, the cobordism hypothesis determines an action of $O(d+1)$ on the $(d+1)$-groupoid of objects in $\Sigma^d\mathrm{Vect}_{\mathbb{C}}$. Using the usual topology on $\mathbb{C}$, this $(d+1)$-groupoid is in fact naturally a topological groupoid. Our conjecture is that the $SO(d+1)$-action is canonically homotopy-trivial on this topological groupoid (but not on the groupoid that only uses the algebraic structure of $\mathbb{C}$). [3] (Added in response to comments discussion.) We work with weak $n$-categories, and manipulate them "synthetically", meaning model independently. Weak $n$-categories are not the same as $(\infty,n)$-categories, but there is an adjunction between the two worlds. The cobordism hypothesis is not known to be satisfied in any model of weak $n$-categories. [4] (Added in response to Henriques' comment below.) One could also imagine looking at categories of boundary conditions which are not gapped. For instance, the category of all boundary conditions, or the category of conformal boundary conditions. My understanding is that this is how to understand the TQFT aspects of the work of Bartels–Douglas–Henriques culminating with Conformal nets V: dualizability (arXiv:1905.03393).<|endoftext|> TITLE: Overconcentration of Poles on the Circle of Convergence of a Power Series with Bounded Coefficients QUESTION [5 upvotes]: Let $V$ be an arbitrary set of infinitely many positive integers, and let: $$\varsigma_{V}\left(z\right)\overset{\textrm{def}}{=}\sum_{v\in V}z^{v}$$ Let $T_{V}$ denote the set of all $t\in\left[0,1\right)$ for which the limit: $$c_{V}\left(t\right)\overset{\textrm{def}}{=}\lim_{x\uparrow1}\left(1-x\right)\varsigma_{V}\left(e^{2\pi it}x\right)$$ exists and is non-zero. Is $T_{V}$ necessarily finite (that is, would an "overconcentration" of radii on which $\varsigma_{V}\left(z\right)$ grows like $\frac{1}{1-\left|z\right|}$ as $z$ tends radially to the unit circle result in a contradiction against the boundedness of $\varsigma_{V}\left(z\right)$'s power series coefficients)? Or do there exist $V$ for which $T_{V}$ is infinite? Part of the problem is that there is so much literature about the boundary behavior and singularities of power series that trying to find information about something this specific is like looking for the needle in the proverbial haystack. Any assistance would be much appreciated. REPLY [8 votes]: The set can be infinite (but only countable). For an example choose any $t_j$ linearly independent over $\mathbb Q$ and let $V$ be the set of all $v$ such that $vt_j\mod 1 \notin (\frac 12-a_j,\frac 12+a_j)$ for all $1\le j\le J(v)$ where $a_j$ decrease so fast that $\sum_j a_j<+\infty$ and the function $J$ increases to $+\infty$ so slowly that by the moment $v_0$ it reaches $j_0$ you may say that $e^{2\pi ivt_j}$ for $v=1,\dots,v_0$ model the joint distribution of $j_0$ independent variables equidistributed over the unit circle with high precision. That (with some small extra care and effort) will give you the existence of non-zero limits even for Cesaro means and the Poisson summation method is stronger. Edit (about the sum of squares). Let us assume that the limit $L_j=c(t_j)$ exists for some finite collection of $t_j$. Consider $$ L=\lim_{x\to 1}\sum_j \bar L_j(1-x)\sum_{v>0}\chi_V(v)x^ve^{2\pi i t_j v}\,. $$ On the one hand, $L=\sum|L_j|^2$. On the other hand, by Cauchy-Schwarz, $$ L^2\le \limsup_{x\to 1}\left[(1-x)\sum_{v>0}\chi_V(v)^2x^v\right] \left[(1-x)\sum_{v>0}\left|\sum_j\bar L_j e^{2\pi i t_j v}\right|^2 x^v\right] $$ The expression in the first brackets is at most $1$, while, if you open the absolute value and take into account that $$ (1-x)\sum_{v>0}e^{2\pi i (t_j-t_k) v} x^v\to 0 $$ for $j\ne k$, you'll find that the limit of the sum in the second brackets is $L$. That gives $L^2\le L$, so $L\le 1$. Apologies to everybody for endless edits. Here comes another one. Consider a very fast increasing sequence $n_k$, $k\ge 1$ of positive integers (as usual, that means that we will choose them inductively). Let $N_k=\prod_{m=1}^k(2^{n_m}-1)$. Define $V_k$ as the set of $v\ge 1$ congruent to $N_{k-1}2^\alpha$ modulo $N_k$ with some $\alpha\in\{0,1,\dots,n_k-1\}$ (we define $N_0=1$). It is clear that we just take $n_k$ possible residues modulo $N_k$, so the density of $V_k$ is $n_k/N_k$. Put $V=\cup_{k\ge 1} V_k$ and $t_k=N_k^{-1}$. That's the whole construction. Now it remains to explain why it works. First, $V_k$ are disjoint. Indeed, if $mN_m$, but $$ L_k(z)=N_k^{-1}\sum_{\alpha=0}^{n_k-1}e^{2\pi i\frac{2^\alpha}{2^{n_k}-1}}\ne 0 $$ (all terms are in one half-plane) and $L_m(z)$ with $m>k$ cannot change that because they are ridiculously small (at most $n_m/N_m$) compared to anything depending on $n_1,\dots, n_k$ only (it is this condition that forces the fast growth of $n_k$). Thus $L(z)\ne 0$.<|endoftext|> TITLE: Slowest initial state for convergence of finite birth-and-death Markov chains QUESTION [7 upvotes]: Consider the continuous-time birth-and-death Markov chain on $\{1,\cdots,n\}$ with all rates equal to $1$. Is it true that the convergence to equilibrium, in total variation distance, is slowest when the initial state is an endpoint? Here is a concrete reformulation: consider the matrix $$ L = \begin{pmatrix} -1 & 1 & 0 & \cdots & 0 \\ 1 & -2 & 1 & \cdots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 1 & -1 \end{pmatrix}. $$ The question is whether for every $t>0$, among the rows of $\exp(tL)$, the $\ell_1$-distance to $(\frac{1}{n}, \cdots ,\frac{1}{n})$ is maximal for the first (and last) row. This is true for small enough $t$ (consider the Taylor expansion of the exponential) and for large enough $t$ (see the answer by Mateusz Kwaśnicki) but I would like an argument working for every $t>0$. It looks obvious but I have no idea how to prove it. REPLY [3 votes]: We denote $\mu = (\frac{1}{n},\cdots,\frac{1}{n})$. First remark : because the convexity of $\ell^1$, for any $t$ the maximum of $\|\exp(tL)\nu-\mu\|_{\ell^1}$ is obtain when $\nu = \delta_x$, $x\in \{1,\cdots,n\}$ so we only have to compare these measures. We now consider the following equivalent system: The markov chain is on $\{1,\cdots,2n\}$ with periodic boundary conditions. ie $$\tilde{L}=\begin{pmatrix} -2 & 1 & 0 & \cdots & 0 &1 \\ 1 & -2 & 1 & 0 & \cdots & 0 \\ 0 & 1 & \ddots & & & \vdots \\ \vdots \\ 0 &&&&\ddots &1\\ 1 & & & &1 & -2\end{pmatrix} $$ with symetric initial condition : $\nu(2n+1-y)=\nu(y)$. As the symetry is conserved : for all $t$ $\nu_t := \exp(t\tilde{L})\nu $ stay symetric. This system is indeed similar to the first one. (with the transformation $\phi : \mathbb{P}(\{1,\cdots,2n\})\rightarrow \mathbb{P}(\{1,\cdots n\})$ $\phi (\nu_t)(k) = \nu_t(k)+\nu_t(2n+1-k)$ As the $\tilde{L}$ is translation invariant for all $t$ we have a kernal $K_t$ $$[\exp(t\tilde{L})\delta_x](y) = K_t(x-y)$$ with $K_t(-k)=K_t(k)=K_t(2n-k)$ for all $k$. Moreover we claim that $K_t(k)$ is decreasing for $k$ in $0,\cdots,(n-1)$. Let $1\leq i\leq n$ and $\nu^i = \frac{1}{2}(\delta_i+\delta_{2n+1-i})$. $$ \|\nu_t^i-\tilde{\mu}\|_{\ell^1} = \sum_{x\leq 2n} \big|\frac{1}{2}(K_t(x-i)+K_t(x+i-1))-\frac{1}{2n}) \big|$$ We divide the sum in two part $$X_1 = [x: K_t(x-i)\leq \frac{1}{2n} \text{,} K_t(x+i-1)\leq \frac{1}{2n}]\cup [x: K_t(x-i)\geq \frac{1}{2n} \text{,} K_t(x+i-1)\geq \frac{1}{2n}]$$ and $X_2 = \{1,\cdots 2n\}/X_1 $. We have then $$ \|\nu_t^i-\tilde{\mu}\|_{\ell^1} = \frac{1}{2} \sum_{x\in X_1} \big|K_t(x-i)-\frac{1}{2n}\big|+\big| K_t(x+i-1)-\frac{1}{2n}\big|+\frac{1}{2} \sum_{x\in X_2} \big| |K_t(x-i)-\frac{1}{2n}|-| K_t(x+i-1)-\frac{1}{2n}|\big| \\ = \|K_t - \tilde{\mu}\|_{\ell^1} - \sum_{x\in X_2} \min(| K_t(x+i-1)-\frac{1}{2n}|,| K_t(x-i)-\frac{1}{2n}|) $$ As $K_t$ is monotone in the distance $|x-i|$, $X_1$ and $X_2$ are an union of two segment in $\{ 1,\cdots , 2n\}$. In the particular case of $i=1$, there exists $k_0$ such that $K_t(x-1)\geq \frac{1}{2n}$ for all $1 \leq x\leq k_0$ and $K_t(x-1)< \frac{1}{2n}$ for all $k_0 TITLE: When does rationalization commute with homotopy fixed points? QUESTION [22 upvotes]: Let $X$ be a $G$-space. There are a number of places in the literature where one can find the claim that under certain conditions rationalization and taking homotopy fixed points with respect to a finite group action commute, that is, the map $X^{hG} \to (X_{\mathbb{Q}})^{hG}$ is a rationalization or equivalently $(X^{hG})_{\mathbb{Q}} \to (X_{\mathbb{Q}})^{hG}$ is a weak equivalence. I know how to prove a number of special cases of this (e.g. when $X$ is based with $G$-fixed base point and nilpotent with finitely many non-zero homotopy groups), but a quite general result is claimed in the 1989 thesis of Goyo (he uses $(-)^{(G)}$ as notation for homotopy fixed points): However, I don't understand his proof (it seems to use the claim that an infinite limit of rationalizations is a rationalization, which is not true). So I am looking for a reference which gives an answer to the following question: When exactly does rationalization commute with homotopy fixed points? REPLY [19 votes]: The statement appears to me to be false. The difficulty, from some point of view, is that in the spectral sequence going from $H^{-i}(G;\pi_j(M))$ to $\pi_{i+j}(M^{hG})$ an infinite number of torsion groups can conspire to make something rationally non-trivial. As I understand it, the statement you are asking about says that if a finite group $G$ acts on a space $M$, and if both $M$ and the homotopy fixed point space $M^{hG}$ are simply connected, and if $M\to M_{\mathbb Q}$ is a rationalization of $M$ that is also a $G$-map for some action of $G$ on $M_{\mathbb Q}$, then the induced map $M^{hG}\to (M_{\mathbb Q})^{hG}$ is also a rationalization. Let's make a counterexample in which $G$ acts trivially on $M$ (and on $M_{\mathbb Q}$). In this case $M^{hG}$ is the function space $Map(BG,M)$. I claim that it is enough if we can find a simply connected space $M$ and a finite group $G$ such that the based function space $Map_\ast(BG,M)$ is simply connected but not rationally trivial. If so, then the fibration sequence $$ Map_\ast(BG,M)\to Map(BG,M)\to M $$ shows that $Map(BG,M)$ is simply connected, and also in the fibration sequence $$ Map_\ast(BG,M)_{\mathbb Q}\to Map(BG,M)_{\mathbb Q}\to M_{\mathbb Q} $$ the right hand map is not a weak equivalence. On the other hand, in the fibration sequence $$ Map_\ast(BG,M_{\mathbb Q})\to Map(BG,M_{\mathbb Q})\to M_{\mathbb Q} $$ the right hand map is a weak equivalence because $H^j(BG;V)=0$ for $j>0$ and $V$ a rational vector space. It follows that $$ Map(BG,M)_{\mathbb Q}\to Map(BG,M_{\mathbb Q}) $$ is not a weak equivalence. To come up with such an $M$ and $G$ we can use the Atiyah-Segal completion theorem. The basic idea is to take $M$ to be $BU$, but I have to modify this a little to make $Map_\ast (BG,M)$ simply connected. Start with $BU$, whose homotopy groups are $\pi_{2k}\cong \mathbb Z$ with complex conjugation acting by $+1$ when $k$ is even and by $-1$ when $k$ is odd. Localize it by inverting $2$, and split the result as a product of two factors according to that conjugation action. $M$ will be the factor corresponding to $-1$, so its homotopy groups are $\pi_j\cong \mathbb Z[1/2]$ if $j$ congruent to $2$ mod $4$. Let $G$ be the dihedral group of order $6$. Then $H^i(BG;\mathbb Z[1/2])$ is trivial when $i$ is not a multiple of $4$ and is of order $3$ if $i>0$ is a multiple of $4$. It follows that $Map_\ast(BG,M)$ is simply connected and that $\pi_2Map_\ast(BG,M)$ is the inverse limit of larger and larger finite $3$-groups. By Atiyah-Segal (which describes the homotopy groups of the related space $Map_\ast(BG,BU)$) there is actually a copy of the $3$-adic integers here, so the group is rationally nontrivial.<|endoftext|> TITLE: What is the homotopy fiber of $X \to X_{hG}$, where this is a pointed homotopy orbit? QUESTION [9 upvotes]: The unpointed version is easy: the model $X = EG \times X \to (EG \times X)/G = X^{un}_{hG}$ is a fibration with fiber $G$. But when we go pointed, $X = EG_+ \wedge X \to (EG_+ \wedge X) / G = X_{hG}$ is no longer a fibration: its fiber changes from $G$ over non-basepoints to $\ast$ over the basepoint. Of course, these two cases are still closely related. So perhaps there's some hope of understanding the pointed case. Questions: Let $G$ be a finite group (or perhaps something a bit more general), and let $X$ be a pointed $G$-space. Is there a good way to understand the homotopy fiber $F$ of the map $X \to X_{hG}$ (where these are pointed homotopy orbits)? In particular, does the homotopy fiber sequence $F \to X \to X_{hG}$ deloop to a homotopy fiber sequence $X \to X_{hG} \to BF$ (as it does in the unpointed case)? REPLY [7 votes]: Here is a special case which gives a partial answer: (i). Suppose $G$ acts in a homotopically trivial way on $X$. This means that there is a trivial $G$-space $Y$ and a pair of $G$-equivariant maps $X \overset\sim\leftarrow X' \overset\sim\to Y$ each which is a weak homotopy equivalence of underlying spaces. (ii). Assume further that $X \simeq \Sigma Z$ has the homotopy type of a suspension of a based space $Z$ (note: $Z$ does not need to have a $G$ action). Using (i), the reduced homotopy orbits $X_{hG}$ is identified with the cofiber of the inclusion $\ast \times BG \to X\times BG$ and the latter is just $X\wedge (BG_+)$. Using (ii), since $X$ is a suspension, $X\wedge (BG_+) \simeq X \vee (X\wedge BG)$. Set $U = X\wedge BG$. The map $X\to X_{hG}$ with respect to this identification corresponds to the inclusion $X\to X \vee U$. So we have to identify the homotopy fiber of the latter. It is straightforward to check that this homotopy fiber is the loop space of the homotopy fiber of the first summand projection $X\vee U \to X$. But the homotopy fiber of the latter is just $U\wedge(\Omega X_+)$. So the homotopy fiber of the map $X\to X_{hG}$ in our special case is then $$ \Omega (X\wedge BG \wedge (\Omega X_+)) \simeq \Omega ((X\wedge BG) \vee (X\wedge BG \wedge \Omega X) ) $$ Answering your question (1) in this instance. Speculation on your Question (2): For the fiber sequence $F\to X\to X_{hG}$ to be induced by a map $X_{hG} \to BF$ in our example, you would need to know that the fiber sequence $U\wedge (\Omega X)_+ \to U \vee X \to X$ is fiber homotopically trivial. The latter is equivalent to finding a retraction to the map $U\wedge (\Omega X)_+ \to U \vee X$. Denote the latter map by $\iota$. This seems unlikely to be true in general: a retraction to $\iota$ is given by a pair of maps $a:U\to U\wedge (\Omega X)_+$ and $b:X\to U\wedge (\Omega X)_+$ such that $\iota \circ a: U \to U\vee X$ and $\iota\circ b:X \to U \vee X$ are the summand inclusions. The map $a$ is clear, but there is no obvious candidate for $b$.<|endoftext|> TITLE: Luxemburg norm as argument of Young's function: $\Phi\left(\lVert f \rVert_{L^{\Phi}}\right)$ QUESTION [8 upvotes]: Let $\Phi$ be a Youngs's function, i.e. $$ \Phi(t) = \int_0^t \varphi(s) \,\mathrm d s$$ for some $\varphi$ satifying $\varphi:[0,\infty)\to[0,\infty]$ is increasing $\varphi$ is lower semi continuous $\varphi(0) = 0$ $\varphi$ is neither identically zero nor identically infinite and define the Luxemburg norm of $f:\Omega\to\mathbb{R}$ as $$ \lVert f \rVert_{L^{\Phi}} := \inf \left\{\gamma\,\middle|\,\gamma>0,\,\int_{\Omega} \Phi\left(\frac {\lvert f(x)\rvert}{\gamma} \right)\,\mathrm{d}x\leq 1\right\}.$$ Question: What can we say about $\Phi\left(\lVert f \rVert_{L^{\Phi}}\right)$? In particular, I'd like to know, if $$\Phi\left(\lVert f \rVert_{L^{\Phi}}\right) \leq C \int_{\Omega}\Phi(\lvert f(x)\rvert) \,\mathrm d x$$ holds for some $C$ independent of $f$. Any idea or hint for a reference is welcome! Notes: The above inequality trivially holds for $\Phi(t) = t^p$, where $p>1$ Maybe it's appropriate to consider this question in the more general framework of Musielak-Orlicz spaces. However, e.g. in Lebesgue and Sobolev Spaces with Variable Exponents I was unable to find an appropriate result. I have asked this question on Math.Stackexchange without luck, so I'm trying here. REPLY [5 votes]: The conjectured inequality does not hold. For a counterexample, consider $\Phi(t)=\max(t^2,t^3)$ and $\Omega=(0,1)$. Let $f=a\chi_{(0,b)}$ for $a,b\in (0,1)$. It can be calculated that $\|f\|_{L^\Phi}= a b^{1/3}$. Then the inequality can be written as \begin{equation*} a^2 b^{2/3} \leq C a^2 b \end{equation*} which is not possible for a constant $C$ independent of $b$. Intuitively, this is because the left-hand side is mostly determined by the values of $\Phi$ for large $t$, whereas the right-hand side is (for small $a,b$) independent of the values of $\Phi$ for large $t$.<|endoftext|> TITLE: $\mathcal A^{\mathcal A}\sim\mathcal B^{\mathcal B}\implies\mathcal A\sim\mathcal B\ ?$ (Does A^A ~ B^B imply A ~ B? --- A, B categories) QUESTION [14 upvotes]: I asked this question on Mathematics Stackexchange, but got no answer. Let $\mathcal A$ and $\mathcal B$ be nonempty categories whose categories $\mathcal A^{\mathcal A}$ and $\mathcal B^{\mathcal B}$ of endofunctors are equivalent. Are $\mathcal A$ and $\mathcal B$ necessarily equivalent? Implicit assumption: we are working in ZFC, and we assume that ZFC is consistent. If my understanding is correct, this post of Joel David Hamkins implies that one cannot prove that the answer is Yes, so that either the answer is No, or the question is undecidable. (I think that the answer is No.) [Reminder 1: Categories are generalized sets in the following sense. Given a set $S$ let $\mathcal C(S)$ be the category whose objects are the elements of $S$ and whose only morphisms are the identity morphisms. Then the assignment $S\mapsto\mathcal C(S)$ commutes with exponentials in the obvious sense.] [Reminder 2: Infinite cardinals $\kappa$ satisfy $\kappa^\kappa=2^\kappa$. Indeed $\kappa^\kappa\le(2^\kappa)^\kappa=2^{\kappa\kappa}=2^\kappa\le\kappa^\kappa$.] REPLY [3 votes]: Here is another counterexample, also using groupoids. For a group $G$ I will use the notation $G$ also for the corresponding one-object groupoid. If $C$ is the disjoint union of groups $G_i$ and $D$ is the disjoint union of groups $H_j$ then $C^D$ is the product over $j$ of the disjoint union over $i$ of the groupoid $G_i^{H_j}$. If $G$ is an abelian group then $G^H$ is the disjoint union, over all homomorphisms $H\to G$, of $G$. Putting all of this together, one can work out a description of $A^A$ when the groupoid $A=\infty 1\coprod \infty\mathbb Z$ is the disjoint union of countably infinitely many trivial groups and countably infinitely many infinite cyclic groups. It comes out be the disjoint union of the following groups, each occurring a continuum's worth of times: free abelian groups of all finite ranks, and a countably infinite product of $\mathbb Z$'s. Now let $B=\infty 1\coprod \infty\mathbb Z\coprod \mathbb Z^2$ be the disjoint union of $A$ with one copy of $\mathbb Z^2$. Then $B^B$ comes out to be isomorphic to $A^A$.<|endoftext|> TITLE: Linear systems separating points QUESTION [6 upvotes]: Is it easy to find an example of a complete linear system on a smooth projective curve (say over $\mathbb C$) which separates points but which is not an embedding? (for just a linear system, one can take the linear system induced by a linear projection (in an embedding) from a point which is not on any secant line of the curve but lies on a tangent line of the curve). REPLY [4 votes]: Take for $X$ a trigonal curve of genus $\geq 5$; that is, $X$ carries a unique degree 3 pencil $P$. Suppose some divisor of $P$ is of the form $p+2q$. Consider the linear system $|K-p|$. If $r,s$ are two distinct points of $X$, we have $h^0(p+r+s)=1$ by unicity of the $g^1_3$, hence $h^0(K-p-r-s)=h^0(K-p)-2$ by Riemann-Roch; thus $|K-p|$ separates $r$ and $s$. But $h^0(K-p-2q)=h^0(K-p)-1$, which means that the associated map is not an embedding at $q$. REPLY [3 votes]: I believe that on any non-rational and non-hyperelliptic curve , the complete linear system $L = K_X(2p)$ will work. In such a case $ 2p - p_1 -p_2$ will always be a non-trivial divisor of degree zero unless $p = p_1 = p_2$ and hence $h^0(L) = g+1$ and $ h^0(L(-p_1 -p_2) ) = g-1 $ unless as stated $p = p_1 = p_2$.<|endoftext|> TITLE: Capturing the $\omega_1^{\mathrm{CK}}$-th stage of Gödel's constructible hierarchy QUESTION [5 upvotes]: For an ordinal $\alpha$, let $L_\alpha$ be the $\alpha^{th}$ set of Gödel's constructible hierarchy and let $\omega_1^{\mathrm{CK}}$ be the first non-recursive ordinal or the first admissible ordinal after $\omega$. Question : does there exists a $\Sigma_2$ formula able to capture that we reached, in Gödel's construction, the $\omega_1^{\mathrm{CK}}$-th stage? That is, is there some $\Sigma_2$ formula $\phi$ such that For all $\alpha < \omega_1^{\mathrm{CK}}$, $L_\alpha \not\models \phi$ $L_{\omega_1^{\mathrm{CK}}} \models \phi$ If no, is $\Pi_2$ powerful enough? REPLY [6 votes]: This couldn't be achieved even by $\Sigma_3$ sentences. First note that $L_{\omega_1^{CK}}$ (as any other model of $\mathsf{KP}\omega+L=V$) satisfies the scheme of $\Sigma_3$-reflection: $$\varphi(\vec{p})\to \exists a \;(\mathsf{Trans}(a)\land \vec{p}\in a\land (\varphi(\vec{p}))^a),\text{ where $\varphi$ is $\Sigma_3$}.$$ And note that there is a $\Pi_2$ sentence $F$ such that for each transitive set $a$, the sentence $F$ is true in $a$ iff $a$ is of the form $L_{\omega(1+\alpha)}$. Henceforth for any $\Sigma_3$ sentence $\varphi$, if $L_{\omega_1^{CK}}\models\varphi$ then there is a transitive set $a\in L_{\omega_1^{CK}}$ such that $a\models \varphi\land F$ which means that $a$ is of the form $L_{\alpha}$, where $\alpha<\omega_1^{CK}$ and $L_{\alpha}\models \varphi$. Note that $\omega_1^{CK}$ is the least $\alpha$ such that $L_{\alpha}$ is a model of $\mathsf{KP}\omega-\mathsf{Foundation}$. And the theory $\mathsf{KP}\omega-\mathsf{Foundation}$ could be axiomatized by a single $\Pi_3$ sentence (the only axiom that isn't $\Pi_2$ is the axiom of $\Sigma_1$-collection). Note that everywhere in this answer the classes $\Pi_n$ were understood as consisting of formulas that start with an unbounded quantifier prefex $\vec{\forall}\vec{x}_1\ldots \vec{Q}\vec{x}_n$ followed by a $\Delta_0$ formula. However if we switch to the classes $\hat\Pi_n$ defined in terms of alternation depth of unbounded quantifiers (bounded quantifiers could appear anywhere) the answer changes. The axiom of $\Sigma_1$-collection is a $\hat\Pi_2$-sentence and hence $\mathsf{KP}\omega-\mathsf{Foundation}$ is $\hat\Pi_2$-axiomatizable. However I don't know whether there is a $\hat\Sigma_2$ sentence that "captures" $L_{\omega_1^{CK}}$. Non-existence of $\hat\Pi_1$ sentence is trivial due to downward-absoluteness. And non-existence of $\hat\Sigma_1$ sentence follows from the fact that $\mathsf{KP}\omega$ proves $\Sigma$-reflection (the class $\Sigma$ is exactly $\hat\Sigma_1$).<|endoftext|> TITLE: Why does MAGMA claim that the automorphism group of an elliptic curve is order 24 when it is order 12? QUESTION [7 upvotes]: I am trying to get the hang of the available software for computing automorphism groups of plane curves over finite fields. I am using this Magma code to test it out on $y^2 = x^3 - x$ over $\mathbb{F}_3$, which we know is of order 12 by (pg. 410, Prop 1.2(c)) Silverman's Arithmetic of Elliptic Curves. The following Magma code gives me as output that the order is 24. Moreover, it also does this for seemingly every other plane curve I can think to put in. I wrote this code according to the exposition here on the Magma webpage. Something terribly wrong is going on, likely I am misunderstanding something fundamental. A := AffineSpace(FiniteField(3),2); f := y^2 - x^3 + x; C := Curve(A,f); G := AutomorphismGroup(C); Order(G); REPLY [16 votes]: Magma is computing the automorphism group of the associated projective curve $E$ defined over the base field (according to the link you gave). You are thinking of the automorphism group of $E$ as an elliptic curve i.e. with $\infty$ fixed. The group magma is computing will also include the translations which are defined over the base, so there is a short exact sequence: $$0 \rightarrow E(k) \rightarrow \mathrm{AutAsProjectiveCurve}(E,k) \rightarrow \mathrm{AutAsEllipticCurve}(E,k) \rightarrow 0.$$ Your curve has $4$ points defined over $\mathbf{F}_3$, but in fact only a subgroup of order $6$ of the order $12$ group $\mathrm{AutAsEllipticCurve}(E,\overline{\mathbf{F}}_3)$ of geometric automorphisms fixing $\infty$ are defined over $\mathbf{F}_3$. Thus $24 = 6 \cdot 4$ is the correct anser. The extra automorphism of the elliptic curve is defined over $\mathbf{F}_9 = \mathbf{F}_3[i]$. Also, we have $|E(\mathbf{F}_9)| = 16$, because the the zeta function is equal to $$\frac{(1 + 3 T^2)}{(1-T)(1-3T)} = 1 + 4 T + 16 T^2 + \ldots = 1 + 4 T + \frac{1}{2} \left(\frac{4^2}{2} + 16\right) T^2 + \ldots $$ And indeed: A := AffineSpace(FiniteField(9),2); f := y^2 - x^3 + x; C := Curve(A,f); G := AutomorphismGroup(C); Order(G); Gives the answer $192 = 12 \cdot 16$. This is also why magma will refuse to do the same computation over $\mathbf{Q}$ --- that would require computing all the rational points! To give some other examples, you can take a random elliptic curve with no automorphisms (as an elliptic curve) besides $\pm 1$ and compare the answer to the number of points modulo p: A := AffineSpace(FiniteField(41),2); f := y^2 - (x^3 + x + 1); C := Curve(A,f); G := AutomorphismGroup(C); Order(G); returns $70$, and indeed $|A(\mathbf{F}_{41})| = 1 + 41 - (-7) = 35$, and $70 = 2 \cdot 35$.<|endoftext|> TITLE: Realizing Morse functions on $S^2$ as height functions QUESTION [7 upvotes]: Let $h: \mathbb{R}^3 \to \mathbb{R}$ be the usual height function (i.e. $h(x,y,z) = z$). One way that Morse functions on $S^2$ are often described is by picking an embedding $i: S^2 \to \mathbb{R}^3$ and then considering $h \circ i$ which, for a generic embedding will be a Morse function. Does there exist a Morse function $f : S^2 \to \mathbb{R}$ so that there is no embedding $i: S^2 \to \mathbb{R}^3$ with $f = h \circ i$? I think that this shows that every Morse function on $S^2$ can be factored through an immersion, but I am interested in embeddings. REPLY [8 votes]: Any Morse function on $S^2$ may be realized by an embedding $S^2\hookrightarrow \mathbb{R}^3 \to \mathbb{R}$. For a Morse function $F:S^2\to \mathbb{R}$, take the equivalence relation with equivalence classes given by the components of the level sets of the Morse function $F$. The quotient of the sphere by this equivalence relation is a cubic tree $\mathcal{T}$, with inducted function $f:\mathcal{T}\to\mathbb{R}$ on the quotient. This was used more generally by Hatcher and Thurston to obtain a presentation of the mapping class group (the figure is from their paper). Hatcher, Allen E.; Thurston, William P., A presentation for the mapping class group of a closed orientable surface, Topology 19, 221-237 (1980). ZBL0447.57005. Now we may embed the tree $\mathcal{T}\hookrightarrow \mathbb{R}^3 \to \mathbb{R}$ so that the composite is $f$ (essentially by general position). Then taking a regular neighborhood of $\mathcal{T}$, we get an embedding $S^2 \hookrightarrow \mathbb{R}^3\to \mathbb{R}$ that induces the Morse function $F$.<|endoftext|> TITLE: Confusion about good reduction QUESTION [8 upvotes]: I am confused about the notion of good reduction. Let $R$ be a DVR, let $K$ be its fraction field. If we have a smooth proper $K$-scheme $V$, then I believe $V$ is said to have good reduction at the unique non-zero prime ideal if there exists a smooth proper $R$-scheme whose generic fiber is $V$. I tend to dislike the word "exists". I think for abelian varieties, a condition equivalent to good reduction can be formulated in terms of the 1st cohomology. Can a condition not involving the existential quantifier and equivalent to good reduction be given for general smooth proper $K$-schemes? A second question, if $F$ is a number field, then what is the right notion of good reduction modulo a non-zero prime ideal of the ring of integers of $F$ for smooth proper $F$-schemes? The issue for me is the compatibility between different prime ideals: if a smooth proper $F$-scheme has "good reduction everywhere", does it mean that there is a single integral model that has smooth fibers over every prime ideal, or just that for any prime ideal you can find a (proper flat, or I don't know what should be required really) model that has smooth fiber over that prime ideal? Third question: given a smooth proper scheme, is there some functorially constructed "best" integral model so that all questions of reduction can be just answered using that particular model? I have heard something about Neron models but I think they only work for abelian varieties. I apologize for these naive questions but all references I found so far refer to good reduction without giving a definition. If there is a reference addressing the above questions I will gladly study it. REPLY [10 votes]: (1) Can a condition not involving the existential quantifier and equivalent to good reduction be given for general smooth proper $K$-schemes? Not in terms of their "topology" in general. For abelian varieties, there's the Neron-Ogg-Shafarevich criterion: $X$ has good reduction iff $H^1(X_{\bar K}, \mathbf{Q}_\ell)$ is unramified (there is also a $p$-adic Hodge theory variant if $R$ has mixed characteristic $(0,p)$, due to Coleman-Iovita). This has been extended to K3 surfaces recently by Liedtke-Matsumoto and Chiarellotto-Lazda-Liedtke, but even there the answer is quite subtle if the residue field is not algebraically closed (roughly speaking, one can only detect good reduction after an unramified extension using $H^2$, and to detect good reduction one has to do some hard work). Already curves give an example where cohomology is insufficient: there exist "curves of compact type" i.e. with bad reduction but whose Jacobian has good reduction (e.g. a curve whose model has a special fiber whose dual graph has no loops). Andreatta-Iovita-Kim provide a criterion in terms of the Galois action on the geometric fundamental group. One could wonder whether looking at the Galois action on the etale homotopy type can see good reduction. I don't know the answer, but it might be not too difficult to find a counterexample. EDIT. I recalled the example due to Cynk and van Straten to the effect that already for Calabi-Yau threefolds, trivial monodromy does not imply (potential) good reduction. Moreover, Chiarellotto-Lazda-Liedtke (see Theorem 4.5) show that for Enriques surfaces one cannot detect good reduction using cohomology of local systems (equivalently, the cohomology of the K3 double cover). (2) If $F$ is a number field, then what is the right notion of good reduction modulo a non-zero prime ideal of the ring of integers of $F$ for smooth proper $F$-schemes? To me, this would mean that for every maximal ideal $\mathfrak{p} \subseteq\mathcal{O}_K$, the base change of $X$ to the henselian (or complete, shouldn't matter) local ring of $\operatorname{Spec}\mathcal{O}_K$ at $\mathfrak{p}$ has good reduction in the sense of the definition over a dvr you gave. An alternative definition would be that $X$ has a smooth proper model over $\mathcal{O}_K$. For abelian varieties the two notions coincide because of Neron models. (3) Given a smooth proper scheme, is there some functorially constructed "best" integral model so that all questions of reduction can be just answered using that particular model? I think the answer is no. Again the most studied case beyond abelian varieties is curves (where one has the Deligne-Mumford compactification) and K3 surfaces (where one has so-called Kulikov models, whose existence is conjectural in general, and which are not unique). If, however, you are interested with smooth and proper models, and you allow yourself to fix a polarization, then the theorem of Mumford and Matsusaka might be useful. See the answers to this question.<|endoftext|> TITLE: Conjectures of Peter Scholze about q-de Rham complex: examples QUESTION [16 upvotes]: Peter Scholze formulated several conjectures about $q$-de Rham complex in the paper Canonical $q$-deformations in arithmetic geometry, Ann. Fac. Sci. Toulouse Math. (6) 26 (2017), no. 5, pp 1163–1192, doi:10.5802/afst.1563, arXiv:1606.01796 Especially intriguing to me is Conjecture 3.3: Betti-de Rham comparison isomorphism. In the paper the author emphasized the lack of nontrivial examples and suggests to do a computation for the universal family of elliptic curves. Question 1: Has anyone done this computation (or any other interesting computations)? I will be especially interested in those involving basic hypergeometric functions. Question 2: What does Conjecture 3.3 mean in practice? Naively one might hope that it leads to some $q$-deformation of periods. REPLY [12 votes]: Good question! I wished I understood what this conjecture really means, concretely. First, I should say that in my paper with Bhatt on prismatic cohomology, much of the content of these conjectures has been proved, although the connection is not made very clear, and we only work after $p$-adic completion for some prime $p$; it would be possible to go back and get full integral statements by pasting together the rational information and the $p$-adic information. The main problem in getting something very concrete out of this formalism --- like explicit $q$-periods, or explicit $q$-difference equations deforming the Picard--Fuchs differential equation --- is that it seems very hard to construct explicit vectors in $q$-de Rham cohomology. For getting say the Picard--Fuchs equation, one uses that de Rham cohomology has a Hodge filtration, and one can write down the differential equation satisfied by a vector in the $1$-dimensional subspace. This subspace does not deform to $q$-de Rham cohomology, so one gets no canonical $q$-difference equation. On the other hand, weight filtrations should still be defined, and maybe one can use those in some cases to get interesting explicit $q$-difference equations. Still, Question 1 seems to be largely answered in this paper by Ryotaro Shirai. If I understand it right, he restricts to the ordinary part of the moduli space of elliptic curves, in which case there is a canonical "ordinary part" of the $q$-de Rham cohomology, giving a canonical $1$-dimensional subspace, for which one can write down the $q$-difference equation. Shirai then shows that this is indeed a hypergeometric $q$-difference equation. For Question 2: This isomorphism is only canonical after going to one of Fontaine's period rings like $A_{\mathrm{inf}}$. Doing this, one essentially ends up with the usual $p$-adic periods, living in Fontaine's field $B_{\mathrm{dR}}$ (or subrings thereof).<|endoftext|> TITLE: Estimating the number of functions which are at most $c$-to-$1$ for some constant $c \ge 2$ QUESTION [7 upvotes]: Notation: $[m] := \{1, 2, \dots, m \}$. How many functions are there $f: [a] \to [b]$? The answer is easily seen to be $b^a$. How many $1$-to-$1$ functions are there $f: [a] \to [b]$? Again the answer is well known, and it is sometimes called the falling factorial: $$b(b-1) \dots (b-a+1).$$ How many functions are there $f: [a] \to [b]$ that are no more than $c$-to-$1$? I don't expect that there is an exact formula, and I am more interested in the asymptotics. For example, can we give "reasonable" upper and lower bounds, in the case that $c \ge 2$ and $|A| / |B|$ are fixed, and $|A| \to \infty$? For a concrete example, roughly how many functions are there $[5n] \to [n]$ that are at most $8$-to-$1$? Call this function $g(n)$. Clearly we have $$\frac{(5n)!}{5!^n} \le g(n) \le n^{5n}.$$ The function ${(5n)!}/{5!^n}$ counts functions that are exactly 5-to-1 (which all satisfy the criterion that they are at most 8-to-1), and the function $n^{5n}$ counts all functions. Applying Stirling's approximation to the first function gives something like $$ \alpha^n n^{5n} \le g(n) \le n^{5n},$$ for some small constant $\alpha > 0$. It seems like there is room for improvement. Is it true, for example, that $$\log g(n) = 5n \log n + C n + o(n) $$ for some constant $C > 0$? REPLY [2 votes]: This is closely related to Lucia's answer. For parameter $t$, let $X_t$ be the random variable with probability generating function $$ p_t(x) = \sum_{i=0}^c \mathrm{P}(X_t=i)\, x^i = \sum_{i=0}^c \frac{x^it^i}{i!}\biggm/ \sum_{i=0}^c \frac{t^i}{i!}. $$ Adjust $t$ so that the expectation of $X_t$ is $b/a$, and let $\sigma^2$ be its variance for that $t$. Then the number of desired functions is asymptotically $$\frac{a!\,t^{-b}}{\sigma\,\sqrt{2\pi a}} \biggl(\sum_{i=0}^c \frac{t^i}{i!}\biggr)^a.$$ This assumes that $X_t$ is suitable for applying the central limit theorem to the sum of $a$ independent copies.<|endoftext|> TITLE: Connected components of real Lie groups QUESTION [7 upvotes]: (This is a follow-up to this question of mine.) Is there an example of a connected reductive algebraic group $G$ over $\mathbb{R}$ such that: $G$ is not isomorphic to a product $G_1 \times G_2$ of smaller groups (isogenous to a product is OK) $G$ is not a torus, the quotient of $G$ by a maximal compact-mod-centre subgroup has a complex structure, $Z_G(\mathbb{R})$ is not contained in the identity component of $G(\mathbb{R})$? The condition $Z_G(\mathbb{R}) \subseteq G(\mathbb{R})^\circ$ is vacuously satisfied if $G$ is adjoint, because then $Z_G = \{1\}$; but it is also vacuously satisfied if $G$ is semisimple and simply-connected, because then $G(\mathbb{R})$ is connected as a Lie group by a theorem of Cartan. So any example would have to lie somewhere in between the two (which makes me wonder if there are any examples at all). PS: Of course $GL_3$ is an example if the "complex structure" condition is dropped. REPLY [6 votes]: There is NO such example. Note that any semisimple algebraic ${\mathbb{R}}$-group $H$ of Hermitian type has a compact (anisotropic) maximal torus. Indeed, by a definition of a group of Hermitian type (see, e.g., Deligne, Travaux de Shimura, condition (1.5.3) on page 128), $H$ is an inner form of a compact algebraic $\Bbb R$-group $K$, namely, $H=\,_\sigma K$, where $\sigma={\rm inn}(x)\in {\rm Aut}(K)$, $x^2=1$, $x\in K^{\rm ad}(\Bbb R)$, $K^{\rm ad}=K/Z_K$. Let $T_K\subset K$ be a maximal $\Bbb R$-torus such that $T_K^{\rm ad}(\Bbb R)$ contains $x$, where $T_K^{\rm ad}=T_K/Z_K$. Then $T_K=\,_\sigma T_K\subset \,_\sigma K=H$ is a compact maximal torus of $H$. Theorem. Let $G$ be a (connected) reductive ${\mathbb{R}}$-group. Write $G^{\rm der}=[G,G]$. Assume that $G^{\rm der}$ has a compact maximal torus $T^{\rm der}$. If the image of $Z_G({\mathbb{R}})$ in $\pi_0(G({\mathbb{R}}))$ is nontrivial, then there exists a nontrivial split ${\mathbb{R}}$-subtorus $T'\subset Z_G$ and a reductive ${\mathbb{R}}$-subgroup $G''\subset G$ such that $G=T'\times_{\mathbb{R}} G''$. Proof. Write $T=Z_G\cdot T^{\rm der}$; then $T$ is a maximal torus of $G$. We have maps $$ Z_G({\mathbb{R}})\to \pi_0(T({\mathbb{R}}))\to \pi_0(G({\mathbb{R}})).$$ Since the image of $Z_G({\mathbb{R}})$ in $\pi_0(G({\mathbb{R}}))$ is nontrivial, we have $ \pi_0(T({\mathbb{R}}))\neq 1$. Write $T=T_0\times_{\mathbb{R}} T_1\times_{\mathbb{R}} T_2$, where $T_0$ is a compact ${\mathbb{R}}$-torus, $T_1$ is a split ${\mathbb{R}}$-torus, and $T_2\simeq (R_{{\mathbb{C}}/{\mathbb{R}}}{\mathbb G}_{m,{\mathbb{C}}})^{n}$. We have $\pi_0(T_0({\mathbb{R}}))=1$, and $\pi_0(T_2({\mathbb{R}}))=1$. Since $\pi_0(T({\mathbb{R}}))\neq 1$, we conclude that $T_1\neq 1$. Note that the ${\mathbb{R}}$-torus $T/Z_G$ is isogenous to $T^{\rm der}$, and hence, compact. It follows that $T_1\subset Z_G$. Set $T'=T_1$ and $T''=T_0\times_{\mathbb{R}} T_2$; then clearly $T=T'\times_{\mathbb{R}} T''$. Set $G''=T''\cdot G^{\rm der}$. Then $$T'\cdot G''=T'\cdot T''\cdot G^{\rm der}=T\cdot G^{\rm der}=G\quad \text{and}\quad T'\cap G''=T'\cap T''=1.$$ Since $T'\subset Z_G$, it commutes with $G''$. Thus $G=T'\times_{\mathbb{R}} G''$, as required.<|endoftext|> TITLE: Splitting low-dimensional $p$-local CW complexes for large $p$ QUESTION [8 upvotes]: Fix a prime $p$. I have a sketch of a proof that if $X$ is a finite simply-connected CW complex with $\mathrm{dim}(X) < p$ then for some $t\in \mathbb{N}$, the $p$-localization $\Sigma^t X_{(p)}$ is a wedge of Moore spaces. (Basically, the idea is that all the interesting attaching maps are Whitehead products, hence stably trivial.) Questions: Does anyone know a reference for this? If it's not true, I'd love to know that too! EDIT: Perhaps this is just a theorem of tame homotopy theory? REPLY [6 votes]: This result appears in the PhD thesis of Hans-Werner Henn, and in this paper: @article {MR884630, AUTHOR = {Henn, Hans-Werner}, TITLE = {Classification of {$p$}-local low-dimensional spectra}, JOURNAL = {J. Pure Appl. Algebra}, FJOURNAL = {Journal of Pure and Applied Algebra}, VOLUME = {45}, YEAR = {1987}, NUMBER = {1}, PAGES = {45--71}, ISSN = {0022-4049}, MRCLASS = {55P42 (55Q70)}, MRNUMBER = {884630}, MRREVIEWER = {Frederick Cohen}, DOI = {10.1016/0022-4049(87)90083-1}, URL = {https://doi.org/10.1016/0022-4049(87)90083-1}, } (For some reason it seems that this took several years to be published, and so is not Henn's earliest paper, contrary to my previous comment.)<|endoftext|> TITLE: $P(x)=P(y)$ has infinitely many integer solutions QUESTION [17 upvotes]: Determine all polynomials $P(x)$ with integer coefficients such that $P(x)=P(y)$ has infinitely many integer solutions in integer $x$ and $y$ with $x \neq y$. Choose $P(x)=a_n(x-k)^{2n}+a_{n-1}(x-k)^{2n-2}+...+a_0$, with integers $k,n,a_n,a_{n-1}, ...,a_0$ Then $P(x)=P(2k-x)$ for every integer $x \neq k$, thus satisfy the condition. Now, I have a gut feeling that if $P(x)=P(y)$ and $x+y=M$, then for every integer $x$, $P(x)=P(M-x)$, but I cannot analyze any futher. So my question is: Are there any other types of polynomials $P(x)$ that satisfy the orange question above? (Any answers or comments will be appreciated!) (If this question should be closed or off topic, please let me know. If this site cannot answer this question, let me know, I will delete this question immediately) REPLY [11 votes]: Here's a more abstract proof: If $P(x) - P(y) =0$ but $x \neq y$ then $ (P(x) - P(y) )/(x-y)=0$. Because this is a polynomial of degree $n-1$, when this identity is satisfied its leading terms $ (x^{n} - y^{n} )/ (x-y)$ must be equal to minus its remaining terms, and thus must be $O( \max(x,y)^{n-2})$. Now if $n$ is odd, $ (x^{n} - y^{n} )/ (x-y)$ is a homogeneous polynomial of degree $n-1$ with no nozero real roots. Because it has no nonzero real roots, its absolute value has some minimum value $C$ on the boundary of the unit square. Then homogeneity gives $ | (x^{n} - y^{n} )/ (x-y)| \geq C \max(x,y)^{n-1}$ and thus can be $O( \max(x,y)^{n-2})$ for only finitely many $x,y$. If $n$ is even, $(x^n-y^n)/(x^2-y^2)$ is a homogeneous polynomial of degree $n-2$ with no real roots. By the same logic, we have $| (x^{n} - y^{n} )/ (x^2-y^2)| \geq C \max(x,y)^{n-2}$. Thus if $| (x^{n} - y^{n} )/ (x-y)| = O ( \max(x,y)^{n-2})$ then $C |x+y| = O(1)$. So this can happen for only finitely many values of $|x+y|$. Thus if it happens infinitely often, it happens infinitely often for one particular value of $x+y$, say $M$, thus $P(x) - P(M-x)$ vanishes for infinitely many $x$ and thus is zero.<|endoftext|> TITLE: Reference request about “internal language of categories” QUESTION [10 upvotes]: I've tried to become familiar with the so-called "internal language of a category" for the last months. However, I'm still not confident enough when, for instance, I find a subobject (of a given object) which is defined through a formula of the internal laguage of the category that I'm considering. In detail, if $C$ is a pretopos, $A$ is an object of $C$ and $\phi$ is a formula in the internal language of $C$, I'm not completely able to understand the following two things: Which is the actual subobject $B$ of $A$, represented by the expression $\{ x\in A:\phi(x)\}$? Meaning, how can I recover $B$ in terms of "categorical operations" in $C$? How can I work with $\{ x\in A:\phi(x)\}$? That is, for instance, how can I verify through a completely syntactical procedure that $\{ x\in A:\phi(x)\}$ is the object that I was looking for? Of course, I'm not asking you to answer points (1) and (2), as they are too generic. I would rather you to suggest me a self-contained chapter of a book or some lecture notes where this subject is fully explained. In my opinion, what I in particular need is a collection of basic examples and exercises regarding its usage. Thanks in advance. P.S. I asked the same question in Math Stack Exchange (https://math.stackexchange.com/questions/3262479/reference-request-about-internal-language-of-categories). REPLY [2 votes]: Here are some resources: Carsten Butz.Regular categories and regular logic, notes, 1998. P. Freyd and A. Scedrov. Categories, Allegories. North–Holland, Amsterdam 1990. S. Mac Lane and I. Moerdijk. Sheaves in Geometry and Logic. Springer–Verlag, New York 1992. B. Jacobs, Categorical Logic and Type Theory, Studies in Logic and the Foundations of Mathematics 141, North Holland, Elsevier, 1999. The first one covers the fragment of logic with $\top$, $\land$, $\exists$ and $=$. The second one has first-order categorical logic, and the third one higher-order logic. Jacobs's book is very thick and intimidating, I am mentioning it because it contains rules for $\lbrace x : A \mid \phi(x)\rbrace$, which you're asking about.<|endoftext|> TITLE: Existence of regular factors in dense graphs QUESTION [5 upvotes]: All graphs here are finite and simple. A $d$-factor of a graph is a spanning regular subgraph of degree $d$. Where can I find theorems of this nature, for constants $a,b,c\gt 0$: If $G$ is a graph with all degrees in $[an,bn]$, and $d$ is an even integer in $[0,cn]$, then $G$ has a $d$-factor? REPLY [4 votes]: To give at least a possible answer, and for those interested in doing a comprehensive literature search for an actual answer, it seems that the somewhat related survey http://web.mat.bham.ac.uk/D.Kuehn/bcc09dkdo3.pdf might be a helpful starting point.<|endoftext|> TITLE: On equation $\Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$ on a Riemannian manifold QUESTION [6 upvotes]: Assume that $M$ is a compact Riemannian manifold whose Laplacian is denoted by $\Delta$. Assume that the Euler characteristic of $M$ is zero. Does $M$ admit a non vanishing vector field $X$ which satisfy $$(*) \qquad \Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$$ What can be said about the structure of the Lie algebra of all vector fields $X$ with the property $(*)$? As a second question: Every vector field $X$ on $M$ defines a second order differential operator on $C^{\infty}(M)$ with $$D(f)=\Delta(X.f)-X.\Delta(f)$$ This is a second order operator since the third order terms cancel each others. What is the principal symbol of this operator , precisely? Can this PDE be an elliptic operator when $M$ is a compact manifold?(I mean:is there an example of this situation in compact case?) Does every compact manifold admit a vector field $X$ for which this PDE would be an elliptic operator?What would be a dynamical interpretation for the index of this PDE. This is a dynamical motivation for the later question.. REPLY [6 votes]: The following formula is known among the experts but hard to find in the literature, so I figure I will document it here. Throughout $(M,g)$ denote an arbitrary pseudo-Riemannian manifold, and $\nabla$ its Levi-Civita connection. Definition Given a vector field $X$, its corresponding 0th order deformation tensor is defined to be ${}^{(X,0)}\pi := \mathcal{L}_X g$, where $\mathcal{L}_X$ is Lie differentiation with respect to $X$. The corresponding 1st order deformation tensor is defined using a formula similar to that of Christoffel symbols: $$ {}^{(X,1)}\pi_{ab}{}^c := \frac12 g^{cd} \left[ \nabla_a ( {}^{(X,0)}\pi_{bd}) + \nabla_b ({}^{(X,0)}\pi_{ad}) - \nabla_d ({}^{(X,0)}\pi_{ab}) \right] $$ Lemma Let $\Xi$ be an arbitrary $k$-covariant tensor field. And let $X$ be a vector field. The following formula holds for the commutation: $$ [ \nabla_a, \mathcal{L}_X ] \Xi_{b_1\cdots b_k} = \sum_{j = 1}^k {}^{(X,1)}\pi_{a b_j}{}^c \Xi_{b_1 \cdots b_{j-1} c b_{j+1} \cdots b_k} $$ With the aid of these formulas, we have immediately that, writing $\triangle_g$ for the Laplace-Beltrami operator, first $$ [ \nabla_X, \triangle_g] f = [\mathcal{L}_X, \triangle_g ] f $$ because Lie derivation and covariant differentiation act identically on scalars, and then $$ [\mathcal{L}_X, g^{ab}\nabla_a\nabla_b] f = \mathcal{L}_X (g^{ab}) \nabla^a \nabla_b f + g^{ab} [\mathcal{L}_X, \nabla_a] \nabla_b f + g^{ab} \nabla_a [\mathcal{L}_X, \nabla_b ]f $$ The first factor we can compute to get $$ \mathcal{L}_X(g^{ab}) = - {}^{(X,0)}\pi^{ab} $$ using that $g^{ab} g_{bc} = \delta^a_c$. The third factor vanishes because Lie differentiation commutes with exterior differentiation. And we use our Lemma for the second term. We get, finally $$ [\nabla_X, \triangle_g] f = - {}^{(X,0)}\pi^{ab} \nabla_a\nabla_b f - g^{ab} ~{}^{(X,1)}\pi_{ab}{}^c \nabla_c f. $$ Remarks: Notice that the first order deformation tensor is defined in terms of the 0th order one. So that when $X$ is Killing, automatically both the ${}^{(X,0)}\pi$ and ${}^{(X,1)}\pi$ vanish, and differentiation with $X$ commutes with the Laplacian. In the case ${}^{(X,0)}\pi = \phi g$ for some scalar function $\phi$ (so $X$ is conformally Killing), one can check that the formula reduces to the one I gave in a comment above. When the function $\phi$ in the previous item is a non-zero constant (which some people refer to as $X$ being a homothetic vector field) one gets the special case $$ [ \nabla_X, \triangle_g] f = \phi \triangle_g f $$<|endoftext|> TITLE: Is there some sort of formula for $\tau(S_n)$? QUESTION [6 upvotes]: Let $G$ be a finite group. Define $\tau(G)$ as the minimal number, such that $\forall X \subset G$ if $|X| > \tau(G)$, then $XXX = \langle X \rangle$. Is there some sort of formula for $\tau(S_n)$, for the symmetric group $S_n$? Here $XXX$ stands for $\{abc| a, b, c \in X\}$. Similar problems for some different classes of groups are already answered: 1) $\tau(\mathbb{C}_n) = \lceil \frac{n}{3} \rceil + 1$, where $\mathbb{C}_n$ is cyclic of order $n$; 2) Gowers, Nikolov and Pyber proved the fact that $\tau(\mathrm{SL}(n, p)) \leq 2|\mathrm{SL}(n, p)|^{1-\frac{1}{3(n+1)}}$ for prime $p$. However, I have never seen anything like that for $S_n$. It will be interesting to know if there is something... REPLY [5 votes]: Here is a lower bound: let $H$ be a subgroup of $S_{n}$ containing $(12)$ and let $\sigma$ be the $n$-cycle $(12 \ldots n)$. Let $X = H \cup \{\sigma \}$ and note that $\langle X \rangle = S_{n}$ since we already have $S_{n} = \langle (12),\sigma \rangle.$ Suppose that we have chosen $H$ so that $|X| > \tau(S_{n}).$ Then we must have $S_{n} = XXX$. But $XXX = H \cup H\sigma \cup H\sigma H \cup H\sigma^{2} \cup \sigma H \cup \sigma H \sigma \cup \sigma^{2}H \cup \sigma^{3}$ has cardinality at most $|H|^{2} + 6|H| + 1,$ so we must have $|H| + 3 > \sqrt{n!}.$ Hence we may choose $k$ minimal so that $k! \geq \tau(S_{n}).$ Then taking $H$ to be the natural copy of $S_{k}$ inside $S_{n}$, we must have $(k! + 3)^{2} > n!$. Then we must have $\tau(S_{n}) > \frac{\sqrt{n!} - 3}{k} \geq \frac{\sqrt{n!} - 3}{n}.$<|endoftext|> TITLE: Visualizing a Whitehead product: the attaching map $S^3\to S^2\vee S^2$ QUESTION [7 upvotes]: There are informative and easily accessible images and videos that illustrate the Hopf fibration $S^3\to S^2$ by describing what happens to the fibers in the unit cube $(0,1)^3\approx S^3\backslash \ast$, e.g. those by Niles Johnson. Is there a similar way to visualize the map $S^3\to S^2\vee S^2$, which is the attaching map of the $4$-cell of $S^2\times S^2$ and allows one to the construct the first interesting higher whitehead product? I understand that this map is not a fibration - it's the restriction of the product of the characteristic maps of the $2$-cells of $S^2\vee S^2$ but, if possible, I think it'd be interesting to be able to understand this visually in a $3$-cube as some kind of higher analogue of conjugation. Note: this was originally a post on MSE that received no answers. REPLY [6 votes]: The natural generality is as follows. For any finite-dimensional inner-product spaces $U$ we have a unit sphere $S(U)\subset U$ and a one-point compactification $S^U\simeq S(U\oplus\mathbb{R})$. If $V$ is another finite-dimensional inner-product space, then we can ask about the attaching map of the top cell in the product $S^U\times S^V$. This naturally arises as a map $$ w\colon S(U\oplus V)\to S^U \vee S^V = S(U\oplus\mathbb{R}) \vee S(V\oplus\mathbb{R}) $$ which can be described like this: $$ w(u,v) = \begin{cases} v^{-2}(2u\sqrt{v^2-u^2},2u^2-v^2) \in S(U\oplus\mathbb{R}) & \text{ if } \|u\| \leq \|v\| \\ u^{-2}(2v\sqrt{u^2-v^2},2v^2-u^2) \in S(V\oplus\mathbb{R}) & \text{ if } \|v\| \leq \|u\|. \end{cases} $$ (Here $u^2$ means $\|u\|^2=\langle u,u\rangle$.) The fibre of $w$ over the basepoint is homeomorphic to $S(U)\times S(V)$. The fibre over all other points of $S(U\oplus\mathbb{R})$ is $S(V)$, and the fibre over all other points of $S(V\oplus\mathbb{R})$ is $S(U)$.<|endoftext|> TITLE: Does there exist another form of the derivative for polynomials? QUESTION [5 upvotes]: Let $F : \mathbb{R}[X] \rightarrow \mathbb R[X]$ be a linear map and let $H \in \mathbb{R}[u,x,y,z]$ be a polynomial. Suppose that $$ F(P \cdot Q) = H(F(P),F(Q),P,Q)$$ for all $P, Q \in \mathbb{R}[X]$. Two obvious solutions for $F$ and $H$ are $F = I$, the identity function, and $H(u,x,y,z) = \alpha u x + \beta u z + \gamma yx + \delta yz$ where $\alpha,\beta,\gamma,\delta \in \mathbb{R}$ and $\alpha +\beta+\gamma+\delta = 1$; $F = D$, the derivative, and $H(u,x,y,z) = u z + x y$, from the Leibniz rule. Do there exist solutions $F$ and $H$ in which $F$ is not a linear combination of $I$ and $D$? REPLY [8 votes]: It's not hard to see that $H$ must be of the form $\alpha ux + \beta uz + \gamma yx + \delta yz$ by $\mathbb{R}$-linearity of $F$ (see Jan-Cristoph Schlage-Puchta's answer for a fuller explanation). By using symmetry of multiplication, we can assume without loss of generality that $\beta = \gamma$. We have the equation $$F(P \times Q) = H(F(P), F(Q), P, Q) = \alpha F(P) F(Q) + \beta F(P) Q + \gamma P F(Q) + \delta PQ$$ Setting $P = Q = 1$, we get $F(1) = \alpha F(1)^2 + \beta F(1) + \gamma F(1) + \delta$. Degree-checking tells us that if $F(1)$ isn't constant, then $\alpha = 0$. More careful checking tells us that $\beta + \gamma = 1, \delta = 0$. This leads to the solution $F_{M, R}(P) = P \times R$, where $R = F(1)$. That answer aside, we assume $F(1)$ is constant. I'm going to take a detour for a moment, to talk about an action of the two-dimensional nonabelian Lie group on the space of pairs $(F, H)$ that satisfy $F(P \times Q) = H(F(P), F(Q), P, Q)$. This action comes from the map $g_{(a, b)} F = aF + b \operatorname{Id}$. We want the equation to remain the same, which leads us to: $$(gH)(u, x, y, z) = aH(\frac{u - by}{a}, \frac{x - bz}{a}, y, z) + b yz$$ $$ = \frac{\alpha}{a} ux + (\beta - \frac{b}{a} \alpha) uz + (\gamma - \frac{b}{a} \alpha) yx + (a \delta - b \beta - b \gamma + \frac{b^2}{a} \alpha + b) yz$$ Back from the detour: we now set just $Q = 1$, getting $F(P) = \alpha F(P) F(1) + \beta F(P) + \gamma P F(1) + \delta P$. Rearranging, we get $(1 - \alpha F(1) - \beta) F(P) = (\gamma F(1) + \delta) P$. The obvious solutions have $F(P) = \lambda P$ for some $\lambda \in \mathbb{R}$, which have been discussed in other answers; otherwise, we must have $1 - \alpha F(1) - \beta = \gamma F(1) + \delta = 0$. We now split into two cases: either $\alpha = 0$, or $\alpha \neq 0$. If $\alpha = 0$, then $\beta = \gamma = 1$. Using the action of the group, we can reduce to the case that $\delta = 0$, implying that $F(P \times Q) = F(P) Q + P F(Q)$, i.e. that $F$ is a derivation. This can only happen when $F(P) = R \times \partial P$ for some polynomial $R$. Undoing the group action, we get $F_{D, R, \lambda}(P) = \lambda P + R \times \partial P$. If $\alpha \neq 0$, then we can use the action of the group to reduce to the case $\alpha = 1, \beta = \gamma = 0$. But then $\delta = 0$, giving us the equation $F(P \times Q) = F(P) F(Q)$ - in other words, $F$ is a homomorphism. Homomorphisms from $\mathbb{R}[X]$ are compositions: $F(P) = P \circ R$ for some polynomial $R$. Undoing the group action, we get $F_{H, R, \lambda} = \lambda P + P \circ R$. So all solutions $(F, H)$ are of the forms: 1) $F(P) = \lambda P, \frac{1}{\lambda} \alpha + \beta + \gamma + \lambda \delta = 1$ 2) $F(P) = \lambda P + R \times \partial P, \alpha = 0, \beta = \gamma = 1, \delta = \lambda$ 3) $F(P) = \lambda P + c (P \circ R)$ with coefficients that aren't difficult to determine using the group action. 4) $F(P) = P \times R, \alpha = 0, \beta = \gamma = \frac{1}{2}, \delta = 0$<|endoftext|> TITLE: On fundamental groupoid of fundamental groupoid QUESTION [7 upvotes]: Given a topological space $X$, we have the notion of the fundamental groupoid $\Pi_1(X)$. Here, the fundamental groupoid $\Pi_1(X)$ is made into a topological groupoid giving a topology on the morphism set. One can then talk about $\Pi_1(\Pi_1(X))$. Is this related to (same as) the fundamental $2$-groupoid? REPLY [4 votes]: The fundamental groupoid $\Pi_1(M)^{top}$, when equipped with the 'usual topology' when the space $M$ permits it, is weakly/Morita equivalent to the same groupoid equipped with the discrete topology, call it $\Pi_1(M)^\delta$. This is essentially Proposition 4.42 in my thesis. Even better, the identity functor $\Pi_1(M)^\delta \to \Pi_1(M)^{top}$ is a weak equivalence. If a functor $X\to Y$ of topological groupoids is a weak equivalence, then $\Pi_1(X) \to \Pi_1(Y)$ is an equivalence of groupoids. And lastly, for a topological groupoid $X$ that has the discrete topology, the canonical functor $X\to \Pi_1(X)$ is an equivalence. So for a suitable topological space $M$, there is a canonical functor $\Pi_1(X)^\delta \to \Pi_1(\Pi_1(M)^{top})$ and it is an equivalence.<|endoftext|> TITLE: Do (3+1)-dimensional Lorentzian manifolds admit unique smoothings? QUESTION [12 upvotes]: Of course, 3-dimensional topological manifolds admit unique smoothings while 4-dimensional topological manifolds generally do not. A (3+1)-dimensional topological Lorentzian manifold (definition below) is a topological 4-manifold, but with extra structure which locally fibers it over 3-dimensional manifolds, so perhaps there's some hope that such a space can be uniquely smoothed. Questions: Let $(M,g)$ be a $C^k$ Lorentzian manifold of dimension $d+1$. Does there exist a refinement of the $C^k$ structure on $M$ to a $C^\infty$ structure? If so, can this refinement be chosen such that $g$ is $C^\infty$? If so, is the $C^\infty$ structure unique subject to requirement (2)? Notes: I'm thinking that the case $d=3$ is the most interesting, but I'm not sure. I think the answer must be no for $d\geq 4$. The answer to (1) is yes for $d \leq 2$, but I'm not sure about (2) or (3) in this case. On the face of it, the question only makes sense for $k \geq 1$. Thus, I will include some speculative remarks about the case $k=0$, i.e. the notion of a $C^0$ Lorentzian manifold. Definitions: Here is a speculative definition of the notion of a chronological $C^0$ Lorentzian manifold ("chronological" refers to the absence of closed timelike curves, enforced by the existence of a distance function.) Following Noldus (Definition 1), a Lorentzian distance on a set $X$ is a function $d: X \times X \to [0,\infty]$ which is reflexive, antisymmetric, and satisfies the reverse triangle inequality. The Lorentzian length of a function $\gamma: [0,1] \to X$ is $$L(\gamma) = \limsup_{0 = t_0 < t_1 < \dots < t_n = 1 \\ \quad |t_{i+1} - t_i| \to 0} \sum_{i=1}^n d(t_i,t_{i+1})$$ A function $\gamma: [0,1] \to X$ is timelike if $s < t \Rightarrow d(f(s),f(t)) >0$. $(X,d)$ is a Lorentzian length space if for all $x,y \in X$, $$d(x,y) = \sup_{\gamma: [0,1] \to X \, \text{timelike} \\ ~~ \gamma(0) = x,\, \gamma(1) = y} L(\gamma)$$ $(X,d)$ is a chronological $C^0$ Lorentzian manifold if it is a Lorentzian length space and, in the coarsest topology such that $d$ is separately continuous in each variable, $X$ is a topological manifold and $d$ is continuous. I might go on to define a notion of "(not necessarily chronological) $C^0$ Lorentzian manifold" by asking for the local structure of a chronological $C^0$ Lorentzian manifold, but perhaps what I've already written is speculative enough. Question: Do $(3+1)$ dimensional chronological $C^0$ Lorentzian manifolds admit unique smoothings? REPLY [6 votes]: Concerning your first three question: depending on what you exactly mean by "refinement", any $C^k$-manifold $M$ with $k\geq 1$ possesses a unique $C^\infty$-structure that is $C^k$-compatible with the given $C^k$-structure on $M$, see Thm. 2.9 in Hirsch, Morris W., Differential topology, Graduate Texts in Mathematics. 33. New York - Heidelberg - Berlin: Springer-Verlag. x, 221 p.(1976). [ZBL0356.57001] So it is no loss of generality to work on smooth manifolds if you want at least $C^1$-regularity of your manifold. If you just want a topological manifold that is of course a complete different story. Moreover, the regularity of metric can in general be not improved. The notion of a $C^0$ Lorentzian manifold / spacetime makes perfect sense (if the $C^0$ refers to the regularity of the metric) and can be quite useful: Chruściel, Piotr T.; Grant, James D. E., On Lorentzian causality with continuous metrics, Classical Quantum Gravity 29, No. 14, Article ID 145001, 32 p. (2012). ZBL1246.83025. Sbierski, Jan, The (C^0)-inextendibility of the Schwarzschild spacetime and the spacelike diameter in Lorentzian geometry, J. Differ. Geom. 108, No. 2, 319-378 (2018). ZBL1401.53058. Galloway, Gregory J.; Ling, Eric; Sbierski, Jan, Timelike completeness as an obstruction to (C^{0})-extensions, Commun. Math. Phys. 359, No. 3, 937-949 (2018). ZBL1396.53095. Sämann, Clemens, Global hyperbolicity for spacetimes with continuous metrics, Ann. Henri Poincaré 17, No. 6, 1429-1455 (2016). ZBL1342.83014. However, if you just have a continuous metric several pathologies in the causal structure can occur, see Chrusciel, Grant above and also https://arxiv.org/abs/1901.07996 Also, we introduced the notion of "Lorentzian length spaces", in the same spirit as you outline above: Kunzinger, Michael; Sämann, Clemens, Lorentzian length spaces, ZBL06970105. Grant, James D. E.; Kunzinger, Michael; Sämann, Clemens, Inextendibility of spacetimes and Lorentzian length spaces, ZBL07030953. This framework allows you for example to define (synthetic) timelike/causal curvature bounds via triangle comparison, analogously to Alexandrov and CAT(k) spaces. Of course, all this does not answer your original question but several of the sides questions and hopefully gives you some context...<|endoftext|> TITLE: Lecture notes by Mahowald and Unell QUESTION [9 upvotes]: I'm trying to find lecture notes of Mahowald and Unell, titled "Lectures on Bott periodicity in stable and unstable homotopy at the prime 2". Does anyone happen to know if a copy exists online (and if it does, where I can find it)? It is cited in Peter May's "Applications and generalizations of the approximation theorem", and in Mahowald's "Some homotopy classes generated by $\eta_j$". Both sources cite specific sections from the document, and in the latter reference, it is said to have been submitted to Springer Lecture Notes Series. Charles Rezk pointed out to me that it's also referenced in footnote 1 of Kuhn's "Geometry of James-Hopf maps", where it's said that the document dates from 1977. REPLY [8 votes]: This is Alan. I'm glad you kept them. My last copy was sent to a student in Indiana years ago. I. glad to have an electronic version...Alan<|endoftext|> TITLE: A mixed preservation theorem for two-sorted structures? QUESTION [6 upvotes]: A well known model theory fact is that for any first-order theory $T$ the collection of universal consequences of $T$, written $T_\forall$, is a precise axiomatization of the class of substructures of models of $T$. This is related to the fact that a sentence is preserved under passing to substructures if and only if it is logically equivalent to a universal sentence. Dually a sentence is preserved under passing to superstructures if and only if it is logically equivalent to an existential sentence. The characterization of models of $T_\exists$ is less clean, specifically it axiomatizes the class of elementary substructures of superstructures of models of $T$. Suppose $\mathcal{L}$ is a two-sorted language with sorts $A$ and $B$. Given $\mathcal{L}$-structures $\mathfrak{M}$ and $\mathfrak{N}$, we'll say that $\mathfrak{N}$ is an $A$-super-$B$-substructure of $\mathfrak{M}$ if $A(\mathfrak{N}) \supseteq A(\mathfrak{M})$ and $B(\mathfrak{N}) \subseteq B(\mathfrak{M})$ with the interpretations of all relation and function symbols agreeing on the common substructure $A(\mathfrak{M}) \cup B(\mathfrak{N})$. Intuitively speaking we've allowed $A$ to grow and $B$ to shrink when passing from $\mathfrak{M}$ to $\mathfrak{N}$. This relationship is obviously transitive. We'll say that a sentence $\varphi$ is preserved under passing to $A$-super-$B$-substructures if for any $\mathcal{L}$-structures $\mathfrak{M}$ and $\mathfrak{N}$, if $\mathfrak{M}\models \varphi$ and $\mathfrak{N}$ is an $A$-super-$B$-substructure of $\mathfrak{M}$, then $\mathfrak{N}\models \varphi$. We may also requires that $\mathfrak{M}$ and $\mathfrak{N}$ be models of some particular theory. A natural example of a sentence with this property is extensionality. Suppose that $B$ is a sort of sets of elements of $A$ and consider the extensionality axiom: $$(\forall x,y:B)(\exists z : A)((z\in x \leftrightarrow z \in y)\rightarrow x=y)$$ Adding more elements to $A$ cannot spoil extensionality, regardless of how you extend $\in$, and removing sets from $B$ cannot spoil extensionality. This sentence has a particular syntactic form, specifically it is prenex and has the property that all $A$-quantifiers are $\exists$ and all $B$-quantifiers are $\forall$. Let's call the collection of sentences of this form $\mathcal{L}_{A\exists B\forall}$ (note that there is no restriction on the number of alternations). An easy argument shows that any sentence in this form is preserved under passing to $A$-super-$B$-substructures. My question is about the converse: Question 1: Fix a theory $T$. Suppose that $\varphi$ is a sentence that is preserved under passing to $A$-super-$B$-substructures provided that both structures are models of $T$. Does it follow that $\varphi$ is equivalent to a sentence in $\mathcal{L}_{A\exists B\forall}$ modulo $T$? Question 2: Let $T_{A\exists B\forall}$ be the set of consequences of a theory $T$ that are sentences in $\mathcal{L}_{A\exists B\forall}$. Are the models of $T_{A\exists B\forall}$ precisely the class of elementary substructures of $A$-super-$B$-substructures of models of $T$? REPLY [2 votes]: $\let\fii\varphi\def\fk{\mathfrak}\def\cL{\mathcal L}\def\aeba{A\exists B\forall}\let\TO\Rightarrow$Here is a quick and dirty proof that Q1 is true for countable languages, using an approach to preservation theorems suggested by §1.5 of Barwise & Schlipf, An introduction to recursively saturated and resplendent models. I have no doubts it holds for uncountable languages as well, though this may require a more elaborate argument. So, assume that $\fii$ is not equivalent over $T$ to an $\cL_{A\exists B\forall}$ sentence. This implies $T+(T+\fii)_{\aeba}\nvdash\fii$, thus there exists $$\fk N\models T+\neg\fii$$ such that $\fk N\models(T+\fii)_{\aeba}$. The latter means that $T+\fii+\mathrm{Th}_{A\forall B\exists}(\fk N)$ is consistent, hence there exists $$\fk M\models T+\fii$$ such that $$\fk M\TO_{\aeba}\fk N,$$ by which I mean that $\fk M\models\psi\implies\fk N\models\psi$ for all $\psi\in\cL_{\aeba}$. Using the Löwenheim–Skolem theorem and standard results on the existence of recursively saturated models, we may assume that $\fk M$ and $\fk N$ are countable, and the joint 4-sorted model $(\fk M,\fk N)$ is recursively saturated. Let us fix enumerations $A(\fk M)=\{a_n:n\in\omega\}$ and $B(\fk N)=\{b_n:n\in\omega\}$. By induction on $n$, we will construct sequences $\{c_n:n\in\omega\}\subseteq A(\fk N)$ and $\{d_n:n\in\omega\}\subseteq B(\fk M)$ such that $$(\fk M,\{a_i:i TITLE: Chromatic number of $C_4$-free graphs QUESTION [7 upvotes]: How large can the chromatic number of an $n$-vertex $C_4$-free graph be? If the maximum degree of the graph $G$ is $\Delta$, is there a bound of the form $\chi(G) \leq O(\Delta/\log(\Delta))$ as in the case of triangles? What happens if $e(G)$ is close to $ex(n,C_4)$, say $e(G) \geq n^{3/2-\alpha}$; is there a better bound (depending on $\alpha$) in this case? REPLY [4 votes]: For $G$ an $n$ vertex graph which is $C_4$-free, $\chi(G)=O(\sqrt{n})$, follows from Kővári–Sós–Turán by the argument found here for instance. Before Johannson proved the chromatic number bound for triangle free graphs, the inequality appeared in a paper of Kim as a conjectured improvement to the girth $>4$ case. In that case, the inequality is due to Kim and takes the form $$\chi(G)≤[1 +o(1)]\frac{\Delta}{\log \Delta},$$ where the $o(1)$ is taken as $\Delta(G)\rightarrow\infty$. For the case that $G$ is $C_4$ saturated, or nearly so, there is less which appears immediately in a quick search.<|endoftext|> TITLE: Embedding metric spaces into Hilbert ones QUESTION [5 upvotes]: Does every compact metric space continuously embed into a Hilbert space (possibly with large distortion)? REPLY [5 votes]: You may also send $X$ into a space $L_2(X,\mu)$ via the Fréchet-Kuratowski isometry $x\mapsto d(\cdot,x)\in C^0(X),\|\cdot\|_\infty$, followed by the bounded linear inclusion $C^0(X)\to L_2(X,\mu)$, where $\mu$ is a probability measure on $X$. If $\text{supp}(\mu)=X$ (e.g. $\mu$ is a series of deltas $\sum_{k=1}^\infty2^{-k}\delta_{q_k}$ for a dense set $\{q_k\}_{k\ge1}$), then $C^0(X)\to L_2(X,\mu)$ is injective, and the composition $j:X\to L_2(X,\mu)$ is a continuous embedding (a homeo of $X$ with $j(X)$).<|endoftext|> TITLE: A generalization of bicharacters QUESTION [5 upvotes]: Let $G$, $H$ be groups and let $F$ be a field. A bicharacter $A: G \times H \to F^\times $ is a map that satisfies $$ A(xy, z) = A(x,z)A(y,z),\quad A(x,zw)=A(x,z)A(z,w),\quad x,y\in G,\, z,w\in H. $$ I came across a map $B: G \times H \to F^\times$ that satisfies a weaker condition: \begin{multline*} \tag{*} B(xy, zw) B(x,z)B(x,w)B(y,z)B(y,w) =\\ B(xy, z) B (xy,w) B(x, zw) B(y, zw) \end{multline*} as well as $B(x, 1)= B(1,z)=1$. A bicharacter satisfies (*) but not vice versa (e.g., for $G=H= \mathbb{Z}_2=\{1,x\}$ one can have $B$ with $B(x,x)$ $=$ a $4$th root of $1$). Do maps satisfying (*) occur somewhere? Is there a name for such maps? Is there a good description of the quotient of the group of such maps by the group of bicharacters? REPLY [5 votes]: Let $ß_F(G,H)$ denote your (abelian) group, where the operation is the convolution product. Let's just run with the recasting I gave in the comments. We will compute some facts about the order $ß_F(G,H)$ (which turns out to always be finite) by relating it to several other groups of morphisms. This won't ultimately resolve exactly what this group and the desired quotient look like, but hopefully it is illuminating and possibly can be developed further to provide the desired answers. We can equivalently rewrite $B$ as a linear map $FG\to F^H$. We can note that any $B$ which is multiplicative in the first or second argument (while the other is held fixed) trivially satisfies your identity. These correspond to the unital coalgebra and augmented algebra morphisms $FG\to F^H$. So your group must be providing some sort of generalization of both of these concepts (for $G,H$ groups) simultaneously. Written in the linear form, your identity is then equivalent to $$\Delta(B(gh))(B(g)B(h)\otimes B(g)B(h)) = (B(gh)\otimes B(gh))\Delta(B(g))\Delta(B(h)),$$ where $FG$ and $F^H$ are given their usual structure as Hopf algebras. The condition $B(g,h)\neq 0$ is equivalent to requiring the linearized version to send group-like elements to invertible elements. The condition $B(x,1)=1=B(1,y)$ is equivalent to requiring that the linearized version respect the unit and counit maps. The invertibility of $B(g),\ \forall g\in G,$ is the amusing thing here, as we can then rewrite your identity as $$\Delta(B(g))(B(g)^{-1}\otimes B(g)^{-1}) \cdot \Delta(B(h))(B(h)^{-1}\otimes B(h)^{-1}) = \Delta(B(gh))(B(gh)^{-1}\otimes B(gh)^{-1}).$$ This means that the map $T(B)\colon FG\to F^H\otimes F^H$ given by $$g\mapsto \Delta(B(g))(B(g)^{-1}\otimes B(g)^{-1})$$ is in fact an algebra homomorphism. Informally speaking the assignment $B\to T(B)$ satisfies: $T(B)$ is an algebra homomorphism if and only if $B$ satisfies your identity. $T(B)$ is a morphism of augmented algebras if and only if $B$ satisfies your identity and $B(g,1)=1$ for all $g\in G$. $T(B)$ is trivial (identically the identity of $F^H\otimes F^H$) if and only if $B$ is a morphism of coalgebras (equiv, $B$ sends group-likes to group-likes). $B$ is a morphism of unital coalgebras if and only if $T(B)$ is trivial and $B(1)$ is the identity of $F^H$. This is informal because I'm more or less acting as if $$T\colon \text{Hom}_{F}(FG,F^H)\to \text{Hom}_{F}(FG,F^H\otimes F^H),$$ but the formula is only well-defined on those $B$ sending group-likes to invertible elements. But by adjusting the domain we can resolve that issue. We only need those $B$ that map elements of $G$ to normalized units of $F^H$. Define $U(A)$ to be the multiplicative group of normalized units of the Hopf algebra $A$. We may then replace the domain with $\text{Hom}_{\text{Set}}(G,U(F^H))$, which is equivalently $\text{Hom}_{\text{coalg}}(FG,FU(F^H))$. Note that using normalized units automatically encodes the relation $B(x,1)=1$ for all $x\in G$. Since $F^H$ is commutative, we can in fact endow $\text{Hom}_{\text{Set}}(G,U(F^H))$ with the structure of an abelian group via the convolution product: $(f_1*f_2)(g) = f_1(g) f_2(g)$ for all $g\in G$. Moreover, the convolution product also endows $\text{Hom}_F(FG,F^H\otimes F^H)$ with the structure of an abelian group. We can then view $T$ as a well-defined group homomorphism $$\text{Hom}_{\text{Set}}(G,U(F^H))\to \text{Hom}_{F}(FG,F^H\otimes F^H).$$ We can adjust the domain further. We also wish to impose the condition that $B(1)$ is the identity, or equivalently that $B(1,y)=1$ for all $y\in H$. As such we may instead think of $T$ as a well-defined group homomorphism $$\text{Hom}_{\text{Set}}(G\setminus\{1_G\},U(F^H))\to \text{Hom}_{F}(FG,F^H\otimes F^H).$$ The augmented algebra morphisms form a subgroup of the codomain, so it follows that $$ß_F(G,H)=T^{-1}(\text{augmented algebra morphisms}).$$ By dualizing, an augmented algebra morphism $FG\to F^H\otimes F^H$ becomes a unital coalgebra morphism $F(H\times H)\to F^G$, which is equivalent to a set map $H\times H\to \widehat{G}$ sending the identity to the identity, where $\widehat{G}$ is the character group of $G$ over $F$. We can then collect a number of facts about $ß_F(G,H)$ and $T$. $\ker(T) = \text{Hom}_{\text{Set}}(G\setminus\{1_G\},\widehat{H})$, which is an abelian group of order $$|\widehat{H}|^{|G|-1}.$$ $\ker(T)\subseteq ß_F(G,H)$ is a (normal) subgroup and $ß_F(G,H)/\ker(T)$ is isomorphic to a subgroup of $\text{Hom}_{\text{Set}}((H\times H)\setminus\{(1_H,1_H)\},\widehat{G})$. This parent group of the quotient has order $$|\widehat{G}|^{|H|^2-1}.$$ $ß_F(G,H)$ is finite and has order dividing $$|\widehat{G}|^{|H|^2-1}|\widehat{H}|^{|G|-1}.$$ As your identity is trivially satisfied when any of $x,y,z,w$ are an identity element, it is easy to verify that $$ß_F(\mathbb{Z}_2,\mathbb{Z}_2)\cong \widehat{\mathbb{Z}_4},$$ so that the bound can be strict. Indeed, $T(ß_\mathbb{C}(\mathbb{Z}_2,\mathbb{Z}_2))$ has order $2$ while the augmented algebra morphisms in question are a group of order $8$, and $\ker(T)$ is precisely the bicharacters. So in general the image of $T$ need not contain all possible augmented algebra homomorphisms. This also shows that your group may contain maps which are not convolution products of unital coalgebra morphisms and augmented algebra morphisms. I currently leave it unresolved as to what, exactly, the image of $T$ is in general. Determining that would in principle resolve your questions, though doesn't seem much easier at first glance. To get to a characterization of $B$ being a group homomorphism you can observe that we could have equivalently rewritten it as a linear map $FH\to F^G$, and this map will be precisely the linear dual of the first one. We then have the obvious analogue to $T$, call it $\check{T}$, and by the preceding case we'd have that $B$ is a group homomorphism if and only $B\in \ker(T)$ and $B^*\in\ker(\check{T})$. Here we're identifying $B\colon FG\to F^H$, so that $B^*\colon FH\to F^G$ is the (equivalent) dualized map. So the group homomorphisms are precisely $\ker(\check{T})^*\cap \ker(T)$. We can also exploit this to apply our results about $T$ to $\check{T}$ as well, which combine to give a better set of divisors that bound the order of $ß_F(G,H)$. $\ker(\check{T}) = \text{Hom}_{\text{Set}}(H\setminus\{1_H\},\widehat{G})$, which is an abelian group of order $$|\widehat{G}|^{|H|-1}.$$ $\ker(\check{T})\subseteq ß_F(G,H)$ is a (normal) subgroup and $ß_F(G,H)/\ker(\check{T})$ is isomorphic to a subgroup of $\text{Hom}_{\text{Set}}((G\times G)\setminus\{(1_G,1_G)\},\widehat{H})$. This latter group has order $$|\widehat{H}|^{|G|^2-1}.$$ $ß_F(G,H)$ is finite and has order dividing $$|\widehat{H}|^{|G|^2-1}|\widehat{G}|^{|H|-1}.$$ Combining with the previous results using $T$, and the fact that $\ker(T)\ \cap\ker(\check{T})^*$ is the subgroup of bicharacters, we have $|ß_F(G,H)|$ is divisible by $\text{lcm}(|\widehat{G}|^{|H|-1}|,\widehat{H}|^{|G|-1})$ and $$\frac{|\widehat{G}|^{|H|-1}|\widehat{H}|^{|G|-1}}{|\text{Hom}_{\text{group}}(G,\widehat{H})|}$$. $|ß_F(G,H)|$ divides $$\gcd(|\widehat{G}|^{|H|^2-1}|\widehat{H}|^{|G|-1}, |\widehat{H}|^{|G|^2-1}|\widehat{G}|^{|H|-1})$$. When $G=H$ this is of limited help, but if $\widehat{G},\widehat{H}$ have coprime orders it tells us the full order of the group: If $\widehat{G},\widehat{H}$ have coprime orders then $$|ß_F(G,H)| = |\widehat{G}|^{|H|-1}|\widehat{H}|^{|G|-1}.$$ So $|ß_\mathbb{C}(\mathbb{Z}_2,\mathbb{Z}_3)| = 12$, for example. Indeed, it is generated by the unital coalgebra morphisms and the augmented algebra morphisms, which intersect trivially.<|endoftext|> TITLE: Large-n limit of the distribution of the normalized sum of Cauchy random variables QUESTION [13 upvotes]: What is the large-n limit of a distribution of the following sample statistic:$$x\equiv\displaystyle\frac{\sum^{n}X_{i}}{\,\sqrt{\,\sum^{n}X_{i}^{2}\,}\,}$$ when sampling the Cauchy(0,1) distribution? Monte Carlo simulation indicates that convergence to this limit is quite fast, and that the resulting (symmetric) PDF has very sharp cusps at -1 and +1, but of course does not yield an analytic expression for this PDF - would anyone be able to find it? Here is how the positive half of the PDF looks like: REPLY [15 votes]: This desired large-$n$ limit of the distribution $P_n(x)$ is calculated in Limit Distributions of Self-normalized Sums (1973). The Cauchy distribution is the case $\alpha=1$ on page 798. The singularity in $\lim_{n\rightarrow\infty}P_n(x)$ at $x=\pm 1$ is logarithmic $$P_n(x)\rightarrow-\pi^{-2}\log|1-x^2|,$$ see equation (5.12) and figure 4 (reproduced below), while for large $|x|$ the decay is Gaussian.<|endoftext|> TITLE: On a morphism from the Brauer group to the Picard group QUESTION [9 upvotes]: Suppose that $k$ is a commutative ring and that $A$ is an Azumaya $k$-algebra. Then there is a well-known morphism from $Aut_{Alg_k}(A)$, the group of algebra automorphisms, to the Picard group $Pic(k)$. One way to phrase it is as follows. Every automorphism $\phi$ determines a $k$-linear autoequivalence $$ \phi^*: Mod_A \to Mod_A $$ However, since $A$ is Azumaya, every $k$-linear autoequivalence of the category of (right) $A$-modules is of the form $M \mapsto M \otimes_k J$ for some $J$ in the Picard group of $k$, unique up to isomorphism. Moreover, this assignment takes composition to tensor product. This assignment is part of the Rosenberg-Zelinsky exact sequence $$ 0 \to k^\times \to A^\times \to Aut_{Alg_k}(A) \to Pic(k). $$ It is possible to express $J$ as $HH^0(A,A^\tau)$ where $A^\tau$ is $A$, but with half of its bimodule structure pulled back along $\phi$. In particular, given an Azumaya $k$-algebra $Q$, let $A = Q \otimes_k Q$. Then A is also an Azumaya $k$-algebra with automorphism $\phi(a \otimes b) = b \otimes a$, and thus $Q$ determines a (2-torsion) element in $Pic(k)$. This assignment turns out to be Morita invariant and it respects the tensor product; this makes it a (2-torsion) homomorphism from the Brauer group $Br(k)$ to the Picard group $Pic(k)$. Are there examples of rings $k$ for which this homomorphism is nontrivial? REPLY [4 votes]: Here is another way to see that the given map should be zero. It corresponds to a map of sheaves of grouplike $\mathbb{E}_\infty$-spaces $K(\mathbb{G}_m,2)\rightarrow K(\mathbb{G}_m,1)$. All such maps are nullhomotopic (for example by delooping to sheaves of spectra and using a $t$-structure argument). I would also guess that the map is given by $A\mapsto\mathrm{HH}(A/k)$. Since Hochschild homology is symmetric monoidal, $\mathrm{HH}(A/k)$ is a line bundle. The argument above shows that it is trivial. On the other hand, I suspect there are derived Azumaya algebras where this is non-trivial. If we look at the derived Brauer sheaf (on the étale site of an ordinary commutative ring) it is an extension $$K(\mathbb{G}_m,2)\rightarrow\mathbf{dBr}\rightarrow K(\mathbb{Z},1).$$ Taking Hochschild homology gives a map of sheaves of grouplike $\mathbb{E}_\infty$-spaces $\mathbf{dBr}\rightarrow\mathbf{dPic}$, where $\mathbf{dPic}$ fits into a fiber sequence $$K(\mathbb{G}_m,1)\rightarrow\mathbf{dPic}\rightarrow K(\mathbb{Z},0).$$ The map $\mathbf{dBr}\rightarrow\mathbf{dPic}$ thus canonically factors through a map $K(\mathbb{Z},1)\rightarrow K(\mathbb{G}_m,1)$ of grouplike $\mathbb{E}_\infty$-spaces. I believe this is the map that sends $1$ to $-1\in\mathbb{G}_m(\mathbb{Z})$. Indeed, the copy of $\mathbb{Z}$ is coming from suspension in the derived category. To make this concrete, one needs a ring $k$ where the induced map $H^1(k,\mathbb{Z})\rightarrow\mathrm{Pic}(k)$ is non-zero. I don't have an example, but I'd be surprised if this didn't exist.<|endoftext|> TITLE: Reference request: norm topology vs. probabilist's weak topology on measures QUESTION [6 upvotes]: Let $(X,d)$ be a metric space and $\mathcal{M}(X)$ be the space of regular (e.g. Radon) measures on $X$. There are two standard topologies on $\mathcal{M}(X)$: The (probabilist's) weak topology and the strong norm topology, where the norm is the total variation norm. Surprisingly, I have found very little discussion in the literature comparing these two topologies rigourously, besides the oft-cited claim that the norm topology is much stronger than the weak topology. I am looking for a reference that discusses and compares these topologies, esp. things like convergence, boundedness, open sets, projections, etc. I am mostly concerned with probability measures $\mathcal{P}(X)\subset\mathcal{M}(X)$, but I am not sure how much of a difference this makes wrt topological concerns. REPLY [4 votes]: A very good treatment of the subject can be found in "Topological Vector Spaces" by Helmut H. Schaefer. If you have no previous experience at all with functional analysis, it may be a bit harsh in the beginning though. You can also learn first some more basic functional analysis, for example from the first 3 chapters of the classical book "Functional Analysis, Sobolev Spaces and Partial Differential Equations" by Haim Brezis (this is the route I have personally followed for example). I expand on my answer in response to the comment Note that first that, if $X$ is compact, then all continuous functions on $X$ are bounded. Therefore in this case the "probabilist's weak" topology on the space of finite Radon measures $\mathcal{M}(X)$ is just the weak* topology (in the usual sense of functional analysis) on $C(X)^*$ via the isomorphism $\mathcal{M}(X)\cong C(X)^*$. Therefore the comparison between the two topologies is clear in this case. This does not work anymore if $X$ is not compact, and here is where the more specific approach of the probabilist comes into play. Let $\mathcal{P}(X)$ be the space of Borel probability on $X$. Then you can define a distance on $\mathcal{P}(X)$ by $$ d_P(\mu,\nu):=\inf\{\alpha>0\,:\,\mu(A)\le \nu(A_{\alpha})+\alpha,\, \mu(A)\le \nu(A_\alpha)+\alpha\},$$ where $$ A_\alpha=\begin{cases} \{x\in X\,:\,d(x,A)<\alpha\} & A\ne \emptyset,\\ \emptyset & A=\emptyset. \end{cases} $$ This is called Prokhorov distance, and you can prove that, provided $X$ being separable, the "probabilist's weak" topology is induced by the metric $d_P$. If fact, in order to make the comparison you are interested in, it suffices to compare this metric with the one inducing the strong topology. What makes things more complicated from a functional analitic point of view is that the metric $d_P$ is in general not induced by norm. It follows that the functional analysis you most commonly learn in your studies does not apply. Schaefer's book is all about extending those tools to the case in which a norm to start with is not available.<|endoftext|> TITLE: Sufficient sets of colimits in small categories QUESTION [10 upvotes]: Let $C$ be a small category, and consider the class of diagrams $G:D\to C$, with $D$ a small category, that have colimits in $C$. This is a proper class even when $C$ is very small, e.g. whenever $D$ has a terminal object $t$, any functor $G:D\to C$ has a colimit $G(t)$, and there is a proper class of small categories with a terminal object. However, those colimits feel kind of "trivial"; in some cases at least we can find a small set of diagrams that "carry all the nontrivial information" about colimit diagrams in $C$. For instance, if $C$ is a poset, then it suffices to consider injective functors $G$ (and we may as well take $D$ to be discrete as well), and these form an essentially small set. For a non-posetal $C$ we can't restrict to injective functors, since coproducts are not idempotent, but maybe there is some other restriction that works. Note that by Freyd's theorem, a non-posetal small category does have a bound on the cardinality of coproducts that it can admit; but this doesn't quite answer the question itself, since a particular colimit can exist even if the coproducts that would be necessary to construct it from coproducts and coequalizers do not. Here are two ways to make the question precise: Given a small category $C$, is there a small set $L$ of diagrams $G:D\to C$ with colimits such that for any diagram $G':D'\to C$ with a colimit, there is a $(D,G)\in L$ and a final functor $F:D\to D'$ such that $G = G' \circ F$? (Edit: As pointed out by Dylan in the comments, this version is impossible. Take $C$ terminal and let $D$ vary over all ordinals; no small set of categories can be cofinal in all ordinals.) Given a small category $C$, is there a small set $L$ of diagrams $G:D\to C$ with colimits such that if a functor $H:C\to E$ preserves colimits of all diagrams in $L$, then it preserves all colimits that exist in $C$? Any solution to question 1 is also a solution to question 2, but I'm not sure whether the converse holds. The mention of Freyd's theorem above suggests that a solution might require classical logic — I would find it more surprising if such a set existed for a non-posetal small complete category, although I don't immediately see an argument that it cannot. One can of course also ask similar questions for enriched categories, internal categories, $\infty$-categories, and so on. Bonus points go to an answer that applies more generally in such contexts. REPLY [6 votes]: I think the answer to (2) is affirmative under Vopenka's Principle. That is, Claim: Let $C$ be a small category, and assume Vopenka's Principle. Then there exists a small set of limit cones $L$ in $C$ such that for any category $D$ and any functor $F: C \to D$, $F$ preserves limits if and only if $F$ preserves the limit cones in $L$. Notes: I have no idea if VP is necessary, or if the statement has large cardinal strength at all. I'm a little suspicious of the claim because of its reliance on the Lemma below. The reason I'm suspicious of the lemma is that it follows very easily from the work of Adamek and Rosicky and I'm surprised I haven't seen it stated before. The proof shows the more general claim, as discussed in the comments, that under VP any limit-sketch on a small category has a small sub-sketch (on the same category) with equivalent models. Proof: As noted in the comments, it suffices to consider the case $D = Set$, because $F: C \to D$ preserves limits (resp. preserves limits in $L$) if and only if $D(d,F-): C \to Set$ preserves limits (resp. preserves limits in $L$) for every $d \in D$. Note that the category $Lim(C)$ of limit-preserving functors $C \to Set$ is the intersection in $Fun(C,Set)$ of the cateogries $Lim_\mu(C)$ of functors preserving $\mu$-small limits in $C$ for each $\mu$. So by the following lemma, we may take $L= L_\mu$ for some $\mu$. Lemma: Let $K$ be a locally presentable category, and assume Vopenka's Principle. Then every decreasing Ord-indexed sequence of reflective categories of $K$ stabilizes after a small number of steps. Proof: Let $(L_\mu)_{\mu \in Ord}$ be such a sequence of localization functors. The proof of Thm 6.22 in [1] shows that even under weak Vopenka's Principle, the sequence $L_\mu(X)$ stabilizes after a small number $\mu_X$ of steps for any fixed $X$ (briefly -- if it doesn't stabilize, we get an embeding of $Ord^{op}$ into $X \downarrow K$). So we just need to show that the number of steps is bounded independent of $X$. Here we use Thm 6.24 of [1], which tells us that by virtue of Vopenka's Principle, $L_\infty = \varinjlim_\mu L_\mu$ reflects onto an accessible, accessibly-embedded subcategory, and so $L_\infty$ is a $\lambda$-accessible functor for some $\lambda$ (viewed as an endofunctor of $K$). Thus, we can take a small subcategory $C \subseteq K$ which generates $K$ under $\lambda$-filtered colimits. Then if $\mu = \max(\lambda, \sup_{c \in C} \mu_c)$, then the colimit defining $L_\infty(X)$ stabilizes after $\mu$-many steps for every $X$. Reference: [1] Adámek, J., and J. Rosickỳ. Locally Presentable and Accessible Categories. London Math. Soc. Lect. Notes Ser. 189. Cambridge Univ Pr, 1994.<|endoftext|> TITLE: Identity involving sum over permutations QUESTION [7 upvotes]: In some work on QFT the following identity has come up: $$ \sum_{\sigma \in S_n}\sum_{j=1}^n \left(\sum_{l=1}^j w_{\sigma_l}\right)\prod_{i=1,i\neq j}^{n}\frac{1}{\sum_{l=1}^j z_{\sigma_l}-\sum_{l=1}^i z_{\sigma_l}}=0 $$ for $2 TITLE: Measure on cosets in a group? QUESTION [6 upvotes]: If $\mu$ is the normalized counting measure on a finite group $G$, then $\mu(G)=1$ and $\mu(C)=1/n$ for every coset $C$ of a subgroup of index $n$. Let's ask for the same for infinite groups: Question: When $G$ is any group, is there a finitely additive measure $\mu$ on $G$ such that for every positive integer $n$ one gets $\mu(G)=1$ and $\mu(C)=1/n$ for every coset $C$ of every subgroup of index $n$? (Question edited for clarity.) REPLY [10 votes]: This should be possible for any group, assuming I have facts about extending measures correct. The idea is to first construct such a measure on a smaller Boolean algebra of subsets of $G$, and then use general measure theory facts to extend this measure to $\mathcal{P}(G)$. So first, let $\mathcal{B}\subseteq\mathcal{P}(G)$ be the Boolean algebra generated by cosets of finite index subgroups of $G$. Define a measure $\mu$ on $\mathcal{B}$ as follows. Suppose $A\in\mathcal{B}$. Then there is some finite index subgroup $H$ of $G$ such that $A$ is a union of cosets of $H$. If $n$ is the index of $H$, and $m$ is the number of cosets in the union, then define $\mu(A)=m/n$. It can be directly checked that $\mu$ is well-defined. Moreover, if $C$ is a coset of subgroup of index $n<\infty$, then $\mu(C)=1/n$ by construction. Note that $\mu$ is also translation-invariant. Now one can (non-uniquely) extend $\mu$ to some finitely additive measure on $\mathcal{P}(G)$ (which won't be translation-invariant necessarily). I believe this follows from Section 457 of Fremlin's "Measure Theory". Specifically, any finitely additive probability measure on a Boolean algebra (of subsets of some fixed set $X$) can be extended to such a measure on any larger Boolean algebra. Remark 1: The initial measure $\mu$ is in fact the unique $G$-invariant finitely additive probability measure on $\mathcal{B}$, and can be constructed from the Haar measure on the profinite completion of $G$. In particular, if $\mathcal{N}$ denotes the collection of finite index normal subgroups of $G$, then the profinite completion is $\hat{G}=\varprojlim_{\mathcal{N}}G/N$. We can write elements of $\hat{G}$ as $(C_N)_{N\in\mathcal{N}}$, where $C_N$ is a coset of $N$. Given a set $A$ in $\mathcal{B}$, define $X_A$ to be the set of $(C_N)_{N\in\mathcal{N}}\in \hat{G}$ such that $C_N\cap A\neq\emptyset$ for all $N\in\mathcal{N}$. Then $X_A$ is closed, and it can be shown that $\mu(A)$ is the Haar measure of $X_A$. Remark 2: Perhaps it's also worth mentioning that if $G$ is amenable (e.g., abelian) then there is a translation-invariant finitely additive probability measure on $\mathcal{P}(G)$, which must satisfy the desired conditions outright by finite additivity and invariance.<|endoftext|> TITLE: A list of all irreducible 4-dimensional real representations QUESTION [11 upvotes]: I need a reference to a complete list of all faithful real 4-dimensional irreducible representations of real Lie algebras. The list itself is not very hard to obtain. Using the Levi decomposition, it's possible to see that the Lie algebra $\mathfrak g \subset\mathfrak{sl}(4, {\Bbb R})$ which does not fix a proper subspace has to be reductive. Then (up to a center acting by constants and/or by rotations with constant angle), $\mathfrak g$ is a semisimple subalgebra of $\mathfrak{sl}(4, {\Bbb R})=\mathfrak{so}(3,3)$. The list of such subalgebras (I think) is $\mathfrak{so}(1,2)=\mathfrak{sl}(2, {\Bbb R})$, $\mathfrak{so}(3)$, $\mathfrak{so}(2,2)= \mathfrak{so}(1,2)\times \mathfrak{so}(1,2)$, $\mathfrak{so}(4)=\mathfrak{so}(3)\times \mathfrak{so}(3)$, $\mathfrak{so}(1,3)=\mathfrak{sl}(2, {\Bbb C})$, $\mathfrak{so}(2,3)=\mathfrak{sp}(4, {\Bbb R})$, $\mathfrak{sl}(4, {\Bbb R})=\mathfrak{so}(3,3)$. It can be (probably) obtained by removing some vertices of the Dynkin diagram of $A_3$ and using the arcane technique of coloring the remaining vertices to obtain the real forms. All in all, it's a pain. I am sure there is a reference somewhere to this list (also I might have missed some algebras). I would be much grateful for all pointers. REPLY [5 votes]: A complete list of semisimple subalgebras of $\mathfrak{sl}(4,\mathbb{R})$ seems to be available here: http://downloads.hindawi.com/journals/jmath/2016/2570147.pdf<|endoftext|> TITLE: Does WRT invariant detect hyperelliptic involution on the genus 2 surface? QUESTION [5 upvotes]: The Witten-Reshetikhin-Turaev invariant cannot detect the hyperelliptic involution on the genus 1 surface, and that if $M_U$ is the mapping torus for a mapping class group element $U\in \mathrm{Mod}(\Sigma_1)$, then $$Z(M_U) = Z(M_{-U})$$ where $-I\in \mathrm{Mod}(\Sigma_1)$ is the hyperelliptic involution. What about the genus 2 case? If $-I\in \mathrm{Mod}(\Sigma_2)$ is the hyperelliptic involution on the genus 2 surface, then is there any $U\in \mathrm{Mod}(\Sigma_2)$, for which the following does not hold? $$Z(M_U) \stackrel{?}{=} Z(M_{-U})$$ REPLY [2 votes]: Unless I'm misunderstanding what's meant by hyperelliptic involution, WRT invariants do, in fact, detect the hyperelliptic involution in genus 1. Let $C$ be the modular tensor category (MTC) corresponding to the WRT theory. The Hilbert space of a torus has a basis indexed by the simple objects $\{a\}$ of $C$, and the hyperelliptic involution sends $a$ to $a^*$. It is not true that $a \cong a^*$ for all MTCs C and simple objects $a$, so WRT invariants detect the hyperelliptic involution in genus 1. (Note that this argument works regardless of how we lift the hyperelliptic involution to a morphism of framed manifolds.) A similar argument works in higher genus. One can choose a "spine" basis for the Hilbert space such that the basis vectors are permuted by the involution, and it's easy to check that this permutation is non-trivial for an MTC which contains a simple object $a$ such that $a \not\cong a^*$.<|endoftext|> TITLE: Does Chowla's conjecture on the Liouville function imply the Riemann hypothesis? QUESTION [8 upvotes]: A paper see here on arXiv claims that Chowla's conjecture (applied to the Liouville function instead of the Mobius function), i.e., that $$ \lim_{N\rightarrow \infty} \sum_{n\leq N} \lambda(n+a_1) \lambda(n+a_2) \cdots \lambda(n+a_k)=o(N), $$ implies the Riemann hypothesis. I have been unable to find any references to this claim after some research. Is this claim new? Any pointers, references appreciated. REPLY [3 votes]: The comment by D Karagulyan (Remark 1, page 9) “ However the multiplicativity property of the Liouville function is not used in the proof.” does not seem to be accurate. Indeed, the multiplicativity of the Möbius or Louville function is used to establish that the RH hypotheses is equivalent to the property defined in his paper as RH. This property should be named after Littlewood (the property says: the partial sum up to $x$ is $o(x^{1/2+\epsilon})).$ The reason for that is due to the fact that this property is valid for Möbius and Louville by Littlewood criterion which use of-course that the Möbius or Louville are multiplicatives. Let us notice furthermore that his example is not multiplicative! On the other hand, as proved by Denjoy in Comptes Rendus Acad. Sci. Paris 192 (1931), 656–658, RH holds almost surely for model which is more close to the zeta.<|endoftext|> TITLE: Lipschitz constant of a function of matrix QUESTION [6 upvotes]: The function is given by $f(X) = (AX^{-1}A^\top + B)^{-1}$ where $X$, $A$, and $B$ are $n \times n$ positive definite matrices. I'm trying to find the Lipschitz constant such that $\| f(X)-f(Y) \| \leq L \|X-Y\|$ where $X \geq 0$ and $Y \geq 0$. Motivated by Lemma 3.1 in Nonlinear Systems(H. Khalil, 3rd Ed.), I tried to find the derivative of $f(X)$ (i.e. $\| \frac{ \partial f(X)}{\partial X} \|$) but it's not easy to find the derivative of a function of a matrix over a matrix. How can I find the Lipschitz constant? or Please let me know if there exists the way to calculate the derivative of a function of a matrix. REPLY [3 votes]: I assume $X\ge0$ means $u^\top X u\ge0$, and that $B$ is definite positive $$\inf_{\|u\|=1} u^\top B\, u:=\beta>0.$$ I also assume matrix norms are the Euclidean operator norms. Compute the differential by the chain rule, as suggested in comments by F.Poloni: $$Df(X)H=(AX^{-1}A^{\top}+B)^{-1}AX^{-1}\cdot H\cdot X^{-1}A^{\top}(AX^{-1}A^{\top}+B)^{-1}$$ $$=(A^{\top}+XA^{-1}B)^{-1}\cdot H \cdot(A+BA^{-\top}X)^{-1}=$$ $$=\big(B^\top+Y\big)^{-1}B^\top A^{-\top}\cdot H\cdot A^{-1}B^\top\big(B^\top+Z)^{-1}, $$ where $Y:=B^{\top}A^{-\top} X A^{-1}B\ge0 $ and $Z:=BA^{-\top}XA^{-1}B^\top\ge0$, conjugated to $X\ge0$. Thus for any unit norm vector $u\in\mathbb{R}^n $ $$\big\|\big(B^\top+Y\big)u\big\|\ge u^\top\big(B^\top+Y\big)u\ge u^\top B u \ge\beta$$ and $$\big\|\big(B^\top+Z)u\big\|\ge u^\top\big(B^\top+Z)u \ge u^\top B u \ge\beta.$$ Hence $$\big\|\big(B^\top+Y\big)^{-1}\big\|\le \beta^{-1}$$ and $$\big\|\big(B^\top+Z)^{-1}\big\|\le \beta^{-1}.$$ Therefore $$\|Df\|_\infty\le\|B\|^2\|A^{-1}\|^2\beta^{-2}$$ which is also a Lipschitz constant for $f$, since its domain is convex, $\{X\ge0\}$. $$*$$ Rmk. The above bounds on the $L_2$ operator norms (or others matrix norms) could be improved, but not up to $$\big\|\big(A^{\top}+XA^{-1}B\big)^{-1}\big\|\le\big\|A^{-1}\big\|,$$ even for symmetric definite positive matrices. Take e.g. $n=2$ and $$A=:I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\quad X:=\begin{bmatrix} 5/2 & 1 \\ 1 & 1/2 \end{bmatrix}\quad B:=\begin{bmatrix} 1 & -1/2 \\ -1/2 & 1/2 \end{bmatrix}$$ Then $$\big(A^{\top}+XA^{-1}B\big)^{-1}=\big(I+XB\big)^{-1}= \begin{bmatrix} 4/15 & 4/15\\ -4/15 & 16/15 \end{bmatrix}$$ whose maximum singular value is $\displaystyle{2\over 15}\sqrt{29}+{2\over 5}>1=\|A^{-1}\|$. The same holds for the Frobenius, and other common entry-wise norms (due to the coefficient $16/15>1$).<|endoftext|> TITLE: Singularities of curves that are moving QUESTION [5 upvotes]: Let $k$ be an algebraically closed field, let $d\ge 2$ be an integer and let $f,g\in k[x,y,z]$ be two homogeneous polynomials of degree $d$ without common factor. We want to know what are the singularities of the curve $C_{[\lambda:\mu]}$ given by $\lambda f+\mu g=0$, for a general $[\lambda:\mu]\in \mathbb{P}^1$. If a point $p\in\mathbb{P}^2$ is a singular point of the curves $C_{[1:0]}$ and $C_{[0:1]}$ given by $f=0$ and $g=0$ then it is of course singular for each $C_{[\lambda:\mu]}$. If $\mathrm{char}(k)=0$, then by Bertini there are no other singularities. If $\mathrm{char}(k)=p>0$, it is false: take for instance $f=x^p$ and $g=y^p$. Are all counterexamples of this type ? One can of course replace $p$ by a power of $p$ and maybe do some more general examples. For instance, if $d<2p$, is the case $f=x^p$ and $g=y^p$ the only possibility (up to change of coordinates)? REPLY [5 votes]: I am just writing an answer summarizing the counterexamples from the comments and adding one positive result for small degree. Let $k$ be an algebraically closed field. Let $X$ and $Y$ be quasi-projective, connected, smooth $k$-schemes. Let $f:X\to Y$ be a flat $k$-morphism. Problem. Is $f$ smooth over a dense open subset of $Y$? By Generic Smoothness / Sard's Lemma, this is true if $\text{char}(k)$ equals $0$. However, if $\text{char}(k)=p$ is positive, this can fail. As the OP explains, the induced map of function fields might be inseparable, in which case $f$ is nowhere smooth. However, there are other examples, such as the ones from the comments. In particular, for $$\mathbb{P}^1_k=\text{Proj}\ k[\lambda,\mu],\ \ \mathbb{P}^2_k = \text{Proj}\ k[x,y,z],$$ $$f = x^dz^{p-d} + (y-x)^p, \ \ g = x^dz^{p-d} + (y-z)^p, \ \ 1\leq d \leq p-1,$$ $$Y=\mathbb{P}^1_k \setminus\{[1,1]\}, \ \ U = \mathbb{P}^2_k \setminus\{[1,0,1]\},$$ $$ X\subset Y\times_{\text{Spec}\ k} U, \ \ X =\text{Zero}(\lambda f - \mu g),$$ the projection $f$ from $X$ to $Y$ is a flat, surjective morphism of quasi-projective, connected, smooth $k$-schemes that is smooth on a dense open of $X$, yet the singular locus of $f$ surjects to $Y$. There is a positive result. Let $Y$ be a quasi-projective, connected, smooth $k$-scheme. Let $X$ be a locally closed subscheme of $Y\times_{\text{Spec}\ k} \mathbb{P}^N_k$ that is connected and smooth. Assume that the projection $f$ from $X$ to $Y$ is flat. Denote the dimension of the generic fiber of $f$ by $n$. Denote the projective degree of the (closure) of the generic fiber by $e$. In the PhD thesis of Jan Gutt, there is the following result. Jan Gutt Hwang-Mok rigidity of cominuscule homogeneous varieties in positive characteristic https://arxiv.org/pdf/1305.5296.pdf Lemma 4.2.5 If the singular locus of the generic fiber of $f$ has dimension $d$, then the projective degree of the associated $d$-cycle is $\leq e(e-1)^{n-d}$. In particular, if $p> e(e-1)^n$, then the generic fiber of $f$ is smooth. I recall that Will Sawin showed me examples demonstrating the sharpness of the inequality. If Will wants to add those examples, that would be great. Otherwise, I will try to add those myself in a few days.<|endoftext|> TITLE: Complete folds and one cut QUESTION [5 upvotes]: The fold-and-cut theorem states that any shape with straight sides can be cut by a single complete straight cut if the paper is the folded flat in the right way. Here is an example from an answer on another MO question:       Note that the folds are not complete, i.e. they are not along lines that run through the whole paper but stop somewhere in the middle. This makes it a little cumbersome to fold the pattern. My questions are: Is there any shape with straight edges that can't be obtained by complete folds and one complete cut? If yes, which kind of shapes can be obtained by complete folds and one complete cut? REPLY [4 votes]: Not an answer. The obvious place to start is to fold-and-$1$-cut an irregular triangle, whose natural method of cuts are incomplete in @Dirk's notation, not simple folds in other notation1:               Figure from How To Fold It: The Mathematics of Linkages, Origami, and Polyhedra.         Copied from MESE post Examples of Mathematical Beauty in School Mathematics . 1 Arkin, Esther M., Michael A. Bender, Erik D. Demaine, Martin L. Demaine, Joseph SB Mitchell, Saurabh Sethia, and Steven S. Skiena. "When can you fold a map?." Computational Geometry 29, no. 1 (2004): 23-46. Journal link.<|endoftext|> TITLE: If $\mathcal C^{\mathcal C}$ is equivalent to $\mathcal C$, is $\mathcal C$ necessarily equivalent to a category with one object and one morphism? QUESTION [19 upvotes]: Let $\mathcal C$ be a category which is equivalent to the category $\mathcal C^{\mathcal C}$ of its endofunctors. Is $\mathcal C$ necessarily equivalent to a category having exactly one object and one morphism? A particular case of this question was asked here. REPLY [5 votes]: This is a partial answer. I tried to mimic the proof of Theorem 3 in [1] Complete lattices and the generalized Cantor theorem by Roy O. Davies, Allan Hayes and George Rousseau, published in Proc. Amer. Math. Soc. 27 (1971), 253–258. (The notation used in this answer is different from the one in the question. I'll stick as much as possible to the notation and terminology of [1].) All categories in this post are small. Let $A$ and $B$ be categories, and let $b_0,b_1$ be two objects of $B$ such that there is a morphism $b_0\to b_1$ but no morphism $b_1\to b_0$. We claim (1) $B^B$ is not equivalent to $B$, (2) there is no full functor $B^A\to A$ and no essentially surjective functor $A\to B^A$. It suffices to prove (2). For any category $C$ write $C_0$ for the set of objects of $C$. Let $C$ and $D$ be categories. Say that a map $F:C_0\to D_0$ is a weak monomorphism if the existence of morphisms $c\to c'$ and $F(c')\to F(c)$ implies that of a morphism $c'\to c$. It is clear that the map $C_0\to D_0$ induced by a full functor is a weak monomorphism. It is also clear that the existence of an essentially surjective functor $D\to C$ implies that of a weak monomorphism $C_0\to D_0$. We claim (3) There is no weak monomorphism $(B^A)_0\to A_0$. It suffices to prove (3). Let $2$ be the ordinal $\{0,1\}$ viewed as a category. We claim (4) There is no weak monomorphism $(2^A)_0\to A_0$. Let us show that (4) implies (3). Let $F:(B^A)_0\to A_0$ be a weak monomorphism, let $J:2\to B$ be a functor mapping $i$ to $b_i$ for $i=0,1$ (such a functor exists by assumption), and define $F':(2^A)_0\to A_0$ by $F'(G):=F(J\circ G)$. It is straightforward to check (using the assumption that there is no morphism $b_1\to b_0$) that $F'$ is a weak monomorphism (details left to the reader). This proves that (4) implies (3). It suffices to prove (4). Say that a subset $R$ of $A_0$ is a right ideal if the conditions $a\in R$ and there is a morphism $a\to a'$ imply $a'\in R$. The right ideals form a complete lattice $\mathcal R$ order isomorphic to $(2^A)_0$. (The order on $\mathcal R$ is given by inclusion.) Thus it suffices to show that there is no weak monomorphism $\mathcal R\to A_0$. Let $\phi:\mathcal R\to A_0$ be a map. Define the map $f:\mathcal R\to \mathcal R$ by letting $f(R)$ be the right ideal generated by $\phi(R)$. By the corollary to Theorem 1 in [1] there is an $R$ in $\mathcal R$ such that $$ f(R)\le\bigcup_{S>R}f(S). $$ As we have $\phi(R)\in f(R)$, this implies $\phi(R)\in f(S)$ for some $S$ in $\mathcal R$ with (5) $R TITLE: How can I prove that the suspension of an Anosov diffeomorphism is an Anosov flow? QUESTION [5 upvotes]: Suppose we have an Anosov diffeo $f$ on $M$. Define $g_t : M\times\mathbb R\to M\times\mathbb R$ by $(x,s)\mapsto (x,s+t)$. Take a quotient of $M\times\mathbb R$ under the relation $(x,s)\sim (f(x),s-1)$. The flow descends to a flow $f_t$ on the quotient manifold. My question is : As the flow is locally a translation, how can it contract the stable direction? I see that a contraction does take place at regular intervals but is that enough? REPLY [3 votes]: The short answer, as pointed out in the comments, is that yes, contraction at regular intervals is enough, because the definition of Anosov requires that there are $C>0$ and $\lambda\in (0,1)$ such that $\|Df^n|_{E^s}\| \leq C \lambda^n$ for all $n\geq 0$, and similarly for $\|Df^{-n}|_{E^u}\|$. The constant $C$ allows the possibility that there are some points at which $E^s$ does not contract and/or $E^u$ does not expand under one iterate (or even several iterates) of $f$. However, if you take $n$ large enough that $C\lambda^n < 1$, then $n$ iterates is always enough to see the hyperbolicity, no matter where you start. There's a subtlety in your question that is worth pointing out, though. The definition of Anosov involves a Riemannian metric (although for a compact manifold the property of being Anosov is in fact independent of the choice of metric). If we write $M_1$ for the quotient manifold, then in order to ask "is the flow $f_t$ Anosov on $M_1$?" we need to first choose a Riemannian metric to use on $M_1$. Even though $M_1$ is a quotient of $M\times \mathbb{R}$, which carries a natural Riemannian metric, you can't simply push the metric down under the quotient map $\pi \colon M\times \mathbb{R}\to M_1$, because $f$ is not an isometry, so pushing the metric from $y\in \pi^{-1}(x) \subset M\times \mathbb{R}$ to $x\in M_1$ may give a different result depending on which point $y\in \pi^{-1}(x)$ you choose. And if you try to make a single global choice, say by identifying $M_1$ with $M\times [0,1)$ so that $\pi$ becomes a bijection (which I suspect is the picture in your head when you say that the flow is "locally a translation"), then you wind up with a discontinuity at the points where you glue $(x,1)$ to $(f(x),0)$. Thus it's not immediately obvious what Riemannian metric to use. One way to proceed is to write $h_0 \colon TM\times TM \to \mathbb{R}$ for the original metric on $M$, put $h_1 = h_0 \circ Df$, connect $h_0$ to $h_1$ by a family of metrics $h_t$ that varies smoothly in $t$, and then extend periodically to $\{h_t \}_{t\in\mathbb{R}}$, so that $h_{t+n} = h_t \circ Df^n$ for all $n\in\mathbb{Z}$. Using $h_t$ as the metric on $M\times \{t\}$ and declaring $T_{(x,t)}(M\times\{t\})$ to be orthogonal to $T_{(x,t)}(\{x\}\times \mathbb{R})$ gives a metric on $M\times\mathbb{R}$ on which $(x,t) \mapsto (f(x),t-1)$ acts isometrically, so this metric does descend to a metric on $M_1$. Then you can check that the Anosov condition holds.<|endoftext|> TITLE: Number of solutions mod p and Betti numbers QUESTION [9 upvotes]: Suppose $X$ is proper flat scheme over Sepc$\mathbb{Z}$. If $p$ is a large prime, then by Weil conjectures one can recover the Betti numbers of $X$ from the size of $X(\mathbb{F}_{p^n})$ for all $n$. My question is, what if instead of fix a prime and vary $n$, we fix an $n$ and vary $p$? More precisely, let $N$ be a fixed number, suppose we know the size of $X(\mathbb{F}_{p^n})$ for all $n$ between 1 and $N$, and (almost) all prime $p$. Can we recover the Betti numbers of $X$? REPLY [5 votes]: Apologies; my first reading of the question (which confused an $n$ for a $p$) assumed you were only given $\#X(\mathbf{F}_p)$ for $p$ in a certain range. If you are given (almost) all the $\#X(\mathbf{F}_p)$ (the case $N = 1$) then you can determine the Betti numbers, as mentioned in the first version of this answer, given with slightly more detail below. The point counts determine, by the Lefschetz trace formula, the trace of Frobenius at $p$ (for good primes $p$) acting on the virtual Galois representation $$[V_l]:=\sum (-1)^i [H^i(X,\mathbf{Q}_l)]$$ (in the Grothendieck group of Galois representations over $\mathbf{Q}_l$) for any fixed $l$. By the Chebotarev density theorem, as long as you include almost all primes $p$ these Frobenii are dense, and hence this determines the trace any any element in the Galois group on this virtual representation $[V_l]$. By the Brauer-Nesbitt Theorem, the set of traces determines the virtual representation (in the Grothendieck group). By purity, you can then determine (the semisimplifications of) the Galois representations $H^i(X,\mathbf{Q}_l)$ completely, and hence also the Betti numbers.<|endoftext|> TITLE: Concurrent normals conjecture QUESTION [15 upvotes]: It is conjectured that if $K$ is a convex body in $n$-dimensional Euclidean space, then there exists a point in the interior of $K$ which is the point of concurrency of normals from $2n$ points on the boundary of $K$. This has been proved for $n=2$ and $3$ by E. Heil. For $n=4$, a proof appeared (under a smoothness assumption on the boundary) in Pardon, John, Concurrent normals to convex bodies and spaces of Morse functions, Math. Ann. 352, No. 1, 55-71 (2012). ZBL1242.52006. but a reviewer’s remark in zbMATH says that in this paper, the proof of the first theorem "does not seem to be quite correct". So my question is: What is the current status of this conjecture for $n=4$? REPLY [27 votes]: Let me address the specific complaint of that review. The situation is the following. Our (bounded, open) convex set is denoted $K\subseteq\mathbb R^n$ with closure $\overline K$, and we consider the "distance to $p$" function $d_p:\partial K\to\mathbb R$ for $p\in\overline K$. Let $V\subseteq\overline K$ be the set of $p\in\overline K$ for which $d_p$ has exactly one local minimum. I claimed in my paper that "it is clear that $V$ is closed". As the reviewer correctly points out, this is false (counterexample: $K$ the unit ball). But here is a corrected version: "if $d_p$ has finitely many local minima for every $p\in\overline K$, then $V$ is closed". Indeed, if $d_p$ has finitely many local minima, then if we perturb $p$ the number of local minima can only increase. This corrected version of the statement is sufficient for the proof given in the paper, since if $d_p$ has infinitely many local minima for some $p$, this exactly means that there are infinitely many normals to $\partial K$ which are concurrent at $p$. So the reviewer's remark doesn't invalidate the argument. I have to admit, this is not a well written paper, and I would be surprised if it did not contain a number of other equally badly presented arguments. I can say that I reread it a couple of years ago and was more or less convinced by the proof. However, I have not discussed it in detail with anyone.<|endoftext|> TITLE: What system suffices to show the strength of PRA is $\omega^\omega$? QUESTION [11 upvotes]: Russell O'Connor wrote in 2009 (link): PRA has consistency strength equivalent to the well-foundness of $\omega^\omega$, which can be stated again as the termination of some other program on all inputs. Presumably this equivalence is proved in a still weaker system. Is this true? What is the weakest system that suffices to prove that well-foundedness of $\omega^\omega$ implies consistency of PRA? And is it truly weaker than PRA? REPLY [10 votes]: First, this is not quite the right question. The implication as such is going to be provable in a trivial base theory; the most important question is what “well-foundedness of $\omega^\omega$” means in the statement. Well-foundedness is not directly expressible in the language of first-order arithmetic, it has to be approximated by the transfinite induction schema, and then the strength of the hypothesis is gauged by the complexity of formulas allowed in the induction schema. Having said that, let $\mathrm{PA}^-$ be the theory of nonnegative parts of discretely ordered rings (see e.g. Wikipedia), which has the proof theoretic strength of Robinson’s $Q$ (much weaker than PRA). By [1], $\mathrm{PA}^-$ is capable of basic sequence coding; let $L^+$ denote the usual language of arithmetic $(+,{\cdot},0,1,{\le})$ augmented with a function symbol $(w)_i$ for extracting the $i$th element of a sequence coded by $w$, and a relation symbol expressing “$w$ and $w'$ encode the same sequences, except possibly for the $0$th element”. Note that in the encoding scheme used in [1], fixed-length sequences $(x_0,\dots,x_k)$ are also computable by $L^+$-terms. Let $\mathrm{Open}^+$ denote the set of quantifier-free (= open) $L^+$-formulas, and let $\mathrm{Open}^+\text-\mathrm{TI}_{\omega^\omega}$ denote the transfinite induction schema $$\forall x\,\bigl(\forall y\,(y\prec x\to\phi(y))\to\phi(x)\bigr)\to\forall x\,\phi(x)$$ for $\phi\in\mathrm{Open}^+$. Here $\prec$ is the ordering relation of the natural definition of $\omega^\omega$ in arithmetic, whereby an ordinal with Cantor normal form $$\omega^mn_m+\omega^{m-1}n_{m-1}+\dots+\omega^0n_0$$ is represented by (a code of) the sequence $(n_0,\dots,n_m)$. Theorem: $\mathrm{PA}^-+\mathrm{Open}^+\text-\mathrm{TI}_{\omega^\omega}\vdash\mathrm{Con_{PRA}}$. In fact, $\mathrm{PA}^-+\mathrm{Open}^+\text-\mathrm{TI}_{\omega^\omega}\equiv I\Sigma_1+\Pi_1\text-\mathrm{TI}_{\omega^\omega}.$ The first assertion follows from the second, because $I\Sigma_1+\Pi_1\text-\mathrm{TI}_{\omega^\omega}$ is more than enough to comfortably formalize the standard proof of the consistency of PRA (or $I\Sigma_1$) by cut elimination, or alternatively, to carry out a model-theoretic proof of ordinal analysis, as presented in [2]. Now, to prove the second claim: first, since we can translate from natural numbers to the corresponding ordinals $<\omega$ and back by $L^+$-terms, it is easy to see that $\mathrm{PA}^-+\mathrm{Open}^+\text-\mathrm{TI}_{\omega^\omega}$ proves ordinary induction for $\mathrm{Open}^+$ formulas, and in particular, it includes the theory $\mathrm{IOpen}$. But we can do much better. The theory actually proves the least number principle for existential formulas, $L\exists_1$, and therefore also the induction schema for existential formulas, $I\exists_1$: to see this, let $\phi(x)$ be an existential formula of the form $$\exists y\,\theta(x,y)$$ where $\theta$ is open. Define an open formula $\psi(\alpha)$ so that $$\psi(\omega^1n+\omega^0m)\iff\theta(n,m).$$ By $\mathrm{TI}_{\omega^\omega}$ for the formula $\neg\psi$, or equivalently, the least element principle (wrt $\omega^\omega$) for $\psi$, if $\exists x\,\phi(x)$, then there exists a least $\alpha=\omega n+m$ such that $\psi(\alpha)$. Then $\phi(n)$, but $\neg\phi(n')$ for all $n' TITLE: Progress on Bondal–Orlov derived equivalence conjecture QUESTION [8 upvotes]: In their 1995 paper, Bondal and Orlov posed the following conjecture: If two smooth $n$-dimensional varieties $X$ and $Y$ are related by a flop, then their bounded derived categories of coherent sheaves are equivalent as triangulated categories, i.e. we have $D^b(\mathsf{Coh}(X)) \cong D^b(\mathsf{Coh}(Y))$. Selected 'particular' cases: Tom Bridgeland proved that this holds in the $n=3$ case by showing that birational smooth projective Calabi-Yau threefolds are derived equivalent. This follows from the fact that any birational transformation between two $3$-dimensional Calabi–Yau varieties can be decomposed into a sequence of flops. Ed Segal has also constructed an example in the $n=5$ case, and Daniel Halpern-Leistner has sketched a proof of the conjecture for the case of Calabi-Yau manifolds which are birationally equivalent to a moduli space of Gieseker semistable coherent sheaves (of some fixed primitive Mukai vector) on a K3 surface. As Sasha mentions in the comments, Yujiro Kawamata has also proved the conjecture in the toric case. There are also other cases in which the conjecture holds which I have not mentioned - thank you in advance for any comments/answers highlighting these. The general case? I have heard however that a proof of this conjecture in general seems rather far off at this moment in time. I am interested in whether any progress has been made with regards to the general case, and what approach/techniques may be involved in a potential proof? REPLY [6 votes]: A lot of work has been done to prove the Bondal-Orlov conjecture for (stratified)-Mukai flops. To quote a few names : Namikawa (Mukai flops), Kawamata (one stratified Mukai flop), Cautis and collab (all stratified Mukai flops type A), Halpern-Leistner (reproves Cautis and collab results with other techniques). In all these examples however, the bases of the contrated locus are "the same" on the two sides of the flop. The example exploited by Segal comes after all these examples. Abuaf indeed noticed it is an interesting example because the bases of the contracted locus are very different on the two sides of the flop. More such examples : https://arxiv.org/abs/1812.10688 Note also that Wierzba and Wisniewski proved that any two symplectic resolutions of the same symplectic singular four-dimensional variety are connected by a chain of Mukai flops. In particular, the Bondal-Orlov conjectures is true for symplectic flops in dimension 4 since we know it is true for Mukai-flops in dimension of type A.<|endoftext|> TITLE: Geometric realization of combinatorial self-duality in polytopes QUESTION [11 upvotes]: Let's say I have a combinatorially self-dual polytope $P\subseteq\Bbb R^d$, i.e., its face lattice is isomorphic to its dual (you reverse the direction of the lattice order). Question: Is it always possible to realize $P$ geometrically, so that $P$ and its polar $$P^\circ := \{x\in\Bbb R^d \mid \langle x,s\rangle \le 1 \text{ for all $s\in P$}\}$$ are geometrically related in some sense, e.g. via orthogonal, linear, affine or projective transformations? In other words: so that they are geometrically self-dual? I would prefer to have a realization of the duality under the weakest possible transformations (i.e. orthogonal), but I wonder whether more general are necessary. REPLY [3 votes]: Alathea Jensen defines "self-polar": Self-polar polytopes are convex polytopes that are equal to an orthogonal transformation of their polar sets. and writes some interesting things about self-polar polytopes here: https://arxiv.org/abs/1902.00784 Your questions are partially answered there: For all two-dimensional polytopes (a.k.a. polygons) we have (obviously): All self-dual polytopes have self-polar realizations. For all 3-dimensional polytopes the answer is a we have (less obviously, using Koebe-Andreev-Thurston): All self-dual polytopes have self-polar realizations. (This is Theorem 4.6) For arbitrary dimensions this question is still open: "Question 9.2. Does every self-dual polytope have a self-polar realization?" The paper does not consider affine or projective transformations in general, but does consider the notion "negatively self-polar".<|endoftext|> TITLE: Can an exotic diffeomorphism of the 4-ball change the isotopy class of an embedded surface? QUESTION [9 upvotes]: Let $W$ be $B^4$ or $S^3 \times I$. Let $Y$ be a properly embedded surface in $W$. Let $f : W \to W$ be a diffeomorphism which is the identity near $\partial W$. Very little is known about $\pi_0(\mbox{Diff($W$) rel $\partial W$})$ (see page 4 of this survey by Hatcher), so $f$ might not be isotopic to the identity (or, in the $S^3\times I$ case, isotopic to a standard twist diffeomorphism). Nevertheless, I believe it is true that $f(Y)$ is isotopic rel boundary to $Y$. In other words, exotic diffeomorphisms cannot do anything exotic to embedded surfaces. The proof I have in mind (based on conversations with an expert) uses a "push-pull" type argument (and, in the $S^3\times I$ case, straightening along an arc) to concentrate any potential exoticness of $f$ in the neighborhood of one or two points. We can take these points to be far from $Y$, and it follows that $f(Y)$ is isotopic to $Y$. Rather than write down the details of this proof, I would prefer to just cite a published result. So my question is: Is there a citable reference for the above claim that $f(Y)$ is isotopic rel boundary to $Y$? Is there a citable reference for the claim that diffeomorphisms of simple 4-manifolds (like $B^4$ or $S^3 \times I$) can be "tamed" in the complement of a finite number of points? REPLY [10 votes]: Isn't this relatively obvious for $W=B^4$ (i.e. it only took me several hours to realize it was trivial)? Isotope $Y$ to $Y'$ by an isotopy $g_t$ into a small collar neighborhood of $\partial W$ (which we can do by general position), then $f(Y')=Y'$ since you've assumed $f$ is the identity in a neighborhood of $\partial W$. But also $f(Y)$ is isotopic to $f(Y')=Y'$ by the conjugate isotopy $fg_tf^{-1}$, hence $f(Y)$ is isotopic to $Y$. For $S^3\times I$ one can probably also achieve this by the lightbulb trick, but I haven't thought it through carefully.<|endoftext|> TITLE: Background needed to understand modern research on knot homology theories QUESTION [5 upvotes]: I am a student of mathematics, and have some background in Algebraic Topology (Hatcher, Bott-Tu, Milnor-Stasheff), Differential Geometry (Lee, Kobayashi-Nomizu), Riemannian Geometry (Do Carmo), Symplectic Geometry (Ana Cannas da Silva) and Differential Topology (Hirsch, Minor (Morse Theory), Milnor (h-cobordism theorem)). I would like to learn Floer homology and Khovanov homology. Q1. Is my background sufficient to learn these topics right now? What are the standard first-level and second-level sources for these topics? Q2. At the research level, will I need to know any other areas to work on these topics? In particular, would I need to know algebraic geometry and to what extent? To give you an idea of my present knowledge, I currently know absolutely no algebraic geometry, and even my commutative algebra background has several gaps, especially in parts about DVRs. Q3. Once I finish books recommended in Q1, what are some good papers to start reading on these areas? I would greatly appreciate reasoning behind any comments that aren't strictly factual and are opinions of the writer. Thank you in advance! :) REPLY [4 votes]: I'm not an expert in Floer homology or Khovanov homology, but if that's your goal I don't think you need quite as wide a background as suggested in the other current answer (though admittedly that answer was written when the title of the question was much broader). For example, in the talks I've seen about these things, I don't think much in the way of spectra comes up, or even homotopy theory at the level of May. My more modest suggestion is first to acquaint yourself with some more classical knot theory. Rolfsen is a very good start for the really classical stuff, while there are now many places to learn about the more modern skein-type invariants. My personal recommendation would probably be Lickorish, but there are several good books and surveys now about that material. After that, some symplectic topology might be useful, though you say you've already got some of that. Beyond that background, I think that this is an area that does not yet have many secondary textbook sources, but there are very many survey articles and I think the primary material tends to be written fairly well compared to some other areas of mathematics. Just Googling "Introduction to Khovanov homology" and "Introduction to Floer homology" brings up a lot of surveys and lecture notes. I would suggest diving into that stuff and then referring outward once you hit something specific that you don't understand or want more background about. Based on what you wrote in your question, I think your broad topological background is already in pretty good shape to get started.<|endoftext|> TITLE: Can someone suggest a path to study Mordell-Weil theorem for someone studying on their own? QUESTION [7 upvotes]: So, I want to read the proof of Mordell-Weil theorem and so, I picked up the book 'Arithmetic of Elliptic Curves' by J. Silverman and J.S. Milne's Elliptic Curves book. But after going through both the books as well as Anthony Knapp's Elliptic curve book....I noticed up one thing, that Silverman takes way longer than Milne or Knapp to reach to Mordell-Weil theorem. My question is- since I can't read all three books at the same time, can someone point out the differences between the approaches taken by these three texts. I know that Milne's has used group cohomology to prove Mordell-Weil but looking at Silverman I don't think he has used the exact same approach. Also, what about Knapp's text? I'm self studying and for this summer, my goal is to read up the proof of a big theorem like Mordell-Weil. But looking at the different books is just spinning my mind. And if Milne's shorter o more readable than Silverman than I would maybe read from it and not Silverman which I'm reading through right now. In short, can someone also suggest me a path that I should follow to read the proof of Mordell-Weil? I don't want it to be unnecessarily long because at the moment, my focus is the big theorem(Mordell-Weil) and not other things, but I will of course come back later to read them according to the need. Thank a lot and please feel free to add appropriate tags as I'm not sure if I've added the correct ones. EDIT: Going through 'Rational points on elliptic curves' by Tate and Silverman, it also discusses Mordell-Weil Theorem in its chapter 3. I guess its proof is not as 'rigorous' as the one in Silverman's bookand is described with far less Algebraic Geometry that is the core of proofs in Silverman. Can someone also comment on the difference between two approaches? I mean, if 'Rational points....' also has a good proof then why do we need to explain everything in Algebraic-geometric language in Silverman's text? EDIT: I'm sorry if this question is not appropriate for this site as it's not exactly a research question but I posted it few days ago on math stack exchange and it has been unanswered there since. REPLY [9 votes]: I've just finished teaching a Master's course on elliptic curves, where we assume no knowledge of number fields and even avoid Galois theory as far as possible. This makes it hard to consider proving Mordell–Weil in full generality. However, the proof over the rational numbers in the case where the curve has a rational 2-torsion point is accessible without any sophisticated tools. I'd suggest understanding this proof first, even if you later want to understand the full proof. I like Cassels' treatment of this, though some might find his style a bit old-fashioned (so my students tell me). I've written my own notes based on Cassels, available here. Of course, Galois cohomology, and the extra tools from algebraic number theory needed for finiteness of the Selmer group in the general case, are excellent topics and I'd encourage you to learn them. But sometimes the structure of a proof is easier to see in a special case.<|endoftext|> TITLE: How dangerous are set-size assumptions? QUESTION [28 upvotes]: Many students' first introduction to the difference between classes and sets is in category theory, where we learn that some categories (such as the category of all sets) are class-sized but not set-sized. After working with such structures, we discover that it is still worthwhile to sometimes treat them as if they were set-sized. But we can't always do so, without running into contradictions. A natural solution to this problem is the "Axiom of Grothendieck Universes". We then work not with the category of all sets, but the category of sets inside some given universe. This type of framework has been used by many notable mathematicians and seems very ubiquitous. For instance, it was an assumption in my universal algebra textbook. I would venture to say that most modern set theorists look at such an axiom as a very tame assumption. My first question concerns the possible dangers of such an assumption, which are not often raised when first learning about this axiom. One obvious danger is that this axiom might lead to an inconsistency. In other words, ZFC might be consistent but ZFC+Universes might be inconsistent. I don't personally subscribe to this belief, but it is certainly a possibility (without further, even stronger, assumptions). What really concerns me is the possibility of another danger: Could such a system proves false things about the natural numbers? In other words, even if we assume that ZFC+Universes is consistent, could it be the case that it proves false arithmetic statements? The motivation for this question came from reading some of the work of Nik Weaver at this link, which argues for a conceptualist stance on mathematics. In particular, if we reject the axiom of power set, we are led to situations where all power sets of infinite sets are class-sized. Nik puts forth the idea that ZFC might prove false things about the natural numbers. Is this a real possibility? I suppose so, since it is possible that ZFC is inconsistent but the natural numbers aren't. But is it still possible even if ZFC is consistent? More generally: Could treating the power set of the natural numbers as a set-sized object, rather than a class, force us to conclude false arithmetic statements (even if such a system is consistent)? Even if the answer to this question is yes, I'm having a hard time seeing how we could recognize this fact, since according to the answers to this linked question it is difficult to state precisely what we mean by the natural numbers. A second motivation for this question comes from what I've read about the multiverse view of set-theory. When creating a (transitive) model $M$ of ZFC, the set $P(\mathbb{N})$ can often be thought of as some "bigger" power set of the natural numbers intersected with the model. Moreover, via forcing, it seems that one can (always?) enlarge the power set of $\mathbb{N}$. This does seem to suggest that $P(\mathbb{N})$ is not completely captured in any model. Thus, in its entirety, perhaps $P(\mathbb{N})$ should be treated as a class-sized object. Added: I believe that the current proof we have of Fermat's Last Theorem uses the existence of a(t least one) Grothendieck universe. However, my understanding is that this dependence can be completely removed due to the fact that Fermat's Last Theorem has small quantifier complexity. I imagine that proofs of statements with higher quantifier complexity that use Grothendieck universes, do not necessarily have a way of removing their dependence on said universes. How would we tell if such arithmetic statements are true of the natural numbers? [Some of this was incorrect, as pointed out by David Roberts and Timothy Chow.] 2nd addition: There are some theories that we believe prove false arithmetic statements. Assuming the natural numbers can consistently exist (which we do!), then both PA+Con(PA) and PA+$\neg$Con(PA) are consistent, but the second theory proves the false arithmetic sentence $\neg$Con(PA). My question then might be rephrased as: What principles lead us to believe that "Universes" is a safe assumption, whereas "$\neg$Con(PA)" is not safe, regarding what we believe is "true" arithmetic? (Next, repeat this question regarding the axiom of power set.) Is any theory that interprets PA "safe", as long as it is consistent with PA, and PA+Con(PA), and any such natural extension of these ideas? Another way of putting this might be as follows: Is the assumption Con(PA) a philosophical one, and not a mathematical one? This ties into my previous question that I linked to, about describing the "real" natural numbers. REPLY [11 votes]: This answer repeats some of the material in other answers but I think it is not entirely redundant. As you seem to have realized, the assertion that a theory is consistent is a much weaker statement than the assertion that the theory does not prove false arithmetical statements (or is "arithmetically sound", to use the standard terminology). So the answer to your question about whether it could be the case that ZFC + universes is consistent but not arithmetically sound is yes, but for perhaps a trivial reason: ZFC + universes could be consistent and yet ZFC itself could fail to be arithmetically sound. Probably what you really wanted to ask was, if ZFC is arithmetically sound, and ZFC + universes is consistent, does it follow that ZFC + universes is arithmetically sound? The answer is no. For example, it is conceivable that under these hypotheses, ZFC + universes could prove $\neg$Con(ZFC). (EDIT: This was a typo. I meant that ZFC+universes could prove $\neg$Con(ZFC+universes). Thanks to Noah Schweber for catching this—see the comments.) Any feeling that universes are a "safe" assumption must come from a direct assessment of the universes axiom itself. The reasons are similar to the reasons that we have for believing that ZFC is arithmetically sound. The ability to remove Grothendieck universes from the proof of Fermat's Last Theorem cannot be deduced from some simple abstract fact like "Fermat's Last Theorem has low quantifier complexity." No meta-theorem of this sort is known. Eliminability must come from careful examination of the nitty-gritty details of the proof. Your question about whether Con(PA) is philosophical or mathematical needs clarification. Offhand, it sounds like a false dichotomy to me, predicated on dubious assumptions about the meaning of the words "philosophical" and "mathematical." Can you define precisely what you mean by a "mathematical" assumption? Are you implicitly assuming that "mathematical" assumptions are rock-solid while "philosophical" ones are not? If so, that in itself is a dubious assumption IMO.<|endoftext|> TITLE: Stronger version of Besicovitch covering theorem QUESTION [6 upvotes]: I'm wondering if the following strengthening of the Besicovitch covering theorem holds: Suppose $A\subset\mathbb R^n$ is a bounded subset and suppose $x\mapsto r_x$ is a function $A\to(0,\infty)$. Is it possible to choose constants $0<\lambda<1$ and $N\in\mathbb N$ (both depending only on the dimension $n$) so that there exists a countable subset $C\subset A$ such that the following two statements hold? (1) the collection of balls $\{B(x,\lambda r_x)\}_{x\in C}$ covers $A$ and, (2) we have the pointwise inequality $\sum_{x\in C} \chi_{B(x,r_x)}\le N$ (where $\chi_E$ denotes the indicator function of a set $E$). In words, no point of $\mathbb R^n$ is contained in more than $N$ of the balls $B(x,r_x)$ (with $x\in C$). If we replaced $\lambda$ by $1$, this statement would be true as a consequence of the usual Besicovitch covering theorem. I'm somewhat stuck trying to modify that proof to get this stronger statement (I would just like $\lambda$ to be strictly smaller than $1$, it can be as close to $1$ as we like). Is this stronger version with some $0<\lambda<1$ true? If not, is there an easy counterexample? If it is true, a proof (or sketch/hint/reference) would be appreciated! If it helps to answer the question, I don't mind adding the additional assumption that $A$ is actually a compact set (i.e., its closed in addition to being bounded) and $x\mapsto r_x$ is a continuous (or even Lipschitz) function. REPLY [3 votes]: I think the statement is false. Consider $n = 1$. Let $A = \{x_i\}_{i \in \mathbb{N}}$ where $x_i = 2^{-i}$. Define $r_i$ as follows: if $k^2 \leq i < (k+1)^2$, let $r_i = 2^{-i} - 2^{-(k+1)^2}$. Notice that $x_{k^2}$ only belongs to $B(x_{k^2}, r_{k^2})$ and no other ball. So any cover of $A$ must include all of $B(x_{k^2}, \lambda r_{k^2})$. Let $\eta = \lceil-\log_2(1 - \lambda)\rceil$, for $\lambda < 1$. Note that $\eta \in \mathbb{N}$. Observe that $B(x_i, \lambda r_i) \subset B(x_i, \lambda x_i)$, which means that $x_{i + \eta} \not\in B(x_i, \lambda r_i)$. Now, suppose we have a cover of $A$. It must include the point $x_{k^2}$. The argument above implies that $x_{k^2+\eta}$ is not included. By construction if $x_{k^2 + \eta}$ belong to $B(x_i, \lambda r_i)$ for some $i$, then $i \leq k^2 + \eta$. This implies that $x_{k^2 + 2\eta}$ cannot belong to whichever ball that covers $x_{k^2+\eta}$. This argument can be iterated $\sigma$ times, where $\sigma$ is the biggest integer such that $k^2 + \sigma \eta < (k+1)^2$, or that $\sigma = \lfloor (2k+1) / \eta\rfloor$. Therefore for any fixed $\lambda < 1$, choosing a starting $k$ sufficiently large we see that the cover for $A$ must include at least $\lfloor (2k+1) / \eta \rfloor$ many $x_i$ with $i$ between $k^2$ and $k^2 + 1$. The corresponding balls for all these $x_i$ have non-empty total intersection, on which set the sum of the characteristic function is at least $\sigma$, which can be made arbitrarily large. Compactness of $A$ will not help. I can add to the set $A$ the origin which we label as $x_\infty$. For any fixed $\lambda, N$ I can choose a sufficiently large $k$ such that the corresponding $\sigma > N$. Then if I take the corresponding $r_\infty \ll 2^{-(k+2)^2}$ we would have a counterexample with a compact $A$. As $A$ is discrete, obviously $x_i \mapsto r_i$ is continuous; even in the compact case we can modify the definitions of $r_i$ for $i > (k+2)^2$ where $k$ is the critical value above so that the radius function is continuous at $x_\infty$.<|endoftext|> TITLE: Is the category of racks semi-abelian? QUESTION [7 upvotes]: I wonder whether the category of (pointed) racks is semi-abelian. Any comments and references would be appreciated. REPLY [13 votes]: The category $\mathbf{Rack}$ of racks is Barr-exact since it is a variety of universal algebras, but it is not protomodular. Indeed, the category of sets is equivalent to the category of racks satisfying the identity $a\triangleleft b =a$, so it is a full epireflective subcategory of $\mathbf{Rack}$. In particular, there is an inclusion functor $\mathbf{Set}\to \mathbf{Rack}$ which preserves limits and reflects isomorphisms; then if $\mathbf{Rack}$ was protomodular $\mathbf{Set}$ would also be protomodular, which is false. I found this argument in the paper "A Galois-Theoretic Approach to the Covering Theory of Quandles" by Valérian Even. It also shows that $\mathbf{Rack}$ cannot be Mal'tsev, or even congruence-permutable.<|endoftext|> TITLE: A coincidence between the Lambert cube, Lobell polyhedron, and hyperbolic 3-manifolds? QUESTION [7 upvotes]: I. Lambert cube $\mathfrak L(\alpha_1,\alpha_2,\alpha_3)$ In this paper (p.8), we find the volume $V$ of the hyperbolic Lambert cube for the special case $\alpha=\alpha_1 = \alpha_2 = \alpha_3$ as $$V_\alpha = \int_x^\infty \ln\left(\frac1{t^2}\left(\frac{t^2-\tan^2\alpha}{1+\tan^2\alpha}\right)^3\right)\frac{dt}{t^2+1}$$ where $x$ is the positive root of $x^4-(1+3\tan^2\alpha)x^2-\tan^6\alpha =0$. Hence, $$\begin{array}{|c|c|c|} \hline \alpha &x& V_{\alpha} & 8V_{\alpha}\\ \hline \pi/3 &\left(\frac{1+\sqrt{13}}2\right)^{3/2}& 0.324423 & \color{brown}{2.59538}\\ \pi/4 &\left(\frac{1+\sqrt{5}}2\right)^{3/2}& 0.538275 & \color{blue}{4.30620}\\ \pi/5 &\left(\frac{5-\sqrt{5}}2\right)^{3/2}& 0.658081 & 5.26465\\ \hline \end{array}$$ II. Lobell polyhedron $L(n)$ In this paper (p.33), we find the volume (using a different formula) of $L(5)$ as $$\begin{array}{|c|c|c|} \hline n & L(n) & V_n\\ \hline 5 & L(5) & \color{blue}{4.30620}\\ 6 & L(6) & 6.02304\\ \hline \end{array}$$ III. Closed hyperbolic 3-manifolds On a hunch, I checked those volumes in the Hodgson & Weeks census and found, $$\begin{array}{|c|c|c|} \hline \text{Dehn filling}& \text{Symmetry} & \text{Volume}\\ \hline m160(-3, 2) & D6 & \color{brown}{2.59538}\\ \hline s648(-5, 1) & D4 & \color{blue}{4.30620}\\ s921(-3, 1) & D2 & 4.30620\\ m400( 4, 1) & Z/2 & 4.30620\\ \hline \end{array}$$ IV. Questions Why are the volumes for $\alpha =\pi/3, \pi/4$ and $L(5)$ found in the census? Conversely, why for $\alpha =\pi/5$ and $L(6)$ are they NOT present? REPLY [5 votes]: Let me make a remark about the first question and then try to answer the second. For the first question, using snappy's arithmetic computations (available when using snappy as a sage module), it looks like m400(4,1), s648(-5,1), s921(-3,1) are likely 60 fold covers of the quotient of $H^3$ by an arithmetic tetrahedral group $\Gamma(3,3,5,2,2,2)=\langle x,y,z| x^3,y^5,z^3, (yz^{-1})^2,(zx^{-1})^2,(xy^{-1})^2\rangle$ (see Maclachlan, Colin; Reid, Alan W., The arithmetic of hyperbolic 3-manifolds, Graduate Texts in Mathematics. 219. New York, NY: Springer. xiii, 463 p. (2003). ZBL1025.57001. (Especially pages 144, 415 for more information on tetrahedral groups.) For an exact representation of this group, you can use Grant Lakeland's notes available from his website: Matrix group computations. It's possible this group (or its index 2 supergroup $\Gamma(3,2,5,2,2,4)$) is the symmetries of the dodecahedron corresponding to $L(5)$, but I don't know if that is written down anywhere. For the second question, $\alpha=\pi/5$ and $(6)$ are likely not in the Hodgson-Weeks census because they are too complicated. The Hodgson-Weeks census is a set of 11,031 fillings of ideal triangulations with 7 or fewer tetrahedra such that (experimentally) the injectivity radius is sufficiently large (I think above .3). So if $L(6)$ can only be obtained from filling an ideal triangulation with 8 or more tetrahedra then it won't appear here. This appears to be the case. Or more precisely, as the geodesic cut-off doesn't seem to be a restriction for $L(6)$, it's seems the answer is that nothing with a fundamental domain with a volume like that of $L(6)$ can be obtained from filling a cusped manifold with 7 or fewer tetrahedra. I would conjecture that the same could be said for $\alpha=\pi/5$, but I don't know what its shortest geodesic is in that case.<|endoftext|> TITLE: Lie algebras with unique invariant scalar product QUESTION [6 upvotes]: Every 1-dimensional or simple complex Lie algebra admits an invariant, symmetric and non-degenerate bilinear form. This form is unique up to multiplication by a nonzero constant (which in Yang-Mills theories plays the role of the coupling constant). Are there other Lie algebras with such a unique scalar product? REPLY [3 votes]: I. Bajo and S. Benayadi already gave a complete answer for your question in their paper entitled "Lie algebras admitting a unique quadratic structure" in 1997, there you can find much more detail.<|endoftext|> TITLE: Algorithm to sample from unknown probability distribution, given its projections? QUESTION [5 upvotes]: The Cramér–Wold theorem states that, if $X$ is a random variable living in $\mathbb{R^d}$, then the distributions of all one-dimensional projections of $X$ uniquely determine the distribution of $X$. But can this be made algorithmic? I have an algorithm that approximately finds the distribution of any one-dimensional projection of a certain random variable $X$. Can I use this algorithm to efficiently approximately sample from $X$? If yes, then -- hopefully this is not asking too much -- are there error bounds on how good the sampling is? Ideally I would want a statement similar to "If one can approximate the distribution of any projection of $X$ to within $\ell_1$ distance at most $\varepsilon$ from its true distribution, then one can efficiently sample from a distribution with at most $\ell_1$ distance [some function of $d$ and $\varepsilon$] from $X$." REPLY [5 votes]: This is too long for a comment. As noted above, this is Radon inversion problem, which can be reduced to the Fourier inversion problem. As is noted in Wikipedia, the inverse problem is ill-posed, which roughly speaking means that no error bounds of the type you want are possible. Let me try and give a simple explanation why. Suppose we know the density of $X\cdot v$ for any unit-norm vector norm $v$, with $L_2(\mathbb{R})$ error $\varepsilon$. This means that we know its Fourier transform $t\mapsto\mathbb{E}(\exp(it(X\cdot v)))$, also with $L_2(\mathbb{R})$ error $\varepsilon$ (maybe up to $2\pi$ factor). This means that we know the d-dimensional Fourier transform $\hat{X}:v\mapsto \mathbb{E}(\exp(i(X\cdot v)))$ with an error of order $\varepsilon$ in the weighted $L^2(\mathbb{R}^d,\omega)$ space, with weight $\omega(v)=1/|v|^{d-1}$ (the Jacobian factor coming from passing to spherical coordinates). But this can correspond to arbitrarily large error in the usual $L_2(\mathbb{R}^d)$ sense, if the error is on high frequencies. When we recover $X$ from its Fourier transform, this arbitrarily large error is preserved. Update: The following argument gives an easy error bound involving Wasserstein distance. Let $\hat{Y}$ be the Fourier transform recovered from the data, and $\hat{X}$ the actual Fourier transform, so that $||\hat{X}-\hat{Y}||_{L^2(\mathbb{R}^d,\omega)}\leq c_d \varepsilon$. Multiply $\hat{Y}$ by $\exp(-\delta|v|^2/2)$ for a small $\delta$ and take the inverse Fourier transform. What we will get is $f_{X+\sqrt{\delta}Z}+f_{\text{err}}$, where $f_{X+\sqrt{\delta}Z}$ is the density of $X+\sqrt{\delta}Z$, with $Z$ standard Gaussian, and $f_{\text{err}}$ is the inverse Fourier transform of $\exp(-\delta|v|^2/2)(\hat{X}-\hat{Y})$. Now, $X+\sqrt{\delta}Z$ is within $C\sqrt{\delta}$ in the $L_1$ Wasserstein distance from $X$. On the other hand, the maximum of $\exp(-\delta|v|^2)|v|^{d-1}$ is of order $\delta^{-(d-1)/2}$. Therefore, $$||e^{-\frac{\delta|v|^2}{2}}(\hat{X}-\hat{Y})||_{L_2(\mathbb{R}^d)}\leq C\delta^{(1-d)/4}||(\hat{X}-\hat{Y})||_{L_2(\mathbb{R}^d,\omega)}\leq C\delta^{(1-d)/4}\varepsilon.$$ So, if we know the densities of all the projections up to $L_2$ error $\varepsilon$, then we can recover a distribution that is within $C\delta^{(1-d)/4}\varepsilon$ in $L_2$ distance from a distribution that is within $C\cdot \sqrt{\delta}$ in the $L_1$ Wasserstein distance from the actual distribution, where $\delta$ is in our disposal.<|endoftext|> TITLE: Vanishing of certain coefficients coming from Coxeter groups QUESTION [6 upvotes]: Let $\left(W\text{, }S\right)$ be a Coxeter system. For every $w\in W$ let us write $\left|w\right|$ for the length of $w$. Set $\lambda\left(e\right)=1$ where $e\in W$ denotes the neutral element of the group and define coefficients $\lambda\left(w\right)\in\mathbb{Z}$ recursively via \begin{eqnarray} \sum_{v\in W \text{: }\left|v^{-1}w\right|=\left|w\right|-\left|v\right|}\lambda\left(v\right)=0 \end{eqnarray} for any $w\in W$. For example this gives us $\lambda\left(e\right)=1$, $\lambda\left(s\right)=\left(-1\right)$ for all $s\in S$, $\lambda\left(st\right)=1$ if $m_{st}=2$ and $\lambda\left(st\right)=0$ if $m_{st}\neq2$ (for the notation see Wikipedia) for all $s\text{, }t\in S$ with $s\neq t$, and so on. My question is: Does there always exist some $l\in\mathbb{N}$ such that $\lambda\left(w\right)=0$ for all $w\in W$ with $\left|w\right|\geq l$? In the case that $\left(W\text{, }S\right)$ is right-angled (i.e. $m_{st}\in\left\{ 2\text{, }\infty\right\}$ for all $s\text{, }t\in S$ with $s\neq t$) this is true and we can take $l=\left|S\right|$. I'm wondering if in the general case this property also holds. If no, is it possible to characterize the property in an instructive way? REPLY [8 votes]: The answer is yes. I claim there are at most $2^n$ elements of $W$ for which $\lambda(w) \neq 0$. Specifically, let $\vee$ be the join in weak order, which is a semi-lattice. If $\{ s_1, s_2, \ldots, s_j \} \subseteq S$ and $s_1 \vee s_2 \vee \cdots \vee s_j$ is defined, then I claim that $\lambda(s_1 \vee s_2 \vee \cdots \vee s_j)=(-1)^j$, and otherwise I claim that $\lambda(w)=0$. Thus $N$ can be taken to be the greatest length of $s_1 \vee s_2 \vee \cdots \vee s_j$, restricting ourselves to cases where this join is defined. The condition $|v^{-1} w| = |w| - |v|$ says that $v \leq w$ in weak order. Therefore, this is the recursion defining the Mobius function of weak order. The Mobius function of a finite lattice $L$ can be computed by Rota's crosscut theorem. (The first online reference I could find was Theorem 1.3 here.) Namely, let $A$ be the set of minimal elements of $L$ -- in this case, this is the set $S$. Then $$\mu(x) = \sum_{B \subseteq A,\ \bigvee B = x} (-1)^{|B|}.$$ So $\mu(w)=0$ if $w$ is not a join of elements of $S$. If $w$ is a join of elements of $S$, then it is so in only one way (this is a way that weak order is simpler than a general lattice) so we get the description from the first paragraph.<|endoftext|> TITLE: When does $BG \to BA$ loop to a homomorphism? QUESTION [8 upvotes]: If I have a compact connected Lie group $G$ and a (relatively nice) simply-connected topological abelian group $A$, when is it the case that a given $f\colon BG \to BA$ loops to a (continuous) homomorphism $G\to A$? This seems like it can't always be true, and also should be some kind of classical result, but I'm not sure where to look. REPLY [13 votes]: If $G$ is a compact connected topological group and $A$ is a locally compact abelian topological group, then for any map $f:BG\to BA$ the looped map $\Omega f:\Omega BG\to \Omega BA$ is homotopically equivalent to a homomorphism $\phi:G\to A$. This follows from the main result of Scheffer, Wladimiro, Maps between topological groups that are homotopic to homomorphisms, Proc. Am. Math. Soc. 33, 562-567 (1972). ZBL0236.22008. I'm not sure if this answers your question, however, which seems to be about when $f$ is in the image of the classifying space functor $B:\mathsf{TopGrp}\to\mathsf{Top}$.<|endoftext|> TITLE: An expansion from Ramanujan related to birthday problem QUESTION [6 upvotes]: A friend designed a drinking game with a lucky wheel of 30 distinct icons. When playing, each one takes turn to spin the wheel, and write down the items until the first one who gets the item that has already appeared then he drinks. And he asked what the average (expectation) of turns is. The formula is not so difficult to get but it is quite unpractical to compute the value. So I was thinking about an asymptotic formula. Luckily, I found this here: See section 5.6, "average number of people". My question is how to prove the Ramanujan's formula given there, $$ Q(M)=\sum_{k=1}^{M} \frac{M!}{(M-k)! M^k}\sim\sqrt{\frac{\pi M}{2}}-\frac{1}{3}+\frac{1}{12}\sqrt{\frac{\pi}{2M}}-\frac{4}{135M}+\cdots$$ Any good reference available? REPLY [8 votes]: A crystal clear derivation using only elementary calculus is given by Donald Knuth on page 116-121 of The Art of Computer Programming (volume 1). I reproduce the first and the last paragraphs of Knuth's exposition: REPLY [7 votes]: We have $$Q(n)=\sum_{k=0}^{n-1} (1-1/n)(1-2/n)\ldots(1-k/n).$$ Write $1-x/n=e^{-x/n-x^2/(2n^2)-\ldots}$, then $$Q(n)=\sum_{k=0}^{n-1} \exp\left(-\frac{k(k+1)}{2n}-\frac{k(k+1)(2k+1)}{12n^2}-\ldots\right).$$ If $k\geqslant n^{1/2+0.00001}$, the corresponding summands are super-polynomially small and do not rely on the asymptotics. For $k TITLE: Co-spectral fractional isomorphic graphs with different Laplacian spectrum QUESTION [7 upvotes]: I am looking for two undirected graphs $G$ and $H$ of the same order (i.e., they have the same number of vertices) such that $G$ and $H$ are cospectral (i.e., their adjacency matrices $A_G$ and $A_H$ have the same multiset of eigenvalues) fractional isomorphic (i.e., there exists a doubly stochastic matrix $S$ such that $A_G\cdot S=S\cdot A_H$, or equivalentlty, $G$ and $H$ have a common equitable partition) but $G$ and $H$ are not cospectral with regards to their Laplacians (signed or otherwise) or Seidel matrix. The context for this question can be found in an earlier post Orthogonal similarity of adjacency matrices of graphs which are cospectral and have a common equitable partition. REPLY [2 votes]: EDIT: According to your link, if two graphs are cospectral with a common equitable partition, then they have cospectral complements. But this implies that they have cospectral Seidel matrices (see Theorem 3 in the first reference from your link). So the best you can do is to not be cospectral with respect to the Laplacian nor signless Laplacian. The two graphs below satisfy this. The equitable partitions are according to the degrees. Verification If $G$ is the top graph and $H$ the bottom graph: Eigenvalues$(G)$=Eigenvalues$(H)\approx\{-2,-2,-1,-1,-0.73205081,1,1,2,2.73205081\}$ Their common coarsest equitable partitions are $\{\{2,3,8\},\{0,1,4,5,6,7\}\}$ for $G$ and $\{\{1,3,8\},\{0,2,4,5,6,7\}\}$ for $H$. Eigenvalues$(L_G)\approx\{4.73\times 10^{-16},1, 1.26, 1.26,3, 4, 4, 4.73,4.73\}$ and Eigenvalues$(L_H)\approx \{-2.08\times 10^{-16},0.585,1.26,2,3,3.41,4,4.73, 5\}$ So $G$ and $H$ indeed satisfy the required conditions.<|endoftext|> TITLE: Formal character of local cohomology groups with support in Schubert cells QUESTION [6 upvotes]: Let $k$ be a field of characteristic zero, $G$ a connected semi-simple algebraic group over $k$ and $B$ a fixed Borel subgroup of $G$ with maximal torus $T$. Also denote by $W$ the Weyl-group of $G$. Let $X$ be the (complete) flag variety of $G$ dimension $n$, hence we can assume that $X=G/B$. Furthermore let $x \in X$ be the unique fixed point of $B$ under the natural left action of $G$ on $X$. Then denote for $w \in W$ by $X_w=Bwx$ the corresponding schubert cell in $X$ with dimension $l(w)$. According to p. 51 in Brylinskis paper "Differential operators on the flag varieties" we have that the formal character of $H^{n-l(w)}_{X_w}(X, \mathscr{O})$ is given by $\frac{e^{-w(\rho)-\rho}}{\prod_{\alpha \in R(B)}(1-e^{-\alpha})}$. Here $R(B)$ is the set the roots of $B$, hence the positive roots of $G$ with respect to $B$ and $\rho$ is the half sum of all positive roots of $G$ with respect to $B$. Finally he concludes that $H^{n-l(w)}_{X_w}(X, \mathscr{O})$ has highest weight $-w(\rho)-\rho$. He refers to Kempfs "The Grothendieck-Cousin Complex of an Induced Representation" for a proof of this result. But when I look it up there, concretely Lemma 12.8, I would rather say that the formal character of $H^{n-l(w)}_{X_w}(X, \mathscr{O})$ is given by $\frac{e^{w(\rho)+\rho}}{\prod_{\alpha \in R(B)}(1-e^{\alpha})}$. So the sign changed everywhere. Hence it would have lowest weight $w(\rho)+\rho$. Can anyone solve my confusion? REPLY [2 votes]: The formula 12.8 in Kempf follows by plugging in 6.5 into 11.10 (see comment by CJS under the question). In the statement of 11.10 Kempf defines $[\chi]$ to denote the isomorphism class of $V(\chi).$ Now $V(\chi)$ is the dual of $\textbf{V}(\chi)$ (see page 313 or 380) and the torus $T$ acts on $\textbf{V}(\chi)$ by $(g,v) \mapsto \chi(g) v$ (see p. 313) which makes the weight/character of $V(\chi)$ to be equal to $-\chi.$ This reconciles Kempf's result/notation with Brylinski's statement.<|endoftext|> TITLE: Importance of the principal bundle in Chern-Simons theory QUESTION [11 upvotes]: This is a very basic beginners question about Chern-Simons theory. The configurations that we sum over to get the partition function are given by a Lie-algebra valued 1-form $A$ on a topological 3-manifold, called the connection. More precisely, $A$ is a connection of a principal Lie-group bundle over the manifold. What is the role of this principal bundle? I didn't find this spelled out explicitly anywhere, but I assume one does not sum over different choices of principle bundles when calculating the partition function? Or are different choices of principle bundle already encoded in different choices of $A$? So is Chern-Simons theory not only a topological field theory that one can put on any topological manifold, but one that we can additionally put on any topological manifold with a Lie-group principle bundle? Can one restrict to trivial principle bundles without loosing any of the physical interpretation? If one talks about the "3-manifold invariants" of the Chern-Simons theory, does that refer to the partition function on the trivial principle bundle over those manifolds? Or is the partition function independent of the choice of bundle? REPLY [11 votes]: In quantum Chern-Simons theory with gauge group $G$ (compact Lie), a field on a 3-manifold $M$ is a principal $G$-bundle with a connection $A$. The partition function/path integral associated to $M$ is supposed to be the integral over the (generally infinite-dimensional) space ("stack") $\mathcal{F}$ of all principal $G$-bundles $P$ over $M$ equipped with a choice of connection $A$ (modulo gauge equivalence) of $\exp(iS(A))$, where $S$ is the classical Chern-Simons action. Perhaps it is easier for pedagogical purposes to explore the simpler case of Chern-Simons theory when $G$ is finite; this is Dijkgraaf-Witten theory. In this case, one can make the notion of an integral over $\mathcal{F}$ perfectly rigorous. If $G$ is finite, then each bundle has a unique connection, so $\mathcal{F}$ is precisely $[M, BG]$, where $BG$ is the classifying space of $G$. The Lie algebra of $G$ is also trivial, so the action vanishes, and we're left with integrating (summing) $1$ over the space $[M,BG]$ with respect to some measure; the weight of a principal $G$-bundle $P$ in this case is $1/|\mathrm{Aut}(P)|$. In other words, the finite-group version of the Chern-Simons partition function is $$Z(M) = \sum_{P\in [M,BG]} \frac{1}{|\mathrm{Aut}(P)|}.$$ This is the literal analogue of Chern-Simons theory for a finite gauge group; however, a more interesting analogue is twisted Dijkgraaf-Witten theory, which might be what you were trying to get at. Recall that the action associated to the field $(P,A)$ over $M$ is $S(A) = \int_M q(A)$; what matters is that the integrand $q(A)$ is a $3$-form. Since a Chern-Simons field on $M$ is a principal $G$-bundle with a choice of connection $A$, one might attempt to "canonically" associate to each principal $G$-bundle a $3$-form on $M$ in the finite group case; this $3$-form would be the replacement of $A$. Note that in this story, $A$ plays a somewhat different role. Think of the $3$-form as a singular cochain on $M$, so integration is pairing the cochain with the fundamental class of the manifold $M$ (assume it's closed and oriented). Since a field on our manifold is still a principal $G$-bundle, determined by a map $f_P:M\to BG$, we can associate to each bundle a $3$-dimensional cohomology class if we fix a choice of $\alpha\in \mathrm{H}^3(BG;\mathbf{C}^\times)$; then, the $3$-form "$q(A)$" associated to $P$ is $f_P^\ast(\alpha)\in \mathrm{H}^3(M;\mathbf{C}^\times)$. (Really, one should fix a $3$-cocycle in $Z^3(BG;\mathbf{C}^\times)$.) The action associated to $P$ is then $\langle [M], f_P^\ast(\alpha)\rangle$, and to obtain the quantum theory, one can now integrate over the space of all $G$-bundles (with the same measure as the untwisted case). Note that $A$ by itself doesn't appear in this story; only the analogue of the associated $3$-form does.<|endoftext|> TITLE: Smooth Julia set for quadratic polynomials QUESTION [7 upvotes]: This question is related to a classification of rational maps in terms of properties of their Julia set. Let $f= z^2 + c$, with $c\in \mathbb{C}$ be a quadratic polynomial such that its Julia set $J(f)$ is connected. Q1: If there exists a relatively open set of $J(f)$ that is (the support of) a smooth curve. Is $f$ conjugate (resp equal) to a Tchebychev polynomial or power map $z^2$? Q2: Is the answer to Q1 yes under the additional assumption that $J(f)$ is also locally connected? Q3: If the answer to Q1 is no, can one describe the set of such counterexamples in terms of the parameter $c$? Thanks a lot. REPLY [7 votes]: The answer to a) is yes, and this was proved by Fatou in 1919. Sur les équations fonctionnelles Bulletin de la S. M. F., tome 48 (1920), p. 208-314. There are many generalizations of this fact. For one generalization, and further references you may look to Meromorphic functions with linearly distributed values and Julia sets of rational functions, Proc. AMS. 137 (2009), 2329-2333.<|endoftext|> TITLE: Does ZF+AD have any unusual arithmetic consequences? QUESTION [10 upvotes]: Motivation: This question is motivated by wondering to what extent "natural" theories are linearly ordered (or at least ordered in a directed manner) by their (first-order) arithmetic consequences, in analogy to the phenomenon that "natural" theories seem to be linearly ordered by consistency strength. In an answer to this question, Joel David Hamkins gave several examples illustrating how these two hierarchies -- arithmetic consequence versus consistency strength - may differ, and indeed his examples also illustrate that the ordering by arithmetic consequence is not linear, even for arguably "natural" theories. But it so happens that these examples largely involved playing games with consistency statements, so the departure from linearity feels "small". One way of making this a bit more precise is that it seems, as far as I can tell, to still be possible that the partial ordering of theories is still directed, at least for "natural" theories. That is, what I know about the matter is consistent with the idea that there really is one "true arithmetic" with which all of our "natural" or "serious" theories (apart from things like $T + \neg Con(T)$ for various $T$) are consistent. In order to have a hope of challenging this view, it seems one needs an alternate "coherent" picture of the mathematical universe. The only example that comes to mind for me is $ZF + AD$, which is inconsistent with ZFC, but nonetheless its consistency strength is well-calibrated (and nontrivial), and seems to have some sort of "inner logic" to it which could potentially yield a different picture of the world even at the level of arithmetic. AD paints a different picture of the universe at least if we include non-arithmetic statements in our scope. But I'm not sure it does anything unusual at the level of arithmetic. Questions: Are the arithmetic consequences of $ZF+AD$ consistent with standard theories like $ZFC + L$ for various large cardinal axioms $L$? Are the arithmetic consequences of $ZF+AD$ implied by large cardinals? More broadly, is there any good candidate out there for a theory $T$ whose arithmetic consequences are inconsistent with (or at least not implied by) ZFC + large cardinals, such that $T$ has some kind of "inner coherence" (in analogy to how large cardinals are said to have "inner coherence" by virtue of inner model theory -- as opposed to theories of the form $T + \neg Con(T)$ and the like)? REPLY [4 votes]: I just want to record here a few things I've learned lately concerning the motivational question about the empirical linearity of "plausible" theories of arithmetic. This empirical fact -- that all "plausible" theories of (first-order -- or even second-order) arithmetic appear to be linearly (or at least directed-ly) ordered by their direct implications -- seems to be a pretty famous fact among logicians. I recently came across some further discussion of it in Woodin's Notices article. To see why one might expect this, one can contemplate the $\omega$-rule, i.e. the (infinitary) deduction rule which says that if you can deduce $\phi(n)$ for all numerals $n$, then you can deduce $\forall x(\phi(x))$ (where we work in the language of first-order arithmetic). It's easy to show that when first-order logic is augmented with the $\omega$-rule, the axioms of $PA$ (or even $Q$) generate a complete theory (though I found it surprisingly hard to find a source bothering to both (i) spell out what the $\omega$-rule is and (ii) explicitly state that it leads to $PA$ being complete!), and that indeed this theory coincides with true arithmetic. Therefore, any theory $T$ in the language of arithmetic extending $PA$ (or even $Q$) must either be a weakening of true arithmetic, or else fail to validate the $\omega$-rule. In particular, if $T$ is complete [1], then $T$ must be $\omega$-inconsistent, i.e. there must be a formula $\varphi(x)$ such that $T \vdash \varphi(n)$ for each numeral $n$, and yet $T \vdash \exists x (\neg \varphi(x))$. Consequently, if $T$ is any theory in the language of arithmetic extending $PA$ (or even $Q$), then either $T$ is a weakening of true arithmetic, or else in every model of $T$ there is an element which is not a numeral (i.e. not of the form $1 + 1 + \dots + 1$). This seems like a pretty good reason to think that any such $T$ is "implausible". At any rate, it means that "alternate arithmetic" can't be thought of as just an alternate way of assigning truth values to formulas with the same underlying set of numbers as we have in true arithmetic -- any alternate arithmetic can have only nonstandard models, with new elements which are not numerals. Of course, this argument requires our metatheory to reassure us that there is a complete theory of "true arithmetic". I'm no expert, but the only way out I can see to entertain the possibility of truly "plausible" alternate theories of arithmetic to exist is if one changes one's metatheoretical assumptions. It seems to me that as long as one assumes classical logic in the metatheory, one has to concede that there is a complete theory of true arithmetic, and I don't see anywhere where choice is used in showing that the $\omega$-rule completes $PA$. (Some form of choice is required to ensure that every consistent theory $T$ of the sort considered above has a model, but if $T$ doesn't have a model, then it again looks "implausible".) So even if the metatheory is $ZF$, I think one will still be led to the conclusion that only true arithmetic is "plausible". Probably, then, one needs to weaken at least to an intuitionistic metatheory in order to contemplate the possibility of "plausible" alternatives in the theory of arithmetic. [1] An earlier version of this post made this claim without the hypothesis that $T$ be complete. Thanks to Emil Jerabek in the comments below for pointing out the error.<|endoftext|> TITLE: Existence of radial limits of products of certain power series and $1-x$ QUESTION [6 upvotes]: Let $V$ be an arbitrary set of infinitely many positive integers, and let: $$\varsigma_{V}\left(z\right)\overset{\textrm{def}}{=}\sum_{v\in V}z^{v}$$ Let $T_{V}$ denote the set of all $t\in\left[0,1\right)$ for which the limit: $$c_{V}\left(t\right)\overset{\textrm{def}}{=}\lim_{x\uparrow1}\left(1-x\right)\varsigma_{V}\left(e^{2\pi it}x\right)$$ exists. Suppose $0\in{T_{V}}$ and $c_{V}\left(0\right)\neq0$ (that is, $V$ has a positive, well-defined natural density). Are there then any $t\in\left(0,1\right)$ which must be in $T_{V}$? More generally, what, if anything, can be said about $T_{V}$ and/or $\left[0,1\right)\backslash T_{V}$? (Ideally, all of $\mathbb{Q}$ (or all but some exceptional set) would have to be a subset of $T_{V}$.) Conversely, what (if anything) does $\mathbb{Q}\subseteq T_{V}$ then force to be true about $V$? In particular, if $\varsigma_{V}\left(z\right)$ is a rational function, then $\mathbb{Q}\subseteq T_{V}$ necessarily holds. Thus, another way of phrasing the question is: does the set $\left\{ \varsigma_{V}\left(z\right):T_{V}\supseteq\mathbb{Q}\right\}$ contain functions which are not rational, and, if such functions do exist, what can be said about them and their associated $V$s? REPLY [3 votes]: As for probabilistic approaches... let's just say they're not my cup of tea Well, that certainly has to be fixed if one "thinks of himself as an analyst". Let me give you a quick overview of the things most relevant to your current project as a series of exercises. 1) Read https://en.wikipedia.org/wiki/Hoeffding%27s_inequality 2) Look up the proof of the Bernstein inequality $$ \|P'\|_\infty\le Cn\|P\|_\infty $$ for trigonometric polynomials $P(t)=\sum_{k=-n}^n c_ke^{ikt}$ of degree $n$. You do not need the sharp value of $C$. 3) Derive from it that $\|P\|_\infty\le C\max_{k=0}^{100n}|P(e^{(2\pi i k)/(100n)}|$ 4) Using Hoeffding and 3), show that if $\xi_j$ are mean zero independent Bernoulli that are $p_j-1$ with probability $p_j$ and $p_j$ with probability $(1-p_j)$, then $$ P(\|\frac 1N\sum_{j=1}^N\xi_jz^j\|_\infty\ge N^{-0.1})\le N^{-0.1} $$ for large $N$. 5) Using Borel-Cantelli and the fact that Chesaro means of a bounded sequence can be read from any subsequence $N_k$ with $N_{k+1}/N_k\to 1$ as $k\to\infty$, show that with probability $1$, we have $$ \lim_{N\to\infty}\frac 1N \sum_{j=1}^N\xi_jz^j=0 $$ for all $z\in\mathbb T$ simultaneously. 6) Consider now the random set $V$ to which each particular integer $j$ belongs independently with probability $a_j\in[0,1]$. Show that with probability $1$, for every $t$, $$ \lim_{N\to\infty}\frac 1N\left|\sum_{v\in V, v\le N}z^v-\sum_{v=1}^N a_vz^v\right|=0 $$ Thus, it does not matter if you consider sums over sets or arbitrary sums with coefficients in $[0,1]$, but with the latter you have way more freedom in various constructions. For instance, 7) Construct a series that gives you a non-zero limit at $0$, $z$, $\bar z$ but $0$ limit everywhere else. 8) Construct a series that gives you non-zero limits at any prescribed self-conjugate set of "rational points" containing $t=0$ and zero limits everywhere else. Etc.<|endoftext|> TITLE: Can the defining rep of $E_7$ split over a finite subgroup while the adjoint remains simple? QUESTION [7 upvotes]: Does the (simply connected compact) Lie group $E_7$ contain a finite subgroup $G \subset E_7$ such that the $56$-dimensional irrep of $E_7$ splits over $G$ as $28 \oplus \overline{28}$, but the $133$-dimensional adjoint representation remains simple when restricted to $G$? By "$28 \oplus \overline{28}$" I mean of course a $28$-dimensional complex irrep plus its dual. Such a subgroup is definitely Lie primitive (meaning it doesn't fit inside a proper Lie subgroup $L \subset E_7$), since an imprimitive subgroup would split $133 = \mathfrak{l} \oplus (\dots)$. According to Griess and Ryba - Finite simple groups which projectively embed in an exceptional Lie group are classified!, of the (quasi)simple subgroups of $E_7$, only three are primitive, and none of them work. ($133$ splits over $SL_2(29)$ and $SL_2(37)$, and $56$ remains simple over $PSU_3(8)$.) But I don't know about nonsimple subgroups. This feels like a good homework problem, but in fact came up in my research: such a subgroup would allow me to build a superconformal field theory with some nice properties. REPLY [4 votes]: I know this is an old question, but I can answer it in the negative. First, I have mostly completed a list of the Lie primitive subgroups of $E_7(k)$ for all $k$, including $k=\mathbb C$, and there is no such example. But even before that, a paper of Liebeck and Seitz, entitled 'Subgroups of exceptional algebraic groups which are irreducible on an adjoint or minimal module', appearing in J. Group Theory, shows that only simple subgroups act irreducibly on the adjoint. In fact, for $p=5$ there are examples, namely the sporadic groups $M_{22}$, $HS$ and $Ru$. But there is none over $\mathbb{C}$.<|endoftext|> TITLE: For what graph does the following algebraic property hold? QUESTION [5 upvotes]: Let $G=(V,E)$ be a simple graph. My question: For what graph $G$, does there exist a permutation $\sigma$ on $V$ such that $$\prod_{uv\in E}(x_{\sigma(u)}-x_{\sigma(v)})=-\prod_{uv\in E}(x_u-x_v)?$$ For example, when $G$ is the complete graph $K_n$, any transposition on $V$ satisfies the above property. Is there any known result about my problem? Any idea is welcome! REPLY [3 votes]: I don't know about the literature on your specific problem but I know this arises as a trivial subquestion of a more general one for which there is literature, most of it from the 19th century. Let $P_G(x)$ denote your polynomial. If the graph is $k$-regular then you can make a homogenized version $$ HP_G(x,y)= \prod_{uv\in E}(x_uy_v-x_vy_u) $$ and then a symmetrized version of the latter $SHP_G$ where you sum over permutations of the pairs $(x_u,y_u)$. What you get is an $SL_2$ invariant of binary forms. In fact you can do this identification in two ways: 1) by seeing the pairs as homogeneous roots and so the invariant is of degree $k$ in the coefficients of a binary form of degree $|V|$, 2) by seeing the pairs as symbolic letters and so the invariant is of degree $|V|$ in the coefficients of a binary form of degree $k$. The precise relationship between the two interpretations is the isomorphism of $SL_2(\mathbb{C})$ modules $$ {\rm Sym}^{k}({\rm Sym}^{|V|}(\mathbb{C}^2))\simeq {\rm Sym}^{|V|}({\rm Sym}^{|k|}(\mathbb{C}^2)) $$ known as Hermite reciprocity. This also works for nonregular graphs by completing the polynomial with extra factors $(x_u\times 0 - x_v\times 1)$ involving the "difference with respect to the point at infinity", in which case you get a covariant instead of an invariant. The above Hermite reciprocity covers this more general case too. Now you see there is a composition of maps: $$ {\rm graph}\ G\ \rightarrow\ SHP_G\ \rightarrow {\rm invariant\ or\ covariant} $$ The second is injective and also surjective if one takes linear spans. The first map however is completely mysterious and highly noninjective. This poses the fundamental question (that every classical invariant theorist has confronted at some point): How to tell if $SHP_G=0$ simply by looking at the graph? The simplest reason for which $SHP_G=0$ is when there exists a permutation as in this MO question. However, this is not the only possibility, as shown by the (multi)graph with $$ P_G=(x_1-x_2)^2(x_1-x_3)(x_2-x_3)\ . $$ For pointers to the literature on the wider question I formulated above, see my answer to the MO question Symmetric polynomial from graphs and in particular the reference to the papers by Sabidussi.<|endoftext|> TITLE: Are the logarithms of the integer polynomials discrete in $L^1$ of the unit circle? QUESTION [13 upvotes]: Tautologically, the integer polynomials form a discrete set in $L^1$ of the unit circle. On the other hand, a set of logarithms ordered by norm becomes generally rather denser than the original set. Is the set $$ \big\{ \log{|P|} \, : \, P \in \mathbb{Z}[X] \setminus \{0\} \big\} \subset L^1(\mathbb{T}) $$ of functions on the complex unit circle $\mathbb{T} = \{ z \mid |z| = 1 \}$ discrete in $L^1$, or does it have an accumulation point? I am equally happy with the $L^2$ norm, if it makes a difference. REPLY [9 votes]: It is still discrete though not uniformly. Since $\log|P|=\frac 12\log|p|$ where $p=P\bar P$ is a real non-negative trigonometric polynomial with integer coefficients, it is enough to work with $p$ instead of $P$. We have $$ |\log p-\log q|\ge \frac{|p-q|}{\max(p,q)}. $$ Now consider the outer function $f$ with $|f|=\frac 1{\max(p,q)}$, so $$ |f(0)| = \exp\Big(-\int_{\mathbb T}\log{\max(p,q)}\Big) \ge \exp\Big(-\int_{\mathbb T}[|\log p|+|\log q|]\Big). $$ Then, denoting by $r$ the difference $p-q$ multiplied by an appropriate power of $z$ so that $r(0)\in\mathbb Z\setminus\{0\}$ and $r$ is analytic, we get $$ \int_{\mathbb T}|\log p-\log q|\ge \int_{\mathbb T}\frac{|p-q|}{\max(p,q)} =\int_{\mathbb T}|r||f|\ge |r(0)||f(0)| \\ \ge \exp\Big(-\int_{\mathbb T}[|\log p|+|\log q|]\Big)\,, $$ finishing the story.<|endoftext|> TITLE: Definition of geometric monodromy QUESTION [5 upvotes]: Consider a polynomial $f \in \mathbb C[x_1,\dots ,x_n]$. An atypical value of $f$ is a complex number about which $f:\mathbb C^n\to \mathbb C$ is not a topological fiber bundle. Writing $\mathrm{Atyp}(f)$ for the atypical values of $f$ we thus have a fiber bundle $$\mathbb C^n\setminus f^{-1}(\mathrm{Atyp}(f))\to \mathbb C\setminus \mathrm{Atyp}(f).$$ A book I'm reading now says the following. The fundamental group $\pi_1(\mathbb C\setminus \mathrm{Atyp}(f))$ acts therefore on the generic fiber $f^{-1}(t_0)$ up to isotopy. The image of this action is called the geometric monodromy group. The pullback of $f$ along an admissible loop (image doesn't hit atypical values) is a fiber bundle which yields an automorphism of the fiber, called the geometric monodromy. I am confused by this. We have for any Hurewicz fibration $E\to B$ the transport functor $\pi_1(B)\to \mathsf{hTop}$ which lands in the homotopy category. This is the only meaning I can think of for "acts up to isotopy". On the other hand, this does not give an automorphism of the fiber in the topological category. Question. Is the geometric monodromy defined as the homotopy type of the continuous map between fibers given by the homotopy lifting property? Is the geometric monodromy group(oid) given by the image $\pi_1(B)$ in $\mathsf{hTop}$? REPLY [3 votes]: As you say, a Hurewicz fibration defines a map from $\pi_1(B)$ to the automorphisms of the fiber in the homotopy category. But a topological fiber bundle has more structure than a Hurewicz fibration. In fact, a topological fiber bundle defines a map from $\pi_1(B)$ to the mapping class group of the fiber (i.e. the component group of the group of topological automorphisms of the fiber). This is what "acts up to isotopy" means. To see this, maybe one way is to observe that the set of pairs of a point in $B$ and an isomorphism between the fiber over that point and $F$ is an an $\operatorname{Aut}(F)$-torsor, and modding out by the identity component gives a $\pi_0(\operatorname{Aut}(F))$-torsor, which gives a homomorphism $\pi_1(B) \to \pi_0(\operatorname{Aut}(F))$.<|endoftext|> TITLE: Finite index subgroup of $\mathrm{GL}_n(\Bbb C)$ and Chevalley groups QUESTION [7 upvotes]: I'm trying to show that if $G$ is a Chevalley group, then every finite index subgroup of $G(\Bbb Z)$ is Zariski dense in $G(\Bbb C)$. ($G(\Bbb Z)$ is the Chevalley group over $\Bbb Z$ and similarly for $G(\Bbb C)$) But I'm struggling to understand some basic stuff, It seems to me that there can't be finite index subgroup in $G(\Bbb C)$ how can I show that? also is there a good source that covers Chevalley groups over $\Bbb Z$? REPLY [5 votes]: (Essentially copied from comments) It's just Borel-Harish-Chandra + Borel's density theorem (+ definition of Chevalley group). More precisely, $G$ is a $\mathbf{Q}$-defined group without nontrivial rational characters (since it has no nontrivial character at all), so Borel-Harish-Chandra implies $G(\mathbf{Z})$ is a lattice in $G(\mathbf{R})$, and hence so are its finite index subgroups. Then Borel's density theorem says that $G(\mathbf{Z})$ is Zariski-dense in $G(\mathbf{R})$ (and hence in $G(\mathbf{C})$ by Rosenlicht and connectedness of $G$). As regards the question about finite index subgroups: this argument probably appears several times on this site: any connected real Lie group has no proper finite index subgroup, i.e., each homomorphism to a finite group is trivial: this follows from being generated by 1-parameter subgroups (which satisfy the given property, by divisibility).<|endoftext|> TITLE: Independence number of $4$-uniform regular hypergraph QUESTION [7 upvotes]: Let $H$ be a $4$-uniform hypergraph on $[1..n]$, i.e. $H$ is a collection of $4$-element subsets of $[1..n]$. The elements of $H$ are called edges. A hypergraph is regular if every element of $[1..n]$ are in the same number of edges. An independent set $I$ of $H$ is a subset of $[1..n]$ such that $I$ does not contain any edge. The independence number of $H$ is the maximal cardinality among independent sets of $H$. Question: Does there exist two positive numbers $c$, $d$ such that every regular 4-uniform hypergraph on $[1..n]$ with size $ TITLE: Derivative of an algebraic power series in positive characteristic QUESTION [11 upvotes]: Let $K$ be a field. It is easy to see that if the characteristic of $K$ is $0$ and $f(T)=\sum_{n\ge0}a_nT^n$ is a power series algebraic over $K(T)$, then $f'$ belongs to $K(T)(f)$. Indeed let $P(X)=\sum_{i=0}^lP_i(T)X^i$ be the minimal polynomial of $f$ over $K(T)$. By differentiating $P(f(T))$, one has $$Q(f(T))+f'(T)R(f(T))=0$$ with $Q(X)=\sum_{i=0}^lP_i'(T)X^i$ and $ R(X)=\sum_{i=1}^niP_i(T)X^{i-1}$. Since $R(f(T))$ can not be zero (otherwise $R$ would vanish $f$ with a smaller degree than $P$) one has $$f'(T)=Q(f(T))/R(f(T))$$ But this proof does not work if the characteristic of $K$ is positive ($R(f(T))$ could be zero). So here is my question. Does $f'$ still belong to $K(T)(f)$ if the characteristic of $K$ is positive? REPLY [6 votes]: Let $p=\operatorname{char}(K)$. By minimality of $P$, if $R(f(T))$ was zero, then $R(X)=0$. Thus, $p$ divides the degree of each nonzero coefficient of $P(X)$, so $P$ is not separable. But the extension $K((T))/K(T)$ is separable, see: Why is $K_{\upsilon}|K$ separable for a global field $K$? So this situation cannot occur, and your argument goes through.<|endoftext|> TITLE: Does the 3875-dimensional rep of $E_8$ have a solution to $x\star x=0$? QUESTION [22 upvotes]: Consider the compact Lie group $E_8$. Its second-smallest fundamental representation is $3875$-dimensional and admits a symmetric invariant form, and so is real: $E_8 \curvearrowright \mathbb{R}^{3875}$. Furthermore, this irrep admits a unique (up to scale) $E_8$-invariant symmetric $3$-tensor, studied for example in Garibaldi and Guralnik, Simple groups stabilizing polynomials, 2015. Using this inner product, I can think of this invariant tensor as a commutative but non-associative multiplication $\star : \mathbb{R}^{3875} \otimes \mathbb{R}^{3875} \to \mathbb{R}^{3875}$. Question: Does $\mathbb{R}^{3875}$ contain any nonzero vectors $x$ such that $x \star x = 0$ for this multiplication? Let me mention two ways that one could try to answer this question. I wasn't able to carry either out to completion, and there might be other approaches. First, pick a random vector $y \in \mathbb{R}^{3875}$, and consider the symmetric 2-tensor $x_1 \otimes x_2 \mapsto \langle x_1 \star x_2, y\rangle$, where of course $\langle,\rangle$ denotes the $E_8$-invariant inner product. This 2-tensor is the symmetric bilinear form corresponding to $\| x\|^2 = \langle x \star x, y\rangle$. Suppose that you had access to a multiplication table for $\star$. Then you could write down this inner product, and diagonalize it — diagonalizing an inner product is fast on the computer — and see if there are any null vectors. If you for some $y$ this inner product is definite, then there are no solutions, and if on the other hand a couple different $y$s have the same null vector, then probably there is a solution. However, I was unable to build a multiplication table for $\star$. Note that it would suffice to write down a set of generators for the $\mathrm{Lie}(E_8)$-action on $\mathrm{Sym}^3(\mathbb{R}^{3875})$, since finding a common eigenvalue is pretty fast, and for that, it would suffice to write down generators for the action on $\mathbb{R}^{3875}$, which is to say it would suffice to construct a crystal basis. But my computer timed out when I asked it to do that. Second, consider the cubic function $f(x) = \langle x \star x, x\rangle$ corresponding to the symmetric $3$-tensor. A solution to $x\star x = 0$ is the same as a critical point of $f$. We may restrict to the unit sphere $S = S^{3874} \subset \mathbb{R}^{3875}$; then a solution to $x\star x = 0$ is the same as a critical point of $f|_S$ at which $f$ vanishes. One could hope that perhaps $f$ is a Morse–Bott function on $S$. It definitely is not Morse because it is $E_8$-invariant, and I expect but haven't proved that $\mathrm{Lie}(E_8)$ acts freely on $S$. Furthermore, one could hope that the critical points at which $f$ vanishes have the same number of attracting and repelling directions — $1813 = (3874 - 248)/2$ of each. Finally, one could hope that perhaps the cells in this Morse(–Bott) complex carry some symplectic or complex structure forcing them to be even-dimensional? This happens for example for flag manifolds. One should be careful a bit: $E_8$, and hence the groupoid $S/E_8$, has torsion in its homology, which is consistent with even-dimensional cells and freeness of the $\mathrm{Lie}(E_8)$-action only if the stabilizers are nontrivial finite groups. Conversely, perhaps $S/E_8$ has so much homology that there must be a degree-$1813$ critical point. REPLY [17 votes]: Yes. The basic representation of $E_8$ has character $j(\tau)^{1/3} = q^{-1/3}(1+248q+4124q^2 + \cdots)$, and the 4124 decomposes as $1+248+3875$. By Frenkel-Kac-Segal, the basic representation has an $E_8$-lattice vertex algebra structure. The tensor on the 3875-dimensional subspace of Virasoro-primary vectors comes from restricting and projecting the $-\cdot_1-$-product on the 4124-dimensional weight 2 subspace, so it suffices to produce a primary vector $v$ such that $v_1v = 0$. The $E_8$-lattice vertex algebra has a grading by the $E_8$-lattice, and the weight 2 subspace has lattice-degree supported by the lattice vectors of norm at most 4. Let $v$ be a nonzero element of weight 2 that is homogeneous with respect to lattice-grading, and whose lattice-degree has norm 4. Then $v$ is primary, and $v_1v = 0$ because it has lattice-degree of norm 16. This construction works over the standard self-dual $\mathbb{Z}$-form as well as the $\mathbb{R}$-form you consider.<|endoftext|> TITLE: Creating many big sets of small numbers QUESTION [6 upvotes]: There are $n$ numbers $a_1,\ldots,a_n\in [0,1]$. Their sum is $\sum_{i=1}^n a_i = s$, where $s$ is some integer. We want to group them into sets so that the sum of each set is at least $t$, where $t$ is some integer. Let $F(n,s,t)$ be the largest number of sets that we can always create (for any $a_i$). What is $F(n,s,t)$? Example. $F(n=8,s=7,t=1)=4$: Proof that $F(8,7,1)\geq 4$: We can always create 4 sets by dividing the $8$ numbers arbitrarily into $4$ pairs. The sum of each pair is at most $2$, and the sum of all pairs is $7$, so the sum of each pair is at least $1$. Proof that $F(8,7,1)\leq 4$: We cannot always create 5 sets. Suppose for all $i$, $a_i=7/8$. In any $5$ sets, at least one set is a singleton so its sum is less than $1$. Similarly, whenever $n$ is even, $F(n,n-1,1)=n/2$. What else is known on the function $F$? Currently I am particularly interested in the case $t=2$, but I will be happy for any more general references. UPPER BOUND: $F(n,s,t)\leq \lfloor {s+1\over t+1}\rfloor$. Proof. Suppose that $s+1$ numbers equal $s/(s+1)$ and the other $n-s-1$ numbers equal $0$. To create a set with sum at least $t$, we need $t+1$ nonzeros. So we can create at most $\lfloor {s+1\over t+1}\rfloor$ such sets. REPLY [5 votes]: For $n\in\mathbb N$ and $s,t\in\mathbb R$ with $0\lt t\le s\le n$, let $F(n,s,t)$ be the greatest integer $m$ such that any family of $n$ numbers $a_1,\dots,a_n\in[0,1]$ with $a_1+\cdots+a_n=s$ can be partitioned into $m$ subfamilies, each with sum $\ge t$. Lemma 1. If $k\in\mathbb N$ and $s\le k\le n$, then $F(n,s,t)\le\left\lfloor\frac k{\lceil kt/s\rceil}\right\rfloor$. Lemma 2. If $n\gt s$ then $F(n,s,t)\le\left\lfloor\frac{\lfloor s+1\rfloor}{\lfloor t+1\rfloor}\right\rfloor$. Proof. Put $k=\lfloor s+1\rfloor$ in Lemma 1. Lemma 3. $F(n,s,t)\ge\left\lfloor\frac{s+1}{t+1}\right\rfloor$. Proof. Let $m=\left\lfloor\frac{s+1}{t+1}\right\rfloor\lt s+1$, so that $t\le\frac{s+1}m-1=\frac sm-\frac{m-1}m$. We may assume that $m\ge2$. Lat $a_1,\dots,a_n\in[0,1]$ be given, $a_1+\cdots+a_n=s$. For notational convenience we assume that $a_1,\dots,a_p\gt0$ while $a_{p+1}=\cdots=a_n=0$. Partition the interval $[0,s]$ into $m$ equal subintervals $J_1,\dots,J_m$, indexed from left to right; that is, $J_i=[c_{i-1},c_i]$ where $c_i=\frac{is}m$. Then $|J_i|=\frac sm\gt1-\frac1m$. Also partition $[0,s]$ into subintervals $A_1,\dots,A_p$ of respective lengths $|A_i|=a_i$. Let $\mathcal A=\{A_1,\dots,A_p\}$. Each interval $A\in\mathcal A$ will be assigned to at most one of the intervals $J_1,\dots,J_m$, and (some of) the numbers $a_1,\dots,a_p$ will be assigned correspondingly to $m$ groups. Namely, an interval $A\in\mathcal A$ is assigned to the interval $J_i=[c_{i-1},c_i]$ if it satisfies one of the following three conditions: $$A\subseteq J_i;$$ $$i\gt1,\ \ c_{i-1}\in A,\ \ \frac{|A\cap J_i|}{|A|}\gt\frac{i-1}m;$$ $$i\lt m,\ \ c_i\in A,\ \ \frac{|A\cap J_i|}{|A|}\gt\frac{m-i}m.$$ It is important to note that no interval $A\in\mathcal A$ is assigned to more than one $J_i$. Now the set of intervals assigned to $J_i$ covers $J_i$, except possibly for an interval at the left of length $\le\frac{i-1}m|A|\le\frac{i-1}m$, and an interval at the right of length $\le\frac{m-i}m|A|\le\frac{m-i}m$. Therefore, the sum of the lengths of intervals assigned to $J_i$ is $\ge\frac sm-\frac{i-1}m-\frac{m-i}m=\frac sm-\frac{m-1}m\ge t$. Theorem. If $t\in\mathbb N$ and $n\gt s$, then $F(n,s,t)=\left\lfloor\frac{s+1}{t+1}\right\rfloor$. Proof. Lemmas 2 and 3.<|endoftext|> TITLE: What if $\mathbb{R}$ is in bijection with the cardinals less than $\frak{c}$? QUESTION [6 upvotes]: I was wondering whether it is consistent to have $\frak{c} = \aleph_{\frak{c}}$ where $\frak{c} = 2^{\aleph_0}$ is the cardinality of the reals (over ZFC). If so, what interesting consequences of this statement are known (besides ¬CH)? I was curious about this because in some sense $\frak{c}$ is the largest possible number of cardinals below $\frak{c}$, and this is partly motivated by the idea that $\frak{c}$ may be so large as to be 'unreachable' via approximation by fewer smaller cardinals, which seems similar in nature to an opinion of Cohen on CH. From what I have read, I think that it is consistent (relative to ZFC) for $\frak{c}$ to be the $ω_1$-th fixed-point of $\aleph$, which would be one possibility satisfying $\frak{c} = \aleph_{\frak{c}}$. But can $\frak{c}$ be the $\frak{c}$-th fixed-point of $\aleph$, and does this yield even more interesting consequences? REPLY [8 votes]: Yes. Start with a model of $\sf CH$, then take the least fixed point with uncountable cofinality. Call that $\kappa$. Now add $\kappa$ Cohen reals. Since fixed points form a club of ordinals, you can iterate the fixed points enumeration. Repeat that $\omega_1$ times, then take the least one of cofinality $\omega_1$ in that club. Now call that $\kappa$, and add that many Cohen reals. Assuming no large cardinals get involved, that means that $\frak c$ is singular. This by itself implies that Martin's Axiom fails, and that Cichon's diagram is not trivial, since some of the cardinal characteristics are provably regular. Other than that, I don't believe we can say a lot more without adding more assumptions on the universe.<|endoftext|> TITLE: Variations on Kaplansky Density QUESTION [5 upvotes]: Let $A$ be a $C^*$-algebra and $\pi:A\rightarrow B(H)$ a faithful $*$-representation, so $M=\pi(A)''$ is a von Neumann algebra and $A\rightarrow M$ is an inclusion. von Neumann's Bicommutant Theorem tells us that $A=\pi(A)$ is weak$^*$-dense in $M$, and the Kaplansky Density Theorem says that further, the unit ball of $A$ is weak$^*$-dense in that of $M$. Suppose now I have $a\in A$ and $k\geq 0$ fixed, and there is $x\in M$ with $$ \|a-x\|\leq k, \quad \|x\|\leq 1. $$ Is $x$ is the weak$^*$-closure of the set $\{ b\in A : \|a-b\|\leq k, \|b\|\leq 1 \}$? If $a=0$ this is just Kaplansky density. Let's weaken this, and just ask: is there $b\in A$ with $\|a-b\|\leq k$ and $\|b\|\leq 1$? This follows from just the triangle-inequality, because it is easy to see that $$ \inf\{k\geq 0 : \exists b, \|a-b\|\leq k, \|b\|\leq 1\} = \max(0, \|a\|-1). $$ So if $\|a-x\|\leq k$ and $\|x\|\leq 1$ then $\|a\|\leq k+1$ and so $\|a\|-1\leq k$. So, let's make this new problem harder. Let $a_0\in A^+$ (so $a_0$ is positive: this is motivated by other considerations) and ask the following: Suppose there is $x\in M$ with $\|a-a_0x\|\leq k$ and $\|x\|\leq 1$. Is there $b\in A$ with $\|a-a_0b\|\leq k$ and $\|b\|\leq 1$? One could also consider more general maps $T:A\rightarrow A$ which extend to $M\rightarrow M$; here $T(b) = a_0b$. REPLY [4 votes]: The answer to the first question is YES (and probably the second as well ). First of all, we may replace the original representation $\pi(A)\subset B(H)$ with a universal representation $\pi\oplus\sigma$, by replacing $x \in \pi(A)''$ with $x\oplus \frac{1}{k+1}\sigma(a) \in A^{**}$. Then, Kaplansky's density theorem is upgraded to the following (which is an easy consequence of the Hahn--Banach separation theorem). Lemma 1: Let $z \in A^{**}$, $w \in A$, and a net $(z_i)_i$ in $A$. If $\| z - w \| \le 1$ and $z_i \to z$ weak*, then $$\lim_j\mathrm{dist}(w,\mathrm{conv}\{ z_i : i \geq j\})\le1$$ The advantage of using convex combination is that it can be iterated without destroying the previously obtained approximation estimate. From Lemma 1, one immediately obtains Lemma 2: Let $x\in A^{**}$ and $a\in A$ be such that $\|x\|\le1$ and $\| x - a \|\le k$. Then for any $\epsilon_1>0$, the element $x$ is weak*-approximated by $y_1\in A$ such that $\|y_1\|\le 1+\epsilon_1$ and $\| y_1 - a \|\le k+\epsilon_1$. We are done once we show the approximant $y_1$ in Lemma 2 is norm-close to an element that satisfies the exact norm inequalities: Lemma 3: Let $y_1 \in A$ and $a \in A$ be such that $\|y_1\|\le1+\epsilon_1$, $\|y_1-a\|\le k+\epsilon_1$, and $\|a\|\le k+1$. Then, there is $y\in A$ such that $\|y\|\le1$, $\|y-a\|\le k$, and $\|y-y_1\|\approx_{\epsilon_1}0$. Here $\approx_{\epsilon_1}$ means that the difference is at most $h(\epsilon_1)$ for some explicit continuous function $h\geq0$ such that $h(0)=0$. Now Lemma 3 is proved by iterating the following approximate version and finding a suitable convergence sequence $(y_n)_n$: Lemma 4: Let $y_1 \in A$ and $a \in A$ be such that $\|y_1\|\le1+\epsilon_1$, $\|y_1-a\|\le k+\epsilon_1$, and $\|a\|\le k+1$. Then, for any $\epsilon_2>0$, there is $y_2\in A$ such that $\|y_2\|\le1+\epsilon_2$, $\|y_2-a\|\le k+\epsilon_2$, and $\|y_2-y_1\|\approx_{\epsilon_1}0$. Proof of Lemma 4: By Lemma 1, it suffices to find $y_2$ in $A^{**}$ (as opposed to in $A$). Put $\alpha=\beta=(2\epsilon_1)^{1/2}\approx_{\epsilon_1}0$. Let $a=v|a|$ be the polar decomposition, $p:=1_{[k+1-\alpha,k+1]}(|a|)$, and $q:=vpv^*$. Since $ap\approx_{\epsilon_1}(k+1)vp$, $\|y_1p\|\approx_{\epsilon_1}1$, $\| y_1p - ap \|\approx_{\epsilon_1}k$, and $vp$ is a partial isometry, one has $y_1p \approx_{\epsilon_1}vp$ and $y_1p \approx_{\epsilon_1} qy_1$. Thus for $a':=ap^\perp=q^\perp a p^\perp$ (which has $\|a'\|\le k+1-\alpha$) and $$y_2:= qvp + q^\perp((1-\beta)\frac{y_1}{\|y_1\|}+\beta\frac{a'}{\|a'\|})p^\perp$$ one has $\| y_2 \|\le 1$ and $y_2\approx_{\epsilon_1}y_1$. Moreover, since \begin{align*} \|q^\perp(y_2-a)p^\perp\|&\le\|y_1-\frac{y_1}{\|y_1\|}\|+\|(1-\beta)q^\perp y_1p^\perp+\frac{\beta}{\|a'\|}a' - a'\|\\ &\le \epsilon_1+(1-\beta)(k+\epsilon_1)+\beta(\|a'\|-1)\\ &\le k+2\epsilon_1-\alpha\beta = k, \end{align*} one has $\|y_2-a\|\le k$ (assuming $k>\alpha$).<|endoftext|> TITLE: (Higher) extensions of mixed Hodge structures QUESTION [5 upvotes]: Mixed Hodge structure is introduced by Deligne and it's very useful for studying complex algebraic varieties. We know $\text{MHS}$, the category of mixed Hodge structures is an abelian category. Where can I find a good reference for the computation of the (higher) extension groups inside MHS (and it's variations)? Carlson's paper "Extensions of mixed Hodge structures" is helpful, but I hope a more general reference. For instance, let $\text{MHS}^+_{\mathbb R}$ denote the category of mixed Hodge structures equipped with an involution $\phi_{\infty}$ preserving the weight filtration and such that $\phi_{\infty} \otimes c$ preserves the Hodge filtration (here $c$ means complex conjugation). Is there no $\operatorname{Ext}^{2}$ in this category? REPLY [6 votes]: Beilinson, Notes on absolute Hodge cohomology is another reference. To answer your last question, the cohomological dimension of the category of mixed Hodge structures is $1$, i.e. there is no $Ext^i$ for $i>1$. See 1.10 of Beilinson's paper.<|endoftext|> TITLE: Does Anosov geodesic flow imply asphericity? QUESTION [6 upvotes]: Let $(M, g)$ be a closed smooth Riemannian manifolds with Anosov geodesic flow, does it implies that $M$ is an aspherical manifold? I am thinking it is not known yet? REPLY [12 votes]: W. Klingenberg in [Riemannian Manifolds With Geodesic Flow of Anosov Type, Annals of Mathematics, Vol. 99, No. 1, 1974, pp. 1-13] proved among other things that a closed Riemannian manifold with Anosov geodesic flow does not have conjugate points, and hence the exponential map at any point is a covering map. In particular, the manifold is aspherical.<|endoftext|> TITLE: Computing the sum of an infinite series as a variant of a geometric series QUESTION [11 upvotes]: I came across the following series when computing the covariance of a transform of a bivariate Gaussian random vector via Hermite polynomials and Mehler's expansion: $$ S = \sum_{n=1}^{\infty} \frac{\rho^n}{n^{1/6}} $$ for $\vert \rho \vert < 1$. We know that $S$ must be finite and satisfy $$ S \le \rho (1-\rho)^{-1} $$ since the original series is dominated by $\sum_{n=1}^{\infty} \rho^n$. However, there is a catch if we use for $S$ the upper bound $\rho (1-\rho)^{-1}$, which tends to $\infty$ when $\rho \to 1-$. This happens when the two marginal random variables in the Gaussian vector are almost surely, positively linearly dependent (asymptotically). So, the target is to obtain a good upper bound, much better than $\rho (1-\rho)^{-1}$ when we restrict $\rho$ to be away from $1$, to reduce the effect of $\rho \to 1-$. In other words, let $1-\rho = \delta$ for some fixed $\delta \in (0,1)$, what is a better upper bound for $S$? Because of the scaling term $n^{-1/6}$ that induces a divergent series $\sum_{n=1}^{\infty} n^{-1/6}$, probably not much improvement should be expected. I have Googled but did not find an illuminating technique for this. Any pointer or help is appreciated. Thank you. REPLY [13 votes]: Let $r:=\rho$, $S(r):=S$, and $a:=1/6$. Let us show that \begin{equation*} S(r)< Cr(1-r)^{a-1}\quad\text{and}\quad S(r)\sim C(1-r)^{a-1},\quad\text{where}\quad C:=\Gamma(1-a). \tag{0} \end{equation*} Everywhere here, the equalities and inequalities are for $r\in(0,1)$, and the limit relations are for $r\uparrow1$. Indeed, we have \begin{equation*} \frac1{n^a}=\frac1{\Gamma(a)}\,\int_0^\infty u^{a-1}e^{-nu}\,du \end{equation*} and hence \begin{equation*} S(r)=\frac1{\Gamma(a)}\,\int_0^\infty du\, u^{a-1}\,\sum_{n=1}^\infty r^n e^{-nu} =\frac r{\Gamma(a)}\,I(r), \end{equation*} where \begin{align*} I(r)&:=\int_0^\infty du\, \frac{u^{a-1} e^{-u}}{1-r e^{-u}} \\ & =\int_0^\infty du\, \frac{u^{a-1}}{e^u-r} \\ & <\int_0^\infty du\, \frac{u^{a-1}}{1+u-r} \\ & =(1-r)^{a-1}\int_0^\infty dv\, \frac{v^{a-1}}{1+v} \\ &=\Gamma(1-a)\Gamma(a)(1-r)^{a-1}, \end{align*} where we used the substitutions $u=(1-r)v$ and then $\frac1{1+v}=t$. So, the inequality in (0) is proved. Also, \begin{align*} (1-r)^{1-a}I(r)& =\int_0^\infty dv\, \frac{v^{a-1}}{1+(e^{(1-r)v}-1)/(1-r)} \\ &\to\int_0^\infty dv\, \frac{v^{a-1}}{1+v} , \end{align*} by dominated convergence. So, the asymptotic relation in (0) is proved. As seen from the above proof, the results in (0) hold for any $a\in(0,1)$. REPLY [6 votes]: Your function is the Polylogarithm function $Li_{1/6}(\rho)$. Mathematica indicates the correct asymptotic (at least for real $\rho$) is $\Gamma(5/6)/(1-\rho)^{5/6}+O(1)$.<|endoftext|> TITLE: How did Lefschetz do mathematics without hands? QUESTION [58 upvotes]: If people think this is the wrong forum for this question, I'll cheerfully take it elsewhere. But: How did Solomon Lefschetz do mathematics with no hands? Presumably there was an amanuensis to whom he dictated his papers, and then dictated his revisions. Does history record that person's identity? Was it someone trained in mathematics? Someone we might have heard of independently? Or maybe a series of graduate students? And what about the research phase, when most of us make a lot of idle notes, come back, revisit, cross things out, etc. Was Lefschetz also dictating all these idle thoughts to someone --- or did he learn to hold them in his mind until he was ready to write a paper? Or something else? Or did he have some sort of prosthetics that allowed him to write? I expect there are people alive --- and perhaps active on this site --- who were at Princeton during the time Lefschetz was active, or who have had direct contact with such people. Maybe one of them can answer? Edited to add: If anybody is tempted to answer that "he used a Lefschetz Pencil, of course!", I've just pre-empted you. REPLY [31 votes]: Surely his wife, nee Alice Berg Hayes, deserves much credit. She graduated with a master's degree in mathematics at Clark, with a thesis on "Reduction of Power Determinants". This was where she met Lefschetz, and she received her M.A. on the same day that he received his Ph.D. Albert Tucker and Frederik Nebeker said in the Dictionary of Scientific Biography: She helped him to overcome his handicap, encouraging him in his work and moderating his combative ebullience. Lefschetz himself said: of debts which I may never succeed in liquidating to the full...the first is my enormous debt to my wife Alice, my Clark companion. Without her constant and unfailing encouragement through 59 years, 56 as my wife, I would have long since ceased to operate.<|endoftext|> TITLE: What are orbifolds with corners? QUESTION [6 upvotes]: What is the geometric definition of orbifolds with corners? Here “geometric" means that there is a definition in chapter 8 of the draft of Dominic Joyce's book D-manifolds and d-orbifolds: a theory of derived differential geometry (book website, direct pdf link), but that is too technical and abstract, and so I am trying to find a definition parallel to the definition of orbifolds in the normal way (locally it is an open set acted by a smooth group, etc.) or more intuitive than the definition in that book. If it is defined as an open set with boundary (its closure) acted by a group keeping the boundary invariant seems does not work, as the boundary point of a manifold with boundary is itself a point in an orbifold (a manifold with boundary is an ordinary orbifold). REPLY [5 votes]: Orbifolds with corners are defined by the same axioms as manifolds with corners and ordinary orbifolds: A topological $n$-dimensional orbifold with corners is a topological space $X$ (2nd countable and Hausdorff) equipped with a (maximal) "orbifold atlas" $\{U_i \ldots : i\in I\}$ consisting of open subsets $U_i\subset X$, open subsets $V_i\subset [0,\infty)^n$ and finite affine groups $\Gamma_i$ preserving $V_i$'s, together with homeomorphisms $\phi_i: V_i/\Gamma_i\to U_i$, satisfying a long list of compatibility conditions which are identical to the ones for ordinary orbifolds with one important addition: Gluing maps $\psi_{ij}: V_i\to V_j$ preserve the boundary stratifications of $V_i$ and $V_j$ given by the product structure of $[0,\infty)^n$ (just as in the case of manifolds with corners). As a special case, a good orbifold with corners is the quotient $M/\Gamma$ of a topological manifold with corners $M$ by a properly discontinuous group action $\Gamma\times M\to M$, which is locally linearizable and is by automorphisms of the manifold with corners $M$.<|endoftext|> TITLE: Using Stiefel-Whitney class to build new principal bundles QUESTION [5 upvotes]: I'm reading this paper and at the beginning of the second section, he states many results that aren't clear to me. Consider a principal $SO(3)$-bundle $P\rightarrow R^2\times \Sigma$, where $\Sigma$ is a Riemann surface. If $\mathcal{w}_2(P)=0$, then $P$ is covered by a principal $SU(2)$-bundle to which we may associate a rank-2 vector bundle $V$, with $\mathcal{c}_1(V)=0$. My questions: 1.1) What does he mean by saying "$P$ is covered by another bundle"? 1.2) Why $\mathcal{w}_2(P)=0$ implies the existence of such a cover? If $\mathcal{w}_2(P)\neq 0$, then there is a principal $U(2)$ bundle $\hat P$ to which $P$ is associeted via homomorphism $U(2)/Z(U(2))\simeq SO(3)$. Associeted to $\hat P$ is a rank-2 vector bundle $V$ with $\mathcal{c}_1(V)$ is odd. Fixing a connection $A_0$ on $\wedge^2V$, we find that a connection $A$ on $\hat P$ lifts to one on $P$, whose curvature is $F(A)+\frac{1}{2}F(A_0)1$. More questions: 2.1) Again, why $\mathcal{w}_2(P)=0$ implies the existence of such a cover? 2.2) How can I find this connection $A$? Sorry for all these questions, but I'm pretty new in this subject and I have no idea of where can I find these results. So, any explanation and reference are very welcome. REPLY [4 votes]: This is an answer only to the question 1.2 (1.1 is already answered in a comment by Ulrich Pennig). Consider the adjoint representation of $SU(2)$ on its Lie algebra. Noting that the real dimension of $SU(2)$ is 3, this gives a homomorphism $SU(2)\to SO(3)$. One can prove that it is surjective with the kernel isomorphic to $Z/2$. Thus we get a fibration sequence $$Z/2 \to SU(2) \to SO(3) \to BZ/2\to BSU(2)\to BSO(3) \to K(Z/2,2).$$ Suppose now $P$ is a principal $SO(3)$ bundle over $X$, classified by a map $f:X\to BSO(3)$. Then $w_2(P)$ is nothing but the composition $X\stackrel{f}{\to}BSO(3)\to K(Z/2,2)$. Thus its vanishing is equivalent to existence of $f$ to $BSO(3)$. Take the corresponding $SO(3)$-principal bundle, and you get the desired cover.<|endoftext|> TITLE: Comparing $\mathcal C$ and $\mathcal C^{\mathcal C}$ (where $\mathcal C$ is a category) QUESTION [5 upvotes]: This is a followup to this question. (Matt Feller also mentioned this followup in a comment to the question linked to above.) For any category $\mathcal C$ write $\mathcal C^{\mathcal C}$ for the category of endofunctors of $\mathcal C$, and write $[\operatorname{Ob}({\mathcal C})]$ for the collection of isomorphism classes of objects of $\mathcal C$. (Note that $[\operatorname{Ob}({\mathcal C})]$ is not necessarily a set.) Let $\mathcal C$ be a category which is not equivalent to a category having exactly one object and one morphism. Are the following statements necessarily true? (1) There is no surjection $[\operatorname{Ob}(\mathcal C)]\to[\operatorname{Ob}(\mathcal C^{\mathcal C})]$. (2) There is no injection $[\operatorname{Ob}(\mathcal C^{\mathcal C})]\to[\operatorname{Ob}(\mathcal C)]$. A positive answer to at least one of the above questions would also answer the question linked to above. A negative answer to at least one of the above questions would also answer this older question. [Emil Jeřábek noticed this mistake.] (We assume that we are working in ZFC.) REPLY [5 votes]: If $\mathcal C$ is the category attached to the ordered set $(\mathbb R,\le)$, then $[\operatorname{Ob}(\mathcal C)]$ coincides with the set $\mathbb R$ and $[\operatorname{Ob}(\mathcal C^{\mathcal C})]$ coincides with the set of weakly increasing functions from $\mathbb R$ to $\mathbb R$. But it is well known that these two sets are equipotent: see for instance this answer of Brian M. Scott or this answer of mechanodroid.<|endoftext|> TITLE: Image of Comultiplication on Finite Quantum Groups/Hopf Algebras QUESTION [8 upvotes]: Let $A=:F(G)$ be the algebra of functions on a finite quantum groups aka a finite dimensional C*-Hopf Algebra. Suppose that $F(G)$ is neither commutative nor cocommutative. In their 1966 paper Kac and Paljutkin, show that when we write $F(G)$ as a multimatrix algebra, one of which must be one-dimensional (to account for the counit), that we can decompose into one-dimensional factors and $n_j>1$ dimensional factors: $$F(G)=\left(\bigoplus_{g\in G_1}\mathbb{C}\delta_g\right)\oplus\left( \bigoplus_{j=1}^mM_{n_j}(\mathbb{C})\right):=A_1\oplus B,\qquad (\star)$$ where $G_1$ is a finite group. Kac and Paljutkin show that: $$\Delta(A_1)\subseteq A_1\otimes A_1+B\otimes B.$$ Assumption 1: Kac and Paljutkin then assume that $m=1$ so that $B=M_{n_1}(\mathbb{C})$ contains a single summand. They show that: $$\Delta(B)\subseteq A_1\otimes B+B\otimes A_1+B\otimes B.$$ Let $f\in B$ and write: $$\Delta(f)=\underbrace{T(f)}_{\in A_1\otimes B+B\otimes A_1}+\underbrace{K(f)}_{\in B\otimes B}.$$ Assumption 2: Then they assume further that $|G_1|=n_1^2$. From there they show that $K(f)=0$ so that in fact $$\Delta(B)\subseteq A_1\otimes B+B\otimes A_1.$$ I am not sure if the assumptions were made to write down a quicker/easier proof for the purposes of illustration... or if there is a finite quantum group such that this inclusion does not hold. I am wondering in the 50 odd years hence has anyone working in Finite Quantum Groups (finite dimensional Hopf *-algebras $H$ with $f^*f\neq 0$ for all $f\in H$) showed that with respect to the decomposition $(\star)$ that $$\Delta(B)\subseteq A_1\otimes B+B\otimes A_1?$$ Question 1: Does this inclusion hold? If not, what is a counterexample? Question 2: Is it possible that $\Delta(M_{n_j}(\mathbb{C}))$ never falls into $M_{n_j}(\mathbb{C})\otimes M_{n_j}(\mathbb{C})$? It might be easier to show this more specific result (that must also be useful for me). Some Efforts: Let $E^j_{mn}$ be a matrix unit in $M_{n_j}(\mathbb{C})$. We want to show that $$(I_{n_j}\otimes I_{n_j})\Delta(E^j_{mn})=0.$$ This would be sufficient for my needs. I have the following assumption about how the antipode relates to the multi-matrix structure but I cannot find any suggestion that it is true. The only thing that seems remotely consistent with it is that $S^2=I_{F(G)}$. The assumption is that on the $m$ non-commutative factors, the antipode is the transpose. This allows a partial result to be proved using the antipodal property. Take a matrix unit. We have that $\varepsilon(E_{mn}^j)=0$ and this implies further that $$m\circ (S\otimes I_{F(G)})\circ\Delta(E^j_{mn})=0=m\circ (I_{F(G)}\otimes S)\circ \Delta(E^j_{mn}).$$ Think about the $A_1\otimes B$ and $B\otimes A_1$ parts of $\Delta(E^j_{mn})$ (there is no $A_1\otimes A_1$ part by Kac and Paljutkin). We know that $S(A_1)\subset A_1$, $S(B)\subset B$, and moreover $A_1B=0$, and so when we multiply $m\circ (A_1\otimes B)$ we get zero. Now choose lots of structure constants (for a fixed $j$) $\alpha^{mn}_{xy,zw}\in\mathbb{C}$: $$(I_{n_j}\otimes I_{n_j})\Delta(E^j_{mn})=\sum_{x,y,z,w=1}^{n^j}\alpha_{xy,zw}^{mn}(E^{j}_{xy}\otimes E^j_{zw}).$$ There are probably lots of conditions on the $\alpha$, but under the assumption that the antipode is the transpose, and the linear independence of the matrix units, it is the case that for any $x,y,w=1,\dots,n_j$, $\alpha_{xy,wy}=0$ and $\alpha_{xy,xw}=0$. I am hoping that the homomorphism property of $\Delta$ might 'finish off' the other structure constants. REPLY [3 votes]: I guess what you are looking for is the inclusion matrix for the unital inclusion of finite dimensional ${\rm C}^{\star}$-algebras $\Delta(A) \subset A \otimes A$. It is given by the fusion rules for $Rep(A)$, see Proposition 7.4 in my preprint arXiv:1704.00745v5. Example: consider the finite dimensional Hopf ${\rm C}^{\star}$-algebra $A=\mathbb{C}S_3$ with the symmetric group $S_3$. Then as an algebra $$A \simeq \mathbb{C} \oplus \mathbb{C} \oplus M_2(\mathbb{C}).$$ Now, $Rep(A)=Rep(S_3)$ and the fusion rules $(n_{ij}^k)$ are given by the following matrices: $$\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right), \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right), \left(\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{matrix} \right) $$ This provides a counter-example for your questions because $n_{33}^3 = 1 \neq 0$. [If you want, you can find the explicit formulas for the comultiplication computed here.] You can get many counter-examples from some non-abelian finite groups. See the following fusion rules for $Rep(A_5)$: $$\left(\begin{smallmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1 \end{smallmatrix} \right), \left(\begin{smallmatrix}0&1&0&0&0\\1&1&0&0&1\\0&0&0&1&1\\0&0&1&1&1\\0&1&1&1&1 \end{smallmatrix} \right), \left(\begin{smallmatrix}0&0&1&0&0\\0&0&0&1&1\\1&0&1&0&1\\0&1&0&1&1\\0&1&1&1&1 \end{smallmatrix} \right), \left(\begin{smallmatrix}0&0&0&1&0\\0&0&1&1&1\\0&1&0&1&1\\1&1&1&1&1\\0&1&1&1&2 \end{smallmatrix} \right), \left(\begin{smallmatrix}0&0&0&0&1\\0&1&1&1&1\\0&1&1&1&1\\0&1&1&1&2\\1&1&1&2&2 \end{smallmatrix} \right)$$ Counter-examples (for your questions) which are neither commutative nor cocommutative Let $\mathcal{C}$ be the class of non-abelian finite groups $G$ having an irreducible complex representation $V$ with $\dim(V)>1$ and which is an irreducible component of $V \otimes V$ (the groups $S_3$ and $A_5$ are of class $\mathcal{C}$, as shown above). Let $G$ be of class $\mathcal{C}$, and let $H$ be any non-abelian finite group. Let $A,B$ be the finite dimensional Hopf ${\rm C}^{\star}$-algebras $\mathbb{C}G$ and $\mathbb{C}H$ (respectively), then the usual tensor product of Hopf algebras $A \otimes B^*$ is a non-commutative and non-cocommutative counter-example of your questions. In addition, for any twist $J$ of $A$, $A_J$ has the same algebra structure than $A$. Moreover, $J$ is an invertible element of $A \otimes A$ and $\Delta_J = J^{-1} \Delta J$, so the inclusions $\Delta(A) \subset A \otimes A$ and $\Delta_J(A_J) \subset A_J \otimes A_J$ have the same inclusion matrices. Thus, $Rep(A_J)$ and $Rep(A)$ have the same fusion rules. Take a group $G$ of class $\mathcal{C}$, which admits a non-trivial twisting $(\mathbb{C}G)_J$ as for $G=A_5$ (by a result of Nikshych). It follows that $(\mathbb{C}G)_J$ is also a non-commutative and non-cocommutative counter-example of your questions.<|endoftext|> TITLE: Simplicial resolutions of varieties QUESTION [7 upvotes]: Let $X$ be a projective variety (over $\mathbb{C}$). Is there a simplicial object $X_{\bullet}$ in the category of smooth projective varieties and a morphism of simplicial varieties $X_{\bullet} \rightarrow X$ inducing a weak homotopy equivalence (of spaces) $|X_{\bullet}| \rightarrow X$? This (I believe) is claimed by Donu Arapura in a comment on the following question: Deligne's Mixed Hodge Theory The statements I've seen about such resolutions (e.g. in Deligne's Hodge III) are all assertions about sheaf cohomology, and not homotopy. I'd be happy with just a reference, or an argument that this is a direct consequence of the statements I have alluded to. REPLY [12 votes]: I'm converting my comment to an answer. Carlson, Polyhedral resolutions of algebraic varieties, shows to each projective variety $X$ there exists an augmented smooth (strict or semi) simplicial projective variety $X_\bullet\to X$ such that the induced map $|X_\bullet|\to X$ is a homotopy equivalence. You can formally add degeneracy maps to make it an actual simplicial object, if you require that. Alternatively, you can look at Deligne's original construction in Hodge III or SGA4 (exp Vbis). This yields a hypercover $X_\bullet\to X$, i.e. $$X_{n+1}\to (cosk_n X_\bullet)_{n+1}$$ is proper and surjective for all $n$. This should imply the map $|X_\bullet|\to X$ is surjective with contractible fibres, as explained in section 8.5 of Artin-Mazur Etale homotopy. I leave to you to flesh out the details.<|endoftext|> TITLE: Knizhnik-Zamolodchikov equation is a connection on "affine slice" QUESTION [6 upvotes]: The question is - what is the precise meaning of the phrase in the title? I heard it from Andrey Okounkov during one of his lectures. The problem is that he didn't really specified which slice is meant, if it is about the Slodowi one, then it is too mysterious. If it will help: the topic of his talk was about equivariant cohomology and intersection theory. REPLY [2 votes]: So, it turns out that we should take affine transversal slice to closure of an $G(\mathcal{O})$ orbit in affine grassmanian $G(K) / G(\mathcal{O})$, then there is "some" $X$ of smallest dimension in this slice, which is of our interest. $X$ turns out to be a singular variety (e.g. $\mathbb{C}^{2} / (\mathbb{Z} / n\mathbb{Z})$). Let $\widetilde{X} \to X$ be it's minimal resolution. We take quantum cohomology of $\widetilde{X}$ and see how basis forms of $H^{2}_{q}(X)$ act on the whole $H^{*}_{q}$ and these guys give rise to a connection which is Trigonometric Knizhnik-Zamolodchikov connection. Yeah, rather complicated story which i surely incorrectly rewrote (and can't find reference). Okounkov said that this was proven in PhD thesis of one of his students.<|endoftext|> TITLE: Explicit evaluation of the derivatives of $p$-adic Gamma function at 0 QUESTION [6 upvotes]: The definition of the $p$-adic Gamma function $\Gamma_p(x)$ for an odd prime number $p$ can be found in the book "A Course in $p$-adic analysis" by A. M. Robert. While the construction of $\log \Gamma_p(x)$ is also included in the book. The function $\log \Gamma_p(x)$ is odd, and has a series expansion in the variable $x$. For more details, see Chapter 7 of the book. \begin{equation} \log \Gamma_p(x)=(\log \Gamma_p)^{(1)}(0)x+\frac{1}{3!}(\log \Gamma_p)^{(3)}(0)x^3+\frac{1}{5!}(\log \Gamma_p)^{(5)}(0)x^5+\cdots. \end{equation} Since $\log \Gamma_p(x)$ is an odd function, we have \begin{equation} (\log \Gamma_p)^{(2n)}(0)=0. \end{equation} I am wondering whether there is an explicit method to evaluate the coefficient $(\log \Gamma_p)^{(s)}(0)$, at least the first several terms $(\log \Gamma_p)^{(3)}(0)$, $(\log \Gamma_p)^{(5)}(0)$, and $(\log \Gamma_p)^{(7)}(0)$? One helpful observations is that since we have \begin{equation} \Gamma_p(0)=1, \end{equation} so $(\log \Gamma_p)^{(s)}(0)$ can be computed by \begin{equation} \Gamma_p^{'}(0),\Gamma^{''}_p(0),\Gamma^{'''}_p(0), \cdots,\Gamma^{(s)}_p(0). \end{equation} So this question is equivalent to evaluate $\Gamma_p^{'}(0),\Gamma^{''}_p(0),\Gamma^{'''}_p(0),\cdots$. Does anyone know an explicit formula to compute them? REPLY [5 votes]: This is very classical, and the formula is very similar to the complex case: the coefficients of the Taylor expansion of $\log(\Gamma_p(x))$ are essentially $p$-adic Bernoulli numbers, or equivalently values at integers of the Kubota--Leopoldt $p$-adic zeta function. This is done in complete detail in particular in my book GTM 240, Chapter 11, specifically Propositions 11.3.15 and 11.5.19, Section 11.6.5, and Exercise 49 of Chapter 11.<|endoftext|> TITLE: Gelfand's trick (Gelfand's lemma) in positive characteristic? QUESTION [10 upvotes]: I came across this preprint that claims in Lemma 1.1 that Gelfand's trick (also known as Gelfand's lemma) only works in characteristic zero: Let $H < G$ be finite groups. Suppose we have an anti-involution $\sigma : G \rightarrow G$ that preserves all H double-cosets. Then over algebraically closed fields of characteristic zero (G, H) is a Gelfand pair. It is not obvious to me where the characteristic of the ground field being zero is used in the proofs from Lang's $SL_2(\mathbb{R})$ book (Theorem 1 and Theorem 3 in Chapter IV), the introduction in this preprint, the last slides in these slides, or anything else that I've seen. What causes Gelfand's trick to fail in positive characteristic? In this setting, the groups are finite but I would also like to know the answer for the more general versions of Gelfand's trick (i.e. also for locally compact groups with compact subgroups or even reductive groups over local fields with closed subgroups). REPLY [5 votes]: Let me try to muddle my way through. In positive characteristic the Gelfand trick will still tell you that the Hecke algebra of $H$-biinvariant functions $A=Func(H\backslash G/H,k)$ is commutative. Then, by general nonsense, the Hecke algebra is isomorphic to $End_G(P)$, where $P=k(G/H)=Ind_H^G (k_{triv})$ is the permutation representation. Since $P$ is not semisimple, its structure can still be complicated. The sort of information you need to conclude that it is a Gelfand pair is that $Soc(P)$ or $P/Rad(P)$ is multiplicity-free. There is no way you can conclude that about $Soc(P)$. But, here, Doc, is a happy ending for your opera: it follows from Nakayama Lemma that $$End(P/Rad(P))\cong A/Rad(A),$$ forcing the semisimple module $P/Rad(P)$ to be multiplicity free. Let $V$ be an irreducible $G$-module. By Frobenius-Nakayama-Fudd Reciprocity, $$ Hom_H(k_{triv},V)=Hom_G(P,V)=Hom_G(P/Rad(P),V) $$ must be at most one-dimensional! And so is $$ Hom_H(V,k_{triv})=Hom_H(k_{triv},V^\ast). $$<|endoftext|> TITLE: Eigenvalues and Domain of the Laplace-Beltrami Operator QUESTION [5 upvotes]: Assume $(M,g)$ is a compact Riemannian manifold without boundary, where $g$ is the Riemannian metric. Let $L:=-\Delta$ be the Laplace-Beltrami operator on $M$ defined by $\Delta \cdot = \text{div}(\nabla \cdot)$. I am reading in a lot of books/papers that the Laplace-Beltrami operator on a closed Riemannian manifold has positive, discrete spectrum whose eigenvalues accumulate at infinity. Here are many questions: What is the domain and range of $\Delta$ in order to have the spectrum described above? Does one treat $\Delta$ as a densely defined unbounded operator from $L^{2}(M)$ to $L^{2}(M)$ with domain $W^{1,2}(M)$ (or $W^{2,2}(M)$?) ? Or, does one think of $\Delta$ as a bounded operator? For example, from $W^{2,2}(M)$ to $L^{2}$. If one defines weak solutions of $\Delta$ by using the Green's identities $$ \int_{M}u\Delta v \text{Vol}_{g} = - \int_{M}g(\nabla u,\nabla v) \text{Vol}_{g} = \int_{M} v\Delta u \text{Vol}_{g},$$ hasn't the target to be some dual space then? Something like $(W^{1,2}(M))^{*}$? What is the precise formulation of spectrum and eigenvalues of $\Delta$, provided one knows the correct domain and range? Do you know any reference, where this is fully discussed? By that I mean, some reference where the domain, range, spectrum of $\Delta$ on $(M,g)$ is discussed? Cheers, Martin REPLY [5 votes]: There are various approaches. First one: consider $\Delta$ as an unbounded operator over $L^2(M)$ with domain $W^{2,2}(M)$. It is closed, densely defined and $-\Delta$ is self-adjoint, positive. There is a well-defined spectral theory for this class, which you should find somewhere in Reed & Simon. That the spectrum is discrete and accumulates at infinity follows from the fact that $-\Delta+1$ has a compact inverse. Second one: the eigenvalues of $-\Delta$ are the critical points of the functional (Rayleigh ratio) $$I[u]=\frac{\int_M|u|^2\,{\rm Vol}_g}{\int_Mg(\nabla u,\nabla u)\,{\rm Vol}_g},$$ which is well-defined over $W^{1,2}(M)\setminus\{0\}$. Once again, you must use the compactness of the embedding $W^{1,2}(M)\subset L^2(M)$. All in all, the way you obtain the eigenvalues is not important in the end, because once you have $-\Delta u=\lambda u$, elliptic regularity plus a bootstrap argument tell you that $u$ is $C^\infty$, hence is an eigenfunction in every sense that you might imagine.<|endoftext|> TITLE: Spherical Bessel functions. Sum of squares QUESTION [6 upvotes]: In (1) there is a property of spherical Bessel functions, which's derivation I can not find in the literature. ${\mathsf{j}_{n}^{2}}\left(z\right)+{\mathsf{y}_{n}^{2}}\left(z\right)=\sum_{k=% 0}^{n}\frac{s_{k}(n+\frac{1}{2})}{z^{2k+2}},$ where $s_{k}(n+\tfrac{1}{2})=\frac{(2k)!(n+k)!}{2^{2k}(k!)^{2}(n-k)!}.$ I tried to obtain it by myself with the use of 10.49.1 $a_{k}(n+\tfrac{1}{2})=\begin{cases}\dfrac{(n+k)!}{2^{k}k!(n-k)!},& k=0,1,\dotsc% ,n,\\ 0,&k=n+1,n+2,\dotsc.\end{cases}$ and 10.49.6 $h_{n}(z)\displaystyle=e^{iz}\sum_{k=0}^{n}i^{k-n-1}\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}},$ but I haven't managed with sums rearrangement. Can someone tell me where to find the derivation? REPLY [5 votes]: From the definition 10.47.10, it follows that $$\mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) = h_n^{(1)}(z)\cdot h_n^{(2)}(z).$$ So, by the expansions 10.49.6 and 10.49.7, \begin{split} \mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) &= \sum_{k=0}^n I^{k-n-1} \frac{a_k(n+\frac{1}{2})}{z^{k+1}}\cdot \sum_{l=0}^n (-I)^{l-n-1} \frac{a_l(n+\frac{1}{2})}{z^{l+1}} \\ &= \sum_{s=0}^{2n} \frac{(-I)^s}{z^{s+2}} \sum_{k=\max\{0,s-n\}}^{\min\{n,s\}} (-1)^k a_k(n+\frac{1}{2})a_{s-k}(n+\frac{1}{2}). \end{split} From the definition 10.49.1, the inner term can be restated as $$(-1)^k\frac{s!}{2^s} \binom{s}{k}\binom{n+k}{s}\binom{n+s-k}{s}$$ and it naturally nullifies when $k$ is outside the summation range. Furthermore, if $s$ is odd then the terms for $k=k'$ and $k=s-k'$ cancel each other. So, we can set $s=2t$ and obtain $$ \mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) = \sum_{t=0}^{n} \frac{(-1)^t (2t)!}{z^{2t+2}2^{2t}} \sum_{k\geq 0} (-1)^k \binom{2t}{k}\binom{n+k}{2t}\binom{n+2t-k}{2t}. $$ It remains to show that $$\sum_{k\geq 0} (-1)^k \binom{2t}{k}\binom{n+k}{2t}\binom{n+2t-k}{2t} = (-1)^t\frac{(n+t)!}{t!^2(n-t)!},$$ which at very least can be done with the WZ method. But perhaps it's just a consequence from something well-known. P.S. This identity has a neat representation in terms of hypergeometric functions, which I posted in a follow-up question.<|endoftext|> TITLE: When does $\sqrt{a_1} + \cdots + \sqrt{a_k} \in K$ imply $\sqrt{a_1}, \ldots, \sqrt{a_k} \in K$? QUESTION [8 upvotes]: Consider $R = \mathbb{Z}[X_1, \ldots, X_k]$, the polynomial ring in $k$ variables over $\mathbb{Z}$, and let $S = \mathbb{Z}[\sqrt{X_1}, \ldots, \sqrt{X_k}]$. Then $S/R$ is an integral extension of commutative ring; let $f \in R[T]$ be the minimal polynomial of $\alpha = \sqrt{X_1} + \cdots + \sqrt{X_k} \in S$. One sees (for example by passing to the fraction fields and invoking Galois theory) that $\deg(f) = 2^k$. Some examples of $f$ for small values of $k$: $k = 1$: $f = T^2 - X_1$ $k = 2$: $f = T^4 - 2(X_1 + X_2)T^2 - (X_1 - X_2)^2$ $k = 3$: $ f =T^8 + (-4(X_1 + X_2 + X_3))T^6 + (6(X_1^2 + X_2^2 + X_3^2) + 4(X_1X_2 + X_1X_3 + X_2X_3))T^4 + (-4(X_1^3 + X_2^3 + X_3^3) + 4(X_1^2X_2 + X_1^2X_3 + X_2^2X_1 + X_2^2X_3 + X_3^2X_1 + X_3^2X_2) - 40X_1X_2X_3)T^2 + (X_1^2 + X_2^2 + X_3^2 - 2(X_1X_2 + X_1X_3 + X_2X_3)^2$ Now consider some arbitrary field $K$ of characteristic different from $2$ and some fixed homomorphism $\varphi : R \to K$. I want to know when the existence of a root of $f(\varphi(X_1), \ldots, \varphi(X_k), T)$ in $K$ implies that $\varphi(X_1), \ldots, \varphi(X_K)$ are squares in $K$, i.e. that $\varphi$ extends to a homomorphism $S \to K$. My understanding is that this will somehow 'generically' be the case by the fact that $K(\sqrt{X_1}, \ldots, \sqrt{X_k}) = K(\sqrt{X_1} + \cdots + \sqrt{X_k})$ for any field $K$ of characteristic different from $2$, but one needs to exclude some values of $t$, possibly depending on the $\varphi(X_i)$. For example, if $k = 2$, $\varphi : R \to K$ any morphism and $t \in K$ such that $f(\varphi(X_1), \varphi(X_2), t) = 0$, then one can verify the identities $$ \varphi(X_1) = \left(\frac{t^2 - \varphi(X_2) + \varphi(X_1)}{2t}\right)^2 \enspace \text{and} \enspace \varphi(X_2) = \left(\frac{t^2 - \varphi(X_1) + \varphi(X_1)}{2t}\right)^2 $$ under the assumption that $t \neq 0$. Hence, any morphism $\varphi : R \to K$ such that there exists a $t \in K^\times$ with $f(\varphi(X_1), \varphi(X_2), t) = 0$ extends to a morphism $S \to K$. The condition $t \neq 0$ cannot be dropped: consider for example $\varphi : R \to \mathbb{R}$ defined by $\varphi(X_1) = \varphi(X_2) = -1$, then $0$ is a root of $f(-1, -1, T) = T^4 + 4T^2$, but $-1$ is not a square in $\mathbb{R}$. How should I approach the problem of finding the values of $t$ one has to exclude for larger values of $k$ - say, for starters, for $k = 3$? Any pointers towards books or articles on related subjects are very welcome. REPLY [4 votes]: No hope, even for $k=3$ and for an arbitrary firld $K$ conaining a non-square (is it non-quadratically-closed?). Let $K$ be such field and $\alpha\in K$ be a non-square; set $L=K[\sqrt\alpha]$. Take any $t\in K$ and define $\psi\colon S\to L$ by $\psi(\sqrt{X_1})=t$, $\psi(\sqrt{X_2})=\sqrt\alpha$, and $\psi(\sqrt{X_3})=-\sqrt\alpha$. Then the image of $\varphi=\psi\big|_R$ is contained in $K$, and $\psi(\sqrt{X_1}+\sqrt{X_2}+\sqrt{X_3})=t\in K$, so $t$ satisfies the required equation; but $\sqrt\alpha\notin K$ It seems that, in order to have a real hope, the weakest property you need is the $\sqrt{X_i}$ be linearly independent over $\varphi(R)$, or at least smewhat close to that (as a model example over $\mathbb Q$ shows). And this is hard to reach just by imposing some condition on $t$.<|endoftext|> TITLE: Talagrand's inequality for the discrete cube QUESTION [8 upvotes]: Talagrand showed that if $f$ is a convex $1$-Lipschitz function on $\mathbb{R}^n$, and if $\mu$ is a product of probability measures supported over the interval, then $f$ has Gaussian concentration w.r.t. to $\mu$: $$ P(\vert f - E_{\mu}f\vert > \epsilon) < C \exp(-c\epsilon^2) $$ for some absolute constants $c$ and $C$. I would like to understand to what extent convexity is necessary in the special case where $\mu$ is the uniform measure on the discrete hypercube $\{0,1\}^n$. What are "representative" counterexamples if convexity is not assumed? REPLY [5 votes]: there is no convexity assumption needed there. Nevertheless here is a counterexample of $1$-Lipschitz function on $\mathbb{R}^{N}$ which fails to satisfy concentration inequality on the Hamming cube $\{-1,1\}^{N}$ as $N$ goes to infinity. Take $N$ to be even. Let $$ A = \left\{ (x_{1}, \ldots, x_{N}) \in \{0,1\}^{N}\, : x_{1}+\ldots+x_{N}\leq \frac{N}{2}\right\}. $$ Next, define a function $$ f(x) = \inf_{y \in A} \| x-y\|_{\mathbb{R}^{N}} $$ Since $f$ is the distance function to a nonempty subset it follows that $f$ is $1$-Lipschitz (an exercise). On the other hand notice that $f(x) = \sqrt{\max\{x_{1}+\ldots+x_{N} - \frac{N}{2},0\}}$ on $\{0,1\}^N$. Then as $N$ goes to infinity we have $$ \begin{aligned} P(|f & -\mathbb{E}f| > N^{1/4}) \\ & = P\biggl(\biggl|\sqrt{\max\{\frac{(2x_{1}-1)+\ldots+(2x_{N}-1)}{\sqrt{N}},0\}} \\ & \qquad\qquad - \mathbb{E}\sqrt{\max\{\frac{(2x_{1}-1)+\ldots+(2x_{N}-1)}{\sqrt{N}},0\}} \biggr|>\sqrt{2} \biggr) \\ & \to P\left( | \sqrt{\max\{\xi, 0\}} - \mathbb{E} \sqrt{\max\{\xi, 0\}}|>\sqrt{2}\right)>10^{-10}, \end{aligned} $$ where we used the central limit theorem, and $\xi$ is the standard normal Gaussian $\xi\in N(0,1)$.<|endoftext|> TITLE: Divergent Series & Continued Fraction (from Gauss' Mathematical Diary) QUESTION [18 upvotes]: I've asked that question before on History of Science and Mathematics but haven't received an answer Does someone have a reference or further explanation on Gauß' entry from May 24, 1796 in his mathematical diary (Mathematisches Tagebuch, full scan available via https://gdz.sub.uni-goettingen.de/id/DE-611-HS-3382323) on page 3 regarding the divergent series $$1-2+8-64...$$ in relation to the continued fraction $$\frac{1}{\displaystyle 1+\frac{\strut 2}{\displaystyle 1+\frac{\strut 2}{\displaystyle 1+\frac{\strut 8}{\displaystyle 1+\frac{\strut 12}{\displaystyle 1+\frac{\strut 32}{\displaystyle 1+\frac{\strut 56}{\displaystyle 1+128}}}}}}}$$ He states also - if I read it correctly - Transformatio seriei which could mean series transformation, but I don't see how he transforms from the series to the continued fraction resp. which transformation or rule he applied. The OEIS has an entry (https://oeis.org/A014236) for the sequence $2,2,8,12,32,56,128$, but I don't see the connection either. My question: Can anyone help or clarify the relationship that Gauss' used? Torsten Schoeneberg remarked rightfully in the original question that the term in the series are $(-1)^n\cdot 2^{\frac{1}{2}n(n+1)}$ and Gerald Edgar conjectures it might be related to Gauss' Continued Fraction. REPLY [23 votes]: The entry from May 24, 1796 is worked out in a more general form on February 16, 1797 [reproduced below from this scan] $$1-a+a^3-a^6+a^{10}+\cdots=\frac{1}{\displaystyle 1+\frac{\strut a}{\displaystyle 1+\frac{\strut a^2-a}{\displaystyle 1+\frac{\strut a^3}{\displaystyle 1+\frac{\strut a^4-a^2}{\displaystyle 1+\frac{\strut a^5}{1+\cdots}}}}}}$$ so the coefficients alternate between $a^{2n+1}$ and $a^{2n}-a^n$. Latin text: Amplificatio prop[ositionis] penult[imae] p[aginae] 1, scilicet $\cdots$ Unde facile omnes series ubi exp[onentes] ser[iem] sec[undi] ordinis constituunt transformantur. Translation: Expanding on the proposition 1 from the next-to-last page $\cdots$ From here one can easily transform every series the exponents of which form a series of the second order. These continued fractions of series of the form $$1+\sum_{n=1}^\infty a^{n(n+1)}-\sum_{n=1}^\infty a^{n^2}=\sum_{n=0}^\infty (-1)^n a^{n(n+1)/2}$$ are related to theta functions, see chapter 29 of "Series and Products from the Fifteenth to the Twenty-first Century". Apparently the series originated from Jakob Bernoulli (1690), see History of Continued Fractions and Padé Approximants.<|endoftext|> TITLE: Smallest set such that all arithmetic progression will always contain at least a number in a set QUESTION [12 upvotes]: Let $S= \left\{ 1,2,3,...,100 \right\}$ be a set of positive integers from $1$ to $100$. Let $P$ be a subset of $S$ such that any arithmetic progression of length 10 consisting of numbers in $S$ will contain at least a number in $P$. What is the smallest possible number of elements in $P$ ? Denote $|P|$ as the number of elements in $P$. We shall find the smallest possible value of $|P|$. For $|P|=16$, we have the answer by @RobertIsrael below. However, for $|P|<16$, I can neither find such set $P$ nor prove that $|P|$ cannot be less than $16$. So my question is: Is it true that $|P| \geq 16$? How can I prove it? If not, what is the minimum amount of elements in $P$ ? Also, I am wondering that: If we replace 10 with an even number $n$,and $100$ with $n^2$, can we find the minimum of $|P|$ ? Any answers or comments will be appreciated. If this question should be closed, please let me know. If this forum cannot answer my question, I will delete this question immediately. REPLY [11 votes]: Using a tabu search procedure, I have found a solution for $|P|=17$, namely ${1, 11, 18, 25, 31, 32, 33, 36, 44, 51, 58, 65, 69, 70, 77, 84, 91}$. I don't know if this is optimal. EDIT: Found a solution for $|P|=16$, namely $$10, 15, 22, 29, 36, 43, 53, 55, 56, 57, 58, 68, 73, 74, 84, 91$$<|endoftext|> TITLE: On the geometric Hahn-Banach theorem QUESTION [7 upvotes]: Let $ X \subset \ R ^ n $ be a closed convex set and let $ L $ be a straight line such that $ X \cap L = \emptyset $. Does there exist a hyperplane containing $ L $ that does not intersect $ X $ ? In the classical Hahn-Banach theorem we require $X$ to be open. REPLY [11 votes]: Not necessarily: consider a 3-D space, let $L$ be the $x$ axis, and let the convex set be $$X=\bigl\{(x,y,z):z\geqslant (\max\{0,y + e^x\})^2\bigr\}.$$ This is the region above the graph of a convex function; here's a 3-D plot. After projecting on the $yz$ plane, the convex set becomes $$X' = \{(y, z) : y < 0 , \, z \geqslant 0\} \cup \{(y, z) : y \geqslant 0, z > y^2\}.$$ There is no line in the $yz$ plane that contains the origin and does not intersect $X'$, and hence there is no hyperplane that contains the $x$ axis and does not intersect $X$.<|endoftext|> TITLE: Origin of the convolution theorem QUESTION [7 upvotes]: I am a chemist, with some interest in signal processing. Sometimes, we use the deconvolution process to remove the instruments response from the desired signals. I am looking for the earliest reference which proposed the convolution theorem which is often utilized in signal processing (i.e., convolution becomes a multiplication in the Fourier domain). The Earliest Known Uses of the Word of Mathematics websites gives lot of details on the word convolution, but who was the first person to specifically show the above mentioned property- the connection of Fourier transforms with convolution? Here is the history of Convolution Operation: https://pulse.embs.org/january-2015/history-convolution-operation/ However, a mathematician privately disagreed with that historical account given in this article. Thanks. REPLY [15 votes]: The general theorem that Fourier’s $\mathscr F:L^2(G)\to L^2(\hat G)$ maps convolution to product and vice versa is in Weil (1940, p. 113), for any locally compact abelian $G$ with dual $\smash{\hat G}$. But the special cases of the real line, the circle group (with $(f,g,h)(t)=$ $\sum (a_n,b_n,c_n)e^{int}$), and the integers: $$ \begin{align} &\mathbf{(R)}\qquad h(t)=\textstyle\int_{-\infty}^\infty f(s)g(t-s)\,ds &&\Rightarrow &&\textstyle\mathscr F(h)=\mathscr F(f)\mathscr F(g)\\ &\mathbf{(T)}\qquad h(t)=\smash[t]{\textstyle\int_0^{2\pi} f(s)g(t-s)\,ds} &&\Rightarrow &&c_n=a_nb_n\\ &\mathbf{(Z)}\qquad h(t)=f(t)g(t) &&\Rightarrow &&\textstyle c_n=\sum_k a_kb_{n-k}\\ \end{align} $$ were known earlier: Dieudonné (1981, p. 195) and Mackey (1980, p. 628) attribute $\mathbf{(R)}$ to Lyapunov (1900, 1901?); it’s also in Hausdorff (1901, p. 169), Poincaré (1912, p. 207), Lévy (1925, p. 184; 1928, p. 79), etc. (They state results in terms of added independent random variables, not repeating the step that $h(t)dt$ is the image of the product measure $f(s)ds\times g(t)dt$ under addition $\mathbf{R\times R\to R}$.) Kahane and Lemarié-Rieusset (1995, p. 19) attribute $\mathbf{(T)}$ to Fourier (1822, p. 259) where it is rather implicit; it’s explicit in Young (1912, p. 32). Burkhardt (1901, p. 84; 1914, p. 947) finds $\mathbf{(Z)}$ in Cauchy (1844, p. 1125) and earlier in Euler (1760), for both trigonometric polynomials (“product-to-sum”, pp. 176-186) and some series (p. 200); it’s also in Pringsheim (1886, p. 158), Hurwitz (1902, p. 369), Lebesgue (1906, §52), etc. (And indeed, the history in Domínguez (2015, p. 46) cheerfully ignores all of the above.) Added: Adams (2009) contains translations of Lyapunov’s (1900, 1901). This should be consulted, as e.g. Fischer (2011, p. 201) contradicts Dieudonné and Mackey by saying: “Lyapunov never used general concepts such as inversion formula or correspondence between convolution of distributions and products of characteristic functions” (and I am indeed not really seeing $\mathbf{(R)}$ in (1900)).<|endoftext|> TITLE: Characteristic classes of symmetric group $S_4$ QUESTION [15 upvotes]: For the symmetric group $S_3$, it is classically known that \begin{equation} H^*(S_3;\mathbb{Z})\cong \mathbb{Z}[x,y]/(2x,6y,x^2-3y), \end{equation} where $|x|=2$ and $|y|=4$. Moreover, $x$ can be identified with the first (and top) Chern class of the 1-dimensional sign representation of $S_3$, and likewise $y$ can be identified with the top Chern class of the standard representation. Question 1: Is something similar true for the integral cohomology of $S_4$? Namely, is there an explicitly computed presentation of $H^*(S_4;\mathbb{Z})$ whose generators $x,y,z$ are the respective top Chern classes of (1) the sign representation, (2) the 2-dimensional representation $S_4$ given by composing the projection $S_4\rightarrow S_3$ with the standard representation of $S_3$, and (3) the standard representation of $S_4$? Question 2: Is there an analogous situation for $H^*(S_4; \mathbb{Z}_2)$ in terms of Stiefel-Whitney classes of the corresponding real representations? REPLY [8 votes]: For Q2, my collaborators and I show that all mod-two cohomology of symmetric groups is generated by Stiefel-Whitney classes of standard representations, if you allow both cup product and transfer (induction) product. It is better, however, to take other Hopf ring generators, in which case all cohomology of symmetric groups is a free divided powers Hopf ring on classes in $H^{2^k - 1}(S_{2^k})$. See http://front.math.ucdavis.edu/0909.3292, in particular Section 10 for discussion of Stiefel-Whitney classes. For Q1, the integral groups are known, but not ring structure. (I happen to be working out the Bocksteins now, at the prime two.) I haven't considered the characteristic class question.<|endoftext|> TITLE: A finite 2-group containing the dihedral group of order 16? QUESTION [17 upvotes]: The dihedral group $D_{16}$ of order 16 has a presentation $D_{16}= \langle a,t \ | \ a^2=t^8=atat=e\rangle$. Question: Does there exist a finite 2-group $G$ containing $D_{16}$ as a subgroup, and an element $g \in G$ such that $gag^{-1}=t^4$? Bonus pats-on-the-back if $G$ has order 64. An obvious reduction: one can assume that $G =\langle D_{16},g\rangle$. An obvious constraint: $D_{16}$ cannot be normal in $G$ (so $G$ can't have order 32). [This question has come up in investigations of the Balmer spectrum of $G$-equivariant stable homotopy for finite $p$-groups $G$. Like Dr. Frankenstein, I am looking for interesting subjects to experiment on, and my student Chris Lloyd is serving as the able assistant to the mad scientist.] REPLY [23 votes]: No. We can prove this by induction. Let $G$ be the smallest $2$-group in which this situation occurs. Then $G$ has a normal central subgroup $N$ of order $2$. If $N$ has trivial intersection with the subgroup $\langle a,t \rangle = D_{16}$, then the same situation occurs in $G/N$, contradicting the minimality of $G$. So that intersection must be nontrivial, and hence $N \le \langle a,t \rangle$, and then we must have $N = Z(\langle a,t \rangle) = \langle t^4 \rangle$. But then $t^4 \in Z(G)$, contradicting the assumption that it is conjugate in $G$ to $a$. The situation you describe can occur in finite groups, such as in simple groups ${\rm PSL}(2,q)$ for prime powers $q$ with $q \equiv 15$ or $17 \bmod 32$, ($q=17$ for example), but not in finite $2$-groups.<|endoftext|> TITLE: Hyperbolic groups and spaces of negative curvature QUESTION [7 upvotes]: Mikhail Gromov states that he "tried for about 10 years to prove that every hyperbolic group is realizable by a space of negative curvature" in his interview with Martin Raussen and Christian Skau (page 6, link below). Is there any information available about the current status of this conjecture? https://www.ihes.fr/~gromov/wp-content/uploads/2018/08/rtx100300391p.pdf REPLY [9 votes]: As already mentioned in the comments, it is still unknown whether hyperbolic groups are CAT(-1) or even CAT(0). A related question is: Let $G$ be a hyperbolic group (endowed with a finite generated set). If $d$ is sufficiently large, does there exist a $G$-equivariant CAT(-1) or CAT(0) metric defined on the Rips complex $P_d(G)$? This question is also open, but a counterexample is given here (Corollary 5.10) for $P_d(X)$ where $X$ is a specific (locally infinite) quasi-tree. An alternative approach is to replace Rips' complex with the injective hull of the group. See the last paragraph of the introduction of Lang's paper. Actually, even the following question is open: Let $G$ be a hyperbolic group acting geometrically on a CAT(0) cube complex. Does $G$ act geometrically on a CAT(-1) cube complex? The question appears for instance here. Also relevant is this article of Brady and Crips constructing a hyperbolic group with CAT(0) dimension two but CAT(-1) dimension three.<|endoftext|> TITLE: A hypergeometric identity related to Bessel functions QUESTION [7 upvotes]: The identity in my recent answer can be stated in a particularly neat form: $${}_2F_0\left({-n, n+1\atop{}};\frac{x}{2}\right) ~\cdot~ {}_2F_0\left({-n, n+1\atop{}};-\frac{x}{2}\right) ~=~ {}_3F_0\left({\frac{1}{2}, -n, n+1\atop{}};x^2\right).$$ Is this by any chance a partial case of something more general and/or has a straightforward proof? REPLY [5 votes]: The coefficient of $x^n$ in $_2F_0(\alpha,\beta;z) {}_2F_0(\alpha,\beta;-z)$, multiplied by $n!$, is $$\begin{aligned} \sum_{k=0}^n (-1)^{n-k}\binom nk &(\alpha)_k(\beta)_k (\alpha)_{n-k}(\beta)_{n-k}\hfill\\ &=(-1)^n (\alpha)_n(\beta)_n\, {}_3F_2\left({-n,\atop}\!{\alpha,\atop 1-\alpha -n,}\, {\beta\atop1-\beta-n}\biggm | 1\right). \end{aligned} $$ The right side can be evaluated by Dixon's theorem to give the identity cited by Johannes Trost.<|endoftext|> TITLE: What are the topological phases of quantum Hall systems? QUESTION [10 upvotes]: (Fractional) quantum Hall systems are $2+1$-dimensional models which are said to possess topological order. One (maybe even complete) set of invariants of topological phases in $2+1$ dimensions is given that anyon statistics, or in mathematical terms, the (2-extended) axiomatic TQFT of the phase. Both are labelled by modular fusion categories. Question: Which are the modular fusion categories associated to quantum Hall systems? Is there a table somewhere in the literature that associates to each value of the "filling fraction" (and whatever other parameters there are in quantum Hall models) a modular fusion category? Also: Quantum Hall systems involve fermionic degrees of freedom. Are they proper fermionic phases? Or are the fermions somehow restricted to local islands of even parity, such that the resulting phases are still bosonic? In the former case, the anyon statistics are not described by modular fusion categories but their fermionic analogue, something like "modular super-fusion categories". I guess this is written down at a lot of places, I'm just having a hard time finding it without having to skim through lots of condensed matter literature that I don't understand. REPLY [3 votes]: We have a paper that contains lists of simple fermionic topological orders in 2+1D: https://arxiv.org/abs/1507.04673 . For fermionic topological without symmetry, there is no filling fraction. So our lists are based on the number of anyon types, together with their quantum dimensions and topological spins. Each entry in the table corresponds to a sequence of fermionic topological orders, differ by stacking of $p+ip$ invertible fermionic topological orders. FQH states have additional U(1) symmetry and correspond to SET states, which do have filling fraction. Filling fraction do not determine the topological order. For a given filling fraction, there can be many different topological orders. Every modular tensor category describes a bosonic topological order (up to $E_8$ invertible topological order). However, fermionic topological orders are not classified by modular tensor category. They are classified by a special type of braided fusion category, with modular extension. This is the main point of our work.<|endoftext|> TITLE: A finite p-group question: can this happen? QUESTION [6 upvotes]: Let all groups here be finite $p$--groups. Given $K TITLE: What structure do natural isomorphisms preserve? QUESTION [10 upvotes]: My understanding from model theory is that, given groups A and B, the statement $ A \cong B $ implies that any for any first order statement $P$ in the language of groups, $P(A) \iff P(B) $. Can an analogous statement be made for naturally isomorphic functors F and G? IE is it true that $F \cong G \implies P(F) \iff P(G)$ for some collection of propositions about functors? I've been told that isomorphisms are the "right" notion of equivalence for algebraic structures because they "preserve structure," and I'm struggling greatly to see why natural isomorphisms are the "right" notion of equivalence of functors. I've seen explainations like: Set functions $X \times C \rightarrow D $ can be identified with functions $X \rightarrow D^C $. Morphisms in a category $X$ can be identified with functors $2 \rightarrow X$, so morphisms in the functor category $D^C$ should be identified with functors $2 \rightarrow D^C$. But in analogy with set functions, $2\rightarrow D^C$ 'equals' $2 \times C \rightarrow D$ from which the definition of natural transformations can be derived. The above explanation intuitively relates to a basic property of set functions and products, but doesn't on the surface tell me about what specifically is true/preserved about isomorphic functors. The definition of natural transformation is forced if we mandate that $Cat$ be cartesian closed, but again I fail to see the relation to preservation of 'structure.' I've seen references to the 'principle of equivalence' which seems to support the idea that isomorphic objects should be be indistinguishable in some language. From ncatlab: Michael Makkai proposed the Principle of Isomorphism, “all grammatically correct properties of objects of a fixed category are to be invariant under isomorphism” How does the definition of natural isomorphisms make this statement true in the category $D^C$ with functors as objects and morphisms as natural transformations? What's invariant? If natural transformations represent a "change in perspective" analogous to conjugation, what is preserved between the perspectives? Can anyone give a formalization of the ideas behind the responses at https://math.stackexchange.com/questions/1077895/functorial-properties-preserved-by-natural-isomorphism? I'm really stuck here. I feel like I'm missing something obvious. Further reading: https://math.stackexchange.com/questions/1432782/how-different-can-equivalent-categories-be, https://math.stackexchange.com/questions/1685227/why-not-just-define-equivalence-relations-on-objects-and-morphisms-for-equivalen, and Can skeleta simplify category theory?. REPLY [6 votes]: The simplest case is that for a fixed (small) category $C$, there is a (multi-sorted) first-order theory whose models are functors $C\to \rm Set$: it has one sort for every object of $C$, and one function symbol for every morphism of $C$, plus axioms saying that these functions preserve identities and composition. So the same reasoning as in the case of groups implies that for functors $F,G:C\to \rm Set$, if $F\cong G$ then $F$ and $G$ satisfy all the same first-order statements in the language of this theory. Moreover, as pointed out in the comments, the first-order qualification is unnecessary; they satisfy all the same statements even in higher-order or infinitary logic. To deal with functors $C\to D$ with $D\neq\rm Set$ (where $C$ and $D$ are small), we can embed them representably in the category of profunctors, i.e. functors $D^{\rm op}\times C\to \rm Set$. A functor $F:C\to D$ thus corresponds to the profunctor $\hat F(d,c) = D(d,F(c))$. This is a full embedding, so in particular two functors are isomorphic if and only if their corresponding profunctors are. Since profunctors, being set-valued functors on a small category $D^{\rm op}\times C$, are a model of a multi-sorted first-order theory as above, it follows that two isomorphic functors satisfy all the same statements in the language of this theory (even in higher-order or infinitary logic). One may object that the language of profunctors does not include all the reasonable things one may want to say about a functor. For instance, how can we talk about "$F(c)$" for some $c\in C$? The answer is that the profunctors $H$ arising in this way are those that are representable, i.e. such that for every $c\in C$ there exists a $d\in D$ and an element $h\in H(d,c)$ such that for any $d'\in D$ and $x\in H(d',c)$ there is a unique morphism $f:d'\to d$ such that $H(f,1)(h) = x$. Representability is not a first-order axiom in this theory, but it can be expressed in infinitary logic: $$ \bigwedge_{c\in C} \bigvee_{d\in D} \exists h\in H(d,c)\, \bigwedge_{d'\in D} \forall x\in H(d',c) \,\Big(\Big(\bigvee_{f:d'\to d} H(f,1)(h) = x\Big)\wedge \Big(\bigwedge_{g:d'\to d} (H(g,1)(h) = x) \Rightarrow (f\equiv g) \Big)\Big) $$ where $(f\equiv g)$ denotes $\top$ if in fact $f=g$ and $\bot$ otherwise. The bit of this formula after the first $\bigwedge\bigvee$ is then a characterization of when "$d=F(c)$", so that we can use infinitary logic to talk about "$F(c)$" and thereby say everything we would want to say about a functor. (To be more precise, an $h$ as in the first $\exists$ is a "witness" that $d=F(c)$ and has to be kept track of.) Finally, so that the HoTT fans are not disappointed, let me add a bit about that perspective. In dependent type theory, the naive way to define a category is with a type $C_0:\rm Type$ of objects, a dependent type $C_1 : C_0 \times C_0 \to \rm Type$ of morphisms, and composition and identity operations satisfying axioms. Similarly, a naive functor consists of a function $F_0: C_0 \to D_0$ and a family of functions $F_1 : \prod_{c,c':C_0} C_1(c,c') \to D_1(F_0(c),F_0(c'))$ satisfying axioms. Now, there is an interpretation of dependent type theory in which types correspond to groupoids (and, in fact, even ∞-groupoids, but that's not necessary here). The identity type "$x=_A y$", for $x,y:A$, is interpreted by the set of isomorphisms $x\cong y$ in the groupoid $A$. Since identity types in dependent type theory satisfy a strong "indiscernibility of identicals" rule, anything in the theory that is true about $x$ must then also be true about $y$. When the naive notion of category is interpreted in this model, it gives something more general than the ordinary notion of category, but there are two ways to embed ordinary categories in such things. Given an ordinary category $C$, we could take the types $C_0$ and $C_1(c,c')$ to be the discrete groupoids corresponding to the set of objects of $C$ and the homsets of $C$. Or, we could take $C_0$ to be the underlying groupoid of $C$ consisting of all its objects and isomorphisms, and $C_1(c,c')$ as before to be the discrete groupoid corresponding to the homset. The former kind of category is called strict, the latter univalent. (There are also ways to pick out these two kinds of category inside the type theory, and a strong argument that the univalent ones are those that should be called simply "categories" for practical purposes.) Finally, if $C$ and $D$ are univalent categories, then the functor category $D^C$ is also univalent. Thus, if $F,G:C\to D$ are ordinary functors such that $F\cong G$, and we embed $C$ and $D$ in the groupoid model of type theory as univalent categories, then $F$ and $G$ become objects of the univalent category $D^C$ that are actually equal as elements of its type of objects $(D^C)_0$, i.e. we have an inhabitant of $F=_{(D^C)_0}G$. Thus, any statement expressible in type theory that is true about $F$ must also be true about $G$. (This is arguably a fancier version of Freyd's argument for categories, which IIRC also uses dependent type theory to characterize the properties of categories that are invariant under equivalence.)<|endoftext|> TITLE: On minimal Kan simplicial sets having finite number of simplexes in each dimension QUESTION [9 upvotes]: What are the examples of “tame” minimal Kan simplicial sets having finite number of simplexes in each dimension besides simplicial point and $B(G)\approx K(G,1) $ for a finite group $G$? I believe that Alain Connes’ simplicial circle is also minimal Kan. REPLY [2 votes]: A tame minimal Kan complex has finite homotopy groups. The converse is also true: if homotopy groups of a minimal Kan complex are finite, then there are finitely many $n$-simplices. The proof is by induction on $n$. First, $X_0 = \pi_0(X)$ is finite. If $n > 0$, there are finitely many possible choices of boundaries for $n$-simplices by induction hypothesis. For every such boundary, the set of its fillers is either empty or isomorphic to $\pi_n(X,x_0)$ for some $x_0$. This completes the proof. This fact implies that, for every Kan complex with finite homotopy group, there is an equivalent tame minimal Kan complex.<|endoftext|> TITLE: Continuity of a differential of a Banach-valued holomorphic map QUESTION [6 upvotes]: Originally posted on MSE. Let $U$ be an open set in $\mathbb{C}^{n}$ let $F$ be a Banach space (in my case even a dual Banach space), and let $\varphi:U\to F$ be a holomorphic map. I seem to be able to prove that the differential map $D\varphi:U\times\mathbb{C}^{n}\to F$ defined by $$D\varphi (z,v)= \lim\limits_{t\to 0}\frac{\varphi(z+tv)-\varphi(z)}{t}$$ is holomorphic. Is there a reference for this assertion? (Or at least for continuity) I tried to look into some sources on infinite-dimensional holomorphicity and could not find such a statement, but some of those sources are rather complicated, and so it is likely I missed it. REPLY [3 votes]: You have a complete detailed treatment of the question of holomorphic functions (there called "analytic") with values in a Banach space in A. Dieudonné, Foundations of Modern Analysis, Acad. Press (1960, enlarged and corrected printing) Chapter 9, in particular 9.6.3 and 9.9.4 for Cauchy formulas for several (complex) variables Banach-valued functions. (indeed 9.9.4 is the weak form of Hartogs theorem for Banach-valued functions, see also 9.9 exercise 3 and the subsequent remark). This implies that, calling $\mathcal{H}(U;F)$ the space of holomorphic functions on $U$ with values in $F$, the operator $D\varphi$ is holomorphic. This property extends to functions with values in any Hausdorff locally convex complete TVS $F$. Of course, this needs a definition. We can take the most natural one which is that A function $f:\ U\to F$ is said holomorphic if for every $a\in U$, there exists an open polydisk $P\subset U$ of center $a$ in which $f(z)$ is the sum of an absolutely convergent power series in the variables $z_k-a_k$. Considering a complete system of seminorms $p_i$ of $F$, and the Banach spaces $B_i=\widehat{F/ker(p_i)}$ we see that this is equivalent to say that, for every Banach space and continuous morphism $\ell:\ F\to B$, the composition $\ell\circ f$ is holomorphic. In other words, this comes from the fact that any locally convex complete TVS $F$ is a dense subspace of the projective (or inverse) limit of Banach spaces.<|endoftext|> TITLE: Sufficient criterion for a simplicial sphere to be polytopal QUESTION [5 upvotes]: Are there any purely combinatorial criteria which allow you to deduce that a spherical simplicial complex is polytopal (i.e., there exists a simplicial polytope whose boundary is isomorphic to it)? For example, I was kind of hoping that shellability might be enough, but this is not true. All 3-spheres on up to ten vertices are shellable, but not all are polytopal, as I learned from this paper by Frank Lutz. This question is not the same as the question What is the best way to test if a sphere is a polytope since that question (at least as it was answered) considers how to carry out the test on any input, while I am interested in a sufficient criterion (which might work only rarely, but hopefully on nice examples). REPLY [3 votes]: I can think of a few purely combinatorial criteria, that allow to deduce realizability as a polytope. All d-polytope with at most d+2 vertices is realizable Stacked polytopes. (It can be easily combinatorially checked if a simplicial complex is stacked.) Those and some more obscure criteria a mentioned for example in a paper by Stefan Felsner and Sarah Kappes: https://arxiv.org/abs/math/0602063v1<|endoftext|> TITLE: Automorphisms and epimorphisms of finite groups QUESTION [9 upvotes]: All groups in this question are finite, and epimorphism means surjective group homomorphism. Suppose I have two epimorphisms $f,g\colon G\to H$. This implies that $\ker(f)$ and $\ker(g)$ have the same composition factors, but they need not be isomorphic. I'll say that $f$ and $g$ are compatible if $g=fh$ for some automorphism $h$ of $G$. This would imply that $\ker(g)\simeq\ker(f)$, so it is not always true. I ask: does there always exist $K$ and an epimorphism $p\colon K\to G$ such that $fp$ and $gp$ are compatible? If $G$ is nilpotent we can reduce to the case where it is a $p$-group, then I think we can take $K$ to be the initial example of an $k$-generator group of exponent $p^n$ and nilpotence class $c$, for sufficiently large $k$, $n$ and $c$. In particular, if $G$ is an abelian $p$-group I think we can take $K=C_{p^n}^k$ for sufficiently large $k$ and $n$. But I am not sure what to do when $G$ is not nilpotent. REPLY [13 votes]: You could start with $K$ being the free group with the elements of $G$ as its free basis, with $p$ the obvious map onto $G$. Then the required automorphism $h$ of $K$ to make the two composite maps compatible is just a permutation of the free basis elements of $K$. Of course this $K$ is infinite, and you are looking for a finite group. But to get that, we could replace $K$ by $K/N$, for any characteristic subgroup $N$ of finite index in $K$, and with $N \le \ker p$. You could take $N$ to be the intersection of the kernels of all epimorphisms from $K$ to $G$, of which there are finitely many.<|endoftext|> TITLE: Examples of independent $\Sigma_4^1$ statements QUESTION [5 upvotes]: As in the title, I'm looking for examples of $\Sigma^1_4$ (preferably complete) sentences which are independent from ZFC in both ways, namely given a model $V$ we can extend it to $V'$ where such a sentence holds, but also extend it to a model $V''$ where such sentence fails. By Shoenfield Theorem $\Sigma_4^1$ (or $\Pi^1_4$) is the lowest available complexity of such a formula, and that's why I'm looking for such examples. REPLY [7 votes]: As a starting point, think about the sentence "There is a nonconstructible real." This is $\Sigma^1_3$ and clearly not downwards-absolute. However, it is upwards-absolute. To get the desired situation, we "relativize" and consider the sentence There is some real $r$ such that every real $s$ is constructible relative to $r$. (That is, for some $r\in\mathbb{R}$ we have $\mathbb{R}=\mathbb{R}\cap L[r]$.) This is $\Sigma^1_4$, and is neither downwards nor upwards absolute.<|endoftext|> TITLE: What do absolute neighborhood retracts look like? QUESTION [15 upvotes]: In the course of filling in my map of non-pathological topology, I'd like to understand the class of ANRs (Absolute Neighborhood Retracts) as a sort of "neighborhood" of the class of CW complexes. This seems warranted by some of the nice properties of ANRs: Every ANR has the homotopy type of a CW complex. Every ANR is locally contractible, and as a partial converse, any locally contractible finite-dimensional metric space is an ANR. But there are also important infinite-dimensional examples of ANR's: The Hilbert cube is an ANR. Many function spaces are ANRs. This leaves me with some Questions: What is a good example of a finite-dimensional ANR which is not a CW complex? Are (finite-dimensional) ANRs an appropriate setting to study either (a) fractals (wikipedia seems to define a fractal to be a subset of Euclidean space whose topological and Hausdorff dimensions differ) or (b) the limit sets of dynamical systems on CW complexes? I think my sense is that both (a) and (b) are generally wilder than ANRs, but I'm not really sure -- perhaps there's some overlap but no strict containments? Do ANRs admit some kind of "generalized cell structure" like CW complexes do? Or is there some other sense in which ANRs can be "classified"? Is there at least a "classification" of what ANRs can look like locally? Less precisely, my feeling is that when somebody says "Let $X$ be a CW complex", I sort of know what they mean. But when somebody says "Let $X$ be an ANR", I don't -- I don't know what to think of as a "typical example", nor do I know what kinds of "typical pathologies" to watch out for. It would be nice if there were a book out there entirely devoted to the topology of ANRs but surprisingly I haven't been able to find one . I found a book Theory of Retracts by Sze-Tsen Hu, but I haven't yet found an example in it of a finite-dimensional ANR which is not a CW complex. EDIT: Another reference is Borsuk's The Theory of Retracts. This contains more examples in the later chapters, though I'm still struggling to piece together a coherent picture of the diversity of ANRs. An important piece of context regarding (1): according to Thm V.10.1 of Borsuk, the (compact) finite-dimensional ANRs coincide with the retracts of (finite) polyhedra. Thus in finite dimensions, the question is: How wild can an idempotent on a polyhedron be?. REPLY [2 votes]: For Euclidean neighborhood retracts, there is the nice characterization of being locally connected (and locally compact). Unfortunately, in the infinite-dimensional case, locally connectedness is necessary but only "almost" sufficient, though there is the nice theorem that being an ANR is a local property.<|endoftext|> TITLE: Is the injectivity radius (semi) continuous on a non-complete Riemannian manifold? QUESTION [13 upvotes]: Let $\mathcal{M}$ be a Riemannian manifold, and let $\operatorname{inj} \mathrel\colon \mathcal{M} \to (0, \infty]$ be its injectivity radius function. It is known that if $\mathcal M$ is connected and complete, then $\operatorname{inj}$ is a continuous function: see for example [Lee, Introduction to Riemannian Manifolds, 2018, Prop. 10.37]. What is known in the case where $\mathcal M$ is not complete? Is $\operatorname{inj}$ also continuous? If not, is there a known counter-example? Would $\operatorname{inj}$ still be semi-continuous? This question is similar to the question "The continuity of Injectivity radius", but the discussion there focuses on compact or complete manifolds. REPLY [5 votes]: The proofs provided here by Stephen M and Jack Lee now appear in my book about Riemannian optimization: An introduction to optimization on smooth manifolds. See Section 10.8. The answer is: yes, the injectivity radius is a continuous function even if the Riemannian manifold $\mathcal{M}$ is not complete. Thank you both!<|endoftext|> TITLE: Cobordism Theory of Topological Manifolds QUESTION [8 upvotes]: Unfortunately, due to my ignorance, my present knowledge is limited to the cobordism Theory of Differentiable Manifolds. Cobordism Theory for DIFF/Differentiable/smooth manifolds However, there are Topological Manifolds which are not Differentiable Manifolds. So my question here for experts is that what do I need to beware and pay attention in order to master a cobordism theory of Topological Manifolds? What are the main differences of the computations of the bordism groups for the given following structures: Say, Cobordism Theory of TOP/topological manifolds Cobordism Theory for PDIFF/piecewise differentiable manifolds Cobordism Theory for PL/piecewise-linear manifolds p.s. Are there Spin, Pin$^+$, and Pin$^-$ versions of these cobordism theories of Topological Manifolds computed in the literature explicitly? REPLY [8 votes]: My memory may be a bit faulty, but I think I should say something since I know quite a lot about calculations in the Top case. There is a 1966 paper "Cobordism of combinatorial manifolds'' of Williamson that works out the PL case geometrically (as in Thom). As follows from a 1966 paper "Cobordism theories" by Browder, Liulevicius, and Peterson, the problem of computing PL or Top cobordism is essentially cohomological at the prime 2, as is worked out in the cited Annals study of Madsen and Milgram. At odd primes, it follows from Kirby-Siebenmann that the Top and PL cases coincide. The calculation there can be tackled by the Adams spectral sequence. I have an unpublished preprint (with Ligaard, Mann, and Milgram) that goes quite a long way towards this calculation, but certainly without full information. I hope to get a paper out eventually. The calculation relies on my paper ''The homology of $E_{\infty}$ ring spaces'' in Springer Volume 533 "The homology of iterated loop spaces'' (with Cohen and Lada) that gives calculations of the homology of $BTop$ and related spaces that are the essential starting point. A short paper "The Bockstein and the Adams spectral sequence'' by Milgram and myself compares the two cited spectral sequences quite generally. That comparison in the case of $MTop$ is surprisingly helpful. But I should apologize for the decades long delay in getting our calculations published.<|endoftext|> TITLE: Mathematical Techniques to Reduce the Width of a Gaussian Peak QUESTION [5 upvotes]: In the chemical analysis by instruments, the signals of several molecules are overlapped which makes it difficult to determine the true area of each peak, such as those shown in red. I simulated this as a sum of six Gaussians (with some tailing elements) One of the simplest technique is to raise the discrete signal values to any positive power (n>0). The standard deviation of the Gaussians becomes smaller and smaller (C, in blue). The big drawback is that we lose all the original peak area information. The transformed data is highly resolved now at the cost of losing true area information. Alternatively, we can add a first derivative of the signal and subtract the second derivative from the original signal i.e., Sharpened signal= Original signal +K (first derivative) - J(second derivative) K and J are small positive real numbers. This neat "trick" maintains the true area because area under the derivatives is negligible (zero in ideal cases). Do mathematicians use any other transformations which can make each overlapping peak very narrow, yet maintain the original peak areas. I am not interested in curve fitting techniques at this moment. Any pointers to some similar "peak sharpening" transformations would be appreciated which can resolve overlapping signals. Thanks. REPLY [5 votes]: OK, here are my 2 cents. I'll try to outline the logic and then make a conclusion. We'll start with the Gaussian case. Suppose that we have a Gaussian peak $f(t)=e^{-t^2/2}$. We want to de-convolve it to the Dirac delta-measure. Since the Fourier transform $\widehat f(s)=\int f(t)e^{-ist}\,dx$ satisfies $\widehat{(f*g)}=\widehat f \widehat {g{,}}$ and the Fourier transform of the standard Gaussian is (up to a constant factor) the standard Gaussian again, the naive but natural idea is just to multiply $\widehat f$ by $e^{s^2/2}$ and to take the inverse Fourier transform of the resulting identically $1$ function. Unfortunately this is unfeasible as written because of the noise, finite computational precision, etc., so we have to truncate the natural Fourier mulitplier $e^{s^2/2}$ somehow. The idea is to multiply it by some even positive cutoff function $\Psi$ with bounded support. Then the noise will be amplified (on the worst frequency) by $$ M=\max_s e^{s^2/2}\Psi(s) $$ but instead of the Dirac point mass, we will get the inverse Fourier transform of $\Psi$. We want that inverse Fourier transform to be as concentrated near the origin as possible. We also want to make it a non-negative function rather than some sign-changing wave. Since we must have $\Psi(0)=1$ to preserve the integral of $f$, we come to the problem of choosing an even function $\Psi$ with bounded support (say $[-\pi,\pi]$; we can always scale later) such that $\Psi(0)=1$, the inverse Fourier transform $\Psi^\vee\ge 0$, and $-\Psi''(0)=\int\Psi^\vee(t)t^2\,dt$ is minimized (other measurements of concentration give slightly different optimizers but pretty much the same results). This problem is solvable and the solution is the (properly normalized) convolution of $\chi_{[-\pi/2,\pi/2]}(s)\cos s$ with itself, i.e., $$ \Psi(s)=\frac 1\pi[(\pi-|s|)\cos s+\sin|s|] $$ (I'll skip the derivation). Thus, after various scalings, the final family of multipliers is $$ H_{\alpha,K}(s)=e^{\alpha^2 s^2/2}\Psi(K^{-1}\alpha s)\,. $$ and the "peak sharpening formula" is $f\mapsto (H_{\alpha,K}\widehat f)^\vee$. It turns out that $K=1$ does not noticeably change the picture (but filters out the high frequency noise from the measurements). The noise amplification grows pretty fast with $K$. The corresponding maxima are about $$ \begin{matrix} K &\max \\ 1.0 & 1.09 \\ 1.1 & 1.55 \\ 1.2 & 2.65 \\ 1.3 & 5.33 \\ 1.4 & 12.4 \\ 1.5 & 33.2 \end{matrix} $$ (with noise at $1$ to $10\%$ of the signal, the higher values of $K$ are pretty useless), so I would recommend setting $K$ between $1.3$ and $1.4$. The whole game is choosing $\alpha$. The legitimate choice is up to the width of the narrowest peak but somewhat higher values are also sort of admissible except you'll get peaks exchanging energy a bit. The rule of thumb is to watch for noticeable negative values in the transformed signal: once they appear, you certainly went too far. Before that moment you are reasonably safe (but not entirely foolproof). The only difference for EMG is that you want to add one more factor $1+i\beta s$ to your multiplier, i.e., to play with the family $$ H_{\alpha,\beta, K}(s)=e^{\alpha^2 s^2/2}(1+i\beta s)\Psi(K^{-1}\alpha s)\,. $$ If $\beta>0$ gets large, it may force you to make $K$ smaller. The effect of going too far in $\beta$ is the same: you'll see noticeable negative values in the transform plus the peaks will noticeably shift to the left. Playing with $2$ parameters is harder than with one and I have no clear advice on in which order to adjust them to get the best results. Just try and let me know what you think. . The original signal is in green, the dots are the actual locations at which EMG's are "centered". Something like that is possible IF you can guess one particular parameter right. I still cannot figure out how to teach the machine that guessing. You'll, probably, like this one too: Or this one: The current version of the code. This is a dirty homemade contraption, of course, but it is totally automatic. import graph; access settings; //settings.outformat="png"; size(500,300,IgnoreAspect); int sec=seconds(); srand(sec); //Just generates a new picture every time //srand(1562245957); //was really bad, fixed //srand(1562264140); //was bad, fixed //srand(1562265837); //still bad bool ON=false; //Set to true if you want to see suppressed low values real beta=0.04; //Noise level (maximum; you need to adjust the code a bit to use the average) real Beta=0.12; //Cutoff level (approximately 3 times maximal noise) real K=1.7, L=3, TOL=2.5; //Don't change these UNLESS int res=8, dirr=5;//you understand what you are doing int M=4; //The number of peaks. Set to 1 to see the individual response. int n=4096; //The sampling range (4096 time ticks with no signal before and after) real h=1/n; real QQ=0.07*unitrand(); //Average lambda; set to 0 to see pure Gaussians real AA=0.15*sqrt(unitrand()); //Average sigma; set to zero to see pure exponentials real det=unitrand(); //Peak spacing from equidistant det=1 to random det=0 real[] A,Q,X,Y; for(int m=0;mpi) return 0; else //return sin(s)/s*(1-s/pi); return ((pi-s)*cos(s)+sin(s))/pi; } pair R(pair x) //Suppression of low values. { if (ON) return x; real y=max(x.x-Beta,0); if(x.x<0) return (0,0); else return ((x.x<4*Beta? y*4/3:x.x),x.y)/(1-4*Beta^2); } real expp(real x) { return 1+x+x^2/2+x^3/12;//A bit more stable than the true exp return exp(min(x,15)); } struct T {pair[] p; real u; real q;} int count=0; real[] pM; pM[n]=0; for(int k=n-1; k>-1; --k) pM[k]=max(p[k].x,pM[k+1]); pair[] pp=copy(p), p1; T U(pair D, real qq) { ++count; write("U "+string(count)); real u=0,v=1; void proc() { for(int kk=0; kk<10;++kk) { real w=(u+v)/2; real a=w*D.x, q=w*D.y; real b=min(max(a*sqrt(2*pi)/K,2*pi*q/L),1/5); pair[] G=fft(pp)/n, mult; p1=copy(pp); for(int k=0;kTOL) mult[k]=TOL*unit(mult[k]); G[k]*=mult[k]; } pair[] ppp=fft(G); ppp=reverse(ppp); ppp.cyclic=true; bool flag=true; //for(int k=0;kTOL) flag=false; for(int k=floor(n*qq-n/res);k-1 && kBeta) ) flag=false; } if(flag && abs(G[floor(n/2)-1])<1/10^10) {u=w; p1=copy(ppp);} else v=w; } } proc(); p1.cyclic=true; if(p1[n-1].x>Beta) {write(p1[n-1].x,pM[n-1]); pause();} real q=0; for(int k=floor(n*qq-n/res);k-1; --m) { D=dir(90*m/dirr); T ttt=U(D,k/res); if(ttt.q>W) {W=ttt.q; tt[k]=ttt;} } //write (tt[k].u, tt[k].q); pp=copy(tt[k].p); } for(int k=0;k0$ really matter. A simple choice is the $L^2$ norm for the residual and the variation norm (aka Radon norm) for the measure $\mu$. This provably leads to a spiky reconstruction and the process is stable.<|endoftext|> TITLE: Trace in the category of propositional statements QUESTION [10 upvotes]: By the result in this paper, there exists a categorification of the trace of a linear operator that generalizes to any endomorphism of a dualizable object in a symmetric monoidal category ($\textbf{Vect}^\text{Fin Dim}_k$ obviously satisfies these requirements, with every object dualizable). Consider the category $\textbf{Prop}$ of propositional statements where the objects are statements with a well-defined truth value, and the morphisms are implication (in particular, $\operatorname{Hom}_{\textbf{Prop}}(A,B)$ is either a singleton set, when $A$ implies $B$, or empty when $A$ does not imply $B$). The categorical product of $A$ and $B$ is $A \land B$, the coproduct is $A \lor B$. The initial object of the category is $0$, the canonical false statement, which implies everything by the principle of explosion, and the terminal object is $1$, the tautology. $\textbf{Prop}$ can be made symmetric monoidal with either $\land$ or $\lor$ as the "tensor product," with $1$ as the identity in the former, and $0$ in the latter. Objects in $\textbf{Prop}$ are dualizable in the obvious way, using $\lnot$. Every object has exactly one endomorphism, its identity. What is the trace of the identity operator of a given statement in either case? I've tried to work this out from the definitions in the paper, but haven't had any luck. Is there some issue with the definition of this category? It's entirely possible that $\textbf{Prop}$ is not well-defined as I've defined it. REPLY [15 votes]: To start with, I just want to make sure no one gets the impression that the categorical notion of trace was introduced by the paper you linked to; however "semi-famous" it might or might not be, it's only an exposition of material that's been well-known for some time. As to your question, your category $\bf Prop$ is also known as the Lindenbaum-Tarski algebra of the underlying logic. Assuming classical logic, it is a Boolean algebra. However, it has no interesting traces relative to either $\wedge$ or $\vee$, since it has no interesting dualizable objects: the only $\wedge$-dualizable object is $1$ and the only $\vee$-dualizable object is $0$. More generally, as noted in Example 3.6 of the paper you linked to, the only dualizable object in a cartesian monoidal category is the terminal object, and dually the only dualizable object in a cocartesian monoidal category is the initial object. In particular, $\neg A$ is not a dual of $A$ in this categorical sense. There is a categorical sense in which $\neg A$ is a dual of $A$, but it requires treating $\bf Prop$ not as a monoidal category with respect to $\wedge$ or $\vee$ separately, but as a linearly distributive category that involves both $\wedge$ and $\vee$ together. This more general sort of dualizability is also important, but unfortunately it doesn't come along with a corresponding notion of trace.<|endoftext|> TITLE: Other homotopy invariants? QUESTION [8 upvotes]: The idea of using maps from a sequence of simple standard objects into a topological space $X$ as $probes$ to explore its topology is ubiquitous. One organizes these maps into equivalence classes in such a way that the collection of classes acquires a nice algebraic structure. These algebraic invariants then serve to recognize $X$ or distinguish it from others. One such sequence is, of course, pointed $n$-spheres, homotopy classes of maps from which yield homotopy groups, $\pi_n (X)$. Has it been useful to consider other sequences of simple spaces for construction of invariants, e.g., homotopy classes of maps from $n$-tori, or from genus $n$ tori? Or can these always be simply expressed in terms of homotopy groups, and are, therefore, redundant? Or too hard to compute? Or lack good properties? Or ... REPLY [16 votes]: A hot topic for 20 years starting in the mid 1980's was the exploration of spaces by `probing' them with the spaces $BV$ where $V$ is a group of the form $(\mathbb Z/p)^n$ with $p$ a prime. It is a theorem of Jean Lannes, building on work of Haynes Miller, that, under remarkably mild hypotheses, the set of maps $[BV,X]$ is a very computable functor of $H^*(X;\mathbb Z/p)$, viewed as an algebra equipped with Steenrod operations. Even better: $H^*(Map(BV,X);\mathbb Z/p) = T_VH^*(X;\mathbb Z/p)$, where $T_V$ is a wonderful algebraic functor discovered by Lannes. This had many applications to a wide range of problems, ranging from the classification of polynomial rings that can be realized as the cohomology of a space, to the theorem that, if $H^*(X;\mathbb Z/p)$ has infinite total dimension as a $\mathbb Z/p$ vector space, then it must also be infinitely generated as a module over the Steenrod algebra.<|endoftext|> TITLE: Is there a connection between representation theory and PDEs? QUESTION [13 upvotes]: As a PhD student, if I want to do something algebraic / linear-algebraic such as representation theory as well as do PDEs, in both the theoretical and numerical aspects of PDEs, would this combination be compatible and / or useful? Is it feasible? I'd be grateful for an online resource to look into. Thanks, REPLY [6 votes]: The book "D-Modules, Perverse Sheaves, and Representation Theory " by Ryoshi Hotta, Kiyoshi Takeuchi and Toshiyuki Tanisaki is the perfect source for this topic. The introduction gives a very nice (and elementary) explanation how representation theory of D-modules and symstems of partial differential equations are related. I just give a very nice excerpt from the introduciton of the book. Let $X$ be an open subset of $\mathbb{C^n}$ and $\mathcal{O}$ the commutative ring of complex analytic functions defined on $X$. Let $D$ be the set of partial differential operators with coefficients in $\mathcal{O}$, whose elements are thus of the form $\sum\limits_{i_1,...,i_n}^{\infty}{f_{i_1,...,i_n} (\frac{\delta}{\delta x_1})^{i_1} ... (\frac{\delta}{\delta x_n})^{i_n}} $. Let $P \in D$ and consider the partial differential equation $Pu=0$ and $M$ the D-module $M=D/DP$. We then have $Hom_D(M,\mathcal{O}) \cong \{f \in \mathcal{O} | Pf=0 \}$. This shows that the set of analytic solution of $Pu=0$ is isomorphic to a $Hom$-space, which are the natural objects of study of representation theory (representation theory can be summarized more or less as the study of representations of rings and their Hom-spaces).<|endoftext|> TITLE: Multiplication and division by a morphism under the “inner composition” in closed monoidal categories QUESTION [7 upvotes]: I asked this a week ago at math.stackexchange, without success, so I hope it will be appropriate here. Let ${\mathcal C}$ be a symmetric closed monoidal category, and let me denote the internal hom-functor by a fraction $$ (X,Y)\mapsto\frac{Y}{X}, $$ so that we have an isomorphism of functors $$ \operatorname{Mor}(A\otimes B,C)\cong \operatorname{Mor}\left(A,\frac{C}{B}\right). $$ As is known, ${\mathcal C}$ is an enriched category over itself. For each objects $A,B,C$ let me denote by $\bullet_{A,B,C}$ the "inner composition" in ${\mathcal C}$ as in an enriched category, i.e. the morphism $$ \bullet_{A,B,C}:\frac{C}{B}\otimes\frac{B}{A}\to\frac{C}{A} $$ with the corresponding properties. I wonder if the following identity always holds $$ \bullet_{A,C,D}\circ\left(1_{\frac{D}{C}}\otimes\frac{\varphi}{1_A}\right)= \bullet_{A,B,D}\circ\left(\frac{1_D}{\varphi}\otimes1_{\frac{B}{A}}\right) $$ (for arbitrary objects $A,B,C,D$ and for arbitrary morphism $\varphi:B\to C$). This is strange, I can prove this only in the case when the unit $I$ is a separating object in ${\mathcal C}$ (what does not always hold). Is it possible that there is a counterexample? REPLY [7 votes]: Noam’s answer is already good and complete, but here is alternate answer phrased more elementarily in terms of the symmetric monoidal category structure. We want to show $$\bullet_{A,C,D}\circ\left(1_{[C,D]}\otimes[1_A,\varphi]\right)= \bullet_{A,B,D}\circ\left([\varphi,1_D]\otimes 1_{[A,B]}\right) : [C,D] \otimes [A,B] \to [A,D] $$ It’s enough to show their corresponding exponential transpose maps $[C,D] \otimes [A,B] \otimes A \to D$ agree. But for these, we have: $$ \newcommand{\ev}{\operatorname{ev}} \ev_{A,D} \circ \left( (\bullet_{A,C,D} \circ(1_{[C,D]}\otimes[1_A,\varphi])) \otimes 1_A \right) \\ = \ev_{A,D} \circ (\bullet_{A,C,D} \otimes 1_A)\circ\left(1_{[C,D]}\otimes[1_A,\varphi] \otimes 1_A\right) \\ = \ev_{C,D} \circ (1_{[C,D]} \otimes \ev_{A,C})\circ \left(1_{[C,D]}\otimes[1_A,\varphi] \otimes 1_A\right) \\ = \ev_{C,D} \circ (1_{[C,D]} \otimes (\ev_{A,C} \circ ([1_A,\varphi] \otimes 1_A))) \\ = \ev_{C,D} \circ (1_{[C,D]} \otimes (\varphi \circ \ev_{A,B})) \\ = \ev_{C,D} \circ (1_{[C,D]} \otimes \varphi) \circ (1_{[C,D]} \otimes \ev_{A,B}) $$ and a similar calculation shows $$ \ev_{A,D} \circ \left( (\bullet_{A,B,D}\circ([\varphi,1_D]\otimes 1_{[A,B]})) \otimes 1_A \right) = \ev_{C,D} \circ (1_{[C,D]} \otimes \varphi) \circ (1_{[C,D]} \otimes \ev_{A,B}).$$ Each step is just using either the monoidal category laws, or the fact that the maps $\bullet_{A,C,D}$, $[1_A,\varphi]$ are defined by their exponential transpose maps. (Please excuse my using the notation $[A,B]$ for mapping objects, since I’m more familiar with it than the fraction notation $\frac{B}{A}$.)<|endoftext|> TITLE: Classify 2-dim p-adic galois representations QUESTION [7 upvotes]: Recently I have known how to classify 1-dim p adic Galois representations $\phi$. The p-adic Galois representations mean that a representation $G_K$ on a p-adic field $E$, where $K$ is also a p-adic field. And a p-adic field means a finite extension of $Q_p$. (1) $\phi$ is hodge-Tate if and only if $\phi$ is de Rham if and only if $\phi\chi^n$ is potentially unramified for some integer $n$. (2)$\phi$ is crystalline if and only if $\phi$ is semi-stable if and only if $\phi\chi^n$ is unramified for some integer $n$. Symbols: $\chi$ is the cyclotomic character of $G_K$. I want to know how to classify 2-dim p-adic Galois representations. When it is Hodge-Tate or de Rham or crystalline or semi-stable? Is there some reference on this? I prefer English reference to Frence reference because I am a beginner on French... Thanks for any answers. REPLY [3 votes]: As Loeffler points out, the classification of $2$-dimensional $p$-adic crystalline Galois representations is complicated. There has been considerable interest in classifying (in a way that is explicit and thus what you are probably seeking) the residual representation $\bar{\rho}_{\restriction D_p}^{ss}$ as $\rho=\rho_f$ ranges over the crystalline Galois representations associated to a cuspidal Hecke-eigenform $f$. Let me quote Buzzard and Gee from "Explicit Reduction Mod $p$ of Certain $2$-dimensional Crystalline Representations" "Question: If $f=\sum_n a_n q^n$ be a normalized cuspidal level $N$ eigenform and $p$ is a prime, and if $\bar{\rho}_f$ is the associated semisimple representation, then can one explicitly read off $\bar{\rho_f}_{\restriction D_p}^{ss}$ from the weight character and $q$-expansion of $f$?" This question is interesting when $f$ ranges over eigenforms which are supersingular at $p$ and has proved to be a whole lot more tractable and yielded interesting results though is not yet settled. Let $\lambda\in \mathbb{Q}_{>0}$ be the slope of $a_p$ (with respect to the normalization with respect to which the slope of $p$ is 1 and a chosen embedding of the field of fourier coefficients of the eigenform). As far as I''m aware the classification is for $\lambda<3$ and the number of possibilities increases considerably as $\lambda$ exceeds $3$. This is a culmination of the results of Berger-Li-Zhu, Buzzard-Gee, Ganguli-Ghate and Pande. Thus in response to what you probably have in mind, the most natural step to pursue would be to try and extend this classification to some slopes $\lambda>3$.<|endoftext|> TITLE: On some analytic property of the Riemann zeta function QUESTION [6 upvotes]: Denote by $\zeta$ the Riemann zeta function. For $\Re(s)=\sigma>0$, it is well known that $$\sum_{n\leq x} n^{-s} = \zeta(s) + \frac{x^{1-s}}{1-s}+ O(x^{-\sigma}).$$ But do there exist infinitely many $x$ such that $$ \Bigg|\sum_{n\leq x} n^{-s} - \zeta(s) - \frac{x^{1-s}}{1-s} \Bigg| \gg x^{-\sigma} ?$$ ADDENDUM: Since the left-hand side is equal to $\Big|s\int_{x}^{\infty} \lbrace u \rbrace u^{-s-1}\mathrm{d}u\Big|$, the problem amounts to finding a lower bound for this integral, where $\lbrace y \rbrace$ denotes the fractional part of $y$. REPLY [7 votes]: $$\int_x^\infty \{ u\} u^{-s-1} du = \int_x^\infty \left( \{ u\}-\frac{1}{2}\right) u^{-s-1} du + \frac{1}{2} \frac{x^{-s}}{s}$$ and by integration by parts, $$\int_x^\infty \left( \{ u\}-\frac{1}{2}\right) u^{-s-1} du= - \int_x^\infty \left(\int_x^u \left( \{ t\}-\frac{1}{2}\right)dt \right) (-s-1) u^{-s-2} du=$$ $$= O\left( |s+1| \int_x^\infty u^{- \sigma -2 } \right) = O\left( |s+1| x^{-\sigma-1} / |\sigma +1| \right)$$ so the $\frac{1}{2} \frac{x^{-s}}{s}$ term dominates and you get a lower bound.<|endoftext|> TITLE: Equivalence of $\sigma$-weak topology to another topology QUESTION [6 upvotes]: Let $\mathcal H$ be a Hilbert space. Define a topology $\tau_1$ on $B(\mathcal H)$ by the family of seminorms $x\mapsto |Tr(xa)|,$ $a\in L^1(B(\mathcal H)).$ Here $B(\mathcal H)$ denotes the set of all bounded linear maps on $\mathcal H$ and $L^1(B(\mathcal H))$ denotes the trace class operators. Again define the $\sigma$-WOT topology $\tau_2$ on $B(\mathcal H)$ by pulling back the weak operator topology of $B(\mathcal H\otimes\ell_2)$ to $B(\mathcal H)$ via the map $x\mapsto x\otimes 1.$ How to show that $\tau_1=\tau_2$? In many books and lecture notes in von Neumann algebras they have just mentioned that this is true. But I could not find a solid proof. REPLY [6 votes]: I think this is just a definition chase, and use of the standard representation of a trace-class operator. Starting with the later, following Chapter II in the first volume of Takesaki, for example, we have that any trace-class operator $a$ can be written as $$ a(\xi) = \sum_n a_n (\xi|\xi_n) \eta_n \qquad (\xi\in H), $$ where $(\xi_n),(\eta_n)$ are orthogonal sequences (perhaps finite) in $H$ and $(a_n)$ is a sequence of positive reals with $\sum_n a_n<\infty$. Further, the trace-class norm of $a$ is $\sum_n a_n$. Given such a representative, let $\alpha_n = a_n^{1/2}\xi_n$ and $\beta_n = a_n^{1/2}\eta_n$ we find that $a(\xi) = \sum_n (\xi|\alpha_n)\beta_n$ and $\sum_n \|\alpha_n\| \|\beta_n\| = \sum_n a_n < \infty$. Conversely, if $(\alpha_n), (\beta_n)$ are any sequences in $H$ with $\sum_n \|\alpha_n\| \|\beta_n\| < \infty$ we have $$ B_0(H)\ni x \mapsto \sum_n (x\alpha_n|\beta_n) $$ defines a bounded linear functional on the compact operators $B_0(H)$, and hence there is a trace-class operator $a:\xi\mapsto \sum_n (\xi|\alpha_n)\beta_n$ with trace-class norm at most $\sum_n \|\alpha_n\| \|\beta_n\|$. We conclude that the $\sigma$-weak topology, $\tau_1$, is given by the seminorms $$ B(H)\ni x \mapsto \Big|\sum_n (x(\alpha_n)|\beta_n)\Big| \text{ where } \sum_n \|\alpha_n\| \|\beta_n\|<\infty. $$ The WOT is defined by seminorms $x\mapsto |(x\xi|\eta)|$. If $B(H)$ acts on $H\otimes\ell_2$ via $x\mapsto x\otimes 1$ then consider the respresentation of vectors in $H\otimes\ell_2$ which is $\xi=\sum_n \xi_n\otimes e_n$ and $\eta=\sum_n \eta_n\otimes e_n$ where $(e_n)$ is the standard unit vector basis of $\ell_2$ and $\sum_n \|\xi_n\|^2, \sum_n \|\eta_n\|^2<\infty$. Hence the induced seminorm on $B(H)$ is $\big| \sum_n (x\xi_n|\eta_n)\big|$. By Cauchy-Schwarz, $\sum_n \|\xi_n\| \|\eta_n\| \leq \Big(\sum_n \|\xi_n\|^2, \sum_n \|\eta_n\|^2\Big)^{1/2}<\infty$. Conversely, if we have $(\alpha_n),(\beta_n)$ with $\sum_n \|\alpha_n\| \|\beta_n\|<\infty$ then by rescaling, we may suppose that $\|\alpha_n\|=\|\beta_n\|$ for each $n$; notice that this does not change the seminorm $x\mapsto \big|\sum_n (x(\alpha_n)|\beta_n)\big|$. Then $\sum_n \|\alpha_n\|^2 = \sum_n \|\beta_n\|^2 < \infty$. The equivalent between $\tau_1$ and $\tau_2$ follows.<|endoftext|> TITLE: Verbal description, or terminology, for the ${\mathcal L}_p$-spaces of Lindenstrauss and Pelczynski QUESTION [10 upvotes]: This question is intended for Banach-space specialists and so I will not repeat all the definitions here. My aim is to find out how the Banach space community refers to such spaces in discussions, and how they go about looking up information on such spaces, given that it does not seem easy to get focused results when searching online. To be a little more precise: are such spaces referred to by anyone as "approximate $L_p$-spaces" or "local $L_p$-spaces"? It seems less than ideal to have an important notion described only by literal typography. By the way: I am aware of the original papers of Lindenstrauss–Pelczynski and Lindenstrauss–Rosenthal, so that is not my question. My question is about the terminology or description that specialists in Banach space theory would use to refer to these spaces, when asking each other questions or giving each other outlines of proofs. REPLY [6 votes]: Just so that this question can be marked as answered: Bill Johnson states in comments that those in the know call these Banach spaces "script $L_p$-spaces", and I trust his awareness/judgement of the norms in the community. I was hoping that in the years since the foundational work, other terminology might have emerged; but so it goes.<|endoftext|> TITLE: Deformations of Vertex Algebras QUESTION [7 upvotes]: As the title suggests, I'm interested in deformation theory of vertex algebras and their representations. In the paper https://arxiv.org/abs/1806.08754, the authors construct, for a vertex algebra $V$ with a module $M$, a family of cohomology spaces $H^{i}_{VA}(V, M)$. Naturally, they in fact construct a complex computing the above and call it vertex algebra cohomology (of $V$ with coefficients in $M$). Further they prove that low degree cohomology groups can be interpreted in the usual way (a version of singular vectors, derivations, extensions etc). In the case of the adjoint representation I believe we obtain a complex controlling the deformation theory of the vertex algebra, as one would hope. I find the calculus of vertex algebras somewhat daunting at times and so I find the construction hard to follow. I'd like to understand it in a simple case, hopefully not so simple as to be completely degenerate. Let $V$ then he a holomorphic vertex algebra, so that for all $v\in V$ the field $v(z)$ is an element of $End(V)[[z]]$. It is not hard to show that such a $V$ is equivalent to the data of a commutative algebra with a derivation. Switching to this language I'll write $(A, \delta)$ for such an object. What is the vertex algebra cohomology of $(A, \delta)$ with cohomolgy in the adjoint representation? If I'm not mistaken, square zero deformations of $(A, \delta) $ are Hochschild cohomology classes $\gamma\in HH^{2}(A)$ such that $Lie_{\delta} (\gamma) =0$. Perhaps this generalizes in the obvious way to other cohomology groups, whatever they are. (Note that the answer should have the structure of a dgla and $Ker(Lie_{\delta})$ indeed has such a structure, in fact that of a Gerstenhaber algebra I believe.) REPLY [5 votes]: The complex considered on that article is a linear algebraic version of a complex constructed by Tamarkin in his ICM address https://arxiv.org/abs/math/0304211 As he points out, in the case of deformations of vertex algebras, instead of a dgla, the complex that controls said deformations is a differential graded Lie conformal algebra. The construction of that complex in the linear algebraic setting is very similar to what we wrote in the article you cite, however there are some technical difficulties that one has to deal with and they have to do with finiteness conditions on the vertex algebra V as a C[T] module. In the case of Tamarkin he considered finitely generated D-modules (which no interesting vertex algebra is). The construction of this complex in the linear algebraic setting will eventually come out. Another comment related to your question (the only one I see is about commutative vertex algebras) is the following. Given a filtered vertex algebra V such that its associated graded A is a Poisson vertex algebra, in particular it is a commutative algebra with a derivation. One can study vertex algebras that deform this given Poisson vertex algebra structure A. This computation was carried by Tamarkin in the case when A is a symmetric algebra. A commutative vertex algebra can be viewed as a Poisson vertex algebra with the zero Poisson bracket (the OPE is regular, or zero). If you look at deformations preserving this Poisson structure you'd get the Hochschild cohomology as you hint. However the second cohomology has more structure corresponding to cocycles deforming the OPE.<|endoftext|> TITLE: Polynomials that share at least one root QUESTION [9 upvotes]: This is a generalization of an MSE question, Polynomials that share at least one root. Let $P(x)$ be a specific polynomial of degree $d$, with given real coefficients $A_i$ ($A_d=1$), and real roots: $$P(x) = x^d + A_{d-1}x^{d-1} + A_{d-2}x^{d-2} + \cdots + A_{0}\;. $$ Q. What does the set of polynomials, with real coefficients $a_i$, $$p(x) = x^d + a_{d-1}x^{d-1} + a_{d-2}x^{d-2} + \cdots + a_{0} \;,$$ look like (geometrically) in $\mathbb{R}^d$, when each $p(x)$ shares at least one root with $P(x)$? I am seeking a description of this set in the space of the $d$ coefficients, $(a_0,\ldots,a_{d-1})$. The reason there is hope for a nice description, is that it makes a pretty picture for $d=2$. Let $P(x) = x^2 + A_1x + A_0$, with $A_1=3$ and $A_0=-1$. The plot in the $a_0 a_1$-plane of all other $p(x)= x^2 + a_1x + a_0$ that share a root with $P(x)$, looks like this:                     Lines intersect at $(a_0,a_1)=(-1,3)$. The discriminant is $a_1^2 = 4 a_0$. All those $(a_0,a_1)$ on the two lines share a root with $x^2 + 3x -1$. The lines are tangent to the discriminant parabola. REPLY [2 votes]: I tried to make a 3D image for $P(x)=x^3+3 x^2-2 x-1$. The set consists of three planes, each tangent to the discriminant surface. But it became too visually complex, partly because the discriminant is complicated. For what it's worth:                     Discriminant surface: blue. Perhaps someone can do better...<|endoftext|> TITLE: How reliable is arXiv to use as a reference in a paper? QUESTION [13 upvotes]: I'm writing a small and simple paper to finish my graduation, I never wrote or published anything before. Is about some irrational/transcendental numbers. I was mentioning some numbers that it's not know to be irrational or not, famous ones are: $e\pi, e+\pi$ and other not so famous like $\ln \pi, e^e, \pi^\pi$. So I was looking where I could find a reference to these statements. And I found on arXiv Some transcendence results from a harmless irrationality theorem that states all that. So I thought: "perfect". However... on arXiv is also: The Product $eπ$ Is Irrational The Zeta Quotient $ζ(3)/π^3$ is Irrational Notice that the two papers above are by the same person (I think this person solved so many open problems that Fields Medal should change the $4$ years rule and give him a medal every week). And I found in arXiv, but can't find the link, that all odd zeta values are irrational. Well, one simple way to check if the paper is trustworthy or not is to see if it has been published somewhere. The first link was published, so everything is ok. But, for example, the proof of the weak goldbach conjecture was posted in arXiv but never published anywhere else. From the wikipedia page Goldbach's weak conjecture : "In 2013, Harald Helfgott published a proof of Goldbach's weak conjecture. As of 2018, the proof is widely accepted in the mathematics community, but it has not yet been published in a peer-reviewed journal." So how much can one trust arXiv to use it as a reference? I think there is a lot there that is true, but is also not published somewhere else, like the weak Goldbach conjecture proof. But also a lot there is wrong and obvious not published. I would appreciate improved tags because I don't know the proper ones. Thanks for the answers! REPLY [35 votes]: In general it is poor scholarly practice to make a final decision about how trustworthy something is solely on the basis of where it appears, whether it's the arXiv or a published journal. As Vladimir Voevodsky eloquently explained some years ago, famous results by famous people in famous journals can still be wrong. In your case, if all you want a reference for is the fact that some constants are not known to be irrational or transcendental, then presumably nothing important in your paper logically depends on this claim being true, so it does not matter all that much how reliable the reference is. If your claim is wrong, the worst that can happen is that someone will be annoyed that their great paper solving the problem was overlooked by you, or someone might work on a solved problem thinking that it is unsolved. So in my opinion it's fine to cite the arXiv paper in this case.<|endoftext|> TITLE: A finite group $G$ all of whose reps are defined over $\mathbb{Z}$ and yet $Rep(G)$ is not generated by permutation representations QUESTION [10 upvotes]: Let $G$ be a finite group and let $Rep(G)$ be its representation ring (as a group it is the free $\mathbb{Z}$-module on the irreducible complex reps). The collection of permutation representations $\mathbb{C}[\mathcal{O}]$ for $\mathcal{O}\cong G/H$ a $G$-orbit generate a $\mathbb{Z}$-subalgebra which we will denote by $Per(G) \subset Rep(G)$. If $Per(G) = Rep(G)$ then it follows that all complex representations of $G$ are defined over $\mathbb{Z}$. To see this note that under our assumption for every representation $V$ there exists a representation $U$ defined over $\mathbb{Z}$ s.t. $U \oplus V = W$ is defined over $\mathbb{Z}$. We can now take $V_{\mathbb{Z}} = W_{\mathbb{Z}} / U_{\mathbb{Z}}$ as a $\mathbb{Z}$-form for $V$ (this is not really a precise proof, maybe a better arguemnt would be to explicitly write the projection operator which projectss onto the isotypic component of $V$ inside $\mathbb{Z}[G]$). Is the converse true? Question: Suppose every $\mathbb{C}$-representation of $G$ has a $\mathbb{Z}$-form, does it follow that $Per(G) = Rep(G)$? If not what's a counter example? As an example when $G=S_n$ both of the properties are satisfied and this is in fact the only non-trivial example I know of. REPLY [16 votes]: The answer is no. Counterexamples include the Weyl groups of types E6, E7, and E8. For a proof that the representations of these Weyl groups are indeed all realisable over $\mathbb{Q}$ (equivalently over $\mathbb{Z}$), see M. Benard, On the Schur Indices of Characters of the Exceptional Weyl Groups, Annals of Mathematics, Vol. 94, No. 1 (Jul., 1971), pp. 89-107 (MSN). For a proof of the statement that $Per(G)\neq Rep(G)$ for these groups, see D. Kletzing, Structure and Representations of Q-Groups, Lecture Notes in Mathematics 1084 (MSN).<|endoftext|> TITLE: Two queries on triangles, the sides of which have rational lengths QUESTION [7 upvotes]: Let us define a "rational triangle" as one in the Euclidean plane, with lengths of all sides rational. We are aware that a positive integer is called "congruent" only if it is the area of a right triangle with rational length sides; so not every rational number is the area of some right rational triangle. However, is every positive rational number the area of some (not necessarily right) rational triangle? for two given rational numbers $A$ and $P$, among the infinitely many general triangles with area $A$ and perimeter $P$ (there are infinitely many such triangles if $A$ and $P$ are within a suitable range), is there a guarantee that there are any (or infinitely many) rational triangles? REPLY [6 votes]: If for the equation derived by Chris Wuthrich, we introduce the new variables $p = 2P$, $q = x - P/2$, $r = y-P/2$, we obtain the more symmetric relation $$ pqr(p+q+r) = A^2. $$ Now I knew I recognized this equation from somewhere, and after a little searching I hit upon some slides by Noam Elkies where your Question 1 is resolved completely. (Elkies further credits Franz Lemmermeyer with bringing this problem under his attention.) It seems that the credit for this goes to Euler. The surface described by the equation above has a K3 surface with maximal Picard rank ($\rho=20$) as its smooth model. Euler presumably used the elliptic fibrations and the presence of many rational curves on the surface to produce a $1$-parameter family of solutions (as suggested by Will Sawin).<|endoftext|> TITLE: Curvature and asphericity of cube complexes QUESTION [5 upvotes]: Let $K$ be a connected cube complex (one may assume that its a cellulation of a smooth, closed manifold). Such a $K$ comes equipped with a length metric (one assumes that each edge is of unit length). It is a well known result of Gromov that if $K$ is non-positively curved (in the sense of Aleksandrov) then its universal cover is contractible; equivalently $K$ is aspherical. Gromov also gave the following criterion to check the curvature: If the link of every vertex in $K$ is flag then $K$ is non-positively curved. Recently I discovered a cube complex structure on some smooth, closed manifolds such that there are vertices whose link is the boundary of a simplex of appropriate dimension. However, some of these manifolds are aspherical (since tori, positive genus surfaces are on that list). So I was wondering whether there are any other (combinatorial) criterion to decide the asphericity of a cube complex? I would also like to know about the papers that prove a certain manifold is aspherical using the combinatorial properties of a regular CW structure on that manifold. REPLY [6 votes]: Regarding your second question, I'm aware of (at least) two combinatorial conditions (beyond Gromov's) for 3-manifolds to be aspherical. Cannon-Floyd-Parry have a combinatorial construction of 3-manifolds by "twisted face pairings". They showed that the 3-manifolds arising from "ample twisted face pairings" are aspherical and have word-hyperbolic fundamental group. However this paper was never published. Elder-McCammond-Meier show that 3-manifolds with a triangulation in which every edge has degree 5 or 6, and every triangle has at most one degree 5 edge, admits a CAT(0) metric, answering a question of Thurston. Regarding your first question about cube complexes, I'm not immediately aware of some combinatorial criterion in the literature. However, there ought to be more general criteria based on Gromov's condition. Suppose one has a vertex in a cube complex with link an $n$-simplex. Then one can glue an $n+1$-cube to this corner to get a simple-homotopy equivalent complex. If this complex satisfies Gromov's condition (links of the vertices are flag), then it is CAT(0) and the original complex is aspherical. For example, this works for a 2-complex which is a union of 3 squares glued cyclically around a vertex. One can try to continue in this way, gluing on higher-dimensional cubes to get rid of "positive" curvature at vertices. If this succeeds, then it gives a combinatorial condition for asphericity. The motivation for this condition comes from Sageev's theory, but I won't delve into this here. Another attempt at a description: suppose one has a cube complex with flag links of vertices (locally CAT(0)), and there are cubes with free faces. Then one can collapse along these faces to obtain a simple-homotopy equivalent cube complex which might no longer be locally CAT(0). One would expect a construction of Sageev to construct cube complexes with higher dimensional cubes which collapse to a manifold associated to e.g. a surface with filling curves, or to a hyperbolic 3-manifold with filling quasi-convex surfaces. Reversing this process gives a sequence of expansions of the cube complex to a CAT(0) complex that one might be able to locate combinatorially, and hence give a combinatorial criterion for asphericity. I haven't thought about whether this might be functorial though - in some situations it might be.<|endoftext|> TITLE: Homotopy type of non-Cohen-Macaulay complexes QUESTION [9 upvotes]: Most interestingly defined (pure) simplicial complexes that occur in topological combinatorics are Cohen-Macualay. Some of these are even shellable (or have the homotopy type of a wedge of spheres; of same dimension). I would like to know examples of pure simplicial complexes (whose simplices are combinatorially defined) that are not shellable. Moreover, can anybody point out papers in which author(s) deal with simplcial complexes that do not have homotopy type of a wedge spheres? REPLY [2 votes]: Here are some examples: There are many papers that deal with topological properties of these complexes and they are rarely Cohen-Macaulay. 1) Triangulated manifolds are very interesting combinatorial objects, and usually, unless the manifold is a sphere or a ball they are not Cohen-Macaulay. Many aspects of the enumerative theory of face numbers of spheres have interesting generalization to manifolds. There is also rich examples of triangulations of specific manifolds with few vertices. 2) Within triangulations of manifolds there is a special role to $d/2+1$ neighborly manifolds when $d$ is even. Their existence for $d=2$ is the Heawood conjecture. For larger values of $d$ only a few examples are known like the Kuhnel 9-vertex triangulation of $CP^2$. 3) Triangulated pseudomanifolds (every co-dimension 1 face is included in two maximal faces + some connectivity assumption) can also be fascinating combinatorial objects. Here are three more interesting examples. 4) Triangulations of algebraic varieties (with singularity) are very interesting combinatorial objects. 5) Quotients of buildings. 6) High dimensional expanders.<|endoftext|> TITLE: Non-associative deformation quantization QUESTION [5 upvotes]: Several physicists consider non-Poisson bivectors but still apply Kontsevich formality in order to get deformation quantization type results: see e.g. Szabos's review An introduction to nonassociative physics. The result they get is a `non-associative star product’. What is the algebraic structure that they get, or, how to state correctly the deformation quantization problem in this non-associative and non-Poisson context? REPLY [3 votes]: This is probably not really an answer to this question, but there are two different context I know where deformation quantization produces something not exactly associative, but associative in a larger sense. This doesn't encompass all cases covered by the question, but might give some hints. a) First, there are so-called quasi-Poisson $G$-manifolds for which the bivector doesn't satisfy the Jacobi identity. Here is the setup. A quasi-Poisson $G$-manifold ($G$ being a Lie group) is a $G$-manifold $M$ together with a bivector $\pi$ and an element $Z\in\wedge^3(\mathfrak{g})^G$ ($\mathfrak{g}$ being the Lie algebra of $G$) and such that $[\pi,\pi]=\vec{Z}$ (here $x\mapsto\vec{x}$ send an element of $\mathfrak{g}$ to the associated fundamental vector field of the action). In other words, the default of the Jacobi identity is determined by $Z$. In this context one can chose an associator $\Phi=1+\hbar Z+\cdots\in(U(\mathfrak{g})^{\otimes3})^G[[\hbar]]$, and consider the monoidal category $\mathcal C_\hbar$ of $\mathfrak{g}[[\hbar]]$-modules whith non-trivial natural associativity isomorphism being given by $\Phi$. A quantization of $\pi$ is then a product on $C^\infty(M)[[\hbar]]$ deforming the usual product of fonctions as an associative monoid in $\mathcal C_\hbar$. Since the forgetful functor $\mathcal C_\hbar\to Vect_\hbar$ is not monoidal, then the quantization doesn't seem associative. But secretely, it is. b) The second situation I have in mind in which one gets non-associative deformation is probably more in the spirit of the what is alluded to in the question. It is the one of twisted Poisson manifolds, where the default of satisfying the Jacobi identity is rather controlled by a closed $3$-form $H$ on $M$ (sometimes refered to as a magnetic charge in more physics papers). Here is what happends: locally, this $3$-form is exact, and thus the twisted Poisson structure is isomorphic to a genuine non-twisted one. on intersections, one can prove that we do have compatibilities... which unfortunately don't glue well (i.e. they don't satisfy the expected cocycle identity). it turns out that things do glue to higher order (getting a gerby version of a sheaf of Poisson algebras). The glueing data is determined by a Cech $3$-cocycle, which represents the same cohomology class in $H^3(M,\mathbb{R})$ as $H$. These gadgets can be naturally deformed to so-called algebroid stacks, which are a generalization of sheaves of algebras, satisfying a higher cocycle condition (sometimes called tetrahedron equation). They can also be seen as linear versions of Giraud's gerbes. A lot of people worked on that in the algebaic and holomorphic contexts: Kontsevich, Yekutieli, Van den Bergh, Halbout and myself, Bressler--Gorokhovsky--Nest--Tsygan, Kashiwara--Schapira, etc... In these contexts these gadgets are necessary for proving existence theorems for deformation quantization of algebraic/holomorphic Poisson structures (indeed, glueing property are known to be harder to handle in these contexts than in the differentiable case). This approach is less known in the differentiable context, though there are references (I guess it would be fruitful to revisit this paper in the light of all the huge work that has been done in the algebraic and holomorphic cases). I'm probably biaised, but it seems to me that all the non-associative structures appearing in deformation quantization are secretely associative in a way: either associativity is satisfied up to homotopy (like in the case of deformation quantization with branes, after Cattaneo--Felder), or it is satisfied in an appropriate monoidal category (like for quasi-Poisson manifolds), or it is satisfied locally but glue in a non-trivial way (like in the twisted Poisson situation), etc... I don't pretend at all that this is all understood. But I would be tempted to say that a good strategy could be to search for hidden associativity whenever one encounters non-associative products of observables in deformation quantization problems.<|endoftext|> TITLE: Dualizable object in the category of locally presentable categories QUESTION [23 upvotes]: The bicategory of locally presentable categoires, and left adjoint functor between them, is monoidal closed for the Kelly tensor product. My question is what are the dualizable objects for this monoidal structure ? Just to be clear, I'm talking about the monoidal stucture for which the internal hom $[A,B]$ is given by the category of left adjoint functor $A \rightarrow B$; or equivalently for which morphismw $A \otimes B \rightarrow C$ are the functors $A \times B \rightarrow C$ that commutes to colimits in each variables. A typical example of such a dualizable object are the presheaves categories. If $C$ is small then the presheaf category $\widehat{C}$ is dualizable with dual $\widehat{C^{op}}$. The evaluation $$\widehat{C} \otimes \widehat{C^{op}} \simeq \widehat{C \times C^{op}} \rightarrow Sets $$ Is given by the coend, and the counit: $$ Sets \rightarrow \widehat{C} \otimes \widehat{C^{op}} \simeq \widehat{C \times C^{op}} $$ Is the unique left adjoint functor sending the singleton to $Hom_C( \_ ,\_ )$. The usual unit/counit relation being essentially the coend formulation of the Yoneda lemma. I suspect there are some dualizable object not of this form, but this is not totally clear to me, and I wonder if it is possible to give a nice description of the dualizable objects in general... REPLY [6 votes]: The following is a completement to the answer by Yonatan. I managed to work it out after I read what he had done. It is basically about proving the converse to the implication he showed in his answer: If I'm correct one can prove that for a locally presentable category $A$, the following are equivalent: $(1)$ $A$ is dualizable in $Pres^L$ $(2)$ $A$ is a retract of a presheaf category in $Pres^L$ $(3)$ $A$ can be constructed from an ideal $C_0 \subset C$ as described in Yonatan answer. This being said, I'm not excluding that a better result can be obtained (for example a more canonical presentation of dualizable objects), though the example given by Yonatan seem to suggest that this is already quite close to be optimal. $(3) \Rightarrow (2) \Rightarrow (1) $ is discussed in Yonatan answer, I'll prove $(1) \Rightarrow (2)$ and $(2) \Rightarrow (3)$ separately. For $(1) \Rightarrow (2)$. Let $A$ be a dualizable object in $Pres^L$, its dual $A^*$ is isomorphic to the category of left adjoint functors $[A,Set]$. The fact that $A$ is dualizable means that there is a coevaluation map $Set \rightarrow A \otimes A^*$, which is fully described by the image of the singleton, giving a specific object $c \in A \otimes A^*$. As any object of $A \otimes A^*$, $c$ can be written as a colimits of pure tensor: $$ c = colim_{i \in I} a_i \otimes \chi_i $$ where $a_i \in A$ and $\chi_i \in A^* = [A,Set]$. The evaluation-coevaluation relation translate into: $$ colim_{i \in I} \chi_i (u) \times a_i \simeq u $$ functorially in $ u \in A$ (where $\times$ above denote the cotensoring objects by sets, i.e. the coproduct of several copies of the same object). It means that the $\chi_i$ together form a cocontinuous functor $A \rightarrow Set^I$, And $F \mapsto colim F(i) \times a_i$ form a cocontinuous functor from $Set^I$ to $A$, whose composite $A \rightarrow Set^I \rightarrow A$ is naturally isomorphic to the identity, hence $A$ is a retract of $Set^I$. I now give a sketch of proof for $(2) \Rightarrow (3)$: Assume that one has two morphisms in $Pres^L$, $f:A \rightarrow \widehat{C}$ and $g:\widehat{C} \rightarrow A$ with $g \circ f \simeq_{\Theta} Id$. Then $P = f \circ g : \widehat{C} \rightarrow \widehat{C}$, is idempotent $P^2 \simeq_{f \theta g} P $, and moreover the isomorphism is an associative multiplication $P^2 \rightarrow P$ (I'm goging to say that $P$ is a non-unital monad), and the category $A$ identifies with the algebras for this multiplication (in the non unital sense), whose structure maps is are isomorphisms $PX \overset{\sim}{\rightarrow} X$. Using that $P$ commutes to coproduct, this makes $P \coprod Id$ into a cocontinuous (unital) monad on $\widehat{C}$, whose algebras are the $P$-algebra (as a non-unital monad), and hence one has a category $C_P$, whose objects are these of $C$ and morphism from $c$ to $c'$ are morphisms from $c$ to $(Id \coprod P) c'$, i.e. either morphism from $c$ to $c'$, or morphism from $c$ to $P c'$ (composition of the morphism of the second type involving the use of the multiplication maps $P^2 \rightarrow P$ The map of the form $c \rightarrow P c'$ form an ideal in $C_P$, and because of the relation $P^2 \simeq P$ they satisfies the additional condition in Yonatan answer. Now presheaves over $C_P$ are the same as presheaves over $C$ with an algebra structure for the monad $Id \coprod P$, i.e. algebra for the non-unital monade $P$. One can then check that the additional condition in Yonatan answer (with respect to the ideal mentioned above) single out the $P$-algebra for which the structure map is an isomorphism, and hence it corresponds exactly to the category $A$ we started from.<|endoftext|> TITLE: Expected value of biggest distance of adjacent points uniformly picked in $[0,1]$ QUESTION [8 upvotes]: We pick $n\ge 2$ points in $[0,1]$ with uniform distribution. What is the expected value of the largest distance of $2$ adjacent points? REPLY [16 votes]: The expected value is asymptotic to $(\log n)/n$ as $n$ tends to infinity (By "asymptotic" I mean that the ratio tends to 1). One way to see this is to use the representation of order statistics of uniform points as the first $n$ points of a Poisson process, normalized by the $n+1$ Point. Since the sum of $n+1$ exponential variables is concentrated, the question reduces to the distribution of the maximum of $n-1$ Exponential variables. Usually one considers the maximum $M_n$ of the $n+1$ gaps including the gap between zero and the first point and between 1 and the last point, but that does not change the asymptotics. The known properties of $M_n$ are surveyed in the introduction of Devroye's paper that Brendan McKay mentioned: https://projecteuclid.org/download/pdf_1/euclid.aop/1176994313 In particular the fact about Poisson processes I noted above is Lemma 2.1 there, and lemmas 2.4, 2.5 and 2.6 describe more precise asymptotics for the maximal gap.<|endoftext|> TITLE: Fibers of continuous maps of $\mathbb{R}^n$ which are injective at dense points QUESTION [7 upvotes]: Question. Suppose that $f\colon\mathbb{R}^n \to \mathbb{R}^n$ is a continuous map and there is a dense subset $D \subset \mathbb{R}^n$ such that $f^{-1}(f(x)) = \{x\}$ for all $x \in D$. Is every fiber of $f$ connected? When $n =1$, it is easy to see that $f$ must moreover be a strictly increasing or decreasing function. I think that it it true in general, but I am not sure. An idea to prove: Suppose, to the contrary, that $f(x)^{-1}$ has two connected components. Then pick an arc, in the domain, with endpoints in different connected components fo $f^{-1}(x)$. Then $f$ identifies the endpoints of the arc, and it seems unlikely that the resulting loop in the target is null homotopic, which would be a contradiction. REPLY [6 votes]: There is a continuous map $f:\mathbb{R}^2\to\mathbb{R}^2$ such that the fiber of the origin $0$ is the disconnected set $f^{-1}(0)=\mathbb{R}\times\{-1,+1\}$, and for any $x\in D:=\mathbb{R}\times\big (\mathbb{R} \setminus \{-1,1\}\big)$, there holds $f^{-1}(f(x))=\{x\}$. I didn't bother to write an equation (I'll do if compelled), but here is a picture: the loops with self-intersection at the origin are lines $f(\{a\}\times\mathbb{R})$, the short segments are lines $f(\mathbb{R}\times\{b\})$, which degenerates to a single point, the origin, for $b=\pm1$. [rmk] For a concrete realization of this picture, we may use the rational parametrization of the Conchoid of Nicomedes. One can even fill the hole in the picture.<|endoftext|> TITLE: Cyclic vectors in irreducible representations of simple Lie algebras QUESTION [6 upvotes]: Is there a notion of "cyclic element" in a simple Lie algebra? In particular, is it independent of the irreducible representation chosen? Explanation. An endomorphism A is called cyclic if there is a vector v which by the action of A generates the whole vector space (v is also called a cyclic vector for A). Let g be a simple Lie algebra. Call an element x cyclic if its matrix in the adjoint representation is cyclic. Let r be an irreducible representation (irrep) of g and x a cyclic element of g. Is it true that r(x) is cyclic? More generally, for two irreps r_1 and r_2 of g, is it true that r_1(x) cyclic implies r_2(x) cyclic? Remarks. 1) It is important to consider only irreducible representations since for a reducible one, no r(x) is cyclic. It is also important to consider simple Lie algebras since for a semi-simple one, you can easily construct counter-examples. 2) The concept of a "cyclic element" as defined above seems new. I would be happy to get references if it is not. REPLY [5 votes]: Summary The answer to the first question is affirmative and to the second question is negative, but for rather mundane reasons. In the simple Lie algebra case, cyclicity of ${\rm ad}\, a$ for some $a$ implies that $\frak{g}$ has rank $1$, in which case every non-zero element $x\in\frak{g}$ is cyclic in every simple module. On the other hand, for any $\frak{g}$, every $x\in\frak{g}$ acts cyclically on any one-dimensional module (e.g. the trivial representation). Thus in the higher rank case, all elements (even $x=0$) are trivial-cyclic and none are ad-cyclic. Moreover, all modules $V$ which admit a semisimple cyclic $x\in\frak{g}$ are weight multiplicity-free (WMF), which have been classified, and the corresponding cyclic elements can be explicitly described. Details Assume that the ground field is not of characteristic $2$. It is easy to see that a simple Lie algebra $\frak{g}$ contains an ad-cyclic element only if $\frak{g}$ has rank $\ell=1$, so that ${\frak g}$ is a form of ${\frak sl}_2$. Indeed, by the standard properties of endomorphisms, the condtion "$A$ is cyclic" implies that the dimension of $\ker{A}$ is at most one, whereas for $A={\rm ad}\, x$, this dimension is at least $\ell$ by the definition of rank; thus $\ell\leq 1$. Conversely, every non-zero element $x$ of a simple rank $1$ Lie algebra acts cyclically in every finite-dimensional simple module: extend the scalars to an algebraically closed field and consider separately the cases of semisimple and nilpotent $x$. A semisimple endomorphism (or a diagonal matrix) is cyclic iff its eigenvalues are distinct. Let $h\in{\frak h}$ be a semisimple element of a split semisimple Lie algebra $\frak{g}$ with a Cartan subalgebra ${\frak h}$ and $V$ an $N$-dimensional $\frak{g}$-module with weights $\lambda_1,\ldots,\lambda_N\in {\frak h}^*$. Then the matrix of $h$ relative to a weight basis of $V$ is diagonal with eigenvalues $\lambda_i(h)$ and $h$ acts cyclically on $V$ iff $\lambda_i(h)$ are pairwise non-equal. In particular, this is possible only if $V$ is weight multiplicity-free (WMF), and all such modules have been classified (consistent with the above, the adjoint module is WMF iff ${\frak g}$ is a direct sum of several copies of ${\frak sl}_2$). For example, in characteristic $0$, if ${\frak g}={\frak sl}_n$, the simple WMF modules are exhausted by the symmetric powers $S^k W$ of the defining module $W$, their duals, and the exterior powers $\Lambda^k W$ of $W$, and the necessary condition for $h$ to act cyclically on $V$ is that $h$ is regular (equivalently, $h$ acts cyclically on $W$, i.e. has distinct eigenvalues). Of course, this condition is not sufficient in general. For example, if $V=S^2 W$ and $h$ is diagonal with eigenvalues $a_i$ that sum to $0$, $h$ is cyclic on $V$ iff all $a_i+a_j, 1\leq i\leq j\leq n$ are pairwise distinct. For $n\geq 3$, this is clearly stronger than just all $a_i$ being pairwise distinct. Note, however, that reducible WMF modules also exist: for example, let $\frak{g}={\frak sl}_n$ and $V=W\oplus\Lambda^2 W$, then $V$ is WMF and every diagonal $h$ with $n$ distinct eigenvalues $a_i$ satisfying appropriate additional inequalities acts cyclically on $V$. Thus contrary to OP's Remark 1, an element of ${\frak g}$ can act cyclically on a reducible module. If $e$ is a nilpotent element acting cyclically on a $\frak{g}$-module $V$ then $V$ has to be simple with respect to the corresponding ${\frak sl}_2$-subalgebra, so that the weights of $V$ belong to a single line. This places severe constraints on the possible data. For a simple classical Lie algebra ${\frak g}\ne{\frak sl}_2$ and a non-trivial module $V$, this leaves only the cases of ${\frak g}={\frak sl}_n$ and $V$ the defining module or its dual, and a symplectic or odd orthogonal ${\frak g}$ and the defining ("vector") module $V$, with the element $e$ being a principal nilpotent in all cases. Finally, if $x=s+n$ is the Jordan decomposition of a general element acting cyclically on a $\frak{g}$-module $V$ then the centralizer $Z(s)$ of the semisimple part $s$ is a reductive subalebra containing $x$, so that $V$ is a $Z(s)$-module with a cyclic action of $x$. This provides an approach to classifying general cyclic pairs $(V,x)$.<|endoftext|> TITLE: When is one 'ready' to make original contributions to mathematics? QUESTION [62 upvotes]: This is quite a philosophical, soft question which can be moved if necessary. So, basically I started my PhD 9 months ago and have thrown myself into learning more mathematics and found this an enjoyable and rewarding experience. However, I have come to realise how much further I still have to go to reach a point where I could even think about publishing original contributions in the literature given how intensively everything has already been studied and the discoveries already made. For example, I have just finished a 600 page textbook on graduate level mathematics. Although it took me a while to understand everything in it, I learned from this and enjoyed doing the exercises, but realised by the end that I still basically know nothing and that it is really intended as a springboard to slightly more advanced texts. I picked up another book which starts to delve more into one of the specific aspects in the book and again, it is 500 pages long. Do I have to read another 500 page book to get a sense of something more specific which I can contribute? At this rate, it will be years and years before I am ever able to publish anything. Later: I am reading this a few years later and realise the question could be hard to answer, as depends on many things (there are some problems where one could contribute decisively without knowing any math at all). However, I will leave the question as I think it's something that many students ask themselves and there is some useful generic advice in the answers. REPLY [9 votes]: Serre, recounting a discussion with Andre Weil, has said that "mathematics is not made by people with long experience, with a lot of knowledge, and so on, no. New ideas come without that." Moreover, he has to be really interested in something. And if you are really interested in a question, and you begin reading what people have done around it, very often you discover that they have not done anything. They always talk on something else, or they made a hypothesis which is not true in your case, they have very rarely done something useful. So you reduce the literature to very little. So you have to find new ideas. But, of course, if you can find connections with something else, it helps a lot. - J.P. Serre (2003). https://www.youtube.com/watch?v=NTZh6cuezv4&t=1612s&ab_channel=TheAbelPrize<|endoftext|> TITLE: Are proper subspaces of Banach spaces which are isomorphic to the ambient Banach space necessarily complemented? QUESTION [6 upvotes]: I had the following little question pop up, but I cannot seem to find any reference to it. Let $X$ be a Banach space and $E\subseteq X$ a proper subspace with $E$ isomorphic to $X$ itself. Is the subspace $E$ necessarily complemented in $X$? Composing the isomorphism with the inclusion map is a linear surjection of $X$ onto $E$, but this map is not necessarily an idempotent on $X$. So this map will require some tweaking to become a projection if it is actually the case that such subspaces are always complemented. It is not entirely clear if this can always be done, so the answer might be no in general. Is there an example? There are some (non-Hilbert space) examples for which the answer is yes. Gowers' solution to the Hyper plane problem [1] is one, but just because that space is not isomorphic to any of its proper subspaces. Another example is one from [2] where all subspaces $E$ that are isomorphic to $X$ are exactly those of finite codimension, and hence are complemented. [1] Gowers, W.T., A Solution to Banach's Hyperplane Problem. Bulletin of the London Mathematical Society, Volume 26, Issue 6, November 1994, Pages 523–530. [2] Gowers, W. & Maurey, B. Banach spaces with small spaces of operators Math Ann (1997) 307: 543. REPLY [8 votes]: The first counterexample that comes to my mind, which is probably overkill (memory is easier than thought!) is J. Bourgain, A counterexample to a complementation problem. Compositio Mathematica, Volume 43 (1981) no. 1, p. 133-144 NUMDAM link which constructs a subspace $E$ of $L_1$ that is isomorphic (non-isometrically) to $L_1$, but which is not complemented inside $L_1$. Some years ago, in discussions with Banach-space theorists, one of them sketched to me how one can bootstrap to get uncomplemented copies of $\ell_1$ inside $\ell_1$ but I don't recall the details right now.<|endoftext|> TITLE: Reference: Hajlasz-Sobolev Spaces with Values in a Metric Space QUESTION [6 upvotes]: Let $(X,d,\mu)$ be a separable metric measure space on which every ball has positive but finite measure. I've come across the definition of a homogeneous Fractional Hajlasz-Sobolev spaces $M^{s,p}(X,d,\mu)$ which are defined as the collection of functions $f \in L^p_{loc}(X)$ for which there exists some $g \in L^p(X)$ satisfying $$ |f(x)-f(y)|\leq d(x,y)^s[g(x)+g(y)], $$ outside a set of $\mu$-measure $0$. These spaces are Banach spaces with the norm $$ \|f\|\triangleq \inf_{g \in D^s(f)} \|g\|_{L^p(X)}, $$ where $D^s(f)$ is the class of functions $g$ satisfying the first inequality. If now $(Y,\rho,\nu)$ is another separable metric measure space satisfying the same conditions as $(X,d,\mu)$ then is there a separable Frechet space of Borel-measurable functions from $X$ to $Y$ generalizing the fractional Hajlasz-Sobolev space? REPLY [7 votes]: The question is not stated in a very clear manner, but nevertheless, the answer is: no. Separability. The space $M^{1,p}(X,d,\mu)$ is not separable even if $X$ is the standard ternary Cantor set, $d$ is the Euclidean metric $d(x,y)=|x-y|$ and $\mu$ is the natural Hausdorff measure. This was proved in [R]. Therefore, there is no reason to expect separability in a more general case of mappings between metric spaces. The space of mappings from $X$ to $Y$ is, in general, not linear so it cannot be a Banach or Frechet space. As I understand MrMMS wants to investigate the Frechet manifold structure, but it is even not clear how to define a metric and hence the topology in the space of $M^{s,p}$ mappings from $X$ to $Y$. A natural way to define a metric is to assume that $Y$ is isometrically embedded in a Banach space, say $Y\subset \ell^\infty$ and then we define the space $M^{s,p}(X,Y)$ as the space of mappings $u\in M^{s,p}(X,\ell^\infty)$ such that $u(x)\in Y$ for $\mu$ almost all $x\in X$. Since $M^{s,p}(X,\ell^\infty)$ is a Banach space, we have a natural metric in $M^{s,p}(X,Y)$ inherited from the norm of $M^{s,p}(X,\ell^\infty)$. The problem is that very little is known about the metric and the topological structure of the space $M^{s,p}(X,Y)$, even in a very classical case when we have Sobolev mappings from a comapact manifold into a compact metric space $W^{1,p}(M,Y)$. Why is it so difficult? The proof that the space of smooth mappings between manifolds has a (infinitely dimensional) manifold structure steams from the fact that mappings can be localized in coordinate charts on manifolds. Since Sobolev mappings are not necessarily continuous, this approach is not available and basically, the only way to define the metric (and topology) in $W^{1,p}(M,Y)$ is through the isometric embedding of $Y$ into $\ell^\infty$ (or another Banach space). This approach does not provide any natural way to equip the space $W^{1,p}(M,Y)$ with a structure of a Banach or a Frechet manifold. Let me just mention one result that show how sensitive the definition is. Let $Y\subset \mathbb{R}^{n+2}$ be compact. There are two different ways to define $W^{1,p}(M,Y)$, where $M$ is a compact Reiamnnian manifold. One way is as $$ W^{1,p}(M,Y)=\{u\in W^{1,p}(M,\mathbb{R}^{n+2}:\, u(x)\in Y\, a.e.\} $$ and another is through the (Kuratowski) isometric embedding $\kappa:Y\to\ell^\infty$ as described above. Denote the space defined through the Kuratowski embedding by $W^{1,p}_\kappa(M,Y)$. Let $S^n$ be the standard sphere. The next result was proved in [H]. Theorem. There is a compact and connected set $Y\subset \mathbb{R}^{n+2}$ such that Lipschitz mappings $\operatorname{Lip}(S^n,Y)$ are dense in $W^{1,n}(S^n,Y)$, while if $\kappa:Y\to\ell^\infty$ is the Kuratowski embedding, then Lipschitz mappings $\operatorname{Lip}(S^n,Y)$ are not dense in $W^{1,n}_\kappa(S^n,Y)$. This result shows that the metric and topology of the space of Sobolev mappings into metric spaces heavily depends on what isometric embedding we choose: with one choice Lipschitz mappings are dense, but with another one they are not dense. [H] P. Hajlasz, Sobolev mappings: Lipschitz density is not an isometric invariant of the target. Int. Math. Res. Not. IMRN Vol. 2011, no.12, 2794-2809. [R] J. Rissanen, Wavelets on self-similar sets and the structure of the spaces $M(E, \mu)$, Ann. Acad. Sci. Fenn. Math. Diss. 125 (2002).<|endoftext|> TITLE: Random Walk on Pentagonal Tiling QUESTION [12 upvotes]: I’ve recently been looking at closed walks on tilings of the plane in which the “player” can move from one tile to any of its edge-adjacent neighbors. In particular, I’m trying to find asymptotic formulae for various tilings giving the number of n-step closed paths starting and ending at an arbitrary tile. For an infinite square checkerboard we have $$\sim \frac{4^n}{n}\cdot\frac{2}{\pi}$$ and for an infinite hexagonal honeycomb we have $$\sim \frac{6^n}{n}\cdot \frac{\sqrt{3}}{2\pi}$$ and for an infinite tiling of equilateral triangles: $$\sim \frac{3^n}{n}\cdot\frac{3}{\pi}$$ where the first and third formulae are applicable only for even $n$, since there are no closed paths of odd length on the square and triangular tesselations. All of these can be derived using the method employed in this answer. QUESTION: Can anybody derive analogous formulae for less “nice” tilings, like the cairo pentagonal tiling, the floret pentagonal tiling, the tetrakis square tiling, the snub square tiling, or any others? Clearly they should be in the form of a constant times $c^n/n$, where $c$ is a constant depending on the tiling that is more easily calculated ($c=5$ for cairo and floret, $c=3$ for tetrakis square, and $c=\frac{1+\sqrt{33}}{2}$ for the snub square). WHY I CAN’T SOLVE IT: The method I used to derive these 3 formulae relied heavily on the fact that all tiles in the tiling had the same orientation, and that the same “moves” could be made no matter what tile is occupied (for the square tiling, the player can always move N,E,S,W, and for the hexagonal tiling, the player can always move in the same 6 directions). This isn’t actually true for the triangular tiling, but luckily the set of moves that the player can make alternates from one move to the next, so I was able to use the same method with a slight modification. This is because there are two orientations of equilateral triangles in this tiling, and all triangles edge-adjacent to a triangle of one orientation are of the opposite orientation. This problem has been slowly killing me for a long time, partly because it seems like it should be easy but isn’t (for me, at least). Help? NOTE: Since posting the question, I have come up with the formula for another tiling that uses only regular hexagons and equilateral triangles. The formula applies to walks starting in a hexagon and is valid only for even $n$: $$\sim \frac{(3\sqrt{2})^n}{n}\cdot \frac{9}{4\pi}$$ REPLY [6 votes]: Another standard approach is to use a multivariate local central limit theorem, which can be found e.g. in Petrov's book "sums of independent random variables" and the paper "Local Limit Theorems for Lattice Random Variables" by A. B. Mukhin https://doi.org/10.1137/1136086 I especially like "An elementary proof of the local central limit theorem B Davis, D McDonald - Journal of Theoretical Probability, 1995 " (see http://www.stat.purdue.edu/research/technical_reports/pdfs/1993/tr93-41.pdf (These references are from a recent discussion at Fair partitioning of a set - Weighted sums of Bernoullis). If all the tiles have the same orientation, the local CLT can be applied directly. Otherwise, by examining only the times when a translate of the original tile is visited, a random walk is obtained so the local CLT applies- the remaining issue is to compute the covariance matrix. In the tiling http://www.math.cmu.edu/~bkell/21110-2010s/3-6-3-6.png one just has to go in skips of two. In the general case, there will be k tile types, and the random walk defines a Markov chain on these tile types. The covariance matrix can then be obtained by averaging on the starting tile (according to the stationary distribution) the covariance matrix of the step from that tile. See also Keilson, Julian, and D. M. G. Wishart. "A central limit theorem for processes defined on a finite Markov chain." In Mathematical Proceedings of the Cambridge Philosophical Society, vol. 60, no. 3, pp. 547-567. Cambridge University Press, 1964.<|endoftext|> TITLE: Algorithm to list all Kostant partitions QUESTION [6 upvotes]: Let $\Phi_+$ be the set of positive roots in some root system, and let $Q_+$ be the positive part of the root lattice, i.e., the set of elements of the form $\sum_{\beta\in \Phi_+}m_\beta\beta$ with $m_\beta \in \mathbb{Z}_{\geq 0}$. For $\theta\in Q_+$, a Kostant partition of $\theta$ is a sequence $(\beta_1^{(m_1)}, \ldots, \beta_t^{(m_t)})$ such that $\beta_i\in \Phi_+$, $m_i\in \mathbb{Z}_{>0}$, $m_1\beta_1 + \cdots + m_t\beta_t = \theta$, and $\beta_1 > \cdots > \beta_t$ according to some fixed total order on $\Phi_+$. For example, for the root system of type $A_3$ with simple roots $\alpha_1, \alpha_2, \alpha_3$, the Kostant partitions of $\theta = \alpha_1+2\alpha_2+\alpha_3$ are $$(\alpha_1+\alpha_2+\alpha_3, \alpha_2), \ (\alpha_2+\alpha_3, \alpha_1+\alpha_2), \ (\alpha_3, \alpha_2, \alpha_1+\alpha_2), \ (\alpha_2+\alpha_3, \alpha_1, \alpha_1), \ (\alpha_3, \alpha_2^{(2)}, \alpha_1).$$ Is there an algorithm for listing or iterating through all Kostant partitions of a given $\theta\in Q_+$ (for an arbitrary root system)? An inefficient/recursive attempt is as follows: Phi := the set of positive roots Q := the positive part of the root lattice def KostantPartitions(theta): L = [] for beta in Phi: if theta-beta is in Q: for kp in KostantPartitions(theta-beta): new = kp with beta inserted if new not in L: L += [new] return L Is there a more efficient algorithm? I realize this question is ill-posed (what does "efficient" mean?); I am really just looking for something better than the above. I would also be happy with partial answers, e.g., an algorithm for root systems of type $A$. REPLY [3 votes]: For fixed $\theta$, the set of $(m_{\beta})$ giving a Kostant partition is evidently the set of lattice points in some polyhedron given in terms of hyperplane inequalities and equalities: they are exactly the integer points satisfying the inequalities $m_{\beta} \geq 0$ and equalities coming from $\theta = \sum_{\beta} m_{\beta} \beta$. There are a lot of software out there which can solve the general problem of finding all the lattice points of a given polyhedron: e.g., general math software like Sage can do this (see the documentation at http://doc.sagemath.org/html/en/reference/discrete_geometry/sage/geometry/lattice_polytope.html), but also there is specialized software like "LattE" (https://www.math.ucdavis.edu/~latte/). Incidentally, this kind of reasoning is what lets you conclude that the Kostant partition function (i.e., the number of Kostant partitions) is a piecewise quasipolynomial, a fact which you may be interested in if you were not aware of it already. And in the particular case of the Type A root system, which you mentioned you were specially interested in, unimodularity implies that the Kostant partition function is a piecewise polynomial; see https://www.sciencedirect.com/science/article/pii/S0021869304000055.<|endoftext|> TITLE: Functoriality of infinite loop space machines? QUESTION [10 upvotes]: If $C$ is a symmetric monoidal category, then $BC$ is canonically an algebra over a certain $E_\infty$ operad, but if $F: C \to D$ is a symmetric monoidal functor then (as far as I can see) $BF: BC \to BD$ is not a map of algebras over that operad (unless all the morphisms $(Fx) \otimes (Fy) \to F(x \otimes y)$ are identities). Because of this I struggle to associate a spectrum map between the two spectra arising from $BC$ and $BD$ by infinite loop space theory, I can only see how to get a zig-zag where the wrong-way map is a weak equivalence. For most practical purposes that's just as good, but nevertheless I wonder: are any of the "well known" infinite loop space machines functorial on the nose, with respect to symmetric monoidal functors? REPLY [8 votes]: Here is a nice gentle old-fashioned answer. Symmetric monoidal categories are functorially equivalent as symmetric monoidal categories to permutative (symmetric strict monoidal) categories, and those are functorially equivalent (essentially the same as) algebras over a certain $E_{\infty}$ operad $\mathcal{P}$, known nowadays as the categorical Barratt-Eccles operad, in Cat. Since $B$ is product preserving it gives a functor from $\mathcal P$-algebras in Cat to $B\mathcal P$-algebras in Top. That gives $B$ as a functor from symmetric monoidal categories to algebras over an $E_\infty$ operad. That goes back, at least in outline, to my 1974 paper ``$E_{\infty}$ spaces, group completions, and permutative categories". Again in outline, by two recent papers, the same argument works equivariantly for (genuine) symmetric monoidal $G$-categories, which give genuine $G$-spectra for finite groups $G$ via infinite loop $G$-space machines. See Equivariant iterated loop space theory and permutative $G$-categories http://www.math.uchicago.edu/~may/PAPERS/GM3.pdf and Symmetric monoidal $G$-categories and their strictification http://www.math.uchicago.edu/~may/PAPERS/AddCat1.pdf}.<|endoftext|> TITLE: A simple cardinal characteristic associated with $\omega^\omega$ QUESTION [12 upvotes]: We can define a very simple cardinal characteristic in the following way. Recall the relation $\leq^*$ on $\omega^\omega$ defined by $x\leq^* y$ iff $x(i)\leq y(i)$ for all but finitely many $i$. For $x,y\in\omega^\omega$, say that $x$ and $y$ are comparable, denoted by $x\parallel y$, if either $x\leq^* y$ or $y\leq^* x$. Let $\omega^{*\omega}$ denote the set of sequences of natural numbers that diverge to $\infty$. Define $\mathfrak{cp}=\min\{|C|:C\subseteq\omega^{*\omega},\ (\forall x\in\omega^{*\omega})(\exists y\in C)\, x\parallel y\}$. It is not hard to see that $\mathfrak{b}\leq\mathfrak{cp}\leq\mathfrak{d}$. On the other hand, since Cohen forcing can add a real in $\omega^{*\omega}$ incompatible with all ground model reals in $\omega^{*\omega}$, we can prove that $\mathrm{cov}(\mathcal{M})\leq\mathfrak{cp}$. Now, to the questions: 1) Is $\mathfrak{cp}=\mathfrak{d}$? 2) What happens to $\mathfrak{cp}$ in Miller's model? Perhaps this cardinal invariant is something trivial, but I haven't been able to figure it out so far. REPLY [9 votes]: I claim that $\mathfrak{c}\mathfrak{p}= \mathfrak{d}$. For any $x\in \omega^{*\omega}$ define its "inverse" $x'$ by $x'(n) = \min \{k\mid\forall j\ge k : x(j)>n\}$. If $x$ grows very fast, then $x'$ grows very slowly. In particular we have $x\le ^* y \Rightarrow y'\le^* x'$, and $x\le^* x''$. Define $x^+$ as follows: $x^+(2n)=x(n)$, and $x^+(2n+1)=x'(n)$. So if $x$ grows very fast, then $x^+$ grows very fast on the even numbers, but very slowly on the odd numbers. Now assume that $C$ is a witness for $\mathfrak c\mathfrak p\le \lambda$. Let $D$ be the closure of $C$ under some natural operations (such as $x\mapsto x'$), then the set $D$ will witness $\mathfrak d\le \lambda$. Proof: Let $x\in \omega^{*\omega}$ be strictly increasing. Find $y$ in $C$ such that $x^+ \|y$. Case 1: $x^+\le^* y$. Then let $x(n)=x^+(2n) \le y(2n)$ for almost all $n$, so the function $n\mapsto y(2n)$ dominates $x$. Case 2: $y \le^* x^+$. Then for almost all $n$ we have $y(2n+1)\le x^+(2n+1)=x'(n)$. Consider the function $z(n)=y(2n+1)$: We have $z\le^* x'$, so $x\le x''\le^* z'$. So all we need to do here is to ensure that $D$ is closed under composition with $n\mapsto 2n+1$ and under $z\mapsto z'$.<|endoftext|> TITLE: The Koch snow flake, Holder exponents of conformal mappings QUESTION [11 upvotes]: The Koch snow flake $K$ is a domain of $\mathbb{C}$, complex plane. Though, I do not state the precise definition, you can see the picture in wikipedia Koch snow flake. The Koch snow flake $K$ is a quasidisk. Let $\mathbb{D}$ be an open unit disk and let $\phi:\mathbb{D} \to K$ be a conformal mapping. It is known that $\phi$ and the inverse map $\phi^{-1}$ are Holder continuous: there exist $\alpha \in (0,1]$, $\beta \in (0,1]$, and $L_1,L_2 \in (0,\infty)$ such that \begin{align*} |\phi(z_1)-\phi(z_2)| &\le L_1 |z_1-z_2|^{\alpha},\quad z_1,z_2 \in \mathbb{D}, \\ |\phi^{-1}(w_1)-\phi^{-1}(w_2)| &\le L_2 |w_1-w_2|^{\beta},\quad w_1,w_2 \in K. \end{align*} My question Is there a study for quantitative estimates on the Holder exponents $\alpha$ and $\beta$? I think that there is such study because the Koch snow flake is a famous fractal set. ADD By an argument by Benoît Kloeckner, $\alpha$ must be less than or equal to $\log3 / \log 4$. Is there a reasonable lower bound for $\alpha$ and $\beta$? ADD2 Let $C$ be a closed Jordan curve. Lasley considers the following condition on $C$. Definition. Let $w_1$ and $w_3$ be points on $C$ and let $w_2$ be on the arc of small diameter between $w_1$ and $w_3$. Then, $C$ is said to be a $c$-quasiconformal curve if there exist positive constants $c$ and $\delta$ such that \begin{align*} \frac{|w_1-w_2|+|w_2-w_3|}{|w_1-w_3|} \le c \end{align*} for any such $w_1,w_2,w_3$ with $|w_1-w_3| \le \delta$. Lasley prove Theorem. Suppose that $f$ maps $\mathbb{D}$ conformally onto the interior $\Omega$ of a $c$-quasiconformal curve $C$. Then, $f$ is $\alpha$-Holder continuous. Here, \begin{align*} \alpha=\frac{2 (\text{arcsin}(1/c))^2}{\pi^2 -\pi \text{arcsin}(1/c)}. \end{align*} If $\Omega=K$, the Koch snowflake, $c=\cdots$. REPLY [5 votes]: U.R Freiberg and M.R. Lancia, Energy Form on a Closed Fractal Curve (2004): The Koch snow flake is the union of three Koch curves of Hausdorff dimension $D=\ln 4/\ln 3$ and Hölder exponent $\beta=\log 2/\log 3$. See Proposition 2.2, attributed to a 1999 paper which I did not find online. (For reference, Pietro Majer's argument for $\alpha=\log 3/\log 4$ is in this 2016 MO answer.)<|endoftext|> TITLE: Ramseyan property of structure QUESTION [5 upvotes]: A binary relation $R$ on $\mathbb{N}$ is $(n,k)$-Ramseyan iff for every coloring $f:[\mathbb{N}]^n\rightarrow l$, there exists a subset $G$ of $\mathbb{N}$ such that $(G,R)$ is isomorphic to $(\mathbb{N},R)$ and $|f([G]^n)|\leq k$ (i.e., there exists a 1-1 mapping $h:G\rightarrow \mathbb{N}$ such that $R(x,y)$ if and only if $R(h(x),h(y))$). Question: Which binary relation is $(n,k)$-Ramseyan? Is there any known result? We know some special example. Say when $R$ is trivial ($R=\emptyset$), then it is $(n,1)$-Ramseyan (which is just the classic Ramsey theorem). When $(\mathbb{N},R)$ is a linear dense order structure then it is $(2,2)$-Ramseyan, $(3,12)$-Ramseyan and for every $n$ there exists $k_n$ such that it is $(n,k_n)$-Ramseyan. REPLY [2 votes]: This question is closely related to the problem of proving the existence of finite big Ramsey degrees. Namely, for a first-order language $L$ and an infinite $L$-structure $\textbf{B}$, we say $\textbf{B}$ has finite big Ramsey degrees if for every finite substructure $\textbf{A}$ there is an integer $T_{\textbf{A}}$ such that, whenever you colour all isomorphic copies of $\textbf{A}$ in $\textbf{B}$, there is a substructure $\textbf{B}'$ of $\textbf{B}$ which is isomorphic to $\textbf{B}$ with the property that the copies of $\textbf{A}$ in $\textbf{B}'$ take no more than $T_{\textbf{A}}$-many colours. Of course, your property is slightly stronger since you're colouring all possible $n$-tuples rather than just the copies of a particular substructure $\textbf{A}$ of cardinality $n$. That said, not much is known even for the above version of the problem, but there are some positive results when $L$ consists of a single binary relation $R$. When $R$ is interpreted as a linear order then $\mathbb{N}$ has finite big Ramsey degrees by Ramsey's theorem, as you alluded to. $\mathbb{Q}$ with its usual order has finite big Ramsey degrees by a theorem of Denis Devlin (a proof of which can be found in Todorcevic's book Introduction to Ramsey Spaces). There are also known results in the case where $R$ is interpreted as an edge relation on a countably infinite graph. For instance, a paper of Laflamme, Sauer, and Vuksanovic shows that the Rado graph has finite Ramsey degrees, while a recent article of Dobrinen shows that the Henson graphs have finite big Ramsey degrees. The introduction of the latter paper also contains a nice overview of the literature on the subject.<|endoftext|> TITLE: Elliptic Regularity with Gibbs Measure Satisfying Bakry-Emery Condition QUESTION [6 upvotes]: Consider $\mathbb{R}^d$ with Gibbs measure $d\mu=Z^{-1}\exp(-V(x))dx$, where the potential $V(x)$ is strongly convex ($\nabla^2 V(x) \ge \lambda Id $). We can assume the regularity of $V$ is as good as we need, and the Hessian is also bounded above. Now, for “Laplace equation” in $(\mathbb{R}^d, d\mu)$: $$\nabla^* \nabla u=f,$$ Here $\nabla^*$ is the $L^2(d\mu)$-adjoint of $\nabla$. Assume $f\in L^2(d\mu)$ and solvability condition $\mathbb{E}_{\mu} f=0$, do we always have well-posedness of the equation in the space of mean-zero functions? If yes, do we have $u \in H^2(d\mu)$, and $\| D^2 u\|_{L^2(d\mu)} \le C \|f\|_{L^2(d\mu)}$? Finally, if yes, what is the constant $C$? Thanks! REPLY [2 votes]: For the estimate of the Hessian you might use Bochner's formula: on such weighted $R^d$ for any smooth compactly supported function $u$ it holds $$ \nabla^*\nabla\frac{|\nabla u|^2}2=|D^2u|^2+\langle\nabla u,\nabla\nabla^*\nabla u\rangle+D^2V(\nabla u,\nabla u) $$ Now using the assumption $D^2V\geq \lambda Id$, $\nabla^*\nabla u=f$ and integrating (the left hand side integrates to 0), after rearrangement we get $$ \int |D^2u|^2\leq \int |f|^2-\lambda |\nabla u|^2 $$ In particular, if $\lambda\geq 0$ then your desired inequality holds with $C=1$. Once you got this for smooth and compactly supported $u$'s, the general case follows by approximation.<|endoftext|> TITLE: "Uniformly continuous" environment sum of a bijection $\varphi:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}$ QUESTION [5 upvotes]: Given any function $f: \mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}$ we define the environment sum of $(x,y)\in\mathbb{Z}\times \mathbb{Z}$ with respect to $f$ by $$\text{es}_f(x,y) = \sum\{f(x', y'): |(x',y') - (x,y)|=1\}.$$ Question. Is there a bijection $\varphi:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}$ such that there is an integer $K\geq 0$ such that for all $(x,y)\in \mathbb{Z}\times\mathbb{Z}$ the following holds? whenever $(x',y')\in \mathbb{Z} \times \mathbb{Z}$ with $|(x',y') - (x,y)|=1$ then $|\text{es}_\varphi(x,y) - \text{es}_\varphi(x',y')| TITLE: Second derivative of integral function QUESTION [6 upvotes]: Let $f: \mathbb R^2 \rightarrow \mathbb R$ be a smooth strictly convex function with unique minimum at $0$ such that all level sets $A_x:=\left\{z ; f(z) \le x \right\}$ are compact. Imagine something like $f(z)=\Vert z \Vert^2.$ Define the integral function $$F(x):=\int_{A_x} g(z) dz$$ where $g$ is as smooth as you like. Question: Is there a way to analytically determine an expression for $F''(x)$? REPLY [4 votes]: Let $B_u:=\{z\colon f(z)=u\}$ for $u>u_0:=\min f$. Let $[0,2\pi)\ni t\mapsto(x_u(t),y_u(t))$ be any smooth parametrization of $B_u$, so that $B_u=\{(x_u(t),y_u(t))\colon t\in[0,2\pi)\}$. (For instance, one may take $(x_u(t),y_u(t))=(\rho_u(t)\cos t,\rho_u(t)\sin t)$, where $\rho_u(t):=f_t^{-1}(u)$ and $f_t^{-1}\colon(u_0,\infty)\to(0,\infty)$ is the function inverse to the function $f_t\colon(0,\infty)\to(u_0,\infty)$ defined by the formula $f_t(r):=f(r\cos t,r\sin t)$.) Then \begin{equation} F'(u)=\int_0^{2\pi}dt\,\sqrt{x'_u(t)^2+y'_u(t)^2}\frac{g(x_u(t),y_u(t))}{|(\nabla f)(x_u(t),y_u(t))|}, \tag{$\ast$} \end{equation} and hence \begin{equation} F''(u)=\int_0^{2\pi}dt\,\frac d{du}\Big(\sqrt{x'_u(t)^2+y'_u(t)^2}\frac{g(x_u(t),y_u(t))}{|(\nabla f)(x_u(t),y_u(t))|}\Big). \end{equation} Here, $dt\,\sqrt{x'_u(t)^2+y'_u(t)^2}$ is the infinitesimal length element of the curve $B_u$, and $\frac{du}{|(\nabla f)(x_u(t),y_u(t))|}$ is the infinitesimal distance between the curves $B_u$ and $B_{u+du}$ near the point $(x_u(t),y_u(t))$. We can verify formula $(\ast)$ for e.g. $f(x,y)=x^2+y^4$ and $g(x,y)=y^2$, in which case we have the following closed-form expressions: for $u>0$ \begin{equation} F(u)=\frac{2 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{3 \Gamma \left(\frac{9}{4}\right)}\,u^{5/4} ,\quad F'(u)=\frac{5 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{6 \Gamma \left(\frac{9}{4}\right)}\,u^{1/4}, \end{equation} and the latter expression coincides with the integral in $(\ast)$, with the parametrization $(x_u(t),y_u(t))=(u^{1/2}\cos t,u^{1/4}(\sin t)^{[1/2]})$ of $B_u$, where $w^{[1/2]}:=|w|^{1/2}\,\text{sign}\, w$.<|endoftext|> TITLE: How complicated can a finite double complex over a field be? QUESTION [16 upvotes]: A finite complex over a field $k$ is pretty simple: it's the direct sum of its homology with a split-exact complex. How complicated can a finite double complex be? Does it make a difference if $k$ is algebraically closed? It seems as though quiver representation theory may be relevant here, but since a double complex is a representation of a quiver with relations, I'm not really sure where to start looking. So my question is: Question: Let $k$ be a field. How complicated is the category of finite-(total) dimensional double complexes over $k$? REPLY [4 votes]: The representation theory of those representations concentrated in the $n \times m$-grid are just representations of the tensor product of the radical square zero algbras $A_n$ and $A_m$ with underlying quiver a linear oriented line. For $n=m$, they have been also classified in https://arxiv.org/pdf/1703.08377.pdf (see proposition 2)and this then also gives the classificaiton for arbitrary $n$ and $m$ when looking at the indecomposables which lie only in the $n \times m$ grid for $m \leq n$. Here another proof using some theorems. Let $C=A_n \otimes_K A_m$. By Theorem 4.5. a) in chapter IX. of the book "Elements of the representation theory of associative algebras" by Assem, Simson and Skowronski the algebra $C$ has a postprojective component. Thus one just have to show that for every indecomposable projective $C$-module $P$ one has $\tau^{-l}(P)=0$ for $l$ large enough to conclude that there are only finitely many indecomposable modules and all are isomorphic to $\tau^{-r}(P)$ for some $r$ and some indecomposable projective module $P$. (starting from this one can get the Auslander-Reiten quiver of $C$ easily) Seeing those complexes as tensor products also shows that in any dimension higher than 2, there are infinitely many indecomposable representations in general as already $A_2 \otimes_K A_2 \otimes_K A_2$ is representation-infinite.<|endoftext|> TITLE: ASD connection for Line bundle over $4$-manifold QUESTION [7 upvotes]: Let $(M,g)$ be an oriented closed Riemannian $4$ manifold. Let $L\to M$ be a complex line bundle. Q Under what condition, we can find an ASD connection of $L$, i.e. a connection $A$ such that $F^+_A=0$. PS: Any reference is welcome. REPLY [3 votes]: By Chern-Weil theory, $c_1(L)=\frac{i}{2\pi}[F_A]$ for any $U(1)$-connection $A$. For an ASD connection ($\ast F_A=-F_A$) we have $$c_1(L)^2=\frac{-1}{4\pi^2}\int_M F_A\wedge F_A=\frac{1}{4\pi^2}\int_M F_A\wedge \ast F_A\equiv-\frac{1}{4\pi^2}||F_A||^2\le0$$ So $c_1(L)^2>0$ is an obstruction. Exercise: Find such an $L$ on $M=S^2\times S^2$. On the other hand, the trivial connection on the trivial bundle is ASD. For an "iff" statement, read sections 1.1.6 and 4.3.3 of Donaldson-Kronheimer's book (The geometry of four-manifolds). Although they talk about ASD connections on $SU(2)$ bundles, you can consider reducible $SU(2)$ connections which correspond to what we care about on $U(1)$ bundles. What we observe is: 1) We really care about cohomology thanks to Hodge-deRham theory, noting that ASD connections have harmonic curvature. 2) Main point: Given a metric $g$, we have projections $\pi_\pm:H^2(M;\mathbb R)\to H^2_\pm(M;\mathbb R)$ and the condition for $L$ to admit an ASD connection is that $\pi_+(c_1(L))=0$. 3) Remark: If $b^2_+(M)>0$ then for generic $g$ we have $\pi_+$ nonvanishing on the lattice $H^2(M;\mathbb Z)$ (which contains $c_1(L)$) minus the origin, because the codimension of $H^2_-(M;\mathbb R)$ is $b^2_+(M)$ and the dimension of the lattice is 0. So we won't have nontrivial line bundles with ASD connections for generic metrics.<|endoftext|> TITLE: A conjecture about the submatrix of orthogonal matrix QUESTION [5 upvotes]: Let $U$ be an $n\times n$ orthogonal matrix, i.e. $U\in\mathbb{R}^{n \times n}$. For any non-empty ordered sets $S_1,S_2\subset\{1,2,...,n\}$, define $U_{S_1S_2}$ to be an $|S_1|\times|S_2|$ submatrix of $U$ which consists of the intersection entries of rows in $S_1$ and columns in $S_2$. Let $\odot$ be the Hadamard product (element-wise product). Here $|S|$ is the cardinality of set $S$. More precisely, an ordered set of cardinality $k$ can be written as a $k$-vector with distinct entries $(i_1,...,i_k)$. Therefore $$U_{(i_1,...,i_k),(j_1,...,j_k)}=\begin{pmatrix} U_{i_1j_1} & \cdots & U_{i_1j_k}\\ \vdots & \ddots & \vdots\\ U_{i_kj_1} & \cdots & U_{i_kj_k} \end{pmatrix}$$ Then is the following conjecture true? For any unit vector $v=(v_1,...,v_k)$ in $\mathbb{R}_+^k, k\leq n$ and any orthogonal matrix $U\in O(n)$, there exists ordered subsets $S_1,S_2\subset\{1,2,...,n\}, |S_1|=k,|S_2|=k$, such that: $$\sum_{j=1}^n\left(\sum_{i=1}^kv_iU_{ij}^2\right)^2\leq v^T[U_{S_1S_2}\odot U_{S_1S_2}]v$$ If this is not true, is the following weaker conjecture true? $$\sum_{j=1}^k\left(\sum_{i=1}^kv_iU_{ij}^2\right)^2\leq v^T[U_{S_1S_2}\odot U_{S_1S_2}]v$$ The conjecture can be easily verified when $k=1$. The weaker conjecture can also be verified when $v=(1/\sqrt{k},...,1/\sqrt{k})$. Maybe we can start from the particular case that $v=(1/\sqrt{k},...,1/\sqrt{k})$. REPLY [3 votes]: Neither conjecture is true in general, as seen from the following counterexample for $n = 4$, $k = 2$, and \begin{align} U = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \ \ \ \ \ \ \ \ \ \ \ \mathrm{and} \ \ \ \ \ \ \ \ \ \ \ \ v = \frac{1}{2}\begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} \mathrm{.} \end{align} Clearly $U$ is orthogonal and $v$ is a unit vector, as required, and for which \begin{align} \sum_{j=1}^n\left(\sum_{i=1}^kv_iU_{ij}^2\right)^2 = \sum_{j=1}^k\left(\sum_{i=1}^kv_iU_{ij}^2\right)^2 = 1 \mathrm{.} \end{align} However, \begin{align} \mathrm{max}_{S_1, S_2} \ v^T[U_{S_1S_2}\odot U_{S_1S_2}]v = \frac{1}{4}\left(2 + \sqrt{3}\right) < 1 \mathrm{,} \end{align} where the maximizing subsets are $S_1 = S_2 = (3, 4)$. Notice that taking $U_{(1,2)(1,2)}$ from $U$ above would suffice as a counterexample as well (for the same $v$), but the above also shows that the conjectures are false when $k$ is strictly less than $n$. The example $v = (1/\sqrt{k},\dots, 1/\sqrt{k})$ may be somewhat pathological, since $U \odot U$ is an orthostochastic matrix (all of its rows and columns sum to 1), and so it is guaranteed to have the all-ones vector as an eigenvector. An arbitrary submatrix of $U \odot U$ will not have this eigenvector in-general, however.<|endoftext|> TITLE: Definition of infinitary regular category QUESTION [7 upvotes]: In Makkai, A theorem on Barr-exact categories, with an infinitary generalization a definition of infinitary regular category is given: a complete regular category $C$ with the additional requirement (DC) for every diagram $F : \alpha^{\text{op}} \to C$, $\alpha$ an ordinal, such that each $F(\beta+1) \to F(\beta)$ is regular epi and $F(\lambda) = \lim_{\beta < \lambda} F(\beta)$ for each limit ordinal $\lambda$, then each projection $\lim F \to F(\beta)$ must be regular epi. In Carboni-Vitale, Regular and exact completions another definition (called "completely regular") is given: a complete regular category such that products of regular epis are regular epi. The first definition implies the second. Are the two definitions equivalent? REPLY [4 votes]: The two definitions are not equivalent: The following counter-example might not be the simplest, and I mostly learned it from Christian Espindola. It is a nice counterexample to quite a lot of similar questions... Let $I$ be the poset of rational number $0 \leqslant q \leqslant 1$, seen as a category with a unique morphism $q \rightarrow q'$ when $q \leqslant q'$, and put the topology on $I$ so that the only non-trivial covering sieve on $q$ is the set of all $q' TITLE: Is it possible that the GHKK canonical basis for cluster algebras is the Lusztig/Kashiwara dual canonical basis? QUESTION [19 upvotes]: Gross-Hacking-Keel-Kontsevich (https://arxiv.org/abs/1411.1394) constructed a canonical basis (the so-called “theta basis”) for a cluster algebra, at least assuming it satisfies a certain combinatorial condition. This condition holds for the homogeneous coordinate ring of the Grassmannian (see e.g. https://arxiv.org/abs/1712.00447 or https://arxiv.org/abs/1803.06901). On the other hand, the $m$th homogeneous component of the coordinate ring is isomorphic to the $GL$ representation $V(m \omega)^*$ for the appropriate minuscule weight $\omega$, hence has a basis (the dual canonical basis of Lusztig/Kashiwara) coming from the theory of quantized enveloping algebras. Could these bases be the same? There are some vague statements alluding to this possibility in GHKK, but nothing definitive. I know the two bases share some interesting properties: e.g., the twisted cyclic shift symmetry of the Grassmannian permutes both bases according to promotion of semistandard tableaux (for the dual canonical basis this was proved by Rhoades, see https://arxiv.org/abs/1809.04965; for the theta basis this was proved by Shen and Weng https://arxiv.org/abs/1803.06901). Most naive bases (e.g. the standard monomial basis) do not satisfy this property. REPLY [10 votes]: I think there is good reason to think the answer is "no". In rank 2, the theta basis agrees with the greedy basis (arXiv:1508.01404). Greedy basis elements are indecomposable positive elements (see arXiv:1208.2391) (i.e., they cannot be written as a sum of two positive elements). If the indecomposable positive elements form a basis, this is what is now called the atomic basis. Consider the cluster algebra associated to the Kronecker quiver. Sherman and Zelevinsky showed arxiv:math/0307082 that the Kronecker-type cluster algebra has an atomic basis. (Note that, confusingly, the term they used for what is now called the atomic basis is the "canonical basis".) If $x$ and $y$ are any two cluster variables from a common cluster, the atomic basis element associated to the imaginary root $\delta$ is $z=x/y + y/x + 1/xy$. The atomic basis element associated to $2\delta$ is $z^2-2$. There is a cluster algebra with a dual canonical basis which is of type $\widetilde A_1$. This is a quantum cluster algebra, so when we talk about the dual canonical basis of the corresponding classical (commutative) cluster algebra, we mean that $q$ has been set to 1. Inconveniently, this cluster algebra also has coefficients. If we simply set them to 1 as well, then the dual canonical basis element corresponding to $2\delta$ is $z^2-1$. This follows from formula (31) in arXiv:1002.2762. (It is a formula for the basis element associated to $n\delta$. Note that $p_0$ and $p_1$ are coefficients, which we are ignoring, while $u_1$ and $u_2$ are what I was calling $x$ and $y$.) Even without the Sherman-Zelevinsky result, it is clear that $z^2-1$ cannot be in the greedy basis, since it is not an indecomposable positive element. It is easy to check that $z^2-1=(z^2-2)+1$ expresses it as a sum of two positive elements. The thing I really wanted to mention, though, is the recent paper by Fan Qin, arXiv:1902.09507, which gives a conceptual explanation for the existence of many good bases which disagree. He shows that there is in a sense a moduli space of good bases. The theta basis is a particular point in this moduli space, and the dual canonical basis is another.<|endoftext|> TITLE: The list chromatic number of some special toroidal grid graphs QUESTION [5 upvotes]: A list-assignment $L$ to the vertices of $G$ is the assignment of a list set $L(v)$ of colours to every vertex $v$ of $G$; and a $k$-list-assignment is a list-assignment such that $|L(v)|\geq k$, for every vertex $v$. If $L$ is a list-assignment to $G$, then an $L$-colouring of $G$ is a colouring (not necessarily proper) in which each vertex receives a colour from its own list; The graph $G$ is $k$-list-colourable, or $k$-choosable, if it is properly $L$-colourable for every $k$-list-assignment $L$ to $G$. The chromatic number $\chi(G)$ of $G$ is the smallest number $k$ such that $G$ is $k$-colourable. The list chromatic number $\chi_l(G)$, is the smallest number $k$ such that $G$ is $k$-list-colourable or $k$-choosable. It is evident that $\chi_l(G)\geq \chi(G)$, since if $k < \chi(G)$ then $G$ is not $L$-colourable when every vertex $v$ of $G$ is given the same list $L(v)$ of $k$ colours. Consider the graph $S_n$ which has as vertex set the $n^2$ cells of our $n\times n$ array with two cells adjacent if and only if they are in the same row or column. The graph $S_n$ Since any $n$ cells in a row are pairwise adjacent we need at least $n$ colors. Furthermore, any coloring with $n$ colors corresponds to a Latin square, with the cells occupied by the same number forming a color class. Since Latin squares, as we have seen, exist, we infer $\chi(S_n) = n$, and the Dinitz problem can be stated as $$\chi_l(S_n) = n? $$ Now we know that the equality holds for all $n$. By the solution of Dinitz's problem, we know that the list chromatic number of $C_3\Box C_3$ is 3, i.e. $\chi_l(C_3\Box C_3)=3$. The method of attack for the Dinitz problem is : We have to find an orientation of the graph $S_n$ with outdegrees $d_+(v) ≤ n−1$ for all $v$ and which ensures the existence of a kernel for all induced subgraphs. I want to know the $\chi_l(C_3\Box C_5)$ and $\chi_l(C_5\Box C_5)$. I conjecure both of them are 3. In the above oriented graph $C_3\Box C_5$, for all vetex $V$, it is easy to see that $$ deg_-(v)=deg_+(v)=2.$$ I verified some of induced subgraph which indeed have a kernel. But I have no idea how to prove that all induced subgraph have a kernel. If someone can give any suggestions or comments, I will appreciate it. Thanks. REPLY [3 votes]: I will encode orientations of $C_3\square C_5$ with outdegress equal to two by labeling vertices with numbers $0,1,2$, such that vertex with label $x$ has exactly $x$ outgoing horizontal arcs. For example, your picture encodes as $$ 21110\\ 02021\\ 10202 $$ One can restore the orientation from such encoding - each horizontal cycle breaks up into an even number of paths oriented from $2$ to $0$, and each vertical cycle breaks up into two paths oriented from $0$ to $2$. There is only one orientation of $C_3\square C_5$ satisfying conditions on outdegrees (modulo cyclic shifts and reflections). To see that first note that in any such orientation there will be one vertex of each label in any vertical cycle. Each horizontal cycle must have an equal amount of $0$'s and $2$'s. That means that two of horizontal cycles will have multisets of labels $\{0,0,1,2,2\}$, and the remaining horizontal cycle will have multiset $\{0,1,1,1,2\}$. We can assume it is the first cycle. Modulo cyclic shifts and reflections, this cycle is encoded either as $21110$, or $21101$. In the first case, the only possible orientation is yours (again, modulo horizontal reflection). In the second case, no such orientation exists. In both cases I use the fact that in each cycle $0$'s and $2$'s must interchange each other, if you remove $1$'s. Finally, the following induced subgraph in the orientation from the picture does not have a kernel, because it is an odd cycle: $$ 2\color{red}{1110}\\ 02021\\ \color{red}{10}20\color{red}2 $$ So unfortunately it looks like the kernel method will not help in this problem.<|endoftext|> TITLE: What can the approximation of a group by some class be used for? QUESTION [6 upvotes]: Recall the following concept due to Malcev and Gromov. Let $C$ be some class of groups. A group $G$ is said to be approximable by the class $C$ if for every finite symmetric subset $F\subset G$ containing 1, there is a group $H$ in $C$ and an injection $f:F\to H$ such that $f(1)=1$ and $f(x)f(y)=f(z)$ for every $x,y,z\in F$ with $xy=z$. My question is: why is this concept useful? I.e. what are real application of it in group theory (say, for establishing some properties of $G$ via its approximation by some class)? REPLY [3 votes]: Let $\mathcal{F}_k$ be the class of free groups that can be generated by at most $k$ elements. In your terminology, limit groups are the groups that can be approximated by the groups in $\mathcal{F}_k$. This article by Champetier--Guirardel develops the theory of limit groups from this point of view. It contains many examples of the kinds of applications that you're looking for.<|endoftext|> TITLE: Is a cofinite topology for a set with cardinality between $\aleph_{0}$ and $2^{\aleph_{0}}$ path-connected? QUESTION [8 upvotes]: It is easy to show that $\mathbb{N}$ with the cofinite topology is not path connected and that any set with cardinality $\geq 2^{\aleph_0}$ equipped with the cofinite topology is in fact path connected. what about cardinalities $\aleph_0<\alpha<2^{\aleph_0}$ (under the assumption that such exist obviously)? If $\alpha $ is path connected then any cardinality $\geq \alpha$ is also, so an interesting direction would be trying and checking whether $\aleph_1$ is path connected (I have no clue about how one can even start checking this). REPLY [5 votes]: A continuous non-constant function from $[0,1]$ into $X$ with the cofinite topology exists iff $[0,1]$ has a partition into $\le |X|$ many disjoint closed non-empty subsets. This question discusses the options for the cardinality of such a partition. One of the conclusions is that under $\textrm{MA}(\omega_1)$ we have that a set of size $\aleph_1$ in the cofinite topology is (connected and ) not path-connected, while of course under CH such a set is path-connected. So the case $\aleph_1$ is undecidable under ZFC.<|endoftext|> TITLE: Wavefront set of characteristic function of rough set QUESTION [5 upvotes]: It is a standard exercise to show that if $X\subseteq\mathbb{R}^n$ has smooth boundary, then the characteristic function $1_X$ has wavefront set $$\{(x,\xi)\in\partial X\times\mathbb{R}^n\setminus\{0\}:\xi\text{ is normal to }\partial X\}.$$ This is shown by locally "flattening" to reduce to the case of the upper half space. My question is what happens when $X$ does not have smooth boundary, such as the interior of a fractal curve. For example, if $X$ is the interior of the Koch snowflake, how would one compute the wavefront set of $1_X$? REPLY [2 votes]: Let me first begin with an elementary example, taking $X=[0,1]^2$ in $\mathbb R^2$. It is then easy to see directly that the wave-front-set of $\mathbb 1_X$ is everywhere the conormal bundle except at the four corners $c_j$: $$ WF\mathbb 1_X=\cup_{1\le j\le 4}(c_j;\mathbb R^2\backslash {(0,0)})\cup\text{conormal bundle at the smooth points}. $$ To prove this, it is enough to check the wave-front-set of $H(x_1)H(x_2)$ with $H=\mathbb 1_{[0,+\infty)}$ which amounts to check the wave-front-set of $\delta_0(x_1)\otimes\delta_0(x_2)$ which contains at $(0,0)$ every direction since the Fourier transform is 1. There are of course (much) more refined results. If you take a look at Lars Hörmander's ALPDO first volume, Springer Grundlehren 256 on page 300, you will find the definition of the normal set to any closed subset of $\mathbb R^n$ and Theorem 8.5.6' asserts that the analytic wave-front-set of a distribution $u$ contains the normal set to the support. For the example you give, I believe that the $C^\infty$ wave-front-set is simply the product of the boundary with $\mathbb R^2\backslash {(0,0)}$. However for $s$ real, we may define the $H^s$ wave-front-set and I guess that the Hausdorff dimension of the fractal set you consider should be linked to the $H^s$ wave-front-set, in the sense that it should be trivial at some threshold $s_0$ and that $s_0$ should be linked to the Hausdorff dimension. The paper MR1277392, by Falcone, is addressing a related question.<|endoftext|> TITLE: Obstruction to homotopy, cohomology operations and Dold-Whitney theorem QUESTION [8 upvotes]: I am reading the famous paper by Dold and Whitney "Classification of Oriented Sphere Bundles Over A 4-Complex". I'll state their theorem for the case of SO(3) bundles Classification Theorem:Let $B_1, B_2$ be principal $SO(3)$ bundles over a complex $K$ of dimension at most 4. let $h_i:K\to G_n\simeq BSO(3)$ be the classifying map for $B_i$. Assume they have the same second Stiefel-Whitney class $w_2(B_1)=w_2(B_2)=: w_2$ (then we can assume that $h_1$ and $h_2$ agree on the 3-skeleton $K^3$). Let $d(h_1,h_2) \in H^4(K;\pi_4(G_n))$ be the difference cocycle cohomology class. The bundles $B_1,B_2$ are equivalent iif there exists a cohomology class $x\in H^1(K,\mathbb{Z}/2)$ such that $$(1) \quad \quad d(h_1,h_2) = \beta x \cup \beta x + \beta (x\cup w_2)$$ where $\beta$ is the Bockstein homomorphism associated with the coefficient sequence $0\to \mathbb{Z} \overset{2\cdot}{\to} \mathbb{Z}\to \mathbb{Z}/2\to 0$ and $\pi_3(G_n)$ has been identified with $\mathbb{Z}$. Usually we require the cohomology class of the cocycle to $d(h_1,h_2)$ to be $0$, why in (1) they require that condition instead? Explanation of the question I have seen stated in others book on obstruction theory that $h_1,h_2:K^n\to Y$ are homotopic rel to $K^{n-2}$ iif $[d(h_1,h_2)]=0$ (under the hypothesis of $Y$ being simple, I don't know if this is the point) this seems to be very different from $(1)$. Cohomology operations. Dold & Whitney appeal to a theorem from Eilenberg MacLane's paper "On the groups $H(\pi,n),\quad III:$ Operations and Obstructions", theorem 14.2. I looked into it and indeed they don't require the difference cocycle to be null-cohomologous. I only explained myself that the RHS of $(1)$ comes from the cohomology operation induced by the $k$-invariant (I guess Postnikov invariant) of the target space. Can someone explain better the setup and theorem 14.2 in Eilenberg MacLane's paper? REPLY [9 votes]: Let me explain why the condition is not the triviality of the difference cocycle. Maybe an important point to note is that the condition with the difference cocycle is not about the vanishing of an obstruction, it is about the enumeration of possible lifts of maps from one Postnikov truncation to the next. If we denote by $\tau_{\leq n} {\rm BSO}(3)$ the Postnikov truncation which kills all homotopy above degree $n$, then we have the usual fiber sequence of obstruction theory $$ K(\pi_4{\rm BSO}(3),4)\to \tau_{\leq 4}{\rm BSO}(3) \to \tau_{\leq 3}{\rm BSO}(3). $$ This gives rise to an exact sequence of pointed sets of homotopy classes $$ [X_+,\Omega\tau_{\leq 3}{\rm BSO}(3)]\to [X_+,K(\pi_4{\rm BSO}(3),4)]\to [X_+,\tau_{\leq 4}{\rm BSO}(3)]\to [X_+,\tau_{\leq 3}{\rm BSO}(3)]. $$ Note that ${\rm BSO}(3)$ is simply connected (hence also simple), so there is no difference between pointed and unpointed maps here. Since $\tau_{\leq 3}{\rm BSO}(3)=K(\mathbb{Z}/2,2)$, we can rewrite that sequence $$ {\rm H}^1(X,\mathbb{Z}/2)\to {\rm H}^4(X,\mathbb{Z})\to [X_+,{\rm BSO}(3)]\to {\rm H}^2(X,\mathbb{Z}/2). $$ The set $[X,{\rm BSO}(3)]$ gives the isomorphism classes of oriented rank 3 bundles, the last map takes a bundle to its second Stiefel-Whitney class. The additional invariant lives in a quotient group of $H^4(X,\mathbb{Z})$ and the first map ${\rm H}^1(X,\mathbb{Z}/2)\to {\rm H}^4(X,\mathbb{Z})$ takes a class $x$ to $\beta x\cup \beta x+\beta(x\cup w_2)$. So the difference cocycle is a class in ${\rm H}^4$, but the bundle is still trivial if the cocycle lies in the image of ${\rm H}^1(X,\mathbb{Z}/2)$ which explains the condition.<|endoftext|> TITLE: Natural examples of $(\infty,n)$-categories for large $n$ QUESTION [19 upvotes]: In Higher Topos Theory, Lurie argues that the coherence diagrams for fully weak $n$-categories with $n>2$ are 'so complicated as to be essentially unusable', before mentioning that several dramatic simplifications occur if we step up to $\infty$-categories but require all higher morphisms to be invertible. What are some examples in nature (topological quantum field theory, string theory etc.) of notions that are best understood as a weak $(\infty,n)$-category, or simply a weak $n$-category, for 'large' $n$? (something which is naturally a 'fully weak' $\infty$-category would be interesting too) Bordisms between $n$-dimensional manifolds have the former structure naturally and serve as the canonical example coming from TQFTs; are there any others? (I believe the present question differs from this one, although the answers may overlap, as I'm looking for examples of $\infty$-categories that are 'motivated by nature' in the above sense and already believe higher categories are useful.) I am also interested in natural occurrences of $\infty$-categories or weak $n$-categories for 'large' $n$ in areas of mathematics outside of category theory, so please share any good ones that occur to you. The use of $2$-category theory for the Galois theorem of Borceux and Janelidze wouldn't quite qualify, but if there is any area outside of category theory that seriously uses weak $3$-categories I would be interested to hear about it (invariants of $3$-manifolds?). This is why 'large' is in quotations; examples with $n>>1$ would be great, but I consider $3$ 'large' in certain contexts where $2$ isn't. REPLY [13 votes]: $\newcommand{\Vect}{\mathrm{Vect}} \newcommand{\Mod}{\mathrm{Mod}} \newcommand{\cc}{\mathbf{C}}$Here are three (related) examples. The first one is simple (although not really related to physics): an $\mathbf{E}_n$-space is "just" an $(\infty,n)$-category with one object, one $1$-morphism, ..., and one $(n-1)$-morphism. This can be generalized: an $\mathbf{E}_k$-monoidal $(\infty,n)$-category can be thought of as an $(\infty,n+k)$-category with one object, ..., and one $(k-1)$-morphism. For the second example (whose details I don't know), let $\Vect(\cc)$ denote the category of vector spaces, regarded as an $(\infty,1)$-category by taking its nerve. This admits a symmetric monoidal structure, and one can look at the $(\infty,2)$-category $\mathrm{Mod}_{\Vect(\cc)}$ of modules over it, taken in presentable $(\infty,1)$-categories. (This is related to the category of $\cc$-algebras, bimodules, and intertwiners.) This is again supposed to be a symmetric monoidal $(\infty,2)$-category, so one can look at "modules" over it to get an $(\infty,3)$-category, and so on. I think that the $(\infty,n)$-category you get this way is supposed to be the natural target of "physical" $n$-dimensional extended TQFTs, and that its Picard space is supposed/conjectured to be $\Omega^\infty \Sigma^{n+1} I_{\cc^\times}$, where $I_{\cc^\times}$ is the Brown-Comenetz dualizing spectrum (although I'm not sure of these statements, and would appreciate if someone confirmed/elaborated upon them!). For the third example, let $\mathcal{C}$ denote a symmetric monoidal $(\infty,1)$-category; then you can inductively construct an $(\infty,n+1)$-category $\mathrm{Mor}_n(\mathcal{C})$ (the "higher Morita category") for any $n$ by defining its objects to be $\mathbf{E}_n$-algebras in $\mathcal{C}$, and such that the $(\infty,n)$-category of morphisms from $R$ and $S$ is the category $\mathrm{Mor}_{n-1}(_R\mathrm{BMod}_S)$, where $_R\mathrm{BMod}_S$ is the $(\infty,1)$-category of $(A,B)$-bimodules. If $n=1$, this is the category of associative algebras (in $\mathcal{C}$), bimodules, and bimodule homomorphisms (so the equivalences are Morita equivalences). You then get a TQFT $Z_A:\mathrm{Bord}_{n+1}\to \mathrm{Mor}_n(\mathcal{C})$ for every $\mathbf{E}_n$-algebra object $A$ of $\mathcal{C}$ which is fully dualizable in $\mathrm{Mor}_n(\mathcal{C})$ (this is a rather strong condition); this TQFT sends $M^d$ to the factorization/topological chiral homology $\int_{\mathbf{R}^{n-d}\times M^d} A$.<|endoftext|> TITLE: Do the Odd Cycles of a Graph Define a Matroid? QUESTION [5 upvotes]: An Odd Cycle Transversal is a set of vertices that, when removed from a graph, renders it bipartite. Question: does the collection of "critical" sets of vertices, whose removal renders a graph bipartite, resemble a Matroid, i.e. will a greedy strategy yield a critical vertex set of minimal cardinality? I am looking for a proof deciding the matroid property that either affirm the success of the greedy strategy or for a proof of at least the existence of instances, where the greedy strategy doesn't yield a minimal set of vertices whose removal renders the graph bipartite; a concrete counter example would be more than I dare to hope for in that case. By a critical set of vertices, in this context, I mean a set of vertices whose removal renders a graph bipartite, but none of its proper subsets has that property. By the greedy strategy I mean repeatedly removing a vertex that is on a maximal number of odd cycles in the graph resulting from previous greedy vertex deletions. Please note that the question is not as to whether it is efficiently possible to determine the number of odd cycles on which a vertex; it is rather assumed that that information comes from a kind of oracle or whatever celestial being you prefer. REPLY [2 votes]: Greedy algorithm doesn't do that well in the worst case, even provided the odd cycle counting oracle. Take large $n$, and consider complete bipartite $K_{n, n}$ with parts $V_0, V_1$ together with an additional disjoint triangle $T$, and connect all vertices of $T$ with two vertices $v, u \in V_0$ pairwise. Odd cycles in this graph consist of paths in $K_{n, n}$ between $v$ and $u$ extended by a two-vertex path in $T$, together with $O(1)$ odd cycles confined to $\{v, u\} \cup T$. One can see that: $v$ and $u$ are contained in all odd cycles except for $O(1)$; vertices of $T$ and other vertices of $K_{n, n}$ are contained in at most $2 / 3 + o(1)$ fraction of all odd cycles. A smallest OCT consists of two vertices of $T$, yet the greedy algorithm WLOG will first take $v$, and then will proceed to delete at least two other vertices (since the remaining graph contains $K_4$, namely, the union of $u$ and $T$).<|endoftext|> TITLE: Is $|\{(j,k):\ 1\le j(k^{16}\ \text{mod}\ p)\}|$ even for each prime $p\equiv1\pmod {16}$? QUESTION [6 upvotes]: In my paper http://arxiv.org/abs/1809.07766, I determined the parity of $$\left|\left\{(j,k):\ 1\le j(k^2\ \text{mod}\ p)\right\}\right|$$ for any prime $p\equiv3\pmod4$, where $(a\ \text{mod}\ p)$ denotes the least nonnegative residue of $a$ modulo $p$. Based on my computation, I have the following three conjectures. Conjecture 1. For any prime $p\equiv1\pmod4$, we have $$\left|\left\{(j,k):\ 1\le j(k^4\ \text{mod}\ p)\right\}\right|\equiv\left\lfloor \frac{p-1}8\right\rfloor\pmod2.$$ Conjecture 2. For any prime $p\equiv1\pmod8$, we have $$\left|\left\{(j,k):\ 1\le j(k^8\ \text{mod}\ p)\right\}\right|\equiv\left|\left\{1\le k<\frac p4:\ \left(\frac kp\right)=1\right\}\right|\pmod2.$$ Conjecture 3. For any prime $p\equiv1\pmod{16}$, we have $$\left|\left\{(j,k):\ 1\le j(k^{16}\ \text{mod}\ p)\right\}\right|\equiv0\pmod2.$$ QUESTION. How to prove Conjectures 1-3? The conjectures might not be very difficult. Maybe some of you could prove them. Your comments are welcome! REPLY [7 votes]: Start with Conjecture 1. Two other look similar, but possibly require additional ideas (UPDATE: they do not actually). Write $(x)_p\in \{0,\ldots,p-1\}$ for the remainder of integer $x$ modulo $p$. Denote $p=2m+1$. We have to calculate the sign of $$ A:=\prod_{1\leqslant j TITLE: A ring of generalized power series QUESTION [7 upvotes]: Let $\Bbbk$ be a field; I am interested in the following ring (which I suspect is a field). Its elements are formal expressions that look like $$ \sum_{n=0}^{\infty} a_n x^{b_n} $$ where $a_n\in \Bbbk$ and $b_n\in \mathbb{R}$, with $b_n$ strictly increasing, and $\lim_{n\to\infty} b_n = \infty$. (Technically, we should quotient these by some relation allowing us to insert and remove terms where $a_n=0$. Or we could require all the $a_n$'s to be nonzero, but then we'd have to include finite sums as well. Or we could represent them by functions $a : \mathbb{R} \to \Bbbk$ assigning a coefficient to each exponent, whose support is left-finite.) We can add and multiply these expressions in fairly evident ways; the condition on $b_n$ ensures that multiplication works (i.e. the resulting set of exponents can again be enumerated with order type $\omega$ and limit $\infty$). This ring of power-series-like-objects is closely related to some others. Specifically, if $\Bbbk=\mathbb{R}$ then it contains the Levi-Civita field as the elements for which each $b_n\in\mathbb{Q}$, while it is contained in the Hahn series field $\Bbbk[[x^{\mathbb{R}}]]$. Note that the set of all Hahn series with order type $\omega$ is not closed under multiplication; this is a natural subset thereof that is. I believe that it is also the set of Hahn series with order type $\omega$ that converge to themselves in the valuation topology of the Hahn series field, and also that it is the closure of the field $\Bbbk(x^{\mathbb{R}})$ of generalized rational functions inside the Hahn series field, and the Cauchy completion of $\Bbbk(x^{\mathbb{R}})$ in its valuation uniformity. Does this field have a standard name and/or a notation? Is it an instance of some more general construction (e.g. replacing $\mathbb{R}$ by something more general, which would presumably then also include the Levi-Civita field as the case of $\mathbb{Q}$)? REPLY [10 votes]: I think your ring looks similar to the Novikov ring (see topology papers).<|endoftext|> TITLE: Closed form $\int_{0}^{\frac{r}{2}} {\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}^{-1}} \ \text{d}p$ QUESTION [5 upvotes]: Note: This is exact copy of my Math.SE question, which I am reposting here, as despite bounty it did not receive any answers. Let there be $n$ pairs of shoes in a box. The the probability that from the $r \le n$ shoes I am taking out of the box there are exactly $p$ pairs is given by \begin{equation*} \mathbb{P}_{n}^{(r)}(p) = \frac{\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}}. \end{equation*} For $n = 15$ and $r \in \{6,8,10\}$. The function (assuming the continuous factorial equivalents) looks like this: I am interested in finding the area under that curve, namely $$\int_{0}^{\frac{r}{2}} \frac{\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}} \ \text{d}p$$ I consulted this question but could derive how that would help me. I also thought about writing the first product of binomial coefficients as $$ \binom{n}{n - p}\binom{n - p}{r - 2p} $$ which is similar to the form $\binom{f(x)}{f(y)} \binom{f(y)}{f(x)}$ mentioned in this question. REPLY [3 votes]: I suspect some form of CLT for large $n, r$, which may possibly be proved by adopting Laplace's method of approximating the integrand by an appropriate gaussian kernel. Interestingly enough, we can prove that: $$ \int_{-\infty}^{\infty} \frac{\binom{n}{p}\binom{n-p}{r-2p}2^{r-2p}}{\binom{2n}{r}}\,\mathrm{d}p = 1. $$ To show this, we may apply the Legendre duplication formula to write $$ \frac{\binom{n}{p}\binom{n-p}{r-2p}2^{r-2p}}{\binom{2n}{r}} = \frac{n!}{\binom{2n}{r}} \frac{\sqrt{\pi}}{\Gamma(1+p)\Gamma(n-r+1+p)\Gamma(\frac{r}{2}+\frac{1}{2}-p)\Gamma(\frac{r}{2}+1-p)}$$ and then apply the Ramanujan's beta integral (see the formula (5.3.14) of DLMF: 5.13).<|endoftext|> TITLE: Generalized Behrend version for Grothendieck-Lefschetz trace formula QUESTION [6 upvotes]: [MOVED HERE FROM MSE.] The statement of the Grothendieck-Lefschetz fixed point theorem is well-known. For a proper algebraic variety $X$ over $\mathbb F_q$, $$\#X(\mathbb F_q) =\sum_i (−1)^i Tr(Fr_X, H^i_c(X, \mathbb Q_l)).$$ Also known is the version for general constructible l-adic sheaves $\mathcal F$: $$\sum_{x\in X(\mathbb F_q)} Tr(Fr_x,\mathcal F_x)=\sum_i (−1)^i Tr(Fr_X, H^i_c(X, \mathcal F)).$$ Thirdly, K. Behrend proved an analog for the first formula in the context of algebraic stacks (replacing the scheme $X$ by a Noetherian algebraic stack $\mathcal X$). Now my question is: is there a version of the second formula for an algebraic stack $\mathcal X$ (with nice hypotheses if necessary)? It would seem natural, since the second formula is a generalization of the first, and the first is true in the context of algebraic stacks by Behrend's work. However, the second formula does not follow directly from the first in the case of schemes (as far as I know: I would be glad if it were true!), so I am not in the position to easily extend the proof of the second formula in the more general context of stacks. Thank you in advance. ADDED QUESTION: Moreover, why is the sum on the left of the second formula finite when the scheme (or stack) is not of finite type? Behrend speaks about this problem, but I do not find where he solves it, if he does. REPLY [10 votes]: This is Theorem 4.2 of Shenghao Sun's paper $L$-Series of Artin stacks over finite fields, Algebra & Number Theory 6 (2012) pp 47–122, doi:10.2140/ant.2012.6.47, arXiv:1008.3689. Let $f:\mathscr X_0\to\mathscr Y_0$ be a morphism of $\mathbb F_q$-algebraic stacks, and let $K_0\in W^{-,stra}_m(\mathscr X_0,\overline{\mathbb Q}_{\ell})$ be a convergent complex of sheaves. Then (i) (Finiteness) $f_!K_0$ is a convergent complex of sheaves on $\mathscr Y_0,$ and (ii) (Trace formula) $c_v(\mathscr X_0,K_0)=c_v(\mathscr Y_0,f_!K_0)$ for every integer $v\ge1.$ Here $c_v$ is the sum of the trace of the $v$th power of Frobenius. Applying this in the case where $\mathscr Y_0$ is a point gets you what you want.<|endoftext|> TITLE: Edges of the contact polytope of the Leech lattice QUESTION [6 upvotes]: Let $P\subset\Bbb R^{24}$ be the contact polytope of the Leech lattice, that is, $P$ is the convex hull of the 196,560 shortest vectors of $\Lambda_{24}$. Question: What are the edges of $P$? Let's say the norm of any vertex is $4\sqrt 2$. I am pretty sure that any two vertices at distance $4\sqrt 2$ form an edge. But what about vertices farther apart? If I am not mistaken, then for any specific vertex, there are further vertices at distance $8$ and $4\sqrt 6$ (and, of course, the antipodal vertex at distance $8\sqrt2$, but there is no edge obviously)? I am aware of Sikirić, Mathieu Dutour, Achill Schürmann, and Frank Vallentin. The contact polytope of the Leech lattice. but it only discusses the facets. In fact, I found no other publication discussing this polytope. REPLY [5 votes]: Using the unimodular scaling of the Leech lattice, the length of each minimal vector is $\sqrt{4}$. Fixing a particular minimal vector $u$, the remaining minimal vectors $v$ are: 1 vector $v$ with $\langle u, v \rangle = 4$ (namely $v = u$); 4600 vectors $v$ with $\langle u, v \rangle = 2$; 47104 vectors $v$ with $\langle u, v \rangle = 1$; 93150 vectors $v$ with $\langle u, v \rangle = 0$; 47104 vectors $v$ with $\langle u, v \rangle = -1$; 4600 vectors $v$ with $\langle u, v \rangle = -2$; 1 vector $v$ with $\langle u, v \rangle = -4$ (namely $v = -u$); and the Conway group $Co_2$ (the pointwise stabiliser of $\{0, u\}$) acts transitively on each of the seven sets of vertices mentioned above. Now, suppose that $u, v, x, y$ are distinct minimal-norm lattice vectors, and further suppose that we have: $$ \lambda u + (1 - \lambda)v = \mu x + (1 - \mu) y $$ where $0 < \lambda, \mu < 1$. Then the line segment with endpoints $u,v$ intersects the line segment with endpoints $x,y$, so neither of these can be edges of the contact polytope $P$. The special case where $\lambda = \mu = \frac{1}{2}$ is particularly helpful: it states (when you double both sides of the equation) that if $u + v = x + y$, then neither of the line segments are edges. So if $u, v$ does form an edge, then all vectors of the form $x + y$ (where $x, y$ is an arbitrary pair of minimal-length vectors satisfying $\langle x, y \rangle = \langle u, v \rangle$) must be pairwise distinct. But the Leech lattice is additively closed, and we know the number of vectors of each norm, so we can apply the pigeonhole principle: There are $\frac{1}{2} \times 196560 \times 1$ sums of pairs of antipodal minimal vectors, but only one Leech vector of length 0, so there are no edges between antipodal vertices. There are $\frac{1}{2} \times 196560 \times 4600 = 452088000$ sums of pairs of minimal vectors $u, v$ with $\langle u, v \rangle = -2$, but only $196560$ Leech vectors of length $\sqrt{4}$, so there are no edges of this form. There are $\frac{1}{2} \times 196560 \times 47104 = 4629381120$ sums of pairs of minimal vectors $u, v$ with $\langle u, v \rangle = -1$, but only $16773120$ Leech vectors of length $\sqrt{6}$, so there are no edges of this form. There are $\frac{1}{2} \times 196560 \times 93150 = 9154782000$ sums of pairs of minimal vectors $u, v$ with $\langle u, v \rangle = 0$, but only $398034000$ Leech vectors of length $\sqrt{8}$, so there are no edges of this form. On the other hand, this stops working for any closer pairs of vectors: There are $\frac{1}{2} \times 196560 \times 47104 = 4629381120$ sums of pairs of minimal vectors $u, v$ with $\langle u, v \rangle = 1$, which is exactly the same as the number of length-$\sqrt{10}$ vectors in the lattice! Assuming that $Co_0$ acts transitively on these vectors (this has probably been proved somewhere?), it means that they're all uniquely expressible as the sum of two minimal vectors. So we've ruled out edges between pairs of vertices $u,v$ when $\langle u, v \rangle \leq 0$. The closest pairs of vertices clearly form edges, so we know that there's an edge whenever $\langle u, v \rangle = 2$ (i.e. the angle is 60 degrees). This leaves the delicate case of $\langle u, v \rangle = 1$, where $u - v$ is a vector of length $\sqrt{6}$. To show that these do indeed form an edge, it suffices to show that there are no other vertices $w$ with $\langle w, u + v \rangle \geq \langle u, u + v \rangle = 5$; that way, the midpoint $\frac{1}{2}(u + v)$ cannot be expressed as a combination of any vertices other than $u$ and $v$ (because all other vertices lie on the same side of the perpendicular bisecting hyperplane of $u + v$ and $0$). Now, if $\langle w, u + v \rangle \geq 5$, then one of $\langle w, u \rangle$ or $\langle w, v \rangle$ must be at least $\frac{5}{2}$. But the inner product of two distinct minimal Leech vectors is at most $2$, so this is only possible if $w = u$ or $w = v$. This proves the desired result, and also implies unconditionally that every length-$\sqrt{10}$ Leech lattice vector is uniquely expressible as the sum of two length-$\sqrt{4}$ lattice vectors. In conclusion, $P$ has $196560$ vertices, $452088000$ 'short' edges of length $\sqrt{4}$, and $4629381120$ 'long' edges of length $\sqrt{6}$.<|endoftext|> TITLE: Do we know SL(2,C) subgroups (not only finite ones)? QUESTION [19 upvotes]: The list of all closed subgroups of O(3) (orthogonal transformation group) or SO(3) (rotation group) is known, and several references exist that give an explicit approach. In the case of SL(2,C) subgroups: I know that any finite subgroup is conjugated to a subgroup of SU(2). Surely it can be extended to a compact subgroup. My questions are: 1) Is the list of SL(2,C) subgroups (with conjugation accuracy) known? 2) Otherwise, is the list of closed subgroups of SL(2,C) known, and is it reduced to subgroups combined with subgroups of SU(2). Thank you, in your answers, for specifying references. REPLY [27 votes]: I'll interpret your question to be regarding closed subgroups of $SL(2,\mathbb{C})$, since you mention in the first paragraph closed subgroups of $O(3)$. Otherwise very little is known (see e.g. Calegari-Dunfield for non-discrete examples). Moreover, usually people work with classification of closed subgroups of $PSL(2,\mathbb{C})$. Then the $SL(2,\mathbb{C})$ case follows by taking a central extension of the discrete subgroup of $PSL(2,\mathbb{C})$ (or a lift avoiding $-I$ if it exists). As you mention, the compact subgroups of $SL(2,\mathbb{C})$ are conjugate into $SU(2)$. For closed subgroups of positive dimension, the identity component is a Lie subgroup, such as $SU(2)$, $SL(2,\mathbb{R})$, and some solvable (upper triangular) examples. These may be classified by considering the fixed points or invariant subsets of the action on $\mathbb{CP}^1$. I believe that this classification is fairly straightforward (if tedious), but I haven't considered the case-by-case analysis (it's possible that it exists in the literature, but I haven't done a search added: see the reference to Greenberg given by Misha in the comments below). Otherwise, one has the discrete subgroups of $PSL(2,\mathbb{C})$. Here there is a big difference in what is known between the finitely-generated and infinitely generated cases. The finitely-generated case (as noted by Yves de Cornulier in the comments) is the classification of Kleinian groups (historically stronger assumptions were made on such groups, and some do not require them to be finitely generated). In a strong sense, this classification was carried out culminating in the Ending Lamination Theorem. There are some caveats here though: the proof has not appeared in the case of decomposable (free product) groups (or more generally groups amalgamated over finite subgroups). Moreover, the classification gives these groups indirectly. It states that the group is the fundamental group of a compact hyperbolic orbifold (possibly with boundary) together with some ending data, which is a set of conformal structures and ending laminations associated to subsets of the boundary. Once one is given such data (which can be encoded combinatorially together with a weighted train track and a point in an appropriate moduli space), it is still a non-trivial step to actually compute the group (e.g. in terms of a set of matrix generators). So this theorem is really giving just a 1-1 correspondence between conjugacy classes of Kleinian groups and some topological and conformal data. As an example, the 2-parabolic case has a well-understood classification. In this case, the free product forms the "Riley slice" of Schottky space (the rainbow shaded region in the picture). The interior contains many more 2-parabolic generated discrete groups which are not free products, such as 2-bridge knot groups. In principle, these all have a classification (although there are still some issues regarding the complete classification of the parabolic generating sets). For the non-finitely-generated case, there is not even a conjectured classification. Tommaso Cremaschi is beginning to consider examples (although there were previous examples known to experts).<|endoftext|> TITLE: Abstract transverse measure theory QUESTION [5 upvotes]: After reading Noncommutative Geometry book (see here) I came across the notion of the so called abstract transverse measure theory which is a generalization of standard measure theory well adapted to deal with singular examples like the space of leaves of the foliation. All the relevant information for me can be found on pages 78 and 79 of the above mentioned book. However this exposition is rather brief and I have several questions regarding this topic. I will divide them into more than one post in order to avoid asking too many questions in one post. So the context is as follows: $X,Y$ are standard Borel spaces and $p:Y \to X$ is a Borel map such that for each $x \in X$ the preimage $p^{-1}(x)$ is at most countable. The first claim is as follows: one defines $F(x)$ to be the cardinality of $p^{-1}(x)$. In this way we obtain a function $F:X \to \mathbb{N} \cup \{0\} \cup \{ \infty \}$ and the claim is that There exists a Borel bijection $\psi: Y \to \{ (x,n): 0\leq n < F(x) \}$ My guess is that $\psi$ should be defined as follows: let $Y=\sqcup_{n=0}^{\infty}Y_n$ be a partition of $Y$ with $Y_n:=\{y \in Y: card(p^{-1}(p(y))=n\}$ and $\psi(y):=(p(y),n)$ for $y \in Y_n$. However Q1 How to prove that $\psi$ is Borel? According to the discussion here the function $F$ need not no be measurable (and if we had defined $F$ as $F(y)=card(p^{-1}(p(y))$ still the same argument seems to work). Now the second claim is the following: suppose that we have two pairs $(Y_1,p_1), (Y_2,p_2)$ as above and let $F_1,F_2$ be as the functions as above constructed with respect to $p_1,$ and $p_2$. The claim is that: $F_1=F_2 \iff$ there exists Borel bijection $\psi:Y_1 \to Y_2$ such that $p_2 \circ \psi=p_1$. One direction of the above claim is clear: if we have $\psi$ then $F_1=F_2$. However I don't see how to prove the converse: so I'd like to ask Q2 How to prove that if $F_1=F_2$ then there exists a Borel bijection $\psi:Y_1 \to Y_2$ such that $p_2 \circ \psi=p_1$? After stating that the author drops the assumption for $X$ being standard Borel space and regard the second condition, i.e. the existence of $\psi$ as the definition of ,,having the same functions $F_i$''. I don't see what is the point here: -First of all there is no need for $F_i$ to be Borel so even if $X$ is a standard Borel space and $p$ is Borel map still $F_i$ may not be Borel so in some sense (from the point of view of $F_i$) there is no difference whether $X$ was standard Borel or not -Second thing: even if $F_i$ are not Borel functions they still make perfect sense even if $X$ is no more a standard Borel space: thus we can still ask whether $F_1=F_2$ -and on the top of that: why are we interested in this strange notion of ,,having the same functions'' i.e. $F_1=F_2$? If we are interested in pairs $(Y,p)$ where $Y$ is standard Borel and $p$ is a map $p:Y \to X$ where $X$ is fixed then the natural notion of equivalence is the existence of $\psi$ and ,,having the same funtions'' seems artificial to me. So to summarize: Q3 Why should we care whether $F_1=F_2$? REPLY [2 votes]: Your proposed $\psi$ can't be bijective, since two $y,y'\in Y$ with $p(y)=p(y')$ are sent to the same pair. But indeed, the sets $Y_n$ are measurable (and hence the functions $F$ are Borel). This follows from Exercise 18.15 in Kechris' Classical Descriptive Set Theory, which states exactly what you need that but for a subset $P\subseteq X\times Y$ with countable sections $P_x$ (you can take $P:=(\mathrm{graph}(p))^{-1}$). Let $A_n= \{x\in X : \mathrm{card}(p^{-1}(x)) = n\}$ for $n=0,\dots,\infty$. Then 18.15 says that for each $n$, $A_n$ is Borel and there are Borel functions $f_i^{(n)}:A_n\to Y$ with disjoint graphs such that for $x\in A_n$, $p^{-1}(x) = \{f_i^{(n)}(x) : i TITLE: Tricategorical coherence QUESTION [6 upvotes]: Why does coherence begin to matter at the tricategorical level? It is well known that every weak $2$-category is equivalent to a strict $2$-category, with the equivalence essentially given by (unless I'm mistaken) the $2$-Yoneda embedding for an arbitrary $2$-category into its strict $2$-category of $2$-presheaves into $\mathfrak{Cat}$ -- this means we can effectively ignore the coherence diagrams for associativity and unitarity in $2$-categories and work with strict ones where composition is associative on the nose and units vanish under composition. This is not so for $3$-categories; Gordon, Power and Street proved that all weak $3$-categories are equivalent to Gray categories, where associativity and unitarity still hold on the nose, but we have an additional coherence notion in play called 'interchange' for $2$-morphisms which introduces fundamental differences between $3$-categories with strict interchange and weak interchange laws. Intuitively speaking, why does this type of coherence 'matter' more than the associative and unital coherence that appears at the $2$-categorical level? The notions of '$n$-equivalence' become more varied as we move into higher $n$ as I understand it, so perhaps the 'correct' notion of $3$-equivalence is sensitive to interchange in some way that $2$-equivalence isn't sensitive to associativity for $1$-cells? I am currently working through the linked Gordon-Power-Street paper on coherence for tricategories and suspect the answer is buried in their proof of the tricategorical coherence theorem, but I thought someone here might already be familiar with it and able to help a newcomer -- any assistance is appreciated. REPLY [6 votes]: The short answer is that what matters is the combination of strict interchange and strict units, because interchange and units are what go into the Eckmann-Hilton argument, and dimension 3 is the first dimension in which you can have weak commutativity (e.g. braided monoidal categories are degenerate tricategories). GPS choose to strictify the units and therefore not the interchange, but you can also choose to strictify the interchange and not the units: André Joyal, Joachim Kock, Weak units and homotopy 3-types, in: Street Festschrift: Categories in algebra, geometry and mathematical physics, Contemp. Math 431 (2007), 257-276, doi:10.1090/conm/431/08277, arXiv:math/0602084.<|endoftext|> TITLE: When is the non-negative derived category compactly generated? QUESTION [13 upvotes]: This question is strongly related to this question. However it seems to me sufficiently distinct to warrant asking it separately. Let $X$ be a quasi-compact, quasi-separated scheme. When is the ∞-category $\mathcal{D}_{\ge0}(X)$ of connective sheaves of complexes compactly generated? Equivalently, when is every connective element of $\mathcal{D}(X)$ expressible as a filtered colimit of connective perfect complexes? (It is not hard to show that the compact objects of $\mathcal{D}_{\ge0}(X)$ are exactly the connective perfect complexes). Note that I am using homological grading, so $\mathcal{D}_{\ge0}(X)$ is a colocalization of $\mathcal{D}(X)$ and not a localization. Thanks to corollary C.6.3.3 in Lurie's Spectral Algebraic Geometry it's enough to show that every connective complex over $X$ receives a nonzero map from a connective perfect complex. For example, every connective complex $C$ such that $R\Gamma(C)$ is not concentrated in negative degree satisfied the condition. In particular the statement is true if $X$ has an ample family of line bundles This is different from this question, because despite the similarities the derived category of a scheme is not the same thing as the derived category of the underlying ringed space. For example, if $U\subseteq X$ is an open subscheme, the restriction map $j^*:\mathcal{D}(X)\to \mathcal{D}(U)$ does not have a left adjoint. REPLY [6 votes]: For qcqs algebraic spaces (non-spectral), this is lemma 2.7 in Algebraization and Tannaka Duality by Bhatt.<|endoftext|> TITLE: Misspelling my name on my mathematical publications QUESTION [18 upvotes]: Perpetuating a mistake by my thesis advisor, I misspelt my name on my mathematical publications so far -- any advice on what to do? As it happens, my thesis advisor, with whom I co-authored my first paper, misspelt my last name on that paper. He then approved the page proofs. For this reason, on that paper, my name reads Leyli Jafari Taghvasani instead of Leyli Jafari Taghvastani. In the sequel, in order to have all my publications under the same name, he advised me to stick to the wrong spelling of my name, and publish all my papers under the name Leyli Jafari Taghvasani instead of Leyli Jafari Taghvastani. I followed his advice, and published already some more papers under my misspelt name: JafariTaghvasaniZarrin2017, JafariTaghvasaniZarrin2018, JafariTaghvasaniMarzangZarrin2018, JafariTaghvasaniKohl2019. My question is: what is the best way to deal with this situation? REPLY [11 votes]: I think the concern about having all your papers be listed together on professional databases like MathSciNet and ZBMath, while real, is more minor than the current answers make it seem. The key point to remember is: 99% of people looking up your publications will use Google or another search engine to do so. Since search engines are extremely good at correcting these sorts of minor typos, the solution is therefore quite simple. The main things you need to do are: Start publishing under your name as it is correctly spelled. Set up a personal home page where you have an easily accessible and well-formatted list of publications, with links to the papers’ final versions and/or arXiv versions, and keep that home page up to date and alive (if you move institutions, set up a redirect from the older page, etc). Add a small note on that page explaining that your first few papers were published with a variant spelling of your name. Steps 1 and 2 are actually things that any early-career mathematician should do, and are much more important than fidgeting over how MathSciNet thinks your name is spelled. The advice to contact MathSciNet and other databases and ask them to link the publications listed under the different spellings of your name is generally good, but, if you follow the steps I am suggesting, whether they will do what you are asking or not will, in my opinion, have an altogether negligible effect on your career success.<|endoftext|> TITLE: About different cohomology theories used to study Shimura varieties QUESTION [6 upvotes]: The classical theory of Eichler-Shimura realizes the space of cusp forms of certain weights and levels in the parabolic cohomology of modular curves, which is the image of the cohomology with compact support in the ordinary cohomology. In Langlands’ Antwerp paper, he also uses parabolic cohomology (of the “Shimura variety” version of modular curve $G(\mathbb{Q})\backslash \mathcal{H}^\pm\times G(\mathbb{A}_f)/K$) and obtain a Matsushima type decomposition of it. But the Matsushima’s formula (also in Borel-Wallach’s book and J.Arthur’s 1981 Invent paper) uses $L^2$-cohomology of the locally symmetric spaces. In Blasius & Rogawski’s paper in Motives, they use intersection cohomology of (the Baily-Borel-Satake compactification of) certain Shimura varieties and use the Matsushima type decomposition because in this case the $L^2$ cohomology is the same as the intersection cohomology. More recently in Sophie Morel’s work, she also uses intersection cohomology. I am not familiar with algebraic topology and the various cohomology theories mentioned above. The only intuition I have is, some Shimura varieties are non-compact, and their compactifications are probably singular, so one has to put some effort on the cohomology theory used. So my questions are: 1: What is the original motivation for using parabolic cohomology in the early works of Eichler-Shimura, since this seems strange at first. Is there any conceptual reason that it works? 2: Are the $L^2$ and parabolic cohomologies of the locally symmetric space isomorphic? 3: It seems that intersection cohomology is the most powerful one, so can one override the theory of Eichler-Shimura by using intersection cohomology? If so, is there any reference on this (or it is too trivial to be wrote down in any reference? ) 4: In the early works of Eichler-Shimura and Langlands, they used different decompositions of the parabolic cohomology, say of Hodge type and Matsushima type respectively. Are they essentially different approaches (to the construction of $\ell$-adic Galois representations) or they are related in some way? REPLY [5 votes]: For Eichler-Shimura, I am not a historian, but I have a guess. I think the first papers of Eichler and Shimura handle only the correspondence between modular forms of weight 2 and the Jacobians of (compactified) modular curves. This is the most natural case to think about because it doesn't require any cohomology at all to define. It's also related to the zeta function, which you can see from the title of Shimura's paper "Correspondances modulaires et les fonctions $\zeta$ de courbes algébriques". If you want to generalize this to higher weight, you might realize that the higher weight modular forms should be related to $H^1$ of a local system $L$. But $L$ has singularities on the cusps, which means there are multiple ways to extend it to the cusps, or equivalently, multiple cohomology theories for the noncompact modular curve. The first thing to do would be to try the two obvious cohomology theories, ordinary and compactly-supported, and realize that they don't work already in the case of constant coefficients that you understand. In fact, the desired group ($H^1$ of the compactification) is the quotient of compactly supported $H^1$ by some group, and is a subgroup of the ordinary $H^1$. From this it's not too hard to come up with the definition of parabolic cohomology (or to recognize it if you already knew it). With regards to Matsushima's formula, his formula relates the cohomology of $G/\Gamma$ to the unitary representations of $G$ appearing in $L^2(G/\Gamma)$, and is generalization of work of Cartan and Hodge in the compact case. When you're looking at unitary representations, an $L^2$ space is a very natural place to look for them, and for obvious reasons an $L^2$ space will be easier to relate to $L^2$-cohomology than to other cohomology theories. With regards to intersection cohomology, using intersection cohomology is very natural if you know (1) the results in $L^2$-cohomology, (2) the relation between $L^2$-cohomology and intersection cohomology, (3) that intersection cohomology exists in characteristic $p$ and has Galois representation structure while $L^2$-cohomology does not. (2) was conjectured by Zucker in 1982, based on some examples where he could prove it, and the existence of the arithmetic structure on intersection cohomology, and its importance, was proved by BBD in 1983. Donu has essentially answered your question 3 - yes, intersection cohomology can be used to do Eichler-Shimura theory. I don't understand your question 4.<|endoftext|> TITLE: What happens with large singular cardinals on the far side of the HOD dichotomy? QUESTION [7 upvotes]: Woodin's HOD Dichotomy Theorem says that if an extendible cardinal exists, then either $V$ and $HOD$ are rather close or rather far apart. My question is whether the "far" case can be strengthened in analogy to Jensen's Covering Lemma to say something more about large singulars. Suppose $\delta$ is extendible and every regular $\kappa \geq \delta$ is measurable in $HOD$. Does there exist a singular $\lambda > \delta$ which is singular in $HOD$? EDIT: The comment by Gabe Goldberg gives an easy "yes" answer. Here's a harder version of the question. Suppose $\delta$ is extendible and $\lambda$ is the least cardinal above $\delta$ such that $V_\lambda \models ZFC$. Are all singular cardinals in the interval $(\delta,\lambda]$ singular in $HOD$? REPLY [6 votes]: This is independent relative to the failure of the HOD hypothesis in the presence of large cardinals. We first give a positive answer under GCH. (Note that if it is consistent for the HOD Hypothesis to fail in the presence of an extendible, then this is consistent with GCH.) More generally, we show that under GCH, for any singular cardinal $\lambda$ that is regular in $\text{HOD}$, $V_\lambda\vDash \text{ZFC}$. Suppose towards a contradiction that $V_\lambda$ does not satisfy ZFC. Then there is a singularization of $\lambda$ definable over $V_\lambda$ from a parameter $a\in V_\lambda$. We may code $a$ by a set of ordinals $A\subseteq \alpha$ for some $\alpha < \lambda$. Thus $\lambda$ is singular in $\text{HOD}_A$. But $\text{HOD}_A$ is a generic extension of $\text{HOD}$ for a forcing of size less than $\lambda$ in $\text{HOD}$. (This bound falls out of Vopenka's theorem given GCH: recall that the Vopenka algebra is in bijection with the OD powerset of $P(\alpha)$.) One cannot destroy the $\text{HOD}$-inaccessible cardinal $\lambda$ by small forcing over $\text{HOD}$, so we have reached a contradiction. On the other hand, the GCH assumption is necessary, which is proved easily by forcing. Start with a model $M$ of the failure of the HOD Hypothesis and an extendible cardinal $\delta$. Let $\lambda$ be the least singular cardinal of $M$ above $\delta$ that is regular in $\text{HOD}$. Force preserving cardinals, preserving the extendibility of $\delta$, and without increasing HOD to blow up $2^\delta$ above $\lambda$. This gives us a generic extension $N$. In $N$, the least $\lambda' > \delta$ such that $V_{\lambda'}$ is a model of ZFC is such that $\lambda' > 2^\delta > \lambda$. But $\lambda$ is still a singular of $N$ that is regular in $\text{HOD}$.<|endoftext|> TITLE: Literature about solitons and Hirota derivatives QUESTION [6 upvotes]: This summer I'm going to learn a mini-course about soliton theory ("Soliton equations and symmetric functions" in LHSM (Russian summer school in mathematics). The web-page of this course is https://mccme.ru/dubna/2019/courses/rozhkovskaya.html (warning: it's in Russian, so here's my translation of its program: Examples of soliton equations, their solutions. Properties of solitons. Definition and properties of Hirota derivatives. KdV and KP equations in terms of Hirota derivatives. Bilinear form of the KP hierarchy. Symmetric functions: main definitions, properties of elementary, complete and power sum symmetric functions. Interpretation of the bilinear form of KP hierarchy in terms of symmetric functions. (If it'll be enough time) Some words about the action of fermions on the symmetric functions and solutions of KP hierarchy). This program has intrigued me (despite my main interest is representation theory and related topics) and now I'm looking for a book in which the soliton theory will be outlined according to this program (i.e. using Hirota derivatives from very beginning), maybe even without any mentions of symmetric functions.. I saw Kasman's book but he doesn't say too much about Hirota derivatives and his style seems simplistic to me... I'll be grateful for anyone who'll give me some references. It'll be great if these books will contain an "algebraic approach" to the subject or some connections with representation theory... But any other reference will also be greeted with admiration! My question's already been posted on math.SE: https://math.stackexchange.com/questions/3286654/literature-about-solitons-and-hirota-derivatives. But I'd no response, so I decided to post it here, because I thought that it's applicable for this site. If it isn't so, please tell me. I'll delete it. REPLY [4 votes]: The canonical reference is The Direct Method in Soliton Theory, Cambridge UP, 2009, by Ryogo Hirota himself.<|endoftext|> TITLE: The rigidity of the countable product of free groups QUESTION [10 upvotes]: For a natural number $n$ let $F_n$ be the free group with $n$ generators. The group $F_n$ is endowed with the discrete topology. Given an increasing sequence $\vec p=(p_k)_{k\in\omega}$ of prime numbers, consider the Polish group $F_{\vec p}=\prod_{k\in\omega}F_{p_k}$. Problem. Let $\vec p,\vec q$ be two increasing sequences of prime numbers such that the Polish groups $F_{\vec p}$ and $F_{\vec q}$ are topologically isomorphic. Is $\vec p=\vec q$? REPLY [13 votes]: Yes, and even in the group category. More generally, I claim that if $\prod_{i\in I}G_i$ and $\prod_{j\in J}H_j$ are isomorphic groups, for two sets $I,J$ and two families $(G_i)$, $(H_j)$ of groups that are center-free and directly indecomposable, then there is a bijection $f:I\to J$ such that $G_i$ and $H_{f(i)}$ are isomorphic for all $i$. (Recall that a group is directly indecomposable if it's nontrivial and not direct product of two nontrivial subgroups.) To show this, it is enough to recognize the subgroups $G_i$ in the product $G=\prod_i G_i$, purely relying on the structure of the group $G$. Indeed, $G_i$ is directly indecomposable, and $G$ is direct product of $G_i$ and its centralizer (using that $G_i$ has trivial center). I claim that this characterizes the subgroups $G_i$. Claim if $H$ is a directly indecomposable subgroup of $G$ such that $G$ is direct product of $H$ and its centralizer, then $H=G_i$ for some unique $i$. Trivial lemma: Let $H$ be a subgroup of $G=\prod G_i$ ($G_i$ arbitrary groups), and $H_i$ its projection on $G_i$. Let $K_i$ be the centralizer of $H_i$. Then the centralizer of $H$ is $\prod_i K_i$. $\Box$ Now to prove the claim, let $H$ be a subgroup with these properties. Then $H\subset\prod H_i$, which has trivial intersection with the centralizer $K=\prod K_i$ of $H$. Since by assumption $G=H\times K$, we deduce that $H=\prod H_i$. Since $H$ is directly indecomposable, $H_i$ is nontrivial for a single $i$. So $H\subset H_i$, and $G_i=H_i\times K_i$. Since $G_i$ is directly indecomposable, we deduce that $H=H_i=G_i$.<|endoftext|> TITLE: Low-degree polynomial approximation of the piecewise-linear function $x \mapsto \max(x, 0)$ on an interval $x \in [-R,R]$ QUESTION [9 upvotes]: For $R > 0$, consider the piecewise-linear function $\sigma_R: [-R,R] \rightarrow \mathbb R^+$, defined by $\sigma_R(x) := \max(x,0)$. Question Given $\epsilon> 0$, find a "low-degree" polynomial $P_\epsilon$ (the smaller the degree, the better!) which approximates $\sigma_R$ within $\epsilon$ w.r.t the sup-norm, ie. $$\|\sigma_R-P_\epsilon\|_\infty := \max_{x \in [-R,R]}|\sigma_R(x)-P_\epsilon(x)| \le \epsilon. $$ Observations Here is a graphical illustration of such an approximation. REPLY [12 votes]: Subtracting $x/2$, and rescaling, the problem is reduced to the best uniform approximation of $|x|$ on $[-1,1]$ by polynomials, a problem already considered by Bernstein. According to Chebyshev's theory, the polynomial of degree $\le n$ that minimizes the uniform distance on $[-1,1]$ from the absolute value function is unique and is characterized by Chebyshev's equioscillation theorem. It is also even, by symmetrization argument. More details should be easily available in the literature; for instance, these notes devote the whole chapter 3 exactly to the problem.<|endoftext|> TITLE: Homotopy pullbacks and pushouts in stable model categories QUESTION [6 upvotes]: There are lots of similar questions that have been answered on this topic (particularly Homotopy limit-colimit diagrams in stable model categories), but I have a specific question that I do not believe has been answered. I suspect there is a simple answer that I'm just not seeing, so apologize in advance if this is the case. Given a commutative square $\require{AMScd}$ \begin{CD} A @>{}>> X\\ @V{f}VV @V{g}VV \\ B @>{}>> Y \end{CD} we say it is a homotopy pullback square if the canonical map $A \to holim(B \to Y \leftarrow X)$ is a weak equivalence and that it is a homotopy pushout square if the canonical map $hocolim(B \leftarrow A \rightarrow X) \to Y$ is a weak equivalence. I'm hoping to understand the proof that in a stable category, the two notions coincide. As pointed out in the reference above, in the 2007 version of Hovey's ``Model Categories,'' an argument is given in Remark 7.1.12 that goes roughly as follows. One can show that the square above is a homotopy pullback square if and only if $$ hofib(f) \overset{\sim}{\to} hofib(g) $$ is a weak equivalence. Similarly, it is a homotopy pushout square if and only if $$ hocofib(f) \overset{\sim}{\to} hocofib(g) $$ is a weak equivalence. One then concludes the result by claiming that $$ \Sigma hofib(f) \simeq hocofib(f) \text{ and } \Sigma hofib(g) \simeq hocofib(g) $$ This is the part on which I am stuck. Hovey claims that it follows from Thereom 7.1.11 (which is the same in both the 1999 and 2007 version) but I do not see this. It seems to me (and here is where I must be going wrong) that Hovey is claiming that the fiber in the homotopy category is the homotopy fiber -- because from what I understand, it is the homotopy category which is triangulated (to which 7.1.11 would apply), not the original stable model category. Since the fiber in the homotopy category is not (usually) the homotopy fiber, I must be misinterpreting what Hovey is saying. So my question is: how do we know that the suspension of the homotopy fiber is weakly equivalent to the homotopy cofiber (or, equivalently, that loops of the homotopy cofiber is weakly equivalent to the homotopy fiber). Am I misunderstanding what Hovey means by "fiber sequence?" Are the distinguished triangles actually homotopy cofiber sequences in the original model category? As is probably clear by now, I do not know much about triangulated categories. Any help/reference would be appreciated! REPLY [2 votes]: Many thanks and much credit to Dmitri Pavlov for clearing up my confusions. In the interest of having a definitive answer, I've organized his comments and my understanding into the following answer. Let $\mathcal{C}$ be a stable model category; then (by definition, for Hovey) $Ho(\mathcal{C})$ is a triangulated category. Note: By definition 6.2.6 (same in both the 1999 and 2007 version of Hovey's "Model Categories") when we say $A \to B \to C$ in $Ho(\mathcal{C})$ is a cofiber sequence, we mean that there is a natural zig-zag of weak equivalences in $\mathcal{C}$ to a cofibration sequence of cofibrant objects $A' \hookrightarrow B' \to C'$. Dually, saying that $X \to Y \to Z$ in $Ho(\mathcal{C})$ is a fiber sequence means that there is a natural zig-zag of weak equivalences in $\mathcal{C}$ to a fibration sequence of fibrant objects $X' \to Y' \twoheadrightarrow Z'$. Claim: Given a map $X \overset{g}{\to} Y$ in $\mathcal{C}$, we have that $\Sigma hofib(g) \simeq hocofib(g)$, where $\Sigma$ is derived suspension (i.e. the suspension of a cofibrant replacement). Proof: Without loss of generality (replacing if needed) we may assume $g$ is a fibration between fibrant objects. Take the (strict) fiber of $g$ (which will also be the homotopy fiber) to obtain $$ F \to X \overset{g}{\twoheadrightarrow} Y $$ Then replace the map $F \to X$ with a fibration and again take the fiber to obtain the following fibration sequence of fibrant objects in $\mathcal{C}$ $$ \Omega Y \to G \twoheadrightarrow X \twoheadrightarrow Y $$ By definition, this descends to a fiber sequence $$ \Omega Y \to G \to X \to Y $$ in $Ho(\mathcal{C})$. By Remark 7.1.9 (same in both versions) the following is also then a fiber sequence in $Ho(\mathcal{C})$ (note our notation clashes with Hovey's here, i.e. our $G$ is his $X$, etc.). $$ \Omega\Sigma G \to X \to Y \to \Sigma G $$ which means (by 6.2.6) that $hofib(X \to Y) \simeq \Omega\Sigma G$. Now apply 7.1.11 (same in both versions) in $Ho(\mathcal{C})$ (taking $Z = \Sigma G$) to obtain a cofiber sequence $$ \Omega \Sigma G \to X \to Y \to \Sigma(\Omega\Sigma G) $$ in $Ho(\mathcal{C})$ which means (by 6.2.6) that $hocofib(X \overset{g}{\to} Y) \simeq \Sigma(\Omega\Sigma G)$. Since $\Sigma(\Omega\Sigma G) \simeq \Sigma hofib(X \overset{g}{\to} Y)$, this completes the proof.<|endoftext|> TITLE: Immersions of surfaces in $\mathbb{R}^3$ QUESTION [15 upvotes]: Stephen Smale famously proved in [Trans. Amer. Math. Soc. 90 (1959), 281-290] that any two $C^2$ immersions $S^2\to\mathbb R^3$ are regularly homotopic. This is how we knew that one can do a sphere eversion before ever constructing such a thing, for example. Later, Morris Hirsch wrote [Trans. Amer. Math. Soc. 93 (1959), 242-276], where he moves from the problem of immersing spheres to immersing general manifolds. His paper is based on Smale's, he says, in roughly the same way that obstruction theory is based on the theory of homotopy groups. In his paper Hirsch uses his results to solve various problems in immersion theory, but I cannot find in it, nor after a google search, an answer to this: What is the classification up to regular homotopy of immersions $M\to\mathbb{R^3}$ from a closed oriented surface? In particular, is there just a single regular homotopy class of immersions of a closed orientable surface of positive genus? Can one everse a sphere with handles? Later. Looking at the explicit construction of representatives of regular homotopy classes of immersions given in the paper by Joel Hass and John Hughes that Danny Ruberman mentions in his answer, the following question asks itself. In some sense, it is closer to the eversion problem, as when turning spheres inside-out we start and end with embeddings. What is the classification up to regular homotopy of embeddings $M\to\mathbb{R^3}$ of a closed oriented surface? Of course, along the homotopy the embedding may degrade into just an immersion. Corollary 3.3 in the paper states that if $M$ has genus $g$, then of the $4^g$ regular homotopy classes of immersions $M\to M\times[0,1]$ which are homotopic to the the obvious map $\iota:p\in M\mapsto (p,0)\in M\times [0,1]$ only one can be represented by an embedding, the one which contains $\iota$. So the second question above is what happens when we replace $M\times[0,1]$ by $\mathbb{R^3}$. REPLY [3 votes]: Regarding the classification up to regular homotopy of embeddings, U. Pinkall proves in [Topology 24 (1984), 421–434] that if $f$, $g:M\to\mathbb R^3$ are two embeddings of a compact orientable surface, then there is a diffeomorphism $h:M\to M$ such that $f$ and $g\circ h$ are regularly homotopic. This is not quite as good as «one can everse a compact orientable surface», though: that the usual embedding $f:S^2\to\mathbb R^3$ and $g=-f$ are related as in Pinkall's theorem is obvious. Pinkall notes that it is not true that two embeddings $f$, $g:M\to\mathbb R^3$ are necesarily regularly homotopic, but does not say anything how often they are.<|endoftext|> TITLE: Do the higher levels of the Borel hierarchy correspond to absolute topological properties? QUESTION [8 upvotes]: It is well known that a subset $Y$ of a Polish space $X$ is completely metrisable iff it is a $G_\delta$ subset. This relates a relative topological property of the subspace $Y \subset X$ to an absolute topological property of $Y$. I wonder are there more general versions of this result where $G_\delta$ is replaced by some other level of the Borel hierarchy and complete metrisability is replaced by some more complex topological property? The first fact can be used to prove $\mathbb Q^\omega \not \cong \mathbb P^\omega$ where $\mathbb P = \mathbb R - \mathbb Q$ is the set of irrational numbers. It is well-known $\mathbb P$ is completely metrisable and this extends to the countable power. Next we find a $G_\delta$ subset $G \subset \mathbb R^\omega$ with $\mathbb P^\omega \subset G \subset \mathbb R^\omega - \mathbb Q^\omega$. $$G = \bigcap_{i \in \omega} \{ x \in \mathbb R^\omega: \text{ all } x_i \ne q_n\}$$ where $\{q_1,q_2,\ldots\}$ is an enumeration of $\mathbb Q$. Since both spaces are dense in $\mathbb R^\omega$ and any two dense $G_\delta$ sets intersect it follows $\mathbb Q^\omega$ is not $G_\delta$ hence not completely metrisable. Here is a harder problem: Suppose instead of $\mathbb Q^\omega$ and $\mathbb P^\omega$ we're interested in the spaces $\mathbb Q^{ \oplus \, \omega}$ and $\mathbb P^{ \oplus \, \omega}$ of choice functions $f: \omega \to \mathbb R$ with all but finitely many coordinates equal to zero and the rest in $\mathbb Q$ or $\mathbb P$ respectively. In this case $G_\delta$ sets are no use. To see $\mathbb P^{ \oplus \, \omega}$ is not $G_\delta$ first observe it is dense in $\mathbb R^{ \omega}$. Then prove $\{x \in \mathbb P^{ \omega}: \text{ all } x_i \ne 0\}$ is $G_\delta$. Then observe the two sets are disjoint so the first cannot be $G_\delta$. In particular $\mathbb P^{ \oplus \, \omega}$ is not $G_\delta$ so cannot be completely metrisable. Edit: Actually the space $\mathbb Q^{\oplus \omega}$ is countable since you can surject $\mathbb Q^{1} \cup \mathbb Q^{2} \cup \ldots$ onto it. So here is a harder question. What if we are interested in the spaces of choice functions with infinitely many coordinates equal to zero and the rest in $\mathbb Q$ or $\mathbb P$ respectively? Getting back to my original question, suppose we could find some level in the Borel Hierarchy that contains $\mathbb P^{ \oplus \, \omega}$ but not $\mathbb Q^{ \oplus \, \omega}$ or vice-versa. Would this translate to some absolute topological property that distinguishes the spaces? I'd also appreciate any other ideas to prove the spaces are (not) homeomorphic. REPLY [2 votes]: You can use complete Borel sets for this. Recall that for Polish spaces $X,Y$ and subsets $A\subseteq X$ and $B\subseteq Y$, $A$ is Wadge-reducible to $B$ if there is a continuous $f:X\to Y$ such that $f^{-1}(B) = A$. It can be proved (for instance, see Theorem 22.10, and Exercises 22.11 and 24.20 in Kechris' Classical Descriptive Set Theory) that $B\in \boldsymbol\Sigma^0_\xi\setminus\boldsymbol\Pi^0_\xi$ if and only if $B$ is complete, in that every $A\in\boldsymbol\Sigma^0_\xi$ is Wadge-reducible to it (for $\xi\geq 1$). This immediately gives that such a $B$ is never homeomorphic to an $A\in \boldsymbol\Pi^0_\zeta$, with $\zeta\leq\xi$, in some other Polish space $X$. Otherwise, take $f:A\to B$ a homeomorphism; by Lavrentiev's theorem, $f$ extends to homeomorphism $\bar f: G \to H$ where $G,H$ are $G_\delta$ subsets such that $A\subseteq G \subseteq X$ and $B\subseteq H \subseteq Y$. But then $B$ reduces to $A$ through $\bar f^{-1}$, a contradiction.<|endoftext|> TITLE: Identification problem: Does this group have a name? QUESTION [5 upvotes]: I've encounter a group with properties that are very familiar, but I cannot say what group is it. Consider the variables $(t,x,y,z)$, an affine transformation $M \in A(3)$ on the last three variables can be represented as $$ M = \begin{pmatrix} 1 & 0 \\ \vec{a} & {R} \end{pmatrix},$$ such that $$ \begin{pmatrix} t' \\ \vec{x}' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ \vec{a} & {R} \end{pmatrix} \begin{pmatrix} t \\ \vec{x} \end{pmatrix} = \begin{pmatrix} t \\ R\vec{x} + \vec{a} \end{pmatrix}. $$ The group I mentioned is composed by elements which are like the transpose of the above $M$, i.e. $$ \begin{pmatrix} t' \\ \vec{x}' \end{pmatrix} = \begin{pmatrix} 1 & \vec{a} \\ 0 & {R} \end{pmatrix} \begin{pmatrix} t \\ \vec{x} \end{pmatrix} = \begin{pmatrix} t + \vec{a} \cdot \vec{x} \\ R\vec{x} \end{pmatrix}. $$ If $d = \dim(\vec{x})$, the above is a $d$-parametric affine group of dimension one! but we still have the possible mixing on the $x$ variables (i.e. $R$). Question: Does this group have a name? Is it a semi-direct product of other groups? REPLY [5 votes]: The group is a (homogeneous) Carroll group: see eq. 2.7 in https://arxiv.org/abs/1405.2264<|endoftext|> TITLE: Classification of minimal sets of properties proving a group is Abelian QUESTION [6 upvotes]: Let $S$ be a non-empty, possibly infinite, set of integers, all of which are greater than $1$. For a given group $G$, let $S[G]$ denote the collection of statements $$ \forall (n \in S, a \in G, b\in g) \,\,(ab)^n=a^nb^n $$ For some sets $S$, $S[G]$ is sufficient to prove $G$ is Abelian. For example, the set $\{2\}$ is sufficient, since $(ab)^2 = a^2b^2\implies ba = ab$. Define a minimal abelian forcing (maf) set as any $S$ such that $S[G]$ proves $G$ is Abelian, but for no proper subset $T \subset S$ does $T[G]$ prove $G$ is Abelian. Has anybody studied the question of classifying all minimal abelian forcing sets? And has anybody addressed the question of whether there exists an infinite maf? For example, some mafs are: $S = \{2\}$ $S = \{3,5\}$ $S = \{n+3,n+4,n+5\}, n\in \Bbb Z^+$ The line of attack I have tried is to exploit the fact that $(ab)^n = a^nb^n$ implies that $n$-th powers commute with $(n-1)$-th powers and that $(ab)^n = a^nb^n \implies (ab)^{n-1} = b^{n-1} a^{n-1} $. This easily confirms the examples listed above, but it is hard to get a systematic test of whether a given set $S$ is maf. I realize that using the word "forcing" has jargon implications. But I felt it was better than minimal abelian proving since the abbreviation of that would be map and that would be more confusing. REPLY [10 votes]: See the Monthly article: Joseph A. Gallian and Michael Reid, Abelian Forcing Sets, Amer. Math. Monthly, Vol. 100, No. 6 (Jun. - Jul., 1993), pp. 580-582. Gallian and Reid show that a set $S$ of integers is abelian forcing if, and only if, the greatest common divisor of the numbers $n(n−1)$, for $n\in S$ is equal to $2$. They show how this makes it particularly easy to see that standard exercise examples, such as $S = \{2\}$, $S = \{-1\}$ and $S = \{ k, k+1, k+2 \},$ (for $k$ an integer) are all abelian forcing sets, and note that those first two are the only singletons $S$ that are abelian forcing.<|endoftext|> TITLE: The tensor product of two monoidal categories QUESTION [9 upvotes]: Given two monoidal categories $\mathcal{M}$ and $\mathcal{N}$, can one form their tensor product in a canonical way? The motivation I am thinking of is two categories that are representation categories of two algebras $R$ and $S$, and the module category of the tensor product algebra $R \otimes_{\mathbb{C}} S$. Also, if $\mathcal{M}$ and $\mathcal{N}$ are assumed to be braided monoidal, can we tensor the braidings? REPLY [4 votes]: If you're acquainted with $\infty$-categories you might also be interested in the tensor product of presentable $\infty$-categories: it's a real tensor product, in that it provides an equivalence $$ \begin{array}{cr} C_1\times\dots\times C_n \to D & \text{cocont.}\\\hline C_1\otimes\dots\otimes C_n \to D \end{array} $$ between the colimit-preserving functors from $C_1\times\dots\times C_n$ to $D$ and the functors from the tensor product to $D$. There's plenty of reasons why "multilinear" gets categorified in "cocontinuous"; also, presentable categories all are cartesian (because they admit finite products, and all other shapes of limits for that matter), so they are not the most general monoidal categories, but their tensor product is fairly well understood. Hope this helps! REPLY [4 votes]: No, there is not tensor product. You need to make further assumptions that could give you several potentially distinct tensor products. You can always form the product category $M\times N$. It is a monoidal category. Any sensible tensor product will be a quotient category of $M\times N$. For instance, in the Deligne tensor product, mentioned by McRae, you assume that $M$ and $N$ are abelian, $k$-linear. Then $M\times N$ is $k$-bilinear and you can ask for the universal quotient $M\otimes_k N$ such that any $k$-bilinear monoidal functor $M\times N\rightarrow P$ factors via a $k$-linear monoidal functor $M\otimes_k N\rightarrow P$. Such thing clearly exists: you dont even need to assume that the categories are abelian but you need $k$-linearity! However, $M\otimes_k N$ is not necessarily abelian, even if $M$ and $N$ are such. If it is abelian, then it is Deligne tensor product. Overall, you need some "balancing" conditions to form a tensor product. In the case of $k$-linear categories, the balancing is performed by $Vec(k)$, the category of vector spaces that acts on both $M$ and $N$. There are several ways you can do "balancing" but only some of them preserve monoidal and/or braided structure. Others will lose it. For instance, the following wabbity balancing is cool. For this I need a third monoidal category $C$ and two monoidal functors $C\rightarrow M$ and $C\rightarrow N$. By a $C$-balanced functor I understand a bifunctor $M\times N \rightarrow D$, together with an equivalence of two resulting trifunctors $M\times C \times N \rightarrow D$. I claim that there exists a monoidal category $M\otimes_CN$ with the universal $C$-balanced bifunctor $M\times N \rightarrow M\otimes_C N$, with two caveats: It is not longer braided, you need some conditions for that. It is no longer a category :-)) The issue is that your hom-s will be proper classes, unless you put some smallness condition, for instance, $C$ being small will do.<|endoftext|> TITLE: Tannaka duality for closed monoidal categories QUESTION [5 upvotes]: I asked this some time ago at mathstackexchange, and people there explained to me the mathematical part of what I was asking, but the question about references remains open. In my impression, people here at MO are more active, so I hope, specialists in the category theory could clarify the remaining part of my question. (At the same time I foresee that somebody can suggest another idea of how this can be proved, and I would welcome this as well.) The nLab article on the Tannaka duality says that this theory was generalized to monoids in arbitrary closed monoidal category (symmetric and complete in some sense): if we take a monoid $A$ in such a category $\mathcal V$ and consider the category $_A{\mathcal V}$ of all left modules over $A$, then $A$ can be recovered from $_A{\mathcal V}$ as the object of enriched endomorphisms of the forgetful functor $F: {_A{\mathcal V}}\to {\mathcal V}$. (As far as I understand, this result is called a reconstruction theorem.) Some details in this construction are however explained at nLab vaguely, in particular, how is the structure of the enriched category introduced on $_A{\mathcal V}$? Can anybody give me a link to a text where this result is explained accurately so that I could refer to it with a clear conscience? (Perhaps I am missing something, I do not see where this is written. From what I see, I have the impression that this is folklore, but I have no confidence, and I do not dare to write this in my paper.) REPLY [2 votes]: I think a good reference for your question would be R. Street, Quantum Groups, A Path to Current Algebra, Chapter 16: Tannaka Duality (see here).<|endoftext|> TITLE: Constructive proof of existence of free algebras for infinitary equational theories QUESTION [18 upvotes]: Is it constructively true that all (not necessarily finitary) equational theories $T = (\Sigma, E)$ have an initial model? The usual proof for finitary equational theories I know constructs first from the signature $\Sigma$ the set $P$ of syntax trees/preterms. This set is by construction the initial model of the theory $(\Sigma, \emptyset)$, i.e. will usually not satisfy equations $E$. One then considers the congruence $R \subseteq P \times P$ generated by (all interpretations of) the equations in $E$, and proves that $Q = P / R$ is a model of $T$ and then that it is the initial one. If $\Sigma$ contains an operation symbol $f$ of non-finitary arity $A$ then I struggle with defining the operations on the quotient $Q$. The interpretation of $f$ for $Q$ should be a function $f_Q : Q^A \rightarrow Q$, and should be defined in terms of the function $f_P : P^A \rightarrow P$ on syntax trees. If $A$ was finite, then any given map $x : A \rightarrow Q$ could be lifted along $P \twoheadrightarrow Q$ to a map $x' : A \rightarrow P$, and then $f_Q(x)$ could be defined as the residue class of $f_P(x')$ in $Q$. But if $A$ is not a choice object/set, then the proof is stuck here. Is there a way to get around this issue without assuming choice, or is it maybe known that the existence of certain initial algebras implies some version of the axiom of choice? EDIT: The reference pointed out by Valery Isaev contains the answer to my questions. There are models of ZF (without C) in which there is no initial algebra for a certain equational theory, in particular it cannot be proved to exist using just constructive logic. On the other hand, initial algebras exist for all theories in all Grothendieck toposes provided that AC holds in the metatheory, so all choice principles that fail to hold in some Grothendieck topos don't follow from the existence of initial algebras. REPLY [8 votes]: As was already pointed out by Valery Isaev, even in the presence of excluded middle initial algebras for equational theories need not exist. I would like to explain a bit what is needed from a constructive point of view. Suppose $T = (\Sigma, E)$ is an equational theory where $\Sigma$ is a family $\Sigma = (A_\mathrm{op})_{\mathrm{op} \in I}$ of sets $A_\mathrm{op}$ indexed by a set $I$. We think of the elements of $I$ as the operation symbols, and $A_\mathrm{op}$ as the arity of the operation symbol $\mathrm{op}$. (Normally arities are natural numbers, but since we allow infinitary operations it is better for arities to be general sets.) A $T$-algebra $C$ is given by a carrier set $|C|$ and, for each $\mathrm{op} \in A$, a map $\mathrm{op}_C : |C|^{A_\mathrm{op}}| \to |C|$, such that the equations $E$ are satisfied. A natural way of constructing the initial $T$-algebra is as follows: Construct the set of well-founded trees $W_T$ whose branching types are $\Sigma$, i.e., the initial algebra for the polynomial functor $X \mapsto \Sigma_{\mathrm{op} \in I} X^{A_\mathrm{op}}$. This is also known as a $W$-type. Quotient $W_T$ by the (interpretations of) equations $E$ to obtain a candidate for the initial algebra. We cannot get either step for free, but in general the first step is the easier one, as it is well understood what it takes to have $W$-types in a constructive setting. For the second step to go through, one needs to resolve the question posed by the OP, namely, how do we lift operations from the quotient $W_T/E$ to $W_T$? It looks like we need choice. Indeed, it suffices for all the arities $A_\mathrm{op}$ to satisfy choice (to be choice sets, also called projective objects), but is that necessary? I do not know of any way of avoiding choice if one attempts to construct the initial algebra as a quotient of an inductively defined set. Homotopy type theory offers an alternative. We avoid stratifying the construction of the initial algebra into an inductive construction followed by a quotient. Instead, we make a purely inductive construction: the initial $T$-algebra is the higher-inductive type $X$ with the following constructors: for each $\mathrm{op} \in I$, there is a point constructor $\overline{op} : X^{A_\mathrm{op}} \to X$; for each equation $\ell_i(x_1, \ldots, x_n) = r_i(x_1, \ldots, x_n)$ in $E$ there is a path constructor $e_i : \prod (x_1, \ldots, x_n : X)\,.\, \overline{\ell}_i(x_1, \ldots, x_n) =_X \overline{r}_i(x_1, \ldots, x_n)$, set-truncation: for all $x, y \in X$ and all paths $p, q : x =_X y$ there is a path $\tau_{p,q} : p =_{x =_X y} q$. For further reference, look at the HoTT book chapter on the real numbers, where a variant of such a construction is used to present the Cauchy completion of rational numbers in an inductive fashion.<|endoftext|> TITLE: Integration with values in a topological vector space QUESTION [7 upvotes]: Is there a general theory of integration of functions with values in a topological vector space (not necessarily locally convex)? Browsing through mathoverflow posts, I came across a discussion regarding Grothendieck's work on integration with values in a topological group L'intégration à valeurs dans un groupe topologique (Reference request : Grothendieck's topological space valued integral ). Unfortunately the requested reference is never provided. Does anyone know of a rough sketch of the approach used? Just to emphasize, I am interested in a general theory and not in notions such as the Bochner integral, or Gelfand-Pettis integral which use local convexity of the topological vector space. REPLY [2 votes]: One reference ... Sion, Maurice, A theory of semigroup valued measures, Lecture Notes in Mathematics. 355. Berlin-Heidelberg-New York: Springer-Verlag. V, 140 p. DM 16.00; $ 6.60 (1973). ZBL0312.28016.<|endoftext|> TITLE: Simplicial set of permutations QUESTION [14 upvotes]: Let $S_k$ be the set of all permutations of $k+1$ elements $0,1,...,k$. introduce boundary maps $d_i : S_k \rightarrow S_{k-1}$ by deleting from permutation $\eta$ element $\eta(i)$ and monotone reordering and degeneracy $s_i :S_k \rightarrow S_{k+1} $ by adding 1 to all elements with $\eta(j)>\eta(i)$ and incerting into the result a new element $\eta(i)+1$ right after $\eta(i)$ on $i+1$ place. It is a simplicial set, contractible and classifies reorderings of simplicial sets. Is it known? May be in higher symmetric something? (Update) Boris Tsygan pointed the right direction in Facebook duscussion The object is classical and it has a name "Symmetric crossed simplicial group”. It was introduced almost simultaneously in Appendix A10, page 191 “Symmetric objects” B. L. Feigin and B. L. Tsygan “Additive K-theory” 1987 K-theory, arithmetic and geometry, Semin., Moscow Univ. 1984-86 LNM 1289 Krasauskas, R. "Skew-simplicial groups", Lithuanian Mathematical Journal, Jan 1987 vol 27 issue 1 p. 47--54 And independently Zbigniew Fiedorowicz and Jean-Louis Loday “Crossed simplicial groups and their associated homology” Trans. Amer. Math. Soc. 326 (1991), 57-87 It has big value in everything symmetric. Geometric realization $|S_\bullet|$ is the topological group structure on infinite dimensional sphere. REPLY [3 votes]: I think something equivalent (or at least closely related) to this has been studied in the combinatorics literature. A CW complex of course has a poset of faces. In this case, this poset is obtained by ordering permutations by subword inclusion up to deletion and monotone reordering. The keyword used in the combinatorics literature for this sort of subword inclusion is permutation patterns. Now, if a CW complex is regular, then the order complex of the face poset is homeomorphic to the complex. As you point out in the comments, the simplicial set has $n!$ faces of dimension $n$, and in particular has a single vertex. So the simplicial set isn't regular, as I'd initially thought it might be, and your question doesn't reduce directly to this poset. It certainly seems like the two objects should be closely related, however. In any case, the lattice of permutations ordered by pattern containment has been studied by Jason Smith. See, for example, the paper Smith, Jason P., A formula for the Möbius function of the permutation poset based on a topological decomposition, Adv. Appl. Math. 91, 98-114 (2017). ZBL1370.05227. That paper cites also his earlier papers on the topic.<|endoftext|> TITLE: Representation-finite quivers over dual numbers QUESTION [6 upvotes]: Given a Dynkin quiver $Q$ and a field $K$. Question 1: For which such $Q$ are there only finitely many indecomposable representations over the dual numbers $K[x]/(x^2)$? Note that those representations are exactly those of $KQ \otimes_K K[x]/(x^2)$. This is for example true for $Q$ being of type $A_1, A_2$ or $A_3$. Question 2: For which combinations of $Q$ and $n$ is $KQ \otimes_K K[x]/(x^n)$ representation-finite? I think question two has an easy answer for $n \geq 3$, namely only when $Q$ is of type $A_1$ for arbitrary $n$ or $A_2$ for $n=3$, but is there an elementary argument? REPLY [3 votes]: You can find this result in [Geiss, Leclerc, Schröer: Quivers with relations for symmetrizable Cartan matrices I: Foundations] as Proposition 13.1. They consider a more general class of algebras. The class of algebras you are considering is obtained by setting their parameters $(c_1,\dots,c_m)$ to all the same number, which you call $n$. Translating to your setup, their result reads: The algebra $KQ\otimes_K K[x]/(x^n)$ for $Q$ a Dynkin quiver is representation-finite if and only if you are in one of the following cases: $n=1$ and $Q$ arbitrary, in which case you just recover $KQ$. $Q=A_1$, the quiver with only one vertex, in this case $n$ is arbitrary. $Q=A_2$ and $n=2$ or $n=3$. $Q=A_3$ and $n=2$. The proof they give (unfortunately) uses the Bongartz-Gabriel and the Happel-Vossieck list. The question for a more elementary argument thus remains open. For this one should try to construct an infinite family of modules.<|endoftext|> TITLE: $(-2)$-curves in complex $3$-folds QUESTION [6 upvotes]: Let $X$ be a smooth complex $3$-fold, and let $C \subset X$ be an embedded smooth rational curve whose normal bundle $N_{C/X}$ is isomorphic to $\mathscr{O}(-1) \oplus \mathscr{O}(-1)$. Is it true that a neighborhood of $C$ in $X$ is biholomorphic to some neighborhood of $C$ in $N_{C/X}$? Could you please give a reference? REPLY [2 votes]: Yes, this result appeared in "Uber Modifikationen und exzeptionelle analytische Mengen" (Mathematische Annalen, vol. 146, n.4) by H. Grauert. See the Corollary on p. 363, which also treats a much more general case.<|endoftext|> TITLE: Reference request: A knot is tame if and only if it has a tubular neighbourhood QUESTION [9 upvotes]: Definitions: A knot is an embedding $\kappa:S^1\hookrightarrow S^3$ (we do not require smooth or polygonal). Two knots $\kappa,\,\lambda:S^1\hookrightarrow S^3$ are equivalent if one of the following equivalent conditions is satisfied: They are ambient isotopic, i. e. there is an isotopy $h:S^3\times I\rightarrow S^3$ with $h_0=\operatorname{id}$, $h_1\circ\kappa=\lambda$. There is an orientation preserving homeomorphism $h:S^3\rightarrow S^3$ such that $h\circ\kappa=\lambda$. Claim: Let $\kappa:S^1\hookrightarrow S^3$ be a knot. The following are equivalent: $\kappa$ is equivalent to a polygonal knot. $\kappa$ can be lifted to an embedding of a full torus $\overline{\kappa}:S^1\times D^2\hookrightarrow S^3$. This claim appears without proof in the following places: Louis H. Kauffman: Knots and Functional Integration, Ch. 1.1 (p. 5). homepages.math.uic.edu/~kauffman/KFI.pdf Unless otherwise specified I shall deal only with tame knots and links. In a tame knot every point on the knot has a neighborhood in 3-space that is equivalent to the standard (ball, arc-diameter) pair. Tame knots (links) can be represented up to ambient isotopy by piecewise linear knots and links. A link is piecewise linear if the embedding consists in straight line segments. Thus a piecewise linear link is an embedding of a collection of boundaries of $n$-gons (different $n$ for different components and the $n$’s are not fixed). A piecewise linear knot (link) is made from “straight sticks.” Wikipedia (en): Wild knot. en.wikipedia.org/wiki/Wild_knot In the mathematical theory of knots, a knot is tame if it can be “thickened up”, that is, if there exists an extension to an embedding of the solid torus $S^1\times D^2$ into the 3-sphere. A knot is tame if and only if it can be represented as a finite closed polygonal chain. REPLY [4 votes]: The existence of PL tubular neighborhoods, plus a uniqueness statement, may be found in Wall, Locally flat PL submanifolds with codimension two, (Proc. Cambridge Philos. Soc. 63 (1967), 5–8.). Is that what you are looking for?<|endoftext|> TITLE: Hopf structure on the universal enveloping of a super Lie algebra QUESTION [10 upvotes]: The universal enveloping algebra of a Lie algebra has a canonically defined Hopf algebra structure. Is the same true of the universal enveloping of a super Lie algebra? A presentation in terms of the universal properties would be of most interest. REPLY [3 votes]: The notion of hopf algebras has slowly emerged from the work of topologists in the late '30's and '40's on the cohomology of compact Lie groups and their homogeneous spaces. Initially the term had been used in the "graded" or "signed" sense (for example, Milnor and Moore's seminal paper refers to $\mathbb{Z}$-graded hopf algebras and I think this is the case in Mac Lane's "Homology" book as well). As far as I can understand, the "present day definition" (i.e. the "unsigned" and "ungraded" as presented in most hopf algebra textbooks after the '70's - that is after Sweedler's textbook) has been formulated during the early '60's and is mainly due to works of Cartier and Dieudonne. (the introduction of A primer of Hopf algebras includes interesting details on that point) However, lots of people from different schools have contributed to the evolution of this notion; lots of authors have kept using the term for signed or graded objects long after that. So it does not come as a surprise that there are still discussions on the "correct" form of the definition. It is not my purpose -in this answer- to argue on what is the "correct" definition but rather to try to provide some terminology, unifying the above descriptions: That is the language of braided groups, in the sense this notion has been introduced and used since the mid'90's after the Majid's school of hopf algebras and quantum groups. Imo, one of the advantages of this language is that it helps in clarifying that the "grading" and the "signs" are not the same thing (the same grading may refer to different sign rules). $\bullet$ Some introductory remarks on graded Lie structures, their "signs", their colors and their universal properties: Let $G$ a countable, abelian group and a function $\theta: G \times G \rightarrow \mathbb{C}^{*}$ satisfying (for all $a, b, c \in G$) \begin{equation} \begin{array}{c} \theta(a+b, c) = \theta(a,c) \theta(b,c) \\ \theta(a, b+c) = \theta(a,b) \theta(a,c) \\ \theta(a,b) \theta(b,a) = 1 \end{array} \end{equation} The function $\theta$ is called a color map on $G$ or a commutation factor for $G$. It is a symmetric bicharacter on $G$. Given such a function and a $G$-graded, complex v.s. $L = \oplus_{g \in G} L_{g}$, then $L$ will be called a $\theta$-colored, $G$-graded Lie algebra or a $(G, \theta)$-Lie algebra, for short, If $L$ is further equipped with a bilinear, non-associative multiplication $\langle .., .. \rangle$ which repsects the grading, is $\theta$-antisymmetric and satisfies the $\theta$-Jacobi, identity i.e. $$ \begin{array}{c} \langle L_{a}, L_{b} \rangle \subseteq L_{a+b} \\ \langle x, y \rangle = - \theta(a,b) \langle y, x \rangle \\ \theta(c,a) \langle x, \langle y, z \rangle \rangle + \theta(b,c) \langle z, \langle x, y \rangle \rangle + \theta(a,b) \langle y, \langle z, x \rangle \rangle = 0 \\ \end{array} $$ for all $x \in L_{a}$, $y \in L_{b}$, $z \in L_{c}$ and for all $a, b, c \in G$. Similarly to the ordinary Lie algebra case, the universal enveloping algebra (UEA) of the $(G, \theta)$-Lie algebra $L$ is a pair $(\mathbb{U}(L), i_{\mathbb{U}})$, where $\mathbb{U}(L)$ is a $G$-graded, associative algebra and $i_{\mathbb{U}}$ an homogeneous linear map of zero degree (i.e.: a $G$-graded v.s. homomorphism) $i_{\mathbb{U}}: L \rightarrow \mathbb{U}(L)$, which -by definition- satisfies $$i_{\mathbb{U}}(\langle x, y \rangle) = i_{\mathbb{U}}(x)i_{\mathbb{U}}(y) - \theta(a,b)i_{\mathbb{U}}(y)i_{\mathbb{U}}(x)$$ $\mathbb{U}(L)$ is defined to be the quotient of the tensor algebra $\mathbb{T}(L)$ with the homogeneous -wrt to the $G$-grading- ideal $Ι(L)$ generated from all elements of the form $\langle x, y \rangle - xy + \theta(a,b) yx$ for all homogeneous elements $x,y$ of $L$ and $i_{\mathbb{U}}$ is the composition $L \hookrightarrow \mathbb{T}(L) \twoheadrightarrow \mathbb{U}(L)$. As a consequence of the above $\mathbb{U}(L) = \mathbb{T}(L)/Ι(L)$ has the following universal property: If $\mathcal{Α}_{gr}$ is an associative, $G$-graded algebra and $f_{L} : L \rightarrow \mathcal{Α}_{gr}$ is a $G$-graded v.s. homomorphism (equivalently: an homogeneous linear map of zero degree) which furthermore satisfies $$ f_{L}( \langle x, y \rangle ) = \langle f_{L}(x), f_{L}(y) \rangle = f_{L}(x)f_{L}(y) - \theta(a,b) f_{L}(y)f_{L}(x) $$ for all homogeneous elements $x \in L_{a}$, $y \in L_{b}$, then there is a unique homomorphism of associative $G$-graded algebras (equivalently: homogeneous assoc algebra homomorphism of zero degree) $f : \mathbb{U}(L) \rightarrow \mathcal{Α}_{gr}$ which extends the linear map $f_{L}$, such that $$f \circ i_{\mathbb{U}} = f_{L}$$ $f$ is fully defined by its values on the generators of $\mathbb{U}(L)$, i.e. from the values of $f_{L}$ on the elements of $L$. The usual Poincare-Birkhoff-Witt theorem generalizes as well. $\bullet$ On the Hopf structure of the UEA $U(L)$, of the $\theta$-colored, $G$-graded Lie algebra: If we equip the v.s. $\mathbb{U}(L) \otimes \mathbb{U}(L)$ with an associative product defined by \begin{equation} (x \otimes y)(z \otimes w) = \theta(a,b) xz \otimes yw \end{equation} for all homogeneous elements $y \in L_{a}$ and $z \in L_{b}$, then the $G$-graded v.s. $\mathbb{U}(L) \otimes \mathbb{U}(L)$ becomes an associative, $G$-graded algebra. We will denote this by $\mathbb{U}(L) \underline{\otimes} \mathbb{U}(L)$ and call it ($G$-graded), $\theta$-braided tensor product algebra or $(G, \theta)$-tensor product algebra. The UEA $\mathbb{U}(L)$ of the $(G, \theta)$-Lie algebra $L$, is not a hopf algebra -well at least not in the "ordinary" (ungraded) sense (here "ordinary" should be taken to mean the modern day definition of hopf algebras as this is presented in most of the hopf algebra textbooks which have appeared after the '70's). Let me try to shed some light in this point: $\mathbb{U}(L)$ is equipped with a "comultiplication" $$ \underline{\Delta} : \mathbb{U}(L) \rightarrow \mathbb{U}(L) \underline{\otimes} \mathbb{U}(L) $$ which is a homomorphism of assoc., $G$-graded algebras (equivalently: a homogeneous homomorphism of assoc algebras of degree $0$) i.e. $$ \underline{\Delta}(ab) = \sum \theta\big(deg(a_{2}), deg(b_{1})\big) a_{1}b_{1} \otimes a_{2}b_{2} = \underline{\Delta}(a) \underline{\Delta}(b) $$ for all $a,b \in \mathbb{U}(L)$, with $\underline{\Delta}(a) = \sum a_{1} \otimes a_{2}$, $\underline{\Delta}(b) = \sum b_{1} \otimes b_{2}$, and $a_{2}$, $b_{1}$ homogeneous. The product $\underline{\Delta}(a) \underline{\Delta}(b)$ in the rhs of the above is understood to be in the $\mathbb{U}(L) \underline{\otimes} \mathbb{U}(L)$ algebra. Due to the universal property of the UEA, $\underline{\Delta}$ is uniquely defined by its values on the elements of $L$ i.e. on the generators of $\mathbb{U}(L)$ $$ \underline{\Delta}(x) = 1 \otimes x + x \otimes 1 $$ Similarly, $\mathbb{U}(L)$ is equipped with the "antipode" $\underline{S} : U(L) \rightarrow U(L)$ which is no more an algebra antihomomorphism (as in the "ordinary" hopf algebra case) but a twisted or braided antihomomorphism of $G$-graded algebras, in the sense that: $$ \underline{S}(ab) = \theta\big(deg(a), deg(b)\big) \underline{S}(b)\underline{S}(a) $$ and $deg(a) = deg(\underline{S}(a))$ for all homog elements $a,b \in \mathbb{U}(L)$. Again the universal property of the UEA, ensures us that the antipode is uniquely defined by its values on the elements of $L$ i.e. on the generators of the UEA $\mathbb{U}(L)$: $$ \underline{S}(x) = -x $$ If the above are complemented with the counit $\underline{\varepsilon}(x) = 0 $ for all $x \in \mathbb{U}(L)$ then we get a $G$-graded, $\theta$-braided Hopf algebra or -for short- a $(G,\theta)$-hopf algebra. $\bullet$ The relation with ordinary Hopf algebras and super-Hopf algebras: The notion of $G$-graded, $\theta$-braided Hopf algebras as defined above, generalizes the definition of ordinary hopf algebras and the various $\mathbb{Z}$ or $\mathbb{Z}_2$-graded (super) Hopf algebras: If $G=\mathbb{Z}_{2}$ and the color function $\theta$ is taken to be $\theta(a,b) = (-1)^{ab}$, then a $(G, \theta)$-Lie algebra is a lie superalgebra (or a $\mathbb{Z}_2$-graded Lie algebra) and its UEA is a $(\mathbb{CZ}_{2},\theta)$-hopf algebra which is nothing different than the hopf superalgebra (known under this name mainly in the mathematical physics literature dealing with SUSY algebras) referred to in the OP and also in Bugs Bunny's answer. If $G$ and $\theta$ are trivial we get the ordinary definition of Hopf algebras. An interesting remark here has to do with the fact that the same grading group $G$ may give rise to different $G$-graded, $\theta$-braided Hopf algebras (for different choices of the color function $\theta$). (imo this imposes interesting and tractable classification problems for such structures). According to present day terminology (here i am mainly following the terminology as used by the Majid's school) such structures are called braided groups or hopf algebras in the braided monoidal Category ${}_{\mathbb{CZ}_2}\mathcal{M}$ of $\mathbb{CZ}_2$-modules (remember that the $\mathbb{CZ}_2$-modules are exactly the $\mathbb{Z}_2$-graded v.s. or "super"-v.s.). The braiding $\theta$ can be shown to be "generated" through a "1-1" correspondence, with the non-trivial, quasitriangular structure (i.e. the non-trivial $R$-matrix) of the $C\mathbb{Z_2}$ group hopf algebra: $$R=\frac{1}{2}\big(1\otimes 1+1\otimes g+g\otimes 1-g\otimes g\big)$$ (More generally, under the term braided groups we frequently mean hopf algebras in the braided monoidal Category ${}_{H}\mathcal{M}$ of $H$-modules, where $H$ is any quasitriangular hopf algebra -and not necessarily a group algebra). $\bullet$ Representations of hopf algebras vs "super"-representations of hopf superalgebras: It has been shown that given a $(\mathbb{CZ}_{2},\theta)$-hopf algebra $H$ (or: a "super"-hopf algebra $H$) we can form the smash product hopf algebra $H\star \mathbb{CZ}_2$ by adjoining an extra generator $g$ to $H$. This is an ordinary hopf algebra and the construction is functorial in the sense that there is an equivalence of categories $${}_{H}\underline{\mathcal{M}} \thicksim {}_{H \star CZ_{2}}\mathcal{M}$$ between the Category ${}_{H}\underline{\mathcal{M}}$ of super-reps of $H$ and the Category ${}_{H \star CZ_{2}}\mathcal{M}$ of representations of the (ordinary) smash product Hopf algebra $H \star CZ_{2}$. The converse procedure can also be done: these are the Bosonization and Transmutation techniques developed in the early 90's (and partially based on the previously known idea of Radford's biproduct). Finally -hoping that the above are somewhat helpful for the OP- some references: Foundations of Quantum Group theory, S. Majid (see mainly chapter 9), Hopf algebras and their actions on rings, S. Montgomery (see mainly chapter 10) If you are further interested on the terminology, maybe you can find some interest in section 3.1 of Gradings, Braidings, Representations, Paraparticles: Some Open Problems (and the references therein).<|endoftext|> TITLE: To find a point in Teichmüller space or measured foliation, how many lengths of curves do you need? QUESTION [7 upvotes]: To parametrize Teichmüller space, it suffices to measure the hyperbolic lengths of a finite number of curves. It is well-known that $9g-9$ curves suffice, by a standard pair-of-pants argument given in, for instance Fathi-Laudenbach-Poenaru. I recall that you need exactly $6g-5$ curves: you cannot achieve it by $6g-6$, because the character variety is not an algebraic subset of $\mathbb{C}^{6g-6}$, but one extra curve suffices. Is this correct, and if so, who proved it? Likewise for measured foliations, the cone over the boundary at infinity: can you parametrize measured foliations with $6g-5$ curves, and how to see that you cannot do it with $6g-6$? REPLY [10 votes]: As indicated in my comments, the Teichmüller question is a duplicate of this question. For the measured lamination case, the fact that $6g-5$ curves suffice was shown by Hamenstädt. Hamenstädt, Ursula, Parametrizations of Teichmüller space and its Thurston boundary., Hildebrandt, Stefan (ed.) et al., Geometric analysis and nonlinear partial differential equations. Berlin: Springer (ISBN 3-540-44051-8/hbk). 81-88 (2003). ZBL1044.32005. MR2008332 To see that $6g-6$ curves do not suffice, suppose we have $6g-6$ curves $(a_1,\ldots, a_{6g-6})$ and an embedding $$\mathcal{MF}_g \hookrightarrow \mathbb{R_{\geq 0}}^{6g-6}-\{{\bf 0}\}, \lambda \mapsto (i(a_1,\lambda),\ldots, i(a_{6g-6},\lambda)).$$ Then we would get an embedding $$\mathcal{PMF}_g \hookrightarrow \Delta^{6g-7} \subset \mathbb{RP}^{6g-7}.$$ But this is impossible by invariance of domain since $\mathcal{PMF}_g \cong S^{6g-7}$.<|endoftext|> TITLE: When does glueing affine schemes produce affine/separated schemes? QUESTION [5 upvotes]: Let $X$ be an affine scheme with an open affine subscheme $U\subset X$. Given an automorphism of $U$, we can glue $X$ with itself along $U$ to get a new scheme. Is there a description in terms of commutative algebra of automorphisms such that the resulting scheme is affine/separated? If $U=\mathrm{Spec}\:B$, $X=\mathrm{Spec}\:A$, then $B$ is an $A$-algebra of finite presentation so there's a chance to be explicit. For example, if $X=\mathrm{Spec}\:k[x]$ and $U=\mathrm{Spec}\:k[x, \frac{1}{x}]$ if we take the identity on $U$ the result is non-separated and if we take $x\rightarrow \frac{1}{x}$ the result is separated. I'm especially interested in what happens for $X$ the spectrum of a discrete valuation ring or a PID. Here is a commutative algebra description of open immersions between affine schemes (not a very convenient one I'd guess). REPLY [5 votes]: There is a general criterion that explains when a gluing of two separated schemes is separated. Proposition: Let $X_1, X_2$ be a separated $S$-schemes, $U_i$ open subschemes in $X_i$ (for $i=1, 2$), and $f:U_1 \to U_2$ an $S$-isomorphism. Then the $S$-scheme $X$ obtained as a gluing of $X_1$ and $X_2$ along the isomorphism $f$ is separated if and only if the ``diagonal'' morphism $$ U_1 \to X_1\times_S X_2 $$ is a closed immersion. In the situation $U_1=\operatorname{Spec} A$ and $X_1=X_2=\operatorname{Spec} B$ are affine the criterion says that separtedness of $X$ is equivalent to surjectivity of the map $$ \phi:B\otimes_{\mathbf Z} B \to A $$ defined by $\phi(a\otimes a')=af^*(b)$.<|endoftext|> TITLE: Symmetric monoidal category with trivial switch morphisms QUESTION [6 upvotes]: Is there a specific terminology for a symmetric monoidal category in which for any object $x$ the switch map $x\otimes x\to x\otimes x$ is the identity ? (Or alternatively the action of the symmetric group $\mathfrak{S}_n$ on $x^{\otimes n}$ is trivial.) Is there a paper or book I can cite where basic properties of such categories are derived ? REPLY [3 votes]: In my thesis I have named objects $x$ whose switch map $x \otimes x \to x \otimes x$ is the identity symtrivial (since I could not find any term in the literature). It was then used by others as well, but right now I can only find this example. Now it is reasonable to call a tensor category symtrivial when every object is symtrivial. The property in Noam's answer is much stronger.<|endoftext|> TITLE: Reference request: The unit of an adjunction of $\infty$-categories in the sense of Riehl-Verity is a unit in the sense of Lurie QUESTION [8 upvotes]: I'm looking for a reference (or proof) for the statement given in the title: that when we have an adjunction between quasicategories in the sense of Riehl and Verity (defined e.g. in Section 4 of their paper "The 2-category theory of quasi-categories" or Definition 2.1.1 of their ongoing book project at http://www.math.jhu.edu/~eriehl/elements.pdf), i.e. an adjunction in the homotopy 2-category of quasicategories, that then its unit is a unit transformation in the sense of Lurie, defined in Definition 5.2.2.7 of "Higher Topos Theory". It is shown in Appendix F of the above mentioned book that their definitions of adjunction agree, but, as far as I could tell, no comparison between the two notions of unit. I also couldn't easily piece it together from other statements. There is Corollary 4.1.3 that goes in that direction, but that still needs an identification of the notions of mapping spaces and maps induced on them, and is less explicit than I would like. REPLY [11 votes]: Let $\mathcal{C},\mathcal{D}$ two $\infty$-categories and $f:\mathcal{C}\to\mathcal{D}$ and $g:\mathcal{D}\to \mathcal{C}$ two functors. Recall (HTT.5.2.2.7) that a natural transformation $u:1_{\mathcal{C}}\to gf$ is a unit transformation if the natural transformation $$(\ast)\qquad\mathrm{Map}_{\mathcal{D}}(f-,-)\xrightarrow{g} \mathrm{Map}_{\mathcal{C}}(gf-,g-)\xrightarrow{u^*}\mathrm{Map}_{\mathcal{C}}(-,g-)$$ is an equivalence. Theorem A natural transformation $u:1_{\mathcal{C}}\to gf$ is a unit transformation if and only if there is a natural transformation $e:fg\to 1_{\mathcal{D}}$ satisfying the triangular identities. Proof: Let us first show that if such an $e$ exists, then the natural transformation $(\ast)$ is an equivalence. Indeed we will show that the natural transformation $$(\ast\ast)\qquad\mathrm{Map}_{\mathcal{C}}(-,g-)\xrightarrow{f} \mathrm{Map}_{\mathcal{D}}(f-,fg-)\xrightarrow{e_*}\mathrm{Map}_{\mathcal{D}}(f-,-)$$ is an inverse. Indeed, if $l:fc\to d$ is an arrow in $\mathrm{D}$, the transformation $(\ast)$ sends it to $gl\circ u_{c}$. Then, $(\ast\ast)$ sends it to $$e_{d}\circ f(gl\circ u_{c})\cong e_{d}\circ fgl \circ fu_{c}\cong l\circ e_{fc}\circ fu_c \cong l$$ where the second equivalence is the naturality of $e$ and the third is one of the triangular identities. So $(\ast\ast)$ is a left inverse of $(\ast)$. The dual proof (using the other triangular identity) implies that it is a right inverse too, so that $(\ast)$ is an equivalence. Let us now prove the other direction. if $(\ast)$ is an equivalence, in particular, the map $$\mathrm{Map}_{\mathcal{D}}(fg-,-)\cong\mathrm{Map}_{\mathcal{C}}(g-,g-)$$ is a natural equivalence. By taking ends, there is an equivalence of the space of natural transformations $$\mathrm{Nat}(fg,1_{\mathcal{D}})\cong \mathrm{Nat}(g,g)\,.$$ Let $e:fg\to 1_{\mathcal{D}}$ be the natural transformation corresponding to $1_g:g\to g$ under this equivalence. I claim that this satisfies the triangular identities. In fact, for every $d\in\mathcal{D}$, the arrow $e_d$ has been chosen so that there is an equivalence $$(ge_d)\circ u_{gd}\cong 1_{gd}$$ naturally in $d$. So one triangular identity is satisfied. On the other hand, arguing as in the previous direction, this implies that $(\ast\ast)$ is a right inverse of the map $(\ast)$. Since that was an equivalence, it must be a left inverse too, Choosing $l=1_{fc}$, this shows $e$ satisfies the second triangular identity. $\square$ If I understood the question in the comments, your doubt can be boiled down to the following construction: if $\mathcal{C},\mathcal{D}$ are two $\infty$-categories, $F,G:\mathcal{C}\to\mathcal{D}$ two functors and $e:F\to G$ a natural transformation, there exists a natural commutative diagram of the form $$\require{AMScd} \begin{CD} \mathrm{Map}_{\mathcal{C}}(-,-) @>{F}>> \mathrm{Map}_{\mathcal{D}}(F-,F-)\\ @V{G}VV @V{e_*}VV \\ \mathrm{Map}_{\mathcal{D}}(G-,G-) @>{e^*}>> \mathrm{Map}_{\mathcal{D}}(F-,G-) \end{CD}\,.$$ In order to construct this we notice that the pullback of the right bottom corner of the diagram is just the functor $\mathrm{Map}_{\mathcal{D}^{\Delta^1}}(e-,e-)$ where we interpret the natural transformation $e$ as a functor $\mathcal{C}\to \mathcal{D}^{\Delta^1}$, sending $c$ to $Fc\to Gc$ (this is just the usual formula for mapping spaces in functor categories using ends, in this case applied to the functor category $\mathcal{D}^{\Delta^1}$). But then giving the required square is just equivalent to giving a natural transformation $$\mathrm{Map}_{\mathcal{C}}(-,-)\to \mathrm{Map}_{\mathcal{D}^{\Delta^1}}(e-,e-)\,.$$ And we can simply choose the standard map induced by the functoriality of $e$.<|endoftext|> TITLE: Have new conjectures generated by the Ramanujan machine been proven? QUESTION [9 upvotes]: Recently the following preprint was published with new automatically generated conjectures on generalizes continuous fractions, e.g., for the Euler constant: Raayoni, Pisha, Manor, Mendlovic, Haviv, Hadad, and Kaminer - The Ramanujan machine: Automatically generated conjectures on fundamental constants. Have these conjectures been proven in the meantime? Are there any partial results? REPLY [4 votes]: In the year or so since this question first appeared on MO, there has been further progress on proving some of these identities, for example by Kadyrov and Mashurov and Dougherty-Bliss and Zeilberger. The comments mention a scathing blog post by Persiflage, condemning the authors' self-promotion and failure to consult experts on continued fractions before claiming that their identities were new. I don't know if the blog post encouraged experts to get involved; possibly it had the opposite effect. At any rate, a number of the formulas have now either been identified as known or been proved, but many seem to be new and remain unproved (e.g., the two examples in the abstract, which have been changed since the original arXiv preprint was posted). IMO it would be good for more number theorists to get involved since the new formulas might be an indication of some new theory that remains to be discovered and elucidated. As a side remark—a natural question to ask is whether any of these identities might lead to irrationality proofs. The only general theorem in this direction that I know about is a sufficient criterion due to Legendre (see Chrystal's Algebra, page 512, for a proof), and unfortunately none of the Ramanujan Machine's new identities seem to satisfy this criterion. Of course, maybe I overlooked something, or maybe they can be manipulated into a form that does satisfy the criterion.<|endoftext|> TITLE: Wildness of codimension 1 submanifolds of euclidean space QUESTION [8 upvotes]: This question arose out of this stack exchange post. I am wirting a thesis about the $s$-cobordism theorem and Siebenmann's work about end obstructions. Combined they give a quick proof of the uniqueness of the smooth structure for $\mathbb{R}^n$, $n \geq 6$. Roughly Siebenmann's theorem says that for $n \geq 6$ a contractible $n$-manifold $M$ that is simply connected at infinity embeds (smoothly) as the interior of a compact manifold. Since this compact manifold is contractible, by the $s$-cobordism theorem, it is diffeomorphic to the standard $n$-disk $D^n$ (see Minor's Lectures on the $h$-cobordism theorem for example). It follows that $M = \text{int } D^n$ is diffeomorphic to $\mathbb{R}^n$. The problem is that the case $n = 5$ is not covered. I am aware of Stallings beautifully written On the Piecewise-Linear Structure of Euclidean Space but I am searching for a way to deal with the $n = 5$ case via Siebenmann's end theorem and the proper $s$-cobordism theorem (see link to the mse question). This leads me to the following question, which is interesting in itself Given a smooth properly embedded codimension 1 submanifold $S \subset \mathbb{R}^{n+1}$, is there a self diffeomorphism $\mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$ which carries $S$ into a one dimensional bounded region $\mathbb{R}^n \times (-1, 1)$ ? Now if $M$ is a manifold that is homeomorphic to $\mathbb{R}^5$, the product $M \times \mathbb{R}$ is homeomorphic to $\mathbb{R}^6$, and hence also diffeomorphic. Granted the existence of the diffeomorphism in my question, we could find a diffeomorphism $f : M \times \mathbb{R} \rightarrow \mathbb{R}^6$ that maps $M \times 0$ into $\mathbb{R}^5 \times (-1, 1)$. This would produce a proper $h$-cobordism between $M$ and $\mathbb{R}^5$ by taking the region between $f(M \times 0)$ and $\mathbb{R}^5 \times 1$ in $\mathbb{R}^5 \times \mathbb{R}$. Since $M$ is simply connected, the proper $s$-cobordism theorem applies and shows that $M$ and $\mathbb{R}^5$ are actually diffeomorphic. REPLY [3 votes]: As you probably know, Siebenmann's thesis can be extended to dimension 5 (subject to a fundamenal group condition) by using the 4-dimensional work of Freedman. Unfortunately, one loses smoothness in the final conclusion, so your hope for a diffeomorphism would fall short. An alternative to using Siebenmann's thesis might be to apply his Open Collar Theorem from the later paper "On detecting open collars". The beauty of that approach is that it is valid in the smooth category down to and including dimension 5.<|endoftext|> TITLE: Point distributions in unit square which minimize E[1 / distance] QUESTION [6 upvotes]: Choose $n$ points $p_1,\ldots,p_n$ in the unit square $[0,1]^2\subset\mathbb{R}^2$ such that $D:=\mathop{\sum}\limits_{1\le id$ (in this case $E_{s}(A) \sim C N^{1+\frac{s}{d}}$). You may look at Hardin, D. P.; Saff, E. B., Discretizing manifolds via minimum energy points, Notices Am. Math. Soc. 51, No. 10, 1186-1194 (2004). ZBL1095.49031. and references therein.<|endoftext|> TITLE: Is the conformal compactification of $M \setminus \{ p \}$ unique? QUESTION [6 upvotes]: Let $(M,c)$ be a compact conformal manifold and $p \in M$. $M$ is a conformal compactification of $M \setminus \{ p \}$, because the embedding $M \setminus \{p\} \hookrightarrow M$ is an isometry. Is this conformal compactification unique? I believe I saw a preprint on the Arxiv that gave a positive answer to this question around a month ago, but I can no longer find it despite spending a long time on it. The reason I am interested in this, is that I am wondering whether the initial statement can be proven using the idea from M Eastwood: Uniqueness of the stereographic embedding. REPLY [5 votes]: Theorem 1.4 in C. Frances' preprint "Rigidity at the boundary for conformal structures and other Cartan geometries" asserts that the geodesic compactification is unique (up to conformal diffeomorphism) in this situation. More generally, he proves uniqueness when one removes a closed set of Hausdorff dimension strictly less than $\dim M - 1$.<|endoftext|> TITLE: Minimal assumptions such that the solution of Poisson equation is $C^2$ QUESTION [6 upvotes]: Take a weak solution $u$ of the Poisson equation on $\mathbb{R}^d$ $$ \Delta u = f $$ By standard elliptic regularity theory we have (for some $\alpha\in (0,1]$) $f\in C^{0, \alpha}_{\text{loc}}(\mathbb{R}^d)$, then $u \in C^{2, \alpha}_{\text{loc}}(\mathbb{R}^d)$. Now my (naive) question is: Are there some weaker assumptions that on $f$ that ensure that $u \in C^2(\mathbb{R}^d)$? By weaker assumptions I mean something like $f\in C^0(\mathbb{R}^d)$ plus some condition which is not of Hölder type (I am aware that $f\in C^0(\mathbb{R}^d)$ is not sufficient). REPLY [9 votes]: Dini continuity may be what you are looking for: if $f = \Delta u$ is Dini continuous, that is, $$\int_0^1 \frac{\omega_f(t)}{t} \, dt < \infty,$$ then $u$ is $C^2$. This is a rather old result, but I do not know the reference. A quick Google search leads to Poisson's equation by T. Gantumur, see Theorem 42 and Corollary 43 there.<|endoftext|> TITLE: Legendary extra parameters to simplify a counting problem QUESTION [14 upvotes]: I am reading Proofs and Confirmations, the history behind the alternating sign matrix conjecture, regarding counting $n \times n$ alternating sign matrices. In the introduction, it is written that the people (Mills Robbins and Rumsey) behind the ASM conjecture added an extra parameter (the position of the 1 in the first row), and by using this refined count, managed to guess a beautiful formula. There are other examples where refining the family reveals beautiful extra structure. My current personal favorite is counting crossings in perfect matchings. $$ T_n(q) := \sum_{M \in PM(n)} q^{crossings(M)} = \frac{1}{(1-q)^n} \sum_{j=-n}^{n} (-1)^j q^{\frac{j(j-1)}{2}} \binom{2n}{n+j} $$ see A067311. The formula on the right hand side is not that nice, but by refining crossings by gathering matchings according to which vertices are starting vertices, one can easily prove that $$ T_n(q) = \sum_{a \in Area(n)} \prod_{i=1}^n [a_i+1]_q $$ where the sum is now over Catalan$(n)$ objects, namely area sequences of Dyck paths. This example IMHO illustrates that refining a problem can reveal nicer properties. (From this formula, one can apply general theory by Flajolet and immediately get a nice continued fraction expansion). Another famous example would perhaps be the recent proof of the shuffle conjecture, which really relies on the refined conjecture (the compositional shuffle conjecture). Question: What are some legendary examples where refining the problem made a significant impact on the solution? Your favorite example? REPLY [7 votes]: Something like this is ubiquitous in lattice path enumeration problems, especially for problems like the enumeration of lattice paths restricted to the quarter plane (a focus of a tremendous amount of research in the past 20 years). The basic idea is that instead of studying the generating function $Q(n)$ of lattice paths of length $n$ starting from the origin, you introduce auxiliary parameters $x$ and $y$ and study the generating function $Q(n,x,y)$ of lattice paths of length $n$ starting from the origin and ending at $(x,y)$. This lets you write a functional equation satisfied by $Q(n,x,y)$. This basic idea is the starting point for the so-called "kernel method" (see e.g. https://arxiv.org/abs/0810.4387) which has been tremendously successful at resolving these enumeration problems.<|endoftext|> TITLE: Variously pointed closed sets QUESTION [7 upvotes]: A tree $A\subseteq \omega^{<\omega}$ - possibly with dead ends - is pointed iff every path $p\in[A]$ has $p\ge_TA$. This lifts to two distinct notions of pointedness for closed sets in Baire space: A closed $S\subseteq \omega^\omega$ is pointable if there is some pointed $A$ with $S=[A]$. A closed $S\subseteq \omega^\omega$ is pre-pointable if for each $p\in S$ there is some tree $A_p\subseteq\omega^{<\omega}$ with $p\ge_TA_p$ and $[A_p]=S$. These definitions also make sense for closed sets in Cantor space. Now it's not hard to whip up a closed set in Baire space which is pre-pointable but not pointable ; however, the only argument I have at the moment breaks down for Cantor space. (Specifically, the set in Baire space we get has two elements, but in Cantor space compactness implies that pointed = pre-pointed for finite sets.) So my question is: Is there a closed set in Cantor space which is pre-pointable but not pointable? REPLY [2 votes]: Let $\mu$ be a measure on $2^\omega$ which doesn't have a least Turing degree. This exists by Theorem 4.2 of Day, Adam R.; Miller, Joseph S., Randomness for non-computable measures, Trans. Am. Math. Soc. 365, No. 7, 3575-3591 (2013). ZBL1307.03026.. Trans. Amer. Math. Soc., 365:3575–3591, 2013. Let $S$ be the class of infinite sequences of reals $X_i$ that are Martin-Lof random with respect to the infinite product measure of $\mu$, $$\mu\times\mu\times\dots$$ with a fixed randomness deficiency constant $c$. This $S$ is pre-pointable because by the Law of Large Numbers each element of $S$ can compute a representation of $\mu$. If $S$ is pointable, it implies something close to saying that $\mu$ has a least Turing degree after all: there is a representation of $\mu$ that all the randoms compute. Then we may use a de Leeuw, Moore, Shannon, Shapiro / Sacks Theorem: Computability by Probabilistic Machines K. de Leeuw, E. F. Moore, C. E. Shannon & N. Shapiro Journal of Symbolic Logic 35 (3):481-482 (1970) for $\mu$ to the effect that if a real $A$ [in particular $A$ could be a representation of $\mu$ or $S$] is computed by all sequences of mutually $\mu$-randoms, then $A$ is computed by all representations of $\mu$. This would be a generalization of a result for Bernoulli measures (which is are already infinite product measures) from Kjos-Hanssen, Bjørn, Permutations of the integers induce only the trivial automorphism of the Turing degrees, ZBL06916706. We also want a lemma to the effect that computing a representation of $\mu$ is equivalent to computing a representation of $S$. Computing a representation of $S$ from a representation of $\mu$ is I guess the familiar direction, similar to the case of Bernoulli measures. Computing a representation of $\mu$ from a representation of $S$ runs into the problem that some strings are dead ends and may mislead us. However we may again use a certain effectivity in the LLN to (nonuniformly) bound the rate of convergence and hopefully overcome this.<|endoftext|> TITLE: Exponential objects in the category of measurable spaces QUESTION [8 upvotes]: Let $\text{Meas}$ be the category of measurable spaces with measurable functions as morphisms. Does $\text{Meas}$ have exponential objects? REPLY [6 votes]: It is also possible to show that the category of measurable spaces, $\newcommand{\Mble}{\mathbf{Mble}}\Mble$, is not cartesian closed by using more category theory and less measure theory (though still some). We reason as follows. If $\Mble$ were cartesian closed, then for each measurable space $Y$, the functor $\newcommand{\blank}{\mbox{-}}\blank \times Y : \Mble \rightarrow \Mble$ would be a left adjoint, and therefore preserve coproducts. Therefore we can show that $\Mble$ is not cartesian closed by finding a coproduct that is not preserved. To follow this line of reasoning, we first need to go over how products and coproducts work in $\Mble$. The product $(X,\Sigma_X) \times (Y,\Sigma_Y)$ in $\Mble$ is given by $(X \times Y, \Sigma_X \otimes \Sigma_Y)$, where $\Sigma_X \otimes \Sigma_Y$ is the $\sigma$-algebra generated by rectangles (this is simple to prove and comes down to showing that the projections and universal mapping are measurable). We will also use the fact that for a singleton $\{x\}$ with its unique $\sigma$-algebra, $\{x\} \times Y \cong Y$ (measurably). You can deduce this from singletons being terminal objects, if that's your bag. The category $\Mble$ also admits coproducts of arbitrary arity, which are preserved by the forgetful functor to $\newcommand{\Set}{\mathbf{Set}}\Set$, but all we need is the fact that for each set $Y$, the space $\newcommand{\powerset}{\mathcal{P}}(Y,\powerset(Y))$ is the coproduct of its elements (again, this comes down to proving that the set-theoretic coprojections $\kappa_y : \{y\} \rightarrow Y$ and universal map are measurable). As discussed in this question and its answers, if $|Y| > 2^{\aleph_0}$, the diagonal $\Delta = \{ (y,y) \mid y \in Y \} \subseteq Y \times Y$ is an element of $\powerset(Y \times Y)$ but not of $\powerset(Y) \otimes \powerset(Y)$. A summary of the proof is: Because of how $\sigma$-algebras are generated, there must exist a countable family of rectangles $(S_i\times T_i)_{i \in \omega}$ such that $\Delta$ is in the $\sigma$-algebra $\Sigma$ generated by $(S_i \times T_i)_{i \in \omega}$. The "distinguishability relation" defined by $(T_i)_{i \in \omega}$ on $Y$ has at most $2^{\aleph_0}$ equivalence classes, so $|Y| > 2^{\aleph_0}$ implies that there exist distinct elements $y_1,y_2 \in Y$ such that for all $i \in \omega$, $y_1 \in T_i$ iff $y_2 \in T_i$. (We could also have done this for $(S_i)_{i \in \omega}$, but this is not needed.) Therefore $(y_1,y_2) \in S_i \times T_i$ iff $(y_1,y_1) \in S_i \times T_i$ for all $i \in \omega$. Proceeding inductively, this holds for all elements of the $\sigma$-algebra $\Sigma$ generated by $(S_i \times T_i)_{i \in \omega}$. As $\Delta \in \Sigma$, we have $(y_1,y_2) \in \Delta$, which contradicts $y_1 \neq y_2$. If $\blank \times Y$ preserved coproducts, then we would have $\coprod_{y \in Y}\{y\} \times Y \cong \left(\coprod_{y \in Y}\{y\}\right) \times Y$ in $\Mble$. The $\sigma$-algebra on the left is $\powerset(Y \times Y)$ because $\coprod_{y \in Y}\{y\} \times Y \cong \coprod_{y \in Y} Y \cong \coprod_{y_1 \in Y} \coprod_{y_2 \in Y} \{y_2\}$ (this is where we use the fact about singletons). But on the right we have $\left(\coprod_{y \in Y}\{y\}\right) \times Y \cong Y \times Y$, so the $\sigma$-algebra is $\powerset(Y) \otimes \powerset(Y)$, and so there is no isomorphism if $|Y| > 2^{\aleph_0}$. Of course, Aumann's result, proved using the Baire hierarchy, is stronger and shows that the spaces $[0,1]$, $2^\omega$, $\mathbb{R}$ etc., which we are often more interested in, are not exponentiable.<|endoftext|> TITLE: Representation theory of inner forms QUESTION [5 upvotes]: I once heard something like "inner forms of reductive groups have the same representation theory". Is this assertion misguided? If this assertion is not misguided, then is there a precise statement to this effect (perhaps in Tannakian terms)? REPLY [4 votes]: Well, it is not misguided if by "representation" you mean "algebraic representation". More generally, if $E$ is a right $G$-torsor over $Spec F$ and $X$ is a $G$-variety you can form a ``twisted form'' $E\wedge_G X=E\times X/(e,x)\sim (eg,gx)$ which is ${}_EG$-variety, where ${}_EG$ is the inner twisted form of $G$ corresponding to $E$. This gives an equivalence between the category of $G$-varieties and the category of ${}_EG$-varieties. The same construction applies to algebraic representations (and the equivalence is additive and preserves tensor products).<|endoftext|> TITLE: Possible isometry groups of open manifolds QUESTION [12 upvotes]: Consider a non-compact manifold $M$. Does there always exist a Riemannian metric on $M$ such that the isometry group is non-compact? REPLY [13 votes]: Let $M$ be the triply punctured 2-sphere (i.e. $M=S^2-\{p_1, p_2, p_3\})$); one can also take any noncompact connected surface of finite topological type as long as $\chi(M)<0$ but the proof is a bit more involved. Suppose that $g$ is a Riemannian metric on $M$. To simplify matters, I consider only orientation-preserving isometries. Then $\mathrm{Isom}(M,g)< \mathrm{Conf}(M,g)$. I claim that $\mathrm{Conf}(M,g)$, the group of conformal automorphisms of $(M,g)$, is finite. The mapping class group of $M$ (the group of self-diffeomorphisms of $M$ modulo isotopy) is finite (isomorphic to the dihedral group of order $6$); hence, it suffices to prove that if $f: (M,g)\to (M,g)$ is a conformal automorphism isotopic to the identity then $f=\mathrm{id}$. This is quite standard (most likely, you will find it in Farb-Margalit's book on the mapping class group). Note that $f$ is isotopic to the identity if and only if it induces an inner automorphism of $\pi_1(M)$. By the uniformization theorem, $(M,g)$ is conformal to the quotient of the hyperbolic plane $H^2$ by a discrete group of isometries $\Gamma$ (isomorphic to $F_2$, the free group on two generators). The map $f$ then lifts to an isometry $F$ of $H^2$ (a linear-fractional transformation in, say, the Poincare disk model of $H^2$) which commutes with every element of $\Gamma$ (since $f$ induces an inner automorphism of $\pi_1(M)\cong \Gamma$). In particular, $F$ acts on the boundary circle $S^1$ of $H^2$ fixing fixed points of all elements of $\Gamma$. Since $\Gamma$ is free of rank 2, the set of fixed points of its nontrivial elements is infinite. Hence, $F$ fixes at least three points of $S^1$. Hence (being a linear-fractional transformation), $F=\mathrm{id}$; thus, $f=\mathrm{id}$. To conclude, every Riemannian metric on $M$ has finite group of isometries. Edit: I can prove the same claim for each noncompact complete hyperbolic $n$-manifold of finite volume, but a proof is more difficult. In general, the question can be reformulated in purely topological terms: Which smooth noncompact connected manifolds admit smooth proper actions of noncompact Lie groups? I suspect, this was studied in 1960s... REPLY [8 votes]: The answer is sometimes yes: If $M$ admits a complete vector field $X$, such that its flow $\operatorname{Fl}^X: \mathbb R\times M \to M$ is a proper action, then there exists a Riemannian metric which is invariant under the flow, by the following Theorem. If M is a proper G-manifold, then there is a G-invariant Riemann metric on M. which is due to Palais if I remember correctly. A proof of this theorem can be found in 6.30 of Topics in Differential Geometry. AMS, 2008. Finally, the isometry group then contains a non-compact 1-parameter group (this is not enough: it might be dense) which moves each point towards infinity for $t\to\infty$. So the isometry group cannot be compact (in the compact-open topology). The existence of such a vector field is a nontrivial condition, since then $M$ is the total space of real line bundle, as pointed out by Misha Kapovich. This is actually proved in 29.21 of 1. See his answer for an example where the answer is no. ADDED: The converse is also true: If the (connected component of the) isometry group is not compact, it contains a closed 1-parameter group whose generating vector field then has proper flow.<|endoftext|> TITLE: Are L-functions uniquely determined by their values at negative integers? QUESTION [12 upvotes]: Are L-functions uniquely determined by their values at negative integers? In another words, is there a sequence of integers $a_1, a_2, a_3, \cdots$ such that the corresponding L-function $$L_{\{a_n\}}(s):=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$$ converges well for $\text{Re}(s) > M$ for some $M\in \mathbb{R}$ $L_{\{a_n\}}(s)$ has analytic continuation to a meromorphic function on the whole complex plane $L_{\{a_n\}}(n)=0$ for all negative integers $n$ not all of $a_n$ are zero? Added : It was suggested in the answers that I should have used the term "Dirichlet series of integer sequences" instead of "L-function" as it lacks Euler product. I apologize for the confusion :) REPLY [4 votes]: Let $F(s) =\sum_{n=1}^\infty a_n n^{-s}= \sum_{j=1}^J c_j F_j(s)$ be a linear combination of L-functions $F_j$ of degree $\le d$, that is $(s-1)^r F_j(s)$ entire and its functional equation contains at most $d$ gamma factors $\Gamma(s/2+a)$. Then for $m > d$ or for $m=d,|x| < C$, $\Gamma(ms)F(s)x^{-s}$ decays fast enough to apply the residue theorem $$\begin{eqnarray}f_m(x)&=&\frac{1}{2i\pi} \int_{2-i\infty}^{2+i\infty} F(s) \Gamma(ms) x^{-s}ds \\ &=& \sum Res(F(s) \Gamma(ms) x^{-s}) \\ &=& \sum Res_{s=1}(F(s) \Gamma(ms)x^{-s}) + \sum_{k=0}^\infty \frac{(-1)^k}{k!m} F(-k/m) x^{k/m}\end{eqnarray}$$ (for $m = d$ it is valid only for $|x| < C$ but the exponential decay of $\Gamma(ms)F(s)$ on $\Im(s)=2$ implies $f_m$ is analytic so it is determined by $x \in (0,C)$) And hence $$F(s) = \frac{1}{\Gamma(ms)} \int_0^\infty f_m(x) x^{s-1}dx$$ is fully determined by its principal part at $1$ and its values at $s=-k/m$. In other words your claim holds only for L-functions of degree $d=1$.<|endoftext|> TITLE: Basic example of a formal affine scheme, functorial point of view QUESTION [5 upvotes]: $\let\opn=\operatorname$For my BA thesis I have to describe formal groups from the functorial point of view. I am hence reading Strickland - Formal Schemes and Formal Groups, which is apparently the only article that deals with this topic in that way. He defines (4.1) an formal scheme as a functor $X: \opn{CRings}\to \opn{Set}$ that is a small filtered colimit of affine schemes i.e., $X(R)=\lim\limits_{\rightarrow i}X_i(R)$. The first example (4.2) is given by the functor $\widehat{\mathbb {A}}^{1} $ defined as $\widehat{\mathbb {A}}^{1}(R)\mathrel{:=}\opn{Nil}(R)$. I don't understand why this functor is the colimit over $N$ of the functors $\opn{spec}(\mathbb{Z}[x]/x^{N+1})\mathrel{:=}\opn{Hom}_{\opn{CRing}}(\mathbb{Z}[x]/x^{N+1},\_)$. I would appreciate it if someone could explain it in general and kindly give an illustrating example. Other simple examples of formal schemes are also highly welcome. Many thanks! REPLY [7 votes]: It might be illuminating to first work the example of (ordinary) affine space $\mathbb{A}^1_\mathbb{Z}$ over the integers. As a functor, $\mathbb{A}^1_\mathbb{Z}$ is the forgetful functor $\mathit{Rings}^\mathrm{op}\rightarrow\mathit{Sets}$, sending a ring $R$ to its underlying set. It is representable by $\mathbb{Z}[t]$, as can be seen by the isomorphism $$\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t],R)\cong R.$$ (As a homomorphism $f\colon\mathbb{Z}[t]\rightarrow R$ is determined by its value $f(t)$ at $t$, we can define an isomorphism $\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t],R)\rightarrow R$ by $f\mapsto f(t)$ for all $f\in\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t],R)$.) $\widehat{\mathbb{A}}^1_\mathbb{Z}$ is similar: there is an isomorphism $$\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t]/(t^n),R)\cong\mathrm{Nil}_n(R),$$ where $\mathrm{Nil}_n(R)$ denotes the set of nilpotent elements $r$ of $R$ of order $n$ (i.e. $r^n=0$). (We define an isomorphism as before, but now the element $f(t)$ that we send $f$ to must be nilpotent in $R$, for it to preserve the ring structure: $f(t)^n=f(t^n)=f(0)=0$.) The answer from here own was changed following Dmitri Pavlov's comment: As (co)limits of presheaves are computed objectwise, we see that $\widehat{\mathbb{A}}^1_\mathbb{Z}$ sends $R$ to the colimit $\mathrm{colim}(\mathrm{Nil}_n(R))=\bigcup_{n\geq0}\mathrm{Nil}_n(R)=\mathrm{Nil}(R)$.<|endoftext|> TITLE: Are there currently any plausible approaches to proving the Penrose сonjecture? QUESTION [11 upvotes]: I have recently been reading some of the literature on the Penrose inequality, especially the papers by Bray and by Huisken and Ilmanen. One notices immediately that the existing proofs for the Penrose conjecture focus on a rather special case with special symmetry for the spatial hypersurface of the spacetime: this then means that the Penrose inequality can be translated into a statement of Riemannian geometry (known as the Riemannian Penrose inequality) which can be attacked with all the machinery available in that branch of differential geometry. Although these proofs are quite profound and make interesting uses of different geometric flows, I was wondering if there were any plausible suggestions anywhere in the literature towards actually proving the Penrose conjecture in full generality without any special symmetry assumptions? I have looked through the literature but cannot really find anything. Perhaps someone could point me to the relevant parts of the literature where some plausible suggestions have been made, perhaps also using geometric flows? REPLY [6 votes]: The difficulty of a general proof was discussed in A counter-example to a recent version of the Penrose conjecture (2010): a general existence theorem cannot be expected with boundary conditions compatible with generalized apparent horizons.<|endoftext|> TITLE: Why is the Langlands dual group always taken over $\mathbb{C}$? QUESTION [9 upvotes]: Whenever I read a statement of the Langlands conjectures for a reductive group $G$, they are formulated in terms of the Langlands dual group, which is essentially the reductive group over $\mathbb{C}$ whose root datum is dual to that of $G$. Instead of $\mathbb{C}$ we could use any separably closed field to obtain a "dual group”. Is there a compelling reason why the conjectures are not formulated in this generality? REPLY [8 votes]: The Satake isomorphism gives a relationship between the convolution algebra of $F$-valued functions on $G (\mathcal O) \backslash G(K) / G(\mathcal O)$ and the ring of conjugacy-invariant polynomial functions on the dual group over $F$. This is one example of where you might want the field of definition of your dual group to equal the field that your automorphic representations are defined over. Outside the $p$-adic Langlands programs, and some steps when you are proving cases of the Langlands correspondence by $\ell$-adic cohomology, it is usually most convenient to have the automorphic representations with coefficient field $\mathbb C$ (so you can do analysis). So then it's natural to have the Langlands dual group defined over $\mathbb C$ as well.<|endoftext|> TITLE: Exactness of the semidefinite programming (SDP) relaxation of maximum cut (Max-Cut) QUESTION [5 upvotes]: Currently, what conditions are known to be sufficient for the SDP relaxation of Max-Cut to be exact? REPLY [6 votes]: This question was studied somewhat in the early '90s (before Goemans--Williamson, in fact; note that it was Delorme and Poljak who first gave a poly-time SDP algorithm for Max-Cut, conjecturing that the 5-cycle gave the worst approximation ratio). Graphs for which the Max-Cut value and the SDP relaxation coincided were called 'exact'. As Dima says, there are not too many classes of exact graphs, with bipartite graphs being one of the main cases. A number of results and examples are given in the paper "The performance of an eigenvalue bound on the max-cut problem in some classes of graphs" by Delorme and Poljak; probably the best place to start looking. Note that it is unlikely there is an exact characterization, since it was shown (again by Delorme and Poljak, in "Combinatorial properties and the complexity of a max-cut approximation") that deciding if a given weighted graph is exact is NP-complete. To be fair, they state therein that they do not know the complexity of recognizing unweighted exact graphs.<|endoftext|> TITLE: Studying for primes $q_k \neq 2$ the sets $\{q_1!+q_k,...,q_{k-1}!+q_k\}$ QUESTION [5 upvotes]: For a prime $q_k \neq 2$ we can study the corresponding set $\{q_1!+q_k,...,q_{k-1}!+q_k\}$, where $q_1,...q_{k-1}$ are all primes strictly less than the prime $q_k$. Peter and Mathphile computed that for $q_k=1193$ we have that all the numbers from the corresponding set are composites I think that $1193$ is rather large as an example of the first prime that generates all composites in the corresponding set and I do not see some reasons, armed with strong principles, of why there shouldn´t be some other primes that generate corresponding sets in which not a single number is prime. Suppose that $\{r_1,...,r_m,...\}$ is the set of all the prime numbers for which the corresponding sets consist of only the composite numbers. Are there any reasons and principles for justifying the assertion that $\{r_1,...,r_m,...\}$ is not a finite set? Edit: This question arrived as the result of thinking about Mathphile´s conjecture. REPLY [5 votes]: Here's a heuristic for the likelihood that $q_k$ generates only composite numbers in this way. For given $j TITLE: Exact sequence involving spectral data for Higgs bundles QUESTION [5 upvotes]: In Beauville, Narasimhan, Ramanan's Spectral curves and the generalized theta divisor, Remark 3.7, the following exact sequence is presented: $0 \rightarrow M(-\Delta) \rightarrow \pi^* E \xrightarrow{\pi^*\varphi - x} \pi^*(L \otimes E) \rightarrow \pi^*L \otimes M \rightarrow 0$ where $X$ is a smooth curve, $L$ is a line bundle on $X$, $(E, \varphi)$ is a $L$-twisted Higgs bundle on $X$, $\pi: X_s \rightarrow X$ is the spectral curve associated to the characteristic polynomial of $\varphi$ (supposed to be integral), $x$ is the tautological section of $\pi^*L$, $\Delta = \pi^*L^{\deg\pi-1}$ is the ramification divisor of $\pi$, and $M$ is the unique torsion free sheaf of rank 1 on $X_s$ such that $E=\pi_* M$. Do you have any suggestion about how to prove the exactness of the sequence? The intuition is that $M(-\Delta)$ must be the eigenspace of $\pi^*\varphi$ associated to the eigenvalue $x$, but it is not clear to me why it should be. REPLY [2 votes]: There is an argument in the proof of Proposition 5.17 of https://arxiv.org/abs/2101.08583<|endoftext|> TITLE: Cauchy reals and Dedekind reals satisfy "the same mathematical theorems" QUESTION [58 upvotes]: The succinct question The conjecture of Birch and Swinnerton-Dyer (to take a random example) mentions L-functions and hence the complex numbers and hence the real numbers (because the complexes are built from the reals). Two naive questions which probably just indicate that I don't understand logic well enough: (1) if we regard BSD as a statement about an explicit model of the real numbers (e.g. the one built from Cauchy sequences or the one built from Dedekind cuts), then why is it "obvious" that BSD is true for one iff it's true for the other? (2) Is it "obvious" that BSD can be formulated as a statement BSD(F) which makes sense for an arbitrary complete ordered archimedean field F? If so, is it also "obvious" that BSD in this sense is isomorphism-invariant, i.e. if F1 and F2 are isomorphic then BSD(F1) iff BSD(F2)? I am interesting in learning the techniques behind why mathematicians treat these claims as obvious. The original question(s) Up to unique isomorphism, there is only one complete archimedean ordered field, and mathematicians refer to it as "the real numbers". There are two standard constructions for showing that such a field exists, one using Dedekind cuts and the other using Cauchy sequences. To be even more explicit, let me define "the Cauchy reals" in this question to mean the set of equivalence classes of Cauchy sequences modulo the usual equivalence relation (so if $x$ is a Cauchy real then $x$ is an uncountably infinite set) and let me define "the Dedekind reals" as being Kuratowski ordered pairs $\{\{L\}, \{L,R\}\}$ with $L$ and $R$ a partition of the rationals with every element of $L$ less than every element of $R$ and both non-empty and $L$ having no rational sup (so if $x$ is a Dedekind real then $x$ is a finite set). Because these two constructions give canonically isomorphic objects, mathematicians think of these constructions as giving "the same answer" and never fuss about which version of the real numbers they are using. I guess there must be some underlying logical principle behind why this works, but I now realise I don't know it. What is it? I am hoping that there is some theorem of logic that says that if I formulate a conjecture (in ZFC set theory, say) about all complete archimedean ordered fields and then I prove the conjecture for the Cauchy reals, then I can somehow deduce that it is also true for the Dedekind reals. But as it stands this is not true. For example, a stupid conjecture about all complete archimedean ordered fields is that they are all equal (as sets in ZFC) to the Cauchy reals. This conjecture is false in general, true for the Cauchy reals, and not true for the Dedekind reals. On the other hand, clearly any "sensible" (I don't know a formal definition of this) mathematical question about complete archimedean ordered fields will be true for the Cauchy reals iff it's true for the Dedekind reals. What would a proof look like? Does one need to give some kind of algorithm which changes a certain kind of proof about Cauchy reals to Dedekind reals? In what generality does this sort of thing work? What are the ingredients? Note that I cannot guarantee that my proof treats the Cauchy reals only as a complete archimedean ordered field, even though I "know in my heart" that there is no advantage in actually starting to look at elements of elements. Here is a related question. Take a normal mathematical conjecture which mentions the reals (for example the Birch and Swinnerton-Dyer conjecture, which mentions L-functions, which are functions on the complex numbers, and a complex number is usually defined to be a pair of reals). Every mathematician knows that it doesn't matter at all whether we use the Dedekind reals or the Cauchy reals. So what is the proof that BSD is true for the L-functions built using the Dedekind reals iff it's true for the L-functions built using the Cauchy reals? It seems to me that we could attempt to use the preceding paragraph, but only once we know that some version of BSD can be formulated using any complete archimedean ordered field, and that the resulting formulation is "a sensible maths question". My gut feeling is that this is "obvious"; however I would rather hear some general principle which I can invoke than actually have to say something coherent about why this is true. Background A few years ago I would have found this kind of question extremely confusing to think about, and would have either dismissed it as trivial or just said that the real numbers were unique up to unique isomorphism and there were probably "theorems of logic" which resolved these issues. But I have a better understanding of what mathematics is now, and I realise that I am not quite sure about how all this works. Here is an attempt to explain what I think are the guts of the first question. Let's say I am doing "normal" mathematics, and I come up with a "normal" mathematical conjecture that mentions real numbers in some way, e.g. the conjecture that pi + e is transcendental, or some much more complicated conjecture which mentions the real numbers implicitly, like the Birch and Swinnerton-Dyer conjecture (which mentions the complex numbers, which are built from the real numbers). No mathematician would ask me whether I mean the Cauchy reals or the Dedekind reals in my conjecture. Let's say I decide to offer $1,000,000 for a proof of my conjecture. Now say some wag who is into computer proof formalisation asks me what foundational system I am using when I make my conjecture, so I say "ZFC set theory". And then they remark that the real numbers have two definitions in ZFC set theory, one using Cauchy sequences and one as Dedekind cuts, and which real numbers was my conjecture about? I am a mathematician, so I know it doesn't matter, so I say "the Cauchy reals" just to shut them up. The next day I realise I could have been more clever, so I take the trouble to reformulate my conjecture so that instead of explicitly mentioning "the real numbers" I make it into a conjecture about all complete archimedean ordered fields (the fact that this reformulation is possible could be thought of as a definition of "normal" mathematics in the paragraph above). Of course I "know" that this does not change my conjecture in any substantial way. I decide to get in touch with the wag to tell them my change of viewpoint, so I call them up, but before I can get a word in, they very excitedly tell me that they left their new deep learning AI ZFC computer proof generator system on all night working on my conjecture about the Cauchy reals, and it has managed to come up with a proof which is a billion lines long and incomprehensible, but each line is formally checked to be valid in ZFC, so it must be right, and can they please have the $1,000,000. I explain that I have now changed my conjecture and it's now a statement about all complete archimedean ordered fields, and ask them if their proof works for all such fields. "Definitely not!" they reply. "My AI needs to generate proofs of trivial things like 3 < 5 to prove your conjecture, and it does it by thinking about the definition of < on the Cauchy reals and coming up with a proof of 3 < 5 which is specific to Cauchy reals. My AI also does a bunch of other weird things with Cauchy reals, and some of them I don't understand at all; they are probably just weird ways of proving standard facts about complete archimedean ordered fields but I can't be sure". "Well, does everything you do for the Cauchy reals have some analogue for the Dedekind reals?" I ask. And they reply "I don't know, all I can guarantee is that my proof is valid in ZFC set theory, and therefore I have proved your conjecture in its Cauchy form. You are claiming that the Cauchy form and the complete archimedean ordered field form are obviously equivalent, hence I have proved your more general conjecture." I think the wag must be right, but I do not understand the details of why. REPLY [6 votes]: There is no need to use explicit models of $\mathbb R$ in formulations of Swinnerton-Dyer conjecture. You could extend $\mathbf{ZFC}$ with symbols $(\mathbb R,+_\mathbb R,\times_\mathbb R,<_\mathbb R,0_\mathbb R,1_\mathbb R)$ (in addition to symbols $=,\in$) and add obvious axioms of Reals. After that you obtain new system, say $\mathbf{ZFC}+\mathbb R$ in which you able to formulate any theorem about Reals model-independently using those new symbols. To be sure that $\mathbf{ZFC}+\mathbb R$ is conservative extension of $\mathbf{ZFC}$ you should to justify your axioms by construction of either Dedekind or Cauchy reals. So, you should to proof in $\mathbf{ZFC}$ statement $$\exists \mathbb R_{Cauchy}, +_{\mathbb {R}_{Cauchy}},\times_{\mathbb {R}_{Cauchy}},<_{\mathbb R_{Cauchy}},0_{\mathbb R_{Cauchy}},1_{\mathbb R_{Cauchy}} ({axiom}_1 \wedge {axiom}_2 \wedge ... \wedge {axiom}_n)$$ where ${axiom}_1,...,{axiom}_n$ is axioms of real numbers rewritten in terms of Cauchy reals, for example ${axiom}_1$ is $$0_{\mathbb R_{Cauchy}} \in \mathbb{R}_{Cauchy}$$ ${axiom}_2$ is $$\forall x \forall y (x \in \mathbb R_{Cauchy} \wedge y \in \mathbb R_{Cauchy} \to x+_{\mathbb R_{Cauchy}} y = y +_{\mathbb R_{Cauchy}} x)$$ and so on (of course you additionaly need to state that $+_{\mathbb R_{Cauchy}}$ is well-defined function on $\mathbb {R}_{Cauchy}$ and write $((x,y),t) \in +_{\mathbb{R}_{Cauchy}}$ rather than $x +_{\mathbb{R}_{Cauchy}} y = t$) . After that you could (externally, on metalevel) deduce that your system $\mathbf{ZFC}+\mathbb{R}$ is conservative extension of $\mathbf{ZFC}$ so you could work now inside $\mathbf{ZFC} + \mathbb{R}$ instead of $\mathbf{ZFC}$ and formulate statements about Reals model-independently. Maybe it sounds a little bit artificial but this is exactly how proof-checkers work see Note on definitions on Metamath Proof Explorer. Also, for example, see axiom ax-resscn and construction-dependent theorem axresscn which is justification of axiom ax-resscn and "should not be referenced directly".<|endoftext|> TITLE: Idempotent functions on Sp(1) QUESTION [6 upvotes]: The quaternion group $Sp(1)\simeq S^3$ can be understood as $(z,w)\in\mathbb {C}^2$ with $|z|^2+|w|^2=1$ where multiplication is defined by $(z,w)(t,s)=(zt-\bar{s}w,zs+\bar{t}w)$. Question: How do the nontrivial continuous idempotent functions (wrt convolution) look like? That is, functions $f*f=f$, defined using Haar measure of 3-sphere. REPLY [7 votes]: As Venkataramana says, this is a natural candidate for the Peter-Weyl theorem: Let $G$ be a compact group. Let $\{ V_i \}_{i \in I}$ be the set of isomorphism classes of irreducible complex representations of $G$ (for some index set $I$), and fix a $G$-invariant Hermitian inner product on each $V_i$. Let $\{ e_i^b \}_{b \in B_i}$ be an orthonormal basis of $V_i$ and put $f_i^{ab}(g) = \langle e_i^a, g e_i^b \rangle$. Then the Peter-Weyl theorem tells us that the $f_i^{ab}$ are an orthonormal basis for $L^2(G)$ (with $\int_G 1$ normalized to $1$). This basis identifies $L^2(G)$ with convolution with the Hilbert direct sum of matrix algebras $$\bigoplus_{i \in I} \mathrm{Mat}_{|B_i| \times |B_i|}.$$ Let's write this isomorphism as $f \mapsto \left( \rho_i(f) \right)_{i \in I}$. So a function in $L^2(G)$ is idempotent iff each matrix $\rho_i$ is idempotent. Moreover, $|f|^2 = \sum_{i \in I} |\rho_i(f)|^2$ (right hand side is Frobenius norm) so $f \in L^2(G)$ implies $|\rho_i(f)| \to 0$. The Frobenus norm of a rank $k$ idempotent is $\sqrt{k}$, so only finitely many $\rho_i(f)$ can be nonzero. So such idempotents correspond to choosing finitely many nonzero subspaces $W_i$ in finitely many $V_i$. Explicilty, take an orthogonal basis $(u_i^c)_{c \in C_i}$ for each $W_i$ and form the sum $\sum \langle u_i^c, g u_i^c \rangle$. The simplest such example is to directly take the two dimensional representation $$a+bi+cj+dk \mapsto \begin{bmatrix} a+bi & c+di \\ -c+di & a-bi \\ \end{bmatrix}$$ and project onto the $1$-dimensional subspace spanned by $\left[ \begin{smallmatrix} 1\\0 \end{smallmatrix} \right]$ to get the idempotent $$\left\langle \begin{bmatrix} 1\\0 \end{bmatrix} ,\ \begin{bmatrix} a+bi & c+di \\ -c+di & a-bi \\ \end{bmatrix}\begin{bmatrix} 1\\0 \end{bmatrix} \right\rangle = a+bi.$$<|endoftext|> TITLE: Volume doubling, uniform Poincaré, counterexample QUESTION [6 upvotes]: The Poincaré inequality and the volume doubling property are important notions related to heat kernel estimates. Pavel Gyrya and Laurent Saloff-Coste obtain the two sided heat kernel estimate of Neumann and Dirichlet heat kernels on inner uniform domains by showing the Poincaré inequality and the volume doubling property for the canonical Dirichlet forms on inner uniform domains. Namely, for inner uniform domains, we can check the Poincaré inequality and the volume doubling property. It is easy to check that domains with cusps are not inner uniform domains. For example, $D=\{(x,y) \in \mathbb{R}^2 \mid x>0,\ y<1,\ y>x^{1/2}\}$ is not inner uniform. The closure $\bar{D}$ of $D$ is regarded as a metric space endowed with the shortest path metric $\rho$. I am concerned with whether the Poincaré inequality holds on $\bar{D}$. Namely, there exists $P_0$ such that for any $x \in \bar{D}$, $r>0$, and smooth $f$, \begin{align*} (\ast)\quad \inf_{\xi \in \mathbb{R}} \int_{B(x,r)}|f-\xi|^2\,dm \le P_o r^2 \int_{B(x,r)}|\nabla f|^2\,dm, \end{align*} where $B(x,r)=\{y \in \bar{D} \mid \rho(y,x) TITLE: Permutations, skew-symmetric forms and degeneracy QUESTION [6 upvotes]: Define a skew-symmetric form $(\cdot,\cdot)$ on $\mathbb{R}^{2k}$ by $$(e_i,e_j) = \begin{cases} 1 &\text{if $ij$},\\ 0 & \text{if $i=j$.}\end{cases}$$ Given a permutation $\pi:\{1,\dotsc, 2 k\}\to \{1,\dotsc, 2 k\}$, let $V_\pi$ be the space spanned by $e_{\pi(2i)}-e_{\pi(2i-1)}$ for $1\leq i\leq k$. For which permutations $\pi$ is the restriction $(\cdot,\cdot)|_{V_\pi\times V_\pi}$ non-degenerate? REPLY [4 votes]: I'll give it a try (clearly, comments aren't enough, and the references I found are too much). Given the permutation $\pi$ of $\{1,2,\dots,2k\}$ one can build an orientable surface with boundary as follows. Take a topological disk, say the rectangle $[0,2k+1]\times[0,1]$ in the complex plane, and consider its quotient by the relation which identifies for each $i=1,\dots,k$ a small interval $[a-\delta,a+\delta]$ ($0<\delta<1/2$) around point $a=\pi(2i-1)$ with interval $[b-\delta,b+\delta]$ around $b=\pi(2i)$, with orientation reversed, that is $a+x\simeq b-x$. [added: this is reversed from my comments under the question : here we glue things on the bottom edge of the rectangle. This has the purely cosmetic effect of numbering points in the usual --counterclockwise-- orientation of the boundary]. The result is a compact, connected, orientable surface with boundary, which you can picture in $3$-space by protruding "nearly horizontal" rectangles from the identified intervals, making them meet only in their tips and the prescribed pair (I'm really sorry for not being able to draw that here, and being probably very bad at describing mental pictures too). The orientation comes from projection to the horizontal plane. The resulting surface $S_\pi$ depends only on the partition of $\{1,2,\dots,2k\}$ into the pairs $\{\pi(2i-1),\pi(2i)\}$, $i=1,\dots,k$, that is on the $\pi$-conjugate $\sigma$ of the involution $\sigma_0=(12)(34)\dots(2k-1)(2k)$. There are $k!2^k$ different $\pi$'s for one $\sigma$ (this amounts to what @Ycor said in a comment). Let $b$ be the number of boundary components of $S_\pi$. A book-keeping exercise tells you that $b$ is the number of cycles of the permutation $i\mapsto \sigma(i)+1$ (or its inverse $i\mapsto\sigma(i-1)$, or any of their conjugates, like $i\mapsto\sigma(i)-1$, that's the same number). Now the all powerful surface classification theorem tells us that this (compact, connected, orientable) $S_\pi$ is (homeomorphic to) an (orientable) surface of some genus $g$ with the interiors of $b$ disjoint closed discs removed. Why that ? Well, glue $b$ closed discs to the boundary components of $S_\pi$, and you have a (compact, connected, orientable) surface without boundary, connected sum of $g$ tori, by classification. Then undo. Hence the amazing fact : there is only one model for each pair $(g,b)$ ! [this should be a footnote] That this is amazing may easily be overlooked nowadays, but can perhaps better be sensed by the fact that mapping class groups of surfaces, that is the discrete groups of components of their autoequivalences (homeo/diffeo doesn't matter here) aren't yet understood. For instance they share many properties with subgroups of linear groups, but they are not known to be linear aside from small cases. Yet they are the means to recognize a surface when we meet one.[end footnote] Now in an oriented manifold $M^n$, one can count algebraically intersections of submanifolds or cycles of complementary dimensions, as Poincaré tolds us. Either triangulate and refine or use Thom's transversality, but in the end it's linear algebra, counting intersections by comparing the orientation of the direct sum of tangent spaces with the ambient one. This gives rise to the integer valued intersection product on cycles of complementary dimensions $p,q=n-p$, which depends only on their homology classes and is $(-1)^{pq}$-symmetric. Crucially, it is non-degenerate when the manifold is closed (and oriented), meaning that $H_p(M,\mathbb{Z})$ mod torsion is isomorphic to $Hom(H_{n-p}(M,\mathbb{Z}),\mathbb{Z})$ via the intersection product. This is part of so-called Poincaré duality, expressed more neatly nowadays by adding cohomology to the picture (but we don't need it). In our context of $1$-cycles on oriented surfaces, intersection is skew-symmetric. If $S_g$ is a closed (=without boundary), connected oriented surface of genus $g$, $H_1(S_g,\mathbb{Z})$ is a free abelian group with a non-degenerate skew-symmetric form, with standard "symplectic" basis $(a_1,b_1,\dots,a_g,b_g)$, meaning $$a_i\cdot b_j=\delta_{ij},$$ $$a_i\cdot a_j=b_i\cdot b_j = 0.$$ But when there are $b\geq 1$ boundary components, $H_1(S_{g,b},\mathbb{Z})\simeq \mathbb{Z}^{2g+b-1}$, mapping to $H_1(S_g,\mathbb{Z})\simeq \mathbb{Z}^{2g}$ with kernel of rank $b-1$, preserving (clearly) the intersection form. [added comment] This could be proved using the homology exact sequence of the pair $(S_g \supset S_{g,b})$, but here it is simpler to use homotopy invariance of homology plus the fact that for $b>0$, $S_{g,b}$ is homotopy equivalent to the graph with one vertex and $2g+b-1$ loops. This is also relevant below. Hence the intersection form is non-degenerate on $H_1(S_{g,b},\mathbb{Z})$ iff $b\leq 1$. Concerning the surface $S_\pi$, it should be clear that it is homotopy equivalent to a graph with one vertex and $k$ loops, and that letting $h_i$ denote the class of a closed curve going counterclockwise [NOTE: this is reversed from my comments under the question] through the handle resulting from gluing the intervals $\pi(2i-1)\pm\delta$ to $\pi(2i)\mp\delta$ (in reverse) that the $h_i$, $i=1,\dots, k$ form a basis of $H_1(S_\pi,\mathbb{Z})\simeq\mathbb{Z}^k$ the matrix of their intersection products is the same as your skew-symmetric form on $V_\pi$ (maybe up to sign, I switched to the upper half plane hoping to correct for this). Hope this helps.<|endoftext|> TITLE: Link of a singularity QUESTION [9 upvotes]: I would like to understand the topological type of a link of a singularity in a simple example. Consider for instance the cone ${xy-z^2=0}\subset\mathbb{C}^3$. If we set $x = x_1+ix_2, y = y_1+iy_2, z = z_1+iz_2$ then this corresponds to the $4$-dimensional real subvariety of $\mathbb{R}^6$ given as the complete intersection $\{x_1y_1-x_2y_2-z_1^2+z_2^2 = x_1y_2+x_2y_1-2z_1z_2 = 0\}$. Therefore the link of the singularity is given by $\{x_1y_1-x_2y_2-z_1^2+z_2^2 = 0, x_1y_2+x_2y_1-2z_1z_2 = 0, x_1^2+x_2^2+y_1^2+y_2^2+z_1^2+z_2^2 = 1\}$ What is the topological type of this link? Can we determine the topological type of the link from these three equations? REPLY [7 votes]: To add to the excellent answers already provided, here are some general facts in the case of rational surface singularities (1 and 2) and hypersurface singularities (3). Many interesting singularities are obtained by quotienting $C^2$ by the action of a finite subgroup $G\subset U(2)$. Since $U(2)$ preserves the unit sphere, you can get the link just by quotienting $S^3$ by $G$. The example you gave is equivalent to the singularity you get by taking the quotient of $C^2$ by the subgroup $G=\{I,-I\}$, so that tells you the link is $RP^3$, as Marco pointed out. For cyclic quotient singularities, you'll get a lens space. There are more exotic examples, for example, for the spin double cover of the symmetry group of the icosahedron (under the covering map $SU(2)\to SO(3)$), the link of the quotient singularity will be the Poincaré homology sphere. If you want to understand these singularities as affine varieties, they're obtained by taking Spec of the ring of invariants for the finite group action on the ring of polynomial functions on $C^2$ (for $G=\{I,-I\}$the invariant functions are generated by $u=x^2,v=xy,w=y^2$ with the obvious relation $uw=v^2$, which is where your equation comes into it). Take the minimal resolution of the singularity. Let's suppose the exceptional divisor is a tree T of spheres with negative self-intersection numbers. A neighbourhood of the exceptional divisor is then a plumbing (according to the tree) of disc bundles over spheres with Euler numbers given by these self-intersection numbers. This has a surgery description: you take a bunch (one for each vertex of T) of unknots in $S^3$ such that two of them link like a Hopf link if they are connected by an edge in T (and are pairwise unlinked otherwise). Give each unknot a framing of minus the self-intersection of the corresponding sphere. This is both a Kirby diagram of the minimal resolution and a Dehn surgery presentation of the link (the exceptional curves in the minimal resolution are given by slice discs for the unknots capped off by the cores of the 2-handles you attach). If you have a complex hypersurface singularity (i.e. one defined by a single equation over C) then you can use the ideas of Milnor to get at the topology of the link (see his book on isolated singularities of complex hypersurfaces). This will give you an "open book decomposition" of the link, i.e. you end up finding a collection of knots (called the "binding") in the link of the singularity whose complement fibres over the circle (the fibres are real surfaces called the "pages"). In higher dimensions, the binding is a real codimension 2 contact submanifold of the link and the pages are real codimension 1 Stein domains.<|endoftext|> TITLE: Is there any conditions on a finite abelian group so that it cannot be class group of any number field? QUESTION [10 upvotes]: The Cohen-Lenstra paper says the probability that the odd part of a class group being cyclic is close to 0.98. So I was thinking: can we find any conditions on a finite abelian group so that it cannot be a class group of any number field? REPLY [16 votes]: It follows from the Cohen-Lenstra heuristic that every finite abelian group is expected to be isomorphic to infinitely many class groups of real quadratic fields (even to a positive proportion of real quadratics, which is stronger), but nothing like this is known. However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.12 and the discussion following it on page 20 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v4. We show there that if $G$ is a cyclic group of order $58$, then there are finite $\mathbb{Z}[G]$-modules that cannot be realised as the class group of a $G$-extension. Of course, there are cheap ways of doing that, by considering modules whose fixed submodule is something silly, contradicting the fact that the class group of $\mathbb{Q}$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.<|endoftext|> TITLE: Categorical Significance of Fibrations QUESTION [7 upvotes]: It is well known that the category $\text{Set}$ classifies covering spaces among $1$-categories. That is, for each topological space $X$, there is an equivalence of categories $[ \Pi (X) , \text{Set}]_{\text{Cat}} \cong \text{Cov}(X)$ where $\text{Cov}(X)$ is the category of covering spaces of $X$, and this equivalence is natural in $X$. Moreover, $\text{Top}$ classifies fibrations among $(\infty, 1)$-categories. That is, for each topological space $X$, there is an equivalence of $(\infty, 1)$-categories $[ \Pi_{\infty} (X), \text{Top} ]_{(\infty, 1) \text{-Cat} } \cong \text{Fib}(X)$, where $\text{Fib}(X)$ is the category of fibrations over $X$ (this is natural in $X$, too). This is what I was reading in the excerpt below: This is quite nice because it seems like fibrations fit into the context of descent, so that we might presume to have the Grothendieck construction and all that. My questions are: 1) Does this give a sort of $(\infty, 1)$-descent along fibrations, just like ordinary descent is for a Grothendieck fibration, but for $(\infty, 1)$-categories? 2) Is there some dual situation for cofibrations? Is there some $(\infty, 1)$-category $C$ such that $[C, \Pi_{\infty}(X)]_{(\infty, 1) \text{-cat}} \cong \text{Cof}(X)$? Is there a codescent? 3) Does this give an analogous theory for $\Pi_i(X), i \in \mathbb{N}_{\geq 1}$ to the one that exists for $\Pi_1(X)$? I'm not quite sure how to analogize here. REPLY [2 votes]: Yes. In roughly your language, the forgetful $(\infty,1)$-functor $\rm Fib\to Top$ is an $\infty$-fibration, where the fiber over a space $X$ is the category of fibrations over $X$, and descent in this $\infty$-fibration corresponds to the fact that there is a fibration classifier. This is described in $\infty$-categorical language in section 6.1 of Lurie's Higher topos theory, and in model-categorical language in Rezk's "Toposes and homotopy toposes". Note, though, that once you are talking about $(\infty,1)$-categories there is no difference between fibrations and arbitrary maps, so the above "$\rm Fib$" is actually just the arrow $(\infty,1)$-category of $\rm Top$. Not as far as I know. Yes. The analogue of $\rm Set$ when $n=0$ and $\rm Top$ when $n=\infty$ is $n \rm Typ$, the space of homotopy $n$-types.<|endoftext|> TITLE: List of all known Riesz representation theorems QUESTION [9 upvotes]: Due to the history and development of measure and integration theory and different mathematical schools, there is a huge variety and inconsistency of definitions for concepts like tightness of a measure, regularity of a measure and even the definition for a measure $\mu : \mathcal{R} \to R$ on its own (including domain of definition such as a ring $\mathcal{R}$ (most often $\sigma$-algebra, $\sigma$-ring, $\delta$-ring or algebra) and target space $R$ (such as $[0, \infty]$ or $\mathbb{R}$ or $(-\infty, \infty]$ or a Banach space for vector measures). There is also a huge variety of Riesz representation theorems between certain function spaces (bounded measurable functions $M_b(X)$, bounded continuous functions $C_b(X)$ (with sup-norm or some form of strict topology) and other classes of continuous functions such as $C_0(X)$ and $C_c(X)$ for locally compact domains $X$). Some representation theorems can be found in books like [Dunford Schwartz, "Linear Operators I"], [Bogachev, "Measure Theory II"], [Fremlin, "Measure Theory"] and many others. These representation theorems use some form of tightness or regularity, but the definitions are all different and in general not equivalent (which extremely slows down the "fast look-up" process.) For instance, one should distinguish between properties like (inner) closed-regular, (inner) compact-regular, outer-regular and so on. I would like to know, if there is a list somewhere that fixes a definition for all the various (useful) tightness and regularity properties and upon this summarizes all (or most of) the known Riesz representation theorems. REPLY [3 votes]: Such a list will always be based on subjective criteria but here is one suggestion, from a functional analytic rather than a probabilistic point of view. In my view the ingredients for an extension of the standard Riesz theorem for $C(K)$-spaces are 1) a space with structure (topology, metric, uniformity, $\sigma$-algebra,...); 2) vector spaces $E$ (of functions on the set—-the integrands) and $F$ (of measures). These are provided with topological structures which are compatible with the vector space one and under which they are complete; 3) $E$ and $F$ are in symmetric duality, i.e., each is naturally identifiable with the other‘s dual. One would also expect these assignments of the vector spaces to the set to have natural functorial properties. One can find such result for topological spaces (the duality between bounded, continuous functions and tight measures), uniform spaces (bounded, uniformly continuous functions and uniform measures (Pachl)) and measure spaces, i.e., sets with a $\sigma$-algebra (bounded, measurable functions and $\sigma$-additive measures). In order to do this one has to extend the classical functional analytic structures to the categories of Saks and CoSaks spaces. However, this can be done in a unified manner, using Grothendieck‘s construction of ind and proj categories by applying it to the standard relevant categories (metric spaces, compacta, Banach spaces and the dual category—-Waelbroeck spaces).<|endoftext|> TITLE: Was Jacobi the first to notice the ambiguity in the partial derivatives notation? And did anyone object to his fix? QUESTION [24 upvotes]: In his 1841 article De determinantibus, Jacobi remarked that the notation $\frac{\partial z}{\partial x}$ for partial derivatives is ambiguous. He observed that when $z$ is a function of $x,y$ as well of say $x,u$, then the coefficient $\frac{\partial z}{\partial x}$ appearing in the linear expansion of $dz$ with respect to $dx,dy$, denotes something else than the partial derivative $\frac{\partial z}{\partial x}$ appearing in the expansion of $dz$ wrt $dx,du$. As he himself writes: In order for the partial differentials, of a function which depends on more than one variable, to be completely determined, it does not suffice to provide the function to be differentiated and the variable with respect to which to differentiate; one must moreover express which quantities remain constant during the differentiation. A longer excerpt can be found here. Probably many others have made the same observation since then. Vladimir Arnold for instance makes it in a footnote to his Classical Mechanics. But partial derivatives already existed for 100 years before De determinantibus. (With a slightly different notation that has the same problem.) Q1: Was Jacobi the first mathematician to point out the ambiguity? In the same article, Jacobi also proposed a fix for the problem: he suggested we write $\frac{\partial z(x,y)}{\partial x}$ for the partial derivative when $y$ is held constant, and $\frac{\partial z(x,u)}{\partial x}$ when $u$ is held constant. This introduces a new problem, since it overloads the notation for function application $f(x,y)$, which already had an established and different meaning at that time. Q2: Did no one object to the suggestion of Jacobi? Finally, in many physics textbooks (thermodynamics in particular), the problem is fixed by writing $ \left(\frac{\partial z}{\partial x}\right)_y $ for the derivative of $z$ wrt $x$ when $y$ is held constant. Q3: Who proposed this fix and when? REPLY [34 votes]: An extensive review of the history is given by Florian Cajori, The History of Notations of the Calculus. Q1: Yes, it does appear that Jacobi was the first to explicitly state this ambiguity. Q2: The German mathematician Paul Stäckel objected to the suggestion of Jacobi: "This notation is ambiguous too, for the symbol is used in two different meanings, in as much as $f(x, y)$ is another function of $x$ and $y$ than is $f(x, u)$ of $x$ and $u$." Q3: The Irish engineer John Perry advocated the notation $\left(\frac{dz}{dx}\right)_y$. I reproduce Perry's 1902 Letter to Nature, where he says he learned this notation at school.<|endoftext|> TITLE: Rank mod $p$ of a non-singular matrix with given determinant QUESTION [6 upvotes]: Let $A$ be a non-singular $n$-by-$n$ matrix with integer entries. Assume that $p^r\nmid \det(A)$. Does it follow that $A$ has an $(n-r+1)$-by-$(n-r+1)$ minor that is non-singular modulo $p$? If the answer is no: what if we put some additional conditions -- say, $p$ large compared to $n$, and/or $A$ having bounded entries? REPLY [8 votes]: Let $s = \operatorname{ord}_p(\det(A))$, i.e. $p^s\ \| \det(A)$. Note that $|\det(A)| = \# \operatorname{coker}(A \colon \mathbf Z^n \to \mathbf Z^n)$. Right exactness of the tensor product shows that $\operatorname{coker}(A \otimes \mathbf F_p) \cong \operatorname{coker}(A) \otimes \mathbf F_p$, so $\operatorname{coker}(A \otimes \mathbf F_p)$ has cardinality at most $p^s$, hence dimension at most $s$. This should answer your question (affirmatively). REPLY [6 votes]: I think this is all about Smith Normal Form. Write $XAY = {\rm diag}(d_{1},d_{2},\ldots, d_{n})$ where $X$ and $Y$ are unimodular and where the $d_{i}$ are integers such that $d_{i} | d_{i+1}$ for each $i$. There are various terminologies for the $d_{i}$ sometimes conflicting. Let me call them the determinantal divisors (sometimes they are called elementary divisors sometimes invariant factors). In any case, they are unique up to sign and they determine and are determined by the structure of the Abelian group $\mathbb{Z}^{n}/{\rm Im}A$ when $A$ acts by multiplication on $\mathbb{Z}^{n}$ identified with $n$ long integer column vectors. Your hypotheses imply that $p \not | d_{i}$ for $1 \leq i \leq n-(r-1).$ On the other hand, it is "well-known" that for $1 \leq i \leq m$, the product $d_{1}d_{2} \ldots d_{m}$ is the gcd of all the $m \times m$ minors of $A$. Hence your assumptions imply that there is indeed some $(n-r +1) \times (n-r+1)$ minor of $A$ which is not divisible by $p$.<|endoftext|> TITLE: When is the poset of acyclic orientations of a graph a lattice? QUESTION [11 upvotes]: $\def\inv{\mathrm{inv}}\def\Acyc{\mathrm{Acyc}}$Let $G$ be a graph whose vertices are numbered $\{ 1,2, \ldots, n \}$. Given an orientation $\omega$ of $G$, define the inversions of $\omega$, written $\inv(\omega)$, to be the set of edges $(i,j)$ with $i TITLE: Number of real roots in type $\tilde{E}_8$ QUESTION [5 upvotes]: Let $\Phi_+$ be the set of all positive roots for a Kac-Moody algebra. Denote by $\alpha_i$ the simple root associated with node $i$ by for $i \in \{1, \ldots, n-1\}$ and by $\beta$ the simple root associated with $n$. The Dynkin diagram for $\tilde{E}_8$ is \begin{align} \circ - \circ - & \circ - \circ - \circ - \circ - \circ - \circ \\ & \ | \\ & \ \bullet \end{align} where $\bullet$ corresponds to the simple root $\beta$. The degree of a root is the coefficient of the root at $\beta$. Has the number of degree $d$ real roots in $\Phi_+$, $d \in \mathbb{Z}_{\ge 1}$, been computed for the root system $\tilde{E}_8$? Are there some references about this? Thank you very much. REPLY [9 votes]: I don't believe there is a reference for this, for this follows immediately from the description of real roots in affine root systems. Namely, by Proposition 6.3(a) in "Infinite dimensional Lie algebras" by V. Kac the real roots of $\Delta=\mathsf{E}_8^{(1)}$ are of the following form. Number the simple roots as $\alpha_0,\alpha_1,\ldots,\alpha_8$, where $\mathring\Delta=\langle \alpha_1,\ldots,\alpha_8 \rangle$ is the subsystem of type $\mathsf{E}_8$, so $\alpha_0$ is the affine root. Then $$ \Delta^{\mathrm{re}} = \{ \alpha+n\delta \mid \alpha\in\mathring\Delta,\ n\in\mathbb{Z} \}, $$ where $\delta=\sum_{i=0}^8 c_i\alpha_i$, and the coefficients $c_i$ are as on the following diagram: In what follows all roots are assumed to be real. Now it's a matter of a simple calculation. The degree $0$ roots form a subsystem of type $\mathsf{A}_8$, so there are $36$ positive roots among them. The number of degree $1$, $2$ and $3$ roots in $\mathring\Delta$ is, respectively, $56$, $28$ and $8$. The degree $1$ roots in $\Delta$ are either in $\mathring\Delta$ or of the form $\alpha+\delta$, where $\alpha\in\mathring\Delta$ has degree $-2$, so there are $56+28=84$ of them; the same works for degree $2$ roots. Degree $3$ roots are either in $\mathring\Delta$ (and form an $\mathsf{A}_7$ subsystem); or of the form $\alpha+\delta$, where $\alpha\in\mathring\Delta$ is of degree $0$; or of the form $\alpha+2\delta$, where $\alpha\in\mathring\Delta$ is of degree $-3$. Thus in total there are $56+8+8=72$ roots of degree $3$. Since there are no roots of degree $\geqslant3$ in $\mathsf{E}_8$, one concludes that $$ \#\left\{ \text{positive real roots of degree $d$ in $\mathsf{E}_8^{(1)}$} \right\} = \begin{cases} 36, & \text{if}\ d=0, \\ 72, & \text{if}\ d \equiv 0 \pmod{3},\ d\neq 0, \\ 84 & \text{otherwise.} \end{cases}$$ The formula for the number of all real roots of a given degree (both positive and negative) is even more simple, namely, $72$ for $d\equiv0\pmod3$ and $84$ otherwise.<|endoftext|> TITLE: Uniqueness of countable version of $L[U]$? QUESTION [5 upvotes]: Suppose $\alpha$ is a countable ordinal and $U_0,U_1,\kappa$ are such that $L_\alpha[U_i] \models \mathrm{ZFC} + U_i$ is a normal ultrafilter on $\kappa$. Does $U_0 = U_1$? The argument for uniqueness of $L[U]$ due to Kunen uses an iteration of length a $V$-regular cardinal above $\kappa$. This is not available for countable models. I am not assuming the models are fully iterable from the outside. REPLY [9 votes]: Andrés's comment can be turned into an answer, but there is a slight subtlety. Suppose $M$ is a countable iterable model of ZFC + $V = L[U]$. Let $\kappa$ be its measurable cardinal and $\alpha$ be its ordinal height. Then there is a transitive model $N\neq M$ of height $\alpha$ satisfying ZFC + $V=L[U]$ + $\kappa$ is measurable. The subtlety is that no such model $N$ can lie in $L$: since $\alpha$ and $\kappa$ are uncountable in $L$ (they're Silver indiscernibles), $N$ contains all the constructible reals, and since $N$ thinks there is a measurable cardinal, $N$ contains nonconstructible reals as well. The model $N$ will instead belong to a generic extension of $L$. Let $g\subseteq \text{Col}(\omega,\alpha)$ be $L$-generic. Then by $\Sigma^1_1$-absoluteness, $L[g]$ contains a model $N$ of height $\alpha$ satisfying satisfying ZFC + $V=L[U]$ + $\kappa$ is measurable. Note that $N$ is not equal to $M$ since $M$ does not belong to any generic extension of $L$: indeed $0^\#\in M$.<|endoftext|> TITLE: Real manifolds in a theorem prover? QUESTION [34 upvotes]: Which of the formal computer proof verification systems (like Lean, Coq, Agda, Idris, Isabelle-HOL, HOL-Light, Mizar etc) have a basic theory of real manifolds? Up to, say, the definition of a smooth map between manifolds, and examples such as real projective space or Grassmannians? PS what is the correct tag for questions about proof verification systems? Are they even welcome here? OK so this question has been around for a week, with no answers. The concept of a real manifold has been around for over 100 years and is both a fundamental mathematical object and something taught in any half-decent undergraduate mathematics degree. Almost all these lectures are given by mathematicians. The concept of a formal proof verification system has been around for maybe 40 years, and many mathematical proofs have been formalised in these systems. Almost all the proofs are written by computer scientists. I am not so sure that the concept of real manifold is mentioned in many computer science degrees. And I am not so sure that there are too many "how to use formal proof verification software" courses in maths departments. And because that's where we stand today in 2019, it appears that there are basic undergraduate-level mathematical objects which nobody has even attempted to formalise the definition of -- the mathematicians because on the whole they don't know where to start, and the computer scientists because on the whole they don't know the definitions. Since I started trying, as a mathematician, to figure out how these systems worked, I have learnt a lot of things. But the fact that these systems have been around for decades but still none of them seem to contain all the theorems and proofs in a standard pure mathematics undergraduate degree was in some sense the most surprising thing I've learnt. Hopefully real manifolds will appear in one or more of these systems at some point (or they are there already but nobody posted here yet). REPLY [11 votes]: Andrew Ashworth posted an answer on the Lean discussion forum, and I'll give it here for the sake of having an answer. Fabian Immler and Bohua Zhan have given a formalisation of $C^k$ and $C^\infty$ manifolds at https://www.isa-afp.org/entries/Smooth_Manifolds.html, describing their development as: We formalize the definition and basic properties of smooth manifolds in Isabelle/HOL. Concepts covered include partition of unity, tangent and cotangent spaces, and the fundamental theorem of path integrals. We also examine some concrete manifolds such as spheres and projective spaces. The formalization makes extensive use of the analysis and linear algebra libraries in Isabelle/HOL, in particular its “types-to-sets” mechanism. There's an accompanying paper http://home.in.tum.de/~immler/documents/immler2018manifolds.pdf, which is interesting. At a few points they point out difficulties that seem to stem from the simple type system available in Isabelle/HOL. In particular they complain that they can't define the tangent space at a point as a type itself, but only as a set, and this is what causes so much difficulty with using the existing Isabelle/HOL library. They also at a few points have some difficulty where they apparently can't do arithmetic with the natural number indexing the dimension of one of their manifolds --- so when defining $n$-spheres for all $n$ they have to jump through some awkward hoops.<|endoftext|> TITLE: Infinitary reasoning in Godel's Completeness Proof QUESTION [8 upvotes]: Godel's Completeness Theorem is a straightforward consequence of Skolem 1922 and yet this conclusion was not drawn by Skolem himself. In a letter to Wang (Dec. 7, 1967 in Godel 2003) Godel gives an explanation for this oversight: At that time, nobody (including Skolem himself) drew this conclusion ... I think the explanation is not hard to find. It lies in a widespread lack, at that time, of the required epistemological attitude toward metamathematics and toward nonfinitary reasoning. I know some form of Konig's Infinity Lemma or the "law of infinite conjunction" (Quine) is necessary to prove semantic completeness, but (a) is this what Godel must be referring to as "nonfinitary reasoning", and (b) where/how does this law figure in Godel's original proof? REPLY [9 votes]: The best write-up I know of Godel's proof of the completeness theorem is by Avigad, in his paper Godel and the metamathematical tradition (section $4$). Avigad divides the proof into $5$(ish) steps, and step $2$ crucially uses Konig's lemma: Step $2$: If a set $\Gamma$ of propositional formulas is not refutable, it has a satisfying truth assignment. Write $\Gamma=\{\varphi_0,\varphi_1,\varphi_2,...\}$. Build a finitely branching tree where the nodes at level one are all the truth assignments to variables of $\varphi_0$ that make $\varphi_0$ true; the nodes at level two are all the truth assignments to variables of $\varphi_0\wedge\varphi_1$ that make that formula true; and so on. (The descendants of a node are all the truth assignments that extend it.) If, at some level $k$, there is no satisfying assignment to $\varphi_0\wedge\varphi_1\wedge ...\wedge\varphi_{k-1}$, then, by step $1$, $\Gamma$ is refutable. Otherwise, by Konig’s lemma, there is a path through the tree, which corresponds to a satisfying truth assignment for $\Gamma$. (Emphasis mine.) This is the only way Konig gets used here. Specifically, here's what's going on in the rest of the proof: Step $1$ is just the completeness theorem for propositional logic for individual sentences (which Step $2$ lifts to sets of sentences via Konig). Step $3$ shows completeness for individual $\forall^*\exists^*$-sentences without function symbols or equality, and is entirely straightforward. Step $4$ extends Step $3$ to individual sentences without function symbols or equality to arbitrary-complexity sentences, a la Skolem. Step $???$ extends the result of Steps $3$ and $4$ to arbitrary sets of such sentences. Avigad doesn't mention this step by name, instead relegating it to the end of Steps $3$ and $4$, but I think it's worth stating separately. However, this extension doesn't involve an application of Konig's lemma, just annoying bookkeeping. Step $5$ wraps everything up by observing that function symbols can be replaced by relation symbols and additional axioms saying they behave as graphs of functions, and that equality can be replaced with a binary relation together with new axioms for its handling (we just take an appropriate quotient structure at the end of the day). This is again entirely straightforward. As to whether that's what Godel meant by "nonfinitary reasoning," I'm not sure. Personally Skolem 1922 already feels pretty nonfinitary (in a good way) to me. My personal guess is that it's a bit more subtle than that: the missing philosophical ingredient was perhaps the realization that nonfinitary methods are relevant to finitary conclusions, even in foundational topics. As opposed to an elementary submodel, a proof is a fully finite object, and it was definitely surprising to me at least that infinitary reasoning could be used to demonstrate the existence of a proof (in fact, the completeness theorem itself deeply shocked me when I first learned it).<|endoftext|> TITLE: Proj construction in derived algebraic geometry QUESTION [12 upvotes]: The question My question is easy to state: Is there a Proj construction in derived geometry, that produces a derived stack from a “graded derived algebra”? Given the vagueness of the question, you’re free to interpret derived geometry in your favourite model for affines: (E-infinity/simplicial/dg)-algebras etc. Perhaps there’s some well-known answer. If not, below I’ll write about why I’m confused about existence of such a notion. The struggle A first obstacle (maybe only to me, as I am dumb) is to formulate a notion of derived graded algebras. This seems possible to do directly. It’s also maybe plausible that we can characterise such graded affine schemes as derived affine schemes that receive an action by the multiplicative group scheme, viewed as a 0-truncated derived scheme. I have not thought either of these through though, but for now let’s assume some notion of graded derived rings exist. At this point we can use a functor-of-points approach to define a notion of Proj: either in terms of maps into invertible projective modules, or as quotient prestack by the $G_m$ action. It’s not immediate (at least to me) that in the derived setting that these approaches produce equivalent results. A more major obstacle is that in the classical case, the way we show the functor of points for Proj is representable by a scheme is by constructing an explicit model for it by gluing affines along open subschemes. If we try to reproduce this argument in the infinity-categorical setting, these gluing diagrams requires an infinite amount of coherence data. This is much like the example outlined in introduction to DAG-XIV. Can we circumvent this by appealing to the Lurie-Artin theorem? REPLY [14 votes]: It is instructive to look at the simplest case of Proj: that of a free module, i.e. the projective space. Lurie works these out for us quite carefully in his Spectral Algebraic Geometry tome. Projective spaces in SAG 1. Projective space by gluing: In Section SAG.5.4, more specifically Construction SAG.5.4.1.3, he constructs a spectral algebraic scheme $\mathbf P^n_S$, following the classical gluing construction of homogeneous coordinates. It looks a little bit more funky due to $\infty$-categorical descent requiring specification of what happends on arbitrary intersections, as opposed to the classically story where double intersections, with a compatibility on triple ones, suffice (this is the infinite coherence data for gluing alluded to in the question). This projective space $\mathbf P^n_S$ is flat, base-changes along $S\to R$ to usual projective spaces we know and love over an ordinary ring $R$, and possess a "good theory" of Serre twisting sheaves $\mathscr O(n)$, e.g. Serre's calculation in FAC of their cohomology still holds. Its drawbacks: $\mathbf P^n_S$ is not smooth over $\operatorname{Spec} S$ (more precisely, it is fiber-smooth and is not differentially smooth), and it does not satisfy the expected universal property in terms of line bundles (for other than classical schemes). 2. Projective space by universal property: On the other hand, Subsection SAG.19.2.6 sees Lurie apply the Artin Representability Theorem to obtain the smooth projective space $\mathbf P^n_{\mathrm{sm}}$, that satisfies the expected universal property: a map of spectral schemes $X\to \mathbf P^n_{\mathrm{sm}}$ corresponds to a line bundle $\mathscr L$ on $X$ together with a map of quasi-coherent sheaves $\mathscr L\to\mathscr O_X^{n+1}$, which exhibits the splitting $\mathscr O^{n+1}_X\simeq \mathscr L\oplus\mathscr Q$. To check the requirements of Artin representability, Lurie uses the already-constructed $\mathbf P^n_S$, however the two spectral schemes do not coincide. The smooth projective space is smooth over $\operatorname{Spec} S$ (i.e. differentially smooth), but it is not flat. And while $\mathscr O(-1)$ is the universal bundle on $\mathbf P^n_{\mathrm{sm}}$, the cohomology of it and its twists is not controlled by Serre's computation anymore. 3. Summary: In the world of SAG there are two notions of the projective space, each satisfying some of the nice properties of projective spaces in classical AG. All that said, that really has nothing to do with projective spaces, but instead with affine ones: it is known that SAG admits two inequivalent notions of the affine space $\mathbf A^n$, one of which $\mathbf A^n_{\mathrm{sm}}$ is (differentially) smooth, and the other of which $\mathbf A^n_S$ is flat. The two projective spaces just correspond to using each of the two variants of affine space to build a projective one. It is explained in Lurie's thesis how requiring the two affine lines to coincide produces DAG from SAG, hence the projective space in derived algebraic geometry (e.g. built out of simplicial commutative rings, as opposed to $\mathbb E_\infty$-rings) will be as nice as you expect. Graded derived rings and Proj The question asks for a good notion of a graded derived $R$-algebra, and the suggestion in the comments was to just define them as affine derived $R$-schemes with a $\mathbf G_m$-action. That works, but it is also possible to imitate the usual classical definition of graded rings: 1. Classical definition of graded rings: A graded derived $R$-algebra is a lax symmetric monoidal functor $A:\mathbf Z\to \mathrm{Mod}_R$, where the LHS is the discrete category indexed by the integers (no non-identity morphisms) with the symmetric monoidal operation given by addition, and the RHS carries the symmetric monoidal structure of the derived relative tensor product $\otimes_R$. If we denote by $A_n$ the value of the functor $A$ on the object $n\in \mathbf Z$, then the colimit $\varinjlim A\simeq \bigoplus_{n\in \mathbf Z} A_n$ is the underlying commutative $R$-algebra. The lax symmetric monoidality translates to the usual definition of a commutative graded ring: the map $R\to A$, picking out the unit $1\in \pi_0A$, factors through the inclusion $A_0\to A$, and the multiplication on $A$ takes $A_m\otimes_R A_n\to A_{m+n}$. Phrasing things as an $\infty$-categorical functor just brings all the necessary homotopy-coherence along for the ride. If you wanted non-negatively graded derived $R$-algebras, you could require that $A_n\simeq 0$ for all $n$, but that amounts to the same thing as a symmetric monoidal functor $\mathbf Z_{\ge 0}\to\mathrm{Mod}_R$, everything else same as before. The relationship with the group scheme $\mathbf G_m$ and the monoid scheme $\mathbf A^1$, alluded to in the comments to the question, come from the fact that $\mathbf G_m = \operatorname{Spec} (R[\mathbf Z])$ and $\mathbf A^1 = \operatorname{Spec} (R[\mathbf Z_{\ge 0}])$. 2. Two options for $\mathbb E_\infty$-rings: This also gives another perspective on what "goes wrong" in SAG to produce two versions of Proj. There are two notions of a polynomial algebra over an $\mathbb E_\infty$-ring $R$: the free $R$-algebra $$R\{t\}=\operatorname{Sym}^*_R(R)\simeq R[\coprod_n B\Sigma_n]$$ and the polynomial $R$-algebra $$R[t] = R[\mathbf Z_{\ge 0}] = R[\coprod_n \mathrm{pt}].$$ Here $\coprod_n B\Sigma_n$, also known as the nerve of the category of finite sets with bijections, is the free $\mathbb E_\infty$-space, while its path-connected components $\mathbf Z_{\ge 0}$ only form the free commutative monoid. This leads to the two different affine lines $\mathbf A^1_{\mathrm {sm}}$ and $\mathbf A^1_S$, the difference btween $\operatorname{GL}_1$ and $\mathbf G_m$ over $\mathrm{Spec S}$, and finally to the two projective spaces. So while $\mathbf P^n_{\mathrm{sm}}$ has a universal property in terms of line bundles, i.e. $\operatorname{GL}_1$-torsors, the corresponding universal property for $\mathbf P^n_S$ would be about $\mathbf G_m$-torsors. Conversely, as the gluing construction for $\mathbf P^n_S$ starts from flat affine spaces $\mathbf A^n_S$, so would the one for $\mathbf P^n_{\mathrm{sm}}$ start from the smooth flat spaces $\mathbf A^n_{\mathrm{sm}}$. As quotient stacks, we have $\mathbf P^n_S\simeq (\mathbf A^n_S-\{0\})/\mathbf G_m$ and $\mathbf P^n_{\mathrm{sm}}\simeq (\mathbf A^n_{\mathrm{sm}}-\{0\})/\operatorname{GL_1}$. 3. Proj: Following the above discussion, you can develop two notions of Proj in the SAG setting (both of which will coincide in the DAG setting), depending on what kind of grading you feed in, of which the two projective spaces will be examples. Either could be defined equivalently via a gluing construction (specifying the classical base-affines via homogeneous localization in the classical construction of Proj) or via quotienting by the $\mathbf G_m$- or $\mathrm{GL_1}$-action respectively. The two constructions will agree under the usual condition on the graded derived ring $A$ (phrased purely on $\pi_0A$ as we expect) that elements in graded degree $1$ generate the irrelevant ideal. Note that this is not terrbily restrictive: even the EGA only really works with Proj in this setting. Just to sketch where the equivalence is coming from: fixing some generators $x_1, \ldots, x_n$ for the irrelevant ideal $\pi_0(A)^+$ in degree $1$, determines an open cover $\coprod_j \operatorname{Spec}(A[x_j^{-1}])\to \operatorname{Spec} A - V(A_+)$ of the complement of the closed subsheme of $\operatorname{Spec} A$ cut out by the irrelevant ideal $A_+ =\bigoplus_{n >0} A_n$. Since all the derived rings in sight are graded, this covering map is $\mathbf G_m$-(or $\mathrm{GL_1}$- resp.)equivariant, and passes to a cover of the quotient. Identifying the quotient of $\operatorname{Spec}\big(A[x_j^{-1}]\big)$ by $\mathbf G_m$ with the spectrum of the $0$-th graded part $(A[x_j^{-1}])_0$ (which goes by the name homogeneous localization in classical AG), we recover the "gluing construction" of $\operatorname{Proj}A$ as the Cech nerve of the open cover.<|endoftext|> TITLE: An entire function all whose forward orbits are bounded QUESTION [5 upvotes]: Edit: I revise the question according to the comment of Gabe Conant. What is an example of a non constant entire function $f:\mathbb{C}\to \mathbb{C}$ which satisfy the following?: For every $z\in \mathbb{C}$, the sequence $z,f(z),f^2(z),\ldots,f^n(z),\ldots$ is a bounded sequence but $f$ is not in the form $f(z)=\lambda z,\; |\lambda|\leq 1$. REPLY [10 votes]: Given an entire function $f\colon\mathbb{C}\to\mathbb{C}$, the escaping set, $I(f)$, is the set of $z\in\mathbb{C}$ such that $f^n(z)\to\infty$. Per the Wikipedia article, the escaping set of a non-linear entire function is nonempty. The reference for this is On the iteration of entire functions by Eremenko.<|endoftext|> TITLE: Practical example in using (homotopy) type theory QUESTION [15 upvotes]: I have just read Grayson's introduction on homotopy type theory as a possible foundation for mathematics. It is very enlightening about what all the fuss is about, but I am left with some doubts. Forgive my naiveté, I am not expert in type theory at all. When one uses ZFC as a foundation for mathematics, one evenutally moves to doing informal reasoning, knowing that it is always possible, although slightly tedious, to formalize all arguments. I am not sure about how to perform this step in HTT. Let me take the example of Grayson, who shows how to define a group in HTT. First, he defines a type $P$ is a proposition iff it has at most one element (that is, the type $\prod_{p : P} \prod_{q : P} p = q$ is inhabited) - which, if existent, is called a proof of $P$ a type $S$ ia a set iff for all $s: S$ and $t: S$ the type $s = t$ is a proposition Then a group is a tuple $(G, e, i, m, \alpha, \lambda, \rho, \lambda', \rho', \iota)$ where $G$ is a type $e: G$, $i: G \to G$, $m: G \times G \to G$ $\alpha$ is a proof of the identity stating associativity, and so on for $\lambda, \rho, \lambda', \rho'$ which prove left and right identity of $e$ and existence of left and right inverses $\iota$ is a proof that $G$ is a set (according to Grayson, it is a theorem that this fact is itself a proposition) So far, so good: I am inclined to believe that with a little ingenuity one can in fact define most (all?) usual concepts in mathematics. Now, I would like to go on proving theorems about groups. The simplest non-trivial theorem I can think of is Lagrange theorem: if $G$ is a finite group and $H < G$ a subgroup, then $|H|$ divides $|G|$. The intuitive proof is very short: for every coset $C$ of $H$ and any $c \in C$, multiplication by $c$ is a bijection $H \to C$, and $G$ is partitioned into cosets. It is clear to me how to formalize this in ZFC, but I would be lost trying to do that in HTT. I have two kind of doubts: 1) Do all steps work as well? It seems to me that one uses the notion of subset. Is there such a notion in HTT? I guess there are no subtypes, so it would seem no subsets as well. I guess that one can define a subgroup as a triple $(H, f, p)$ where $H$ is a group, $f \colon H \to G$ and $p$ is a proof that $f$ is injective. But eventually we are doing something like "counting elements" - is this justified? 2) Even logic has changed into this new setting: proof are themselves objects of this mathematical universe. I guess I need to have something like a constructive proof to make this work out? Does it matter that I have to choose arbitrary elements $c \in C$ to find a bijection? Sorry for the vague question: I am trying to understand what it would be like to do mathematics knowning in the back of your head that one can formalize thing into HTT, and it is not obvious to a beginner like me REPLY [12 votes]: The answers and comments on this question show that there is still a ton of misinformation out there about HoTT. The short answer (but much more time-consuming for you) is that you should read the HoTT Book, which was explicitly written to address this sort of question. To address the specific questions you asked, and the points raised in the other answers: HoTT is not a structural set theory. Nor is it a material set theory; it is a type theory, a different kind of beast altogether. For an explanation of the differences, see this blog post of mine. There absolutely is a notion of subset type (or subtype). The rules for subset types are explained in that blog post. Subsets of a type $A$ are usually encoded as their characteristic functions $P:A \to \rm Prop$. Given such a subset, one can always take its $\Sigma$-type $\sum_{x:A} P(x)$ which is a type with an injective function to $A$, but often it's easier and better to work directly with subsets than with such things. Of course, when you want a subset of a group to be itself a group, you'll need to consider it as a type in its own right, and you might find it easier to formulate Lagrange's theorem in terms of an arbitrary injective group homomorphism. Constructive logic is a red herring if what you're wondering about is what mathematics looks like in HoTT versus in ZFC. It's true that HoTT is "by default" constructive due to its inheritance from type theory, but one can simply add the axioms of excluded middle and choice to obtain a mathematics that looks just like classical mathematics in most respects. (If you do want to be constructive, then as others have pointed out you have to be careful about the meaning of "finite" -- but this would also be true if you were trying to use a constructive set theory like IZF, and is irrelevant to doing mathematics in HoTT+LEM+AC.)<|endoftext|> TITLE: A conjecture on partitions QUESTION [9 upvotes]: Let $p>q$ be two prime numbers. Let $\lambda$ be a partition whose $p$-core is $\lambda_p$ and $q$-core is $\lambda_q$. Assume that $|\lambda|>|\lambda_p|+|\lambda_q|$. Is it true that there always exists some partition $\mu\neq \lambda$ with $|\mu|=|\lambda|$, such that $\mu$ also has $p$-core $\lambda_p$ and $q$-core $\lambda_q$? I have verified this conjecture for partitions with small sizes. But I can't prove it in general. Any hint will be appreciated. REPLY [9 votes]: The answer is yes, and it follows from the results in my paper "A generalisation of core partitions", J. Combin. Theory 127. In that paper I define a class of partitions called $[p:q]$-cores. One characterisation of these partitions (Corollary 5.2 in the paper) is that $\lambda$ is a $[p:q]$-core if and only if $\lambda$ is the unique partition with size $|\lambda|$, $p$-core $\lambda_p$ and $q$-core $\lambda_q$ (in other words, if and only if there is no partition $\mu$ as asked for in the question). But in this case Lemma 4.8 of the paper says that the $q$-core of $\lambda_p$ coincides with the $p$-core of $\lambda_q$; call this partition $\lambda_{pq}$. And now Proposition 5.1 in the paper says that $|\lambda|=|\lambda_p|+|\lambda_q|-|\lambda_{pq}|$, contrary to hypothesis.<|endoftext|> TITLE: Abel-Ruffini theorem for systems of polynomial equations QUESTION [5 upvotes]: I know the Abel-Ruffini theorem, which states that a general polynomial equation in one variable with degree $\geq 5$ is not solvable in radicals. I wonder whether there is a similar result for systems of polynomial equations? Say I have a system \begin{cases} f_1(x_1,x_2,\ldots,x_n)=0\\ f_2(x_1,x_2,\ldots,x_n)=0\\ \vdots\\ f_n(x_1,x_2,\ldots,x_n)=0\\ \end{cases} where $f_i(x_1,x_2,\ldots,x_n)=0$ is a polynomial of some degree in the variables $x_1,\ldots,x_n$. (Just to clarify my notation, I am not restricting to homogeneous systems, as $f_i$ may contain also the constant term as well as mixed terms such as $x_1^2x_2^3x_8$ and so on.) My question is, what can I say about the solvability of this system in radicals? I would be tempted to say that if the total degree of the system is $\geq5$ then the system is not solvable in radicals, because a system of degree $\alpha$ corresponds be a single equation in one variable of degree $\alpha$, but I am not sure if it is right. How would you comment this issue? REPLY [2 votes]: (@VictorProtsak points out I was assuming the "total degree" $n$ referred to the maximum of the degrees of the polynomials. I'm not sure what the actual definition of "total degree" is but perhaps the only sensible definition is to take the degree in the scheme theoretic sense, i.e. the number of solutions if it is reduced. In which case the example below still applies:) To show equations are not "generally" solvable in radicals it suffices to show that there exists at least one specialization which is not solvable. So surely you immediately see that these equations cannot be solved for $d \ge 5$ by taking $f_1 = x^5_1 - x_1 - 1$ and then $f_i = x_i$ for $i \ge 1$. If the degree is $\le 4$ then the Galois group acts on the roots via a subgroup of $S_n$ for $n \le 4$ and so it will be solvable in that case. (If you bound the maximal degree) you can do better; if $n \ge 3$, then you can take $$f_1 = x_1 - x^2_2, \ f_2 = x_2 - x^2_3, \ f_3 = x^2_1 - x_3 - 1,$$ and $f_i = x_i$ for $i > 3$. Since this implies that $x^8_3 - x_3 - 1 = 0$, which is not solvable, so there can be no general solution in radicals for $d \ge 2$ if $n \ge 3$. For the remaining cases, $d = 1$ is always solvable using linear algebra. For $n = 2$ there are not general solutions for $d \ge 3$; you use the same idea as above by taking $f_1 = x_1 -x^3_2$ and $f_2 = x^3_1-x_2-1$. The last case is $n = d = 2$ which is solvable as one can see in a number of ways, for example, by using resultants to get equations in each single variable of degree $4$.<|endoftext|> TITLE: Finite-dimensional Hilbert $C^*$-modules QUESTION [6 upvotes]: Does there exist a classification, or characterization, of finite-dimensional Hilbert $C^*$-modules? More generally, does there exist a characterization of countable direct sums of finite-dimensional Hilbert $C^*$-modules Edit: By finite dimensional I mean that the $C^*$-module is finite dimensional as a vector space. REPLY [5 votes]: Let $E$ be a Hilbert $C^*$-module over a $C^\ast$-algebra $A$. (I'll use the right module convention, which basically just makes life difficult for myself.) Then $I$, the closure of $\newcommand{\lin}{\operatorname{lin}} \lin \{ (x|y):x,t\in E \}$ is a (two-sided) closed ideal in $A$. (Because, for example $(x|y)a = (x|y\cdot a)\in I$). Recall that $E$ is full when $A=I$. Now suppose that $E$ is a finite-dimensional vector space. Then $\lin \{ (x|y):x,y\in E \}$ is finite-dimensional, so equals $I$. As $I$ is itself a $C^\ast$-algebra, and is finite-dimensional, it in particular has a unit $e$. Then $I = eA = Ae = AeA$ and $A\rightarrow I, a\mapsto ae = eae = ea$ is a $*$-homomorphism which is the identity on $I$. It follows that $A \cong I \oplus B$ for some $C^\ast$-algebra $B$. We know that $E = E\cdot I$ and so $x = x\cdot e$ for each $x\in E$. We can hence treat $E$ is a full Hilbert $C^\ast$-module over $I$. We now use that $I$ is a direct sum of matrix algebras, say $I = \sum_{k=1}^N M_{n_k}$. Let $1_k$ be the unit of $M_{n_k}$ so $x = x\cdot e = x\cdot(\sum 1_k) = \sum_k x\cdot 1_k$ for $x\in E$, and so $E = \sum_{k=1}^N E_k$ where $E_k = E\cdot 1_k$ is a full module over $M_{n_k}$. Finally, we classify modules $F$ over $M_n$. One can show (I do not know a fast way to show this) that if $F$ is a finite-dimensional vector space with a (left) action of $M_n$ then $F \cong \mathbb C^n \otimes F'$ where $M_n$ acts on $\mathbb C^n$ in the usual way (treat $\mathbb C^n$ as column vectors) and the action on $F$ is as $a\otimes 1$. As $F$ is a right module over $M_n$, we have that $(\xi\otimes x')\cdot a = a^\top\xi \otimes x'$ where $a^\top$ is the transpose of $a\in M_n$. Let $\delta=\delta_1$ be the first unit vector in the standard basis of $\mathbb C^n$, so $a^\top\delta\otimes x'$ is a typical elementary vector in $F$. Now consider the $M_n$-valued inner-product on $F$ which is $$ (a^\top\delta\otimes x'|b^\top\delta\otimes y') = a^\ast (\delta\otimes x'| \delta\otimes y') b = a^\ast \theta(x'|y')b $$ say, for some sesquilinear $\theta:F\times F\rightarrow M_n$. However, notice that if $a^\top\delta=0$ or $b^\top\delta=0$ then $a^*\theta(x'|y')b=0$ for any $x',y'$. Let $\theta_{\xi,\eta}\in M_n$ be the rank-one operator $\alpha\mapsto \xi(\eta|\alpha)$. Further let $J:\mathbb C^n\rightarrow \mathbb C^n$ be the anti-linear operator given by pointwise complex conjugation. Then $a^\top = Ja^\ast J$. Thus, if $(\xi|\delta)=0$ or $(\eta|\delta)=0$, then $\theta_{\xi,\alpha}^\ast \theta(x',y') \theta_{\eta,\beta}=0$ for any $\alpha,\beta$, so $(\xi|\theta(x',y')\eta)=0$. It follows that $\theta(x',y') \in \mathbb C \theta_{\delta,\delta}$. So, there is a Hilbert space structure on $F'$ with $\theta(x',y') = (x'|y') \theta_{\delta,\delta}$. We conclude that the $M_n$-valued inner-product on $F=\mathbb C^n\otimes F'$ is $$ (\xi\otimes x'|\eta\otimes y') = (x'|y') \theta_{\eta,\xi} \in M_n. $$ So, this completely answers the case of a finite-dimensional module over $A$. There was also the question of countable direct sums of such things. I'm not sure what is meant here. If you know that $E$ is a countable direct sum $E\cong \oplus E_n$ each $E_n$ finite-dimensional, then you can piece together the structure for each $E_n$. However, if you want to know which $E$ can occur in this way, then that seems more complicated.<|endoftext|> TITLE: Definition of $E_n$-modules for an $E_n$-algebra QUESTION [8 upvotes]: The category $Mod^{E_n}_A(\mathcal{C})$ of $E_n$-modules for an $E_n$-algebra in a symmetric monoidal $\infty$-category $\mathcal{C}$ is defined in Lurie's Higher Algebra as a special case of a more general definition (just replace the little $n$-cubes operad with your favorite $\infty$-operad). The definition is rather obscure, however, and requires a lot of terminology I'm not entirely comfortable with. I was wondering if there is a more intuitive way to define/think about this category. For example, what's the best way to think of $E_n$-modules for an $E_n$-algebra in the category of spaces? There is also a proof in Lurie that $Mod^{E_1}_A(\mathcal{C})$ for an $E_1$-algebra $A$ is equivalent to the category of bimodules for $A$ in $\mathcal{C}$, but again the proof is quite long and difficult. Is there an intuitive way to think about this result? I'd also be happy if there was a short proof for $A$ an ordinary associative algebra. REPLY [10 votes]: $E_n$ algebras have compatible multiplications for every way of placing a bunch of elements into a collection of balls in $\mathbb{R}^n$. A module for an $E_n$ algebra has an action for every way of placing a bunch of elements of the algebra into balls not at the origin, and an element of the module into a ball at the origin. Here's what this looks like in a few examples to help you gain some intuition. $E_1$ algebras in vector spaces are ordinary (non-commutative) algebras. An $E_1$-module for $A$ is just an $A$-$A$ bimodule. I.e. there's actions for every way of placing another ball not at the origin (so either to the left or right) and you get an associativity axiom for every way of placing two balls not at the origin (both left is the left module axiom, both right is the right module axiom, and one on each side is the bimodule axiom). $E_2$ algebras in vector spaces are commutative algebras (because you can continuously deform two discs in the plane into the opposite order, so $xy = yx$). An $E_2$-module for $A$ is just an $A$-module (which is also automatically a right $A$-module by using commutativity to turn the left action into a right action). $E_1$ algebras in categories are monoidal categories. $E_1$-modules for monoidal categories are bimodule categories in the sense of Chapter 7 of tensor categories. I.e. you have bifunctors $\vartriangleright: A \times M \rightarrow M$ and $\vartriangleleft: M \times A \rightarrow M$ together with natural associators for each triple product (e.g. $\alpha_{a,b,m}: a \vartriangleright (b \vartriangleright m) \rightarrow (a \otimes b) \vartriangleright m)$ satsifying some hexagon axioms for quadruple products. $E_2$ algebras in categories are braided monoidal categories. Both horizontal and vertical composition are the monoidal structure, but the braiding tells you how 180-degree rotation gives a braiding iso $\beta_{x,y}: x \otimes y \rightarrow y \otimes x$ satisfying a hexagon axiom for triple tensor products. Note that you can either rotate clockwise or counterclockwise, yielding the braiding and its inverse, but these need not agree. Now let's think about $E_2$ modules. The action comes from a pair of discs, one of which is a marked disc at the origin. I.e. we want a left module category $M$. The important isotopy here is no longer the swap, but instead a full rotation around the marked disk (i.e. the generator of the affine braid group on one strand). So we need a full-twist isomorphism $\eta_{a,m}: a \vartriangleright m \rightarrow a \vartriangleright m$. But looking at three discs, this needs to satisfy a compatibility axiom related to the defining relation of the 2-strand affine braid group which states that $\eta_{a\otimes b,m}: (a \otimes b) \vartriangleright m \rightarrow (a \otimes b) \vartriangleright m$ is equal to the composite: $$(a \otimes b) \vartriangleright m \rightarrow a \vartriangleright (b \vartriangleright m) \rightarrow a \vartriangleright (b \vartriangleright m) \rightarrow a \vartriangleright (b \vartriangleright m) \rightarrow (a \otimes b) \vartriangleright m \rightarrow (b \otimes a) \vartriangleright m \rightarrow (a \otimes b) \vartriangleright m$$ where the first and fourth maps are associators and the second and third maps are the full-twists $\mathrm{id}_a \vartriangleright \eta_{b,m}$ and $\eta_{a, b\vartriangleright m}$ (or possibly vice-versa), and the fifth and sixth maps are both the braiding (or possibly both the inverse braiding). This structure is called a "braided module category" by Enriquez (see also Brochier) and the translation between $E_2$-modules and braided module categories is given in Theorem 3.11 of Ben-Zvi-Brochier-Jordan. $E_3$ algebras in categories are symmetric monoidal categories. I.e. braided tensor categories where the braiding and the inverse braiding agree. Similarly, for $E_3$-modules the full-twist maps are automatically trivial by untying them in the extra dimension. So an $E_3$-module for a symmetric tensor category is just a module category. (Note that in vector spaces $E_2$-algebras are $E_\infty$-algebras, while in categories $E_3$-algebras are $E_\infty$-algebras. So in both cases we see that $E_\infty$-modules for $E_\infty$-algebras agree with the usual notion of left-module = right-module, but below the stable range the notion is subtler and is a bimodule with the appropriate amount of compatibility between the left and right module structures.)<|endoftext|> TITLE: Zariski closure of set of units in a number ring QUESTION [6 upvotes]: Let $\mathcal{O}$ be a number ring. Letting $r$ and $2s$ be the number of real and complex embeddings of $\mathcal{O}$, the number ring $\mathcal{O}$ is a lattice in $\mathcal{O} \otimes \mathbb{R} \cong \mathbb{R}^{r+2s}$, and thus is Zariski dense in this vector space. Question: What is the Zariski closure of the group $\mathcal{O}^{\times}$ in $\mathcal{O} \otimes \mathbb{R} \cong \mathbb{R}^{r+2s}$? There is one obvious polynomial condition coming from the fact that the norm is $\pm 1$. Are there others? REPLY [5 votes]: Let $x_1,\dots, x_n$ be the embeddings of $\mathcal O$ into $\overline{\mathbb Q}$. Every monomial in $x_1,\dots, x_n$ restricts to a character of $\mathcal O^\times$. If some polynomial in $x_1,\dots, x_n$ vanishes on units, then we get a linear relation among characters of $\mathcal O^\times$, unless two of the monomials in $x_1,\dots, x_n$ appearing in the polynomial give the same character of $\mathcal O^\times$. Thus all relations are generated by the relations of the form $\prod_{i=1}^n x_i^{d_i} - \prod_{i=1}^n x_i^{e_i}$ where these two monomials give the same character or equivalently by $1- \prod_{i=1}^n x_i^{e_i}$ for $e_i \in \mathbb Z$ where this monomial gives the trivial character. Each complex conjugation element of the Galois group defines a permutation of the set $\{1,\dots,n\}$ of embeddings. If, for all units, we have $$ 1= \prod_{i=1}^n x_i^{e_i}$$ then we have $$ 1 = \prod_{i=1}^n x_i^{e_i} \overline{ \prod_{i=1}^n x_i^{e_i} } = \prod_{i=1}^n |x_i|^{ 2e_i } .$$ By Dirichlet's unit theorem, this can only happen if all the exponents on each real place are equal, and all the exponents on each complex place are double that. In other words, for each real place $i$ we must have $e_i=n$ for some constant $n$, and for each complex place $i$ we must have $e_i+ e_{\sigma{i}}=2n$, so altogether we must have $e_i + e_{\sigma(i)}=2n$ for all $i$. This can only be no constant if the graph whose edges are $(i, \sigma(i))$ for all complex conjugations $\sigma$ has a bipartite connected component. In this case, because the graph is Galois-invariant, every component is bipartite, and because each connected graph is bipartite in at most one way, the graph of this set of functions is the number of components plus one. We have to convert this information back into Galois theory, and then number theory. Because each vertex of the graph defines a part of its connected component, the stabilizer of a part contains the vertex stabilizer, hence corresponds to a subfield. Because each complex conjugation sends one part to the other part of the same graph, all complex conjugations in the Galois group of this field are equal, so it is a CM field. Moreover, the number of complex places of this field is the number of components (orbit-stabilizer theorem). So we can see that the number of relations is at most the number of complex places of the maximum CM subfield, plus one. Is this exactly equal to the number of relations? Yes, because in such a CM field $K$ with totally real subfield $L$, a finite index subgroup of the unit group is contained in $L$, and so we have the one relation from $L$ plus $[L: \mathbb Q]$. Then we can pull these relations back to any extension of $K$ by the norm map. Thus, for $\mathcal O$ the ring of integers of a number field $F$, $K$ the maximal CM subfield, $L$ its maximal totally real subfield, the Zariski closure of the units is the inverse image under $N_{F}^K$ of the Zariski closure of the units of $K$, which is a finite union of cosets of the norm one torus of $L$.<|endoftext|> TITLE: Existence of algebraic integers with certain properties QUESTION [5 upvotes]: Is the following statement true? ($\star$) Given integers $n > k > 0$, there exists a monic polynomial of degree $n$ with integer coefficients and constant term $\pm 1$, irreducible over $\mathbb{Z}[x]$, whose roots are all real and simple, with exactly $k$ roots of absolute value $<1$ and $n-k$ roots of absolute value $>1$. If $k=n-1$ then the answer is yes: in this old paper by Vijayaraghavan there is an explicit example of such a polynomial (see the first proof of Theorem 1; detailed calculations are omitted). It seems possible to tweak Vijayaraghavan's construction to other values of $k$, but the calculations are somewhat ugly. I wonder whether statement ($\star$) has a cleaner proof and/or is already known. Remark: If $k=n-1$ then the unique root of absolute value $>1$ is a Pisot–Vijayaraghavan number. REPLY [2 votes]: @orthodontist has already given a clean and educated answer to my question. Nevertheless let me include an elementary answer, tweaking Vijayaraghavan's construction. The calculations are not terribly ugly after all. Fix integers $$ \ell_1>\cdots>\ell_{n-k}>0>\ell_{n-k+1}>\cdots>\ell_n $$ whose sum is $0$ and such that $\ell_i-\ell_{i+1}\ge 2$ for each $i\in\{1,\dots,n-1\}$. Let $e_0:=0$ and $e_i:=\ell_1+\cdots+\ell_i$ for $i\in\{1,\dots,n\}$. Let $b \ge 3$ be another integer. I claim that the polynomial $$ P(x) := \sum_{i=0}^n (-1)^i b^{e_i} x^{n-i} $$ has the required properties $(\star)$ (except perhaps for irreducibility). To begin, note that $e_i \ge 0$ for each $i$ and so $P$ has integer coefficients. Furthermore, it is monic, has constant term $(-1)^n$. In order to locate the roots of $P$, fix integers $c_1, \dots, c_{n-1}$ such that $$ \ell_1>c_1>\ell_2>c_2>\cdots>c_{n-1}>\ell_n \quad \text{and} \quad c_{n-k}=0. $$ Fix $j\in\{1,\dots,n-1\}$. I claim that in the expression $$ P(b^{c_j}) = \sum_{i=0}^n (-1)^i b^{e_i + (n-i)c_j} , $$ the dominant term corresponds to $i=j$. Indeed, the sequence of integers $$i \in \{0,\dots,n\} \mapsto e_i + (n-i)c_j$$ is increasing for $i$ in the interval $\{0,\dots,j\}$, and decreasing afterwards. Using the fact that $2\sum_{r=1}^\infty b^{-r} \le 1$ (i.e. $b \ge 3$) we conclude that the sign of $P(b^{c_j})$ is $(-1)^j$. By the intermediate value theorem, $P$ contains a root in each of the intervals $$ (0, b^{c_{n-1}}), \ (b^{c_{n-1}},b^{c_{n-2}}), \ \dots, \ (b^{c_2},b^{c_1}), \ (b^{c_1},+\infty) \, . $$ Since $b^{c_{n-k}}=1$, we conclude that $P$ has $k$ simple roots on the interval $(0,1)$ and $n-k$ simple roots on the interval $(1,+\infty)$. EDIT: Getting irreducibility (Thanks @GabeConant for pointing the error of the previous "argument".) Let $z_1>\dots>z_n$ be the roots of $P$. The construction actually ensures that: $$ b^{\ell_i - 1} < z_i < b^{\ell_i + 1} \quad \text{for each } i. $$ Obviously, $\prod_{i=1}^n z_i = 1$. Assume that $P$ is reducible. Then there exists a nonempty proper subset $I \subset \{1,\dots,n\}$ such that: $$ \prod_{i \in I} z_i = 1. $$ (Indeed, if $P=QR$ is a non-trivial factorization of $P$ then $\prod_{Q(z)=0} z$ and $\prod_{R(z)=0} z$ must be positive integers and so must equal $1$.) Therefore: $$ \left| \sum_{i \in I} \ell_i \right| < n \, . $$ Now, we have lots of freedom in the choice of the numbers $\ell_i$, and it is certainly possible (though no clean argument occurs to me right now) to choose them so that $\left| \sum_{i \in I} \ell_i \right| \ge n$ for each nonempty proper subset $I \subset \{1,\dots,n\}$. In this way we can guarantee that $P$ is irreducible.<|endoftext|> TITLE: A curious relation between angles and lengths of edges of a tetrahedron QUESTION [89 upvotes]: Consider a Euclidean tetrahedron with lengths of edges $$ l_{12}, l_{13}, l_{14}, l_{23}, l_{24}, l_{34} $$ and dihedral angles $$ \alpha_{12}, \alpha_{13}, \alpha_{14}, \alpha_{23}, \alpha_{24}, \alpha_{34}. $$ Consider solid angles \begin{split} &\Omega_1=\alpha_{12}+\alpha_{13}+\alpha_{14}-\pi \\ &\Omega_2=\alpha_{12}+\alpha_{23}+\alpha_{24}-\pi \\ &\Omega_3=\alpha_{13}+\alpha_{23}+\alpha_{34}-\pi \\ &\Omega_4=\alpha_{14}+\alpha_{24}+\alpha_{34}-\pi \\ \end{split} and perimeters of faces \begin{split} &P_1=l_{23}+l_{34}+l_{24} \\ &P_2=l_{13}+l_{14}+l_{34} \\ &P_3=l_{12}+l_{14}+l_{24} \\ &P_4=l_{12}+l_{23}+l_{13}. \\ \end{split} Then the following cross-ratios are equal to each other: $$ [e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}]=[P_1, P_2, P_3, P_4]. $$ Question: Is it known? I have found a proof of this statement (see here), but it involves quite tricky algebraic geometry. It will be very interesting to me to see a more elementary approach. Addition: Similar statements hold in spherical and hyperbolic geometry. For a spherical tetrahedron $$ [e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}]=[e^{iP_1}, e^{iP_2}, e^{iP_3}, e^{iP_4}]. $$ For a hyperbolic tetrahedron $$ [e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}]=[e^{P_1}, e^{P_2}, e^{P_3}, e^{P_4}]. $$ Addition 2: One can prove a more general statement, which I formulate in the spherical case (but it is true in other geometries after appropriate modifications). There exists a $PSL_2(\mathbb{C})-$ transformation, sending eight numbers $$ 1, e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}, e^{i(\alpha_{12}+\alpha_{23}+\alpha_{34}+\alpha_{14})}, e^{i(\alpha_{12}+\alpha_{24}+\alpha_{34}+\alpha_{13})}, e^{i(\alpha_{13}+\alpha_{23}+\alpha_{24}+\alpha_{14})} $$ to $$ 1, e^{iP_1}, e^{iP_2}, e^{iP_3}, e^{iP_4}, e^{i(l_{12}+l_{23}+l_{34}+l_{14})}, e^{i(l_{12}+l_{24}+l_{34}+l_{13})}, e^{i(l_{13}+l_{23}+l_{24}+l_{14})}. $$ I know neither an elementary proof of this statement nor any interpretation of the coefficients of the $PSL_2(\mathbb{C})-$ transformation. REPLY [17 votes]: Here goes the elementary proof of the claim by Robert Houston that the quadraples $(P_1^{-1},P_2^{-1},P_3^{-1},P_4^{-1})$ and $(\cot \frac{\Omega_1}2,\cot \frac{\Omega_2}2,\cot \frac{\Omega_3}2,\cot \frac{\Omega_4}2)$ are affinely equivalent. In the spherical case the first quadraple should be replaced to $\{\cot\frac{P_i}2\}$, in the hyperbolic case to hyperbolic cotangents. Further the tetrahedron $A_1A_2A_3A_4$ is assumed to be generic (that implies the general case automatically.) I write the solution in Euclidean case, but the changes in spherical/hyperbolic case are almost straightforward, since the basic construction works in all three geometries. At first, we note that if the quadraples $(a_1,a_2,a_3,a_4)$ and $(b_1,b_2,b_3,b_4)$ of reals satisfy $(a_1-a_2):(a_3-a_4)=(b_1-b_2):(b_3-b_4)$ and two other analogous relations, then they are affinely equivalent. Indeed, without loss of generality $a_4=b_4=0,a_3=b_3=1$, $a_1\leqslant a_2$ and we have $a_1-a_2=b_1-b_2$, $1+(a_1-a_2-1)/a_2=(a_1-1)/a_2=(b_1-1)/b_2=1+(b_1-b_2-1)/b_2$, $a_2=b_2$, $a_1=b_1$. So we have to prove (call it equation $(\star)$) that $$ \frac{P_1^{-1}-P_3^{-1}}{\cot \frac{\Omega_1}2-\cot \frac{\Omega_3}2}= \frac{P_3-P_1}{\sin \frac{\Omega_3-\Omega_1}2}\cdot\frac{\sin \frac{\Omega_1}2 \sin \frac{\Omega_3}2}{P_1P_3} $$ equals to $$ \frac{P_4-P_2}{\sin \frac{\Omega_4-\Omega_2}2}\cdot\frac{\sin \frac{\Omega_2}2 \sin \frac{\Omega_4}2}{P_2P_4}. $$ In order to prove it we draw three planes: through $A_2$ orthogonal to the external bisector of $\angle A_1A_2A_3$, through $A_3$ orthogonal to the external bisector of $\angle A_2A_3A_4$, through $A_4$ orthogonal to the external bisector of $\angle A_3A_4A_1$. Let them meet at point $I$, and let $Q_1,Q_2,Q_3,Q_4$ be projections of $I$ onto lines $A_1A_2,A_2A_3,A_3A_4,A_4A_1$ respectively (see the picture). We have $IQ_1=IQ_2=IQ_3=IQ_4=:r$ and also $A_2Q_1=A_2Q_2$, $A_3Q_2=A_3Q_3$, $A_4Q_3=A_4Q_4$ by construction (the segments are directed in the natural sense: either $Q_1,Q_2$ both belong to rays $A_2A_1,A_2A_3$ respectively, or both do not, and so on). Also the right triangles $\triangle A_1IQ_1,\triangle A_1IQ_4$ are equal by hypotenuse and cathetus. Thus $A_1Q_1=A_1Q_4$. It implies that $$A_1A_2+A_3A_4=A_1Q_1+A_2Q_1+A_3Q_3+A_4Q_3=\\A_1Q_4+A_2Q_2+A_3Q_2+A_4Q_4=A_1A_4+A_2A_3,$$ which is not quite what you need but also an interesting relation. Well, to be serious, of course $Q_1Q_4$ is parallel to the internal, not external, bisector of $\angle A_2A_1A_4$: Denoting $x_i=A_iQ_i$ we find (on this picture, in general we should consider directed segments) $x_2-x_1=l_{12},x_2+x_3=l_{23},x_3+x_4=l_{34},x_4+x_1=l_{14}$. Thus $2x_{3}=l_{23}+l_{34}-l_{12}-l_{14}=P_1-P_3$, $2x_1=l_{14}+l_{23}-l_{12}-l_{34}$. Analogously for dihedral angles we have four relations like $\angle(A_1A_2A_3,A_1A_2I)=\angle(A_1A_2A_3,A_2A_3I)=:\beta_4$, $\angle(A_2A_3A_4,A_2A_3I)=\angle(A_2A_3A_4,A_3A_4I)=:\beta_1$, $\angle(A_3A_4A_1,A_3A_4I)=\angle(A_3A_4A_1,A_4A_1I)=:\beta_2$, $\angle(A_1A_2A_4,A_4A_1I)=\pi-\angle(A_1A_2A_4,A_1A_2I)=:\beta_3$ (suppose that $I$ lies inside the tetrahedron, otherwise carefully use directed angles: I am going to divide by 2 that is dangerous for the angels which are considered remainders modulo $\pi$, or consider several cases or use "generic case" abstract nonsense reasoning or whatever). We get $\beta_4+\pi-\beta_3=\alpha_{12}$,$\beta_1+\beta_4=\alpha_{23}$,$\beta_2+\beta_1=\alpha_{34}$,$\beta_3+\beta_2=\alpha_{41}$; $2\beta_1=\alpha_{23}+\alpha_{34}+\pi-\alpha_{12}-\alpha_{41}=\pi+\Omega_3-\Omega_1$, $2\beta_3=\alpha_{23}+\alpha_{14}+\pi-\alpha_{12}-\alpha_{34}$. Let $H$ be a projection of $I$ onto the plane $A_2A_3A_4$, then $A_3H$ is the internal bisector of $\angle A_2A_3A_4$. We get $Q_2H=r\cos\beta_1=r\sin \frac{\Omega_1-\Omega_3}2$, $\frac{P_1-P_3}2=x_3=Q_2H\cot \frac{\angle A_2A_3A_4}2$. Therefore $$ \frac{P_1-P_3}{2\sin \frac{\Omega_1-\Omega_3}2}=r\cot\frac{\angle A_2A_3A_4}2, $$ analogously considering the vertex $A_1$ instead of $A_3$ we get $$ \frac{l_{14}+l_{23}-l_{12}-l_{34}}{2\sin \frac{\alpha_{12}+\alpha_{34}-\alpha_{23}-\alpha_{14}}2}=r\tan \frac{\angle A_2A_1A_4}2. $$ So we may exclude $r$ from the formulae and get $$ \frac{P_1-P_3}{2\sin \frac{\Omega_1-\Omega_3}2}= \frac{l_{14}+l_{23}-l_{12}-l_{34}}{2\sin \frac{\alpha_{12}+\alpha_{34}-\alpha_{23}-\alpha_{14}}2}\cdot \cot \frac{\angle A_2A_3A_4}2 \cot \frac{\angle A_2A_1A_4}2. $$ Analogously $$ \frac{P_2-P_4}{2\sin \frac{\Omega_2-\Omega_4}2}= \frac{l_{14}+l_{23}-l_{12}-l_{34}}{2\sin \frac{\alpha_{12}+\alpha_{34}-\alpha_{23}-\alpha_{14}}2}\cdot \cot \frac{\angle A_1A_2A_3}2 \cot \frac{\angle A_1A_4A_3}2. $$ Therefore $(\star)$ reads as $$ \frac{\sin \frac{\Omega_1} 2\sin \frac{\Omega_3} 2}{P_1P_3}\cot \frac{\angle A_2A_3A_4}2 \cot \frac{\angle A_2A_1A_4}2= \frac{\sin \frac{\Omega_2} 2\sin \frac{\Omega_4} 2}{P_2P_4}\cot \frac{\angle A_1A_2A_3}2 \cot \frac{\angle A_1A_4A_3}2. $$ This immediately follows from the following "sine theorem" for the tetrahedron: the expression $$ \frac{\sin \frac{\Omega_1} 2 \sqrt{\cot \frac{\angle A_2A_1A_3}2 \cot \frac{\angle A_2A_1A_4}2\cot \frac{\angle A_3A_1A_4}2} } {P_1\sqrt{\tan \frac{\angle A_2A_3A_4}2 \tan \frac{\angle A_2A_4A_3}2 \tan \frac{\angle A_3A_2A_4}2}} $$ (denote it $\eta_1$) equals to analogous expressions for other indices. For proving it we denote $S_1$ and $r_1=2S_1/P_1$ the area and inradius of $\triangle A_2A_3A_4$, then we have $$ P_1^2\tan \frac{\angle A_2A_3A_4}2\tan \frac{\angle A_3A_2A_4}2 \tan \frac{\angle A_2A_4A_3}2=P_1^2\cdot\frac{r_1^3}{(P_1/2-l_{24}) (P_1/2-l_{34})(P_1/2-l_{23})}=\\ \frac{(2S_1)^3}{4(P_1/2)(P_1/2-l_{24}) (P_1/2-l_{34})(P_1/2-l_{23})}=\frac{8S_1^3}{4S_1^2}=2S_1 $$ by Heron formula. Next, using the Cagnoli formula (see p. 101 here) $$\sin \frac{\Omega_1}2=\sin \alpha_{12}\frac{\sin \frac{\angle A_3A_1A_2}2\sin \frac{\angle A_4A_1A_2}2}{\cos \frac{\angle A_3A_1A_4}2}$$ we get $$\sin^2 \frac{\Omega_1}2 \cot \frac{\angle A_2A_1A_4}2\cot \frac{\angle A_2A_1A_3}2\cot \frac{\angle A_3A_1A_4}2 =\sin^2 \alpha_{12}\frac{\sin \angle A_3A_1A_2\sin \angle A_4A_1A_2}{2\sin \angle A_3A_1A_4}.$$ So, $\eta_1^2=\eta_2^2$ reads as $$ \frac{\sin \angle A_3A_1A_2\sin \angle A_4A_1A_2}{S_1\cdot \sin \angle A_3A_1A_4}= \frac{\sin \angle A_3A_2A_1\sin \angle A_4A_2A_1}{S_2\cdot \sin \angle A_3A_2A_4}. $$ Substituting $2S_2=l_{13}\cdot l_{14}\cdot \sin \angle A_3A_1A_4$, $2S_1=l_{23}\cdot l_{24}\cdot \sin \angle A_3A_2A_4$ this reduces to a product of two sine laws, and everything is proved. Note that we had the expression $l_{12}-l_{23}+l_{34}-l_{41}$ above, and if you replace the internal bisectors to external bisectors you may also get $l_{12}+l_{23}+l_{34}+l_{41}$ that should answer to your extended question.<|endoftext|> TITLE: Prove that this expression is greater than 1/2 QUESTION [6 upvotes]: Let $0 TITLE: Real orthogonal and sign QUESTION [6 upvotes]: I came across the following conjecture, reading a recent paper in the Monthly, an orthogonal matrix of order $n\neq 0 \pmod 4$ has a nonnegative (up to a scalar) row vector. It should be straight in dimensions $2$ and $3$. REPLY [12 votes]: Here's a simple argument that this is false for $n > 2$. In dimension $n$ there are $2^{n}$ orthants, $2^{n-1}$ if one considers them modulo sign. A pair of antipodal orthants means a pair of orthants whose sign patterns are exactly opposite (e.g. the positive orthant union the negative orthant). The key point is that an orthonormal frame contains $n$ vectors, no two of which lie in the same pair of antipodal orthants. If two nonzero vectors lie in the interior of the same orthant, then their inner product is positive because it is a sum of nonnegative numbers not all of which are zero. (If they lie on the boundary of the same orthant, a small rotation moves them to different orthants). The original "conjecture" supposes that necessarily one of the vectors of an orthonormal frame lies in a particular orthant. However, since $2^{n-1} > n$ if $n > 2$, necessarily some pair of antipodal orthants contains no vector of the frame (pigeonhole principle). Since permutations of the coordinates are orthogonal transformations permuting the orthants (and preserving antipodal orthants), given a particular orthant there is always an orthonormal frame containing no vector of the orthant. (Note that the preceding is a bit vague in that it is not specified whether orthants are open or closed; since the goal is to build a counterexample, this does not matter much.)<|endoftext|> TITLE: Replacing the Fibre of a Fibration QUESTION [5 upvotes]: This was a question I first asked on stack exchange, here. In my head it seems like a fairly reasonable thing to ask for, but I'm not aware of any construction in the literature. Let $p:E\rightarrow X$ be a Serre fibration over a pointed, connected CW complex $X$ with strict fibre a CW complex $F=p^{-1}(\ast)$. Given another space $F'$ and a homotopy equivalence $F\simeq F'$, is it possible to construct a Serre fibration $p':E'\rightarrow X$ with strict fibre $F'=p'^{-1}(\ast)$? If $E'$ exists, is it then possible to extend the homotopy equivalence $F\simeq F'$ to a fibre homotopy equivalence $E\simeq E'$ over $X$? If you assume some extra structure on the fibration $p$ and the homotopy equivalence $F\simeq F'$ then things work out fairly easily, but I'm interested in the more general case. Really I would like both questions to be taken together, since as pointed out in the comments, the projection $X\times F'\rightarrow X$ trivially satisfies the requirements of the first question alone, and such an answer is not exactly what I was looking for. REPLY [5 votes]: You can make the construction as follows. There are three steps. First make a Serre fibration over the unit interval, pi:T->I=[0,1] such that the fibre over 0 (resp. 1) is F (resp. F'). You will find how to make this in the general case; for example, for F=1 point and F'=the interval, here is a proof that the triangle does it. Consider in the plane the triangle T: 0<=y<=x<=1 and the projection pi(x,y)=x. Given a CW-complex A, a continuous map from A to T:a->(x_0(a),y_0(a)) and a homotopy (x_t(a)) (0<=t<=1), consider K={(a,t)\in AxI/x_t(a)=0} and K_0=(Ax0)\cap K. On the complement (Ax0)\K_0 you have the slope function s_0(a):=y_0(a)/x_0(a); extend it to a continuous function s from (AxI)\K to I; the wanted homotopy is (a,t)->(x_t(a),s(a,t)x_t(a)) for (a,t)\notin K and (a,t)->(0,0) for (a,t)\in K. Hence, pi is a Serre fibration. Second, let $f:X\rightarrow X$ be homotopic to the identity and contract some neighborhood $N$ of $*$ onto $*$. Pulling back your fibration $p$ through $f$, you are reduced to the case where $p$ is a projection $F\times N\rightarrow N$ over $N$. Third, consider a function $g: N\rightarrow I$ whose value is $0$ on $\partial N$ and $1$ on $*$; define $E'$ over $N$ as the amalgamated product of $T$ with $N$ over $pi$ and $g$; define $E'$ as $E$ over $(X-N)$. If you have chosen $N$ to retract by deformation on $*$, then this $E'$ will have the property you want. Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes)<|endoftext|> TITLE: Proofs of Young's inequality for convolution QUESTION [20 upvotes]: For $1\leq p,q \leq \infty$ such that $\frac1p +\frac1q\geq 1$, Young's inequality states $\|f\star g\|_r\leq \|f\|_p\|g\|_q$ (we work on $\mathbf{R}^d$ here), where $1+\frac1r = \frac1p+\frac1q$. Equivalently \begin{align*} \|f\|_p=\|g\|_q=\|h\|_{r'}=1\Rightarrow \int_{\mathbf{R}^d}\int_{\mathbf{R}^d}f(x)g(y)h(x+y)\,\mathrm{d}x\,\mathrm{d}x \leq 1. \end{align*} The most elementary proof that I know is based on the (generalized) Hölder inequality on $\mathbf{R}^d\times\mathbf{R}^d$ (for three functions), applied on three "mixing" functions $\varphi(x,y)^a \psi(x,y)^b$ where $\{\varphi,\psi\}$ runs over the possible pairs of $\{(x,y)\mapsto f(x); (x,y)\mapsto g(y) ; (x,y)\mapsto h(x+y)\}$ and $a$ and $b$ are adequately chosen. There is of course a way to guess the correct exponents, but I find this proof a bit tedious and, when it comes to teach it, a bit articial ("consider these three functions and ... the magic happens"). Instead, I am wondering if it is possible to prove it in a different way, remaining at the same level of knowledge. The relation between $p,q,r$ rewrites $\frac{1}{r'} = \frac{1}{p'}+\frac{1}{q'}$. This, together with Hölder inequality, proves that any element in $L^{r'}(\mathbf{R}^d)$ is the (ponctual) product of two elements respectively in $L^{p'}(\mathbf{R}^d)$ and $L^{q'}(\mathbf{R}^d)$. Can we use this to prove (something like) \begin{multline*} \sup_{\|h\|_{r'}=1} \int_{\mathbf{R}^d}\int_{\mathbf{R}^d} f(x)g(y) h(x+y)\,\mathrm{d} x\,\mathrm{d} y \\ \leq \sup_{\|\varphi\|_{q'}=1,\|\psi\|_{p'}=1} \int_{\mathbf{R}^d}\int_{\mathbf{R}^d} f(x)g(y) \varphi(x)\psi(y)\,\mathrm{d} x\,\mathrm{d} y\quad ? \end{multline*} I did not succeed but still feel that the correspondance between the exponents in the convolution and poncutal products is not a coincidence. Note that using (a bit of) interpolation theory (I did not check in details) : Young's inequality can be obtained by Fourier transform (precisely using $\widehat{f\star g}=\widehat{f}\widehat{g}$), at least for exponents in $[1,2]$ and then all the other ones by a duality argument. The case $\{p,q\}=\{1,\infty\}$ is straightforward and by a duality argument it is possible to recover then $\{p,q\}=\{1,r\}$, and then an interpolation argument should recover some intermediate exponents. However, I'd really much appreciate a proof without interpolation. REPLY [2 votes]: Thanks Daniele and Willie for these nice answers. Willie : I tried this doubling variable thing but got stuck : I was writing $|h(x+y)|$ as the product of two elements respectively in $L^\infty_y(L^{p'}_x)$ and $L^\infty_x(L^{q'}_y)$, which was not the good point of view, nice insight that you got there ! Since the answer of Daniele came first and started the whole thing, I vote for him. I found yet another way to present the proof, which is somehow linked to your answers. As I was writing in my original post, the "easy" case $L^\alpha \star L^{\alpha'}$ implies the $L^\alpha\star L^1$ one by duality because of the formula \begin{align*} \int_{\mathbf{R^d}} (f\star g) h = \int_{\mathbf{R^d}} f(g\star \check{h}). \end{align*} Now as usual, w.l.o.g. we can assume $f,g\geq 0$ with $\|f\|_p=\|g\|_q=1$. The case $L^r \star L^1$ shows that $f^p \star g^{q/r}$ has $L^r(\mathbf{R}^d)$ norm less than $1$ and the same thing holds for $f^{p/r}\star g^q$. Thanks to Hölder's inequality, this means that we only need to prove a.e. for some $\theta\in[0,1]$ \begin{align*} f\star g \leq (f^p \star g^{q/r})^\theta (f^{p/r} \star g^{q})^{1-\theta}, \end{align*} and another use of Hölder's inequality allows to reduce this to proving a.e. \begin{align*} f(x-y)g(y) \leq (f(x-y)^p g(y)^{q/r})^\theta (f(x-y)^{p/r}g(y)^{q})^{1-\theta}, \end{align*} and in fact we even have an equality of this type. Indeed $p\in[1,r]$ so we have $\frac1p = \theta + \frac{1-\theta}{r}$ for some $\theta\in[0,1]$ and adding $\frac1q$ on both sides shows that $1-\theta+\frac{\theta}{r} = \frac{1}{q}$.<|endoftext|> TITLE: Injectivity of a class of integral operators QUESTION [7 upvotes]: Given a probability measure $\mu$ on the interval $[0,1]$, the linear operator $$ T_\mu \! f(y) := \int_0^1 f(yx) \, d\mu(x) $$ takes the space of continuous functions $f: [0, \infty) \rightarrow \mathbb{R}$ such that $f(0) = 0$ to itself. Assuming that $\mu$ is not the delta function at the origin, is this operator injective? For some measures this is really easy to show. For instance, if $d\mu(x) = g(x)|dx|$ with $g$ a homogeneous function. Remarks. $T_\mu$ acts like an invertible diagonal matrix on polynomials of a given degree. If a function $g$ is in the kernel of $T_\mu$, then so are the functions $x \mapsto g(\lambda x)$ $(0 \leq \lambda < \infty)$. Although the only requirement on $\mu$ is that it is not the delta at zero, in the applications I have in mind $\mu$ is positive almost everywhere. In the original question (addressed in Robert Israel's answer) the domain of the functions was $[0,1]$ instead of $[0,\infty)$. If there is an advantage in dealing with the Banach space of continuous function on $[0,1]$ (I thought there might be), then just add the hypothesis that $\mu$ is positive almost everywhere. At first I thought this was going to be easy, but I've been blocked for a while. It may still be easy though. REPLY [6 votes]: If the support of $\mu$ is contained in $[0, b]$ for some $b \in (0,1)$, then $T_\mu f = 0$ for any $f$ that is $0$ on $[0,b]$, so it is not injective. EDIT: Another example, where the support of $\mu$ is all of $[0,1]$: $d\mu(x) = g(x)\; dx$ where $g(x) = 5/3$ for $0 \le x < 1/2$, $1/3$ for $1/2 \le x \le 1$. Let $f(x) = x \sin(\pi \log_2(x))$, and note that $f(2x) = - 2 f(x)$. Then $$ \eqalign{\int_0^1 f(xy) g(x)\; dx &= \frac{1}{3} \int_0^1 f(xy)\; dx + \frac{4}{3} \int_0^{1/2}f(xy)\; dx\cr &= \frac{1}{3} \int_0^1 f(xy)\; dx - \frac{1}{3} \int_0^1 f(xy)\; dx = 0}$$<|endoftext|> TITLE: What work can be done to study the solutions of $\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)$? QUESTION [5 upvotes]: For integers $n\geq 1$ I denote the Euler's totient function as $\varphi(n)$ and the divisor function $\sum_{1\leq d\mid n}d$ as $\sigma(n)$, that are two well-known mulitplicative functions. We assume also the theory of odd perfect numbers, see if you want the corresponding section of the Wikipedia with title Perfect number. It is easy to prove the following statement, on assumption that there exists an odd perfect number $x$. Fact. If $x$ is an odd perfect number then $$\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)\tag{1}$$ holds. Computational fact. For integers $1\leq n\leq 5000$, the only solution of $(1)$ is $n=1$. To see it, after some seconds, choose GP as language and evaluate next code (it is just a line written in Pari/GP) in the web page Sage Cell Server for (x = 1, 5*10^3,if (eulerphi(x^(sigma(x))*(sigma(x))^x)==2^(x-1)*x^(3*x-1)*eulerphi(x), print(x))) I believe that the following conjecture holds. Conjecture. The only solution of our equation $(1)$ is the integer $1$. Motivation for the post. My belief is that an interesting way (but my attempts were failed) to study the unsolved problem related to odd perfect numbers (that is if there exist any of them) should be to create intrincated/artificious equations similar than $(1)$ involving the sum of divisors functions and the Euler's totient function with the purpose to invoke inequalitites, asymptotics, heuristics or conjectures for these arithmetic functions (my belief is that the problem of odd perfect numbers is related to the distribution of prime numbers, thus maybe in the equations similar than $(1)$ that previously I've evoked should be required also that arise functions as the radical of an integer $\operatorname{rad}(x)$ or even the prime-counting function $\pi(x)$, both specialized for odd perfect numbers $x$). Question. What work can be done to prove of refute previous conjecture, that the only solution of $$\varphi\left(n^{\sigma(n)}\sigma(n)^n\right)=2^{n-1} n^{3n-1}\varphi(n)$$ should be $n=1$? It is welcome unconditionally statements or heuristics, but also feel free to invoke conjectures if you can get some advanced statement. Many thanks. Thus, as how it is perceived in the title of the post, previous Question is also an invitation to add remarkable statements about the nature of the solutions of $(1)$, if we are in the situation that the Question can not be solved. Last remarks to to emphasize my ideas. What is saying myself thus previous Motivation and Question? That of couse I understand that the equation/characterization for odd perfect nubmers by means the equation $\sigma(x)=2x$ for odd integers $x\geq 1$ is easiest (to understand and study it) than others involving more arithmetic functions, but in my belief is that there exists a chance to get some statement for odd perfect numbers by the method to create more intrincated/artificious equations. I think that my question is interesting, and I think that arises in a natural way when one tries to drop solutions like $2^{2^{\lambda-1}-1}$, that is the sequence A058891 from the On-Line Encyclopedia of Integer Sequences, for equations like this $$\varphi(x^x\sigma(x))=x^x\varphi(x).$$ See if you want the code for (x = 1, 10^4,if (eulerphi((x^x)*sigma(x))==(x^x)*eulerphi(x), print(x))) I would like to refer that certain characterizations of primes are feasibles from answers of next posts (the problem [2] remains as unsolved), these posts are not directly related to this my post in MathOverflow, but maybe can be inspiring for some user of MathOverflow since are similar problems. Thus I justify this last paragraph as a compilation of similar equations for constellations of primes: thanks to the excellence of the user who provides the answer of [1] we've the characterization of Sophie Germain primes and similarly for twin primes; thanks to the excellence of the user who provides the answers of problems [2] and [3] we've a characterization of Mersenne exponents, Fermat primes and near-square primes. [1] From the equation $\sigma(x^{\varphi(y)})=\frac{1}{\varphi(x)}(x^y-1)$ involving arithmetic functions to a characterization of Sophie Germain primes, question 3578715 from Mathematics Stack Exchange (Mar 12 '20). [2] From the equation $\sigma(x^{\sigma(y)-1})=\frac{1}{\varphi(x)}(x^{y+1}-1)$ involving arithmetic functions to a characterization of Mersenne exponents, question 3587159 from Mathematics Stack Exchange (Mar 19 '20). [3] On characterizations for near-square primes and Fermat primes in terms of equations involving arithmetic functions, question 3588192 from Mathematics Stack Exchange (Mar 20 '20). REPLY [3 votes]: Here is a proof that if an odd integer $x>1$ satisfies (1), then $x$ is a perfect number. First, by using the property that $\varphi(nm)=n\varphi(m)$ whenever $\mathrm{rad}(n)\mid\mathrm{rad}(m)$, we can rewrite (1) as $$\big(\frac{\sigma(x)}2\big)^{x-1}\cdot \frac{\varphi(x\sigma(x))}{\varphi(x)} = x^{3x-\sigma(x)}.$$ Since the fractions in the l.h.s. are integer, the r.h.s. is also integer and odd, and thus $\sigma(x)=2y$ for some odd $y$ such that $\mathrm{rad}(y)\mid \mathrm{rad}(x)$. Then we further simplify the above equation to $$y^x = x^{3x-2y}.$$ In other words, $$\big(\frac{y}x\big)^x = x^{2(x-y)}.$$ Now, if $x>y$, then lhs < 1 while rhs > 1. Vice versa, if $x 1 while rhs < 1. Hence, $x=y$, meaning that $\sigma(x)=2x$. QED<|endoftext|> TITLE: Is there a classification of finite simple groups of perfect power order? QUESTION [12 upvotes]: The finite simple group $\operatorname{PSp}(4,7)$ has order $138297600 = 11760^2$. There also seems to be a description of the $q$ such that $\operatorname{PSp}(4,q)$ has square order, see for example here . Some natural questions: Which finite simple groups have order $n^2$? Are there any finite simple groups order $n^k$ for $k \geq 3$? Is there a classification of finite simple groups of order $n^k$ ($k > 1$)? For a sporadic simple group $G$, one can check there always exists a prime $p$ such that $p \mid |G|$ and $p^2 \nmid |G|$, so $|G| \neq n^k$ for $k > 1$. The same is also true for $G = \operatorname{Alt}(n)$ by Bertrand's postulate. The difficult case is then that of the finite simple groups of Lie type. REPLY [8 votes]: Not anything resembling an answer, but too long for a comment. This question seems very hard to me, from a purely diophantine perspective. We have formulas for the orders of all the finite simple groups. Let's take just the example of $PSp(4,q)$, where $q$ is an odd (for convenience) prime power. I believe the order of this group is $$f(q) = \frac{1}{2}q^4(q^4-1)(q^2-1).$$ If we fix some $k$ such as 2 or 3 and ask for the order of the group to equal $n^k$, then a group in this family of order $n^k$ corresponds to an integer point on the curve $$n^k = f(q)$$... and you impose the additional constraint that $q$ is a positive prime power. It sounds like the authors of the paper linked in the MSE post you reference have found some solutions to this equation (and I'm guessing this curve has genus zero if they are conjecturing it has infinitely many), but proving you've found them all sounds challenging. And then to do this for even all groups of the form $PSp(2m,q)$, let alone all FSG's, seems very hard to me. It wouldn't surprise me if you ran across a family of curves which was quite intractable.<|endoftext|> TITLE: Stationary sets and $\kappa$-complete normal ultrafilters QUESTION [5 upvotes]: Let $\kappa$ be a measurable cardinal, and let $u$ be a normal $\kappa$-complete ultrafilter over $\kappa$. It is a standard easy fact that every closed unbounded set must belong to $u$ (notice that normality is key here), and therefore every element of $u$ is stationary. My question is about the converse: Is it the case that, for every stationary set $S\subseteq\kappa$, there exists a $\kappa$-complete normal ultrafilter $u$ over $\kappa$ with $S\in u$? Maybe there's an easy argument, but I don't see it. For example, attempting to take an existing $\kappa$-complete normal ultrafilter $u$ and considering its Rudin--Keisler image $v=f(u)$ under a bijection $f:\kappa\longrightarrow X$ does give us that $v$ is $\kappa$-complete and $X\in v$, but we may lose normality (e.g., if $X$ was not stationary). So I'd like to know if anyone is aware of a way of getting around this difficulty when dealing with a stationary set. REPLY [11 votes]: Asaf's answer is totally right, but let me also point out that you don't even have to go to a special model to see that your conjecture fails. The point is that sets in any normal measure on $\kappa$ must reflect many properties of $\kappa$ itself (since these sets $X$ are exactly the ones for which $\kappa\in j(X)$, where $j$ is the ultrapower map). For example, the set of regular (or inaccessible or Mahlo etc.) cardinals below $\kappa$ will be in any normal measure on $\kappa$. This means that the set $S$ of singular cardinals below $\kappa$ (which is stationary) is omitted from all normal measures on $\kappa$. REPLY [6 votes]: No. Work in $L[U]$, the canonical inner model, then $U$ is the unique normal measure on $\kappa$. Pick any $S$ such that $S$ and $\kappa\setminus S$ are stationary, and then only one of them can be in a normal ultrafilter.<|endoftext|> TITLE: Conceptual explanation for the appearance of entropy in $\frac{d}{dp}\|x\|_p$ QUESTION [10 upvotes]: For $x\in \mathbb{R}^d$, an elementary computation yields that $$\frac{d}{dp}\log \|x\|_p =\frac{1}{p^2}\sum_{i=1}^d \frac{|x_i|^p}{\|x\|_p^p}\log \frac{|x_i|^p}{\|x\|_p^p}=-\frac{1}{p^2}\operatorname{Ent}(\mu_{x,p}),$$ where $\mu_{x,p}$ is the law of a random variable taking values in $\{1,\ldots,d\}$ that takes the value $i$ with probability proportional to $|x_i|^p$ and $\operatorname{Ent}(\mu_{x,p}) = -\sum \mu_{x,p}(i) \log \mu_{x,p}(i)$ is its entropy. I've been curious about the following question: Is there a good/conceptual reason for an entropy to appear here? Obviously the computation is very easy, but could I have predicted this equality without any computation? REPLY [9 votes]: As Von Neumann said "Nobody really know what entropy is." so it is quite difficult to give a conceptual reason. However I think your calculation appears and can be interpreted in the statistical-mechanics setting (so from a physicits' point of view) as the Free Energy https://en.wikipedia.org/wiki/Helmholtz_free_energy and the canonical ensemble https://en.wikipedia.org/wiki/Canonical_ensemble Let $E$ such that $|X|=e^{- E}$. The partition function is $Z = \sum e^{-\beta E} = \|x\|^\beta_\beta$ with $p=\beta=\frac{1}{T}$ the inverse of temperature and $F=T\log(Z)$ is the free energy. Then the physics relation $$ F = \langle E \rangle- TS $$with $S$ the entropy and $\langle E\rangle$ the mean energy and this calculation $$\langle E\rangle =\frac{1}{Z}\sum e^{-\beta E}=\partial_{\beta}\log(Z) = \beta\partial_{\beta}[\frac{1}{\beta}\log(Z)]+\frac{1}{\beta}\log(Z)= \beta \partial_{\beta}[\log(\|x\|_\beta)] +F$$ gives $$\frac{1}{\beta}S = \beta \partial_{\beta}[\log(\|x\|_\beta)$$ which is what you wanted.<|endoftext|> TITLE: Moment map interpretation of Einstein equation QUESTION [5 upvotes]: Einstein's famous equation relates the geometry of a (4-dimensional) manifold to the matter content in that manifold. Is there a way to obtain Einstein's equation as a moment map? More precisely, given a Riemannian 4-fold $M$ (maybe Lorentzian), is there an infinite-dimensional symplectic manifold $X$ with a hamiltonian action of a group $G$ such that the moment map is Einstein's equation? For simplicity take for instance the case of an empty space (no matter). So Einstein's equation just tells that the Ricci tensor vanishes. Since the moment map is a map from $X$ to the dual Lie algebra of $G$, we could ask which group $G$ has a dual Lie algebra formed by Ricci tensors? REPLY [3 votes]: I'm not aware of a momentum map interpretation of the Einsteins's equation, but you can bring Einstein's equations in a Hamiltonian form with momentum map constraints (this is due to Fischer & Marsden). In more detail, you write spacetime $M$ (locally) as the product $\mathbb{R} \times \Sigma$, where $\Sigma$ is a spatial Cauchy surface. Then Einstein's equation split into dynamical equations of a time-dependent Riemannian metric on $\Sigma$ and two constraint equations (called the momentum constraint and the Hamiltonian constraint). The momentum constraints turns out to be of the form $J = 0$ for the momentum map $J$ of the natural $Diff(\Sigma)$ action on the cotangent bundle $T^* Metr(\Sigma)$. The Hamiltonian constraint is also related to the $Diff(M)$-symmetry of Einstein's equation but its interpretation as a momentum map constraint is more complicated (for this have a look at recent work by Blohmann & Weinstein).<|endoftext|> TITLE: Weight spaces of representations of finite dimensional simple Lie algebras QUESTION [5 upvotes]: This question has probably been asked before on this website, but I could not find any solution and neither can I solve this question. So again I am asking the following question: Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$ with Cartan subalgebra $\mathfrak{h}$ and $V$ be an irreducible weight module of $\mathfrak{g}$ with respect to $\mathfrak{h}$, i.e. $V= \oplus_{\lambda \in \mathfrak{h}^*} V_{\lambda}$ where $V_{\lambda} = \{v \in V : h.v = \lambda(h)v$ $\forall h \in \mathfrak{h}\}$. If $V_{\mu}$ is a non-zero finite dimensional weight space for some $\mu \in \mathfrak{h}^*$, then show that $V_{\lambda}$ is finite dimensional $\forall \lambda \in \mathfrak{h}^*$. Any help or reference will be highly appreciated. REPLY [2 votes]: The requisite property follows from the following key proposition: $U_{\lambda}$ is a finitely generated right $U_0$-module. Notation The subscripts denote the grading of the universal enveloping algebra $U=U({\frak g})$ with respect to the adjoint action of the Cartan subalgebra ${\frak h},\, \displaystyle U=\bigoplus_{\lambda\in P}U_{\lambda},$ with the grading abelian group the root lattice $P.$ The subspace $U_0$ is $U^{\frak h}$, the subalgebra of ${\frak h}$-invariants in $U.$ Similarly, $\displaystyle S=\bigoplus_{\lambda\in P}S_{\lambda}$ for the symmetric algebra $S=S({\frak g})$ and $S_0$ is the subalgebra of ${\frak h}$-invariants in $S.$ Proof of the property The action of $U_0$ stabilizes each weight subspace of $V$ and the action of $U_{\lambda}$ increases the weight by $\lambda$. Let $W=UV_{\mu}$, where a weight subspace $V_{\mu}$ is non-zero and finite-dimensional. Since $V$ is simple and $W$ is a non-zero submodule of $V$, $W=V$. Note that $U_{\lambda}V_{\mu}$ is $W_{\lambda+\mu}$, the weight subspace of $W$ of weight $\lambda+\mu$. Therefore $$\displaystyle V=\bigoplus_{\lambda\in P}U_{\lambda}V_{\mu}=\bigoplus_{\lambda\in P}W_{\lambda+\mu}$$ is the weight decomposition of $W=V$. For any $\lambda\in P$, $U_{\lambda}=X_{\lambda}U_0$ with a finite set $X_{\lambda}$, according to the proposition, and $U_{0}V_{\mu}=V_{\mu}$. It follows that each weight subspace $W_{\lambda+\mu}$ is finite-dimensional: $$W_{\lambda+\mu}=U_{\lambda}V_{\mu}=X_{\lambda}U_{0}V_{\mu}=X_{\lambda}V_{\mu}.$$ Proof of the proposition The algebra $U$ is almost commutative (i.e. its associated graded algebra ${\rm gr\,}U=S$ is commutative and generated by degree 1 part) and the adjoint action of $\frak h$ on $U$ is semisimple and preserves the filtration, so that ${\rm gr\,}U_{0}=S_{0}$ and ${\rm gr\,}U_{\lambda}=S_{\lambda}$. Hence it suffices to prove corresponding statement for the associated graded algebra: $S_{\lambda}$ is a finitely generated $S_0$-module. Recall that $S$ is a polynomial ring, the symmetric algebra of ${\frak g}$, and it is graded by $P$, i.e. it is a multigraded ring, and the last statement is a general property of multigraded rings. A good reference for these rings is the Miller-Sturmfels book "Combinatorial Commutative Algebra".<|endoftext|> TITLE: Which failed attempts have there been to find a contradiction in ZFC or ZF? QUESTION [14 upvotes]: Can you provide any failed attempts to prove that ZF or ZFC to be inconsistent? References to articles in the literature if there are any will be much appreciated. Thanks! REPLY [22 votes]: I think the most noticeable one was Nelson's attempt to prove the inconsistency of primitive recursive arithmetic. Terence Tao found a mistake in the proof, but Nelson's attempt was posthumously uploaded to arxiv (https://arxiv.org/pdf/1509.09209.pdf), together with an introduction by Sarah Jones Nelson and an afterword by Sam Buss and Terence Tao himself. Nelson was amongst the very few serious mathematicians who supported the view that arithmetic was indeed inconsistent, based on his ultrafinitist philosophy.<|endoftext|> TITLE: ZFC applications of Shelah's creature forcing QUESTION [11 upvotes]: Shelah's creature forcing is a very powerful method, with wide range of applications. The method also has some applications in ZFC, let's quote a few of them that I am aware of: (1) In A partition theorem Shelah proves a very general infinitary Ramsey theorem in ZFC, which is parallel to the Galvin-Prikry theorem and the Carlson-Simpson theorem. (2) In Partition theorems from creatures and idempotent ultrafilters, creature forcing is used to prove some Ramsey type theorems. As an application of their general method, new proof of Carlson-Simpson theorem is given. See also Creature forcing and topological Ramsey spaces (3) In Ramsey theorems for product of finite sets with submeasures creature forcing is used to prove a parametrized partition theorem on products of finite sets equipped with submeasures. It improves the results of DiPrisco, Llopis, and Todorcevic. Question 1. What other ZFC examples are available whose proofs is given by creature forcing? Question 2. Are there any applications of creature forcing in proving ZFC results, beyond those in Ramsey type theorems? REPLY [7 votes]: Also check: MR3563073 Dobrinen, Natasha Creature forcing and topological Ramsey spaces. Topology Appl. 213 (2016), 110–126; DOI: 10.1016/j.topol.2016.08.008, arXiv: 1509.06402.<|endoftext|> TITLE: What is the intuition for higher homotopy groups not vanishing? QUESTION [33 upvotes]: The homotopy groups of the spheres $S^n$ (see Wikipedia) vanish for the circle $S^1$ as, naively speaking, there are not higher order holes to be grasped by higher order homotopy groups. This intuitions already breaks down for the two sphere $S^2$, e.g. $\pi_3(S^2)$ is non-trivial because of the Hopf fibration. This non-triviality seems to keep on going for all the higher spheres $S^n$. What makes $S^1$ so fundamentally different? REPLY [9 votes]: Here's a very simple argument as to why not all of the spheres have higher homotopy groups vanish. Higher homotopy groups vanishing (along with the lower ones) is equivalent to the sphere representing n-dimensional cohomology. By Yoneda, natural transformations between functors correspond to maps between representing objects. Hence, natural transformations between nth cohomology with integral coefficients and mth cohomology with integral coefficients correspond to maps $S^n \rightarrow S^m$ if these both represent singular cohomology. However, there are nontrivial cohomology operations (say for example, reduction mod p followed by the integral Bockstein), and so this implies that we have a map between the spheres representing them. However, this would imply that the higher homotopy groups of this sphere do not vanish, and so we get a contradiction.<|endoftext|> TITLE: Label distance number and chromatic number of a graph QUESTION [7 upvotes]: Let $n\in\mathbb{N}$ be a positive integer, and let $[n] = \{1,\ldots,n\}$. We set $S_n$ for the set of all bijections $\varphi:[n]\to [n]$. Let $G= ([n], E)$ be a simple, undirected graph, and let $\varphi\in S_n$ We define the label distance number of $G$ in the following way: $$\lambda(G) = \min\big\{\max\{|\varphi(v)-\varphi(w)|:\{v,w\}\in E\}: \varphi\in S_n\big\}.$$ Conjecture: if $n$ is a positive integer, and $G, H$ are graphs with $V(G) = V(H) = [n]$ and $\chi(G) < \chi(H)$, then $\lambda(G)\leq \lambda(H)$. Is this conjecture true? I would also be interested in knowing whether there is an established name for what I call label distance number. REPLY [7 votes]: The cycle on five vertices has chromatic number 3 and bandwidth 2. The complete bipartite graph with blocks 2 and 3 has chromatic number 2 and bandwidth 3.<|endoftext|> TITLE: Can there be a small complete category in $ZF$? QUESTION [11 upvotes]: It's a ZFC theorem of Freyd, any small complete category is a preorder. Freyd's theorem continues to hold in any Grothendieck topos. But Hyland showed it fails in some elementary toposes. I don't actually know, but I suspect that Hyland's topos examples don't readily translate into models of ZF -- toposes don't play well with replacement. So Question: Does Freyd's theorem hold in ZF? That is, in ZF is every small complete category a preorder? REPLY [16 votes]: Yes, Freyd's theorem holds in ZF. The theorem relies on excluded middle, choice does not play a role. Just to be sure, let's work with excluded middle but without choice. Theorem: A small-complete small category is a preorder. Proof. Let $C$ be a small-complete small category, with $C_0$ the set of objects and $C_1$ the set of morphisms. Consider any $x, y \in C_0$. We need to show that the set of morphisms $C(x,y)$ contains at most one element. For this purpose, suppose $r_0 : x \to y$ and $r_1 : x \to y$. By excluded middle it suffices to show that $r_0 \neq r_1$ leads to a contradiction. So assume $r_0 \neq r_1$, and define $z = \prod_{C_1} y$, the $C_1$-fold product of $y$'s. Note that a morphism $h : x \to z$ is given as a $C_1$-indexed family $h = \langle h_f \rangle_{f \in C_1}$ of morphisms $h_f : x \to y$. Define the map $i : \{0,1\}^{C_1} \to C_1$ by $$i(c) = \langle r_{c(f)}\rangle_{f \in C_1} : x \to z.$$ we claim that $i$ is injective, which is impossible because, by excluded middle, it would give us a surjection $C_1 \to \{0,1\}^{C_1}$, contradicting Cantor's theorem (which is constructive). To see that $i$ is injective, suppose $i(c) = i(d)$. Then $r_{c(f)} = r_{d(f)}$ for all $f \in C_1$, but since $r_0 \neq r_1$ it follows that $c(f) = d(f)$ for all $f \in C_1$ (no excluded middle here because equality on $\{0,1\}$ is decidable), hence $c = d$. $\Box$ In the above proof there are tree applications of excluded middle: concluding $r_0 = r_1$ from $\lnot\lnot (r_0 = r_1)$ and concluding that there is no injective map $2^{C_1} \to C_1$. As far as I can tell, both are necessary. And we never used choice.<|endoftext|> TITLE: Good reduction of rational surfaces QUESTION [9 upvotes]: Let us work over $K = \mathbf{C}((t))$ for simplicity. We say that a smooth proper scheme $X/K$ has good reduction if it extends to a smooth and proper algebraic space $\mathcal{X}/\mathcal{O}_K$ where $\mathcal{O}_K = \mathbf{C}[[t]]$, and that it has potentially good reduction if for some finite extension $K' = \mathbf{C}((t'))$ ($t' = t^{1/N}$), the base change $X_{K'}$ has good reduction over $K'$. My question may sound a bit silly: Question. Can you give an example of a (smooth, projective) rational surface $X/K$ which does not have potentially good reduction? On the one hand, the usual "homotopical" obstructions to good reduction vanish. The only nontrivial one I can think of is the action of the Galois group of $K$ on the Neron-Severi group ${\rm NS}(X_{\bar K})$, but that one factors through the action of a finite quotient, and hence becomes trivial over some finite extension $K'$. On the other, it is easy to think of a possible culprit. Take a sufficiently complicated zero-dimensional subscheme $Z_0\subseteq \mathbf{P}^2_{\mathbf{C}}$, and pick a formal curve $$ z\colon \operatorname{Spec} \mathbf{C}[[t]] \longrightarrow \operatorname{Hilb}(\mathbf{P}^2_\mathbf{C}), \quad z(0) = [Z_0]$$ through $Z_0$ and such that $z(\eta)$ ($\eta$ being the generic point) corresponds to a smooth subscheme $Z_\eta \subseteq \mathbf{P}^2_K$. Take $X$ (resp. $\mathcal{X}$) be the blowup of $\mathbf{P}^2_K$ (resp. $\mathbf{P}^2_{\mathcal{O}_K}$) along $Z_\eta$ (resp. the subscheme $Z$ corresponding to $z$). Then $\mathcal{X}$ will not be smooth if $Z$ is complicated enough. Of course, this does not imply that $X$ does not have potentially good reduction, and I do not see how to check this. REPLY [7 votes]: As Ulrich commented, this is true. In fact, if $\mathcal{O}_K$ is a Dedekind domain with fraction field $K$ and $X$ is a rational surface over $K$, then there is a finite field extension $L/K$ such that $X_L$ has a smooth projective model over the normalization $\mathcal{O}_L$ of $\mathcal{O}_K$ in $L$. I just wanted to seize this opportunity to make a couple of additional comments. 1) Suppose that $X$ is a smooth split cubic surface over $\mathbb{Q}$, so that $X$ is the blow-up of $\mathbb{P}^2_{\mathbb{Q}}$ in six points in general position. Then, $X$ has a smooth projective model over $\mathbb{Z}$. However, the smooth cubic surface $X$ does not have a smooth projective model over $\mathbb{Z}$ which is a smooth cubic surface over $\mathbb{Z}$. To prove this, it is good to have a look at Scholl's observation that having a smooth proper model does not imply having a smooth proper model with relatively ample anti-canonical bundle. This is in Remark 4.6 of Scholl's paper here. A variation of Scholl's observation is given in Lemma 4.9 of this paper. The fact that there really are not such smooth cubic surfaces over $\mathbb{Z}$ is proven below. Let me make another remark on what it means (or should mean) to have "good reduction" as a rational surface. 2) The fact that every smooth rational surface potentially has a good model (in your sense) means that the stack of such surfaces satisfies the existence part of the valuative criterion of properness. The comment above shows that the (algebraic) stack of del Pezzo surfaces does not satisfy this property. 3) Here is how you show that there are no smooth cubic surfaces $X$ over $\mathbb{Z}$. First, note that any smooth cubic surface $X$ over $\mathbb{Z}$ is split. This follows from the fact that Hilbert scheme of lines $Lines_{X/\mathbb{Z}}$ is finite étale over $\mathbb{Z}$ and that $\mathbb{Z}$ is simply connected (Hermite). Then, it follows that $X$ is the blow-up of six $\mathbb{Z}$-points of $\mathbb{P}^2_{\mathbb{Z}}$ in general position. The coordinates of such points would (after some appropriate rescaling) be solutions to the unit equation in $\mathbb{Z}$. As the unit equation has no solutions in $\mathbb{Z}$ it follows that there are no smooth cubic surfaces over $\mathbb{Z}$.<|endoftext|> TITLE: Deligne's example of $\deg \pi_{*}\Omega_{X/Y}<0$ QUESTION [5 upvotes]: While reviewing Lang's book on Arakelov theory, I saw the following comment by Paul Vojta: "...Deligne has found an example when $\deg \pi_{*}\Omega_{X/Y}$ can be negative, because Green's functions at infinity. This is of course unlike the functional field case, but this is of no consequence for next section..." (page 159) May I ask is this example ever published? I know people are usually interested in the upper bound, not lower bound for the height functions. But I feel such a result can still be interesting to know. REPLY [2 votes]: This example should be the elliptic curve with j-invariant 0 and can be found in Deligne's paper Preuve des conjectures de Tate et de Shafarevitch (p. 29). If you are more interested in small values for elliptic curves, the article On the essential minimum of Faltings' height by Burgos Gil, Menares and Rivera-Letelier may be interesting for you. An example of negative value in genus 2 can be found in Sur le calcul explicitedes “classes de Chern” des surfaces arithmétiques de genre 2 by Bost, Mestre and Moret-Bailly.<|endoftext|> TITLE: Spectral gaps for spin manifold Laplace spectrum QUESTION [7 upvotes]: For a (compact) spin manifold, we know that the eigenvalues $\lambda_n$ of the Dirac operator are countable, with finite multiplicity, and satisfy $$ |\lambda_n| \to \infty, ~~~ \text{ as } n \to \infty. $$ This can be concluded, for example, from the fact that they have compact resolvent, as established in Friedrich's book on Dirac operators in Chapter 4.2. I am wondering for the gap between the eigenvalues, as we tend to infinity, will it become as large as we want, or at least is there a minimum distance between succesive eigenvalues. (Honestly, I care most about Hermitian manifolds that are spin, so if it is easier in this case, please let me know!) REPLY [6 votes]: Edit: In Eigenvalues of the dirac operator, M. Atiyah, 1984, the following results are stated and proved. Let $M$ be a $d$-dimensional closed spin manifold, and $D$ the Dirac operator associated to a Hermitian bundle $V$ with a connection over $M$. Let $\lambda_k$ be its eigenvalues, ordered by absolute value and increasing. The following results are named after their version in the above paper. Theorem 2. There exists a constant $C$ independent of $V$ such that $$|\lambda_k|\leq Ck^{1/d}.$$ ‎ Theorem 1*. If $d$ is odd, there exists a constant $C$ independent of $V$ such that every interval of length $C$ contains an eigenvalue. ‎ Theorem 2*. If $d$ is odd, there exists a constant $C$ independent of $V$ such that every interval of length $Ck^{1/d}$ contains at least $k$ eigenvalues. About the case of even-dimensional manifolds, Atiyah gives an argument showing that such a constant would have to depend on $V$, but I don't think it rules out similar theorems for fixed $V$. Out of transparency, here is my previous answer, which in the odd case is of course way worse. It seems classical, although I cannot seem to be able to find a reference with a complete proof (Atiyah writes a sentence about it in the above article), that the Dirac operator on a closed spin manifold $M$ of dimension $d$ satisfies a Weyl type law: $$\frac{N(\lambda)}{|\lambda|^d}\to C$$ for some known constant $C>0$ ($C_d\mathrm{Vol}(M)$, for $C_d$ a constant depending only on $d$), where $N(\lambda)$ is the number of eigenvalues with absolute value less than $\lambda$ (counted with multiplicity). I claimed before that it implied that the gap between two consecutive eigenvalues cannot be bounded below. In fact, we get $$\liminf_{k\to\infty}\frac{|\lambda_{k+1}|-|\lambda_k|}{k^{1/d-1}}<\infty$$ so for $d\geq2$ the gaps cannot be bounded below at least on one side. However it does not rule out large gaps; it only rules out relatively large gaps in absolute value in the sense that $$\limsup_{k\to\infty}\frac{|\lambda_{k+1}|-|\lambda_k|}{\lambda_k} =\limsup_{k\to\infty}\frac{|\lambda_{k+1}|-|\lambda_k|}{k^{1/d}}=0.$$ Again, I don't know if there exists Dirac operators on even-dimensional manifolds with arbitrarily large eigengaps. A few references: Weyl laws on open manifolds (S. Moroianu, 2003), claims that this law is well-known in the case of closed manifolds, and that the case of compact manifolds with boundary is treated in Spectral Asymmetry and Riemannian Geometry. I (M. F. Atiyah, V. K. Patodi and I. M. Singer, 1975). It is also possible that the result of Moroianu holds trivially in the compact case, see Theorem 3. The more recent work The Dirac spectrum on manifolds with gradient conformal vector fields (A. Moroianu and S. Moroianu, 2007) states in the introduction that [The eigenvalues of the Dirac operator on a closed spin manifold] grow at a certain speed determined by the volume of the manifold and its dimension. which I read as “the Weyl law holds for the Dirac operator on compact spin manifolds.” I bet someone with more experience with Dirac operators than myself will be able to find the missing pieces.<|endoftext|> TITLE: P-adic functions on annuli QUESTION [7 upvotes]: It is known that a complex analytic function defined on an annulus, say, takes its maximum on the boundary. Does an analogue hold for $p$-adic analytic functions? More precisely suppose we have a doubly infinite power series $f(z) = \sum_{n\in \mathbb{Z}}a_n z^n $ with coefficients $a_n \in K$ where $K$ is a finite extension of $\mathbb{Q}_p$ (the rationals completed by a $p$-adic absolute value). Suppose further that $f$ converges for all $z \in K$ with $r_1 \leq |z| < r_2$. Does it hold that $|f(z)| \leq \max\{|f|_{r_1}, |f|_{r_2}\}$ for $r_1\leq |z| < r_2$? $|f|_r $ denotes the maximum of $|f(z)|$ as $z$ varies over $z \in K$ with $|z| = r$. If it isn't true I would be very grateful for a counterexample. Thanks a lot! As an aside: If true one should be able to take any complete non-archimedean field but I am unsure about whether compactness helps. REPLY [10 votes]: $\def\bQ{\mathbb{Q}}\def\bF{\mathbb{F}}\def\bZ{\mathbb{Z}}$This is false as stated because of the following important difference between $K$ and $\mathbb{C}$: the former is not algebraically closed. For example. consider $K=\bQ_p(p^{1/k})$ with $k>1$ and the polynomial $$f(z)=z\prod\limits_{a\in\bF_p^{\times}}(z-[a])^2$$ with $r_1=1/p,r_2=1$(here $[x]$ denotes the Teichmuller representative of an element $x$ of the residue field of a complete non-archimedean field). Then for any $z\in K$ with $|z|$ equal to $1$ we have $|z-[a]|\leq p^{-1/k}$ for exactly one $a$ because the residue field of $\mathcal{O}_K$ is equal to $\bF_p$, so $|f(z)|\leq p^{-2/k}$. For $z$ with $|z|=p^{-1}$ we just get $|f(z)|=p^{-1}$. However, taking $z=p^{1/k}$ gives $|f(z)|=p^{-1/k}$ which is larger than any value of $f$ on the boundary. The statement becomes true if we assume that the residue field of $\mathcal{O}_K$ is infinite(the counterexample above depends crucially on the finiteness of the residue field). The formation of the Newton polygon is a very convenient way to visualize the behavior of roots of a $p$-adic analytic function and the desired inequality in the case of an algebraically closed field $K$ follows quickly from the fact that the slopes of the Newton polygon of a Laurent series correspond to the valuations of the roots. However, It might be instructive to unpack the proof of this theorem to get a direct argument for our boundary inequality: let $f(z)=\sum\limits_{n\in \bZ}a_nz^n$ be a Laurent series converging for $z$ satisfying $r_1\leq |z|\leq r_2$. Assume that $r_1,r_2$ are in the image of the norm map on $K$. We will prove that for every such $z$ there is $u$ with $|u|$ equal to $r_1$ or $r_2$ such that $|f(z)|\leq |f(u)|$. Lemma. If the residue field of $\mathcal{O}_K$ is infinite, then for every $z\in K$ there exists an element $z'\in K$ with $|z|=|z'|$ and a number $n\in\bZ$ such that $|a_n(z')^n|$ is larger or equal to any $|a_m(z')^m|$ with $m\neq n$ and $|f(z)|=|a_n(z')^n|$. Proof. Since the values $|a_nz^n|$ tend to zero as $n$ tends to $\pm\infty$, there exists a finite set of indices $i_1<\dots< i_k$ such that $|a_{i_1}z^{i_1}|=\dots=|a_{i_k}z^{i_k}|$ and $|a_mz^m|$ is less that this common value for any $m\notin\{i_1,\dots, i_k\}$. We want to find $z'$ such that the norm of the sum of these $k$ summands is precisely equal to the norm of each separate summand. To arrange that, pick $\lambda\in \mathcal{O}_K/\mathfrak{m}_K$ such that $1+\lambda^{i_2-i_1}\rho\left(\frac{a_{i_2}z^{i_2}}{a_{i_1}z^{i_1}}\right)+\dots+\lambda^{i_k-i_1}\rho\left(\frac{a_{i_k}z^{i_k}}{a_{i_1}z^{i_1}}\right)\neq 0$ where $\rho:\mathcal{O}_K\to \mathcal{O}_K/\mathfrak{m}_K$ is the reduction map. Then $z'=[\lambda]z$ will do the job. We can now prove the statement: let $z\in K$ be any element in the annulus $r_1\leq |z|\leq r_2$. There exists $k\in\bZ$ such that $|f(z)|\leq |a_kz^k|$. Assume that $k\geq 0$. Then $|a_kz^k|\leq |a_k|r_2^k$. Using the lemma, find $u$ with $|u|=r_2$ such that $|f(u)|=|a_nu^n|$ and $|a_nu^n|\geq |a_mu^m|=|a_m|r_2^m$ for every $m$. It follows that $|f(u)|\geq |a_k|r_2^k\geq |f(z)|$. If $k<0$, then arguing in the same way with $r_2$ replaced by $r_1$ gives the result.<|endoftext|> TITLE: Simplicial Complex Induced by a Morphism QUESTION [5 upvotes]: Let $(C, \otimes, I)$ be a symmetric monoidal abelian category. For an object $M$ in $C$ with a map $\rho : M \rightarrow I$, we can form the chain complex $M_*$ where $M_n = \otimes_{i = 1}^n M$ and $d : M_n \rightarrow M_{n-1}$ is $$ \sum_{i = 1}^n (-1)^i 1 \otimes \cdots \otimes \rho \otimes \cdots \otimes 1 $$ where $\rho$ is in the $i$th slot (this is a less-than-ideal of writing the map, but I think it should be clear what I mean). This is sort of like a Bar construction, but we got less information to start with than a typical module over a monad. My questions are What is the simplicial object over $C$ corresponding to $M_*$ under the Dold-Khan correspondence? Is there a name for this construction? Can we view this as an instance of the Bar construction? Example: Note that, if we take $C$ to be the symmetric monoidal category of $R$-modules over a commutative ring $R$ and modify the twist map $\tau$ so that $\tau_{M} : M \otimes_R M \rightarrow M \otimes_R M$ sends $a \otimes b$ to $-b \otimes a$, then we get $\Lambda (M)$ above. If $M = R^n$ and we are given a map $\rho : R^n \rightarrow R$, then $M_*$ above is the Koszul complex. So this means that the Koszul complex is an example of the kind of resolution above. Another one along the same vein is De Rham cohomology. Thanks very much! REPLY [4 votes]: Have you looked at Illusie's Complexe Cotangent et Deformations I? Specifically Section 1.3, "The Theorem of Dold-Puppe" where there's a fairly formal formula for the simplicial object to which your complex corresponds. Also, Section 1.5, "The Standard Simplicial Resolution" which is Illusie's name for the Bar construction.<|endoftext|> TITLE: "Bootstrapping" an unbounded class of inaccessible cardinals QUESTION [8 upvotes]: The "richness principle" of set theory asserts roughly that "everything that happens once should happen an unbounded number of times". An example would be the existence of an unbounded class of inaccessible cardinals. Now there are many examples of large cardinals $\kappa$ whose existence guarantees an unbounded class of inaccessibles below $\kappa$. So, my first question: Is there any large cardinal such that "$\kappa$ exists" $\implies$ "There exists an unbounded class of inaccessible cardinals in V" ? An even stronger example of this would be something like: "There exists a large cardinal of Type A" $\implies$ "There exists an unbounded class of cardinals of Type A". Is there any large cardinal with this property? I know that rank-into-rank cardinals satisfy an "upward reflection" property, but that only happens $\omega$-many times. Would a Reinhardt cardinal (in ZF rather than ZFC say) reflect upwards unboundedly many times ? REPLY [4 votes]: The maximality principle implies a sweeping uniform version of the idea you mention. Hamkins, Joel David, A simple maximality principle., J. Symb. Log. 68, No. 2, 527-550 (2003). ZBL1056.03028. Namely, theorem 14 of that paper shows that if MP holds, then there is a proper class of inaccessible cardinals, if any; and similarly for Mahlo cardinals and indeed any large cardinal notion that is downward absolute to small-forcing grounds. To explain, the maximality principle is the principle that any statement that is forceably necessary is already true. That is, if there is a forcing extension such that the statement is true in all further forcing extensions, then the statement is already true. This principle is expressible in modal terms by the scheme $\Diamond\Box\varphi\to\varphi$, and indeed, the modal logic of forcing was introduced in this paper specifically because of the connection with the maximality principle. The point now is that if MP holds and there is an inaccessible cardinal, then there cannot be only one or a bounded number of them, because then we could collapse them all and thereby make it forcing necessarily true that there would be none; so under MP there would have to have been none to begin with. So if there is one, there must be a proper class of them.<|endoftext|> TITLE: Oriented vector bundle with odd-dimensional fibers QUESTION [6 upvotes]: Is it true that for every oriented vector bundle with odd-dimensional fibers, there is always a global section that vanishes nowhere? REPLY [17 votes]: No. Let us consider some vector bundles over $S^4$. Since $S^4$ is simply connected all vector bundles are oriented and rank $k$ vector bundles over $S^4$ are classified by $\pi_{3}(SO(k))$ by the clutching construction (I am glossing over basepoint issues here, but I think it is correct in this setting). Now $\pi_3(SO(1))$ and $ \pi_3(SO(2)=\pi_3(S^1)$ are trivial groups, so there do not exist non-trivial rank $1$ and rank $2$ bundles over $S^4$. What about rank $3$? Well $SO(3)\cong \mathbb RP^3$ and the long exact sequence of homotopy groups of the fiber bundle $$ \mathbb{Z}_2\rightarrow S^3\rightarrow \mathbb{RP}^3 $$ shows that $\pi_3(SO(3))\cong\pi_3(\mathbb{RP}^3)\cong \mathbb{Z}$. Hence there are $\mathbb{Z}$ different rank $3$ vector bundles over $S^4$. Only the trivial one admits a non-zero section. If another one admits a section, then the orthogonal complement (a rank $2$ bundle) must be non-trivial, otherwise the bundle is trivial. But our previous computation shows that there are no-nontrivial rank $2$ bundles over $S^4$.<|endoftext|> TITLE: Is equality of formulas with floor rounding or integer division decidable? QUESTION [8 upvotes]: As far as I know, formulae involving rationals and basic arithmetic ($+$, $-$, $\cdot$ and $/$) have decidable equality. Is this still the case if we add floor rounding (or integer division)? Define a "basic formula" by the following grammar (in Backus-Naur form): $$ \begin{matrix} e &= &q \\ &| &v \\ &| &e + e \\ &| &e - e \\ &| &e \cdot e \\ &| &e / e \\ \end{matrix} $$ Here, $q$ stands for rational literals and $v$ for a set of variables. A valuation is the assignment of a rational number to every variable. The semantics of such a formula for a given valuation is ordinary arithmetics. Two formulae are semantically equal if they are equal for every allowed valuation. (Division by 0 is not allowed, of course.) As far as I understand, this notion of equality is decidable. Given two formulas, we essentially multiply by all denominators, apply the distributive law everywhere and then compare summands. We've basically reduced each expression to a normal form where we can compare. Now, let us add the floor function, which rounds down to the nearest integer. We extend the above definition by: $$ \begin{matrix} e &= &\dots \\ &| &\lfloor e \rfloor \\ \end{matrix} $$ Is equality still decidable here? It's not so obvious to me that a well-behaved normal form even exists, since the floor function satisfies some equations such as $\lfloor \lfloor x \rfloor + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$, Hermite's identities (e.g. $\lfloor x\rfloor + \lfloor x+1/3\rfloor + \lfloor x+2/3\rfloor = \lfloor 3x\rfloor$), and so on. Is there still some simple enough algorithm that decides whether two such formulae are equal? Note: I could have added integer division instead of the floor function as well, but I believe that this wouldn't make a difference. Feel free to add or suggest tags, I'm not so sure whether the question is categorized well. REPLY [11 votes]: Equality of formulas is undecidable. To prove this it is enough to show that an algorithm for equality of formulas would enable us to determine whether or not a multivariate polynomial with integer coefficients vanishes at at least one integer point, because the latter condition is undecidable according to the negative solution to Hilbert's Tenth Problem. But the polynomial $f(x)$, where $x$ is a tuple of variables and $f$ has integer coefficients, does not vanish at any integer point if and only if the following equation holds at all integer points $x$: $$\Bigl\lfloor\dfrac{1}{1+f(x)^2}\Bigr\rfloor=0.$$ I believe this settles it. However it would be nice to know a simple elegant decidability result for formulas excluding division.<|endoftext|> TITLE: Moduli of curves over finite field QUESTION [5 upvotes]: This is related to this question. I learnt about moduli problem mainly with the book Harris and Morrison. Therefore, I have only seen the construction of moduli spaces $M_{g}$ over $\mathbb{C}$. But now I want to change the base field. It does not really make sense to me to take $\mathbb{F}_q$-points as described in the comments of that post, because $M_g$ is a $\text{Spec}(\mathbb{C})$-scheme not a $\text{Spec}(\mathbb{F}_q)$-scheme. So how should I understand $M_{g}$ as a coarse moduli space properly? REPLY [10 votes]: As requested, I am posting my comment as an answer. Mumford discusses this in the preface to the first edition of "Geometric Invariant Theory". Already in "Geometric Invariant Theory" (which was Mumford's thesis), Mumford constructed the coarse moduli scheme $M_g$ over $\text{Spec}\ \mathbb{Z}.$ You can read about what this means in Mumford's book. The great irony of GIT is that, although GIT was largely motivated by this problem, in fact Mumford constructs $M_g$ over $\text{Spec}\ \mathbb{Z}$ without using GIT (precisely because neither had Haboush yet extended GIT to char $p$, nor had Seshadri extended GIT to mixed characteristic).<|endoftext|> TITLE: “Taylor series” is to “Volterra series” as “Padé approximant” is to _________? QUESTION [13 upvotes]: Padé approximants are often better than Taylor series at representing a function. Given a Taylor series, one can use Wynn's epsilon algorithm to easily produce the Padé approximants to it. Volterra series are a generalization of Taylor series that can also model "memory" phenomena. Does there exist a similar generalization of Padé approximants that can model these phenomena, or an algorithm like Wynn's to compute them from the Volterra series? I would be at least happy to know the answer for the discrete Volterra series, which (I think) would be equivalent to something like a multivariate Padé approximant. Originally asked at MSE, but seems too advanced for that site. REPLY [8 votes]: "I would be at least happy to know the answer for the discrete Volterra series, which (I think) would be equivalent to something like a multivariate Padé approximant." Multi-variate versions of Padé approximants got attention in the 1970s, for example a generalization to functions of two variables was published by J.S.R. Chisolm in 1973 (see "Rational Approximates Defined from Double Power Series") and a generalization to functions of countably more variables was studied at around the same time by P.R. Graves-Morris and D.E. Roberts who were both at the University of Kent in Canterbury, leading to this generalization being called the "Canterbury approximant" (see "Calculation of Canterbury approximants"). The original papers about Canterbury approximants never seemed to get many citations, so it's easy to look through the entire list of papers that cited them to find further generalizations (for example, to functions of an uncountably infinite number of variables). Some of the most significant post-Canterbury papers on multivariate versions of Padé approximants that can be found in this list are: Annie Cuyt "How well can the concept of Padé approximant be generalized to the multivariate case?" (May 1999) Philippe Guillaume and Alain Huard. "Multivariate Padé approximation" (September 2020) Finally, one of the original authors of the Canterbury approximant co-wrote a book with George Baker Jr. called "Padé Approximants" (first edition: 1982, second edition: 1996) which provides you with any developments which they considered significant between the aforementioned papers from the 1970s and about the time of the aforementioned papers from 1999-2000. Finally, some steps towards a generalization to functions of complex variables can be found in my analogous question: “Taylor series” is to “Volterra series” as “Laurent series” is to _________?<|endoftext|> TITLE: Formal completion of an elliptic curve along the $0$ sectioin and the formal expansion of functions QUESTION [6 upvotes]: Let $ S = \operatorname{Spec}A $ be an affine scheme, $ f : E \to S $ an elliptic curve and $\mathscr{I}$ the ideal sheaf of the $0$-section. (This is invertible since the section defines the effective relative Cartier divisor.) Assume that $f_* \Omega_{E/S}, f_*\mathscr{I}^n$ are free over $\mathscr{O}_S$. ($n = 1, \cdots , 6$) I want to show $\hat{E} \cong \operatorname{Spf} A[[T]]$. And I don't understand the formal expansion, of a basis $\omega$ of $f_* \Omega_{E/S}$ and a basis of $f_*\mathscr{I}^n$, along the $0$-section. Here is what I have tried: Since the $0$-section is a regular immersion, for any $x \in S$, there exists affine opens $ x \in V \subseteq S$, $ 0(x) \in U \subseteq E$ s.t. $0(V) \subseteq U$ and the diagram $$\require{AMScd} \begin{CD} S @>{0}>> E \\ @VV{1}V @VV{f}V \\ S @>{1}>> S \end{CD}$$ corresponds to $$\require{AMScd} \begin{CD} C @<{0}<< B \\ @A{\text{localization by one element}}AA @AAA \\ A @<{1}<< A, \end{CD}$$ where the kernel $I$ of $B \to C$ is generated by $t \in B$, a regular element. I showed that $\hat{B}$ (the completion of $B$ along the kernel $I$) $\cong C[[t]]$. And $\Omega_{B/A} \otimes_B \hat{B} = dt \hat{B} = dt C[[t]].$ That is, I can show $ \hat{E} \cong \operatorname{Spf}A[[t]]$ locally, and I can expand $\omega$ locally. How can I extend these operation globally? I also showed that the isomorphism $\hat{B} \cong C[[t]]$ is compatible with localization. So I think intuitively that these isomorphisms (at any points) are glued together, and we have $ \hat{E} \cong \operatorname{Spf}A[[t]]$. Any help will be much appreciated! REPLY [4 votes]: Yes, what you are saying is true, at least over an algebraically closed base $k$ of characteristic $0$. In fact, all you need is that $S$ is affine and that the normal bundle $I/I^2$ is trivial (as a bundle over $S$), which in your case is equivalent to $f_*\Omega(E/S)$ being free. The rest follows essentially from deformation theory. There is a standard way to reduce questions about formal thickenings to questions in deformation theory, and this is called deformation to the normal cone. Here's the idea. Let $R = O(\widehat{E})$ be the algebra of function on the formal thickening, viewed as a formal ring over $A$. Then this algebra is filtered by $R_n : = I^n,$ with associated graded ring $$R_{gr}:=\bigoplus_n \frac{I^n}{I^{n-1}}\cong A[I/I^2].$$ Since $I/I^2$ is trivial and one-dimensional over $A$, we have $R_{gr}\cong A[[t]].$ Now the deformation to the normal cone is an algebra over $\text{Spf} (k[[s]])$ interpolating between $R$ and $R_{gr}$, given by the Rees construction $$R_{\text{rees}} : = R[s^{-1}\cdot I]^{\wedge} = \bigoplus_n^\wedge s^{-n}I^n,$$ where the completion is taken along (positive) powers of $s$. The key thing to observe here is that the fiber over $s=0$ of $R_{\text{rees}}$ is $\bigoplus s^{-n}(I^n/I^{n+1})$ (not completed), which is $R_{gr}$ and the generic fiber over $k((s))$ is isomorphic to $R\otimes k((s))$. Now a standard result in the deformation theory of schemes tells us that flat deformations of a smooth scheme $X$ up to isomorphism (as well as of a smooth formal scheme) are classified by (what is noncanonically isomorphic to) a subset of $H^1(X,T)^\infty,$ i.e. infinite sequences of vectors in the first homology of the tangent bundle (deformations of order $n$ would be certain sequences of size $n$). In particular, if $X$ is affine (in your case — $X = \text{Spf}A[[t]],$ the special fiber of the Rees construction), the relevant $H^1$ is trivial, so there is only one deformation — the trivial one. Equivalently, $R_{\text{rees}}\cong R_{gr}[[s]]$. Now you're done! Basechanging to the generic point gives $R\otimes k((s)) \cong A[[t]]\otimes k((s)).$ But since you are over an algebraically closed base of characteristic zero, such an isomorphism implies an isomorphism $R\cong A[[t]]$ over $k$.<|endoftext|> TITLE: Are there infinitely many insipid numbers? QUESTION [6 upvotes]: A number $n$ is called insipid if the groups having a core-free maximal subgroup of index $n$ are exactly $A_n$ and $S_n$. There is an OEIS enter for these numbers: A102842. There are exactly $486$ insipid numbers less than $1000$. Question: Are there infinitely many insipid numbers? Let $\iota(r)$ be the number of insipid numbers less than $r$. The following plot (from OEIS) leads to: Bonus question: Is it true that $\lim_{r \to \infty}r/\iota(r)=2$? REPLY [17 votes]: Almost all $n$ are insipid. In fact, the number of non-insipid numbers at most $n$ grows like $2n/\log n$. See the paper Cameron, Peter J.; Neumann, Peter M.; Teague, David N. On the degrees of primitive permutation groups. Math. Z. 180 (1982), 141–149. doi.org/10.1007/BF01318900<|endoftext|> TITLE: Bounded Torsion, without Mazur’s Theorem QUESTION [11 upvotes]: Mazur’s torsion theorem famously tells us exactly which finite groups can occur as the torsion subgroup of $E(\mathbb{Q})$ for an elliptic curve $E$ defined over $\mathbb{Q}$. In particular, it implies that only finitely many torsion subgroups are possible, which seems like a much weaker result. My question: Is there any way to see that the weak version of bounded torsion (only finitely many groups occur, but nevermind what they are) is true, without recourse to the full proof of Mazur’s theorem? I saw some references to a paper of Demjanenko from 1971 (EDIT: originally said 1975) which claims to prove this, but it’s only available in Russian and other sources don’t seem to think its argument is correct. REPLY [16 votes]: By all means hold out hope, but I don't think that the ideas in Dem'janenko's papers are going to work. I spent a lot of time in grad school looking at them. If I remember correctly, Dem'janenko also claimed to have proven that on $E:y^2=x^3+D$, if a $P\in E(\mathbb Q)$ is a non-torsion point, then $\hat h(P)\ge c\log|D|$ for an absolute constant $c$. But again, no one has managed to decipher his proof. Lang was intrigued enough to conjecture that $\hat h(P)\ge c\log|\Delta_E|$ for all elliptic curves, where $\Delta_E$ is the minimal discriminant. Using quite different techniques, I proved a weaker version of Lang's conjecture, and Hindry and I proved that Lang's full conjecture follows from $ABC$. However, for twists such as in Dem'janenko's paper, there is an alternative easier argument, and it's possible that that is what's lurking in his paper. I mention all of this, because height arguments such as those in Dem'janenko, and in my work with Hindry, tend to lead to statements of the form: Let $P\in E(\mathbb Q)$. Then either $NP=0$ or $\hat h(P)$ isn't too small. So torsion bounds and height bounds come packaged together. As for proofs that just handle torsion points, note that any such proof would be a strong uniform version of the Mordell conjecture for the modular curves $X_1(\mathbb Q)$, which is why it seems unlikely that purely elementary, albeit complicated, algebraic manipulations as in Dem'janenko's paper could give full uniformity. On the other hand, Dem'janenko came up with a very nice way of proving the Mordell conjecture in certain situations. For example, if there are two independent maps $f_1,f_2:C\to{E}$ and if $E(\mathbb Q)$ has rank 1, then $C(\mathbb Q)$ is finite, essentially by comparing $\hat h(f_1(x))$ and $\hat h(f_2(x))$. Manin generalized this to the tower of modular curves $X_1(p^{n+1})\to X_1(p^n)$, and used this to give an elementary (at least, compared to Mazur's work) proof of $p$-power uniformity for torsion points: Theorem (Manin) Fix $p$ and a number field $K$. There is a constant $C=C(p,K)$ such that for every elliptic curve $E/K$, there are no torsion points in $E(K)$ of exact order $p^C$.<|endoftext|> TITLE: Conditionally convergent spectral sequences with exiting and entering differentials QUESTION [6 upvotes]: I have to deal with unbounded filtrations and want to use the conditional convergence of spectral sequences and the results from [1]: J. Michael Boardman, Conditionally Convergent Spectral Sequences, March 1999 (http://hopf.math.purdue.edu/Boardman/ccspseq.pdf) The article uses cohomological spectral sequences derived from the exact couple coming from a cochain complex $C$ and a decreasing filtration $F$ of $C$. The system of inclusions is $$A^s := H(F_s C) \leftarrow A^{s+1}$$ and the pages are denoted by $E^s_r$ for $s\in \mathbb{Z}$ and $r\in \mathbb{N}$ ($r$ is the page number and $s$ the ``filtration degree''). The symbol $A^\infty$ denotes the limit and the symbol $A^{-\infty}$ the colimit. The symbol $RA^\infty$ denotes the right derived module of the limit. I basically work over $\mathbb{R}$. The following are the two theorems (or their parts) from [1] which I am interested in: Theorem 6.1 (p.19): Let $C$ be a filtered cochain complex. Suppose that \begin{equation}\label{Eq:Exit}\tag{C1} E^s = 0\quad\text{for all } s>0.\end{equation} If $A^\infty = 0$, then the spectral sequence converges strongly to $A^{-\infty}$. Theorem 7.2 (p.21): Let $f: C \rightarrow \bar{C}$ be a morphism of filtered cochain complexes and suppose that $E^s$, resp. $\bar{E}^s$ converge conditionally to $A^{-\infty}$, resp. $\bar{A}^{-\infty}$. Suppose, moreover, that \begin{equation}\tag{C2} E^s = \bar{E}^s = 0\quad\text{for all }s<0. \end{equation} If $f$ induces the isomorphisms $E^\infty\simeq \bar{E}^\infty$ and $RE^\infty\simeq R\bar{E}^\infty$, then it induces the isomorphism $H(C)\simeq H(\bar{C})$. Let me introduce the standard (degree shifted) bigrading on $E_r$ and visualize $E_r^{s,d}$ as sitting at the coordinate $(s,d)$ in plane. The differentials are then $$ d_r: E_r^{s,d}\rightarrow E_r^{s+r,d-r+1}. $$ My questions are the following: How does Theorem 6.1 generalize if (C1) is replaced by the following condition of exiting differentials? $$ E_r \text{ sit in a half-plane and if we fix any coordinate }(s,d), \text{ then all but finitely many }d_r\text{ starting at }(s,d)\text{ leave the half-plane.}$$ How does Theorem 7.2 generalize if (C2) is replaced by the following condition of entering differentials? $$ E_r \text{ sit in a half-plane and if we fix any coordinate }(s,d), \text{ then all but finitely many }d_r\text{ ending at }(s,d)\text{ start outside of the half-plane.}$$ The author of [1] addresses the questions as follows: On p.19, Chapter 6 in brackets right before Theorem 6.1: ...The results generalize appropriately, as all arguments can be carried out degreewise; the main difficulty is to find notation that would help rather than hinder the exposition On p.20, Chapter 7 in brackets a couple of paragraphs before Theorem 7.2: ...The results remain valid when appropriately modified, as all arguments can be carried out degreewise; the difficulty is to find notation that helps rather than hinders. How do these theorems generalize precisely? Has it been done anywhere? Thanks! P.S. I come from differential geometry and am not familiar with the proof methods for spectral sequences at all. I use it merely as a black box. REPLY [6 votes]: After a long life in preprint form, Boardman's paper was published in the conference proceedings celebrating his 60th birthday: \bib{MR1718076}{article}{ author={Boardman, J. Michael}, title={Conditionally convergent spectral sequences}, conference={ title={Homotopy invariant algebraic structures}, address={Baltimore, MD}, date={1998}, }, book={ series={Contemp. Math.}, volume={239}, publisher={Amer. Math. Soc., Providence, RI}, }, date={1999}, pages={49--84}, review={\MR{1718076}}, doi={10.1090/conm/239/03597}, } In the $(s,d)$-bigraded case, you can replace Boardman's (left half-plane) condition that $E_1^{s,d} = 0$ for $s > 0$ (or $s > s_0$ for some fixed integer $s_0$) by the (upper half-plane) condition that $E_1^{s,d} = 0$ for all $d < 0$ (or $d < d_0$ for some fixed integer $d_0$). Similarly, you can replace his (right half-plane) condition that $E_1^{s,d} = 0$ and $\bar E_1^{s,d} = 0$ for $s < 0$ (or $s < s_0$ for some fixed integer $s_0$) by the (lower half-plane) condition that $E_1^{s,d} = 0$ and $\bar E_1^{s,d} = 0$ for $d > 0$ (or $d > d_0$ for some fixed integer $d_0$). This adjustment is often enough. Do you need to refer to other half-planes than those bounded by a horizontal or a vertical line? EDIT: The OP added some questions, including one about the case of a right half-plane cohomological bicomplex $(B^{i,j}, d_h, d_v)$, where $B^{i,j} = 0$ for $i<0$, $d_h : B^{i,j} \to B^{i+1,j}$ and $d_v : B^{i,j} \to B^{i,j+1}$. Suppose that $Z^{i,j} \subset B^{i,j}$ is such that $d_h(Z^{i,j}) = 0$, and filter the total complex $$ (C, d) = (\bigoplus_{i,j} B^{i,j}, d_h + d_v) $$ by $$ F^s = \bigoplus_{j-i>s} B^{i,j} \oplus \bigoplus_{j-i=s} Z^{i,j} $$ for all integers $s$. Here $C$ and $F^s$ are graded, with $B^{i,j}$ and $Z^{i,j}$ in degree $i+j$. Then $(F^s, d)$ is a subcomplex of $(C, d)$, and contains $(F^{s+1}, d)$ as a further subcomplex. We get an exact couple in the usual way, with $A^{s,t}$ and $E_1^{s,t}$ equal to the degree $s+t$ parts of $H(F^s, d)$ and $H(F^s/F^{s+1}, d)$, respectively. I claim that $\lim_s F^s = 0$ and $\lim^1_s F^s = 0$. This can be checked one degree at a time, since $F^s = 0$ in degrees less than $s$. It follows (see Boardman's Theorem 9.2) that $\lim_s A^s = 0$ and $\lim^1_s A^s = 0$, so the spectral sequence is conditionally convergent to the colimit $G = H(C, d)$. Furthermore, $F^s/F^{s+1}$ and $E^1_s$ are concentrated in degrees $\ge s$, corresponding to $t\ge0$, so this is an upper half-plane cohomological spectral sequence with exciting differentials. Hence it is strongly convergent to $G$, by the modified form of Boardman's Theorem 6.1 that I mentioned above. To prove the modified form, one does as Boardman says. Let $F^s G$ be the image of $H(F^s, d)$ in $G$. One must check that the filtration $\{F^s G\}_s$ of $G$ is complete Hausdorff and exhaustive, and that the natural inclusion $F^s G/F^{s+1} G \to E_\infty^s$ is an isomorphism. Both claims can be checked one degree at a time, and for each degree the proof of Theorem 6.1(a) carries over. (I do not have Cartan--Eilenberg at hand: I do not recall if they spelled this out.)<|endoftext|> TITLE: Diffeomorphism type of the added sphere in simply connected surgery QUESTION [6 upvotes]: A classical result of simply connected surgery theory, is that if two normal maps $f:M_i\rightarrow X$, $i=0,1$ are normally cobordant and if the dimension of the manifolds is odd, there exists a homotopy sphere $\Sigma$ such that $M_0$ is diffeomorphic to $M_1\#\Sigma$. This was first proved by Novikov in Homotopically equivalent smooth manifolds. I, Izv. Akad. Nauk SSSR Ser. Mat., 28:2 (1964), 365–474. (Theorem 5.1) but can also be found in Browder's Surgery on simply-connected manifolds, Springer-Verlag (1972), Ergebnisse der Mathematik und ihre Grenzgebiete, Band 65 (II.3.7 Theorem) and many other references of surgery theory. My question is the following: is there a way to determine the diffeomorphism type of the added homotopy sphere? REPLY [5 votes]: The ambiguity in the the added homotopy sphere is captured by a suitable inertia group $I(X)$, which in this case is the group of homotopy spheres $\Sigma$ such that $\Sigma$ bounds a parallelizable manifold, and the standard homeomorphism $\Sigma\# X\to X$ is homotopic to the diffeomorphism. There is a related inertial group $\bar I(V)$ that consists of homotopy spheres $\Sigma$ such that $\Sigma\# X$ and $X$ are diffeomorphic. For example, if $X=S^3\times CP^2$, then $\bar I(X)=\Theta_7$, the group of all homotopy $7$-spheres. The group $bP_d$ of homotopy spheres bounding parallelizable $d$-dimensional manifolds is cyclic of known order. More precisely, the order is known except when $d=126$, the only remaining case of the Kervaire invariant problem, see here. Thus to understand the ambiguity in the the added homotopy sphere one needs to compute the index of $I(X)$ in $bP_{d}$ where $\dim(X)=d-1$. Suppose $d$ is divisible by $4$ and $d\ge 8$, and $X$ is a closed oriented manifold of dimension $d-1$. Then $bP_{d}$ has large order given in terms of Bernoulli numbers. For such $d$ a theorem of L.Taylor says that the index of $I(X)$ in $bP_{d}$ is $\ge 2$. This is actually sharp, e.g., the index is $2$ for $X=S^3\times CP^{2m}$. The index is $4$ if $X=S^7\times CP^2$. On the other hand, Browder showed that $I(X)$ is trivial when $d$ is not divisible by $8$, the group $H^1(X;\mathbb Z_2)$ is zero, and $X$ stably parallelizable; thus in this case one can ``determine the diffeomorphism type of the added homotopy sphere". More details and references to the above claims can be found here on p.8.<|endoftext|> TITLE: Ultraproduct of Dividing Lines QUESTION [5 upvotes]: Let $\{M_i\}_{i\in I}$ be a family of $L$-structures such that for each $i\in I$, $T_i=Th(M_i)$ is X where $X\in\{\text{stable, simple, NIP, NSOP}, \dots\}$. Let $U$ be a non-principal ultrafilter on $I$, and $M=\prod_U M_i$. Also, let $T=Th(M)$. Is $T$ an X theory as well? What can we say about $T$ in general? REPLY [4 votes]: I'm going to assume we're talking about $\omega$-incomplete ultrafilters. Others have pointed out that none of the major dividing lines are elementary; that is, these properties are not equivalent to a countable conjunction of sentences. Some dividing lines, like stability and NIP, have the form "for each formula $\phi$, there is a numeric n such that a sentence $\sigma_{n,\phi}$ holds". For instance, NIP holds if each $\phi$ has bounded VC dimension. Such a property holds in an ultraproduct exactly when it holds uniformly in the ground models. That is, $\prod_{\mathcal{U}}M_i$ is NIP if and only if, for each formula $\phi(x;y)$, there is a bound $d$ such that, for most $i$, $\phi$ has VC dimension $d$ in $M_i$. That means an ultraproduct of NIP models can be IP (if the VC dimension is unbounded) and an ultraproduct of IP models can be NIP (if each model has an IP formula, but each individual formula is IP in most models) I'm not sure if every dividing line can be expressed in that form. But (at the risk of self-promotion) there's a more general framework for describing properties of ultraproducts in terms of "higher order uniformity", so it will, in general, be true that a dividing line holds in the ultraproduct if it holds uniformly in the ground models, for a suitable notion of uniformity.<|endoftext|> TITLE: Differential forms of a Lie group giving cohomology of the Lie group QUESTION [7 upvotes]: Consider a manifold $M$. Then, we have the notion of differential forms on $M$ and complex associated to that, denoted by $$\cdots\rightarrow \Omega^{k-1}(M)\rightarrow \Omega^k(M)\rightarrow \Omega^{k+1}(M)\rightarrow \cdots$$ giving de Rham cohomology groups $H^k_{\mathrm{dR}}(M)$ for $k\in \mathbb{N}$. Now consider a Lie group $G$. There is a notion of cohomology of this Lie group. Ignoring the group structure, we can talk about de Rham cohomology of the underlying manifold. Question : Is there a notion of restricted complex of differential forms, that is a sub complex $\{\widetilde{\Omega^k(G)}\}$, of the complex of differential forms $\{\Omega^k(G)\}$ of $G$, whose cohomology groups gives cohomology of the Lie group? All that (of great importance) extra structure coming in Lie group $G$ is the multiplication (and inverse) map $G\times G\rightarrow G$. So, I am expecting this restricted differential forms to show the action of $G$ on itself. By cohomology of the Lie group, I mean the cohomology of the underlying manifold. I do not think there is any notion of cohomology of Lie group. Even Google search does not give anything. REPLY [7 votes]: For every Lie group $G$, you have in your $\Omega_G$ the subcomplex of the left-invariant differential forms. Moreover, $G$ acts on the right on this subcomplex. It has the virtue of being finite-dimensional and isomorphic to a complex defined purely in terms of the Lie algebra, thus the computation of its cohomology is reduced to linear algebra. On the other hand, considering a maximal compact subgroup $K\subseteq G$, since $G/K$ is contractible, by Poincare's lemma, $\Omega_G$ and $\Omega_K$ have the same cohomology.<|endoftext|> TITLE: Deligne’s letter to Kazhdan on $\ell$-adic Fourier transform QUESTION [10 upvotes]: I wonder if anyone has a copy of Deligne's letter to Kazhdan, dated 29 November 1976. The letter is on the $\ell$-adic Fourier transform. In his essay on Deligne's work on the occasion of the Abel prize, Illusie cites Laumon's paper and his own article in Algebraic Geometry - Bowdoin 1985 as auxiliary references, but I am curious to read the 1976 letter. (It is not listed on Deligne's IAS page or Wikipedia article.) Thanks. REPLY [13 votes]: Thanks to Beilinson & Drinfeld I was able to make a copy of this letter, now online at https://dl.dropboxusercontent.com/s/ufc66njc36svfm1/deligne_to_kazhdan.pdf.<|endoftext|> TITLE: Large deviations for discrete uniform distribution QUESTION [6 upvotes]: (Not sure if this belongs on stack-exchange or overflow; let me know if I should switch it). Given a sum of $n$ IID random variables $\{X_i\}_{i=1}^n$, each uniform on the integers $0,1,...,r$ for some (fixed) $r$, I would like to estimate $$ \mathbb{P}[\sum_{i=1}^n X_i = k]$$ for $k$ of order $n$. If $k$ happened to be the mean of the sum, then I know how to handle this via the Local Central Limit Theorem. But, when $k$ is far from the mean of the sum, the error term in the LCLT is larger than the first-order term. This seems like it should be a very standard exercise in large-deviations, but I am not very familiar with that field and am having trouble finding the right tool. Could someone help point me to a theorem (and ideally an example calculation) that might help? REPLY [4 votes]: As Iosif Pinelis mentioned this is quite standard in large deviations theory so let me explain a bit the idea of the theorem he quote. Let $Y$ a random variable defined as $\mathbb{P}(Y=y)=\frac{1}{Z(\alpha)}e^{-\alpha y}$ for $y=0,\cdots,r$ with $Z(\alpha) = \sum_y e^{-\alpha y}$ and choose $\alpha$ such that $\mathbb{E}(Y)=\frac{k}{n}$. Let $(Y_i)_{i\leq n}$ $n$ independant copy of $Y$. Then $$\mathbb{P}[\sum_i Y_i =k]$$ can be estimated with the CLT. Moreover we have $$\mathbb{P}[X_1=y_1,X_2=y_2,\cdots,X_n=y_n]=\mathbb{P}[Y_1=y_1,Y_2=y_2,\cdots,Y_n=y_n]\frac{Z(\alpha)^n e^{\alpha \sum_i y_i}}{(r+1)^n}$$ and then $$\mathbb{P}[\sum_i X_i =k] = \Big(\frac{Z(\alpha) }{r+1}\Big)^n e^{\alpha k} \mathbb{P}[\sum_i Y_i =k] $$ which solve your problem. In order to find $\alpha$ we use that $$\mathbb{E}[Y]=\frac{1}{Z(\alpha)}\sum_{y\leq r}ye^{-\alpha y}=-\frac{d}{d\alpha} \log(Z(\alpha))$$ which gives an easy equation in $\alpha$. Similarly the variance of $Y$ is calculated with the second derivative of $Z(\alpha)$ [To translate with Iosif Pinelis answer : $\alpha=h$, $Z(\alpha)=(r+1)\mathbb{E}[e^{-\alpha X}]=(r+1)R(-\alpha)$]<|endoftext|> TITLE: Compact space $K$ without convergent sequence while $c_0$ is complemented in $C(K)$ QUESTION [5 upvotes]: Let $K$ be a compact Hausdorff infinite topological space and $C(K)$ the Banach space of continuous functions from $K$ in $\mathbb{R}$ with sup norm. It is known that $c_0$ is complemented in $C(K)$ for $K$ that contains infinite convergent sequence. I would appreciate if somebody let me know an example of $K$ without infinite convergent sequences such that $C(K)$ contains a complemented subspace isomorphic to $c_0$. REPLY [3 votes]: Take $K = \beta N \times \beta N$, the square of the Čech--Stone compactification of the integers. The space of continuous functions on that space is not a Grothendieck spaces. A space of continuous functions on a compact space is Grothendieck if and only if it doesn't contain complemented copies of $c_0$. More generally you may take any infinite $K = L \times L$, such that $C(L)$ is a Grothendieck space.<|endoftext|> TITLE: Does the $\overline{\partial}$ operator have closed image? QUESTION [13 upvotes]: Let $X$ be a complex-analytic manifold, not necessarily compact. Does $\overline{\partial} : C^\infty(X) \rightarrow \Omega^{0,1}(X)$ have closed image with respect to the Fréchet topology given by the norms of derivatives on compact sets? This question is related to Serre's paper about Serre duality but does not seem to answered there. If $X$ is Stein, the answer is yes. REPLY [10 votes]: The range of $\bar\partial$ is closed iff $H^{0,1}(X)$ is separable with the induced topology. According to the paper "On the compactification of concave ends" by M. Brumberg and J. Leiterer, the concave end of a $1$-corona $X$ can be compactified iff $H^{0,1}(X)$ is separable. Since there are $2$-dimensional coronas whose concave ends cannot be compactified (these examples are due to Grauert, Andreott, Siu and Rossi), it follows that there are examples where the range of $\bar\partial$ is not closed.<|endoftext|> TITLE: Path of Diffeomorphisms Fixing the Boundary QUESTION [7 upvotes]: Could you please let me know if the following is true. The problem came up while constructing a solution of a PDE. I have browsed through the net for an answer. While I came across some articles regarding the identity component of the diffeomorphism group, with my poor geometry and topology I could not really figure out what really is happening. Any help in this direction is appreciated. QUESTION: Let $n\geqslant 2$, $k\geqslant 1$, $\Omega\subset\mathbb{R}^n$ be open, bounded, smooth, simply connected with $\partial\Omega$ connected. Let $u:\overline{\Omega}\to\overline{\Omega}$ be a diffeomorphism of class $C^k$ satisfying $$\det(\nabla u)>0\text{ in }\overline{\Omega}\text{ and }u(x)=x\text{ on }\partial\Omega.$$ Does there exist $H\in C^k\left([0,1]\times\overline{\Omega};\mathbb{R}^n\right)$ satisfying $H(0,\cdot)=\text{Id}$ in $\overline{\Omega}$. $H(1,\cdot)=u$ in $\overline{\Omega}$. $\det(\nabla H(t,\cdot))>0\text{ in }\overline{\Omega}$, for all $t$. $H(t,x)=x\text{ on }\partial\Omega$, for all $t$. Note that, 3 and 4 imply that $H(t,\cdot)$ is a diffeomorphism of $\overline{\Omega}$. If the result is negative in general, it will be great to have an explicit counterexample. It will also be good to know some cases, if any, when the result is positive. REPLY [13 votes]: A diffeomorphism for which there exists such an H is called an isotopy (relative to the boundary). This has been much studied in the case where Omega is the unit ball in R^n. Your question is well-known as: does pi_0(D^n,\partial D^n)=0? The answer is positive for the unit balls in R^2 (Smale 1958) and R^3 (Cerf's famous "Gamma4=0" in 1969). This answers positively tour question for n=2 and 3, since in these dimensions, every simply connected compact domain with smooth connected boundary is diffeomorphic to the n-ball. For the compact unit ball in R^4, your question is a big open question. The answer to your question is widely negative in large dimensions. The answer to your question is negative for the unit ball in R^6, as discovered by John Milnor in 1959. It is linked to the existence of "exotic spheres". There are many more counterexamples in large dimensions; see for example Hatcher's review "A 50 year-view on diffeomorphism group". He writes "\pi_0 of Diff(D^n, \partial D^n) is not zero for most n ≥ 5. However it is zero for n = 5, 11, 60." It seems that no other exception than 2,3,5,11,60 is known.<|endoftext|> TITLE: Reference request: mod 2 cohomology of periodic KO theory QUESTION [7 upvotes]: The mod 2 cohomology of the connective ko spectrum is known to be the module $\mathcal{A}\otimes_{\mathcal{A}_2} \mathbb{F}_{2}$, where $\mathcal{A}$ denotes the Steenrod algebra, and $\mathcal{A}_2$ denotes the subalgebra generated by $Sq^1$ and $Sq^2$. Where can I find the original calculation for referencing it ? What is the structure as module over the steenrod algebra of the mod 2 cohomology of periodic KO theory? REPLY [10 votes]: Ravenel in his Complex Cobordism and Stable Homotopy Groups of Spheres attributes this result to Stong, in Determination of $H^*(BO(k,⋯,∞),Z_2)$ and $H^∗(BU(k,⋯,∞),Z_2)$, but looking at that paper (which is concerned mainly with the "unstable" cohomology of the various constituent spaces of $ko$), he attributes this further to Adams, J. F., On Chern characters and the structure of the unitary group, Proc. Camb. Philos. Soc. 57, 189-199 (1961). ZBL0103.16001. There you can find indeed the required result as Lemma 4. Regarding your second question, the cohomology $H^*(KO;\mathbb{F}_2)$ is zero for chromatic reasons (the spectrum $KU\wedge H\mathbb{F}_2$ carries an isomorphism of the additive and multiplicative formal group law in characteristic two, so it must be trivial, and then you can use the fact that $KU=KO/\eta$ to conclude, since $\eta$ is nilpotent in $\pi_*KO$), hence the map $H^*(KO;\mathbb{F}_2)→H^*(ko;\mathbb{F}_2)$ is trivial.<|endoftext|> TITLE: Function orthogonal to powers of $1/\left(1+x^2\right)$ QUESTION [11 upvotes]: Does there exist any continuous function $f:\mathbb{R}\to\mathbb{R}$, $f(x)/(1+x^2)\in L^1(\mathbb R)$, such that $f(0)=1$ and $$\int_{-\infty}^{\infty}\frac{f(x)}{\left(1+x^2\right)^p}dx=0$$ for every $1\leq p\leq 2$? REPLY [2 votes]: The connection between this question and uniqueness sets for holomorphic functions has already been mentioned but I think that it could benefit from a more systematic treatment. Let me begin with Müntz’ result concerning when the span of the sequence $(x^{\lambda_n})$ is dense in $C[0,1]$. If one considers the holomorphic function $F(z)=\int x^z dx$ then it can be dualised to the question of sets or sequences of uniqueness for bounded analytic functions in the right half plane. These are completely understood (e.g. by using a Möbius tranformation to reduce to functions on the unit disc and then Blaschke products). This provides a proof of the Müntz result in the special case where the exponents are real and go to infinity but also in many other cases (convergence to zero, complex exponentials which converge slowly to the imaginary small corrections—-axes). In the case in question here one uses the function $F(z)=\int f(x)(1+x^2)^{-z} dx$ which is analytic on a suitable region of the plane. Then sets of uniqueness provide results of this type—-in the case of the original posting the rather crude one of a real interval.<|endoftext|> TITLE: Is there a way to find any non-trivial $\mathbb{F}_p(t)$-point on the given elliptic curve? QUESTION [5 upvotes]: Consider a finite field $\mathbb{F}_p$ (where $p \equiv 1 \ (\mathrm{mod} \ 3)$, $p \equiv 3 \ (\mathrm{mod} \ 4)$) and the elliptic curve $$ E\!:y^2 = x^3 + (t^6 + 1)^2 $$ over the univariate function field $\mathbb{F}_p(t)$. Is there a way to explicitely find any $\mathbb{F}_p(t)$-point on $E$ outside $E[3]$ (i.e., $x \neq 0$)? REPLY [9 votes]: You can observe that the elliptic curve $E_2: y^2=x^3+(t^3+1)^2$ is a generic fibre of a rational elliptic surface. Over the algebraically closed field $k$ the group of $k(t)$ points on the curve has rank equal to 2 (by the Shioda-Tate formula) and from the classification of possible groups of $k(t)$-rational points by Oguiso-Shioda (case 39 because we have three places of reduction type $IV$ away from characteristic 3) the group of $k(t)$-rational points modulo torsion has a lattice structure $A_2^{*}$. So we are looking at the points of height $1/2$ and this can be found only when one hits two singular points in the Weierstrass equation at places of bad reduction. So you obtain points of the form $P_i=((4^{1/3}\cdot \zeta_3^i)(1 - t + t^2), \sqrt{-3}(-1 + t)(1 - t + t^2)$ for i=0,1,2 We have the relation $P_0+P_1+P_2=0$ and th points $P_1,P_2$ span a lattice of type $A_2^*$, hence are free generators. Now you have to apply the map $t\mapsto t^2$ and on the elliptic curve $y^2=x^3+(t^6+1)^2$ you obtain two linearly independent points $P_1'$ and $P_2'$. Notice that $P_1'-P_2'$ is the negative of the point found by Noam Elkies. The points are defined over $\overline{\mathbb{Q}}(t)$. By reduction they will become naturally points over $\overline{\mathbb{F}}_{p}(t)$ and can provide points defined over $\mathbb{F}_{p}(t)$. With a trace formula and point count (which can be done in Magma as pointed out by Jeremy Rouse in his answer) you can check that in several characteristics the rank over $\mathbb{F}_{p}(t)$ can be higher than $2$. This follows from Tate conjecture and is unconditional because the given elliptic curve is a generic fibre of a K3 surface. For example: in characteristic $p=5,11,13,17$ the rank over $\mathbb{F}_{p}$ is equal to $4$. Addendum: One can play the same game with the elliptic curve $E_{tw}=E_{1}^{(t)}: y^2=x^3+(t+1)^2\cdot t^3$, which under the base change $t\mapsto t^6$ becomes isomorphic to the original curve. Curve $E_{tw}$ is a generic fibre of the rational elliptic surface with rank over $\overline{\mathbb{Q}}(t)$ equal to $2$ again (now the singular fibres have different reduction type: $I_{0}^{*}$, $II$ and $IV$.). This is the case $32$ in the Oguiso-Shioda table. We find the following points of required height $Q_{i}=(\zeta_{6}^{i}(1+t)t, \sqrt{-1}t^2(1 + t))$ for $i=1,3,5$. There is the equality $Q_{1}+Q_{3}+Q_{5}=0$ and the points $Q_{1}$ and $Q_{3}$ are lineary independent. Now, we base change again to the elliptic curve $y^2=x^3+(t^6+1)^2$ and obtain the following set of four linearly independent points obtained via a base change $$\{P_{1}',P_{2}', Q_{1}',Q_{3}'\}$$ The Gram matrix is a block matrix of determinant 4. Second addendum: Finally, we consider a curve $F:y^2=x^3+(t^3-3t)^2$. It is isomorphic to $y^2=x^3+(t^6+1)^2$ under the pullback by $t\mapsto t+1/t$. Curve $F$ has three places bad reduction of type $IV$. Hence we find from the Oguiso-Shioda the points of height $1/2$. Those points are $$R_{i}=(\zeta_{6}^{i}(t^2 - 3),\sqrt{3}(-3 + t^2))$$ for $i=1,3,5$. We have $R_{1}+R_{3}+R_{5}$ is zero and we finally obtain the set of 6 linearly independent points $$\{P_{1}',P_{2}', Q_{1}',Q_{3}',R_{1}',R_{3}'\}$$ which have a Gram matrix of determinant $4/3$. The calculations above for several characteristics show that we cannot have more than 6 independent points in the group $E(\overline{\mathbb{Q}}(t))$, so up to finite index we found a basis of points. From this using the Galois action on the points you can work out the exact value of the $\mathbb{F}_{p}(t)$ rank in the cases when the Tate conjecture computation matches this bound. For the primes where the rank over $\overline{\mathbb{F}}_{p}(t)$ is higher than $6$ the extra points might exist for very peculiar reasons, e.g. the Shioda-Inose structure might throw some extra divisors into the picture and the heights of those points can be very large.<|endoftext|> TITLE: Is the 2-сategory of groupoids locally presentable? QUESTION [6 upvotes]: I am wondering if the 2-сategory of groupoids is locally presentable. Locally presentable means the category is accessible and co-complete. It has been pointed out that the category of groupoids is locally presentable. It would be useful if I could find a theorem that says that if your base category, namely groupoids, was LP, then so is the 2-category on that base. I have done a bit of research and I cannot find anything to that effect. I still cannot say if the 2-category of groupoids is LP. REPLY [8 votes]: This is true. Since the $2$-category of groupoids is equivalent to the $2$-category of $1$-truncated spaces, the statement follows immediately from 5.5.1.8 and 5.5.6.21 in Lurie's Higher Topos Theory.<|endoftext|> TITLE: How does one compute the Hecke algebra acting on modular forms? QUESTION [13 upvotes]: I asked this on mathstackexchange, but got no answer. Let $N\geq1$ be an integer, and let $\mathbb{T}$ be the Hecke algebra acting on the cusp forms of weight k and level $\Gamma_0(N)$. Then $\mathbb{T}$ is finite and flat over $\mathbb{Z}$, but the proof I know for this fact is slightly roundabout, by letting it act on homology. How can one compute $\mathbb{T}$ in explicit examples? By which I mean, writing a presentation for it. I am not necessarily looking for an algorithm, but all the examples I have seen are pretty degenerate (i.e. level 1, weight 12 or level 23, weight 2), and I would like to get some better feel for it. Moreover, it seems to me this sort of data cannot be easily found on the LMFDB. Answers to slight variations of this question are also very welcome (i.e. level $\Gamma_1(N)$, or all forms instead of just cusp forms). Thanks! REPLY [3 votes]: This isn't quite an answer, but since I cant comment, I'll do it here. In MAGMA you can ask for HeckeAlgebra of a space of ModularSymbols (or maybe it only works for the cuspidal subspace of such), but in any case, it'll give you some generators and such. You can see more here https://magma.maths.usyd.edu.au/magma/handbook/text/1613 I think one can also do something similar in Sage, see here http://doc.sagemath.org/html/en/reference/hecke/sage/modular/hecke/algebra.html<|endoftext|> TITLE: Diffeomorphic but not isotopic symplectic forms QUESTION [5 upvotes]: Do we know of any closed symplectic manifold $M$ with 2 cohomologous symplectic forms $\omega_1$ and $\omega_2$ such that there exist $\psi \in \text{Diff}(M)$ and $\psi^* \omega_1 = \omega_2$ but $\omega_1$ and $\omega_2$ are not isotopic? Here the word isotopy is supposed to mean that the two forms a joined by a path of cohomologous symplectic forms. REPLY [3 votes]: This answer is an extension of my last comment, which in turn is just a reference to this answer by MO user Petya, and the paper Symplectic Topology and Capacities by McDuff cited therein. First, your condition for two forms to be isotopic is equivalent to the existence of a path of diffeomorphisms $\phi_t$ such that $\phi_{0} = \operatorname{id},\phi_1^*\omega_2 = \omega_1$. One direction is clear; in the other, let $\omega_t$ be the family of symplectic forms and define a time-dependent vector field $X_t$ by $\mathrm d\left(\iota_{X_t}\omega_t\right) = \partial_t\omega_t$ (by assumption, the RHS is exact). The flow of $X_t$, which exists by compactness, is the required family of diffeomorphisms. Take $M = S^2\times S^2\times T^2$, with $T^2 = S^1\times S^1$ the torus, and let $\omega_1 = p_1\omega_{S^2} + p_2^*\omega_{S^2} + p_3^*\omega_{T^2}$ where $\omega_{S^2}$ and $\omega_{T^2}$ are volume forms on $S^2$ and $T^2$, respectively. (Note that we use the same volume form for both factors of $S^2$.) Let $\psi(x,y,u,v) = (x,R(x,v)y,u,v)$, where $R:S^2\times S^1\to SO(3)$ sends $(x,v)$ to the rotation by $v$ around the axis through $x$. For fixed $y_0,u_0$, the map $x,v\mapsto (x,y_0,u_0,v)$ defines an $S^1$-family of $J$-holomorphic spheres for the product complex structure, and the projection of this family to the second $S^2$-factor si constant. Since each of these spheres has minimal positive area, there is no bubbling, so that for any deformation one obtains a compact cobordism between this family and its pushforward under the end of the deformation. In particular, the projection of the pushforward to the second factor is bordant to a constant map. But one can check that the map $f: S^2\times S^1\to S^2,(x,v)\mapsto R(x,v)y_0$ is not bordant to a constant map: Its restriction to $S^2\vee S^1$ is nullhomotopic, and the resulting map $S^2\times S^1/(S^2\vee S^1)\cong S^3\to S^2$ is homotopic to the Hopf map, which is not nullbordant.<|endoftext|> TITLE: The norm of tensor product operator on Lp spaces QUESTION [6 upvotes]: Let $X, Y$ be two $\sigma$-finite measure spaces and $p,q\in [1,\infty]$. Let $T_1, T_2:L^p(X)\rightarrow L^q(Y)$ be two bounded linear operators. Then one can define a linear operator $$T_1\otimes T_2: L^p(X)\otimes L^p(X)\rightarrow L^q(Y)\otimes L^q(Y).$$ Here $L^p(X)\otimes L^p(X), L^p(Y)\otimes L^p(Y)$ are subspaces of the Banach spaces $L^p(X\times X), L^p(Y\times Y)$ respectively and equipped with the norms of the latter. My questions is: would the following equality of operator norms hold? If yes, is there a quick proof? $$\Vert T_1\otimes T_2\Vert=\Vert T_1\Vert \cdot \Vert T_2\Vert$$ (When $p=q=2$ and $X, Y$ are finite sets, one can easily show the equality using the property of normal transformations or singular values of matrices). Edit: When $p\leqslant q$, the above equality is true, as shown in the discussion below. So the question becomes: does the equality also hold when $p>q$, or is there an obvious counter-example? REPLY [2 votes]: The equality does not need to hold for $p>q$, due to the following counterexample, which is based on the counterexample from this article. We choose $q=1$, $p=\infty$, $X=Y=\{1,2\}$ (equipped with the counting measure) and $$T=T_1=T_2= \begin{bmatrix} 1 & -1 \\ 1 & 2 \end{bmatrix} : L^p(X)\to L^q(Y). $$ Then one can show $\|T\|=\|T e_2\|_1 = 3$. We consider $f:=(e_1+e_2)\otimes (e_1+e_2)-e_1\otimes e_1\in L^p(X)$. Then one can calculate $$ (T\otimes T) f = 9 e_2\otimes e_2 - (e_1+e_2)\otimes (e_1+e_2). $$ Due to $\|f\|_\infty=1$ we have $$ \|T\otimes T\| \geq \|(T\otimes T)f \|_1 = 11 > 9 = \| T\| \cdot \|T\|. $$ For continuity reasons, this counterexample works also for small $q>1$ and large $p<\infty$.<|endoftext|> TITLE: unlinking in 5 dimensions QUESTION [6 upvotes]: If I have a linked pair of circles in $\mathbb{R}^3$, they can be unlinked in $\mathbb{R}^4$. Said differently, there is an isotopy in $\mathbb{R}^4$ between two strands which have been twisted, and two untwisted strands. Here is an ascii art picture! a b a b | | | | \ / | | \ | | / \ | | | | | | | | ~~~> | | \ / | | \ | | / \ | | | | | | I'll call this "the unlinking isotopy" (actually I don't know if it's unique, but I mean to use some chosen one throughout). Now if I cross the two strands three times, I have two different isotopies to the once-crossed version. To clarify, let $t$ denote a single crossing, transposing $a$ and $b$. Then the picture above shows the first and last steps of an isotopy between $t^2$ and $t^0$. One can use this in two different ways to make an isotopy from $t^3$ to $t^1$. My question is whether these two isotopies are "the same", and in what sense. Consider the surfaces swept out by these two isotopies. Can one be deformed (isotoped?) to the other, while keeping the boundaries fixed? I think maybe this should be possible in $\mathbb{R}^5$ but not $\mathbb{R}^4$. Maybe this can be answered as some sort of higher-dimensional linking; if so, please explain! Post script My motivation is coming from Baez-Dolan's Tangle Hypothesis. In particular, I've wanted to give a geometric description for sylleptic monoidal structure in 2-categories (and in bicategories). This is an intermediate notion, between braided monoidal and symmetric monoidal, which doesn't appear for monoidal 1-categories. I've been trying to think about this by imagining the unlinking isotopy as a surface in $\mathbb{R}^4$ -- does it intersect itself? The two isotopies from the triple twist to the single twist give two surfaces, and the geometric avatar of sylleptic structure is an isotopy between them. If I'm understanding the relationship correctly, this should be possible in $\mathbb{R}^5$ but not $\mathbb{R}^4$. REPLY [6 votes]: Thanks, I understand your question now. I asked a similar question in an old paper of mine, but it was with knots rather than braids. In short, the answer is yes they are isotopic in $\mathbb R^4$ subject to certain carefulness assumptions. But if you are particularly fussy about the set-up there is a way in which the answer could be no. The unknotting operation you see for braids is an aspect of what Fred Cohen might call the Freudenthal suspension map, for configuration spaces. Specifically, the pure braid group is the fundamental group of $C_n \mathbb R^2$, and you are looking at the natural inclusion $C_n \mathbb R^2 \to C_n \mathbb R^3$. The latter space is simply connected, so you get an unknotting operation. But there's more you can say. This inclusion is null-homotopic in exactly two different ways. The idea is when you think of $n$ points in $\mathbb R^2$ included into $\mathbb R^3$, you can lift those $n$ points to prescribed distinct altitudes (the height function on $\mathbb R^3$ that has $0$-set the inclusion of $\mathbb R^2$. You then do the straight-line homotopy in the $\mathbb R^2$ factor, and push back into the plane. So that converts your inclusion $C_n \mathbb R^2 \to C_n \mathbb R^3$ to a map $$C_n \mathbb R^2 \to \Omega C_n \mathbb R^3.$$ and this map is non-trivial on the fundamental group. I suppose Fred would write the map as $\Sigma C_n \mathbb R^2 \to C_n \mathbb R^3$. By 'converts' I mean you can recover your original inclusion by evaluating the loop at its mid-point. Anyhow, one issue with this is it shows you there are precisely two distinct ways to untangle a given braid canonically in this co-dimension one setting. If you included $C_n \mathbb R^2 \to C_n \mathbb R^4$ there would be a circle of ways of untangling your braids, and they would all (individually) be isotopic, since the circle is connected. But in co-dimension one, the two ways of untangling are not isotopic. Anyhow, back to your question. Basically you are concatenating a braid with an isotopy of a braid, in two different ways and you are asking if these operations commute. To make it a proper question you would have to be a little more careful settings things up -- this is the issue of loop spaces being homotopy-associative and not (usually) strictly associative. But I see where you are going. In your case of three half-twists, if you use the same unknotting operation on the two braids, they are not isotopic. If you use the opposite unknotting operations, then they are isotopic. If on the other hand you concatenated four half-twists and were comparing the same unknotting operation on the top half vs. the bottom half, yes they would be isotopic. The way to see the isotopy (or lack of it) is just to be careful about exactly what you are doing. When you have your 2-stranded braid that is the product of three half-twists, imagine a little red rectangle that engulfs two of the half twists. In that rectangle you can do the unknotting operation as the two points in the plane are the same at the top of the rectangle as at the bottom -- so the unknotting operation is defined. Now imagine sliding that rectangle down. You keep the dimensions of the rectangle the same. Provided your braid started off as a proper helix, regardless of where your rectangle is, the two points from the braid at the top of the rectangle will be in the same position (in $\mathbb R^2$) as the two points at the bottom of the rectangle. So the untangling operation is defined for this entire 1-parameter family. That said, if you have three half-twists there is the issue of which point in the braid gets pushed up into $\mathbb R^4$. For a half-twist the point that gets pushed up switches at the end of the braid, due to the representation $B_n \to \Sigma_n$. So when you concatenate three half-twists the unknotting operation on the top two strands is isotopic to the reverse unknotting operation on the bottom two strands. For the same reason, if this braid were a concatenation of four half-twists, unknotting the first two would be isotopic to unknotting the last two (provided you use the same unknotting isotopy). If you are curious about my analogous question for knots, it's in my "Family of embedding spaces" paper, proposition 5.1 and the discussion immediately after it: https://arxiv.org/pdf/math/0605069.pdf edit: and yes these are all isotopic in $\mathbb R^5$.<|endoftext|> TITLE: Examples of simultaneous independent breakthroughs QUESTION [34 upvotes]: I'm looking for examples where, after a long time with little progress, a simultaneous mathematical discovery, solution, or breakthrough was made independently by at least two different people/groups. Two examples come to mind: Prime Number Theorem. This was proved by Jacques Hadamard and Charles Jean de la Vallée Poussin in 1896. Sum of three fourth powers equals a fourth power. In 1986, Noam Elkies proved that there are infinitely many integer solutions to $a^4 + b^4 + c^4 = d^4$. His smallest example was $2682440^4 + 15365639^4 + 18796760^4 = 20615673^4$. Don Zagier reported that he found a solution independently just weeks later. Can you give other instances? REPLY [3 votes]: The HOMFLY or HOMFLYPT polynomial is a clear example. The polynomial was introduced in "A new polynomial invariant of knots and links", and on the first page there is a footnote: Editor's Note. The editors received, virtually within a period of a few days in late September and early October 1984, four research announcements, each describing the same result—the existence and properties of a new polynomial invariant for knots and links. There was variation in the approaches taken by the four groups and variation in corollaries and elaboration. These were: A new invariant for knots and links by Peter Freyd and David Yetter; A polynomial invariant of knots and links by Jim Hoste; Topological invariants of knots and links, by W. B. R. Lickorish and Kenneth C. Millett, and A polynomial invariant for knots: A combinatorial and an algebraic approach, by A. Ocneanu. It was evident from the circumstances that the four groups arrived at their results completely independently of each other, although all were inspired by the work of Jones (cf. [10], and also [8, 9]). The degree of simultaneity was such that, by common consent, it was unproductive to try to assess priority. Indeed it would seem that there is enough credit for all to share in. Each of these papers was refereed, and we would have happily published any one of them, had it been the only one under consideration. Because the alternatives of publication of all four or of none were both unsatisfying, all have agreed to the compromise embodied here of a paper carrying all six names as coauthors, consisting of an introductory section describing the basics written by a disinterested party, and followed by four sections, one written by each of the four groups, briefly describing the highlights of their own approach and elaboration. The result also appears in work by Józef H. Przytycki and Paweł Traczyk, hence the PT. The reason for simultaneous discovery is quite clear: the HOMFLYPT polynomial is a relatively straightforward generalization of the (at the time) recently introduced Jones polynomial.<|endoftext|> TITLE: Is every connected semisimple linear Lie group the connected component of (the real points of) an algebraic group? QUESTION [10 upvotes]: Is every connected semisimple linear Lie group the identity connected component of (the real points of) an algebraic group? I was told some fact along this line is true but could not find any reference after searching for a while. REPLY [6 votes]: (Comments converted into an answer:) Mostow (to whom I think this is often attributed?) gives a detailed proof in (1949, Lemmas 2.2, 2.3). Borel (2001, pp. 152, 114) notes that algebraicity of perfect (e.g. semisimple) linear Lie algebras was claimed by Cartan in (1897, p. 547), and spells out what “it may not too far fetched to believe” would have been his simple proof.<|endoftext|> TITLE: Converse to Wolpert's Lemma QUESTION [9 upvotes]: Recall Wolpert's lemma: Let X,Y be hyperbolic surfaces and $f:X\to Y$ a $K$-quasiconformal homeomorphism. For any homotopy class of curves $c$ let $\ell(c)$ denote the length of the geodesic in the class. Then $$\frac{\ell(c)}{K}\leq \ell(f(c))\leq K\ell(c)$$ I am wondering: if $f$ is a $C^1$ diffeomorphism between closed hyperbolic surfaces that has this property, is it $K$-quasiconformal? Of course it is quasiconformal for some other constant, by nature of being $C^1$. REPLY [6 votes]: The answer is no, but it's actually a deep question and leads to another metric on the Teichmüller space of surfaces. The minimum quasi-conformal constant of a map in a given homotopy class is the Teichmüller metric on surfaces; it is also equal to the maximal ratio of extremal lengths of any curves. If you look at the ratio of hyperbolic lengths, as in your question, you get a quite different metric on the space of surfaces, studied by William Thurston: Minimal stretch maps between hyperbolic surfaces. That paper effectively uses an asymmetric version of the inequality ($\ell(f(C)) \le K \ell(C)$), but the symmetrized version is also a very different metric from the standard one given by quasiconformal stretch. Both the asymmetric stretch metric was studied recently in detail by Dumas, Lenzhen, Rafi, and Tao. The paper The converse of the Schwarz Lemma is false by Maxime Fortier-Borque gives many relevant examples, although that paper is concerned with the slightly different situation of embeddings between non-compact surfaces and maps that decrease length, effectively the case $K=1$ in your questions.<|endoftext|> TITLE: Is it possible to stab (every rotation of) any four element subset of $\mathbb Z_n$ with less than $n/2$ elements? QUESTION [11 upvotes]: Say that $S\subset \mathbb Z_n$ is stabbed by $X\subset \mathbb Z_n$ if for every $t$ we have $(S+t)\cap X\ne \emptyset$. Is there for every $|S|=4$ an $|X| TITLE: Polynomial defined recursively by a resultant QUESTION [5 upvotes]: Cross posting from MSE. Definition: For any natural number $n\ge 3$, define the polynomial $P_{n}\left(x_1,x_2,...,x_{n-1},x_{n} \right)$, with indeterminates $x_{i}$, where $i\in\{1,2,...,n-1,n\}$, by the following recursive rule: $$ \begin{align} P_{n}\left(x_1,x_2,...,x_{n-1},x_{n} \right) \equiv \begin{cases} \left(x_1+x_2+x_3\right)^2-2 \left(x_{1}^2+x_{2}^2+x_{3}^2 \right), & \text{if } n=3 \\\ \mathrm{Res}_{\mathbf{x}}\left(P_{3}\left(x_1,x_2,\mathbf{x} \right),P_{n-1}\left(x_3,x_4,...,x_{n},\mathbf{x} \right) \right), & \text{if } 4 \le n \end{cases} \end{align} $$ Observe that if $4 \le n$ then $P_{n}\left(x_1,x_2,...,x_n \right)= \mathrm{Res}_{\mathbf{x}}\left(P_{n-1}\left(x_1,x_2,...,x_{n-2},\mathbf{x} \right),P_{3}\left(x_{n-1},x_n,\mathbf{x} \right) \right)$. Example #1: If $n=4$ then $$ \begin{align} P_{4}\left(x_1,x_2,x_3,x_4 \right)=\mathrm{Res}_{\mathbf{x}}\left(P_{3}\left(x_1,x_2,\mathbf{x} \right),P_{3}\left(x_3,x_4,\mathbf{x} \right) \right)=\left(\left(x_1+x_2+x_3+x_4\right)^2-2 \left(x_{1}^2+x_{2}^2+x_{3}^2+x_{4}^2 \right)\right)^2-64x_{1}x_{2}x_{3}x_{4} \end{align} $$ Example #2: If $n=5$ then $$ \begin{align} P_{5}\left(x_1,x_2,x_3,x_4,x_5 \right)=\mathrm{Res}_{\mathbf{x}}\left(P_{3}\left(x_1,x_2,\mathbf{x} \right),P_{4}\left(x_3,x_4,x_5,\mathbf{x} \right) \right)=\left(\left(\left(x_1+x_2+x_3+x_4+x_5\right)^2-2 \left(x_{1}^2+x_{2}^2+x_{3}^2+x_{4}^2+x_{5}^2 \right)\right)^2-64\left( x_{1}x_{2}x_{3}x_{4}+x_{1}x_{2}x_{3}x_{5}+x_{1}x_{2}x_{4}x_{5}+x_{1}x_{3}x_{4}x_{5}+x_{2}x_{3}x_{4}x_{5}\right) \right)^2-2048x_{1}x_{2}x_{3}x_{4}x_{5}\left(8\left(x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{2}x_{5}+x_{1}x_{3}x_{4}+x_{1}x_{3}x_{5}+x_{1}x_{4}x_{5}+x_{2}x_{3}x_{4}+x_{2}x_{3}x_{5}+x_{2}x_{4}x_{5}+x_{3}x_{4}x_{5}\right) -\left(x_1+x_2+x_3+x_4+x_5 \right)\left( \left(x_1+x_2+x_3+x_4+x_5\right)^2-2 \left(x_{1}^2+x_{2}^2+x_{3}^2+x_{4}^2+x_{5}^2 \right)\right)\right) \end{align} $$ As you can see, $P_n$ becomes quite complicated fairly quickly. Question: I want to find a closed form expression for $P_n$. I doubt that this is feasible, so I also look for alternative ways to compute $P_n$ which may be quicker and more efficient. I would also like to know if there are papers or theories which deal with this sort of polynomial objects. Keeping this in mind, I realize that this question may be regarded as "soft". Possible lead #1: The object at hand (might, still) "screams" partition-related symmetric-polynomial objects, like the Schur Polynomials and their generalizations. Unfortunately, playing around with those hasn't produced a hopeful pattern for a general formula, yet. Possible lead #2: Even though $P_n$ is a symmetric polynomial, so it has a canonical representation in Elementary Symmetric Polynomials, and some powerful tools for manipulation, it just seems to me, from playing around, that the complexity of $P_{n+1}$ explodes when compared to $P_n$ when they are represented with symmetric polynomials. Another direction might have something to do with identities involving powers of sums of squares, like those apperaing on this wonderful page. Major Edit (definition changed): Changed the definition of $P_n$ to an equivalent, more simple one, so one doesn't have to check whether $P_n$ is well defined. The original definition can still be seen on the MSE question (link at the top of this question). REPLY [3 votes]: This question is close to a central one in Rational Trigonometry, which is to determine the relations between successive quadrances between a cyclic list of points on a line. Here quadrance refers to what might naively be called "squared distance", but it is better to think of just applying a fixed symmetric bilinear form to successive vectors (it turns out not to matter what the bilinear form is, as long as it is non-degenerate). This is coming from my 2005 book Divine Proportions: Rational Trigonometry to Universal Geometry. The problem stated is more or less Exerise 5.19: Is there a Quintuple quad formula? Generalize. Here is a solution. First observe that the first polynomial $P_3$ is what in the book is called Archimedes function (pg 64) $$ A(a,b,c)=(a+b+c)^2-2(a^2+b^2+c^2). $$ This determines the squared area of a triangle, in the Euclidean case, in terms of the three quadrances of the sides; in fact $A=16 \cdot \text{area}^2$ so this is really a rational version of Heron's formula. Second observe that the second polynomial $P_4$ is what in the book is called the Quadruple quad function (pg 70): $$ Q(a,b,c,d)=((a+b+c+d)^2-2(a^2+b^2+c^2+d^2))^2-64abcd. $$ This is closely related to formulas for areas of quadrilaterals. In Theorem 31 it is shown that if $A_1,A_2,A_3,A_4$ are collinear points and $Q_{ij}=Q(A_i,A_j)$ are the quadrances formed by them, then $$Q(Q_{12},Q_{23},Q_{34},Q_{14})=0.$$ So these polynomials are intimately tied up with the geometry of polygons in the plane, and what happens to them when the polygons become degenerate, that is lie on a line. When they lie on a line, the quadrances can be interpreted, after a normalization, as squares. This is the reason for the following identities which make the pattern hopefully clear: $$ A(x^2,y^2,z^2)=(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$$ (Archimedes formula, Theorem 29) $$ Q(x^2,y^2,z^2,w^2)= (w+x+y+z)(w-x+y+z)(w-x+y-z)(w-x-y+z)(w-x-y-z)(w+x+y-z)(w+x-y+z)(w+x-y-z)$$ (Brahmagupta's identity, Theorem 32) In general we get, up to a sign, that the $n$th polynomial, evaluated on $n$ squares, is the product of all linear factors where one variable has a coefficient of $1$, and the others all have coefficients of $1$ or $-1$. This should make it clear how to write down such polynomials: write down the products of all these linear factors, and then expand the polynomial, and rewrite it as a polynomial in the squares of the original variables. This story is closely connected with many other remarkable formulas, for example due to Robbins on areas of non-convex cyclic polygons and generalizations. For more info, have a look at my YouTube series MathFoundations B starting around video MathFoundations123 and the next ten or so.<|endoftext|> TITLE: O. Leroy's thesis on fundamental groupoids QUESTION [7 upvotes]: Does someone have a copy of the O. Leroy's thesis: Groupoïde fondamental et théorème de van Kampen en théorie des topos or has the ability to make a digitalization ? The thesis was done at Université Montpellier II 1978. REPLY [3 votes]: Jean Malgoire kindly sent me a scan back to 2015, with authorization to transmit it. So here it is: Link: https://plmbox.math.cnrs.fr/f/7ffa366379144dd4bacc/ Direct Download Link: https://plmbox.math.cnrs.fr/f/7ffa366379144dd4bacc/?dl=1 The password is : groupoide Actually, the plan was to type it in $\LaTeX$ and send it to arXiv. As far as I know, this has not be done yet. That being said, if enough people are interested, we could make a collaborative project. Please just leave an answer if you are (especially if you already have some expertise in such projects !). Of course, this could be the opportunity to translate the text into english.<|endoftext|> TITLE: Normal bundle of Whitney embedding QUESTION [10 upvotes]: Let $X$ be a real $n$ dimensional manifold. One knows that it can be embedded into $\mathbb{R}^{2n}$ by the Whitney embedding theorem. The normal bundle for such an embedding will be a rank $n$ real vector bundle. I would be interested in understanding the connection between this embedding, its normal bundle and the rank $n$ bundles on $X$. How does the normal bundle change under the isotopy of this embedding? Is it possible to obtain all the isomorphism classes of rank $n$ vector bundles by such isotopies? If the answer for 2. is negative, is it still possible to find an embedding of $X$ into $\mathbb{R}^{2n}$ for each vector bundle of rank $n$, such that the corresponding normal bundle is isomorphic to it? I was thinking that the answer should have something to do with the self-intersection of $X$ in the vector bundles and its self-intersection in $\mathbb{R}^{2n}$, but don't really see how to use that. Edit: I have realized that it shouldn't be possible to obtain a vector bundle with a non-zero self intersection of the zero section like this. However, I would still like to know if it works for all the other cases. Edit 2: Based on Mike's comment, I have realized I have missed something very obvious. I am just thinking whether one still gets all the rank $n$ vector bundles which complete the tangent bundle to a trivial $2n$ bundle. REPLY [5 votes]: This is to shed some light on Part 3 which asks for a classification of normal vector bundles of a smooth $n$-dimensional submanifold $X$ of $\mathbb R^{2n}$. The normal bundle $\nu$ to $X$ is stably isomorphic to the negative of $TX$, and in particular the Pontryagin and Stiefel-Whitney classes of $\nu$ are completely determined by those of $X$. Also if $X$ is orientable, then the Euler class of $\nu$ vanishes (contactibility of $\mathbb R^{2n}$ allows to push the embedding off itself). On the other hand, if $X$ is non-orientable, the twisted Euler class (i.e., the first obstruction to the existence of a nowhere zero section) of $\nu$ can be nonzero and is an important invariant (see below). For simplicity let us assume that $X$ is closed. Let us also ignore the cases $n\le 3$ where if $X$ is orientable one can use the classification of oriented bundles Dold-Whitney in [Classification of Oriented Sphere Bundles Over A 4-Complex], and presumably with some work one deal with the nonorientable case. Instead of classifying normal bundles to $X$ let us describe isotopy classes of embeddings of $X$ into $\mathbb R^{2n}$. Of course, isotopic embeddings have isomorphic normal bundles. One quick statement is a theorem of Haefliger-Hirsch [On the existence and classification of differentiable embeddings. Topology 2 1963 129–135] that If $X$ is simply-connected and $n\ge 4$, then there is only one isotopy class of embeddings from $X$ into $\mathbb R^{2n}$, and hence, only one normal bundle. More generally, if $X$ is orientable and $n\ge 4$, then the set of isotopy classes of smooth embeddings of $X$ into $\mathbb R^{2n}$ is bijective to $H_1(X)$ if $n$ is odd, and to $H_1(X;\mathbb Z_2)$ if $n$ is even. If $X$ is non-orientable and $n$ is even and $\ge 4$, there is a more complicated classification by Kitada in [Classification of embeddings of a non-orientable manifold. J. Fac. Sci. Univ. Tokyo Sect. IA Math. 18 (1971/72), 435–442]. The conclusion is that the set of isotopy classes is bijective to $\mathbb Z\oplus H^{n-1}(X)/K$ where $H$ is a certain subgroup of $H^{n-1}(X)$. The $\mathbb Z$-factor corresponds to the twisted Euler class of $\nu$, and in particular, if $H^{n-1}(X)=0$, then $\nu$ is completely determined by the twisted Euler class. I don't know what happens if $X$ is non-orientable and $n$ is odd.<|endoftext|> TITLE: monoidality of $ A\otimes (-) $ with $ A $ monoid belonging to the center QUESTION [6 upvotes]: Let $(\mathcal{C}, \otimes)$ a monoidal category, and $(A, m, e)$ a monoid (where $m: A\otimes A\to A$, $e: I\to A$ ecc. ), with $(A, u)$ belonging to the centre of $(\mathcal{C}, \otimes)$: $u: A\otimes (-)\cong (-)\otimes A$, ecc. see [JS] p.38. Consider tha (usual) functor $F_A(-):= A\otimes (-): \mathscr{C}\to \mathscr{C}$, there are the natural morphisms: $\alpha_{X,Y}: F_A(X)\otimes F_A(Y)= A\otimes X\otimes A\otimes Y\xrightarrow{1u1} A\otimes A\otimes X\otimes Y \xrightarrow{m11} A\otimes X\otimes Y=F_A(X\otimes Y)$ $\phi: I\cong I\otimes I\xrightarrow{e1}A\otimes I=F_A(I)$. Question: Is the data above define a monoidal functor? What coherence axioms (between the monoidal and centre object structure) we need? [JS] Braided Tensor Categories, A.Joyal, R.Street https://www.sciencedirect.com/science/article/pii/S0001870883710558 REPLY [5 votes]: I write $XY$ instead of $X\otimes Y$ and ignore canonical isomorphism and morphisms labels (unambiguous). I assume the axiom: Ax) $(AXAY\to AAXY\to AXY) = (AXAY\to XAAY\to XAY\to AXY)$. The second member is an alternative computation of $\alpha$. First I have to show that: $F_A(X)F_A(Y)F_A(Z)\to F_A(XY)F_A(Z)\to F_A(XYZ) =$ $F_A(X)F_A(Y)F_A(Z)\to F_A(X)F_A(YZ)\to F_A(XYZ)$ i.e. that $$AXAYAZ\to AAXYAZ\to AXYAZ\to AXAYZ\to AAXYZ\to AXYZ$$ is equal to: $$AXAYAZ\to AXAAYZ\to AXAYZ\to AAXYZ\to AXYZ$$ proceed: $(AXAYAZ\to AAXYAZ\to AXYAZ)\to (AXAYZ\to AAXYZ\to AXYZ)=^{Ax} =(AXAYAZ\to XAAYAZ\to XAYAZ\to AXYAZ) \to (AXAYZ\to XAAYZ\to XAYZ\to AXYZ)=^{natuality}$ $=AXAYAZ\to XAAYAZ\to (XAYAZ \to XAAYZ\to XAYZ\to AXYZ)=^{Ax}$ $=AXAYAZ\to XAAYAZ\to (XAYAZ \to XYAAZ\to XYAZ\to XAYZ\to AXYZ)=^{Ax}$ $=A_1XA_2YA_3Z\to XA_1A_2YA_3Z\to XA_{1,2}YA_3Z \to XA_{1,2}A_3YZ\to XAYZ\to AXYZ=^{naturality}$ $=A_1XA_2YA_3Z\to XA_1A_2YA_3Z\to XA_1A_2A_3YZ\to XA_{1,2}A_3YZ \to XAYZ\to AXYZ=^{monoid}$ $=A_1XA_2YA_3Z\to XA_1A_2YA_3Z \to XA_1A_2A_3YZ\to XA_1A_{2,3}YZ \to XAYZ\to AXYZ=^{naturality}$ $=A_1XA_2YA_3Z\to A_1XA_2A_3YZ\to XA_1A_2A_3YZ \to XA_1A_{2,3}YZ \to XAYZ\to AXYZ=^{naturality}$ $=A_1XA_2YA_3Z\to A_1XA_2A_3YZ\to (A_1XA_{2,3}YZ\to XA_1A_{2,3}YZ \to XAYZ\to AXYZ)=^{Ax}$ $=A_1XA_2YA_3Z\to A_1XA_2A_3YZ\to A_1XA_{2,3}YZ\to A_1A_{2,3}XYZ \to AXYZ.$ About the unity axioms we prove that $F_A(X)\cong IF_A(X)\to F_A(I)F_A(X)\to F(IX)\cong F(X)$ is the identity, this is:: $AX\cong IAX\to AAX\cong AIAX\cong IAAX\to IAX\cong AX$ for naturality this is: $AX\cong IAX\to AAX\cong AIAX\cong IAAX\cong AAX\to AX$ for the usual monoidal topic this is : $AX\cong IAX\to AAX\to AX=1$. The other unitary axiom is quite similar.<|endoftext|> TITLE: Regularity properties of Turing-invariant and arbitrary sets of reals QUESTION [6 upvotes]: The question whether Turing determinacy implies $AD$ is a well-known open problem. I was wondering if anything is known about the following analogous question: Let $\Gamma$ be a regularity property (e.g. Lebesgue measurability, Baire property, $\mathbb P$-measurability for a reasonable arboreal forcing $\mathbb P$, etc). Assuming $ZF+DC+"\Gamma$ holds for all Turing-invariant sets of reals$"$, does it follow that $\Gamma$ holds for all sets of reals? REPLY [2 votes]: Here is not an answer but an example: Let PSP be the statement that every uncountable set of reals has a perfect subset and TPSP that every uncountable set of Turing degrees has a perfect subset. Then PSP and TPSP are equivalently consistent over $ZF+DC$. Clearly over $ZF+DC$, PSP implies TPSP but I don't know whether the inverse is true.<|endoftext|> TITLE: Equivalence between Lowenheim-Skolem Theorem and Godel Completeness QUESTION [7 upvotes]: In papers published 1920 and 1922, Skolem offered two separate proofs of a result due to Lowenheim. On this basis we can distinguish a strong and a weak version of the Lowenheim-Skolem theorem as follows: The weak (1922) version states that if a closed formula $\phi$ of quantification theory is satisfiable, then it is satisfiable in a countable domain. This version does not require the Axiom of Choice; a model of $\phi$ is built up from below using numerals to instantiate the variables. The (1920) "subdomain" version states that if $\phi$ is satisfiable in an (infinite) domain D, then it is satisfiable in a countable subdomain D' of D, where the predicates retain the same meaning in D' as in D (modulo the restriction). Skolem (1922) gives no formal proof procedure. However, the level-by-level construction of the denumerable model implicitly supplies an effective procedure for refuting a formula in a finite number of steps. This occurs if we reach a level for which no satisfying truth-value assignment exists for the approximation to the formula considered at that level, thus implying the formula's negation. This explains why Godel (Coll. Wrks. Vol 1, p. 52) writes that Skolem's weak theorem implies completeness: "Skolem...could justly claim...that, in his 1922 paper, he implicitly proved: 'Either A is provable or $\neg$A is satisfiable” (“provable” taken in an informal sense).' What about the subdomain version? In this version, Skolem starts with $\phi$ in normal form (i.e. $\forall x \exists y\psi$, where $\psi$ quantifier-free) and uses the axiom of choice to find witnesses $f(x)$ for the existential quantifier, taken from a domain D in which $\phi$ is assumed to be satisfied. Let $a$ be an arbitrary element from D. The proof continues by closing $a$ under the following operation. Consider all the classes $X\subseteq $D such that $a \in X$ and if $x \in X$, then $f(x) \in X$. Skolem then applies a result from Dedekind's chain theory to conclude that the intersection of all such classes $X$ must be denumerable (cf. Dedekind 1888). By omission, Godel implies that this subdomain version cannot be interpreted so as to yield completeness. I assume this is because the method Skolem uses does not, as in the weak version, implicitly describe a refutation procedure for $\phi$ since the assumption of $\phi$'s satisfiability in D is crucial in the description of the submodel. Question: (a) Is that the correct reading? (b) Does it mean that the weak version, in order to provide a refutation procedure, does not make use of the antecedent in its statement? Main references Skolem 1920, Logisch-kombinatorische Untersuchungen über die Erfüllbarkeit oder Beweisbarkeit mathematischer Sätze nebst einem Theoreme über dichte Mengen, Videnskapsselskapet Skrifter, translated in (3) as Logico-combinatorical investigations in the satisfiability or provabilitiy of mathematical propositions: A simplified proof of a theorem by L. Löwenheim and generalizations of the theorem Skolem 1922, Einige Bemerkungen zu axiomatischen Begründung der Mengenlehre, 5th Scand. Math. Congress, translated in (3) as Some remarks on axiomatized set theory Jean van Heijenoort (ed.) 1977, From Frege to Gödel: A Source Book in Mathematical Logic, 1879–1931, Harvard University Press REPLY [6 votes]: On Question (a), yes this is correct. On Question (b), almost yes. If you drop the assumption that ϕ is satisfiable, then Skolem has given a sound and complete procedure for refuting a formula in a finite number of steps. However it is not (and cannot be) also effective. If the formula is in fact not refutable then the procedure may run forever without showing at any finite point that the formula is not refutable. It is easy to prove tree/tableaux methods like this cannot be sound, complete, and effective: some formulas have only infinite interpretations, and the tree cannot construct such in a finite number of steps. What is harder is to show there is no sound, complete, effective procedure for refutability at all.<|endoftext|> TITLE: Extension of Hopf fiber bundle to (an equivariant) 2 dimensional vector bundle QUESTION [5 upvotes]: Let $p:S^3 \to S^2$ be the Hopf fibration which is a result of the standard action of $S^1$ on $S^3$. Is there a $2$ dimensional vector bundle $\tilde{p}:E \to S^2$ such that $S^3\subset E$ and $\tilde{p}_{|S^3}=p$? Is there a vector bundle $E$ as above with the following extra condition:The total space $E$ can be acted by $S^1$ with linear isomorphism and this action would be the extension of the standard action of $S^1$ on $S^3$? Note: One can ask the same question for such extension of an arbitrary principal bundle to an equivariant vector bundle of arbitrary dimension. REPLY [7 votes]: I think both of your questions can by answered positive. Since the Hopf fibration is an $S^1$ bundle it comes as a sphere bundle of complex line bundle over $S^2$. From the Gysin sequence the first Chern class of that line bundle must be a generator of $H^2(S^2;\mathbb Z)$ which determines the isomorphism type of the line bundle. Moreover a circle action on $S^2$ can be lifted to a line bundle if and only if the first Chern class possess an equivariant extension in $H_{S^1}(S^2;\mathbb Z)$ which here is always true. Edit: I would like to remark that lifting actions from the base space to a vector bundle is in general a very difficult problem (which is solved for complex line bundles) but unknown (as far as I know) for complex vector bundle of rank higher than 2. There are some results for oriented vector bundles over spheres.<|endoftext|> TITLE: Is a sign-preserving operator on $L^2$ a multiplication? QUESTION [7 upvotes]: Let $T:L^2(\mu)\to L^2(\mu)$ be a linear and continuous operator, where $L^2(\mu)$ is the (real) $L^2$-space to some $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$. $T$ is assumed to be sign-preserving in the sense that $$ v(x) \cdot (Tv)(x) \ge0 $$ for $\mu$-almost all $x\in \Omega$ and all $v\in L^2(\mu)$. Does this imply that $T$ is a multiplication? That is, does there exist $\phi\in L^\infty(\mu)$ such that $Tv = u\cdot v$? I could show the following property: $$ \chi_{A^c} \cdot (T\chi_A) = 0 $$ $\mu$-almost everywhere for all characteristic functions $\chi_A$ of $A\in\Sigma$. This would prove the question for $\mathbb R^n$ or $l^2(\mathbb N)$. I was not able to prove the question in the general case. REPLY [5 votes]: First, if $uv=0$ then $uTv=0$ a.e.: indeed, applying the hypothesis with the function $\epsilon u+\epsilon^{-1}v$ and evaluating on $\{u\neq 0\}$, we get $$0\le \epsilon^2uTu+\epsilon^{-2}vTv+uTv+vTu=\epsilon^2 uTu+uTv$$ a.e. and deduce the claim (sending $\epsilon\to 0$). This gives the property $(*)$ that you showed. If the space is $\sigma$-finite, using $(*)$ we can easily reduce to the case that the measure is finite. Now take $v:=T1$. We claim that $Tu=uv$ (a.e.) when $u$ is simple: since $T$ is linear, it suffices to show this when $u=1_A$ is a characteristic function. But $$v=T1=T1_A+T1_{A^c}$$ and $T1_{A^c}$ vanishes (a.e.) on $A$, hence $T1_A=v$ (a.e.) on $A$. Since $T1_A$ vanishes (a.e.) on $A^c$, the claim follows. Taking $A:=\{|v|>\|T\|\}$, if $\mu(A)>0$ we see that $\|T1_A\|_{L^2}>\|T\|\|1_A\|_{L^2}$, contradiction. So $v\in L^\infty$. The statement now follows since simple functions are dense. It seems false to me without assuming the space to be $\sigma$-finite: take $\Omega:=[0,1]^2$, with $\mu:=\mathcal{H}^1$ (1-dimensional Hausdorff measure) and the $\sigma$-algebra $\mathcal A$ generated by horizontal and vertical slices ($\{s\}\times[0,1]$ and $[0,1]\times\{t\}$). Now with little work you can show that all elements of $\mathcal A$, up to adding and removing negligible sets, are of the form $$\bigcup\Big(\{s_i\}\times[0,1]\Big)\cup\bigcup\Big([0,1]\times\{t_j\}\Big),$$ where both unions are (at most) countable. Hence, $L^2(\mu)$ splits as a direct sum $V\oplus W$, where $V$ consists of functions of the form $f=\sum_i a_i 1_{\{s_i\}\times[0,1]}$ and $W$ of similar "vertical" functions. Now declare $T$ to act by multiplication by $0$ on $V$ and multiplication by $1$ on $W$. It's easy to see that there is no consistent choice of $v$. If you don't want atoms in the counterexample, take instead the $\sigma$-algebra generated by sets of the form $\{s\}\times E'$ and $E\times\{t\}$, where $s,t$ range in $[0,1]$ and $E,E'$ vary among Borel subsets of $[0,1]$. In this case, measurable sets have the form $$\bigcup(\{s_i\}\times E_i')\cup\bigcup(E_j\times\{t_j\})\cup (E\times E')\cup N,$$ where $E_j,E$ are Borel subsets of $[0,1]\setminus\bigcup\{s_i\}$, $E_i',E'$ are Borel subsets of $[0,1]\setminus\bigcup\{t_j\}$, and finally $N$ is any subset of $\Big(\bigcup\{s_i\}\Big)\times\Big(\bigcup\{t_j\}\Big)$. Once you have this, you can argue as before.<|endoftext|> TITLE: Equivalence between categories of coherent sheaf of codimension p QUESTION [6 upvotes]: Let $X$ be a noetherian and separated scheme and $M(X)$ denote the abelian category of coherent sheaves on $X$. Let $M^{P}(X) = \lbrace \mathcal{F} \in M(X) \hspace{2mm} : Codim(sup(\mathcal{F}), X) \geq p \rbrace$ be the full subcategory of $M(X)$. I want to show that the quotient category $M^{p}(X)/M^{p+1}(X)$ is equivalent to $\bigoplus _{x \in X^{p}} \mathcal{A}(O_{X,x})$, where $\mathcal{A}(O_{X,x})$ denote the category of $O_{X,x}$-module of finite length. Here $X^{p}$ denote the points of codimenison $p$. The natural functor which seems to be equivalence is given as $L : M^{P}(X) \rightarrow \bigoplus _{x \in X^{p}} \mathcal{A}(O_{X,x})$ by $L(\mathcal{F})$ = $\oplus (\mathcal{F}_{x_{i}})$ where $x_{i}$ are codimension $p$ points. Clearly this functor has kernel $M^{p+1}(X)$, so it induces a faithful functor $U : M^{p}(X)/M^{p+1}(X) \rightarrow \bigoplus _{x \in X^{p}} \mathcal{A}(O_{X,x})$. Now I don't know how prove that $U$ is full and essential surjective. Any help would be great. Thanks in advance. REPLY [3 votes]: First, note that the category of finite length modules on a noetherian local ring $(A, \mathfrak{m})$ is equivalent to the direct limit of the categories of finitely generated modules on $A/\mathfrak{m}^n$ as $n \rightarrow \infty$ (this just says that any finite-length module is killed by a power of the maximal ideal). In other words, the natural inclusion maps induce an equivalence $$ \varinjlim_n \mathrm{Coh}(\mathrm{Spec}(A/\mathfrak{m}^n)) \rightarrow \mathscr{A}(A) $$ Next, we will show the statement in the special case where the scheme is irreducible and $p = 0$. Indeed, let $Y$ be an irreducible noetherian scheme with unique generic point $y$. Then the natural restriction map $$M^0(Y)/M^1(Y) = \mathrm{Coh}(Y)/M^1(Y) \rightarrow \mathrm{Coh}(\mathrm{Spec}(\mathscr{O}_{Y,y}))= \mathscr{A}(\mathscr{O}_{Y,y})$$ is an equivalence of categories (the last equality is because $\mathscr{O}_{Y,y}$ is artinian and therefore all finitely generated modules are finite length). Now, by definition $\mathscr{O}_{Y,y} = \varinjlim_{U \ni y} \mathscr{O}_Y(U)$. By "spreading out", this implies that the natural restriction map $$\varinjlim_{U \ni y} \mathrm{Coh}(U) \rightarrow \mathrm{Coh}(\mathrm{Spec}(\mathscr{O}_{Y,y}))$$ is an equivalence of categories. (This is a special case of the direct limit techniques developed in EGA IV$_3$ §8, but can be proved simply by "chasing denominators"). These maps are compatible (as they are all defined by restriction of modules), so we are reduced to showing that the natural functor $$ \mathrm{Coh}(Y)/M^1(Y) \rightarrow \varinjlim_{U \ni y} \mathrm{Coh}(U) $$ is an equivalence of categories. Note that the kernel of the natural map $\mathrm{Coh}(Y) \rightarrow \varinjlim_{U \ni y} \mathrm{Coh}(U)$ consists of those coherent sheaves on $Y$ whose restriction to some non-empty open subset is zero, which is exactly $M^1(Y)$ (as $Y$ is irreducible, so any proper open subset has codimension at least $1$). Thus, the functor is at least faithful. Now, fix some $U \ni y$, define the closed subset $Z := |Y| - |U|$, and consider the restriction map $\mathrm{Coh}(Y) \rightarrow \mathrm{Coh}(U)$. This is an exact functor with kernel equal to the thick Serre subcategory $M_Z(Y)$ consisting of modules whose support (considered as a topological space) is contained in $Z$. We claim that the induced faithful functor $$ \mathrm{Coh}(Y)/M_Z(Y) \rightarrow \mathrm{Coh}(U) $$ is an equivalence. To see that it is essentially surjective, we use Exercise II.5.15 in Hartshorne, or Tag 01PF in the Stacks project: for any quasi-coherent sheaf $\mathscr{F}$ on $Y$ and any coherent subsheaf $\mathscr{G} \subseteq \mathscr{F}|_U$, there is a coherent subsheaf $\widetilde{\mathscr{G}} \subseteq \mathscr{F}$ such that $\widetilde{\mathscr{G}}|_U = \mathscr{G}$. Now, if $\mathscr{G}$ is any coherent sheaf on $U$, we let $j \colon U \hookrightarrow Y$ be the open immersion, and take $\mathscr{F} = j_* \mathscr{G}$, so $\mathscr{F}|_U = \mathscr{G}$. This shows that any coherent sheaf on $U$ is the restriction of a coherent sheaf on $Y$. To see that the functor is full, let $\varphi \colon \mathscr{G}_1 \rightarrow \mathscr{G}_2$ be a morphism of sheaves. Take $\widetilde{\mathscr{G}_1} \subseteq j_* \mathscr{G}_1$ as above, so $\widetilde{\mathscr{G}_1}|_U = \mathscr{G}_1$. We have a morphism $j_* \varphi|_{\widetilde{\mathscr{G}_1}} \colon \widetilde{\mathscr{G}_1} \rightarrow j_* \mathscr{G}_2$. Its image is a coherent subsheaf $\mathscr{H} \subseteq j_* \mathscr{G}_2$. Now, we can apply the above lemma to the sheaf $j_* \mathscr{G}_2/\mathscr{H}$ (whose restriction to $U$ is $\mathscr{G}_2/\mathrm{im } \varphi$) to find a coherent subsheaf $\widetilde{\mathscr{G}_2}$ of $j_* \mathscr{G}_2$ containing $\mathscr{K}$ such that $\widetilde{\mathscr{G}_2}|_U = \mathscr{G}_2$. Thus, $j_* \varphi$ restricts to a map $\widetilde{\varphi} \colon \widetilde{\mathscr{G}_1} \rightarrow \widetilde{\mathscr{G}_2}$, which restricts to $\varphi$. Applying this equivalence, we are reduced to showing that the natural faithful functor $$ \mathrm{Coh}(Y)/M_1(Y) \rightarrow \varinjlim_{U \ni y} \mathrm{Coh}(Y)/M_{|Y| - |U|}(Y) $$ is an equivalence. This functor comes from the identity functor on $\mathrm{Coh}(Y)$ by compatibility of the various restriction maps. But $M_1(Y) = \cup_{U \ni y} M_{|Y| - |U|}(Y)$ is the (full) thick Serre subcategory of $\mathrm{Coh}(Y)$ consisting of objects of $\mathrm{Coh}(Y)$ which are contained in $M_{|Y| - |U|}(Y)$ for some $U \ni Y$. Now, if $\mathcal{A}$ is an arbitrary abelian category and $B_i, i \in I$ are a direct system of thick Serre subcategories, the natural functor $$ A/(\cup_{i \in I} B_i) \rightarrow \varinjlim_{i \in I} A/B_i $$ induced by the identity on $A$ is an equivalence. This follows by comparing the universal properties of the left and right sides (alternatively, this is easy to see from the construction of the quotient categories as localizations). This finishes the proof of the case $p = 0$ for irreducible $Y$. Now, let $Z \subseteq |X|$ be an irreducible closed subset of codimension $p$. Define $M_Z(X)$ to be the thick Serre subcategory consisting of coherent sheaves on $X$ whose (topological) support is contained in $Z$, and $M_Z^1(X) = M_Z(X) \cap M^{p+1}(X)$ to be the further subcategory consisting of coherent sheaves on $X$ whose (topological) support is a proper closed subset of $Z$. Let $z$ be the unique generic point of $Z$, and consider the local ring $(\mathscr{O}_{X, z}, \mathfrak{m}_z)$. Restriction defines an exact functor $M_Z(X) \rightarrow \mathscr{A}(\mathscr{O}_{X, z})$ with kernel $M_Z^1(X)$, so we have a faithful exact functor $$ M_Z(X)/M_Z^1(X) \rightarrow \mathscr{A}(\mathscr{O}_{X, z}) $$ Let $Y_1$ be the closed subscheme of $X$ given by $Z$ with the reduced closed subscheme structure, and $Y_n$ the $n$-th infinitesimal neighborhood of $Y_1$ in $X$. This is the closed subscheme of $X$ cut out by the $n$-th power of the ideal sheaf defining $Y_1$. Each of the $Y_n$'s has unique generic point $z$ and $\mathscr{O}_{Y_n, z} = \mathscr{O}_{X, z}/\mathfrak{m}_z^n$. By the case $p = 0$ for the irreducible $Y_n$'s, we have an equivalence $$ \varinjlim_n \mathrm{Coh}(Y_n)/M^1(Y_n) \rightarrow \varinjlim_n \mathrm{Coh}(\mathrm{Spec}(\mathscr{O}_{Y_n, z})) = \mathscr{A}(\mathscr{O}_{X, z}) $$ Thus, the inclusion functors $\mathrm{Coh}(Y_n) \rightarrow M_Z(X)$ induce a functor $$ \mathscr{A}(\mathscr{O}_{X, z}) \rightarrow M_Z(X)/M_Z^1(X) $$ Chasing the definitions, we can see easily that these functors are quasi-inverse to each other. Finally, we have an obvious inclusion $M_Z(X) \hookrightarrow M^p(X)$ for any $Z \subseteq |X|$ of codimension $p$. Since $M_Z(X) \cap M^{p+1}(X) = M_Z^1(X)$, this induces a faithful exact functor $$ M_Z(X)/M_Z^1(X) \hookrightarrow M^{p}(X)/M^{p+1}(X) $$ Then, taking direct sums of coherent sheaves defines a functor $$ \bigoplus_{x \in X^p} \mathscr{A}(\mathscr{O}_{X, x}) = \bigoplus_{\overline{x}\mid x \in X^p} M_{\overline{x}}(X)/M_{\overline{x}}^1(X) \rightarrow M^p(X)/M^{p+1}(X) $$ We claim that this is the desired quasi-inverse functor to the restriction functor $M^p(X)/M^{p+1}(X) \rightarrow \bigoplus_{x \in X^p} \mathscr{A}(\mathscr{O}_{X,x})$. To see this, use the fact that if $\mathscr{F} = \oplus_{i=1}^n \mathscr{F}_i$ is a coherent sheaf on $X$ with $\mathrm{supp}(\mathscr{F}_i) \subseteq \overline{x_i}$ for some $x_i \in X^p$, we have $\mathscr{F}_{x_i} = (\mathscr{F}_i)_{x_i}$, as $x_i \not \in \overline{x_j}$ for $i \neq j$.<|endoftext|> TITLE: Are radicals dense in the real closure of an ordered field? QUESTION [17 upvotes]: Let $F$ be an ordered field and let $R$ denote its real closure. It is well-known that $F$ is cofinal in $R$, but not necessarily dense. For example, consider $F=\mathbb{R}(\omega)$ with the order determined by $r<\omega$ for all $r\in \mathbb{R}$, and consider the open interval $(\sqrt{\omega}-1,\sqrt{\omega}+1)$ in $R$, which does not meet $F$. Question 1. Is always the case that radicals of elements of $F$ in $R$ are dense in $R$? Equivalently, is it possible to find for all $0<\alpha<\beta$ in $R$ an element $\gamma\in F$ and some $n\in \mathbb{N}$ such that $\alpha^n<\gamma<\beta^n$? It is not difficult to see that a positive answer would imply a positive answer to the next question, which is really what I am after. Question 2. Let $a_0,a_1,\dots,a_t\in R$ be distinct elements. Is there always a polynomial $p\in F[x]$ satisfying $p(a_0)<0$ while $p(a_i)>0$ for all $i\in\{1,\dots,t\}$? The most interesting case is when $a_0,\dots,a_t$ constitute the list of roots of an irreducible polynomial in $F[x]$. REPLY [2 votes]: Update: the following doesn't answer the question, since the element $\omega^\pi$ is transcendental rather than algebraic. Let $K$ be an ordered field, and let $$F = \left\{ \alpha = \sum_{k=k_0}^\infty a_k\omega^{-k/d}\Big\vert k_0\in\mathbb{Z},\,d\in\mathbb{Z}_{\geq 1},\, a_k\in K\right\}$$ be the field of Puiseaux series over $K$ ordered lexicographically (take the convention that $a_{k_0}\neq 0$ if $\alpha\neq 0$ and then $\alpha>0$ iff $a_{k_0}>0$). Note that $F$ is complete with respect to the valuation $v(\alpha) = k_0/d$. We call $-k_0/d$ the order of $\alpha$ and $a_{k_0} \omega^{-k_0/d}$ the leading term of $\alpha$. Let $R = F(\omega^\pi)$ where $\pi\in \mathbb{R}$ is any irrational. Then the valuation and order extend to $R$. In particular, elements of $R$ have leading terms and orders. Clearly if $\alpha,\beta\in R$ are positive of the same irrational order then no elements of $F$ separate them (the order of any positive element of $F$ is rational, hence either smaller or larger than that of $\alpha,\beta$). Now let $p(x) = \sum_{i=0}^n a_i x^i \in F[x]$ be a non-constant polynomial. Then for any $\alpha \in R$ of irrational order, the elements $a_i x^i$ (for $a_i\neq 0$) have distinct orders (because $1,\pi$ are linearly independent over $\mathbb{Q}$). It follows that the order of $p(\alpha)$ is irrational and depends only on the order of $\alpha$ (when that is irrational), and in particular that $p$ cannot separate elements of $R$ of the same irrational order. Remark It is not hard to see that polynomials in $K[x]$ can be used to separate elements of $F$. More generally, if $K$ is an ordered field and $R$ is the field of multivariable Puiseaux series over $K$ in the variables $\omega_1,\ldots,\omega_r$ where we repeatedly order lexicographically then the same holds. In other words, the claim is true if $R=F(\omega_1,\ldots,\omega_r)$ where the extensions are successively non-archimedean in that $\omega_i$ is larger than every element of $F(\omega_1,\ldots,\omega_{i-1})$.<|endoftext|> TITLE: Strong tournaments QUESTION [7 upvotes]: Let $T$ be a strong tournament, and let $N=v_1v_2 \cdots v_n$ be an enumeration of $V(T)$. Let $C$ be a circuit in $T$. We define $i_N(C)=|\{(v_i,v_j) \in E(C); i>j\}|$. Suppose that $N$ is chosen in such a way that $i_N(C_1)+ \cdots + i_N(C_t)$ is minimum, where $C_1, \cdots, C_t$ are all the circuits of $T$. Prove that $\forall i$ such that $1 \le i \le n-1$, $(v_i,v_{i+1}) \in E(T)$ and that $(v_n,v_1) \in E(T)$. My attempt: I already proved that $\forall i$ such that $1 \le i \le n-1$ we have $(v_i,v_{i+1})\in E(T)$. I first assumed that $(v_i,v_{i+1}) \not \in E(T)$, and this gives that $(v_{i+1},v_{i})\in E(T)$, so I took the enumeration $N'=v_1 \cdots v_{i-1}v_{i+1}v_iv_{i+2} \cdots v_n$, and proved that $i_{N'}(C_1)+ \cdots + i_{N'}(C_t)< i_N(C_1)+ \cdots + i_N(C_t)$, which is a contradiction. But for proving $(v_n,v_1) \in E(T)$ I supposed that $(v_1,v_n) \in E(T)$ and tried to take the enumeration $N''=v_nv_1\cdots v_{n-1}$ but I wasn't able to get to a contradiction, since to get to a contradiction from this enumeration I must be sure that the number of forward edges going to $v_n$ was less than that of the backward edges from $v_n$ in the first enumeration, can I prove this?Or do I take 2 cases if the number of forward edges was less or more than that of the backward edges? Or is there another enumeration that can finish it? Please help, and thanks in advance. REPLY [4 votes]: As you suspected, taking a better enumeration suffices. If $(v_n,v_1)\not\in E$ then consider the enumeration $N'=(v_2,v_3,\cdots,v_{n-1},v_1,v_n)$. It is easy to see that for any circuit $C$ such that $(v_1,v_n)\not\in E(C)$ we have $i_N(C)=i_{N'}(C)$. For any circuit $C$ such that $(v_1,v_n)\in E(C)$ (and there is one since $T$ is strong) we get $i_{N'}(C)=i_N(C)-1$, contradicting our assumption on $N$.<|endoftext|> TITLE: Tensoring adjointable maps on Hilbert modules QUESTION [6 upvotes]: Given a right Hilbert $A$-module $E$, and a right Hilbert $B$-module $F$, together with non-degenerate $*$-homomorphism $\phi:A \to \mathcal{L}_B(F)$, we can form the tensor product $$ E \otimes_{\phi} F, $$ by completing the $A$-balanced tensor product of $E$ and $F$ and completing with respect to the obvious norm. For two adjointable maps $L \in \mathcal{L}_A(E)$, and $K \in \mathcal{L}_B(F)$, such that $K$ is in addition a bimodule map, will their tensor product again be adjointable? (I am assuming here, that just as in the Hilbert space case, there is no difficulty with tensoring bounded maps to produce bounded maps.) REPLY [4 votes]: That $K$ is a bimodule map means that $\phi(a)K(x) = K(\phi(a)(x))$ for all $a\in A, x\in F$. That is, $K \in \phi(A)' \subseteq \mathcal{L}_B(F)$. I am here following the ideas of Lance's little Hilbert $C^*$-modules book, Chapter 4. Let $z = \sum_i x_i\otimes y_i \in E\odot F$, so $(z|z) = \sum_{i,j} (y_i|\phi((x_i|x_j))y_j) = (y|\phi_n(X)y)$. Here $X\in M_n(A)$ is the matrix with entries $(x_i|x_j)$, $\phi_n$ is the $n$th amplification (as $\phi$ is a $*$-homomorphism, it is completely positive, and so $\phi_n(X) \in M_n(\mathcal{L}_B(F))$ is positive), and $y$ is column with entries $(y_n)$. Here we have turned $F^n$ (as column vectors) into a Hilbert $C^*$-module over $B$ in the obvious way. Now consider $z'=(\iota\otimes K)z$ which has $(z'|z') = (y|K^*\phi_n(X)Ky)$ where $K$ acts on $F^n$ pointwise. As $K$ commutes with $\phi_n(X)$, $K^*\phi_n(X)K = K^*K \phi_n(X)$. As $\phi_n(X)$ is positive and $\phi_n(M_n(A)) = M_n(\phi(A))$ is a $C^*$-subalgebra of $M_n(\mathcal{L}_B(F)) = \mathcal{L}_B(F^n)$, there is $T\in M_n(A)^+$ with $\phi_n(T^2) = \phi_n(X)$. Then $K^*K\phi_n(T^2) = \phi_n(T) K^*K \phi_n(T)$. It follows that \begin{align*} (z'|z') &= (\phi_n(T)y|K^*K \phi_n(T)y) \leq \|K\|^2 (\phi_n(T)y|\phi_n(T)y) \\ &= \|K\|^2 (y|\phi_n(X)y) = \|K\|^2 (z|z). \end{align*} Hence $(\iota\otimes K)$ is bounded (with norm $\|K\|$). As always $L\otimes\iota$ is bounded (see Lance's book) we can compose to see that $L\otimes K$ is bounded, as required. (Contrary to the original question, I think it is far from obvious that $L\otimes K$ is bounded). Once we have established that $L\otimes K$ exists, it follows from just a calculation that it is adjointable, with adjoint $L^*\otimes K^*$. Indeed, $(x\otimes y| (L\otimes K)(x'\otimes y')) = (y|\phi((x|Lx')) Ky') = (y|K \phi((x|Lx')) y') = (K^*y|\phi((L^*x|x')) y') = ((L^*\otimes K^*)(x\otimes y)|x'\otimes y')$.<|endoftext|> TITLE: Irreducible factors of primitive permutation group representation QUESTION [6 upvotes]: Consider a primitive permutation group $\Gamma\subseteq\mathrm{Sym}(N)$ on the $n$-element set $N=\{1,...,n\}$, that is, $\Gamma$ does not preserve any non-trivial partition of $N$. Consider the linear representation of $\Gamma$ by permutation matrices on $\Bbb R^n$ in the obvious way. What can be said about the decomposition of this representation into irreducible factors? I am especially interested in the multiplicities of the factors: is it known whether any two irreducible factors in such a decomposition are non-isomorphic? REPLY [4 votes]: Peripherally related: In a paper I wrote in 1997 about bases for primitive permutation groups, it is noted that if $G$ is a (faithful) primitive permutation group of degree $n$, and a complex irreducible character $\chi$ of $G$ occurs with multiplicity $m$ in the associated permutation character of degree $n$, then there is a base for $G$ of size at most $\frac{\chi(1)}{m}.$ Recall that a base for the permutation group $G$ acting on $\Omega$ is a subset $\beta$ of $\Omega$ such that only the identity element of $G$ fixes every element of $\beta$. Hence we obtain $|G| \leq n(n-1) \ldots (n+1 - \frac{\chi(1)}{m}) < n^{\frac{\chi(1)}{m}}.$<|endoftext|> TITLE: Groups without factorization QUESTION [8 upvotes]: A group G is said to have a factorization if there exist proper subgroups $A$ and $B$ such that $G = AB = \{ ab \ | \ a \in A, b \in B \}$. The paper Factorisations of sporadic simple groups (by Michael Giudici) provides the classification of all the factorizations of the sporadic simple groups. We observe there that only $11$ sporadic simple groups (among $26$) admit a factorization. So the remaining $15$ (which are $J_1$, $M_{22}$, $J_3$, $McL$, $O'N$, $Co_3$, $Co_2$, $F_5$, $Ly$, $F_3$, $Fi_{23}$, $J_4$, $F_{3+}$, $F_2$, $F_1$) admit no factorization. Obviously, a cyclic group admits no factorization if and only if it is of prime power order. Question: What are all the (other) finite groups without factorization? (Is there an official name for such a group?) Proposition: Let $G$ be a group without factorization. If the intersection of all the maximal subgroups of $G$ is the trivial subgroup then $G$ is simple. proof: Assume $G$ non-simple and let $N$ be a non-trivial proper normal subgroup of $G$. If every maximal subgroup of $G$ contains $N$ then their intersection also, which contradicts the assumption. So there is a maximal subgroup $M$ of $G$ not containing $N$. It follows that $NM=G$, contradiction. $\square$ Sub-question: must a non-simple finite group without factorization be cyclic (of prime power order)? Lemma: A finite group $G$ admits a unique maximal subgroup $M$ iff it is cyclic of prime power order. proof: Let $g \in G \setminus M$ and $H = \langle g \rangle$. If $H \neq G$ then there must exist a maximal subgroup $M'$ of $G$ containing $H$, but $M=M'$ by assumption, so $g \in M$, contradiction. So $G = H$ is cyclic. It is moreover of prime power order by Chinese Remainder Theorem. $\square$ Bonus question: What do we know about the infinite groups without factorization? REPLY [3 votes]: The smallest non-abelian group without factorization is simple of order $1092$: it is $A_1(13)$. Using the answer of Geoff and by browsing the book The maximal factorizations of the finite simple groups and their automorphism groups (by Martin W. Liebeck, Cheryl E. Praeger and Jan Saxl), the finite simple groups without factorization are the following: Every cyclic group of prime order. Alternating: none Chevalley groups: $A_n(q)$ with $n=1$, $q \equiv 1 (\textrm{mod}\ 4)$ and $q\neq 5,9,29$, or with $n \ge 2$ even and $(n+1) | (q-1)$; $E_6(q)$; $E_7(q)$; $E_8(q)$; $F_4(q)$ with $q \neq 2^n$; $G_2(q) $ with $q>4 $ and $q \neq 3^n$. Twisted Chevalley groups: $^2A_{2n}(q^2)$ except $(n,q) = (1,3),(1,5),(4,2)$; $^2B_2(2^{2n+1})$; $^2D_{2n}(q^2)$; $^3D_4(q^3)$; $^2E_6(q^2)$; $^2F_4(2^{2n+1})$; $^2G_2(3^{2n+1})$. Sporadic: $J_1$, $M_{22}$, $J_3$, $McL$, $O'N$, $Co_3$, $Co_2$, $F_5$, $Ly$, $F_3$, $Fi_{23}$, $J_4$, $F_{3+}$, $F_2$, $F_1$. Before any use: this list requires a double checking.<|endoftext|> TITLE: What subalgebras of $\mathfrak{so}(2n)$ or $\mathfrak{sp}(2n)$ are orthogonal to the centre of $\mathfrak{gl}(n)$? QUESTION [11 upvotes]: Consider the Lie algebra inclusions $\mathfrak{gl}(n) \subset \mathfrak{so}(2n)$ and $\mathfrak{gl}(n) \subset \mathfrak{sp}(2n)$. Let $\mathfrak{c} \subset \mathfrak{gl}(n)$ denote the centre. Thinking of it as a subspace of $\mathfrak{so}(2n)$ or $\mathfrak{sp}(2n)$, let $\mathfrak{c}^\perp$ denote the orthogonal complement (with respect to the Killing form). Note that $\mathfrak{c}^\perp$ is not a Lie subalgebra of $\mathfrak{so}(2n)$ or $\mathfrak{sp}(2n)$, but it contains various subalgebras (e.g. $\mathfrak{sl}(n) \subset \mathfrak{c}^\perp$). What are the (maximal) Lie subalgebras of $\mathfrak{c}^\perp$? Are there any for which the induced $2n$-dimensional representation is simple? The specific case I care about is when $n=28$, and I look at $\mathfrak{sp}(56)$. Then I want to know if $\mathfrak{c}^\perp$ contains a subalgebra of type $\mathfrak{e}_7$. I suspect it does not, but I cannot prove it. REPLY [2 votes]: Here is an example of a Lie subalgebra of $\mathfrak{c}^\perp$ for which the $2n$-dimensional remains simple, with $n=16$. Consider the Lie group $\mathrm{Spin}(12)$, and its vector representation, which I will call $\mathbf{12}$, and its two half-spin representations, each of dimension $32$, which I will call $\mathbf{32}_\pm$. So for instance the adjoint representation is $\mathfrak{so}(12) = \operatorname{Alt}^2(\mathbf{12})$. Each half-spin representation supports a $\mathfrak{so}(12)$-invariant symplectic form. Mathieu's group $\mathrm{M}_{12}$ has no 12-dimensional irreps, but its double cover $2\mathrm{M}_{12}$ has one (up to isomorphism) 12-dimensional irrep. It supports a symmetric invariant form, so it defines a (conjugacy class of) map(s) $2\mathrm{M}_{12} \to \mathrm{O}(12)$, and there is only one conjugacy class of irreducible maps like this. However, there are two conjugacy classes of irreducible maps $2\mathrm{M}_{12} \to \mathrm{SO}(12)$, exchanged by the outer automorphism thereof, because any copy of the 12-dimensional $2\mathrm{M}_{12}$-irrep is chiral. Choose one of these maps $2\mathrm{M}_{12} \to \mathrm{SO}(12)$. The choice breaks the symmetry between the half-spin representations $\mathbf{32}_\pm$. Namely, one of them, which I will arbitrarily call $\mathbf{32}_+$, descends from $2\mathrm{M}_{12}$ to $\mathrm{M}_{12}$ and splits as the sum $\mathbf{16} \oplus \overline{\mathbf{16}}$ of a 16-dimensional complex irrep and its dual, and the other, $\mathbf{32}_-$, is nontrivially charged under the centre of $2\mathrm{M}_{12}$ and remains simple upon restriction. Thus we have a commutative square: $$ \begin{matrix} 2\mathrm{M}_{12} & \hookrightarrow & \mathrm{Spin}(12) \\ \downarrow & & \downarrow \\ \mathrm{GL}(16) & \hookrightarrow & \mathrm{Sp}(32) \end{matrix} $$ where the map $\mathrm{Spin}(12) \to \mathrm{Sp}(32)$ is via the representation $\mathbf{32}_+$. (Both horizontal arrows are inclusions, and both downward arrows have kernel of order $2$.) Now look at the adjoint representation $\mathfrak{sp}(32)$, on which $2\mathrm{M}_{12}$ acts orthogonally, and the image therein $\mathfrak{c}$ of the centre of $\mathfrak{gl}(16)$. On the one hand, this centre is fixed by $2\mathrm{M}_{12}$, and in fact is the only fixed subspace. On the other hand, $\mathfrak{so}(12) \subset \mathfrak{sp}(32)$ remains simple when restricted to $2\mathrm{M}_{12}$. It follows that $\mathfrak{so}(12)$ and $\mathfrak{c}$ are orthogonal. Thus $\mathfrak{g} = \mathfrak{so}(12) \subset \mathfrak{c}^\perp$. But by construction the $32$-dimensional defining represenation of $\mathfrak{sp}(32)$ restricts to $\mathfrak{g}$ as the irrep $\mathbf{32}_+$.<|endoftext|> TITLE: Equivalent definitions of Thom spectra QUESTION [12 upvotes]: Background and notations: Recall the classical contruction and definition of Thom spectra. To a spherical fibration $S^{n-1} \to \xi \to B$, we can associate the data of a Thom space $T_n(\xi)$, given by the cofiber: $\require{AMScd}$ \begin{CD} \ \xi @>>> B\\ @VVV @VVV\\ * @>>> T_n(\xi). \end{CD} This construction can be viewed as a (suitably homotopy invariant functor) from the slice category of spaces over $BGL_1S^{n-1}$ (which is the classifying space for $n-1$ spherical fibration) to spaces. We can promote the functor $T_n$ to a spectra valued functor by defining the Thom spectrum $T(\xi)$ to be $$ T(\xi):= \Sigma^{-n}\Sigma^{\infty}T_n(\xi). $$ And again (by filtering the base space of the fibration by compact subspaces) $T$ can be viewed as functor from the slice category over the colimit space $BGL_1\mathbb{S}$ to the category of spectra. There is another definition of Thom spectrum, under the hypothesis that the classifying map $B \to BGL_1\mathbb{S}$ is an infinite loop space map, which i found in these two papers AHR10 (def. 2.6), ABGHR14 (def. 4.2) and in the book "Formal Geometry and Bordism Operations" by E.Peterson Pet (Lemma A.4.1), which present the Thom spectrum in the following way. Write $\mathbb{S}$ for the sphere spectrum and take a map of connective spectra $f: b \to bgl_1\mathbb{S} = \Sigma gl_1\mathbb{S}$, where $gl_1$ is the right adjoint of the stabilization functor $\Sigma^\infty_+\Omega^\infty$. We can view $f$ also as a map of infinite loop space $$ \Omega^\infty f: \Omega^\infty b \to BGL_1\mathbb{S}$$ and hence as a stable spherical fibration over the base space $B=\Omega^\infty b$. Now the Thom spectrum of this spherical fibration is the homotopy pushout in the category of $E_\infty$ spectra of the following diagram $\require{AMScd}$ \begin{CD} \ \Sigma^\infty_+\Omega^\infty gl_1\mathbb{S} @>{\epsilon}>> \mathbb{S}\\ @V{\Sigma^\infty_+\Omega^\infty \lambda}VV @VVV\\ \Sigma^\infty_+\Omega^\infty Cj @>>> T(\Omega^\infty f), \end{CD} where: $\epsilon$ is the counit map in the adjunction $\Sigma^\infty_+\Omega^\infty/ gl_1$ $Cj$ is the cone of the desuspension $\Sigma^{-1}f$ and $\lambda$ is the map of the cofiber sequence $\require{AMScd}$ \begin{CD} \ \Sigma^{-1}b @>{j}>> gl_1\mathbb{S} @>{\lambda}>> Cj \end{CD} Question: Is there any motivation for the "homotopy pushout" definition of Thom spectrum to be equivalent to the classic definition for stable spherical fibration classified by map $B \to BGL_1\mathbb{S}$ that are infinity loop space map? I am facing big difficulties in reading the references that both of the papers and the book gives to motivate the latter definition of Thom spectra and i could not be able to find any other references. Further observations: The only case in which the equivalence is clear to me is the case of the map $* \to bgl_1\mathbb{S}$ in which both definitions produce immediately the sphere spectrum $\mathbb{S}$. The general case is still obscure to me. There should be another (equivalent?) definition (treated in ABGHR09) for the Thom spectrum of stable spherical fibration classified by map $f: BG \to BGL_1\mathbb{S}$, induced by group map $G \to GL_1\mathbb{S}$, in which $T(f)$ is presented as the "space of sections" of the sphere bundle over $BG$ associated to the map $f$. However, either this definition and how it should be related with the others is still not clear to me. REPLY [13 votes]: The details of the comparison are treated in detail in the original ABGHR paper (and then unfortunately split in half across two papers in the updated version), so I'll just try to give a sketch of what's going on. Thom spaces Given any type of bundle $E \to B$ we can view this as a diagram of spaces indexed by $B$. The homotopy colimit of this diagram in the homotopy theory of spaces is just $E$ itself (which makes sense- we are 'summing up over $B$'). If $E \to B$ has a section $\infty: B \to E$, then this supplies each fiber with a basepoint. The homotopy colimit over $B$ of this diagram of pointed spaces can't be $E$, because it has a whole $B$-family of basepoints. So we need to identify those all together and take the cofiber $B \to E \to E/B$. The homotopy colimit is then $E/B$. Example. Given a pointed spherical fibration, $E \to B$, with fibers $S^n$ (pointed!), this procedure gives the Thom space. So we learn that: the Thom space of a pointed spherical fibration is the homotopy colimit taken in the homotopy theory of pointed spaces of the family of spheres indexed over the base. Example. Given an unpointed spherical fibration $E \to B$, with fibers $S^{n-1}$, we can form the fiberwise suspension by taking the cofiber $S^{n-1} \to CS^{n-1}$, fiberwise. This produces a pointed spherical fibration, and we can then take the Thom space as above. It's not hard to see that collapsing the section at $\infty$ of this suspended fibration gives the same answer as taking the mapping cone of the projection for the original fibration, whence the connection with your first definition. Example. If $B = BG$ is the classifying space for a group, then the 'homotopy colimit indexed by $BG$' is also called the 'homotopy orbits for the action of $G$', since we may identify a bundle $E \to BG$ with a homotopy coherent action of $G$ on the fiber over a fixed basepoint of $BG$. So, in this case, the Thom space is of the form $S^n_{hG}$ where $G$ acts in some way on $S^n$. This includes the universal example, when $G = \mathrm{hAut}(S^n)$, the space of homotopy automorphisms of $S^n$ (i.e. the degree $\pm 1$ components of $\Omega^nS^n$ when $n>0$). Thom spectra For bundles of spectra it is nice to take the 'diagrams indexed by $B$' point of view as a definition. So a bundle is given by a map $B \to \{\text{space of spectra equivalent to }S^0\}=\mathrm{BGL}_1(S^0)$. This, in particular, defines a diagram of spectra and its homotopy colimit is the Thom spectrum. Since $\Sigma^{\infty}_+$ commutes with taking homotopy colimits (it's a left adjoint), we see that taking the suspension spectrum of a Thom space gives the Thom spectrum of the bundle obtained by taking fiberwise suspension spectra. Now, in the case that $B = BG$, then we may identify functors $BG \to \mathsf{Sp}$ with modules over the $\mathbb{E}_1$-ring $S^0[G] := \Sigma^{\infty}_+G$. (This is a spectral version of the relationship between $G$-modules and $\mathbb{Z}[G]$-modules for an ordinary group). So we have: spherical fibrations over $BG$ $\iff$ a coherent action of $G$ on the sphere spectrum $S^0$ $\iff$ an $S^0[G]$-module structure on $S^0$. Under this correspondence, taking homotopy colimits corresponds to the construction on modules $M \mapsto M \otimes_{S^0[G]}S^0$ where we use the augmentation. (This is analogous to the statement that the coinvariants of a $G$-module are computed via a similar tensor product in the classical setting). Thus, in the case $B = BG$, we may compute Thom spectra as $S^0 \otimes_{S^0[G]} S^0$ where the left $S^0$ has an interesting module structure, and the second $S^0$ is acted on through the augmentation. Of course, the action of $G$ factors through the largest action of all of $\mathrm{GL}_1(S^0)$, so we can write this as $S^0 \otimes_{S^0[G]}S^0 = S^0 \otimes_{S^0[\mathrm{GL}_1(S^0)]} S^0[\mathrm{GL}_1(S^0)] \otimes_{S^0[G]} S^0$. In the special case when $G \to \mathrm{GL}_1(S^0)$ is 'normal', ie arises as the fiber of an $\mathbb{E}_1$-map $\mathrm{GL}_1(S^0) \to H$ (which happens int he infinite loop space context if we take $H = \Omega^{\infty}Cj$ in your notation), then we may simplify the right hand side of the tensor product as $S^0 \otimes_{S^0[\mathrm{GL}_1(S^0)]} S^0[\mathrm{GL}_1(S^0)] \otimes_{S^0[G]} S^0=S^0 \otimes_{S^0[\mathrm{GL}_1(S^0)]} S^0[H]$. So that's the relationship between the two definitions you write down.<|endoftext|> TITLE: Effect of adding one Hechler real versus adding two on the meager ideal QUESTION [9 upvotes]: In "The Kunen-Miller Chart (Lebesgue Measure, The Baire Property, Laver Reals and Preservation Theorems for Forcing)" by Haim Judah and Saharon Shelah JSL Vol. 55, No. 3 (Sep., 1990), pp. 909-927 ([JdSh308] in Shelah's numbering) the authors remark on the final page that adding a pair of Hechler reals makes the union of the meager sets coded by the ground model meager. My questions are: How does one prove this? (and/or is there a citation with the proof, in the paper none is given that I can find) Does this not happen when one adds only one Hechler real? I suppose more generally my question is as follows: clearly in the Hechler model $\mathrm{add}(\mathcal M) = \mathfrak{c}$ by Miller-Truss theorem that $\mathrm{add}(\mathcal M) = \min\{\mathfrak{b}, \mathrm{cov}(\mathcal M)\}$ since both dominating reals and Cohen reals are added but I don't quite see how this proliferates down to the finite steps. Is there a more direct way to see how Hechler forcing effects $\mathrm{add}(\mathcal M)$? REPLY [9 votes]: If $c$ is Cohen over $V$, and $d$ is dominating over $V[c]$ (not necessarily Hechler-generic), then in $V[c][d]$ there is a meager set covering all meager sets from $V$. Hence 2 successive Hechler reals make the union of all old meager sets meager. (I suspect that this is not true if you just add one Hechler real.) Proof: (This is implicit in Bartoszynski-Judah 2.2, and implicit or perhaps even explicit in other papers, such as Miller or Truss. I seem to remember that Andreas Blass invented a notion of "composition" of Galois-Tukey relations, which gives a general framework for arguments such as the one I give below.) For each meager set $M\subseteq 2^\omega$ coded in $V$ there are functions $f:\omega\to \omega$ and $x\in 2^\omega$ (again in $V$) such that $M$ is contained in $ M_{f,x}:= \{ y\in 2^\omega\mid \forall^ \infty n: x\restriction I_n\not= y\restriction I_n\}$, where $I_n:=[f(n), f(n+1))$. In $V[c]$ there are infinitely many $n$ such that $c\restriction I_n = x\restriction I_n$. (As $c$ is Cohen). So there is an increasing sequence $(n_k)$ such that $x\restriction I_{n_k} = c\restriction I_{n_k}$ for all $k$. (The sequence $(n_k:k\in \omega)$ depends on $x$, of course. But all we will use about it in the next paragraph is that it is an element of $V[c]$.) In $V[c][d]$ we may wlog assume that $d$ strongly dominates $V[c]$, e.g. by first assuming $d(k)>k$ for all $k$, and then replacing $d$ with $k\mapsto d^{(k)}(0)$ ($k$-th iterate). Let $J_\ell$ be the interval $[ d(\ell), d(\ell+1))$. As these intervals are very long (compared with anything from $V[c]$), almost every $J_\ell$ contains at least one intervall of the form $I_{n_k}$. I claim that $M_{f,x} \subseteq M_{d,c}$. So let $y\in M_{f,x}$ (in $V[c][d]$). For notational simplicity assume that $\forall n: y\restriction I_n \not= x\restriction I_n$. Since $x$ and $c$ agree on all the $I_{n_k}$, we also have $y\restriction I_{n_k}\not= c\restriction I_{n_k}$. For any (sufficiently large) $\ell $ we can find $k$ such that $J_\ell$ contains $I_{n_k}$; so we also have $y\restriction J_\ell\not= c\restriction J_\ell$. Hence $y\in M_{d,c}$.<|endoftext|> TITLE: Set of perfect subsets of a Borel set QUESTION [7 upvotes]: Let $\mathbb{P}$ be the set of all perfect (i.e., every node has incomparable successors) subtrees of the full binary tree $2^{<\omega}$. We can endow $\mathbb{P}$ with a Borel structure by considering it as a subspace of $2^{2^{<\omega}}$ with the product topology. If $p\in\mathbb{P}$, let $[p]$ denote the set of infinite branches through $p$, a nonempty perfect subset of $2^{\omega}$. Question: Given a Borel subset $B\subseteq 2^\omega$, is the set $$S_B=\{p\in\mathbb{P}:[p]\subseteq B\}$$ necessarily Borel? Some observations: $S_B$ is $\mathbf{\Pi}^1_1$ (i.e., co-analytic), so this is equivalent to asking if $S_B$ is also $\mathbf{\Sigma}^1_1$ (i.e., analytic). If $B$ is open, then $S_B$ is Borel: Let $\{t_n:n\in\omega\}$ be finite binary strings such that $B$ is the union of the corresponding basic clopen sets $\mathcal{N}_{t_n}$. Then, $p\in\mathbb{P}$ if and only if every node in $p$ is comparable to some $t_n$; this is a Borel condition. It is likewise easy to see that if $B$ is closed, then $S_B$ is Borel. If you consider the collection $\{B\subseteq 2^\omega:S_B\text{ is Borel}\}$, it contains the open sets (by the previous bullet) and it is closed under countable intersections, so it suffices to show that it is closed under complements. One strategy to find a counterexample might be to embed trees $T$ on $\omega$ into $2^{<\omega}$, via some nice map $f$, and find a Borel $B$ such that $T$ is well-founded if and only if $[f(T)]\in B$. REPLY [2 votes]: This appears not to be the case: there is a $F_\sigma$ set $B$ such that $S_B$ is not Borel. This is optimal since the bullets in the question explain how $S_B$ is Borel when $B$ is $G_\delta$. There is surely a better way to explain this. I arrived at this answer by first realizing that the answer is yes if we replace "$[p] \subseteq B$" by "$[p]\cap B$ is comeager in $[p]$" in the question. I then attempted to transform this into a positive answer to the question, and my failure to do so led to what follows. Let $c_0 \subseteq c_1 \subseteq c_2 \subseteq \cdots$ be a sequence of nonempty subtrees of $2^{<\omega}$ with no dead ends. For a (possibly empty) tree $p \subseteq 2^{<\omega}$, define $$p' = \{ t \in p \mid (\forall n)(p_t \nsubseteq c_n)\},$$ where $p_t = \{ s \in p \mid s \subseteq t \lor t \subseteq s \}.$ Topologically: $c_0,c_1,c_2,\ldots$ is a sequence of codes for closed sets $[c_0] \subseteq [c_1] \subseteq [c_2] \subseteq \cdots$ If $U_i$ is the relative interior of $[c_i]\cap[p]$ in $[p]$, then $[p'] = [p] \setminus \bigcup_{i<\omega} U_i.$ By the Baire Category Theorem, if $[p] \cap \bigcup_{i<\omega} [c_i]$ is comeager in $[p]$ then $[p] \cap \bigcup_{i<\omega} U_i$ is open dense in $[p]$ and thus $[p']$ is nowhere dense in $[p]$. So, as a first approximation to determining whether $[p] \subseteq \bigcup_{i<\omega} [c_i]$ we can check that $[p']$ is comeager in $[p]$. This is a $\Pi^0_2$ check: $(\forall t \in p)(\exists u \in p)(t \notin p' \land t \subseteq u).$ This is not enough however as we have merely reduced the question to whether $[p'] \subseteq \bigcup_{i<\omega} [c_i].$ We can iterate this operation: define $p^{(0)} = p,$ $p^{(\alpha+1)} = (p^{(\alpha)})',$ and $p^{(\alpha)} = \bigcap_{\beta<\alpha} p^{(\beta)}$ when $\alpha$ is a limit ordinal. Since $p$ is a countable set, there is always some $\alpha<\omega_1$ such that $p^{(\alpha+1)} = p^{(\alpha)}$ and the sequence stabilizes from then on. Let's call $p^{(\alpha)}$ the core of $p$ and lets call the first such $\alpha$ the core rank of $p$. For convenience, let's write $p^\ast$ for the core of $p$. Note that $[p] \subseteq \bigcup_{i<\omega} [c_i]$ if and only if the core of $p$ is empty. The only if direction is follows from the observation that we always have $[p] \subseteq [p'] \cup \bigcup_{i<\omega} [c_i]$. For the if direction, notice that the core $p^\ast$, when nonempty, has the property that $p^\ast \setminus c_n$ is dense in $p^\ast$ for every $n$. So a generic path through $p^\ast$ witnesses that $[p^\ast] \subseteq [p] \nsubseteq \bigcup_{i<\omega} [c_i].$ For perfect $p$, there are $c_0 \subseteq c_1 \subseteq c_2 \subseteq\cdots$ with empty $p^\ast$ of arbitrarily large core rank with respect to $p$. For simplicity, we take $p = 2^{<\omega}.$ To get started, note that if $$c_i = \{0\}^{<\omega} \cup \{ s \in 2^{<\omega} \mid |s| > i \land (\exists j \leq i, s_j = 1)$$ then $2^\omega = \bigcup_{i<\omega} [c_i]$ and $c_0,c_1,c_2,\ldots$ has core rank $2$ with respect to $2^{<\omega}$. To get increasingly larger core ranks, given $c_{k,0} \subseteq c_{k,1} \subseteq \cdots$ for $k = 0,1,2,\ldots$ with $2^{\omega} = \bigcup_{i<\omega} [c_{k,i}].$ Define $$c_i = z \cup \bigcup\nolimits_{k<\omega} \{ 0^{k-1}1 t \mid t \in c_{k,i} \}.$$ Then $2^{\omega} = \bigcup_{i<\omega} [c_i]$ and a straghtforward inductive calculation shows that this sequence has core rank at least $\alpha+1$ where $\alpha$ is any ordinal for which there are infinitely many $k$'s where $c_{k,0},c_{k,1},c_{k,2},\ldots$ has core rank at least $\alpha$. In particular, if $c_{k,0},c_{k,1},c_{k,2},\ldots$ has core rank $\alpha_k$ and $\alpha_0 \leq \alpha_1 \leq \cdots$ then $c_0,c_1,c_2,\ldots$ has core rank $\sup_{k<\omega} (\alpha_k+1).$ Now a sequence $c_0 \subseteq c_1 \subseteq c_2 \subseteq \cdots$ can be encoded by an $f \in \omega^\omega$ in such a way that $f(n)$ determines all $c_i \cap 2^n$ at once. This is because $c_0 \cap 2^n \subseteq c_1 \cap 2^n \subseteq \cdots,$ so it suffices for $f(n)$ to encode the finitely many subsets of $2^n$ that occur in this sequence along with the first index at which they appear. Define the universal sequence $d_0 \subseteq d_1 \subseteq d_2 \subseteq \cdots$ as follows: $t \in d_i$ if the longest initial segment of $t$ which is of the form $$1^{n_0}0s_01^{n_1}0s_1\cdots1^{n_{k-1}}0s_{k-1},$$ where each $n_j$ appropriately encodes an infinite nondecreasing sequence of subsets of $2^j$ according to the scheme above, is such that $s_0s_1\cdots s_{k-1}$ belongs to the $i$th set coded by $n_{k-1}.$ Note that if $f$ is the code for $c_0 \subseteq c_1 \subseteq c_2 \subseteq \cdots$ then the perfect set $p$ where $$[p] = \{1^{f(0)}0x_01^{f(1)}0x_11^{f(2)}0x_2\cdots \mid x \in 2^\omega\}$$ is such that $p,d_0 \cap p,d_1 \cap p, d_2 \cap p,\ldots$ is isomorphic to $2^{<\omega},c_0,c_1,c_2,\ldots$ It follows that for this sequence $d_0 \subseteq d_1 \subseteq d_2 \subseteq \cdots$ there are $p$ with $p^\ast = \varnothing$ and arbitrarily large core rank with respect to $d_0,d_1,d_2,\ldots$ Let $B = \bigcup_{i<\omega} [d_i]$, which is $F_\sigma.$ For each $\alpha<\omega_1,$ the set $S_\alpha = \{ p \mid p^{(\alpha)} = \varnothing \}$ is Borel, where $p^{(\alpha)}$ is computed with respect to $d_0,d_1,d_2,\ldots$, and if $\alpha \leq \beta < \omega_1$ then $S_\alpha \subseteq S_\beta$. We now have $S_B = \bigcup_{\alpha<\omega_1} S_\alpha$ but $S_B \nsubseteq S_\alpha$ for any $\alpha <\omega_1$. Therefore $S_B$ is not Borel.<|endoftext|> TITLE: $2$-norm distance between square roots of matrices QUESTION [7 upvotes]: Suppose two square real matrices $A$ and $B$ are close in the Schatten 1-norm, i.e. $\|A-B\|_1=\varepsilon$. Can this be used to put a bound on the Schatten 2-norm distance between their square roots. Namely, is there something of the form $\|\sqrt{A}-\sqrt{B}\|_2\leq f(\varepsilon)$? It is important that $f(\varepsilon)$ be independent of the dimension of the matrices. One can assume that these are symmetric, positive definite matrices. I have a proof for the above statement when $A$ and $B$ are taken to be simultaneously diagonal. However, I was wondering if there is a more general proof. REPLY [12 votes]: As you yourself discovered by finding a paper of Audenaert: an upper bound of the form you require is provided by the Powers–Størmer inequality: Theorem (Powers–Størmer, 1970, Lemma 4.1; link) Let $S$ and $T$ be positive Hilbert-Schmidt operators on a Hilbert space. Then $\Vert S - T \Vert_2^2 \leq \Vert S^2-T^2\Vert_1$, where $\Vert \quad\Vert_p$ denotes the Schatten $p$-norm. The starting idea of the proof is to work in an ONB with respect to which $R=S-T$ is diagonal (which is possible by the spectral theorem for compact self-adjoint operators). One then observes that, putting $Q=S+T$, we have $S^2-T^2 = (RQ+QR)/2$ and then one exploits the fact that $Q\geq \pm R$. More general inequalities are known, see the discussion in Section X.1 of Bhatia's Springer GTM book on Matrix Analysis (Springer GTM).<|endoftext|> TITLE: Сlosed formula for $(g\partial)^n$ QUESTION [9 upvotes]: The objective is to obtain a closed formula for: $$ \boxed{A(n)=\big(g(z)\,\partial_z\big)^n,\qquad n=1,2,\dots} $$ where $g(z)$ is smooth in $z$ and $\partial_z$ is a derivative with respect to $z$. I think the first few terms are, \begin{equation} \begin{aligned} A(1) &= g\,\partial\\ A(2)&= g\,(\partial g)\,\partial+g^2\,\partial^2\\ A(3)&= \big[(\partial^2g)g^2+(\partial g)^2g\big]\partial+3(\partial g)g^2\,\partial^2+g^3\partial^3\\ A(4) &= \big[(\partial^3g)g^3+4(\partial^2g)(\partial g)g^2+(\partial g)^3g\big]\partial\\ &\quad +\big[4(\partial^2g)g^3+7(\partial g)^2g^2\big]\partial^2+6(\partial g)g^3\partial^3+g^4\partial^4\\ &\,\,\vdots \end{aligned} \end{equation} and perhaps there is a simple pattern that I'm failing to see. The partitionings of the $\partial$ and $g$ are reminiscent of Bell polynomials but the coefficients are more complicated. Perhaps it is useful to make explicit that the general expansion is of the form: $$ (g\,\partial)^n=g^n\sum_{p=0}^{n-1}a_{n,p}(g)\,\partial^{\,n-p} $$ with, $$ a_{n,p}(g)=\sum_{m_1+2m_2+\dots+pm_{p}=p} C_{n,p}(m_1,\dots,m_{p})\Big(\frac{\partial g}{g}\Big)^{m_1}\Big(\frac{\partial^2 g}{g}\Big)^{m_2}\dots \Big(\frac{\partial^{p} g}{g}\Big)^{m_{p}}\qquad (*) $$ and the latter sum is over all non-negative integers, $\{m_1,\dots,m_{p}\}$, subject to: $$ m_1+2m_2+\dots+pm_{p}=p $$ From this viewpoint the objective is to determine the coefficients $C_{n,p}(m_1,\dots,m_{p})$, which in turn depend on all integers, $n$, $p$ and $\{m_1,\dots,m_p\}$. Any ideas? Many thanks in advance. REPLY [4 votes]: The Ihara reference "Derivations and automorphisms on non-commutative power series" (open archive now) in OEIS A139605 contains an explicit formula for the coefficients you are looking for, obtained from the Comtet ref. "Une formule explicite pour les puissances successives de l'opérateur de dérivation de Lie." See A139605 (also related OEIS A145271) for simple matrix computations for these partition polynomials and numerous other references. The formula section of A139605 contains the matrix formula. Multiply the $n$-th diagonal (with $n=0$ the main diagonal) of the lower triangular Pascal matrix A007318 by $g_n = D_x^n g(x)$ to obtain the matrix $VP$ with $VP_{n,k} = \binom{n}{k}g_{n-k} $. Then $$(g(x)D_x)^n = (1, 0, 0,..) [VP \dot \; S]^n (1, D, D^2, ..)^T,$$ where S is the shift matrix A129185, representing differentiation in the divided powers basis $x^n/n!$. Example: $$(g(x)D_x)^3$$ $$= (1, 0, 0, 0) [VP \dot \; S]^3 (1, D, D^2, D^3)^T$$ $$= \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & g_0 & 0 & 0 \\ 0 & g_1 & g_0 & 0\\ 0 & g_2 & 2g_1 & g_0 \\ 0 & g_3 & 3g_2 & 3g_1 \end{pmatrix}^3 \begin{pmatrix} 1 \\ D \\ D^2 \\ D^3 \end{pmatrix} $$ $$ = [g_0g_1^2 + g_0^2 g_2] D + 3 g_0^2g_1 D^2 + g_0^3D^3 $$ And, the pdf Mathemagical Forests gives a diagrammatic method for creating forests of trees through "natural growth" that represent the partition polynomials.<|endoftext|> TITLE: Texts on moduli of elliptic curves QUESTION [5 upvotes]: I want to study FLT (Fermat's Last Theorem), and now I'm studying moduli of elliptic curves. I've heard that Deligne-Rapoport, Katz-Mazur, Mazur's "Modular curves...", and Katz's "p-adic..." are very good for this topic. But I don't know what's the difference between these papers. For me it seems that these treat almost same topics. So what should I read at first? (Now I'm reading Katz-Mazur. It's easy to read even for me, a beginner of arithmetic geometry. So I think I should read it at first. And glancing through Mazur, it seems to use many results from other 3 papers.) And would you recommend other good papers which I read understand for understanding FLT? Any help will be much appreciated! REPLY [2 votes]: Not having been mentioned before, I would recommend the two books "Fermat's Last Theorem, Basic Tools" and "Fermat's Last Theorem, The Proof" by Takeshi Saito. https://bookstore.ams.org/mmono-243 and https://bookstore.ams.org/mmono-245. In particular, moduli of elliptic curves appear in chapter 2.2 but only over $\mathbb{Q}$. Then the second volume starts with chapter 8 "Modular curves over $\mathbb{Z}$" which covers the topics on the moduli spaces of elliptic curves needed for Fermat's Last Theorem.<|endoftext|> TITLE: A conjectural trigonometric identity QUESTION [30 upvotes]: Recently, I formulated the following conjecture which seems novel. Conjecture. For any positive odd integer $n$, we have the identity $$\sum_{j,k=0}^{n-1}\frac1{\cos 2\pi j/n+\cos 2\pi k/n}=\frac{n^2}2.\tag{1}$$ Using Galois theory, I see that the sum is a rational number. The identity $(1)$ has some equivalent versions, for example, $$\sum_{0\le j TITLE: Prove that $\left(\frac{x^n+1}{x^{n-1}+1}\right)^n+\left(\frac{x+1}{2}\right)^n\geq x^n+1$ QUESTION [27 upvotes]: Let $x>0$ and $n$ be a natural number. Prove that: $$\left(\frac{x^n+1}{x^{n-1}+1}\right)^n+\left(\frac{x+1}{2}\right)^n\geq x^n+1.$$ This question is very similar to many contests problems, but I think it's much more harder than contest problem and it's just impossible to solve this problem during a competition. In my fifth comment I wrote: I took it here: https://math.stackexchange.com/questions/3304808/refinement-of-a-famous-inequality. but I think there is no a chance that it will be solved there. I tried to use $$M_n^2\geq M_{n+1}M_{n-1},$$ where $x>0$ and $M_n=\sqrt[n]{\frac{x^n+1}{2}}$, but without success. I think a perspective way to solve this problem it's something like the following wrong way: https://mathoverflow.net/revisions/337457/1 Thank you! REPLY [12 votes]: This is a supplement (correction) to Peter Mueller's nice solution. As he observed, it suffices to show that, for any fixed $n\geq 1$ and $y\in[0,1]$, the function $$z\mapsto\left(\frac{y^{n+1}+z}{y^{n-1}+z}\right)^n+\left(\frac{zy^{n+1}+1}{zy^{n-1}+1}\right)^n,\qquad z\in(0,1),$$ is increasing. (Indeed, $y:=x^{1/2}$ and $z:=x^{(n-1)/2}$ yields the LHS of the OP's inequality, while $y:=x^{1/2}$ and $z:=0$ yields the RHS of the OP's inequality.) Taking the derivative with respect to $z$, the statement becomes $$\left(\frac{y^{n+1}+z}{zy^{n+1}+1}\right)^{n-1}\ge \left(\frac{y^{n-1}+z}{zy^{n-1}+1}\right)^{n+1},\qquad y,z\in(0,1).$$ Let us now fix $y,z\in(0,1)$ and think of $n\geq 1$ as the variable. Taking the logarithm of both sides and dividing by $(n-1)(n+1)$, it suffices to show that the function $$t\mapsto\frac{1}{t}\log\frac{y^t+z}{zy^t+1},\qquad t>0,$$ is increasing. Making the change of variable $w:=y^t$, it suffices to show that the function $$w\mapsto\frac{\log(w+z)-\log(wz+1)}{\log w},\qquad w\in(0,1),$$ is increasing. Writing $w=:\tanh u$ and $z=:\tanh v$, it suffices to show that the function $$u\mapsto\frac{\log\tanh(u+v)}{\log\tanh(u)},\qquad u>0,$$ is increasing. Taking the derivative with respect to $u$, the statement becomes $$\sinh(u)\cdot\cosh(u)\cdot\log\tanh(u)\geq\sinh(u+v)\cdot\cosh(u+v)\cdot\log\tanh(u+v).$$ That is, it suffices to show that the function $$u\mapsto \sinh(u)\cdot\cosh(u)\cdot\log\tanh(u),\qquad u>0,$$ is decreasing. With the notation $s:=-\log\tanh(u)$, we have $$\sinh(u)\cdot\cosh(u)\cdot\log\tanh(u)=\frac{e^{-s}}{\sqrt{1-e^{-2s}}}\cdot\frac{1}{\sqrt{1-e^{-2s}}}\cdot(-s)=\frac{-s}{2\sinh s},$$ hence it suffices to show that the function $$s\mapsto\frac{\sinh s}{s},\qquad s>0$$ is increasing. However, this is clear, because the Taylor series of this function converges everywhere, and it has nonnegative coefficients.<|endoftext|> TITLE: A question on "$p$-group bounds"? QUESTION [8 upvotes]: Suppose $n_p(G)$ is the number of elements of order $p$ in a group $G$. Does there for every prime $p$ exist a $\epsilon_p > 0$, such that for any group $G$ $n_p(G) > (1 - \epsilon_p(G))|G|$ implies, that $G$ is a $p$-group? This question was inspired by a well known fact, that for any group $G$ $n_2(G) > \frac{3}{4}|G|$ implies $G$ being a $2$-group (or, alternatively, that $\epsilon_2 = \frac{1}{4}$). REPLY [18 votes]: Yes, this is the main theorem of Laffey, Thomas J., The number of solutions of (x^p=1) in a finite group, Math. Proc. Camb. Philos. Soc. 80, 229-231 (1976). ZBL0343.20006. Namely, if a finite group $G$ has more than $\tfrac{p}{p+1} |G|$ elements of order $p$, then $G$ is a $p$-group. This bound is achieved by $C_p \ltimes C_2^k$ where $p = 2^k-1$ is a Mersenne prime. There is a related very frustrating problem: Is there an $\delta_p$ such that, if $G$ has more than $1-\delta_p$ elements of order $p$, then $G$ is $p$-torsion? We can take $\delta_3 = \tfrac{2}{9}$, see Laffey, Thomas, The number of solutions of (x^3=1) in a 3-group, Math. Z. 149, 43-45 (1976). ZBL0314.20020. I thought for a bit that the right bound might be $\tfrac{p-1}{p^2}$ in general, which occurs for $C_p \ltimes \mathbb{Z}[\zeta_p]/(1-\zeta_p)^N$ where $\zeta_p$ is a $p$-th root of unity and $p \leq N \leq 2p-2$. But that's wrong! Wall constructed a $5$-group of exponent $25$ where all the elements of order $25$ were in an index $25$ subgroup, demonstrating that $\delta_5$, if it exists, is $< \tfrac{1}{25}$. Wall, G. E., On Hughes’ $H_{p}$ problem, Proc. Int. Conf. Theory Groups, Canberra 1965, 357-362 (1967). ZBL0189.31701 But, as far as I know, no one knows whether any such $\delta_5$ exists at all! See Havas, George; Vaughan-Lee, Michael, On counterexamples to the Hughes conjecture., J. Algebra 322, No. 3, 791-801 (2009). ZBL1187.20010 for groups achieving $\tfrac{1}{p^2}$ for various $p>5$. I will also note that I talked to Harry Altman about a lot of this material.<|endoftext|> TITLE: Ruelle-Perron-Frobenius theorem for shift of finite type QUESTION [6 upvotes]: I know a version of Ruelle's theorem for expansive transformations in a compact metric space that says there is a single equilibrium state for a potential holder. In this Ruelle-Perron-Frobenius theorem from the Bowen book: Equilibrium States and the Ergodic Theory of Anosov Diffeomorphisms, the same thing is said is basically Ruelle's theorem for shift of finite type, but my question is: what does this limit that I marked green intuitively mean? REPLY [7 votes]: The most intuitive explanation I know is the following: suppose that you have a certain amount of mass (I usually picture a pile of sand) that is distributed over $\Sigma_A^+$ according to the density $g\,d\nu$, where $g$ is an arbitrary continuous function and $\nu$ is the eigenmeasure for the transfer operator $\mathcal{L}$. Then you move this mass around according to the shift map: a grain of sand at $\underline{x}$ gets moved to $\sigma(\underline{x})$, for each $\underline{x} \in \Sigma_A^+$. After doing this $n$ times, the sand is distributed according to some new density $g_n \,d\nu$. Then the new density function $g_n$ is given by $g_n = \lambda^{-n} \mathcal{L}^n g$; in other words, the transfer operator tells you how mass moves around (w.r.t. the underlying reference measure $\nu$). This much can be proved just by manipulating the definition of the transfer operator, and is a good exercise if you didn't see it yet. The intuitive content of the limit you marked in green is that as $n\to\infty$, the density functions $g_n$ converge to the eigenfunction $h$, perhaps multiplied by some constant that corresponds to the initial amount of sand; in other words, mass that is spread out according to any initial distribution absolutely continuous w.r.t. $\nu$ will eventually converge towards a distribution with density function given by $h$.<|endoftext|> TITLE: Proof that geodesics have zero curvature with Cartan's moving frames method QUESTION [5 upvotes]: I tried to use Cartan's moving frames method to prove that any minimizing-length curve in $\mathbb{R}^2$ has zero curvature. Here below is my idea of proof. I am asking for a proof-verification and for some specific doubts. The set of all orthonormal frames could be identified with the euclidean plane group $E(2)$. We have the principal bundle $\pi: E(2) \to \mathbb{R}^2$. We fix two points $A, B \in \mathbb{R}^2$ and suppose that it exists a 1-manifold $C \subseteq \mathbb{R}^2$ such that $\partial C = \{A, B\}$ and $C$ has minimal length. It exists $f: C \to E(2)$ such that $\pi f = \text{id}_C$ and $f^*\omega^2 = 0$. This is the classic Frenet-Serret-Cartan frame associated to a curve and we have two possible choice for the orientation. We define $F = f(C) \cong C$. We have that $\int_C f^*\omega^1$ is minimum because $C$ has minimal length. We consider a variation with fixed end-points in $\mathbb{R}^2$, i.e. a vector field $X \in \mathfrak{X}(\mathbb{R}^2)$ such that $X_A = X_B = 0$. Let be $\phi_t: \mathbb{R}^2 \to \mathbb{R}^2$ the flow of $X$ for $t \in (-\epsilon, \epsilon)$. We define the variations of $C$ as $C_t = \phi_t(C) \cong C$. For each $C_t$ it exists the unique frame $f_t: C_t \to E(2)$ such that $\pi f = \text{id}_{C_t}$, $f_t^*\omega^2 = 0$ and $f_t$ has the same orientiation of $f$. Obviously we have that $C_0 = C$ and $f_0 = f$. Because $C$ has minimal length we have that $\frac{\text{d}}{\text{d}t}\int_{C_t}f_t^*\omega^1|_{t=0} = 0$ and so that $\frac{\text{d}}{\text{d}t}\int_{F}(f_t\phi_t\pi)^*\omega^1|_{t=0} = 0$ after a change of variable. If we define $Y: F \to TE(2)$ such that $Y(p) = \frac{\partial}{\partial t}(f_t\phi_t\pi)(p)\big|_{t=0}$ we have that 0 = $\frac{\text{d}}{\text{d}t}\int_{F}(f_t\phi_t\pi)^*\omega^1|_{t=0} = \int_{F}\frac{\partial}{\partial t}(f_t\phi_t\pi)^*\omega^1|_{t=0} = \int_{F}\mathcal{L}_Y\omega^1 = \int_{\partial F}i_Y\omega^1 + \int_F i_Y\text{d}\omega^1$ I think that the first integral vanishes. In fact I guess that $T\pi \circ Y = X \circ \pi$ so $T\pi \circ Y(\partial F) = X \circ \pi (\partial F) = X (\partial C) = 0$. Hence $0 = \omega^1(Y) = i_Y\omega^1$ on $\partial F$. For the second integral we have that $\text{d}\omega^1 = \omega^1_2\wedge\omega^2$ and so $i_Y\text{d}\omega^1 = Y^1_2\omega^2 - Y^2\omega^1_2$ where $Y^1, Y^2, Y^1_2$ are the local components of $Y$. Because $\omega^2$ on $F$ vanishes we have that $0 = - \int_FY^2\omega^1_2 = - \int_F Y^2 k \omega^1$ where $k: F \to \mathbb{R}$ is the curvature of $C$. Because $Y^2$ depends only of $X^2$ and $X^2$ could be choosen randomly and also because $\omega^1 \neq 0$ always we can conclude that $k = 0$. My really first question is: is this proof correct? My doubts regarding the proof are: Does $f_t$ always exist? More important: is it $\mathcal{C}^\infty$ in $t$? Does Lie derivative definition work for $Y$ even if $Y$ is defined only on $F$ and not on all $E(2)$? Are $X$ and $Y$ $\pi$-related? Is true that $i_Y\omega^1$ vanishes on $\partial F$? Is true and $Y^2$ could be everything and $Y^2(p) = X^2(\pi(p))$? Can we conclude that it is necessary that $k=0$ if the integral vanishes for all $Y^2$? Other questions regard possible generalizations: Can we modify this proof for $C$ not being a $1$-manifold but a parametrized curve? Does this proof work for a 2 Riemannian manifold with principal bundle of orthonormal frames $\pi: O(M) \to M$ and the Levi-Civita connection $\omega^1_2$? Can we avoid using local components of $Y$ in the proof? Because maybe this could be a problem if the topology of $O(M)$ is more complicated of those of $E(2) \cong \mathbb{R}^2 \times \mathbb{S}^1$. REPLY [2 votes]: $\newcommand{\R}{\mathbb{R}}$ Here's another way to write out the calculation (CORRECTED): Let $f_0: [0,1]\rightarrow E(2)$ be the Frenet frame of a curve. In particular, if $s$ denotes the coordinate on $[0,1]$, then \begin{align*} f_0^*\omega^1 &= ds\\ f_0^*\omega^2 &= 0\\ f_0^*\omega^1_2 &= k\omega^1 = k\,ds, \end{align*} where $k$ is the curvature. Define a variation to be a map $f: [0,1] \times (-\delta,\delta) \rightarrow E(2)$ such that $f(\cdot,0) = f_0$. For each $t$, let $f_t = f(\cdot, t)$. Let $(s,t)$ denote the coordinates on $D = [0,1]\times (-\delta,\delta)$, $S = \partial_s$, and $T = \partial_t$. We also assume the frame along $f_t$ has $e_1$ tangent to the curve, i.e., $$ \langle f_t^*\omega^1, S\rangle = 0. $$ Also, note that if $\theta$ is a differential form on $E(2)$, then $$ \langle S, f^*_t\theta\rangle = \langle S, f^*\theta\rangle, $$ and, if $\eta$ is a $1$-form on $D$, $$ \partial_t\eta = \mathcal{L}_T\eta = d\langle T,\eta\rangle + \langle T, d\eta\rangle $$ Therefore, the derivative of the length of the curve parameterized by $f_t$ is \begin{align*} \frac{d}{dt}\int_{s=0}^{s=1} \langle S,f_t^*\omega^1\rangle\,ds &= \frac{d}{dt}\int_{s=0}^{s=1} \langle S, f^*\omega^1\rangle\,ds\\ &= \int_{s=0}^{s=1} \langle S, \partial_t(f^*\omega^1)\rangle\,ds\\ &= \int_{s=0}^{s=1} \langle S, \mathcal{L}_T(f^*\omega^1)\rangle\,ds\\ &= \int_{s=0}^{s=1} \langle S, d\langle T,f^*\omega^1\rangle\rangle + \langle T\otimes S, d(f^*\omega^1)\rangle\,ds\\ &= \int_{s=0}^{s=1} \partial_s(\langle T,f^*\omega^1\rangle) + \langle T\otimes S, f^*(d\omega^1)\rangle\,ds\\ &= \left.\langle f_*T,\omega^1\rangle\right|_{s=0}^{s=1} + \int_{s=0}^{s=1} -\langle T\otimes S,f^*(\omega^1_2\wedge\omega^2)\rangle\,ds\\ \end{align*} If we assume that $f$ fixes the endpoints of the curve, then $f_*T(0,0)= \partial_t f(0,0)= 0$ and $f_*T(1,0)= \partial_t f(1,0)= 0$. Therefore, \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\int_{s=0}^{s=1} \langle S, f_t^*\omega^1\rangle\,ds &= \int_{s=0}^{s=1} -\langle T,f^*\omega^1_2\rangle\langle S,f^*\omega^2\rangle + \langle T,f^*\omega^2\rangle \langle S, f^*(\omega^1_2)\rangle\, ds\\ &= \int_{s=0}^{s=1} -\langle T,f^*\omega^1_2\rangle\langle S,f_0^*\omega^2\rangle + \langle T,f^*\omega^2\rangle \langle S, f_0^*(\omega^1_2)\rangle\, ds\\ &= \int_{s=0}^{s=1} \langle T,f^*\omega^2\rangle \langle S, f_0^*(\omega^1_2)\rangle\, ds\\ &= \int_{s=0}^{s=1} \langle T,f^*\omega^2\rangle \langle S, k\,ds\rangle\, ds\\ &= \int_{s=0}^{s=1} \langle f_*T,\omega^2\rangle k\, ds\\ &= \int_{s=0}^{s=1} \langle X,\omega^2\rangle k\, ds, \end{align*} where $X = \partial_tf(\cdot,0)$. If this vanishes for any variation $f$ that fixes the endpoints, then $k$ must vanish on the curve.<|endoftext|> TITLE: On the integral $\int_0^1\log(x!)dx$ revisited QUESTION [5 upvotes]: I was interested in an integral that I known from [1], it is $$\int_0^1 \log(x!)dx.$$ I tried to get such closed-form using myself ideas and symbolic calculations, also with the help of Wolfram Alpha online calculator. But I don't know how get the sum of certain series involving a special function. My first step was to invoke the formula $(3.13)$ from [2], taking the logarithm and integrating one has that $$\int_0^1 \log(x!)dx$$ is equals to $$\log 2+\frac{1}{24}\sum_{n=1}^\infty\left(-2^{n}(6(6\log A)+\log \pi)-24\psi^{(-2)}(\frac{1}{2}+2^{-n})+\log(32))-12\log\pi\right),$$ where $A$ is the Glaisher-Kinkelin constant and $\psi^{(n)}(s)$ denotes the $n^{th}$ derivative of the digamma function. See the first comment that you can to evaluate in Wolfram Alpha online calculator. I've some computational evidence, for example the following code (it is a line) sum 1/24 (-2^n (6 (6 log(Glaisher) + log(π)) - 24 polygamma(-2, 1/2 + 2^(-n)) + log(32)) - 12 log(π)), from n=1 to 100 and log(2)-0.7742086473552725676369-(1/2log(2pi)-1) but I can not to prove the corresponding closed-form. Of course I know how to get the sum of geometric series but the problem here is different. Question. Please, prove that $$\log 2+\frac{1}{24}\sum_{n=1}^\infty\left(-2^{n}(6(6\log A)+\log \pi)-24\psi^{(-2)}(\frac{1}{2}+2^{-n})+\log(32))-12\log\pi\right)$$ is equals to the closed-form for $\int_0^1 \log(x!)dx$ given in [1]. What I ask is if you can analyze the series to calculate its sum. I've deduced/considered previous expression and I would like to know if it is possible to prove that previous expression equals to $\frac{1}{2}\log(2\pi )-1$ analizing the series to get its sums (without invoking that it is equals to $\int_0^1 \log(x!)dx$). Many thanks. I have no intuition/knowledges to know if it is easy to get the sum of the series. References: [1] Muliplicative integral of $\Gamma(x)$, this MathOverflow (July of 2010). [2] Manuel Benito, Luis M. Navas and Juan Luis Varona, Möbius inversion from the point of view of arithmetical semigroup flows, Biblioteca de la Revista Matemática Iberoamericana, Proceedings of the "Segundas Jornadas de Teoría de Números", (2008), pages 61-81. REPLY [6 votes]: Details of the simple integration by series for $\int_0^1\log(x!)dx$ mentioned above (hopefully yours may be treated analogously, if you wish to try it). Start from the series of the logarithm of the infinite product for $x!$ $$\log(x!)=- \gamma x + \sum_{k=1}^m\bigg(\frac{x}{k}-\log\big(1+\frac{x}{k}\big)\bigg)+o(1),$$ uniformly on $[0,1]$ as $m\to+\infty$. Integrating on $[0,1]$ we get $$\int_0^1\log(x!)dx= -\frac{\gamma}{2}+\frac{1}{2}\sum_{k=1}^m \frac{1}{k}+m-\sum_{k=1}^m\Big((k+1)\log(k+1)-k\log k\Big) +\sum_{k=1}^m \log k +o(1)$$ Simplifying the telescopic sum, and by the definition of the Euler-Mascheroni constant, this is $$= \frac{1}{2}\log m +m -(m+1)\log(m+1)+ \log(m!) +o(1)$$ $$= \frac{1}{2}\log m +m -\Big(m\log m + \log m + 1\Big)+ \log(m!) +o(1)$$ $$=- \frac{1}{2}\log m +m -m\log m -1 + \log(m!) +o(1)$$ Finally by the (logarithm of the) Stirling formula this is exactly $$\frac{1}{2}\log(2\pi)-1+ o(1),$$ ending the computation.<|endoftext|> TITLE: Physicists misuse the term "Kac Moody algebra". Does that bring problems? QUESTION [11 upvotes]: In physics textbooks one frequently sees the name (affine) Kac Moody algebra used to describe the universal (one dimensional) central extension of the loop algebra of a semisimple algebra. But this is not an affine Kac Moody algebra: one would still need a derivation extension to obtain a true (untwisted) affine Kac Moody algebra. Do the properties of the object that physicists call affine Kac Moody algebra resemble the properties of the true Kac Moody algebra so closely that this misuse of the terms is rendered harmless? Meaning, are physicists somehow justified to assume that properties proved for affine Kac Moody algebras hold for "their" affine Kac Moody algebras? REPLY [8 votes]: I can't address all uses by all physicists, but in many contexts, they consider only representations at a fixed level that admit a well-behaved energy grading. That is, sometimes an energy grading is implicit in the environment. Such a grading can be given by the eigenvalues of some semisimple operator, and adjoining such an operator to the centrally extended loop algebra yields the full affine Kac-Moody Lie algebra. In this case, the representations in question naturally extend to representations of the larger algebra, but when energy is implicit, a physicist may feel that references to an explicit operator are unnecessary.<|endoftext|> TITLE: Computing a cone in a $\otimes$-triangulated category QUESTION [5 upvotes]: I have a $\otimes$-triangulated category $\mathcal T$ and two triangles in $\mathcal T$: $$ x_0\to x_1\to c_x\to \Sigma x_0\ \ \ \text{and}\ \ \ y_0\to y_1\to c_y\to \Sigma y_0. $$ Consider the following diagram, where the object $P$ is constructed as a homotopy push-out of the obvious maps $\Sigma^{-1}c_x\otimes y_0 \leftarrow\Sigma^{-1}c_x\otimes\Sigma^{-1}c_y\to x_0\otimes \Sigma^{-1} c_y$ (one can do this either as in Neeman's book, or assuming that there is some enhancement for $\mathcal T$): Is it true that the cone of the map $\varphi$ obtained this way is isomorphic to $x_1\otimes y_1$? The question is relatively natural, so I hope that somebody with a little more working experience with $\otimes$-triangulated categories will be able to give a simple proof or a known counterexample in some concrete situation. If this is not known in general, I am happy to assume that $\mathcal T$ has some kind of enhancement (e.g., $\mathcal T$ is the homotopy category of a stable $(\infty,1)$-category, or $\mathcal T$ is the base of a strong and stable derivator), or even, to start with, to assume that $\mathcal T$ is the derived category of a commutative ring. REPLY [4 votes]: In what follows I am going to assume $\mathcal{T}$ is the homotopy category of a stably symmetric monoidal ∞-category. The argument I am going to give can probably be generalized to any case of interest, but I'll stay in familiar territory (for me). What you are trying to compute is usually called the total cofiber of the square $$\require{AMScd} \begin{CD} \Sigma^{-1}c_x \otimes \Sigma^{-1}c_y @>>> x_0 \otimes \Sigma^{-1}c_y\\ @VVV @VVV\\ \Sigma^{-1}c_x\otimes y_0 @>>> x_0\otimes y_0 \end{CD}$$ Now the total cofiber of a square can be computed by taking the cofiber of the induced map on the cofiber of any pair of parallel arrows (this is a well known result, at the end of this answer of mine you can find a short proof in the case of the total fiber), so we can compute the required object by first taking the cofibers of parallel arrows. Since the tensor product preserves fiber sequences, and the cofiber of $\Sigma^{-1}c_x\to x_0$ is $x_1$, the required object is the cofiber of $$x_1\otimes \Sigma^{-1}c_y\to x_1\otimes y_0$$ Hence it coincides with $x_1\otimes y_1$, as required.<|endoftext|> TITLE: Why would one number theorems, propositions and lemmas separately? QUESTION [24 upvotes]: When it comes to numbering results in a mathematical publication, I'm aware of two methods: Joint numbering: Thm. 1, Prop. 2, Thm. 3, Lem. 4, etc. Separate numbering: Thm. 1, Prop. 1, Thm. 2, Lem. 1, etc. Every piece of writting advice I have encountered advocates the use of 1. over 2., the rationale being that it makes it easier to find the result based on the number. It seems that 1. is more popular than 2., although 2. still exists, especially in books. I can only imagine that people using 2. must have a reason, but I have not yet to encounter one. I hope it is not too opinion-based to ask: What is the rationale for separately numbering theorems, propositions and lemmas, like in 2.?" REPLY [4 votes]: This is a slight elaboration of François Dorais's comment. If you have a small number of theorems/lemmas/propositions—let's say, small enough that readers can reasonably be expected to hold all the theorems in their head at once—then the second method of numbering can help readers grasp the flow of the paper and can even serve as a mnemonic aid. A secondary consideration, similar to what Fedor Petrov said, is that the reader may want to skim through and just look at the main theorems. If you adopt the first method of numbering, then readers might accidentally skip from (say) Theorem 8 to Theorem 17 without realizing that they missed Theorem 14. One famous book that uses the second method of numbering is Serre's Course in Arithmetic. Serre uses the "Theorem" designation very sparsely in that book, and the numbering system helps make the Theorems stand out.<|endoftext|> TITLE: Flat maps and Zariski tangent spaces QUESTION [6 upvotes]: Let $f:A \to B$ be a finite flat local homomorphism of noetherian local rings. Are there some nice conditions on $A$ and $B$ which guarantee that the dimension of the Zariski tangent space of $A$ (at its maximal ideal) is smaller or equal to the dimension of the Zariski tangent space of $B$ (at its maximal ideal)? For example, if $B$ is regular then $A$ is regular, so the above would hold, but I don't want to assume something so strong. I know no example where it fails, but I am particularly interested in the case that $A$ and $B$ are both lci and Artin. REPLY [10 votes]: Lemma. Let $A \to B$ be a flat local homomorphism of Noetherian local rings with the same residue field $k$ such that $A$ and $B$ are both (local) complete intersections. Then $$ \dim(B) - \dim(A) + \dim_k \mathfrak m_A/\mathfrak m_A^2 \leq \dim_k \mathfrak m_B/\mathfrak m_B^2 $$ provided $B$ is essentially of finite type over $A$. Proof. By a theorem of Avramov, the ring map $A \to B$ is a local complete intersection homomorphism! Consider the distinguished triangle $$ L_{B/A} \otimes_B^\mathbf{L} k \to L_{k/A} \to L_{k/B} \to (L_{B/A} \otimes_B^\mathbf{L} k)[1] $$ of cotangent complexes associated to the ring maps $A \to B \to k$. Observe that $H^0(L_{k/A}) = 0$ and $H^{-1}(L_{k/A}) = \mathfrak m_A/\mathfrak m_A^2$ and similarly for $B$. Since $A \to B$ is a flat local complete intersection of relative dimension $\dim(B) - \dim(A)$, the complex $L_{B/A}$ is isomorphic in $D(B)$ to a complex of the form $$ B^{\oplus r} \to B^{\oplus r + \dim(B) - \dim(A)} $$ for some integer $r \geq 0$ with the two terms sitting in degrees $-1$ and $0$ and the map given by a matrix whose coefficients are in $\mathfrak m_B$. To see this we use that $B$ is essentially of finite type over $A$ and hence a localization of an algebra of the form $A[x_1, \ldots, x_{r + \dim(B) - \dim(A)}]/(f_1, \ldots, f_r)$ which is a relative global complete intersection. Putting everything together we see that we get an exact sequence $$ k^{\oplus r} \to \mathfrak m_A/\mathfrak m_A^2 \to \mathfrak m_B/\mathfrak m_B^2 \to k^{\oplus r + \dim(B) - \dim(A)} \to 0 $$ of $k$-vector spaces. We conclude by dimension theory for finite dimensional vector spaces. Remark. The assumption of $B$ being essentially of finite type over $A$ can be removed, but it becomes a bit technical if you want to do this. Conversely, you can spell out what Avramov's theorem means in nontechnical terms and prove the lemma without using the cotangent complex (but since Avramov's theorem is kind of hard you don't gain a lot in doing so I think). Note. I haven't considered what happens if the residue fields aren't the same (for example if the residue field extension is inseparable). I haven't tried to make a counter example if the rings aren't lci.<|endoftext|> TITLE: The use of Schur's lemma for Lie algebras in physics (CFT) QUESTION [7 upvotes]: Anytime a one-dimensional central extension appears in the physics literature, immediately they assume that in any irreducible representation the central charge will be a multiple of the identity, implicitly (and sometimes explicitly) using Schur's Lemma (for Lie algebras). However, the version of the Schur's Lemma which tells us this is only valid for finite-dimensional modules, and physicists are usually interested in representing Hilbert spaces which in general can have infinite dimension. Is it somehow still justified to say that the central charge is a multiple of the identity even for infinite-dimensional irreducible representations? The two examples I have in mind are the Virasoro algebra as one dimensional central extension of the Witt algebra and the universal central extension of a loop algebra, both appearing in CFT texts, and in both cases they use Schur's Lemma as described. REPLY [6 votes]: Let $\mathfrak{g}$ be a complex Lie algebra with a distinguished nonzero central element $x$, and let $V$ be an irreducible representation of $\mathfrak{g}$. The usual proof of Schur's lemma can be adapted to show that if $x$ admits an eigenvector in $V$, then $x$ acts by a scalar: If $v$ is an eigenvector with eigenvalue $\lambda$, then $v$ lies in a submodule, namely the kernel of the central transformation $x - \lambda$, so irreducibility implies this kernel is equal to $V$. Your question becomes: What makes it safe for physicists to assume a central element has an eigenvector? The usual answer is that the representations that appear naturally tend to have gradings with at least one finite dimensional component, so the action of the central element restricts to such finite dimensional parts where one necessarily has an eigenvector.<|endoftext|> TITLE: Is there a way to prove, that $2$-generated groups are rare among finite groups? QUESTION [9 upvotes]: Is there a way to prove, that $\lim_{n \to \infty} \frac{\text{the number of all } 2 \text{-generated groups of order less than }n}{\text{the number of all groups of order less than } n} = 0$? This statement is implied by a well known conjecture: $\lim_{n \to \infty} \frac{\text{the number of all groups of nilpotency class } 2 \text{, exponent } 4 \text{ and order less than }n}{\text{the number of all groups of order less than } n} = 1$ (because the $2$-generated relatively free group of nilpotency class $2$ and exponent $4$ is finite) However, the problem with aforementioned statement is, that it is... ...well, just a conjecture. Is there a way to strictly prove the main statement of the question? REPLY [12 votes]: Lubotzky proved that the number of $2$-generator groups of order at most $n$ (or order exactly $n$) is bounded by $n^{A \log n}$ for some constant $A$. The number of $2$-groups of order $2^m$ with $n \ge 2^m > n/2$ is (corrected, see Will Sawin's comment) $n^{B \log^2 n}$ for some explicit constant $B$. So you win! Using Lubotzky's theorem, you still win even if $2$-generator groups are replaced by $d$-generator groups for any fixed $d$. Here is Lubotzky's paper: Enumerating Boundedly Generated Finite Groups, Journal of Algebra 238 (2001) pp 194–199. doi:10.1006jabr.2000.8650, core.ac.uk version.<|endoftext|> TITLE: Automatically solving olympiad geometry problems QUESTION [50 upvotes]: Warning: I am only an amateur in the foundations of mathematics. My understanding of this Wikipedia page about Tarski's axiomatization of plane geometry (and especially the discussion about decidability) is that "plane geometry is decidable". The 2019 International Maths Olympiad happened recently, and there were two plane geometry questions in it (problems 2 and 6). Their solutions look really intimidating! However even as a student I felt that one should be able to solve these questions, in theory, by just "writing down coordinates of everything and doing the algebra". Tarski's work, which I will freely confess that I do not understand fully, might even vindicate my view. The question: Is there an algorithm for solving these kinds of questions, or have I misunderstood? If so, is this algorithm actually feasible to run in practice nowadays (on a computer say) for IMO-level problems? In other words -- are there computer programs which will take as input a planar geometry question of "olympiad level" (for example problems 2 and 6 in this year's IMO) and actually output a solution? Currently I am not too bothered about whether the solution is human-readable -- it could just be a formal proof in some kind of type theory or something, but the output would be some object that some expert could coherently argue was a solution of some sort. The reason I'm asking is that I was talking to some computer scientists about various goals in the long-term project of getting computers to do mathematics "better than humans", and having a computer program which could solve IMO problems by itself was a suggested milestone. REPLY [5 votes]: There is a pretty general method (although not always sufficient) to apply your intuition that one could translate everything into algebra and then solve it there. Essentially, you introduce coordinates for your points, encode all your hypothesis as polynomial equalities between coordinates, do the same for the thesis, and then try to prove that the thesis is in the ideal generated by the hypotheses (or even its radical) using Gröbner bases. Of course, the issue here is that the classical Nullstellensatz does not hold for $\mathbb{R}$, so the thesis may hold even if it does not lie in the radical of the ideal generated by the hypotheses. Using the real Nullstellensatz, it may be possible to adapt the technique, but I did not give it much thought. To make a concrete example, say you want to prove Heron's formula. Let $T$ be a triangle with side length $a, b, c$ and area $s$. You choose coordinates for the vertices of $T$ so that they are $(0, 0), (a, 0), (x, y)$ (this particular nice choice of coordinates is not necessary on a computer but simplifies the discussion for humans). Then the hypotheses are: $b^2 = x^2 + y^2$ $c^2 = (a - x)^2 + y^2$ $2s = a y$. The thesis is Heron's formula $16 s^2 = (a + b - c)(c + a - b)(b + c - a)(a + b + c)$. What you do is consider the ideal $I \subset \mathbb{R}[a, b, c, x, y, s]$ generated by $b^2 - x^2 - y^2$, $c^2 - (a - x)^2 - y^2$ and $2s - ay$, and use Gröbner bases to check that $16 s^2 - (a + b - c)(c + a - b)(b + c - a)(a + b + c) \in \sqrt{I}$. In fact, since the thesis does not involve $x, y$, one can compute $I \cap \mathbb{R}[a, b, c, s]$ - again using Gröbner bases - and discover that it is generated by the equation expressing Heron's formula. EDIT The above can actually be implemented very efficiently. I used rings, an efficient Scala library to perform polynomial computations, and the following implicit val ring = MultivariateRing(Q, Array("a", "b", "c", "x", "y", "s")) val h1 = ring("b^2 - x^2 - y^2") val h2 = ring("c^2 - (a - x)^2 - y^2") val h3 = ring("2 * s - a * y") val t = ring("16 * s^2 - (a + b - c) * (c + a - b) * (b + c - a) * (a + b + c)") val I = Ideal(ring, Seq(h1, h2, h3)) I.contains(t) gave the answer true is about a second on my laptop.<|endoftext|> TITLE: A criterion for finite abelian group to embed into a symmetric group QUESTION [10 upvotes]: Let $G$ be a finite abelian group. Write $G\approx \mathbb{Z}/p_1^{i_1}\mathbb{Z}\times\dots \mathbb{Z}/p_m^{i_m}\mathbb{Z}$, with $m\ge 0$, $p_1,\dots,p_m$ primes (not necessarily distinct) and $i_k\ge 1$ for all $k$. For $n\ge 0$, is it true that there is an injective homomorphism $G\rightarrow S_n$ if and only if $p_1^{i_1}+\dots+p_m^{i_m}\leq n$? REPLY [13 votes]: Yes, this is true (provided we add the hypothesis $i_1, \ldots, i_n \ge 1$), and not hard to prove by considering centralizers of products of cycles in symmetric groups. According to http://www-history.mcs.st-andrews.ac.uk/Biographies/Povzner.html, the minimum degree of a permutation representation of an arbitrary abelian group was found in A. Ya. Povzner, Finding groups of permutations of least degree which are isomorphic to a given Abelian group, Kharkov Mat. Obshch., 14 (1937) 151–158. (I have not been able to obtain this paper to check.) Theorem 4.1 in Kay Jin Lim, Specht modules with abelian vertices, J. Algebraic Combin. 35 (2012) 157–171 states your conjectured result in the special case when all the primes are the same with a reference to Povzner's paper. REPLY [4 votes]: Yes. Your notation $n$ is used for two different meanings, so let me denote by $m$ the number of summands. A homomorphism $G\to S_n$ correspond to an action of $G$ on a set with $n$ elements, which can be split into orbits $A_1,...,A_\ell$. Each orbit can can be provided with a structure of a group with a surjective homomophism $G\to A_i$. Then, the condition is precisely that the induced map $G\to \oplus_i A_i$ is injective. Note the total number of cyclic summands in the right hand side is at least $n$. If $A_i$ is not cyclic, using $ab\ge a+b$ for $a,b>1$, we see that it only decreases the sum of sizes if we split $A_i$ into two parts corresponding to a decomposition $A_i = B\oplus C$. So we may assume that all the $A_i$-s are cyclic of order a prime power. By comparing each primary component at a time, we may assume that all the $p_i$-s are a single prime $p$. The result now follows from the following standard lemma Lemma: Let $\phi: \oplus_i \mathbb{Z}/p^{a_i} \to \oplus_j \mathbb{Z}/p^{b_j}$ be injective, with $a_1 \ge a_2 \ge ...\ge a_k$ and $b_1 \ge ...\ge b_\ell$. Then $a_i \le b_i$. edit: As mensioned correctly by @Richard Lyons, the proof I gave for it was totally incorrect so I deleted it. Here's an hopefully less wrong proof, im really sorry for that. Let $A\subseteq B$ be abelian finite $p$-groups. Write $B=\mathbb{Z}/p^k \oplus B'$ for $B'$ of exponent smaller than or equal $k$. Let $B\to \mathbb{Z}/p^k$ be the projection. If the composition $A\to B\to \mathbb{Z}/p^k$ is not surjective, then $A$ is contained in $p\mathbb{Z}/p^k \oplus B$ and we can proceed by induction onthe total number of elements of $B$. Otherwise $A$ contains an element of the form $(1,b)$ for $b\in B'$. Since the exponent of $B'$ is no more than $k$, there is a map $\mathbb{Z}/p^k\to B'$ mapping $1$ to $b$, and so by applying the automorphism $(x,c)\mapsto (x,c-xb)$ we may assume that $(1,0)$ is in $A$. It follows then that $A=\mathbb{Z}/p^k \oplus A\cap B'$ and we proceed by induction on the number of summands.<|endoftext|> TITLE: Is the set of escaping endpoints for $e^z-2$ completely metrizable? QUESTION [13 upvotes]: Let $f:\mathbb C \to \mathbb C$ be the complex exponential $$f(z)=e^z-2.$$ It is known that $J(f)$, the Julia set of $f$, is a uncountable collection of disjoint rays (one-to-one continuous images of $[0,\infty)$). It looks something like the figure below. The number $2$ is rather arbitrary; the dynamics of $f$ are practically identical if we replace $2$ with any number greater than $1$. Let $E(f)$ be the set of all $0$-endpoints of these rays, i.e. the "endpoints" of $J(f)$. Let $I(f)=\{z\in \mathbb C:f^n(z)\to\infty\}$. Here $f^2$ is the composition $f\circ f$, etc. Let $\tilde E(f)=I(f)\cap E(f)$ be the set of escaping endpoints of $J(f)$. It is known that $E(f)$ is completely metrizable. Question. Is $\tilde E(f)$ completely metrizable? We may independently consider the set of escaping points. It is easy to show $I(f)$ is an $F_{\sigma\delta}$-subset of the plane using only continuity of $f$. Question. Is $I(f)$ completely metrizable? Alhabib, Nada; Rempe-Gillen, Lasse, Escaping endpoints explode, Comput. Methods Funct. Theory 17, No. 1, 65-100 (2017). ZBL1381.37051. for further information. REPLY [2 votes]: $\tilde E(f)$ and $I(f)$ are first category, so the answer to each question is NO. https://doi.org/10.1017/etds.2019.111<|endoftext|> TITLE: Lower bound for the number of lattice points on high dimensional spheres QUESTION [5 upvotes]: Let $rS^{d-1}$ denote the sphere of radius $r$ in dimension $d$ (centered at the origin). I'm interested in the number of lattice points on the sphere (not inside). More precisely, let $$ N(r,d):=\text{number of lattice points on the sphere of raduis } r=\#\{x\in rS^{d-1}: x\in \mathbb{Z}^d\}. $$ I'm especially interested in the lower bound of $N(r,d)$ for any $d\ge 3$ and large $r$ (with $r^2\in\mathbb{Z}$, of course). For example, I found in the book by F. Fricker Einführung in die Gitterpunktlehre. (German) [Introduction to lattice point theory] that the following result seems to be true (my German is poor): $N(r,d)\gtrsim r^{d-2}$ for $d\ge 4$. So what about $d=3$ case? What is the current best lower bound? The book is in 1982 so I guess there might be a better exponent than $d-2$ now. One can also ask a weaker question: is there a sequence of $r$ tending to $\infty$ such that the above inequality holds with a better lower bound? REPLY [4 votes]: My answer to this MO question contains the answer to your question, especially if you take into account that $L\left(1,\left(\frac{D}{\cdot}\right)\right)$ can be estimated unconditionally (i.e. without GRH): $$|D|^{-\varepsilon}\ll_\varepsilon L\left(1,\left(\tfrac{D}{\cdot}\right)\right)\ll \log|D|.$$ The lower bound is ineffective (i.e. we don't know the implied constant), and it is due to Siegel (1934). The upper bound is effective and older. Both bounds are explained in Montgomery-Vaughan: Multiplicative number theory I.<|endoftext|> TITLE: One more generator needed for a Z/6 elliptic curve QUESTION [7 upvotes]: I am trying to find the next rank 8 curve with the torsion subgroup Z/6 using Kihara's family as described in https://arxiv.org/pdf/1503.03667.pdf. Meanwhile, I came across a curve generated by $t=629/3287$ (or $t=6202/8089$, $t=-8089/1772$, $t=-23009/1258$). Magma Calculator (http://magma.maths.usyd.edu.au/calc/) and mwrank return 6 generators for this curve. SetClassGroupBounds("GRH"); E := EllipticCurve([1, 0, 1, -134523401167995213138670219183146040563810987418811883, 66402369909929526433604564866758135700820111823876373971833120805994125518227306]); MordellWeilShaInformation(E); Sagemath 8.4 returns 7 for the upper bound of analytic rank. E = EllipticCurve([1,0,1,-134523401167995213138670219183146040563810987418811883,66402369909929526433604564866758135700820111823876373971833120805994125518227306]) E.analytic_rank_upper_bound(max_Delta=2.8,root_number="compute") Is there a way to find one more generator? A working piece of any code would be greatly appreciated. Max REPLY [7 votes]: Yes. A 7th generator has $x$-coordinate $$ 181265389257356655988118224516379188326810855287159053664052560/3919647209484520988422390115383428889. $$ Knowing the $6$ generators Magma finds (let's say they are P1, P2, P3, P4, P5, P6), the Magma command twocovers := TwoDescent(E : RemoveTorsion := true, RemoveGens := {P1,P2,P3,P4,P5,P6}); finds the unique $2$-cover which should have a rational point, but on which we haven't yet found one. Then, one can obtain $4$-covers for this via fourcovers := FourDescent(twocovers[1] : RemoveTorsion := true, RemoveGensEC := {P1,P2,P3,P4,P5,P6}); This command takes about 20 minutes to run on my machine, and it returns an intersection of two quadrics on which one needs to find a point. Point searching to a height of $10^{8}$ turns up a point, and this leads to a point P7 on $E$ with $\hat{h}(P7) \approx 171.3$. Subtracting off a linear combination of P1, P2, $\ldots$, P6 one obtains the point with $x$-coordinate above and canonical height $\approx 94.34$. If this hadn't worked, a potentially feasible (but quite time-consuming) strategy is the following. Since $E$ has a rational point of order $6$, the image of the mod $3$ Galois representation is quite small and doing a $3$-descent is feasible. Magma has an implementation of Tom Fisher's algorithm (see his 2008 paper from the Journal of Algebra) of combining $3$-covers and $4$-covers to obtain $12$-covers, and point searching on those can yield rational points on $E$ with canonical height in excess of $1000$.<|endoftext|> TITLE: what do we know about the product $\prod_{n\in \mathbb{N}}(1-\frac{x^m}{n^m})?$ QUESTION [9 upvotes]: My question is what do we know about the product $\prod_{n\in \mathbb{N}}(1-\frac{x^m}{n^m})?$ which is a slightly modified product from the eulerian product. REPLY [2 votes]: The even case $$\prod_{n \ge 1} (1-(x/n)^{2m}) =\prod_{a=1}^m \prod_{n \ge 1} (1- e^{2i \pi a/m}(x/n)^2)= \prod_{a=1}^m \frac{\sin(\pi e^{i \pi a/ m} x)}{\pi e^{i \pi a/ m} x}$$ Otherwise $$\prod_{n \ge 1} (1-(x/n)^m) =\lim_{N \to \infty }\prod_{a=1}^m \prod_{n=1}^N (1- e^{2i \pi a/m}x/n)= \prod_{a=1}^m \frac{-e^{2i \pi a/ m} x}{\Gamma(-e^{2i \pi a/ m} x)}$$<|endoftext|> TITLE: A theorem by Harald Cramér? QUESTION [7 upvotes]: In the paper “On the order of magnitude of the difference between consecutive prime numbers” by Harald Cramér there is the following statement: Suppose $\{X_n\}_{n=2}^\infty$ is a sequence of independent random variables, such that $X_n \sim Bern(\frac{1}{\ln(n)})$. Then $\lim_{n \to \infty} \sup |\frac{\sqrt{\ln(n)}(\Sigma_{i=2}^n X_i - li(n))}{\sqrt{2n \ln(\ln(n))}}| = 1$ However, he does not prove this result there, but rather states, that it is proved in his paper “Prime numbers and probability” (which I could not find) My question is: How can this statement be proved? Probably, it has something to do with the Law of Iterated Logarithm, but I do not know for sure ... REPLY [11 votes]: This is a special case of a general law of the iterated logarithm for non-iid random variables (r.v.'s), which states the following: Suppose that $Y_1,Y_2,\dots$ are independent zero-mean r.v.'s, $S_n:=\sum_1^n Y_i$, $B_n:=Var\, S_n\to\infty$, $|Y_n|\le M_n\in(0,\infty)$, and $M_n=o((B_n/\ln\ln B_n)^{1/2})$. Then $$\limsup_n\frac{S_n}{\sqrt{2B_n\ln\ln B_n}}=1 $$ almost surely. See e.g. V. Petrov, Ch. X, Theorem 1. This theorem is due to Kolmogorov (1929). (Just in case, here is a reference to Cramér's paper: Cramér, H. 1935 Prime numbers and probability. Skand. Mat.-Kongr. 8, 107--115. I found it in Granville's paper.)<|endoftext|> TITLE: Analogue of integration for group cohomology QUESTION [6 upvotes]: Consider some oriented surface $S$ with fundamental group $\pi_1(S)$. The group cohomology of $\pi_1(S)$ with coefficients in $\mathbb{R}$ is isomorphic to the de Rham cohomology of $S$. In degree 2, integration gives a map $\int_S\colon H^2_{dR}(S;\mathbb{R})\to \mathbb{R}$. In terms of group cohomology, with cochains being (inhomogeneous) maps $\pi_1(S)\times \pi_1(S)\to \mathbb{R}$ instead of 2-forms, what is the analogue of integration ? Say differently, is there a natural map $H^2(\pi_1(S);\mathbb{R})\to \mathbb{R}$ that corresponds to integration of differential forms ? REPLY [3 votes]: We always have a pairing $H_k(X, \mathbb{R}) \otimes H^k(X, \mathbb{R}) \to \mathbb{R}$ between chains and cochains. If $X$ is a closed oriented $n$-manifold, the orientation equips it with a class $[X] \in H_n(X, \mathbb{R})$ called the fundamental class, and pairing this class with elements in $H^n(X, \mathbb{R})$ reproduces integration of $n$-forms when $X$ is smooth. When $X$ is furthermore aspherical, so a $K(\pi_1(X), 1)$, then its homology and cohomology can be computed as group homology and cohomology of $\pi_1(X)$ as Ben Wieland says in the comments.<|endoftext|> TITLE: Quotient of arbitrary free involution on $S^n$ QUESTION [9 upvotes]: If we consider arbitrary free involution on $S^n$, then the quotient need not be diffeomorphic to $\Bbb RP^n$ if $n\geq 5$ and a reference for this is "some curious involutions of spheres" by Morris W. Hirsch and John Milnor. In that same paper, it was mentioned as an unsolved problem that whether the quotient space is homeomorphic to $\Bbb RP^n$?. I would like to know if the answer of the above question is known by now and references for that will be very helpful. Thank you so much! REPLY [11 votes]: In every dimension $\ge 4$ there is a fake real projective space, i.e., a manifold that is homotopy equivalent but not homeomorphic to $RP^n$. Here you can find a computation for the topological surgery structure set of $RP^n$. It is stated for $n>4$ but I think it extends to $n=4$ because $\mathbb Z_2$ is good in the sense of Freedman. To get a homeomorphism classification of homotopy $RP^n$'s we also need to know the group of homotopy self equivalences of (any manifold homotopy equivalent to) $RP^n$: it is $\mathbb Z_2$ if $n$ is odd and trivial if $n$ is even (see e.g. p.61 of Rutter's survey ``Spaces of homotopy self-equivalences''). Moreover, if $n$ is odd, then $RP^n$ admits a self-map of degree $-1$, namely, the map induced by reflection in an equator of $S^n$. Thus any homotopy self-equivalence of $RP^n$ is homotopic to a diffeomorphism. Let $f_i: M_i\to RP^n$ be homotopy equivalences representing different elements in the structure set. If $d_{ij}: M_i\to M_j$ is a homeomorphism, then $f_i^{-1}f_j d_{ij}$ is a homotopy self-equivalence of $M_i$. By the previous paragraph $f_i^{-1}f_j d_{ij}$ is homotopic to a homeomorphism, so $f_i$, $f_j$ represent the same element in the structure set. Thus in every dimension there is a fake $RP^n$. A word of caution: the above argument does not show that the fake $RP^n$ is smoothable. For example, the only fake $RP^4$ is not smoothable as I learned from ``Invariant knots of free involutions of $S^4$'' by Ruberman, see here. There do exist many smoothable fake $RP^n$'s, see e.g., Smooth free involutions on homotopy $4k$-spheres by Fintushel-Stern.<|endoftext|> TITLE: Formal mathematical definition of renormalization group flow QUESTION [25 upvotes]: I was watching some lectures by Huisken where he mentioned that one-loop renormalization group flow was in some analogous to mean curvature flow. I have tried reading up the exact definition of what this flow actually is, but could not find anything suitable and was wondering if anyone could explain it to me. I have tried reading texts on QFT, but I don't really want the physics behind it and I don't want vague descriptions. I just want to know what the precise mathematical definition of the flow is, similar to the one for Ricci flow or mean curvature flow. Is there a manifold, what is the PDE involved etc.? I am familiar with the idea of loops in the context of Feynman diagrams and integrals if that helps, so I know what a one-loop Feynman diagram is. REPLY [17 votes]: The renormalization group (RG) as a geometric flow (like the Ricci flow) is a very special case of the RG, namely, the one corresponding to the nonlinear sigma-model (NLSM) in two dimensions with values in a Riemannian manifold. Now the RG is much more general and applies to all sorts of models, not just the NLSM. In order to find satisfactory answers to your questions, I recommend first understanding how the RG works in general by specializing to a simpler model: the scalar field. This is explained below. Then, look at the NLSM and see how in this particular case, the Ricci flow emerges from the RG. For this second part, I think the two references given by Igor are spot on. I should however mention a common source of confusion about the RG. There are two different but related RG's (this distinction is not specific to the scalar field or the NLSM but holds across models): 1) the old Stueckelberg-Peterman-Gell-Mann-Low RG (SPGLRG), 2) the more recent Wilsonian RG (WRG). Below, I provide details which should hopefully explain the relation between the two. The short story is the WRG is a flow on the space of theories with a fixed ultraviolet cutoff (say at unit scale) whereas the SPGLRG only concerns theories in the continuum, after the removal of the ultraviolet cutoff. These continuum theories (points in some space) are parametrized by coordinates (renormalized couplings). When rescaling a theory by some factor, one changes the point and therefore one would like to know how the coordinates change. The answer to this question is the SPGLRG. As far as I can tell, Igor's answer and the second (very nice) reference he gave by Carfora et al. only concerns the SPGLRG, because I did not see the word "cutoff" mentioned once. I think the first reference he gave by Nguyen would be the first place to go to understand the RG vs. Ricci flow connection (immediately after reading my nontechnical explanations below) because it articulates both RG's, the SPGLRG and the WRG, and thus provides a fuller picture. Another reference I could add is also a review written by Carfora for mathematicians about the RG vs. Ricci flow connection, which is available at https://arxiv.org/abs/1001.3595. Below is the answer I gave at https://physics.stackexchange.com/questions/372306/what-is-the-wilsonian-definition-of-renormalizability which has more details than the one in the MO post mentioned by J. C. This is a very good question which, however, shows the extent of the reigning confusion about renormalization even four decades after Wilson's Nobel Prize winning theory on the matter. I essentially answered the OP's question, and much more, about constructing continuum QFTs in Wilson's framework in my expository article "QFT, RG, and all that, for mathematicians, in eleven pages" but in a very condensed fashion (one needs to do computations on the side to follow what is being said). So let me give more details pertaining to the OP's specific question. I should preface this by saying that what follows is a "cartoon" for renormalization. I will oversimplify things by ignoring anomalous dimensions, marginal operators, and nonlocal terms generated by the RG. You will not find technical details but hopefully the conceptual picture and logical structure of renormalization will become clearer. The OP is right to point out that in the setting of ODEs and dynamical systems a first order equation can be run backwards in time. So let me start by recalling some important terminology from that area. Consider a first order nonautonomous ODE of the form $$ \frac{dX}{dt}=f(t,X)\ . $$ It generates a flow (groupoid morphism from time pairs to diffeomorphisms of phase space) I will denote by $U[t_2,t_1]$ which sends the initial value $X(t_1)$ to the value of the solution at time $t_2$. It trivially satisfies $\forall t, U[t,t]={\rm Id}$ and the semigroup property $$ \forall t_1,t_2,t_3,\ \ U[t_3,t_2]\circ U[t_2,t_1]=U[t_3,t_1]\ . $$ This time-dependent situation is to be distinguished from the autonomous ODE case $$ \frac{dX}{dt}=f(X) $$ where $U[t_2,t_1]=U[t_2-t_1,0]=:U[t_2-t_1]$. In Wilson's RG, time is scale or more precisely, $t=-\log\Lambda$ where the UV cutoff is implemented in momentum space by a condition like $|p|\le\Lambda$ or in position space by $\Delta x\ge \Lambda^{-1}=e^t$. The high energy physics literature usually works in a nonautonomous setting while it is essential to translate the equation to autonomous form for a proper understanding of Wilson's RG. The latter imported tools and concepts from dynamical system theory like fixed points, stable and unstable manifolds etc. It is possible to do some contortions to try to make sense of these concepts in the nonautonomous setting, but these truly are notions that are congenial to autonomous dynamical systems. Let $\mu=:\mu_{-\infty,\infty}$ denote the probability measure corresponding to the free Euclidean theory. Its propagator is $$ \int \phi(x)\phi(y)\ d\mu_{-\infty,\infty}(\phi)=\langle \phi(x)\phi(y)\rangle_{-\infty,\infty}= \int\frac{dp}{(2\pi)^D} \frac{e^{ip(x-y)}}{p^{D-2\Delta}} $$ where $\Delta$ is the scaling dimension of the field $\phi$. Normally, $\Delta=\frac{D-2}{2}$ but I will allow more general $\Delta$'s in this discussion. Now let me introduce a mollifier, i.e., a smooth function of fast decay $\rho(x)$ such that $\int \rho(x)\ dx=\widehat{\rho}(0)=1$. For any $t$, let me set $\rho_t(x)=e^{-Dt}\rho(e^{-t}x)$, so in particular $\rho_0=\rho$. Let $\mu_{t,\infty}$ be the law of the field $\rho_t\ast\phi$ where $\phi$ is sampled according to $\mu_{-\infty,\infty}$ and we used a convolution with the rescaled mollifier. In other words, $\mu_{t,\infty}$ is the free cutoff measure at $\Lambda_H=e^{-t}$ and propagator $$ \int \phi(x)\phi(y)\ d\mu_{t,\infty}(\phi)=\langle \phi(x)\phi(y)\rangle_{t,\infty}= \int\frac{dp}{(2\pi)^D} \frac{|\widehat{\rho}_t(p)|^2\ e^{ip(x-y)}}{p^{D-2\Delta}}\ . $$ Note that $\widehat{\rho}_t(p)=\widehat{\rho}(e^t p)$ which we assume to have decreasing modulus with respect to $t$. We have $\widehat{\rho}_{-\infty}=1$ and $\widehat{\rho}_{\infty}=0$ and $|\widehat{\rho}_{t_1}(p)|^2-|\widehat{\rho}_{t_2}(p)|^2\ge 0$ whenever $t_1\le t_2$. One can thus define a more general family of modified free/Gaussian theories $\mu_{t_1,t_2}$ with $t_1\le t_2$ by the propagator $$ \int \phi(x)\phi(y)\ d\mu_{t_1,t_2}(\phi)=\langle \phi(x)\phi(y)\rangle_{t_1,t_2}= \int\frac{dp}{(2\pi)^D} \frac{\left(|\widehat{\rho}_{t_1}(p)|^2-|\widehat{\rho}_{t_2}(p)|^2\right)\ e^{ip(x-y)}}{p^{D-2\Delta}}\ . $$ One has the semigroup property for convolution of (probability) measures $$ \mu_{t_1,t_2}\ast\mu_{t_2,t_3}=\mu_{t_1,t_3} $$ when $-\infty\le t_1\le t_2\le t_3\le \infty$. This means that for any functional $F(\phi)$, $$ \int F(\phi)\ d\mu_{t_1,t_3}=\int\int d\mu_{t_1,t_2}(\zeta)\ d\mu_{t_2,t_3}(\psi)\ F(\zeta+\psi)\ . $$ The other key players are scale transformations $S_t$. Their action on fields is given by $(S_t \phi)(x)=e^{-\Delta t}\phi(e^{-t}x)$ and obviously satisfies $S_{t_1}\circ S_{t_2}=S_{t_1+t_2}$. Using the notion of push-forward/direct image of measures, one has $(S_t)_{\ast}\mu_{t_1,t_2}=\mu_{t_1+t,t_2+t}$, i.e., $$ \int d\mu_{t_1,t_2}(\phi)\ F(S_t\phi)=\int d\mu_{t_1+t,t_2+t}(\phi)\ F(\phi)\ . $$ Since these are centered Gaussian measures, it is enough to check the last property on propagators, i.e., $F(\phi)=\phi(x)\phi(y)$ where this follows from a simple change of momentum integration variable from $p$ to $q=e^{-t}p$ in the above formula for the propagator. This also covers the infinite endpoint case with the conventions $t+\infty=\infty$, $t-\infty=-\infty$ for finite $t$. The high energy physics Wilsonian RG is the transformation of functionals $RG[t_2,t_1]$ for pairs $t_1\le t_2$ obtained as follows. Using the convolution semigroup property $$ \int e^{-V(\phi)} d\mu_{t_1,\infty}(\phi)=\int e^{-V(\zeta+\psi)} d\mu_{t_1,t_2}(\zeta)\ d\mu_{t_2,\infty}(\psi) $$ $$ =\int e^{-(RG[t_2,t_1](V))(\phi)} d\mu_{t_2,\infty}(\phi) $$ after renaming the dummy integration variable $\psi\rightarrow\phi$ and introducing the definition $$ (RG[t_2,t_1](V))(\phi):=-\log \int e^{-V(\zeta+\phi)} d\mu_{t_1,t_2}(\zeta)\ . $$ If $V$ is the functional of $\phi$ corresponding to the bare action/potential with UV cutoff $\Lambda_H=e^{-t_1}$, then $RG[t_2,t_1](V)$ is the effective potential at momentum/mass scale $\Lambda_L=e^{-t_2}$. Trivially (Fubini plus associativity of convolution of probability measures) one has, for $t_1\le t_2\le t_3$, $$ RG[t_3,t_2]\circ RG[t_2,t_1]=RG[t_3,t_1] $$ which is indicative of a nonautonomous dynamical system structure, to be remedied shortly. At this point one can already state the main goal of renormalization/taking continuum limits of QFTs: finding a correct choice of cutoff-dependent potentials/actions/integrated Lagrangians, $(V_t^{\rm bare})_{t\in\mathbb{R}}$ such that $$ \forall t_2,\ \lim_{t_1\rightarrow -\infty} RG[t_2,t_1](V_{t_1}^{\rm bare})\ =:\ V_{t_2}^{\rm eff}\ {\rm exists}. $$ The OP's intuition is correct in seeing this as a backwards shooting problem: choosing the right initial condition at $\Lambda_{H}$ to arrive where we want at $\Lambda_{L}$. A difficulty here (related to scattering in classical dynamical systems) is this involves an IVP at $t=-\infty$ instead of finite time. Note that the continuum QFT, its correlations, etc. should be completely determined by the collection of its effective theories indexed by scales $(V_{t}^{\rm eff})_{t\in\mathbb{R}}$. This is most easily seen when considering correlations smeared with test functions with compact support in Fourier space and with a sharp cutoff $\widehat{\rho}(p)$ given by the indicator function of the condition $|p|\le 1$ (or at least one which satisfies $\widehat{\rho}(p)=1$ in a neighborhood of zero momentum). Switching to an autonomous setting involves some twisting by the scaling maps $S_t$. Given a potential V (bare or effective) which "lives at" scale $t_1$, one has $$ \int e^{-V(\phi)}\ d\mu_{t_1,\infty}(\phi)=\int e^{-V(S_{t_1}\phi)}\ d\mu_{0,\infty}(\phi)= \int e^{-(S_{-t_1}V)(\phi)}\ d\mu_{0,\infty}(\phi) $$ where we now define the action of rescaling maps on functionals by $$ (S_t V)(\phi):=V(S_{-t}\phi)\ . $$ As maps on functionals, one has the identity $$ RG[t_2,t_1]=S_{t_1}\circ RG[t_2-t_1,0]\circ S_{-t_1}\ . $$ Wilson's Wilsonian RG is $WRG[t]:=S_{-t}\circ RG[t,0]$, for $t\ge 0$. It acts on the space of "unit lattice theories" (I put quotes because I am using Fourier rather than lattice cutoffs). Thus the previous identity becomes $$ RG[t_2,t_1]=S_{t_2}\circ WRG[t_2-t_1]\circ S_{-t_1}\ . $$ The identity can be derived as follows (note the orgy of parentheses due to the increase of abstraction from functions to functionals, then to maps on functionals): $$ [(RG[t_2-t_1,0]\circ S_{-t_1})(V)](\phi)=-\log\int d\mu_{0,t_2-t_1}(\zeta) \exp[-(S_{-t_1}V)(\phi+\zeta)] $$ $$ =-\log\int d\mu_{0,t_2-t_1}(\zeta) \exp[-V(S_{t_1}\phi+S_{t_1}\zeta)] $$ $$ =-\log\int d\mu_{t_1,t_2}(\xi) \exp[-V(S_{t_1}\phi+\xi)] $$ where we changed variables to $\xi=S_{t_1}\zeta$. From this one gets $$ [(S_{t_1}\circ RG[t_2-t_1,0]\circ S_{-t_1})(V)](\phi)=[(RG[t_2,t_1,]\circ S_{-t_1})(V)](S_{t_1}\phi) $$ and the identity follows from the trivial fact $S_{t_1}(S_{-t_1}\phi)=\phi$. Note that $(V_t)_{t\in\mathbb{R}}$ is trajectory of $RG$, i.e., $$ \forall t_1\le t_2,\ V_{t_2}=RG[t_2,t_1](V_{t_1}) $$ if and only if $W_t:=S_{-t}V_t$ is a trajectory of $WRG$, i.e., $$ \forall t_1\le t_2,\ W_{t_2}=WRG[t_2-t_1](W_{t_1})\ . $$ The semigroup property for $RG$ readily implies that for $WRG$, namely, $$ \forall t_1, t_2\ge 0,\ WRG[t_1+t_2]=WRG[t_1]\circ WRG[t_2]\ . $$ Now define $W_{t}^{\rm start}:=S_{-t} \circ V_t^{\rm bare}$. Then assuming continuity of all these RG maps one has $$ V_{t_2}^{\rm eff}=\lim_{t_1\rightarrow -\infty} RG[t_2,t_1](V_{t_1}^{\rm bare})=S_{t_2}(W_{t_2}^{\rm eff}) $$ where $$ W_{t_2}^{\rm eff}:=\lim_{t_1\rightarrow -\infty} WRG[t_2-t_1](W_{t_1}^{\rm start})\ . $$ The definiteness of the continuum QFT can also be rephrased as the existence of the potentials $W_{t}^{\rm eff}$. A common source of confusion is the failure to see that while $(W_{t}^{\rm eff})_{t\in\mathbb{R}}$ is (by definition, the semigroup property and continuity) a trajectory of $WRG$, the family of bare potentials $(W_{t}^{\rm bare})_{t\in\mathbb{R}}$ is not. The same statement is true, by undoing the "moving frame change of coordinates", when replacing $W$'s with $V$'s and $WRG$ with $RG$. For concreteness, we need coordinates on the space where the RG acts. Assume the bare potential $V_t^{\rm bare}$ is determined by a collection of coordinates or couplings $(g_i)_{i\in I}$ via $$ V_{t}^{\rm bare}(\phi)=\sum_{i\in I} g_i^{\rm bare}(t)\ \int \mathcal{O}_i(x)\ dx $$ for local operators of the form $$ \mathcal{O}_i(x)= :\partial^{\alpha_1}\phi(x)\cdots \partial^{\alpha_k}\phi(x):_t\ . $$ The Wick/normal ordering is with respect to the free cutoff measure $\mu_{t,\infty}$. More precisely, for every functional $F$, $$ :F(\phi):_t\ \ :=\exp\left[-\frac{1}{2} \int dxdy\ \frac{\delta}{\delta\phi(x)}\ C_{t,\infty}(x,y)\ \frac{\delta}{\delta\phi(y)} \right]\ F(\phi) $$ where we denoted the propagator by $C_{t,\infty}(x,y):=\langle\phi(x)\phi(y)\rangle_{t,\infty}$. Note that changing $-\frac{1}{2}$ to $+\frac{1}{2}$ followed by setting $\phi=0$ is integration with respect to $\mu_{t,\infty}$. For instance $:\phi(x)^2:_t=\phi(x)^2-C_{t,\infty}(x,x)$ and $:\phi(x)^4:_t=\phi(x)^4-6C_{t,\infty}(x,x)\phi(x)^2+3C_{t,\infty}(x,x)^2$. An easy change of variables $y=e^{-t}x$ shows that $$ (S_{-t}V_{t}^{\rm bare})(\phi)=\sum_{i\in I} g_i^{\rm start}(t) \int :\partial^{\alpha_1}\phi(y)\cdots \partial^{\alpha_k}\phi(y):_0\ dy $$ where $g_i^{\rm start}(t):=e^{(D-\Delta_i)t}\ g_i^{\rm bare}(t)$ and I used the notation $\Delta_i=k\Delta+|\alpha_1|+\cdots+|\alpha_k|$ for the scaling dimension of the local operator $\mathcal{O}_i$. The switch $g_i^{\rm bare}\rightarrow g_i^{\rm start}$ corresponds to that from dimensionful to dimensionless coupling constants. The indexing set splits as $I=I_{\rm rel}\cup I_{\rm mar}\cup I_{\rm irr}$, respectively corresponding to the three possibilities for operators: $D-\Delta_i>0$ or relevant, $D-\Delta_i=0$ or marginal, $D-\Delta_i<0$ or irrelevant. $W=0$ is a fixed point of the autonomous dynamical system $WRG$. The behavior near this (trivial/Gaussian/free) fixed point is governed by the linearization or differential at $W=0$, i.e., the maps $\mathcal{D}WRG[t]$ given by $$ [\mathcal{D}WRG[t](W)](\phi):=\int W(S_t\phi+\zeta)\ d\mu_{0,t}(\zeta) $$ as follows from the definition $$ [WRG[t](W)](\phi)=-\log \int e^{-W(S_t\phi+\zeta)}\ d\mu_{0,t}(\zeta) $$ and the crude approximations $e^z\simeq 1+z$ and $\log(1+z)\simeq z$. If $W$ has coordinates $(g_i)_{i\in I}$ (with $:\bullet :_0$ Wick ordering), then one can show (good not so trivial exercise) that $\mathcal{D}WRG[t](W)$ has coordinates given exactly by $(e^{(D-\Delta_i)t}g_i)_{i\in I}$, in the same frame, i.e., with the same $t=0$ Wick ordering. If instead of flows one prefers talking in terms of the vector field $\mathcal{V}$ generating the dynamics, then a trajectory $(W_t)_{t\in\mathbb{R}}$ of $WRG$ satisfies $\frac{dW_t}{dt}=\mathcal{V}(W_t)$ with $\mathcal{V}:=\left.\frac{d}{dt} WRG[t]\right|_{t=0}$ admitting a linear plus nonlinear splitting $\mathcal{V}=\mathcal{D}+\mathcal{N}$. The linear part, in coordinates, is $$ \mathcal{D}(g_i)_{i\in I}=((D-\Delta_i) g_i)_{i\in I}\ . $$ Assume the existence of $W_{\rm UV}:=\lim_{t\rightarrow -\infty} W_{t}^{\rm eff}$, the UV fixed point, and $W_{\rm IR}:=\lim_{t\rightarrow \infty} W_{t}^{\rm eff}$, the infrared fixed point (they have to be fixed points by continuity). The discussion of perturbative renormalizability always refers to the situation where $W_{\rm UV}=0$ corresponding to continuum QFTs obtained as perturbations of the free CFT $\mu_{-\infty,\infty}$. By definition, the QFT or the trajectory $(W_t)_{t\in\mathbb{R}}$ of its "unit lattice"-rescaled effective theories lies on the unstable manifold $\mathcal{W}^{\rm u}$ of the $W=0$ fixed point. In what follows I will assume for simplicity there are no marginal operators so the fixed point is hyperbolic and there are no subtleties due to center manifolds. The tangent space $T\mathcal{W}^{\rm u}$ is then spanned by functionals $\phi\longmapsto \int \mathcal{O}_i$, for $i$ in $I_{\rm rel}$ which is typically finite. Note that, in principle, knowing a QFT is the same as knowing a trajectory $(W_t^{\rm eff})_{t\in\mathbb{R}}$ and thus the same as knowing just one point of that trajectory say $W_0^{\rm eff}$ (if the $t=0$ IVP is well-posed forwards and backwards in time, which is another delicate issue as explained in Arnold's answer). The point $W_0^{\rm eff}$ can be made to sweep the unstable manifold which can be identified with the space of continuum QFTs obtained by perturbing the $W=0$ fixed point. On the other hand our control parameter is the choice of cut-off dependent starting points $(W_t^{\rm start})_{t\in\mathbb{R}}$. These belong to the bare surface $T\mathcal{W}^{\rm u}$. This is why when considering say the $\phi^4$ model only a small finite number of terms are put in the bare Lagrangian, otherwise we would be talking about some other (family of) model(s) like $\phi^6$, $\phi^8$, etc. So after all these explanations, it should be clear that renormaliztion in Wilson's framework can be seen as a parametrization of the nonlinear variety $\mathcal{W}^{\rm u}$ by the linear subspace $T\mathcal{W}^{\rm u}$. If we denote the stable manifold by $\mathcal{W}^{\rm s}$ and its tangent space by $T\mathcal{W}^{\rm s}$ then, assuming hyperbolicity of the trivial fixed point, the full space where the RG acts should be $T\mathcal{W}^{\rm u}\oplus T\mathcal{W}^{\rm s}$. The stable manifold theorem gives a representation of $\mathcal{W}^{\rm u}$ as the graph of a map from $T\mathcal{W}^{\rm u}$ into $T\mathcal{W}^{\rm s}$. The main problem is to find $(W_t^{\rm start})_{t\in\mathbb{R}}$ so that the limit $W_0^{\rm eff}=\lim_{t\rightarrow -\infty} WRG[-t](W_t^{\rm start})$ exists. The stable manifold theorem is the $t=-\infty$ case of a mixed boundary problem where on a trajectory one imposes conditions (on coordinates) of the form $g_i^{\rm start}(t)=0$, $i\in I_{\rm irr}$, and $g_i^{\rm eff}(0)=\lambda_{i}^{\rm R}$, $i\in I_{\rm rel}$. Irwin's proof is a nice way to slove this and it works even if the RG is not reversible. This method can be applied for finite negative $t$, and this should produce a collection $(W_t^{\rm })_{t<0}$ (all that is needed in fact) dependent on the renormalized couplings $\lambda_{i}^{\rm R}$. Let us assume for instance that $I_{\rm rel}=\{1,2\}$ and $I_{\rm irr}=\{3,4,\ldots\}$. Consider the map $P_t$ given by $$ (\lambda_{1}^{\rm B},\lambda_{2}^{\rm B})\longmapsto (g_i\{WRG[-t](\lambda_{1}^{\rm B}, \lambda_{2}^{\rm B},0,0,\ldots)\})_{i=1,2} $$ where $g_i\{W\}$ denotes the $i$-th coordinate of $W$. A possible choice of starting points is thus $$ W_t^{\rm start}:=(P_t^{-1}(\lambda_{1}^{\rm R},\lambda_{2}^{\rm R}),0,0,\ldots)_{i\in I}\ . $$ The above is more like a road map for what needs to be done but it does not quite provide a recipe for doing it. In the perturbative setting, one trades numbers in $\mathbb{R}$ for formal power series in $\mathbb{R}[[\hbar]]$. The propagators of the $\mu$ measures get multiplied by $\hbar$ and there is now $\frac{1}{\hbar}$ in front of the $V$'s or $W$'s in the exponential. All the couplings $g_i$ now also become elements of $\mathbb{R}[[\hbar]]$. The invertibility of $P_t$ in this setting is easy and follows by analogues of the implicit/inverse function theorem for formal power series (e.g. in Bourbaki, Algebra II, Chapters 4-7, Berlin, Springer-Verlag, 1990). All the work is in showing that for $i\ge 3$, the quantities $$ f_i(\lambda_{1}^{\rm R},\lambda_{2}^{\rm R}):=\lim_{t\rightarrow -\infty} g_i\{WRG[-t](P_t^{-1}(\lambda_{1}^{\rm R},\lambda_{2}^{\rm R}),0,0,\ldots)\} $$ converge to finite values. This gives the wanted parametrization $(\lambda_{1}^{\rm R},\lambda_{2}^{\rm R}) \mapsto(\lambda_{1}^{\rm R},\lambda_{2}^{\rm R},f_3(\lambda_{1}^{\rm R},\lambda_{2}^{\rm R}),f_4(\lambda_{1}^{\rm R},\lambda_{2}^{\rm R}),\ldots)$ of $\mathcal{W}^{\rm u}$ by $T\mathcal{W}^{\rm u}$. There are two ways of showing the above convergence statement. Underlying both ways is the fact (see Bourbaki above) the formal power series $P_t^{-1}(\lambda_{1}^{\rm R},\lambda_{2}^{\rm R})\in \mathbb{R}[[\hbar]]^2$ exist and are unique. Fans of combinatorics would prefer a two-step procedure consisting in 1) finding an explicit formula for $WRG[-t](P_t^{-1}(\lambda_{1}^{\rm R},\lambda_{2}^{\rm R}),0,0,\ldots)$ for finite $t$; then 2), with this formula in hand, analyze the limit $t\rightarrow -\infty$. The explicit formula in 1) is Zimmermann's forest formula. See this article by Hairer for a recent take on the delicate analytical estimates needed for step 2). For those who abhor combinatorics, there is another method which avoids explicit formulas. Change the scale $0$ in the mixed boundary problem to an arbitrary scale $s>t$. Namely, impose $g_i(t)=0$ for $i\ge 3$ and $g_i(s)=\lambda_i^{\rm R}$ for $i=1,2$ and study the variation of $s$ from $s=t$ to $s=0$ by ODE techniques. This is the Wilson-Polchinski approach. The best rigorous account that I know for this second approach is in the book "Renormalization: An Introduction" by Salmhofer. Finally, one could ask what would happen if one used $W_{s}^{\rm eff}$, for some fixed $s\neq 0$, to parametrize the QFTs instead of $W_{0}^{\rm eff}$. The answer is obtained by noticing that the maps $W_s^{\rm eff}\mapsto {\rm QFT}$ intertwine the action of $WRG$ on $\mathcal{W}^{\rm u}$ and that of the scaling maps $S_t$ on QFTs (simply rescale correlations, i.e., do $\phi\rightarrow S_t\phi$ inside correlations). This is the relation to the old Stueckelberg-Peterman-Gell-Mann-Low RG (i.e., change of scale can be absorbed in a change of renormalized coupling constants). In other words, the restriction of the nonreversible $WRG$ to the finite dimensional manifold $\mathcal{W}^{\rm u}$ should be reversible since $S_t$'s (on collections of correlations) are, or because of the remark I made about Irwin's proof being applicable even for noninvertible (discrete) dynamical systems.<|endoftext|> TITLE: Does every positive-definite integral lattice admit an angle-preserving homomorphism into $\Bbb Z^n$ for some $n$? QUESTION [9 upvotes]: Some initial clarifications By lattice I mean an additive subgroup of $\mathbb R^n$ which is isomorphic to $\mathbb Z^n$ and has full rank (i.e. spans $\Bbb R^n$ when considered as set of vectors). A lattice $\mathcal L$ is integral if $\langle v,w\rangle\in\mathbb Z$ for all $v,w\in\mathcal L$, where $\langle\cdot,\cdot\rangle$ denotes the standard inner product. Above terms are standard, but some of the terms I will use in the following, like "sublattice" and "lattice isomorphism", are probably not standard. Feel free to correct my terminology, as I simply do not know enough of that subject. The actual question I wonder whether every integral lattice $\mathcal L$ is isomorphic to a sublattice of $\mathbb Z^n$ in the following sense: Consider some vectors $v_1,..., v_k\in\mathbb Z^n$. Then $$\mathcal L(v_1,...,v_k):=\mathrm{span}\{v_1,...,v_k\}\cap \mathbb Z^n$$ is an integral lattice in $\mathrm{span}\{v_1,...,v_k\}\subseteq\mathbb R^n$ (with the inner product inherited from $\mathbb Z^n$), and will be called a sublattice of $\mathbb Z^n$. I care about whether I can find $\mathcal L$ in $\Bbb Z^n$ with the right angles, not necessarily the right scale. So I need the following kind of "isomorphism": two lattices $\mathcal L_1$ and $\mathcal L_2$ are isomorphic, if there is a group isomorphism $\phi:\mathcal L_1\to\mathcal L_2$ and a constant $\alpha\in\mathbb R$ with $$\langle \phi(v),\phi(w)\rangle=\alpha \langle v,w\rangle,\quad\text{for all $v,w\in\mathcal L_1$}.$$ REPLY [24 votes]: Yes, this is true. There's some fancy number theory that one can apply (the Hasse-Minkowski invariant and embedding of quadratic forms), but one can see this directly without number-theoretic machinery. First, notice that one can choose a basis for the lattice which is orthogonal. Just start with any basis and apply Gram-Schmidt orthogonalization. The new orthogonal basis for the lattice may no longer be integral, but we may multiply the inner product by an integer to clear denominators make it integral (and so that the original lattice is a sublattice up to scaling). Now one is left with the problem of embedding the 1-dimensional quadratic form/lattice $\mathcal{L}_m$ given by $\mathbb{Z}v_1, \langle v_1,v_1\rangle = m$ into an integral lattice. This may be done using Lagrange's 4-square theorem. Let $m=w^2+x^2+y^2+z^2$. Then we may embed $\mathcal{L}_m$ into the lattice $\mathbb{Z}^4$ via $$v_1 \mapsto (w,x,y,z).$$ Do this for each basis vector individually, and take the orthogonal direct sum of these $n$ 4-dimensional lattices (so this embeds into the lattice $\mathbb{Z}^{4n}$ up to scaling, which is likely far from optimal in general). Addendum: The answer above shows how to embed as a sublattice up to scaling, whereas the question asks for a saturated sublattice (intersection of $\mathbb{Z}^n$ with a subspace). I don't know how to obtain this in general, but in dimensions $1$ and $2$ one can make a construction by hand without scaling. In dimension $1$, send the basis vector $v_1$ in the lattice $\mathcal{L}_m$ with $\langle v_1, v_1 \rangle = m$ to the vector $(1,\ldots,1)\in\mathbb{Z}^m$. This vector is primitive, and hence the image is a saturated sublattice. In dimension $2$, one may do a variation on this embedding. Suppose that one has a 2-dimensional lattice with $\langle v_1,v_1\rangle =a >0, \langle v_1,v_2\rangle =b\geq 0, \langle v_2,v_2\rangle = d >0$, with $ad-b^2>0$. By reduction theory, we may in fact assume that $a \leq d$ and $ ad \geq 2b^2$ (by taking $v_1$ to have minimal norm, and the acute angle between $v_1$ and $v_2$ to be $\geq \pi/3$). Let $b=(a-1)q+r$, $0\leq r < a-1$. Then send $v_1\mapsto (1,\ldots,1,0,\dots,0)$ and $v_2\mapsto (q,\ldots, q, r, 1, \ldots, 1) \in \mathbb{Z}^n$, where $n=a+d- (a-1)q^2 -r^2$ (so the second vector has $(a-1)$ $q$s) . For this to work, one has to check that $d > (a-1)q^2+r^2$ which can be deduced from the inequalities $ad \geq 2b^2$ and $d\geq a$. Moreover, one may see that this is a saturated sublattice of $\mathbb{Z}^n$: if any of the last $d-(a-1)q^2-r^2$ coordinates is non-zero, then we may subtract off the corresponding multiple of $v_2$ to make these coordinates $=0$. What's left must be a multiple of $v_1$.<|endoftext|> TITLE: Representing a number as a sum of four squares and factorization QUESTION [8 upvotes]: Rabin and Shallit have a randomized polynomial-time algorithm to express an integer $n$ as a sum of four squares $n=a^2+b^2+c^2+d^2$ (in time $\log(n)^2$ assuming the Extended Riemann Hypothesis). I'm wondering why this does not give an efficient factorization algorithm? Here's what one could try: run their algorithm $m$ times, with different random steps. This should give expressions $n=a_l^2+b_l^2+c_l^2+d_l^2, l\leq m$, presumably with many distinct representations as a sum of four squares (cf. Jacobi's theorem). We can think of these as factorizations of $n$ over the Lipschitz integers, so $n=|a+bi+cj+dk|^2=(a+bi+cj+dk)(a-bi-cj-dk)$. The Lipschitz integers do not admit a Euclidean algorithm, but the Hurwitz quaternions do. Hence one should be able to take the $\gcd$ of Hurwitz quaternions efficiently. I.e., for $N,D$ Hurwitz quaternions, there should be an efficient algorithm to find $N=QR, D=PR$, with $|R|< |N|,|D|$. Now, take $\gcd(a_l+b_li+c_lj+d_lk,a_p+b_pi+c_pj+d_pk)$, $1\leq l TITLE: Voronoi formula for the symmetric $L$-function with level $N $ QUESTION [6 upvotes]: Sorry to disturb. Does any experts here know something upon the Voronoi type for the symmetric $L$-functions$$\sum_{n\le X} A_F(1,n)e\left ( \frac{an}{c}\right)=?$$ Here $F$ is a symmetric-lift of a $GL_2$ cusp form of square-free level $N$ (i.e., $F=\text{sym}^2 f$), and $A_F(1,n)$ is the $n$-th coefficient of the symmetric $L$-function $L(s,\text{sym}^2f)$. The denominator $c$ is co-prime with $N$ (i.e., $(c,N)=1$). I checked F. Zhou's paper "The Voronoi formula on $GL(3)$ with ramification" at https://arxiv.org/abs/1806.10786. But I am not sure if his definition of the $GL_3$ cusp form on $\Gamma_{N}$ covers my case. Note that in the paper of Buttcane Jack and Rizwanur Khan's “$L^4$-norms of Hecke newforms of large level'', the authors said it seems challengeable to give a Voronoi type for the symmetric $L$-functions with $(c,N)=1$ as in the present case. If any expert here leans something upon this type of Voronoi formula, please guide some reference or share some your very valuable comments here. Very grateful for your time. Much obliged, and thanks in advance. REPLY [9 votes]: Edit: This is the third update of my initial answer. It includes some more details requested by the OP in the comments as well as some general updates. I apologize for the many edits and the length. As pointed out in the comments such a formula is not explicitly in the literature. However, it can be derived from a general Voronoi type formula due to A. Corbett. Let me try to shed some light on how to do this in the following. For simplicity I will treat the modified sum $$S_F(X) = \sum_{n\in\mathbb{Z}} \frac{A_F(n,1)}{\vert n\vert}e\left(n\frac{a}{c}\right)W\left(\frac{n}{X}\right),$$ for a smooth function $W$ with compact support in $\mathbb{R}_+$. To derive this formula it will be useful to switch to the language of automorphic representations. So let us assume that $f$ and $F$ are newforms and let $\pi_f$ and $\Pi_F=\text{sym}^2(\pi_f)$ be the associated automorphic representations. Before we can come to the summation formula in question we gather some preliminaries. Note that, because $N$ is square free and $f$ has trivial nebentypus, we know that $\pi_{f,p}=\text{St}$ is the Steinberg representation for all $p\mid N$. From this we can derive several useful facts. At the archimedean place we find that $\Pi_{F,\infty}$ has representation parameters $$\lambda=(2\nu,-2\nu,0) \text{ and } \delta=\begin{cases} (0,0,0) &\text{ in the Maa\ss\ case,}\\ (1,0,1) &\text{in the holomorphic case.}\end{cases} $$ In particular, $$ L_{\infty}(s,\Pi_{F,}) = \begin{cases} \Gamma_{\mathbb{R}}(s+2\nu)\Gamma_{\mathbb{R}}(s)\Gamma_{\mathbb{R}}(s-2\nu) &\text{ in the Maa\ss\ case,} \\ \Gamma_{\mathbb{C}}(s+2k-1)\Gamma_{\mathbb{R}}(s+1) &\text{ in the holomorphic case.} \end{cases}$$ See for example Proposition~5.12 in "Summation formulas, from Poisson and Voronoi to the present". The associated gamma-quotient then is \begin{equation} \gamma(s,\Pi_{F,\infty},\psi_{\infty}) = \epsilon(s,\Pi_{F,\infty},\psi_{\infty})\frac{L_{\infty}(1-s,\tilde{\Pi}_F)}{L_{\infty}(s,\Pi_F)}.\nonumber \end{equation} The computation at the finite places is slightly more involved. We start by mentioning that the Steinberg representation corresponds under the Local Langlands Correspondence to $\Vert \cdot \Vert^{-\frac{1}{2}}\otimes \text{sp}(2)$, where $\text{sp}(2)$ is the $2$-dimensional special representation of $W_{\mathbb{Q}_p}$. Further, using Artin reciprocity we can identify a character $\chi$ of $\mathbb{Q}_p^{\times}$ with a character of $W_{\mathbb{Q}_p}$. The upshot is that, we can compute the $L$- and $\epsilon$-factors of $\chi \Pi_{F,p}$ on the Weil-Deligne side. Indeed we find that \begin{equation} \left[\Vert \cdot \Vert^{-\frac{1}{2}}\otimes \text{sp}(2)\right]\otimes \left[\Vert \cdot \Vert^{-\frac{1}{2}}\otimes\text{sp}(2)\right] \cong 1 \oplus \left[ \Vert \cdot \Vert^{-1}\otimes\text{sp}(3)\right]. \nonumber \end{equation} From this we conclude that instead of computing the local factors of $\chi \Pi_{F,p}$ we can compute those of $\chi\Vert \cdot \Vert^{-1}\otimes \text{sp}(3)$ on the Weil-Deligne side of the Local Langlands Correspondence. According to ["Elliptic Curves and the Weil-Deligne group"](https://math.berkeley.edu/~dyott/Elliptic Curves and the Weil-Deligne Group.pdf) by D. E. Rohrlich we have that \begin{multline} a(\chi \Pi_{F,p}) = \begin{cases} 2 &\text{ if $\chi$ is unramified,}\\ 3a(\chi) &\text{ if $\chi$ is ramified,} \end{cases}, L(s,\Pi_{F,p}) = L(s+1,\chi) \\ \text{ and } \epsilon(\frac{1}{2},\chi\Pi_{F,p}) = \begin{cases} 1 &\text{ if $\chi=1$,}\\ \epsilon(\frac{1}{2},\chi)^3 &\text{ if $\chi$ is unitary and ramified.} \end{cases} \nonumber \end{multline} (Here some clarification concerning the epsilon factors is necessary. We are always following Langlands convention and consider the canonical additive character with the corresponding self dual Haar measure as fixed and drop them from the notation.) Our local computation implies the well known fact that the conductor of $\Pi_F$ is $N^2$. Further, we can write \begin{multline} A_F(m,n)=A_F\left(\frac{m}{(m,N^{\infty})},\frac{n}{(n,N^{\infty})}\right) (mn,N^{\infty})\\ \cdot \prod_{p\mid N} W_{F,p}\left(\left(\begin{matrix} (mn,p^{\infty})&0&0\\0&(n,p^{\infty})&0\\0&0&1\end{matrix}\right)\right), \nonumber \end{multline} where $W_{F,p}$ denotes the (suitably normalised) Whittaker new vector of $\Pi_{F,p}$. The Fourier coefficients $A_F(m,d)$ depend only on unramified data of $\pi_F$. In particular we have the following well known relations: \begin{equation*} A_F(m,1) = A_F(1,m) = \sum_{d^2\mid n} a_f(\frac{n^2}{d^4}) \text{ and } A_F(m,n) = \sum_{d\mid (m,n)}\mu(d)A_F(\frac{n}{d},1)A_F(\frac{m}{d},1), \end{equation*} for $(md,N)=1$. Knowing the local $L$-factor we can compute \begin{multline} W_{F,p}\left(\left(\begin{matrix} p^{f_1+f_2} & 0 & 0\\ 0 & p^{f_2} & 0 \\ 0 & 0 &1\end{matrix}\right)\right) = p^{-f_1-f_2}s_{(f_1+f_2,f_2)}(p^{-1},0)W_{F,p}(1) \\= \begin{cases} p^{-2f_1}W_{F,p}(1) &\text{ if }f_2=0, \\ 0& \text{ if } f_2\neq 0. \end{cases} \nonumber \end{multline} This can be deduced from Theorem 4.1 in "Whittaker functions associated to newforms of GL(n)" by M. Miyauchi. These preliminaries show that the case of symmetric square lifts (of forms of square free level) is very special. We now turn towards developing the requested Voronoi formula. Sine our method is flexible enough we will start by working in greater generality. Let $\Pi$ be an automorphic representation for $\text{GL}_3(\mathbb{A})$ of level $Q$ containing the newform $G$ of character $\chi$. (In the setting above we have $\Pi=\Pi_F$, $G=F$, $\chi=1$ and $Q=N^2$.) We now make the first step towards our summation formula. By applying Theorem~1.1 from A. Corbett's paper "Voronoi summation for $\text{GL}_n$: collusion between level and modulus" with $\chi=1$, $c_2=1$, $\phi_{\infty}=W(\frac{\cdot}{X})$, $M=Q$, $l=1$, $q=c$ we get \begin{multline} S_G(X) = \sum_{n\in\mathbb{Z}} \frac{A_G(n,1)}{\vert n\vert}e\left(n\frac{a}{c}\right)W\left(\frac{n}{X}\right)\\ =c\sum_{\substack{m,r\in\mathbb{Z}_{\neq 0},\\ (m,N)=1,\\ r\mid N^{\infty}}}\sum_{d\mid c}\text{KL}(\overline{aQL^3}r;m;c,1,d)\chi(\frac{md}{c})\frac{A_G(d,m)}{\vert md\vert} \\ \cdot \mathcal{B}_{\Pi_{\infty}, \phi_{\infty}}\left(\frac{rmd^2}{c^3QL^3}\right)\cdot \prod_{p\mid N} \mathcal{B}_{\Pi_{p}}\left(\frac{rmd^2}{c^3QL^3}\right), \nonumber \end{multline} where $L$ is the largest square free integer dividing $Q$. Here $\text{KL}$ is a 2-dimensional Kloosterman sum defined by $$\text{KL}_2(x,y;\frac{c}{d})\colon =\text{KL}(x;y;c,1,d) = \sum_{\alpha\in \left(\mathbb{Z}/\frac{c}{d}\mathbb{Z}\right)^{\times}}e\left(-\frac{xd\alpha}{c}+\frac{y\overline{\alpha}}{c/d}\right)$$ and the function $\mathcal{B}_{\Pi_{\infty},\phi_{\infty}}(\cdot)$ is a Bessel-transform of $\phi_{\infty}$ given by \begin{multline} \mathcal{B}_{\Pi_{\infty},\phi_{\infty}}(y) = \frac{1}{4\pi i}\sum_{r=0,1} \text{sgn}(y)^r\int_{(\sigma)}\gamma(1-s,\text{sgn}^r\Pi_{\infty},\psi_{\infty})\vert y\vert^{1-s}\int_0^{\infty}W(\frac{x}{X})\vert x\vert^{-1-s}dxds \\ = \frac{1}{X} [\mathcal{H}_{\Pi_{\infty}} W] \left(\frac{y}{X}\right). \nonumber \end{multline} Here we have rewritten $\mathcal{B}_{\Pi_{\infty},\phi_{\infty}}$ in terms of the transform \begin{equation} [\mathcal{H}_{\Pi_{\infty}} W](y) = \frac{1}{4\pi i}\sum_{r=0,1} \text{sgn}(y)^r\int_{(\sigma)}\gamma(1-s,\text{sgn}^r\Pi_{\infty},\psi_{\infty})\vert y\vert^{1-s}[\mathfrak{M}W](-s) ds, \label{C} \end{equation} where $[\mathfrak{M}W]$ is just the normal Mellin-transform of $W$. The transforms $\mathcal{B}_{\Pi_{p},\phi_{p}}(\cdot)$ are $p$-adic versions of this Bessel transform. They are given by \begin{multline} \mathcal{B}_{\Pi_{p},\phi_{p}}(y)= \frac{\log(p)}{2\pi} \sum_{\substack{\xi\colon F^{\times}\to S^1,\\ \xi(p)=1}}\xi(y)\int_{\sigma-i\frac{\pi}{\log(p)}}^{\sigma+i\frac{\pi}{\log(p)}}\epsilon(\frac{1}{2},\xi\Pi_{p})p^{a(\xi \Pi_{p})(s-\frac{1}{2})}\frac{L(s,\xi^{-1}\tilde{\Pi}_{p})}{L(1-s,\chi\Pi_{p})}\vert y \vert_p^{1-s} \\ \cdot \int_{\mathbb{Q}_p^{\times}} \xi(x)\psi_p(x\frac{a}{c})W_{G,p}\left(\left(\begin{matrix} x&0&0\\0&1&0\\0&0&1\end{matrix}\right)\right)\vert x\vert_p^{-s}d^{\times}x ds. \label{D} \end{multline} We claim that \begin{equation} \int_{\mathbb{Q}_p^{\times}} \xi(x)\psi_p(x\frac{a}{c})W_{G,p}\left(\left(\begin{matrix} x&0&0\\0&1&0\\0&0&1\end{matrix}\right)\right)\vert x\vert_p^{-s}d^{\times}x = \begin{cases} L(1-s,\Pi_p)W_{G,p}(1) &\text{ if }\xi\equiv 1,\\ 0 &\text{ else.} \end{cases} \nonumber \end{equation} To see this we note that since $(c,N)=1$ we have $\psi_p(x\frac{a}{c})=1$ for $x\in \mathbb{Z}_p$. Then according to Lemma 2.2 in "Voronoi summation for $\text{GL}_n$: collusion between level and modulus" we have \begin{align} &\int_{\mathbb{Q}_p^{\times}} \xi(x)\psi_p(x\frac{a}{c})W_{G,p}\left(\left(\begin{matrix} x&0&0\\0&1&0\\0&0&1\end{matrix}\right)\right)\vert x\vert_p^{-s}d^{\times}x \nonumber \\ &\quad = \int_{\mathbb{Z}_p\setminus \{ 0\}} \xi(x)W_{G,p}\left(\left(\begin{matrix} x&0&0\\0&1&0\\0&0&1\end{matrix}\right)\right)\vert x\vert_p^{-s}d^{\times}x \nonumber \\ &\quad = \sum_{k\geq 0}\int_{\mathbb{Z}_p^{\times}} \xi(p^kx)d^{\times}x \cdot W_{G,p}\left(\left(\begin{matrix} p^k&0&0\\0&1&0\\0&0&1\end{matrix}\right)\right)p^{sk}. \nonumber \end{align} In the last step we have used that $W_{G,p}$ is right-invariant under the standard congruence subgroup. If $\xi$ is ramified, then the $x$-integral vanishes by orthogonality. Otherwise, if $\xi\equiv 1$, we can use the identity \begin{equation} \int_{\mathbb{Q}_p^{\times}} \xi(x)\psi_p(x\frac{a}{c})W_{G,p}\left(\left(\begin{matrix} x&0&0\\0&1&0\\0&0&1\end{matrix}\right)\right)\vert x\vert_p^{-s}d^{\times}x = Z(1-s,W_{G,p}), \nonumber \end{equation} which follows from the equation, above and apply Theorem~5.1 in "Whittaker functions associated to newforms of GL(n)". (We use the notation $Z(s,W)$ as in loc. cit. for the local zeta integral.) With this claim being established the Bessel-transform as given above takes the following nice form: \begin{equation} \mathcal{B}_{\Pi_{p},\phi_{p}}(y)= \epsilon(\frac{1}{2},\Pi_p)\frac{\log(p)}{2\pi}\int_{\sigma-i\frac{\pi}{\log(p)}}^{\sigma+i\frac{\pi}{\log(p)}}p^{a( \Pi_{p})(s-\frac{1}{2})}L(s,\tilde{\Pi}_{p})\vert y \vert_p^{1-s}ds\cdot W_{G,p}(1). \nonumber \end{equation} Writing $L(s,\tilde{\Pi}_p) = \sum_{k\geq 0} s_{(k,0)}(\tilde{\alpha})p^{-ks}$ we find \begin{align} \mathcal{B}_{\Pi_{p},\phi_{p}}(y) &= \epsilon(\frac{1}{2},\Pi_p)\frac{\log(p)}{2\pi}\sum_{k\geq 0}s_{(k,0)}(\tilde{\alpha})\int_{\sigma-i\frac{\pi}{\log(p)}}^{\sigma+i\frac{\pi}{\log(p)}}p^{a( \Pi_{p})(s-\frac{1}{2})-ks}\vert y \vert_p^{1-s}ds\cdot W_{G,p}(1) \nonumber \\ &= \delta_{v_p(y)\geq -a(\Pi_p)}\epsilon(\frac{1}{2},\Pi_p)\vert y \vert_p p^{-\frac{a(\Pi_p)}{2}} s_{(a(\Pi_p)+v_p(y))}(\tilde{\alpha})W_{G,p}(1). \nonumber \end{align} According to Theorem~4.1 from "Whittaker functions associated to newforms of GL(n)" we have \begin{equation} W_{\tilde{G},p}\left(\left(\begin{matrix} p^{a(\Pi_p)+v_p(y)} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix} \right)\right) = p^{-v_p(y)-a(\Pi_p)}s_{(a(\Pi_p)+v_p(y))}(\tilde{\alpha})W_{\tilde{G},p}(1).\nonumber \end{equation} Thus we get \begin{equation} \mathcal{B}_{\Pi_{p},\phi_{p}}(y) = p^{\frac{a(\Pi_p)}{2}} \epsilon(\frac{1}{2},\Pi_p) W_{\tilde{G},p}\left(\left(\begin{matrix} p^{a(\Pi_p)+v_p(y)} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix} \right)\right)\frac{W_{G,p}(1)}{W_{\tilde{G},p}(1)}. \nonumber \end{equation} (If $\Pi_p$ is self-dual we can replace $\tilde{G}$ by $G$ in the formula above. This applies in particular to $\Pi_p=\Pi_{F,p}$, where this can be checked more directly using the explicit computations from the beginning.) We use the evaluation of $\mathcal{B}_{\Pi_{p},\phi_{p}}$ to explicate Corbett's formula as follows. Putting $y=\frac{rmd^2}{c^3QL^3}$ we observe that the condition $a(\Pi_p)+v_p(y)\geq 0$ becomes $v_p(\frac{r}{L^3})\geq 0$ (since $v_p(Q)=a(\Pi_p)$ and $(mdc,Q)=1$). The latter is nothing but the condition $L^3\mid r$. We get \begin{multline} S_G(X) = \frac{cQ^{\frac{1}{2}}}{X} \cdot \left[\prod_{p\mid Q}\epsilon(\frac{1}{2},\Pi_p)\right]\cdot \sum_{\substack{(m,Q)=1,\\ L^3\mid r\mid Q^{\infty},\\ d\mid c}} \chi(\frac{md}{c})\frac{A_G(d,m)}{\vert md\vert}\text{KL}_2(\overline{aQL^3}r,m;\frac{c}{d})\\ \cdot \left[\prod_{p\mid Q} W_{\tilde{G},p}\left(\left(\begin{matrix} p^{v_p(r/L^3)} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix} \right)\right)\frac{W_{G,p}(1)}{W_{\tilde{G},p}(1)} \right] \cdot [\mathcal{H}_{\Pi_{\infty}}W]\left(\frac{rmd^2}{c^3QL^3X}\right). \nonumber \end{multline} The condition $L^3\mid r$ comes from the support of the local Whittaker functions. Indeed the denominator $L^3$ appears in a the first place due to a rough estimate of the $p$-adic Bessel-transforms. Our refined analysis of the case at hand has now revealed the true support. This allows us to shift the $r$-sum accordingly to get: \begin{multline} S_G(X) = \frac{cQ^{\frac{1}{2}}}{X} \cdot \left[\prod_{p\mid Q}\epsilon(\frac{1}{2},\Pi_p)\right]\cdot \sum_{\substack{(m,Q)=1,\\ r\mid Q^{\infty}}}\sum_{d\mid c} \chi(\frac{md}{c})\frac{A_G(d,m)}{\vert md\vert}\text{KL}_2(\overline{aQ},rm;\frac{c}{d})\\ \cdot \left[\prod_{p\mid Q} W_{\tilde{G},p}\left(\left(\begin{matrix} p^{v_p(r)} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix} \right)\right)\frac{W_{G,p}(1)}{W_{\tilde{G},p}(1)} \right] \cdot [\mathcal{H}_{\Pi_{\infty}}W]\left(\frac{rmd^2}{c^3QX}\right). \nonumber \end{multline} We now specialise to the case $\Pi=\Pi_F$. (Actually we only use that in this case $\Pi_p$ is self-dual it is easy to obtain a more general formula from the discussion above.) First we write $Q=N^2$, $\epsilon(\frac{1}{2},\Pi_{F,p})=1$ and $\chi=1$ in the formula above. Further, self-duality allows us to write \begin{equation} \frac{A_F(d,m)}{\vert md\vert }\left[\prod_{p\mid Q} W_{F,p}\left(\left(\begin{matrix} p^{v_p(r)} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix} \right)\right) \right] = \frac{A_F(d,mr)}{\vert mdr\vert }.\nonumber \end{equation} Thus the $m$-sum and the $r$-sum can be combined. Thus, we have proved the following result. $\mathbf{Theorem}$: Let $F$ be a newform for $\text{SL}_3(\mathbb{R})$ underlying the symmetric square lift of a newform $f$ of square free level $N$ and trivial character. Further fix a smooth function $W$ of compact support and take $(c,Na)=1$ as well as $X\in \mathbb{R}_{>0}$. Then we have \begin{multline} \sum_{n\in\mathbb{Z}} \frac{A_F(n,1)}{\vert n\vert}e\left(n\frac{a}{c}\right)W\left(\frac{n}{X}\right) \\ = \frac{cQ^{\frac{1}{2}}}{X} \cdot \sum_{m\in\mathbb{Z}_{\neq 0}}\sum_{d\mid c} \frac{A_F(d,m)}{\vert md\vert }\text{KL}_2(\overline{aQ},m;\frac{c}{d}) \cdot [\mathcal{H}_{\Pi_{\infty}}W]\left(\frac{md^2}{c^3QX}\right). \nonumber \end{multline} Here the Bessel transform $[\mathcal{H}_{\Pi_{\infty}}W]$ is defined above. We essentially recover the formula given in equation (2) of "The Voronoi formula on $\text{GL}(3)$ with ramification". Let us mention that the formula provided by A. Corbett is very general and very flexible. Thus, along those lines it is possible to derive many more explicit Voronoi type formulae for $\text{GL}_n$ in particular one can generalize the approach presented above in several directions. In particular, the argument above gives an explicit Voronoi summation formula fo $\text{GL}_n$ in the $(c,Q)=1$ situation. However, for more complicated cases the computational overhead might be far bigger. Let us finish off by making some remarks concerning a follow up question by the OP in the comments. Looking at the $\text{GL}_2$-case one observes that very convenient Voronoi formulae for sums $$\sum_{n\in\mathbb{Z}} e(n\frac{a}{c})a_f(n)W(\frac{n}{X})$$ are available whenever $(c,N)=1$ or $N\mid c$. A for analytic purposes useful completely general formula was only recently found in "Subconvexity for modular form $L$-functions in the $t$ aspect". In particular the case $N=p^{2k}$ and $(N,c)=p^k$ is very hard. Returning to the $\text{GL}_3$-world we can say the following. The Voronoi formula for the sum $S_G(X)$ if $(c,Q)=1$ is discussed in this post (with a little more work one can derive an explicit formula for general $G$ dropping the self-duality assumption). The case $Q\mid c$ is covered in "The Voronoi formula on $\text{GL}(3)$ with ramification" (along the lines of this answer one can also derive the corresponding formula from A.Corbett's work). The case for general $c$ will be very difficult. It is covered by A. Corbett's work but the compute the local Bessel transforms promises to be very hard. To illustrate this let us look at the following toy situation. Let $F$ be the symmetric-square lift of a newform $f$ of level $p$ and trivial character. Then $F$ has level $p^2$. We now take $c=c'p$ for $(c',p)=1$. (We are brief in details and translate everything straight away in the explicit situation at hand!) This hits exactly the hardest case possible. However, our advantage is that we have full information on $\Pi_{F,p}$ as discussed above. Applying A. Corbett's theorem once again we find \begin{equation} S(X) = \frac{c'}{X} \sum_{\substack{(m,p)=1,\\ r\mid p^{\infty}}}\sum_{d\mid c'} \text{KL}_2(\overline{ap^4}r,m;\frac{c'}{d}) \frac{A_F(d,m)}{\vert md\vert} [\mathcal{H}_FW] \left(\frac{rmd^2}{Xp^5c'^3} \right)\cdot \mathcal{B}_{\Pi_{F,p},\phi_p}\left(\frac{rmd^2}{p^5c'^3}\right).\label{A} \end{equation} The job is now to evaluate the $p$-adic Bessel transform. For simplicity we assume that $W_{F,p}(1)=1$. In this case we have \begin{multline} \mathcal{B}_{\Pi_{p},\phi_{p}}(y)= \frac{\log(p)}{2\pi} \sum_{\substack{\xi\colon F^{\times}\to S^1,\\ \xi(p)=1}}\xi(y)\int_{\sigma-i\frac{\pi}{\log(p)}}^{\sigma+i\frac{\pi}{\log(p)}}\epsilon(\frac{1}{2},\xi\Pi_{p})p^{a(\xi \Pi_{p})(s-\frac{1}{2})}\frac{L(s,\xi^{-1}\tilde{\Pi}_{p})}{L(1-s,\chi\Pi_{p})}\vert y \vert_p^{1-s} \\ \cdot \int_{\mathbb{Z}_p\setminus \{0\}} \xi(x)\psi_p(x\frac{a}{c'p})\vert x\vert_p^{2-s}d^{\times}x ds. \nonumber \end{multline} The $x$-integral can be evaluated using Gau\ss\ sums. Indeed one gets \begin{equation} \int_{\mathbb{Z}_p\setminus \{0\}} \xi(x)\psi_p(x\frac{a}{c'p})\vert x\vert_p^{2-s}d^{\times}x = \begin{cases} -\frac{p}{p-1}+L(1-s,\Pi_{F,p}) &\text{ if } \xi=1,\\ (1-p^{-1})^{-1}p^{-\frac{1}{2}}\xi(\frac{c'}{a})\epsilon(\frac{1}{2},\xi^{-1}) &\text{ if }a(\xi)=1,\\ 0&\text{ else.} \end{cases} \nonumber \end{equation} We conclude that in this case the Bessel transform reads \begin{multline} \mathcal{B}_{\Pi_{p},\phi_{p}}(y)=\frac{\log(p)}{2\pi} \int_{\sigma-i\frac{\pi}{\log(p)}}^{\sigma+i\frac{\pi}{\log(p)}} \bigg[p^{2s-1}L(s,\tilde{\Pi}_{F,p})-\frac{p^{2s}}{p-1}\frac{L(s,\tilde{\Pi}_{F,p})}{L(1-s,\Pi_{F,p})} \\ +\frac{p^{3s-1}}{p-1}\sum_{a(\xi)=1} \xi(-y\frac{c'}{a})\epsilon(\frac{1}{2},\xi)^2 \bigg] \vert y\vert^{1-s}ds. \label{B} \end{multline} With enough will-power one can certainly go further in the evaluation of this expression but we stop at this point. Our goal was to highlight some of the difficulties that arise when the additive twist 'colludes' with the level and we hopefully succeeded.<|endoftext|> TITLE: A conjectural formula for the class number of the field $\mathbb Q(\sqrt{-p})$ with $p\equiv3\pmod8$ QUESTION [9 upvotes]: Question. Is my following conjecture new? How to prove it? Conjecture. Let $p>3$ be a prime with $p\equiv3\pmod 8$, and let $h(-p)$ denote the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. Then we have $$h(-p)=\frac1{2\sqrt p}\sum_{k=1}^{(p-1)/2}\csc\left(2\pi\frac{k^2}p\right).$$ I have checked the conjecture numerically for all primes $3 TITLE: Does every functor between Grothendieck categories have adjoints? QUESTION [5 upvotes]: Let $F:\mathcal C\longrightarrow \mathcal D$ be an additive functor that preserves colimits. Suppose that $\mathcal C$ and $\mathcal D$ are Grothendieck categories. Does $F$ have a right adjoint? I know the adjoint functor theorems. But when I checked the exact statements in standard references like Kashiwara Schapira, it is not clear to me whether the results apply to additive functors. Could someone help clarify with references? REPLY [4 votes]: A Grothendieck category is locally (finitely) presentable; see for example Theorem 2.2 in this paper by Positselski and Rosický. And any colimit-preserving functor between locally presentable categories has a right adjoint. One way to see this is by citing Theorem 1.58 in Locally Presentable and Accessible Categories which shows that locally presentable categories are co-well-powered, and then applying the Special Adjoint Functor Theorem. Another way is to observe that locally presentable categories are total categories, and applying the adjoint functor theorem for total categories; see the nLab.<|endoftext|> TITLE: Have the "Bayreuther Mathematische Schriften" ever been digitalized? QUESTION [14 upvotes]: See update below. This is borderline off-topic (certainly not a research question), but the rather successful story of a similar question gives me hope. The Bayreuther Mathematische Schriften were a periodical published by the University of Bayreuth from 1979 till 2011, and included research papers, lecture notes and theses. Its 80 volumes include a surprisingly large amount of (Rota-style) combinatorics (which otherwise is a rare topic in German mathematics), including papers by Dress, Thévenaz, Kohnert, Strehl and Sturmfels as well as a long set of lecture notes on Polya enumeration and representations of symmetric groups. Both German- and English-language works appear in the periodical. By now, the volumes are neither being sold nor seem to be available on the internet, and only a few papers and some amount of theses (quite possibly in a non-final form) can be found on the internet. Hence I am wondering: Have the Schriften ever been digitalized? If yes: What is preventing their open publication? Who has the copyrights? If no: Are good sources for digitalization available? Is this something that a library could do with a bit of crowdfunding? About the last question: I am currently near a full archive and have made a bunch of scans of the parts relating to algebraic combinatorics for myself. However, the binding of the volumes often limits the quality of my scans, as it prevents fully opening the journals. I am also having the impression that the volumes here are not quite the originals; a number of pages are missing due to incorrect printing, and quality is suboptimal in many places. My impression is that someone, somewhere should still be having the originals for most of the content (LaTeX for the later volumes, probably microfiche for the earlier ones). Does anyone know where to look? UPDATE: According to inside sources, there is a license problem: The authors have not granted the Bayreuth University any digital distribution rights (probably the copyright agreement did not include such language). Thus, authors can post their own papers online (if you are an author, please do so!), but the university cannot unless it gets green light from the authors. This seems to be a general problem in Germany (or the EU?), where distribution rights are not understood to include digital media by default. This brings me to the next question: How did other German/European publishers (Springer, e.g.) handle this issue? I'm pretty sure that copyright agreements in the 1970s did not include any language on digital distribution, yet Springer distributes its full back catalogue. REPLY [6 votes]: Q1: Yes, the Bayreuther mathematische Schriften have been digitized by the Hathi Trust Ditigal Library. Q2: The Hathi Trust Library states that "This item is not available online (Limited - search only) due to copyright restrictions." Since the publisher is Universität Bayreuth it is most likely that they hold the copyright. Q3: No library will digitize and provide access to a publication that has copyright protection, but you can find a list of libraries that hold the volumes by following the links in the Hathi Trust database. Incidentally, you mention that "the binding of the volumes often limits the quality of my scans, as it prevents fully opening the journals." Are you using a book scanner? I am a happy user of the CZUR book scanner, which corrects for the curved pages by postprocessing the image. --- UPDATE --- A brief addendum in response to the updated question: In the United States copyright has been broadly defined to include reproduction by "any method now known or later developed" (see footnote 7 in Revising Copyright Law for the Information Age). I would be surprised if the definition would be more restrictive in Europe, and that is presumably why publishers just adapt their content to new reproduction methods, without having to once again secure permission from the original author.<|endoftext|> TITLE: Examples of hyperbolic groups QUESTION [14 upvotes]: What are some other classes of word-hyperbolic groups other than the finite groups, fundamental groups of surfaces with Euler characteristics negative and virtually free groups? REPLY [28 votes]: Below are some sources of hyperbolic groups. Of course, the list is far from being exhaustive. Groups defined by generators and relations: Finitely generated free groups, as their Cayley graphs are simplicial trees. If $\varphi$ is an atoroidal automorphism of a free group $\mathbb{F}_n$, then the extension $\mathbb{F}_n \rtimes_\varphi \mathbb{Z}$ is hyperbolic. One-relator groups with torsion, namely groups admitting a presentation of the form $\langle x_1, \ldots, x_n \mid r^m=1 \rangle$ with $m \geq 2$, are hyperbolic. Coxeter groups not containing $\mathbb{Z}^2$ are hyperbolic. (See Theorem 12.6.1 in Davis' book The geometry and topology of Coxeter groups for a better characterisation.) As a nice particular case, right-angled Coxeter groups defined by finite square-free graphs are hyperbolic. Generalising hyperbolic right-angled Coxeter groups, graph products of finite groups over square-free graphs are hyperbolic. If $\Gamma$ is a topological graph which does not contain two disjoint loops, then the braid group $B_2(\Gamma)$ is hyperbolic. Small cancellation groups C'(1/6) or C'(1/4)-T(4). A very rich source of two-dimensional hyperbolic groups. Random groups in Gromov's density model are hyperbolic and non-elementary below a given density. Fundamental groups: Fundamental groups of compact negatively curved Riemannian manifolds are hyperbolic. Fundamental groups of hyperbolic 3-manifolds. A general construction is to take a hyperbolic 3-manifold of finite volume and to "fill in" its cusps with tori in different ways. The keyword is Dehn fillings. Uniform lattices in real, complex and quaternionic hyperbolic spaces are hyperbolic. Arithmetic constructions may lead to explicit examples. If $f$ is a pseudo-Anosov homeomorphism of a closed surface $S$, then $\pi_1(S) \rtimes_\varphi \mathbb{Z}$ is hyperbolic, where $\varphi$ denotes the automorphism of $\pi_1(S)$ induced by $f$. (A reference is Otal's monograph Le théorème d'hyperbolisation pour les variétés fibrées de dimension 3; also, a generalisation to free groups of pseudo-Anosov homeomorphisms can be found in Farb and Mosher's article Convex cocompact subgroups of mapping class groups.) Constructions Dehn fillings are now defined purely algebraically, and it is possible to create hyperbolic groups by quotienting relatively hyperbolic groups for instance, or even to create exotic hyperbolic groups from classical hyperbolic groups. The family of hyperbolic groups is also stable under various operations, including commensurability, some graphs of groups, and some graph products. For instance, free products (as mentioned by Yves in the comments). I can add a few precise references if some examples interest you. Otherwise, you have some keywords to help you in your research.<|endoftext|> TITLE: Is there some sort of classification of finite groups that force solvability? QUESTION [5 upvotes]: Suppose $G$ is a finite group. We will say, that it force solvability if any finite group $H$, such that $G$ is isomorphic to its maximal proper subgroup, is solvable. Does there exist some sort of classification of such groups? On one hand, all such groups have to be solvable. On the other hand, there are several large classes of such groups known. One of them is yielded by a theorem from “A condition for the solvability of a finite group” by W. E. Deskins: All groups of nilpotency class 2 force solvability The other comes from a theorem by Thomas Browning: If a finite group is nilpotent and all its $2$-subgroups are normal, then it forces solvability. His proof is here: Let $G$ be minimal such that $G$ is not solvable and such that $G$ contains a maximal subgroup $M$ that is nilpotent and whose 2-subgroups are normal. If $M$ contains a nontrivial normal subgroup $N$ of $G$ then $G/N$ contradicts the minimality of $G$. Thus, $M$ does not contain nontrivial normal subgroups of $G$. In particular, $N_G(P)=M$ for all Sylow $p$-subgroups $P$ of $M$. Then $P$ is a Sylow $p$-subgroup of $N_G(P)$ so $P$ is a Sylow $p$-subgroup of $G$. This shows that $M$ is a Hall subgroup of $G$. If $P$ is a Sylow $p$-subgroup of $M$ and if $Q$ is a nontrivial normal subgroup of $P$ then $N_G(Q)=M$ which has a normal $p$-complement. For $p=2$, Frobenius' normal $p$-complement theorem gives that $G$ has a normal $p$-complement. For $p\geq3$, Thompson's normal $p$-complement theorem or Glauberman's normal $p$-complement theorem gives that $G$ has a normal $p$-complement (since you only have to consider characteristic $p$-subgroups). Thus, for each prime $p$ dividing the order of $M$, $G$ has a normal $p$-complement. Then $M$ has a normal complement $N$ in $G$. Since $M$ is solvable but $G$ is not solvable, $N$ is not solvable. In particular, $N$ does not admit a fixed-point-free automorphism of prime order. If $m\in Z(M)$ has prime order then $C_N(m)$ is nontrivial. Then $C_N(m)M$ is a subgroup of $G$ that properly contains $M$ so $C_N(m)M=G$ by the maximality of $M$. Comparing cardinalities shows that $C_N(m)=N$ so $m\in Z(G)$. Then $\langle m\rangle$ is a nontrivial normal subgroup of $G$ contained in $M$ which is a contradiction. It is also known, that if $H$ and $K$ force solvability, then $H \times K$ also does. However, I do not know, whether there is anything else here... REPLY [4 votes]: I make a remark in the opposite direction. Let $p \geq 17$ be a Fermat or Mersenne prime, so that $X = {\rm PSL}(2,p)$ has a dihedral Sylow $2$-subgroup $D$ which is maximal. Let $d >1$ be a power of $2$, and let $Q$ be a transitive $2$-subgroup of $S_{d}$. Let $A= {\rm Aut}(X)$ and let $T$ be a Sylow $2$-subgroup of $A$. Then $T$ is a maximal subgroup of $XT$. Using this action of $Q$ on $d$ points, form the wreath product $XT \wr Q$. Then $T \wr Q$ is a Sylow $2$-subgroup of $(XT) \wr Q$, and is a maximal subgroup of $(XT) \wr Q$.<|endoftext|> TITLE: Generalization of a problem, involving radicals and the floor function, proposed by Ramanujan to the Journal of the Indian Mathematical Society QUESTION [7 upvotes]: The section Solved problems from the Wikipedia Floor and ceiling functions shows several problems proposed by Ramanujan ([1]). The purpose of this post, if possible, is try to get the generalization of some of these identities, for positive integers $n\geq 1$, involving fractions or radicals and the floor function $\lfloor x\rfloor$. I tried to get a generalization of the identity $(iii)$. I don't know if my conjectural identiy is in the literature or has a good mathematical content, these are my previous failed attempts. Counterexamples for different formulas. A counterexample for the (false) identity $$\lfloor \sqrt[k]{n}+\sqrt[k]{n+1}\rfloor=\lfloor \sqrt[k]{2^k n+k}\rfloor$$ is the integer $n=525$ for the case $k=5$. Counterexamples for the (false) identity $$\lfloor \sqrt[k]{n}+\sqrt[k]{n+1}\rfloor= \left\lfloor \sqrt[k]{2^k n+2(k-1)} \right\rfloor$$ are the integers $n=11$ or $n=610$ for the case $k=6$. From this thread of experiments I get the following conjecture. Conjecture. For each integer $k\geq 2$ one has that the identity $$\lfloor \sqrt[k]{n}+\sqrt[k]{n+1}\rfloor=\left\lfloor 2\sqrt[k]{n+\frac{1}{2}}\right\rfloor$$ holds over integers $n\geq 1$. I don't know if it is easy to prove, or if you can to find a counterexample. Question. Do you know if generalizations (thus with a good mathematical meaning, with mathematical significance) of the mentioned problems proposed by Ramanujan are in the literature? In this case, please refer the literature and I try to find and read from the literature these generalizations of $(i)$, $(ii)$ or $(iii)$. In case that aren't in the literature please add yourself generalization, if possible, with its respective proof for some of those identities. In particular, if you know that my conjeture can be proved or can be refuted by finding a counterexample. Many thanks. Please if some professor/user finds a counterexample it is welcome that he/she comment it, many thanks. I add for example the following scripts written in Pari/GP as a proof of concept/toy model of my conjecture for(k=2,10,for(n=1, 1000,if(floor((n)^(1/k)+(n+1)^(1/k))!=floor(2*((n+1/2)^(1/k))),print(k," ",n)))) that you can to evaluate from the web Sage Cell Server, just choose GP as language. Thus aren't showed conunterexamples as outputs. Also you've this one for(k=2,10,for(n=1, 100,print(floor((n)^(1/k)+(n+1)^(1/k))))) or this for(k=2,10,for(n=1, 100,print(floor((n)^(1/k)+(n+1)^(1/k))-floor(2*((n+1/2)^(1/k)))))) I add as reference the PARI/GP Developers group of Université Bordeaux 1. References: I believe that the corresponding reference is [1] Srinivasa Ramanujan, Collected Papers, Question 723 in p. 332, Providence RI: AMS / Chelsea (2000). [2] I've used also the PARI-GP Reference Card (version 2.2.5), by Karim Belabas (2003), based on an earlier version by Joseph H. Silverman. REPLY [4 votes]: I can provide a partial proof of your conjecture. It shows that the statement is true for sufficiently large $n$, which in this case means that it is true for $n\geq A_k$, where $A_k$ is a number depending on $k$. Exact form of $A_k$ will be provided in the proof. We'll need following ingredients: $\left\lfloor\sqrt[k]{2^kn+2^{k-1}-1}\right\rfloor =\left\lfloor\sqrt[k]{2^kn+2^{k-1}}\right\rfloor$ $\sqrt[k]{2^kn+2^{k-1}-1}<\sqrt[k]{n}+\sqrt[k]{n+1}$, for $n\geq \left\lceil A_k\right\rceil$, where $\left\lceil A_k\right\rceil=\left\{\begin{array}{ll} 0 & \text{for }k=1 \\ 2^{k-3} & \text{for }k\geq2 \end{array}\right.$ $\sqrt[k]{n}+\sqrt[k]{n+1}<\sqrt[k]{2^kn+2^{k-1}}$ These three ingredients imply our conjecture. Proofs of individual ingredients provided below. Note that ${2^kn+2^{k-1}-1}$ and ${2^kn+2^{k-1}}$ differ only by 1, so the only possibility for 1. not being true is when $\sqrt[k]{2^kn+2^{k-1}}$ is an integer. If we assume that $$\sqrt[k]{2^kn+2^{k-1}}=q,\quad\text{for }q\in\mathbb{N},$$ then $$2^kn+2^{k-1}=2^{k-1}(2n+1)=q^k.$$ We can notice, that $q^k$ is divisible by $2^{k-1}$, but this implies that $2^k|q^k$, which can't be true since $2n+1$ is odd. Hence 1. is true. Let's rewrite 2. a little bit. We want to prove that $$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}>\sqrt[k]{n+\frac{2^{k-1}-1}{2^k}}.$$ By applying AM-GM inequality we obtain $$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}>\sqrt[2k]{n^2+n}.$$ The inequality $$\sqrt[2k]{n^2+n}\geq\sqrt[k]{n+\frac{2^{k-1}-1}{2^k}}$$ occurs if and only if $n\geq2n\dfrac{2^{k-1}-1}{2^k}+\left(\dfrac{2^{k-1}-1} {2^k}\right)^2$, but this inequality implies that $$n\geq \dfrac{2^{2(k-1)}+1-2^k}{2^{k+1}}=:A_k,$$ so if $n\geq \left\lceil A_k\right\rceil$, then 2. is true. Let's rewrite 3. in the same manner as we've rewritten 2. We want to prove that $$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}<\sqrt[k]{n+\frac{1}{2}}.$$ By Generalized mean inequality for $k$ exponent we obtain $$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}<\sqrt[k]{\frac{n + n + 1}{2}}=\sqrt[k] {n+\frac{1}{2}}.$$ The equality can't occur, because $\sqrt[k]{n}\neq \sqrt[k]{n+1}$ for all $n\in\mathbb{N}$. $$\tag*{$\blacksquare$}$$ The cases, when $n TITLE: Non-small objects in categories QUESTION [10 upvotes]: An object $c$ in a category is called small, if there exists some regular cardinal $\kappa$ such that $Hom(c,-)$ preserves $\kappa$-filtered colimits. Is there an example of a (locally small) category $C$ and an object $c$ of $C$, such that $c$ is not small, i.e. such that $Hom(c,-)$ doesn't preserve all $\kappa$-filtered colimits for any $\kappa$ whatsoever? REPLY [14 votes]: In the opposite category of the category of sets, and of many algebraic categories, the only small objects are the empty set and the singleton. A conceptual reason for this is Freyd's (or Gabriel and Ulmer's?) theorem that it is impossible for a category and its opposite both to be locally presentable, unless they are both posets. Indeed, if $A$ is a set with at least two elements, consider functions $f:\{0,1\}^\kappa\to A$ where $\kappa$ is some infinite cardinal. If $\lambda<\kappa$ then $\{0,1\}^\kappa$ may be viewed as a $\lambda$-cofiltered limit of all products of at most $\lambda$ of the copies of $\{0,1\}$. For $A$ to be $\lambda$-small in $\mathrm{Set}^{\mathrm{op}}$, we would have to be able to guarantee that $f$ depends on at most $\lambda$ coordinates in the domain. Since the opposite of the category of sets is the category of complete atomic Boolean algebras (CABAs), we can also make this argument directly in there, where it amounts to the fact that there are elements in a coproduct of CABAs that do not come from any smaller sub-coproduct, since we can always take a join or a meet of elements from every term in the coproduct. REPLY [12 votes]: In the category $\mathsf{Top}$ of topological spaces and continuous maps the only $\lambda$-presentable objects are discrete spaces. This appears 1.14(6) in Locally presentable and Accessible categories by Adamek and Rosicky. The reason is explained in 1.2(10) in the same reference.<|endoftext|> TITLE: Criterion to construct a $\mathbb{Z}$-basis of a free $\mathbb{Z}$-Lie algebra QUESTION [5 upvotes]: Let $L(\mathbb{Z},n)$ (resp. $L(\mathbb{Q},n)$) be the free Lie algebra over $\mathbb{Z}$ (resp. over $\mathbb{Q}$) with generating set $\{x_1,\dots,x_n\}$. Let $\mathcal B$ be a $\mathbb{Q}$-basis of $L(\mathbb{Q},n)$ consisting of elements of the form $[x_{i_1},[x_{i_2},\dots,[x_{i_{r-1}},x_{i_r}]\dots]]$ for some $i_1,\dots,i_r\in\{1,\dots,n\}$. Is $\mathcal B$ automatically a $\mathbb{Z}$-basis of $L(\mathbb{Z},n)$? REPLY [2 votes]: It turns out my intuition was wrong and the answer is "no". Take for instance $n=3$, and attribute the degree $\alpha_i$ to $x_i$, so that $L(\mathbb{Q},3)$ becomes graded by the free $\mathbb{Z}$-module $\oplus_{i=1}^3\mathbb{Z}\alpha_i$. Then the homogeneous component of degree $2\alpha_1+\alpha_2+\alpha_3$ of $L(\mathbb{Q},3)$ has dimension $3$, and admits the $\mathbb{Q}$-basis $\mathcal B=\{[x_1,[x_1,[x_2,x_3]]], [x_2,[x_1,[x_1,x_3]]],[x_3,[x_1,[x_1,x_2]]]\}$. However, $$[x_1,[x_3,[x_1,x_2]]]=-\frac{1}{2}[x_1,[x_1,[x_2,x_3]]]+\frac{1}{2}[x_2,[x_1,[x_1,x_3]]]+\frac{1}{2}[x_3,[x_1,[x_1,x_2]]].$$<|endoftext|> TITLE: Uncountable disjoint closed coverings of $[0,1]$ QUESTION [8 upvotes]: It is well known that the unit interval $[0,1]$ cannot be decomposed as a countable union of pairwise disjoint closed (nonempty) subsets. See for instance this math.stackexchange question. The proof using Baire category theorem can be trivially adapted to show that, under MA, the unit interval cannot be decomposed as a union of less than $\mathfrak c$ pairwise disjoint closed subsets, but, on the other hand: Is it consistent that $[0,1]$ can be expressed as a union of $\aleph_1<\mathfrak c$ pairwise disjoint (nonempty) closed subsets? REPLY [12 votes]: This is question has a long and interesting history, which is discussed in Arnie Miller's paper cited below. The first construction of a model of ZFC + $\aleph_1 < 2^{\aleph_0}$ where $[0,1]$ can be partitioned into $\aleph_1$ pairwise disjoint nonempty closed sets is due to Jim Baumgartner (unpublished). Early on, Hausdorff showed that one can always write $[0,1]$ as a union of $\aleph_1$ pairwise disjoint nonempty $F_{\sigma\delta}$ sets. Hausdorff, F., Summen von $\aleph_1$ Mengen., Fundam. Math., Warszawa, 26, 241-255 (1936). ZBL62.0228.03. Sierpiński then asked if $F_{\sigma\delta}$ could be improved by $G_\delta$ in Hausdorff's result. Sierpiński, Wacław, Sur deux consequences d’un théorème de Hausdorff, Fundam. Math. 33, 269-272 (1945). ZBL0060.12715. Fremlin and Shelah answered Sierpiński's question by showing that $[0,1]$ can be written as a union of $\aleph_1$ pairwise disjoint nonempty $G_\delta$ (equivalently $G_{\delta\sigma}$) sets if and only if $[0,1]$ can be covered by $\aleph_1$ meager sets. Fremlin, D. H.; Shelah, S., On partitions of the real line, Isr. J. Math. 32, 299-304 (1979). ZBL0413.04002. Then, Miller showed that the Fremlin-Shelah result is false if we replace $G_\delta$ by closed (equivalently $F_\sigma$). Miller, Arnold W., Covering $2^\omega$ with $\omega_1$ disjoint closed sets, The Kleene Symp., Proc., Madison/Wis. 1978, Stud. Logic Found. Math., Vol. 101, 415-421 (1980). ZBL0444.03026. http://www.math.wisc.edu/~miller/res/cov.pdf However, it is still consistent with $\aleph_1 < 2^{\aleph_0}$ that $[0,1]$ can be written as a union of $\aleph_1$ pairwise disjoint nonempty closed sets. Baumgartner was perhaps the first to find such a model (see Theorem 4 in Miller's paper), but this result was rediscovered many times (with different arguments). For example, the earliest published proof is due to Stern, who showed that this is true in any forcing extension of a model of CH by adding $\aleph_2$ random reals. Stern, Jacques, Partitions of the real line into $\aleph_1$ closed sets, Higher Set Theory, Proc. Oberwolfach 1977, Lect. Notes Math. 669, 455-460 (1978). ZBL0393.03038. It is perhaps interesting that even without assuming AC, one can always partition $[0,1]$ into $\aleph_1$ pairwise disjoint nonempty Borel sets. (See this MO answer by Andreas Blass.) Hausdorff's gap construction makes heavy use of choice, so one can ask whether it is provable in ZF that one can always partition $[0,1]$ into $\aleph_1$ pairwise disjoint nonempty Borel sets of bounded rank. This question was investigated by Stern. The answer is negative, assuming the consistency of some large cardinal axioms. For example, the answer is no assuming AD. Stern, Jacques, Effective partitions of the real line into Borel sets of bounded rank, Ann. Math. Logic 18, 29-60 (1980). ZBL0522.03032.<|endoftext|> TITLE: The projective functor $\mathbb{P}^n: \operatorname{CRing} \to \operatorname{Set}$ is not representable: categorical argument QUESTION [7 upvotes]: Using a "geometrical" argument of dimension, like the one here, one can show that the projective space is not affine. I am interested in showing that, but using a categorical argument, i.e. I want to show that $\mathbb{P}^n:\operatorname{CRing} \to \operatorname{Set}$ which sends a ring $R$ to the set of equivalence classes $\mathbb{P}^n(R):= R^{n+1}/R^{\times}$ is not representable. Similar to the example of the $\operatorname{Nil}$ functor, this could be done either by showing that its category of elements has no initial object, or by showing that it does not preserve limits. Any ideas what could work? REPLY [5 votes]: $\def\PP{\mathbb{P}}\def\AA{\mathbb{A}}\def\GG{\mathbb{G}}\def\Spec{\mathrm{Spec}}$This is probably going to sound too classical to satisfy, but it seems straightforward to me. Let $\PP^{n}_{charts}$ be the functor represented by the scheme which is normally called projective $n$-space. In other words, $\PP^{n}_{charts}$ is the co-equalizer of a certain diagram $(\AA^{n-1} \times \GG_m)^{\binom{n+1}{2}} \rightrightarrows (\AA^n)^{n+1}$. As discussed in comments, the correct definition of $\PP^n(R)$ is that $\PP^n(R)$ is the set of rank one direct summands of $R^{n+1}$ (see here). Grothendieck preferred to dualize and work with rank one projective quotients of $R^{n+1}$. I'm not sure if there is a deep reason which this is better; from a shallow perspective, it seem to me to introduce unnecessary duals in the notation. I'll work with the summand version. For $0 \leq j \leq n$, let $X_j$ be the submodule $(r_0, r_1, \ldots, r_{j-1}, 0 , r_{j+1}, \ldots, r_n)$ of $R^{n+1}$. Let $U_n$ be the subfunctor of $\PP^n$ where $U_j(R) = \{ L \subset R^{n+1} : L + X_j = R^{n+1} \}$. Every submodule in $U_n(R)$ is uniquely of the form $R(u_0, u_1, \ldots, u_{j-1}, 1, u_{j+1}, \ldots, u_n)$; the coordinates $(u_0, \ldots, u_{j-1}, u_{j+1}, \ldots, u_n)$ give an isomorphism $U_j \cong \AA^n$. The overlap $U_i \cap U_j$ is the chart $u_i \in R^{\times}$ in $U_j$, so $U_i \cap U_j \cong \AA^{n-1} \times \GG_m$, and the gluing is precisely the classical chart formula. So, by the universal property of co-equalizers, we get a map $\PP^{n}_{charts} \to \PP^{n}$. It shouldn't be hard to show that this is an isomorphism, but we don't need to work that hard to show that $\PP^{n}$ isn't affine. For a field $k$, the map $\PP^{n}_{charts}(k) \to \PP^{n}(k)$ is definitely a bijection. If $\PP^{n}$ were affine, then global functions on $\PP^{n}$ would separate $k$-points, so such functions pulled back to $\PP^{n}_{charts}$ would separate $k$-points. But global functions on $\PP^{n}_{charts}$ don't separate $k$-points, a contradiction. Or, briefer but less intuitively: The module $k[t] (1,t) \subset k[t]^2$ corresponds to a map $\Spec\ k[t] \to \PP^1$, so if $\PP^1$ were $\Spec\ S$, there would be a corresponding map $S \to k[t]$. The module $k[t^{-1}] (t^{-1},1) \subset k[t^{-1}]^2$ would similarly give a map $S \to k[t^{-1}]$. The inclusions of $k[t]$, $k[t^{-1}]$ into $k[t, t^{-1}]$ give us the same module, $k[t,t^{-1}] (1,t) = k[t,t^{-1}](t^{-1},1)$ inside $k[t,t^{-1}]^2$, so the maps $S \to k[t]$ and $S \to k[t^{-1}]$ must become the same after further composing to $k[t,t^{-1}]$. So the image of $S$ would have to lie in $k[t] \cap k[t^{-1}] = k$. But then all the different maps $k[t] \to k$ would have to give to the same $k$-submodule of $k^2$, and they don't.<|endoftext|> TITLE: Is there a Kolmogorov complexity proof of the prime number theorem? QUESTION [27 upvotes]: Lance Fortnow uses Kolmorogov complexity to prove an Almost Prime Number Theorem (https://lance.fortnow.com/papers/files/kaikoura.pdf, after theorem $2.1$): the $i$th prime is at most $i(\log i)^2$. This differs from the bound in the Prime Number Theorem only by a factor of $\log i$. Is there a way to sharpen this proof to a bound of $O(i\log i)$, making good use of complexity theory (Matt. F comment gives a smaller $O((\log\log i)^2)$ factor)? added: If 1. is not possible what additional information do we need to get to the Prime Number Theorem? The only limitation seems to be the creativity needed in prefix free codes. Perhaps there are asymptotically efficient coding or some other reasoning getting to needed. Is there a connection between prefix free codes and multiplicative number Theory? Both try to characterize unique representation in some manner. REPLY [7 votes]: I think it’s not uncommon for arguments in elementary number theory to be “philosophically” information-theoretic in nature. But this is not a very deep fact - at the end of the day it all comes down to the well-known heuristic principle that the events of a “random integer” being divisible by different primes are approximately independent events. It’s making that heuristic notion precise that always requires considerably more subtle arguments involving complex analysis etc, in which information theoretic thinking plays no apparent role. So I‘m reasonably sure the answer to your title question about the prime number theorem is “no”. Nonetheless, one can do some cute things with this idea. Here is an example I noticed many years ago, related to the standard estimate $$ \textrm{(*)} \qquad \sum_{p \le n} \frac{\log p}{p} = \log(n) + O(1) \qquad (n\to \infty) $$ (summation over primes) — a series that is very much related to the sum of the reciprocals of the primes and to estimates of $p_n$ which you and others were discussing here. I will give (*) the following information-theoretic interpretation (which will give a rigorous proof of a bound in one direction): Take a random variable $X$ that is chosen uniformly at random from the numbers $1,\ldots,n$. For each prime $p\le n$, let $V_p$ be the $p$-valuation of X (the exponent of $p$ dividing $X$). A key observation is that knowing all the $V_p$’s, one can recover $X$. That means that there is an inequality of Shannon entropies: $$ \log_2(n) = H(X) \le H(\{V_p, p\le n\}) \le \sum_{p\le n} H(V_p), $$ by well-known properties of the entropy function $H(\cdot)$ (subadditivity and monotonicity with respect to the $\sigma$-field, precisely analogous to the properties of Kolmogorov complexity that Fortnow uses in his notes). Now, what is the entropy $H(V_p)$ of the random variable $V_p$? It’s the expression $$ - \sum_{k\ge 0} \operatorname{Pr}(V_p=k) \log \operatorname{Pr}(V_p=k), $$ which can be bounded from above by the $k=1$ term $$ -\operatorname{Pr}(V_p=1) \log \operatorname{Pr}(V_p=1) = -\left( \frac{1}{n} \lfloor n/p \rfloor \right) \log \left( \frac{1}{n} \lfloor n/p \rfloor \right), $$ plus all the other parts (which I’ll leave as an exercise to show is $O(1)$). And this last quantity is approximately equal to $\frac{\log p}{p}$ as long as $p$ is smaller in order of magnitude than $n$. Thus after some more simple estimates one gets essentially the “$\ge$” side of (*), together with the added insight that the error term in approximations such as (*) says something about the extent to which divisibility properties of a typical number by different primes are correlated with each other. As I was saying at the beginning, this is kind of interesting at a philosophical level, but I’m not aware that anyone has found a way to make these sorts of arguments precise enough to prove something as strong as the prime number theorem, or indeed anything stronger than the most elementary sorts of estimates that already have very short and direct proofs.<|endoftext|> TITLE: Are the exp and log maps of Riemannian geometry conformal QUESTION [7 upvotes]: For any Riemannian manifold are the exp and log maps (from a predetermined base) conformal? If not, are there some manifolds where they are and others where they aren't? REPLY [6 votes]: See Robert Bryant's answer to Complex manifolds in which the exponential map is holomorphic in which he proves that the surfaces for which the exponential map is conformal are precisely the flat ones. If the exponential map is conformal on some Riemannian manifold of dimension $\ge 3$, it is a conformal map on the intersection $P \cap B$ of any 2-plane $P\subset T_m M$ in any tangent space with the ball $B\subset T_m M$ of radius equal to the injectivity radius. So the curvature vanishes on $e^{P\cap B}$, i.e. the sectional curvature vanishes, so the Riemannian manifold is flat.<|endoftext|> TITLE: Equivalence relations in arbitrary categories QUESTION [6 upvotes]: Let $C$ be a category and $A\in\mathrm{ob}(C)$. A relation is a subobject $q:Q\hookrightarrow A^{\times 2}$ and the quotient is defined as the coequalizer $$A/Q:=\mathrm{coeq}\left(Q\stackrel{q}{\hookrightarrow} A^{\times 2}\rightrightarrows A\right)$$ where the two maps are the two projections. Moreover, we can define what it means to be an equivalence relation: The inclusion $q$ should satisfy the following: Reflexivity: Consider the diagonal $\Delta:A\to A^{\times 2}$. Then there should be a map $i:A\to Q$ such that $q\circ i=\Delta$. Symmetry: Consider the flip $t:A^{\times 2}\to A^{\times 2}$. Then there should be an automorphism $s:Q\to Q$ such that $t\circ q=q\circ s$. Transitivity: Consider the map $q\times_A q: Q\times_AQ\to A^{\times 2}\times_AA^{\times 2}$ and the outer projections $p:A^{\times 2}\times_AA^{\times 2}\to A^{\times 2}$. Then there should be a map $j:Q\times_A Q\to Q$ such that $p\circ (q\times_A q)=q\circ j$. Given an arbitrary relation $q:Q\hookrightarrow A^{\times 2}$, we can define $\overline{Q}\stackrel{\overline{q}}{\hookrightarrow} A^{\times 2}$ as the inital equivalence relation having a morphism $i:Q\to \overline{Q}$ such that $\overline{q}\circ i=q$. I claim that the canonical map $$A/Q\to A/\overline{Q}$$ is an isomorphism. In $\mathbf{Set}$, this is clear as building the coequalizer is the same as quotienting out the spanned equivalence relation. Is this also true in other categories? REPLY [6 votes]: Short answer :Yes, assuming $\overline{Q}$ exists and $C$ has kernel pairs (for example if it has finite limits). For more details: The relation $\overline{Q}$ do not always exists, you need some assumption on the underlying categories, and there are various type of assumption that can work. For example if $\mathcal{C}$ is a pretopos with a parametrized natural number object you can mimick the usual construction of the transitive closure of a relation internall to give a construction of $\overline{Q}$. In a completely different style, if $Sub(A \times A)$ is small and order complete then a variant of the small object argument also allows to construct $\overline{Q}$. But, assuming that $\overline{Q}$ exists, it is actually easy to see that assuming $C$ has Kernel pair, if $\overline{Q}$ exists, then the canonical map $A/Q \rightarrow A/\overline{Q}$ is always an isomorphisms. Indeed, let $R$ the kernel pair of $A \rightarrow A/Q$, $R$ is an equivalence relation on $A$ containing $Q$, so $\overline{Q} \subset R$. It follows from the universal property of the quotient that $A \rightarrow A/Q $ factors as $A \rightarrow A/\overline{Q} \rightarrow A/Q$ the uniqueness part of both universal property shows that the map constructed $A/\overline{Q} \leftrightarrows A/Q$ are invers of each other. The same argument also shows that $A/R$ is also isomorphic to these (but $R$ can actually be strictly bigger than $Q$).<|endoftext|> TITLE: Why does every chain complex have a map into its cone? QUESTION [6 upvotes]: In Weibel's An introduction to homological algebra he defines a cone as an explicit chain complex associated to the given one -i.e. for a chain $C=(C_i, d)$ he defines $Cone(C)=\left(C_{i-1} \oplus C_i, \begin{bmatrix} -d & 0 \\ -id & d \end{bmatrix} \right)$. It is good and it is trivial that we have a monomorphism $C\rightarrow Cone(C)$ which gives us a short exact sequance $0\rightarrow C\rightarrow Cone(C)\rightarrow C[-1]\rightarrow 0$, however we can define cone as an universal object as follows below. For a chain complex $C$ let $F$ be a functor from chain complexes to sets given by the formula $F(D)= \left\{(f,s); f:C\rightarrow D, f=ds+sd,\text{f is a morphism}, \text{s is a chain homotopy}\right\}$ then the above cone represents it because $Hom(Cone(C),D)$ is naturally bijective to $F(D)$. Assume now that our cone is defined as an universal object. Why do we have a monomorphism from $C$ to $Cone(C)$? I don't know almost anything about a homotopical algebra but maybe it is a proposition there? REPLY [7 votes]: This is really just rephrasing Simon Henry's comment, but there is a natural transformation $F\to\text{Hom}(C,-)$ given by $(f,s)\mapsto f$, and so if $\text{Cone}(C)$ represents $F$ then by Yoneda's lemma this natural transformation is induced by a map $C\to\text{Cone}(C)$. You can also see that this map is a monomorphism without using the construction of general cones. You need to show that for every complex $B$ and every nonzero map $\alpha:B\to C$, the composition $B\to C\to\text{Cone}(C)$ is nonzero, or equivalently, by Yoneda, that the natural transformation $F\to\text{Hom}(B,-)$ given by $(f,s)\mapsto f\circ\alpha$ is nonzero. Or in other words, that there is some complex $D$ and a map $f:C\to D$ that is homotopic to zero with $f\circ\alpha$ nonzero. But if $\alpha$ is nonzero in degree $n$, then you can take $D$ to be the complex $\dots\to0\to C_n\stackrel{\text{id}_C}{\to}C_n\to0\to\dots$ with nonzero terms in degrees $n-1$ and $n$, with $f_{n-1}=d_{n-1}$ and $f_n=\text{id}_{C_n}$.<|endoftext|> TITLE: Sets A such that A+A contains the largest set [0,1,..,t] QUESTION [10 upvotes]: I look for a reference for the following problem. Given an integer $k$, find a set $A\subset\mathbb{N}$ with $|A|=k$ that maximizes $t$ such that $\left[0,1,..,t\right]\subset A+A$. REPLY [10 votes]: A table of values for these $t$ are given in the introduction Graham and Sloane's On Additive Bases and Harmonius Graphs (your sequence corresponds to $n_\beta(k)$ in their notation). Graham and Sloane also give some references to previous work with this sequence, both under the name of "interval basis" (or Abschnittsbasis), going back to a paper in German from Rohrbach in the 1930's, and under the name of "The Postage Stamp Problem". This is sequence A001212 in the OEIS, which has additional references. REPLY [5 votes]: This is related to ``thin additive bases" of order $2$. Clearly $t$ cannot be larger than $k(k+1)/2$. It is also possible to give examples where $t$ grows quadratically. Take $A=A_0 \cup A_1$ where $A_0$ contains all integers below $t$ with binary expansion $\sum_{j} \epsilon_j 2^j$ with $\epsilon_j= 0$ unless $j$ is even, and $A_1$ consists of numbers with binary digits $\epsilon_j=0$ unless $j$ is odd. Then $A$ has $O(\sqrt{t})$ elements in it; or alternatively $t\ge Ck^2$ for some constant $C>0$. See for example this paper of Blomer which has other references.<|endoftext|> TITLE: Integral of Schur functions over $SU(N)$ instead of $U(N)$ QUESTION [5 upvotes]: Schur functions are irreducible characters of the unitary group $U(N)$. This implies $$ \int_{{U}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\delta_{\lambda\mu},$$ where the overline means complex conjugation. My question is what is the result of the same integral performed over $SU(N)$ instead, $$ \int_{{SU}(N)}s_\lambda(u)\overline{s_\mu(u)}du=?$$ REPLY [6 votes]: The desired integral is given in equation (13) of arXiv:1812.06069: $$\int_{{SU}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\sum_{q=-\infty}^\infty\prod_{i=1}^N\delta_{\lambda_i,\mu_i+q},$$ where $\lambda=(\lambda_1,\lambda_2,\ldots\lambda_N)$ and $|\lambda|=\sum_{i}\lambda_i$, with $\lambda_1\geq\lambda_2\cdots\geq 0$. This is still an orthogonality relation, because the Schur functions $s_\lambda$ and $s_\mu$ are identical in $SU(N)$ iff $\lambda=\mu+(q,q,\ldots q)$ for some integer $q$. The $U(N)$ integral $\int_{{U}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\delta_{\lambda\mu}$ corresponds to the $q=0$ term in the sum over $q$. It follows that the integrals over $SU(N)$ and over $U(N)$ are the same if $|\lambda|,|\mu| TITLE: In a topological space if there exists a loop that cannot be contracted to a point does there exist a simple loop that cannot be contracted also? QUESTION [34 upvotes]: I'm interested in whether one only needs to consider simple loops when proving results about simply connected spaces. If it is true that: In a Topological Space, if there exists a loop that cannot be contracted to a point then there exists a simple loop that cannot also be contracted to a point. then we can replace a loop by a simple loop in the definition of simply connected. If this theorem is not true for all spaces, then perhaps it is true for Hausdorff spaces or metric spaces or a subset of $\mathbb{R}^n$? I have thought about the simplest non-trivial case which I believe would be a subset of $\mathbb{R}^2$. In this case I have a quite elementary way to approach this which is to see that you can contract a loop by shrinking its simple loops. Take any loop, a continuous map, $f$, from $[0,1]$. Go round the loop from 0 until you find a self intersection at $x \in (0,1]$ say, with the previous loop arc, $f([0,x])$ at a point $f(y)$ where $0 TITLE: A non-geodesible foliation of $S^3$ or $S^2\times S^1$ QUESTION [5 upvotes]: Is there a $1$-dimensional foliation of $S^3$ which is not a geodesible foliation? Is there a $1$-dimensional foliation of $S^2\times S^1$ which is not a geodesible foliation? If the answer is affirmative, can one produce an example of a non-geodesible foliation of $S^3$ or $S^2\times S^1$ whose leaves are mapped to the singular foliation of $S^2$ associated to the Poincaré compactification of a planar quadratic vector field? (By "mapped" we mean the Hopf map $S^3\to S^2$ or the projection $S^2\times S^1\to S^2$.) For detailed motivation please see the following post. REPLY [7 votes]: If a 1-dimensional foliation of a 3-manifold contains a Reeb component, then it is not geodesible. This follows from a theorem of Sullivan, see Section 2.5, Obstruction 1. A Reeb component is a saturated annulus embedded in the manifold with induced foliation looking like: One can find a 1-dimensional foliation of any 3-manifold containing such a Reeb component. Foliate $T^2\times I = S^1 \times (S^1\times I)$ by foliating $S^1\times I$ by the Reeb foliation, and extending by the product with $S^1$ to $T^2 \times I$. Now, embed $T^2\times I$ in a manifold $M$ with this foliation, and extend it to a 1-dimensional foliation on the rest of $M$.<|endoftext|> TITLE: If it quacks like a conifold resolution and it waddles like a conifold resolution, $\ldots$ QUESTION [8 upvotes]: Suppose that $X$ is a projective threefold with at worst conifold singularities and suppose $\omega_X$ trivial. Suppose $Y$ is a projective variety with a birational morphism $f: Y\to X$ which is an isomorphism away from the conifold points and such that $f^{-1}(p) = \mathbb{P}^1$ for each conifold point $p \in X$. Can I conclude that $Y$ is smooth? i.e. that $f:Y\to X$ is a conifold resolution? This seems too good to be true, but I was unable to come up with a counterexample and it would be really useful (to me at least) if it were true. Edit: In light of Sasha/Jason's counterexample I would like to impose a normality condition. The most useful version to me would be to let $\bar{Y} \to Y $ be the normalization of $Y$, and ask: is $\bar{Y} \to X$ a conifold resolution? Alternatively, is it true if I assume that $Y$ is normal to begin with? REPLY [5 votes]: Take a conifold resolution $\tilde{X}$ and then "impose a cusp" on its exceptional fiber (in the vertical tangent direction). In other words, in a local chart replace $Spec(A)$ by the spectrum of the subring $$ A' := \{a \in A \mid \partial(a) \in \mathfrak{m} \}, $$ where $\mathfrak{m}$ is the ideal of a point on the exceptional fiber and $\partial$ is a derivation tangent to exceptional fiber. Define $Y$ by gluing $Spec(A')$ with the rest of $\tilde{X}$. This is a counterexample to your question.<|endoftext|> TITLE: Cohn-Vossen rigidity theorem in hyperbolic space QUESTION [5 upvotes]: There is the following rigidity theorem of Cohn-Vossen as stated on p. 86 of these lecture notes: http://www.math.brown.edu/~deigen/chern.pdf Any isometry between two closed smooth convex surfaces in the Euclidean space $\mathbb{R}^3$ is established by an isometry of $\mathbb{R}^3$. Is the same result true if one considers convex surfaces with the same assumptions in the hyperbolic space $\mathbb{H}^3$ instead of $\mathbb{R}^3$? This post is a continuation and a more precise version of Extendability of isometries of convex surfaces REPLY [2 votes]: Hsiung-Liu: Generalization of the rigidity theorem of Cohn-Vossen prove the rigidity theorem for surfaces in hyperbolic 3-space, assuming that their second fundamental form is positive definite and a condition on the directions of normal vectors (these assumptions replacing the convexity assumption one needs in Euclidean 3-space).<|endoftext|> TITLE: Is Multilinear Hilbert's tenth problem version undecidable? QUESTION [16 upvotes]: A multilinear polynomial $f\in\mathbb Z[x_1,\dots,x_t]$ has terms only of form $$b\prod_{i=1}^tx_i^{a_i}$$ where $a_i\in\{0,1\}$ and $b\in\mathbb Z$. Is there no general purpose algorithm for finding integer roots of this class of polynomials? REPLY [3 votes]: This is not an answer but a longish comment. There is no Hasse Principle for multilinear polynomials. Take for example the polynomial equation $$(5x+2)(5y+3)=11.$$ Evidently the equation has no integer solutions. I'll show that it has $p$-adic solutions for every prime $p$. First take $x=0$ in the displayed equation. We then require a $p$-adic integer $y$ such that $2(5y+3)=11$. The latter equation is equivalent to $10y=5$, which has a $p$-adic integer solution for all $p\ne2$. Next, take $y=0$ in the displayed equation. We then require a $p$-adic integer $x$ such that $3(5x+2)=11$. This is equivalent to $15x=5$, which has a p-adic integer solution for all $p\ne3$. Therefore the displayed equation has $p$-adic integer solutions for all $p$. Clearly there are real solutions. So the Hasse Principle fails for multilinear polynomials.<|endoftext|> TITLE: Simple book on model theory QUESTION [6 upvotes]: I was expressed by how Mendelson describes models in his Introduction to mathematical logic. Now I am looking for a nice model theory guide. The book (video source, etc.) must: Include the concrete methods with their proofs and must answer the following questions: 1.1. how to know if a theory has a model 1.2. how to build a model if a theory is consistent 1.3. how to know if a class of structures forms the models of some theory 1.4. how to build a theory for an elementary class 1.5. given a structure, what information can be obtained by logic Be concentrated on finite models and theories (that's why Keisler doesn't fit) Not contain too much algebra (that's why Marker doesn't fit) Not contain complexity theory at all (that's why Ebbinghaus doesn't fit) So I am interested in a simple guide containing necessary proofs. If there is no such a book, is it real to discover the methods above by myself? REPLY [9 votes]: The best book for you is probably A Shorter Model Theory by Hodges. Some comments on your question, though: First, you should be aware that the model theory of finite structures and the model theory of infinite structures have extremely different characters - so much so that finite model theory is essentially a separate subfield of logic, which is much closer to computer science and complexity theory. This can be (partially) explained by the fact that first-order logic is powerful enough to completely describe finite structures, so interesting questions in the first-order model theory of finite structures have to impose some constraints: working with fragments of first-order logic and taking complexity into account. If you're really interested in finite model theory, you can take a look at this question, which has some references in the comments and answers. To my knowledge, the book by Ebbinghaus and Flum is the textbook on the subject which contains the most material not directly related to complexity theory (though there are probably books that I'm not aware of). On the other hand, "ordinary" model theory is primarily concerned with infinite models, and as a result it's hard to avoid some set theory creeping in. If you're really turned off by ordinals and cardinals, I would recommend: (1) learn something about them, set theory is a beautiful subject! (2) in the mean time, concentrate on the model theory of countably infinite structures. This is a domain in which you get to see many of the concepts and techniques of model theory at work without any transfinite inductions in sight (except in more advanced topics: the Scott rank and Morley rank can be useful for studying countable structures, and they are both ordinal-valued). It's also the case that many of the interesting examples in model theory come from algebra. So it's hard to achieve your requirements 2, 3, and 4. But this is why I suggested Hodges: In my experience students without a strong background in algebra and set theory find Hodges's book to be easier to read than Marker's.<|endoftext|> TITLE: In a subset of $\mathbb{R}^2$ which is not simply connected does there exist a simple loop that does not contract to a point? QUESTION [8 upvotes]: I previously asked In which topological spaces does the existence of a loop not contractable to a point imply there is a non-contractable simple loop also? Given the broad scope of this question I propose this special case as an independent question: In a path-connected subset of $\mathbb{R}^2$ which is not simply connected does there exist a simple loop that does not contract to a point? REPLY [10 votes]: One-dimensional metric spaces and planar sets do have the property that you're interested in. To explain why this works out in such generality requires a combination of planar topology, continuum theory, and shape theory. One-Dimensional Case: This is pretty classical, going back to work of Curtis and Fort in the 1950's. Proof Sketch. Suppose $X$ is a one-dimensional metric space. If $X$ contains a simple closed curve, then any loop parameterizing that curve cannot be null-homotopic. A rigorous argument for this requires the shape injectivity of one-dimensional spaces. On the other hand, if $X$ contains no simple closed curves, then the image of any loop in $X$ is a dendrite, which is contractible. Thus either $X$ contains a non-contractible simple closed curve or is simply connected. Planar Case: All the ingredients to prove what you want (and more) can be found in H. Fischer, A. Zastrow, The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655--1676. Proof Sketch. Suppose $X\subseteq \mathbb{R}^2$ is path connected and $\alpha:S^1\to X $ is not null-homotopic. Let $Y=\alpha(S^1)\cup\bigcup\{U\mid U\text{ is a bounded path component of }\mathbb{R}^2\backslash \alpha(S^1)\text{ with }U\subseteq X\}$. Fischer and Zastrow show $Y$ is a Peano continuum homotopy equivalent to a “Sierpinski-like” space, i.e. a copy of the Sierpinski Carpet with some, possibly empty, set of (originally deleted) squares filled back in. Since all we want to do is find some simple closed curve that doesn’t contract, pick any bounded path component $C$ of $\mathbb{R}^2\backslash Y$ and a point $z\in C\backslash X$. While $C$ is homeomorphic to an open disk, its boundary need not be a simple closed curve. However, you basically have a planar Peano continuum $Y$ surrounding a “hole” at $z$. Since $Y\subseteq X$ and $z\notin X$, any simple closed curve in $Y$ with winding number $1$ around $z$ will do the trick. If you want to create one explicitly, use some planar geometry to cover $Y$ with enough arc-wise connected open (in $Y$) sets to create a finite loop of neighborhoods in $Y$ with winding number $1$ around $z$ and build your curve piece-wise. This simple closed curve can't contract in $X$ since then it would contract in $\mathbb{R}^2\backslash \{z\}$, which clearly is impossible.<|endoftext|> TITLE: Geodesic preserving diffeomorphisms of constant curvature spaces QUESTION [10 upvotes]: Let $X$ be either Euclidean space $\mathbb{R}^n$, the sphere $\mathbb{S}^n$, or hyperbolic space $\mathbb{H}^n$. I would like to have a classification of all diffeomorphisms $X\to X$ which map every geodesic line to a geodesic line. In the first two cases, the group of all such transformations is strictly larger than the group of isometries, but for the hyperbolic space I am not sure. REPLY [14 votes]: For $\mathbb{R}^n$: the fundamental theorem of projective geometry (proof: https://www3.nd.edu/~andyp/notes/FunThmProjGeom.pdf) says that the bijections of $\mathbb{R}^n$ taking lines to lines are the affine maps $x\mapsto Ax+b$ for an invertible matrix $A$ and a constant vector $b$. For $S^n$: a theorem by the same name shows that the bijections of projective space taking projective lines to projective lines is a projective transformation. One easily uses this to prove the result for the sphere that such a bijection is a linear transformation defined up to positive rescaling, acting on the sphere as the space of rays in a real vector space. For $\mathbb{H}^n$: Kobayashi defined a Kobayashi pseudometric for projective connections, which determines the usual metric when applied to hyperbolic space, so the unparameterized geodesics determine the metric, and so the diffeomorphisms preserving them are isometries: S. Kobayashi, Intrinsic distances associated with flat affine or projective structures, J. Fac. Sci. Univ. Tokyo IA 24 (1977), 129--135.<|endoftext|> TITLE: Max order of an isogeny class of rational elliptic curves is 8? QUESTION [8 upvotes]: I am looking for a reference for the proof of the following question following Theorem 5 in Mazur's Rational Isogenies of Prime Degree. Theorem 5 There is a constant $C$ such that every elliptic curve $E_{/\mathbb{Q}}$ is isogenous (over $\mathbb{Q}$) to at most $C$ (mutually nonisomorphic) elliptic curves. "Can one take $C=8$?" Has this question been settled? And if so, what is a reference to the proof of the result. REPLY [13 votes]: M. Kenku, On the number of $\mathbf{Q}$-isomorphism classes of elliptic curves in each $\mathbf{Q}$-isogeny class, J. Number Theory 15, 199 (1982): It is shown that there are at most eight $\mathbf{Q}$-isomorphism classes of elliptic curves in each $\mathbf{Q}$-isogeny class.<|endoftext|> TITLE: When were triples called monads for the first time? QUESTION [9 upvotes]: I am fine-tuning a short note on basic category theory; any such course must introduce monads, and I want to give a bit of history of the subject. I soon realized that I don't know the precise series of events that led Mac Lane to create the name "monad" instead of the less creative "triple" or "standard construction". Did he coin the term, or did he simply advertize it? The only reference I can find is a link at the nLab page. REPLY [8 votes]: This is covered in this English.SE question. In short, people were not all very happy about the term "triple", and tried to come up with something better. Jean Bénabou suggested "monad" during lunch at a meeting in 1966, and it was quickly adopted; for example, it appears in the titles of Anders Kock's and Eduardo Dubuc's theses from 1967 and 1969 respectively, and the use of the term is alluded to in the Introduction of the "Seminar on triples and categorical homology theory" from 1969 and in Lawvere's "Ordinal sums and equational doctrines" from that volume. Peter May also convinced Mac Lane to use "monad" in his book, which helped to popularize the term.<|endoftext|> TITLE: On math looking obvious in retrospect QUESTION [35 upvotes]: Admittedly, a soft-question. I, being a very young researcher (PhD student) have personally faced the following situation many times: You delve into a problem desperately. No progress for a very long while. All of a sudden, you get the light, and boom: the result is proven. Looking in retrospect, though, the result looks extremely obvious (to the extent you sometimes are ashamed of not getting till then, or embarrassed sharing/publishing). I wonder people's personal opinion on this matter (would also like to hear several real-stories on it, related to published papers). Note If moderators believe Mathoverflow is not the right venue for my question, I can consider relocating it. REPLY [8 votes]: One issue to keep in mind is that things look obvious after you've been thinking about that thing for weeks, or months, or possibly years. Then when you have a proof, you think about how to present it, and how to generalize the idea in question. After all of that, the basic direction will seem more obvious to you than to someone who hasn't thought about it that much, since you've gone down the same mental paths many times. This also has a consequence in teaching; if there's disconnect for research, it may be even stronger when one is dealing with things one considers basic or obvious because you've used them since you were young.<|endoftext|> TITLE: What is the growth rate of the products of binomial coefficients? QUESTION [6 upvotes]: Question 1: Are the following empirically observed relationships true $$ {n \choose 1^a}{n \choose 2^a}{n \choose 3^a}\cdots {n \choose m^a} \sim \exp\bigg(\frac{2n^{1 + \frac{1}{a}}}{a+3}\bigg) $$ where $a$ is a fixed positive integer and $m = \lfloor n^{1/a}\rfloor$. $$ {n \choose b}{n \choose 2b}{n \choose 3b}\cdots {n \choose mb} \sim \exp\bigg(\frac{n^{2}}{2b}\bigg) $$ where $b$ is a fixed positive integer and $m = \lfloor n/b\rfloor$. Question 2: What is the growth rate of $$ {n \choose 1^ab}{n \choose 2^ab}{n \choose 3^ab}\cdots {n \choose m^ab} $$ For $a = 3$, the $\%$ error between the asymptotic and the actual product is shown below. We observe that the error is small and is decreasing with $n$? Note: Posted in MO since it was unanswered in MSE. REPLY [4 votes]: Partial answer (but the general case should be similar): using Stirling+Euler-MacLaurin+Glaisher's constant we have for $a=1$: $$\prod_{1\le j\le n}\binom{n}{j}\sim n^{-(n/2+1/3)}e^{n^2/2+n(1-\log(2\pi)/2)+K}$$ with $$K=1/12-2\zeta'(-1)-\log(2\pi)/2\;.$$ Thus I presume that $A\sim B$ in the OP's question means that $\log(A)/\log(B)$ tends to 1.<|endoftext|> TITLE: Computing spectra without solving eigenvalue problems QUESTION [14 upvotes]: There is a rather remarkable conjecture formulated in this paper, "Computing spectra without solving eigenvalue problems," https://arxiv.org/pdf/1711.04888.pdf and in this talk by Svitlana Mayboroda at the International Congress of Mathematicians 2018 (towards the end): https://www.youtube.com/watch?v=FhPsWJL9eNQ Namely, she considers the following eigenvalue problem (otherwise known as the Schrödinger equation): $$ [-\Delta + V(x)] \psi(x) = E\psi(x) $$ with $x\in \Omega \subset \mathbb{R}^d$ and $\psi({x})\Bigr|_{\partial \Omega}=0$,and where $V(x)$ is a random potential (in some sense, as defined in the paper). I.e., the potential has many valleys of random location and possibly random depth, but the exact form of randomness appears unimportant. The statement seems to be that if we solve instead the following much simpler problem $$ [-\Delta + V(x)] u(x) = 1, \mbox{with } u(x)\Bigr|_{\partial \Omega}=0 $$ then the $n$-th consecutive minimum of the function, $u^{-1}(x)$, dubbed localization landscape will determine with great accuracy (there is no exact statement) the $n$-consecutive eigenvalue as follows $$ E_n \approx (1 + d/4) \inf_x u^{-1}(x)|_n $$ I wonder if there are experts here in ODE, etc, who could comment on the status of this statement/conjecture and in general this localization landscape perspective. The conjecture seems suspicious to me, because diagonalizing and inverting operators are in different computational complexity classes (the latter - required for finding $u$ - is much simpler). But if it's actually true, it would have important implications for physics (I am a physicist). REPLY [3 votes]: Here is how one might rationalize the computational complexity issue: in a system of size $L$ (for example, a chain of $L$ sites), a direct calculation of $L$ eigenvalues requires of order $L^2 \log L$ operations. The calculation of the localization function $u$ only requires of order $L\log L$ operations. The scaling with $L$ rather than $L^2$ can be understood if the eigenfunctions are localized over a length $\xi\ll L$. One can then divide the entire system into independent segments of length only somewhat larger than $\xi$, and that should be sufficient to find most of the eigenvalues with good accuracy. The computational complexity of a calculation of the eigenvalues of localized eigenfunctions would then scale as $\xi L\log L$ rather than as $L^2\log L$. Victor's question motivated me to write a commentary on this development for the Condensed Matter Journal Club,<|endoftext|> TITLE: How does proof assistant organize knowledge? QUESTION [13 upvotes]: I am reading a paper Ittay Weiss, The QED Manifesto after Two Decades — Version 2.0, Journal of Software, 11 no. 8 (2016) pp. 803–815, doi:10.17706/jsw.11.8.803-815 The paper says Goal 7: Organization of Knowledge. The volume of mathematical knowledge is staggering. Not only is it the entirety of mathematics that is well beyond the scope of mastery for any single person, the same can be said of each of the major areas of mathematics (e.g., topology, algebra, analysis, logic, etc.). Moreover, an overwhelming quantity of new mathematics is accumulated each year. For users of mathematics as well as for novice and expert mathematicians, navigating the ocean of results is becoming an ability that appears to require superhuman capabilities. Both QED 1.0 and QED 2.0 aim to provide invaluable tools for this purpose. The system will have tools to navigate, search, and compare results. By analyzing the interdependencies in the coding of various results the system will be able to automatically locate similar results suspected of being related (or perhaps duplicates), thus identifying areas of mathematics that are more closely related than what appears to be. I think organizing knowledge is really important for interactive theorem proving (ITP). Especially, I want to figure out the status of comparing, analyzing the mathematics represented by ITP. But I can not find any project or paper works on comparing and analyzing mathematics represented by ITP. Could you recommend me some papers or projects related to comparing and analyzing mathematics using ITP? REPLY [11 votes]: Organization of mathematics in computerized form is a somewhat separate topic from automated and assisted theorem proving. Here are some pointers that will get you started: MathWebSearch, a content-based search engine for mathematical formulae OpenMath, standards for representing the semantics of mathematical objects and communicating them between software systems The MMT Language System for representation of mathematical knowledge A typical theme in mathematical knowledge organization is how to represent mathematical knowledge in a platform independent way, agnostic with respect to mathematical foundations (or at least translatable between foundations), and readily searchable. None of these are easy to solve. You might think, for instance, that we could all just agree on some format to represent formulas, but that is not the case. Should we use first-order logic, higher-order logic, or type theory? Which meta-level operations on formal theories should we support? Who governs the naming conventions for lemmas, theorems and definitions? And there are meta-theorems showing that figuring out whether two theorems "state the same thing" is a very hard computational problem, even just for very simple fragments of logic. I personally think that these efforts are immensly important, but also that we should not expect them to result in a grand unification of mathematical knowledge. The accepted view of mathematical foundations is that, surely, all of mathematics can be unified as a coherent whole in a single system. However, if you put a bunch of experts on foundations of math into a room and lock them up until they have agreed on one particular mathematical formalism which will serve for expressing all of mathematics, they will die of thirst before they agree on anything. I don't think this is the fault of logicians or logic. It has something deeper to do with how multi-agent system (as computer scientists like to call collections of intelligent entities) organize knowledge. REPLY [4 votes]: Of course a hard part is to know whether two similar-looking lemmas are really related, and even more whether two superficially very different statements might have a short proof of their equivalence. This ITP (Interactive Theorem Proving) paper looks promising: Learning-assisted theorem proving with millions of lemmas Cezary Kaliszyk and Josef Urban https://doi.org/10.1016/j.jsc.2014.09.032<|endoftext|> TITLE: Durov approach to Arakelov geometry and $\mathbb{F}_1$ QUESTION [31 upvotes]: Durov's thesis on algebraic geometry over generalized rings looks extremely intriguing: it promises to unify scheme based and Arakelov geometry, even in singular cases, as well as including geometry over the tropical semiring. The analytic notions in Arakelov geometry seem to appear naturally, as a consequence of rephrasing the construction of $\mathbb{Z}_p$ in terms of maximal compact subgroups of $GL(n, \mathbb{Q}_p)$, and carrying this construction to $GL(n, \mathbb{R})$. A proposed construction of the field with one element $\mathbb{F}_1$ and its finite extensions also appear in the thesis. Given these premises, I'd want to understand more of Durov's work. But Durov has left mathematics to found Telegram, and I am not sure about the status of his work. Since my background on arithmetic geometry is limited, and I am not currently in the university, I have troubles evaluating the impact of this approach, and I would like to know more before delving into a 568 pages thesis. Has this theory been developed after he left mathematics? Did anyone find applications outside the theory itself? What is the point of view of people working in "classical" Arakelov geometry? EDIT I had incorrectly assumed that after founding Telegram, Durov had left mathematics, but as @FedorPetrov points out, he is still active. Yet the question still stands: what is the status of this approach to Arakelov geometry? Is he (or other people) still developing it? Were there any results unrelated to generalized rings proved using this theory? I do not have access to his recent papers, but judging from the first page, his focus seems to have shifted somehow REPLY [5 votes]: Regarding, Did anyone find applications outside the theory itself? The approaches of Durov (and a number of similar methods) are guided by getting an elegant philosophy, or getting some philosophies "right" already at the level of foundations. Also: Not all these ideas or aims are necessarily informed by Arakelov theory alone. The state of the art is very far from being usable for attacking the kind of questions one could try to use classical Arakelov theory for (or, simpler, one could use Algebraic Number Theory for). One could, for example, demand that such an approach should give a novel (genuinely different) reproof for the finiteness theorems of classical number theory; but this so far has been lacking. People have looked particularly at the $K$-theory because many arithmetic invariants can be extracted ("automatically") from $K$-theory, e.g. intersection theory (this is also the viewpoint frequently used in classical Arakelov theory, e.g. in Gillet-Soule; more classically unit group and class group of a number field also are found in $K$-theory). So if you get $K$-theory right, this should give you a whole supply of "right" definitions for free. To the best of my knowledge, although interesting things have been done, no miraculously useful information (for pragmatic number theorists) has yet been found in any of these viewpoints/computations. Of course K(F_1) being the sphere spectrum is aesthetically very pleasing, but it's still very far from giving you anything useful if you're interested in, say, a concrete Diophantine problem. New ideas are needed. Maybe yours. Future will tell.<|endoftext|> TITLE: Is there a Morita cocycle for the mapping class group Mod(g,n) when n > 1? QUESTION [13 upvotes]: Write Mod(g,n) for the mapping class group of a genus-$g$ surface $\Sigma$ with $n$ boundary components. When $n=0,1$ we define the Torelli group $T$ to be the subgroup of Mod(g,n) which acts trivially on the homology $H = H_1(\Sigma,\mathbf{Z})$. The Johnson homomorphism is a much-studied homomorphism from the Torelli group to $\mathrm{Hom}(H,\wedge^2 H)$ (when n=1) or a quotient of this (when n=0) whose kernel turns out to be the commutator subgroup of Torelli. Morita showed in 1993 that the Johnson homomorphism extends to the whole group Mod(g,1), not as a homomorphism, but as a 1-cocycle in $H^1(\mathrm{Mod}(g,1), \mathrm{Hom}(H,\wedge^2 H))$ where the action is given by the action of Mod(g,n) on $H$. (Thus the Morita cocycle restricts to a homomorphism on Torelli, as claimed.) It can be thought of as keeping track of the action of the mapping class on the quotient of $\pi_1(\Sigma)$ by the third term of its lower central series. All of the above is well-known, or at least well-known to the people who know this kind of thing well. Now here's my question: is there a Morita cocycle on Mod(g,n) when n > 1? Of course, such a cocycle would restrict to a Johnson homomorphism from the Torelli subgroup of Mod(g,n), and even this is subtle; but Church's paper "Orbits of curves under the Johnson kernel," gives a way to define a Torelli group and a Johnson homomorphism for Mod(g,n) which behaves well with respect to inclusion of subsurfaces. So a more specific version of my question would be: when n > 1, does Church's "Johnson homomorphism" extend to a "Morita cocycle" on all of Mod(g,n) which behaves well with respect to inclusion of subsurfaces? REPLY [8 votes]: The answer is "yes" -- in fact one can do better and get a class in $$H^1(\text{Aut}(F_m), \text{Hom}(H, \wedge^2 H)),$$ where $F_m$ is the free group on $m$ generators and $H$ is the Abelianization of $F_m$, if I'm not mistaken. This gives the cocycle you want since there is an obvious map $\text{Mod}_{g, n}\to \text{Aut}(\pi_1(\Sigma_{g, {n-1}}))\simeq F_{2g+n-2}$ for $n\geq 2$, given by the conjugation action of $\text{Mod}_{g,n}$ on the point-pushing subgroup, namely $\pi_1(\Sigma_{g, {n-1}})$. A construction goes as follows. Let $\mathbb{Z}[F_m]$ be the group ring of $F_m$, and let $\mathscr{I}$ be the augmentation ideal. Then $H \simeq \mathscr{I}/\mathscr{I}^2$ canonically (via the map sending $g$ to $g-1$) and $\mathscr{I}^2/\mathscr{I}^3\simeq H^{\otimes 2}$ canonically (via the multiplication map). There is a short exact sequence of $\text{Aut}(F_m)$ modules $$0\to \mathscr{I}^2/\mathscr{I}^3\to \mathscr{I}/\mathscr{I}^3\to \mathscr{I}/\mathscr{I}^2\to 0,$$ which we can think of as an extension of $H$ by $H^{\otimes 2}$, and hence gives a class in $$H^1(\text{Aut}(F_m), \text{Hom}(H, H^{\otimes 2})).$$ But in fact a direct computation of a crossed homomorphism representing this class shows that it lands in $\text{Hom}(H, \text{Alt}^2(H)).$ EDIT: This is now worked out in detail here: https://arxiv.org/abs/2004.06146<|endoftext|> TITLE: Collinear Galois conjugates QUESTION [9 upvotes]: This is inspired by this old question, which may provide a bit more background. But the two present questions seem somewhat more fundamental to me. Let $p$ be an irreducible polynomial with integer coefficients. Is it possible that three of the roots of $p$ are collinear on a line in "general position", i.e. which is neither horizontal nor vertical nor through the origin? Further, for $\alpha\in\mathbb R$, denote by $N_p(\alpha)$ the number of zeros of $p$ with real part $\alpha$. If $N_p(\alpha)>1$, is it true that $N_p(\alpha)$ is always a power of $2$? REPLY [11 votes]: The answer to Q1 is yes. For example, $p(x) = x^6 + 45x^4 + 122x^3 + 504x^2 + 1740x + 2213$ is a polynomial with three roots on the line $y = 2x+3$ (and the other three on the line $y = -2x-3$). Take your favorite irreducible cubic with real roots $f(x)$ (mine is $x^{3} - 3x + 1$) and let $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$ be the roots. Now, choose rational numbers $m$ and $b$ - we'll make a degree 6 polynomial whose roots (thought of in the form $x + iy$) lie on the line $y = mx + b$. (I chose $m = 2$ and $b = 3$.) Let $p(x) = f\left(\frac{x-bi}{1+mi}\right) f\left(\frac{x+bi}{1-mi}\right)$. If $z$ is a root of $p(x)$ coming from the first factor, then $\frac{z-bi}{1+mi} = \alpha$ for one of the real roots $\alpha$ of the cubic $f(x)$ and then $z = \alpha + (m \alpha + b)i$ lies on the line $y = mx+b$. The answer to Q2 is no. You can take an irreducible cubic $f(x)$ with all roots real and negative. Then, $f(x^{2})$ can be an irreducible polynomial with all imaginary roots and so $N_{p}(0) = 6$. (For example, you can take $f(x) = x^{3} + 6x^{2} + 9x + 3$. Then $f(x^{2})$ is irreducible by Eisenstein's criterion.)<|endoftext|> TITLE: Easy example of an infinite simple group with an embedding into a finitely presented group QUESTION [6 upvotes]: I would like an easy example of an infinite simple group along with an embedding into a finitely presented group. I know that there are infinite simple finitely presented group such as Thompson’s group $V$ and I know that any group with solvable word problem embeds into a simple subgroup of a finitely presented group, but I am looking for an example of a more elementary nature. REPLY [7 votes]: I think D. L. Johnson's article Embedding some recursively presented groups should answer your question. The abstract is: We seek to illustrate the Higman Embedding Theorem by finding actual embeddings of various popular recursively presented groups in finitely presented ones, and are successful in at least one case. In particular, Section 3 contains a short and elementary proof of the fact that the group $S_\infty$ of bijections of $\mathbb{N}$ with finite supports (which contains the infinite simple group $A_\infty$ of alternating bijections) embeds into a finitely presented group. A presentation for such a group, simplified in Section 4, is $$\langle a,b,x \mid a^2=1, (xaxa^{-1})^3=1, [x,a^2xa^{-2}]=1, x=[a,b], axa^{-1}=bxb^{-1} \rangle.$$ Actually, the group defined by the presentation coincides with Houghton's group $H_3$ already mentioned by Yves in the comments.<|endoftext|> TITLE: A property about probability distribution QUESTION [6 upvotes]: Suppose $g(x)$ is a pdf function and k is a positive real number. Let $F(\alpha)=\int_{-\infty}^{\infty}\frac{1}{\frac{g(x+\alpha)}{g(x)}+k}g(x)dx$, where $\alpha$ is positive. I feel $F(\alpha)$ is increasing in $\alpha$. But I don't know how to prove it for general $g(x)$. Or maybe it is only right for some $g(x)$. Could anyone provide some ideas on this property? Thanks! Thank Iosif Pinelis for showing it is not true for general distributions. Maybe we should focus on unimodal distributions. REPLY [2 votes]: If $\lim_{x \to \pm \infty} g(x) = 0$ then the assertion is false. Define $h(x) := g(-x)$, $H(\alpha) := \int_{-\infty}^\infty \frac{1}{\frac{h(x+\alpha)}{h(x)}+k} ~ h(x) ~ dx = \int_{-\infty}^\infty \frac{1}{\frac{g(-x-\alpha)}{g(-x)}+k} ~ g(-x) ~ dx = \int_{-\infty}^\infty \frac{1}{\frac{g(x-\alpha)}{g(x)}+k} ~ g(x) ~ dx$. (We only need this since $\alpha \geq 0$.) Then again $h$ is a density and by Lebesgue's theorem on dominated convergence $\lim_{\alpha \to \infty} F(\alpha) = \frac{1}{k} = \lim_{\alpha \to \infty} H(\alpha)$. Define $F(\alpha) := H(-\alpha)$ for $\alpha <= 0$. If the assertion is always true then $\alpha \to F(\alpha)$ is always decreasing for $\alpha \leq 0$ and increasing for $\alpha \geq 0$. But this then also holds true for $F_\beta(\alpha) := \int_{-\infty}^\infty \frac{1}{\frac{g(x+\beta-\alpha)}{g(x+\beta)}+k} ~ g(x+\beta) ~ dx = F(\alpha-\beta)$, $\beta \in \mathbb{R}$. As a consequence any $\beta \in \mathbb{R}$ is a minimum point of $F$, i.e. $F \equiv \frac{1}{k}$. This is only possible if $g \equiv 0$, a contradiction. This sort of reasoning of course holds for unimodal densities too. Edit: This answer is not correct. Actually $F_\beta(\alpha) = F(\alpha)$.<|endoftext|> TITLE: Triangle inequality for Ito integral? QUESTION [5 upvotes]: For Lebesgue integrals one has the triangle inequality saying that for continuous functions let's say $$\left\vert\int_0^t f(s) \ ds\right\vert \le \Vert f \Vert_{\infty} \int_0^t \ ds$$ Now if we consider an Ito integral, then $$\left\vert\int_0^t f(s) \ dW(s)\right\vert \le \Vert f \Vert_{\infty} \vert \int_0^t \ dW(s)\vert$$ does not hold pointwise, but I was wondering whether this one holds probabilitically, i.e. does there exist a constant $c(t)>0$ such that for all deterministic continuous $f$ $$\mathbb P\left(\vert W(t) \vert \Vert f \Vert_{\infty}\ge a\right)\ge c(t)\mathbb P\left(\left\vert\int_0^t f(s) \ dW(s)\right\vert \ge a\right)?$$ REPLY [2 votes]: If the function $f$ is indeed deterministic, with $M:=\|f\|_\infty$ and $\sigma^2:=\int_0^t f(s)^2\,ds$, then $X:=\int_0^t f(s)\,dW(s)\sim N(0,\sigma^2)$, whereas $Y:=MW(t)\sim N(0,M^2t)$, and $k^2:=\sigma^2/(M^2t)\le1$. So, $X$ equals $kY$ in distribution, and so, $P(|Y|\ge a)\ge P(|X|\ge a)$ for all real $a$; that is, your inequality holds with $c(t)=1$.<|endoftext|> TITLE: Starting letters of equivalent infinite geodesic paths of hyperbolic Coxeter groups QUESTION [6 upvotes]: Let $\left(W\text{, }S\right)$ be a Gromov hyperbolic Coxeter system and denote by $\partial W$ the corresponding Gromov boundary. For $z\in\partial W$ let $\alpha$, $\beta$ be infinite geodesic paths with $z=\left[\alpha\right]=\left[\beta\right]$ and assume that there exists $w\in W$ such that $\left|w^{-1}\alpha_{n}\right|=\left|\alpha_{n}\right|-\left|w\right|$ for all large enough $n$ (i.e. $\alpha_n$ starts with $w$). Here $\left|\cdot\right|$ denotes the word metric on $W$ regarding the generating set $S$. Is it true that then $\left|w^{-1}\beta_{n}\right|=\left|\beta_{n}\right|-\left|w\right|$ for all large enough $n$? My intuition for the Gromov boundary is not well-developed yet, so intuitively I would say yes. However, I doubt this is true for general hyperbolic groups. REPLY [4 votes]: This is false. The archetypical family of hyperbolic Coxeter groups are the hyperbolic triangle groups $$ T(l,m,n) = \langle a,b,c \mid a^2=b^2=c^2=(ab)^l=(bc)^m=(ca)^n=1\rangle $$ where $l,m,n\geq 2$ and $(1/l)+(1/m)+(1/n)<1$. Such a group acts as a group of symmetries of a tiling of the hyperbolic plane by congruent triangles with angles of $\pi/l$, $\pi/m$, and $\pi/n$. Here are some basic facts about these groups: If we choose a base triangle $T_0$ (a.k.a. the "fundamental chamber"), then each triangle of the tiling can be expressed uniquely as $gT_0$ for some $g\in T(l,m,n)$. As such, the (right) Cayley graph of $T(l,m,n)$ is precisely the dual graph to the triangular tiling, with the identity vertex lying in the triangle $T_0$. The triangular tiling can be obtained by cutting the hyperbolic plane along a countable family of hyperbolic lines. A path of edges in the Cayley graph is a geodesic if and only if it crosses each line of the tiling at most one time. The Gromov boundary $\partial T(l,m,n)$ is the circle $S^1$, which can naturally be identified with the boundary circle of the hyperbolic plane. An infinite geodesic path in the Cayley graph represents a point on the boundary if and only if it converges to that point in the closed unit disk. Now, let $L$ be the hyperbolic line that separates $T_0$ from $aT_0$, let $z$ be an endpoint of $L$ on the boundary circle, and consider the following two geodesic paths in the Cayley graph: The path $\alpha$ that goes from $T_0$ to $aT_0$, and then "follows along" $L$ in the direction of $z$. The path $\beta$ that starts at $T_0$ and "follows along" $L$ in the direction of $z$. To be precise, $\beta$ is the path in the Cayley graph corresponding to the sequence of triangles passed through by the hyperbolic ray from the identity vertex to $z$, and $\alpha$ consists of the dual edge from $T_0$ to $aT_0$ followed by the reflection of $\beta$ across $L$. Then $[\alpha]=[\beta]=z$, but $\alpha$ starts with $a$ and $\beta$ does not. Edit: Part of the reason I chose the above example was to convey some intuition for hyperbolic Coxeter groups. However, there are simpler examples available. For example, consider the group $$ G = \langle a,b,c \mid a^2=b^2=c^2=(ab)^2=(ac)^2=1\rangle $$ This is the direct product of $\mathbb{Z}_2$ with the infinite dihedral group $\mathbb{Z}_2*\mathbb{Z}_2$, and its Cayley graph is a bi-infinite "ladder" with $a$ edges as rungs and alternating $b$ and $c$ edges along the sides. This group is Gromov hyperbolic since it's virtually cyclic (making it an "elementary" hyperbolic group), and the Gromov boundary $\partial G$ is a two-point set corresponding to the two ends of the ladder. Now, if $\alpha$ is the geodesic path corresponding to the infinite word $a(bc)^\infty = abcbcbc\cdots$ and $\beta$ is the infinite path corresponding to $(bc)^\infty$, then $[\alpha]=[\beta]$ since the two paths go the same direction on the ladder, but $\alpha$ starts with $a$ and $\beta$ does not.<|endoftext|> TITLE: An equality about sin function? QUESTION [9 upvotes]: Empirical evidence suggests that, for each positive integer $n$, the following equality holds: \begin{equation*} \prod_{s=1}^{2n}\sum_{k=1}^{2n}(-i)^k\sin\frac{sk\pi}{2n+1}=(-1)^n\frac{2n+1}{2^n}, \end{equation*} where $i=\sqrt{-1}$. Is it a known equality? If it is true, would you please give me some insights on how to derive this equality? REPLY [26 votes]: We have $$ \sum_{k=0}^{2n}(-i)^k\sin\frac{sk\pi}{2n+1}=\frac{h(s)-h(-s)}{2i},\quad\text{where}\\ h(s)=\sum_{k=0}^{2n}e^{i(-\pi/2+\frac{\pi s}{2n+1})k}=\frac{1-e^{-i\pi(2n+1)/2+i\pi s}}{1-e^{i(-\pi/2+\frac{\pi s}{2n+1})}}=\frac{1+i(-1)^{n+s}}{1+ie^{i\frac{\pi s}{2n+1}}}. $$ The numerators for $s$ and $-s$ are the same, and $$ \frac1{1+ie^{i\theta}}- \frac1{1+ie^{-i\theta}}=\frac{2\sin\theta}{2i\cos \theta}=-i\tan\theta, $$ so the product reads as $$ 2^{-2n}\prod_{s=1}^{2n} (1+i(-1)^{n+s})\tan \frac{s\pi}{2n+1}. $$ The product of $(1+i(-1)^{n+s})$ equals $2^n$, since the product of two consecutive guys equals 2. It remains to prove that $$ \prod_{s=1}^{2n}\tan \frac{s\pi}{2n+1}=(-1)^n(2n+1). $$ This should be well known, and in any case it is standard: using the formula $$ i\tan \theta=\frac{e^{2i\theta}-1}{e^{2i\theta}+1} $$ we get $$ (-1)^n\prod_{s=1}^{2n}\tan \frac{s\pi}{2n+1}=\prod_{s=1}^{2n} \frac{\omega^s-1}{\omega^s+1},\quad\text{where}\, \omega=e^{2\pi i/(2n+1)}. $$ We have $\prod_{s=1}^{2n}(z-\omega^s)=1+z+\ldots+z^{2n}=:P(z)$, therefore $$\prod_{s=1}^{2n} \frac{\omega^s-1}{\omega^s+1}=\frac {P(1)}{P(-1)}=2n+1$$<|endoftext|> TITLE: Who invented Monoid? QUESTION [10 upvotes]: I was trying to find (and failed) the original author of either the concept of Monoid (set with binary associative operation and identity) the name (which sounds french ? and also Dioid (for what seems to be a semiring) is exclusively french wiki article) Question: Is there a text source to attribute the invention of either the name or concept of monoid to a person or a group (pun intended) ? REPLY [14 votes]: The name "monoid" was first used in mathematics by Arthur Cayley [*] for a surface of order $n$ which has a multiple point of order $n-1$. In the context of semigroups the name is due to Bourbaki [source, page 30] It is also worth commenting on the related term monoid, meaning an associative magma with identity. This term is a little more recent than semigroup, and seems to originate with Bourbaki [**]. Before this, Birkhoff (1934) was using the term groupoid for an associative magma with identity. Dov Tamari [source, page 1] argues that Bourbaki "probably intended to abolish the term semigroup for reasons of linguistic taste." [*] A. Cayley, Second and Third Memoirs on Skew Surfaces, Otherwise Scrolls, Phil. Trans. (1863 and 1869). [**] N. Bourbaki, Éléments de Mathématique, Algèbre, Hermann, Paris (1943): Chapter I, §2.<|endoftext|> TITLE: Why "monoidal" transformation? QUESTION [9 upvotes]: In Carlo Beenakker's answer to this recent MO question, it turns out that the name "monoid" was first used in mathematics by Arthur Cayley for a surface of order $$ which has a multiple point of order $−1$. On the other hand, a (old) name for the blow-up is "monoidal transformation". I suspect that there should be a link relating these two classical terminologies, but I was not able to find one. So let me ask the following Question. Why was the blow-up classically known as "monoidal transformation"? Was this related to Cayley's "monoid"? REPLY [10 votes]: As I mentioned in the comments, Semple and Roth (Introduction to Algebraic Geometry) discuss this in "Examples on Chapter VIII" example 12 (p.187 of my version). As far as I can tell, this is what they are saying: Homoloidal webs: Let $V$ be a 4-dimensional vector space and suppose we're working in the projective 3-space $PV$ with homogeneous coordinates $[x_0:\cdots:x_3]$. Pick four homogeneous (degree $d$) polynomials $F_0,\ldots,F_3\in Sym^d(V)$. We get a rational map $PV\to PV$ defined by $$[x_0:\cdots:x_3]\mapsto[F_0(x):\cdots:F_3(x)]$$ How should we think of this? We have a 3-parameter family of surfaces $\sum_i\lambda_iF_i=0$ (parametrised by $[\lambda_0:\cdots:\lambda_3]\in PV^*$, called a "homoloidal web" of surfaces) and the equation $\sum_i\lambda_iF_i(x)=0$ defines a hyperplane in $PV^*$ consisting of surfaces which contain $x$. Our map sends $x$ to the point in $PV$ dual to this hyperplane. Monoids: Now suppose that our surfaces $F_i$ all have a singularity of order $d-1$ at the point $p$. In other words, they're all monoids in the sense of Cayley (because they have degree $d$). They call the corresponding birational maps monoidal transformations. Certainly $p$ will be in the base locus of this system of surfaces, so you need to blow up at $p$ to get a morphism, but I'm still not sure why this is a reasonable definition. They do discuss examples of this immediately afterwards, but I get lost in the notation.<|endoftext|> TITLE: Analog of the Lie Product formula for commutators QUESTION [6 upvotes]: Let $X, Y$ be elements of a Lie algebra. Consider the group $G$ generated by (limits of) arbitrary products of the elements $$ G = \langle{e^{tX},e^{sY}\rangle}$$ for all $t,s$. The Lie product formula tells us that $e^{u(X+Y)} \in G$. What about the commutator? E.g., for all $u$, what conditions are necessary so that $$ e^{u [X,Y]} \in G$$ Clearly we can achieve this for ``infinitesimal'' $u$ by considering something like $e^{t X} e^{t Y} e^{-t X} e^{-t Y}$ for infinitesimal $t$. But what about for finite $u$? Is there a formula like $$ e^{u [X,Y]} = \mathcal{P} e^{\int X t' + Y s' d\tau }$$ for some $t(\tau), s(\tau)$? If so, is there an explicit formula for $t$ and $s$ in terms of $u$? I know of such a formula for SU(2), but I do not know how general group. REPLY [7 votes]: The formula $$ e^{u[X,Y]}=\lim\limits_{n\to\infty}\left(e^{\sqrt{u/n}\,X}e^{\sqrt{u/n}\,Y}e^{-\sqrt{u/n}\,X}e^{-\sqrt{u/n}\,Y}\right)^n \tag{$*$} $$ (Godement 2017, p. 70) shows that $e^{u[X,Y]}\in G$ always, and seems sufficiently “like” “Lie’s” to be what you want. Note added: The origin of this formula was left unresolved before. Howe (1983) and Godement (originally 1982) point to von Neumann (1929, §II.3) who proves $e^{u[X,Y]}\in G$, but apparently not $(*)$. Formula $(*)$ is explicit in e.g. Chorin et al. (1978, p. 207), Goldstein (1970), Nelson (1969, p. 111), Goto (1969, p. 159), Hausner-Schwartz (1968, p. 78), Cohn (1957, p. 112), and somewhat implicit in Gluškov (1957, p. 137), Yamabe (1950, p. 14), Yosida (1936, Remark p. 470). Almost none of whom cite each other!<|endoftext|> TITLE: Shapovalov form on Verma modules QUESTION [8 upvotes]: I will formulate my question first; below is the relevant background information and notation. I have asked this question on stack, but I think the odds are better that I actually get an answere here (but do correct me if I'm wrong to post it here). WHY is the Shapovalov form on a Verma module symmetric? The fact that it's orthogonal with respect to its weight decomposition simplifies things of course, so that we only have to calculate expressions of the form $$ \langle f^{n_1}_1\ldots f^{n_l}_lv_\lambda,f^{m_1}_1\ldots f^{m_l}_lv_\lambda\rangle $$ where $n,m\in\mathbb{Z}^l_{\ge0}$ are such that $\sum^l_{i=1}n_i\beta_i=\sum^l_{i=1}m_i\beta_i$. This seems rather hard to do because one has to commute elements of root spaces of non-simple roots. In principle, any element of the form $f^{n_1}_1\ldots f^{n_l}_l$ for $n\in\mathbb{Z}^l_{\ge0}$ can be expressed as a sum of elements of the form $f^{n_{i_1}}_{i_1}\ldots f^{n_{i_p}}_{i_p}$, where now $p\in\mathbb{Z}_{\ge1}$, and $i_1,\ldots,i_p\in\{1,\ldots,r\}$ (not necessarily distinct!) and $n_{i_1},\ldots,n_{i_l}\in\mathbb{Z}_{\ge1}$. This seems hopeless as well. In general, I don't feel like such a computation-based approach can be very helpful (let alone insightful). An elementary argument would be awesome. Background Let $L$ be a simple Lie algebra (over $\mathbb{C}$), let $L=L_-\oplus H\oplus L_+$ be a root space decomposition with $H$ a Cartan subalgebra of $L$ ($\dim(H)=r$ and $\dim(L_-)=\dim(L_+)=l$), let $\Delta=\{\alpha_1,\ldots,\alpha_r\}$ be a set of simple positive roots, and enumerate the positive roots as $\beta_1,\ldots,\beta_r$ such that the first $r$ are the simple onese. Denote by $B$ the Borel subalgebra $H\oplus L^+$. Let $\{e_1,\ldots,e_r,f_1,\ldots,f_r\}$ be a set of Chevalley generators, and let $\{e_{r+1},\ldots,e_l,f_{r+1},\ldots,f_l\}$ be elements of the root spaces corresponding to the positive, negative roots that are not simple. Let $\lambda\in H^\ast$, denote by $\lambda^+\in B^\ast$ its extension to $B$ (by declaring it to be zero on $L_+$), and denote by $\mathbb{C}_\lambda$ the one-dimensional (over $\mathbb{C}$) $\mathfrak{U}(B)$-module defined by $xv_\lambda=\lambda^+(x)v$ for any $x\in B$ and for any $v\in\mathbb{C}_\lambda$, where $v$ is a nonzero element of $\mathbb{C}_\lambda$. The Verma module is then the induced module $M_\lambda:=\mathfrak{U}(L)\otimes_{\mathfrak{U}(B)}\mathbb{C}_\lambda$. Denote $1\otimes 1\in M_\lambda$ by $v_\lambda$. The Verma module is a direct sum of weight spaces that are finite-dimensional, i.e. $$ M_\lambda=\bigoplus_{k\in\mathbb{Z}^r_{\ge0}}M^k_\lambda $$ where $M^k_\lambda=\{v\in M_\lambda\mid\forall\,h\in H:hv=(\lambda-\sum^r_{i=1}k_i\alpha_i)(h)v\}$ for any $k\in\mathbb{Z}^r_{\ge0}$. It is easy to see that, for any $k\in\mathbb{Z}^r_{\ge0}$, the weight space $M^k_\lambda$ is the linear span of $$ \{f^{n_1}_1\ldots f^{n_l}_l\mid n\in\mathbb{Z}^l_{\ge0}\,\wedge\,\sum^l_{i=1}n_i\beta_i=\sum^r_{i=1}k_i\alpha_i\}. $$ The restricted dual $M^\vee_\lambda$ of $M_\lambda$ corresponding to the Chevalley involution $\omega\colon L\to L$ defined by $$ \forall\,i\in\{1,\ldots,r\}\colon\omega(e_i)=-f_i\,\wedge\,\omega(f_i)=-e_i, $$ is as the $\mathfrak{U}(L)$-module whose underlying abelian group is $$ \bigoplus_{k\in\mathbb{Z}^r_{\ge0}}(M^k_\lambda)^\ast $$ and where the action of $\mathfrak{U}(L)$ is defined for each $k\in\mathbb{Z}^r_{\ge0}$ by setting $$ (xf)(v)=f(-\omega(x)v) $$ for all $x\in L$, for all $f\in(M^k_\lambda)^\ast$, and for each $v\in M^k_\lambda$. The unique vector $v^\ast_\lambda\in(M^0)^\ast_\lambda$ dual to $v_\lambda$ is an element of the weight space $(M^\vee_\lambda)^\lambda$ and it's annihilated by $L_+$, so we have a unique morphism $\phi\colon M_\lambda\to M^\vee_\lambda$ of left $\mathfrak{U}(L)$-modules such that $\phi(v_\lambda)=v^\ast_\lambda$. Now we define the Shapovalov form to be the bilinear form $$ \langle\cdot,\cdot\rangle:M_\lambda\times M_\lambda\to\mathbb{C},(v,w)\mapsto\phi(v)(w). $$ It has the property that $\langle M^k_\lambda,M^{k^\prime}_\lambda\rangle=\{0\}$ for all $k,k^\prime\in\mathbb{Z}^r_{\ge0}$ such that $k\neq k^\prime$. In addition, it satisfies $\langle v_\lambda,v_\lambda\rangle=1$ and $\langle xv,w\rangle=\langle v,-\omega(x)w\rangle$ for any $x\in L$ and for any $v,w\in M_\lambda$. REPLY [6 votes]: There is alternative way to define the Shapovalov form which makes the symmetry easy to see. By the PBW theorem you can write each element $X$ of $\mathfrak{U(g)}$ as $X = f_{i_1} \cdots f_{i_m} \cdot h_{j_1} \cdots h_{j_n} \cdot e_{k_1} \cdots e_{k_o}$. Now denote by $\pi: \mathfrak{U(g)} \to \mathfrak{U(h)}$ the projection map defined by $$X \mapsto h_{j_1} \cdots h_{j_n}.$$ Then the universal Shapovalov form $S: \mathfrak{U(g)} \otimes \mathfrak{U(g)} \to \mathfrak{U(h)}$ is defined by $$S(X, Y) = \pi(\tau(X)\cdot Y),$$ where $\tau: \mathfrak{U(g)} \to \mathfrak{U(g)}$ is anti-automorphism that is identity on $ \mathfrak{U(h)}$ and $\tau(f_{i_1} \cdots f_{i_m}) = e_{i_m} \cdots e_{i_1}$. Since $\pi(\tau(u)) = \pi(u)$ for any $u \in \mathfrak{U(g)}$, the symmetry follows. Now for any $\lambda \in \mathfrak{h}^*$ we can evaluate $\lambda$ on the image of $\pi$ and composing with $S$ we get a $\mathbb{C}$-valued form, call it $S^\lambda$. It is not hard to show that $S^\lambda$ defines a symmetric, $\tau$-contravariant form on any highest weight module of highest weight $\lambda$. For details see Representations of Semisimple Lie Algebras in the BGG Category O by James Humphreys.<|endoftext|> TITLE: turn $\pi/n$, move $1/n$ forward QUESTION [11 upvotes]: start at the origin, first step number is 1. turn $\pi/n$ move $1/n$ units forward Angles are cumulative, so this procedure is equivalent (finitely) to $$ u(k):=\sum_{n=1}^{k} \frac{\exp(\pi i H_{n})}{n}$$ Is the limiting shape formed by a line plot of partial sums a circle or a spiral? Relying on visual intuition with the harmonic numbers seems to be perilous. Where's its center? This is equivalent to "If $u(k)$ converges, what does it converge to?". Naively resumming within exp produces lots of divergent series as a result of the harmonic numbers $H_{n}$. At the cost of being a numerically ill-conditioned sum with lots of cancellation, I have to wonder what this says of arithmetic properties of the harmonic numbers. REPLY [12 votes]: The large-$n$ asymptotics of the harmonic numbers is $H_n\simeq \gamma_E+\log n$, which we use for $n\geq k_0$, replacing the sum $\sum_{n=k_{0}}^k$ by an integral $\int_{k_0}^k dn$. We thus find $$\begin{align} U(k)&=u(k_0-1)+e^{i\pi\gamma_E}\int_{k_0}^k \frac{1}{n}e^{i\pi\log n}\,dn\\ &=u(k_0-1)+\frac{i}{\pi} e^{i \pi\gamma_E } \left(k_0^{i\pi}-k^{i \pi }\right). \end{align}$$ So $U(k)$ traces out a circle in the complex plane, of radius $1/\pi$ and center at $z_0=u(k_0-1)+(i/\pi)e^{i \pi\gamma_E } k_0^{i\pi}$. The plot compares $u(k)$ (gold) and $U(k)$ (blue) for $k_0=50$ and $k$ up to 1000, when $z_0=-0.66- 0.28i$. The agreement is quite satisfactory.<|endoftext|> TITLE: Is the diophantine equation $3x^2+1=py^2$ always solvable for each prime $p\equiv 13\pmod{24}$? QUESTION [6 upvotes]: In Question 337879, I conjectured that for any prime $p\equiv3\pmod4$ the equation $$3x^2+4\left(\frac p3\right)=py^2\tag{1}$$ always has integer solutions, where $(\frac p3)$ is the Legendre symbol. Motivated by this, here I pose the following conjectures. Conjecture 1. For any prime $p\equiv13\pmod{24}$, the equation $$3x^2+1=py^2\tag{2}$$ always has integer solutions. For example, when $p=829$ the least positive integer solution of $(2)$ is $$(x,\,y)=(1778674,\,106999).$$ Conjecture 2. For any prime $p\equiv3\pmod4$, the equation $$2x^2-py^2=\left(\frac 2p\right)\tag{3}$$ always has integer solutions, where $(-)$ is the Legendre symbol. For example, when $p=167$ the smallest positive integer solution of $(3)$ is $$(x,\,y)=(3993882,\,437071).$$ Conjecture 3. For any prime $p\equiv3\pmod4$ and $q\in\{7,11,19,43,67,163\}$, the equation $$qx^2+4\left(\frac pq\right)=py^2\tag{4}$$ always has integer solutions. QUESTION. How to solve the conjectures? Your comments are welcome! REPLY [3 votes]: Here's a second proof of Claim 1 (I guess the others can be taken care of similarly) that perhaps explains a little bit better what is going on. Assume first that $3x^2 + 1 = py^2$ has an integral solution. Then $\eta = x\sqrt{3} + y \sqrt{p}$ is a unit in ${\mathbb Q}(\sqrt{3},\sqrt{p})$ satisfying $\eta^2 = t + u\sqrt{3p}$; thus $\eta^2$ is an odd power of the fundamental unit $\varepsilon$ of ${\mathbb Q}(\sqrt{3p})$. For $p = 829$ we have $\varepsilon = 18982087189657 + 380632678652\sqrt{3p}$, and using the observation that $(2 \cdot 18982087189657 - 2)/3 = 4 \cdot 1778674^2$ (see the proof below) we find $\eta = 1778674 \sqrt{3} + 106999 \sqrt{829}$. For proving the existence of a solution we simply work backwards (essentially this a classical descent on Pell conics). We start with the fundamental solution $(t, u)$ of $t^2 - 3pu^2 = 1$ and write this equation in the form $(t-1)(t+1) = t^2 - 1 = 3pu^2$. The fact that the fundamental unit has norm $+1$ (the discriminant is divisible by $3$) and that $(t,u)$ is fundamental implies that $3$ and $p$ divide different factors. Using elementary congruences and the fact that $(2/p) = -1$ and $(3/p) = +1$ it is easy to show that the only possibility is $$ t-1 = 6a^2, \quad t+1 = 2pb^2, $$ which implies $1 = pb^2 - 3a^2$.<|endoftext|> TITLE: Is there any edge- but not vertex-transitive polytope in $d\ge 4$ dimensions? QUESTION [8 upvotes]: I consider convex polytopes $P\subset\Bbb R^d$. The polytope is called vertex- resp. edge-transitive, if any vertex resp. edge can be mapped to any other by a symmetry of the polytope. I am looking for polytopes which are edge- but not vertex-transitive. There are infinitely many of these for $d=2$, and exactly two for $d=3$ (rhombic dodecahedron and rhombic tricontrahedron, see below). $\quad$$\quad$ $\quad$$\quad$ I do not know a single example for $d\ge 4$. I believe it is easy to see that the edge-graph of such a polytope must be bipartite, and thus, zonotopes might be a good place to start looking. But my constructions fail for $d\ge 4$. REPLY [9 votes]: The answer is No, there are no other such polytopes. The proof is quite laborious in parts, and I wrote it down in this article. Theorem. In dimension $d\ge 4$, an edge-transitive polytope is vertex-transitive. The idea is as follows: first, show that every edge-transitive polytope $P$ that is not vertex-transitive has the following three properties: all edges of $P$ are of the same length, $P$ has an edge in-sphere, and the edge-graph of $P$ is bipartite. Call a polytope with these three properties bipartite. One then tries to classify these polytopes instead. This is easier, because every face of a bipartite polytope is again bipartite (not true for edge- or vertex-transitive polytopes). The second step is to deal with all inscribed bipartite polytopes. It is not hard to see that these are zonotopes. By a result from this article (see also this question), inscribed zonotopes with all edges of the same length are vertex-transitive. We can therefore exclude all the inscribed bipartite polytopes. In the third step one classifies all the 3-dimensional non-inscribed bipartite polyhedra. This is quite tedious. Here is one example of a polyhedron which satisfies 2. and 3., but fails to have all edges of the same length. The deviation is so miniscule, that it cannot be spotted visually. The result is then that there are only two such polyhedra: exactly those that I already mentioned in the question. The final step is then to show that no 4-dimensional non-inscribed bipartite polytope can be built if we can use only these two polyhedra as facets. This uses a straight-forward argument on dihedral angles (see also Nick's answer).<|endoftext|> TITLE: Is Binary Integer Linear Programming solvable in polynomial time? QUESTION [5 upvotes]: The paper Solving the Binary Linear Programming Model in Polynomial Time claims that Binary Integer Linear Programming is in P. However, it seems that no subsequent literature in the mainstream has done any further study on this. I am a bit doubtful regarding the correctness of the claim made in that paper. Therefore, I have put this question here. REPLY [7 votes]: Often called Binary Integer Programming (BIP). Wikipedia: Integer programming is NP-complete. In particular, the special case of 0-1 integer linear programming, in which unknowns are binary, and only the restrictions must be satisfied, is one of Karp's 21 NP-complete problems. Here is a list of those 21 Karp problems. You can also find the claim that BIP $\in$ NPC in many class notes, e.g., this set.<|endoftext|> TITLE: When simple cohomological computations predict ingenious algebro-geometric constructions? QUESTION [21 upvotes]: Classical algebraic geometry is full of ingenious constructions and miraculous coincidences: 27 lines on a cubic surface are related to Weyl lattice of type $E_6,$ lines on an intersection of four-dimensional quadrics are parametrized by a Jacobian of genus two curve and so on. Existence of some constructions of this type may be predicted by simple cohomological computations. For instance, consider a smooth intersection $X$ of two quadrics $Q_1$ and $Q_2$ in $\mathbb{P}^5.$ This example is explained in detail in the beautiful paper The complete intersection of two or more quadrics of Miles Reid. It is easy to see that $h^{p,q}(X)$ equals to $1$ for $$(p,q)\in \{(0,0),\ (1,1),\ (2,2),\ (3,3)\},$$ equals to 2 for $$(p,q)\in \{(1,2),\ (2,1)\}$$ and equals to $0$ in other cases. Generalized Hodge Conjecture predicts that there exists a certain curve $C$ and a correspondence $D\subset C\times X$ which induces a surjective morphism $$H^1(C,\mathbb{Z})\longrightarrow H^3(X,\mathbb{Z})(-3).$$ Explicit construction of this correspondence is based on an identification of the Hilbert scheme of lines on $X$ with Jacobian $J(C).$ Question 1: What other classical constructions in algebraic geometry can be a posteriori motivated by simple computations with (mixed) Hodge structures? Question 2: It will be very interesting to see some concrete examples, when such a construction is expected to exist, but has not been worked out yet. REPLY [6 votes]: Cohomology, with the intersection form, can give interesting lattices. Geometry tells us that they contain many elements of square length $2$: they are given by vanishing cycles in a Lefschetz pencil. This also applies to Milnor fibers of deformations of isolated singularities. This is how one recognizes the $E_n$ in the orthogonal of the canonical class in the $\text{H}^2$ of $\mathbb{P}^2$ with $n$ points blown up ($n\le8$). Here, $n=6$ is the cubic surfaces case; the affine case $n=9$ when one blows up the intersection of two cubic curves is interesting too. When a $\text{H}^{2n+1}$ is of type $\{(2n+1,0),(0,2n+1)\}$, one expects a principally polarized abelian variety is lurking around. A Jacobian? A Prym variety? For complete intersections in $\mathbb{P}^N$ there is a table in SGA 7 telling us when this Hodge level one case occurs. I do not know whether all have been unraveled. Similarly, cubic fourfolds look very much like K3 surfaces (with cohomology one bigger). It follows that there are related Kuga-Satake abelian varieties attached to them (whose $\text{H}^2(-1)$ contains their $\text{H}^4$). This allowed to prove the Weil conjecture for them (as for the K3) early on, but remains quite unexplicit. The Milnor fibers story relates to which quadratic singularities one can have, and when many are imposed, the related abelian variety reduces to a lower dimensional one, possibly easier to see (case of Kummer surfaces among K3's). A different game is guessing periods of differential forms ($\zeta(3)$ related to an extension of $\mathbb{Z}$ by $\mathbb{Z}(3)$, $\ldots$).<|endoftext|> TITLE: What generalizes symmetric polynomials to other finite groups? QUESTION [5 upvotes]: Multivariate polynomial indexed by ${1, \ldots, n}$ are acted on by $S_n$: for $\sigma \in S_n$, define $\sigma(x_i) = x_{\sigma(x_i)}$, etc. Symmetric polynomials are those polynomials which are invariant under this action. If we consider just a subgroup of $S_n$, a polynomial which is not symmetric my be invariant. An example: Let $S_5$ be the symmetric group of degree 5, $\tau = (1\ 2\ 3\ 4\ 5)$, and $C_5 = \{\tau^n : n = 1, ..., 5\}$. The polynomial $$f(x_1, x_2, x_3, x_4, x_5) = x_1x_3 + x_2x_4 + x_3x_5 + x_4x_1+x_5x_2$$ is invariant under the elements of $C_5$, but not under the elements of $S_5$. I was wondering what sort of theory exists around these sets of polynomials? What are they called? Also, Every symmetric polynomial can be expressed as some polynomial evaluated on the elementary symmetric polynomials. Is there some systematic way of generating some set of polynomials which will have this property? REPLY [8 votes]: The name of the body of theory you are asking for is "invariant theory of permutation groups." You will also find relevant papers by searching for "polynomial permutation invariants." It falls under the broader rubric of "invariant theory of finite groups", which is a developed field. See the books by Benson, Smith, Neusel and Smith, Campbell and Wehlau, and Derksen and Kemper. But there is a fair amount of work specifically on permutation groups, i.e., subgroups of $S_n$, acting in the way you describe, dating back to the 19th century. I'll mention the paper by Leopold Kronecker, Grundzüge einer arithmetischen theorie der algebraischen grossen, Crelle, Journal für die reine und angewandte Mathematik 92:1-122, 1881 (reprinted in Werke, vol. 2, 237–387) as an important early contribution. The paper by Garsia and Stanton linked in Richard Stanley's comment is an important contribution to this literature, dating to 1984. The MO question linked in KConrad's comment speaks to the cyclic action you mention in the OP. I summarized a large amount of currently known theory (for general $G\subset S_n$) in this math.SE answer, which I thought belonged on MO to begin with, so I'm reproducing it here: Let $G\subset S_n$ be the permutation group in question. Let $A[\mathbf{t}]:=A[t_1,\dots,t_n]$ as a shorthand. General results: (a) The invariant ring $A[\mathbf{t}]^G$ is generated by the invariants of degree at most $\operatorname{max}(n,n(n-1)/2)$, a result usually attributed to Manfred Göbel (see here), although it was actually anticipated by Leopold Kronecker (see section 12 of his paper Grundzüge einer arithmetischen theorie der algebraischen grossen, Crelle, Journal für die reine und angewandte Mathematik 92:1-122, 1881, reprinted in Werke, vol. 2, 237–387). (b) If the coefficient ring $A$ is a field of characteristic not dividing the group order $|G|$, then $A[\mathbf{t}]^G$ is free as a module over the subring generated by any homogeneous system of parameters (equivalently, $A[\mathbf{t}]^G$ is Cohen-Macaulay). This result is not specific to permutation groups -- it is a consequence of the Hochster-Eagon theorem. (Though again it happens that Kronecker proved it in the case of a permutation group and a field of characteristic 0.) Then any homogeneous system of parameters for $A[\mathbf{t}]^G$ is called a set of primary invariants, and a module basis over the subring they generate is a set of secondary invariants. There are algorithms based on Gröbner bases to compute primary and secondary invariants, again not specific to permutation groups; see the book by Derksen and Kemper. However, in the case of permutation groups, the elementary symmetric polynomials provide a uniform choice for the primary invariants, and there is a method due to Nicolas Borie that aims for more effective computability of the secondary invariants (see here). (c) There is also a method due to Garsia and Stanton that produces secondary invariants from a shelling of a certain cell complex (specifically, the quotient of the barycentric subdivision of the boundary of an $(n-1)$-simplex by the $G$ action on the simplex's vertices), when such exists (see here). When this shelling exists, the assumption that $A$ be a field of characteristic not dividing $|G|$ becomes superfluous, i.e. the secondary invariants produced by the method give a module basis for $A[\mathbf{t}]^G$ over the subring generated by the elementary symmetric polynomials, entirely regardless of $A$. It is not an easy problem to find the shelling in general, but has been done in specific cases (the original paper by Garsia and Stanton handles the Young subgroups, work of Vic Reiner handles $A_n\subset S_n$ and the diagonally embedded $S_n\hookrightarrow S_n\times S_n\subset S_{2n}$, and work of Patricia Hersh handles the wreath product $S_2\wr S_n\subset S_{2n}$). There is a detailed development of Garsia and Stanton's shelling result in my thesis, sections 2.5 and 2.8, along with a discussion of its connection to Göbel's work (see last paragraph) and some speculation about generalizations. (d) From (b) you can see that $A[\mathbf{t}]^G$ has a nice structure of free-module-over-polynomial-subring when $A$ is a field of characteristic not dividing $|G|$, but from (c) you can see that sometimes this nice structure still exists even when $A$ doesn't satisfy this (e.g. perhaps it is $\mathbb{Z}$, or else a field whose characteristic does divide $|G|$). There is a characterization, due to myself and Sophie Marques, of which groups $G\subset S_n$ have the property that this structure in $A[\mathbf{t}]^G$ exists regardless of $A$. It turns out to be the groups generated by transpositions, double transpositions, and 3-cycles. (Our paper is framed in the language of Cohen-Macaulay rings and is focused on the situation that $A$ is a field. To see that my claim about "any $A$" in the previous paragraph follows, one shows that if for a given $G$, the described structure obtains for $A$ any field, then it also obtains with $A=\mathbb{Z}$ -- this is supposedly well-known, but "just in case", it is written down carefully in section 2.4.1 of my thesis -- and then one notes that a free module basis of $\mathbb{Z}[\mathbf{t}]^G$ over the subring generated by the elementary symmetric polynomials will also be a free module basis of $A[\mathbf{t}]^G$, just by base changing to $A$. See this MSE question for why the base change doesn't mess anything up.) (e) As lisyarus stated, the special case of $G=A_n$ is well-understood: the invariant ring is generated by the elementary symmetric polynomials and the Vandermonde determinant. Actually this requires the hypothesis that $2$ is a unit in $A$, as you note in comments. If $2$ is not a unit in $A$, one can still generate the invariant ring with the elementary symmetric polynomials and the sum of the positive terms in the Vandermonde determinant (or, the sum of the negative terms). Certain other cases, e.g. $D_4$, also have explicit descriptions coming from Galois theory. The classical material usually assumes $A$ is a field, but see sections 5.4 and 5.5 in Owen Biesel's thesis for $A_n$ and $D_4$; Biesel is working over general $A$. [The references to "lisyarus" and "you" in this final paragraph make sense in the context of the original post on math.SE.]<|endoftext|> TITLE: Inequality involving tensor product of orthonormal unit vectors QUESTION [8 upvotes]: Let $e_1,...,e_r$ be the first $r$ standard basis of $\mathbb{R}^n, r0$ (for example $0<\delta<\frac{1}{rn}$), then there exists a set $T\subset\{1,2,...,n\}$ with $|T| = r$ satisfying $$1-C\delta\leq\sum_{i\in T}\left|\langle\Lambda,u_i\otimes u_i\rangle\right|^2$$ The conjecture can be verified when $r=1$. Note that $C$ cannot depend on $\delta$, which means the conjecture states that there is a $C>0$ to make the assertion true for all $\delta$ small enough. For example, $C$ can be taken as 2 when $r=1$. REPLY [13 votes]: Without loss of generality, $u_1,\dots,u_n$ can be taken to be the standard basis of ${\bf R}^n$ (so $e_1,\dots,e_r$ is just some arbitrary orthonormal system). We can view $\Lambda$ as a real symmetric $n \times n$ matrix of Frobenius norm $1$ and rank at most $r$. (Conversely, every such matrix has a representation of the desired form for some $e_1,\dots,e_r$ by the spectral theorem, so this reformulation has not lost any information.) If $D = \sum_{i=1}^n \langle \Lambda, u_i \otimes u_i \rangle u_i \otimes u_i$ is the diagonal component of $\Lambda$, the hypothesis is then $\|D\|^2 \geq 1-\delta$. Let $\lambda = \mathrm{diag}(\lambda_1,\dots,\lambda_r,0,\dots,0)$ be the diagonalisation of $\Lambda$ (the ordering of the eigenvalues is irrelevant), thus $\lambda$ is a unit vector supported on a set of $r$ indices in $\{1,\dots,n\}$. By the Schur-Horn theorem, $D$ is a convex combination of the permutations $\sigma(\lambda)$ of $\lambda$, $\sigma \in S_n$, where the symmetric group $S_n$ acts on diagonal matrices in the obvious manner, thus $$ D = \sum_{\sigma \in S_n} c_\sigma \sigma(\lambda)$$ for some non-negative coefficients $c_\sigma$ summing to $1$. (Again, the Schur-Horn theorem is an if-and-only-if statement, so we have still not lost any information so far.) Now we exploit the uniform convexity of the Frobenius norm. We take the norm square $$ 1-\delta \leq \|D\|^2 = \sum_{\sigma,\sigma' \in S_n} c_\sigma c_{\sigma'} \langle \sigma(\lambda), \sigma'(\lambda) \rangle$$ and then apply the cosine rule to conclude $$ \sum_{\sigma,\sigma' \in S_n} c_\sigma c_{\sigma'} \| \sigma(\lambda) - \sigma'(\lambda)\|^2 \leq 2\delta$$ hence by pigeonholing there exists $\sigma_0 \in S_n$ such that $$ \sum_{\sigma \in S_n} c_\sigma \| \sigma(\lambda) - \sigma_0(\lambda)\|^2 \leq 2\delta$$ hence by Cauchy-Schwarz $$ \sum_{\sigma \in S_n} c_\sigma \| \sigma(\lambda) - \sigma_0(\lambda)\| \leq \sqrt{2\delta}$$ hence by the triangle inequality $$ \| \sum_{\sigma \in S_n} c_\sigma \sigma(\lambda) - \sigma_0(\lambda)\| \leq \sqrt{2\delta}$$ thus $$ \| D - \sigma_0(\lambda) \| \leq \sqrt{2\delta}.$$ The diagonal matrix $\sigma_0(\lambda)$ is supported on a set $T \subset \{1,\dots,n\}$ of cardinality $r$, and the above inequality implies that the Frobenius norm of $D$ outside of $T$ is at most $\sqrt{2\delta}$. Thus $$ \sum_{i \not \in T} |\langle \Lambda, u_i \otimes u_i \rangle|^2 \leq 2\delta$$ so that $$ \sum_{i \in T} |\langle \Lambda, u_i \otimes u_i \rangle|^2 \geq 1-3\delta$$ as required.<|endoftext|> TITLE: Bending surfaces in Riemannian manifolds QUESTION [6 upvotes]: Let $S$ be an immersed surface in $\mathbb{R}^3$ (with the flat metric). We will call it flexible if there exists a smooth (or whatever regular) family of immersions $s_t: S\to \mathbb{R}^3$, such that each $s_t$ induces the same metric on $S$ and no $s_{t_1}$ and $s_{t_2}$ are related by an isometry of $\mathbb{R}^3$ (i.e. we rule out trivial deformations by one-parameter subgroups of ambient isometries). 1) Do there exist flexible smooth closed surfaces in $\mathbb{R}^3$? In the polyhedral world the answer is yes, with a well-known example of flexible polyhedron by Connelly. Convex surfaces seem to be rigid, which should be a theorem of Alexandrov or Cauchy. Are there any other global criterions for rigidity/flexibility? 2) Is locally any surface flexible? 3) Can anybody give me an example of a rigid surface with boundary? REPLY [10 votes]: The standard reference, where the state of the art concerning all of your questions is found: Q. Han, J.-X. Hong, Isometric Embedding of Riemannian Manifolds in Euclidean Spaces, Amer. Math. Soc., Providence, R. I., Math. Surveys and Monographs, vol. 130, 2006. There is still no proof that every compact smooth embedded surface in 3-dimensional Euclidean space is rigid. Michael T. Anderson recently produced a proof, but a flaw emerged. The local flexibility of any smooth surface with nonzero or nonnegative Gauss curvature is, I believe, a straightforward consequence of the results in Han and Hong, although they don't state it explicitly. Han and Hong prove (theorem 8.1.2) that any smooth closed surface, with nonnegative Gauss curvature vanishing on a nowhere dense set, is rigid. Another very important work, Open problems in geometry of curves and surfaces (and a very nice one to read) on these questions is the unpublished manuscript of Mohammed Ghomi, which sums up many of the open questions on curves and surfaces in Euclidean space, and particularly the flexibility questions, with a thoughtful collection of references. Rigidity of the top half of a torus is discussed here, but perhaps without a complete proof having become clear yet.<|endoftext|> TITLE: Is $B\mathbb{G}_m$ strongly $A^1$-invariant? QUESTION [5 upvotes]: I have just seen the definition of strongly ${A}_1$ invariance: A sheaf of group $G$ is strongly $A_1$ invariance , if both $H^0(-;G)$ and $H^1(-;G)$ is $A_1$-invariant. I haven't got too much about the point of this definition and just try to think about some examples. Can someone give me some simple examples? Is $B\mathbb{G}_m$ strongly $A_1$-invariant where $\mathbb{G}_m$ is the multiplicative group? REPLY [6 votes]: The point of the notion is that for a strongly $\mathbb{A}^1$-invariant sheaf of groups $G$ the classifying space $BG$ is $\mathbb{A}^1$-local. The characterization in terms of $H^0$ and $H^1$ is equivalent to that since $H^0(-,G)$ and $H^1(-,G)$ are the homotopy presheaves $\pi_1$ and $\pi_0$ of $BG$ in the homotopy theory of simplicial Nisnevich sheaves, with all other homotopy presheaves trivial. The sheaf $\mathbb{G}_m$ is a simple example of a strongly $\mathbb{A}^1$-invariant sheaf of groups (since $\mathbb{G}_m$ and ${\rm Pic}$ are $\mathbb{A}^1$-invariant, assuming we are over a regular base). Over a field, other examples of strongly $\mathbb{A}^1$-invariant sheaves are the sheaves $\mathbf{K}^{\rm M}_n$ of Milnor K-groups. The space $B\mathbb{G}_m$ in the question isn't a sheaf of groups, so strictly speaking the notion doesn't apply. Of course, the tautological examples are the $\mathbb{A}^1$-fundamental group sheaves, by the theory in Morel's book "$\mathbb{A}^1$-algebraic topology over a field". More examples can be found there. Section 7.3 of the book also contains an example of a strongly $\mathbb{A}^1$-invariant sheaf of groups which is not abelian, namely $\pi_1^{\mathbb{A}^1}(\mathbb{P}^1)$. Most of the examples of strongly $\mathbb{A}^1$-invariant sheaves are actually abelian, but that's mostly due to our lack of knowledge of how to compute $\mathbb{A}^1$-fundamental groups of things.<|endoftext|> TITLE: Regular subsets of $\text{PSL}(2, q)$ QUESTION [10 upvotes]: Suppose a group $G$ acts on a set $\Omega$. Call a subset $A \subset G$ regular (or sharply transitive or simply transitive or...) if for every two points $\omega_1, \omega_2 \in \Omega$ there is a unique $a \in A$ such that $\omega_1^a = \omega_2$. This generalizes the concept of a regular subgroup to subsets. Consider the action of $G = \def\PSL{{\text{PSL}}}\def\F{{\mathbf{F}}}\def\projline{{\mathbf{P}^1 \F_q}}\PSL(2, q)$ on the projective line $\Omega = \projline$. It is not hard to prove that $G$ has a regular subgroup if and only if $q \not\equiv 1 \pmod 4$. Indeed if $q$ is even then there is a cyclic regular subgroup, and if $q \equiv 3 \pmod 4$ then there is a dihedral regular subgroup, while if $q \equiv 1 \pmod 4$ then any subgroup of order $q+1$ must contain an element of order $2$, but every element of order $2$ has eigenvalues $\pm i$ and hence two fixed points. But could $G$ yet have a regular subset? Does $G = \PSL(2, q)$ for $ q \equiv 1 \pmod 4$ acting on $\projline$ have a regular subset? The answer is no for $q = 5, 9, 13, 17$, by direct computation. REPLY [4 votes]: I have come across some references for this problem, which amount to a solution very different from the (beautiful) one given by Peter Mueller, and which also goes further. It turns out that it is possible to show directly that a sharply transitive subset of $\mathrm{PGL}_2(q)$ acting on $\mathbf{P}^1$ must be a coset of a subgroup. Since the subgroups of $\mathrm{PGL}_2(q)$ are classified, so are regular subsets. This result is due to Bader, Lunardon, and Thas, who classified flocks of the hyperbolic quadric $\mathcal{Q} \subset \mathbf{P}^3$. The solution was given in the following two papers. Thas, J. A., Flocks, maximal exterior sets, and inversive planes, Finite geometries and combinatorial designs, Proc. AMS Spec. Sess., Lincoln/NE (USA) 1987, Contemp. Math. 111, 187-218 (1990). ZBL0728.51010. Bader, Laura; Lunardon, Guglielmo, On the flocks of (Q^+(3,q)), Geom. Dedicata 29, No. 2, 177-183 (1989). ZBL0673.51010. On the face of it the problem these authors study is different, but it is equivalent. A flock is a partition of a quadric into conics. The quadric $\mathcal{Q}$ can be taken without loss of generality to be the subset $\det A = 0$ of the space $M_2(K) / Z$ of $2\times2$ matrices up to scalars. Each conic can be identified with the plane it spans and hence by duality with a point of $\mathbf{P}^3 = M_2(K)/Z$. The condition that the planes intersect $\mathcal{Q}$ in disjoint subsets is equivalent to saying that the the corresponding points define a regular subset (or maximal exterior set, as they call it). The equivalence is explicit in the introduction here: Bonisoli, Arrigo; Korchmáros, Gábor, Flocks of hyperbolic quadrics and linear groups containing homologies, Geom. Dedicata 42, No. 3, 295-309 (1992). ZBL0756.51009. I took some time to boil off all the geometry, and I wrote up what was left here: https://arxiv.org/abs/2105.02760<|endoftext|> TITLE: Sums of binomial coefficients weighted by incomplete gamma QUESTION [9 upvotes]: I am interested in proving that $$\sum_{k=0}^n\frac{k}{k!}\sum_{l=0}^{n-k}\frac{(-1)^l}{l!}=1 $$ and $$\sum_{k=0}^n\frac{k^2}{k!}\sum_{l=0}^{n-k}\frac{(-1)^l}{l!}=2. $$ I verified it numerically for many values of $n$ and would like now to get to a proof. I rewrote it to make incomplete Gamma functions appear, but it is (apparently) not leading me further. Has somebody ever encountered such an identity ? Is there an easy proof I am just completely missing ? I apologize if the question is trivial, I may lack some combinatorics tools. Note: It was edited and should now be correct REPLY [2 votes]: A combinatorial proof may not be easy, or even considered proper, but it could shed some light on the meaning of these formulae. The sum with alternating signs resembles the definition of derangements: \begin{equation*} d(n) = n! \sum_{i=0}^n \frac{(-1)^i}{i!}. \end{equation*} This is the number of permutations where no element appears in its original position. It is straightforward to rewrite the first equation as \begin{equation*} \sum {n \choose k} \cdot k \cdot d(n-k) = n! \label{r1}\tag{r1} \end{equation*} The role of the factor $k$ is not obvious, so assume for a moment that it is not there (!), and focus on: \begin{equation*} \sum {n \choose k} \cdot d(n-k) = n! \label{r2}\tag{r2} \end{equation*} This new equation allows for a fairly simple combinatorial proof: On the one hand we are counting in how many ways a class of $n$ students can be rearranged, and the result is $n!$ by definition of permutation. On the other hand, we observe that in any given permutation each element can either stay in its original position or not. Let's say that $k$ students keep their place: there are ${n \choose k}$ ways to select those, and $d(n-k)$ ways to scramble the others. By summing over $k$ we cover all permutations without overlap. We found a partition of permutations, and this can be a starting point to attack the original problem. We can prove (\ref{r1}) with a slight variation of the above: We count in how many ways a class of $n$ students can be rearranged, given that one student at random will receive a red hat and will not change place. There are $n$ ways to extract who will receive the red had, and $(n-1)!$ ways to rearrange the others, for a total of $n!$. On the other hand, we select $k$ students that will not change place, then pick at random one of those and give her the hat, finally scramble the rest. The product of the number of way for these three actions gives one addend in LHS of (\ref{r1}). The partition of the permutations in (\ref{r2}) is maybe surprising on its own, but together with (\ref{r1}) is a kind of magic: apparently we can add or drop a factor $k$ with no effect on the result. This until we consider that the sums distribute weights differently. A striking difference is for $k=0$, where $d(n-k)$ is maximum in (\ref{r2}) but nullified in (\ref{r1}). We might now want to go further and attempt the second equation. This can be rewritten as \begin{equation*} \sum {n \choose k} \cdot k^2 \cdot d(n-k) = 2 \cdot n! \label{r3}\tag{r3} \end{equation*} The same reasoning can be adapted: Each student has $2$ lottery tickets, and each can get with the same probability and independently a red hat and a blue scarf. Each winner will stay in its original position, while the others are allowed to move. In how many ways we can do that? There are two mutually exclusive cases, either we have only one winner for both prizes, $n$ ways, and that others are rearranged, $(n-1)!$ ways; or we have two winners, $n\cdot (n-1)$ ways, and the others are rearranged, $(n-2)!$ ways. Two cases but same results, so $2 \cdot n!$. On the other hand, the winners belong to the $k$ students that will not change place. We pick $k$ students who do not move, ${n \choose k}$ ways, assign two prizes independently, $k^2$ ways, and derange the others, $d(n-k)$ ways. This is quite remarkable, and would be a pity to stop here. What if we added more prizes, a pair of green gloves, a yellow whistle, etc. for a total of $m$ gifts? The reasoning above can be applied in a straightforward way to the LHS of (\ref{r3}), simply changing $k^2$ with $k^m$. A bit more work is needed for the RHS, which so far was the easiest. The number of ways to group $m$ prizes is $B(m)$, the $m$-th Bell number. Now a final treat. Assume the prizes are distributed to $r$ winners. We can select the winners in $n\cdot(n-1)\cdot\ldots\cdot(n-r+1)$ ways and rearrange the others in $(n-r)!$ ways. It is very convenient, because we always have $n!$ for each of the possible grouping of prizes regardless of the number of winners. This concludes the combinatorial interpretation of \begin{equation*} \sum {n \choose k} \cdot k^m \cdot d(n-k) = B(m) \cdot n! \label{r4}\tag{r4} \end{equation*}<|endoftext|> TITLE: CG spaces from the perspective of sheaves over compact Hausdorff spaces QUESTION [7 upvotes]: A compactly generated space is a space $X$ such that $f : X \rightarrow Y$ is continuous if and only if $K \rightarrow X \stackrel{f}{\rightarrow} Y$ is continuous for each compact hausdorff space $K$. Let $I$ be the category of compact hausdorff topological spaces with continuous maps. I think being compactly generated is equivalent to the canonical map $\epsilon_X : \int_{K \in I} [K, X]_{\text{Top}} \times K \rightarrow X$ being an isomorphism. $\epsilon_X$ is the counit of an adjunction between functors $L : [ I^{op}, \text{Set}]_{\text{Cat}} \rightarrow \text{Top}$ sending $S$ to $\int_{K \in I} S(K) \times K$ and $R : \text{Top} \rightarrow [ I^{op}, \text{Set}]_{\text{Cat}}$ sending a space $X$ to the presheaf $S : I^{op} \rightarrow \text{Set}$ sending $K$ to $[K, X ]_{\text{Top}}$. $L$ is the left Kan extension of the inclusion $\iota : I \rightarrow \text{Top}$ along the yoneda embedding $Y : I \rightarrow [I^{op}, \text{Set}]_{\text{Cat}}$, as in the diagram below: That $L$ is left adjoint to $R$ is a famous theorem, which was discussed here. If I am not mistaken, this adjunction is related to Peter Scholze and Dustin Clausen's condensed mathematics. Surprisingly, there is no information lost if we take instead of $I$ the smaller category of profinite sets. There is a fully faithful functor from compactly generated spaces to condensed sets (sheaves on the étale site of a point) that arises in this way. My question is, has anyone taken this perspective to get a streamlined, categorical proof that the category of compactly generated spaces is cartesian closed? We might start by observing that the category of presheaves on $I$ has all small limits and colimits, $L$ preserves colimits, and $R$ preserves limits. Might this allow us to carry over the cartesian closed-ness of $[I^{op}, \text{Set}]_{\text{Cat}}$ to the category of compactly generated spaces? REPLY [3 votes]: I found an answer using Ivan Di Liberti's very helpful reference in the comments. I figured I would post it here in case anyone has interest in this approach to compactly generated spaces in the future. For a cardinal $\kappa$, let $I_{\kappa}$ be the category of compact hausdorff spaces with cardinality less than or equal to $\kappa$. The cardinal is to deal with size issues. We make $I_{\kappa}$ into a site where covers are jointly surjective collections of maps. $I_{\kappa}$ induces an adjunction between $\text{Sh}( I_{\kappa}^{op})$, the category of sheaves of sets on $I_{\kappa}^{op}$, and $\text{Top}$. The left adjoint $L$ sends $F$ to $\int_{K \in I_{\kappa}} F(K) \times K$, and the right adjoint $R$ sends $X$ to the sheaf sending $K$ to $[K, X]_{\text{Top}}$. $L$ is a left Kan extension of $I_{\kappa} \rightarrow \text{Top}$ along the inclusion $I_{\kappa} \rightarrow \text{Sh}( I_{\kappa}^{op})$. This adjunction between $L$ and $R$ is idempotent; it factors as $\text{Sh}( I_{\kappa}^{op}) \leftrightarrow \text{CG}_{\kappa} \leftrightarrow \text{Top}$, where $\text{CG}$ is what I call $\kappa$-compactly generated spaces. $\text{CG}_{\kappa}$ is a reflexive subcategory of $\text{Sh}( I_{\kappa}^{op})$ and a coreflexive subcategory of $\text{Top}$. What Ivan Di Liberti realized is that this may allow us to apply proposition 4.3.1 in The Elephant. Let $L' : \text{Sh}( I_{\kappa}^{op}) \leftrightarrow \text{CG} : R'$ be the reflexive adjunction. To show that $\text{CG}_{\kappa}$ inherits a cartesian closed structure from $\text{Sh}(I_{\kappa}^{op})$, it suffices to show that $L'$ preserves products. $L'$ sends a representable sheaf $[-, K]$ to $K$. It preserves products of compact hausdorff spaces; the canonical map $L'([-, K] \times [-, K']) \rightarrow L'([-, K]) \times L'([-, K'])$ is an isomorphism. To see this, note that it is a bijection (one can show that it is an isomorphism after composing with the forgetful functor to set, and this can be made very canonical). Now the famous lemma about a continuous map from a compact space to a hausdorff space being open finishes the job. This result can be extended to work for the general case. $\text{Sh}(I^{op}_{\kappa})$ is a sheaf topos. It is therefore locally presentable (see Borceux, Handbook of Categorical Algebra, prop. 3.4.16, page 220). So there is a small set of $\lambda$-small objects which generates this category under $\lambda$-filtered colimits. Every sheaf is a $\lambda$-filtered colimit of representables here. $L'$ and products commute with $\lambda$-filtered colimits, reducing to the previous paragraph. Now it seems that, if we take the large colimits $\text{colimit } \text{Sh}(I_{\kappa}^{op})$ and $\text{colimit } \text{CG}_{\kappa}$, I think we obtain compactly generated spaces, along with the desired product preserving reflexive adjunction.<|endoftext|> TITLE: Cusp forms have an orthonormal basis of eigenfunctions for all Hecke operators QUESTION [6 upvotes]: I am reading Langlands' pape Euler Products and have a few questions. Let $G$ be a split adjoint semisimple group over $\mathbb Q$. If $p$ is a place of $\mathbb Q$, finite or infinite, let $G_{\mathbb Z_p}$ be a maximal compact subgroup of $G_{\mathbb Q_p}$. Let $K = \prod\limits_p G_{\mathbb Z_p} \subset G_{\mathbb A}$, and let $L$ be the Hilbert space of square integrable functions on $G_{\mathbb Q} \backslash G_{\mathbb A}$ which are right invariant under $K$. Let $L_0 \subset L$ be the subspace of cusp forms (whose definition as given by Langlands does not quite make sense to me). Langlands writes: I have a few questions about this. I would greatly appreciate any explanation or references. Is $H_p$ always commutative? If $p$ is finite, there is an injection of the "spherical Hecke algebra" $C_c^{\infty}(G_{\mathbb Q_p}, G_{\mathbb Z_p})$, the space of locally constant, compactly supported functions on $G_{\mathbb Q_p}$ which are left and right invariant under $G_{\mathbb Z_p}$, into $H_p$, where a function $f \in C_c^{\infty}(G_{\mathbb Q_p}, G_{\mathbb Z_p})$ is associated with the measure $\mu_f \in H_p$ on $G_{\mathbb Q_p}$ defined by $$\mu_f(E) = \int\limits_{E} f(x) d\mu_{\textrm{Haar}}(x)$$Is $f \mapsto \mu_f$ an isomorphism of the spherical Hecke algebra onto $H_p$? Why in the $p$-adic case are all measures in $H_p$ absolutely continuous with respect to Haar measure? Why is there a countable orthonormal basis of $L_0$ consisting of eigenfunctions for all operators $\lambda(\mu)$ over all $\mu \in H_p$ and all $p$? Is this some sort of version of spectral theorem? REPLY [10 votes]: These things are not trivial at all, but by the time Langlands was writing "Euler products" they were known, and quite familiar to many people at Princeton and Yale, even if not so many other places. $H_p$ is certainly mostly commutative. As commented by @GTA, at least for classical groups the Gelfand criterion is easy to verify (from a Cartan decomposition, both $p$-adic and archimedean). I do not know whether there is an intrinsic proof that treats, for example, Galois twists of exceptional groups. About the continuity of that class of measures with respect to Haar measure: the left-and-right (or even one-sided) $G_{\mathbb Z}$-invariance, together with compact support, implies (by some abstract uniqueness-of-invariant-distributions result) that the functional is a finite linear combination of (Haar) integrals over cosets of $G_{\mathbb Z}$. So, yes, barring some technicalities, your map is an isomorphism of Hecke algebras. (At archimedean places, similar things can be said, but since there are derivatives, one cannot purely echo what is true for $p$-adic places.) The existence of an orthonormal basis of spherical-Hecke eigenvectors for the space of cuspforms is highly non-trivial. True, for holomorphic cuspforms, the finite-dimensionality makes things elementary at a given level, for good primes. The general argument (with or without level one, etc.) does indeed use the spectral theorem for families of compact operators closed under adjoints. The difficult piece is proving that the integral operators attached to test functions are compact on cuspforms. The case of compact $\Gamma\backslash G$ is often treated in introductory sources, since the compactness of integral operators follows from their being Hilbert-Schmidt, and proving the latter is essentially elementary. The compactness of integral operators on cuspforms on non-compact quotients is a much bigger production. One sort of argument was sketched in Lax-Phillips "Scattering theory for automorphic forms" (Princeton orange series), although a scrupulous reader will notice many analytical details needing to be filled-in. They treated $SL_2(\mathbb Z)$, proving that spaces of pseudo-cuspforms decompose discretely for (a certain self-adjoint extension of a restriction of) the invariant Laplacian. The ideas admit generalization. Another argument for compactness of suitable integral operators on cuspforms was given by R. Godement in the Boulder Conference from 1965/6, and is visible in the AMS Proc. Symp. Pure Math IX. It, too, has pretty steep functional analysis prerequisites, which may be not obvious to a casual number-theorist reader. Both types of argument are illustrated in several examples in my recent books, which, I hasten to point out, are legally available in a PDF http://www.math.umn.edu/~garrett/m/v/current_version.pdf at my website, published by Cambridge Univ. Press ("Modern Analysis of Automorphic Forms, by Example"). Although the analytical backdrop may not be of great interest to all number theorists, I hope that at least the demonstrable existence of (perhaps tedious or uninteresting) proofs will be comforting. :) Seriously, many of these details plagued me for decades... (Also, Gelfand's criterion is treated in my books, and several other of these apocryphal things...)<|endoftext|> TITLE: Faithfully flat descent for modules from the simplicial point of view QUESTION [7 upvotes]: Let $R \rightarrow R'$ be a faithfully flat ring map, let $M$ be an $R$-module, and let $M_n$ be the base change of $M$ to the tensor product of $n + 1$ copies of $R'$ over $R$. One way to formulate the statement of faithfully flat descent is as the claim that the complex $$ 0 \rightarrow M \rightarrow M_0 \rightarrow M_1 \rightarrow \dotsc$$ is exact, where the maps are the Cech style alternating sums of pullbacks. This is a fairly common way that is discussed further here https://stacks.math.columbia.edu/tag/023M. Another, probably less common, way to formulate faithfully flat descent is to say that $$ M \xrightarrow{\sim} Rlim_{\Delta} M_n$$ where $\Delta$ is the usual simplex category that is used to define simplicial objects. My question is: why are these two formulations equivalent? How does one see that one implies the other? I suspect that one probably uses the Dold-Kan equivalence and then some "standard" but possibly difficult to locate in the literature simplicial calculus. I am guessing that the core of my question is not really about descent for modules, but rather about simplicial or bisimplicial objects and Rlim. I would very much appreciate if someone could explain the equivalence between the two points of view in detail (or give a precise reference to the literature where this is treated). REPLY [6 votes]: Essentially, your question is: for any co-simplicial abelian group $M_{\bullet}$, why is ${\rm R lim}_{\Delta} M_{\bullet}$ given by the complex $$C:= M_0 \to M_1 \to M_2 \to \dots ?$$ (If $A$ is an abelian group $A \to C$ is a quasi-isomorphism if and only if $M \to M_0 \to M_1 \to M_2 \to \dots $ is acyclic). There is a direct lowbrow proof, given by resolving the constant cosimplicial abelian group $\mathbb Z *$ given by $\mathbb Z *_n = \mathbb Z$ and all maps the identity. Since we have a natural isomorphism $${\rm Hom}_{\Delta}(\mathbb Z *, M_{\bullet}) \cong {\rm lim}_{\Delta} M_{\bullet}$$ there is a quasi-isomorphism ${\rm RHom}_{\Delta}(\mathbb Z *, M_{\bullet}) \simeq {\rm R lim}_{\Delta} M_{\bullet}$. To compute it, we resolve $\mathbb Z *$ by free (or if you prefer, representable) co-simplicial abelian groups. The resolution looks like $$\mathbb Z \Delta(0,-) \leftarrow \mathbb Z \Delta (1, -) \leftarrow \mathbb Z \Delta(2, -) \leftarrow \mathbb Z \Delta(3,-) \leftarrow \dots $$ Here, the number $n$ represents the totally ordered set $\{0, \dots, n\}$ and the cosimplicial abelian group $\mathbb Z \Delta(n, -)$ is given by $\mathbb Z \Delta(n, -)_m = \mathbb Z \Delta(n, m)$. The differential $\mathbb Z \Delta(n, -) \to \mathbb Z \Delta(n-1, -)$ is given by the alternating sum of precomposition by the $n+1$ face maps in $\Delta(n-1, n)$. The complex resolves $\mathbb Z *$, because if we look at level $n$, we see that we get exactly the simplicial homology complex for the $n$-simplex! $$\mathbb Z \Delta(0,n) \leftarrow \mathbb Z \Delta(1,n) \leftarrow \dots$$ Since we know the homology of the simplex, we know that the complex is a resolution. Now by the Yoneda lemma, we have that ${\rm Hom}_{\Delta}( \mathbb Z \Delta(n,-), M) = M_n$. In particular, this resolution is projective. Using it to compute ${\rm RHom}(\mathbb Z *, M)$ we get the complex $${\rm Hom}_{\Delta}( \mathbb Z \Delta(0,-), M) \to {\rm Hom}_{\Delta}( \mathbb Z \Delta(1,-), M) \to {\rm Hom}_{\Delta}( \mathbb Z \Delta(2,-), M) \to \dots, $$ and applying Yoneda, this is exactly the complex $C$ we wanted.<|endoftext|> TITLE: The space of cusp forms for $\mathrm{GL}_2$ over ${\mathbf{F}}_q(T)$ QUESTION [16 upvotes]: This question is about automorphic forms for the group $\mathrm{GL}_2$, over a rational function field. Let's say $\mathbf{F}_q$ is a finite field, and $X=\mathbf{P}^1_{\mathbf{F}_q}$ is the projective line, with function field $K=\mathbf{F}_q(T)$. For an effective divisor $D\subset X$ we have the space $S(\Gamma_0(D))$ of cusp forms with $\Gamma_0(D)$ structure. By this I mean: complex-valued smooth functions on $\mathrm{GL}_2(\mathbf{A}_K)$, which are left $\mathrm{GL}_2(K)$-invariant, which have trivial central character, which are right-invariant under the appropriate compact open subgroup (integral matrices which are upper-triangular modulo $D$), and which satisfy the cuspidality condition. This space is finite-dimensional. What is the dimension of $S(\Gamma_0(D))$? I'm happy to have an answer in the case that $D$ is a sum of distinct degree 1 points. As I understand it, there should be no cusp forms when $D$ is empty. I expect to get cusp forms when $D$ has degree at least 4, because then there are non-constant elliptic curves over $\mathbf{F}_q(T)$ with four places of multiplicative reduction; then Langlands' theory predicts cusp forms of this level. What if $D$ has degree $\leq 3$? REPLY [16 votes]: If $D$ has degree $\leq 3$ there won't be any cusp forms. By the Langlands correspondence these correspond to irreducible Galois representations into $GL_2$, unramified away from a degree $3$ divisor, with unipotent local monodromy at $3$ points. Such representations would have Euler characteristic $1$, contradicting the claim that they are irreducible. One can do an automorphic version of this argument - the conductor would be $q^3$, so the constant of the functional equation of the standard $L$-function would make it a rational function of degree $-1$, contradicting the fact that it must be polynomial. However, one can also check this by hand from the definition of an automorphic form. I did this in the language I am most comfortable with (vector bundles), rather than adelically. I worked out once how to think concretely about automorphic forms over function fields. I don't know if this makes sense to anyone else, but I put it below the line. Before giving this argument, let me note that, with four points, there's many more examples than elliptic curves. There's $q+1 -O(1)$ and this was counted in the paper of Deligne and Flicker. This uses automorphic methods (trace formula). The associated Langlands parameters are almost totally opaque, except for the information we can glean about them from Langlands and the few special cases that arise from elliptic curves. Note that, for most sets of four points, there's no elliptic curves at all - only four special sets of four points do the trick. We want to check that any nontrivial function on the set of isomorphism classes of rank two vector bundles on $\mathbb P^1$ with a chosen line-sub-bundle over $D$ has some nonzero constant term map. The constant term map will be associated to the data of two line bundles $L_1$ and $L_2$ and a divisor $D'$ between $0$ and $D$, and will equal the sum over your function over all elements of $\operatorname{Ext}^1 (L_1, L_2)$ plus a choice of a sub-line-bundle of the induced extension $0 \to L_2 \to V \to L_1$ over $D'$ that doesn't intersect $L_2$ (and then take the sub-line bundle $L_2$ over $D-D'$). The group parameterizing pairs of such choices is $\operatorname{Ext}^1 (L_1, L_2(-D'))= H^1 ( \mathbb P^1, L_2 (-D) \otimes L_1^{-1} )$. If $\deg L_2 - \deg L_1 - \deg D' > -2$, this ext group vanishes, and so the sum is over a single vector bundle with extra data, and so if the constant term map vanishes, the function vanishes on all vector bundles arising this way. In particular, if the vector bundle with extra data splits into a sum of two line bundles, then the total degree from ordering the line bundles in both ways is $\deg L_2 - \deg L_1 + \deg L_1 - \deg L_2 - \deg D = -3$ so one must be at least $-1$ so any function with zero constant terms vanishes on such data. It then follows that if we have $\deg L_2 - \deg L_1 - \deg D' = -2, because the ext group is one-dimensional, the sum is over a single vector bundle with extra data plus a single split one, so if all constant term vanishes, the function vanishes on every vector bundle with extra data arising in this way as well. But it is easy to see that one can always choose a sub-line-bundle $L_2$ to make this at least $-2$: If $V = \mathcal O(a) + \mathcal O(b)$, with $a TITLE: Does using continued fractions work to give a homeomorphism $\mathbb{Q}^+ \rightarrow (\mathbb{Q}^+)^2$? QUESTION [5 upvotes]: Let $\mathbb{Q}$ be the topological space of rational numbers (with topology induced by inclusion in the real line) and let $\mathbb{Q}^+$ be the set of positive ($x>0$) rationals. I'm looking for a simple construction of a homeomorphism $\phi: \mathbb{Q} \rightarrow \mathbb{Q}^2$ (not using an abstract result). In an early post, it was suggested to use continued fractions, but this has problems with negative numbers. My idea is we can first make a homeomorphism $f: \mathbb{Q}^+ \rightarrow \mathbb{Q}$ via $f(x) = (x-1)^3/x$ and then we can just worry about $\mathbb{Q}^+$. Then the homeomorphism $\phi: \mathbb{Q}^+ \rightarrow (\mathbb{Q}^+)^2$ would just go like $[a_0, a_1, ..., a_n] \rightarrow ([a_0, a_2, ...], [a_1, a_3, ...])$ this seems to be obviously bijective, and I believe it is bicontinous just based on the idea that two continued fractions are close if and only if enough of their initial terms agree (which sounds plausible). Does this make sense? Or am I missing something? Thanks for your help. REPLY [2 votes]: Here is one way to form a (not very) explicit homeomorphism. As has been mentioned, the function $\phi=(\phi_0,\phi_1)$ from the question applied to the set $I^{>1}$ of irrationals greater than $1$ gives you a homeomorphism $I^{>1}\to I^{>1}\times I^{>1}$. So we will be done if we find a homeomorphism from $\mathbb{Q}$ to a countable subset $A\subseteq I^{>1}$ such that $x\in A$ iff $(\phi_0,\phi_1)(x)\in A\times A$ (equivalently, $A=\phi_0(A)=\phi_1(A)$). We can achieve that if our set $A$ is dense in $I^{>1}$ and we have an enumeration of $A$: then we can follow the proof of Cantor's isomorphism theorem to get an explicit order isomorphism between $A$ and $\mathbb{Q}$ (and thus a homeomorphism, as both sets have their order topologies). For example we can take $A=\mathbb{I}^{>1}\cap (\mathbb{Q}+\mathbb{Q}\sqrt{2})$, the set of irrationals $>1$ with periodic continued fraction. P.S. I know almost nothing about continuous fractions, but this could be made more explicit if there was a set $A$ as above and such that we know a homeomorphism from $\mathbb{Q}$ to $A$ without using Cantor's isomorphism theorem. We don't even need the condition $A=\phi_0(A)=\phi_1(A)$ if we know enumerations of $A,\phi_0(A)$ and $\phi_1(A)$ and explicit homeomorphisms of them with $\mathbb{Q}$.<|endoftext|> TITLE: Simple vector bundle isomorphic to one of its twistings QUESTION [5 upvotes]: Let $V$ be a vector bundle over an algebraic curve $C$, ad assume that that $V \cong V \otimes L$ for some line bundle $L$. If $V$ is decomposable this is clearly possible, for example take $V \cong \mathcal{O}_C \oplus L$ with $L^2 \cong \mathcal{O}_C$. Q. Does $V \cong V \otimes L$ imply $L \cong \mathcal{O}_C$ if $V$ is simple? REPLY [5 votes]: The answer is no, as shown by the following classical example due to Atiyah. Take an elliptic curve $E$, choose a point $p \in E$ and consider the unique non-split extension $$0 \longrightarrow \mathcal{O}_E \longrightarrow V_p \longrightarrow \mathcal{O}_E(p) \longrightarrow 0.$$ Then $V_p \simeq V_p \otimes L$ for every $2$-torsion line bundle $L$ on $E$, see Remark p. 35 of [F]. Note that $V_p$ is stable and thus simple, see [F], Theorem 9 p. 89. References. [F] R. Friedman: Algebraic surfaces and holomorphic vector bundles, Springer 1998.<|endoftext|> TITLE: Number of matrices with bounded products of rows and columns QUESTION [11 upvotes]: Fix an integer $d \geq 2$ and for every real number $x$ let $M_d(x)$ be number of $d \times d$ matrices $(a_{ij})$ satisfying: every $a_{ij}$ is a positive integer, the product of every row does not exceed $x$, and the product of every column does not exceed $x$. I'm looking for a good upper bound for $M_d(x)$ as $x \to +\infty$. If we forget about the condition on the columns, since it is well known that the number of $d$-tuples $(b_1, \dots, b_d)$ of positive integers satisfying $b_1 \cdots b_d \leq x$ is $\ll_d x (\log x)^{d - 1}$ (a generalization of Dirichlet divisor problem), we get the upper bound $$M_d(x) \ll_d x^d (\log x)^{d(d-1)}.$$ In the special case $d = 2$, we have that $a_{12}, a_{21} \leq \min(x / a_{11}, x / a_{22})$ and consequently $$M_2(x) \leq \sum_{a_{11}, a_{22} \leq x} \min\left(\frac{x}{a_{11}}, \frac{x}{a_{22}}\right)^2 \ll \sum_{a_{11} \leq a_{22} \leq x} \left(\frac{x}{a_{22}}\right)^2 \ll x^2 \log x ,$$ which is a better upper bound than the general one given in the above paragraph. However, I have no idea of how to generalize this trick to $d \geq 3$ (if possible). Has this problem been studied before? Do you have any idea/suggestion about it? REPLY [14 votes]: This problem was considered in passing in the proof of Theorem 4.1 in Granville and Soundararajan, see the argument starting at the bottom of page 17. They show (in your notation) that $M_d(x)$ is of order $x^d (\log x)^{(d-1)^2}$. You should also look at work of Harper, Nikeghbali and Radziwill which shows an asymptotic formula for closely related objects (see Theorem 3 there). REPLY [6 votes]: Nice question. Some thoughts on lower bounds: for $d=2$, the order of magnitude $x^2\log x$ is correct—here is an argument giving such a lower bound. For any real numbers $U,V$ such that $UV=x$ (and say $U,V\ge2$), any $2\times 2$ matrix such that $a_{11},a_{22}\in\big(\frac U2,U\big]$ and $a_{21},a_{12}\in\big(\frac V2,V\big]$ is counted by $M_2(x)$, and the number of such matrices is $\gg U^2V^2 = x^2$. We can sum this dyadic-interval lower bound over $U=2^k$ for a suitable range of $k$ to obtain the lower bound $M_2(x) \gg x^2\log x$. A similar argument gives a lower bound for general $d\ge3$. If $U_1U_2\cdots U_d=x$, with the convention that $U_{d+j}=U_j$, then the matrices such that $a_{ij} \in \big( \frac{U_{i+j}}2, U_{i+j}\big]$ for all $1\le i,j\le d$ contribute $\gg U_1^d \cdots U_d^d = x^d$ to $M_d(x)$. We then sum over all $U_1=2^{k_1},\dots,U_d=2^{k_d}$ such that $k_1+\cdots+k_d\le(\log x)/\log 2$; the number of such $d$-tuples of positive integers is $\gg_d(\log x)^d$, giving the lower bound $M_d(x) \gg_d x^d(\log x)^d$. REPLY [6 votes]: Let $M_d(x_1,y_1,\dotsc,x_d,y_d)$ be the number of matrices $(a_{ij})$ with positive integer entries such that the product of the $i$-th row is at most $x_i$, and the product of the $i$-th column is at most $y_i$. I claim that $$M_d(x_1,y_1,\dotsc,x_d,y_d)\ll_d\sqrt{x_1y_1}\prod_{i=2}^d\sqrt{x_iy_i}\log(2x_iy_i)^{2i-3}.\tag{$\ast$}$$ In particular, this implies the bound in Lucia's response: $$M_d(x)=M_d(x,x,\dotsc,x,x)\ll_d x^d\log(2x)^{(d-1)^2}.$$ We can prove $(\ast)$ by induction on $d$. For $d=1$ the statement is clear (the empty product means $1$). So let us assume that $d\geq 2$, and $(\ast)$ holds with $d-1$ in place of $d$. For a matrix $(a_{ij})$ to be counted, let us abbreviate $$s_i:=a_{id},\qquad t_j:=a_{dj},\qquad a:=a_{dd}.$$ Then we see that \begin{align*}M_d(x_1,y_1,\dotsc,x_d,y_d) &\ll_d\sum_{a\leq\sqrt{x_dy_d}}\sum_{\substack{s_1\dots s_{d-1}\leq y_d/a\\t_1\dots t_{d-1}\leq x_d/a}}M_{d-1}\left(\frac{x_1}{s_1},\frac{y_1}{t_1},\dotsc,\frac{x_{d-1}}{s_{d-1}},\frac{y_{d-1}}{t_{d-1}}\right)\\ &\ll_d\sum_{a\leq\sqrt{x_dy_d}}\sum_{\substack{s_1\dots s_{d-1}\leq y_d/a\\t_1\dots t_{d-1}\leq x_d/a}}\sqrt{\frac{x_1y_1}{s_1t_1}}\prod_{i=2}^{d-1}\sqrt{\frac{x_iy_i}{s_it_i}}\log(2x_iy_i)^{2i-3}. \end{align*} On the right hand side, we observe that \begin{align*} \sum_{\substack{s_1\dots s_{d-1}\leq y_d/a\\t_1\dots t_{d-1}\leq x_d/a}}\frac{1}{\sqrt{s_1t_1\dots s_{d-1}t_{d-1}}} &=\left(\sum_{m\leq x_d/a}\frac{\tau_{d-1}(m)}{\sqrt{m}}\right) \left(\sum_{n\leq y_d/a}\frac{\tau_{d-1}(n)}{\sqrt{n}}\right)\\ &\ll_d\frac{\sqrt{x_dy_d}}{a}\log(2x_dy_d)^{2d-4}. \end{align*} Therefore, we can bound $M_d(x_1,y_1,\dotsc,x_d,y_d)$ as \begin{align*}M_d(\cdots) &\ll_d\sqrt{x_1y_1}\left(\prod_{i=2}^{d-1}\sqrt{x_iy_i}\log(2x_iy_i)^{2i-3}\right)\sum_{a\leq\sqrt{x_dy_d}}\frac{\sqrt{x_dy_d}}{a}\log(2x_dy_d)^{2d-4}\\ &\ll_d\sqrt{x_1y_1}\prod_{i=2}^d\sqrt{x_iy_i}\log(2x_iy_i)^{2i-3}. \end{align*} The proof of $(\ast)$ is complete.<|endoftext|> TITLE: Structure of the module of derivations on the space of Holomorphic functions QUESTION [8 upvotes]: Maybe this is well-known, maybe not. Let $\Omega\subset \mathbb{C}$ be connected open and non-empty. It can be shown that if $d\in\mathfrak{Der}(\mathcal{H}(\Omega))$ (i.e. $d$ is a derivation of the algebra $\mathcal{H}(\Omega)$) and is continuous for the topology of compact convergence then $d$ is of the form $d=\varphi(z)\frac{d}{dz}$. My questions are the following Can the continuity be withdrawn ? If yes, for which $\Omega$ ? Can $\Omega$ be replaced by a one dimensional complex manifold ? (in particular, for these manifolds, is there a principal derivation like $\frac{d}{dz}$ ?) In particular the third question is trivial in the case of a compact connected manifold (because $\mathcal{H}(\Omega)=\mathbb{C}1_{\Omega}$ by maximum principle.). REPLY [7 votes]: If you have a smooth manifold $M$, then you have a linear map $\mathfrak{X}(M) \to \text{Der}(C^{\infty}(M))$ from the space of smooth global vector fields to the space of derivations of the algebra of smooth global functions. It sends a vector field $X$ to the operator of differentiation of functions along $X$. It is a standard fact from the smooth manifold theory that this map is an isomorphism. Now, let $M$ be a complex manifold. You still have a similar map $\mathfrak{X}^{hol}(M) \to \text{Der}(\mathcal{O}(M))$ from global holomorphic vector fields to derivations of the algebra of global holomorphic functions. As I understand, your question is about the surjectivity of this map. First of all, as you mentioned, if $M$ is compact, then there are no nonzero derivations of the algebra of holomorphic functions, but there maybe plenty nonzero holomorphic vector fields, so the map is surjective but not injective in this case. Now, we need the following fact: If $M$ is Stein, then this map is an isomorphism. Here is a proof of a similar result for real-analytic manifolds (real-analytic vector fields are exactly derivations of the algebra of real-analytic functions) by David Speyer. He uses the fact that every real-analytic manifold can be properly embedded into a Euclidean space, but the same is true for Stein manifolds (see the Wikipedia article on Stein manifolds; in fact Stein manifolds are exactly those complex manifolds that admit a proper holomorphic embedding into $\mathbb{C}^n$ for $n$ sufficiently large). After that, as far as I see, the proof goes without any changes. Now, it is known that every non-compact connected Riemannian surface is Stein (see the same Wikipedia article for references). In particular, every domain in $\mathbb{C}$ is Stein. So the answers to your questions are: Yes. For any $\Omega$. For any Riemannian surface $X$, every derivation of $\mathcal{O}(X)$ comes from some holomorphic vector field. It comes from a unique vector field if and only if $X$ has no compact components. Remark. It is unknown to me if there are any examples of complex manifolds where our map is not surjective. If there is one (connected, WLOG), it must be of dimension greater than 1, non-Stein, and noncompact. Besides, I am interested if there are non-Stein complex manifolds with this map still bijective. By the way, one can prove that every open subset of $\mathbb{C}$ is Stein without using the deep theorem that noncompact connected Riemannian surfaces are Stein. If you want, you may look for a proof in this wonderful survey into the Levi problem by Harry J. Slatyer.<|endoftext|> TITLE: Subgroup generated by a subgroup and a conjugate of it QUESTION [6 upvotes]: Let $H\leq G$ be groups, and $a\in G$ so that $\langle H,a\rangle=G$. Does it follows that $\langle H\cup aHa^{-1}\rangle$ is a normal subgroup of $G$? My hope is that this is true, and my guess is that it is not. It might be easy. REPLY [4 votes]: Too long for a comment. The answer is no. Just take the free product $G=H*\langle a \rangle$, where $a\notin H$ is of infinite order. By definition, $H$ and $a$ generates $G$. Then $\langle H\cup aHa^{-1}\rangle $ is not normal. For example, letting $h\in H$, $g:=a^2ha^{-2}\notin \langle H\cup aHa^{-1}\rangle$. Indeed by contradiction, write $g=h_1...h_n$, where $h_k\in H$ or $h_k\in aHa^{-1}$. Then, the first letter needs to be an $a$ so that $h_1\in aHa^{-1}$ but the second letter of $h_1$ is in $H$, whereas the second letter of $g$ is again $a$. It seems to me that the problem actually comes from the powers of $a$. For example, assume that $a^2\in H$, then, the answer is yes. Indeed you then have that $G/H$ is of order 2, so that $H$ is normal in $G$ and so $\langle H\cup aHa^{-1}\rangle =H$ is indeed normal. So if you have a particular group in mind, maybe you should first look carefully as the relations between powers of $a$ and $H$. Also quick comment: try to find more suited titles. REPLY [3 votes]: Work in $GL_2(\mathbb{Q})$. Let $H$ be the subgroup generated by $$ \left(\begin{array} 01 & 1\\ 0 & 1\end{array}\right). $$ Let $G=\langle H,a\rangle$ where $$ a=\left(\begin{array} 02 & 0\\ 0 & 1\end{array}\right). $$ Then $aHa^{-1}$ is a proper subgroup of $H$. In particular, $H$ is not normal in $G$, and $H=\langle H\cup aHa^{-1}\rangle$. REPLY [3 votes]: As other have said, the answer is no in general. However if $a$ has finite order $n$, then it is true that $\langle H \cup aHa^{-1} \cup \ldots \cup a^{n-1}Ha^{-(n-1)} \rangle$ is a normal subgroup of $G$. This is because the group in question is normalized by $H$ (since it contains $H$ ) and is normalized by $a$, since the given generating set is stable under conjugation by $a$ ( and hence the group it generates is stable under conjugation by $a$). Hence the given group is normalized by $\langle H, a \rangle = G$, so is a normal subgroup of $G$.<|endoftext|> TITLE: Existence of a certain set of 0/1-sequences without the Axiom of Choice QUESTION [8 upvotes]: Is there a set $\mathcal X\subset\{0,1\}^{\Bbb N}$ of 0/1-sequences, so that For any two 0/1-sequences $x,y\in\{0,1\}^{\Bbb N}$ for which there is an $N\in\Bbb N$ with $$x_i=y_i,\;\;\text{for all $i< N$},\qquad x_i\not=y_i,\;\;\text{for all $i\ge N$},$$ exactly one of these belongs to $\mathcal X$. $\mathcal X$ can be proven to exist without using the Axiom of Choice. REPLY [14 votes]: A set $\mathcal X$ with the first of the two properties you want cannot have the Baire property (in the space $\{0,1\}^\omega$ with the product topology). Proof: Suppose it had the Baire property, so it differs from an open set $U$ by a meager set. Suppose for a moment that $U$ is nonempty, and consider a basic open subset of $U$, say the set $B$ of all $0/1$-sequences extending a certain finite $0/1$-sequence $s$. Then, $\mathcal X\cap B$ is a comeager subset of $B$. But then so is its image under the self-homeomorphism of $B$ that switches all $0$'s and $1$'s beyond the end of $s$. Your assumption says that this switching maps $\mathcal X$ to its complement, so we have two disjoint comeager subsets of the complete metric space $B$, which is absurd. So $U$ can't be nonempty. But if $U$ is empty, then $\mathcal X$ is meager and therefore so is its image under the self-homeomorphism of $\{0,1\}^\omega$ that switches $0$ with $1$ in all components. Then, by your assumption, $\{0,1\}^\omega$ is covered by two meager sets, again an absurdity. This completes the proof that $\mathcal X$ cannot have the Baire property. It is consistent, relative to ZF, that all subsets of $\{0,1\}^\omega$ have the Baire property (and that dependent choice holds, so that the Baire category theorem still works). So it is consistent with ZF that no $\mathcal X$ as in your question exists.<|endoftext|> TITLE: Convolution in K-Theory via an Example (From StackExchange) QUESTION [8 upvotes]: I've spent lots of time in Chriss and Ginzburg's "Complex Geometry and Representation Theory" and despite convolution (in Borel-Moore homology or K-theory) being very central, I feel like I'm still lacking a little understanding. I'd like some help with the following example. On the representation theory side, I'd like to consider the following simple example. Let $G=SL_2(\mathbb{C})$, and let $T \subset B$ be the toral subgroup of diagonal matrices and the Borel subgroup consisting of upper triangular matrices respectively. Let $V_{\Lambda_1}$ denote the irreducible representation of highest weight $\Lambda_1$. Taking the tensor square of this representation yields the following decomposition into irreducibles: $V_{\Lambda_1} \otimes V_{\Lambda_1} \simeq V_{2 \Lambda_1} \oplus V_0$. I'd like to geometrize this a la Ginzburg. Via Borel-Weil, we know that $H^{0}(G/B, L_{\Lambda_1}) \simeq V_{\Lambda_1}$, where $L_{\Lambda_1}$ is the associated bundle $G \times_{B} \mathbb{C}^{-\Lambda_1}$. What I would like is an operation on $G$-equivariant sheaves which corresponds to the tensor product of representations, so that $H^0(G/B, L_{\Lambda_1} * L_{\Lambda_1}) \simeq V_{2 \Lambda_1} \oplus V_0$. Note that the operation cannot be the tensor product. To see this, remember that $G/B = \mathbb{P}^1$, and $L_{\Lambda_1}$ is isomorphic to $\mathscr{O}_{\mathbb{P}^1}(1)$; if I tensor this sheaf with itself and take global sections I will get the irreducible 3-dimensional representation $Sym^2(V_{\Lambda_1})=V_{2\Lambda_1}$. Here is where I know that $*$ is supposed to be convolution, as defined by Ginzburg. (If anyone would like the definition, I can provide it, but that would lengthen this post even more). Question 1: Is it correct to expect that $\mathscr{O}_{\mathbb{P}^1}(1) *\mathscr{O}_{\mathbb{P}^1}(1) \simeq \mathscr{O}_{\mathbb{P}^1}(2) \oplus \mathscr{O}_{\mathbb{P}^1}$? This is the only way I can see the global sections giving me the correct representation. Question 2: If this is indeed the case, is there an explicit description in terms of global sections $T_1, T_2$ of $\mathscr{O}_{\mathbb{P}^1}(1)$, if the coordinates on $G/B=\mathbb{P}^1$ are $[T_1:T_2]$? It is easy to get the global sections $T_1^2, T_1T_2, T_2^2$ as a basis for $V_{2 \Lambda_1}$, but I cannot see how to get the basis for $\mathscr{O}_{\mathbb{P}^1}$. It also occurs to me that I have been working with $G-$equivariant sheaves here instead of their $K$-theory, and maybe that is incorrect. I've got more thoughts, but this is already quite long for a post. Please let me know if I can provide any additional information. REPLY [2 votes]: If I understand your question correctly, you are trying to compute the convolution product on $K^G(G/B)$. Since $G/B$ is smooth, in this case the convolution product $\ast$ coincides with the tensor product $\otimes$ of $G$-equivariant sheaves (see Corollary 5.2.25. in Chriss & Ginzburg). Since $K^G(G/B)\simeq R(T)$ (Lemma 6.1.6. in Chriss & Ginzburg), the tensor product (and therefore the convolution product) of Borel-Weil line bundles with dominant weights will have the effect of adding the corresponding weights (as you computed in your question). To get a geometric construction of $R(G)$ (the representation ring of $G$ rather than that of $T$), you might want to look at the Geometric Satake correspondence that establishes that this category is equivalent (as a tensor category) to $P_{G^\vee_\mathcal{O}}(Gr_{G^\vee})$, the category of $G^\vee_\mathcal{O}$-equivariant perverse sheaves on the affine Grassmannian $Gr_{G^\vee}$ of the Langlands dual group $G^\vee$.<|endoftext|> TITLE: Survey article model theory research QUESTION [6 upvotes]: I've taken a graduate course in model theory and I like it so much that I can imagine doing research in this area. Are there survey articles or review papers on the current research topics in model theory? Where can I find them? Also, I wish there is literature about the common proof techniques and tricks one uses in the current research. (Sometimes I have the feeling that everybody just writes down their proofs and nobody writes down the essential ideas or an overview of techniques used.) Do you know of any such text? REPLY [6 votes]: This answer is partly an answer to your questions, and also partly a response to the conversation unfolding in the comments. So I am just going to list a somewhat disconnected list of resources. One thing I'll say first though is in response to the query "or are they just the polished versions of the results of the last century" made in a comment. As a "stand alone" field, model theory is not even a century old (barely more than half a century by some accounts). So, yes, textbooks such as Marker's book on model theory are relevant for learning about current research. Other good texts are: A Course in Model Theory, by K. Tent and M. Ziegler A Course in Model Theory, by B. Poizat Stable Groups, by B. Poizat An Introduction to Stability Theory, by A. Pillay Model Theory, by W. Hodges Another remark is that, in my opinion, it could be difficult to compile a list of common techniques and "tricks" used in model theory research. Perhaps it's because of how new the field is, and so there hasn't been enough time for such things to congeal. It could also be because a major portion of current research in model theory focuses on applications and interactions with other areas of math, and so the "techniques and tricks" are widely varied. At any rate, here are more papers, books, and notes. These aren't in any particular order, and certainly this list is incomplete. But I think the list covers a lot, and provides starting points into several active areas. The Forking and Dividing website also has many definitions and examples from model theory with citations. Notes and survey articles. Survey articles on recent research in model theory can be scarce, but you should look through recent volumes of the Bulletin of Symbolic Logic. That being said, here are some survey-type articles that I like. a. Stability theory and its variants, by B. Hart b. Seminar and talk notes by E. Casanovas (you can find many topics of current research). c. Diophantine geometry from model theory, by T. Scanlon d. Approximate groups (after Hrushovski, and Breuillard, Green, Tao), by L. van den Dries (more "applied" model theory) e. Notes on the model theory of finite and pseudo-finite fields, by Z. Chatzidakis f. Unpublished notes of H. Adler Papers and books on specific research topics. a. A Guide to NIP Theories, by P. Simon. b. Simple Theories, by B. Kim c. Tame Topology and O-minimal Structures, by L. van den Dries d. Model Theory of Fields, by D. Marker, M. Messmer, and A. Pillay. e. Vapnik-Chervonenkis density in some theories without the independence property, I and II, by M. Aschenbrenner, A. Dolich, D. Haskell, D. Macpherson, and S. Starchenko f. Theories without the tree property of the second kind, by A. Chernikov g. A geometric introduction to forking and thorn-forking, by H. Adler h. Model theory for metric structures, by I. Ben Yaacov, A. Bernstein, C. W. Henson, and A. Usvyatsov i. Topological dynamics of definable group actions, by L. Newelski Lastly, one can read Classification theory and the number of non-isomorphic models by S. Shelah.<|endoftext|> TITLE: Geometry of a manifold after Dehn filling, in terms of geometry pre-filling QUESTION [5 upvotes]: First time posting, so sorry if this is an uninteresting or overly long post! The inspiration for this question was sparked by this answer given by Bruno Martelli in response to a question about horizontal surfaces in Seifert fibered spaces as the fiber of a fiber bundle over the circle with periodic monodromy. When seeing Seifert fibered space and a surface bundle over the circle with periodic monodromy, torus knots in $S^3$ fit into this class and are well behaved. In Martelli's response however, he restricts to Seifert fibered spaces without boundary, so let's Dehn fill the torus knot, say $T(p,q)$ along boundary slope $\frac{r}{s}$. Now if I remember correctly, we can't fill along $\frac{pq}{1}$ to get a closed Seifert fibered space, because it results in a connect sum of lens spaces from capping off the cabling annulus. Every other slope should yield a Seifert fibered space over $S^2$ with 3 singular fibers, a rather tricky class of Seifert fibered spaces. The surgered manifold should be a Seifert fibered space over the sphere as the underlying topological space for the torus knot's base orbifold was a disk, so Dehn filling the boundary should cap off the base orbifold's underlying topological space boundary to an $S^2$. Another way to see this is perhaps simply from the classification/construction of Seifert fibered spaces. If $T(p,q)$'s exterior was $M_k$ we can give an easy van Kampen argument to yield a presentation for $\pi_1(M_k)$ as $$ where $a$ and $b$ are the classes of the singular fibers in the Seifert fibering of $M_k$. From here we should be able to give another Van Kampen argument to yield a presentation for the fundamental group of the the surgered manifold $M_k(\frac{r}{s})$, where we just add a relation to the previous presentation based on $\frac{r}{s}$. (EDIT, my original post had calculated $\pi_1(M_k(\frac{r}{s})) = $, which I believe is wrong. I think mixed up the Heegard torus from the previous calculation with the knot's boundary torus so the new relation is wrong.) I suppose geometrization in all its high powered glory may be able to tell us something about the geometry of $M_k(\frac{r}{s})$ based on this fundamental group calculation, if this presentation isn't too unpleasant. A more pedestrian understanding would be appreciated though, as I understand little about geometrization and how the geometry looks near the decomposing tori, and know nothing of how the geometries glue together coherently to produce whatever it is they produce. So, I'm wondering if we can track through the construction of these Seifert fibered spaces through Dehn filling a trivial circle bundle over a pair of a pants with slopes $\frac{p}{q}$, $\frac{q}{p}$, and $\frac{r}{s}$, how the geometry changes, if we imagine that we start with a hyperbolic pair of pants $P$ with geodesic boundary. So onto the actual questions: 0.) Is there a way to see the geometry evolve as we fill $P \times S^1$? 1.) Of particular interest is the $\mathbb{H}^2 \times \mathbb{R}$ geometry that might arise, given that we can start with an $\mathbb{H}^2 \times \mathbb{R}$ geometry. Can the $\mathbb{H}^2 \times \mathbb{R}$ geometry ever show up in filled torus knots? 2.) It is known that torus knots are L-space knots, and so for every filling slope $\frac{r}{s} \geq pq-(p+q) = 2g-1$, where $g$ denotes the knot genus of $T(p,q)$, the surgered manifolds are L-spaces, is the class of possible geometries for the L-space slopes different from the class of geometries possible for the non-L-space slopes? REPLY [3 votes]: The general principle is that a generic filling of a geometric manifold belongs to the same geometry of the original manifold. This holds notably in hyperbolic geometry by Thurston's Dehn filling theorem. In the other geometries, this principle is somehow also true, but some care is needed to interpret it correctly: the manifold $P\times S^1$ can be equipped with two geometries, either $\mathbb H^2 \times \mathbb R$ or $\widetilde{SL_2}$, and both may appear in a filling - the latter being much more generic than the former. The geometry of the fillings of $P\times S^1$ can be described easily. Use the (oriented) boundaries of $P$ and the fiber $S^1$ as meridians and longitudes. Represent the fillings with respect to this basis as coprime pairs $$(p_1,q_1), \qquad (p_2,q_2), \qquad (p_3,q_3).$$ If $p_i=0$ for some $i$, you get a sum of lens spaces. If not, you get a Seifert manifold, whose geometry is nicely determined by the invariants $$\chi = -1 + \frac 1{p_1} + \frac 1{p_2} + \frac 1{p_3}$$ $$e = \frac{q_1}{p_1} + \frac{q_2}{p_2} + \frac{q_3}{p_3}$$ If $p_i=1$ for some $i$, you get a lens space, that has the $S^3$ geometry. Here $\chi$ is the Euler characteristic of the base orbifold. Generically you get $\chi < 0$, and in this case you get the geometry $\mathbb H^2 \times \mathbb R$ if $e=0$ and $\widetilde{SL_2}$ if $e\neq 0$. You get a torus knot by filling the first two components with parameters $(p_1,q_1)$ and $(p_2,q_2)$ such that $p_1q_2+p_2q_1=1$. In this case $$e = \frac{1}{p_1p_2} + \frac{q_3}{p_3}.$$ Therefore every torus knot has precisely one Dehn filling with $e=0$, obtained by taking $(p_3,q_3)=(-p_1p_2,1)$. This gives a manifold of geometry $\mathbb H^2 \times \mathbb R$ as soon as $\chi < 0$. The latter condition should be satisfied by all torus knots except the trefoil knot, where $p_1,p_2,p_3 = 2,3,6$ and you get a flat manifold. Conclusion: Every torus knot except the trefoil has precisely one filling with geometry $\mathbb H^2 \times \mathbb R$. Generically, its Dehn fillings are of the geometry $\widetilde{SL_2 \mathbb R}$, but as noted by Marco there are also infinitely many fillings that are lens spaces, obtained by taking $p_3=1$: these have $\chi>0$, and are not generic in some natural sense.<|endoftext|> TITLE: Are there categories whose internal hom is somewhat 'exotic'? QUESTION [30 upvotes]: The internal hom (or exponential object) is basically a reification of the 'external' hom. It can be defined in any cartesian (or even monoidal, more on this later) category as the right adjoint of the (monoidal) product. My question is: are there some categories whose internal hom behaves quite unexpectedly? Or even some interesting examples where internalization really deforms the external hom in a non-trivial way. The question arises from the observaton that the 'external' hom can be assumed to be a set (assume the category to be locally small, or wave your hand hard enough), and thus is already somewhat reified. We expect some 'structure' on it, namely elements and perhaps even subobjects. In a sufficiently rich category (say a topos), this structure is internalizable as well. So it might be the case that some exotic behavior emerge, or some collapse happens. I expect this to happen mainly for non-Cartesian monoidal closed category, because the monoidal product can be quite convoluted. On the other hand, Section 3 of the internal hom page of the nLab seems to prove the internal hom shares some strong properties of the 'external' hom, which might hint to the fact they are really 'the same'. REPLY [7 votes]: I've seen many great positive answers to this question, providing examples of exotic internal homs. In this answer, I would like to show obstructions to the exoticism of the monoidal closed structure. I think that these examples will make this question and all the provided answers even more interesting. Nothing shapes like a boundary. The most classical result in this direction is the monoidal structure on the category $\mathsf{Top}$ of topological spaces. Prop. 7.1.1 Handbook of categorical Algebra 2, Borceux shows that the internal hom of any simmetric monoidal structure in Top has to match with the cartesian structure at least on the level of the underlying set. This result highly depends on the fact that whatever monoidal closed structure $(\mathsf{I}, \otimes, [\_,\_])$ you have on a category $\mathcal{A}$, the $\mathsf{I}$-points of the internal hom recover the external hom, $$\mathcal{A}(\mathsf{I}, [A,B]) \cong \mathcal{A}(A,B).$$ This observation spots an entanglement between the internal and the external logic of the category that unveil some rigidities of the monoidal structure. Along those lines some research has been developed in the direction of showing that there exists some obstruction in admitting a monoidal biclosed structure. Topological categories with many symmetric monoidal closed structures, by Kelly and Rossi, show that there exist topological categories which admit a proper class of symmetric monoidal closed structures. Algebraic categories with few monoidal biclosed structures or none by Kelly, Foltz and Lair, goes in the other direction, proving (among other stuff) the two following theorems. Prop. If an equational variety admits a monoidal biclosed structure, every idempotent algebra is self-commuting. Prop. The categories of magmas, of semigroups, of magmas with identity, of monoids, of groups, of rings, and of commutative rings, admit no monoidal biclosed structures whatsoever; the category of abelian groups admits none but the classical one, and similarly for abelian monoids; and that the category of small categories admits exactly two, each symmetric, one being the classical Cartesian closed structure. The last paper is very much inspired by the Czech school, among the very influential papers let me mention A. Pultr, Extending tensor products to structures of closed categories, Comm. Math. Univ. Carolinae 13 (1972) 599-616. A. Pultr, Closed categories of models of Gabriel theories (manuscript, Charles Univ. Prague, 1973). J. Rosický, One obstruction for closedness, Comm. Math. Univ. Carolinae 18 (1977). 311-318. If you have other examples that prove how having a closed monoidal structure imposes some rigidity on the underlying category, please contribute to this question with a comment.<|endoftext|> TITLE: Why doesn't mathematics collapse even though humans quite often make mistakes in their proofs? QUESTION [192 upvotes]: To begin with, I am aware of these questions, which seems to be related: How do I fix someone's published error?, Examples of common false beliefs in mathematics, When have we lost a body of mathematics because errors were found?, etc... My background: I am a senior undergraduate student in mathematics. Recently, I got a nice chance in a REU program, and started to read some journal articles. My impression was: any result in modern mathematics critically depends on another result, and that result depends on some other result, and ad infinitum. On the other hand, some graduate students and professors in my university, who stand in quite intimate relations to me, say that, they do not check every details of proofs when they read mathematical monographs and research articles. They simply do not have enough time to read all the details and fill in the lines. (Clearly, I also do not read all the proofs in detail, if it seems to be so difficult or not much relevant to what I am interested in.) Finally, I've been heard of some stories on fatal mathematical errors. To be honest, I do not understand what the errors precisely are. What I've been heard about are some "urban legends". (I intentionally didn't write down the details of these urban legends, since if I write down everything I've heard, maybe someone working in the mentioned field may feel insulted...) For the above reasons, recently I am afraid of the situation where a field in mathematics collapse down because of a single, fatal, but very subtle error in the foundations of that field. In mathematics, everything seems to be so much intertwined, and it seems that no one actually checks every single detail in every mathematical articles. But the mathematics community seems to be very sound. Maybe at least one of the followings are true: Actually, a typical mathematical result does not depend that much on other results. So whenever if possible, a mathematician can check the details of every results which is of interest to him/her. Strictly speaking, rigor is actually not that important. Even if a mathematical result turns out to be false, there is still something true in the statement. Therefore, only minor changes will be needed, and all the results depending on the turned-out-to-be-false result remains sound. Here are my questions. Why the whole mathematics remains so sound, even though humans are imperfect and quite often produce errors? Are my explanations above correct? If a theorem turns out to be wrong, then mathematicians will try to correct (if possible) all the results depending on that theorem. How hard is this job? Isn't it very tedious and frustrating? I want to hear some personal stories. As an undergraduate student, I want to know if anybody who is much wiser, older, or experienced, had the same fear as mine. (Again, I want to hear some personal stories.) As an undergraduate student who will get into a graduate school in the near future, I want to get some advice. Should I stop worrying and believe the authors of the books and articles I read? When should I check all the details, and when should I just accept the theorem as given? Thanks to everyone for reading my question. REPLY [5 votes]: An interesting paper on this topic is Explosive Proofs of Mathematical Truths by Scott Viteri and Simon DeDeo. Roughly speaking, they propose that the high probability of any particular sequence of deductive steps from A to B being wrong is compensated for by the large number of distinct paths from A to B. They support their hypothesis by examining 48 Coq proofs and 4 human proofs.<|endoftext|> TITLE: On subsets of $\mathbb{N}$ reciprocally summable to $1$ QUESTION [7 upvotes]: Let $\mathbb{N}$ denote the set of positive integers. If $A\subseteq \mathbb{N}$ is finite, we say that $A$ is reciprocally summable to $1$ ("rs1") if $\sum_{a\in A} \frac{1}{a} = 1$. If $A\subseteq \mathbb{N}$ is finite and $\sum_{a\in A} \frac{1}{a} < 1$, is there a finite rs1 set $A'$ with $A\subseteq A'$? REPLY [12 votes]: Yes, several algorithms for Egyptian fractions suggested by Gerhard Paseman works. We want to represent the number $r=1-\sum_{a\in A}$ as a sum of distinct Egyptian fractions with denominators not in $A$. This may be done by many ways, for example we may use Lemma. For any positive integers $a,n$ the number $1/a$ is representable as a sum of distinct Egyptian fractions with denominators greater than $n$. Proof. Start with $1/a=1/a$. If it does not work (i.e., $a\leqslant n$), replace $1/a$ to $1/(a+1)+1/(a^2+a)$. After that do the same with both $1/(a+1),1/(a^2+a)$. We get four fractions which sum up to $1/a$, then eight fractions and so on. Stop when we get, say, $2^k$ fractions which are all less than $1/n$. They are all distinct (unless $a=1$, in this case use $1=1/2+1/3+1/6$), that proves Lemma. Why distinct? Assume that two of them coincide, choose the first step when this happens, say $b+1=c^2+c$ where $b,c$ were denominators on the previous step. Then the total number of steps is less than $c$, and $b=c^2+c-1$ was obtained from $c^2+c-2$, this guy in turn from $c^2+c-3$ and so on. Thus initial number $a$ was not less than $c^2$ and could not generate $c$. Lemma is proved. Now start with the representation $r=1/N+\ldots+1/N$ for large $N>\max(A)$. Then perform the replacement algorithm: while two fractions $1/a,1/a$ in the representation are equal, replace one of them using lemma by a sum with denominators greater than anything already used. After finitely many steps we remove all repetitions.<|endoftext|> TITLE: Do regular (but non-smooth) conics over a discretely valued field of characteristic $2$ admit a regular model over the valuation ring? QUESTION [5 upvotes]: Let $K$ be a non-perfect field of characteristic $2$. Let $T \subseteq K$ be a discrete valuation ring. Assume there exist $a,b \in K^{\times}$ such that the projective conic $C$ defined by $$aX^2 + b Y^2 + Z^2=0$$ contains no $K$-rational point (in particular $a,b$ and $ab$ are not squares in $K$). Then $C$ is non-smooth over $K$ at every point. However, $C$ is regular (see Exercise 4.3.22 (d) of Qing Liu's book Algebraic Geometry and Arithmetic Curves). Question: Does $C$ have a regular projective model over $T$ (i.e. a regular fibered projective surface over $T$ with generic fiber isomorphic to $C$) ? Every smooth curve over $K$ does admit a regular projective model over $T$, as is shown (for example) in Corollary 8.3.51 of Liu's aforementioned book. This seems to be based on Lipman's resolution of singularities of $2$-dimensional excellent Noetherian schemes, applied to a normal model of the curve over $T$, or rather some base change to the completion of $T$ (excuse the sloppy description, this may not be a very accurate description of what is really going on). In any case, it seems to be crucially used that the generic fiber of the normal model is not only regular, but smooth over $K$. Nevertheless, this is a general statement, and it is a priori possible that in concrete cases of regular projective curves (such as all integral regular conics over $K$, which I am interested in) there do exist regular projective models over $T$, regardless of whether the regular curve is smooth over $K$ or not. REPLY [5 votes]: In general this isn't possible. The material that follows is a bit technical, but I do not have the time to explain all the details here now. Suppose that $a, b \in T$ and that $a, b$ become squares in the completion of $T$ (see example below). Then the projective scheme $S$ defined by $a X^2 + b Y^2 + Z^2 = 0$ in $\mathbf{P}^2_T$ is $2$-dimensional with generic fibre $C$ and has the following property: for every point $s \in S$ lying over the closed point of $\text{Spec}(T)$ the local ring $\mathcal{O}_{S, s}$ is a domain with non-reduced completion (this is where we use that $a, b$ become squares in the completion of $T$). If $C$ had a regular projective model $S'$ over $T$, then we could eleminate indeterminacies in the rational map from $S'$ to $S$ by a finite sequence of blowing ups in closed points of $S'$. Note that the blowing up of $S'$ in a closed point produces another regular surface. Thus we may assume there is a morphism $S' \to S$ over $T$ which is proper and birational. In particular $S' \to S$ is finite over an open $U \subset S$ which contains all codimension $1$ points. Let $s \in U$ be the generic point of an irreducible component of the closed fibre of $S$. The scheme $S' \times_S \text{Spec}(\mathcal{O}_{S, s})$ is affine, say equal to the spectrum of $A$. Then the inclusion $\mathcal{O}_{S, s} \subset A$ is finite and $A$ is regular. It follows that the $\mathfrak m_s$-adic completion $A^\wedge$ of $A$ is a finite product of dvrs in particular reduced. Since completion is exact on finite modules, we get $\mathcal{O}_{S, s}^\wedge \subset A^\wedge$ which is a contradiction with the non-reducedness seen above. A reference for some of the things discussed above is the chapter on resolution of surfaces in the Stacks project, especially the section on dominating by quadratic transformations and the section on implied properties. Remark. It seems possible to me that for "most" choices of $a$ and $b$ the curve $C$ does have a regular model, but I didn't try to prove or disprove it. Example. Let $k = \mathbf{F}_2(s_1, s_2, s_3, \ldots, t_1, t_2, t_3, \ldots)$ be the purely transcendental extension of $\mathbf{F}_2$ on elements $s_1, s_2, s_3, \ldots, , t_1, t_2, t_3, \ldots$. Let $T \subset k[[x]]$ be the set of power series $\sum a_i x^i$ such that $[k^2(a_0, a_1, a_2, \ldots) : k^2] < \infty$. Then $T$ is the standard example of a non-Nagata dvr in char $2$. Let $a = 1 + \sum_{i \geq 1} s_i^2 x^{2i}$ and $b = 1 + \sum_{i \geq 1} t_i^2 x^{2i}$. Note that $a, b, ab$ are not squares in $T$ but become squares in the completion of $T$. We omit the verification that $C(K)$ is empty.<|endoftext|> TITLE: Who proved that the Mandelbrot set's Julia sets are locally connected? QUESTION [5 upvotes]: I'd be greatly interested in a reference to the respective article. Was it Douady? Julia? Hubbard? Fatou? Bonus question: Who gave the proof that can be found in the Orsay notes? EDIT: The question was based upon a misconception on the asker's part: There are points in the Mandelbrot set whose Julia sets are NOT locally connected. Yet, all interior points of the Mandelbrot set have a locally connected Julia set. (There are boundary points of the Mandelbrot set whose Julia set is locally connected.) REPLY [9 votes]: Nobody. This is the principal unsolved problem in the area, which is called MLC (That the Mandelbrot set is locally connected). Two Fields medals were awarded for partial progress in this problem. About Julia sets, some of them are locally connected, others are not. See The deep significance of the question of the Mandelbrot set's local connectedness? for more detail. Remarks. All proofs in the Orsay notes are due to Douady and Hubbard, unless stated otherwise. That Julia sets corresponding to the interior of the Mandelbrot set are locally connected is also not known. It is only known for parts of this interior: for the main hyperbolic component this was proved by Fatou, and for the rest of hyperbolic components by Douady-Hubbard. But the existence of components of the interior, other than hyperbolic components is not known.<|endoftext|> TITLE: Averaging Chebotarev's density theorem over families of number fields QUESTION [5 upvotes]: The Chebotarev density theorem is one of the most celebrated and important results in number theory. We state the following version: for a number field $K$, Galois over $\mathbb{Q}$ with Galois group (isomorphic to) $G$ and a conjugacy class $C$ of $G$, put $\pi(x; C, K)$ to be the number of prime ideals $\mathfrak{p}$ in $\mathcal{O}_K$ which is unramified in $K$ and for which the Frobenius $\sigma_{\mathfrak{p}} = C$. Then $\pi(x; C, K)$ satisfies $$\displaystyle \pi(x; C, K) = \frac{|C|}{|G|} \text{Li}(x)(1 + o(1)).$$ The original proof given by Chebotarev is ineffective, so that the determination of the error term $o(1)$ above is not tenable. Various effective improvements have been proved over the past several decades. Conditioned on GRH, one can obtain the expression $$\displaystyle \pi(x; C, K) = \frac{|C|}{|G|} \text{Li}(x) + O \left(x^{1/2} [K : \mathbb{Q}] \log(|\Delta_K x|)\right),$$ where $\Delta_K$ is the discriminant of $K$. Thus, assuming GRH and running over a family $\mathcal{F}$ of fields $K$ of equal degree and the same Galois group with the property that for any integer $n$ there is a uniformly bounded number of fields $K$ in $\mathcal{F}$ with discriminant $n$, one has (1) $$\displaystyle \sum_{\substack{n \leq Q \\ |\Delta_K| = n \\ K \in \mathcal{F}}} \sum_C \left \lvert \pi(x; C, K) - \frac{|C|}{|G|} \text{Li}(x) \right \rvert = O \left(Q x^{1/2} \log(xQ) \right).$$ Thus, the error term remains acceptable (i.e., $o(x)$) even when $Q$ is as large as $x^{1/2} (\log x)^{-A}$, for $A > 1$. The latter statement is similar to a famous result of Bombieri and Vinogradov. For a positive integer $q$ and an integer $a$ such that $\gcd(a,q) = 1$, put $\pi(x; a, q)$ for the number of (rational) primes $p \leq x$ such that $p \equiv a \pmod{q}$. Then Bombieri-Vinogradov theorem asserts that $$\displaystyle \sum_{q \leq Q} \max_{y < x} \max_{\substack{1 \leq a \leq q \\ \gcd(a,q) = 1}} \left \lvert \pi(y; a, q) - \frac{1}{\phi(q)} \text{Li}(y) \right \rvert = O \left(Q x^{1/2} (\log x)^5\right),$$ so we may take $Q$ to be as large as $x^{1/2} (\log x)^{-A}$ for $A > 5$. The Elliott-Halberstam conjecture is the statement that one can take $Q$ as large as $x^{1 - \varepsilon}$ for any $\varepsilon > 0$ in the statement of Bombieri-Vinogradov. Can one prove (1) without assuming GRH? Further, can one expect that the 'level of distribution' in (1) can in fact be taken as large as in the Elliott-Halberstam conjecture? The most basic example of where this question may apply is the family of (real or imaginary) quadratic fields. REPLY [4 votes]: Let us consider the related problem of finding a suitable $\delta>0$ such that $\displaystyle\sum_{\substack{q\leq x^{\delta-\epsilon} \\ K\cap \mathbb{Q}(e^{2\pi i/q}) = \mathbb{Q}}}\max_{(a,q)=1}\Big|\sum_{\substack{p\leq x \\ p\equiv a\pmod{q} \\ [\frac{K/\mathbb{Q}}{p}]=C}}1 - \frac{|C|}{|G|}\frac{\mathrm{Li}(x)}{\varphi(q)}\Big|\ll \frac{x}{(\log x)^A}$. Let $H$ be an abelian subgroup of $G$ such that $H\cap C$ is nonempty, and let $E$ be the fixed field of $H$. Ram Murty and Kumar Murty proved that this holds when $\delta = \frac{1}{\max\{2,[E:\mathbb{Q}]-2\}}$. The strong Artin conjecture for the irreducible representations $\rho$ of $G$ would imply that one can replace $[E:\mathbb{Q}]$ with $\max \rho(1)$. Here, the averaging is much simpler than you propose in your initial problem. We are simply averaging over Dirichlet characters and have an optimal version of the large sieve. We have a level of distribution $\delta = 1/2$ when $[E:\mathbb{Q}]\leq 4$, which includes the cases where $K/\mathbb{Q}$ is abelian or dihedral. If $K/\mathbb{Q}$ is "sufficiently nonabelian", then the level of distribution is quite small. Ultimately, this is related to the lack of strong bounds for Artin $L$-functions in the critical strip; we're using the convexity bound and Phragmen-Lindelof. So the level of distribution will be proportional to the reciprocal of the largest degree of all of the $L$-functions (when viewed over $\mathbb{Q}$) under consideration. I find it hard to believe that a proper analogue of Bombieri-Vinogradov (with a level of distribution equal to $1/2$) is possible without some serious advance toward Lindelof for Artin $L$-functions twisted by Dirichlet characters. If one cannot establish a proper BV in this setting, I don't see how one could do it in the more complicated setting you propose. This also ignores many subtleties in averaging over number fields with a fixed nonabelian Galois group, many of which are catalogued very nicely in the work of Pierce, Turnage-Butterbaugh, and Wood arxiv.org/abs/1709.09637 (as Alison already mentioned). Also, a small point: You don't necessarily have that $|\{K\in \mathcal{F}\colon |\Delta_K|\leq Q\}|\ll Q$. So the RHS of your equation (1) might be off, depending on what $\mathcal{F}$ is.<|endoftext|> TITLE: Every riemannian length structure on $\mathbb{R}^n$ is induced by a continuous function $f:\mathbb{R}^n\to \mathbb{E}^n$, to the euclidean space QUESTION [8 upvotes]: This question is a cross post from Math.SE. Unfortunately the migration of the question is not possible after two months of posting. I have been reading about length spaces in the (great) book Metric Geometry by Y. Burago, D. Burago and S. Ivanov. They define what an induced length structure is and they do the following claim without details or references; they just say that it is not easy to prove. It is the Example 2.2.3. I am looking for a proof of the following claim: Every riemannian length structure on (the differential manifold) $\mathbb{R}^n$ is induced by a continuous function $f:\mathbb{R}^n\to \mathbb{E}^n$, where $\mathbb{E}^n$ is the euclidean space of dimension $n$. An answer could be (a sketch of) a proof or a reference contaning it. It is not clear for me if we can choose the set of admissible paths in $\mathbb{E}^n$ or it is part of the claim that we have to take every continuous path as admissible. At the moment I have no progress in the solution so any remark or comment is also very welcome. Thanks in advance! Relevant definitions: A length structure on a Hausdorff space $X$ is a pair $(\mathcal{C},\mathcal{L})$, where $\mathcal{C}$ is a set of continuous paths (with closed intervals as domains) in $X$ and $\mathcal{L}$ is a function $\mathcal{C}\to \mathbb{R}_{\geq0}\cup \{+\infty\}$, satisfying the following axioms. 1) The set $\mathcal{C}$ is closed under restriction (to closed intervals), concatenation and linear reparameterization. 2) The function $\mathcal{L}$ is additive and invariant under linear reparameterizations. It also depends continously on the path in the following sense: If $c:[a,b]\to X$ is in $\mathcal{C}$ then the function $t\in [a,b]\mapsto \mathcal{L}(c|_{[a,t]})\in \mathbb{R}_{\geq0}\cup\{+\infty\}$ is continuous. 3) For every $x$ in $X$ there exists an open neightborhood $U_x$ and a positive real number $R_x$ such that every path $c$ in $\mathcal{C}$ with $x$ in its image and with image not contained in $U_x$ verifies that $\mathcal{L}(c)\geq R_x$. We call the elements of $\mathcal{C}$ admissible paths and the value of $\mathcal{L}$ on such a path is the length of the path. A length structure induces a metric on $X$ in which the distance between two points is the infimum of the lengths of admissible paths joining them. If $f:Y\to X$ is a continuous function between two Hausdorff spaces and $X$ has a length structure $(\mathcal{C},\mathcal{L})$ we define an induced structure $(\mathcal{C}',\mathcal{L}')$ on $Y$ such that a continuous path $c:[a,b]\to Y$ is admissible if and only if its composition with $f$ is admissible and the length of $c$ is equal to the length of such composition. The axiom 3) can fail for this induced structure. If the structure satisfies axiom 3) we call it the induced length structure. A riemannian length structure on a differential manifold $M$ is a length structure such that there exists a riemannian metric on $M$ with the same induced metric. Suggestions and progress: @HKLee has suggested to look in the chapter 6 of Petrunin and Yashinski. From there we can extract a proof of the fact that every non-expanding map $g:X\to Y$ between riemannian manifolds of the same dimension can be approximated (maybe uniformly or uniformly on compact sets) by length-preserving maps. REPLY [5 votes]: Well, as far as I can tell, this follows from the Corollary in Section 2.4.11 (p. 216) of Gromov's book Partial Differential Relations. I quote it here: Corollary: Let $V$ be an $n$-dimensional stably parallelizable manifold. Then $V$ admits an isometric map $V\rightarrow \mathbb{R}^n$. However, I'm not an expert on the subject, so you should check whether this is really what you need.<|endoftext|> TITLE: A second course in the representation theory QUESTION [12 upvotes]: I've read Etingof's and then Fulton-Harris' books about the representation theory ("Intrdouction to representation theory" and "Representation theory. A first course" respectively) and found their subject very exciting! Can someone, please, recommend me a textbook containing a kind of the "second course" in this branch of the mathematics? (It'd be great if someone recommended a "russian-style" book.) (I heard a lot about the "Representation theory and complex geometry" written by Chriss and Ginzburg. Is it really so wonderful? And what other good references do you know?) I'm going to learn the advanced representation theory for its own sake. But I'll be glad to see interesting intersections with the algebraic or complex geometry! UPD: User Vincent's answer was really great! But it was only about the Lie groups. I'm also interested in such topics as Kac-Moody algebras, (double affine) Hecke algebras, category $\mathcal O$, Soergel (bi)modules, quantum groups and other things like those (maybe, quivers)... Can some of them be a part of the second (not third, fourth, etc.) course? And are there (less or more) introductory textbooks covering a part of this material? REPLY [13 votes]: The best textbook that covers a wide range of subjects is for me "A Tour of Representation Theory " by Lorenz. Of course it is close to a first course but I would say a little more advanced than the two books that you mentioned. Another similar one is "A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras" by Gruson and Serganova. Here is a small collection in specific directions: Modern (modular) group representation theory: "The Block Theory of Finite Group Algebras" Volume 1 and 2 by Linckelmann . Representation theory of finite dimensional algebras with some indepth look into some special classes such as hereditary, Hecke or Hopf algebras: "Frobenius Algebras " part 1 and 2 (3 will come soon) by Skowronski and Yamagata. Representation theory of the symmetric group (including modular representation theory): "The Representation Theory of the Symmetric Group" by James and Kerber. Homological methods in representation theory (mostly with application for group algebras): "Representation Theory: A Homological Algebra Point of View " by Zimmermann. For an application of representation theory (of quivers) to data science: "Persistence Theory: From Quiver Representations to Data Analysis " by Oudot<|endoftext|> TITLE: Can a group have a cyclical derived series? QUESTION [18 upvotes]: Given any group $G$, one can consider its derived series $$G = G^{(0)}\rhd G^{(1)}\rhd G^{(2)}\rhd\dots$$ where $G^{(k)}$ is the commutator subgroup of $G^{(k-1)}$. A group is perfect if $G=G^{(1)}$ and thus has constant derived series, and solvable if its derived series reaches the trivial group after finitely many steps. Is it possible for a group’s derived series to be cyclical, i.e. that $G \cong G^{(n)}$ for some $n>1$ and $G\not\cong G^{(k)}$ for all positive $k TITLE: Is there a generalization of Furstenberg theorem from SL(2,R) to SL(2,C) matrices? QUESTION [6 upvotes]: I learnt from a talk that consider a random product of i.i.d. matrices, randomly chosen from SL(2,R): $T_n=A_n \cdots A_2 A_1$, where the random matrices $A_i$ are i.i.d. A classical Furstenberg theorem then implies, that under some very mild nondegeneracy conditions (no finite common invariant set of lines, no common invariant metric) for the law of $A_i$'s the norm of such a product almost surely grows exponentially. Now I wonder if this conclusion also holds if the matrices are chosen from SL(2,C)? For simplicity, I checked the case with two matrices $A=\left[\begin{array}{cccc}i &0\\0 &-i\end{array}\right]$ and $B=\left[\begin{array}{cccc}1+i &-i\\i &1-i\end{array}\right]$. Now I multiply the matrices which are randomly chosen from $A$ and $B$ with probability $1/2$ and $1/2$, and find that indeed the norm of such a product grows exponentially. But I am not sure if there is a proof for this. Thanks! REPLY [3 votes]: The original work of Furstenberg "Non-commuting random products" (1963) actually contains an answer to your question in Theorem 8.6 which states the positivity of the top Lyapunov exponent for any random walk on $SL(d,\mathbb R)$ under the natural first moment condition and the irreducibility condition your mention. So in your case you just have to embed $SL(2,\mathbb C)$ into a higher order real matrix group.<|endoftext|> TITLE: Injectivity radius of manifolds with boundary QUESTION [9 upvotes]: This question stems from the discussion in: how to define the injectivity radius of manifolds with boundary? Suppose $(M,g)$ is a compact Riemannian manifold with boundary. In this context, let the injectivity radius of a point $x$ be the minimum distance from $x$ at which there is a point $y$ with more than one length-minimizing geodesic connecting $x$ to $y$. Is it true that the injectivity radius as defined this way is bounded below by some nonzero value? If so, is there a standard reference for this fact? In the discussion linked above, the following theorem is referenced: Corollary 2. If for a complete Riemannian manifold with boundary, M, the sectional curvatures of the interior and the outward sectional curvatures of the boundary are no greater than $K$, then $N(p,\frac{\pi}{2K})$ is open in M and the distance function from p is convex on $N(p,\frac{\pi}{2K})$. Where $N(p,\frac{\pi}{2K})$ is the set of points connected to $p$ by a unique geodesic of length $\frac{\pi}{2K}$ or less. (Reference Paper) https://www.ams.org/journals/tran/1993-339-02/S0002-9947-1993-1113693-1/S0002-9947-1993-1113693-1.pdf This seems like it is close to the result I am looking for. Earlier in the paper, it is also stated that there are no conjugate points in $N(p,\frac{\pi}{K})$. Is there a simple step from this result that proves that the injectivity radius is nonzero? REPLY [11 votes]: Yes, they show that any compact Riemannian manifold with boundary is locally $\mathrm{CAT}(\kappa)$ for some $\kappa\in\mathbb{R}$. In particular the injectivity radius is positive.<|endoftext|> TITLE: De Finetti-style theorem for Point Processes QUESTION [5 upvotes]: I am new to point processes. I know there are a number of theorems along the lines that if a point process $\eta$ satisfies: Complete independence (the random variables $\eta(B_1), \ldots, \eta(B_n)$ are independent for pairwise disjoint bounded measurable $B_1, \ldots, B_n$) and Some regularity conditions like being simple and uniform $\sigma$-finite on a Borel subset of a complete separable metric space, then $\eta$ is a Poisson process. It seems that something like the following must be known. I think (I haven't worked through all the details) that if $\eta$ satisfies: (a) A condition one might call "complete exchangeability": for any disjoint bounded measurable $B_1, \ldots, B_n$, there are $A_i \subseteq B_i$, with equality for at least one $i$, such that $\eta(A_1), \ldots, \eta(A_n)$ are exchangeable random variables; or equivalently (b) For any disjoint bounded measurable $B_1, \ldots, B_n$ with $\mathbb{E}\eta(B_1) = \ldots = \mathbb{E}\eta(B_n)$, $\eta(B_1),\ldots, \eta(B_n)$ are exchangeable random variables; as well as Similar regularity conditions including $\mathbb{E}\eta(B)<\infty$ for all bounded measurable $B$; and $\mathbb{E}\eta(\text{whole space}) = \infty$ then there exists a nonnegative-valued scalar random variable $G$ with $\mathbb{E}G = 1$ such that conditioned on $G$, $\eta$ is a Poisson process with intensity measure $G\mathbb{E}\eta$. Without (3) there are simple counterexamples, e.g. $\eta = \delta_x$ where $x$ is distributed according to a nonatomic probability distribution. Could anyone provide a reference for such a point process de Finetti theorem? REPLY [2 votes]: This is Theorem 3.34 in Kallenberg, Olav, Random measures, Berlin: Akademie-Verlag. London - New York - San Francisco: Academic Press. 104 p. M 28.00 (1976). ZBL0345.60032.<|endoftext|> TITLE: Homotopy in $X$ and homology in $X \times I$ QUESTION [7 upvotes]: Suppose $X^n$ and $M^{n-2}$ are manifolds, and $f_1,f_2 : M \to X$ to two homotopic embeddings of $M$ into $X$. We can then embed $M$ into both boundary components in $X \times I$ using $f_1$ and $f_2$, respectively. I am wondering if there will always be some submanifold $N^{n-1} \subset X \times I$ with boundary equal to these two embeddings of $M$ into $X \times I$. If we choose a homotopy between $f_1$ and $f_2$, say $H$, then we can look at the track of the homotopy $M \times I \to X \times I$, but this will not be an embedded $N$. This then shows that the two different images of $M$ are homologous in $X \times I$ - is there some sort of Steenrod realization thing that allows we to conclude that there is such a desired $N$? Maybe I need assumptions on $n$... For what its worth, I'm really interested in the case where everything is connected and oriented. REPLY [8 votes]: You are talking about the notion of L-equivalence, studied by Thom in his seminal paper Thom, René, Quelques propriétés globales des variétés différentiables, Comment. Math. Helv. 28, 17-86 (1954). ZBL0057.15502. (Nowadays some people, myself included, might call this cobordism of embeddings.) The discussion starts with the definition on page 71. At the bottom of page 74 you find the statement that (assuming everything is oriented) in codimensions $1$ or $2$, two submanifolds are $L$-equivalent if and only if they are homologous. This is slightly stronger than what you ask (since homotopic implies homologous, but not the other way round). In general it should be possible to find embeddings which are homotopic but not $L$-equivalent. Note that $L$-equivalence classes of codimension $k$ embeddings in $X$ are in one-to one correspondence with homotopy classes of maps from $X$ to the Thom space $MSO(k)$ (in the oriented case, or $MO(k)$ forgetting orientations) so finding such examples involves studying the unstable homotopy of Thom spaces. Edit: I must have learnt this from Thom's paper, as I can't think of a canonical reference in English. This is the basic Pontrjagin-Thom construction, so it should be in many textbooks. An expository paper (in English) is here. It should be noted that the statement that two codimension $2$ oriented embeddings in an oriented $n$-manifold $X$ are cobordant if and only if they are homologous is down to the (somewhat miraculous, in my opinion) fact that $MSO(2)\simeq \mathbb{C}P^\infty$ is an Eilenberg-Mac Lane space $K(\mathbb{Z},2)$. Then there are isomorphisms $$ [X,MSO(2)]\cong [X,K(\mathbb{Z},2)]\cong H^2(X;\mathbb{Z})\cong H_{n-2}(X;\mathbb{Z}), $$ the last being Poincaré duality.<|endoftext|> TITLE: A list of proofs of "Coherent topoi have enough points" QUESTION [7 upvotes]: For my research I would like to read all the known proofs of the very classical result "Coherent topoi have enough points", by Deligne. Ref 1: D3.3.13 in Sketches of an Elephant provides a very logic-rooted proof of the statement, I would like to see a more geometric or a more category theoretic proof. Suggested by Christian Espindola Ref 2: 7.44 in Topos Theory by Johnstone. Ref 3: 9.11.3 in Sheaves in Geometry and Logic. REPLY [7 votes]: The proof you cite is certainly based in the completeness theorem for coherent logic, but that book contains a fully categorical proof of such a theorem, based on ideas of Joyal, so it can qualify as category-theoretic. It makes use of previous categorical constructions from other parts of the book. For a more geometric proof you can refer to Deligne's original proof, which is actually exposed in detail in Johnstone's other book "Topos theory", section 7.4. It is actually essentially the same argument, though you can distinguish the geometric flavour here. And there is of course the proof given in Maclane-Moerdijk "Sheaves in geometry and logic" which is based on Barr's theorem and is also quite geometric.<|endoftext|> TITLE: Which homotopy types can be realized as the classifying space of a right-cancellative discrete monoid? QUESTION [5 upvotes]: McDuff showed that every connected homotopy type can be realized as the classifying space of a discrete monoid, but the monoid she constructs has lots of idempotents. Question: Which homotopy types are realized as the classifying space of a right-cancellative discrete monoid? In the commutative case, my guess would be that $BM \simeq B(M[M^{-1}])$, so that the classifying space is aspherical. But I'm less confident that this happens in the noncommutative case. REPLY [5 votes]: By universal properties, we have that $BM$ is the classifying space of the homotopy localization $B(M[M^{-1}]^h)$. Thus $BM$ is aspherical if and only if the homotopy localization is discrete. Further, Dwyer-Kan showed that if $(M,W)$ admits a calculus of fractions, then the homotopy localization agrees with the ordinary localization. When $M$ is cancellative, $(M,M)$ admits a calculus of fractions if and only if it satisfies the Ore condition: $$\forall m_1, m_2 \in M, \exists n_1, n_2 \in M, ~ n_1 m_1 = n_2 m_2.$$ So in this case $M[M^{-1}]^h \simeq M[M^{-1}]$ and the homotopy localization is aspherical. In general, I do not know what happens.<|endoftext|> TITLE: Dividing a chocolate bar into any proportions QUESTION [12 upvotes]: Suppose I have a chocolate bar of integer length $L$, and there are $m\leq L$ people that are going to share it. We do not know ahead of time how much each person should receive, all we know is that they will receive an integer amount. Our goal is to chop the bar into the smallest number of pieces possible, such that, no matter what the desired amounts are for each person, there exists a way of assembling the pieces together so that each person receives that amount. It is easy to verify that if $m=2$, then it is possible to accomplish this with $ \lceil \log_{2}(L+1)\rceil $ pieces. I also verified computationally, for example, that for $L=19$ and $m=3$, then one possible optimal solution is to divide the bar into the following $7$ pieces: $7,4,3,2,1,1,1$. Is this a well-known problem? REPLY [4 votes]: Thanks to Fedor Petrov for an advice how to simplify the estimates. We present a simple algorithm which finds an optimal partition, and then find some estimates on the number of parts in it. Call a partition satisfying the requirements $m$-universal. Step 1. Consider any partition of $L$. For every $k$, denote by $S_k$ the sum of all parts not exceeding $k$. We claim that a partition is $m$-universal if and only if $(*)$ $S_k\geq (m-1)k$ whenever $(m-1)k\leq L$. The `only if' part is easy: if $m-1$ persons want to get $k$ units each, all of them should get a combination of pieces accounted for $S_k$. For the `if' part, distribute the pieces to the people, starting from the largest ones. When we want to give to someone a piece of length $k$, the total length of non-distributed pieces is at least $S_{k-1}+k\geq (m-1)(k-1)+k=m(k-1)+1$, so there is a person who needs at least $k$; give them the piece and continue. The claim is proved. Step 2. Now we present an algorithm which finds one of optimal $m$-universal partitions. Define the sequence $a_1,a_2,\dots$ inductively as follows. On the $i$th step, when $a_1,\dots,a_i$ are already defined, find the minimal $k$ such that $\Sigma_i:=a_1+\dots+a_i<(m-1)k$ and set $a_{i+1}=k$ (thus, in partivular, $a_1=1$). In other words, $$ a_{i+1}=\left\lfloor\frac{\Sigma_i}{m-1}\right\rfloor+1. $$ For any $i$ we have $$ (a_i-1)(m-1)\leq \Sigma_{i-1}<\Sigma_ia_i$ for some (minimal) $i$. By the definition of $a_i$, we have $a_1+\dots+a_{i-1}<(m-1)a_i$ and hence $c_1+\dots+c_{i-1}<(m-1)a_i$ as well. But then in $\mathcal Q$ we have $S_{a_i}<(m-1)a_i$, so $\mathcal Q$ violates $(*)$. Thus, our algorithm indeed provides an optimal partition. Step 3. Notice that $\Sigma_n$ is the largest value of $L$ for which we can survive with $n$ pieces. Recall that $a_n=\lfloor \Sigma_{n-1}/(m-1)\rfloor+1$, so $$ \frac {m\Sigma_{n-1}+1}{m-1}\leq \Sigma_n\leq\frac m{m-1}\Sigma_{n-1}+1. $$ This rewrites as $$ \frac m{m-1}(\Sigma_{n-1}+1)\leq \Sigma_n+1 \quad\text{and}\quad \Sigma_n+(m-1)\leq \frac m{m-1}\left(\Sigma_{n-1}+(m-1)\right). $$ Therefore, $$ \left(\frac m{m-1}\right)^{n-t}(\Sigma_t+1)-1\leq \Sigma_n \leq \left(\frac m{m-1}\right)^{n-t}(\Sigma_t+(m-1))-(m-1). $$ Now, we may choose an appropriate value of $t$. Setting $t=m$ (where $\Sigma_m=m+1$), we get $$ \left(\frac m{m-1}\right)^{n-m}(m+2)-1\leq \Sigma_n\leq 2m\left(\frac m{m-1}\right)^{n-m}-(m-1). $$ This, in view of $\Sigma_{s-1} TITLE: Mathematical uses of string theory QUESTION [33 upvotes]: It is widely believed that correctness of string theory as a physical theory will not be decided in the near future. Regardless whether this will turn out to be correct or not, mathematical concepts derived from string theory may have proved theorems whose correctness (and relevance for mathematics) is undisputed. What are (important in mathematics) mathematical theorems which wouldn‘t have been proved without the development of string theory? REPLY [5 votes]: I recall that Richard Wentworth's first paper on precise constants in bosonization formula (which is part of his PhD thesis?) extensively used computational methods from bosonic string theory. I am not sure if the methods have been justified, since his second paper used completely different methods to derive the same result. Later I was informed that Jorgenson proved the same result using much more classical methods like asymptotic expansion of heat kernel, construction and estimate of the paramatrix, etc. To me I feel the fact that path integral and $\zeta$-function regularization methods "coincide" in actual computation for topics related to Polyakov measure is not a mere coincidence. I do not really know string theory, but this observation striked me as something deep and subtle connecting physics to mathematics.<|endoftext|> TITLE: Which mathematician sampled published proofs and found one third of them to have errors? QUESTION [10 upvotes]: A recent question about whether/how we can trust mathematics in the face of human fallibility reminded me of a paper or article I read probably more than twenty years ago about a mathematician who was working at Bell Labs (I think) that had developed a novel proof mechanism. (He might have called it a "lucid proof"?) As I recall, it consisted of taking every single concept in the proof that wasn't blindingly obvious and giving it its own "appendix" where the proof of that bit was expanded until it was blindingly obvious that said part was true, possibly with its own appendices, etc, until every claim of the proof was fully exhausted in that manner. Once he had the mechanism working, he tested it against some of his previous papers. To his horror, he found out that a bunch of his previous results were wrong. When he forced himself to eliminate every last shred of doubt about every claim, it turned out that many of his papers had claims--which had seemed obvious enough to not go through in excruciating detail at the time of writing the paper--which were, in fact, actually incorrect. The way I remember it was that his initial reaction was something along the lines of "holy crap, I'm an awful mathematician!". Then it occurred to him to check the published work of other authors. From a random sampling (I doubt this was a statistically rigorous sample, I don't think that was the point) of published works, he found that a third of the results he tested failed to prove out when attacked with this method. I have occasionally tried searching for this article to no avail, although in preparation for asking this question I tried again, and found this from Leslie Lamport which might refer to it: Anecdotal evidence suggests that as many as a third of all papers published in mathematical journals contain mistakes—not just minor errors, but incorrect theorems and proofs. How to Write a Proof (1993) [EDIT: Maybe Lamport is the person, this paper describes the proof mechanism, and that "anecdotal evidence" he cited was from his own investigation. If you read the linked PDF, you will see that many of the parts of the story are there. It might well be that I mixed up Bell Labs with DEC, for example...] The copy of the paper that I read was downloaded as a .ps file from some website in the 90s if I remember correctly. I remember wondering if anyone paid attention to this result, if not, why not, etc, but I have not been able to locate it since. Does anyone know who the mathematician was, or where I can find the paper? I would also be happy to find out what Lamport is referring to in the quoted section of the linked paper, if it isn't this. Or anything that will help me pick up this trail. REPLY [4 votes]: I don't want anyone wasting time chasing this down for me, now that I've actually read the rest of the document I linked to in the question, I'm just going to assume that the mathematician I'm looking for is in fact Leslie Lamport or one of the people he mentions collaborating with. He refers to this proof mechanism by the term "structured proof". The assertions or guesses about how much published literature might be incorrectly proven, in addition to the one-third number quoted above (and which @ToddTrimble mentions having heard from Lamport in a comment) are alluded to here (references in the original, emphasis mine): The style was first applied to proofs of ordinary theorems in a paper I wrote with Martín Abadi 1. He had already written conventional proofs—proofs that were good enough to convince us and, presumably, the referees. Rewriting the proofs in a structured style, we discovered that almost every one had serious mistakes, though the theorems were correct. Any hope that incorrect proofs might not lead to incorrect theorems was destroyed in our next collaboration [3]. Time and again, we would make a conjecture and write a proof sketch on the blackboard—a sketch that could easily have been turned into a convincing conventional proof—only to discover, by trying to write a structured proof, that the conjecture was false. Since then, I have never believed a result without a careful, structured proof. My skepticism has helped avoid numerous errors. I'm not sure if this document is the exact one that I read, but it's certainly close enough.<|endoftext|> TITLE: External tensor product of irreducible representations is not irreducible? QUESTION [12 upvotes]: I'm writing up some notes, and I realize I don't have a counterexample for something I suspect is false. Dubious claim: If $(\pi, V)$ and $(\rho, W)$ are irreducible representations of two groups $G$ and $H$, respectively, then the "external" tensor product $\pi \boxtimes \rho$ is an irreducible representation of $G \times H$. Of course this is true and well-known in the usual cases, e.g., when $G$ and $H$ are finite groups. The proof I know uses the converse to Schur's Lemma, or something similar. Is there a nice counterexample for complex representations of some infinite groups? Published somewhere? REPLY [13 votes]: You can generate examples from standard counterexamples to (generalisations of) Schur's lemma. Let E/F be a field extension. Let $G=H=E^\times$, acting on the F-vector space E. Then the external tensor product is not irreducible, for example the kernel of the multiplication map $E\otimes_F E\to E$ is a submodule.<|endoftext|> TITLE: Root lattices and (resolutions of) singular cubic surfaces QUESTION [5 upvotes]: (Cross-posted from math.SE since I'm not sure what platform is suitable -- see https://math.stackexchange.com/questions/3331104/root-lattices-and-resolutions-of-singular-cubic-surfaces) Given a smooth cubic surface $X$ (say over $\mathbb{C}$) considered as a blowup of $\mathbb{P}^2$ at $6$ points) with Neron-Severi group $\operatorname{NS}(X)$ and canonical divisor $K_X$, the subset $R = \{ \alpha \in \operatorname{NS}(X) : \alpha\cdot K_X = 0, \alpha\cdot\alpha = -2\}$ is the root system of the Weyl group of $E_6$. I understand that there are previous questions on MSE related to this lattice (e.g. https://math.stackexchange.com/questions/1477220/27-lines-on-a-cubic-surface and https://math.stackexchange.com/questions/82199/automorphism-group-of-the-configuration-of-lines-on-a-cubic-surface-and-quadrati), but I haven't been able to find an explanation about what happens for singular cubic surfaces (on this site or other sources). More specifically, suppose that $\varphi : X' \longrightarrow X$ is a "minimal" resolution of singularities of a singular cubic surface $X$ with rational double points as singularities. Suppose that $X \subset \mathbb{P}^3$ has rational double points as singularities. For simplicity, we take $X$ to have only one singular point (which we take to be $(1 : 0: 0 : 0)$) with singularity type $A_1$ or $A_2$. Then, what sublattice of the root system described above for $X'$ does the components of the exceptional divisor of this resolution generate? Under the assumptions above, we can write $X = (f = 0)$ with $f(t_0, t_1, t_2, t_3) = t_0 g_2(t_1, t_2, t_3) + g_3(t_1, t_2, t_3)$ for some homogeneous polynomials $g_2$ and $g_3$ of degrees $2$ and $3$ respectively (listed in many sources, such as section 9.2 of Dolgachev - Classical algebraic geometry). In the $A_2$ case, I've tried looking at $(-2)$-curves coming from two consecutive blowups of three collinear points (blow up one point, and then some point on the exceptional divisor in the next blowup). We repeat this for two collinear triples of points. It isn't clear to me how to obtain a sublattice of $R$ (constructed for $X'$) that keeps track of the information above. I think this is what I need to better understand "usual" generators for the root lattice in the smooth case. For example, what do classes in $X'$ coming from classes of lines passing through the singular point in $X$ look like? Are there any suggestions on how to proceed in this situation (or even in the $A_1$ case or other singularity types)? REPLY [7 votes]: I think you will find everything you want in Demazure, Pinkham, Teissier (eds.), Séminaire sur les singularités des surfaces, Springer LNM 777. Over non-closed fields there's also an article by Coray and Tsfasman, "Arithmetic on singular del Pezzo surfaces", Proc. LMS 57 (1988). Briefly, the answer is as nice as you might hope. The resolution is what's called a generalised del Pezzo surface of degree 3, that is, a smooth surface obtained by blowing up the projective plane in 6 points in "almost general position". Its Picard lattice is the same as that of a smooth cubic surface; what differs is which classes are effective. It turns out that the effective $-2$-curves form a set of positive roots in a sub-root system of $R$. The possible configurations of rational singularities on a cubic surface correspond exactly to the sub-root systems of $E_6$. (I believe this all holds more generally for del Pezzo surfaces of degree at least 3; there's an article by Urabe showing that in degrees 1 and 2 there are a few root systems that aren't realised by del Pezzo surfaces.)<|endoftext|> TITLE: A characterisation of faces of rational polyhedral cones QUESTION [5 upvotes]: This is about a (seemingly) basic lemma about rational polyhedral cones that is sometimes used when working with toric varieties and is usually "left to the reader". Unfortunately, I could neither prove it myself nor find a complete proof in the literature. So, I am looking for either a proof or a proper reference. Lemma. Let $V$ be an $\mathbb{R}$-vector space of finite dimension, and let $N$ be a $\mathbb{Z}$-structure on $V$ (i.e., a free abelian group with $N\otimes_{\mathbb{Z}}\mathbb{R}=V$). Let $\sigma$ and $\tau$ be $N$-rational polyhedral cones in $V$, and suppose that $\tau$ is a subset of $\sigma$. Then, the following statements are equivalent: (i) $\tau$ is a face of $\sigma$; (ii) If $x,y\in\sigma$ with $x+y\in\tau$, then $x,y\in\tau$; (iii) If $x,y\in\sigma\cap N$ with $x+y\in\tau$, then $x,y\in\tau$. Showing that (i) and (ii) are equivalent and imply (iii) is clear. My problem is to show that (iii) implies (i) or (ii), i.e., that it suffices to consider only rational points. Note 1. Trying to show (iii)$\Rightarrow$(i) analogously to (ii)$\Rightarrow$(i) leads to the question whether $(\sigma-\tau)\cap N=(\sigma\cap N)-(\tau\cap N)$, which has a negative answer in general. Note 2. Condition (iii) is sometimes expressed by saying that the monoid $\tau\cap N$ is a face of the monoid $\sigma\cap N$, and similarly for (ii). REPLY [4 votes]: To prove that (iii) implies (i), assume w.l.o.g. that $\tau\neq\sigma$. We first need to show that $\tau\subset \partial \sigma$. If this is not the case then either there exists a hyperplane $H$ containing $\tau$ such that $H$ is not a supporting hyperplane for $\sigma$, or $\tau$ is full-dimensional. If $H$ exists then there are $x,y\in\sigma$ on different sides of $H$, which may be chosen in $N$ as $\sigma$ is rational, so that $x+y\in\tau$, contadicting (iii). If $H$ does not exist, then $\dim \tau=\dim\sigma$, and there exists $x\in\sigma\setminus\tau$ ($x$ may be chosen to be a generator for an extreme ray of $\sigma$) and $y\in\tau\setminus\partial\tau$ (again, it's possible to choose $x,y\in N$) so that $x+y\in\tau$, again contradicting (iii). Thus $\tau\subset \partial \sigma$. Let $\sigma'$ be the minimal face of $\sigma$ containing $\tau$. Note that $\sigma'$ is a rational polyhedral cone, and we are basically in the situation as above, with $\sigma$ replaced by $\sigma'$, except that we don't have any more dimensions to spare, and so either $\tau=\sigma'$, as required, or we contradict (iii) as above.<|endoftext|> TITLE: Examples of topoi that are not ordinary spaces QUESTION [14 upvotes]: In [SGA6] we find: Mais nous lui conseillons néanmoins, de préférence, de s'assimiler le langage des topos, qui fournit un principe d'unification extrêmement commode. (DeepL translate: However, we nevertheless advise him, preferably, to assimilate the language of toposes, which provides an extremely convenient principle of unification.) Lets motivate this advice following some examples of topoi that are not “ordinary” spaces: Comme autres exemples remarquables de topos qui ne sont pas des espaces ordinaires, et pour lesquels il ne semble pas y avoir non plus de substitut satisfaisant en termes des notions “admises”, je signalerai : les topos quotients d’un espace topologique par une relation d’équivalence locale (par exemple des feuilletages de variétés, auquel cas le topos quotient est même une “multiplicité” i.e. est localement une variété) ; les topos “classifiants” pour à peu près n’importe quelle espèce de structure mathématique (tout au moins celles “s’exprimant en termes de limites projectives finies et de limites inductives quelconques”). Quand on prend une structure de “variété” (topologique, différentiable, analytique réelle ou complexe, de Nash, etc. … ou même schématique lisse sur une base donnée) on trouve dans chaque cas un topos particulièrement alléchant, qui mérite le nom de “variété universelle” (de l’espèce envisagée). Ses invariants homotopiques (et notamment sa cohomologie, qui mérite le nom de “cohomologie classifiante” pour l’espèce de variété envisagée) devraient être étudiés et connus depuis longtemps, mais pour le moment ça n’en prend nullement le chemin…. [ReS] DeepL translation: As other remarkable examples of topos that are not ordinary spaces, and for which there also seems to be no satisfactory substitute in terms of “accepted” notions, I would point out: the topos quotients of a topological space by a local equivalence relation (e.g. foliations of manifolds, in which case the topos quotient is even a “multiplicity” i. e. is locally a manifold); “classifying” topos for almost any species of mathematical structure (at least those “expressed in terms of finite projective limits and any inductive limits”). When we take a “manifold” structure (topological, differentiable, real or complex analytical, Nash, etc. … or even smooth schematic on a given base) we find in each case a particularly attractive topos, which deserves the name of “universal variety” (of the species considered). Its homotopic invariants (and in particular its cohomology, which deserves the name of “classifying cohomology” for the species of variety considered) should have been studied and known for a long time, but for the moment it does not take any way…. What are some other examples of topoi that are not “ordinary” spaces? [ReS] Récoltes et Semailles, A. Grothendieck [SGA6] SGA6 Théorie des intersections et théorème de Riemann-Roch, 1966–1967. Séminaire de Géométrie Algébrique du Bois Marie doi:10.1007/BFb0066283 REPLY [10 votes]: This is a translation of Dmitri Pavlov's answer into a more intrinsic, more geometric, and more elementary language. In particular, I will show that the étale topos of a positive-dimensional variety is never the topos of a topological space. If $X$ is a topological space, then the associated topos $E = \mathbf{Sh}(X)$ is generated by subobjects of the final object $\mathbf 1_X$. In fact we may take the sheaves $h_U = j_{U,!}(\mathbf 1_U)$ for $U \subseteq X$ open: if two maps $\mathscr F \rightrightarrows \mathscr G$ are different, there is an open $U \subseteq X$ such that $\mathscr F(U) \rightrightarrows \mathscr G(U)$ don't agree, so we get a map $j_{U,!}(\mathbf 1_U) \to \mathscr F$ such that the compositions $j_{U,!}(\mathbf 1_U) \to \mathscr F \rightrightarrows \mathscr G$ differ. On a general site $\mathscr C$, the same argument shows that the $j_{U,!}(\mathbf 1_U)$ for $U \in \mathscr C$ generate the topos $E = \mathbf{Sh}(\mathscr C)$. But the difference is that we cannot do this with subobjects of $\mathbf 1$. For example, if $G$ is a discrete group, then the terminal object $\mathbf 1$ of $BG = G\text{-}\mathbf{Set}$ is a point with a trivial $G$-action, which does not have any subobjects except $\varnothing$ and $\mathbf 1$. More generally¹, let $\mathscr C$ be a subcanonical site with fibre products and a terminal object $X$, such that every object $V$ covers a subobject $W \subseteq X$ (for example, the (small) étale site $X_{\operatorname{\acute et}}$ on a scheme $X$, since $W = \operatorname{im}(V \to X)$ is open and $V \to W$ a covering). Assume $\mathscr C$ has an object $U$ that does not admit a map from a subobject of $X$ (for example, the étale site of any positive-dimensional variety $X$ over a field $k$, or of $\operatorname{Spec} k$ when $k$ is not separably closed). Then I claim that $h_U = j_{U,!}(\mathbf 1_U)$ does not admit any map from a nonempty subobject of $\mathbf 1_X$. Indeed, suppose $\mathscr F \subseteq \mathbf 1_X$ maps to $h_U$. Let $V \in \mathscr C$, and choose a covering $V \to W$ of a subobject $W \subseteq X$. By assumption, $h_U(W) = \varnothing$, so $\mathscr F(W) = \varnothing$. But since $V \to W$ is a covering and $\mathscr F \subseteq \mathbf 1_X$, for any section in $\mathscr F(V)$ the restrictions $\mathscr F(V) \rightrightarrows \mathscr F(V \times_W V)$ vacously agree, hence they glue to a section of $\mathscr F(W)$, which is impossible. We conclude that $\mathscr F(V) = \varnothing$, so $\mathscr F = \varnothing$ as $V$ was arbitrary. $\square$ So the étale topos on a reasonable scheme $X$ is basically never a topos of sheaves on a topological space. ¹For example, the case $\mathscr C = (\operatorname{Spec} \mathbf R)_{\operatorname{\acute et}}$ recovers the example $B(\mathbf Z/2)$ above.<|endoftext|> TITLE: Minimizing the number of segments in drawings of planar graphs QUESTION [6 upvotes]: Every planar graph has at most $3n-6$ edges, where $n$ is the number of vertices. Moreover, every planar graph can be drawn with straight-line edges in the plane, without crossings. For example, for the complete graph $K_4$ one actually need 6 segments, so its segment number in 6. For any path its segment number is 1. Dujmović, Eppstein, Suderman and Wood (2006) have shown that every planar graph can be drawn with at most $\frac52 n$ segments (where $n$ is the number of vertices). Can this be improved? Remark. There are examples of planar graphs that need $2n\pm O(1)$ segments. The problem was asked on 24.03.2019 by Alex Ravsky and Alexander Wolff on pages 94-95 of Volume 2 of the Lviv Scottish Book.                     Fig.1 in Dujmović et al. (a) shows a path has segment number $1$. REPLY [3 votes]: Recently, Durocher and Mondal improved the upper bound for plane triangulations to $7n/3$: https://doi.org/10.1016/j.comgeo.2018.02.003<|endoftext|> TITLE: On sums of independent random variables in Banach spaces QUESTION [14 upvotes]: Let $(\xi_n)_{n\ge 1}$, $(\eta_n)_{n\ge 1}$ be independent mean-zero random variables with values in a Banach space $X$ such that $$\sum_n\mathbb P(\xi_n\in A)\le\sum_n\mathbb P(\eta_n\in A)$$for any Borel set $A\subset X\setminus\{0\}$. Let $1\le p<\infty$. Is there a constant $C$ (perhaps depending on $p$ and $X$) such that $$\textstyle\mathbb E\|\sum_n\xi_n\|^p\le C\,\mathbb E\|\sum_n\eta_n\|^p?$$ This problem was posed on 28.06.2019 on page 133 of Volume 2 of the Lviv Scottish Book. REPLY [3 votes]: First of all, many thanks to @LvivScottishBook for putting the problem online (that was indeed very surprising for me) and to @fedja for finding a counterexample. While writing the problem I was sure that it has a positive answer for any Banach space (for some reason this was kind of intuitive for me) with $C$ independent of $X$, so @fedja answered the original question. Nonetheless, as @fedja noticed, it is not clear for which Banach spaces such an inequality holds true. The only thing that I can guarantee so far is that the necessary assumption on the Banach space is having a finite cotype thanks to @fedja counterexample and the Maurey–Pisier theorem (see Corollary 7.3.14 in Analysis in Banach spaces, Volume II by Hytönen, van Neerven, Veraar, and Weis). It remains open whether finite cotype is a sufficient condition as well.<|endoftext|> TITLE: Remarkable applications of Dickson's lemma QUESTION [6 upvotes]: Dickson's lemma states that, for a fixed $k \in \mathbf N^+$, every set of $k$-tuples of natural numbers has finitely many elements that are minimal with respect to the product order induced on $\mathbf N^k$ from the integers. The lemma has many and diverse applications: E.g., it was used by L.E. Dickson in his work on odd perfect numbers (that's where the lemma takes its name from); and by P. Gordan as part of his 1899 proof of Hilbert's basis theorem (though it is probably more common to use Hilbert's basis theorem to prove Dickson's lemma, and hence to regard the former as a generalization of the latter). Further applications include finiteness results in factorization theory, where the lemma is used, e.g., to show that every atomic, cancellative, commutative monoid that is finitely generated up to units, has accepted elasticity (with all the nice things this implies for a wide spectrum of arithmetical invariants including local elasticities, distances, unions of sets of lengths, etc.). Question. What are some striking applications of Dickson's lemma you have encountered in your research? One answer per post, and please provide some context (if possible). This post should be community wiki, but I don't know how to make it so. REPLY [3 votes]: In theoretical computer science, Dickson's lemma (or one of its equivalent forms) is frequently used to show that various problems involving the behavior of structured transition systems are decidable. One example is the coverability problem for vector addition systems: given an alphabet of vectors $\{ v_1, \ldots , v_k \}\subset \mathbb{Z}^n$ and starting/target vectors $v_0, u\in \mathbb{Z}_{\ge 0}^n$, we're looking for a sequence $v_{i_1},\ldots , v_{i_k}$ such that $v_0 + v_1 + \ldots + v_{i_j}\ge 0$ for all $j$ and $v_0 + v_1 + \ldots + v_{i_k}\ge u$---or a certificate that no such sequence exists. Rackoff proved that, if such a sequence exists, we may take $k$ to be doubly exponential in $n.$ [1] The proof technique can be understood in terms of ascending chains of monomial ideals---in another language, these are descending chains of order ideals. [2] The complexity of this and related problems are connected to the following effective variant of Hilbert's basis theorem: how long can a strictly ascending chain of ideals $$I_1\subsetneq I_2 \subsetneq \cdots \subsetneq I_l$$ in $k[x_1,\ldots , x_n]$ be? To make some sense of the question we should control the degrees of the generators of each $I_i$---assume we have some explicit function $f: \mathbb{N} \to \mathbb{N}$ such that each $I_i$ is generated in degree $f(i).$ For each fixed $n,$ if the function $f$ is primitive recursive, then there exists a bound (depending primitive-recursively on $f$) on the length of such a chain. [3] However, the dependence of this bound on $n,$ even when $f(i)$ grows linearly in $i,$ is generally not primitive recursive! [4] This is somewhat remarkable, given that we "expect" doubly-exponential bounds for similar-looking questions. This problem was also studied in the context of polynomial dynamical systems. [5] In the case of interest to the authors, they are able to determine better bounds (doubly exponential) by exploiting the fact that the dimensions of successive colon ideals $\dim (I_k:I_{k+1})$ are non-increasing. Some personal context is that these problems came up in my master's research---part of which was incorporated into the paper [6]. [1] Rackoff, Charles. "The covering and boundedness problems for vector addition systems." Theoretical Computer Science 6.2 (1978): 223-231. [2] Lazić, Ranko, and Sylvain Schmitz. "The ideal view on Rackoff’s coverability technique." International Workshop on Reachability Problems. Springer, Cham, 2015. [3] Seidenberg, A. "On the length of a Hilbert ascending chain." Proceedings of the American Mathematical Society 29.3 (1971): 443-450. [4] Socias, Guillermo Moreno. "Length of polynomial ascending chains and primitive recursiveness." Mathematica Scandinavica (1992): 181-205. [5] Novikov, Dmitri, and Sergei Yakovenko. "Trajectories of polynomial vector fields and ascending chains of polynomial ideals." Annales de l'institut Fourier. Vol. 49. No. 2. 1999. [6] Benedikt, Michael, et al. "Polynomial automata: Zeroness and applications." Proceedings of the 32nd Annual ACM/IEEE Symposium on Logic in Computer Science. IEEE Press, 2017.<|endoftext|> TITLE: Odd differential forms QUESTION [8 upvotes]: In de Rham's classical book "Variétés Différentiables" de Rham, Georges, Variétés différentiables. Formes, courants, formes harmoniques. 3e éd. revue et augmentée, Publications de l’Institut de Mathématique de l’Université de Nancago III. Actualités scientifiques et industrielles 1222 b. Paris: Hermann. X, 198 p. (1973). ZBL0284.58001. it is defined the concept of an differential form of "odd type" that carries a sign and makes sense in non-oriented(able) manifolds. Since we usually only encounter so called forms of "even type" in modern standard textbooks, I was wondering what further developments these odd forms had in the past decades. Then my question is: are these forms of odd type still studied? Do they have any further applications? REPLY [4 votes]: A form of odd type is the same thing as a pseudoform, as mentioned by Mike Shulman. This is a form twisted by the pseudo-scalar bundle $\Psi$. The issue with the term "density" is that it has multiple meanings. It can refer to forms twisted by $\Psi$ or by $\bigwedge^{\!n}T^*\!X$ (where $X$ is an $n$-dimensional manifold). These bundles are isomorphic, but not canonically so, because the isomorphism depends on a choice of Riemannian metric. I think that pseudo-forms are more natural objects to integrate and to multiply. As described in de Rham, we get a system of multiplication for even and odd forms, where the parity of multiplication looks like addition in $\mathbb Z/2\mathbb Z$. This is because $\Psi\otimes \Psi$ has a canonical trivialization, so the product of two $\Psi$-twisted forms can be identified with a form twisted by the trivial line bundle, i.e. our usual notion of a differential form. Unfortunately, I don't have the knowledge/expertise to point to various present applications, so this answer is somewhat incomplete. But I can say how I stumbled upon it: I am reading de Rham in order to understand currents, so that I can read Harvey and Lawson - Finite volume flows and Morse theory. Currents are a generalization of forms of even and odd type, allowing us to integrate "Schwartz distributions" on manifolds. The paper I mentioned shows that pulling back a differential form by a nice enough flow will limit towards a current as $t\rightarrow \infty$. The authors use this notion to provide a beautiful description of Morse homology.<|endoftext|> TITLE: An extrapolation method QUESTION [9 upvotes]: I've stumbled upon a method of extrapolation that I haven't seen before. We are trying to approximate $f(0)$ for a certain function $f$, which we have only measured at points $x_0, \ldots, x_N$ in an interval $[a,b]$ that does not contain $0$. We have reason to believe that $f$ is analytic in a neighbourhood of a region of $\mathbb C$ (containing both $0$ and $[a,b]$) bounded by a simple positively oriented closed contour $\Gamma$. Suppose we can approximate $1/z$ on $\Gamma$ by a linear combination of $1/(z-x_j)$, say $$ \left| \frac{1}{z} - \sum_{j=0}^N \frac{a_j}{z-x_j} \right| \le \varepsilon \ \text{for}\ z \in \Gamma $$ Then using Cauchy's formula, $$ \left|f(0) - \sum_{j=0}^N a_j f(x_j) \right| \le \frac{M\; \text{length}(\Gamma) \varepsilon}{2\pi}$$ Rather than uniform approximation, it is more convenient to use an $L^2$ approximation. This will let us find the $a_j$ by minimizing a quadratic form. In the case where $\Gamma$ is a circle of radius $r$ centred at $0$, I get a nice closed form: $$ a_j = \frac{r^2 - x_j^2}{r^{2N+2}} \prod_{k \ne j} \frac{x_k (r^2 - x_j x_k)}{x_k - x_j}$$ I can't believe I'm the first to think of this idea. Has anyone seen something like this? REPLY [5 votes]: There are extrapolation schemes in use that take essential recourse to analyticity, for example, the "z-expansion fits" described in arXiv:1008.4619 (it's sufficient to read the first two pages to see what the method is). In the details, however, these fits differ substantially from the scheme you are proposing. I have not seen anything that is very similar to your scheme. Having said that, I worry about the stability of your scheme. You say that you "measure" the function $f$, which I assume means that, generically, you don't have exact data on $f$, but they come with statistical or systematic error. In that case, it is important that the scheme be robust against these errors. Let's look at an example that, I think, by no stretch of the imagination can be dismissed as contrived: 10 equally spaced data points at $x$ ranging from $x=0.1$ to $x=1$, with $r=1.5$, i.e., in your notation, $N=9$, $x_j = 0.1(j+1)$, $r=1.5$. Then, your coefficients $a_j $ are: $a[0]=7.80096236306711$ $a[1]=-27.161848927028$ $a[2]=55.5523167543375$ $a[3]=-73.8548460044706$ $a[4]=66.6361882416331$ $a[5]=-41.2843129034312$ $a[6]=17.3237609884424$ $a[7]=-4.70612791676158$ $a[8]=0.746272144521954$ $a[9]=-0.0523647403501327$ The coefficients are large and alternating! This means they are prone to magnify fluctuations of the data. Let's take the simplest possible true functional dependence, $f(x)=1$. As long as you have no errors in the data, $f(x_j )=1$, you predict $f(0)=1$ with high accuracy. However, as soon as I put small random fluctuations on the data $f(x_j )$, the prediction fluctuates wildly. Here is a sequence of 10 predictions for $f(0)$ when I put random fluctuations on the $f(x_j )$ that are smaller than 1% (i.e., I'm adding $0.01 (2 \cdot rand()-1)$ in perl): $1.978$ $1.114$ $0.529$ $1.286$ $1.032$ $-0.179$ $1.384$ $0.102$ $0.371$ $1.656$ Thus, 1% errors in the data are amplified to 100% errors in the predictions of $f(0)$, which is of course not at all surprising in view of the $a_j $. So, unfortunately, I'm not sure whether the fact that we're having a hard time finding someone who has seen a scheme of the type you describe used means that no one has thought of it before ...<|endoftext|> TITLE: Papers on arXiv solving the same problem at the same time QUESTION [29 upvotes]: I am little bit curious about the following examples at least two papers appeared on arxiv at the same day solving one and the same problem. Have you ever seen such a coincidence? If yes, can you provide the links to that papers? of course the papers maybe solving one and the same problem in a slightly different generalities, for example, one solved a certain case of the problem, and the other one solved the problem completely. the authors should be different, for example, it is not allowed that both papers have one and the same person as an author/co-author. REPLY [3 votes]: This happened in 2013 in the context of finding $\epsilon$-approximate solutions to the maximum $s-t$ flow problem in near-linear-time (taking day to mean any 24 hour period). The two references are https://arxiv.org/abs/1304.2338 (Jonathan A. Kelner, Yin Tat Lee, Lorenzo Orecchia, Aaron Sidford) and https://arxiv.org/abs/1304.2077 (Jonah Sherman). As usual, the timing was not a coincidence.<|endoftext|> TITLE: Piecewise linear Poincaré conjecture QUESTION [10 upvotes]: Let $M$ be a PL-manifold that is a homotopy sphere (PL stands for Piecewise Linear). Does it follow that $M$ is PL-homeomorphic to the sphere $S^n$ (with the usual PL-structure)? Here is the background: Zeeman (1962 [ 2 ]: The Poincaré conjecture for $n\geq 5$) writes: [Smale] assumed a differentiable structure upon the manifold instead of a combinatorial structure, and because of the triangulation theorem for differentiable manifolds, this was at first a weaker version than Stallings' result. Then, however, he used a constructed structed differentiable structure to prove a stronger combinatorial version (which we state without proof): ``THEOREM 3 (Differential: Smale). Let $M^n$ be a connected closed combinatorial manifold, $n\ne 4,5,7$. If $M^n$ is a homotopy sphere then it is a combinatorial sphere.'' I cannot find a proof for this statement. The best I could find is the following: Stallings (Polyhedral homotopy spheres [ 1 ]) says that it is combinatorially equivalent to $\mathbb{R}^n$ away from a point. [ 1 ] J. Stallings. Polyhedral homotopy-spheres. Bull. Amer. Math. Soc. 66 (1960), 485-488. ProjetEuclid link. [ 2 ] E. Zeeman. The Poincaré conjecture for $n\ge 5$. 1962 Topology of 3-manifolds and related topics (Proc. The Univ. of Georgia Institute, 1961) pp. 198-204 Prentice-Hall, Englewood Cliffs, N.J. REPLY [10 votes]: For spheres of dimension $n>5$ the PL Poincare conjecture follows from the s-cobordism theorem. Indeed, removing disjoint two small open disks one gets an s-cobordism (this uses excision in homology, Poincare duality and Hurewicz theorem). The s-cobordsim is PL trivial (if the cobordsim has dimension is $>5$) from where a PL homeomorphism to a sphere should be obvious. This is done in detail eg in Rourke-Sanderson PL topology book, pp 8-10. The 5-dimensional case involves a trick. By a surgery theoretic argument Kervaire proved that every PL homology $n$-sphere $\Sigma$ with $n\neq 3$ bounds a contractible PL manifold, see p.71 in [Smooth homology spheres and their fundamental groups. Trans. Amer. Math. Soc. 144 (1969), 67–72]. Removing a small open disk from the interior of that manifold gives a cobordism $W$ between $\Sigma$ and $S^5$ which is an $s$-cobordism since $\Sigma$ is simply-connected. Since $\dim(W)>5$ the s-cobordism is trivial, so $\Sigma$ is PL homeomorphic to $S^5$. It seems that the result was first discovered through the smoothing theory as follows: Step 1 is to show that any homotopy $5$-sphere has a smooth structure. Munkres and Hirsch developed an obstruction theory for putting a smooth structure on a PL manifold. The obstruction groups are cohomology groups of $M$ rel boundary (if any) with coefficients in the groups $\Gamma^k$ of isotopy classes of diffeomorphism of $S^{k-1}$ moduli those that extend to the $k$-disk. Now $\Gamma^k=0$ for $k\le 4$ where the hardest case $k=4$ is a famous theorem of Cerf. This shows that any PL $5$-manifold has a smooth structure. Step 2 is to show that any smooth homotopy $5$-sphere $\Sigma$ bounds a smooth compact contractible manifold $W^6$. This is done by a surgery theoretic argument staring with some $6$-manifolds with boundary $\Sigma$ and the simplifying it by surgery. This is due to Milnor-Kevaire and Wall (independently). See theorem 9.1 of Milnor's "Lectures on the h-cobordsim theorem". Step 3 is to remove a small open ball from the interior of $W^6$ to get a smooth $h$-cobordism between $\Sigma$ and $S^5$, which is trivial by the smooth $h$-cobordism theorem. Thus $\Sigma$ is diffeomorphic and hence PL homeomorphic to $S^5$.<|endoftext|> TITLE: Is transfinite composition of weak equivalences of simplicial presheaves a weak equivalence? QUESTION [5 upvotes]: In a left Bousfield localization of the projective model structure on the category of simplicial presheaves, what is the condition that transfinite composition preserves weak equivalences? How about for transfinite composition of fibrant simplicial presheaves? I apologize if the question doesn't make sense. My question comes from the proof here for Theorem 4.27 showing that the $A^1$ localization functor $L_{A^1}L_{Nis}$ is equivalent to the transfinite composition of $L_{Nis}Sing^{A^1}$, where they show that $L_{Nis}Sing^{A^1}$ takes values in $A^1$-fibrant objects and preserves $A^1$-weak equivalence then conclude the following. So I think the reason for this is that transfinite composition of $A^1$-weak equivalences between fibrant objects are $A^1$-weak equivalence. Is this true and how to prove it? Edit: The idea of the proof comes as following: Let $\Phi=L_{A^1}L_{Nis}$, they first show that $\Phi^{\circ \mathbb{N}}X$ is fibrant in $L_{A^1}L_{Nis} sPre(Sm_S)$ for any $X$. Then they show that $\Phi$ preserves $A^1$-local weak equivalences. Since $X\simeq \Phi (X)$ when $X$ is $A^1$-fibrant, applying $\Phi$ tranfinitely to this equivalence should give the following $\Phi^{\circ \mathbb{N}}(X)\simeq \Phi ^{\circ\mathbb{N}}(L_{A^1}L_{Nis} X)\simeq L_{A^1}L_{Nis} X$. The first equivalence comes from that $\Phi^{\circ \mathbb{N}}$ preserves weak equivalences if $\Phi$ does, the second should come from applying $\Phi$ tranfinitely to $X\simeq \Phi (X)$. But if weak equivalences are already closed under transfinite composition. Why do they need to show that $\Phi^{\circ \mathbb{N}}X$ is $A^1$-fibrant? REPLY [3 votes]: Yes, weak equivalences are always closed under transfinite compositions in this model category. The standard set of generating cofibrations of (a left Bousfield localization of) the projective model structure has compact domains and codomains. In any model category with such a property, weak equivalences are closed under transfinite compositions. See, for example, Lemma 2.0.3(iv) or Proposition 6.1.3(i) in the paper https://arxiv.org/abs/1510.04969, but this is a fairly standard fact.<|endoftext|> TITLE: Process quicker than Fourier for squares of polynomials QUESTION [5 upvotes]: FFT is a quick algorithm for multiplying two polynomials, but given it's a square (i.e. multiplying the polynomial with itself) can we find something better? REPLY [5 votes]: As requested, here is an answer. $$2(AB + BA) = (A+B)^2 - (A-B)^2$$ Is an identity that holds in rings in general. When the ring multiplication is commutative and one can divide by 4 nicely, this gives a means of multiplying $A$ and $B$ in terms of adding, subtracting, and two squaring operations. So any fast routine for squaring in the right kind of ring leads to a fast multiplication algorithm. It should be clear how fast multiplication leads to fast squaring. Back in another lifetime, I looked at implementing the above using programmable logic chips to produce a fast multiplier. (The basic idea was that minimizing the logic for squaring binary integers might lead to a smaller circuit depth.) The result did not gain much for 16 bit integers. Nowadays 64 bit arithmetic is "kept under the hood" meaning multiplication is seen as fast enough, so this approach of reducing circuit depth by squaring is unlikely to be practical. In the case that your domain is restricted, so that you are squaring a small class of polynomials, you might find a more optimal circuit/method, especially if the polynomials are multiplicatively generated, or if you have a fast transform like a logarithm/exponential pair available. At present, I am not aware of anything that in practice is much faster than FFT for squaring. Gerhard "It's All About The Speed" Paseman, 2019.08.24.<|endoftext|> TITLE: How many small dots can be drawn in a region such that no three are "collinear"? QUESTION [10 upvotes]: When people draw dots on paper, they are actually not points, but small regions filled with ink. Suppose that each dot has disc-shape with fixed radius $r\ll 1$ and must be drawn inside (1) a square region with side length $1$; (2) a circle region with radius $1$. How many dots can be drawn such that no three are collinear? Three dots are collinear if one of them intersects with the strip determined by the other two at more than one point. (I think it is equivalent to: there exists a line that intersects all the dots at more than one point.) Comment. I was working on an Olympic level problem that requires $100$ points in a circle without any three being collinear. It is a very difficult task with usual pen and paper. So I came up with this problem and hopefully someone has done similar problems before, or can give an algorithm to estimate the bound when $r$ is small. REPLY [4 votes]: For a circle, suppose we distribute $n$ dots evenly near the circumference. One dot has a center at $(1-r,0)$ and the adjacent dots have centers at $$((1-r)\cos(2\pi/n),\ \pm (1-r)\sin(2\pi/n))$$ The middle one goes as far to the left as $1-2r$, and the neighbors go as far to the right as $(1-r)\cos(2\pi/n)+r$ To avoid collinearity, we need \begin{align} (1-r)\cos(2\pi/n)+r &< 1-2r\\ n &< 2\pi/\arccos\left(\frac{1-3r}{1-r}\right) \end{align} Asymptotically, this means we can choose $n$ as large as $$\frac{\pi}{\sqrt{r}}\left(1-\frac{2r}{3}+O(r^2)\right)$$ which is a reasonable lower bound for the problem. For a unit square, $[-1/2,1/2]^2$, putting all the dots on one circle still seems to work well. We can try an alternative, putting dots on four circular arcs (black, blue, purple, red) each of which stays closer to one of the four sides. The black arc will have a center at $(-h,0)$, a radius of $k=1/2+h-r$, and a dot with a center at $(1/2-r,0)$. The dashed lines show radii of the black arc; we call the smallest central angle $\alpha$ and the biggest central angle $\beta$. Then the total number of dots on that arc will be twice the floor of $\beta/\alpha$ where \begin{align} -h+k\cos(\beta)&=k\sin(\beta)\\ -h+k\cos(\alpha)&=1/2-3r \end{align} Empirically, we maximize $\beta/\alpha$ at $h=0$, which is to say that we might as well take the four circular arcs to be part of the same circle. In the diagram, that corresponds to spacing the dots evenly outside the light gray circle instead. REPLY [2 votes]: Well one observation is that if three dots are not "collinear" then the triangle formed by all of the centers has the lengths of its altitudes being at least $r$. This implies that the area is at least $\frac{r^2}{\sqrt{3}}$. This relates the problem to the Heilbronn triangle problem which has some bounds.<|endoftext|> TITLE: Slicing up monads on categories with pullbacks: what are these mysterious "zerosumfree" monads" QUESTION [10 upvotes]: Introduction I'll describe a way of taking a monad on a category $\mathcal{E}$ with pullbacks, and obtaining a monad on each slice category. I'll show that this construction is always lax-natural in $\mathcal{E}$. However, for many monads on $\mathbf{Set}$ this construction is actually strongly natural, most interestingly in the case of the powerset monad and the probability distributions monad. For others, it's not natural, e.g. for the free abelian group monad $\mathbb{Z}[-]$. The condition seems to characterize monads with a certain sort of ``zerosumfreeness'' condition, e.g. $\mathbb{Z}[-]$ fails because zero can arise as the result of non-trivial linear combinations. The zerosumfree monads $M$ admit a certain strengthening of the usual notion of ``lax monoidality'' for the cartesian product monoidal structure: the products in the usual coherence maps $M(X_1)\times M(X_2)\to M(X_1\times X_2)$ can be replaced by fiber products $$M(X_1)\times_{M(X)}\; M(X_2)\to M(X_1\times_X X_2).$$ For example, for any functions $X_1\xrightarrow{f_1} X\xleftarrow{f_2} X_2$ in $\mathbf{Set}$, there is a natural map \begin{align*} \mathsf{Dist}(X_1)\times_{\mathsf{Dist}(X)}\;\mathsf{Dist}(X_2)&\to \mathsf{Dist}(X_1\times_X X_2)\\ (p_1,_pp_2)&\mapsto (x_1,_x x_2)\mapsto \frac{p_1(x_1)\cdot p_2(x_2)}{p(x)} \end{align*} where $\mathsf{Dist}(f_1)(p_1)=p=\mathsf{Dist}(f_2)(p_2)$ is the distribution on $X$ induced by $p_1$ and $p_2$, and $f_1(x_1)=x=f_2(x_2)$. Questions: Where else do these "zerosumfree" monads arise and have they been studied? Have these ``stronger lax monoidal conditions" for categories with pullbacks been studied? (Edit: what I'm now calling "zerosumfree" was called "non-cancellative" in my original post; I'm now following the naming suggested by Tobias Fritz in his answer below.) Slicing a monad in a category with pullbacks Suppose $\mathcal{E}$ is a category with pullbacks and we're given an adjunction $L:\mathcal{E}\rightleftarrows\mathcal{E}':R$. Then for any object $e\in\mathcal{E}$, there is an induced adjunction between slice categories $$L_e:\mathcal{E}/e\rightleftarrows\mathcal{E}'/L(e):R_e$$ Indeed the left adjoint sends $x\to e$ to $Lx\to Le$, and the right adjoint sends $y\to L(e)$ to the pullback $$ \begin{array}{ccc} R_e(y) & \to & Ry\\ \downarrow & & \downarrow\\ e & \underset{\eta_e}{\to}& RLe \end{array} $$ In particular, if $(M,\eta,\mu)$ is a monad on $\mathcal{E}$, we can consider the canonical adjunction with the Eilenberg-Moore category $\mathcal{E}^M$ of $M$-algebras: $$ \hat{M}\colon\mathcal{E}\rightleftarrows\mathcal{E}^M:U $$ Here $\hat{M}(e)$ is the free algebra on $e$, and $U$ is the forgetful functor. Then for every object $e\in\mathcal{E}$ the above construction gives us an adjunction $(\hat{M}_e\vdash U_e)$, and hence a monad on the slice $\mathcal{E}/e$. We denote the underlying functor of this sliced monad by $$ M_e\colon\mathcal{E}/e\to\mathcal{E}/e $$ Example: Take $\mathcal{E}:=\mathbf{Set}$ and $M:=\mathsf{Dist}$ the finite probability distributions monad. Then for any function $g\colon x\to e$, the map $\mathsf{Dist}_e(g):\mathsf{Dist}_e(x)\to e$ is given by the pullback $$ \begin{array}{ccc} \mathsf{Dist}_e(x) & \to & \mathsf{Dist}(x)\\ \downarrow & & \downarrow\\ e & \underset{\eta_e}{\to} & \mathsf{Dist}(e) \end{array} $$ In other words, the sliced monad $\mathsf{Dist}_e(x)$ is the set of fiberwise distributions on $x$: an element consists of an element of $e$ and a distribution on the fiber of $x$ over it. Naturality All of the above is natural in $e\in\mathcal{E}$. Indeed, for any morphism $f\colon d\to e$ there is an adjunction between slice categories: $$f_!\colon\mathcal{E}/d\rightleftarrows\mathcal{E}/e:f^*$$ and it is easy to check that the both the following diagram of left adjoints (shown) and that of right adjoints commutes: $$ \begin{array}{ccc} \mathcal{E}/d & \overset{L_d}{\rightleftarrows} & \mathcal{E}'/Ld\\ \scriptsize f_!\normalsize \downarrow\uparrow~~~ & & \scriptsize (Lf)_!\normalsize\downarrow \uparrow~~~~\\ \mathcal{E}/e & \overset{L_e}{\rightleftarrows} & \mathcal{E}'/Le \end{array} $$ Thus there is an induced "mate" natural transformation $L_d\circ f^*\Rightarrow (Lf)^*\circ L_e$. In the case of monads, where $L$ is the free monad functor to the Eilenberg-Moore category, the Beck-Chevalley condition seems to rarely hold. However, something interesting happens when we compose with the right adjoints $R_d$ and $R_e$. "Zerosumfree monads"? Composing the above mate natural transformation with $R_d$ and using the commutativity of the right adjoint square above, we obtain: $$ R_d\circ L_d\circ f^*\Rightarrow R_d\circ (Lf)^*\circ L_e\cong f^*\circ R_e\circ L_e $$ Returning to the monad case, we take $L_d:=U_d$ and $R_d:=\hat{M}_d$ to be the adjoint functors $\mathcal{E}/d\leftrightarrows\mathcal{E}^M/Md$ to which $M_d=R_d\circ L_d$ is the associated monad, and we obtain a natural transformation $$f^M\colon M_d\circ f^*\Rightarrow f^*\circ M_e.$$ This is a kind of lax naturality condition for the sliced monads. Claim: Let $\mathcal{E}=\mathbf{Set}$ and $f\colon d\to e$ any function. For the following examples of monads, the lax naturality map $f^M$ is—or is not—an isomorphism as indicated $$ \begin{array}{l | l} \text{Monad } M&f^M\text{ is always iso?}\\\hline \text{identity}&\text{Yes}\\ \mathsf{Dist}&\text{Yes}\\ \mathsf{P},\text{ powerset}&\text{Yes}\\ \mathsf{P}^{\mathsf{fin}},\text{ finite powerset}&\text{Yes}\\ \mathsf{P}_+,\text{ nonempty powerset}&\text{Yes}\\ \mathsf{P}_+^{\mathsf{fin}}&\text{Yes}\\ -^S,\text{any set $S$}&\text{Yes}\\ (-\times S)^S,\text{any set $S$}&\text{Yes, but reduces to $-^S$}\\ \mathsf{List},\text{free monoid}&\text{Yes, but reduces to identity}\\ -\times G,\text{ $G$ a monoid}&\text{Yes, but reduces to identity}\\ \mathbb{N}[-],\text{free comm. monoid}&\text{Yes, but reduces to identity}\\ \mathbb{R}_+[-],\text{free semi-module over nonneg. reals}&\text{Yes, but reduces to identity}\\ \mathbb{Z}[-],\text{free abelian group}&\text{No!}\\ k[-],\text{free vector space}&\text{No!} \end{array} $$ The reason that $\mathbb{Z}[-]$ and $k[-]$ fail is roughly that there are expressions in the free algebra that "sum to zero" and are thus "locally indetectable". Question: Has this phenomenon been studied before? Strengthening lax monoidality of monad In the case that the above-mentioned maps $f^M$ are invertible, we have natural transformations $$ (f^M)^{-1}\colon f^*\circ M_e\Rightarrow M_d\circ f^* $$ But stepping back a bit, these maps actually arise by composing other natural transformations with $U_d$ and $U_e$ as above. That is, there is a natural transformation as shown: $$ \begin{array}{ccccc} \mathcal{E}/d && \overset{\hat{M}_d}{\rightarrow} && \mathcal{E}^M/Md\\\\ f^*\uparrow & &\Uparrow & & \uparrow\scriptsize (Mf)^*\\\\ \mathcal{E}/e && \overset{\hat{M}_e}{\rightarrow} && \mathcal{E}^M/Me \end{array} $$ The above maps $(f^M)^{-1}$ arise—though not a priori as inverses to $f^M$—by composing the map $(Mf)^*\circ\hat{M}_e\Rightarrow \hat{M}_d\circ f^*$ with the functor $U_d\colon\mathcal{E}^M/Md\to\mathcal{E}/d$ adjoint to $\hat{M}_d$. Anyway, what this says is that for any maps $d\to e\leftarrow y$, there is a natural map $$\varphi_{d,_ey}: M(d)\times_{M(e)}M(y)\to M(d\times_ey).$$ (This will automatically be a section of the universal map going the other way.) I have not checked that these maps satisfy the axioms for being lax monoidal coherence maps in the case that $e=1$ is a terminal object, but it seems likely. Question: Has this phenomenon been studied before? Are there any cases you know of where the maps $\varphi_{d,_ey}\;$ exist but do not lead to $f^M$ being an isomorphism? REPLY [6 votes]: Here are some remarks and pointers to the literature which are too long for a comment. The map $$ \mathsf{Dist}(X_1) \times_{\mathsf{Dist}(X)} \mathsf{Dist}(X_2) \longrightarrow \mathsf{Dist}(X_1 \times_X X_2) $$ is a well-known construction called conditional product. While this has been rediscovered many times over, the first axiomatics that I am aware of were proposed in a paper by Dawid and Studený. More recently, one categorical axiomatization based on the resemblance with sheaf-theoretic gluing was given in this paper. Another categorical axiomatization has been given by Simpson in this paper, where conditional products are called local independent products. Note that the first two references only consider conditional products of the special type $\eqref{laxproduct_special}$ below, but I believe this to be only a minor restriction. An important caveat is that although the conditional product is natural in the left and right arguments, it is not natural in the middle one. This plays a prominent role in such categorical axiomatizations. A general construction of conditional products for suitable monads can be extracted from my latest paper. As described in Section 3, the Kleisli category of any symmetric monoidal affine monad $M$ on a category with finite products is a Markov category. If this Markov category has conditionals in the sense of Definition 11.1, then Definition 12.8 shows that one can form conditional products in the special form $$\tag{1}\label{laxproduct_special} M(A \times B) \times_{M(B)} M(B \times C) \longrightarrow M(A \times B \times C), $$ and I suspect that one can arrive at the general form as in the OP with a bit more fiddling. However, I do not know how to translate the assumption for the Kleisli category to have conditionals into an assumption on the monad, and this may end up being related to your $f^M$. I also do not know whether the existence of conditionals is related to any sort of "noncancellativity"; I have investigated one such condition in Definition 11.19 and after, but it does not seem to play any role in the context of conditional products. (By the way, what is called "noncancellativity" in the OP should arguably be called positivity or zerosumfreeness, since cancellativity usually means more something like $a + c = b + c \: \Rightarrow \: a = b$.) Finally, I agree with Todd Trimble that the question also triggers strong associations with weakly cartesian monads. Replacing the title of the second column by "has weakly cartesian underlying functor and weakly cartesian multiplication" indeed gives the exact same yes/no answers. This can be seen by applying the results of this paper by Clementino, Hofmann and Janelidze, where this property is called condition (BC). Properties like zerosumfreeness—i.e. what the OP calls "noncancellativity"—indeed play a major role in characterizing monads on $\mathsf{Set}$ which satisfy condition (BC).<|endoftext|> TITLE: Homologous quotient of fundamental groupoid QUESTION [5 upvotes]: Let $X$ be a connected space and $\Pi_1(X)$ be its fundamental groupoid. We consider the homologous relation $\mathcal R$ on every morphism space: $f,g\in \Pi_1(X)(p,q)$ are related if the singular one-chain $g-f= \partial S$ for a two-chain $S$. It seems this is a congruence relation: if we further have $f'-g'=\partial S'$ for $f',g'\in \Pi_1(X)(q,r)$ then $f'\circ f-g'\circ g= \partial(S+S')$. So we can consider the quotient category $$ \Pi'_1(X):=\Pi_1(X)/\mathcal R $$ Is this stuff well-studied and is there a good reference? For example, It is known that the fundamental groupoid can be viewed as a functor $\Pi_1: \mathcal{Top}\to \mathcal{Grpd}$ from the category of topological spaces to the category of groupoids. Does this property also holds for our quotient $\Pi_1'$? The fundamental group $\pi_1(X,p)$ is the object group of $\Pi_1(X)$ at $p\in X$; in analogy, it seems $H_1(X)$ is simply the object group at any point $p\in X$ of our new $\Pi'_1(X)$. Is this correct? REPLY [2 votes]: The resulting groupoid is equivalent to the disjoint union of groupoids $B(H_1(X_i))$ taken over all connected components $X_i$ of $X$. This answers both 1 and 2 in the positive. To see this, observe that maps in both directions can be constructed using the corresponding universal properties. The fundamental group maps to the first homology group via the Hurewicz homomorphism. Vice versa, any 1-cycle can be converted to a continuous loop (in a nonunique way), and any two such choices will differ by a 2-boundary, which yields the desired map to the quotient under consideration.<|endoftext|> TITLE: Higman's lemma and a manuscript of Erdős and Rado QUESTION [11 upvotes]: Motivated by a problem in factorization theory, I've recently proved the following: Theorem. If $X$ is a non-empty finite alphabet and $\mathcal W$ an infinite subset of the free semigroup, $X^\ast$, over $X$, then there exists a sequence $(w_n)_{n \ge 1}$ with values in $\mathcal W$ such that $w_n$ is a proper subword of $w_{n+1}$ for every $n \in \mathbf N^+$. The theorem implies at once Higman's lemma. The proof is elementary and self-contained (the most advanced thing one is using, is the pigeonhole principle), but I wouldn't call it trivial: The basic idea is to introduce a non-standard factorization of the elements of $X^\ast$ that is well suited to an induction on $|X|$, and then distinguish two cases depending on a certain invariant associated with this factorization. Unfortunately, it turned out that the result is nothing new and comes down to a special case of Theorems 2.1 and 4.3 of G. Higman's paper Ordering by divisibility in abstract algebras [Proc. Lond. Math. Soc., III. Ser. 2 (1952), 326-336]. In particular, Theorem 2.1 in Higman's paper states that the following conditions are equivalent for a quasi-ordered set $(A, \preceq)$: (a) Every sequence of elements of $A$ has a subsequence that is strictly increasing wrt $\preceq$. (b) If $(a_n)_{n \ge 1}$ is a sequence of elements of $A$, there exist $i,j \in \mathbf N^+$ such that $i < j$ and $a_i \preceq a_j$. For the proof of the equivalence, Higman cites an unpublished manuscript of P. Erdős and R. Rado. Question. Which manuscript of Erdős and Rado does Higman refer to? Has the manuscript been eventually published? If not, is there a book, article, etc. with the details of a proof of the equivalence between (a) and (b)? I browsed through the joint papers of Erdős and Rado listed at https://www.renyi.hu/~p_erdos/Erdos.html, but I couldn't find what I'm looking for. REPLY [7 votes]: This is an answer not for the original question but for how to get the equivalence of (a) and (b) from the infinite Ramsey theorem (as requested in a comment). The implication from (a) to (b) is trivial. For the converse, I'm going to assume that "sequence" means one-to-one sequence (i.e., no repetitions), because otherwise a one-element set $A$ is a counterexample (because of the "strictly" in (a) and the non-strict $\preceq$ in (b)). So assume (b) and let $(a_i)_{i\in\mathbb N}$ be a sequence of distinct elements of $A$. Partition the set $[\mathbb N]^2$ of $2$-element subsets $\{i TITLE: How many minimum generating sets are there in a finite group? QUESTION [8 upvotes]: Let $G$ be a finite group of order $n$. A generating set in $G$ is said to be minimum if it has minimal size. Is there a known lower bound on number of minimum generating sets in a group of order $n$? For cyclic groups I know the answer. REPLY [2 votes]: Some fairly general lower bounds are given in Igor Pak's "What do we know about the product placement algorithm" [2]. Let $d(G)$ denote the minimal number of generators of $G$. By $\varphi_{k}(G)$ we denote the number of generating $k$-tuples of $G$. The function $\varphi_k$ is usually referred to as the $k$-th Eulerian function of the group $G$ [1]. OP's Question. Let $d = d(G)$. Is there a non-trivial lower bound for $\frac{\varphi_{d}(G)}{d!}$ given as a function of the cardinal $\vert G \vert$ of $G$? The probability that $k \ge d(G)$ uniformly and independently chosen elements in $G$ generate the whole group is: $$\overline{\varphi}_k(G) \Doteq \frac{\varphi_k(G)}{{\vert G \vert}^k}.$$ Here is the most general (and the weakest) lower bound extracted from [2]: [2, Theorem 1.1.7]. Let $G$ be a finite group, let $m \Doteq \lceil\log_2(\vert G \vert)\rceil$ and let $k \ge d(G)$. Let $0 < \epsilon < 1$. Then we have: $\overline{\varphi}_k(G) \ge \overline{\varphi}_k((\mathbb{Z}/2 \mathbb{Z})^m)$. $\overline{\varphi}_k(G) > 1 - \varepsilon$ if $k \ge m + 2 + \log_2(1/\epsilon)$. There are also interesting general bounds for nilpotent groups. [2, Proposition 1.1.4]. Let $G$ be a finite $p$-group with $p$ a prime number and let $d = d(G)$. Then we have $$\overline{\varphi}_d(G) \ge 1 -\frac{1}{p} - \frac{1}{p^2}.$$ For an arbitrary finite nilpotent group, we have: [2, Proposition 1.1.4] Let $G$ be a finite nilpotent group and let $d = d(G)$. Then we have $$ \overline{\varphi}_d(G) > \frac{1}{5 \log(\log(G))}. $$ If $(G_n)$ is sequence of pairwise non-isomorphic finite simple groups then $\overline{\varphi}_2(G_n)$ tends to $1$ as $n$ tends to infinity [2, Theorem 1.1.1]. More detailed asymptotics are known for several specific sequences [2, Theorem 1.1.2]. [1] P. Hall, "The Eulerian functions of a group", Quarterly Journal of Mathematics 7 (1936), 134–151. [2] I. Pak, "What do we know about the product replacement algorithm", 2000.<|endoftext|> TITLE: Minimum number of relations that must be added to make a group abelian QUESTION [12 upvotes]: Let $G$ be a finitely generated group and let $c(G)$ denote the minimal number of relations that we must add to a presentation fro $G$ in order to make $G$ abelian. I would like to find examples of groups where $c(G)$ is arbitrarily large but I do not know how to get lower bounds on $c(G)$. Do you know of such groups? There is no algorithm to compute $c(G)$ since $c(G) = 0$ implies $G$ is abelian - but I just want some examples where $c(G)$ is big. Actually, for an application I have in mind, I would like to find 3-manifold groups (or in particular knot complement groups) $G_i$ where $c(G_i) \to \infty$. I imagine there is a relation between $c(G)$ and some other invariants of groups and I am hoping someone can point these out to me. Maybe $c(G)$ has a name that I am totally unaware of... REPLY [12 votes]: The question has been answered in the comments by SashaP and Derek Holt, taking the following definition of $c(G)$: Definition 1. Let $G$ be a finitely generated group. We denote by $c(G)$ the minimal number of elements of $G$ required to normally generate a subgroup of $G$ that contains $G'$, the derived subgroup of $G$. The invariant $c(G)$ is related to both the weight $w(G)$ and the deficiency $\text{def}(G)$ of $G$. The weight $w(G)$ of a finitely generated group $G$ is the minimal number of elements required to generate $G$ as a normal subgroup. Clearly, we have $$d(G/G') \le w(G) \le d(G)$$ where $d(G)$ denotes the minimal number of generators of $G$. Neil Hoffman has suggested another possible definition of OP's invariant. Definition 2. For $G$ a finitely generated group, we define $c_{ab}(G)$ as the minimal number of elements required to normally generate $G'$ as a normal subgroup of $G$. I will address Definition 1 first by wrapping up the hints provided in the comments attached to the question. Definition 2 will be addressed later in Claim 3. The following is a combination of Derek Holt's and Neil Hoffman's observations. Claim 1. Let $G$ be a finitely generated group. Then we have $$c(G) \le \min(w(G), d(G) - 1).$$ In particular, $c(G) = 1$ if $G$ is a non-Abelian knot group. The deficiency $\text{def}(P)$ of a finite group presentation $P = \langle x_1, \dots , x_n \vert \, r_1, \dots, r_m \rangle$ on $n$ generators and $m$ relators is the integer $n - m \in \mathbb{Z}$. The deficiency $\text{def}(G)$ of a finitely presented group $G$ is the maximum of the deficiencies of its finite presentations. The following summarizes SashaP's observations. Claim 2. If $G$ is a finitely generated Abelian group, then $\text{def}(G) \le 1$. In addition, we have $\text{def}(G) = 1$ if and only if $G \simeq \mathbb{Z}^r$ for some $r \in \{1, 2\}$. $\text{def}(G) = 0$ if and only if $G \simeq \mathbb{Z}/n\mathbb{Z}, \mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ with $n > 0$ or $\mathbb{Z}^3$. Proof. Let $r = \dim(G \otimes_{\mathbb{Z}} \mathbb{Q})$. Therefore we can write $G = \mathbb{Z}^r \times H$. By [2, Lemma 1.2], we have $\text{def}(G) \le \dim\left(H_1(G, \mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{Q}\right) - d(H_2(G, \mathbb{Z}))$. Actually, equality holds since $G$ is efficient by [2, Lemma 1.3]. As $d(H_2(G, \mathbb{Z})) = d(\Lambda^2(G)) = d(\Lambda^2(\mathbb{Z}^r \times H)) \ge \frac{r(r-1)}{2}$, we deduce that $\text{def}(G) \le r - \frac{r(r-1)}{2} \le 1$. The remaining statements are easily inferred from the efficiency of finitely generated Abelian groups. We are now in position to answer the main question, that is, to provide a sequence of groups $(G_n)$ with arbitrarily large values of $c(G_n)$. Corollary 1. Let $G$ be a finitely presented group. Then $c(G) \ge \text{def}(G) - 1$. Proof. Consider a presentation $P$ of $G$ with deficiency $\text{def}(G)$. By adding $c(G)$ relators to $P$ we obtain the presentation $P'$ of an Abelian group. Thus $\text{def}(P') = \text{def}(G) - c(G) \le 1$ by Claim 2. The result immediately follows. Note that the free group on $n$ generator is the fundamental group of a 3-dimensional handlebody of genus $n$. It is also the fundamental group of the connected sum of $n$ copies of $S^1 \times S^2$. Corollary 2. Let $F_n$ be the free group on $n$ generators. Then $c(F_n) = n - 1$. Proof. By Claim 1, we have $c(F_n) \le n - 1$. It follows from Corollary 1 that $c(F_n) \ge n - 1$, hence the result. In order to address Definition 2, further definitions are required. If $M$ is a finitely generated module over a unital ring $R$, we denote by $d_R(M)$ the minimal number of generators of $M$. Let $G$ be a group. Then the conjugation action of $G$ on the Abelian group $M \Doteq G'/G''$ induces an action of $G_{ab}$ on $M$ so that $M$ is naturally a $\mathbb{Z}[G_{ab}]$-module. Claim 3. Let $G$ be a finitely generated group and let $n = d(G)$. Then the following holds $$d_{\mathbb{Z}[G_{ab}]}(G'/G'') \le c_{ab}(G) \le c_{ab}(F_n).$$ In addition, we have $$c_{ab}(F_n) = \frac{n(n - 1)}{2}.$$ Proof. The first two inequalities follow easily from the definitions. Clearly, we have $c_{ab}(F_n) \le \frac{n(n-1)}{2}$. The reverse inequality follows from [3, Theorem 3]. Indeed, set $M \Doteq F_n'/F_n''$, $(F_n)_{ab} = \mathbb{Z}^n$ and let $I$ be the augmentation ideal of $\mathbb{Z}[(F_n)_{ab}]$. Then by [3, Theorem 3] the group $M$ is a $\mathbb{Z}[(F_n)_{ab}]$-module generated by $\frac{n(n-1)}{2}$ elements and we have $M/IM \simeq \mathbb{Z}^{\frac{n(n-1)}{2}}$. As $(F_n)_{ab}$ acts trivially on $M$, this module cannot be generated by less than $\frac{n(n - 1)}{2}$ elements. A similar reasoning applies to the integral Alexander module $A(K)$ of a classical $1$-knot $K \subset \mathbb{S}^3$. Recall that the Alexander module $A(K) \Doteq G'/G''$ is a module over $\mathbb{Z}[t^{\pm 1}] \simeq \mathbb{Z}[G_{ab}]$ where $G = \pi_1(\mathbb{S}^3 \setminus K)$. Claim 4. Let $K$ be the knot sum of $n$ copies of an oriented knot with a non-zero Alexander module. Let $G = \pi_1(\mathbb{S}^3 \setminus K)$. Then we have $c_{ab}(G) \ge n.$ Claim 4 is an easy consequence of Lemma 1. Let $K_1$ and $K_2$ be two oriented $1$-knots. Then $A(K_1 \# K_2) = A(K_1) \times A(K_2)$. Proof. This is a straightforward consequence of the two following facts: If $V$ is a Seifert matrix of a $1$-knot $K$, then $V - tV^{T}$ is a presentation matrix of $A(K)$ where $V^T$ denotes the transpose of $V$. If two $1$-knots $K_1$ and $K_2$ admit $V_1$, resp. $V_2$, as a Seifert matrix, then the block-diagonal matrix $\text{diag}(V_1, V_2)$ is a Seifert matrix of $K_1 \# K_2$. We are now in position to prove Claim 4. Proof of Claim 4. Write $K = K_0 \# K_0 \# \cdots \# K_0$ with $A(K_0) \neq \{0\}$. By Lemma 1, the Alexander module $A(K)$ of $K$ is the direct product of $n$ copies of $A(K_0)$. Considering a maximal ideal $\mathfrak{m}$ of $\mathbb{Z}[t^{\pm 1}]$ such that $A(K_0)$ surjects onto the field $\mathbb{Z}[t^{\pm 1}]/\mathfrak{m}$, we see that $A(K)$ cannot be generated by less than $n$ elements. [1] E. Rapaport, "On the commutator subgroup of a knot group", 1960. [2] D. Epstein, "Finite presentations of groups and 3-manifolds", 1961. [3] S. Bachmuth, "Automorphisms of free metabelian groups", 1965.<|endoftext|> TITLE: Minimum of $z:\mathbb{R}^n \to \mathbb{R}$ along paths implies local minimum of $z$ QUESTION [6 upvotes]: Suppose we are given a smooth function $z: \mathbb{R}^n \to \mathbb{R}$, a point $x_0 \in \mathbb{R}^n$ and a set $\mathcal{F}$ consisting of certain paths in $\mathbb{R}^n$, i.e. $f: [0,1] \to \mathbb{R}^n$ for $f \in \mathcal{F}$. Assume that for each $f \in \mathcal{F}$, there exists a $t \in [0,1]$ such that $f(t) = x_0$, and that the function $z \circ f: [0,1] \to \mathbb{R}$ has a local minimum in $t$. Under what conditions on $\mathcal{F}$ does it follow that $z$ has a local minimum at $x_0$? Is is easily seen that it is enough if $\mathcal{F}$ consists of all continuous paths going through $x_0$. However, the set of straight lines going through $x_0$ is not enough (consider $x_0 = (0,0)$ and $z(x,y) = (y^2 - 2x)(y^2 - x)$, by Peano). The standard argument to see that continuous paths are sufficient, is to assume not; i.e. there are points $(x_k)_{k\in\mathbb{N}}$ such that $x_k \to x_0$ and $z(x_k) < z(x_0)$. Now construct a continuous path along going through $(x_k)$ and you are done. So one may also consider which $\mathcal{F}$ are such that for each converging sequence of points $(x_k)$ one can find a subsequence $(x_{k_l})$ and an $f \in \mathcal{F}$ such that $f$ goes through $(x_{k_l})$. Can we classify sets $\mathcal{F}$ for which the local minimality of $z$ holds? More specifically, I am interested in paths for which each component is twice differentiable. REPLY [4 votes]: Answer: $C^\infty$ curves suffice for arbitrary functions $z: \mathbb{R}^n \to \mathbb{R}$. Suppose we are given any function $z: \mathbb{R}^n \to \mathbb{R}$, a point $x_0 \in \mathbb{R}^n$ and let $\mathcal{F}$ consisting of $C^\infty$ (i.e. infinitely differentiable) $f: [0,1] \to \mathbb{R}^n$ such that $f(0)=x_0$ and all derivatives of $f$ vanish at $x_0$ (these functions are all restrictions of $C^\infty$ maps from $\mathbb{R}$ to $\mathbb{R}^n$.) Claim: If for every $f \in \mathcal{F}$ the function $z \circ f: [0,1] \to \mathbb{R}$ has a local minimum at $0$, then $z$ has a local minimum at $x_0$. Proof: We may assume that $x_0=0$. If $z$ does not have a local minimum at $x_0=0$, then as noted by the OP, there are points $(x_k)_{k\in\mathbb{N}}$ such that $x_k \to 0$ and $z(x_k) < z(0)$ for all $k \ge 1$. Passing to a subsequence, we may assume that the Euclidean norm $| \cdot|$ satisfies $|x_k|<2^{-k}$. We will construct $f \in \mathcal{F}$ such that $f(1/k)=x_k$ for all $k>1$ and this will prove the claim. Let $\psi_0$ be a $C^\infty$ bump function on $[0,1]$ such that $\psi_0$ and all its derivatives vanish at 0 and 1 yet $\psi_0$ is positive on $(0,1)$. e.g., $\psi_0(t)=\exp(-1/[t(1-t)])$. Define $\psi_1(t)=\int_0^t \psi_0(s) ds$ and $\psi(t)=\psi_1(t)/\psi_1(1)$. Then $\psi(0)=0$ and $\psi(1)=1$ and all the derivatives of $\psi$ vanish at 0 and 1. For $k \ge 2$, write $d_k= 1/(k-1)-1/k$ and $y_k=x_{k-1}-x_{k}$. Define $f: [0,1] \to \mathbb{R}^n$ as follows: First $f(0)=0$. Second, if $k\ge 2$ and $1/k TITLE: Open subspaces of CW complexes QUESTION [12 upvotes]: I am looking at the paper Covering homotopy properties of maps between CW complexes or ANRs by Mark Steinberger and James West and a claim is made in the proof of their first main theorem that (slightly rephrased) since $U$ is a contractible subspace of the CW complex $B$, $U$ "is" a CW complex Question: Is it generally true that an open subspace of a CW complex can be given the structure of a CW complex? Is it true in general only for contractible subspaces? Why? Is there a reference? REPLY [19 votes]: It is not generally true that each open subset of a CW complex admits the structure of a CW complex. Counterexamples were given in \bib{MR1157891}{article}{ author={Cauty, Robert}, title={Sur les ouverts des CW-complexes et les fibr\'{e}s de Serre}, language={French}, journal={Colloq. Math.}, volume={63}, date={1992}, number={1}, pages={1--7}, issn={0010-1354}, review={\MR{1157891}}, doi={10.4064/cm-63-1-1-7}, } in response to the paper you mention by Steinberger and West. Here is the abstract: M. Steinberger et J. West ont prouvé dans [7] qu'un fibré de Serre p:E → B entre CW-complexes a la propriété de relèvement des homotopies par rapport aux k-espaces. Malheureusement, leur démonstration contient une légère erreur. Ils affirment que certains ensembles (notés U et p−1U×U) sont des CW-complexes car ce sont des ouverts de CW-complexes. Ceci est généralement faux, et notre premier objectif dans cette note est de donner des exemples d'ouverts de CW-complexes n'admettant aucune décomposition CW. Malgré cela, le théorème de Steinberger et West est vrai, et notre deuxième objectif est de montrer comment leur démonstration peut être rectifiée.<|endoftext|> TITLE: Applications of number theory in dynamical systems QUESTION [19 upvotes]: I am looking for references (or ways to find references) on significant and/or recent applications of techniques in number theory to problems in the areas of dynamical systems and nonlinear dynamics. While there may be overlap with arithmetic dynamics (see, for instance Current Trends and Open Problems in Arithmetic Dynamics by Benedetto, DeMarco, Ingram, Jones, Manes, Silverman & Tucker), I would like examples leaning more towards traditional dynamical systems, in other words differential equations or discrete dynamical systems over the reals or complex numbers. Note that Lagarias writes in The Unreasonable Effectiveness of Number Theory, Proceedings of Symposia in Applied Mathematics, Volume 46, 1992, American Mathematical Society in the chapter Number Theory and Dynamical Systems: Number theoretic problems have occurred repeatedly in dynamical systems. This initially seems surprising, since number theory deals with discrete objects. A citation search using this reference was unrewarding. Other nonrecent papers which might yield a successful citation search would be welcome. Also, useful search terms would be appreciated. REPLY [2 votes]: Some more references (which i think have not been mentioned in the other posts): A couple of collections: LMS 134: Number Theory & Dynamical Systems Workshop on Combinatorics, Number Theory and Dynamical Systems an article: NUMBER THEORY, DYNAMICAL SYSTEMS AND STATISTICAL MECHANICS, ANDREAS KNAUF, Reviews in Mathematical PhysicsVol. 11, No. 08, pp. 1027-1060 (1999) and a (possibly) related journal: Communications in Number Theory and Physics<|endoftext|> TITLE: Curves with isogenous Jacobians QUESTION [5 upvotes]: Suppose that $C_1, C_2$ are two curves of genus $g \geq 2$ defined over a number field $K$. Let $J_1, J_2$ respectively be their Jacobians. Suppose that $J_1, J_2$ are isogenous over $K$ and $C_1(K), C_2(K)$ are both non-empty, can $C_1(K), C_2(K)$ have different cardinalities? For $g = 1$ and without the assumption that $C_i(K) \ne \emptyset$, the conclusion is obviously false. Take any elliptic curve $E$ over $\mathbb{Q}$ of positive rank such that the $2$-Selmer group of $E$ is non-trivial, so that there is a genus one curve $C$ which represents a non-trivial $2$-Selmer element of $E$. Then $E$ is isomorphic to the Jacobian of $C$, and the Jacobian of $E$ is itself, so that $C,E$ have isogenous Jacobians but $E(\mathbb{Q})$ is by assumption infinite but $C(\mathbb{Q}) = \emptyset$. REPLY [13 votes]: Yes it is possible, already for $(K,g) = ({\bf Q},2)$, and already with the first example of isogenous $C_1,C_2$ listed in the LMFDB: curve 249.a.249.1, $y^2 + (x^3+1) y = x^2 + x$, has one rational Weierstrass point, while curve 249.a.6723.1, $y^2 + (x^3+1) y = -x^5 + x^3 + x^2 + 3x + 2$, has two, so the rational-point counts don't even have the same parity. (The counts are known to be $5$ and $6$ respectively.) Further examples are isogeny class 277.a (two curves, each with a unique rational Weierstrass point, but the total counts are $1$ and $5$), and isogeny class 644.a (two curves, one with two rational points, the other not solvable over $\bf R$ --- so this violates your assumption that each $C_i(K)$ is nonempty). The first example where the counts do agree is the two curves in isogeny class 294.a, each with $4$ rational points. By the way, for $g=1$ there are examples even with each $C_i(K)$ nonempty; for example, take $K=\bf Q$, let $C_1$ be either $X_0(11)$ or $X_1(11)$, and let $C_2$ be the third elliptic curve of conductor $11$. Then $C_1$ has five rational points and $C_2$ has only one.<|endoftext|> TITLE: Fourier coefficients of Siegel Eisenstein series QUESTION [5 upvotes]: I am looking for reference about Fourier coefficients of Eisenstein series. Currently I am mainly interested Eisenstein series given by Siegel parabolic subgroup of $SP_{2n}$ and $U(n,n)$. Let's consider symplectic case for now. Given a symmetric matrix T, we can define a Fourier coefficient with respect to T: $E_T(g,s,\Phi)$=$\int_{[N(A)]}$$E(n(b)g,s,\Phi )$$\psi (-tr(Tb)) db$. If $det(T)\neq 0$, then by unfolding, only one term is nonzero, which is $\int_{sym(A)}\Phi(\omega^{-1}n(b)g,s)\psi(-tr(Tb))db$ where $\omega$ is:$$\begin{pmatrix} 0 & 1_n \\ -1_n & 0 \\ \end{pmatrix}$$ Then if $\Phi$ is factorizable, we can decompose the Fourier coefficient into product of Whittaker integral to study the Fourier Coefficient. Further study will be about local density, Siegel-Weil formula, Arithmetic Siegel-Weil formula etc. But if T is singular, then more terms can survive. I guess we can relate the terms with sum of nonsingular Fourier coefficients of Eisenstein series comes from $SP_{2r}$ where $r TITLE: Dual of a bimodule QUESTION [9 upvotes]: For a noncommutative ring $R$, and an $R$-$R$-bimodule $B$, is there a "correct/natural" notion of a dual bimodule? I am interested, really, when $B$ is projective as a left $R$-module. Note: Switched from Stackexchange, since no answers REPLY [13 votes]: As explained in more detail in this blog post linked by Jakob in the comments, every $(A, B)$-bimodule $M$ has two natural duals: If $M$ is finitely generated projective as a left $A$-module, it has a left dual given by the $(B, A)$-bimodule $\text{Hom}_A(M, A)$. If $M$ is finitely generated projective as a right $B$-module, it has a right dual given by the $(B, A)$-bimodule $\text{Hom}_B(M, B)$. These duals come from thinking of an $(A, B)$-bimodule as a 1-morphism in the Morita 2-category whose objects are rings 1-morphisms are bimodules 2-morphisms are bimodule homomorphisms and applying the general equational definition of dual or adjoint 1-morphisms in a 2-category given by the zigzag identities (the one which, applied to the 2-category of categories, produces left and right adjoints).<|endoftext|> TITLE: Failure of SVC in Grothendieck toposes QUESTION [5 upvotes]: The axiom SVC (for "small violations of choice") asserts that there is a set $S$ such that for every set $X$ there is a choice set $A$ such that $X$ is a subquotient of (i.e. admits a surjection from a subset of) $A\times S$. The original formulation of Blass (Injectivity, projectivity, and the axiom of choice) took $A$ to be an ordinal and the subquotient to be a surjection, which are equivalent in the presence of excluded middle; but the above version seems arguably more natural in intuitionistic logic. Blass showed that many models of set theory satisfy SVC, including permutation and symmetric models, $L(T)$, and ${\rm HOD}(T)$, but that SVC is not a consequence of ZF because it can fail in a class-forcing model. In Avoiding the axiom of choice in general category theory, Makkai remarks (p171) that "direct topos-theoretic translates" of SVC do not hold in all Grothendieck toposes. This is intriguing because Grothendieck toposes are a rough category-theoretic correspondent of (ordinary, set-based, intuitionistic) forcing models; and if SVC can be violated in an ordinary forcing model, why did Blass have to recourse to a class-forcing model to prove its independence? Unfortunately, Makkai gives no examples or explanation. So: what is an example of a Grothendieck topos (defined over a base model of ZFC) in whose internal logic the SVC fails? (An ordinary forcing model in which it fails would be interesting enough, although since a forcing model doesn't necessarily coincide exactly with its corresponding topos it might not quite answer the question yet.) REPLY [5 votes]: It seem to me that the problem that Makkai has in mind is that the existence of non-trivial choice objects is in conflict with non-booleaness. The core of the arguement, is the following lemma, which essentially follows from Diacunescu's proof that $AC \Rightarrow LEM$: Lemma: Let $A$ be a choice object in a topos, then internally: $$ \forall x,y \in A, \forall U \in \Omega, (U \Rightarrow (x=y)) \cup U $$ Remark: that is basically a rephrasing of lemma D4.5.11 of P.T.Johnstone sketches of an elephant. Proof: Let $A$ be a choice object and $x, y \in A$. Let $U$ be any proposition, consider the set $1 \coprod_U 1$ and the relation that send the first component to $x$ and the second to $y$ (and to both for element that are in both component). It is an entire relation to $A$, so there is function $1 \coprod_U 1 \rightarrow A$ included in that relation, in particular which takes values in $\{x,y\}$. I'm calling $a$ and $b$ the "two" elements of $1 \coprod_U 1$, I have four case to deal with regarding the values of $a$ and $b$ by this functions, but in each of them you can either prove $U$ or $U \Rightarrow (x,y)$: $a \mapsto x$ and $b \mapsto y$ then $U \Rightarrow (x=y)$ $a \mapsto y$ then $U$ $b \mapsto x$ then $U$ $\square$ Geometrically speaking, this is a pretty rough restriction on what can choice objects be ! it means that if $A$ is a choice object in a topos $\mathcal{T}$, and $a,b$ are two section of $A$ on $X$, then the closed complement of $(x=y)$ in $\mathcal{T} / X$ is Boolean. So either $A$ is very close to be subterminal, or the slices of $\mathcal{T}$ have some large boolean closed subtoposes. (this is the conflict I was referring to at the begining) But one can do better: Proposition: Let $\mathcal{T}$ be a Grothendieck topos which: Is nowhere boolean, i.e. no non-degenerate slice of $\mathcal{T}$ is boolean, satisfies SVC, then $\mathcal{T}$ is degenerate. For example, $Sh([0,1])$ do not satisfies SVC. Note that a nowhere boolean topos is a topos where LEM is internally false, in the sense that the interpretation of "$\forall U, U \cup \neg U$" in the internal logic is $\bot$ (the initial object). We start with another lemma: Lemma: In a nowhere boolean topos, if $A$ is a choice object and $D$ is decidable object (i.e. internally $\forall x,y \in D, x=y \cup x \neq y$) then one can internally show that: $$ \texttt{Any partial map $A \rightarrow D$ is constant} $$ Indeed, consider internally a partial map $f:A \rightarrow D$, for each $a , a'$ in the domain of definition of $f$, either $f(a)=f(a')$ or $f(a) \neq f(a')$, but $f(a) \neq f(a')$ implies $a \neq a'$, which internally implies LEM by the previous lemma, which is false as $\mathcal{T}$ is nowhere boolean. Hence the result. One can now prove the proposition: If one assumes further that $\mathcal{T}$ satisfies SVC, it means that every decidable object of $\mathcal{T}$ is a subquotient of the object $S$ (as the lemma above implies that the partial map $S \times A \rightarrow D$ is constant in the $A$-direction). But if $\mathcal{T}$ is non degenerate, for $\kappa$ large enough (larger than the size of the site and the size of $S$), the object $p^* \kappa$ is locally decidable and of size $\kappa$, so it cannot be a subquotient of $S$. Edit: My apologies for the needlessly non mathematical and complicated first answer. I think I found the theorem I was after.<|endoftext|> TITLE: How wide is the Birkhoff Polytope? QUESTION [6 upvotes]: This question is migrated from MSE where it turned out to be much harder than I thought. I still cannot figure this out. Does anyone have any ideas? Define the width of a polytope $P \subset \mathbb R^d$ as the minimum length of the interval $\{v \cdot p:p \in P\}$ for $v$ in the unit sphere. In other words the width is the smallest number $W$ such that you can sandwich $P$ between two hyperplanes distance $W$ apart. Here's a picture: More generally suppose the polytope $P \subset \mathbb R^d$ has affine hull $A + x$ for $A \subset \mathbb R^d$ a hyerplane. Define the relative width as the smallest length of $\{v \cdot p:p \in P\}$ as $v$ ranges over the unit sphere in $A$. In other words translate the affine subspace to contain the origin and then ignore the perpendicular directions. Equivalently the width is the minimiser of $$F(v) = \max\{v \cdot (p_1 - p_2) :p_1,p_2 \in P \text{ are vertices}\}.$$ Note $F$ is the maximum of a bunch of linear functions so is convex, and we are looking to minimise a convex function. The problem is the domain is a sphere rather than a convex region. The Birkhoff polytope $\mathcal B$ is defined as the convex hull of the $n!$ permutation matrices. That means the $n \times n$ matrices with all zeros except for exactly one $1$ in each row and column. Equivalently $\mathcal B$ is the set of nonnegative matrices with all row and column sums equal to $1$. In this case the affine subspace is defined as $$\left \{x \in \mathbb R^d: \sum_j x^i_j =1, \sum_i x^i_j =1\right \}.$$ and $$A= \left \{x \in \mathbb R^d: \sum_j x^i_j =0, \sum_i x^i_j =0\right \}.$$ This just says the row and column sums equal $1$. Within that subspace the polytope is defined as the intersection with the first quadrant. I am having trouble computing or estimating the height of $\mathcal B$. I would imagine the $v$ that minimises the projection is something like $$ v_1 = \left( {\begin{array}{cccc} 1/4 & -1/4 & 1/4& -1/4\\ -1/4 & 1/4 & -1/4 & 1/4\\ 1/4 & -1/4 & 1/4 & -1/4\\ - 1/4 & 1/4 & - 1/4 & 1/4\\ \end{array} } \right)\\[30pt] v_2 = \left( {\begin{array}{cccc} 1/2 & -1/2 & 0& 0\\ -1/2 & 1/2 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0& 0 & 0 \end{array} } \right)$$ In these cases we can choose the correct permutations (vertices) to force the interval to have length 2. Other choices like $$ v_3 = \left( {\begin{array}{cccc} 1/4 & -1/4 & 0& 0\\ -1/4 & 1/4 & 0 & 0\\ 0 & 0 & \sqrt{3/16} & -\sqrt{3/16}\\ 0 & 0& -\sqrt{3/16} & \sqrt{3/16}\\ \end{array} } \right) $$ You can use to get interval greater than 1. My intuition for why $v_1,v_2$ are optimal is along the lines of "If you try to shift mass to ruin some choice of vertices, others choices will become better." Here are some things I am able to prove: The vectors $v_1$ and $v_2$ are local minima of the function $F(v) = \max\{v \cdot( \sigma - \rho): \sigma - \rho \text{ vertices of } \mathcal B\}$. However we do not have a local minimum over the ball, or any guarantee this is a global minimum. At $v_1$ and $v_2$ then $F$ has a subgradient normal outwards to the sphere. This means moving along the sphere will have a small influence on $F$ compared to moving towards the centre. If we add a perturbation $\epsilon^i_j$ to $v= v_1,v_2$ such that $\|v + \epsilon\| = 1$ and $v + \epsilon \in A$ then we have $$\sum_{i+j \ \text{even}} \epsilon^i_j \le 0 \qquad \qquad \sum_{i+j \ \text{odd}} \epsilon^i_j \ge 0$$ This is because otherwise you push $v$ out of the unit ball. From this I can show there is either a positive diagonal $\sigma$ with $\epsilon^1_{\sigma(1)} + \ldots+ \epsilon^1_{\sigma(n)} \ge 0$ or a negative diagonal $\rho$ with $\epsilon^1_{\rho(1)} + \ldots+ \epsilon^1_{\rho(n)} \le 0$. Here positive diagonal means all $v^i_{\sigma(i)} >0$. If I could prove both exist at once I'd be done. Partial converse to 1: If at some some $w$ in the sphere the subgradient to $F$ contains $w$ itself then for each positive entry $w^i_j$ there is a diagonal $\sigma$ with $w^1_{\sigma(1)} + \ldots+ w^1_{\sigma(n)} = \max\{w \cdot \rho : \rho \text{ a vertex of } \mathcal B\}$ and likewise for each negaive entry $w^i_j$ there is a diagonal $\sigma$ with $w^1_{\sigma(1)} + \ldots+ w^1_{\sigma(n)} = \min\{w \cdot \rho : \rho \text{ a vertex of } \mathcal B\}$. If I could probe deeper into 4. and somehow categorise all vectors similar to $v_1,v_2$ then I could check them case by case and determine the minimiser. But so far I am stuck and imagine the correct proof is a big more elementary than what I'm trying. Any ideas? REPLY [8 votes]: For $n$ even, the width is exactly $\frac{2}{\sqrt{n-1}}$. For $n$ odd, I can prove this as a lower bound and $\frac{2n}{(n-1) \sqrt{n+1}} = \frac{2}{\sqrt{n-1} \sqrt{1-1/n^2}}$ as an upper bound. Upper bound To start, let $n$ be even. Let $$\vec{j} = (1,1,\ldots, 1)^T$$ $$\vec{u} = \frac{1}{\sqrt{n}} (1,1,\ldots,1,-1,-1,\ldots,-1)^T$$ $$\vec{v} = \frac{1}{\sqrt{n(n-1)}} (n-1, -1,-1,\ldots,-1)^T$$ where $\vec{u}$ has equally many $1$'s and $-1$'s. We note that $|\vec{u}| = |\vec{v}|=1$ and $\vec{j} \cdot \vec{u} = \vec{j} \cdot \vec{v} = 0$. Let $X$ be the $n \times n$ matrix $\vec{v} \vec{u}^T$. We have $X \vec{j} = \vec{v} (\vec{u}^T \vec{j}) = 0$ and $\vec{j}^T X = (\vec{j}^T \vec{v}) \vec{u} = 0$, so the rows and columns of $X$ sum to $0$. We also have $\mathrm{Tr}(X^T X) = \mathrm{Tr}(\vec{u} \vec{v}^T \vec{v} \vec{u}^T) = \mathrm{Tr}( \vec{v}^T \vec{v} \vec{u}^T \vec{u} ) = \mathrm{Tr}(1 \cdot 1) = 1$. So $X$ has length $1$. Now, consider the linear functional $\mathrm{Tr}(X\ \underline{\quad } )$ on the Birkhoff polytope. For any permutation matrix $\sigma$, we have $\mathrm{Tr}(X \sigma) = \mathrm{Tr}(\vec{v} \vec{u}^T \sigma) = \mathrm{Tr}(\vec{u}^T \sigma \vec{v}) = \vec{u} \cdot \sigma(\vec{v})$. If $\sigma$ maps the first coordinate into one of the first $n/2$ coordinates, the dot product of $\vec{u}$ and $\sigma(\vec{v})$ is $$\frac{1}{n\sqrt{n-1}} {\Big(} (n-1) - (n/2-1) + n/2 {\Big)} = \frac{n}{n \sqrt{n-1}} = \frac{1}{\sqrt{n-1}}.$$ If $\sigma$ maps the first coordinate into one of the last $n/2$ coordinates, then we get negative this. So $\mathrm{Tr}(X\ \underline{\quad } )$ ranges from $\tfrac{1}{\sqrt{n-1}}$ to $- \tfrac{1}{\sqrt{n-1}}$ on the Birkhoff polytope, and the Brikhoff polytope has width $\leq \tfrac{2}{\sqrt{n-1}}$. For the case where $n$ is odd, replace $\vec{u}$ by the vector $$\frac{1}{\sqrt{n^3-n}} (n+1,n+1,\ldots,n+1,-n+1,-n+1,\ldots,-n+1)$$ where there are $\tfrac{n+1}{2}$ negative terms and $\tfrac{n-1}{2}$ positive ones. Lower bound: Here is the key lemma: Lemma Let $X$ be an $n \times n$ matrix with row and column sum $0$, and $\sum_{ij} X_{ij}^2 = 1$. Then $$\sum_{\sigma \in S_n} \left( \mathrm{Tr}(\sigma X) \right)^2 = n (n-2)!.$$ Here the sum runs over all permutation matrices. Proof Expanding the sum gives $$(n-1)! \sum_{ij} X_{ij}^2 + (n-2)! \sum_{i_1 \neq i_2,\ j_1 \neq j_2} X_{i_1 j_1} X_{i_2 j_2}.$$ Letting $J$ denote the $n \times n$ matrix which is all $1$'s, we have $$\sum_{i_1 \neq i_2,\ j_1 \neq j_2} X_{i_1 j_1} X_{i_2 j_2} = \mathrm{Tr}{\Big(} (J - \mathrm{Id}) X^T (J - \mathrm{Id}) X {\Big)}.$$ But $JX=XJ=0$ since the rows and columns of $X$ sum to $0$. So $$\mathrm{Tr}{\Big(} (J - \mathrm{Id}) X^T (J - \mathrm{Id}) X {\Big)} = \mathrm{Tr}(X^T X) = 1.$$ Our sum in total is thus $(n-1)! + (n-2)! = n (n-2)!$. $\square$ Also, $\sum_{\sigma \in S_n} \mathrm{Tr}(\sigma X) = (n-1)! \sum X_{ij} =0$. So, if $\sigma$ ranges uniformly over $S_n$, then $\mathrm{Tr}(\sigma X)$ has expected value $0$ and standard deviation $\sqrt{\tfrac{n(n-2)!}{n!}} = \tfrac{1}{\sqrt{n-1}}$. So the range between its greatest and least value is at least $\tfrac{2}{\sqrt{n-1}}$.<|endoftext|> TITLE: What numbers are not represented by $5xy+2x+2y$? QUESTION [14 upvotes]: What numbers are not represented by $5xy+2x+2y$? Do they have a positive density? This came up for me while investigating some cases here. Here's what I've found: All evens are represented with $x=0$, and all $3m+1$ are represented with $x=-1$. There are infinitely many $n$ not represented, e.g. any $n$ for which $5n+4$ is prime. If $5xy+2x+2y=n$, then either $|x|$ or $|y|$ is less than $|n|/5+2$, so for each $n$ this is decidable. Of numbers with absolute value less than 6000, about 80% are represented by this polynomial. I'd expect a nice characterization for these numbers, but I haven't found it. REPLY [34 votes]: $n = 5xy + 2x + 2y$ if and only if $5n+4 = (5x+2)(5y+2)$. So a necessary and sufficient condition is that $5n+4$ have a factor congruent to $2 \bmod 5$ --- or $3 \bmod 5$ since you're allowing negative $x,y$ such as $x = -1$. This makes it easy to decide whether a given $n$ is so represented. In particular, the numbers not represented by $5xy + 2x + 2y$ are those for which $5n+4$ is the product of primes all congruent to $\pm 1 \bmod 5$ (you already found the special case of prime $5n+4$). Such numbers have density zero, but convergence is slow: the density in $|n| < X$ is asymptotically proportional to $1 \left / \sqrt{\log X} \right.$. P.S. Here's some quick gp code to count such $n$ up to $N$: f(v) = sum(n=1,#v,(v[n]%5!=1)&&(v[n]%5!=4)) F(n) = f(factor(n)[,1])==0 S(N) = sum(n=1,N,F(5*n+4)) For $N=6000$ the count $S(N)$ is $1204$, which agrees with Matt F.'s calculation that "about 80%" of $n \leq 6000$ are represented by $5xy+2x+2y$. Taking $N=10^5, 10^6, 10^7, 10^8$ finds $S(N) = 17992$, $166612$, $1557892$, $14680787$ (the last count took about 5.5 minutes to compute); this is quite close to $CN \left / \sqrt{\log(5N)} \right.$ for $C$ somewhere between $0.65$ and $0.66\,$.<|endoftext|> TITLE: Prominent examples of $q$-analogs without known cyclic sieving QUESTION [8 upvotes]: The cyclic sieving phenomenon is nicely summarized in the following AMS Notices "What is...?" article: https://www.ams.org/notices/201402/rnoti-p169.pdf. In that article, Reiner, Stanton, and White explain some desiderata (conditions (i)-(vi)) for "very nice" $q$-analogs $X(q)$ of cardinalities $\#X$ of combinatorial sets. Question: What are some "very nice" $q$-analogs (especially, those with simple product formulas) for which there is no known cyclic action which gives rise to a CSP? REPLY [5 votes]: I think I can provide one example, from this paper, The Cone of Cyclic Sieving Phenomena by N. Amini and I. We look at a principal specialization of Schur polynomials, $$ s_{n \lambda}(1,q,\dotsc,q^k) $$ for some partition $\lambda$, and integers $n$ and $k$. Here, $n\lambda = (n\lambda_1,\dotsc,n\lambda_\ell)$. This polynomial evaluates to non-negative integers at $n$th roots of unity, and I think one should be able to prove without effort that there should in principle be a cyclic group action which gives CSP for this family. The results by Rhoades show that (a suitable chosen power of) promotion works as group action whenever $\lambda$ is a rectangle. As a side note, I would be happy to compile a list of suspected instances of the cyclic sieving phenomenon on my web page. At the moment, it lists (most?) published instances of CSP. I also have a private list with suspected instances, but I prefer to save these for student projects, and a few others I am working on myself.<|endoftext|> TITLE: Elementary lower bounds for the number of primes in arithmetic progressions QUESTION [15 upvotes]: Some version of the Prime Number Theorem provides the asymptotic behavior of the number of primes in arithmetic progression $qn+a$ with $(q,a)=1$, $n \ge 1$. I was wondering there are Chebyshev-type arguments (using the binomial coefficients or variants thereof) that establish the existence of at least $$c x/\log x$$ primes in such arithmetic progressions up to $x$ for some even small $c>0$ and some values of $a, q>1$. For instance, is there such an argument for primes of the form $ 4k \pm 1$? This is mostly a curiosity triggered by thinking about a number theory course I will teach soon. REPLY [4 votes]: Here is another, more Chebyshev like, approach. It is possible that this is the proof Terry was sketching and I just didn't understand it. I think this should generalize to AP's for any modulus, but I'll stick to $4$. Let $\chi$ be the quadratic character modulo $4$. To prove PNT in AP's mod 4, we must show $$\sum_{n \leq N} \Lambda(n) = N + o(N),\ \sum_{n \leq N} \chi(n) \Lambda(n) = o(N).$$ Chebyshev style arguments prove bounds of the form $a N < \sum_{n \leq N} \Lambda(n) < b N$; I will analogously prove bounds of the form $\left| \sum_{n \leq N} \chi(n) \Lambda(n) \right| < cN$. This will only be interesting if I manage to get $c<1$. For any positive real number $x$, define $$\langle x \rangle = \sum_{1 \leq k \leq x} \chi(k).$$ So $\langle x \rangle$ is $1$ if $\lfloor x \rfloor \equiv 1,2 \bmod 4$ and $0$ if $\lfloor x \rfloor \equiv 3,4 \bmod 4$. Put $$S = \sum_{n \leq N} \chi(n) \log n.$$ Start by noticing that $$\left| S \right| \leq \log N$$ because each term switches the sign of the partial sum. Expand the sum as $$S = \sum_{n \leq N} \sum_{d|n} \chi(d) \chi(n/d) \Lambda(d) = \sum_d \chi(d) \Lambda(d) \sum_{m \leq N/d} \chi(m) = \sum_d \chi(d) \Lambda(d) \ \langle\! N/d \rangle. \quad (\ast)$$ Now, $\langle N/d \rangle$ is $1$ for $N/3 < d \leq N$, is $0$ for $N/5 < d \leq N/3$ and is between $0$ and $1$ for all $d$, so $$\left|\sum_{N/3< d \leq N} \chi(d) \Lambda(d) \right| \leq |S| + \sum_{d \leq N/5} \Lambda(d).$$ Invoking Chebyshev's bound $\sum_{d \leq M} \Lambda(d) \leq 1.11 M$, we have $$\left|\sum_{N/3< d \leq N} \chi(d) \Lambda(d) \right| \leq \log N + 1.11 \times \frac{1}{5} N .$$ Summing up this equation for $N$, $N/3$, $N/9$, ..., we get $$\left|\sum_{d \leq N} \chi(d) \Lambda(d) \right| \leq \frac{(\log N)^2}{\log 3} + 1.11 \times \frac{3}{10} N.$$ Since $0.333<1$, we win. I think there is substantial room to improve this argument. If we sum up equation $(\ast)$ at $N$ and $N/3$, we get $$\sum_d \chi(d) \Lambda(d) {\Big(} \langle N/d \rangle + \langle N/(3d) \rangle {\Big)} = O(\log N).$$ We have ${\Big(} \langle N/d \rangle + \langle N/(3d) \rangle {\Big)} = 1$ for $N/7 < d \leq N$, then $=0$ for $N/9 < d \leq N/7$, and is always at most $2$. So similar arguments give $$\left| \sum_{N/7 < d \leq N} \chi(d) \Lambda(d) \right| \leq 1.11 \times \frac{2}{9} N + O(\log N)$$ and thus replace $1.11 \times \tfrac{3}{10}$ with $1.11 \times \frac{16}{63}$ in the final analysis. I think this constant could be chased much lower if we took smarter linear combinations. More precisely, we have $\sum_{t=1}^T \chi(t) \mu(t) \langle x/t \rangle = 1$ for $x \geq T$. So, for fixed $T$, we have $$\sum_{t=1}^T \mu(t) \chi(t) \sum_d \chi(d) \Lambda(d) \ \langle\! N/(td) \rangle =$$ $$\sum_{N/T < d \leq N} \chi(d) \Lambda(d) + \sum_{d \leq N/T} \chi(d) \Lambda(d) \left( \sum_{t=1}^T \mu(t) \chi(t) \langle\! N/(td) \rangle \right) = O(\log N).$$ If we could get the parenthesized quantity to be $o(T)$, independent of $N$, we would win.<|endoftext|> TITLE: Which polygons can be turned inside out by a smooth deformation? QUESTION [35 upvotes]: Take a non-degenerate polygon with side lengths $\{a_1,\dots,a_n\}$ in a convex configuration. What is the condition on the $a_i$'s so that the polygon can be turned inside out by a continuous motion that preserves the side lengths. For example any triangle cannot be, but the unit square and unit hexagon can be, since you can collapse both to form a degenerate shape with zero area. A four sided figure can sometimes be inverted, sometimes not if the smallest side is too small. Just working out the four sided figure could be interesting I think. I suspect that it is possible iff you can find a configuration of zero area, where the area is defined as $\frac{1}{2}(x_1y_2-x_2y_1+x_2y_3-x_3y_2+\dots+x_ny_1-x_1y_n)$ where the $(x_i,y_i)$ are vertices of the polygon with i increasing from 1 to n as you progress around the polygon in order. It is trivial that if a deformation exists the area must change sign and hence be zero but the reverse implication may be more challenging. (Of course this doesn't solve the problem as this is not a condition on the side lengths but might be a useful observation). Update: My conjecture that you can invert a polygon iff you can deform it into a configuration of zero area is incorrect due to the following counterexample based on the figure included in Joseph O'Rourke's answer: The polygon on the left has 3 side lengths $\frac{1}{\sqrt 3}$ and $1+\frac{2}{\sqrt 3}$ and hence does not satisfy the criteria in the paper because $$2(1+\frac{2}{\sqrt 3})>(1+\frac{2}{\sqrt 3})+\frac{3}{\sqrt 3}.$$ However you can deform the left configuration into the one on the right which has zero area. In addition you can slightly increase the short sides AB, CD, EF to create a negative area on the right configuration but still not satisfy the criteria for inversion. Therefore we cannot modify the conjecture to the obvious one that there must be mutually deformable configurations of both strictly positive and strictly negative area. Therefore the two equivalence classes of configurations for a polygon that doesn't satisfy the criteria must sometimes contain a mixture of positive and negative area configurations. REPLY [41 votes]: This question was explored here: Lenhart, William J., and Sue H. Whitesides. "Reconfiguring closed polygonal chains in Euclidean $d$-space." Discrete & Computational Geometry 13, no. 1 (1995): 123-140; DOI: 10.1007/BF02574031, eudml. From the Abstract: "It is shown that in three or more dimensions, reconfiguration is always possible, but that in dimension two this is not the case. Reconfiguration is shown to be always possible in two dimensions if and only if the sum of the lengths of the second and third longest links add to at most the sum of the lengths of the remaining links."<|endoftext|> TITLE: The real numbers as a wreath product? QUESTION [11 upvotes]: In Faltin-Metropolis-Ross-Rota's [FMRR] paper The Real Numbers as a Wreath Product [Adv. Math. 16(3), 278-304 (1975)], the real numbers are constructed as a quotient of a certain subset of the ring of formal Laurent series $\mathbb{Z}((T))$, emphasizing the digit-expansion aspect of elements of $\mathbb{R}$ by formalizing the idea of "infinite carrying". The paper by itself is not hard to follow, and my question is actually more about the title of the paper, for even after reading it completely, I did not understand what exactly was being referred to as a wreath product. (The term wreath product is mentioned exactly once in the entire paper: in the title!) Here's a (very) quick summary of the construction: Let $b\geq 2$ be a natural number (base). Write $\mathbf{C}_b \subseteq \mathbb{Z}((T))$ for the ring of convergent sequences: $$ \mathbf{C}_b = \left\{ \sum_{n\in\mathbb{Z}} a_n T^n \in \mathbb{Z}((T)) ~\bigg|~ |a_n| = o(b^n) \right\}. $$ Write $\mathfrak{K}_b := 1 - bT$ for the carry constant. Then, denote the ring of bounded sequences and the carry ideal by, resp., $$ R_b := \left\{ \sum_{j\in\mathbb{Z}} a_j T^j \in \mathbb{Z}((T)) ~\bigg|~ \sum_{j=1}^{n} |a_j| b^{n-j} = O(b^n) \right\}, \quad I_b := (\mathfrak{K}_b \mathbf{C}_b) \cap R_b, $$ so that $I_b \subseteq R_b \subseteq \mathbf{C}_b$. Then: Theorem (FMRR). $R_b/I_b \simeq \mathbb{R}$ as ordered topological fields, for every $b \geq 2$. Remark. The big-$O$ notation is just a shorthand for the appropriate condition (which, although similar, is defined solely in terms of $\mathbb{Z}$). Both the order and the topology turn out to be the "obvious ones" (of course); while a topology may be defined in terms of the magnitude of $\sum_j |\alpha_j - \beta_j| b^{n-j}$ as $n$ grows, the order depends on a canonical representation of strings (referred to as clearing in [FMRR]). My question is: what exactly in this construction is being referred to as a wreath product? I see that $\prod_{j\in\mathbb{Z}} \mathbb{Z}/(b)$ could be acting as a basis for some potential wreath product, but I see no obvious group action that makes it substantially different from the generalized Lamplighter group $\mathbb{Z}/(b) \,\wr\, \mathbb{Z}$. REPLY [3 votes]: Consider the finite analogue where we model $\mathbb{Z}/b^n \mathbb{Z}$ as the ring $R = \mathbb{Z}[T]/\langle T^n \rangle$ modulo the ideal generated by $1-bT$. Note that $1$ in $R$ corresponds to $b^{n-1}$. Reversing sequences, so we get the 'little-endian' representation, we take the abelian group $\mathbb{Z}^n$ and quotient by the subgroup generated by all $$ (0,\ldots, 0, -b, 1, 0, \ldots, 0) .$$ A set of representatives for the cosets is the $b$-ary words of length $n$. These are in bijection with the leaves of the rooted $b$-ary tree with $n$ levels. The automorphism group of the tree is the iterated wreath product $S_b \wr \ldots \wr S_b$, with $n$ factors. Since addition respects congruence modulo powers of $b$, addition of a constant is an automorphism of the tree. This already gives some connection. To bring in carries, consider the binary case. Now addition of $1$ is the automorphism which swaps the two halves of the tree, and then continues on the (new) right-hand subtree, by swapping its two halves, and then continues working further down the tree. (For instance, after the first swap, an even numbered leaf $2m$ is replaced with the odd numbered leaf $2m+1$, after the second, an odd numbered leaf $4m+1$ is replaced with $4m+2$, and so on.) I think this repeated swapping is essentially the carry relation. To make this explicit take $b=2$ and $n=3$. Then the leaves of the tree are labelled $0,4,2,6,1,5,3,7$, corresponding to the binary words $$000, 001, 010, 011, 100, 101, 110, 111.$$ Addition of $1$ is the permutation $(0,1)(4,5)(2,3)(6,7)$ followed by $(0,2)(4,6)$ followed by $(0,4)$. (Confusingly the elements in these permutation come from successive left sub-trees, because the labels change as we apply the automorphism.) Chasing $3$, we get $3 \mapsto 2 \mapsto 0 \mapsto 4$. This corresponds to $$110 \mapsto 210 \mapsto 020 \mapsto 001$$ in the carries model. For $7$ the carries model requires four steps $$111 \mapsto 211 \mapsto 021 \mapsto 002 \mapsto 000$$ rather than the three predicted by the permutations, but this should be expected because we truncated the elegant infinite construction in the paper at $n=3$.<|endoftext|> TITLE: Do the operators in $B(E,F)$ separate points on the projective tensor product $F' \mathop{\tilde\otimes_\pi} E$? QUESTION [7 upvotes]: Let $E$ and $F$ be Banach spaces, and let $\mathfrak L_{co}(E,F)$ denote the space of bounded linear operators $E \to F$ equipped with the topology of uniform convergence on the absolutely convex compact subsets of $E$. Defant and Floret [DF93, §5.5] point out that there is a natural linear map $$ D_F : F' \mathop{\tilde\otimes_\pi} E \to (\mathfrak L_{co}(E,F))' $$ which is surjective. (No continuity claims are made.) Furthermore, they prove [DF93, §5.7] that $D_F$ is injective if $F$ is reflexive or if $F'$ or $E$ has the approximation property, so in that case one has an isomorphism $F' \mathop{\tilde\otimes_\pi} E \cong (\mathfrak L_{co}(E,F))'$ of vector spaces. They conclude with: We do not know whether or not $D_F$ is always injective. Question. Has this since been settled? Can someone provide (a reference to) a proof or a counterexample? Some background: $D_F$ is defined by taking the natural map $$ \Phi_F : \mathfrak L(E,F) \stackrel{1}{\hookrightarrow} \mathfrak L(E,F'') \stackrel{1}{\cong} (E \mathop{\tilde\otimes_\pi} F')' \stackrel{1}{\cong} (F' \mathop{\tilde\otimes_\pi} E)'. $$ It is shown that $\Phi_F$ is continuous as a map $\mathfrak L_{co}(E,F) \to [(F' \mathop{\tilde\otimes_\pi} E)',\text{weak-$*$}]$, and then $D_F$ is defined as the adjoint of this map. (I guess one could deduce some continuity property of $D_F$ from this, but the authors steer clear of that, presumably to focus on questions related to Banach spaces.) Another way to interpret the question is this: the map $D_F$ (or $\Phi_F$) gives rise to a bilinear map $(F' \mathop{\tilde\otimes_\pi} E) \times \mathfrak L(E,F) \to \mathbb{F}$. One readily verifies that this is simply the natural map $$ \left(\sum_{n=1}^\infty y_n' \mathop{\otimes} x_n \, , \, T\right) \mapsto \sum_{n=1}^\infty y_n'(Tx_n). $$ Since we have $\mathfrak L(E,F) \stackrel{1}{\hookrightarrow} (F' \mathop{\tilde\otimes_\pi} E)'$, it is clear that $F' \mathop{\tilde\otimes_\pi} E$ separates points on $\mathfrak L(E,F)$. The question whether $D_F$ is injective is equivalent to the question whether $\mathfrak L(E,F)$ separates points on $F' \mathop{\tilde\otimes_\pi} E$ (so that the bilinear map becomes a dual pairing). Yet another equivalent formulation: is the image of $\Phi_F$ weak-$*$ dense in $(F' \mathop{\tilde\otimes_\pi} E)'$? References. [DF93] A. Defant, K. Floret, Tensor Norms and Operator Ideals (1993), Mathematics Studies 176, North-Holland. REPLY [2 votes]: This question was settled in 2012 by Petr Hájek and Richard J. Smith [HS12]. They prove the following beautiful result (reformulated here to match the notation from the question). Theorem (cf. [HS12, Theorem 2.5]). Let $F$ be a Banach space with the AP. Then the following conditions are equivalent: $F'$ has the AP. For every Banach space $E$, the map $F' \mathbin{\tilde\otimes_\pi} E \to (\mathfrak L_{co}(E,F))'$ is injective. The map $F' \mathbin{\tilde\otimes_\pi} F'' \to (\mathfrak L_{co}(F'',F))'$ is injective. As there are known examples of Banach spaces with the AP whose dual fails the AP, the question is settled in the negative. References. [HS12]: Petr Hájek, Richard J. Smith, Some duality relations in the theory of tensor products, Expositiones Mathematicae, volume 30 (2012), issue 3, pages 239–249. https://doi.org/10.1016/j.exmath.2012.08.004<|endoftext|> TITLE: Is there a name for this "stack" of graphs? QUESTION [5 upvotes]: Let $G_1,\ldots,G_m$ be a sequence of graphs, all having the same number $n$ of vertices. For each pair $(G_i, G_{i+1})$ we add $n$ edges that connect the vertices of $G_i$ and $G_{i+1}$ bijectively. My question: Is there an established name for this "stack" of graphs? REPLY [3 votes]: In the case that you only have two graphs, and they are the same, say $G$, then the edges between the two copies of $G$ can be described by a permutation, and your graphs are precisely the permutation graphs defined by Chartrand and Harary in their 1967 paper "Planar permutation graphs" [Ann. Inst. H. Poincaré Sect. B (N.S.), 3, pp. 433–438]. A cute example of such a "permutation graph" is the Petersen graph, which we can obtain with $G=C_5$. (I cannot stress strongly enough that this is not the standard notion of permutation graphs!)<|endoftext|> TITLE: A reformulation of Erdős conjecture on arithmetic progressions QUESTION [12 upvotes]: Erdős conjecture on arithmetic progressions states that if $S$ is a set of positive integers such that $c(S):=\sum_{n \in S} \frac{1}{n} = \infty$ (large set), then $ \forall \ell \ge 3$ the set $S$ contains an arithmetic progression of length $\ell$. In other words, this conjecture states that for all set $S$ of positive integers, if $\exists \ell \ge 3$ such that $S$ contains no arithmetic progression of length $\ell$, then $c(S) < \infty$. Let $\ell \ge 3$ and $S_{\ell}$ be the lexicographically earliest (strictly) increasing sequence of positive integers that contains no arithmetic progression of length $\ell$. We identify $S_{\ell}$ with its set of terms. According to Erdős conjecture, we should have $c_{\ell}:=c(S_{\ell}) < \infty$. But intuitively, $S_{\ell}$ is the worst set with no arithmetic progression of length $\ell$ to deal with Erdős conjecture. This leads to: Main Question: Is Erdős conjecture equivalent to $c_{\ell} < \infty$ for all $\ell \ge 3$? Let $\ell \ge 3$ and $S$ be a set of positive integers with no arithmetic progression of length $\ell$. Stronger Question 1: Is it true that $c(S) \le c_{\ell}$? Answer: No, see the comment of Andrés E. Caicedo citing this paper of J. Wróblewski (see below). Stronger Question 2: Is it true that $c(S) < c_{\ell}$ or $c(S) \sim c_{\ell}$? Note that OEIS includes the sequences $S_{\ell}$ for all $\ell \in \{3,4,\dots,10 \}$: $S_3$: $1, 2, 4, 5, 10, 11, 13, 14, 28, 29, 31, 32, 37, 38, 40, \dots$ (A003278) $S_4$: $ 1, 2, 3, 5, 6, 8, 9, 10, 15, 16, 17, 19, 26, 27, 29, 30, \dots$ (A005837) $\vdots$ $S_{10}$: $1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, \dots$ (A020664) Ralf Stephan put the following formula in A003278: If $S_3(n)=b(n+1)$ then $b(0)=1$, $b(2n)=3b(n)-2$ and $b(2n+1)=3b(n)-1$. It follows that $S_3(n)$ grows like $n^{\alpha}$ with $\alpha:=\frac{\log(3)}{\log(2)}$, so that $c_3<\infty$. Moreover, $c_3 = \sum_{n=1}^{N}S_3(n)^{-1}+ O(N^{1-\alpha})$. We computed (after the comment of Bullet51) $∑_{n=1}^{10^8}S_3(n) = 3.007886\dots$, whereas $10^{8(1-\alpha)} = 0.0000209\dots$. Now, Wróblewski's paper states that $ 3.007933$, the graph seems to lose this flavor (see here), so that we can no more expect such a recursive formula. Question of Sylvain Julien: Is $c_{\ell}$ an increasing function of $\ell$? More strongly: Is it true that $ℓ≤ℓ′$ implies $S_ℓ(n)≥S_{ℓ′}(n)$ for all $n≥1$? Wróblewski's paper provides a set $S$ of positive integers with no arithmetic progression of length $3$ and with $c(S)>3.00849$. That's why the answer to Stronger Question 1 is no, but the Stronger Question 2 and Main Question are still open (to me). Let $\mathcal{A}_{\ell}$ be the set of sets of positive integers with no arithmetic progression of length $\ell$, and let $a_{\ell}:=\sup_{S \in \mathcal{A}_{\ell}}(c(S))$. Now my belief is that $a_{\ell} < \infty$, that $a_{\ell} \sim c_{\ell}$, but also that there is no $S \in \mathcal{A}_{\ell}$ with $c(S) = a_{\ell}$. I believe that a proof of $a_3<\infty$ should be hard but reachable, on the other side, I believe that the exact value of $a_3$ or even a good approximation (with say 10 digits) is not reachable with the current knowledge/technics. What could be reachable is to break records, starting by the above old record (July 1984) of Wróblewski. Is Wróblewski's record broken today? If yes: What is the largest $c(S)$ known today, with $S \in \mathcal{A}_3$? (same question for every fixed $\ell > 3$) Thomas' answer suggests a set $S$ of Moser (1953), but we don't even know if $c(S)>c_3$. REPLY [2 votes]: I asked Prof. Jaroslaw Wróblewski by email, below is his answer (reproduced with his authorization): I do not know of any new results regarding searching a set $B$ in $\mathcal{A}_3$ [the set of sets of positive integers with no arithmetic progression of length $3$] with large $c(B)$. However since I wasn't interested in the subject for years, I may not be a good source of information. I am sure however, that such a set can be constructed easily with help of nowadays computers - my 35 year old construction was made virtually by hand and a limited access to a computer was used only to get a good estimate of the sum of reciprocals. I have been collecting results regarding large (i.e. having many elements) non-avereging finite sets and the best results known to me are here: http://www.math.uni.wroc.pl/~jwr/non-ave.htm As for $a_3$ [$:=\sup_{S \in \mathcal{A}_{3}}(c(S))$], I am afraid it is impossible to have an educated guess on that. Note that Szekeres's sequence seems to be the best possible before you see a construction which can improve it. So I may give only my speculations which have value of tea leaves reading: I feel that most likely $a_3$ is finite, but only slightly larger than $3$. Less likely, but reasonable is that $a_3 = \infty$. I feel that it is rather unlikely that $a_3$ is finite, but significantly larger than $3$. The reason for that is the following: Szekeres's sequence is the best one for a quite long run. There are sets that are better, but they are better far away (on large numbers). If they give finite sum of reciprocals, then this is sum of reciprocals of large numbers, so it is likely to be very small (assuming it is finite) - most accidental sets of large numbers have small or infinite sum of reciprocals - a set of large numbers with large but finite sum of reciprocals can be easily constructed (must have a very specific density of terms), but is rather unlikely to appear accidentally as a set arising in a problem not direcly related to such a sum.<|endoftext|> TITLE: How can I improve my formal definitions? QUESTION [27 upvotes]: I am a Software Architect and not very familiarized with standard notation in mathematics. Nonetheless, I would like to write a paper explaining a normalization of a computing model for expert systems. It has a very deep background on geometry, logic and group theory, so I have to define some [new] unusual mathematical objects, and in order to get it accepted by the reader (some of them scientists of different disciplines) I would like to be as clear and correct as possible. How can I improve these definitions? Does anyone or any company offer this "help" as a service? I have about 20 definitions like the following (only for instance) to be improved. The $m$-crown of a set $S$, denoted by $S^m$, is the family of sets of every subset of its index set of cardinality $m$ not containing its index, such that $$\forall(X_i \in S^m : i \in S) \rightarrow X_i = \{ x \subset S :(i\notin x \land|x|=m) \}$$ (Note: I am sure that it is a correct definition but may be not easy to understand with a not very standarized notation) REPLY [62 votes]: I don't know about a definition-checking service, but I can give some general advice which I think will help. Let me begin by rewriting your definition (hopefully correctly!): Suppose I have a set $S$ and a natural number $m$. For $i\in S$, let $$X_i=\{a\subseteq S: \vert a\vert=m\mbox{ and }i\not\in a\}.$$ We then let the $m$-crown of $S$ be the indexed set (with indexing set $S$ itself) $$(X_i)_{i\in S}.$$ For example, if $S=\{1,2,3,4\}$ and $m=2$ then e.g. $$X_2=\{\{1,3\}, \{1,4\}, \{3,4\}\},$$ and the whole $m$-crown of $S$ is $$($$ $$X_1=\{\{2,3\}, \{2,4\}, \{3,4\}\},$$ $$X_2=\{\{1,3\}, \{1,4\}, \{3,4\}\},$$ $$X_3=\{\{1,2\}, \{1,4\}, \{2,4\}\},$$ $$X_4=\{\{1,2\}, \{1,3\}, \{2,3\}\}$$ $$).$$ Now, what's behind this? Let's start with what I have done. First, AND MOST IMPORTANTLY, I've broken the definition into separate pieces and worked from the inside out. My version of the definition ends by introducing the $m$-crown, and starts by introducing the much simpler $X_i$s. Moreover, my version breaks into multiple separate sentences. Think of this all as a kind of cognitive piton: it lets the reader digest the definition in discrete steps, without ever having to be kept guessing what things mean. It also helps prevent errors on your end, by making you think carefully about what exactly is going on at each step along the way - when you write an "outside-in" definition, it's easy to experience a kind of "precision fatigue" and wind up finishing with something unclear or garbled. Finally, it winds up helping you use natural language in a precise way, for the same reason. Second, I've provided nontrivial explicit examples of everything. Moreover, I haven't tried to mix notations to be more efficient - the only thing I've done is throw in some spacing for readability (which is actually quite useful in some situations, and sets-of-sets is one of those since the curly braces can blur together). This helps the reader both understand your definition and repair it if you have made any errors. Besides all that, examples also provide a more high-level kind of cognitive piton: in the course of a slew of definitions, they help me keep straight what each thing is and how the various things differ from each other. Really, you should always give examples of everything. Now let me point out two things I haven't done. First, I avoided using too many symbols. In particular I avoided quantifiers. There's a positive and a negative aspect to this. The positive aspect is that precise natural language is easier to read than symbolic expressions; it's almost always better to focus on making your explanation precise than using symbols to shorten everything (and the "piton" stuff above is a big help here). The negative aspect is that unless one has some experience with the relevant formalism, it's easy to misuse - and this is exceptionally true of quantifiers for whatever reason. In the case of your definition, when blindly rendered into English your expression $$\forall(X_i \in S^m : i \in S) \rightarrow X_i = \{ x \subset S :(i\notin x \land|x|=m) \} $$ translates to "If for all [missing bound variable], $X_i$ is in $S^m$ [grammatically incorrect symbol - maybe "such that"?] $i$ in $S$, then $X_i$ is the set of subsets of $S$ of size $m$ not containing $i$." The second half of that is what you want, but the first half is thoroughly garbled. Indeed, my first guess at rewriting this would be "If $X_i$ is in $S^m$ for every $i\in S$, then [rest]," which is very much not what you want. And reading not being much easier than writing, even if you'd gotten the symbols right your readers might still have trouble following the definition. Second, I didn't put motivation inside the definition proper. This isn't something you did either, but it is a common issue (and like everything else makes it much easier to mess up natural-language definitions) so it's worth mentioning here. You should definitely include motivation, but put it before or after the definition (or both); keep the definition itself nice and clean. Here's an example of the sort of thing I'm railing against: Let FOO be the tensor product of BLAH, which by Theorem 11.36 characterizes the FLEEN completely, with (in order to de-VORP the resulting algebra) BLEELG. It would be much better to write this as: Let FOO = BLAH $\otimes$ BLEELG. Remember that BLAH completely characterizes the FLEEN (Thm. 11.36); meanwhile, bringing BLEELG prevents VORPiness. Note that I've also broken the motivation itself into separate pieces: why the BLAH?, and why the BLEELG?, are separate issues and I've treated them as such. Now it's important to note that these rules don't always have to be followed. But I think one should follow them very closely if one isn't already rather experienced with this sort of writing, especially if the intended readership isn't necessarily either.<|endoftext|> TITLE: NP-completeness of a covering problem QUESTION [6 upvotes]: I was wondering about the complexity of the following covering problem. Let $B_i,\,i=1,\ldots,n$ be a set of unit disks in $\mathbb{R}^2$. The problem is to decide whether there exists $C\subset\{1,\ldots,n\}$ with $|C|\leq k$ such that $\bigcup_{i\in C} B_i = \bigcup_{i=1}^n B_i$. This could be considered as a continuous version of the geometric set cover problem. There is a paper that provides a constant approximation algorithm (Basappa et al, "Unit disk cover problem in 2D") but no proof of NP completeness has been provided. In this paper, the problem is referred to as the "rectangular region cover" problem. Also, in one version of the paper, they claim that the problem is NP-complete and they refer to Garey & Johnson but I could not find a proof there either. Computational complexity is far to my area of expertise so I do not want to reinvent the wheel by reproducing some well-known result. Any help will be greatly appreciated. REPLY [4 votes]: The typical presentation of this problem is more like the following: Given: A finite set of points $P$ and a set of unit disks $C$ in the plane Question: Does there exist a subset $C' \subseteq C$ such that $|C'| \leq k$ and $P \subseteq \cup_{S \in C'} S$? In this formulation, this problem is commonly known as DISCRETE UNION DISK COVER (DUDC). I believe the canonical reference for demonstrating that this problem is NP-complete is [1]. Note that in DUDC, $P$, the set of points to be covered, is given as finite. In the version proposed in this question, the set of points to be covered is given as $\cup_{S\in C} S$. In order to show that this version is NP-complete, we will need to demonstrate how some known NP-complete problem (either DUDC or otherwise) can be translated into this version. [A previous version of this answer had the conversion the wrong way around.] EDITED TO ADD: One other problem that seems related but not quite on the nose is DOMINATING SET IN DISK GRAPHS. It can be paraphrased as follows: Given: A set of unit disks $C$ in the plane Question: Does there exist a subset $C' \subseteq C$ such that $|C'| \leq k$ and for each $c \in C$ there exists $c' \in C'$ such that $c'$ intersects $c$? I believe the original reference for the NP-completeness of this problem is [2] where it is referred to as "EUCLIDEAN m-CENTER ON POINTS". Fowler, Robert J., Michael S. Paterson, and Steven L. Tanimoto. "Optimal packing and covering in the plane are NP-complete." Information processing letters 12.3 (1981): 133-137 Masuyama, Shigeru, Toshihide Ibaraki, and Toshiharu Hasegawa. "The computational complexity of the m-center problems on the plane." IEICE TRANSACTIONS (1976-1990) 64.2 (1981): 57-64.<|endoftext|> TITLE: Spelling König's Lemma QUESTION [11 upvotes]: I was surprised to learn here that the man responsible for "König's Lemma" was Hungarian, and spelled his last name Kőnig (with a different accent on the o), presumably with the same accent that occurs in Paul Erdős' last name. I still remember being taught that this accent was not an umlaut, and had a special LaTeX command. Now, in the Konig-case, what adds more to the confusion is that we mathematicians historically have used the (incorrect) umlaut spelling when referring to the the tree lemma. Moreover, Kőnig's father apparently spelled his name with an umlaut (according to a comment by Asaf Karagila here). Should we as a math community correct the spelling of "König's Lemma" to match the name of the one honored by it? Edited to add: The story gets even weirder. Kőnig sometimes spelled his name (or, at least, the editors spelled his name) as König in some of his publications. For the 1927 publication in question, the marking on his name, on the first page, is hard to make out, but appears to be the Hungarian marking. But on the running header throughout the paper, the markings are clearly the German umlaut! REPLY [4 votes]: In A tale of three eras: The discovery and rediscovery of the Hungarian Method Harold W. Kuhn adresses the issue under point 4. Who was Dénes Konig? and clarifies that the correct spelling is with the Hungarian "double acute" and not the German "umlaut" that appears in the German word for "king".<|endoftext|> TITLE: A conjecture on primitive tenth roots of unity QUESTION [9 upvotes]: QUESTION. How to solve my following conjecture involving primitive tenth roots of unity? Conjecture. Let $\zeta$ be any primitive tenth root of unity. Then $$\prod_{k=1}^{(p-1)/2}(\zeta-e^{2\pi ik^2/p})=(-1)^{|\{1\le k\le\frac {p+9}{10}:\ (\frac kp)=-1\}|}$$ for each prime $p\equiv21\pmod{40}$, and $$\prod_{k=1}^{(p-1)/2}(\zeta-e^{2\pi ik^2/p})=(-1)^{ |\{1\le k\le\frac {p+1}{10}:\ (\frac kp)=-1\}|}\zeta^2$$ for any prime $p\equiv29\pmod{40}$, where $(\frac kp)$ is the Legendre symbol. Remark. The conjecture was motivated by Questions 337879 and 338876 on MathOverflow, and I have checked it numerically via Mathematica. For related materials, see Conjectures 5.4 and 5.5 of my preprint arXiv:1908.02155. REPLY [5 votes]: Not a complete solution. Let $p$ be (1 mod 4) and $r,n$ runs over quadratic residues/non residues mod $p$ in $[1,p-1]$ and let $R_p(x)=\prod_r(x-\zeta_p^r),\;\;\; N_p(x)=\prod_n(x-\zeta_p^n).$ It is known (see for example, Daveport's Multiplicative number theory, p.19), that $R_p(x)=[Y_p(x)-\sqrt{p}Z_p(x)]/2, \;\; N_p(x)=[Y_p(x)+\sqrt{p}Z_p(x)]/2$ where $Y_p(x),Z_p(x) \in \mathbb{Z}[x]$, so $Z_p(x)=[N_p(x)-R_p(x)]/\sqrt{p}$ and $N_p(x)R_p(x)=(x^p-1)/(x-1)$. If one compute a few terms of $Z_p(x)$, one can discover immediately that $Z_p(x)$ is divisible by the 10th cyclotomic polynomial $\Phi_{10}(x)$ if and only if $p=21, 29$ mod 40. This explain why one gets nice result as conjectured only for these two congruence classes. Assuming this, (we are still missing a proof that $N_p(\eta)=R_p(\eta)$), it will follow that for any primitive 10th root of unity $\eta$, $R_p(\eta)^2=R_p(\eta)N_p(\eta)=(\eta^p-1)/(\eta-1)$, so $R_p(\eta)=\pm 1$ if $p$ is (21 mod 40), and $R_p(\eta)=\pm \eta^2$ if $p$ is (29 mod 40). It remains to determine the sign. This is reminiscent of the determination of the sign of the Gauss's sum. We now observe that the very useful trick of writing $(1-e^{i\theta})=(e^{-i \theta/2}-e^{i \theta/2})e^{i \theta/2}=(-2i)\sin( \theta/2) e^{i \theta/2}$ can be extended to the case of two roots of unity, namely $R_p(\zeta_m^a)=\prod_r(\zeta_m^a-\zeta_p^r)(\zeta_m^{-a/2}\zeta_p^{-r/2})\prod_r(\zeta_m^{a/2}\zeta_p^{r/2}) $ $= \zeta_m^{a(p-1)/4}\zeta_p^{-p(p-1)/8}(2i)^{(p-1)/2} \prod_r \sin \pi (\frac{a}{m}-\frac{r}{p}). $ Now set $m=10,a=1,p=40k+21$, we get $R_p(\zeta_{10})=(-1)2^{(p-1)/2} \prod_r \sin \pi (\frac{p-10r}{10p}).$ Clearly $-1 < \frac{p-10r}{10p}<1$, so the number of negative $\sin$ factors in the product is the number of $r$ with $r>p/10$ which equals $(p-1)/2- \#\{1 \le r \le p/10\}=(p-1)/2-\#\{1 \le r \le (p+9)/10\},$ since $c=(p+9)/10$ is always a non residue mod $p$ as $10c=9$ mod 10 is a residue while 10 is nonresidue. So the number of negative $\sin$ factors is $(p-1)/2-( (p+9)/10-\#\{n : n \le (p+9)/10 \})$. Since $1+(p-1)/2-(p+9)/10$ is even for $p=40k+21$, $R_p(\zeta_{10})=(-1)^{\#\{n : 1 \le n \le (p+9)/10 \}}$. Now $Y_p(x)=x^{(p-1)/4}W(x+1/x)$, since $Y(x)$ is known to be reciprocal, we have $2R_p(\zeta_{10})=Y_p(\zeta_{10})=Y_p(\zeta_{10}^9)$. Similarly, we have $R_p(\zeta_{10}^3)=R_p(\zeta_{10}^7)$. We still need to show $R_p(\zeta^3)=R_p(\zeta)$. The case for $p=29$ mod 40 is similar. (Added) The congruence condition $p=21,29$ mod 40 is determined by $-1$ and $5$ are quadratic residue while $2$ is a non residue. The sign depends on whether $5$ is a quartic residue mod $p$, ie $5^{(p-1)/4} = \pm 1$ mod $p$.<|endoftext|> TITLE: Surprising appearances of Painlevé transcendents QUESTION [15 upvotes]: What are some of your favorite examples of enumerative problems whose answer ended up being (related to) a solution to one of the Painlevé equations? I have seen examples from enumeration of classes of graphs, to Hodge integrals and Hurwitz numbers and I'm sure there are many more out there. Bonus points if there is an explanation/heuristic that demystifies or explains why one should expect coefficients of a Painlevé transcendent to appear. REPLY [5 votes]: The OP likely has surprising appearances in the mathematical literature in mind, but perhaps the physical observation of a Painléve transcendent is noteworthy? In Universal Fluctuations of Growing Interfaces: Evidence in Turbulent Liquid Crystals (2001) the growth of a liquid crystal cluster is observed in time. The time-dependent probability distribution $\rho$ of the radius $R$ follows (upon rescaling) the solution of the Painléve II equation (dashed curve in the right plot). The origin of the correspondence is described by Craig Tracy in Integrable Probability and the Role of Painlevé Functions. I am not aware of any other real-world observation of a Painlevé function. REPLY [2 votes]: I have seen Painlevé transcendents appear a few times in the context of field theory. For example, they appear in model solutions of Hitchin's equations for Higgs fields. (See [MSWW16] and references therein.) I have also discovered the Hastings-McLeod solution in a certain rescaling limit of the Ginzburg-Landau equations. (Unfortunately this is unpublished.) Both instances occurred in the context of radially-equivariant solutions to a non-linear elliptic PDE in two dimensions. The resulting ODE is unstable, and there is a distinguished solution which is well-behaved at both $r=0$ and $r=\infty$. If you have a reasonably-natural nonlinear ODE which is not Painléve, you might be able to add some parameters, take a limit, get lucky, and find something which is Painléve. I also have a vague intuition that their appearance is related to their nice asymptotics. [MSWW16] Mazzeo, Rafe; Swoboda, Jan; Weiss, Hartmut; Witt, Frederik, Ends of the moduli space of Higgs bundles, Duke Math. J. 165, No. 12, 2227-2271 (2016). ZBL1352.53018.<|endoftext|> TITLE: Heuristic argument for the Riemann Hypothesis QUESTION [33 upvotes]: Is there a heuristic argument that supports the validity of the Riemann hypothesis or are we just relying on numerical evidence? Moreover, what is the strongest theorem that supports the validity of RH? REPLY [13 votes]: I am not sure that this is a proper answer for your question. Sorry in advance if it is not. Other answers provided some good examples of heuristic arguments supporting the RH. But I want to point at a heuristic result that puts some serious doubts on it. Since I am not so familiar with the technical details, it is better to quote directly from this answer: The De Bruijn-Newman constant $\Lambda$ was defined and upper bounded by $\Lambda\le\frac{1}2$ in 1950. After 58 years of work, in 2008 this upper bound was finally improved to... $\Lambda<\frac{1}2$ (a 0% improvement) in a 26-page paper. The best known upper bound is currently $\Lambda<0.22$. The Riemann hypothesis was known to be equivalent to $\Lambda\le 0$, so if it's true then we've got quite a ways to go. But Terrence Tao and Brad Rogers recently proved that $\Lambda\ge 0$. So in Tao's words (actually, Newman's words quoted by Tao): If the Riemann hypothesis is true, then it's 'just barely' true.<|endoftext|> TITLE: $K_3(\mathbb{Z})$ and $\pi ^S_3$ QUESTION [9 upvotes]: This is an afterthought on this MO question, and also on Gannon's book mentioned there, about $K_3(\mathbb{Z})=\mathbb{Z}/48$. Neither the question nor the book mentions a possible connection with the third stable homotopy group of spheres, $\pi ^S_3=\mathbb{Z}/24$. Does anyone know about such a connection? REPLY [9 votes]: We have $\pi_3(\mathbb{S}) \cong \mathbb{Z}/24\{ \nu\}$ and $\pi_3K(\mathbb{Z}) \cong \mathbb{Z}/48\{ \lambda \}$. As Achim suggested, the unit map $\mathbb{S} \to K(\mathbb{Z})$ induces on $\pi_3$ the injection sending $\nu$ to $2\lambda$. See the first paragraph of Section 2 of 'Divisibility of the Dirac magnetic monopole as a two-vector bundle over the three-sphere' by Ausoni, Dundas, and Rognes.<|endoftext|> TITLE: Optimization with weaker oracle than projection QUESTION [5 upvotes]: I'm looking to solve the optimization problem $$ min_{x \in C} ~ f(x), $$ where $C \subset R^n$ is a closed, convex, bounded set and $f : R^n \to R$ a Lipschitz differentiable (nonconvex) function. In my problem, $C$ is the solution set of a difficult convex optimization problem, so the projection onto $C$ and also a linear minimization oracle are intractable to compute in closed-form, thus projected gradient or Frank-Wolfe methods are not applicable. However, I can efficiently compute a separating hyperplane between a point $\bar x$ and the set $C$. My question is whether iterations of the type $$ \bar x^{t+1} = x^t - \alpha_t \nabla f(x^t), $$ $$ x^{t+1} = \text{proj}_H(\bar x^{t+1}), $$ have been analyzed in literature or have hope of converging to a stationary point. Here $\{ \alpha_t \}$ is a suitable vanishing step-size sequence, and $proj_H$ the projection onto a separating half-space to the set $C$ at point $\bar x^{t+1}$. REPLY [2 votes]: I would guess that the method is going to converge (weakly), even with constant stepsizes. Off the top of my head I don't know a precise reference. The method is close in spirit to the "hybrid projection proximal point method" by Solodov and Svaiter, but you have a gradient step instead of a proximal step.<|endoftext|> TITLE: Norms as Points in $C(X)$ QUESTION [7 upvotes]: $\newcommand\abs[1]{\lvert{#1}\rvert}$Let $X$ be a compact hausdorff space, and put $C(X)$ for the $\mathbb{R}$-algebra of continuous maps from $X$ to $\mathbb{R}$. For each point $x$, there is a multiplicative semi-norm $\abs-_x$ on $C(X)$, where $\abs f_x = \abs{f(x)} \in \mathbb{R}_{\geq 0}$. That is, $\abs{f}_x \geq 0$. $\abs{cf}_x = \abs c \cdot \abs f_x$. $\abs{fg}_x = \abs{f}_x\cdot\abs{g}_x$. $\abs{f + g}_x \leq \abs f_x + \abs g _x$. Let $Y$ be the set of multiplicative semi-norms on $C(X)$ mod the equivalence relation where $\abs- \sim \abs-'$ when each is bounded by the other up to a constant. We have just constructed a function $\nu : X \rightarrow Y$. My questions are: How might we put a topology on $Y$ using only the information of $C(X)$? Is $\nu$ above a bijection? In light of the last question, is it a homeomorphism? So in this view, multiplicative seminorms are like points or generalized points. Interestingly, the submultiplicative norm $\abs- : C(X) \rightarrow \mathbb{R}$ sending $f$ to $\operatorname{sup}_x \abs f_x$ seems like it could be the supremum of multiplicative norms in $Y$. That way, multiplicative norms are the "local" versions of submultiplicative norms, which would be "global". REPLY [8 votes]: A slight improvement to Terry Tao's answer: It is not necessary to assume a priori that the multiplicative seminorm $\|\cdot\|$ is continuous with respect to the sup-norm on $C(X)=C(X\to \mathbb R)$ to conclude $\|f\|=|f(x_0)|$ for some $x_0\in X$. This follows from a theorem of Mazur [Sur les anneaux linéaires, C. R. Acad. Sci. Paris, 2 0 7 (1938), 1025-1027] which says that the only real algebras with a multiplicative norm are (isometrically isomorphic to) the real numbers, the complex numbers, or the quaternions. We will apply this to the the quotient algebra $C(X)/I$ with the ideal $I=\{f\in C(X): \|f\|=0\}$. Denoting by $\tilde f$ the equivalence class of $f\in C(X)$ one easily sees that $\tilde f \mapsto \|f\|$ is a well-defined multiplicative norm on $C(X)/I$ to which Mazur's theorem applies. Since this quotient algebra is commutative it cannot be the quaternions and we show that it neither can be ismorphic to $\mathbb C$. Indeed, assume that $\pi: C(X)\to \mathbb C$ is an algebra morphism onto $\mathbb C$ with kernel $I$. If $g\in C(X)$ satisfies $\pi(g)=i$ we get $\pi(g^2+1)=0$ so that $g^2+1 \in I$ which is absurd because $g^2+1$ is invertible in $C(X)$ and thus does not belong to the (proper) ideal $I$. We conclude that $C(X)/I$ is isomorphic to $\mathbb R$ so that $I$ is a maximal ideal in $C(X)$ and hence of the form $I=\{f\in C(X): f(x_0)=0\}$ for some $x_0\in X$. Finally, given $f\in C(X)$ write $f=f-f(x_0)+f(x_0)$ (where the number $f(x_0)$ is identified with the constant function $f(x_0)\mathrm 1)$ to get $$\|f\|=\|f(x_0)\mathrm 1\|=|f(x_0)| \|\mathrm 1\|=|f(x_0)|.$$<|endoftext|> TITLE: $|(A+B)(X)|=o(X)$ if $|A(X)|=O(X^{1/2})$ and $|B(X)|=O(X^{1/2})$? QUESTION [7 upvotes]: Sorry if this is trivial: it is well-known that the number of sums of two squares less than $X$ is asymptotic to $CX/\log(X)^{1/2}$ for some $C$. Is this a general phenomenon ? More precisely, if $A$ and $B$ are subsets of the natural numbers whose counting functions $|A(X)|$ and $|B(X)|$ are $O(X^{1/2})$, it it true that $|(A+B)(X)|$ is $o(X)$ (or perhaps always $O(X/\log(X)^{1/2})$ ? REPLY [11 votes]: Here are four approaches to find counter examples, three of them even work in the special case $A=B$. Together all this means that cases like sums of two squares (or more generally quadratic forms in the integers create an algebraic reason that those integers that can be written in this form have a representation number (say $r_2(n)/4$ for sums of two squares) which is much larger than for random sets. One could ask vice versa, which conditions on $A$ with counting function $\gg \sqrt{X}$ allow that $(A+A)(X)=o(X)$? Examples would be $A$: contains large subsets which come from linear or quadratic polynomials. What other examples? Now the four types of counter examples: A) The elements lie close to squares: Atkin proved that it is possible to use elements $s_j=j^2+O(\log j)$ such that the sumset of these pseudo squares has positive density. (See references 1) and 2).) B) Another approach is from (infinite) Sidon sets, which have by definition the property that all sums $a_i+a_j$ are distinct, or more generally $B_2(g)$ sequences, with the number of representations as $a_i+a_j$ bounded by $g$. (Then $A(X)=O(\sqrt{X})$ follows.) Erdos constructed (see for example Krückeberg or Cilleruelo and Trujillo below): an infinite Sidon set $A$ with counting function $A(X)$ and $\lim \sup_{x \rightarrow \infty} \frac{A(X)}{\sqrt{X}}\geq \frac{1}{2}$. By the Sidon property the sumset $A+A$ must have positive density at least infinitely often, which contradicts $A(X)=o(X)$. (To do this with $\lim \inf$ is much harder and unknown, see 5) below.) (Krückeberg improved the constant of Erdos, Cilleruelo and Trujillo generalized it to $B_2(g)$ sequences. C) A third related approach are "thin bases". A set $A\subset\mathbb{N}_0$ is said to be a basis of order 2 if every positive integer can be written as a sum of 2 (not necessarily distinct) elements of $A$, and is called "thin" if $A(X)=O(\sqrt{X})$. Stöhr, Raikov and many others proved the existence of such bases. Moreover, they actually have so called UR-bases, where $A=B \cup C$, every element can be written in a unique way as $b+c, b \in B, c\in C$. (The Stöhr-Raikov constructions are digit based, and thus related to Gjergji Zaimi's solution. Nathanson (and others) have also non-digit based constructions.) D) It follows by work of Sarnak, https://mathscinet.ams.org/mathscinet-getitem?mr=1472786 and Eskin, Margulis and Mozes https://mathscinet.ams.org/mathscinet-getitem?mr=2153398 that many quadratic forms with irrational coefficients and integer variables generate a positive density of the integers, in particular this works with $\lfloor x^2+\alpha y^2\rfloor$, if $\alpha$ is not extremely well approximable. (Here algebraic irrationals such as $\alpha=\sqrt 2$ are admissable.) Choose $A$ as the set of squares and $B=\{\lfloor \alpha y^2\rfloor ; y \in \mathbb{N} \}$. See also this MO post: Distribution of $a^2+\alpha b^2$ References: 1) A.O.L. Atkin, On Pseudo-squares, Proc. LMS 14 (1965), 22-27 Atkin's theorem on pseudo-squares, R. Balasubramanian and D.S. Ramana PUBLICATIONS DE L'INSTITUT MATHÉMATIQUE (BEOGRAD) (N.S.) Vol. 63 (77), pp. 21-25 (1998) https://www.emis.de/journals/PIMB/077/3.html Krückeberg. B2-Folgen und verwandte Zahlenfolgen, Journal für die reine und angewandte Mathematik (1961) Volume: 206, page 53-60 https://eudml.org/doc/150472 Javier Cilleruelo, Carlos Trujillo. Infinite $B_2[g]$ sequences Israel Journal of Mathematics, December 2001, Volume 126, Issue 1, pp 263–267. https://doi.org/10.1007/BF02784156 Gowers's webblog (2012): https://gowers.wordpress.com/2012/07/13/what-are-dense-sidon-subsets-of-12-n-like/ A. Stöhr Eine Basis h-ter Ordnung für die Menge aller natürlichen Zahlen, Mathematische Zeitschrift (1937) Volume: 42, page 739-743 https://eudml.org/doc/168747 D. Raikov Über Basen der natürlichen Zahlenreihe Mat. Sb. N.S., 2 (44) (1937), pp. 595-597. http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=5584&option_lang=eng M.B. NathansonNathanson, Thin bases in additive number theory Discrete Math. 312 (2012), no. 12-13, 2069–2075. https://mathscinet.ams.org/mathscinet-getitem?mr=2920867 Eskin, Margulis, Mozes, Quadratic forms of signature (2,2) and eigenvalue spacings on rectangular 2-tori. Ann. of Math. (2) 161 (2005), no. 2, 679–725.<|endoftext|> TITLE: Is the equational theory of groups axiomatized by the associative law? QUESTION [8 upvotes]: Consider the class of groups in the signature {*}. Is the equational theory of that class axiomatized by the associative law? I asked this on math stack exchange but I didn't receive a satisfactory answer? REPLY [4 votes]: The title should read " Is the equational theory of (the class of those semigroups which are groups) axiomatized by the associative law?", and the answer given by Bjorn Kjos-Hanssen makes even more sense (and seems more immediate): Take any free group on a set X of generators, and consider its semigroup reduct. The set X generates a subsemigroup S of this reduct, which is free (in semigroups) on X. (If it weren't, the words in X would be identified in the free group.) So any hope of finding a nontrivial semigroup identity in the equational theory of this class is toast. Gerhard "Please Pass The Butter Knife" Paseman, 2019.08.31.<|endoftext|> TITLE: What is this sequence counting? QUESTION [5 upvotes]: While solving (a system of) a system of linear equations level-by-level recursively, I am finding some redundant equations for level $n\geq5$. The reason why the redundancies arise is because $P(n)\neq P(n-1)+P(n-2)$ for $n\geq5$. The redundancies are given by the sequence : $$ 0,0,0,0,1,1,3,4,7,10,16,21,32,43,60,80,110,\dots~. $$ Here, $P(n)$ is the number of partitions of the integer $n$. This is the sequence is given by $P(n-1)+P(n-2)-P(n)$ for $n\geq1$. A generating function for the above sequence is $$ 1 - (1-q-q^2)\prod_{n=1}^\infty {1\over (1-q^n)}~. $$ Is this sequence (or any closely related one) encountered in some context in combinatorics? What is being counted? (I have searched this in OEIS; there are a few sequences matching till the $16$ above but disagrees thereafter.) REPLY [3 votes]: Notice that $P(n-1)$ counts the number of partition of $n$ that contain $1$, while $P(n-2)$ counts the number of partition of $n$ that contain $2$. It follows that $P(n-1)+P(n-2)-P(n)$ equals the difference between the number of partitions of $n$ that contain $\{1,2\}$ and the number of partitions of $n$ that contain neither $1$ nor $2$.<|endoftext|> TITLE: To minimize the Hausdorff distance between convex polygonal regions QUESTION [7 upvotes]: Definition: The Hausdorff distance is the greatest of all the distances from a point in one set to the closest point in the other set. Question: Given two convex polygonal regions P1 and P2 on the plane, give an efficient algorithm that locates and orients P2 with respect to P1 (the two regions are allowed to overlap) so that the Hausdorff distance between P1 and P2 is minimized. REPLY [6 votes]: A polynomial-time algorithm was presented in this paper: Chew, L. Paul, Michael T. Goodrich, Daniel P. Huttenlocher, Klara Kedem, Jon M. Kleinberg, and Dina Kravets. "Geometric pattern matching under Euclidean motion." Computational Geometry 7, no. 1-2 (1997): 113-124. Elsevier journal link. If I read their results correctly, "the minimum (bidirectional) Hausdorff problem under Euclidean motion" for polygons can be solved in time $O((m + n)^6 \log^2 mn)$, for polygons of $n$ and $m$ vertices. They did not specialize to convex polygons. These are primarily theoretical results, and I doubt have been implemented. Other work focuses on approximation algorithms, e.g., Alt, Helmut, Bernd Behrends, and Johannes Blömer. "Approximate matching of polygonal shapes." Annals of Mathematics and Artificial Intelligence 13, no. 3-4 (1995): 251-265. Springer link.                     Fig.5. A more practical algorithm specialized to convex polygons is described here: Danilov, Dmitry I., and Aleksei Stanislavovich Lakhtin. "Optimization of the algorithm for determining the Hausdorff distance for convex polygons." Ural Mathematical Journal 4, no. 1 (2018): 14-23. doi.<|endoftext|> TITLE: An example of a morphism of rigid analytic spaces with affinoid base which is proper but does not satisfy $(\dagger)$ QUESTION [6 upvotes]: Let $k$ be a complete non-archimedean field and let $\varphi \colon X \to Y$ be a morphism of rigid analytic spaces over $k$, where $\newcommand{\Sp}{\operatorname{Sp}}Y = \Sp(B)$ is affinoid. Consider the following condition: $(\dagger)$ The morphism $\varphi$ is separated and there exist two finite admissable affinoid coverings $\mathfrak{U} = (U_i)_{i \in I}$, $\mathfrak{V} = (V_i)_{i \in I}$ of $X$ such that $V_i \Subset_Y U_i$ (i.e. $V_i$ lies relatively compact in $U_i$ w.r.t. $Y$) for all $i \in I$. A morphism $\varphi \colon X \to Y$ (with $Y$ not necessarily affinoid) is called proper if it is separated and there is an admissable affinoid covering $(W_j)_{j \in J}$ of $Y$ such that for each $j \in J$ the morphism $\varphi^{-1}(W_j) \to W_j$ satisfies condition $(\dagger)$. If $Y$ is affinoid, then obviously a morphism satisfying condition $(\dagger)$ is proper, whereas the converse is not clear. In fact, Bosch in his book Lectures of Formal and Rigid Geometry writes at the beginning of section 6.4 that $(\dagger)$ is slightly stronger than properness. Is there an example of a proper morphism with affinoid base which does not satisfy condition $(\dagger)$? REPLY [3 votes]: These two notions are actually equivalent (at least if $k$ is the fraction field of a dvr $R$), but I do not know any direct way to see this. The proof I know heavily uses the theory of formal schemes. The three main results we need are Lemma 2.5, Lemma 2.6 and the statement after Corollary 3.2 from Lutkebohmert's paper ``Formal-algebraic and rigid-analytic geometry''. Lemma 2.5: Let $\mathfrak W \to \mathfrak V=\text{Spf}(A)$ be a morphism of admissible formal schemes whose generic fibers are affinoid and let $\mathfrak U \subset \mathfrak W$ be an open formal subscheme. Then $\mathfrak U_K$ is relatively compact in $\mathfrak W_K$ over $\mathfrak V_K$ if and only if the Zariski-closure $\overline{\mathfrak U}_0$ of $\mathfrak U_0$ in $\mathfrak W_0$ is proper over $\mathfrak V_0$. Lemma 2.6: Let $f\colon \mathfrak X \to \mathfrak Y$ be a morphism of admissible formal schemes and let $f_K\colon \mathfrak X_K \to \mathfrak Y_K$ be its generic fiber. If the rigid map $f_K$ is proper, the formal map $f$ is proper. Claim on page 350: Let $f\colon \mathfrak X \to \mathfrak Y=\text{Spf} B$ be a proper morphism of admissible formal schemes. Let $\mathfrak U$ be a formal open subscheme of $\mathfrak X$ which is affine. Then there exists an admissible blowing-up $\mathfrak X \to \mathfrak X'$ with a center contained in the complement of $\mathfrak U_0$ such that there exists an open subscheme $\mathfrak U'$ of $\mathfrak X'$ whose associated rigid subspace $\mathfrak U'_K$ is affinoid and such that the Zariski-closure $\overline{\mathfrak U_0}$ of $\mathfrak U_0$ in $\mathfrak X'_0$ is contained in $\mathfrak U'_0$. These results easily imply the desired claim. Indeed, suppose that $g\colon X \to \text{Sp}(A)$ is a proper map. Choose some topologically finitely generated ring of definition $A_0 \subset A$. Then using the standard machinery on formal models, we can find a morphism of admissible formal schemes $f\colon \mathfrak X \to \text{Spf } A_0$ such that the generic fiber $f_K$ is equal to $g$. Then Lemma 2.6 reads that $f$ is proper as a map of formal schemes. Now comes the key step: we use the last cited claim to find a good covering of $X$. Namely, we choose any covering of $\mathfrak X$ by open affine formal schemes $\mathfrak U_i$. We apply that claim to find an admissible blow-up $\mathfrak X'_i \to \mathfrak X$ with an open formal subscheme $\mathfrak U'_i\subset X'_i$ such that $(\mathfrak U'_i)_K$ is affinoid and $\overline{\mathfrak U_0}$ (that is proper as a closed subscheme of $\mathfrak X'_0$) is contained in $(\mathfrak U'_i)_0$. Lemma 2.5 implies that $(\mathfrak U_i)_K$ is relatively compact in $(\mathfrak U'_i)_K$. The last thing to observe is that $(\mathfrak U_i)_K$ form a finite covering of the rigid space $X$. P.S. Lutkebohmert's paper is written entirely in the noetherian setup. So if one wants to give a proof along these lines without noetherianness assumption on $\mathcal O_K$, then one needs to reprove these results in the set-up of arbitrary (complete) rank-$1$ valuation rings. My recollection is that it should not be that difficult given the recent finiteness results of Fujiwara and Kato (Fujiwara, Kato ``Foundations of Rigid Geometry I'')<|endoftext|> TITLE: Is $\sum_{n=1}^\infty\frac{S(n)}{n!}$ an irrational, where $S(n)$ denotes the sum of remainders function? QUESTION [12 upvotes]: For each integer $n\geq 1$ we consider the arithmetic function $$S(n)=\sum_{k=1}^n n\text{ mod }k,\tag{1}$$ the sum of remainders function, the arithmetic function A004125 from the OEIS. Example. We've that for $n=6$ $$S(6)=0+0+0+6\text{ mod }4+6\text{ mod }5+0=2+1=3.$$ This arithmetic function was studied for example in [1]. I wondered about a type of problems that are in the literature, that is in our case what about the irrationality of the real number $$\sum_{n=1}^\infty\frac{S(n)}{n!}.\tag{2}$$ I don't know if this previous example is in the literature or has good mathematical content. Question. Is it possible to deduce that $$\sum_{n=1}^\infty\frac{S(n)}{n!}$$ is irrational? Or well, is it possible to discard it as an irrational? I am asking about if it possible to do or provide some work, reasonings or heuristics, about it. Then I should to accept an answer. If it is in the literature refer the article and I try to search and read the statement and proof. References: [1] Michael Z. Spivey, The Humble Sum of Remainders Function, Mathematics Magazine, Vol. 78, No. 4 (Oct., 2005). REPLY [4 votes]: Well, let me elaborate Ilya Bogdanov's argument. First of all, $$S(n)=\sum_{k=1}^n \left(n-k\lfloor n/k\rfloor\right)=n^2-\sum_{k,d:kd\leqslant n} k= n^2-\sum_{d=1}^n (1+2+\ldots+\lfloor n/d\rfloor). $$ We have $1+2+\ldots+\lfloor n/d\rfloor=\frac1{2d^2}n^2+O(n/d)$, thus $$S(n)=\beta n^2+O\left(n\sum_{i=1}^n \frac1d+n^2\sum_{d=n+1}^\infty\frac1{2d^2}\right)=\beta n^2+O(n\log n),$$ where $\beta=1-\frac12\sum_{d=1}^\infty \frac1{d^2}=\frac12-\frac{\pi^2}{12}$. Assume that $\alpha:=\sum_{k=1}^\infty \frac{S(k)}{k!}$ is rational. Choose large $n$. Then $$(n-1)!\alpha=(n-1)!\sum_{k=1}^\infty \frac{S(k)}{k!}=\text{integer}+\frac{S(n)}n+\frac{S(n+1)}{n(n+1)}+o(1)=\\ \text{integer}+\frac{S(n)}n+\beta+o(1),$$ thus $S(n)/n=k_n-\beta+o(1)$ for certain integer $k_n$. We get $$ k_{n+1}-\beta+o(1)=\frac{S(n+1)}{n+1}=\frac{S(n)+(2n+1)-\sigma(n+1)}{n+1}=\\ \frac{S(n)}{n+1}+\frac{(2n+1)-\sigma(n+1)}{n+1}= \frac{S(n)}n-\frac{S(n)}{n(n+1)}+\frac{(2n+1)-\sigma(n+1)}{n+1}=\\ k_n-\beta+o(1)-\beta+\frac{(2n+1)-\sigma(n+1)}{n+1}. $$ If, say, $n+1$ is large prime, this is not possible.<|endoftext|> TITLE: Which right square pyramids are scissors congruent to a cube? QUESTION [17 upvotes]: Consider a right square pyramid whose base has side length $2r$ and whose height is $h$. Let the dihedral angle between the base and each triangular side be $\theta$, and the dihedral angle between adjacent triangular sides be $\phi$. We have $$\cos(\theta)=\frac{r}{\sqrt{r^2 + h^2}}$$ and $$\cos(\phi)=\frac{-r^2}{r^2+h^2}$$ so in particular $$\cos^2(\theta)=-\cos(\phi).$$ This square pyramid is scissors congruent to a cube iff its Dehn invariant is zero, for which it is necessary that $\pi$ be expressible as a rational linear combination of $\theta$ and $\phi$. If $r=h$ then this condition is satisfied, and indeed such a square pyramid is scissors congruent to a cube. There is no other nontrivial way to satisfy this condition with both $\theta$ and $\phi$ being rational multiples of $\pi$, as shown by this answer on MSE. It seems unlikely that there are any other solutions at all, but I have no idea how to even begin to prove such a thing. To ask a precise Question: Are there any pairs $\theta$, $\phi$ such that $\cos^2(\theta)=\cos(\phi)$ and $\pi$ is expressible as a rational linear combination of $\theta$ and $\phi$, except where $\cos(\phi)\in\{0,\frac12,1\}$? Equivalently, are there any pairs $\theta$, $\phi$ such that $\cos^2(\theta)=\cos(\phi)$, and $\pi$ is expressible as a rational linear combination of $\theta$ and $\phi$, but $\theta$ and $\phi$ are not rational multiples of $\pi$. Added later: I asked the question above in the expectation that the answer would be no, which would then imply the nonexistence of other right square pyramids that are scissors congruent to a cube. Unfortunately it is quite easy to see that the answer is actually yes, as Daniil Rudenko points out below. This shows that I asked the wrong precise question. The right question is more complicated, and is described – and answered – by user145307. REPLY [14 votes]: This is not the question you want to ask. (The actual question asked is easy by a continuity argument.) If the sides of the pyramid have length $2r$ and the height is $h$, then the other side lengths of the pyramid have length $\sqrt{2r^2 + h^2}$. You want to ask whether the element $$\xi = (2r \otimes \theta) + (\sqrt{h^2 + 2 r^2} \otimes \phi)$$ is trivial in the Dehn group. For this to be true, either: $\theta$ and $\phi$ are both rational multiples of $\pi$ (case already done). $r$ is a rational multiple $0 < v < 1/\sqrt{2}$ of $\sqrt{h^2 + 2 r^2}$, and so $$\xi = \sqrt{h^2 + 2 r^2} \otimes (2 v \theta + \phi);$$ then one moreover demands $2 v \theta + \phi$ is a rational multiple of $\pi$. This is a much more stringent requirement. In fact, it never happens, by the following elementary but tedious argument. By scaling, we can assume that $h = 1$. Hence it follows that $r^2$ is a rational multiple of $1 + 2 r^2$, which certainly implies that $r^2$ is rational. So let $r^2 = t$. Thus we require that $$\frac{r}{\sqrt{1 + 2 r^2}} = \sqrt{\frac{t}{1 + 2 t}} = v$$ is rational. It follows that we have $$r^2 = t = \frac{v^2}{1 - 2 v^2}$$ for some rational $v$. Thus we can rephrase the problem as follows: Find all rational $0 < v < 1/\sqrt{2}$ with $$\cos(\theta) = \frac{r}{\sqrt{r^2 + 1}} = \frac{v}{\sqrt{1 - v^2}},$$ $$\cos(\phi) = \frac{-r^2}{r^2 + 1} = \frac{-v^2}{1 - v^2},$$ and such that $$(2 v \theta + \phi) \in {\mathbf{Q}} \pi.$$ Let us do an example to explain how one can eliminate any specific $v$. Take the case of $v = 1/2$. We find that $$\cos(\theta) = 1/\sqrt{3}, \quad \cos(\phi) = -1/3,$$ from which we deduce (for example) that $$\cos(2 v \theta + \phi) = \cos(\theta + \phi) = - \frac{5}{3 \sqrt{3}}.$$ If $\alpha = 2v \theta + \phi$ is a multiple of $\pi$, then $$\cos(\alpha) = \frac{\zeta + \zeta^{-1}}{2},$$ where $\zeta = e^{i \alpha}$ is a root of unity. But knowing $\cos(\alpha)$ one can solve for $\zeta$ and then determine if it is a root of unity or not from its minimal polynomial. In the example above, we win immediately because $2 \cos(\alpha)$ should be an algebraic integer and it is not. The cases $v = 1/3$ and $v = 2/5$ can be handled in a very similar way: if $v = 1/3$, then $\cos(\theta) = 1/(2 \sqrt{2})$ and $\cos(\phi) = -1/8$, and, with $\alpha = 3 (2v \theta + \phi)$, $$\cos(\alpha) = \cos(2 \theta + 3 \phi) = \frac{87}{256},$$ and if $v = 3/5$, then $\cos(\theta) = 3/5$ and $\cos(\phi) =9/25$, and with $\alpha = 5 (2v \theta + \phi)$, $$\cos(\alpha) = \cos(6 \theta + 5 \phi) = \frac{3617721}{4194304}.$$ In both cases, the corresponding $\zeta$ is manifestly not a root of unity because $2 \cos(\alpha)$ is not an algebraic integer. Returning to the original problem, the first thing we will do is prove that ${\mathbf{Q}}(e^{2 v i \theta})$ is an abelian extension. We may write $$e^{2 v i \theta} = e^{(2 v \theta + \phi) i} \cdot e^{- \phi i}.$$ Since $2 v \theta + \phi \in {\mathbf{Q}} \pi$, the first factor is a root of unity and so lies in an abelian extension. On the other hand, the second term is simply $$\cos(\phi) - i \cdot \sin(\phi).$$ Since $\cos(\phi) \in {\mathbf{Q}}$, it follows that $i \cdot \sin(\phi) = \sqrt{\cos^2(\phi) - 1}$ lives in an imaginary quadratic extension of ${\mathbf{Q}}$. In particular, $e^{- i \phi}$ clearly lies inside an abelian extension. Taken together, we deduce: The extension ${\mathbf{Q}}(e^{2 v i \theta})$ is abelian. We also have the explicit formulae $$e^{i \theta} = \cos(\theta) + i \sin(\theta) = \frac{v + i \sqrt{1 - 2 v^2}}{\sqrt{1 - v^2}},$$ and then squaring: $$e^{2 i \theta} = \frac{ 3 v^2 - 1 + 2 v \sqrt{2 v^2 - 1}}{1 - v^2}.$$ Let $v = a/b$ with $(a,b) = 1$, this becomes $$e^{2 i \theta} = \frac{ 3 a^2 - b^2 + 2 a \sqrt{2 a^2 - b^2 }}{(b^2 - a^2)}.$$ Let $E = {\mathbf{Q}}(\sqrt{2 a^2 - b^2 })$, which is an imaginary quadratic extension of ${\mathbf{Q}}$. (The condition that $a/b = v < 1/\sqrt{2}$ implies that $b^2 > 2 a^2$.) Let us write $$x = 3 a^2 - b^2 + 2 a \sqrt{2 a^2 - b^2 } \in \mathcal{O}_E.$$ Note that $N(x) = (b^2 - a^2)^2$. Secondly, note that $$e^{2 v i \theta} = (e^{2 i a\theta})^{1/b}$$ By Galois theory, $E(\alpha^{1/b})$ can only be an abelian extension of ${\mathbf{Q}}$ (or even of $E$) under the following conditions: $\alpha$ is a perfect $b$th power in $E$. $b$ is even and $\alpha$ is a perfect $b/2$th power in $E$. (Added explanation: You can work prime by prime on the divisors $p$ of $b$. Suppose that $\alpha$ is not a perfect $p$th power for a prime $p > 3$. Then the Galois closure of $E(\alpha^{1/p})$ is $E(\alpha^{1/p},\zeta_p)$ and contains the automorphism $\tau: \alpha^{1/p} \rightarrow \zeta_p \alpha^{1/p}$. But then Galois group of $E(\zeta_p)$ over $E$ contains an element $\sigma$ which fixes $\alpha$ and sends $\zeta_p$ to $\zeta^i_p \ne \zeta_p$. Then $\tau$ and $\sigma$ do not commute. The same argument works if $p = 3$ as long as $E(\zeta_p) \ne E$, which can only happen if $E = \mathbf{Q}(\sqrt{-3})$. But $2 a^2 - b^2 \ne - 3 c^2$ by $3$-adic considerations. The same argument works for "$p = 4$" as well, as long as $E(\zeta_4) \ne E$, which can only happen if $E = \mathbf{Q}(\sqrt{-1})$. But $2 a^2 - b^2 = - c^2$ can't happen when $b$ is even (which is the only relevant case) because then $a$ is odd and $2 a^2 - b^2$ is $2 \mod 4$.) Moreover, since$(a,b) = 1$, if $e^{2 i a \theta}$ is a perfect $b$ or $b/2$th power, then so is $e^{2 i \theta}$. In particular, $e^{4 i \theta}$ is a $b$th power in $E$. Note also that $$e^{4 i \theta} = \frac{x^2}{N(x)} = \frac{x^2}{x \overline{x}} = \frac{x}{\overline{x}},$$ so $x/\overline{x}$ is a perfect $b$th power in $E$. Our goal is now to prove that the ideal $(x)$ is (almost) a $b$th power, and deduce that $N(x) = (b^2 - a^2)^2$ is (almost) a $b$th power. Suppose that $\mathfrak{p}$ is a prime ideal of $\mathcal{O}_E$ which divides both $x$ and $\overline{x}$. I claim that $\mathfrak{p}$ divides $2$. Note that $N(x) = (b^2 - a^2)^2$, so so $a^2 \equiv b^2 \mod \mathfrak{p}$. But then $$x + \overline{x} = 2(3 a^2 - b^2) \equiv 4 a^2 \mod \mathfrak{p}.$$ If $a \in \mathfrak{p}$ then also $b \in \mathfrak{p}$ contradicting that $(a,b) = 1$. Hence $(x,\overline{x})$ is only divisible by primes above $2$. Since $x/\overline{x}$ is a $b$th power, and $(x,\overline{x})$ is supported at primes above $2$, we deduce that $(x) = I^b \cdot J$ where $J$ is supported at primes above $2$. We deduce that $$N(x) = (b^2 - a^2)^2 = n^b \cdot 2^k.$$ where $n$ is an odd integer. Let us first consider the case when $a$ and $b$ are not both odd. In this case $k$ is trivial, and $(b^2 - a^2)^2 = n^b$. Since $b^2 - a^2 = 1$ has no solutions in positive integers, it follows that $b^2 - a^2 > 1$, but then $$b^4 > (b^2 - a^2)^2 \ge 2^b,$$ from which we deduce that $b < 16$. Now suppose that $a$ and $b$ are both odd. Since $(b-a,a+b) = 2$ in this case, it must be the case that $$b-a = r^b 2^u, b+a = s^b 2^v,$$ where one of $u$ and $v$ must be equal to $1$. Case 1: $u = 1$. If $b-a=2$, then from the inequality $v < 1/\sqrt{2}$ and $a < b/\sqrt{2}$, we deduce that $$2 = b - a > b(1 - 1/\sqrt{2}),$$ and thus $b < 7$. If $b-a=2 \cdot r^b$ with $r > 1$, then $$b \ge b - a \ge 2^{b+1},$$ which is impossible. Case 2: $v = 1$. We have $a+b = 2 \cdot s^b$, and now $$2b > a + b \ge 2^{b+1},$$ which once again is impossible. Putting this together, we deduce that $b < 16$. Checking all the cases with $b < 16$ (noting that $a/b < 1/\sqrt{2}$ and $(a,b) = 1$) we find that the only possible pairs are $$(a,b) = (1,2), (1,3), (3,5), \ \text{or} \ v =1/2, 1/3, 3/5.$$ But these cases have already been considered previously.<|endoftext|> TITLE: Restriction of irreducible representations from $G(\mathbb Q_p)$ to $[G, G](\mathbb Q_p)$ QUESTION [7 upvotes]: Let $\pi_p$ be a smooth irreducible representation of $G(\mathbb Q_p)$, where $G$ is a connected reductive group over $\mathbb Q_p$. Consider the restriction of $\pi_p$ to $[G, G](\mathbb Q_p)$, how does it decompose? Can we determine the multiplicities in term of some data of $\pi_p$? In the case that $\pi$ "comes from geometry" (this is kind of vague), can we determine the decomposition using the involved geometric objects? I am interested in some examples like general unitary groups, general orthogonal groups, general symplectic groups... How about the global theory? Example: if $\pi$ is a smooth irreducible admissible representation of $GL(2,\mathbb Q_p)$ then its restriction to $SL(2,\mathbb Q_p)$ is either irreducible, or splits as a direct sum of 2 non-isomorphic representations, or, occasionally, as a direct sum of 4. For instance, see Overview of automorphic representations for $SL(2)/{\mathbf{Q}}$?. Example: the admissible $GL_2(\mathbb Q_p)$ representation attached to an elliptic curve $E$ with good reduction is reducible upon restriction to $SL_2(\mathbb Q_p)$ if and only if $E$ has supersingular reduction for $p＞3$. REPLY [6 votes]: I won't say anything here about the number of components in your restricted representation, but here is some information about multiplicities. Kwangho Choiy and Dipendra Prasad have (independently) formulated a conjecture that expresses multiplicities in terms of the enhanced Langlands parameter attached to your representation $\pi$. Choiy proves the conjecture for tempered representations under the assumption that the Langlands correspondence exists and has lots of expected properties. Prasad and I reduce the conjecture to the tempered case, give a heuristic for why it should be true, and compute some examples of multiplicity. In some sense, both papers are thus trying to answer your question. However, I don't know if the answer provided is useful for you, since one doesn't usually know the Langlands correspondence in an explicit way. So let me outline a few examples, and you can find details for most of them in the papers above or here. Let $G'$ be a subgroup of $G$ that contains the derived group, and $F$ the local field. I'll assume that $F$ is nonarchimedean, but won't assume $F=\mathbf{Q}_p$. Let $Z$ denote the center of $G$. If the quotient $G(F)/Z(F)G'(F)$ is cyclic, then the multiplicity is $1$ for elementary reasons. However, this doesn't happen all that often, so one must use other techniques. Here are some pairs $(G,G')$ for which restriction from $G(F)$ to $G'(F)$ is always multiplicity free: $(GL_n,SL_n)$, $(GO_n,O_n)$ (using any nondegenerate symmetric form, and assuming $p\neq 2$), $(GSO_n,SO_n)$ (ditto), $(GSp_{2n},Sp_{2n})$, $(GU_n, U_n)$ (any nondegenerate hermitian form), $(U_n, SU_n)$ if $p$ is coprime to $n$, or $F=\mathbf{Q}_p$ with $p\neq 2$, $(GU_n,SU_n)$ if $p$ is coprime to $n$ and $n$ is odd. One can also show that multiplicity $1$ holds for certain kinds of representations, rather than for all representations of certain groups. For example, it easily holds for generic representations, due to the uniqueness of Whittaker models. Monica Nevins has shown that multiplicity one always holds for certain kinds of supercuspidal representations. More generally, Manish Mishra and I can show that multiplicity one holds for Kaletha's ``regular'' supercuspidal representations that satisfy an additional regularity condition, which is equivalent to the components of the restricted representation remaining regular. Some or all of the examples above are predicted by the conjecture. It has long been known that higher multiplicities can occur if $G= SL_1(D)$ for $D$ a division algebra (I apologize for not having the reference at hand), so one should expect similar behavior for many other groups involving division algebras. See one of the papers above for an example of higher multiplicity that can occur for the pair $(GU_n, SU_n)$, even when our unitary groups are quasi-split. The example involves a regular supercuspidal representation whose restriction breaks up into components that are not regular.<|endoftext|> TITLE: (Barely) linearly independent vectors over $\mathbb{Z}/2\mathbb{Z}$ QUESTION [11 upvotes]: Let $V$ be a vector space over $\mathbb{Z}/2\mathbb{Z}$. Can there be a set $S$ of $2 n$ vectors in $V$ such that any $n$ vectors in $S$ span a space of dimension exactly $n-1$, but no $n$ vectors $v_1,\dotsc ,v_n\in S$ satisfy $v_n = v_1 + v_2 + \dotsc + v_{n-1}$? REPLY [20 votes]: Since every $n$ vectors span a space of dimension $n-1$, the whole set $S$ spans a space of dimension $n-1$, and we can assume that $V$ has dimension $n-1$. Choose $n-1$ linearly independent vectors $T=\{t_1,t_2,\dots,t_{n-1}\}\subseteq S$ and assign the standard basis to them, i.e. $t_k=(0,0,\dots,0,1,0,\dots,0)$ where the $1$ is at index $k$. Let $x,y\in S\setminus T$ be distinct. For each $k\in\{1,\dots,n-1\}$, $x_k\neq 0$ or $y_k\neq 0$, otherwise there are $n$ vectors in $S$ spanning a $n-2$ dimensional space: $x$, $y$, and all the vectors in $T$ excluding $t_k$. By the pigeonhole principle, there are at most $n-1$ zero indices in $S\setminus T$. So there are at most $n$ vectors in $S\setminus T$ (otherwise there would be two all-$1$ vectors). So the size of $S$ is at most $2n-1$. As a result, there is no set $S$ of size $|S|=2n$ such that any $n$ vectors from $S$ span an $(n-1)$-dimensional subspace.<|endoftext|> TITLE: Ideal characterization of almost convergence QUESTION [9 upvotes]: $\bullet$ A real sequence $x=(x_n)_n$ is called convergent to $\alpha$ in usual sense if for any $\epsilon>0$ the set $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}$ is finite. $\bullet$ A real sequence $x=(x_n)_n$ is called statistically convergent to $\alpha$ if for any $\epsilon>0$ the set $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}$ has natural density $0$. The natural density $d$ of $A\subset\mathbb N$ is defined by $d(A)=\lim\limits_{n\to\infty}\frac{|A\cap\{1,2,\dots,n\}|}{n}$, (provided the limit exists) where $|A|$ denotes the cardinality of $A$. $\bullet$ A bounded real sequences $x=(x_n)_n$ is said to be almost convergent to $\alpha$ if all the Banach limit functionals give an unique value for the sequence $x$. A family $\mathcal I$ of subsets of $\mathbb N$ is said to be an ideal in $\mathbb N$ if (i) $A,B\in \mathcal I$ $\implies $ $A\cup B\in \mathcal I$ (ii) $A\in \mathcal I$ and $B\subset A$ $\implies$ $B\in \mathcal I$ $\bullet$ A real sequence $x=(x_n)_n$ is called $\mathcal I$-convergent to $\alpha$ in usual sense if for any $\epsilon>0$ the set $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}\in \mathcal I$. $$\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots$$ $\mathcal I_f=\{A\subset\mathbb N: A \text{ is finite}\}$ and $\mathcal I_d=\{A\subset\mathbb N: d(A)=0\}$ become ideals in $\mathbb N$. Moreover, $\mathcal I_f$-convergence and $\mathcal I_d$-convergence coincide with usual convergence and statistical convergence respectively. But, what is the ideal in the case of almost convergence? My Question : Find the ideal $\mathcal I$ for which $\mathcal I$-convergence coincides with the almost convergence. Is it available in literature? REPLY [2 votes]: A slightly different argument using the sequence $x=(1,0,1,0,1,0,\dots)$.$\newcommand{\I}{\mathcal I}\newcommand{\Ilim}{\operatorname{\I-lim}}\newcommand{\Flim}{\operatorname{\mathcal F-lim}}\newcommand{\Glim}{\operatorname{\mathcal G-lim}}$ This sequence is almost convergent to $1/2$. At the same time, it is not difficult to show that if this sequence has $\I$-limit of some ideal $\I$, then the $\I$-limit can only be $0$ or $1$. We can use the fact that $\I$-limit of a sequence is a cluster point of that sequence. (This holds for any admissible ideal, i.e., for any ideal which contains all finite sets. If we allow also non-admissible ideals, then we can get cluster points or terms of the sequence as limits.) For $\I$-convergence we have multiplicativity, i.e., $\Ilim (x_ny_n)=\Ilim x_n\cdot\Ilim y_n$. In particular, for our sequence $x$ we have $x^2=x$. Consequently, if $L$ is an $\I$-limit, then we get $L^2=L$. This is basically just a reformulation of Lorenz's criterion for almost convergent sequence, but since you're looking at connection between almost convergent ideals, I'll mention that a sequence is almost convergent to $L$ if and only if $$\Flim_n \Glim_k \frac{x_k+\dots+x_{k+n-1}}n=L$$ for any free ultrafilters $\mathcal F$, $\mathcal G$. You can check the section on almost convergent sequences of the paper Jerison, Meyer, The set of all generalized limits of bounded sequences, Can. J. Math. 9, 79-89 (1957). ZBL0077.31004, MR83697. I have notes on some part of the paper where the results are expressed using ultrafilters (rather than using nets in Stone-Čech compactification) available here: notes (Wayback Machine), slides (Wayback Machine).<|endoftext|> TITLE: Are manifolds admitting a circle foliation covered by manifolds with a (non-trivial) circle action? QUESTION [13 upvotes]: More precisely, is there a criterion that decides the above question? I am particularly interested in the smooth setting: is a smooth manifold with a smooth regular foliation by circles covered by a smooth manifold with a smooth non-trivial circle action? I am mostly interested in finite covers. REPLY [11 votes]: Dennis Sullivan constructed a non-vanishing Lispschitz vector field on a closed $5$-manifold such that all orbits are periodic but they amazingly have unbounded lengths (!). An addendum by Nicholas Kuiper shows that the flow can be chosen smooth, and another by William Thurston gives another construction which is real analytic.<|endoftext|> TITLE: If $S_\mathbb N$ is partitioned into finitely many pieces, must one piece contain a "skew copy" of every countable group? QUESTION [20 upvotes]: Suppose $G$ and $H$ are groups. $C \subseteq H$ is called a skew copy of $G$ in $H$ if $C = hK$ for some $h \in H$ and some subgroup $K$ of $H$ with $K \cong G$. Question 1: Suppose the infinite symmetric group $S_\mathbb N$ is partitioned into finitely many pieces. Must one of these pieces contain a skew copy of every countable group? Given two groups $G$ and $H$, let us write $H \rightarrow G$ to mean that whenever $H$ is partitioned into finitely many pieces, one of the pieces contains a skew copy of $G$. Question 2: If the answer to Question 1 is negative, is it nonetheless true that there is some (possibly very large) group $H$ such that $H \rightarrow G$ for every countable group $G$? What I know so far about this question is: $(1)$ The answer to Question 1 is positive if and only if $S_\mathbb N \rightarrow G$ for every countable group $G$. $(2)$ $S_\mathbb N \rightarrow G$ for every finite group $G$. In fact, something a little stronger is true: for every finite group $G$ and every $r \in \mathbb N$, there is a finite group $H$ such that if $H$ is partitioned into $r$ pieces, then one of them contains a skew copy of $G$. $(3)$ If $S_\mathbb N$ is partitioned into finitely many Borel pieces (or, a little more liberally, pieces with the Property of Baire), then one of the pieces contains a skew copy of $S_\mathbb N$, and therefore contains a skew copy of every countable group. I do not know whether the Property of Baire can be replaced with Lebesgue measurability. REPLY [6 votes]: The answer to both questions is no. For every group $G$, by induction on $|G| = \kappa$, we'll construct a partition $G = A \sqcup B$ such that for every $g, h \in G$, if $g$ has infinite order, then $h\langle g\rangle$ meets both $A, B$ on an infinite set -- Call such a partition of $G$ good. If $\kappa = \aleph_0$, this is clear so assume $|G| = \kappa \geq \aleph_1$ and fix a continuously increasing sequence $\langle H_{\alpha}: \alpha < \kappa \rangle$ of subgroups of $G$ such that $|H_{\alpha}| = |\alpha + \omega|$ and $G = \bigcup \{H_{\alpha}: \alpha < \kappa\}$. For each $\alpha < \kappa$, let $H_{\alpha} = A'_{\alpha} \sqcup B'_{\alpha}$ be a good partition of $H_{\alpha}$. For $\alpha \leq \kappa$, define $A_{\alpha}, B_{\alpha}$ as follows. $A_0 = A'_0$ and $B_0 = B'_0$, $A_{\alpha + 1} = A_{\alpha} \cup (A'_{\alpha + 1} \setminus H_{\alpha})$ and $B_{\alpha + 1} = B_{\alpha} \cup (B'_{\alpha + 1} \setminus H_{\alpha})$ and if $\gamma \leq \kappa$ is limit, then $A_{\gamma} = \bigcup \{A_{\alpha}: \alpha < \gamma\}$ and $B_{\gamma} = \bigcup \{B_{\alpha}: \alpha < \gamma\}$. Let us check, by induction on $\alpha \leq \kappa$, that $A = A_{\alpha}$, $B = B_{\alpha}$ form a good partition of $H_{\alpha}$. If $\alpha = 0$ or a limit, this is clear. So assume $A_{\alpha}, B_{\alpha}$ form a good partition of $H_{\alpha}$ and we'll show that $A_{\alpha+1}, B_{\alpha+1}$ form a good partition of $H_{\alpha + 1}$. Fix $g, h \in H_{\alpha + 1}$ such that $g$ is torsion free. We have the following cases. Case 1: $g \in H_{\alpha}$. If $h \in H_{\alpha}$, this is clear. If $h \notin H_{\alpha}$, then note that $h \langle g\rangle \subseteq H_{\alpha + 1} \setminus H_{\alpha}$ and $A'_{\alpha + 1}, B'_{\alpha + 1}$ form a good partition of $H_{\alpha + 1}$. Case 2: Not Case 1. If $|h\langle g\rangle \cap H_{\alpha}| \leq 1$, then this is clear as $A'_{\alpha + 1}, B'_{\alpha + 1}$ form a good partition of $H_{\alpha + 1}$. If not, then for some $n \neq m$, $hg^n, hg^m \in H_{\alpha}$ and so $g^{n - m} \in H_{\alpha}$. Now apply Case 1.<|endoftext|> TITLE: Can matrices over $F[G]$ have infinite order when $G$ is periodic? QUESTION [12 upvotes]: Let $G$ be a group, $F$ a field and $d \geq 1$. When does the ring $M_d(F[G])$ of $d$-by-$d$ matrices over the group ring $F[G]$ admit an element of infinite multiplicative order? When does it admit such a (left and right) invertible matrix? This easily reduces to the case where $G$ is an infinite f.g. torsion group, and (as noted by grok) we may also assume $F$ has positive characteristic. I imagine that typically there are such matrices (perhaps always), but I have no nontrivial examples when $G$ is a f.g. infinite torsion group. If $d = 1$, the question is whether the group ring $F[G]$ contains elements (resp. invertible elements) of infinite order under multiplication. In particular, I am interested in the following variant. Let $G$ be a f.g. infinite torsion group. Can you find a finite field $F$, $d \geq 1$ and a matrix $A \in M_d(F[G])$ such that $A$ has infinite multiplicative order, i.e. $\{A^n \;|\; n \in\mathbb{N}\}$ is infinite. edit: I reverted the second question to something closer to the original form, because while basically logically equivalent, it was probably not clear from the rewording what is a useful partial answer. Finding a single element of $M_d(F[G])$ of infinite order, for some fixed triple $(G,F,d)$ is already interesting to me, and finding some $F, d$ and $A \in M_d(F[G])$ for every f.g. infinite torsion group $G$ is really what I'm looking for in my application. Changed the title as well, though maybe it is now less catchy. REPLY [5 votes]: With Ilkka Törmä, we proved that the Grigorchuk group $G$ admits infinite order matrices over any field $K$ and for every dimension $d$. For all claims about the Grigorchuk group and omitted definitions, see e.g. the paper "Topological full groups of minimal subshifts with subgroups of intermediate growth" by Nicolás Matte Bon. If $\mathrm{char} \; K \neq 2$ then $p = a+b+c+d \in K[G]$ has infinite multiplicative order. Proof. Recall the form of the Schreier graphs of the Grigorchuk group for its natural action on $\{0,1\}^{\mathbb{N}}$, with generators $a, b, c, d$. On the orbit of $111...$, ignoring the labels, this is the following infinite $4$-regular undirected (multi)graph $S$: and we have some group action $V(S) \curvearrowleft G$ of $G$ on the nodes of $S$, which just follows the edges. This action of course depends on the omitted edge labels, but the argument that follows does not. Label the nodes $V(S)$ of the graph $S$ with nonnegative integers, in order from left to right. Let $\phi : M_4 \to G$ be the natural map from the free monoid on generators $a,b,c,d$ to $G$. Define $$ T_k = \{g \in G \;|\; 0 \cdot g = k \}, $$ and $$ V_n = \mathbb{S}_{M_4}(2n) \cap \phi^{-1}(T_{2n}) \subset M_4 $$ where $\mathbb{S}_{M_4}(2n)$ is the Cayley sphere, i.e. words of length exactly $2n$. So $V_n$ is just the set of all words of length $2n$ that map $0$ to $2n$ in the action on the Schreier graph. We clearly have $|V_n| = 2^n$ from the form of the graph. Since $q \nmid 2^n$ and $$ V_n = \bigsqcup_{h \in T_{2n}} \mathbb{S}_{M_4}(2n) \cap \phi^{-1}(h) $$ not all of $|\mathbb{S}_{M_4}(2n) \cap \phi^{-1}(h_n)|$ can be divisible by $q$, that is, $$ \exists h_n \in T_{2n}: |\mathbb{S}_{M_4}(2n) \cap \phi^{-1}(h_n)| \not\equiv 0 \bmod q. $$ That means the number of words $w \in M_4$ of length exactly $2n$ which represent $h_n$ is not divisible by $q$. Because $p^{2n}$ is precisely the sum of (elements represented by) all words of length $2n$, we have that the coefficient of $h_n$ in $p^{2n}$ is not zero. Since $h_n \in T_{2n}$, $h_n$ must have word norm exactly $2n$ in $G$, and thus the support of $p^{2n}$ is not globally bounded. QED The argument doesn't work for $a+b+c+d$ in characteristic $2$ (because a geodesic branches over $\{b,c\}$-bridges into an even number of geodesics), and currently we don't know if $a+b+c+d$ has infinite order over $\mathrm{char} \; 2$. For the same reason, the approach doesn't work for Mark Sapir's suggestion $1+a+b+c$ in characteristic $2$. Maybe it's not so hard to count geodesics directly, we'll try that next. In any case, the problem can be solved this way: if we pick another polynomial the Schreier graph changes in an entirely local way, and with a bit of experimenting you can find ones where the argument goes through. A particularly nice Schreier graph is obtained for $ada+dad+c$. It has a unique infinite geodesic (and some dead ends which die in a few steps). This gives: For any field $K$, $ada + dad + c \in K[G]$ has infinite order. The proof of this is simpler than the above in that you don't need to think about parities (since, as mentioned, geodesics are essentially unique), but you need to do a bit of local analysis on the Schreier graph (which I think is easier to reproduce than to write). The first argument works directly for all groups $G_\omega$ in the Grigorchuk family. For the second, you can use one of $ada + dad + c$, $aca + cac + b$ or $aba + bab + d$ depending on the first symbol of $\omega$, but otherwise the argument is the same. I don't know if these polynomials are invertible, but it seems you can make the matrices invertible in a standard way as soon as $d \geq 2$: suppose $p \in K[G]$ has infinite multiplicative order and $K$ is finite (this is the case for our $p$, as we can consider $p$ to be over the prime field). Then $p, p^2, p^3, ...$ must be linearly independent. Consider then $A = \begin{pmatrix} 0 & 1 \\ 1 & p \end{pmatrix}$ (which has inverse $A^{-1} = \begin{pmatrix} -p & 1 \\ 1 & 0 \end{pmatrix}$). By induction $A^n = \begin{pmatrix} O(p^{n-1}) & O(p^{n-1}) \\ O(p^{n-1}) & p^n + O(p^{n-1}) \end{pmatrix}$ (where $O(p^{n-1})$ just means linear combinations of powers of $p$ up to and including $n-1$) so all these matrices are distinct by linear independence of $p^n$ from the previous powers. For any field $K$, any $d \geq 1$ and any group $G = G_\omega$ in the Grigorchuk family, there exists a matrix $A \in M_d(K[G])$ of infinite order. If $d \geq 2$, there is such an invertible matrix.<|endoftext|> TITLE: Looking for a simple one-dimensional noetherian domain whose regular locus is not open QUESTION [8 upvotes]: In the wikipedia webpage for "excellent ring", one finds the following. If R is the subring of the polynomial ring k[x1,x2,...] in infinitely many generators generated by the squares and cubes of all generators, and S is obtained from R by adjoining inverses to all elements not in any of the ideals generated by some xn, then S is a 1-dimensional Noetherian domain that is not a J-1 ring as S has a cusp singularity at every closed point, so the set of singular points is not closed, though it is a G-ring. This ring is also universally catenary, as its localization at every prime ideal is a quotient of a regular ring. I am not sure what the "elements not in any of the ideals generated by some xn" are, because no xn lies in R. Also I am not able to prove noetherianity. In fact, I am not sure that the example has all the claimed properties. In Exposé XIX of the volume "Travaux de Gabber" in Astérisque 363-364, there is an example of a one-dimensional noetherian domain whose regular locus is not open, with ordinary double points as singularities at closed points. I understand this latter example, but it is much more complicated than the former and I'd really like to find an example simple enough to be presented in a colloquium-style talk. Can anyone help me understand the wikipedia example, or find an example in the same vein? REPLY [7 votes]: This example appears as Example 1 in the following reference: Melvin Hochster, Non-openness of loci in Noetherian rings, Duke Math. J. 40 (1973), 215–219. MR311653. ZBL0257.13015. DOI: 10.1215/S0012-7094-73-04020-9. In fact, the example in question is a special case of the general result quoted below. […] $K$ denotes a field and all otherwise unspecified tensor products are taken over $K$. If $R$ is a $K$-algebra, we say that $R$ is absolutely Noetherian over $K$ if for every overfield $L \supset K$, $L \otimes R$ is Noetherian. If $R$ is a localization of a finitely generated $K$-algebra, then $R$ is absolutely Noetherian over $K$. $R$ is absolutely a domain over $K$ if, likewise, each $L \otimes R$ is a domain, and $P$ is absolutely prime if, equivalently, either $R/P$ is absolutely a domain or each $L \otimes P$ is prime (in $L \otimes R$). […] Proposition 1. Let $\{R_i\}_{i \in I}$ be a family of absolutely Noetherian $K$-algebras indexed by an infinite set $I$, and for each $i \in I$, let $P_i$ be a nonzero absolutely prime ideal of $R_i$. Assume that each $R_i$ is absolutely a domain. Let $R' = \bigotimes_{i \in I} R_i$. Then $R'$ is a domain and for each $i$, $P_iR'$ is prime. Moreover, if $S = R' - (\bigcup_i P_iR')$ and $R = S^{-1}R'$, then $R$ is a Noetherian domain whose maximal ideals are in one-to-one correspondence with $I$ via the map $i \mapsto P_iR$. In addition, each nonzero element of $R$ belongs to only finitely many maximal ideals, and for any maximal ideal $P_iR$ of $R$ $$R_{P_iR} \cong (L_i \otimes R_i)_{P_i^e},$$ where $L_i$ is a certain extension field of $K$ and $P_i^e$ is $P_i(L_i \otimes R_i)$. In particular, if for each $i \in I$, $R_i$ is a subring of a polynomial ring over $K$ and is generated by a finite nonempty set of forms of positive degree and $P_i$ is the ideal generated by these forms, then the hypotheses of the first paragraph are satisfied, and the conclusions of the second paragraph hold. Moreover, the local rings of $R$ are algebro-geometric in this case. In Example 1, Hochster applies the Proposition to the situation where $I$ is the set of positive integers and $K$ is an arbitrary field, in which case he sets $R_i = K[x_i^2,x_i^3]$ and $P_i = (x_i^2,x_i^3)R_i$ for each $i$. Then, $R_{P_iR} \cong L_i[x_i^2,x_i^3]_{P_i^e}$ for every $i$, which is a non-normal local domain of dimension one. By the Proposition these are the localizations of $R$ at nonzero prime ideals, and the only regular point in $\operatorname{Spec}R$ is the generic point corresponding to the zero ideal. It is useful to note the following consequence of Proposition 1, also in Hochster's paper: Proposition 2. Let $\mathscr{P}$ be a property of local rings and suppose there is an algebro-geometric local ring $(R_1,P_1)$ over a field $K$ such that $R_1$ is absolutely a domain, $P_1$ is absolutely prime, and for every overfield $L \supset K$ $$(L \otimes R_1)_{P_1^e}$$ fails to have property $\mathscr{P}$. Suppose that each field $L \supset K$ has $\mathscr{P}$. Then there is a locally algebro-geometric Noetherian domain $R$ over $K$ in which the $\mathscr{P}$ locus is not open. This gives a nice way to construct rings that are locally excellent but not excellent, for instance.<|endoftext|> TITLE: Coend calculus as a deductive system QUESTION [14 upvotes]: I've always had in the back of my head the feeling that co/end calculus, regarded as a set of rules allowing to mechanically prove nontrivial statements as initial and terminal points of a chain of deductions, can indeed be regarded as a deductive system. I must admit I have no grip on the issue, but as far as I can justify this analogy with the only deductive system I've seen in my ife (Gentzen natural deduction), I'd say that elimination and introduction rules resemble very much elimination and introduction of integrals in formulas like $$ Nat(F,G) \cong \int_C \hom(FC,GC) $$ or $$ \text{Lan}_GF(A) = \int^X \hom(GX,A)\times FX $$ These equalities allow one to deduce the validity of more complex statements, like for example the fact that profunctor composition is associative, or that pointwise right extensions in the bicategory of profunctors are indeed exhibited by a certain end. How far am I from being right? I there a way to turn this intuition into something more formal? Is there a way to actually prove that {co/end calculus}, by which I mean the set of elementary theorems regarding natural transformations, the Yoneda lemma,etc. as suitable co/ends, all taken as "manipulation rules", indeed forms a deductive system? REPLY [5 votes]: This kind of question has been taken very seriously in the past, and I am surprised to discover it only now. You should definitely check Sec. 3 in: Cáccamo and Winskel, A Higher-Order Calculus for Categories. Boulton R.J., Jackson P.B. (eds) Theorem Proving in Higher Order Logics. TPHOLs 2001. Lecture Notes in Computer Science, vol 2152. Springer, Berlin, Heidelberg. Is it any meaningful to wonder what a Haskell implementation of this paper would look like?<|endoftext|> TITLE: Higher categorical analogue of the equivalence between the category of representations of a monoid and the category of the monoid ring of the monoid QUESTION [6 upvotes]: In classical algebra, there is a notion of "monoid rings" such that the functor taking monoids to the monoid rings is the left adjoint to the forgetful functor from the category of commutative rings to that of commutative monoids forgetting the addition. Moreover, there is an equivalence between the category of representations of a monoid and the category of the monoid ring of the monoid. Question 1: Let a functor $F$ from the $\infty$-category of $E_\infty$-rings of spectra to the $\infty$-category of $E_\infty$-spaces be the functor post-compositing infinitely looping functor $\Omega^\infty$(we see algebras as $\infty$-operad maps and $\Omega^\infty:Sp \rightarrow S$ is a lax monoidal functor). Then, is there a right adjoint functor to $F$? ($Sp$ is the $\infty$-category of spectra and $S$ is the $\infty$-category of spaces.) Question 2: If the Question 1 is true, then we will denote the functor by $\mathbb{S}[-]$. Let $M$ be an $E_\infty$-space. Is there an equivalence between the $\infty$-category $Fun(BM, Sp)$ of repesentations of $M$ and the $\infty$-category $Mod_{\mathbb{S}[M]}(Sp)$ of modules of the monoid ring $\mathbb{S}[M]$? ($BM$ is an $\infty$-category which has a single object $*$ and $Map(*, *)$ = $M$ as $A_\infty$-spaces.) REPLY [4 votes]: I think you made a sign mistake, and asked for a left adjoint to $\Omega^\infty$ since the monoid ring is a left along to the forgetful functor. If so then the answer is yes. $\Sigma^\infty_+:\mathrm{Space}→\mathrm{Spectra}$ is a symmetric monoidal left adjoint, so it can be extended to an adjunction $$\Sigma^\infty_+:\mathrm{CAlg}(\mathrm{Space})→\mathrm{CAlg}(\mathrm{Spectra})\dashv \Omega^\infty:\mathrm{CAlg}(\mathrm{Spectra})→\mathrm{CAlg}(\mathrm{Space})$$ (see Higher Algebra, 7.3.2.13). The functor $\Sigma^\infty_+$ is what you want to call $\mathbb{S}[-]$. The answer to your question 2 is also true, since $\mathrm{Fun}(BM,\mathrm{Spectra})$ has a compact generator ($\mathbb{S}[M]$ with the obvious $M$-action, which represents simply the evaluation at the basepoint of $BM$) and then you can just apply Schwede-Shipley Morita theory.<|endoftext|> TITLE: A rather curious equality: is this true? QUESTION [5 upvotes]: I came across (coincidentally) two integral evaluations, which seem to agree according to numerical tests. It did not seem easy to convert one into the other. QUESTION. Is this true? $$\int_0^1\left(\frac{\arcsin x}x\right)^3dx =\frac34\pi\int_0^1\left(2\,\text{arctanh}\, x +\frac{\log(1-x^2)}x\right)dx.$$ REPLY [16 votes]: The proposed equality is true. Details: To find $$l:=\int_0^1\left(\frac{\arcsin x}x\right)^3\,dx =-\frac{1}{16} \pi \left(\pi ^2-24 \ln2\right), $$ make the substitution $t=\arcsin x$ and repeatedly integrate by parts to kill the powers of $t$ and reduce this integral to $$\int_0^{\pi/2}\ln\sin t\,dt=-\frac\pi2\,\ln2, $$ by formula 4.225.3, page 531, of Gradshteyn--Ryzhik. To find $$r_1:=\int_0^1 2\,\text{arctanh}\, x\,dx =\ln4, $$ integrate by parts to find an antiderivative of $\text{arctanh}$. Alternatively, expand $\text{arctanh}\, x$ into the Maclaurin series (using $\text{arctanh}'x=\frac1{1-u}=1+u+u^2+\dots$ with $u=x^2$) and integrate termwise, to get $$r_1/2=\sum_1^\infty\frac1{(2j-1)2j} =\sum_1^\infty\Big(\frac1{2j-1}-\frac1{2j}\Big) =\sum_1^\infty\frac{(-1)^{k-1}}k=\ln2. $$ To find $$r_2:=\int_0^1 \frac{\ln(1-x^2)}x\,dx =-\frac12\,\sum_1^\infty\frac1{j^2}=-\frac{\pi^2}{12}, $$ expand $\ln(1-x^2)$ into the Maclaurin series (by using the Maclaurin series for $\ln(1-u)$) and integrate termwise. Now one can see that $l=\frac{3\pi}4\,(r_1+r_2)$, that is, the equality holds.<|endoftext|> TITLE: Strange formulas that gave rise to Koszul duality QUESTION [10 upvotes]: According to p.8 of the note KOSZUL DUALITY AND APPLICATIONS IN REPRESENTATION THEORY by Geordie Williamson. Let $M(\eta)$ be the Verma module of weight $\eta$, $L(\eta)$ be its unique simple quotient and $w_0$ be the longest element in $W$. The strange formula in our notation is $$[M(x\cdot 0):L(y\cdot 0)]=\sum_{i\ge 0}\dim\mathrm{Ext}_{\mathcal{O}}^i(M(w_0x\cdot 0):L(w_0y\cdot 0))$$ for $x,y\in W$. Let $\mu$ be an integral, antidominant weight, $\Delta$ be the set of simple roots, $\Sigma=\{\alpha\in\Delta:\langle\mu+\rho,\alpha^\lor\rangle=0\}$ and $W^{\Sigma}=\{w\in W:w TITLE: applications of finding least quadratic nonresidue mod $p$? QUESTION [5 upvotes]: I saw some papers from famous mathematicians (assuming GRH or without it) which are devoted to finding bound for least quadratic nonresidues modulo prime number $p$. My question is that why it is so important and what are the applications of it? Thanks. REPLY [2 votes]: Let me expand my comment into a short answer. The explicit knowledge of a quadratic nonresidue $a$ modulo $p$ allows us to construct the irreducible quadratic polynomial $x^2-a$ in $\mathbb{F}_p[x]$, and so we have a very explicit description of the field with $p^2$ elements as $$\mathbb{F}_{p^2}=\mathbb{F}_p[x]/(x^2-a).$$<|endoftext|> TITLE: Is every metric uniformly close to a metric with negative scalar curvature? QUESTION [7 upvotes]: Let $M$ be a smooth manifold with non-empty boundary. Let $g$ be a smooth Riemannian metric on $M$. Is the following true? For every $\epsilon >0$ there exist a Riemannian metric $g_{\epsilon}$ with non-positive scalar curvature such that $\|g-g_{\epsilon}\|_{C^0}<\epsilon$? (Here $\|\cdot\|_{C^0}$ is the $C^0$ norm w.r.t $g$, but I guess one can use $g_{\epsilon}$ as well). I ask whether the metrics of non-positive scalar curvature are dense in the space of metrics. On other hand, I heard that the space of metrics with non-negative scalar curvature is closed in the uniform topology. Where can I find references for these two facts? (I am not sure if the boundary creates problems. I might have heard this about closed manifolds). If it matters, I am fine with assuming that $M$ is contractible. REPLY [7 votes]: About the $C^0$-closure of metrics with nonnegative scalar curvature, the results actually holds without any completeness assumptions and thus can be applied locally, so the fact that your manifold has boundary doesn't make any difference here. There are now two proofs available : The theorem was first proved by Gromov in Dirac and Plateau Billiards in Domains with Corners, it proceeds by showing that non-negative scalar curvature is equivalent to the non-existence of cubical domains with : $(1)$ mean convex faces. $(2)$ angles between adjacent faces less than $\pi/2$. This is accomplished by gluing several (regflected) copies of such a cube and smoothing the resulting metric to get a metric with positive scalar curvature on the $n$-torus, which do not exist thanks to previous work (Gromov-Lawson, or alternatively Schoen-Yau in dimension $n\leq 7$). To prove $C^0$ stability from this, one consider a sequence $g_i\xrightarrow{C^0} g$, and assume that $g$ has negative scalar curvature at a point $x$. Around such an $x$ one can build a small cubical domain $C$ with the two properties above, the hard work is then to show that for $i$ big enough, one can slightly deform $C$ to get a cubical domain $C_i$ which has the same properties $(1)$ and $(2)$ with respect to $g_i$, which is hard because the positive mean curvature condition at first glance looks like a $C^2$ condition on the metric. The second proof is (for me) much simpler and uses Ricci flow, it was given by Bamler in A Ricci flow proof of a result by Gromov on lower bounds for scalar curvature.<|endoftext|> TITLE: Iterated derivative and rectangular standard Young tableaux QUESTION [5 upvotes]: We first make a few definitions, seemingly out of the blue (they are introduced/defined in this paper). Let $F^0_{a}(z) = (1-z)^{-1}$ and define recursively $$ F^{k+1}_{a}(z) = z^{a-1} \frac{d^a}{dz^a} F^{k}_{a}(z), \qquad k\geq 0. $$ Let $G_{a,b}(z)= \left(\prod_{k=0}^{b-1}\frac{k!}{(a+k)!} \right) z^{1-a}(1-z)^{ab+1} F^{b}_{a}(z)$. Here are the first few values of $G_{a,b}(z)$ as $a=1,2,3$, $b=1,\dotsc,4$. \begin{matrix} 1 & 1 & 1 & 1 \\ 1 & z+1 & z^2+3 z+1 & z^3+6 z^2+6 z+1 \\ 1 & z^2+3 z+1 & z^4+10 z^3+20 z^2+10 z+1 & z^6+22 z^5+113 z^4+190 z^3+113 z^2+22 z+1 \end{matrix} The combinatorialist might then be inclined to make the following conjecture (verified for $1\leq a,b\leq 4$): $$ G_{a,b}(z) = \sum_{T \in \mathrm{SYT}(a^b)} z^{des(T)-b+1}. $$ Here, $\mathrm{SYT}(a^b)$ denotes the set of standard Young tableaux of rectangular shape $(a,a,\dots,a)$, and $des(T)$ is the number of descents. Question: What is going on? Can this be proved? I have never seen anything like this, but the paper above introduces these as generalizations of Eulerian polynomials and Narayana polynomials, so the first two columns (and rows) of the table above agrees (by finding the appropriate pages in Stanley's EC2). If the conjecture is true, then $G_{a,b}(z)$ can be shown to be $h^*$-polynomials for some nice family of polytopes. Mathematica code for the function $G_{a,b}(z)$ is as follows: GG[a_, b_] := (Product[(k)!/(a + k)!, {k, 0, b - 1}]) z^(1 - a) (1 - z)^(a b + 1) Nest[Simplify[z^(a - 1) D[#, {z, a}]] &, 1/(1 - z), b]; REPLY [8 votes]: The formula follows from a result in EC2 (Stanley's "enumerative Combinatorics" Vol. 2) -- Chapter 7, equation (7.96), which is a result from the expansion of Schur functions in terms of fundamental quasisymmetric functions. The equation reads: $$\sum_{m\geq 0} s_{\lambda/\mu} (1^m) z^m = \frac{\sum_T z^{des(T)+1}}{(1-z)^{n+1}},$$ where $T$ ranges over all SYTs of shape $\lambda/\mu$. In the case of rectangular shape $(a^b)$, the hook-content formula (equivalent to MacMahon's here) gives $$s_{a^b} (1^m) = \prod_{i=1}^{b}\prod_{j=1}^a \frac{m+j-i}{i+j-1},$$ and we notice that the product of hook-lengths in the denominator is exactly the constant $\prod_{k=0}^{b-1} \frac{k!}{(a+k)!}$. So after canceling the common factors on both sides, Per's identity becomes equivalent to: $$\sum_{m\geq 0} \prod_{i=1}^b \prod_{j=1}^a (m+j-i) z^{m+a-b-1} = F_a^b(z)$$ Now this follows easily by induction on $b$, since $$z^{a-1}\frac{d^a}{dz^a} z^{m+a-(b-1)-1} = \prod_{j=1}^a(m+j-b) z^{m+a-1-b}.$$<|endoftext|> TITLE: Equivalent notions of congruence for elliptic curves over $\mathbb{Q}$ QUESTION [8 upvotes]: Let $E_1$ and $E_2$ be elliptic curves over $\mathbb{Q}$ and $f_i$ the eigencuspform of weight $2$ attached to $E_i$. Express $f_1=\sum a_i q^i$ and $f_2=\sum b_i q^i$. Suppose that the residual Galois representations $E_1[p]\simeq E_2[p]$. Note that $E_1$ and $E_2$ have bad reduction at the same primes. It is easy to see that $a_l\equiv b_l\mod{p}$ at a prime $l$ of good reduction for both elliptic curves (since the residual representations are unramified at $l$). Does it follow that $a_l\equiv b_l\mod{p}$ at a prime $l$ of bad reduction for both elliptic curves. If so how do we show this? Perhaps this comes down to showing that $E_1$ has the same kind of bad reduction at $p$ as $E_2$. I know that if $E_1$ has additive reduction then so does $E_2$, but I'm not sure about distinguishing between split multiplicative and non-split multiplicative reduction. REPLY [3 votes]: Note that a conjecture attributed to Frey and Mazur says that for each elliptic curve $E_1$, there exists a constant $C$ depending on $E_1$, such that for every prime $p>C$ and for every elliptic curve $E_2$ defined over $\mathbb{Q}$ with $E_1[p]\simeq E_2[p]$ as Galois modules, one has $E_1$ isogenous to $E_2$. In particular, $E_1$ and $E_2$ have the same $L$-functions and hence the same coefficients $a_\ell$ (including at the bad primes). It is even conjectured that $C$ can be taken independently of $E_1$. Moreover there is no known pair of non-isogenous elliptic curves $(E_1,E_2)$ such that $E_1[p]\simeq E_2[p]$ for some prime $p>17$. If you want to learn more about this conjecture you can have a look at the recent preprint by Cremona and Freitas "Global methods for the symplectic type of congruences between elliptic curves": https://arxiv.org/abs/1910.12290<|endoftext|> TITLE: Every possible number of partitions by restricting parts? QUESTION [9 upvotes]: Write $p(n)$ for the number of integer partitions of $n$. For $S \subseteq \{1, \ldots, n\}$, let $p_S(n)$ be the number of partitions of $n$ with all parts in $S$. So $p(n) = p_{\{1,\ldots,n\}}(n)$. Conjecture: Given positive integers $n$ and $k$ with $0 \le k \le p(n)$, there is an $S \subseteq \{1, \ldots, n\}$ such that $p_S(n)=k$. I have verified this through $n=20$. Some very loose intuition supporting the conjecture is that the number of subsets grows much faster than $p(n)$. This conjecture is inspired by David Newman's question where he wonders about $S$ with $p_S(n)=p(n)/2$ or $(p(n)\pm1)/2$. My non-answer there suggests that a greedy algorithm is enough to find the $S$ he wants, verified for even $n$ up to 50. REPLY [2 votes]: The conjecture seems to hold. For brevity, denote $[k,n]=\{k,k+1,\dots,n\}$ and $[n]=[1,n]$. Start with $S = [n]$. Algorithm: Given $0\leq k\leq p(n)$, consider the numbers $1,2,\dots,n$ one by one, removing the number if the remaining set $S$ satisfies $p_S(n)\geq k$. After all $n$ numbers are considered, we are done. Let $S_m$ be the set obtained after considering the numbers up to $m$. Hence, the algorithm reads as follows: $S_m=S_{m-1}\setminus\{m\}$ if $p_{S_{m-1}\setminus\{m\}}(n)\geq k$, and $S_m=S_{m-1}$ otherwise. We claim that at each step, the inequality $$ p_{S_m\cap[m]}(n)\leq k \qquad(*) $$ holds. If $(*)$ is proved, we will get $k\leq p_{S_n}(n)\leq k$, as required. We prove $(*)$ by induction on $m$. Clearly, it holds for $m=0$. Assume now that $p_{S_{m-1}\cap[m-1]}(n)\leq k$. If $S_m=S_{m-1}\setminus\{m\}$, then $S_{m-1}\cap[m-1]=S_m\cap[m]$, and $(*)$ holds. Otherwise, $S_m=S_{m-1}$, which means that $p_{S_{m-1}\setminus\{m\}}(n)\leq k-1$. So, in order to finish this case, it suffices to show that $$ p_{S\cap[m]}(n)\leq p_{S\setminus\{m\}}(n)+1 \qquad(**) $$ for every $S\supseteq[m,n]$ (and apply it to $S=S_{m-1}$). If $m=n$, the inequality is obvious. Otherwise, we will show even the version without `$+1$'. We present an ``injective'' proof, after taking the complements. Namely, to each partition with parts in $S$, at least one of which equals $m$, we assign a partition with parts in $S$, at least one of which exceeds $m$, so that this correspondence is injective. This will prove $$ p_S(n)-p_{S\setminus\{m\}}(n)\leq p_S(n)-p_{S\cap[m]}(n), $$ which yields $(**)$. The partitions containing both $m$ and $>m$ correspond to themselves. Now take any partition with maximal element $m$. If it contains a unique copy of $m$, let $a(1$ copies of $m$, replace them by $dm$. Clearly, this correspondence is injective, as required.<|endoftext|> TITLE: Is the Hilbert–Pólya intuition vindicated in the function field case? QUESTION [21 upvotes]: The Hilbert–Pólya conjecture is the name given to the idea that the "reason" or "explanation" for the collinearity of the non-trivial zeros of the Riemann zeta function $\zeta(s)$ is that they are the spectrum of some self-adjoint operator. The empirical evidence for the Montgomery pair-correlation conjecture provides some support for the idea that the zeros of $\zeta(s)$ are "spectral." In the function field case, the Riemann hypothesis has been famously proved by Deligne. My question is, is the intuition behind the Hilbert–Pólya conjecture vindicated in the function-field case? As I understand it, in the function field case, there is an interpretation of the Riemann zeros as the zeros of a linear operator (a Frobenius action on cohomology). However, it's not clear to me that this necessarily answers my question affirmatively. It seems to me that for the (original) Riemann hypothesis to be "explained" by the spectrum of a self-adjoint operator, it isn't enough to demonstrate the mere existence of such an operator; there should be some "natural reason" for the operator to be self-adjoint. To put it another way, suppose someone were to prove the Riemann hypothesis by some kind of seemingly ad hoc computation that just happened to come out right, and then after the fact, we were to artificially construct a self-adjoint operator to fit the zeros. This would prove the Hilbert–Pólya conjecture in some sense, but it would seem not to satisfy the original desire to "explain" the Riemann hypothesis spectrally. In the function field case, does Deligne's proof seem to give a satisfying "spectral explanation" of the Riemann hypothesis, or does it feel more like a calculation that just happens to work out? I realize that this is a vague question that may be largely opinion-based, but I'm holding out hope that people with a thorough understanding of Deligne's proof will find this to be a meaningful question. REPLY [6 votes]: In the function field setting, the most natural spectral explanation for the Riemann hypothesis might be expressing the eigenvalues of Frobenius as the eigenvalues of a unitary operator on a finite-dimensional vector space (times $\sqrt{q}$). This would be related to the Hilbert-Polya picture by a logarithm. There are a formidable number of proofs of the Riemann hypothesis over finite fields. For curves, there are the two proofs of Weil, and one of Bombieri-Stepanov. For higher-dimensional varieties there are Deligne's proof in Weil I and his proof in Weil II, as well as later simplifications of these. I think the short answer is that none of these proofs really construct a unitary operator prior to proving these inequalities. In particular, for Deligne's method, note that all variants of Deligne's proof use crucially his inductive argument where a series of stronger bounds for a Frobenius eigenvalue is proved, converging on the correct one. It's hard to see how such a proof could be converted into a reasonable construction of a vector space and operator. Stepanov's method has no linear algebra in characteristic zero at all, and works entirely with algebra over the finite field and inequalities. For Weil's proofs this is the most subtle, as he works with the algebra of endomorphisms of the Jacobian variety of a curve (in the first proof) or the divisor class group of a surface (in the second proof). Both of these are integer lattices and so can be naturally embedded into a complex vector space, unlike anything in Deligne's proof, which is entirely $\ell$-adic. But both of these complex vector spaces do not play the role of the vector space, but rather of the algebra of operators on that vector space. I am not too sure of this but it might be said that Weil's first proof naturally produces a representation of Frobenius as a unitary element of a $C^*$ algebra.<|endoftext|> TITLE: Koszulness of some DG-algebras and a paper by Kohno and Oda QUESTION [8 upvotes]: This is a follow-up of my previous question Formality of the 2nd ordered configuration space of a closed Riemann surface. At page 131 of [B], R. Bezrukavnikov states Proposition 4.1, in which he shows (among other things) that a certain enveloping algebra $U(L)$, related to the topology of the $n$th configuration space $X_n$ (where $X$ is a smooth compact Riemann surface of genus $g \geq 1$) is a Koszul quadratic algebra if and only if $n \leq 2$. Then he makes this cryptic remark (see p. 133): An equivalent form of the incorrected statement "the graded algebra $U(L)$ is a Koszul quadratic algebra" was stated $($and "proved"$)$ in [KO], saying nothing more on the subject. I would like to use these results in my research, but I am not a specialist in the field. So let me ask the following Q. What is the "equivalent statement" in [KO] Bezrukavnikov refers to, and what are (if any) the mistakes in its proof? Bibliography. [B] R. Bezrukavnikov: Koszul DG-algebras arising from configuration spaces, Geometric and functional analysis 4 (1994), 119-135. [KO] T. Kohno, T. Oda: The lower central series of the pure braid group of an algebraic curve, Advances in Pure Math 12 (1987), 201-219. REPLY [7 votes]: (Just for notation, I let $\mathrm{Conf}_n(X) = \{ x \in X^n \mid \forall i \neq j,\, x_i \neq x_j \}$. This is what Bezrukavnikov calls $X_n$ and what Kohno–Oda call $F_{0,n}(X)$.) The issue is in §3, on page 208, when they claim that the monodromy representation $\rho : \pi_1(\mathrm{Conf}_n(\Sigma_g) \to \mathrm{Aut}(\pi_1(\Sigma_g - \{p_1, ..., p_n\})^{ab})$ is trivial. They claim that $\pi_1(\mathrm{Conf}_{n+1}(\Sigma_g)) \to \pi_1(\mathrm{Conf}_n(\Sigma_g))$ has a section. Unfortunately, this is not correct if $g \ge 2$ and $n \ge 2$. (For the torus it exists because it's a Lie group so you can define one explicitly, see "Configuration spaces" by Fadell–Neuwirth.) Here I should give some credit as I didn't find the issue on my own: see the last remark in "Braids on surfaces and finite type invariants" by Bellingeri and Funar, and a remark in Hain's "Infinitesimal presentations of the Torelli groups". To see which statement is equivalent to the Koszulity of the universal enveloping algebra, I'll try to rephrase everything in slightly more "modern" language (at the very least, language that I am more familiar with). Bezrukavnikov uses a convention (the same as in the book Quadratic Algebras of Polishchuk–Positselski) that a dg-algebra $A$ is quadratic if it is generated by $A^1$ and the kernel of $T(A^1) \to A$ is generated by elements in degree $2$. If $A$ is graded commutative, then the Koszul dual $A^!$ is the enveloping algebra of $\mathcal{L}$, where $\mathcal{L}$ is the free Lie algebra on the dual $(A^1)^\vee$ modded out by the image of the dual of the product $(A^2)^\vee \to (A^1)^\vee \otimes (A^1)^\vee$. In the case where $A = H^*(X)$, this is exactly the holonomy Lie algebra $\mathfrak{g}_X$ defined by Kohno–Oda. There are various equivalent definitions of Koszulness of a quadratic DGA $A$. One of them is that the Koszul complex $(A \otimes A^¡, d)$ is acyclic, where $A^¡$ is dual to $A^!$. The Koszul complex of $U(\mathfrak{g}_X)$ is exactly the one called $R(X)$, and Kohno–Oda prove that it is acyclic in the proof of Lemma 4.1, using the section claimed above. (Maybe a quick remark: nowadays, algebras which aren't $1$-generated can also be considered Koszul. Instead of asking $A$ to be generated by $A^1$ with relations in degree $2$, you can ask $A$ to be generated by some set of generators and ask for relations of weight $2$ with respect to the word length. Then whenever you see an $\operatorname{Ext}$ or something, it is weight graded, and whenever there is a condition of the type "generated in degree $1$" you can replace it by "generated in weight $1$".)<|endoftext|> TITLE: Are buttons really enough to bound validities by S4.2? QUESTION [16 upvotes]: Joel Hamkins recently claimed on twitter that buttons suffice to bound the validities of a potentialist system to the modal logic S4.2 (see here), and that switches are not necessary. We have been trying to reproduce this result here in Amsterdam, and encountered a few problems, so we decided to seek help here. For the sake of this post, let me stick to the standard terminology and notation used in this field (see, for example, these slides of Joel Hamkins). Our aim is to prove the following claim. Claim. If a potentialist system has infinitely many independent buttons, then its modal validities are contained in the modal logic S4.2. Let me now describe two problems we encountered while trying to reprove this claim, and related questions. 1. A possible counterexample? Consider the potentialist system on $P = (\mathcal{P}(\mathbb{N}),\subseteq)$ in the language of propositional logic: every world in $P$ is a structure for propositional logic (i.e. this is just a Kripke model for modal propositional logic). We define the valuations of these worlds as follows for every $x \in \mathcal{P}(\mathbb{N})$: $x \vDash p_i$ if and only if $i \in x$. It is now easy to verify that $\{ p_i | i \in \mathbb{N} \}$ constitutes a collection of infinitely many independent buttons. Consider now the so-called $\mathsf{Top}$-Axiom, i.e. $$ \Diamond((\Box p \leftrightarrow p) \wedge(\Box \neg p \leftrightarrow \neg p)). $$ This axiom is not a consequence of $\mathsf{S4.2}$ (it fails on every $\mathsf{S4.2}$-frame whose top cluster contains at least two different elements), but it is easy to verify that it holds in the potentialist system $P$ defined above (exactly because it has a unique maximal point $\mathbb{N}$ in which all buttons are pushed). Question. This counterexample shows that the Claim needs some additional assumptions, possibly on the language or the kind of structures we're dealing with. What are these additional assumptions? What is the exact statement that we can prove? 2. A possible proof? Let me now explain how we think the Claim should be proved, and which problems we encounter here. Of course, this will only be a sketch. Let $P$ be a potentialist system with infinitely many buttons. Given a modal formula $\phi$ such that $\mathsf{S4.2} \not \vdash \phi$, we can find a Kripke model $(K,R,V)$ of $\mathsf{S4.2}$ and a node $v \in K$ such that $(K,R,V), v \not \vdash \phi$. The idea is to unfold the model $K$ for $n$ steps, where $n$ is the modal depth of the formula $\phi$. This yields a finite tree Kripke model $T = (T,R_T,V_T)$ whose root node is $n$-bisimilar to the node $v$ of the original model $K$. By $n$-bisimilarity, it follows that $T, v \not \Vdash \phi$. We can now use the infinitely many buttons to find a labelling that imitates the behaviour of the tree in the potentialist system. As the original model $K$ is transitive, it follows that the resulting model $T$ cannot possibly transitive and it cannot possibly be reflexive as otherwise, the notion of $n$-bisimilarity is just usual bisimilarity. In particular, the relation $R_T$ of the tree $T$ connects a node of the tree only to its direct successor, and to no other nodes. Moreover, if the resulting trees $T$ were transitive and reflexive, the above would show that $\mathsf{S4.2}$ is complete with respect to the finite trees, but that is not the case. In particular, the Kripke model $T$ does not satisfy $\mathsf{S4.2}$ anymore. If we want to continue the proof from the non-reflexive and non-transitive tree $T$, we have to find a labelling of the potentialist system $P$ that imitates the non-reflexive and non-transitive tree $T$. It does not seem that this can be done with usual buttons: At any world $v$ of the potentialist system corresponding now to a node $t$ of the tree $T$, we can press several buttons at once and reach a world $w$ that corresponds to a node $t_1$ of $T$ that is not an immediate successor of the node $t$ we started with, a contradiction. This suggests that the standard way of labelling worlds of the potentialist system does not seem to work. Question. Given a non-reflexive and non-transitive tree $T$ and a potentialist system $P$ with infinitely many buttons, can we find a labelling of the worlds of $P$ using the buttons to imitate the modal behaviour of $T$ in the potentialist system? Related is the following: Question. The above argument could also suggest that the notion of $n$-bisimilarity does not work for this situation. How can this proof be salvaged? In some sense, we could simplify the question and just ask: Question. How can we prove (a possibly strengthened version of) the Claim? I am very curious about this result, and hope that someone here can clarify these questions. REPLY [13 votes]: Unfortunately, the argument I had had in mind seems broken to me now, and I retract the claim. I wonder what of it can be salvaged? Let me explain what I had had in mind. The main idea was this: it seems we don't need to flip the switches infinitely often, but rather only finitely many times for any given formula, based on the modal depth of the formula. And so I wanted to mimic that by weakening the notion of independence to $k$-independence. Specifically, define recursively a notion of $k$-independence for a family of buttons $b_j$ and statements $s_i$ to be used as weak switches, as follows: for $k=0$, every such family is $0$-independent at every world; the family is $(k+1)$-independent at a world, if you can operate the buttons and switches to achieve a new configuration (pushing any desired additional buttons and setting the switches as desired) by moving to a world where the family remains at least $k$-independent. Basically, a family is $k$-independent at a world, when you can operate the controls as you like $k$ times. The observation was that if one has a lot of independent buttons, then you can easily make $k$-independent families of buttons and switches. Simply form disjoint groups of $k$ buttons, and for each group, let the corresponding assertion $s$ be the parity count of the number of them that are pushed. With such a family, you can flip the parity count by pushing one button, and so this family will be $k$-independent at the original world. My idea then was to adapt the usual button+switch proof, by folding in a $k$-independent requirement: the simulation lemma would work for formulas of depth $k$, provided the family remained $k$-independent. But that part seems broken to me now. I had thought at first one could rule out your counterexample by insisting that every node has independent unpushed buttons. This does in fact rule our your example, but it doesn't seem actually to enable the simulation lemma to go through. If the weak switches are not actually switches, then they could be possibly necessary in the potentialist system, and this will not necessarily be simulated in the Kripke model.<|endoftext|> TITLE: Internal $2$-categories QUESTION [8 upvotes]: Has the notion of an internal $2$-category been studied, or more generally an internal $n$-category? Do we have any examples of naturally occurring internal $2$-categories/$n$-categories? The motivation here is understanding internalization at higher levels. In particular, it seems the 'nuts and bolts' set theoretical definitions of higher $n$-categories as a class of objects, a class of $1$-cells, a class of $2$-cells, ..., a class of $n$-cells, together with functions between them for domains, codomains, and identities, and functions out of pullbacks for compositions (see below for an example), is more naturally internalized at higher levels than the 'high-tech' definitions relying on component $n-1$ categories for each pair of objects together with $n-1$ functors for composition, $n-1$ natural transformations for associative coherence, $n-1$ modifications for coherence of associative coherence, $\dots$. This seems counterintuitive (even ironic) in light of the goal of internal category theory -- internalization is supposed to allow us to eschew any use of sets in our categorical reasoning, but we have to use a very set theoretical definition of higher categories in order to correctly internalize them? Perhaps I'm just not seeing a simple internalization of the latter definition. To make the question more precise, here's one possible internalization of the 'nuts and bolts' definition of a bicategory on the nlab: Let $\mathfrak{C}$ be a $2$-category with $1$-cell pullbacks. An internal $2$-category $\mathrm{C}$ in $\mathfrak{C}$ consists of An object of objects ${\bf Ob}_\mathrm{C}\in{\bf Ob}_\mathfrak{C}$. An object of $1$-cells ${\bf 1-Hom}_\mathrm{C}\in{\bf Ob}_\mathfrak{C}$. An object of $2$-cells ${\bf 2-Hom}_\mathrm{C}\in{\bf Ob}_\mathfrak{C}$. $1$-cells as below for domains, codomains and identities: $1$-cells as below for vertical and horizontal composition: where we pull back ${\bf 2-Hom}_\mathrm{C}$ over ${\bf 1-Hom}_\mathrm{C}$ using ${\sf dom}$ and ${\sf cod}$ for vertical composition, and over ${\bf Ob}_\mathrm{C}$ using ${\sf dom\circ dom}$ and ${\sf cod\circ cod}$ for horizontal composition. A $2$-cell $\gamma$ as below for associative coherence of composition: $2$-cells $\iota_\ell$ and $\iota_r$ as below for left and right unital coherence: With all the above data satisfying appropriate coherence diagrams (in the spirit of bicategories), this definition seems to shake out correctly -- realized in the discrete $2$-category ${\bf Set}$ this yields a strict $2$-category, since all $2$-cells will be identities. Realizing it in the $2$-category of categories yields something else, a 'triple category' of sorts, but perhaps this discrepancy has to do with the fact that we've internalized a set-theoretical definition of $2$-categories rather than a $1$-categorical definition. Has this idea been explored, or is it trivialized by some higher notion of internal categories already in the literature? REPLY [4 votes]: Bicategories internal to 2-categories with pullbacks are defined in §3 of Douglas–Henriques's Internal bicategories. They have two motivating examples described in §1: a bicategory $\mathrm{Bord}_0^2$ of bordisms internal to the 2-category $\mathrm{CAT}$; and a bicategory of conformal nets internal to the 2-category $\mathrm{SYMMONCAT}$.<|endoftext|> TITLE: Zermelo's proof for unique factorisation QUESTION [7 upvotes]: In Peter Bundschuh's "Einführung in die Zahlentheorie" I came across a possibly well-known but to me rather peculiar proof of unique factorisation, which is attributed to Ernst Zermelo. The proof bypasses Euclid's lemma to prove that $ \mathbb{Z}$ is a UFD. It seems to me that the deepest property of the ring of integers used is some properties of the order on $ \mathbb{Z}$! An account of this proof can be seen here: https://planetmath.org/inductionproofoffundamentaltheoremofarithmetic Can this idea be applied or has it been applied to any other situations? Or is it simply a tricky proof for the ring of integers? REPLY [6 votes]: In Pete Clark's notes on factorization, he calls this the "Lindemann–Zermelo proof" of the fundamental theorem of arithmetic, and he uses a similar idea to prove the well-known fact that if $R$ is a unique factorization domain, then so is $R[t]$. See Theorem 27.<|endoftext|> TITLE: Reduction to Lie algebra version of fundamental lemma? QUESTION [5 upvotes]: Ngo famously proved the Langlands-Shelstad fundamental lemma for Lie algebras using the geometry of the Hitchin fibration. For the purposes of the trace formula, one actually needs the fundamental lemma for the group. In his ICM notes, Ngo states that Waldspurger shows how to reduce the group statement to the Lie algebra statement, but the reference is not clear (to me at least). My question is Where is this reduction discussed? A precise reference would be great. In lieu of the reference, I would also gladly accept a sketch of the reduction: How does this reduction work? REPLY [4 votes]: For the purpose of this answer let us say that "fundamental lemma" means "fundamental lemma for the unit of the unramified Hecke algebra". I do not think that "FL for Lie algebras => FL for groups" was proved in "Le lemme fondamental implique le transfert". This implication is however proved in greater generality (twisted endoscopy) in the paper "L'endoscopie tordue n'est pas si tordue", also by Waldspurger (see Conjecture 2 there; I think the non-standard FL for Lie algebras was also proved by Ngo). Just below Conjecture 2 Waldspurger writes that Hales "A simple definition of transfer factors for unramified groups" gives the implication in the case of standard endoscopy, but I do not think that this is completely obvious. In Hales' paper there is a reduction to the case of topologically unipotent elements, and I think that for these elements you can deduce the FL from the case of Lie algebras via the exponential map. There are detailed proofs in Waldspurger's paper for the twisted case; presumably they can be slightly simplified for the ordinary case. Note that the actual reduction in section 5.12 is not very long, so you can probably get a more precise idea of the proof than "use the exponential" without reading the whole paper. Of course Ngo's theorem is about equicharacteristic local fields, whereas the version of FL for Lie algebras needed here is for characteristic zero local fields (and indeed the exponential map is used!). There are two ways to "switch fields": a paper of Waldspurger, or using model theory (I imagine you can find all the relevant references in Ngo's paper).<|endoftext|> TITLE: Examples where "thin + thin = nice and thick" QUESTION [57 upvotes]: I'm interested in examples where the sum of a set with itself is a substantially bigger set with nice structure. Here are two examples: Cantor set: Let $C$ denote the ternary Cantor set on the interval $[0,1]$. Then $C+C = [0,2]$. There are several nice proofs of this result. Note that the set $C$ has measure zero, so is "thin" compared to the interval $[0,2]$ whose measure is positive. Goldbach Conjecture: Let $P$ denote the set of odd primes and $E_6$ the set of even integers greater than or equal to 6. Then the conjecture states is equivalent to $P + P = E_6$. Note that the primes have asymptotic density zero on the integers, so the set $P$ is "thin" relative to the positive integers. Are there other nice examples? REPLY [7 votes]: I know you asked for examples of the "thin + thin = nice and thick" phenomenon but, since Palindrome Week is all the rage these days, I can't avoid mentioning the following example of "thin + thin + thin = nice and thick". A couple of years ago, J. Cilleruelo(†), F. Luca, and L. Baxter proved that every natural number $n$ can be written as a sum of three palindromic numbers. Since the natural density of the set of palindromic numbers is $0$, if we agree to regard $\mathbb{N}$ as "nice" and "thick", we do have here--as promised--an example of the "thin + thin + thin = nice and thick" phenomenon.<|endoftext|> TITLE: Numbers divisible by precisely the same set of primes QUESTION [5 upvotes]: Say two positive integers are "peers" if they are divisible by precisely the same set of primes, such as 12 and 18 (both divisible by 2 and 3), or 70 and 350 (both divisible by 2, 5 and 7). What are the best estimates known for the number of pairwise non-peers not greater than an arbitrary positive integer N? REPLY [8 votes]: Your count equals the number of square-free numbers up to $N$. This is because a set of positive integers are pairwise "non-peers" if any only if their radicals are distinct. This is a well-studied problem in analytic number theory, see in particular Walfisz's estimate here.<|endoftext|> TITLE: What are the solutions of this Diophantine equation? QUESTION [7 upvotes]: Besides $(x, y, z)=(0, 0, 0)$ and $(1, 1, -2)$ (and their permutations) are there any other integer solutions to the equation $$3(x^{3}+y^{3}+z^{3})+3(x^{2}+y^{2}+z^{2})+(x+y+z)=0 $$ ? REPLY [17 votes]: A simple transformation renders this equivalent to $(3x + 1)^3 + (3y + 1)^3 + (3z + 1)^3 = 3$; any solutions would give solutions to $a^3 + b^3 + c^3 = 3$ (and vice versa, as pointed out by Emil Jeřábek in a comment). According to a recent arxiv article, the only known solutions to the latter are $(1, 1, 1), (4, 4, -5)$ and its permutations, but the existence of other solutions remains open, so this question is also open. Edited to add: This question seems to have been asked at a very opportune time; Booker and Sutherland have just found another solution: $569936821221962380720^3−569936821113563493509^3−472715493453327032^3 = 3$. That corresponds to $x = 189978940407320793573, y = -189978940371187831170, z = -157571831151109011$ for your question.<|endoftext|> TITLE: Are constant $\infty$-sheaves constant on connected components? QUESTION [5 upvotes]: Let $C$ be an $\infty$-category endowed with a Grothendieck topology $J$ and consider the $\infty$-topos $\infty\text{Sh}(C, J)$. There is a natural geometric morphism to $\infty\text{Grpd}$ whose left adjoint is the constant sheaf functor $\Delta : \infty \text{Grpd} \to \infty\text{Sh}(C, J)$. In the 1-topos case, $\Delta$ acts as follows: Take a set $S$ and take the constant presheaf on $S$, then sheafify. The resulting sheaf has the property that $\Delta(S)(U) = S$ for $U$ connected. Is this also true in the $\infty$ case, or do we need $U$ to be contractible? What even is $\Delta$?? REPLY [4 votes]: You need $U$ "contractible". In general $\Delta(S)$ is defined exactly as in the $1$-topos case: "take the constant presheaf valued at $S$ and sheafify it" (i.e. applies the left adjoint to the forget full functor from sheaf to presheaf). This theorem, when one takes $U$ contractible, is indeed also true for $\infty$-topos, but as far as I'm concerned this is the definition of a "contractible $\infty$-topos". To put in another way, if you want to see a proof it, it highly depends on what you mean by "U is contractible". To give an example that should show you that 'connected' is not enough, consider sheaves on a nice topological space $X$ (typically a CW-complex), and take $S$ to be some Eilenberg-Mac lane space $S = K(\pi,n)$ (though any space would do). Then in general, $\Delta(S)(U)$ for $U \subset X$ an open subspace is the space of maps from $U$ to $S$. So as soon as as $H^n(U,\pi)$ is non zero, $\Delta(S)(U)$ has several connected component (corresponding exactly to elements of $H^n(U,\pi)$) while $S$ itself only has one connected component. Note that in this special case (when $X$ is a nice topological space) it is easy to see that the formula $\Delta(S)(U) = S$ holds for all $S$ exactly when $U$ is contractible as a topological space in the usual sense. But, as mentioned before, for a more general topos you might want to clarify what "contractible" should mean in the first place... and depending on what definition you go for, it makes the question trivial or not.<|endoftext|> TITLE: Fast Bourgain embedding (or similar embeddings)? QUESTION [5 upvotes]: Currently I am working on applications of Bourgain Embedding (or similar embeddings of finite metric spaces to $l_2$) to automatic feature engineering for machine learning/data science ( http://www.orges-leka.de/automatic_feature_engineering.html, https://github.com/orgesleka/bourgain_embedding/blob/master/bourgain.py ). The method works quite well, but the only drawback is its runtime $O(N^2)$ where $N$ is the number of data points to be embedded. I have seen ( https://arxiv.org/abs/1805.07674 ) and it is a natural idea to chose $n < N$ random data points and to do the Bourgain embedding for those points. But the only proven preserved quantity is the distance distribution. My question is, if there is a way to speed up the Bourgain embedding while preserving the low distortion? This would be potentially useful for large datasets and I think it would have applications in data science / machine learning. Thanks for your help! REPLY [4 votes]: There is a way to speed-up Bourgain's embedding in case if the original metric space has low "intrinsic dimension". The resulting algorithm will have a theoretical runtime $O(CN\log^2(N))$, where $C$ is a constant depending only on "intrinsic dimension". And also will show speedups over the standard Bourgain's embedding varying between one and several orders of magnitude on natural machine learning datasets. The idea is to use Nearest Neighbor search in metric spaces. When you construct the standard Bourgain Embedding you do the following. For each coordinate you fix a random subset $A$ of you metric space $X$. Then for each point $x \in X$ you compute $d(x,A)$ and this is the value of you coordinate. You need $O(|X||A|)$ operations to do this if you use the brute force method. If $|A| \approx |X| = n$ you have $O(n^2)$ just from this step. But let's do this in a sneaky way instead. We will use [1] "Cover trees for nearest neighbor" by Beygelzimer, Alina, Sham Kakade, and John Langford. https://www.cc.gatech.edu/~isbell/reading/papers/cover-tree-icml.pdf First we construct the "cover tree data structure" on $A$. Which can be done in $O(C|A|\log|A|)$ time. (All constants $C$ in this text are different constants not related to each other.) This data structure allows to compute nearest neighbors in $O(C\log|A|)$ time. Thus, computing all $n$ required distances will take $O(Cn\log|A|)$ operations. Since we have $C\log(n)$ coordinates the runtime of the whole algorithm will be $O(CN\log^2(N))$. I assume that computations of $d(x,A)$ is the only computationally heavy part of Bouragin's Algorithm. Thus, we probably will have the same speed up on natural machine learning datasets as in "nearest neighbor via cover trees" vs brute force, see [1]. PS:I'm currently working on writing some kind of academic paper out of this observation. But struggling since I'm a mathematician and don't really know how to write CS papers. I implemented Beygelzimer-Kakade-Langford in Java. But now I'm not sure what to do next. It would be cool to write some kinda of package, maybe people will try to use it. But I'm so noob at coding. Probably I should choose the language wise and not stick with Java out of habit. Probably should use parallelization. Also I'am only familiar with CPU computations but I heard that practitioners mostly use GPU/TPU. So If you can give me an advice or want to collaborate with me please email me at paranuel@mail.ru.