diff --git "a/stack-exchange/math_overflow/shard_17.txt" "b/stack-exchange/math_overflow/shard_17.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_17.txt" +++ /dev/null @@ -1,28227 +0,0 @@ -TITLE: Generalised Isospectrality of Graphs -QUESTION [6 upvotes]: Q: Is there a graph matrix-representation (not necessarily an $n \times n$ matrix for an $n$-graph) such that isospectrality implies graph-isomorphism? For instance, would the simple distance-matrix do the job? -Background: The 'can you hear the shape of a drum?' question can be answered of undirected graphs in the negative, but I do not know if the results are only for the adjacency and/or Laplacian matrix representations. - -REPLY [2 votes]: Distance matrix is not going to help, as examples of non-isomorphic co-spectral strongly regular graphs would tell you. -The smallest examples like this exist on 16 vertices: Shrikhande graph. -In general, there seem to be logic-related obstacles for non-isomorphism certificates of this sort, related to 1st order logics with counting. This has started with the famous paper -An Optimal Lower Bound on the Number of -Variables for Graph Identification by Cai, Furer and Immerman.<|endoftext|> -TITLE: Faces of the intersection of convex sets -QUESTION [8 upvotes]: Let $V$ be a normed real vector space and let $K_1, K_2\subseteq V$ be closed convex subsets such that the intersection $K_1\cap K_2$ is non-empty. Assume that $F_1$ is a face of $K_1$ and $F_2$ is a face of $K_2$ (face $F$ of a convex set $K$ is a convex subset such that $a=tb+(1-t)c$, for $a\in F$, $b, c\in K$, $0 -TITLE: Category theory without axiom of choice -QUESTION [8 upvotes]: I'm looking for references on the development of (some of) Category theory without the axiom of choice. One possible axiom system (that, to me, seems the natural setting) is ZF + there are arbitrarily large inaccessible cardinals. -I found a link to these notes by Solovay somewhere in mathoverflow, that I can't seem to find now. It gives a notion of inaccessible cardinals ("v-inaccessible" in the text) that can be formulated in ZF. -And the existance of arbitrary large such cardinals does not imply AC (this is the first example in the notes), but is still equivalent (over ZF) to axiom of "every set belongs to a Grothendieck universe", which means that, theoretically [1], category theory can be done in such set theory. -Are there any such texts/papers? -[1] What I mean is that there are no foundational obstructions. - -REPLY [11 votes]: As was pointed out in a comment (that should have been an answer), Makkai's work on anafunctors gives a way to replace the notion of "functor" by a more general notion that doesn't require the axiom of choice in order to prove, for instance, that every fully faithful essentially surjective functor is an equivalence (in the sense of having an inverse up to isomorphism), or that if every pair of objects have a cartesian product then there is a product-assigning functor. However, as was also pointed out in a comment, this doesn't work perfectly inside ZF: in particular the category of anafunctors between two small categories may no longer be small, unless you assume extra axioms. It's also kind of tedious to work explicitly with anafunctors everywhere, which may be why no one has tried to push Makkai's work much further. -Another approach to choice-free category theory is to decide to live with the fact that not every fully faithful essentially surjective functor is an equivalence (i.e. not every "weak equivalence" is a "strong equivalence"), and define a category to (for instance) "have products" if there is a functor assigning a product to every pair of objects rather than merely that a product exists for every pair of objects. This generally works quite well too; although in general, objects with universal properties are only determined up to isomorphism, in practice it seems that (at least when doing category theory based on set theory) nearly all objects with universal properties can in fact be constructed functorially without assuming choice. For instance, in the category of sets there is a cartesian product functor defined by using "the" definition of cartesian products (e.g. Kuratowski ordered pairs), and almost any way of constructing a new category from others that inherits limits or colimits will preserve their functorial nature without using choice. A reference that uses this approach pretty systematically is Peter Johnstone's Sketches of an Elephant, which conveniently also includes one of the rare sort of cases where it doesn't work (section D1.5). -A third way of dealing with this problem is to use "saturated" or "univalent" categories in homotopy type theory; see chapter 9 of the homotopy type theory book. In this context, one can prove that if "essentially unique" objects (e.g. with universal properties) exist, then there is always a functor selecting them, even without assuming the axiom of choice: it's a higher-categorical version of the "unique choice principle" which doesn't depend on the axiom of choice. -Finally, this is not actually the only use of choice in category theory. In particular, transfinite induction is sometimes used to construct free algebras, left adjoints, cofibrant replacements, and so on, especially when working with locally presentable categories. This is actually a "real" use of choice, as such free algebras may not exist at all in ZF; this was shown by Andreas Blass in his paper Words, free algebras, and coequalizers. I don't know of a good way to deal with this in ZF, but in homotopy type theory one can often use higher inductive types to construct such things.<|endoftext|> -TITLE: $L^1$ and $L^4$ norms of trigonometric polynomials -QUESTION [7 upvotes]: Let $p(x) = \sum_{n=1}^N e^{2 \pi i a_n x}$ be a trigonometric polynomial, where $a_n$ are distinct positive integers. There is a classical trick which (using Hölder's inequality) allows to give a lower bound for the $L^1$ norm of $p$ in terms of the $L^4$ norm of $p$. One obtains -$$ -\|p\|_1 \geq \frac{\|p\|_2^{3}}{\|p\|_4^{2}} = \frac{N^{3/2}}{\|p\|_4^{2}}. -$$ -So roughly speaking a small $L^4$ norm implies a large $L^1$ norm. -Question: Is the opposite also true? That is, does a large $L^4$-norm imply a small $L^1$-norm? (And if "yes", is there a quantitative estimate?) -(This might be a stupid question, but still I am grateful for an answer.) -(PS: For a reference to the trick mentioned above, see for example A. A. Karatsuba, "An estimate of the L1-norm of an exponential sum", Mathematical Notes, 1998, 64:3, 401-404.) - -REPLY [12 votes]: No this need not be the case. Take $f(x) = \sum_{n=1}^{N/2} e(nx)$ and $g(x) = \sum_{k=N/2}^N e(2^kx)$. Then the $L^4$ norm of $f$ is big -- of size $N^{\frac 34}$ -- and its $L^1$ norm is very small -- of size $\log N$. On the other hand the $L^4$ norm of $g$ is small -- of size $\sqrt{N}$ -- and its $L^1$ norm is correspondingly large -- of size $\sqrt{N}$. But now the triangle inequality shows that the $L^4$ norm of $f+g$ is big (of size $N^{\frac34}$), whereas the $L^1$ norm of $f+g$ is also big (of size $N^{\frac 12}$). -Even if you want the coefficients $a_n$ to be small, you could arrange this by making the first half of the coefficients be all natural numbers in $[1, N/2]$ and then choosing $N/2$ integers randomly from $[N/2, 2N]$.<|endoftext|> -TITLE: Making compact subsets "parallel" -QUESTION [9 upvotes]: Let $X$ be a compact metric space. Say that two compact subsets $E,F\subset X$ are parallel if -$$ dist(x,F) = dist(y,E)$$ -for all $x\in E$ and $y\in F$. Here $ dist(y,E) = \inf\{d(y,z):z\in E\}.$ -The overall question I would like to understand is the following: - -Let $X$ be a compact (Hausdorff, second countable, hence metrizable) topological space with a collection of disjoint compact subsets $\{E_t\}_{t\in T}$. When is there a metric on $X$ that makes every pair of sets in $\{E_t\}$ parallel? - -I generally think of $T$ as uncountable and $\{E_t\}$ as a partition of $X$, but I don't see the need to assume this. -There is at least one clear necessary condition that $\{E_t\}$ must satisfy: Namely, if $\{x_n\in E_{t_n}\}$ is a sequence converging to $x\in E_t$, then every element $x'\in E_t$ is the limit of a sequence $\{x'_n\in E_{t_n}\}$, and every sequence $\{x''_n\in E_{t_n}\}$ accumulates only at points in $E_t$. -Call this condition (*). (The part after the and in condition (*) was added after MTyson's example.) -For example, if $X=[0,1]$ and $\{E_t\}$ consists of $[0,1/2]$ as well as all the singletons $\{x\}$ for each $x\in (1/2,1]$, the collection fails condition (*) and can't be made parallel. -So I refine the question: - -Let $X$ be a compact (Hausdorff, second countable, hence metrizable) topological space with a collection of disjoint compact subsets $\{E_t\}_{t\in T}$, satisfying (*). Is there a metric on $X$ that makes every pair of sets in $\{E_t\}$ parallel? - -As I don't know if I believe this, I also ask: - -Are there conditions, in addition to or instead of (*), that ensure the sets $\{E_t\}$ can be made parallel by the appropriate choice of a metric on $X$? - -I'd be interested in counterexamples, other possible obstructions, references to similar things in the literature, or general sufficient conditions (even if they assume more about $X$ or $\{E_t\}$). -One could imagine cases in which $T$ is a topological space and $t\mapsto E_t$ is continuous in some topology on the space of compacta of $X$ (e.g., that induced by the Hausdorff metric coming from some metric on $X$.) Maybe this is equivalent to my formulation. Regardless, if that makes the problem easier or more well-known then by all means assume it. - -REPLY [5 votes]: The answer to this problem is affirmative (at least for covers). -Definition. A family $\mathcal C$ of subsets of a topological space $X$ is called -$\bullet$ lower semicontinuous if for any open set $U\subset X$ its $\mathcal C$-star $St(U;\mathcal C):=\bigcup\{C\in\mathcal C:C\cap U\ne\emptyset\}$ is open in $X$; -$\bullet$ upper semicontinuous if for closed set $F\subset X$ its $\mathcal C$-star $St(F;\mathcal C)$ is closed in $X$; -$\bullet$ continuous if $\mathcal C$ is both lower and upper semicontinuous. -Theorem. For a cover $\mathcal C$ of a metrizable space $X$ by compact subsets the following conditions are equivalent: - -The topology of $X$ is generated by a metric $d$ making all sets in the cover $\mathcal C$ parallel. -The cover $\mathcal C$ is disjoint and continuous. -The cover $\mathcal C$ has property $(*)$. - -Proof. The implications $(1)\Rightarrow(2)\Leftrightarrow(3)$ are simple and are left to the reader. -To prove that $(2)\Rightarrow(1)$, assume $\mathcal C$ is disjoint and continuous. Fix any admissible metric $\rho\le 1$ on $X$. -Let $\mathcal U_0(C)=\{X\}$ for every $C\in\mathcal C$. -Claim. For every $n\in\mathbb N$ and every $C\in\mathcal C$ there exists a finite cover $\mathcal U_n(C)$ of $C$ by open subsets of $X$ such that -(i) each set $U\in\mathcal U_n(C)$ has $\rho$-diameter $\le\frac1{2^n}$; -(ii) if a set $A\in\mathcal C$ meets some set $U\in\mathcal U_n(C)$, then $A\subset\bigcup\mathcal U_n(C)$ and $A$ meets each set $U'\in\mathcal U_n(C)$. -Proof of Claim. Using the paracompactness of the metrizable space $X$, choose an open locally finite cover $\mathcal V$ of $X$ consisting of sets of $\rho$-diameter $<\frac1{2^n}$. -For every compact set $C\in\mathcal C$ consider the finite subfamily $\mathcal V(C):=\{V\in\mathcal V:V\cap C\ne \emptyset\}$ of the locally finite cover $\mathcal V$. Since the cover $\mathcal C$ is upper semicontinuous, the set $F_C=St(X\setminus \bigcup\mathcal V(C);\mathcal C)$ is closed and disjoint with the set $C$. Since $\mathcal C$ is lower semi-continuous, for any open set $V\in\mathcal V(C)$ the set $St(V;\mathcal C)$ is open and hence $W(C):=\bigcap_{V\in\mathcal V(C)}St(V;\mathcal C)\setminus F_C$ is an open neighborhood of $C$. -It can be shown that the family $\mathcal U_n(C):=\{W(C)\cap V:V\in\mathcal V(C)\}$ satisfies the conditions (i), (ii). -Continuation of the proof of Theorem. -Given two points $x,y\in X$ let $$\delta(x,y):=\inf\big\{\tfrac1{2^n}:\mbox{$\exists C\in\mathcal C$ and $U\in\mathcal U_n(C)$ such that $x,y\in U$}\big\}.$$ -Adjust the function $\delta$ to a pseudometric $d$ letting $$d(x,y)=\inf\sum_{i=1}^m\delta(x_{i-1},x_i)$$where the infimum is taken over all sequences $x=x_0,\dots,x_m=y$. -It can be shown that the metric $d$ generates the topology of $X$ and $\rho\le d$. -Let us prove any sets $A,B\in\mathcal C$ are parallel in the metric $d$. We need to show that $d(a,B)=d(A,B)=d(A,b)$ for any $a\in A$, $b\in B$. Assuming that this equality is not true, we conclude that either $d(a,B)>d(A,B)>0$ or $d(A,b)>d(A,B)>0$ for some $a\in A$ and $b\in B$. -First assume that $d(a,B)>d(A,B)$ for some $a\in A$. Choose points $a'\in A$, $b'\in B'$ such that $d(a',b')=d(A,B) -TITLE: Petersson square of twist -QUESTION [8 upvotes]: I should know this, but can't find a reference. Let $F$ be a cusp form, -not necessarily an eigenform, on some congruence subgroup. It seems experimentally that the ratio of the Petersson square of a twist of $F$ -(say by some quadratic character) to the Petersson square of $F$ belongs -to $\mathbb Q(F)$. Is this true, and what is the proof ? Or are there conditions on the twist ? Is this directly related to Manin's rationality -theorem, i.e., are the periods $\omega^{\pm}$ of the twist related to -those of $F$ ? - -REPLY [2 votes]: For an eigenform, the product of the periods associated to $f$ ($\omega^{\pm}$ in your notation) is equal modulo $\mathbf{Q}(f)^\times$ to $i^{1-k} G(\chi) \langle f, f\rangle$, where $G(\chi)$ is the Gauss sum of the nebentypus character of $f$, and $k$ is the weight. This is Theorem 1 (iv) of Shimura's On the periods of modular forms (Math. Ann. 229, 1977). -Since the periods of the twist $f_\varepsilon$ are $G(\varepsilon)$ times the periods of $f$, and $G(\varepsilon^2 \chi) / [G(\chi) G(\varepsilon)^2]$ lies in the field of values of $\chi$ and $\varepsilon$, it follows that the ratio $\langle f_\varepsilon, f_\varepsilon\rangle / \langle f, f \rangle$ lies in $\mathbf{Q}(f, \varepsilon)^\times$ (and hence in $\mathbf{Q}(f)^\times$ if $\varepsilon$ is quadratic). -(Are you sure the statement is true for non-eigenforms? It seems very unlikely, somehow.)<|endoftext|> -TITLE: Computational complexity of finding the smallest number with n factors -QUESTION [11 upvotes]: Given $n \in \mathbb{N}$, suppose we seek the smallest number $f(n)$ with -at least $n$ distinct factors, -excluding $1$ and $n$. -For example, for $n=6$, $f(6)=24$, -because $24$ has the $6$ distinct factors $\{2,3,4,6,8,12\}$, -and $24$ is the smallest integer with $6$ factors. -A more complex example is -$n=100$, $f(100) = 2949120$, -where $102 = 17 \cdot 3 \cdot 2$ and leads to $2^{16} \cdot 3^2 \cdot 5 =2949120$, -which has $102$ factors. -Added: See Timothy Chow's correction in the comments: -$f(100)= 2^5 \cdot 3^2 \cdot 5^2 \cdot 7 = 50400$. -All this is known; the sequence is OEIS A061799. -E.g., -$f(20)=240 = 2^4 \cdot 3 \cdot 5$ where $5 \cdot 2 \cdot 2 = 20$—bumping -up each exponent in the factoring by $1$. -My question is: - -Q. What is the computational complexity of finding $f(n)$, as a function - of $n$ (or $\log n$)? - -Is this known? - -Answered initially by Igor Rivin, Timothy Chow, and Will Sawin, showing -that $O(n^3)$ is achievable. Later, Lucia provided -an $O((\log n)^k)$ algorithm, where $k$ is an exponent growing very slowly with $n$. - -REPLY [13 votes]: The problem asks for the least number $N$ such that the number of divisors of $N$ is at least $n+2$. Since all numbers below $N$ must have fewer divisors, clearly $d(N) > d(m)$ for all $1\le m < N$. Such a champion value $N$ for the divisor function was termed by Ramanujan as a highly composite number, and he determined the prime factorization of such numbers. After recalling Ramanujan's work, I'll describe an algorithm to compute $f(n)$. It executes in time -$$ -O((\log n)^{C\log \log \log n}), -$$ -for some constant $C$. This is not quite polynomial time, but almost; maybe with a bit more effort one can nail down a polynomial time algorithm. -Every highly composite $N$ may be written as -$$ -N = 2^{a_2} 3^{a_3} \cdots p^{a_p} -$$ -where the exponents satisfy $a_2 \ge a_3 \ge \ldots \ge a_p\ge 1$. Apart from $4$ and $36$, the last exponent $a_p =1$. Ramanujan's main result concerns the exponents $a_\ell$ for primes $\ell \le p$. He works out detailed estimates for these exponents; roughly they satisfy -$$ -a_\ell \approx \frac{1}{\ell^{\alpha}-1}, -$$ -with $\alpha= \log 2/\log p$, in keeping with the example in Will Sawin's answer. -The numbers produced in Will Sawin's answer are what Ramanujan calls "superior highly composite numbers." These numbers $N$ are characterized by the property that for some $\epsilon >0$ one has -$$ -\frac{d(N)}{N^{\epsilon}} > \frac{d(n)}{n^{\epsilon}}, -$$ -for all $n >N$, and -$$ -\frac{d(N)}{N^{\epsilon}} \ge \frac{d(n)}{n^{\epsilon}} -$$ -for all $n\le N$. The "superior highly composite numbers" are strictly a subset of the highly composite numbers. -The table on pages 110-112 of Ramanujan's paper lists all the highly composite numbers (with superior highly composite numbers marked with an asterisk) with number of divisors up to $10080$ (that is, Ramanujan computes your $f(n)$ for all $n\le 10078$). Ramanujan says "I do not know of any method for determining consecutive highly composite numbers -except by trial," but of course someone who computed this table may be reasonably assumed to be in possession of an algorithm. -Now for the algorithm and its complexity. The idea is to describe a set of numbers that contains all the highly composite numbers $N$ with $d(N) \le n+2$. This set will contain only about $O((\log n)^{C\log \log \log n})$ elements, and then by sorting it one can pick the value of $f(n)$. -We are looking for numbers $N=p_1^{e_1} \cdots p_{k}^{e_k}$ where $p_i$ is the $i$-th prime, and the exponents are in descending order $e_1 \ge e_2 \ge \ldots \ge e_k\ge 1$. Now we can assume that $k\le [\log_2 (n+2)] +1=K$, else $d(N)$ is already larger than $n+2$. Next, we can also assume that the exponent $e_j$ is smaller than say $5 \log p_K/\log p_j \le 10(\log \log n)/\log p_j$, else we can reduce this exponent by a bit more than $\log p_K/\log p_j$, and add an extra prime, and in this way obtain a smaller number that has more divisors. -Now the idea is simply to list all numbers (together with their prime factorizations) that satisfy the above conditions on the exponents. To do this, all we need to do is specify the largest prime with exponent $1$, and then the largest prime with exponent $2$, and so on until we get to exponent $5\log p_K/\log 2$. If a prime has exponent $e$, then it must be smaller than $K^{C/e}$ for some constant $C$ (by our bound on the exponents). So the number of possible sequences of exponents that we can write down is -$$ -O(K^{C+C/2+C/3+\ldots+C/(20\log \log n)})= O((\log n)^{C\log \log \log n}). -$$ -That finishes our running time analysis. - -          - - -          - -The beginning of Ramanujan's table.<|endoftext|> -TITLE: What ordinal corresponds to the T(3)? -QUESTION [5 upvotes]: Let's play a game. You start with the ordinal $\alpha$ and I start with the empty sequence. Each turn, you decrease your ordinal, and I add a tree (where each node can have one of three labels), subject to the rule that no previous tree I choose can homeomorphically embed into the new one. Whoever moves last wins! What $\alpha$ should we choose to make the game fair, meaning whoever moves first will lose (so both players make the same number of moves)? -Note we can generalize this any number of labels $n$, or indeed, any well-quasi-ordered set instead of trees. - -Well, note that the game can not go on forever, for two reasons. There is no infinite decreasing sequence of ordinals, so you must eventually hit $0$ and be out of moves. Likewise, due to Kruskal's tree theorem, I too can not go on forever. Indeed, we could have replaced trees with any other well-quasi-ordered set. -Now, what do we mean by fair? Well, since the tree picking game correspond to a game in the CGT sense, there is some ordinal $\alpha$ such that $T(3) - \alpha = 0$ (implying $T(3) = \alpha$, hence the title), where $T(3)$ is the tree picking game. $T(3)-\alpha$ is therefore the game described in the first paragraph. Since $T(3) - \alpha = 0$, whoever moves first in that game will lose (this is why we say whoever moves first will lose). - -REPLY [5 votes]: If Player 1 moves first, Player 2 wins the game if and only if $\alpha\leq o(T(3))$, where $o$ denotes the maximal order type of the well-quasi-order $T(3)$. This ordinal was studied by Diana Schmidt in her 1979 Habilitation thesis Well-partial orderings and their maximal order types, which might not be so easy to find online. The introduction of the 2015 Ph.D. thesis of Van der Meeren Connecting the two worlds: Well-partial-orders and ordinal notation systems does a nice quick survey of these questions in Section 1.2; the answer to the question is $\vartheta(\Omega^\omega\cdot 3)$, assuming you are talking about finite ordered unranked trees.<|endoftext|> -TITLE: How to compute the Gromov-Hausdorff distance between spheres $S_n$ and $S_m$? -QUESTION [18 upvotes]: Can we compute the Gromov-Hausdorff distance $d(\mathbb{S}_n,\mathbb{S}_m)$ for two different spheres $\mathbb{S}_n$ and $\mathbb{S}_m$, $m\neq n$? We consider the spheres with the metrics induced by the embedding -$\mathbb{S}_n \to \mathbb{R}^{n+1}$. -For $n=2$, $m=3$, this is evaluating $$d(\mathbb{S}_2,\mathbb{S}_3)=\inf_{M,f,g}d_{M}(\mathbb{S}_2,\mathbb{S}_3)$$ where $M$ ranges over all possible metric spaces and $f:\mathbb{S}_2\to M$ and $g:\mathbb{S}_3\to M$ range over all possible distance-preserving embeddings. As an upper bound, $d(\mathbb{S}_2,\mathbb{S}_3)\leq \sqrt{2}$. -More generally for $0 \leq n \leq m$, -$$d(\mathbb{S}_n,\mathbb{S}_m)\leq d(point,S_n)+d(point,S_m)\leq 2$$ -But I find it difficult to control the lower bound with the inf over all possible metric spaces $M$. -I conjecture that for all $0 \leq n \leq m$: -$$d(\mathbb{S}_n,\mathbb{S}_m)\geq \lambda_{m,n}\frac{m-n}{m}, \text{ where } \liminf_{m,n\to \infty}\lambda_{m,n}>0$$ -I only know the Gromov-Hausdorff theory from Petersen's Riemannian Geometry, which does not give enough information to compute this distance. I will appreciate any pointers. - -REPLY [9 votes]: Hu Xiyu. Even though yours is a question from four years ago, I want to bring to your attention my recent paper “Gromov-Hausdorff distance between spheres”( https://arxiv.org/abs/2105.00611 ) coauthored with Facundo Memoli and Zane Smith, since it is very closely related to your question. -In this paper, we compute/bound the Gromov-Hausdorff distance between two spheres with different dimension (each with geodesic metric). We use topological methods in order to obtain lower bounds: more precisely, we resort to a certain version of Borsuk-Ulam Theorem for discontinous functions. On the other hand, we design specialized optimal correspondences in order to estimate upper bounds. In particular, we were able to compute precise value of the Gromov Hausdorff distance for $\mathbb{S}^1$ vs $\mathbb{S}^2$, $\mathbb{S}^1$ vs $\mathbb{S}^3$, and $\mathbb{S}^2$ vs $\mathbb{S}^3$. -Finally, the last section of the paper deals with the case of spheres with Euclidean metric. Even though we could not give a full answer, I believe you will be able to find some useful observations there.<|endoftext|> -TITLE: Tracing the word “form” -QUESTION [17 upvotes]: Today the word form can refer to (at least) three different kinds of mathematical object: - -A homogeneous polynomial. This was apparently started by Gauss (1801), renaming what others had called formulasa. (See e.g. Bachmann 1922, p. 17.) -A scalar-valued linear or multilinear map. Apparently started by Kronecker (1866) / Weierstrass (1868), rather out of the blue. -A field of forms in the sense 1 or 2. Apparently started by Christoffel (1869) / Lipschitz (1869), renaming what others called differential formulasb or expressionsc. (See e.g. Weitzenböck 1922, p. 29.) - - -Question: Has anyone anywhere ever discussed these choices and switches in terminology? - -$\ $ - -References: e.g. -a) -Euler (1770, 1774, 1827), Lagrange (1773, 1774), Liouville (1852). -b) -Bernoulli (1712), -Euler (1755, 1768), -Agnesi (1775), Cousin (1777), -Lagrange (1786), Bossut (1798), Poisson (1811), Abel (1826), Liouville (1852, 1856). -c) -Gauss (1815), Jacobi (1845), Riemann (1867), Sturm (1877), Frobenius (1879), Darboux (1882), Cartan (1899). - -REPLY [6 votes]: The evolution of the concept of a form from arithmetic to algebra is discussed on pp. 20, 21, 27 of F. Brechenmacher (arXiv:0712.2566; revised version published in 2016): - -Whereas such terms as “forms” and “transformations” had been given an explicit mathematical definition in the arithmetic of quadratic forms in relation to the notion of equivalence relation that had been introduced by Gauss’ 1801 Disquisitiones arithmeticae, they pointed to various and mostly implicit meanings within the algebraic framework of the discussion. (...) -Kronecker had been implicitly referring to the legacy of the works of Gauss and Hermite on the arithmetic of quadratic forms in 1866 — [when] he preferred to make use of the term “form” to name what others would designate as a function ([Weierstrass, 1858]) or as a “polynom” ([Jordan, 1873]) (...) -Kronecker blamed algebraic methods [notably by Jordan] for their tendency to think in term of the “general” case with little attention given to the arithmetic difficulties that might be caused by assigning specific values to the symbols (...) [He] appealed to the tradition of Gauss on behalf of his claim that the theory of forms should be considered as belonging to arithmetic and should consequently focus on the characterisation of equivalence classes in establishing arithmetical invariants thanks to some effective procedures such as g.c.d.s computations.<|endoftext|> -TITLE: Immersions of manifolds with boundary (regular homotopy classes, h-principle) -QUESTION [6 upvotes]: Have regular homotopy classes of immersions of manifolds with boundary been studied? i.e. immersions $(M, \partial M) \looparrowright (N, \partial N)$. The references I've looked at from this question state theorems for target manifolds without boundary. -Does the $h$-principle apply, or something like it? -Simple example to keep in mind: $M$ is $D^1$ with boundary $S^0$, $N$ is $D^1 \times \mathbb{R}$ with boundary $S^0 \times \mathbb{R}$, and let's say each boundary point of $M$ goes to a different component of $\partial N$. It seems to me at a glance that there is a regular homotopy class for each integer, corresponding to the number of counterclockwise turns made along the way from one component to the other. How is this proved? (and generalized?) - -REPLY [4 votes]: There are two questions you could ask: about the space of immersions $Imm((M, \partial M), (N, \partial N))$ of $M$ in $N$ taking the boundray to the boundary, where the boundary is allowed to move, or you could fix an immersion $i : \partial M \looparrowright \partial N$ and consider the subspace $Imm((M, \partial M), (N, \partial N); i)$ of those immersions which restrict to $i$ on the boundary. -These fit into a fibration sequence -$$Imm((M, \partial M), (N, \partial N); i) \to Imm((M, \partial M), (N, \partial N)) \to Imm(\partial M, \partial N)$$ -so the answers are closely related. The space $Imm(\partial M, \partial N)$ can certainly be studied by the Smale--Hirsch $h$-principle for immersions, but the usual proof of that theorem is an induction over handles, where handles are immersed with given bundary conditions: this proof applies equally well to $Imm((M, \partial M), (N, \partial N); i)$, too, and gives a bundle-theoretic description of this space. If you like, you can then combine these results to get a bundle-theoretic description of $Imm((M, \partial M), (N, \partial N))$. This will be: the space of bundle monomorphisms $\hat{f}: TM \to TN$, covering a map $f : (M, \partial M) \to (N, \partial N)$, which preserve the inward-pointing normal vector over the boundary.<|endoftext|> -TITLE: What is the name of this kind of equivalence between two topological spaces? -QUESTION [9 upvotes]: I apologize if my question is trivial. I am a group theorist with a minor knowledge of topology. Suppose $(X, T_X)$ and $(Y, T_Y)$ are two topological spaces and there is an inclusion-reversing (inclusion preserving) bijection $\varphi:T_X\to T_Y$. What is the name of this kind of equivalence? What can be said about the relation of these two spaces? - -REPLY [11 votes]: As far as I can see there are two possibilities where we can say more: - -The bijection is order preserving. As Qiaochu Yuan said in a comment, this is the same as asking that their associated locales are isomorphic. Thanks to the equivalence of sober spaces with spatial locales, this is the same as asking that their soberifications are homeomorphic. In particular, if $X$ and $Y$ are already sober (a very common condition, sober is strictly weaker than Hausdorff), this is the same as asking that $X$ and $Y$ are homeomorphic. -The bijection is order reversing. Then the lattice of opens of both spaces need to have arbitrary meets distribute over finite joins (i.e. each frame of opens must have the property that its opposite lattice is a frame too). This is a very rare situation, that happens mainly for Alexandrov spaces,although as მამუკაჯიბლაძე notes in the comments there might be other examples. - -REPLY [8 votes]: Edited after realizing that my preceding answer was false. -By elementary tools (i.e., without soberifications) it can be shown that for $T_1$-spaces this equivalence coincides with the homeomorphness. -Theorem 1. For $T_1$-spaces $X,Y$ the following conditions are equivalent: - -$X$ and $Y$ are homeomorphic; -The topologies of the spaces $X,Y$ are order isomophic. - -Proof. The implication $(1)\Rightarrow(2)$ is trivial. -To prove that $(2)\Rightarrow(1)$, assume that there exists an order isomorphism $i:\tau_X\to \tau_Y$ between the topologies of the spaces $X$ and $Y$. -Since $X$ is a $T_1$-space, for every $x\in X$ the open set $X\setminus\{x\}$ is a maximal element of the poset $\tau_X\setminus\{X\}$. Since $i$ is an isomorphism of the posets $\tau_X$ and $\tau_Y$, the set $i(X\setminus\{x\})$ is a maximal element of the poset $\tau_Y\setminus\{Y\}$ and hence $i(U_x)=Y\setminus \{f(x)\}$ for some point $f(x)\in Y$. The injectivity of $i$ implies that the map $f:X\to Y$ is injective. To see that $f$ is surjective, observe that for every $y\in Y$ the set $i^{-1}(Y\setminus \{y\})$ is a maximal element of the poset $\tau_X\setminus\{X\}$ and hence it coincides with the set $X\setminus\{x\}$ for some - $x\in X$. It follows that $Y\setminus\{y\}=i(X\setminus\{x\})=Y\setminus \{f(x)\}$ and hence $y=f(x)$. -Therefore the map $f:X\to Y$ is bijective. To show that $f$ is continuous, it suffices to check that for every open set $U\in\tau_Y$ we get $f^{-1}[U]=i^{-1}(U)$. -To show that $f^{-1}[U]\subset i^{-1}(U)$, observe that for any $x\in f^{-1}[U]$ we get $f(x)\in U$ and hence $(Y\setminus \{f(x)\})\cup U=Y$. Taking into account that order isomorphisms of lattices preserve lattice operations, we conclude that $$(X\setminus\{x\})\cup i^{-1}(U)=i^{-1}(Y\setminus \{f(x)\})\cup i^{-1}(U)=i^{-1}(Y)=X$$ and hence $x\in i^{-1}(U)$. Therefore $f^{-1}[U]\subset i^{-1}(U)$. -To see that $i^{-1}(U)\subset f^{-1}[U]$, observe that for any $x\in i^{-1}(U)$ we get $X=(X\setminus \{x\})\cup i^{-1}(U)$ and hence $(Y\setminus\{f(x)\})\cup U=i(X\setminus\{x\})\cup U=Y$, which implies $f(x)\in U$ and hence $x\in f^{-1}[U]$. -This shows that the preimage $f^{-1}[U]=i^{-1}(U)$ of the open set $U$ is open and $f$ is continuous. By analogy we can check that the map $f^{-1}:Y\to X$ is continuous. -So, $f$ is a homeomorphism. - -Remark. The theorem does not generalize to $T_0$-spaces as the $T_0$-space $X=[0,1)$ with topology $\tau_X=\{[0,a):0\le a\le 1\}$ and its subspace $Y=(0,1)$ are not homeomorphic but have order isomorphic topologies. -Also there exist a $T_1$-space $X$ and a sober space $Y$ which are not homeomorphic but have order-isomorphic topologies. -Example 1s. Take any infinite set $X$ such that $X\notin X$. On the set $Y= X\cup\{X\}$ consider the topology $\tau_Y=\{\emptyset\}\cup\{Y\setminus F:F\subset X$ is finite$\}$. The space $Y$ is sober but not $T_1$ and its subspace $X$ is $T_1$ but not sober. Nonetheless the spaces $X,Y$ have order-isomorphic topologies. - -Added after a comment of მამუკა ჯიბლაძე: -Theorem 1 can be generalized to $T_{Constructible}$-spaces. We recall that a set $A$ in a topological space $X$ is constructible (resp. Borel) if it belongs to the smallest algebra (resp. $\sigma$-algebra) of subsets of $X$, containing all open subsets of $X$. -A topological space $X$ is defined to be $T_{Constructible}$-space (resp. $T_{Borel}$-space) if for each $x\in X$ the singleton $\{x\}$ is a constructible (resp. Borel) subset of $X$. -By Theorem 2.1 of this paper, a topological space $X$ is $T_{Borel}$ (resp. $T_{Constructible}$) if and only if each singleton $\{x\}\subset X$ can be written as the intersection $F\cap U$ of a closed set $F\subset X$ and an (open) $G_\delta$-set $U$. -This characterization implies that the class of $T_{Constructible}$-spaces coincides with the class of $T_D$-spaces (i.e. spaces in which each singletion is an intersection of an open and a closed sets). -So, we have the implications -$$T_1\Rightarrow T_D\Leftrightarrow T_{Constructible}\Rightarrow T_{Borel}\Rightarrow T_0.$$ -Observe that if the set $X$ in Example 1s is countable, then the spaces $X,Y$ from this example are $T_{Borel}$. This shows that Theorem 1 does not generalize to $T_{Borel}$-spaces. -On the other hand, it does generalize to $T_{Constructible}$-spaces (and to $T_D$-spaces as was suggested by მამუკა ჯიბლაძე). -Theorem 2. Two $T_{Constructible}$-spaces are homeomorphic if and only if their topologies are order-isomorphic. -Proof. Let $(X,\tau_X)$ be a $T_{Constructible}$-space. -A pair $(V,W)$ of open sets in $X$ is called point-determining if - -$V\subset W$ and $V\ne W$; -for any open set $U\in\tau_X$ with $V\subset U\subset W$ we get $U=V$ or $U=W$; -for any open set $U\subset X$ with $V\subset U$ and $W\not\subset U$ we get $V=U$. - -Using the fact that each $T_{Constructible}$-space is a $T_D$-space, one can easily check: -Lemma 1. For any point $x\in X$ the pair $(V_x,W_x)$ of the open sets $V_x=X\setminus\overline{\{x\}}$ and $W_x=V_x\cup\{x\}$ is point-determining. -Lemma 2. Each point-determining pair $(V,W)$ is equal to $(V_x,W_x)$ for a unique point $x\in X$. -Lemma 3. An open set $U\in\tau_X$ is a neighborhood of a point $x\in X$ if and only if $W_x\subset V_x\cup U$. -Now theorem 2 easily follows from Lemmas 1-3 and the observation that point-determining pairs of open sets are preserved by order-isomorphisms of topologies. -Added in an edit. The reconstruction theorems for continuous maps and homeomorphisms on sober and $T_D$-spaces can be found in this book, which also contains Example 1s (see Example 3.1 in this book).<|endoftext|> -TITLE: Bound on probabilities of the sum of uniform order statistics -QUESTION [6 upvotes]: Let $X_1,...,X_n$ be i.i.d. random variable with a uniform distribution on [0,1]. Denote by $X_{(1)}\leq X_{(2)} \leq \ldots \leq X_{(n)}$ their order statistics. -Given $k\geq 1$ and $u\in[0,k]$, I want a simple formula for -$$ -p_k(u):=\mathbb{P}[X_{(1)}+\ldots + X_{(k)}\leq u], -$$ -or at least a simple lower bound on $p_k(u)$. (By simple, I mean that I don't want a k-dimensional integral...) -Note: -Of course, the case $k=1$ is easy: -$$p_1(u) = 1-(1-u)^n.$$ -And, by using the representation -$$U_{(1)},U_{(2)} = 1-Y^{1/n},1-Y^{1/n}Z^{1/(n-1)},$$ -where $X,Y$ are iid $\sim U([0,1])$, -I was able to compute $p_2(u)$ (with the help of Maple): -$$p_2(u) = 1-2(1-\frac{u}{2})^n+ \big(\max(1-u,0)\big)^n$$ - -REPLY [5 votes]: As pointed out by user63957, the conditional distribution of $T_k:=X_{(1)}+\ldots + X_{(k)}$ given $X_{(k+1)}=x$ is that of $xS_k$, where $S_k:=U_1+\dots+U_k$ and the $U_i$'s are iid random variables (r.v.'s) uniformly distributed on $[0,1]$. So, we have the key relation -$$T_k\overset D=X_{(k+1)}S_k,$$ -assuming that the $U_i$'s and hence $S_k$ are independent of $X_{(k+1)}$, where $\overset D=$ denotes the equality in distribution. So, the distribution of $T_k$ is the product-convolution of the distributions of $X_{(k+1)}$ and $S_k$. -The cdf of $S_k$ is given by the Irwin--Hall formula: -\begin{equation} - P(S_k\le x)=\frac1{k!}\sum_{j=0}^k(-1)^j\binom kj(x-j)_+^k -\end{equation} -for real $x$, where $z_+:=0\vee z$ and $z_+^k:=(z_+)^k$. -The distribution of $X_{(k+1)}$ is the beta one with parameters $k+1,n-k$. So, for $t>0$ we have -\begin{align*} - P(T_k\le t)&=\int_0^1 P(S_k\le t/y)P(X_{(k+1)}\in dy) \\ - &=\frac1{k!}\sum_{j=0}^k(-1)^j\binom kj \int_0^1 (t/y-j)_+^k\frac{n!}{k!(n-k-1)!}y^k(1-y)^{n-k-1}dy \\ - &=\frac{n!}{(k!)^2(n-k-1)!}\sum_{j=0}^k(-1)^j\binom kj J_{n,k,j}\big(t,(1-t/j)_+\big), \tag{1} -\end{align*} -where $(1-t/0)_+:=0$ and (using the substitution $y=1-x$) -\begin{align*} - J_{n,k,j}(t,u)&:=\int_u^1 (t-j+jx)^k x^{n-k-1}dx - =\sum_{r=0}^k\binom kr(t-j)^{k-r}j^r \int_u^1 x^{r+n-k-1}dx \\ - &=\sum_{r=0}^k\binom kr(t-j)^{k-r}j^r \frac{1-u^{r+n-k}}{r+n-k}.\tag{2} -\end{align*} -Thus, by formulas (1)--(2), the cdf of $T_k$ is expressed as a double sum of products of powers and binomial coefficients. In particular, for $k=1,2$ this yields the expressions obtained by the OP. -One can also obtain some asymptotics, as follows. -Suppose that $k\wedge(n-k)\to\infty$. Then both $X_{(k+1)}$ and $S_k$ are asymptotically normal, and we have -\begin{align*} -T_k\overset D=X_{(k+1)}S_k&=\Big(\frac kn+(Z_1+o_P(1))\sqrt{\frac{k(n-k)}{n^3}}\Big) -\Big(\frac k2+(Z_2+o_P(1))\sqrt{\frac{k}{12}}\Big) \\ -&=\frac{k^2}{2n}+(Z_1+o_P(1))\frac k2\sqrt{\frac{k(n-k)}{n^3}} -+(Z_2+o_P(1))\frac kn\sqrt{\frac{k}{12}} \\ -&+(Z_1+o_P(1))(Z_2+o_P(1))O\Big(\frac kn\Big) \\ -&\overset D=\frac{k^2}{2n}+(Z+o_P(1))\frac{k\sqrt k}{2n}\sqrt{\frac13+\frac{n-k}n}, -\end{align*} -where $Z_1,Z_2,Z$ are iid standard normal r.v.'s, and $o_P(1)$ denotes any r.v.'s $Y_{n,k}$ that go to $0$ in probability as $k\wedge(n-k)\to\infty$. -We see that the distribution of $T_k=X_{(1)}+\ldots + X_{(k)}$ is concentrated near $\frac{k^2}{2n}$, which was easy to predict. -One can refine this asymptotics by using bounds on the errors of the normal approximations for $X_{(k+1)}$ and $S_k$.<|endoftext|> -TITLE: What are the uses of coefficient systems for arithmetic cohomology theories? -QUESTION [8 upvotes]: In topology when studying a space with non-trivial fundamental group it becomes important to consider homology and cohomology with coefficients in representations of the fundamental group, i.e. local systems. The general Serre spectral sequence for a fibration with non-simply connected base provides an example. Another use of local systems is to get a version of Poincaré duality for non-orientable manifolds. -I am wondering about similar instances when it comes to e.g. algebraic de Rham cohomology, where the coefficient systems are given by quasi-coherent sheaves with integrable connections. - -What does the ability to construct the de Rham complex with coefficients in a module with integrable connection buy us? Or for crystalline cohomology; what are some uses of the full theory of crystals (rather than just the obvious crystal given by the structure sheaf)? - -(note the related question: de Rham cohomology and flat vector bundles was more focused on analytic and topological situation. Also, the question: Coefficients of Weil Cohomology Theories focused on conceptualizations of these coefficient systems, whereas I am more interested in applications.) - -What are some of the uses of coefficient systems in algebraic geometry, especially in arithmetic contexts? - -I guess one still needs these coefficient systems to prove duality statements? -I also think that Deligne's proof of the Weil conjectures relied heavily on using constructible $l$-adic sheaves. The same goes for Kedlaya's proof of the Weil conjectures using rigid cohomology, with F-isocrystals as coefficients. What are some reasons why one needs such coefficient systems in these proofs? What are other instances where this extra flexibility is needed? -Thanks! - -REPLY [6 votes]: Well when you say: - -The general Serre spectral sequence for a fibration with non-simply connected base provides an example. - -you've already pretty much got it - except in algebraic geometry we usually use the Leray spectral sequence instead of the Serre spectral sequence. -A lot of the foundational proofs of etale cohomology (notably duality, as you state) are proved or can be proved by fibering your variety over a curve and using a Leray spectral sequence argument to reduce the statement in dimension $n$ to the statement in dimension $n-1$ plus a statement for curves, with coefficients in a local system. The first hypothesis can be handled by induction and the second by doing explicit calculations on curves. -Then of course these foundational theorems are used in the proof of the Weil conjectures. But that is probably not what you're thinking of. Deligne has two proofs of the Weil conjectures, there are later simplifications, and then of course the $p$-adic version. -However, as far as I know, the core of all these arguments is Deligne's version of Rankin's squaring trick. Fundamentally, we take some very simple observation - that for a local system with some positivity properties, the size of the largest Frobenius eigenvalue at any point is at most the position of the last pole in the zeta function, which is the size of the largest Frobenius coinvariant times q - and calculate what it implies about tensor powers of a fixed local system. -I think if you read Katz's very simple version of the proof for curves and hypersurfaces you will agree that this idea is essentially all that is needed to get the bound for some varieties, though of course more ideas are needed to get the result in full generality. -So you need local coefficient systems because you need to take high tensor powers of them to enhance this bound, which is always trivial if you try to apply it directly to bound something, into something nontrivial. -But one thing that Deligne's second proof of the Weil conjecture demonstrates, and a lesson that arithmetic geometers have taken it is that the Riemann hypothesis is really better understood as a statement about local systems the whole time. The "real" version of the statement is that the $i$th cohomology of proper variety with coefficients in a sheaf mixed of weight $\leq w$ is mixed of weight $\leq i+w$, and the original Weil conjecture is just a corollary.<|endoftext|> -TITLE: How does Riemann hypothesis implies estimates? -QUESTION [10 upvotes]: In Iwaniec, Luo and Sarnak article (precisely (4.23)), it is said that GRH for $L(s, \mathrm{sym}^2(f))$, for a holomorphic cusp newform $f$ of level $N$ and weight $k$, implies -$$\sum_{p \nmid N} \frac{\lambda_f(p^2) \log p}{p} \ll \log\log kN$$ -Why is that true? -It is not the first time that I use this "blackbox" of GRH giving bounds, is there a general intuition to have behind it, or a systematic formal justification (like using the explicit formula to relate zeroes and eigenvalues)? - -REPLY [13 votes]: We have that -\[-\frac{L'}{L}(s,\mathrm{sym}^2 f) = \sum_{n = 1}^{\infty} \frac{\Lambda_{\mathrm{sym}^2 f}(n)}{n^s},\] -where $\Lambda_{\mathrm{sym}^2 f}(n)$ is equal to $\lambda_f(p^2) \log p$ if $n = p$ with $p \nmid N$, is essentially a bounded multiple of $\log p$ if $n = p^k$, and vanishes otherwise. (There are some minor issues at $p \mid N$ that are no big deal). In particular, this shows that the value at $1$ of this sum is basically equal to the desired sum up to a negligible error term. -Now use Theorem 5.17 of Iwaniec and Kowalski, which states that for $s = \sigma + it$ with $1/2 < \sigma \leq 5/4$ and assuming RH for $L(s,\mathrm{sym}^2 f)$ (as well as the Ramanujan conjecture, which is known via the work of Deligne), -\[-\frac{L'}{L}(s,\mathrm{sym}^2 f) = \frac{r}{s - 1} + O\left(\frac{1}{2\sigma - 1} (\log \mathfrak{q}(\mathrm{sym}^2 f, s))^{2 - 2\sigma} + \log \log \mathfrak{q}(\mathrm{sym}^2 f, s)\right),\] -where $r$ is the order of the pole of $L(s,\mathrm{sym}^2 f)$ at $s = 1$ and $\mathfrak{q}(\mathrm{sym}^2 f, s)$ is the analytic conductor. Note that $N$ is squarefree and $f$ has trivial nebentypus, so that $r = 0$. Taking $s = \sigma = 1$ and noting that $\log \mathfrak{q}(\mathrm{sym}^2 f, s) \ll \log kN$ yields the result.<|endoftext|> -TITLE: Classification of unitary modular tensor categories (UMTCs) -QUESTION [12 upvotes]: Context/background: -I'm approaching this topic from the perspective of anyonic systems. -In the study of anyons, one works with fusion categories. Of course, for physicality, we demand that -i) The category is unitary (all $F$ and $R$ symbols are unitary) i.e. a UTC. -ii) The category is modular ($S$ matrix is unitary - this ensures nondegenerate braiding) i.e. an MTC. -Therefore, we are interested in classifying all UMTCs (which may play host to a physical theory of anyons in Nature). The advantage of working with MTCs is that demanding $S$ unitary provides a nice mechanism for discarding fusion rules that are unphysical (a technique that is used by Wang et al. in https://arxiv.org/abs/0712.1377). -Questions: -1) Given modular fusion rules $\mathcal{N}$, how do we count the number of corresponding UMTCs? -The obvious way seems to be to count the no. of distinct unitary solutions of $F$ and $R$ symbols (upto gauge freedom). However, it is conjectured that the $S$ and $T$ matrices uniquely define a UMTC, and so we might count the number of 'modular symbols' $(\mathcal{N}; S, T)$ corresponding to $\mathcal{N}$ - only this increases the count due to symmetries $(\mathcal{N}; -S, T)$, $(\mathcal{N}; S^{\dagger}, T^{\dagger})$ of the symbol. Which way is the more appropriate method for counting UMTCs w.r.t $\mathcal{N}$? The latter method seems stronger yet based on conjecture (though I suppose this is harmless in practice). -2) Are there any known examples of modular $\mathcal{N}$ with no unitary solutions for $F$ and $R$ symbols? -3) Are there examples of fusion rules $\mathcal{N}$ that yield UTCs which aren't modular? - -REPLY [12 votes]: Recently, Mignard and Schaunberg found a counter example to the conjecture (https://arxiv.org/abs/1708.02796). Counting $(S,T)$ pairs is still just a shorthand way of counting monoidal equivalence classes, which is not exactly the same as counting gauge classes (it's close though). Taking a note along that line though, it is known that for whatever fusion rules you want to study there exist polynomial invariants (https://arxiv.org/abs/1509.03275) which will get the job done at all levels from fusion to modular. -Eric Rowell showed (https://arxiv.org/abs/math/0503226) that the braided tensor categories coming from $\mathcal C(\mathfrak{so}_{2r+1},l,q)$ for $l$, $2(2r+1) -TITLE: Graded rings with compatible S_n actions -QUESTION [8 upvotes]: Does the following mathematical gadget have a standard name? Let $R$ be an $\mathbb{N}$-graded ring together with an $S_n$ action on each $R_n$ which are compatible in the following sense. Let $i:S_a \times S_b \rightarrow S_{a+b}$ be the standard inclusion. If $x$ is in grade $a$ with $\sigma \in S_a$ and $y$ is in grade $b$ with $\tau$ in $S_b$, then $$\sigma(a) \cdot \tau(b) = i(\sigma,\tau)(a\cdot b).$$ -The main example I have in mind is the ring of G-invariant tensors. -The point is to have the right language to say that such a gadget has certain generators and relations, where I think of the symmetric group actions as unary operations. - -REPLY [6 votes]: Steven Sam and Andrew Snowden and their other collaborators call these `twisted commutative algebras' and have been having fun writing papers about properties of generators and similar. -But predating this, any topologist of a certain sort would call this a monoid in the category of symmetric abelian groups. If you do the same construction in simplicial sets (or topological spaces), an example is the sphere spectrum, with nth space $S^n$. Modules over this object are what are called symmetric spectra, and serve as one of the main models of modern day stable homotopy theory. (This was an observation by Jeff Smith in the mid 1990s.) - -REPLY [4 votes]: As a warmup, an $\mathbb{N}$-graded ring is a monoid object in the symmetric monoidal category of $\mathbb{N}$-graded abelian groups under the convolution tensor product, which you can think of as Day convolution from the usual addition on $\mathbb{N}$. -Similarly, this thing is a monoid object in the symmetric monoidal category of species in abelian groups (presheaves on the category $S$ of finite sets and bijections valued in abelian groups) under the convolution tensor product, which you can again think of as Day convolution from disjoint union on $S$. -So I might be inclined to call such a thing an $S$-graded ring, and you can feel free to replace $S$ with your favorite name for $S$, maybe $\text{FinSet}^{\times}$ or something.<|endoftext|> -TITLE: A quantity associated with a smooth groupoid -QUESTION [10 upvotes]: Assume that $(G,G^0,r,s)$ is a smooth groupoid such that $G$ is a compact connected manifold. -The graph of "source" and "range" maps $s, r: G \to G^0$ are compact submanifolds $S$ and $R$ of $G\times G$. To our smooth groupoid we associate the quantity -$$q= S\ \#\ R,$$ -the intersection number of $S$ with $R$. -This quantity vanishes for the particular case that $G$ is a Lie group and $G^0=\{e\}$, the neutral element. -On the other hand, for a given compact manifold $M$, this quantity for the trivial groupoid structure $(M,M, Id, Id)$ is equal to the Euler characteristic of $M$. Hence this quantity may be non zero. -But I search for some nontrivial smooth groupoids for which this quantity is non-zero. What are some examples of this situation? -Furthermore can this quantity be realised and be reintroduced via some quite algebraic formulation that would be defined for more abstract (not necessarily smooth) groupoids? -In the other word, and with some abuse of terminology, how can one "Noncommutativize" the above quantity "q"? - -REPLY [2 votes]: Here are a few more examples and thoughts. - -first of all, it seems to me that it does not necessarily vanish for the particular case of a Lie group $G$, unless $G$ is assumed to be connected. Indeed, in the case of a finite group $G$ I think that the intersecion number is the cardinal of $G$ (unless I clearly misunderstood something). -Then one can consider the pair groupoid $G_1=M\times M$, $G_0=M$, $s$ and $r$ are the two projections, and $e$ is the diagonal embedding. Then it seems to me that the intersection number is again the Euler characteristic of $M$. -Consider the action groupoid of a finite group $G$ on $M$: $G_1=G\times M$, $G_0=M$, $s(g,m)=m$ and $r(g,m)=g.m$. You might then get the sum over $g\in G$ of the Euler characteristic of $M^g$. Note that if $M=pt$ then one recovers Example 1 above. If $G$ is trivial one then recovers your example of $(M,M,id,Id)$. - -Concerning your last question about non-smooth issues, I would recommend to look at derived intersections instead (this is somehow already what you've already done by saying that the intersection number of the self-intersection of the diagonal is the Euler characteristic). -As for the meaning of this quantity, I would first check if this is a Morita invariant one (i.e. it is an invariant associated with the underlying quotient stack of the groupoid). I would say it's not, unless I've made a stupid mistake with Example 2. Indeed, the quotient stack of the pair groupoid of $M$ is a point... and the Euler characteristic of $M$ doesn't seem to be an invariant of the point.<|endoftext|> -TITLE: General term formulas for nonlinear recurrence sequences -QUESTION [5 upvotes]: It seems to be a well known question: in which cases will there be general term formulas for sequences like $p_n=a p_{n-1} ^2 +b p_{n-1} +c$ where $a, b, c$ are real or complex numbers and n is greater or equal to 1? I think there are something called Julia theorem (maybe I spell it wrong) can explain this, but I never studied dynamic systems, so can anyone tell me something about that? - -REPLY [6 votes]: Polynomial maps $f(z)$ for which there is a general formula for the $n$-th iterate -are called integrable. Besides polynomials of degree $1$, there are two types of them: a) those which are conjugate (by an affine map) with $z^d$ and b) those which are conjugate with $T_d$, where $T_d$ is the Chebyshev -polynomial, defined by -$$\cos dx=T_d(\cos x).$$ -For example $f(z)=2z^2-1$, the general formula for the $n$-th iterate -is $\cos(2^n\arccos z)$. -The result can be extended to rational functions of one variable. -Then we have the third class c) consisting of Lattes functions $L_m$, defined by -equations like -$$\wp(mx)=L_m(\wp(x)).$$ -(These are not all. For a complete list of Lattes functions, see -he paper by -A. Douady and J. Hubbard, A proof of Thurston’s topological characterization of rational functions. Zbl 0806.30027 -Acta Math. 171, No.2, 263-297 (1993).) -This is not a theorem, because it was never rigorously defined what an "explicit formula" is, but this is a well-extablished "fact". (Instead a map is usually called "integrable" if here is a "non-trivial" family of other maps of the same class which commute with it, see, for example -A. P. Veselov -What is an integrable mapping? Zbl 0733.58025 -What is integrability, Springer Ser. Nonlinear Dyn., 251-272 (1991). -Remark. In fact "a formula" for the $n$-th iterate exists for every polynomial, -namely $f^{*n}=\phi^{-1}\circ(\phi^{d^n})$, where $f^{*n}$ is he $n$-th iterate, -and $d$ is the degree of $f$. Here $\phi$ is the so-called Boettcher function -of $f$. But Boettcher function is "explicit" exactly in the examples of polynomials listed above as a), b). -Remark 2. The result for polynomials is indeed due to P. Fatou and G. Julia (independently) and the result for rational functions to J. F. Ritt.<|endoftext|> -TITLE: Why are free objects "free"? -QUESTION [16 upvotes]: What is the origin/motivation for the adjective "free" in the term "free object"? -Does it refer to them coming "for free" (as being constructed from a set in a straight-forward manner) or does it refer some type of freedom they enjoy? - -REPLY [34 votes]: Jakob Nielsen's 1921 paper in Danish, with the title "Om Regning med ikke-kommutative Factorer og den Anvendelse i Gruppetoerien", is where he proved that subgroups of free groups are free. I don't know about the original Danish, but the English translation by Anne W. Neumann is available in Nielsen's collected works, with translated title "On calculation with Non-commutative Factors and its Application to Group Theory", and it contains the following sentences - -In our case we will call it the 'free group $G_n$' generated by the generators $a_1,...,a_n$ and will denote it $[a_1,...,a_n]$. -... -In $G_n$, two distinct products of the $a_i$, irreducible in the $a_i$, always describe two different elements. It is for this reason that $G_n$ is called a free group. - -One finds free groups described but not named in earlier works, such as Dehn's 1912 paper "Über unendliche discontinuierliche Gruppen", with English translation by John Stilwell titled "On infinite discontinuous groups" in Dehn's collected works: - -Consider, say, the group which is given by two generators $S_1$ and $S_2$ without defining relations... - -There is a telling passage later in this same paper. Dehn's focus in this paper is to understand groups in which there is a presentation with a single relator, and every generator appears exactly twice in that relator (he proves that this characterizes surface groups). Along the way he considers variations on this property, such as groups in which "each generator appears in the relation either twice or not at all", and then he introduces some terminology: - -... we arrive at the following normal form of the group: among the generators are $w$ generators which appear in no relation ("free generators")...<|endoftext|> -TITLE: Freely adding finite limits preserves some colimits? -QUESTION [6 upvotes]: Let $\mathcal{K}$ be a category and $\mathcal{K}_{\text{fin}}$ its free completion with finite limits. - - -Does the embedding $\mathcal{K} \hookrightarrow \mathcal{K}_{\text{fin}}$ preserve some colimits? - - -I am especially interested in directed colimits. - -REPLY [8 votes]: Yes, all the colimits that exist in ${\cal K}$. Indeed, ${\cal K}_{{\rm fin}}$ can be identified with the smallest full subcategory of ${\rm Fun}({\cal K},{\rm Set})^{{\rm op}}$ which contains the representable functors ${\rm Hom}(x,-)$ and is closed under finite limits, and the composition ${\cal K} \to {\cal K}_{{\rm fin}} \hookrightarrow {\rm Fun}({\cal K},{\rm Set})^{{\rm op}}$ is the opposite of the Yoneda embedding ${\cal K}^{{\rm op}} \to {\rm Fun}({\cal K},{\rm Set})$. The result you need now follows from the fact that the Yoneda embedding preserves all limits which exist in ${\cal K}^{{\rm op}}$.<|endoftext|> -TITLE: Derivative of the flow for ODEs on manifolds -QUESTION [12 upvotes]: Let $\mathbf V \colon [0,T] \times \mathbb R^d \to \mathbb R^d$ (for $T>0$) be a given, bounded smooth vector field and let $\mathbf X=\mathbf X(t,x)$ be its flow, i.e. the unique solution to the initial-value problem -\begin{equation} -\begin{cases} -\frac{\partial}{\partial t} \mathbf X(t,x) = \mathbf V(t,\mathbf X(t,x)) & \text{ in } (0,T) \times \mathbb R^d \\ -\mathbf X(0,x) = x \quad \text{ for all } x \in \mathbb R^d.   -\end{cases} -\end{equation} -A well-known result in standard ODE's theory says that -$$\tag{1} -\nabla_x \mathbf X(t,x) = \exp\bigg( \int_0^t \nabla \mathbf V(s,\mathbf X(s,x))\,ds\bigg). -$$ - -Is there an analogous formula to (1) for ODEs driven by (smooth) vector fields on Riemannian manifolds? In particular, does this formula involve somehow the geometry of the Riemannian manifold? - A rather precise question could be: consider the $C^1$ norm of $\mathbf X$ (or even its Lipschitz constant) w.r.t. space variable $x$: does it depend on some known tensors on the manifold (e.g. curvature)? - -I have gone through books in differential geometry/differential topology (e.g. Lee, Lang) and they prove that $\mathbf X$ is smooth but do not compute explicitly the derivative. -References are very much welcome. Thanks. - -REPLY [3 votes]: I think you may be interested in this remarkable, recent paper by E. Brué and D. Semola (see in particular Theorem 3.11 which answers to your question in a much more general setting).<|endoftext|> -TITLE: $\zeta^{(k)}(s) < 0$ for $s\in (0,1)$ -QUESTION [12 upvotes]: A bit of plotting suggests that $\zeta^{(k)}(s) < 0$ for all $s\in (0,1)$ and all integers $k\geq 0$. (Or, what is the same: $\zeta^{(k)}(s)$ has no zeroes on $(0,1)$.) Is there a brief, clean proof of this apparent fact (and/or a reference for it)? - -REPLY [14 votes]: The coefficients computed in the comments appear to imply that the Taylor expansion at $s=0$ of $\zeta(s)+\frac1{1-s}-\frac12$ has very small coefficients, which would imply the result. -Following section 2.1 in Titchmarsh Theory of the Riemann zeta function, -by integration/summation by parts (or one step of Euler-Maclaurin), -$$\zeta(s)=\frac1{s-1}+\frac12+s\int_1^{\infty}\frac{1/2-\{x\}}{x^{s+1}}dx$$ -absolutely convergent for $\Re(s)>0$ where $\{x\}$ denotes the fractional part of $x$. -Integrating by parts again (or two steps of Euler-Maclaurin), -$$\zeta(s)=\frac1{s-1}+\frac12+\frac{s}{12}-(s+1)\int_1^{\infty}(\{x\}^2-\{x\}+\frac16)x^{-s-2}dx$$ -Consider the integrand $(\{x\}^2-\{x\}+\frac16)x^{-s-2}dx$ as a function of $s$. Its Taylor coefficients are alternating in sign and dominated by the coefficients of $\frac16x^{-s-2}$, so the Taylor coefficients of the absolutely convergent integral are dominated by those of -$$\int_1^{\infty}\frac16x^{-s-2}dx = \frac1{6(s+1)}$$ -and the result follows.<|endoftext|> -TITLE: What are the minimal local models for Riemannian manifolds? A local question about isometric embeddings -QUESTION [5 upvotes]: There are many results about isometric embeddings of Riemannian manifolds but I haven't been able to find one that quite answers this question (which I believe must have some kind answer in the literature). - -Question: For $n \in \mathbb{N}$ what is the minimal integer $r$ (depending on $n$) such that every point in every $n$-dimensional Riemannian manifold $M$ has an open neighborhood which admits a smooth (resp. analytic) locally closed isometric embedding into $\mathbb{R}^r$? - -REPLY [6 votes]: The Nash-Kuiper theorem says that for $C^1$ isometric embeddings, $r$ just needs to be $n+1$, so in what follows smooth means $C^2$ or better. -These results are all in chapter 1 of Han and Hong's book "Isometric Embedding of Riemannian Manifolds in Euclidean Spaces," which is a great source for these sorts of results. -Let $s_n=\binom{n+1}{2}=\frac{n(n+1)}{2}$. -Theorem 1.1.6 (Janet-Cartan) states: - -Any $n$-dimensional analytic Riemannian manifold admits a local analytic isometric embedding in $\mathbb{R}^{s_n}$. - -This is proved using the Cauchy-Kowalevski theorem for analytic solutions of PDE in section 1.1. -Note that this is in some sense best possible since $s_n$ is the number of equations in the system $\sum_{k=1}^q\partial_i u^k\partial_j u^k = g_{ij}$ where $g_{ij}$ are the components of the metric on $M^n$ in some local coordinate system and $u:M^n\rightarrow\mathbb{R}^q$ is the embedding. -Theorem 1.2.4 (Gromov and Rokhlin, Greene) implies: - -Any smooth $n$-dimensional Riemannian manifold admits a smooth local isometric embedding in $\mathbb{R}^{s_n+n}$. - -It seems that it is an open question whether $r$ can be reduced to $s_n$ for smooth local isometric embeddings. In the notes to chapter 1 of their book, Han and Hong state that this is open even for $n=2$ and cite a 1982 list of problems by Yau. -Note that for some $n$, there are better results; e.g. when $n=2$, $r$ may be taken to be 4.<|endoftext|> -TITLE: Must any continuous odd map from $\mathbb{S}^2$ to $\mathbb{R}$ have a path of zeros between antipodal points? -QUESTION [15 upvotes]: Let $f : S^2 \to \mathbb{R}$ be a continuous map such that $f(-x) = -f(x)$. Consider the set $Z = f^{-1}(0)$. Must $Z$ contain some path from some point to its antipode? Indeed, must $Z$ contain a continuous loop intersecting each "meridian", passing through antipodal points on antipodal meridians? -[I see this often asserted as a step in intuitive arguments for the Borsuk-Ulam theorem, but do not see why rigorously it must be so. Indeed, as a child, I once attempted precisely this argument for the Borsuk-Ulam theorem in a math camp, and was chided for asserting this unsupported lemma; thus it has stuck with me always. Certainly, $Z$ must contain a point on each meridian, but that these points must line up into a continuous path is not obvious to me.] -Update: My original question turned out to have already been answered at How bogus is the glitzy proof of Borsuk-Ulam?, as well as by Loïc Teyssier in the same way below, but a new question with same motivation is: must Z contain two antipodal points in the same connected component? (If so, then the Borsuk-Ulam proof via this approach can still be salvaged; if not, we may write it off as a lost cause) - -REPLY [3 votes]: The answer to the connected component version is yes, though the only proof I can come up with seems like more trouble than it would've been worth in math camp homework! Two steps: - -If there there are finitely many components of $S^2 \smallsetminus Z$, produce a finite-diameter bipartite tree on which the antipodal map on $S^2$ induces an involution fixing a point. -Approximate any $f$ by functions satsifying step 1. - -Most of the work seems to be a part of Step 1 that vaguely sounded like the Jordan curve theorem: - -Key Fact. Let $X$ be a connected, compact, locally path-connected metric space with zero first homology. For an open subset $U$, inclusion induces a bijection between the connected components of $\partial U = \bar{U} \smallsetminus U$ and those of $X \smallsetminus U$. - -Proof of Key Fact: Mayer-Vietoris and some point-set debris -Given $\epsilon > 0$, let $U_\epsilon$ be the set of points less than $\epsilon$ away from the closure $\bar{U}$ of $U$; and let $V_\epsilon$ be the set less than $\epsilon$ away from $X \smallsetminus U$. The Mayer-Vietoris sequence in reduced homology ends with -$$ H_1(X) \to \tilde{H}_0(U_\epsilon \cap V_\epsilon) \to \tilde{H}_0(U_\epsilon) \oplus \tilde{H}_0(V_\epsilon) \to \tilde{H}_0(X) , $$ -which gives a bijection between the path components of $U_\epsilon \cap V_\epsilon$ and those of $V_\epsilon$. Pass to connected components of their closures using: - -A connected open set $W$ in a locally path-connected space is path-connected. -Proof. Each point of $W$ is has a path-connected open neighborhood in $W$; so the path components form a partition of $W$ by open sets. -The closure of a connected open set $W$ is connected. -Proof. In a partition of $\bar{W}$ by clopen subsets, whichever one contains $W$ contains $\bar{W}$. - -Then send $\epsilon$ to $0$. That is, since -$$ -\begin{align*} -X \smallsetminus U &= \bigcap_{\epsilon > 0} \bar{V}_\epsilon & &\text{and} & -\partial U &= \bigcap_{\epsilon > 0} \overline{U_\epsilon \cap V_\epsilon} , -\end{align*} -$$ -the Key Fact is proven if we can biject connected components of $X \smallsetminus U$ with nested sequences of components $C_n$ of $\bar{V}_{1/n}$ (and similarly for $\partial U$): - -Lemma. In a compact space $X$, the intersection $C$ of nested closed connected sets $C_1 \supseteq C_2 \supseteq \cdots$ is connected. - -Proof. Given disjoint open subsets $U$ and $V$ of $X$ that cover $C$, their union and the sequence $X \smallsetminus C_n$ form an open cover of $X$—which has a finite subcover, so some $C_n$ lies in $U \cup V$. Since $C_n$ is connected, it lies entirely in one of $U$ or $V$; and so does $C$. -Step 1: Turn connectivity data into a tree - -Claim. Let $Z$ be the zero locus of a continuous odd $f: S^2 \to \mathbb{R}^2$. If $S^2 \smallsetminus Z$ has finitely many connected components, then the antipodal map on $S^2$ sends some component of $Z$ to itself. - -Let $G$ be the graph where: - -Vertices are connected components of $Z$ and $S^2 \smallsetminus Z$, the sets of which we respectively denote $V_0$ and $V_{\neq 0}$. -Two vertices are joined by an edge if their union is a connected set in $S^2$—equivalently, if $K \in V_0$ meets the closure of $U \in V_{\neq 0}$. - -$G$ is bipartite with parts $V_0$ and $V_{\neq 0}$, and the assumption of Step 1 is that $V_{\neq 0}$ is finite; so—since $S^2$ is connected—$G$ is connected with finite diameter. By the Key Fact, every vertex in $V_{\neq 0}$ is a cut vertex, which makes $G$ a tree. -Every involution of a finite-diameter tree preserves either an edge or a vertex; and since the antipodal map can't exchange the parts of $G$ or fix any member of $V_{\neq 0}$, it has to fix exactly one vertex in $V_0$. This is the connected component of $Z$ we're after. -Step 2: Infinitely many components in $S^2 \smallsetminus Z$ is okay -Approximate $f$ by a sequence $f_n$ where $f_n$ is equal to $f$ on the connected components of $\{f \neq 0\}$ admitting an open ball of radius $\pi/n$ and zero everywhere else. -The area of an open ball of radius $\pi/n$ on $S^2$ is at least $4\pi/n^2$ (the area of a sphere of circumference $2\pi/n$). So each $f_n$ has no more than $n^2$ components to its nonzero locus; and running them through Step 1 yields a sequence of nested closed connected sets $C_1 \supseteq C_2 \supseteq \cdots$ preserved by the antipodal map. -Then the intersection of the sets $C_n$ is preserved by the antipodal map, in $Z$ (every component of $S^2 \smallsetminus Z$, being open, contains an open ball), and connected (by the Lemma before Step 1).<|endoftext|> -TITLE: Set theoretical multiverse and truths -QUESTION [22 upvotes]: In J.D. Hamkins' multiverse view of set theory, every universe has an ill-founded $\mathbb{N}$ from the perspective of another universe. Does this mean that every proof in our universe can be seen as a nonstandard length proof from the perspective of some other universe, so that there is no real "truth" in the Multiverse? -Could this happen even for a simple concrete proof, like the fundamental theorem of arithmetic? Or are some some proofs standard-length in all universes? - -REPLY [28 votes]: Thank you for your interest in my views on the set-theoretic -multiverse. -Yes, indeed, the well-foundedness mirage axiom you mention is -probably the most controversial of my multiverse axioms, and so -allow me to explain a little about it. -The axiom expresses in a strong way the idea that we don't actually -have a foundationally robust absolute concept of the finite in -mathematics. Specifically, the axiom asserts that every universe of -set theory is ill-founded even in its natural numbers from the -perspective of another, better universe. Thus, every set-theoretic -background in which we might seek to undertake our mathematical -activity is nonstandard with respect to another universe. -My intention in posing the axiom so provocatively was to point out -what I believe is the unsatisfactory nature of our philosophical -account of the finite. -You might be interested in the brief essay I wrote on the topic, A -question for the mathematics -oracle, -published in the proceedings of the Singapore workshop on Infinity -and Truth. For an interesting and entertaining interlude, the -workshop organizers had requested that everyone at the workshop -pose a specific question that might be asked of an all-knowing -mathematical oracle, who would truthfully answer. My question was -whether in mathematics we really do have a absolute concept of the -finite. -To explain a bit more, the naive view of the natural numbers in -mathematics is that they are the numbers, $0$, $1$, $2$, and so -on. The natural numbers, with all the usual arithmetic structure, -are taken by many to have a definite absolute nature; arithmetic -truth assertions are taken to have a definite absolute nature, in -comparison for example with the comparatively less sure footing of -set-theoretic truth assertions. -To be sure, many mathematicians and philosophers have proposed a -demarcation between arithmetic and analysis, where the claims of -number theory and arithmetic are said to have a definite absolute -nature, while the assertions of higher levels of set theory, -beginning with claims about the set of sets of natural numbers, are -less definite. Nik Weaver, for example, has suggested that -classical logic is appropriate for the arithmetic realm and -intuitionistic logic for the latter realm, and a similar position -is advocated by Solomon Feferman and others. -But what exactly does this phrase, "and so on" really mean in the -naive account of the finite? It seems truly to be doing all the -work, and I find it basically inadequate to the task. The situation -is more subtle and problematic than seems to me to be typically -acknowledged. Why do people find their conception of the finite to -be so clear and absolute? It seems hopelessly vague to me. -Of course, within the axiomatic system of ZFC or other systems, we -have a clear definition of what it means to be finite. The issue is -not that, but rather the extent to which these internal accounts of -finiteness agree with the naive pre-reflective accounts of the -finite as used in the meta-theory. -Some mathematicians point to the various categoricity arguments as -an explanation of why it is meaningful to speak of the natural -numbers as a definite mathematical structure. Dedekind proved, -after all, that there is up to isomorphism only one model -$\langle\mathbb{N},S,0\rangle$ of the second-order Peano axioms, -where $0$ is not a successor, the successor function $S$ is -one-to-one, and $\mathbb{N}$ is the unique subset of $\mathbb{N}$ -containing $0$ and closed under successor. -But to my way of thinking, this categoricity argument merely pushes -off the problem from arithmetic to set theory, basing the -absoluteness of arithmetic on the absoluteness of the concept of an -arbitrary set of natural numbers. But how does that give one any -confidence? -We already know very well, after all, about failures of -absoluteness in set theory. Different models of set theory can -disagree about whether the continuum hypothesis holds, whether the -axiom of choice holds, and so with innumerable examples of -non-absoluteness. Different models of set theory can disagree on -their natural number structures, and even when they agree on their -natural numbers, they can still disagree on their theories of -arithmetic truth (see Satisfaction is not -absolute). -So we know all about how mathematical truth assertions can seem to -be non-absolute in set theory. -Skolem pointed out that there are models of set theory $M_1$, $M_2$ -and $M_3$ with a set $A$ in common, such that $M_1$ thinks $A$ is -finite; $M_2$ thinks $A$ is countably infinite and $M_3$ thinks $A$ -is uncountable. For example, let $M_3$ be any countable model of -set theory, and let $M_1$ be an ultrapower by a ultrafilter on -$\mathbb{N}$ in $M_3$, and let $A$ be a nonstandard natural number -of $M_1$. So $M_1$ thinks $A$ is finite, but $M_3$ thinks $A$ has -size continuum. If $M_2$ is a forcing extension of $M_3$, we can -arrange that $A$ is countably infinite in $M_2$. -No amount of set-theoretic information in our set-theoretic -background could ever establish that our current conception of the -natural numbers, whatever it is, is the truly standard one, since -whatever we assert to be true is also true in some nonstandard -models, whose natural numbers are not standard. -The well-foundedness mirage axiom asserts that this phenomenon is -universal: all universes are wrong about well-foundedness. -In defense of the mirage axiom, let me point out that whatever attitude toward it one might harbor, nevertheless the axiom cannot be seen as incoherent or inconsistent, because Victoria Gitman and I have proved that all of my multiverse axioms are true in the multiverse consisting of the countable computably saturated models of ZFC. So the axiom is neither contradictory nor incoherent. See A natural model of the multiverse axioms. -I have discussed my multiverse views in several papers. - - -Hamkins, Joel David, The set-theoretic multiverse, Rev. Symb. Log. 5, No. 3, 416-449 (2012). Doi:10.1017/S1755020311000359, ZBL1260.03103. -Hamkins, Joel David, A multiverse perspective on the axiom of constructibility, Chong, Chitat (ed.) et al., Infinity and truth. Based on talks given at the workshop, Singapore, July 25--29, 2011. Hackensack, NJ: World Scientific (ISBN 978-981-4571-03-6/hbk; 978-981-4571-05-0/ebook). Lecture Notes Series. Institute for Mathematical Sciences. National University of Singapore 25, 25-45 (2014). DOI:10.1142/9789814571043_0002, - ZBL1321.03061. -Gitman, Victoria; Hamkins, Joel David, A natural model of the multiverse axioms, Notre Dame J. Formal Logic 51, No. 4, 475-484 (2010). DOI:10.1215/00294527-2010-030, ZBL1214.03035. -Hamkins, Joel David; Yang, Ruizhi, Satisfaction is not absolute, to appear in the Review of Symbolic Logic. - - -But finally, to address your specific question. Of course, there -are specific finite numbers that will be finite with respect to any -alternative set-theoretic background. As Michael Greinecker points -out in the comments, the number 35253586543 has that value -regardless of your meta-mathematical position. So of course, there -are many proofs that are standard finite with respect to any of the -alternative foundations. -Meanwhile, I find it very interesting to consider the situation -where different foundational systems disagree on what is provable. -In very recent work of mine, for example, we are looking at the -theory of set-theoretic and arithmetic potentialism, where -different foundational systems disagree on what is true or -provable. -For example, recently with Hugh Woodin, I have proved that there is -a universal finite set $\{x\mid\varphi(x)\}$, a set that ZFC proves -is finite, and which is empty in any transitive model of set -theory, but if the set is $y$ in some countable model of set theory -$M$ and $z$ is any finite set in $M$ with $y\subset z$, then there -is a top-extension of $M$ to a model $N$ inside of which the set is -exactly $z$. The key to the proof is playing with the non-absolute -nature of truth between $M$ and its various top-extensions.<|endoftext|> -TITLE: Characterizing the Plancherel measure -QUESTION [5 upvotes]: If $A$ is an abelian locally compact group, the Plancherel measure on $\hat A$ is a Haar measure, so, up to scaling it is the unique invariant Radon measure. -Now for a nonabelian locally compact group $G$. Can the Plancherel measure on the unitary dual $\hat G$ be characterized by any (invariance) property? -In a way that it is the unique (up to scaling) measure with property XYZ, where XYZ should not be the property that defines the Plancherel measure. -Just to make it clear: I do know the definition of the Plancherel measure. I am looking for a different characterization. - -REPLY [7 votes]: If $G$ is a unimodular second countable Type I group, then the Plancherel measure is the unique measure $\mu$ such that -$$\|f\|_2^2 = \int_{\widehat{G}} \|\pi(f)\|_{\mathrm{HS}}^2 \mathrm{d}\mu(\pi).$$ -for every $f \in \mathrm{L}^1(G) \cap \mathrm{L}^2(G)$. This appears as Theorem 18.8.2 in Dixmier's book on $C^*$-algebras. -When $G$ is not unimodular, the question becomes more complicated, because the Plancherel measure needs to be twisted by a section of a line bundle; see the paper of Duflo-Moore on the subject for the gory details. When $G$ is not second countable, I do not know of a published result; the technical details of direct integral theory are more difficult in this case and not standard. When $G$ is not Type I, the decomposition of the left regular representation into irreducibles is no longer unique, and some of the operators on the right-hand side of the formula will fail to have finite Hilbert-Schmidt norm. -Edit: The question has been clarified to indicate that it is not asking for this characterization of the Plancherel measure, but rather a characterization that is closer to the description in the abelian case of the Plancherel measure as a Haar measure on $\widehat{G}$. Since there is usually no algebraic structure on $\widehat{G}$ for which to consider invariance, a direct generalization is impossible. -The closest analogue is the characterization of the Plancherel measure as a unique co-invariant trace (or weight) on the von Neumann algebra $\mathcal{M}$ generated by the left-regular representation of $G$. Suppose $G$ satisfies the same hypotheses as above and $\Delta : \mathcal{M} \to \mathcal{M} \overline{\otimes} \mathcal{M}$ is the comultiplication on $\mathcal{M}$ given by $\lambda(s) \mapsto \lambda(s) \otimes \lambda(s)$. Then the Plancherel trace is the unique normal semifinite trace $\tau$ on $\mathcal{M}$ such that -$$\tau((\varphi \otimes \mathrm{id}) (\Delta(a))) = \tau(a)$$ -for all $a \in \mathcal{M}_\tau^+$ and $\varphi \in \mathcal{M}_*$. A similar characterization holds for the Plancherel weight of an arbitrary locally compact group, or for the Haar weight of a locally compact quantum group. For proofs, see volume 2 of Takesaki or any of the literature on von Neumann algebraic quantum groups.<|endoftext|> -TITLE: Reference on Highest Weight Module of Kac-Moody Algebra -QUESTION [6 upvotes]: I am trying to understand this paper. The construction requires the understanding of the following concepts in the representation theory of simple and affine Lie algebras: - -The construction of Verma module for a general (not necessarily integral) highest-weight state; -The character for these modules (the Weyl-Kac character formula cannot be applied for a generic non-integral highest-weight module); -BRST reduction of affine Lie algebras; -Quantum Drinfeld-Sokolov reduction; -... - -I am looking for some references that explain these concepts or some detailed examples of the construction for simplest cases. I appreciate any comment. - -REPLY [4 votes]: I highly recommend the book - Affine Lie Algebras, Weight Multiplicities, and Branching Rules -by Kass, Moody, Patera, and Slansky. -https://www.ucpress.edu/op.php?isbn=9780520067684<|endoftext|> -TITLE: Presentation of the covariant power set functor -QUESTION [8 upvotes]: How to write the covariant power set functor (restricted to finite sets for simplicity) -$$P : \mathsf{FinSet} \to \mathsf{Set}$$ -concisely as a colimt of representable functors? There is an epimorphism $$\coprod_{n \geq 0} \hom(\{1,\dotsc,n\},-) \to P,$$ -mapping $f \in \hom(\{1,\dotsc,n\},X)$ to $\mathrm{im}(f) \in P(X)$. This already provides a generating set of $P$. Compare this with the contravariant power set functor, which is already a representable functor. -This question is motivated by the exercise to find all morphisms of functors $P \to P$ without too much calculations. - -REPLY [9 votes]: It helps a lot that in this case $\hom(\{1,...,n\},-)\cong\hom(\{1\},-)^n$. Denote $\hom(\{1\},-)$ by $X$ and take more generally in any category with finite products and countable colimits distributing over each other the free monoid on $X$, that is, $1\sqcup X\sqcup X^2\sqcup X^3\sqcup\cdots$. Now in addition factor out by the following: identify, for any surjection $\pi:\{1,...,m\}\twoheadrightarrow\{1,...,n\}$, the summand $X^n$ with its image under $X^\pi:X^n\rightarrowtail X^m$ (note that in particular this forces quotienting $X^n$ by the action of the $n$th symmetric group, so we get the free commutative monoid on $X$ as an intermediate step). -I believe what we get is the free internal semilattice on $X$, and for $X=\hom(\{1\},-)$ one obtains the covariant powerset. This answers the question since the diagram we start from consists of representables since, as already said, $X^n$ is isomorphic to $\hom(\{1,...,n\},-)$. -As HeinrichD notes in the comment below, we probably only need quotients by $X^k\times\text{switch}\times X^l:X^k\times X^2\times X^l\to X^k\times X^2\times X^l$ and by $X^n\times\text{diagonal}:X^n\times X\to X^n\times X^2$, although I fail to organize this into a honest diagram for the moment. -In any case, seems like the resulting description of natural transformations $P\to F$ is as follows: they are in one-to-one correspondence with families -$$ -\left(\xi_n\right)_{n=0,1,2,...}\in\prod_{n=0}^\infty F(n)^{\Sigma_n} -$$ -satisfying -$$ -F(\pi)(\xi_{n+1})=\xi_n -$$ -for all $n>0$, where $\pi:\{1,...,n+1\}\to\{1,...,n\}$ is given by $1\mapsto1,...,n\mapsto n,n+1\mapsto n$. -Later: -Here is a sketch of an alternative calculation for the above $\hom$, using the fact that $P\cong i_!(1)$, where $i:\mathsf{Finepi}\hookrightarrow\mathsf{FinSet}$ is the embedding of the subcategory with the morphisms surjections only, while $i_!$ is the left adjoint to the restriction $i^*:\mathsf{Set}^{\mathsf{FinSet}}\to\mathsf{Set}^{\mathsf{Finepi}}$. -Indeed, using the left Kan extension formula,$$i_!(1)(n)=\varinjlim\left(i/n\xrightarrow{\text{(take domain)}}\mathsf{Finepi}\xrightarrow{\text{constant $1$}}\mathsf{Set}\right);$$ -now every object $f:i(m)\to n$ of $i/n$ admits a morphism to the object $\operatorname{image}(f)\hookrightarrow n$, so that we may restrict from $i/n$ to its cofinal subcategory whose objects are inclusions of subsets into $n$. From this it is easy to see that indeed $P=i_!(1)$. -Using this then, $\hom(P,F)$ $=$ $\hom(i_!(1),F)$ $=$ $\hom(1,i^*(F))$ $=$ $\varprojlim(i^*(F))$, which more or less amounts to the same expression as in the first version above.<|endoftext|> -TITLE: Is there an information exchange in this game? (Bell's inequality) -QUESTION [5 upvotes]: This question concerns quantum mechanics experiment. But I believe it belongs here, on MathOverflow. -So, we have two players. They play a simple game and either both win or both loose, so they cooperate. -A fair coin is thrown in front of each player. (Each player sees only "his" coin). Having seen the result of the throw, each player tells a number: either 0 or 1. The outcome of the game is determined by the following rules: if both coins are "heads", players win if they named different numbers. Otherwise players win if they named the same number. Otherwise players loose. -Of course, if players can communicate (send some information to each other), they can easily win each game. But each player can't see the other coin and the game master made his best not to allow any information exchange. -In this case it's easy for players to achieve 75% win rate. Just always tell 1. (Players can discuss the strategy before the game starts). -The question is, can players achieve higher win rate? -It's "obvious" they can't. I can prove that if they use strategy like "if outcome the the coin throw is this, I tell number 0 with probability X, etc..." they can't beat 75% result. In other circumstances I would be satisfied, because no other strategy comes to my mind. But things are more complicated. -As I mentioned the question relates to quantum mechanics and physics. -One of the best ways to ensure that players can't exchange information is to put them quite far away from each other and require that they name the number fast enough. Relativity theory says that if "speed of light" * "given time" < "distance between players" they have absolutely no way to send each other information about the coin throw result. Sending information faster than light would make time-machine possible. -But players can "cheat" using quantum mechanics. The strategy looks like this: they prepare a pair of entangled particles and each player takes one of these particles. After the coin is thrown, a player makes an experiment with his particle: he positions the measurement device in either of two ways (depending the coin throw result) he measures the state of the particle: the outcome of measurement is either 0 or 1 and the player names the outcome. Quantum mechanics predicts win rate of about 85% in this case. -My description of the strategy is not at all complete. You can easily find full description if you want. My point is that the strategy exists, and it was experimentally confirmed (!). -So, there are three statements: - -75% is a maximum win rate if there is no information exchange between players -there IS no information exchange between players in experiments -win rate is above 75% in experiments - -These statements contradict each other. At least one of them must be wrong. I am pretty sure the second and third statements are correct. I think the first statement is wrong. Could you please confirm this and explain why exactly is it wrong? - -REPLY [2 votes]: The protocol you describe satisfies 2 and 3 but not 1, so when that protocol is adopted, 1 is wrong. -The correct form of statement 1 is that 75% is a maximum win rate if each player must choose a strategy that is contingent on the realization of some classical random variable (so that in particular there exists a joint distribution for the set of all random variables that are available for at least one player to use). -In a world governed by classical physics, this correct form of statement 1 is equivalent to the statement you've given about information exchange. In that world, all observables can be modeled as classical random variables. -But for some sets of quantum observables, that's no longer true, so in a world governed by quantum mechanics the two forms of Statement 1 are not equivalent. In your example, if we take four observables --- the outcome of Player $i$'s measurement with the device in position $j$, where $i$ and $j$ each have two possible values --- there is no joint probability distribution for the outcomes (i.e. they cannot be modeled as classical random variables satisfying Kolmogorov's axioms). -Edited to add: For a general formalism that specializes both to the case of classical game theory with mixed strategies (where your version of statement 1 holds) and to the quantum case, see here.<|endoftext|> -TITLE: Eisenstein series for quadratic number fields -QUESTION [6 upvotes]: I am familiar with the theory of modular forms and weight k Eisenstein series, and I am wondering if such a theory exists when the base field is not $\mathbb{Z}$. -Is there a theory of modular forms over $SL_2(\mathcal{O_k})$ where $k$ is a real or imaginary quadratic number field? Moreover, is there a nice Fourier series expansion of their Eisenstein series (if they exist)? Please provide references. - -REPLY [9 votes]: When $k$ is a real quadratic field (or more generally a totally real number field) the short answer is Hilbert modular forms. The corresponding Eisenstein series are called Hecke-Eisenstein series and are quite easily defined, -and yes they have a very nice Fourier expansion. For instance, this is what allowed Siegel to show that $\zeta_K(1-k)$ is always a rational number for -$k\ge2$. -The case of imaginary quadratic fields is a little more complicated: the modular forms are here called Bianchi modular forms. -There is a vast literature on Hilbert mf, a little less on Bianchi. -If I may, you can look at the last chapter of my recent book with Fredrik Str\"omberg published by the AMS.<|endoftext|> -TITLE: Is the homology of $\Omega^2\Sigma^2X$ free as a Gerstenhaber algebra? -QUESTION [8 upvotes]: Let $X$ be a connected space. According to Getzler BV-algebras and two-dimensional topologcial field theories , page 271, we have and isomorphism -$ -H_*(\Omega^2\Sigma^2X) \cong {\cal G}( \widetilde{H}_* X ) -$ -where ${\cal G}( V)$ means the free Gerstenhaber (Getzler calls it "braid") algebra over the graded space $V$ and $\widetilde{H}$ is the reduced homology. -Getzler credits Cohen's results in The homology of iterated loop spaces for this isomorphism. The closest thing I can find there is Cohen's theorem 3.2, in his chapter "The homology of $C_{n+1}$-spaces, $n\geq 0$", which sounds like it, but I'm having some problems to deduce Getzler's claim. -First of all, Getzler says to be working with complex coefficients, and Cohen with $\mathbb{Z}_p$ ones. Is it clear that the result should be true no matter which coefficients? Rational coefficients too? -Secondly, Cohen's result for $n=1$ would be, I guess, Getzler's case: -$ -H_*(\Omega^2\Sigma^2X) \cong GW_1(H_*X) \ . -$ -But here the free algebra functor is this $GW_1$ which I'm having some troubles to identify with ${\cal G}$. -Any hints or other references will be greatly appreciated. - -REPLY [11 votes]: Over $\mathbb{Z}_p$ it is not true that $H_*(\Omega^2\Sigma^2X)$ is the free Gerstenhaber algebra. Instead, Cohen proves that $H_*(\Omega^n\Sigma^nX)$ is a free object in a more elaborate category involving some Dyer-Lashof operations. In the case $n=2$ there is only one Dyer-Lashof operation but it still creates a lot of complexity. However, the Dyer-Lashof operations are controlled by the homology of symmetric groups. If we use rational coefficients then the homology of any finite group is zero in positive degrees, and because of this, all the Dyer-Lashof operations are zero. You therefore expect to get a free Gerstenhaber algebra, but I do not know where that is spelled out. There will not be any interesting difference between $\mathbb{Q}$ and $\mathbb{C}$ here; there is a natural isomorphism $H_*(X;\mathbb{C})=H_*(X;\mathbb{Q})\otimes\mathbb{C}$ for all spaces $X$.<|endoftext|> -TITLE: Uniqueness of dualizing objects -QUESTION [10 upvotes]: One definition of (symmetric) star-autonomous category is as a closed symmetric monoidal category $(C,\otimes,I,\multimap)$ equipped with an object $\bot$ such that all double-dualization maps $A \to ((A\multimap\bot)\multimap \bot)$ are isomorphisms. It follows that the functor $A \mapsto (A\multimap \bot)$ is a contravariant autoequivalence of $C$. -Such an object is sometimes called a dualizing object, although sometimes that name only requires double-dualization to be an isomorphism when $A$ is suitably "finite". Here I'm interested in the case where it is an isomorphism for all objects $A\in C$. -Can a given closed symmetric monoidal category admit more than one star-autonomous structure, i.e. can there be more than one such object $\bot$? - -REPLY [15 votes]: If a dualizing object exists, there is a bijection between isomorphism classes of dualizing objects and isomorphism classes of $\otimes$-invertible objects (i.e. the Picard group), given by tensoring your favorite dualizing object by a $\otimes$-invertible object. So the groupoid of $\ast$-autonomous structures, if nonempty, is equivalent to the groupoid of $\otimes$-invertible objects (canonically as soon as one chooses a basepoint -- it's a torsor over the grouplike symmetric monoidal groupoid of $\otimes$-invertible objects). -One direction: -First let's check that if $D$ is dualizing and $L$ is $\otimes$-invertible (with inverse $L^\vee$), then $L\otimes D$ is dualizing. Write $[,]$ for the internal hom (sorry, I seem to be changing all of your notation :). Then -$\begin{align*} -[[A, L\otimes D], L\otimes D] -&= [L^\vee \otimes [A, L\otimes D], D] \\ -&= [L^\vee \otimes L \otimes [A,D],D] \\ -&= [[A,D],D] \\ -&= A -\end{align*}$ -I suppose I should verify that the above isomorphism is the canonical morphism $A \to [[A, L\otimes D], L\otimes D]$, but since all the isomorphisms used were canonical, maybe I'll just wave my hands and ask rhetorically, "what else could it be?". -The converse: -In fact, it's always the case that any two dualizing objects differ by tensoring by a $\otimes$-invertible object. Here's a proof. - -First note that if $D$ is dualizing, then $[D,D] = I$. To see this, it suffices to check that $[[D,D],D] = [I,D]$ because $[-,D]$ is a contravariant equivalence of categories. But both sides are $D$, so this is the case. -Now if $D,D'$ are both dualizing, I claim that $[D,D']$ is $\otimes$-invertible, with inverse $[D',D]$. To see this, it suffices by symmetry to show that $[D,D'] \otimes [D',D] = I$. To check this, it suffices to check that $[[D,D'] \otimes [D',D],D] = [I,D]$. The lefthand side simplifies to $[[D,D'], [[D',D],D]] = [[D,D'],D'] = D$ where we have curried, and then used the dualizing property of both $D$ and $D'$. Of course, this is the same as the righthand side. -Finally, I claim that $D \otimes [D,D'] = D'$. To see this, it suffices to check that $[D \otimes [D,D'], D'] = [D',D']$. We've already seen that the righthand side is $I$ in the first bullet. And the lefthand side simplifies to $[D, [[D,D'],D']] = [D,D] = I$. - -So $D$ and $D'$ differ by tensoring by the $\otimes$-invertible object $[D,D']$. -And of course, since tensoring with a $\otimes$-invertible object is an equivalence of categories, the action map $L \mapsto L \otimes D$ is fully faithful; we've just seen it's essentially surjective, so it's an equivalence of groupoids. - -Endnote: -More than once I've found myself questioning the equation $[A,L\otimes D] = L \otimes [A,D]$ used in the second line of the forward direction so let me just record the proof here for my own benefit: -$\begin{align*} -Hom(X,[A,L\otimes D]) &= Hom(X \otimes A, L\otimes D) \\ -&= Hom(X \otimes A \otimes L^\vee, D) \\ -&= Hom(X \otimes L^\vee, [A,D]) \\ -&= Hom(X, L \otimes [A,D]) -\end{align*}$ -and conclude by Yoneda. So this isomorphism holds for any dualizable $L$ and arbitrary $A,D$.<|endoftext|> -TITLE: Double sum of negative powers of integers: a direct approach? -QUESTION [9 upvotes]: Let $\alpha,\beta\in (0,1\rbrack$, $\alpha\ne \beta$. I wish to estimate $$\sum_{m\leq x} \frac{1}{m^\alpha} \sum_{n\leq x/m} \frac{\log(x/mn)}{n^\beta}.$$ There is an obvious approach, namely, to estimate the inner sum first (the second- and third-order terms will be proportional to $\zeta(\beta)$ and $\zeta'(\beta)$; there is a connection with Ramanujan summation) and then input that estimate into the outer sum, which gets estimated in much the same way. -An improved version of the approach consists essentially in putting the longer of the two sums always inside, by means of a bit of combinatorial manipulation. -I have to wonder - is there a more elegant, less piecemeal approach, considering both sums in one go? If $\alpha$ were equal to $\beta$, the answer would be yes - we would get an estimate based on $(\zeta^2)'(s)$. Of course, that's precisely the case we aren't in. -(You may assume $\alpha = 2 \beta$, since that is the particular case I am most interested in.) -Update: what I get by a careful version of the above (continuous partition) and some pain is that the double sum equals $$\begin{aligned}&\frac{y^{1-\alpha}}{(1-\alpha)^2} \zeta(1-\alpha+\beta) + \frac{y^{1-\beta}}{(1-\beta)^2} \zeta(1-\beta+\alpha)\\ &+ \zeta(\alpha) \zeta(\beta) \log y + \zeta'(\alpha) \zeta(\beta) + \zeta(\alpha) \zeta'(\beta)\end{aligned}$$ plus an error term of size $O\left(y^{\frac{1}{2} - \frac{\alpha+\beta}{2}}\right)$, where the implied constant is explicit (and fairly small). I do not know whether the error term ought to be of a lower order of magnitude. Let me see what I can do with Lucia's approach below. - -REPLY [3 votes]: Let me carry out matters using a complex-analytical approach, as Lucia suggests, and then say where the difficulty lies. -Let $0<\beta<\alpha\leq 1$. First of all, as Lucia says, -$$\sum_{m\leq x} \frac{1}{m^\alpha} \sum_{n\leq x/m} \frac{\log(x/mn)}{n^\beta} = \frac{1}{2πi} \int_{c-i\infty}^{c+i\infty} \zeta(s+\alpha) \zeta(s+\beta) \frac{x^s}{s^2} ds$$ -for $c>1$. We shift the contour of integration to the left of $\Re(s)=0$, picking up the main terms $$\begin{aligned}&\frac{y^{1-\alpha}}{(1-\alpha)^2} \zeta(1-\alpha+\beta) + \frac{y^{1-\beta}}{(1-\beta)^2} \zeta(1-\beta+\alpha)\\ &+ \zeta(\alpha) \zeta(\beta) \log y + \zeta'(\alpha) \zeta(\beta) + \zeta(\alpha) \zeta'(\beta)\end{aligned}$$ plus an error term of size $O\left(y^{\frac{1}{2} - \frac{\alpha+\beta}{2}}\right)$ along the way. We are left with the task of estimating an error term $$\frac{1}{2πi} \int_R \zeta(s+\alpha) \zeta(s+\beta) \frac{x^s}{s^2} ds,$$ where the integral is over a contour $R$ of our choice going from $-r-i\infty$ to $-r+i\infty$, say, and satisfying $\Re s\leq -r$ at all points. The error will be clearly bounded by $O(K x^{-r})$, where -$$K = \frac{1}{2πi} \int_R \frac{|\zeta(s+\alpha)| |\zeta(s+\beta)|}{|s|^2} ds.$$ -The problem does reduces to estimating $K$. -Now, there are rigorous-numerics packages that include integration and the possibility to compute the zeta function $\zeta(s)$. (I currently use ARB.) However, (a) computations must obviously be finite (at least assuming mortal mathematicians), and (b) computing $\zeta(s)$ is never a walk in the park, and rigorous integration only adds to the overhead. Integrating an expression such as above from $-1/2 - i T$ to $1/2 + i T$ takes 15 minutes for $T = 10000$ (says a better programmer than I), but we should not expect to go much further than $T = 100000$ programming casually on our laptops. -The problem that remains, then, is how to bound a tail $$\frac{1}{2\pi i} \left(\int_{-r-i\infty}^{-r-i T} + \int_{-r+i T}^{-r + i \infty} \frac{|\zeta(s+\alpha)| |\zeta(s+\beta)|}{|s|^2} ds\right).$$ -The most obvious approach is to use Backlund's explicit bounds (1918) on $\zeta(\sigma + it)$ (see http://iml.univ-mrs.fr/~ramare/TME-EMT/Articles/Art06.html#Size). They are of the quality $$|\zeta(\sigma + i t)| = (1+o(1)) (t/2\pi)^{(1-\sigma)/2} \log t$$ for $0\leq \sigma\leq 1$ and $$|\zeta(\sigma + i t)| = (1+o(1)) (t/2\pi)^{1/2-\sigma} \log t$$ for $-1/2\leq \sigma\leq 0$. The problem here is that convergence is painfully slow. If, say, $\alpha = 1$, $\beta=1/2$ and $r =-1/4$ (reasonable values all around), the tails will be bounded by a constant times $(\log T)^2/\sqrt{T}$. For $T=10000$, $(\log T)^2/\sqrt{T} > 0.848\dotsc$ - not exactly small; for $T=100000$, the same equals $0.419\dotsc$ - barely an improvement. -Notice, however, that why Backlund's bounds are essentially tight for $\Re s<0$, that is not the case for $0<\Re(s)<1$. Of course, they are convexity bounds, so improving on them explicitly would amount to translating into explicit terms rather non-trivial material. However, as long as we are satisfied with $r>-\beta$, what we can do instead is give $L_2$ bounds for the tails, that is, bound -$$\int_{r-i \infty}^{r-i T} \frac{|\zeta(s)|^2}{|s|^2} ds + \int_{r+iT}^{r+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds.$$ -(The integral $\int_{r-i \infty}^{r-i T}$ is obviously the same.) Then we use Cauchy-Schwarz to bound the tail of the integral we were discussing. -This is non-trivial, and takes us further afield, so I will make it into a separate question: $L_2$ bounds for tails of $\zeta(s)$ on a vertical line .<|endoftext|> -TITLE: What is known about ideal and divisibility lattices of GCD domains and their generalizations? -QUESTION [11 upvotes]: The divisibility relation "$a$ divides $b$", or concisely, $a \vert b$ defined over a commutative integral domain $R$ with identity induces a partial order on the multiplicative semigroup $R/R^{\times}$ where $R^{\times}$ is the unit group of $R$. The inclusion relation $I \subset J$ is a partial order on the set of principal ideals of $R$ and the two corresponding partially ordered sets are anti-isomorphic in a natural sense. -If $R$ is a GCD domain, then $D(R) \Doteq (R/R^{\times}, \vert)$ is a bounded distributive lattice, the joint $a \vee b$ being the lowest common multiple of $a$ and $b$ and the meet $a \wedge b$ being the greatest common divisor of $a$ and $b$. We call $D(R)$ the divisibility lattice of $R$. -If $R$ is any ring (commutative or not, with identity or not and possibly with zero divisors), the set $\mathcal{I}(R)$ of ideals of $R$ is partially ordered by the inclusion and the operations $I \vee J \Doteq I + J$ and $I \wedge J \Doteq I \cap J$ turns $L(R) \Doteq (\mathcal{I}(R), \subset)$ into a complete and modular lattice. We call $L(R)$ the ideal lattice of $R$. -If $R$ is a principal ideal domain, then $D(R)$ and $L(R)$ are naturally anti-isomorphic. More generally, if $R$ is a Bézout domain, then $D(R)$ is anti-isomorphic to the lattice $L_{\text{principal}}(R)$ of its principal ideals. - - -Question 1. Are there remarkable results that bind the structure of $R$ to lattice properties of $D(R)$ and $L(R)$? In particular, what can be said about $R$ if one of the two lattices is - $(i)$ complete or, $(ii)$ distributive or, $(ii)$ complemented or, $(iii)$ boolean? - - -Here are the few things I know. The lattice $D(R)$ is complete if and only if $R$ satisfies the ascending chain condition on principal ideals (ACCP). Moreover, there exist non-atomic Bézout rings, e.g., the ring of entire functions (think about $\sin(\frac{z}{2^k})$ for $k \ge 0$). Hence there exists a ring $R$ which is a GCD domain and such that $D(R)$ is not complete. If the lattice $L(R)$ is distributive for a commutative ring $R$, then $R$ is called arithmetical and this is equivalent to say that the localization $R_{\mathfrak{m}}$ of $R$ at $\mathfrak{m}$ is a uniserial ring for every maximal ideal $\mathfrak{m}$ of $R$. (An arithmetical domain is the same as Prüfer domain). The Jaffard-Ohm-Kaplansky Theorem states that the positive cone of every lattice-ordered Abelian group arises as the semigroup of divisibility of a Bézout domain. -The answers of Sridhar Ramesh and Luc Guyot show that for a unique factorization domain $R$, the lattice $D(R)$ is complemented if and only if $R$ is a field, and $D(R)$ is Boolean in this case. In addition, $D(R)$ is totally ordered if and only if $R$ is a discrete valuation ring or a field. -My second question is - - -Question 2. Have $D(R)$ and $L(R)$ been studied or classified for specific classes of rings such as principal ideal domains (PIDs), unique factorizations domains (UFDs), Dedekind domains and rings of integers in particular? - - -The answers of Sridhar Ramesh and Luc Guyot show that the situation is extremely tamed for UFDs, since the cardinality of the set of primes of $R$ (up to multiplication by units of $R$) is a complete invariant of isomorphism for $D(R)$. Indeed, $D(R)$ identifies with the lattice of $\mathbb{N}$-valued functions with finite support on the set of primes (modulo units), plus an additional top element corresponding to $0$. Any cardinal can be achieved as the cardinality of the set of primes of some UFD (think about $k[X_i; i \in I]$ for $k$ a field and any set $I$) and actually it is even true for a PID (think of $k[X]$ for a $k$ an arbitrary algebraically closed field). -Any references, specifically comprehensive surveys or textbooks, will be appreciated! - -REPLY [8 votes]: A PID is a unique factorization domain, so the divisibility lattice (identifying associate elements, of course) has the same ordering as the set of $\mathbb{N}$-valued functions with finite support on some (possibly infinite) set of primes, plus an additional top element corresponding to 0. Actually, I suppose it is the reverse of this in your convention, where $a \leq b$ corresponds to "$b$ divides $a$". -It is thus distributive, but not Boolean (except in the trivial case where the set of primes is empty; i.e., when the PID is a field); the only complemented elements are the top and bottom element, and indeed, if $a \wedge b = 0$ (in the sense that $lcm(a, b) = 0$), then at least one of $a$ and $b$ is itself $0$.<|endoftext|> -TITLE: Name for the action of a bialgebra on an algebra -QUESTION [8 upvotes]: Give an algebra $A$, a bialgebra $B$, and an action, that is, a bilinear map $\triangleright: B \times A \to A$ such that -$$ -(b_1b_2) \triangleright a = b_1\triangleright(b_2 \triangleright a). -$$ -When it also holds that -$$ -b \triangleright(ac) = (b_{(1)} \triangleright a)(b_{(2)}\triangleright c) -$$ -what do we call this kind of action? - -REPLY [10 votes]: According to nLab, such an action is called a Hopf action and your data specify a left $B$-module algebra. Such a structure is also referred to in the literature as an algebra in the category (of left $B$-modules). -Note also, that, if instead of the bialgebra $B$ we consider a hopf algebra $H$ acting on $A$ and satisfying your condition supplemented by: -$$ -h\triangleright 1_A=\varepsilon(h)1_A -$$ -where $\varepsilon:C\rightarrow k$ is the counity map of $H$, then under such an action, $A$ is called a $H$-module algebra or a Hopf-module algebra or an algebra in the category (of left $H$-modules). -(You can also check section 3.1 of this article, for a review on this and other similar kinds of actions and coactions on algebras and coalgebras). -Edit: If we drop the demand for the bilinear map $\triangleright: H \times A \to A$ to be an action, i.e., if we drop $(h_1h_2) \triangleright a = h_1\triangleright(h_2 \triangleright a)$ and simply require $h \triangleright(ac) = (h_{1} \triangleright a)(h_{2}\triangleright c)$ and $h\triangleright 1_A=\varepsilon(h)1_A$, then we say that the bilinear map $\triangleright$ is a measuring or that $H$ measures $A$. If, furthermore: $1_H\triangleright a=a$ for all $a\in A$ then we are speaking about a weak action of $H$ on $A$ (see def.1.1, p.674 of the linked article). The same terminology is used for the bialgebra case as well.<|endoftext|> -TITLE: Is there a simple system that has $\text{SU}(3)$ symmetry? -QUESTION [15 upvotes]: The buckle at the end of a belt has $\text{SU}(2)$ symmetry, if the rotations around the three coordinate axes are taken as generators. See, for example, the paper by Hart, Francis and Kauffman, https://www.researchgate.net/publication/2389584_Visualizing_Quaternion_Rotation . You can also use the hand at the end of your arm to visualize SU(2), including its double cover of SO(3). -Is there something similar for $\text{SU}(3)$ symmetry? Is there a simple system, maybe from everyday life, that realizes this symmetry? -Alternatively: how do you picture $\text{SU}(3)$ in your head? - -REPLY [9 votes]: One of the simplest mechanical systems having an SU(3) dynamical symmetry is the three dimensional isotropic harmonic oscillator. Its phase space is $\mathbb{R}^6$ parametrized by 3 position coordinates $x_i$ and 3 momenta $p_i$, $i=1,2,3$. -A fixed energy hypersurface in phase space is given by the constraint (in units where the natural frequency is 1): -$$\sum_i x_i^2 + \sum_i p_i^2 = E $$ -This a the $5$-dimensional sphere $S^5$, and the set of classical trajectories (the reduced phase space corresponding to the Hamiltonian action) on this hypersurface is the complex projective space $CP^2$ , obtained from $S^5$ by a Hopf fibration and from the original phase space by Marsden-Weinstein symplectic reduction with respect to the action of the total energy function: -$$CP^2 = S^5/U(1) = \mathbb{C}^3//U(1)$$ -Being a $SU(3)$ homogeneous space $CP^2$ has a natural $SU(3)$action: -The functions representing $SU(3)$ in the $CP^2$ Poisson algebra have simple geometric expressions: - -First we have the three angular momentum components: -$$L_{ij} = x_ip_j-x_jp_i$$ -In addition we have 5 quadrupole moments built from the (traceless)symmetric combinations of the coordinates and momenta: -$$Q_{ij} = \frac{3}{2} (x_ip_j+x_jp_i) - \delta_{ij} \sum_k x_kp_k$$ -Together they generate the $SU(3)$ Lie algebra. The quadrupole moment space is closed under the $SU(2)$ action generated by the angular momenta. They transform according to the spin-2 representation. - -Physically, the quadrupole moments interact with gradients of an external magnetic field. If the harmonic oscillator is charged, then its interaction energy with an external magnetic field has the form: -$$E_{int} = \sum_{ijk} \epsilon_{ijk}L_{ij}B_k+ \sum_{ij} Q_{ij} \partial_{i}B_j$$ -Please, see the following Melih Sener lecture note by for a detailed exposition.<|endoftext|> -TITLE: Have you seen my matroid? -QUESTION [28 upvotes]: Let $M(n,k)$ be the matroid on the ground set $\{\pm 1,\ldots,\pm n\}$ for which a set is independent if and only if it contains at most $k$ pairs $\pm i$. Note that the signed permutation group (the Coxeter group of type $B_n$) acts on this matroid. Questions: - -Does this matroid have a name? -Has it been studied before? -Is there a nice formula for its characteristic polynomial? - -Here are some boring special cases: - -$M(n,n)$ is the Boolean matroid on $2n$ elements. -$M(n,n-1)$ is the uniform matroid of rank $2n-1$ on $2n$ elements. -$M(n,0)$ is the direct sum of $n$ copies of the uniform matroid of rank 1 on 2 elements. - -The first interesting case is $M(3,1)$, which has rank 4 and characteristic polynomial -$$q^4 - 6q^3 + 15q^2 - 17q + 7$$ -I am also interested in truncations of this matroid. That is, let $M(n,k,d)$ be the matroid on the ground set $\{\pm 1,\ldots,\pm n\}$ for which a set is independent if and only if it contains at most $k$ pairs $\pm i$ and has size at most $d$. All of the same questions apply! -Remark: I would like to regard these matroids as type B analogues of uniform matroids. Uniform matroids are the permutation-invariant matroids on the ground set $\{1,\ldots,n\}$, while these are the signed-permutation-invariant matroids on the ground set $\{\pm 1,\ldots,\pm n\}$. - -REPLY [14 votes]: Let $U$ be the uniform matroid of rank $k$ on $n$. Since $U$ is orientable one can consider the Lawrence oriented matroid $\Lambda(U)$ associated with any orientation of $U$ (the Lawrence construction doesn't care about which orientation you take). Then $M(n,k)$ is precisely the underlying unoriented matroid $\underline{\Lambda(U)}$ of $\Lambda(U)$. -Also, the dual matroid $M^*(n,k)$ is a symplectic matroid, which explains why the group $B_n$ acts on the primal.<|endoftext|> -TITLE: Can we classify all finite 2-generated groups $G$ such that if $x,y$ generate $G$, then so does $x,yxy^{-1}$? -QUESTION [5 upvotes]: Can we classify finite 2-generated groups $G$ satisfying the following property: -For any pair $x,y$ which generate $G$, the pair $x,yxy^{-1}$ also generates $G$. -By the comments, no nontrivial abelian group can satisfy this property, so I suppose the first question is: Do such groups $G$ exist? - -REPLY [10 votes]: Theorem The only finite group satisfying the condition is the trivial group. -Proof. Let $G$ be a nontrivial finite group and $S$ be a simple quotient of $G$, which satisfies the condition by Gaschutz's lemma. Then $S$ is non-abelian as mentioned above. If $x\in S$ is any involution, then by well-known result of Guralnick and Kantor in Probalistic generation of finite simple groups, there exists an element $y\in S$ such that $S=\langle x,y\rangle$ while $\langle x,yxy^{-1}\rangle$ is a dihedral group, that is, $S\neq\langle x,yxy^{-1}\rangle$, a contradiction.<|endoftext|> -TITLE: Averaging $2^{\omega(n)}$ over a region -QUESTION [10 upvotes]: Let $R(X)$ be the region defined by -$$\displaystyle R(X) = \{(a,b,c) \in \mathbb{R}^3 \colon |b| \leq a \leq c, \, a,c \geq 1, \, a(4ac-b^2) \leq X\}.$$ -I want to know how to estimate the sum -$$\displaystyle \sum_{(a,b,c) \in R(X) \cap \mathbb{Z}^3} 2^{\omega(4ac-b^2)},$$ -where $\omega(n)$ denotes the number of prime divisors of $n$. An upper bound suffices. -One can show that -$$\displaystyle \sum_{(a,b,c) \in R(X) \cap \mathbb{Z}^3} 1 = O (X \log X),$$ -and it's known that the average of $2^{\omega(n)}$ over $n \leq X$ is of order $\log X$. Therefore, the expected upper bound should be about $O(X (\log X)^2)$. - -REPLY [8 votes]: Your guess is correct. Since $2^{\omega(n)} = \sum_{d | n} \mu^2(d)$ one has -$$ -S = \sum_{(a,b,c) \in R(X)} 2^{\omega(4ac-b^2)} = \sum_{d \leq X} \mu^2(d) |R_d(X)| -$$ -where $R_d(X)$ is the set of integer triples $(a,b,c) \in R(X)$ such that $4ac \equiv b^2 \pmod d$. -Let $d$ be a squarefree integer. Let $a,b$ be such that $|b| \leq a$ and let $s = \mathrm{gcd}(4a,d)$. Then for $(a,b,c)$ to be in $R_d(X)$ one must have $s | b^2$, or equivalently $s |b$ since $s$ is squarefree, and also $c \equiv (4a/s)^{-1} b^2 \ \mathrm{mod} \ d/s$. The number of integers $c$ like that, with $a \leq c \leq b^2/(4a) + X/(4a^2)$ is -$$ -\leq \frac{s}{d}( \frac{b^2}{4a} + \frac{X}{4a^2} - a) + 1 \leq \frac{s}{d}( \frac{X}{4a^2} - \frac{3 a}{4}) + 1\leq\frac{sX}{4 a^2d} + \mathrm{1}_{as \leq 4d/3}. -$$ -Moreover the number of integers $b$ such that $|b| \leq a$ and $s | b$ is -$$ -\leq \frac{2a}{s} + 2. -$$ -Since the constraint $X \geq a(4ac-b^2)$ yields $ad \leq X$, we have -$$ -|R_d(X)| \ll \sum_{ad \leq X} (\frac{sX}{a^2d} + \mathrm{1}_{as \leq 4d/3})(\frac{a}{s} + 1) \\ -\ll \sum_{ad \leq X} ( \frac{X}{ad} + \frac{sX}{a^2d} + \frac{a}{s} \mathrm{1}_{as \leq 4d/3} + 1 ) -\ll \frac{X \log X}{d} + \frac{X}{d} + \sum_{ ad \leq X} ( \frac{sX}{a^2d} + \mathrm{1}_{as \leq d} \frac{a}{s} ). -$$ -Summing over $d$ we get -$$ -S \ll X \log^2(X) + \sum_{s} \sum_{ad \leq X \\ s|4a , s|d} ( \frac{sX}{a^2d} + \mathrm{1}_{as \leq 4d/3} \frac{a}{s} ). -$$ -For each $s$, the inner sum is -$$ -\ll \sum_{ad \leq X/s^2 } ( \frac{X}{s^2 a^2d} + \mathrm{1}_{as \leq 16d/3} a ) -$$ -The first sum is $\ll X s^{-2}$, while in the second we have the constraint $a^2 \ll X s^{-3}$ which yields -$$ -\ll \sum_{a \ll X^{1/2} s^{-3/2}} a \ll X s^{-3} + X^{1/2} s^{-3/2}. -$$ -So the global sum over $s$ is $\ll X$ and we are done. -Remark: one should be able to extract from this proof an asymptotic formula of the form $c X \log^2(X) + O(X \log X)$.<|endoftext|> -TITLE: $L^p$-estimates for elliptic pseudodifferential operators -QUESTION [5 upvotes]: Assume we have an pseudodifferential operator -$P:\mathcal{S}(\mathbb{R}^n) \rightarrow \mathcal{S}(\mathbb{R}^n), Pf(x) = (2\pi)^{-n/2}\int\mathrm{d}\xi\; p(x,\xi)\,\hat{f}(\xi)e^{i\xi x}$ -acting on Schwarz-functions. Assume it is of degree $m$, that is we have $|\partial_\xi^\alpha \partial_x^\beta p(x,\xi)| \leq C_{\alpha,\beta} \, (1 + |\xi|)^{m - |\alpha|}$ for some real $m$ and multiindices $\alpha,\beta$. Now, assume that P is elliptic, that is $c(1+|\xi|)^m \leq |p(x,\xi)|$ for some constant. -Now it is a standard result, that one has the following elliptic $L^2$-estimate: -$\lVert f \rVert_{W^{s,2}} \le C (\lVert Pf\rVert_{W^{s-m,2}} + \lVert f\rVert_{W^{s-m,2}})$ -My question: Is there a $L^p$-generalization of this estimate? If yes, where can I find a reference to it? - -REPLY [3 votes]: Yes you do have a generalization of your elliptic inequality to the $L^p$ case for $p\in (1,+\infty)$. In fact the operators with symbols in the class $S^0_{1,0}$ (as in your question) are bounded on $L^p$ ($p$ in the same range as above). You can find a proof of this in the Astérisque book by R. Coifman & Y. Meyer "Au delà des opérateurs pseudo-différentiels".<|endoftext|> -TITLE: Irrational rotation - recurrence times -QUESTION [10 upvotes]: I consider the irrational rotation -$T_\alpha(x) = x + \alpha \text{ mod } 1$ for given irrational $\alpha \in [0,1]$. For a given open interval $A \subset [0,1]$ with length $|A|>0$, I consider the recurrence times $I = \{n\in \mathbb{N}: T^n(0) \in A \}$. I want to show that -$\sum_{i \in I} p\cdot(1-p)^i \to |A|$ as $p \to 0$. -My very informal motivation for this is that the above sum should be equal to $\sum_{n\in \mathbb{N}} p \cdot (1-p)^{n\cdot\frac{1}{|A|}}$ "give or take" "a few" $(1-p)$-factors (which tend to $1$ as $p \to 0$), and the latter sum can be shown to converge to $|A|$ as $p \to 0$. -I have obtained a somewhat similar (but obviously not identical) result for the case of rational $\alpha$ (happy to add details, but I'm not sure if it's useful), and tried to derive the above using a rational series converging to $\alpha$, but wasn't successful. -Unfortunately, I have virtually no background in ergodic theory, numbers theory and similar subjects which apparently treat irrational rotations, and thus don't quite know where to start. - -REPLY [8 votes]: You can recover this result in two steps: - -a variation on Birkhoff's ergodic theorem yields the Cesàro convergence of the sums; -Cesàro convergence implies Abel convergence. - -First step: $T_\alpha$ preserves the Lebesgue measure and is uniquely ergodic. Hence, for all $f \in \mathcal{C} (\mathbb{S}_1, \mathbb{R})$, -$$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} f \circ T_\alpha^k = \int_{\mathbb{S}_1} f(x) \ dx,$$ -where the limit is uniform. By approximating $\mathbf{1}_A$ from above and from below by continuous functions, we get that: -$$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} \mathbf{1}_A \circ T_\alpha^k = |A|,$$ -where the limit is again uniform. I had to use uniform ergodicity to be able to tell something about a specific starting point ($0$) instead of a generic one. -Second step: let $p \in (0,1)$. Let $(a_k)_{k \geq 0}$ be a bounded real sequence. Then: -$$\sum_{n=0}^{+ \infty} (1-p)p^n a_n -= \sum_{n=1}^{+ \infty} \frac{n(1-p)^2 p^n}{p} \frac{1}{n} \sum_{k=0}^{n-1} a_k$$ -Now, putting $b_n^p := n(1-p)^2p^{n-1}$, for all $p \in (0,1)$, the sequence $(b_n^p)_{n \geq 1}$ defines a probability measure on the positive integers. In addition, for all $n$, $b_n^p$ converges to $0$ as $p$ goes to $1$. Hence, -$$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} a_k = \ell \quad \implies \quad \lim_{p \to 1} \sum_{n=0}^{+ \infty} (1-p)p^n a_n = \ell.$$ -Finally, take $a_n := \mathbf{1}_A \circ T_\alpha^n (0)$, and apply the first step.<|endoftext|> -TITLE: Categories with meaningful "core" objects -QUESTION [6 upvotes]: Let $G$ be a graph. A graph $C$ is a core of $G$ if: - -There is a homomorphism $G \to C$. -There is a homomorphism $C \to G$. -$C$ is smallest possible with these properties. - -A core graph is well-defined up to isomorphism for any finite graph $G$. One can see that under these restrictions $C$ is a subgraph of $G$ (since otherwise the image of $C \to G$ must be a smaller graph with all these properties), and is its own core. Further, we may equivalently replace 3 with: -3'. All endomorphisms of $C$ are isomorphisms. -3' is one of the main reasons why core graphs are a useful notion (see references in the article above), so it is interesting to seek for similar objects in other categories. Let us define a core of an object $G$ as any $C$ satisfying 1, 2 and 3'. -In any category with an object $X$ both initial and terminal, $X$ is trivially a core of any object (as in $\mathbf{Grp}$, $\mathbf{Mon}$, or $\mathbf{Vect}_K$). In $\mathbf{Field}$ any object is a core of itself since any homomorphism is injective. In both of this situations the cores are kinda trivial. We could say that in the first case objects have too 'admissive' internal structure, while in the latter case objects have too 'restrictive' internal structure. -$\mathbf{Graph}$ is an example of a category with "meaningful" cores, that is, different graphs may have different cores, and there are graphs that are not cores of themselves. In $\mathbf{Ring}$ cores may be meaningful too: for example, the core of $\mathbb{Z}_n \times \mathbb{Z}_m$ is $\mathbb{Z}_{\mathrm{lcm}(n, m)}$. -Questions: In which (concrete) categories cores are meaningful? In which categories cores are defined up to isomorphism? -Late edit: Another aspect of how graph cores are interesting is that they are hard to compute (see above reference). Naturally, if cores of a category are computationally hard, then they must be meaningful in the above sense, and it feels like a good way to encapsulate their 'interestingness'. Can we say something about categories with computationally hard cores? - -REPLY [3 votes]: I think of the core phenomenon as a special case of a basic fact about finite transformation monoids. Suppose we have a concrete category of finite structures such that bijective endomorphisms are isomorphisms and the image of each endomorphism is a substructure. Then for each object $X$ there is a subject $Y$ which is a retract of $X$, any endomorphism of $Y$ is an isomorphism and $Y$ is unique up to isomorphism. Moreover, $Y$ is of minimal cardinality among images of a morphism from $X$ to itself. -Why? Let $M$ be the endomorphism monoid of $X$. Define the rank of an element of $M$ to be the cardinality of its image. Let $m\in M$ have minimal rank $r$. Then all powers of $m$ have the same rank and hence there is an idempotent $e$ achieving the rank $r$. Let $Y=eX$. Then $Y$ is a retract of $X$. If $f$ is an endomorphism of $Y$, then $ife$ (where $i$ is the inclusion) is an endomorphism of $X$ and so by minimality of the rank of $e$, we have that $|eX|=|Y|\leq |fY|=|feX|\leq |eX|=|Y|$ and thus $f$ is an automorphism of $Y$ by our hypotheses. -Suppose $Z$ is another retract of $X$, say by an idempotent $e'$, such that every endomorphism of $Z$ is an isomorphism. Then $e'|_Y\colon Y\to Z$ and $e|_Z\colon Z\to Y$ are morphisms by restriction and hence $e|_Ze'|_Y$ is an automorphism of $Y$ and $e'|_Ye|_Z$ is an automorphism of $Z$. It easily follows that $e'_Y$ is an isomorphism from $Y$ to $Z$.<|endoftext|> -TITLE: What is the status of the 4-dimensional Smale Conjecture? -QUESTION [20 upvotes]: 4-dimensional Smale conjecture claims the following: -The inclusion $SO(5)$ → $SDiff(S^4)$ is a homotopy equivalence. -or Does $Diff(S^4)$ have the homotopy-type of $O(5)$ ?. -The inclusion $SO(n + 1$) → $SDiff(S^n)$ is a homotopy equivalence for n = 1 (trivial proof), n = 2 [1004,Smale,1959,Proc. Amer. Math. Soc.], n = 3 [464,Hatcher, 1983,Ann. of Math.], and is not a homotopy equivalence for n ≥ 5 [41,Antonelli, Burghelea, & Kahn,1972,Topology] and [164,Burghelea & Lashof,1974,Trans. Amer. Math. Soc.]. -I looked everywhere but I could not find anything. Is this problem still open? Thanks. - -REPLY [10 votes]: Tadayuki Watanabe has a preprint for the disproof here: https://arxiv.org/abs/1812.02448<|endoftext|> -TITLE: Why is Oka's coherence theorem a deep result? -QUESTION [16 upvotes]: This is a very naive question. -Let $X$ be a complex manifold. Let $\mathcal{O}_X$ be the structure sheaf of $X$, a sheaf of rings whose sections over opens $U\subset X$ are just the holomorphic functions $U\rightarrow\mathbb{C}$. -A sheaf of $\mathcal{O}_X$-modules $F$ is coherent if: - -It is locally finitely generated: Ie, there is an open cover $\{U_i\}$ of $X$ such that $F|_{U_i}$ admits surjections $\mathcal{O}_{U_i}\rightarrow F|_{U_i}$ for each $i$. -For any open $V\subset X$, and any morphism $f : \mathcal{O}_V^s\rightarrow F|_V$, $\text{Ker}(f)$ is a locally finitely generated sheaf on $V$ (ie, satisfies condition 1). - -Then, Oka's coherence theorem states that the structure sheaf $\mathcal{O}_X$ is coherent. -When $X$ is a scheme, this statement is almost tautological. What is it about the setting of complex manifolds that makes this theorem deep? - -REPLY [23 votes]: In scheme theory applied to complex geometry one usually does not encounter coherent rings which are not noetherian as well. -However if $X$ is (for example) a Stein manifold then the ring $R = \mathcal{O}(X)$ of global holomorphic functions is typically non-noetherian, which makes remarkable the fact that $R$ is nonetheless coherent.<|endoftext|> -TITLE: A question on connected sum of compact manifolds -QUESTION [13 upvotes]: Let $M$ be a compact orientable manifold which is homeomorphic to its connected sum with itself $M\# M$. Must $M$ be homeomorphic to a sphere? -I will explain why I am interested (at the risk of being outright foolish or overambitious). The 'equation' $M = M\#M$ is the simplest conceivable equation in the category of compact topological manifolds, just as $x+x=x$ characterises $0$ in an abelian group. Although I am suspicious if $M$ could be a homology sphere, I hope to get a characterisation of spheres. -Thanks for any comments or suggested references. - -REPLY [18 votes]: Let us suppose that $dim(M)\geq 3$ then we have that: - -$\pi_1(M \# M)\cong \pi_1(M)*\pi_1(M)$, -$H_*(M;\mathbb{Z})\cong H_*(M;\mathbb{Z})\oplus H_*(M;\mathbb{Z})$ when $*< dim(M)$. - -As $M$ is compact this implies that $\pi_1(M)$ is finitely presented thus that $M$ is simply connected (*). Together with the second point it implies that $M$ has the homotopy type of a sphere. And you conclude using the Poincaré conjecture that $M$ is homeomorphic to a sphere. -Edit (*): we have to use a non-trivial theorem in group theory about the minimal number of generators $m(G)$ of a group $G$. Namely we have $m(G*H)=m(G)+ m(H)$ (conjectured by Levi and proved independently by Gruschko and Neumann).<|endoftext|> -TITLE: On statistical bases in Banach spaces -QUESTION [9 upvotes]: Let $K$ be a subset of the positive integers $\mathbb{N}$. For each $n\in \mathbb{N}$, $K_{n}$ denotes the set $\{k\in K: k\leq n\}$ and $|K_{n}|$ denotes the number of the elements in $K_{n}$. The natural density of $K$ is defined by $$\delta(K)=\lim_{n\rightarrow \infty}\frac{|K_{n}|}{n}.$$ -A sequence $(x_{k})_{k}$ in a Banach space $X$ is said to be norm statistically convergent to $x\in X$ if $\delta(\{k\in \mathbb{N}:\|x_{k}-x\|\geq \epsilon\})=0$ for every $\epsilon>0$. -It is natural to define a series $\sum_{k=1}^{\infty}x_{k}$ to be norm statistically convergent to $x\in X$ by $(s_{n})_{n}=(\sum_{k=1}^{n}x_{k})_{n}$ to be norm statistically convergent to $x$. -We say that a sequence $(x_{k})_{k}$ in a Banach space $X$ is a statistical basis for $X$ if, for each $x\in X$, there exists a unique $(a_{k})_{k}$ such that the series $\sum_{k=1}^{\infty}a_{k}x_{k}$ norm statistically convergent to $x$. -Clearly, statistical basis is a generalization of the notion of Schauder basis. -Question: For each $n$, we can define a linear functional $f_{n}$ in $X$ by $f_{n}(x)=a_{n}$. If $(x_{k})_{k}$ is a Schauder basis, we know that $f_{n}$ is continuous. However, if $(x_{k})_{k}$ is a statistical basis for $X$, is $f_{n}$ continuous? - -REPLY [5 votes]: Edit 26.04.2022. The problem has been recently solved in ZFC. arxiv.org/abs/2203.15123. - -In a recent note with J. Swaczyna (arXiv:2005.04873), we proved that assuming the existence of certain large cardinals (for example, the existence of a super-compact cardinal) that filter bases whose underlying filter is a projective subset of the Cantor set, have continuous coordinate functionals. The role of large cardinals is to make the heuristic proof I outlined in my other answer work. -The filter of statistical convergence is actually $F_{\sigma \delta}$ (hence Borel, hence projective), so in a theory stronger than ZFC, the answer to your question is affirmative. We still expect that, at least for Borel filters, the question about continuity of coordinate functionals should have a positive answer in ZFC. However, our proof method does not give a chance to invoke Schoenfield's absoluteness theorem in that context.<|endoftext|> -TITLE: Does there exist a Haken manifold where all its incompressible surfaces are non-separating? -QUESTION [6 upvotes]: Every non-zero element in $H_2(M,\mathbb Z)$ corresponds to an incompressible surface. So these surfaces are non-separating. But I'm interested in knowing about separating incompressible surfaces. A result of Peter Shalen guarantees that a compact, connected, orientable, irreducible 3-manifold $M$ [such a manifold is known as a Haken manifold] will have a non-trivial separating incompressible surface provided that $H_1(M;\mathbb Q)$ is carried by the boundary of M and that some boundary component of M has genus $> 1$. -Question 1: Does there exist any closed Haken 3-manifold with all its incompressible surfaces are non-separating. -My intuition is NO. If we consider $S^1\times S^1\times S^1$, then all its incompressible surfaces has to be torus since this is the only surface with abelian fundamental group. And then any incompressible torus transverse to the fiber (thinking it as a torus bundle over circle) has to intersect it with a non-trivial simple closed curve. And by some more argument we can prove that it is non separating since it can't separate any fiber. -Question 2: What if the fundamental group is non-abelian? Moreover what if we consider a hyperbolic Haken manifold, then can anyone give me an explicit example or reference of such a manifold whose all incompressible surfaces are non-separating? - -REPLY [10 votes]: There exist closed orientable hyperbolic 3-manifolds that are surface bundles such that the fiber is the only incompressible surface in the manifold (up to isotopy). Such manifolds can be obtained by Dehn surgery on certain 2-bridge knots, using the classification of incompressible surfaces in 2-bridge knot complements in a paper of Bill Thurston and myself in Inventiones math. 79 (1985), 225-246. Here are a few more details: -It is a well-known general fact that if one does Dehn filling of slope $p/q$ on the exterior $X$ of a knot $K\subset S^3$ to produce a closed manifold $M=M_{p/q}(K)$, then each incompressible surface in $M$ comes from an incompressible surface in $X$ that is either closed (and not the peripheral torus) or has boundary consisting of curves of slope $p/q$ in $\partial X$. -Suppose $K$ is a fibered knot, so its exterior $X$ is fibered with Seifert surface fibers. Then if one does Dehn filling with slope $p/q=0/1$ the resulting manifold $M$ will also be fibered with fibers obtained by capping off the Seifert surface fibers with disks. We would like to choose a fibered knot $K$ whose fiber is the only incompressible surface (up to isotopy) in $X$ of boundary slope $0$, and with no closed incompressible surfaces in $X$. All 2-bridge knots have the latter property, as shown in the paper cited above where it is also shown that all Dehn surgeries on nontrivial 2-bridge knots produce irreducible manifolds. The paper gives an algorithm for computing all the incompressible surfaces in $X$ and their boundary slopes. The incompressible surfaces correspond to certain continued fraction expansions of the rational number $\beta/\alpha$ classifying the 2-bridge knot, where the terms in the continued fractions are allowed to be negative as well as positive. Incompressibility corresponds to no terms being $\pm 1$. There are finitely many incompressible surfaces for each continued fraction. Seifert surfaces correspond to the unique continued fraction for $\beta/\alpha$ with all terms even, and the ones that are fibers of surface bundle structures on $X$ correspond to the case that all the terms in the continued fraction are $\pm 2$. In these cases the fiber surface is the unique surface associated to the continued fraction. There is a simple formula for the boundary slope of each surface in terms of the continued fraction. -A few examples are worked out on page 231 of the paper, including the knots $6_2$ and $6_3$ which correspond to $\beta/\alpha = 4/11$ and $5/13$, respectively. These are fibered knots, and although they have other incompressible surfaces, only the fiber surface has boundary slope $0$. This means that when we do the slope $0$ Dehn surgery we obtain a fibered closed manifold $M$ with a unique incompressible surface, the fiber. For these two knots the fiber has genus $2$ (its Euler characteristic is easy to compute from the continued fraction) so $M$ has a good chance of being hyperbolic. (It is irreducible with $H_1={\mathbb Z}$ and contains no incompressible tori.) Using SnapPea for example one can check that $M$ is actually hyperbolic in the cases of $6_2$ and $6_3$, but perhaps there's a general reason for this. The monodromy of the bundle structure should be computable and maybe it can be checked that this has infinite order in the mapping class group of the fiber, which is all that is necessary since the monodromy would have finite order if $M$ was a Seifert manifold. -It seems likely that many more choices for $\beta/\alpha$ should also work, with fiber surfaces of arbitrarily large genus.<|endoftext|> -TITLE: Loop space of a Simplicial Abelian group -QUESTION [5 upvotes]: Let $X$ be a simplicial abelian group. Let $NX$ be its normalised chain complex denoted -...$\rightarrow NX_{K}$ $\rightarrow$ $NX_{k-1}$ $\rightarrow$... -Now define a new chain complex $Y$ by shifting the normalised chain complex; i.e. $Y_{k} = NX_{k+1}$. Now applying the functor from Dold Kan correspondence on $Y$ we get a simplicial abelian group $\Gamma(Y)$ whose homotopy groups are isomorphic to those of $\Omega X$, since simplical abelian groups are cofibrant-fibrant one can say $\Gamma(Y)$ is isomorphic to $\Omega X$ in homotopy category (using Whitehead's theorem) provided there is a map $\Omega X$ $\rightarrow$ $\Gamma(Y)$ inducing an isomorphism of homotopy groups. -My question is , is there such a canonical map $\Omega X$ $\rightarrow$ $\Gamma(Y)$ . - -REPLY [8 votes]: You can find such a map, but it goes the other way: $\Gamma(Y)\to \Omega X$. It is always wise to keep one's right adjoints on the right hand side.t -But first, let me note that every simplicial abelian group $Z$ is equivalent to $\prod_{k\ge0} K(\pi_kZ,k)$ (this is just a restatement of the classical fact that every chain complex of $\mathbb{Z}$-modules is quasi-isomorphic to its homology), so just the observation about homotopy groups suffices to conclude that $\Gamma(Y)$ and $\Omega X$ are equivalent. -Now, for the construction of the map, note that there is a canonical diagram -$$\require{AMScd} -\begin{CD} - Y @>>> 0\\ - @VVV @VVV\\ - 0 @>>> N(X) -\end{CD} -$$ -equipped with a chain homotopy of the zero map $Y\to N(X)$ with itself (the latter is just the canonical isomorphism $Y[1]\to N(X)$). -Since Dold-Kan sends chain homotopies to simplicial homotopies, after applying $\Gamma$ we get a homotopy commutative square -$$\require{AMScd} -\begin{CD} - \Gamma(Y) @>>> *\\ - @VVV @VVV\\ - * @>>> X -\end{CD} -$$ -which again induces the required map from $\Gamma(Y)$ to the homotopy pullback of the rest of the diagram, that is $\Omega X$. -There is a more concrete (although messier) way of obtaining the above map, and it is via constructing the adjoint map $\Sigma \Gamma(Y)\to X$. This is because you can get the suspension via the Kan suspension functor ([1], III.5) -$$(\Sigma K)_n = K_n \vee K_{n-1}\vee \cdots \vee K_0$$ -so to get map $\Sigma \Gamma(Y)\to X$ you just need to give maps -$$(\Gamma(Y))_l=\bigoplus_{[l]\twoheadrightarrow [i]} Y_i = \bigoplus_{[l]\twoheadrightarrow [i]} (NX)_{i+1}\to X_n =\bigoplus_{[n]\twoheadrightarrow [i]} (NX)_{i}$$ -for every $l\le n$ satisfying some compatibilities. Spoiler alert: these will just be the canonical inclusions. -In fact what we are doing is just identifying $X$ with $\bar W \Gamma(Y)$, where $\bar W$ is the Eilenberg-MacLane classifying space for a simplicial group ([1] Remark III.5.6). -[1] Goerss, Jardine Simplcial Homotopy Theory, 2006<|endoftext|> -TITLE: The possible degrees of $\mathbb{Q}(a,b)$ in terms of the degrees of $a$ and $b$ -QUESTION [17 upvotes]: Given two positive integers $n,m$, which positive integers $d$ appear as the degree of $\mathbb{Q}(a,b)$ for two algebraic numbers $a$ and $b$ of degrees $n$ resp. $m$? - -Two necessary conditions are $\mathrm{lcm}(n,m) \mid d$ and $d \leq nm$. In particular, if $n,m$ are coprime, only $ d=nm$ is possible. Are they sufficient? -I have chosen $\mathbb{Q}$ just to fix ideas. Maybe the same analysis works for any field, or at least any field of characteristic zero. Answers treating more general cases are appreciated as well. - -REPLY [8 votes]: The necessary condition is not sufficient. In particular, it is impossible for $|\mathbb{Q}(a) : \mathbb{Q}| = |\mathbb{Q}(b) : \mathbb{Q}| = 5$ and $|\mathbb{Q}(a,b) : \mathbb{Q}| = 15$. This is because there is no group $G$ with subgroups $H$ and $K$ with $|G : H| = |G : K| = 5$ and $|G : H \cap K| = 15$. (So the argument works for arbitrary separable extensions.) I'm not happy with the proof I have of this, and I welcome cleaner arguments. -We may assume without loss of generality that $G$ is finite. Consider first the case in which $H$ and $K$ are conjugate in $G$. The action of $G$ on the cosets of $H$ gives a homomorphism from $G$ into $S_{5}$ with kernel contained in $H \cap K$, so we can assume that $G$ is a transitive subgroup of $S_{5}$. The only such groups of order a multiple of $15$ are $A_{5}$ and $S_{5}$ which are both doubly transitive and so the intersection of two point stabilizers has index $20$ in $G$, not $15$. -Now, suppose that $H$ and $K$ are not conjugate and consider the partition of $G$ into double cosets $HxK$ for $x \in G$. The double coset $HK$ has size $|H| |K|/|H \cap K| = \frac{3}{5} |G|$. The size of the double coset $|HxK| = |K| |H : H \cap xKx^{-1}| = |H| |K : K \cap x^{-1} Hx|$. Since $H \ne xKx^{-1}$, then $|HxK| = |K| |H : H \cap xKx^{-1}| \geq 2|K| = \frac{2}{5} |G|$. Thus, there are exactly two double cosets in the partition: $HK$ and $HxK$, with sizes $\frac{3}{5} |G|$ and $\frac{2}{5} |G|$. -It follows that $|K : K \cap x^{-1} Hx| = |H : H \cap xKx^{-1}| = 2$ and so $M_{1} = K \cap x^{-1} Hx$ is normal in both $K$ and $x^{-1} Hx$. Likewise $M_{2} = H \cap xKx^{-1}$ is normal in both $H$ and $xKx^{-1}$. Since $H$ and $K$ are maximal, it follows that $M_{1}$ and $M_{2}$ are both normal in $G$. Thus, $xM_{1}x^{-1} = x(K \cap x^{-1}Hx)x^{-1} = xKx^{-1} \cap H = M_{2}$, but since $M_{1}$ is normal in $G$, $M_{1} = M_{2}$. However, $M_{1} \subseteq H$ and $M_{2} \subseteq K$ and thus, $M_{1} = M_{2} = H \cap K$ actually has index $2$ in $H$ and $K$, which contradicts the assumption that $|G : H \cap K| = 15$. -EDIT: Here's a more general claim taking into account YCor's approach. -Claim: Suppose that $m = n$ is prime. Then $d = kn$ where -$\bullet$ $k = n$ or -$\bullet$ $k | n-1$ or -$\bullet$ if $n = \frac{p^{q}-1}{p-1}$ for primes $p$ and $q$, then $k = p^{q-1}$ or -$\bullet$ $n = 11$ and $k = 6$. -Proof: Letting $G$ act on the cosets of $H$ we get a homomorphism $\phi : G \to S_{n}$. Let $N = \ker \phi$. Then $NK$ is a subgroup of $G$ containing $K$. If $NK = G$ then $|G : N \cap K| = n^{2}$ but $N \subseteq H$ and this implies that $|G : H \cap K| = n$ or $n^{2}$. -In the case that $NK = K$, then we have that $N \subseteq K$. Replacing $G$ with $G/N$ we have that $G$ is a transitive subgroup of $S_{n}$. A theorem of Burnside from 1911 implies that either $G$ is solvable (and hence contained in the normalizer of a Sylow $p$-subgroup) or $G$ is doubly-transitive. If $G$ is solvable, then $G \cong (\mathbb{Z}/n\mathbb{Z}) \rtimes (\mathbb{Z}/k\mathbb{Z})$ where $k | n-1$ and $H$ and $K$ are conjugate in $G$ (since they are Hall $\pi$-subgroups). The orbits of a point stabilizer in such a group have size $1$ and $k$ and so $|H \cap K| = 1$ which has index $nk$. -In the case that $G$ is doubly-transitive, then for any minimal normal subgroup $N$ of $G$, $N$ is transitive and this implies that $C_{G}(N) = 1$. It follows that $G \subseteq {\rm Aut}(N)$. It therefore suffices to find the finite simple groups that are doubly-transitive. This determination follows from the classification of finite simple groups (and is given by Peter Cameron in "Finite Permutation Groups and Finite Simple Groups" published in the Bulletin of the London Mathematical Society in 1981; see the table on page 8). -The only such groups that can have prime degree are $A_{n}$, ${\rm PSL}_{q}(\mathbb{F}_{p})$ of degree $\frac{p^{q}-1}{p-1}$, ${\rm PSL}_{2}(\mathbb{F}_{11})$, $M_{11}$ and $M_{23}$. Moreover, if the subgroups $H$ and $K$ are conjugate the double transitivity implies that $|G : H \cap K| = n(n-1)$. Therefore, examples where this doesn't occur must arise from simple groups with more than one conjugacy class of subgroups of index $n$. This occurs only for ${\rm PSL}_{q}(\mathbb{F}_{p})$ with $q > 2$ and ${\rm PSL}_{2}(\mathbb{F}_{11})$. -In the case of ${\rm PSL}_{q}(\mathbb{F}_{p})$, one class of subgroups comes from the stabilizers of one-dimensional subspaces of $\mathbb{F}_{q}^{d}$ and the other class comes from stabilizers of codimension one subspaces. The stabilizer of a codimension one subspace has two orbits on the one-dimensional subspaces: one of size $\frac{p^{q-1} - 1}{p-1}$, and the other of size $p^{q-1}$. The former is a divisor of $n-1$. -Finally, in the case of ${\rm PSL}_{2}(\mathbb{F}_{11})$ there are two classes of subgroups of index $11$. If $H$ comes from the first class, the orbits of $K$ have sizes $5$ and $6$, and this gives rise to the possibility $m=n=11$ and $k = 6$. QED<|endoftext|> -TITLE: Topos properties from coverage conditions -QUESTION [8 upvotes]: For any category $C$ and coverage $J$ on it, let $\mathcal{E}:=\mathsf{Shv}(C,J)$ denote topos of sheaves on the site $(C,J)$. What sorts of results are known about the relationship between properties of the topos $\mathcal{E}$, and conditions on the 'diagrammatic shape' of the covering families in $J$? -My interest includes any sort of property of $\mathcal{E}$, for example logical properties. To that end, let $\mathtt{Prop}$ denote the subobject classifier in the internal language of $\mathcal{E}$. -Here are some first examples of the kinds of relationships I'm talking about: - -if all covering families in $J$ are empty, then $\mathcal{E}$ is terminal. -if all covering families in $J$ consist of the identity, then $\mathcal{E}$ is of presheaf type. -if all covering families in $J$ are filtered, then constant objects satisfy a dual Frobenius property. - -By #3, I mean the following. Say that a sheaf $K\in\mathcal{E}$ satisfies the dual Frobenius property if, for any proposition $P:\mathtt{Prop}$, and predicate $P':K\to\mathtt{Prop}$, the following logical statement is sound in $\mathcal{E}$: -$$\big(\forall(k:K)\ldotp P\vee P'(k)\big)\implies\big(P\vee\forall(k:K)\ldotp P'(k)\big).$$ -It is relatively straightforward—though seems to require classical reasoning—to prove that constant sheaves satisfy this property when every covering family in $J$ is filtered. -Questions: - -Are there more examples of such relationships? Obviously the three above are not very thorough (they don't have much 'coverage,' so to speak). -Is there a name for toposes of type 3 above? - -REPLY [10 votes]: This is a very broad question, we have a huge numbers of such characterization. -But part C of "Sketches of an elephant" contains most of those I know (especially C3). Basically all the notion introduced have such a "site characterization", i.e. a properties of a topos or a morphism is characterized by the existence of a Site description having some properties. -And I would like to add that Part C is in my opinion the better written and easiest to read part of the elephant. -From memory, you can find there at the very least conditions for: - -Atomic toposes, as sheaves for a topology where all non-empty sieve are covering (called an atomic topology). -Coherent toposes by combining the fact that every covering is finitely generated with existence of finite limits in the site. ( it is a bit more subtle than that, I refer you to the elephant for the precise statement) -A characterization of proper geometric morphisms. (but maybe not the simplest possible ? ) - -and a lots of other examples (open geometric morphisms, locally connected morphisms and so one...) but I do not remember all of them. -This dual frobenius property that you sate seem related to properness (maybe be not equivalent, but that is what it makes me think about...) you should have a look to the section of the elephant on properness.<|endoftext|> -TITLE: Bialynicki-Birula Decomposition and moment polytopes/graphs -QUESTION [6 upvotes]: Let $X$ be a possibly singular projective scheme which admits a torus $T$ action and has finitely many $T$ fixed points and one-dimensional $T-$orbits. There are many such schemes in the Grassmannian/flag variety for an algebraic group/Kac-Moody group. Then some theorems by GKM(Goresky-Kottwitz-MacPherson) allow us to compute the (equivariant)cohomology of $X$ from the moment graph of $X$, which consists of the zero and one dimensional $T-$orbits. -Based on the Bialynicki-Birula Decomposition (or equivalently on Morse Theory), we can give $X$ a "cell decomposition", and the cells are the "attracting sets" of the fixed points. -My questions are: how is the information of this cell decomposition reflected in the moment graph defined above? In particular, given a $T-$fixed point $p$, how is its attracting set/cell related to the set of one-dimensional edges connected to $p$ in the moment graph for $X$? For example, can we get a bound on the dimension of this cell by counting the number of the one-dimensional edges connected to $p$ in the moment graph? Any references would be appreciated. Thanks! - -REPLY [4 votes]: The moment graph comes with an additional structure: every edge is labelled by a character of $T$. More precisely, every edge $e$ corresponds to a 1-dimensional $T$-orbit closure $C_e$. Its normalization is isomorphic to $\mathbf P^1$. An orientation of $e$ determines which fixed point in $C$ corresponds to $0$ or $\infty$, respectively. Thus, each oriented edge determines a character,namely the one with which $T$ acts on the affine line $\mathbf P^1\setminus\{\infty\}$. If $-e$ equals $e$ with the reverse orientation then $\chi_{-e}=\chi_e^{-1}$. -Now let $\lambda:\mathbf G_m\to T$ be a 1-parameter subgroup and let $X^\lambda$ be its fixed point set. If $\dim X^\lambda>0$ then $X^\lambda$ would contain one of the curves $C_e$ which means $\langle\chi_e|\lambda\rangle=0$. -So assume $\langle\chi_e|\lambda\rangle\ne0$ for all $e$. Then $X^\lambda=X^T$. Let $v\in X^T$ considered as a vertex of the moment graph. Let $e$ be an edge with outgoing orientation form $v$. Then the curve $C_e$ lies in the Bialinicki-Birula cell $X(v)$ if and only $\langle\chi_e|\lambda\rangle>0$. -With this observation it should be possible to glean a lot of information about the BB-decomposition. For example using a Noether normalization argument one can show that $X(v)$ contains at least $\dim X(v)$ $T$-stable curves. This yields an upper bound for $X(v)$. If $X$ is smooth in $v$ this becomes even an equality.<|endoftext|> -TITLE: Vandermonde matrix is totally positive -QUESTION [14 upvotes]: A totally positive matrix $M\in \mathcal{M}_{n\times m}(\mathbb R)$ is such that all of its minors of all sizes are positive. It is true that any Vandermonde matrix (with well-ordered positive entries) is totally positive. It seems that this fact should be classic. Although I can prove it by a variational argument, I cannot find a reference (in books I can think of or on the Internet) and I would like to know whether this is the "standard" way of proving the result, or if there is another (algebraic?) method known to the community. -Any pointer to a reference or direct proof would be very much appreciated ! - -REPLY [9 votes]: The minors of a Vandermonde matrix are known as alternants. Their positivity follows from the fact that they are products of the Vandermonde determinant of the variables involved with a Schur function in these variables. For a short and self-contained proof of this fact, see "Corollary (The Bi-Alternant Formula)" in John R. Stembridge, A Concise Proof of the Littlewood-Richardson Rule, The Electronic Journal of Combinatorics 9 (2002), Note #N5. This note itself is the distillate of several years of algebraic combinatorics (ideas of Lindstrom, Gessel, Viennot, Gasharov, Bender and Knuth are all in there), and longer proofs have been found before (e.g., in Macdonald's book).<|endoftext|> -TITLE: Good book for measure theory and functional analysis -QUESTION [15 upvotes]: I have taken advanced courses both in measure theory and also in functional analysis (Banach and Hilbert spaces, spectral theory of bounded and unbounded operators, etc.) -The connections between the two arises in several theorems: - -Riesz theorem showing that under some conditions a continuous functional can be represented as integral with respect to some measure. -Spectral measure and functional calculus for the bounded/unbounded self-adjoint operators. - -I have also seen some other results that state that the dual of specific Banach spaces are the same than those of finitely additive measures. -In spite of having advanced course, the connection between measure theory and functional analysis is still really mysterious to me. -I would like to learn more about the connection between the two subjects in a more systematic fashion. I have already seen several related books but the connection is discussed only superficially. -I was wondering if anyone has a suggestion for a rigorous book that focuses specifically on the connection between measure theory and functional analysis. - -REPLY [6 votes]: You have to go fairly far with measure theory and functional analysis in order to use one to understand the other better. There are some intersections such as $L_p$-spaces and integral representations. Books that will teach you about happy marriages of measure theory and functional analysis are "Lectures on Choquet's theorem" by Robert Phelps and the first few chapters of "Optimal Transport: Old and New" by Cedric Villani. There is also the rich topic of vector measures; the book "Vector Measures" by Diestel and Uhl is surprisingly readable. -On a deeper level, measure theoretic results rely heavily on order structure and this is where measure theory and functional analysis have deep connections. One can, for example, obtain the Hahn decomposition from a general result valid in all vector lattices, but vector lattices are not a topic usually taught in introductory functional analysis courses. A not so gentle read on the connection between vector lattices and measure theory is "Topological Riesz Spaces and Measure Theory" by David Fremlin. -As an aside: In the heyday of Bourbaki, it used to be popular to reduce measure theory to integration theory and integration theory to a study of dual spaces via the Riesz representation theorem. This works reasonably well when doing topological measure theory on locally compact, but works in general only via some clumsy constructions using sophisticated compactification arguments. In probability theory, the most cheerful importer of measure theory, one regularly has to deal with measures on function spaces that are not locally compact, so the approach via the Riesz representation theorem is of rather limited usefulness.<|endoftext|> -TITLE: Undecidable easy arithmetical statement -QUESTION [11 upvotes]: Is there a basic arithmetic statement which is known to be undecidable ? -By basic arithmetic statement I do mean an easy statement in the spirit of the Collatz conjecture . By the way is there some reasons to believe that the Collatz conjecture is undecidable ? - -REPLY [15 votes]: The "mortal matrix" problem: Given a set of $n\times n$-matrices with integer entries, decide whether they can be multiplied, in any order and possibly with repetition, to give the $0$-matrix. If I remember correctly, it is already undecidable for $n\geq 3$. -I decided to put this here, since matrix multiplication is a sequence of simple arithmetical operations.<|endoftext|> -TITLE: Minimal number of n/2-subsets of [n] that contains every d-subset -QUESTION [8 upvotes]: Let $d , n$ be positive integers such that $d < n/2$. Consider collections $\mathcal{F}$ consisting of subsets of $[n] = \{1,2,\ldots, n\}$ of size $n/2$. Question: what is the minimal size of a collection $\mathcal{F}$, such that every size-$d$ subset of $[n]$ is contained in \textit{at least} one set in $\mathcal{F}$ ? -I did searches in literature and if the "at least" above is changed to "exactly one" then it is called Steiner designs and is considered a hard problem. Just wondering if the above version is easier... -I had some initial idea but got stuck on how to make it work: say $n = 2^k$, then we can regard $[n]$ as $\mathbb{F}_2^k$, and select all hyperplanes. But I don't know if such collection satisfies the condition ... any ideas are appreciated! - -REPLY [4 votes]: These are called covering designs. See https://www.ccrwest.org/cover.html for references and tables of known values and bounds.<|endoftext|> -TITLE: About weak convergence of probability measure -QUESTION [6 upvotes]: Suppose $\mu_j$ is a sequence of measures on $\mathbb{R}$. By the definition of weak convergence of measures, $\mu_j$ weak converges to $\mu$ means that for any bounded continuous function $f$, there holds that $$\int_{\mathbb{R}} f\mu_j \to \int_{\mathbb{R}} f\mu.$$ My question is whether we can just check the equality $\int_{\mathbb{R}} f\mu_j \to \int_{\mathbb{R}} f\mu$ for all smooth functions $f$ with compact support. -I come across this problem while I try to prove the following problem: -If $\phi_j$ are a sequence of smooth convex functions defined on $\mathbb{R}$ with uniformly bounded second order derivative and $\phi_j$ converges to the convex function $\phi$ in $L^\infty(\mathbb{R})$, then $\phi_j''dx$ weakly converges to $\phi''dx$ as measures. -Can this be obtained just from the fact that the equality $$\int_{\mathbb{R}}f \phi_j''dx\to\int_{\mathbb{R}}f \phi''dx$$ holds for all smooth functions $f$ with compact support? - -REPLY [7 votes]: If the measures are probability measures, then yes you can; it's kind of a standard exercise. -The argument I've seen goes something like this: fix $\epsilon$ and choose a smooth compactly supported cutoff function $g$ with $0 \le g \le 1$ and $\int g\,d\mu \ge 1-\epsilon$ (possible by monotone convergence); let $K$ be the support of $g$. Then by assumption $\int g\,d\mu_j \to \int g\,d\mu \ge 1-\epsilon$ so we see that $\limsup_{j \to \infty} \mu_j(K) \ge 1-\epsilon$. In other words, $\{\mu_j\}$ is tight. Hence after passing to a subsequence, $\mu_j$ converges weakly to some measure $\nu$. Now we notice that $\int f\,d\mu = \int f\,d\nu$ for all smooth compactly supported $f$, and it follows from a monotone class type argument that $\mu=\nu$. Finally use the "double subsequence" trick to conclude that the original sequence $\mu_j$ also converges weakly to $\mu$. -If you don't assume they are probability measures then this can be false; let $\mu_j$ be a point mass at $j$ and let $\mu$ be the zero measure.<|endoftext|> -TITLE: Are Pointwise conditions studied?, do they make sense?, do they have any applications? -QUESTION [5 upvotes]: In weakly formulated PDE (or even ODE), we seem to be interested in solutions that satisfy or take desired values at some boundary points of a domain we are interested in. For example, Dirichlet boundary conditions impose the function to take zero or certain value on the boundary of the domain, and so are other types like periodic boundary conditions and Neumann boundary conditions, Robin boundary conditions, etc., -My question is, can we also impose condition like, not only the function staisfy one of the above mentioned conditions, but also a new condition that function should take a so and so values at so and so points (finite in number) that lie in the interior of the domain. -Writing down as equation, let $P = \{p_i\}$ be a finite set of points in the interior of the domain, and condition is that $f(p_i) = d_i$, $d_i\in\mathbb{R}$ $d_i$ are the desired values the function should take at the points $p_i$ -I want to know the questions in the title of this post, for the above mentioned conditions. The above pointwise conditions make sense only for weakly formulated PDE, so I am not interested in strong PDE. - -REPLY [2 votes]: Such interpolatory conditions are better treated by solving your PDE $Af=\sum_ic_i\delta_{p_i}$ as distributions in the whole domain, with the $c_i$s to be determined so that $f(p_i)=d_i$. For elliptic $A$ this is exactly what splines (in the variational sense) do, typically for $A=\partial_x^2$ (linear interpolation) or $A=\partial_x^4$ (cubic splines), in one dimension, or $A=\Delta^2$ in two dimensions (thin plate splines).<|endoftext|> -TITLE: Prove an anti-concentration inequality for a martingale -QUESTION [7 upvotes]: My problem can be described easily: -I have a sequence $(X_l)_{l \in \mathbb{N}}$ of r.v. adapted to some filtration $(\mathcal{F}_l)_{l \in \mathbb{N}}$, such that - -$\left|X_{l+1}-X_l\right|\le R$ a. s. -$\mathbb{E}[X_{l+1}-X_l| \mathcal{F}_l]\le \delta$ a. s. -$X_0 = x_0$ a.s. where $x_0 \in \mathbb{R}$ is fixed -$\operatorname{Var}[X_{l+1}-X_l |\mathcal{F}_l] \ge v$ where $v>0$. - -I want to prove $\mathbb{P}[X_l \le x_0-c\sqrt{vl}+\delta l]\ge p $ for some $p>0$ and some constant $c>0$. -By using 2) one can easily reduce the problem to a martingale -$$\bar{X}_l:= \left(\sum_{k=1}^l D_k - \mathbb{E}[D_k| \mathcal{F}_{k-1}]\right)+x_0$$ where $D_k:=X_k-X_{k-1}$. The martingale again has bounded increments, $\bar{X}_l=x_0$ and $\operatorname{Var}[\bar{X}_l]\ge vl$. The task then is to prove $\mathbb{P}[\bar{X}_l \le x_0-c\sqrt{vl}]\ge p$. -Sadly, the proof in the original paper (Lemma 6.5) was wrong and as I've found out one can't adapt it after having corrected the wrong inequality (Jensen's inequality used in the wrong direction). -The only possibility I see so far is using the Martingale Central limit theorem and then try to control the speed of convergence for my original process using the inequalities described in the beginning of this paper. This seems pretty hard to check for my concrete considered process. -May I miss any obvious proof using simple martingale theory? Can anybody prove this result using only the above assumptions (or maybe some small additional assumptions, e.g. $L^{2+\delta}$ moments) ? - -REPLY [7 votes]: Basically, the proof goes along the following lines: -(1) Take a small $\varepsilon>0$ and show that the expected exit time from the interval $[-\varepsilon\sqrt{vl},\varepsilon\sqrt{vl}]$ is less than $\varphi l$ (this is standard, using the fact that your martingale squared becomes a submartingale with uniformly positive drift, see e.g. Example 7.1 of Section 4.7 of Durrett/Probability) with small $\varphi$. Chebyshev's inequality then will show that you martingale will go out of that interval with probability close to 1 until time $l$. -(2) By the Optional Stopping Theorem, with probability bounded away from 0 (in fact, even close to $1/2$) it will exit the above interval through $(-\varepsilon\sqrt{vl})$. -(3) now, you only need to show that the walker will remain to the left of (say) $(-\frac{1}{2}\varepsilon\sqrt{vl})$ till time $l$. -(4) For this, first note that the Optional Stopping Theorem implies that, starting at $(-\varepsilon\sqrt{vl})$, the probability that the walker exits the interval $[-M\sqrt{vl},-\frac{1}{2}\varepsilon\sqrt{vl}]$ through the left side is at least constant (depending on $M$). -(5) Using the Doob's inequality, we then observe that the process is unlikely to reach $(-\frac{1}{2}\varepsilon\sqrt{vl})$ in time $l$. -You may find it interesting to look at the proof of Lemma 2.1 in https://arxiv.org/abs/1201.6089, it contains all these ideas.<|endoftext|> -TITLE: Extension problem for Seiberg-Witten solutions -QUESTION [7 upvotes]: Let $X$ be a compact $4$-manifold, possibly with boundary. -Theorem 17.1.2 of Kronheimer-Mrowka's book "Monopoles and Three-Manifolds" states - -Let $X' \subset X$ be a codimension-zero submanifold with boundary, contained in the interior of the manifold $X$. The restriction map $L^2_k(X) \to L^2_k(X')$ is a surjective map with continuous left inverse. - -I was wondering if such an extension statement for solutions to the Seiberg-Witten equations could be found anywhere in the literature. -Specifically, can we fix boundary conditions $P$, $P'$ on $X$, $X'$ such that: - -There is a well-defined restriction map from $\mathcal{M}(X, P)$ to $\mathcal{M}(X', P')$, the spaces in question being the gauge-equivalence classes of solutions to the Seiberg-Witten equations satisfying the boundary conditions. -This restriction map is surjective and admits a continuous left inverse. - -REPLY [2 votes]: Here is a counterexample. -Consider a flat ball $B^4$ embedded in a closed manifold $X$ on which there are irreducible solutions to the Seiberg-Witten equations (e.g. a symplectic manifold). Suppose your conjecture is true for some boundary conditions $P_B$ on $\partial B^4$ (there are no boundary conditions on $X$, obviously). Thus $\mathcal{M}(B^4,P_B)$ is nonempty and, specifically, contains an irreducible solution. -Now embed $B^4$ in any closed connected $4$-manifold $X'$ which has nonnegative, somewhere positive scalar curvature. By the Weitzenbock formula and a unique continuation theorem for the Dirac equation, there are no irreducible solutions on $X'$, so we cannot extend the irreducibles from $\mathcal{M}(B^4,P_B)$ to $\mathcal{M}(X',\varnothing)$. -Another (less rigorous) argument: the Seiberg-Witten equation may blow up in finite time on a cylinder $[a,b] \times Y$. Your conjecture would imply existence of some boundary conditions on the cylinder which would guarantee that the solutions do not blow up on $[a-t,b+t] \times Y$ for any $t$. I suppose there should be embeddings of $[a,b] \times Y$ into closed $4$-manifolds $X$ such that restricting the solutions on $X$ to the cylinder yields solutions which blow up in finite time on $\mathbb{R} \times Y$.<|endoftext|> -TITLE: LMS Lectures on Geometric Langlands -QUESTION [15 upvotes]: Everybody knows how insightful are David Ben-Zvi talks (and comments/answers here on mathoverflow). I was trying to watch the LMS 2007 Lecture Series on Geometric Langlands by David, supposedly made available via GRASP Lectures, but the server is down or no longer exists. -Some of the GRASP lectures are available on the iTunes Podcasts of the University of Texas Austin, but unfortunately not these ones. -Does anybody know where else to find them online or did anyone downloaded them back then and could, please, make them available to the community again? - -REPLY [10 votes]: The videos from the LMS lectures and all of the GRASP videos are now available again from the links you gave (for download, not streaming). Many apologies for their long hiatus offline and many thanks for your enthusiasm and persistence!! Please do email me for broken links etc., and I will update the site with some more recent materials.<|endoftext|> -TITLE: Are finite nilpotent groups the only finite groups with abelian Frattini quotient? -QUESTION [6 upvotes]: It is obvious that if the Frattini quotient of a finite group $G$ is abelian, then $G$ is abelian by nilpotent and that finite nilpotent groups have abelian Frattini quotient. -I wonder if there is an example of a non-nilpotent finite group with abelian Frattini quotient ? -A better thing would be to have a nice characterization of such finite groups. - -REPLY [4 votes]: Here is my attempt to entertain those who knew the answer. - - -Question 1. Let $G$ be a finitely generated group such that $G/\Phi(G)$ is Abelian. Is $G$ nilpotent? - - -The answer is yes if we assume moreover that $G$ is linear (e.g., $G$ is finite). But the answer is no in general, as the the first Girgorchuk group $G_1$ is such that $[G_1, G_1] \subset \Phi(G_1)$. For more examples, see this MO post. -Let $W(G)$ denote the $N$-Frattini subgroup of $G$, that is the intersection of the maximal normal subgroups of $G$ when defined, $G$ otherwise. (See this post for results related to $W(G)$.) Note that $G/W(G)$ is Abelian for any soluble group $G$. - - -Question 2. Let $G$ be a finitely generated group such that $G/W(G)$ is Abelian. Is $G/\Phi(G)$ Abelian? - - -The answer is no, because some Sunic group of intermediate growth is a counter-example, see this preprint. - - -Question 3. Let $G$ be a finite group such that $G/W(G)$ is Abelian. Is $G$ soluble? - - -The answer is no because of the symetric group $S_n$ with $n \ge 5$. One may try to get a classification though. Possibly useful: - - -Lemma (Baer). Let $G$ be a group and let $N$ be a normal subgroup of $G$. Then $W(N)$ is a normal subgroup of $G$ and $W(N) \subset W(G)$.<|endoftext|> -TITLE: a comparison between LS and cohomological dimension -QUESTION [6 upvotes]: Let $X$ a simply connected elliptic space. Assume $\pi_\star(X)\otimes\Bbb{Q}$ is concentrated in odd degrees. So, we have $dim~\pi_\star(X)\otimes\Bbb{Q}=TC(X_\Bbb{Q})=catX_\Bbb{Q}$ (ie) the topological complexity and the Lusternik-Schnirelmann category. I want to compare the LS category or the $TC$ with $dim~H^\star(X;\Bbb{Q}).$ I just want an idea if it's possible - -REPLY [2 votes]: The Hilali conjecture is a well-known conjecture in rational homotopy theory, which asks whether the inequality -$$ -\dim \pi_*(X)\otimes\mathbb{Q}\leq H^*(X;\mathbb{Q}) -$$ -always holds for $1$-connected rationally elliptic spaces $X$. You ask about the oddly generated case. I believe the Hilali conjecture is still unkown in this generality, but is known to hold if in addition the space is coformal, meaning the differential in the minimal model is purely quadratic.<|endoftext|> -TITLE: When are these definitions of "toric variety" equivalent? -QUESTION [7 upvotes]: Let $k$ be an algebraically closed field. Let $X$ be an integral $k$-scheme, separated and of finite type over $k$. Let $d := \dim X$, let $T := (\mathbb{G}_{m,k})^{d}$ be the $d$-dimensional torus, and suppose we have an action of $T$ on $X$ over $k$ given by the action morphism $\sigma : T \times_{k} X \to X$. Let $$ \mathrm{stab}_{\sigma} : Z_{\sigma} \to X $$ be the stabilizer of $\sigma$, namely the pullback of the morphism $(\sigma,p_{2}) : T \times_{k} X \to X \times_{k} X$ via the diagonal $\Delta_{X/k} : X \to X \times_{k} X$. For a $k$-point $x \in X(k)$, let $$ \alpha_{x} : T \to X $$ denote the orbit morphism of $x$, namely the composition $\mathrm{id}_{T} \times x : T \to T \times_{k} X$ with $\sigma$, and we denote by $T \cdot x \subset X$ the (set-theoretic) image of $\alpha_{x}$. By the closed orbit lemma, we have that $T \cdot x$ is a locally closed subset of $X$. -Depending on the source, the scheme $X$ is called a toric variety if it satisfies one of the following conditions (ordered roughly from strongest to weakest): - -The scheme $X$ is given by a construction involving fans (e.g. [4, Theorem 1.4] or [1, Section 2.1]). -There is a $T$-equivariant open immersion $j : T \to X$, where $T$ acts on itself by left multiplication. -The action $\sigma$ has generically trivial stabilizers, i.e. there exists a dense open subscheme $U \subseteq X$ such that the restriction $\mathrm{stab}_{\sigma}^{-1}(U) \to U$ is an isomorphism. -The action $\sigma$ has generically finite stabilizers, i.e. there exists a dense open subscheme $U \subseteq X$ such that the restriction $\mathrm{stab}_{\sigma}^{-1}(U) \to U$ is a quasi-finite morphism. -There is a $k$-point $x \in X(k)$ such that $T \cdot x$ is an open subset of $X$. -The action $\sigma$ is faithful, i.e. the corresponding morphism of group sheaves $T \to \underline{\mathrm{Aut}}_{(\mathrm{Sch}/k)}(X)$ on the category of $k$-schemes $(\mathrm{Sch}/k)$ is injective. - - -Do the above conditions imply any others (possibly with mild assumptions about $X$)? - -At the moment, I am most interested in whether (4) implies (2). -Here is what I have so far: In (1), the scheme $X$ is obtained by equivariantly gluing certain affine $T$-schemes which satisfy (2), hence (1) implies (2). If $X$ is normal, then (2) implies (1) by Sumihiro's theorem [4, Theorem 1.5]. For (2) implies (3), take any $k$-point $x \in X(k)$ lying in the image of $j$. Clearly (3) implies (4). For the equivalence of (4) and (5), use the orbit-stabilizer theorem for algebraic groups (i.e. that $\dim (T \cdot x) + \dim (x^{-1}(\mathrm{stab}_{\sigma})) = \dim T$, see [2, Proposition 3.20]) and the assumption that $T$ and $X$ have the same dimension (since $T \cdot x$ is locally closed, saying that $T \cdot x$ is open is the same as saying that $\dim (T \cdot x) = d$). For (2) implies (5), either compose previous implications or take $U = j(T)$. For (2) implies (6), restriction via $j$ gives a map $\underline{\mathrm{Aut}}_{(\mathrm{Sch}/k)}(X) \to \underline{\mathrm{Aut}}_{(\mathrm{Sch}/k)}(T)$ such that the composition $T \to \underline{\mathrm{Aut}}_{(\mathrm{Sch}/k)}(X) \to \underline{\mathrm{Aut}}_{(\mathrm{Sch}/k)}(T)$ is injective, hence the first arrow is injective. -References: -[1] Elizondo, Lima-Filho, Sottile, Teitler, "Arithmetic Toric Varieties", Mathematische Nachrichten, vol. 287, no. 2-3 (2014) link -[2] Hoskins, "Moduli problems and geometric invariant theory", online course notes (2015) -[3] Kempf, Knudsen, Mumford, Saint-Donat, "Toroidal Embeddings 1", Springer Lecture Notes in Mathematics, vol. 339 (1973) -[4] Oda, "Convex Bodies and Algebraic Geometry", Ergebnisse der Mathematik und ihrer Grenzgebiete, vol. 3 (1985) - -REPLY [5 votes]: As noticed by Dave Anderson, (4) does not imply (2) because the generic stabilizer might be finite but non-trivial. But still, let $E$ be the stabilzer subgroup scheme of $x$ as in (5). Because $T$ is abelian, $E$ is normal in $T$. Hence $E$ acts trivially on the open set $Tx$. The fixed point scheme of $E$ being closed this implies that $E$ acts trivially on all of $X$. Hence the action of $T$ on $X$ factors through the torus $T_0=T/E$ for which then (2) is satisfied. -More interesting is the relationship of (6) and the other conditions. It is known that an effective torus action has trivial generic stabilizers. Therefore (6) implies the other conditions except for (1) where one needs normality. -To see the claim one can e.g. use that subgroup schemes of a torus are rigid. More precisely, let $U\subseteq X$ be open, dense over which $Z_\sigma$ is flat. Then one can show that it is a constant subgroup scheme of $T\times U$. Therefore any fiber $E$ acts trivially on $U$, hence on $X$. Thus $E=1$ by (6).<|endoftext|> -TITLE: Inequalities on elementary symmetric polynomials -QUESTION [14 upvotes]: I have recently come across the following result. - -Let $0 < d \leq n$. Given any vector $x \in \mathbb{R}^n$ that satisfies $e_{d-1}(x) = 0$, show that $$|x_1 \cdots x_d| \leq |e_d(x)|$$ where $e_k$ is the $k$-th elementary symmetric polynomial. - -I think the inequality is tight exactly when $x_{d+1} = \cdots = x_{n} = 0$. I believe this result is true, but I am having problems making progress. Any references would be appreciated. - -REPLY [5 votes]: It looks false to me and the counterexample is the second one that comes to one's head: three $1$'s and plenty (say, $m$) of $-x$ where $x$ is a small number. -$e_2=3-3mx+\frac{m(m-1)}{2}x^2=0$ means that $x=y/m+o(1/m)$ where $y$ is the root of $3-3y+y^2/2=0$, i.e., $y=3-\sqrt 3$. -Then -$e_3=1-3mx+3\frac{m(m-1)}{2}x^2-\frac{m(m-1)(m-2)}{6}x^3 -\\ -\approx 1-3y+3y^2/2-y^3/6\approx -0.73$, -which is smaller than $1\cdot 1\cdot 1=1$ in absolute value.<|endoftext|> -TITLE: What rectangles can a set of rectangles tile? -QUESTION [8 upvotes]: (I asked this question first on math.stackexchange, but did not get any responses so I thought I would try here.) -If we have a set of $p_i \times q_i$ rectangles ($p_i, q_i \in \mathbf{N}$), which $m \times n$ rectangles can be tiled with copies from the set? (No rotation allowed.) -I am particularly interested in the algorithm that realizes Theorem 4 below. -What I know so far: -Theorem 0 - -We need $mn = \sum p_iq_ic_i$ for some $c_i \geq 0$. -Considering how rectangles form the border, we need at least $m = \sum a_ip_i$ and $n = \sum b_iq_i$ for some $a_i \geq 0$ and $b_i \geq 0$. - -Theorem 1 -For two rectangles with $\gcd(p_1, p_2) = \gcd(q_1, q_2) = 1$, a tiling exists if and only if one of the following holds [2]: - -$p_1 \mid m$ and $q_1 \mid n$ -$p_2 \mid m$ and $q_2 \mid n$ -$p_1q_1 \mid m$ and $ap_2 + bq_2 = n$ for some integers $a, b$ -$p_2q_2 \mid n$ and $ap_1 + bq_1 = n$ for some integers $a, b$. - -Theorem 2 -For any number of rectangles, if any side of all rectangles share a common factor, then they can only tile a larger rectangle if one side has the same common factor [3]. -(Between these first theorems dealing with sets of two rectangles is easy.) -Theorem 3 A set of three or more rectangles satisfying $\gcd(p_i, p_j) = \gcd(q_i, q_j) = 1$ , for $i \neq j$ there exist some $C$ such that the set will tile any rectangle with $m, n > C$ [4, 5]. -How to find such a $C$ is given in [3]. Unfortunately, this $C$ can be quite large, and is not generally tight (there is a smaller $C$ that also works). So there is a whole bunch of rectangles for which it says nothing. -In addition, it says nothing about rectangles that do not satisfy the conditions. For example, it is hard to know much about which rectangles can be tiled by -a set with a $6\times 6, 10\times 10$ and $15 \times 15$ rectangle. In this example, pairs of squares share a common factor, but we cannot reduce the tiling problem because there is not a common factor among all tiles. -Theorem 4 For every finite set of rectangular tiles, the -tileability problem of an $m\times n$ rectangle can be decided in $O(\log mn)$ time. -This result is mentioned in [4] (and some others), but unfortunately it references a mysterious unpublished manuscript, and there is no details available; no proof, and no hint at what the algorithm might be. -(The unpublished manuscript is Tiling rectangles with rectangles by Lam, Miller and Pak. I also saw a reference to "Packing boxes with boxes", also unpublished, by the same authors, which I suspect is the same. I could find neither one :-/) - -I also wrote a computer program to investigate some examples. My own optimized-but-still-exponential-time algorithm starts becoming unpractical around for $m, n >80$ with a set of only three tiles, so I have not been able to get much insight from it. - -[2] Bower, Richard J.; Michael, T.S., When can you tile a box with translates of two given rectangular bricks?, Electron. J. Comb. 11, No. 1, Research paper N7, 9 p. (2004). ZBL1053.05027. -[3] de Bruijn, N.G., Filling boxes with bricks, Am. Math. Mon. 76, 37-40 (1969). ZBL0174.25501. -[4] Labrousse, D.; Ramírez Alfonsín, J.L., A tiling problem and the Frobenius number, Chudnovsky, David (ed.) et al., Additive number theory. Festschrift in honor of the sixtieth birthday of Melvyn B. Nathanson. New York, NY: Springer (ISBN 978-0-387-37029-3/hbk; 978-0-387-68361-4/ebook). 203-220 (2010). ZBL1248.11022. -[5] Pak, Igor; Yang, Jed, Tiling simply connected regions with rectangles, J. Comb. Theory, Ser. A 120, No. 7, 1804-1816 (2013). ZBL1314.05034. - -REPLY [4 votes]: This is not a complete answer, but another piece of the puzzle. This all follows from @Aaron Meyerowitz's idea of extending the range of Theorem 3 by building rectangles from your tile set that satisfy the conditions of Theorem 3, so you can prove the set will tile any sufficiently large rectangle. -Theorem 5 -(The numbering is just to keep track; it really is a more general version of Theorem 2.) -If we can partition a set of rectangles into two partitions, where the first has a common factor $r$ among their widths, and the second has a common factor $s$ among their heights, then any tileable rectangle either has width with factor $r$, or height with factor $s$. -The proof follows easy from the tiling by rectangles of integer side theorem (Theorem 2 as stated earlier also follows from this) which states that any rectangle which is tileable by rectangles with at least one side an integer, has at least one side an integer. (There are 14 proofs of this fact in [1].) -To prove Theorem 5, convert a tiling problem to a new one with widths and heights divided by $r$ and $s$ respectively. In the new problem, partition 1 rectangles have integer widths, and partition 2 rectangles have integer heights. Together, they can only tile a rectangle with either integer width or integer height. Translating it back to the original problem, we see the original tiles can only tile a rectangle whose width has factor $r$ and height has factor $s$. - -Applying this to my example problem using the tile set $6\times 6$, $10 \times 10$, and $15 \times 15$, we can form partitions several ways; one way is $\{ 6\times 6, 10\times 10\}$ and $\{15 \times 15\}$, giving us tileable rectangles have either width with a factor of 2, or height with a factor of 16. $r = 2, s = 15$. Other ways of partitioning gives us other conditions; there are 6 in total (here, $m$ and $n$ are the width and height of the rectangle we wish to tile): - -$2\mid m$ or $15\mid n$ -$3\mid m$ or $10\mid n$ -$5\mid m$ or $6\mid n$ -$15\mid m$ or $2\mid n$ -$10\mid m$ or $3\mid n$ -$6\mid m$ or $5\mid n$ - -Now a tileable rectangle must satisfy all 6 conditions. So, for example, if it has a width $m = 6$, then it must, to comply with conditions 3, 4 and 5, a height with a factor of $6$. It is also not hard to see that a tileable rectangle must have one side with factor 6, 10, or 15; and when one side is prime the other has factor of 30 (which corresponds with my observations from my experiments). - -Looking a bit more broadly, for all sets of three rectangles, we have one of these cases: - -$\gcd(p_i, p_j) = \gcd(q_i, q_j) = 1$ for $i \neq j$ (so we can tile any rectangle "sufficiently large"). -Each rectangle has one side with a factor $r$, so we can only tile rectangles with one side with a factor $r$. -We can partition the rectangle as in Theorem 5 (and so we can only tile rectangles which either has width with factor $r$ and height with factor $s$). - -When we have the last case, we cannot build 3 rectangles from our set such that no pairs have common factors among either their widths or heights, since for every three rectangles 2 share a factor along either width or height. So we can never apply Theorem 3. -(However, we can say slightly more. I have not worked out the details of this exactly, but sometimes you can tile every rectangle with width of a certain factor for sufficiently large heights.) - -For larger sets of rectangles, things become interesting again. There are 4 cases, the three above and an additional case where none of the others apply. Currently, my suspicion is that when we are in this case, then we can build bigger rectangles that satisfy the conditions of Theorem 3. I have managed to do this with a handful of examples, but have not yet looked at a general proof. [I will update this answer if I find anything.] - -Update -This is the situation for tile sets with 4 or more rectangles: -Theorem 6 -For a set of 4 or more rectangles, one of the following is true: - -We can select from the set 3 rectangles such that $\gcd(p_i, p_j) = \gcd(q_i, q_j)$ for $i \neq j$. -We can partition the set as in Theorem 5. -We select four rectangles that can tile any sufficiently large rectangle. - -For case 1 and 3 we can therefor tile any sufficiently large rectangle, and for case 2 at least one of the sides must have a certain factor (and therefor there are some rectangles, however large, we cannot tile). -The proof of this is a bit tedious. We can use induction on the number of rectangles in the tile set. The base case for $n = 3$ is is discussed above. The rest is just confirming that adding a rectangle to a set that satisfy one of the three cases will lead to a set that will also follow one of these three cases. (It's tedious because the new rectangle can share factors in various ways with the existing set). -The only tricky bit is dealing with case 3. The basic idea is, supposing the other cases don't hold, that there are four rectangles $R_1, \cdots, R_4$, that satisfy: - -$\gcd(p_1, p_2) = r > 1$ -$\gcd(p_3, p_4) = s > 1$ -$\gcd(q_i, q_j) = 1$, for $i, j = 1, 2, 3, 4$, $i \neq j$ -$\gcd(r, s) = 1$ - -(OR, symmetrically, all $q$s and $p$s swapped.) -$\DeclareMathOperator{\lcm}{lcm}$ -Now let $u = \lcm(p_1, p_2)$ and $v = \lcm(p_3, p_4)$. We can then build these rectangles: - -$u \times q_1$ -$u \times q_2$ -$v \times q_3$ -$v \times q_4$ - -Form the first two, we can then build $u \times x$ for large enough $x$, and from the second two $v \times y$ for large enough $y$. Furthermore, if $x = y$, since $\gcd(u, v) = 1$, from these two rectangles we can build $z \times x$ rectangles for any large enough $z$. - -This completes the "for sufficiently large" and "has a factor" type characterization; of course there is still what happens if the rectangles we wish to tile is not sufficiently large, or they do have the required factors (since these does not guarantee a tiling exists). - -[1] Wagon, Stan, Fourteen proofs of a result about tiling a rectangle, Am. Math. Mon. 94, 601-617 (1987). ZBL0691.05011.<|endoftext|> -TITLE: On non-isotriviality of families -QUESTION [6 upvotes]: Let $f:X\to S$ be a smooth proper morphism of schemes with geometrically connected fibres. Assume $S$ is a smooth irreducible variety over $\mathbb{C}$. Assume that there is a sequence of closed points $(s_i)_{i=1}^\infty$ in $S$ such that the fibres $X_{s_i}$ are pairwise non-isomorphic over $\mathbb{C}$. -Is there a curve $C\subset S$ such that the restricted family $f|_{X_C}:X_C\to C$ is non-isotrivial? (That is, are there infinitely many points $c_i$ such that the fibres $X_{c_i}$ are pairwise non-isomorphic?) -I think the answer is yes, and that one can construct $C$ by taking a "general" curve. But how does one make this precise? - -REPLY [5 votes]: Let $k$ be an algebraically closed field that is uncountable. Let $S$ be finite type $k$-scheme. Let $f:X\to S$ be a proper, flat morphism of schemes. -Proposition. For every closed subset $T\subset S$, if there exists an infinite subset of $T(k)$ such that the corresponding fibers of $f$ are pairwise non-isomorphic, then there exists an irreducible curve $C\subset T$ such that $X_C\to C$ is non-isotrivial. In particular, for $T$ equals $S,$ if $f$ has pairwise non-isomorphic fibers over an infinite subset of $S(k)$, then there exists an irreducible curve $C\subset S$ such that $X_C\to C$ is non-isotrivial. -Proof. This is proved by Noetherian induction on the closed subset $T$ of $S.$ If $T$ is reducible, then the infinite subset of $T$ has infinite intersection with one of the finitely many irreducible components of $T.$ Thus, to prove the proposition, it suffices to assume that $T$ is irreducible. -Denote $X\times_S T$ by $X_T.$ On $T\times T,$ form the two pullback families $X_1=X_T\times T$ and $X_2 = T\times X_T.$ Edit. Thanks to Laurent Moret-Bailly who pointed out the projectivity issue. First assume that $X$ is projective. Then by the existence of the Hilbert scheme, there is a locally finite type morphism of schemes $\rho:I\to T\times T$, whose domain $I =\text{Isom}_{T\times T}(X_1,X_2)$ is a countable increasing union of quasi-compact opens $I_j$, and there exists a universal $I$-isomorphism of the pullback families, $$ \phi: I\times_{T\times T} X_1 \to I\times_{T\times T} X_2.$$ In case $f:X\to S$ is only proper, flat (and locally finitely presented), then Artin's paper Algebraization of Formal Moduli, I (reference to follow) proves that there is such $I$ that is an algebraic space. Every algebraic space has an fppf representable morphism $I'\to I$ such that $I'$ is a scheme, which means that $\phi'$ over $I'$ will not be universal, but for every isomorphism $\phi_J$ after base change by a morphism $J\to T\times T$, there will exist an fppf morphism of schemes $J'\to J$ and a morphism $J'\to I'$ such that the pullback of $\phi_J$ to $J'$ equals the pullback of $\phi'$ to $J'$. In both cases, the image of $I\to S\times S$, resp. of $I'\to S\times S$, is a countable increasing union of constructible subsets $Z_j=\rho(I_j).$ -Since $k$ is uncountable, if the union contains all $k$-points of a nonempty Zariski open (or even just a nonempty analytic open when $k=\mathbb{C}$), then one of the $Z_j$ contains a dense Zariski open. In this case, the union of all opens contained in the image of $\rho$ is a dense open of the form $U\times U$ for $U\subset T$ a dense Zariski open. Thus, the infinite subset of $T$ must be in the complement $T'=T\setminus U.$ By Noetherian induction, the claim holds in this case. -Thus, assume that the image of $\rho$ contains no nonempty open, i.e., every closure $\overline{Z}_j$ is a nowhere dense closed subset of $T\times T.$ Since $k$ is uncountable, there exists a point $p\in T(k)$ such that $\{p\}\times T$ has nowhere dense intersection with every $\overline{Z}_j.$ For a very general complete intersection curve $C\subset T$ that contains $p,$ also $\{p\}\times C$ has nowhere dense intersection with every $\overline{Z}_j.$ Thus, $C\times C$ has nowhere dense intersection with every $\overline{Z}_j.$ Therefore, the restriction of $X$ over $C$ is non-isotrivial. Thus, the result is proved by Noetherian induction on $T.$ QED -Example. There are examples where you need to remove a dense open $U$ as in the proof. For instance, let $S$ be the Hilbert scheme of length $4$, zero-dimensional closed subschemes of $\mathbb{P}^2$ that are reduced. Let $X\to S$ be the blowing up of $S\times \mathbb{P}^2$ along the universal closed subscheme of $S\times \mathbb{P}^2.$ This is isotrivial over a dense Zariski open subset of $S$, but it is non-isotrivial over the locus parameterizing linearly degenerate closed subschemes.<|endoftext|> -TITLE: Are the pure genus zero mapping class groups residually torsion-free nilpotent? -QUESTION [6 upvotes]: Are the pure genus zero mapping class groups residually torsion-free nilpotent? - -They are --a quotient of the pure braid groups (which are residually torsion-free nilpotent). --torsion-free. - -REPLY [7 votes]: The answer is yes. This is because these groups are fundamental groups of complements of fiber-type hyperplane arrangements; the fact that such groups are residually torsion free nilpotent goes back to Falk and Randell. -Indeed we are considering the fundamental group of the moduli space $M_{0,n}$ of genus zero Riemann surfaces with $n$ distinct ordered marked points. Since the group of Möbius transformations acts 3-transitively on the Riemann sphere, we can take the first three markings to be $0,1,\infty$. This identifies $M_{0,n}$ with the space $$\{(x_1,\ldots,x_{n-3}) \in \mathbf C^{n-3} : x_i \neq x_j, x_i \neq 0, x_i \neq 1 \text{ for all }i,j\},$$ -which is of the required form.<|endoftext|> -TITLE: Are these large cardinals properties equivalent? -QUESTION [9 upvotes]: Consider the three following large cardinal axioms: - -there exists a nontrivial elementary embedding $j:V\to V$. -there exists a n.e.e. $j:V\to M$ such that $M^{j^\omega(crit(j))}\subseteq M$. -there exists a n.e.e. $j:V_{\lambda+2}\to V_{\lambda+2}$ for some $\lambda$. - -Kunen's inconsistency theorem states that, in ZFC (with replacement extended to formulas containing $j$), these three statements are false. However there is no such (known) result within ZF (with replacement sufficiently extended) alone. -Clearly (1) implies (2). I remember seeing somewhere that (2) implies (1), but I don't remember where, so I might be mistaken. Does anyone knows how these three statements relates to each other? -EDIT: this question already deals with the case (1)=>(3) - -REPLY [10 votes]: No. -This is a very recent work in progress of myself with Juan Aguilera. - -Definition. We say that $\kappa$ is a Kunen cardinal if there is a nontrivial elementary embedding $j\colon V_{\lambda+2}\to V_{\lambda+2}$ with $\lambda=\sup j^n(\kappa)$. - -(Sometimes it is easier to talk about the critical points of the sequence, and sometimes on $\lambda$, like in the case of $I0$ and such.) - -Theorem. If $\kappa$ is a Reinhardt cardinal, i.e. the critical point of an elementary embedding $j\colon V\to V$, then $\kappa$ is the limit of Kunen cardinals. - -Proof. Note that being a Kunen cardinal is a first-order property in the language of set theory. And note that a Reinhardt cardinal itself is Kunen by a fairly easy verification. -Let $j\colon V\to V$ be a nontrivial elementary embedding, and $\kappa$ its critical point. Now, it is easy to see that the set $A\subseteq\kappa$ defined as $\{\mu<\kappa\mid\mu\text{ is a Kunen cardinal}\}$ satisfies $\kappa\in j(A)$. Therefore it follows that $A$ is in the normal measure derived from $j$, and therefore it is stationary. -In particular, a Reinhardt cardinal is the limit of Kunen cardinals. $\square$<|endoftext|> -TITLE: References for Langlands classification -QUESTION [17 upvotes]: I kindly ask about some references concerning the representation theory of the Langlands dual of a compact Lie group, and how it relates to things related to the original compact Lie group. -My background: I know some basic facts about Lie groups/algebras, such as their root systems, Weyl groups etc. I am not familiar yet with the Langlands program, and related things. I am mostly interested for now in working over $\mathbb{R}$ and $\mathbb{C}$. -Edit 1: after some research, I realize that what I want is a reference on the Langlands classification, done by Langlands himself. So I will start by reading that article by Langlands ("On the classification of irreducible representations of real algebraic groups"). -Edit 2: I found some introductory notes on endoscopy by J-P Labesse, which look very promising to me! http://www.math.utah.edu/~ptrapa/src2006/labesse.pdf -Edit 3: Knapp's article, suggested by Desiderius Severus, is indeed a really good introduction to the Langlands classification, which is part of the local Langlands program in the Archimedean case. In some of my comments, one can see that I was confused between the Langlands classification, and the geometric Satake isomorphism, which plays a role in the Geometric Langlands program. I apologize for this confusion. It took me some time to get used to some of the jargon of the Langlands program (and even now, I cannot claim to have mastered the jargon, but I have improved a little). - -REPLY [13 votes]: The first source in which I really discovered quite explicitly the archimedean local Langlands classification is in this beautiful article of Knapp, reviewing it in some pages. Moreover, it has the appeal to give a short historical motivation, to deal with the $SL_2$ case, and then to turn to the general one in an explicit manner. - -A. Knapp, The Local Langlands Correspondence: The Archimedean Case, - Proc. Symp. Pure Math, Volume 55 (1994), Part II<|endoftext|> -TITLE: Submanifolds of $\mathbb{R}^N$ whose local charts have uniformly bounded derivatives -QUESTION [9 upvotes]: Working on a problem in Differential Geometry, which is quite far away from my area of expertise, I was recently led to consider the class of those smooth, $n$-dimensional embedded submanifolds $M \subset \mathbb{R}^N$ such that the following condition is satisfied: - -the manifold $M$ possesses an oriented atlas $\{(U_{\alpha}, \, \phi_{\alpha})\}$, where $\phi_{\alpha} \colon U_{\alpha} \stackrel{\cong}{\longrightarrow} V_{\alpha} \subset \mathbb{R}^{n}$, such that, denoting by $i_{\alpha} \colon U_{\alpha} \hookrightarrow \mathbb{R}^N$ the inclusion map, the composition $$i_{\alpha} \circ \phi_{\alpha}^{-1} \colon \, V_{\alpha} \longrightarrow \mathbb{R}^N$$ - has bounded partial derivatives up to order $k$, uniformly on $\alpha$. - -My question is now the following: - -Does the condition above have a name? In this case, is there any characterization of the manifolds $M$ satisfying it in terms of the usual Riemannian invariants (for instance, curvature)? - -Any answer or reference to the relevant literature will be greatly appreciated. - -REPLY [4 votes]: Posting this a an answer since it's too long for a comment. -I have a feeling that the question as formulated might be ill posed, or at least not quite what you'd expect in the sense that all embedded manifolds $M \subset \mathbb{R}^N$ satisfy it, as already noted by Deane Yang and Fan Zheng: you didn't fix an atlas on $M$, so you can always stretch charts. -Possibly a better condition would be to consider if the embedding $\iota: M \to \mathbb{R}^N$ is a $C^k$ bounded map in the sense of bounded geometry, e.g. with Definition 2.9 from my book (that Thomas Rot already linked to) and with $M$ having its metric induced by the Euclidean metric of $\mathbb{R}^N$. Modulo some issues with loss of smoothness due to curvature conditions in (that Rbega seems able to address) this should be equivalent to $C^k$ boundedness of $M$ in terms of local graphs. -The fact that your original condition would be satisfied by any $C^k$ embedding $M$ seems to match with the fact that any smooth manifold possesses a metric of bounded geometry (see Greene, "Complete metrics of bounded curvature on noncompact manifolds", Arch. Math. (31) 1978), as mentioned by Deane Yang.<|endoftext|> -TITLE: Homotopy groups of smooth part of moduli space -QUESTION [6 upvotes]: Let $M_g$ be the moduli space of Riemann surfaces, as described for example in the book of Harris and Morrison - Moduli of curves. As a topological space, or better as orbifold, it has smooth points and singular points. Let $S\subseteq M_g$ be the subset of smooth points. In the article by Cornalba ("On the locus of curves with automorphisms") this subset is identified, in the case $g>3$, with the set of those curves which have just the trivial automorphism. -Is there any idea (or any reference) of how to compute the homotopy groups -$$\pi_2(M_g,S) \qquad \mbox{and} \qquad \pi_1(S)?$$ -Edit: -I should have included a base point above, but I intentionally didn't because I wouldn't know which one to choose. - -REPLY [9 votes]: I'm not sure why you're talking about codimension $1$ components of $S$ since the smooth locus $S$ is actually dense in $\mathcal{M}_g$. -Anyway, the fundamental group of the locus $S$ of smooth points is the mapping class group $Mod_g$ of the surface $\Sigma_g$. To see this, recall that $\mathcal{M}_g$ is the quotient of Teichmuller space $\mathcal{T}_g$ by the action of $Mod_g$. Let $\widetilde{S}$ be the preimage of the smooth locus in $\mathcal{M}_g$ under the projection $\mathcal{T}_g \rightarrow \mathcal{M}_g$. The action of $Mod_g$ on $\mathcal{T}_g$ preserves $\widetilde{S}$, and moreover the restriction of the action of $Mod_g$ to $\widetilde{S}$ is free (the only fixed points of the action of $Mod_g$ on $\mathcal{T}_g$ come from curves with automorphisms). It follows that $\widetilde{S} \rightarrow S$ is a regular cover with deck group $Mod_g$. To prove that the fundamental group of $S$ is $Mod_g$, it is enough to prove that $\widetilde{S}$ is $1$-connected. For this, recall that the locus of curves with automorphisms has complex codimension $2$ in $\mathcal{M}_g$ as long as $g > 3$. This implies that the complement in $\mathcal{T}_g$ of $\widetilde{S}$ also has complex codimension $2$ and thus real codimension $4$, which implies in particular that the inclusion $\widetilde{S} \hookrightarrow \mathcal{T}_g$ induces an isomorphism on fundamental groups. The desired result now follows from the fact that $\mathcal{T}_g$ is contractible.<|endoftext|> -TITLE: Stability of root-finding near the unit circle -QUESTION [5 upvotes]: It is stated in several sources on numerical analysis that the general problem of polynomial root-finding is ill-conditioned, but that it is well-conditioned if the roots are near the unit circle. (e.g. see page 133 of Approximation Theory and Approximation Practice by Trefethen.) What is the underlying reason for this? Thank you very much. - -REPLY [8 votes]: The issue is explained nicely in Six Myths of Polynomial Interpolation and Quadrature. It is not a stability problem of polynomial root finding, but a problem of finding the proper representation of the polynomial. If the roots are on or near the unit circle, you want to express the polynomial in an orthogonal basis on the unit circle, which is the basis of monomials $x^k$. If the roots are near the real axis, close to the interval $[-1,1]$, you want instead to use a basis that is orthogonal on that interval (Chebyshev polynomials). If you use the wrong basis the algorithm is unstable, but that is not a problem with the algorithm per se, but with the choice of basis. -The cited reference gives an example how root finding in the interval $[-1,1]$ is highly robust if you represent the polynomial in terms of the orthogonal Chebyshev polynomials [using the Chebfun algorithm], while a representation in terms of monomials is highly sensitive to small errors in the coefficients.<|endoftext|> -TITLE: Non-metric topological continua -QUESTION [7 upvotes]: What important results hold for non-metric continua, or where can I find a survey of such results? -There are three definitions of a continuum around: a non-empty topological space that is -(1) connected compact metric, or -(2) connected compact Hausdorff [e.g., General Topology by Willard], or -(3) connected compact [ProofWiki]. -I am interested in non-trivial properties commonly known for definition (1) that have been found to also hold for definitions (2) or even (3). -I have asked this question on math.stackexchange but did not get any answer. - -REPLY [2 votes]: The non-cut point existence theorem is a good one. -Wilder proved sometime around the 1930's that every type (1) continuum has at least two non-cut points. This was later generalized for type (2) continua sometime around the 1960's by Whyburn, and finally, for all (type (3)) continua in 1999 by Honari and Brahmanpour. -Most results do not proceed this way, I must warn you. -Usually either (a) there is a non-metric counterexample, or (b) the original proof for metric continua "obviously" does not require metrizability and so applies to more general continua.<|endoftext|> -TITLE: Volume of $-K_X$ for a weighted projective variety -QUESTION [5 upvotes]: Let $X:=\mathbb P(a_0,a_1, \ldots, a_n)$ be a well formed weighted projective variety. Let $-K_X$ be its anticanonical divisor, then how to express its volume ${\rm vol}(-K_X)=(-K_X)^n$ in terms of $a_0,\ldots, a_n$? -In principle, this could be computed by toric geometry, but the data seems too complicated to compute (especially to write the dual lattice). Besides, I was wondering if there is a generally formula for the self intersection $D^n$ of the torus invariant divisor -$D = \sum_{i=0}^n c_i D_i$? - -REPLY [7 votes]: I am just rewriting my comment above as an answer. -Let $S=k[x_0,\dots,x_n]$ be the $\mathbb{Z}_{\geq 0}$-graded $k$-algebra with every $x_i$ homogeneous of degree $a_i.$ Denote by $X$ the associated projective $k$-scheme, $X =\text{Proj}\ S.$ Denote by $a$ the least common multiple of $(a_0,\dots,a_n).$ For every $i,$ denote by $b_i$ the integer such that $a_i\cdot b_i$ equals $a.$ Denote by $b$ the product $b_0\dots b_n.$ -Let $R=k[y_0,\dots,y_n]$ be the $\mathbb{Z}_{\geq 0}$-graded $k$-algebra with every $y_i$ homogeneous of degree $a.$ Denote by $Y$ the associated projective $k$-scheme, $Y=\text{Proj}\ R.$ This is $k$-isomorphic to $\mathbb{P}^n_k.$ There is a unique homomorphism of $\mathbb{Z}_{\geq 0}$-graded $k$-algebras, $$f^*:R \to S, \ \ y_i\mapsto x_i^{b_i}.$$ The inverse image of the irrelevant (prime) ideal is primary for the irrelevant (prime) ideal. Thus, there is an induced morphism of $k$-schemes, $$f:X\to Y.$$ This morphism is finite, and it is flat over an open subscheme that includes the open $Y_*=D_+(y_0\dots y_n).$ In fact, because of the well-formedness hypothesis, the restriction of the morphism over $Y_*$ is naturally a torsor for the finite, flat, commutative group scheme $\Gamma=\mu_{b_0}\times \dots \times \mu_{b_n}$ acting by $$(\zeta_0,\dots,\zeta_n)\cdot [x_0,\dots,x_n] = [\zeta_0\cdot x_0,\dots, \zeta_n\cdot x_n].$$ -Assume that $(a_0,\dots,a_n)$ is well-formed, i.e., the greatest common divisor of any $n$ of the $n+1$ weights equals $1$. In particular, this implies that the smooth locus $X^o$ of $X$ is a dense open subscheme whose complement has codimension $\geq 2$. Moreover, the Picard group of $X^o$ is generated by the ample invertible sheaf $\mathcal{O}_X(1)|_{X^o}$ that is the restriction of the rank $1$, reflexive, coherent sheaf $\mathcal{O}_X(1) = \widetilde{S[1]}.$ In particular, for every integer $d\geq 0,$ $$H^0(X^o,\mathcal{O}_X(d)|_{X^o}) = H^0(X,\mathcal{O}_X(d)) = S_d.$$ -The Picard group of $Y$ is generated by an ample invertible sheaf $\mathcal{O}_Y(1)$ whose vector space of global sections is the free $k$-vector space with basis $y_0,\dots,y_n.$ Since $f^*(y_i)$ has degree $a,$ the pullback $f^*\mathcal{O}_Y(1)$ is an ample invertible sheaf on $X$ whose restriction to $X^o$ equals $\mathcal{O}_X(a)|_{X^o}.$ In particular, $f^*\mathcal{O}_Y(-(a_0+\dots+a_n))|_{X^o}$ is isomorphic to $\omega_{X^o/k}^{\otimes a}.$ Thus, the $n$-fold self-intersection on $X$ of $c_1(f^*\mathcal{O}_Y(a_0+\dots +a_n))$ equals $(a_0+\dots+a_n)^n$ times the $n$-fold self-intersection on $X$ of $f^*c_1(\mathcal{O}_Y(1)).$ -The $n$-fold self-intersection on $Y$ of $c_1(\mathcal{O}_Y(1))$ is the unique class whose cap product with $[Y]$ equals the class of every $k$-point of $Y$. Thus, the $n$-fold self-intersection on $X$ of $f^*c_1(\mathcal{O}_Y(1))$ equals the class of every fiber of $f$ over any element of $Y_*.$ Since the morphism is a torsor for the finite, flat, commutative group scheme $\Gamma$ of length $b = b_0\cdots b_n,$ it follows that the $n$-fold self-intersection $X$ of $f^*c_1(\mathcal{O}_Y(1))$ equals $b.$ -Putting the pieces together, there is a unique invertible $\mathcal{O}_X$-module, $f^*\mathcal{O}_Y(a_0+\dots + a_n)$, whose restriction to $X^o$ equals $(\omega_{X^o/k}^\vee)^{\otimes a}.$ The $n$-fold self-intersection of $c_1(f^*\mathcal{O}_Y(a_0+\dots+a_n))$ equals $(a_0+\dots+a_n)^n b.$ Thus, considered as a rational number, the $n$-fold self-intersection of $c_1(\omega_{X/k}^\vee)$ equals $(a_0+\dots+a_n)^nb/a^n.$ Finally, using the fact that $b_i/a$ equals $1/a_i,$ this gives, $$\left( c_1(\omega_{X/k}^\vee) \right)^n_X = \frac{(a_0+\dots+a_n)^n}{a_0\cdots a_n}.$$<|endoftext|> -TITLE: How to understand the Deligne' tensor product of finite abelian category -QUESTION [6 upvotes]: In the sec 1.11. "Delignes' tensor product of locally finite abelian categories" of the book "Tensor Categories" of EGNO, the deligne's tensor -product $C \boxtimes D$ of two k-linear locally finite abelian categories $C$ ad $D$ is defined by the universal property: -$$ -\matrix{ -C \times D & \xrightarrow{\boxtimes} & C \boxtimes D\\ -& F \searrow & \downarrow{ \exists ! G} \\ -& & A -} -$$ -i.e., $\boxtimes: C \times D \rightarrow C \boxtimes D$ is a bifuntor which is right exact in both variables and is such that -for any right exact bifunctor $F: C \times D \rightarrow A$, where $A$ is k-linear locally finite abelian category, there exists a unique right exact functor $G: C \boxtimes D \rightarrow A$ such that -$$ -G \circ \boxtimes = F. -$$ -Let $Vec$ be the category of finite dimensional $k$-vector spaces. By the proof provided in the book, $Vec \boxtimes Vec \cong Vec$ and $\boxtimes$ is give by the tensor product of vector spaces. To my understanding, if $B$ is a category which is -equivalent to $C \boxtimes D$, then $B$ can also be viewed as the Delignes tensor product of $C$ and $D$. -Let $sVec$ be the skeleton of $Vec$. Then there should exists a functor -$G: sVec \rightarrow Vec$ such that the following diagram commutes: -$$ -\matrix{ -Vec \times Vec & \xrightarrow{\otimes} & sVec \\ -& \otimes \searrow & \downarrow{ G} \\ -& & Vec -}. -$$ -However, this is clearly impossible. So my question is if we can -view $sVec$ as $Vec \boxtimes Vec$? - -REPLY [7 votes]: The universal property isn't a characterization of $C \boxtimes D$, per se: the universal property is a property of the pair $(C \boxtimes D, \boxtimes)$ as an object of the coslice 2-category whose objects are pairs consisting of a category $B$ and a functor $C \times D \to B$. Now, $\mathsf{Vec}$ and $\mathsf{sVec}$ are equivalent just as categories, but there's no way to make them equivalent in this coslice 2-category (being equivalent here is a stronger condition). Does this help?<|endoftext|> -TITLE: Valuative criterion to extend morphism of schemes -QUESTION [5 upvotes]: Let $k$ be an algebraically closed field, $X, Y$ integral $k$-schemes and $Y$ proper over $k$. Let $U$ be a non-empty open subset $U \subset X$ -and $f:U \to Y$ a morphism of finite-type. Suppose that for any closed point $x \in X \backslash U$, any DVR $R$ with fraction field $K$, residue field $k$ and any morphism $\phi:\mathrm{Spec}(R) \to X$ with $\phi(\mathrm{Spec}(K)) \in U$ and $\phi(\mathrm{Spec}(k))=x$ we have that the unique morphim $\phi_f:\mathrm{Spec}(R) \to Y$ lifting the morphism $$\mathrm{Spec}(K) \xrightarrow{\phi} U \xrightarrow{f} Y$$ satisfies the property $y(x):=\phi_f(\mathrm{Spec}(k))$ does not depend on the choice of the DVR $R$ or the morphism $\phi$ (i.e., $y(x)$ depends only on the choice of $x$). Does this imply that the morphism morphism $f$ extend to a morphism $\tilde{f}:X \to Y$ such that $\tilde{f}(x)=y(x)$ for all closed $x \in X\backslash U$? - -REPLY [7 votes]: No. Take $X$ to be a projective curve with one cusp at $x$, $U:=X\smallsetminus\{x\}$, $Y=$ the normalization of $X$, $f=$ the section of $Y\to X$ over $U$. -[Edit after Ron's comment] The answer is yes if $X$ is normal. Let $X'\subset X\times Y$ be the closure of the graph of $f$. Then $\pi:X'\to X$ induced by the first projection is proper, and is an isomorphism above $U$. The assumption implies that $\pi$ is bijective on closed points. In particular, $\pi$ is finite birational, hence an isomorphism if $X$ is normal.<|endoftext|> -TITLE: Covering compactness in the weak sequential topology -QUESTION [5 upvotes]: Let $X$ be a real Banach space. Apart from the norm topology, we can consider the following weak topologies on $X$: - -the weak toplogy, defined as the initial topology with respect to $X^*$. In other words, it is the coarsest topology for which all $f\in X^*$ are continuous. -the weak sequential topology, which is essentially the topology induced by weak convergence. More precisely, we call a set closed if it is weakly sequentially closed, and this induces a topology. - -It is easy to see that the weak topology is the weaker of the two (since every $f\in X^*$ is weakly sequentially continuous). Moreover, it is well known that a weakly sequentially closed set is not necessarily weakly closed. However, the picture is not so clear when it comes to compactness: -By the Eberlein-Smulian theorem, weak compactness coincides with weak sequential compactness. However, it is important to note that weak sequential compactness means sequential compactness, not compactness in the weak sequential topology (!!). In particular, this raises the following question: - -What does ordinary compactness (i.e., covering compactness) look like in the weak sequential topology? Is it equivalent to weak (sequential) compactness? - -(Also: if these are not equivalent in general, what about the case of convex sets?) - -REPLY [2 votes]: Taras' answer seems perfectly fine to me, but I'm adding a second answer which is a little more "elementary" (despite being fundamentally the same). -The key observation is that, for subsets of weakly (sequentially) compact sets, the notions of weak closedness and weak sequential closedness coincide. To see this, let $C$ be weakly compact and $A\subseteq C$ weakly sequentially closed. Then $A$ is weakly sequentially compact (since $C$ is), hence weakly compact (by the Eberlein-Smulian theorem), hence weakly closed. The converse implication (weak closedness $\Rightarrow$ weak sequential closedness) holds trivially. -It follows that, if $C$ is weakly compact, then the weak and weak sequential topologies (or, more precisely, their induced subspace topologies) coincide on $C$. This yields the result.<|endoftext|> -TITLE: Do all symmetries of a Kähler quotient come from the original space? -QUESTION [8 upvotes]: For a Kähler manifold $M$, let $\operatorname{Iso}_{\mathbb{C}}(M)$ denote the group of holomorphic isometries. -Suppose that $K$ is a compact subgroup of $\operatorname{Iso}_{\mathbb{C}}(M)$ and there is a moment map $\mu:M\to\mathfrak{k}^*$ for the $K$-action such that the Kähler quotient $M_0:=\mu^{-1}(0)/K$ is smooth. Then, if $N_{\operatorname{Iso}_{\mathbb{C}}(M)}(K)$ denotes the normalizer of $K$ in $\operatorname{Iso}_{\mathbb{C}}(M)$, there is a natural group homomorphism -$$N_{\operatorname{Iso}_{\mathbb{C}}(M)}(K)/K\to\operatorname{Iso}_{\mathbb{C}}(M_0).$$ -Are there instances where this map is not surjective? I.e. $M_0$ posseses more symmetries than those comming from $M$? - -REPLY [2 votes]: Let $f : M \to E$ be a line bundle over an elliptic curve such that $\deg(M)<0$ - and let $G = \mathbf{C}^*$ act on $M$ (on the left) by scalar multiplications. The maximal compact subgroup $K$ of $G$ is $U(1)$ and let $M$ be endowed with the $K$-invariant Kähler metric $h$ constructed as follows. -There exists a unitary representation of $\pi_1(E)$ of a hermitian vector space $L$ of dimension 1 such that $M = (L \times \tilde{E})/\pi_1(E)$ where $\tilde{E} = \mathbf{C}$ is the universal cover of $E$. This identification is compatible with the $G$-actions where on the right hand side, $G$ acts on $L$ on the left by scalar multiplications. Let $L \times \tilde{E}$ be endowed with the product metric $\tilde{h}= (h_L,h_{\mathrm{std}})$ where $h_L$ is the constant hermitian metric coming from the hermitian structure of $L$ and $h_{\mathrm{std}}$ the standard hermitian metric on $\mathbf{C}$. The metric $\tilde{h}$ is Kähler and is at the same time $K$-invariant and $\pi_1(E)$-invariant. So $\tilde{h}$ descends to a $K$-invariant Kähler metric $h$ on $M$. -The Kähler quotient $M_0 = M // G$ is isomorphic to the elliptic curve $E$ and the induced Kähler metric on $M_0$ is invariant under translations. So the group of translations of $E$ (still denoted by $E$) is contained in $\mathrm{Iso}_\mathbf{C}(M_0)$. We shall show that not every element of $E$ can be lifted to $N_{\operatorname{Iso}_{\mathbf{C}}(M)}$. -The $N_{\operatorname{Iso}_{\mathbf{C}}(M)}$-action on $M$ preserves fibers of $f$. As $H^0(E,M)= 0$, the action also preserves the $0$-section. Thus for each $g \in N_{\operatorname{Iso}_{\mathbf{C}}(M)}$, there exist a biholomorphic map $\phi : E \to E$ and an isomorphism $i:\phi^*M \simeq M$ such that $g = i\circ \phi^*$, and the image of $g$ under the map $N_{\operatorname{Iso}_{\mathbf{C}}(M)} \to \operatorname{Iso}_{\mathbf{C}}(M_0)$ is $\phi$. Finally since $\deg(M) \ne 0$, the line bundle $M$ is not invariant under translations, which shows that $N_{\operatorname{Iso}_{\mathbf{C}}(M)} \to \operatorname{Iso}_{\mathbf{C}}(M_0)$ is not surjective.<|endoftext|> -TITLE: Is every continuous endomorphism of the Schwartz space a pseudo-differential operator? -QUESTION [11 upvotes]: Let $\mathcal{S}:= \mathcal{S}(\mathbb{R}^n)$ be the Schwartz space of smooth functions with rapid decay. The question is pretty simply stated in the title. Pseudo-differential act continuously on the space $\mathcal{S}$, it is therefore natural to wonder whether they are the only ones: - -Is there a continuous operator $T:\mathcal{S} \to \mathcal{S}$ which is not a pseudo-differential operator? i.e. not of the form: -$$T(f)(x) = \int_{\mathbb{R}^n}a(x,\xi)\hat f(\xi)e^{ix\xi}d\xi$$ -For some appropriate symbol $a(x,\xi)$ (of class $S^m$ for some $m$) - -EDIT: For a more interesting follow up version of this question (which is not trivially false) see here. - -REPLY [15 votes]: No, the most obvious example is the reflection operator: $Rf(x) = f(-x)$ this is not pseudolocal (in fact the $\xi$-compotent of the wavefront set gets a sign flip). Also the Fourier transform. -More generally, every compactly supported Fourier integral operator with non-trivial Lagrangian (not the co-normal to the diagonal) is not a pseudodifferential operator, but preserves the Schwartz-space.<|endoftext|> -TITLE: Comparison of the absolute value of an operator with its positive parts -QUESTION [10 upvotes]: It is well known that the absolute value on operators does not satisfy the triangle inequality. -My question is whether for all positive operators $P,Q \in B(\mathcal H)$ is there a universal constant $C$ such that -$$ |P + iQ| \leq C(P + Q)? $$ -I am doubtful that this is true but cannot find a counterexample. - -REPLY [7 votes]: OK, here goes. -We start with a reformulation of the problem. Assume that everything is invertible and $P+iQ=UBB^*$ where $U$ is unitary and $BB^*=|P+iQ|$. We can also re-parameterize $P$ and $Q$ as $BPB^*$ and $BQB^*$. Then the question reduces to whether for two positive operators $P,Q$, the relation $P+iQ=B^{-1}UB$ implies $P+Q\ge\delta I$ for some universal $\delta>0$. In finite dimension, this condition merely means that all eigenvalues of $P+iQ$ equal $1$ in absolute value (and, say, are distinct for the counterexample purposes to avoid boring discussions about Jordan blocks and approximation arguments). -If the dimension $n$ is fixed, then it implies that $|\operatorname{Tr}(P+iQ)|=|\operatorname{Tr}P+i\operatorname{Tr}Q|\le n$, so, since the traces are positive, we can bound them both by $n$, which, in turn, implies (due to positive definiteness) that $\|P\|,\|Q\|\le n$. Thus $\|P+iQ\|\le 2n$ but $|\det(P+iQ)|=1$, so $\|(P+iQ)^{-1}\|\le C_n$. In particular, for every $x$, we have $\|Px\|+\|Qx\|\ge\|(P+iQ)x\|\ge C_n^{-1}\|x\|$. On the other hand, if we have a vector $x$ with $\langle Px,x\rangle+\langle Qx,x\rangle\le\delta\|x\|^2$, then, by Cauchy-Schwarz, for any unit vector $y$, we have -$$ -|\langle Px,y\rangle|\le\langle Px,x\rangle^{1/2}\langle Py,y\rangle^{1/2}\le \sqrt{\delta n}\|x\|\,, -$$ -i.e., $\|Px\|\le \sqrt{\delta n}\|x\|$ -and similarly for $Qx$, so if $\delta>0$ is too small, we get a contradiction. -This is a terribly crude bound but it suffices to explain why Christian's approach with $2\times 2$ matrices had no chance to work. -Now the example in high dimension. We need to know that the $n\times n$ real antisymmetric matrix $H$ with diagonal entries $0$ and $H_{i,j}=\frac 1{i-j}$ for $i\ne j$ (the truncated discrete Hilbert transform) has norm bounded by some universal constant $\Psi>0$ regardless of $n$ while the $n\times n$ real symmetric matrix $L$ with $0$ diagonal entries and $L_{i,j}=\frac 1{|i-j|}$ for $i\ne j$ has norm $\Phi_n\approx\log n$. Now put -$$ -P=a[D_1-(\Phi_n+\Psi)^{-1}(L-iH)], Q=a[D_2-(\Phi_n+\Psi)^{-1}(L+iH)] -$$ -where $D_1, D_2$ are diagonal with distinct positive entries slightly above $1$ so that $D_1^2+D^2_2=a^{-2}I$ with $a$ just a tiny bit less than $1/\sqrt 2$. -Since $(1-i)+i(1+i)=0$, we conclude that $P+iQ$ is triangular (zero entries for $i>j$) with the diagonal $a(D_1+iD_2)$, so the spectral condition is satisfied. On the other hand, $P+Q$ is norm close to the degenerate matrix $2a(I-\Phi_n^{-1}L)$. -This construction shows that in dimension $n$ the constant in the original problem should be at least of order $\log n$. I believe it is the right order of growth, but I'll leave it to the "people who know all that stuff well" to comment on that and to provide relevant references.<|endoftext|> -TITLE: Upper bounds for lattice points in orbits, and representations of binary quadratic forms -QUESTION [11 upvotes]: Write $\mathbb{Z}^{a\times b}$ for the $a\times b$ integer matrices. Let $n\geq 3$ and $Q\in\mathbb{Z}^{n\times n}$. Let $G=O(Q)$ be the orthogonal group of $Q$. For $X_0\in \mathbb{Z}^{2\times n}$, set -$$ -R'(T,Q,X_0) = \{ X\in X_0G(\mathbb{Z}) : |X_{ij}|\leq T\}. -$$ -Dubious heuristic: If $Q$ is indefinite then -$$ -\#R'(T,Q,X_0) \ll_{Q,\epsilon} -T^{\max\{0,2n-6\}+\epsilon} -\tag{$\star$} -$$ -for all $\epsilon>0$, with an implicit constant depending only on $Q$ and $\epsilon$. -Question: For which $Q$ and which $X_0$ is this true? How about simple (?) examples like $Q=\operatorname{diag}(1,1,-1)$ or $Q=\operatorname{diag}(1,1,-1,-1)$? -Update 2018/4/11: I wonder if the work of Gorodnik and Nevo allows one to prove upper and lower bounds of the form $R'(T,Q,X_0)\asymp_Q T^{\text{something}}$ for all $X_0$ with entries $\ll T^{\delta}$, for some small $\delta$. In particular this would answer the question for all small $X_0$. I will try to look into this further. -Update 2018/4/11: As he explains in his answer below, Emilio Lauret gave in his PhD thesis a general strategy to answer questions of this type. This uses the result of Gorodnik and Nevo mentioend above. There is nontrivial work still to be done to implement this strategy in any particular case, more specifically one must compute the constant $\delta$ from page 45, line -5 of his thesis. (See his answer for a link). Unless and until someone comes up with an ever better approach to the problem, I accept his answer! - - - -Background -Let $B\in\mathbb{Z}^{2\times 2}$ be nonsingular and symmetric. I am interested in the number of solutions -$$ -R(T,Q,B) = \{ X\in \mathbb{Z}^{2\times n} : XQX^T =B, \, |X_{ij}|\leq T\} -$$ -with height up to $T$, when $Q$ is indefinite. A naive guess based on the circle method would be that -$$ -\#R(T,Q,B) \ll_Q 1+T^{2n-6} -\tag{$\ast$} -$$ -with an implicit constant depending at most on $Q$. The set $R(T,Q,B)$ is the union of finitely many $G(\mathbb{Z})$-orbits $R'(T,Q,X_0)$, so the conjecture above is a weak version of this. -Note that ($\ast$) is true for large $n$. For instance when $Q$ is indefinite and $n\geq 12$ this (and much more) follows from the asymptotic result of Brandes. (The condition $n\geq 12$ follows from the comment at the end of section 1 in the subsequent paper.) On the other hand ($\ast$) is certainly false for small $n$, though I don't know of a good example in the indefinite case. -If $B$ is positive definite and $Q$ has signature $(p,q)$ with $p\geq q$ then it is a special case of Theorem 1.4 of Eskin, Mozes and Shah that -$$ -\# R(T,Q,B) \sim C T^{\min\{2,q\}(n-1-\min\{2,q\})} -\tag{$\dagger$} -$$ -for some $C= C(Q,B)$, unless a certain volume is infinite. In the latter case, the comments after formula (1.4) of Duke, Rudnick and Sarnak suggest that the right-hand side of ($\dagger$) just changes by a factor of $\log T$. If this is correct then ($\star$) can be false in the case when $(p,q)=(2,1)$ and $B=X_0Q X_0^T$ is positive definite, since $\# R(T,Q,B)\gg T$ for all $T\gg_{Q,B} 1$. -Perhaps a proof or disproof of ($\star$) in other cases is implicit in the work cited in the previous paragraph, but I have not been able to verify this. - -REPLY [2 votes]: This is not an answer, but a long comment providing related known results. -In the particular case $Q=I_{n,1}= diag(I_n,-1)=diag(1,\dots,1,-1)$ (called the Lorentzian case), Ratcliffe and Tschantz gave in 1 an asymptotic formula for the number of $x=(x_1,\dots,x_{n+1})^t\in \mathbb Z^{n+1}$ satisfying that $x^tI_{n,1}x=k$ ($k\in\mathbb Z$) and $\|x\|^2\leq t$, as $t\to\infty$. -When $k$ is negative, they applied (in a clever way) the lattice point theorem by Lax and Phillips, obtaining a formula with an error term. -In my Ph.D. thesis (available at my web page, though it is written in Spanish) I extended the above formulas for more general indefinite quadratic or hermitian forms $Q$ of signature $(n,1)$ (see [2]). Furthermore, I also worked on counting certain solutions of $X\in \mathbb Z^{(p+q)\times q}$ satisfying $I_{p,q}[X]=X^*I_{p,q}X=-L$, where $L$ is a positive definite integral matrix by using a lattice point theorem by Gorodnik and Nevo [3] or [4]. Unfortunately, concerning the last case, it is not easy to compute explicitly the constants involved in the asymptotic formulas. -I hope similar technics as above may help to find upper bounds like those requested in the question. -1 Ratcliffe, John G.; Tschantz, Steven T., On the representation of integers by the Lorentzian quadratic form, J. Funct. Anal. 150, No. 2, 498-525 (1997). ZBL0883.11017. -[2] Lauret, Emilio A., An asymptotic formula for representations of integers by indefinite Hermitian forms, Proc. Am. Math. Soc. 142, No. 1, 1-14 (2014). ZBL1329.11029. -[3] Gorodnik, Alexander; Nevo, Amos, The ergodic theory of lattice subgroups, Annals of Mathematics Studies 172. Princeton, NJ: Princeton University Press (ISBN 978-0-691-14185-5/pbk; 978-0-691-14184-8/hbk). 136 p. (2009). ZBL1186.37004. -[4] Gorodnik, Alexander; Nevo, Amos, Counting lattice points, J. Reine Angew. Math. 663, 127-176 (2012). ZBL1248.37011.<|endoftext|> -TITLE: supercompactness measure projections -QUESTION [5 upvotes]: Suppose $\kappa$ is $\kappa^{+\omega}$-supercompact. If $U$ is a normal measure on $\mathcal P_\kappa(\kappa^{+\omega})$, and $j : V \to M$ is the derived embedding, then it is easy to see that $j[\mathcal P_\kappa(\kappa^{+\omega})] \in M$, and thus $M$ is closed under $\kappa^{+\omega+1}$-sequences by the cardinal arithmetic. We can then derive a normal measure $U'$ on $\mathcal P_\kappa(\kappa^{+\omega+1})$ by putting $X \in U'$ iff $j[\kappa^{+\omega+1}] \in j(X)$. Let $i : V \to N$ be the ultrapower embedding by $U'$. There is an embedding $k : N \to M$ such that $j = k \circ i$. Define $k([f]) = j(f)(j[\kappa^{+\omega+1}])$. -Question 1: Is $k$ the identity? -Now let $\pi : \mathcal P_\kappa(\kappa^{+\omega+1}) \to \mathcal P_\kappa(\kappa^{+\omega})$ be $\pi(z) = z \cap \kappa^{+\omega}$. Let $U'' = \{ X \subseteq \mathcal P_\kappa(\kappa^{+\omega}) : \pi^{-1}[X] \in U' \}$. -Question 2: Does $U'' = U$? - -REPLY [4 votes]: I think about this a little more abstractly using what I call seed theory. See for example my article, -Hamkins, Joel David, Canonical seeds and Prikry trees, J. Symb. Log. 62, No.2, 373-396 (1997). ZBL0890.03024, which provides an elementary development of this theory. I tried hard in that paper to find a nice, careful treatment of the ideas, which I view as providing a useful framework for answering all these kinds of questions; so I recommend taking a look at the paper, particularly sections 1, 2 and 3. (The paper was adapted from a chapter in my dissertation.) -If you have an elementary embedding $j:V\to M$, then the seed hull of an object $a\in M$ is the class $X_a=\{j(f)(a)\mid f\in V\}$, and this is an elementary substructure of $M$, which is isomorphic to the ultrapower by the measure $\mu_a=\{X\mid a\in j(X)\}$ by the association $[f]_{\mu_a}\mapsto j(f)(a)$. It is a basic fact that if two seeds $a$ and $b$ generate each other, then the measures $\mu_a$ and $\mu_b$ are isomorphic. In particular, any seed $a$ generating all of $M$ will have $j_{\mu_a}=j$. -In your case, for a $\lambda$-supercompactness ultrapower, the seed $j"\lambda$ and the seed $j"\lambda^{<\kappa}$ easily generate each other, for the reasons you gave in your post. Namely, from $j"\lambda$ we can easily construct $j"P_\kappa(\lambda)$, and from $j"\lambda^{<\kappa}$ we can project to $j"\lambda$. It follows that these seeds generate each other and therefore all of $M$, and so the measures they give are isomorphic, the ultrapowers are in each case $j$ itself and the factor embedding $k$ is the identity. The same is true whenever we have two elements $a$ and $b$ in $M$ that can define each other in $M$ using parameters in $\text{ran}(j)$; the corresponding measures $\mu_a$ and $\mu_b$ will be isomorphic and the factor embeddings identical.<|endoftext|> -TITLE: Modular forms for different groups than $SL(2,\mathbb Z)$ -QUESTION [7 upvotes]: I know some theory of "classical" modular forms, that is functions in the complex upper-half plane satisfying - -$f(\frac {az+b} {cz+d})=(cz+d)^kf(z)$ - -I know one can study modular forms on finite-index subgroups of $SL(2,\mathbb Z)$. But I have not seen much theory of modular forms on arbitrary Fuchsian groups. Which are the most interesting cases of such groups? Can somebody recommend a good reference? -I have also come across Hilbert and Siegel modular forms, but I don't have these in mind as an answer to this question. I wonder whether one can use arbitrary Lie group instead of $SL(2,\mathbb R)$, which is what the Wikipedia page about automorphic forms suggests, but I am not on a level to tackle the theory. - -REPLY [4 votes]: Yes, as you surmise, since about 1950, there has developed a general theory of automorphic forms on (semi-simple or reductive, mostly, "Jacobi forms" are a sort of exception) real Lie groups... and also on the corresponding adele groups when the group is defined over (some localization of) $\mathbb Z$. The general development is due to Harish-Chandra, Borel, Gelfand-PiatetskiShapiro, Godement, Langlands, and many others subsequently. -One of the earliest overviews of various aspects is the 1965 Boulder Conference, which appeared in 1966 as AMS Proc. Symp. Pure Math 9. The next iconic source is the 1977 Corvallis conference, which occurred in two volumes as AMS Proc Symp Pure Math 33. -There were and are many more sources... -I note that, apart from the relatively isolated studies on Siegel and Hilbert modular forms, and Maass' waveforms, by Maass, Siegel, Shimura, Klingen, and a few others, until 1960 in the U.S. "automorphic forms" exactly meant "with respect to (suitable) Fuchsian subgroups of $SL(2,\mathbb R)$, and/or intense examination of ratios of products of the weight $1/2$ modular form $\eta(z)$, for purposes of examining partition functions... -The renaissance of the general theory was perhaps due to Selberg-Roelcke's study of spectral theory in the 1950's, and Shimura's study of arithmetic consequences throughout the 1960's (and later), and then Langlands' generalizations and abstractions.<|endoftext|> -TITLE: Is there a complex which computes Cech cohomology? -QUESTION [17 upvotes]: Suppose $X$ is a (paracompact, Hausdorff) topological space and we want to define its Cech cohomology with coefficients in $\mathbb Z$. Here is the way I have seen this constructed. For each open cover $\mathcal U$, we define the complex ${\check{\mathcal C}}^\bullet(\mathcal U,\mathbb Z)$ of Cech cochains relative the open cover with the usual Cech differential. Now, if $\mathcal V$ is another open cover of $X$ which is a refinement of $\mathcal U$, then we have a map ${\check{\mathcal C}}^\bullet(\mathcal U,\mathbb Z)\to {\check{\mathcal C}}^\bullet(\mathcal V,\mathbb Z)$ for each "presentation" of $\mathcal V$ as a refinement of $\mathcal U$. But the chain homotopy class of this map is well-defined and thus, we get a well-defined directed system of groups $\check{H}^\bullet(\mathcal U,\mathbb Z)$ indexed by the open covers $\mathcal U$ of $X$, and their direct limit is defined to be $\check{H}^\bullet(X,\mathbb Z)$. On the other hand I have heard people speak of the complex $\check{\mathcal C}^\bullet(X,\mathbb Z)$ of Cech cochains on a space $X$, which computes the Cech cohomology of $X$. -I am interested in knowing the definition of this complex which computes the Cech cohomology. It can't simply be the direct limit of the complexes ${\check{\mathcal C}}^\bullet(\mathcal U,\mathbb Z)$ since these do not form a directed system of complexes (as the maps are well-defined only up to homotopy). - -REPLY [19 votes]: This is the kind of thing sieves are good for. For an open cover, let $S$ denote the sieve it generates, so $S$ is poset of open subsets $V$ such that $V$ is contained in some element of the cover. A quasi-isomorphic model for the Cech complex is given by the "homotopy limit over the sieve", so the $0$th term in the complex is the product of the $F(U)$ for all $U$ in $S$, the $1$st term is the product of the $F(U)$ for all $U \subset U'$ in $S$, etc. Basically, instead of modeling "a section on each element of the open cover, which agree on intersections", you model "a section on each element of the sieve, which agree on restriction". The point now is that a refinement of open covers gives an honest inclusion of sieves, and the sieve model for the Cech complex is just plain functorial in inclusions of sieves. So you can just pass to the filtered colimit.<|endoftext|> -TITLE: Why are there so many fractional derivatives? -QUESTION [31 upvotes]: I have been interested in fractional calculus for some time now, and I have seen "lots" of definitions of the $\frac {d^\alpha} {dx^\alpha}$ operator. -I started with the book The Fractional Calculus by Oldham and Spanier, and it comes as no surprise that I favor the Grünwald-Leitnikov derivative. It seems to me a great definition, because it directly generalizes the basic definition of the derivative $\frac {df} {dx}=\lim_{h \rightarrow 0} \frac {f(x)-f(x-h)} {h}$. And it also produces the integral when $\alpha$ is set to be a negative number. -Another (which I think is the Liouville definition, but I'm not sure) generalizes the property of differentiating an exponential $\frac {d^k} {dx^k} e^{rx} = r^ke^{rx}$ and thus if $\frac {d^\alpha} {dx^\alpha}f(x)=\sum A_ne^{nx}$ then $f(x)=\sum A_n n^\alpha e^{nx}$. -A definition, which is used really often for some reason, is the Caputo derivative. Lot of people find it natural that $\frac {d^{\frac 1 2}} {dx^{\frac 1 2}} [1]=0$, but I think it is "evident" that it should be proportional to $x^{-\frac 1 2}$. -Now comes the actual question. Why are there so many definitions of the fractional derivative? Are some of them "better" than the others in some sense? And lastly, is there a general framework, wherein "functions" of differential operators, maybe more general than (fractional) powers, can be given an explicit meaning? - -REPLY [2 votes]: I will give an answer concerning definitions of fractional\nonlocal derivatives that are Markovian generators of stochastic processes with jumps. I will briefly argue that - -Different definitions arise naturally, -there is a clear interpretation of many properties (like nonlocality or killing/not-killing constants), and -generalizations are natural and meaningful for applications. - -It is useful to look at the most simple stochastic jump process and its corresponding generator. Take a Markov chain $P=\{p_{i,j}\}_{i,j\in \text{State space}}$ (which is intrinsically jumpy) and write out its generator -$$ -\mathcal G f(x):=(P-I)f(x)=\sum_{y\in\text{ State space}}(f(y)-f(x))p_{x,y},\quad x\in\text{ State space}. -$$ -Here the intuition is clear: the infinitesimal jump (working with unit time in this case) from $x$ to $y$ is assigned intensity/probability $p_{x,y}$. The operator $\mathcal G$ is non-local. If we modify the process (impose boundary conditions), say by forcing the process to be absorbed at $a\in\text{ State space}$ once it tries to jump to a state $y\notin \Omega\subset \text{State space},$ we obtain a new generator -$$ -\mathcal G^{\text{abs}} f(x):=(P^{\text{abs}}-I)f(x)=\sum_{y\in\Omega}(f(y)-f(x))p_{x,y}+(f(a)-f(x))\sum_{y\notin\Omega}p_{x,y},\quad x\in\Omega. -$$ -If we instead decide to kill it (by testing against functions with $f(a)=0$, for example), the new generator will be -$$ -\mathcal G^{\text{kill}} f(x):=(P^{\text{kill}}-I)f(x)=\sum_{y\in\Omega}(f(y)-f(x))p_{x,y}-f(x)\sum_{y\notin\Omega}p_{x,y},\quad x\in\Omega. -$$ -So from one single process we can obtain many different generators/fractional derivative (as mentioned in a comment above, the boundary conditions are reflected in the representation of the operator away from the boundary due to the non-locality of $\mathcal G$). -Let us now move to the Riemann-Liouville and Caputo derivatives of order $\beta\in(0,1)$. Consider the three fractional derivatives for $x -TITLE: Compactification of open manifolds in the form of a manifold( with zero Euler characteristic) -QUESTION [7 upvotes]: Edit: According to the interesting comments of Michael Albanese and Nick L we revise the question as follows: -By manifold compactification of a manifold $M$ we mean a compact manifold $\tilde{M}$ which contains $M$ as an open dense subset. -Assume that $M$ is an open connected manifold which admits a manifold compactification. -Does $M$ necessarily admit a manifold compactification with zero Euler characteristic? - -REPLY [8 votes]: First note that odd dimensions the question of Euler characteristic $0$ is automatic, $M$ will embed in the orientable double cover of $\tilde{M}$, which will have $\chi = 0$ by Poincare Duality. -In even dimension = $2n$ (we assume $n > 1$), we recall the following fact. If $M_{1},M_{2}$ are compact connected manifolds then $\chi(M_{1} \# M_{2}) = \chi(M_{1}) + \chi(M_{2}) - \chi(S^{2n}) = \chi(M_{1}) + \chi(M_{2}) - 2$. -To prove that some embedding into any manifold implies embedding in to Euler characteristic $0$ manifold, it is sufficient to show that any integer is equal to the Euler characteristic of some manifold of dimension $2n$, since we can do connect sums on the complement of the embedding $M \hookrightarrow \tilde{M}$ (It is easy to see that the embedding can be changed so that this complement contains an open set) to shift the Euler characteristic of $\tilde{M}$ to the correct value. -In dimension $4$ we have $\chi(\mathbb{R}\mathbb{P}^{2} \times \mathbb{R}\mathbb{P}^{2}) =1$ and $\chi(\mathbb{C} \mathbb{P}^{2} ) = 3$. So for any $4$-manifold $N$ connect summing with $\mathbb{R}\mathbb{P}^{2} \times \mathbb{R}\mathbb{P}^{2}$ subtracts 1 from $\chi(N)$, connect summing with $\mathbb{C} \mathbb{P}^{2} $ adds one to $\chi(N)$, hence there is a $4$-manifold with Euler characteristic equal to any integer. In higher even dimension taking appropriate products with $\mathbb{R}\mathbb{P}^{2}$ will give the same result. -Edit 1 As Misha points out we can now ensure the image is dense by using the fact that every connected n-manifold is a compactification of an open n-cell. -Edit 2. Note that the above solution holds in even dimensions atleast $4$. I will give details a counterexample in the dimension $2$ case (which was pointed out Tom Goodwillie). -Let $S$ be an orientable surface of genus $g \geq 2$. I will show there is no embedding $I: S \setminus \{p\} \hookrightarrow S'$ (for some $p \in S$). Where $S'$ is a compact surface of Euler characteristic $0$ (i.e. a torus or a Klein bottle). -We argue by contractiction, suppose such an embedding $I$, exists. Let $C$ be the boundary of a small neighbourhood of $p$. -Suppose there is an embbeding $I: S \setminus \{p\} \hookrightarrow S' $, then image of $I(C)$ decomposes $S'$ as the connect sum of $S$ and compact surface $\Sigma$. In symbols: $S' = S \# \Sigma$. Hence by the formula for Euler characteristic of a connect sum given above $\chi(S') \leq -2$. -To see this note by the classification of surfaces $\chi(\Sigma) \leq 2$ and by direct computation $\chi(S) = 2-2g \leq -2$, the fact that $\chi(S') \leq -2$ now follows directly from the formula for Euler characteristic of a connect sum. This is the desired contradiction.<|endoftext|> -TITLE: Galois twist of a variety -QUESTION [8 upvotes]: Suppose $X$ is a variety over $\mathbb{Q}$ and it has a Galois twist $X'$, i.e. $X$ is isomorphic to $X'$ over $\overline{\mathbb{Q}}$. The set of isomorphism classes of twists of $X$ is classified by $H^1(\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), \text{Aut}(X_{\overline{\mathbb{Q}}}))$. -If $X'$ corresponds to a class $c \in H^1(\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), \text{Aut}(X_{\overline{\mathbb{Q}}}))$, what is the relation between the Galois representations $H^q_{et}(X_{\overline{\mathbb{Q}}},\mathbb{Q}_{\ell})$ and $H^q_{et}(X'_{\overline{\mathbb{Q}}},\mathbb{Q}_{\ell})$? - -REPLY [5 votes]: Just follow your nose. Functoriality of cohomology gives a map $\operatorname{Aut}(X_\bar{\mathbb{Q}})\to \operatorname{Aut}\big(H^q_{\text{ét}}(X_{\bar{\mathbb{Q}}}, \mathbb{Q}_\ell)\big)$ (actually, the functoriality is contravariant, so gives a map to the opposite group -- but for groups, $G \cong G^{op}$ via $g\mapsto g^{-1}$). This map is compatible with Galois action, hence you get (again by functoriality) a map $$H^1\big(\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}),\operatorname{Aut}(X_{\bar{\mathbb{Q}}})\big)\to H^1\big(\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}),\operatorname{Aut}(H^q_{\text{ét}}(X_{\bar{\mathbb{Q}}}))\,\big).$$ The RHS once again classifies Galois twists (now of a vector space). -There is a topological way of seeing why this functoriality should hold. Very informally, you should think of an equivariant object (e.g. $X_{\bar{\mathbb{Q}}}$) of any category with $\Gamma$ action (for $\Gamma$ the Galois group) as a local system of objects in a local system of categories over a topological space $S$ with $\pi_1(S) = \Gamma.$ The object over the basepoint $s_0\in S$ is $X_{\bar{\mathbb{Q}}},$ and the Galois action $\Gamma\to \operatorname{Aut}(X_{\bar{\mathbb{Q}}})$ is encoded by the monodromy; a twist in $H^1(S, \operatorname{Aut})$ is some class over $S$ which modifies the local glueing data between the copies of $X_{\bar{\mathbb{Q}}}$ on intersections locally by some elements of the relevant automorphism group. Now any functor of $\Gamma$-equivariant categories gives maps of all the relevant local data over $S$, and in particular takes a twist to a twist.<|endoftext|> -TITLE: $L_2$ bounds for tails of $\zeta(s)$ on a vertical line -QUESTION [8 upvotes]: Let $0<\sigma\leq 1$. Let $T$ be large. How can we give good explicit $L^2$ bounds on the tails of $\zeta(\sigma+it)$? That is, we want to bound the quantity $$\int_{\sigma-i\infty}^{\sigma-iT} + \int_{\sigma+iT}^{\sigma+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds.$$ -Morally, we should expect something about as good as $1/T$ for $\Re(s)\geq 1/2$, and presumably $1/T^{2\sigma}$ for $0<\Re(s)\leq 1/2$ (or maybe I ate a $\log$). In particular, we should certainly expect to do better than what we can get from convexity bounds on individual values of $\zeta(s)$. (Such bounds are, incidentally, explicit (Backlund)). - -How to go about this? One inviting approach is to use the fact that the Mellin transform is an isometry. (That is, if $F$ is the Mellin transform of $f$, then $\frac{1}{2\pi} \int_{-\infty} |F(\sigma+it)|^2 dt=\int_0^\infty |f(x)|^2 x^ {2\sigma-1}$.) Recall that $\zeta(s)/s$ is the Mellin transform of $x\mapsto \lfloor 1/x\rfloor$. An obvious (even crass) sub-approach would be to let $$g(x) = \frac{1}{2 \pi i} \int_{\sigma-iT}^{\sigma+iT} \frac{\zeta(s)}{s} x^{-s} ds,$$ and then note that the Mellin transform of $\lfloor 1/x\rfloor - g(x)$, evaluated at $\sigma+it$, equals $0$ when $|t|< T$ and $\zeta(s)/s$ when $|t|>T$. (I'm papering over the fact that we should check that there is no pole at $s=1$, so that we can analytically continue the Mellin transform to the left of $\Re s=1$.) It follows, by isometry, the quantity in the first displayed equation above (the integral over the tails) equals $$\int_0^\infty |\lfloor 1/x\rfloor - g(x)|\cdot x^{2\sigma-1} dx.$$ We can write down $g(x)$ explicitly, though it is not completely trivial; if I understand correctly (see Theorem 5.2 in Montgomery-Vaughan), $$g(x) = \frac{1}{\pi} \sum_{n\geq 1} \mathrm{si}(T \log n x),$$ where $\mathrm{si}$ is the sine integral function. Of course the problem here is that $g(x)$ is not a particularly pleasant function to work with. -My instinct is to work instead with a function of the form $$\frac{1}{2\pi i} \int_{\sigma-iT}^{\sigma+iT} \frac{\zeta(s)}{s} \phi(s) x^{-s} ds,$$ where $\phi(s)$ is close to $0$ for $s = \sigma+i t$, $|t|\geq T$, and is close to $1$ on most of the segment $\lbrack \sigma-iT,\sigma+iT\rbrack$. I haven't hit on the right choice of weight $\phi$, though. It should give some sort of advantage over $g$ above, or else it is pointless. -Any other approaches? Is this all known (explicitly)? - -REPLY [3 votes]: Thanks, GH! Let me have another go. I think the following is the right way to go about things, at least if one wants something self-contained and with good, explicit constants. (The latter more or less implies the former, given that almost all of the literature is non-explicit.) -We want to estimate -$$\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+i \infty} \left|\frac{1}{s} -G(s)\right|^2 |\zeta(s)|^2 ds,$$ -where $G(s)$ is the Mellin transform of a well-chosen function $g:\lbrack 0,\infty)\to \mathbb{R}$. It is easy to see that $G(s) \zeta(s)$ is the Mellin transform of $x\mapsto \sum_n g(n x)$. We will choose $g$ so that (a) $G(\sigma+it)$ is small for $|t|\geq T$, (b) the "physical-space" estimation we are about to do is easy. -By Plancherel, -$$\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+i \infty} |1-G(s)|^2 \frac{|\zeta(s)|^2}{|s|^2} ds = \int_0^\infty |h(x)|^2 x^{2\sigma-1} dx,$$ -where $$h(x) = \lfloor 1/x\rfloor - \sum_n g(n x).$$ (We will make sure that $G(1)=1$, so that there is no pole at $s=1$; in this way, the equation above will hold for $\Re(s)>0$, and not just for $\Re(s)>1$.) -First, let us show that $h(x)$ is bounded for all $x$. (This part is the same as what I had before.) By second-order Euler-Maclaurin, -$$\sum_n g(n x) = \frac{1}{x} -\int_0^\infty g(t) dt - \frac{g(0)}{2} - \frac{g'(0)}{12} + \textrm{err},$$ -where $|\textrm{err}| \leq \frac{x}{24} \int_0^\infty |g''(t)| dt$. We will work with $g$ such that $g(0)=1$, $g'(0)=0$ and $\int_0^\infty g(t) dt = 1$. Then $$|h(x)| = \left|\lfloor 1/x\rfloor - \left(\frac{1}{x} - \frac{1}{2} + \text{err}\right)\right|\leq \frac{1}{2} + \textrm{err}$$ for all $x$. -We will now choose $g$ so that we can give a much better estimate for $x$ not too small. For starters, we will have $g(x)=1$ for $x\leq 1-\delta$ and $g(x)=0$ for $x\geq 1+\delta$. Then $$h(x) = 0$$ unless $n x \in \lbrack 1-\delta,1+\delta\rbrack$. Moreover, for each $x>2\delta$, there is at most one $n$ such that $n x$ is in that interval, since $\frac{1+\delta}{x} - \frac{1-\delta}{x} = \frac{2 \delta}{x} < 1$. Hence $$\begin{aligned}\int_0^\infty |h(x)|^2 x^{2\sigma-1} dx &\leq \int_0^{2\delta} c^2 x^{2\sigma-1} dx + \sum_{n\leq \frac{1}{2 \delta}} \int_{\frac{1-\delta}{n}}^{\frac{1}{n}} |1-g(n x)|^2 x^{2\sigma-1} dx \\ &+ \sum_{n\leq \frac{1+\delta}{2\delta}} \int_{\frac{1}{n}}^{\frac{1+\delta}{n}} |g(n x)|^2 x^{2\sigma-1} dx,\end{aligned}$$ where $c =1/2 + \frac{2\delta}{24} \int_0^\infty |g''(t)| dt$. Obviously $$\int_0^{2\delta} c^2 x^{2\sigma-1} dx = c^2 \frac{(2\delta)^{2\sigma}}{2\sigma}.$$ To estimate the two other integrals, we have to choose a convenient $g$ obeying our conditions. I will simply take $$g(x) = \begin{cases} 1 &\text{if $x\leq 1 - \delta$,}\\ \frac{1+\delta-x}{2\delta} &\text{if $1-\delta 1/2$,}\\ \log(2 y + 1) &\text{if $\sigma=1/2$,}\\ \frac{y^{1-2\sigma}}{1-2\sigma} & \text{if $\sigma<1/2$.}\end{cases}$$ -Taking totals, we conclude that $$\int_0^\infty |h(x)|^2 x^{2\sigma-1} dx \leq c^2 \frac{(2\delta)^{2\sigma}}{2\sigma} + \begin{cases} \frac{\zeta(2\sigma)}{6} \delta& \text{if $\sigma>1/2$,}\\ \frac{\delta}{6} \cdot \log\left(\frac{1}{\delta} + 2\right) &\text{if $\sigma=1/2$,}\\ \frac{\eta (1+\delta)^{1-2\sigma}}{6\cdot 2^{1-2\sigma} (1-2\sigma)} \cdot \delta^{2\sigma} & \text{if $\sigma<1/2$.}\end{cases}$$ -Now it just remains to estimate how much of the tail we captured. A quick calculation shows that we then have $$G(s) = \frac{(1+\delta)^{s+1} - (1-\delta)^{s+1}}{2\delta (s+1) s}.$$ Not unexpectedly, this is close to $1/s$ for $\Im(s)$ small. Even more to the point, for any $s$, $$|G(s)| \leq \frac{(1+\delta)+(1-\delta)}{2\delta |s+1| |s|} = \frac{1}{\delta |s+1| |s|},$$ and so $$|1-G(s) s|^2 \geq \left(1-\frac{1}{\delta |s+1|}\right)^2 \geq \left(1 - \frac{1}{\delta T}\right)^2$$ for every $s=\sigma+it$ with $|t|\geq T$. Hence $$\int_{\sigma-i\infty}^{\sigma-iT} + \int_{\sigma+iT}^{\sigma+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds \leq \frac{1}{\left(1 - \frac{1}{\delta T}\right)^2} \int_0^\infty |h(x)|^2 x^{2\sigma-1} dx .$$ -Minimizing what will be the main term, we choose $\delta = 3/T$ if $\sigma\geq 1/2$, and $\delta = \frac{1+1/\sigma}{T}$ if $0<\sigma<1/2$. Then $\frac{1}{\left(1 - \frac{1}{\delta T}\right)^2}$ equals $9/4$ for $\sigma\geq 1/2$ and $(1+\sigma)^2$ for $0<\sigma<1/2$. -We conclude that $$\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma-iT} + \int_{\sigma+iT}^{\sigma+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds \leq \begin{cases} \frac{9 \zeta(2\sigma)}{8} \cdot \frac{1}{T} + \frac{9 c^2 6^{2\sigma}}{8\sigma} \cdot \frac{1}{T^{2\sigma}} & \text{if $\sigma>1/2$,}\\ \frac{9}{8} \frac{\log\left(\frac{T}{3}+2\right)+12 c^2}{T} &\text{if $\sigma=1/2$,}\\ \left(\frac{ \frac{1}{2} \left(\frac{1+\delta}{1-\delta}\right)^{1-2\sigma} + \frac{1}{2}}{6\cdot 2^{1-2\sigma} (1-2\sigma)} + c^2 \frac{2^{2\sigma}}{2\sigma}\right)\cdot \frac{(1+\sigma)^{2\sigma+2}}{\sigma^{2\sigma}}\cdot \frac{1}{T^{2\sigma}} & \text{if $\sigma<1/2$.}\end{cases}$$ -Oh, by the way, $\int_0^\infty |g''(x)| = 1/\delta$, so $c = 7/12$. -No doubt the constants can still be improved, and any suggestions on how to simplify things further are very welcome. I think I can be happy now, though.<|endoftext|> -TITLE: Is every continuous microlocal operator a pseudo-differential operator? -QUESTION [12 upvotes]: Let $\mathcal S'=\mathcal S'(\mathbb R^n)$ be the Schwartz distribution space. -Suppose $A\colon\mathcal S'\to\mathcal S'$ is linear, continuous and microlocal. -By being microlocal I mean that the wave front sets satisfy $WF(Af)\subset WF(f)$ for all $f$. -(For another version, one could consider the singular supports instead of wave fronts, but I assume the answer wouldn't be different.) -Does it follow that $A$ is a pseudo-differential operator? -This is a variation on this question about continuous endomorphisms on the Schwartz space. -There one only assumed linearity and continuity, and the answer was negative. -The other question was about $\mathcal S$ instead of $\mathcal S'$, but I need distributions to allow singularities. -I realize that there are many classes of pseudo-differential operators. -The question is whether such a microlocal operator is always a ΨDO of some kind. -Any details on what kind would be very interesting, of course. - -REPLY [3 votes]: This is more like a longish series of comments somewhat complementing Ilya Zakharevich's answers rather than an answer by itself. First of all, notice that since $\mathscr{S}(\mathbb{R}^n)$ embeds continuously (and densely) into $\mathscr{S}'(\mathbb{R}^n)$, any continuous linear map $A:\mathscr{S}'(\mathbb{R}^n)\rightarrow\mathscr{S}'(\mathbb{R}^n)$ induces by restriction a unique continuous linear map $A|_{\mathscr{S}(\mathbb{R}^n)}:\mathscr{S}(\mathbb{R}^n)\rightarrow\mathscr{S}'(\mathbb{R}^n)$, which we also denote by $A$ not to overburden our notation. By Schwartz's kernel theorem, we must have $$(A\phi_1)(\phi_2)=K_A(\phi_1\otimes\phi_2)\ ,\quad\forall\phi_1\ ,\,\phi_2\in\mathscr{S}(\mathbb{R}^n)$$ for a unique $K_A\in\mathscr{S}'(\mathbb{R}^{2n})$. -One sees from the calculus of wave front sets that microlocality of $A$ $$\mathrm{WF}(Au)\subset\mathrm{WF}(u)\ ,\quad\forall u\in\mathscr{S}'(\mathbb{R}^n)$$ is entailed by the condition $$\mathrm{WF}(K_A)\subset\{(x,x;\xi,-\xi)\ |\ (x;\xi)\in\mathbb{R}^{2n}\smallsetminus 0\}\ ,$$ which all pseudodifferential operators with (say) smooth symbols must satisfy by the stationary phase formula, but the converse (as far as I know) is not necessarily true. -To have a better idea of what the would-be symbol of $A$ should look like, we have that $$\sigma_A(x,\xi)=A(e^{-i\langle x-\cdot,\xi\rangle})(x)$$ defines for each $\xi\in\mathbb{R}^n$ a tempered distribution $\sigma_A$ in $x$ which is smooth and polynomially bounded in $\xi$ together with all its $x$- and $\xi$-derivatives. By microlocality, $\sigma_A$ should be smooth in $x$ as well. However, this alone does not guarantee that $\sigma_A$ fits into any decent symbolic calculus since the polynomial bounds on its derivatives in $x$ and $\xi$ may vary too wildly. -One can say more if we know that $A$ is translation invariant in the sense that $$A(u(\cdot+x))=(Au)(\cdot+x)$$ for all $x\in\mathbb{R}^n$, $u\in\mathscr{S}(\mathbb{R}^n)$. Due to microlocality, this can only be the case if $A$ is a convolution operator, that is, there is a unique $F_A\in\mathscr{S}'(\mathbb{R}^n)$ such that $$Au=F_A*u\ ,\quad\forall u\in\mathscr{S}'(\mathbb{R}^n)\ .$$ Equivalently, $K_A$ is the convolution kernel associated to $F_A$: $$K_A(x,y)=F_A(x-y)\ .$$ The Fourier transform $\sigma_A$ of $F_A$ is in this case the obvious candidate for the symbol of $A$ (in this case, $\sigma_A$ no longer depends on $x$, of course). The image of $\mathscr{S}(\mathbb{R}^n)$ under $A$ consists of smooth functions with at most polynomial growth together with all their derivatives, but that is pretty much it - the convolution with any $F\in\mathscr{S}'(\mathbb{R}^n)$ will have the same property. Microlocality of $A$ follows if in addition $\mathrm{WF}(F_A)\subset\{(0;\xi)\ |\ \xi\in\mathbb{R}^n\smallsetminus 0\}$, which is equivalent to the above condition on $\mathrm{WF}(K_A)$ in the translation invariant case, but again this (as far as I know) is not a necessary condition. -If we could show that: - -$(1+\|\xi\|^2)^{-k}\sigma_A\in L^\infty(\mathbb{R}^n)$ for some $k\in\mathbb{N}$, which amounts to saying that $(1-\Delta)^{-k}A$ is a bounded operator in $L^2(\mathbb{R}^n)$ by the Plancherel formula ($\Delta$ is the Laplacian in $\mathbb{R}^n$), and -The commutators of $(1-\Delta)^{-k}A$ with (multiplication by) polynomials are also bounded operators in $L^2(\mathbb{R}^n)$, - -one could appeal to the Beals-Cordes commutator characterization of pseudodifferential operators of order zero since $A$ commutes with derivatives in the translation invariant case (notice that (2.) and the latter form of (1.) still make sense in the non-translation-invariant case). In the general (non-translation-invariant) case, we need as well to show that - -The commutators of $(1-\Delta)^{-k}A$ with derivatives are also bounded operators in $L^2(\mathbb{R}^n)$ - -in order to use the Beals-Cordes criterion, which precisely guarantees the kind of polynomial bounds on the derivatives of $\sigma_A$ needed for it to be a symbol (of order $2k$). I have no idea if microlocality is enough to yield (1.)-(3.) (or even (1.)-(2.) in the translation-invariant case), but it seems a rather tall order to me.<|endoftext|> -TITLE: Formal vector fields vs. (standard) vector fields -QUESTION [9 upvotes]: Given a smooth manifold $M$, one can consider the Lie algebra $\mathcal{X}(M)$ of vector fields equipped with the standard Lie bracket. This is a standard machinery of differential geometry. Gelfand and Fuchs defined the Lie algebra of formal vector fields at $0 \in \mathbb{R}^n$ as linear combinations -$$\sum_{j=1}^np_j(x_1,...,x_n)e_j$$ -where $e_j$ is a standard basis of $\mathbb{R}^n$ and $p_j$ are formal power series in variables $x_1,...,x_n$. This definition is at the given point $0$ in $\mathbb{R}^n$, so formal vector fields are not ,,globally'' defined. - -Is it possible to define formal vector fields globally on a given manifold $M$? - -Also I would like to know - -What is the significance of formal vector fields? For example, do they naturally arise as a Lie algebra of some natural (infinite dimensional) group? - -REPLY [3 votes]: The Lie algebra of all formal vector fields of a real or complex manifold germ $(M,p)$ (where $p\in M$ is any point), does arise as the Lie algebra of the Lie group of all formal power series automorphisms of $(M,p)$, which is the same as the group of all invertible infinite jets of local diffeomorphisms of $(M,p)$, smooth or holomorphic respectively. -Also, formal vector fields of $(M,p)$ can be defined as derivations of the algebra of all formal power series at $p$. -Yet alternatively, the formal vector field algebra can be seen as the quotient of the algebra of all germs of smooth vector fields at $p$ modulo all germs of "flat" vector fields, i.e. ones vanishing of infinite order. For a complex manifold, that definition needs to be appropriately modified by allowing germs of smooth vector fields that are holomorphic of infinite order at $p$. -From that perspective, all these notions are fundamentally local. Of course, you can always consider subalgebras of formal vector fields extendible to global ones on $M$, which would be only interesting on a complex manifold. On a real manifold, of course, any formal vector field would arise that way.<|endoftext|> -TITLE: Action of upper triangular matrices -QUESTION [5 upvotes]: Let $M,N$ be two $n\times m$ matrices with $n\leq m$ and coefficients in an algebraically closed field of characteristic zero $K$, both of full rank $n$. -Do there exist two upper triangular matrices $A\in SL(n)$ and $B\in SL(m)$ such that $A\cdot M \cdot B^{T} = \lambda N$ for $\lambda\in K\setminus\{0\}$ ? - -REPLY [7 votes]: I think the answer to your question is negative. Consider for instance -$$M = \begin{pmatrix} -1 & 0 \\ -0 & 1 -\end{pmatrix}, \quad N = \begin{pmatrix} -0 & 1 \\ -1 & 0 -\end{pmatrix}$$ -Assume that there exist -$$A = \begin{pmatrix} -a_{11} & a_{12} \\ -0 & a_{22} -\end{pmatrix}, \quad B = \begin{pmatrix} -b_{11} & b_{12} \\ -0 & b_{22} -\end{pmatrix}$$ -with $\det(A) = \det(B) = 1$ and such that -$$A\cdot M\cdot B^{T} = N$$ -Then -$$A\cdot B^{T} = \begin{pmatrix} -a_{11}b_{11}+a_{12}b_{12} & a_{12}b_{22}\\ -a_{22}b_{12} & a_{22}b_{22} -\end{pmatrix} = \begin{pmatrix} -0 & 1\\ -1 & 0 -\end{pmatrix}$$ -and hence either $a_{22} = 0$ or $b_{22}=0$ which contradict $\det(A) = \det(B) = 1$. -More generally, your action stabilizes the locus of matrices of the form $\begin{pmatrix} -m_{11} & m_{12}\\ -m_{21} & 0 -\end{pmatrix}$.<|endoftext|> -TITLE: Minimum degree of a variety with $H^i(X,\mathcal{O}_X)\neq 0$ for some $i$ with $0 -TITLE: Can the methods of classical algebraic geometry be made rigorous with a synthetic approach? -QUESTION [13 upvotes]: There are approaches to real analysis that use an axiomatization of nilpotent infinitesimals to enable rigorous synthetic reasoning about infinitesimals, which is arguably closer to the reasoning employed by mathematicians prior to the arithmetization of analysis. -While I am far from an expert, it seems that the conventional narrative is that increasing doubts about the validity of some of the results of classical algebraic geometry led to a similar arithmetization of algebraic geometry by Zariski and Weil. The gap between the resulting methods and the underlying geometric reasoning seems a bit wider than the gap between $\epsilon$-$\delta$ arguments and infinitesimals, although that is entirely personal opinion and may only be due to my familiarity with the latter. -Is it possible to do algebraic geometry in a synthetic manner that enables rigorous reasoning but is closer to the style of argument employed by classical algebraic geometers? - -REPLY [2 votes]: Is it possible to do algebraic geometry in a synthetic manner that enables rigorous reasoning but is closer to the style of argument employed by classical algebraic geometers? - -I sure hope so. You can have a look at notes of mine which develop the basics of a synthetic account of algebraic geometry. Especially relevant is Section 20, which presents a couple of case studies, including computing the cohomology of Serre's twisting sheaves. -To give a short teaser: Synthetically, we can define the projective space $\mathbb{P}(V)$ associated to a vector space $V$ simply as the set of one-dimensional subspaces of $V$. The twisting "sheaf" $\mathcal{O}(-1)$ is then simply the family $(\ell)_{\ell \in \mathbb{P}(V)}$ of vector spaces. Its dual sheaf is the family $(\ell^\vee)_{\ell \in \mathbb{P}(V)}$. The scheme structure is automatically taken caren of. -However there is still much more to be done. The most pressing concerns are maybe: - -There should be an easy and intuitive synthetic description of proper morphisms. Right now we do have a synthetic description, but it's very close the usual non-synthetic description and doesn't exploit the unique possibilites of the synthetic context. -We have to develop a synthetic account of cohomology, derived categories and intersection theory.<|endoftext|> -TITLE: Are all infinite graphs $3$-weak-edge colorable? -QUESTION [5 upvotes]: Let $G=(V,E)$ be a simple, undirected graph such that every vertex has degree at least $2$. Given $n\in\mathbb{N}$, a map $c:E \to \{1,\ldots, n\}$ is said to be a weak coloring if for every $v\in V$ the edges adjacent to $v$ do not all have the same color. (More formally, we want the restriction $c|_{E(v)}$ to be non-constant, where $E(v) = \{e\in E: v\in e\}$.) -These two nice posts by Mikail Tikhomirov and Brendan McKay respectively show that for every finite graph there is a weak edge coloring with $3$ colors. I tried to carry through their arguments with transfinite induction to infinite graphs - without success. -Question. If $G=(V,E)$ is an infinite simple undirected graph, is there a weak edge coloring $c:E \to \{1,2,3\}$? - -REPLY [2 votes]: Wlog $G$ is connected. Let $T\subseteq G$ be a spanning tree. Choose a vertex $r$ in $T$ with dgeree at least 2, make it the root. Split the neighbors of $r$ into two nonempty sets, $A$ and $B$. Color all tree edges between $r$ and $A$, blue, those between $r$ and $B$ red. Then continue coloring the edges of $T$: those from $A$ red, from $B$ blue, etc. Eventually all edges of $T$ are colored with red and blue. Finally color the remaining edges with green. -This is good: all nonleaf vertices of $T$ have colors red and blue, the leaf vertices necessarily have a red/blue colored edge in $T$, and another in $G-T$, colored green.<|endoftext|> -TITLE: Primes of the form $4x+1$ -QUESTION [5 upvotes]: For a given $n$, i am trying to find a constant value $c$ such that you can always find a prime $p$ of the form $4x+1$ for some $x \in \mathbb{Z}_+$ and $n < p < cn$. I want to find the smallest such $c$ for which the above property holds for all values of $n$. -I searched in literature, i could only find that $(n!)^2+1$ has a prime divisor $p$ of the form $4x+1$ and $n < p < (n!)^2+1$. But $n!$ upper bound is huge for my need. -Also i found that we can find a prime $p$ of the form $4x+1$ such that $n< p < (p_1...p_k)^2+1$ where $p_1,...,p_k$ are all the prime numbers of the form $4x+1$ and less than $n$. This upper bound of $(p_1...p_k)^2+1$ is also large for my need. -Also i found some asymptotic formulas but i want formulas which holds for all numbers. -It would be great if i can find a constant $c$ such that there is always a prime of the form $4x+1$ between $n$ and $cn$ and the property holds for all $n$. -Thanks in advance. - -REPLY [3 votes]: I feel bad posting this as an answer, but: using Felipe's reference, the last number for which there is not a prime equal to $1 \mod 4$ between $n$ and $2n-1$ (inclusive) is $6$. For $6$ the first such number is $13,$ for $1$ it is $5,$ and for $2$ it is also $5.$ So, if you care about $1,$ the best constant is $6,$ if you care about $2$ but not $1,$ it is $3,$ and if you care about $6$ but neither $1$ nor $2,$ the best is $7/3.$<|endoftext|> -TITLE: Linear equations with absolute values -QUESTION [7 upvotes]: Assume we have a set of equations in $x \in \mathbb{R}^n$ -$$|a_i\cdot x|=b_i$$ -where $a_i \in \mathbb{R}^n$ and $b_i>0$ are given. -Could such a system be solved efficiently? - -In a theoretical machine storing reals with perfect accuracy. -An approximate solution taking rounding errors into account. - -This is equivalent to having $2$ possible values for each linear combination, which is a valid question over any field. Does this equivalent problem over a fixed finite field have an efficient solution? Or is it perhaps known to be NP-complete? -So far, I concluded that squaring both sides of the equation we get linear equations in $y_{ij} = x_i x_j$. However, this helps only if the system is quadratically overdetermined. - -REPLY [3 votes]: If you square your equations to get $|\langle x,a_i\rangle|^2 = b_i^2$ your problem is the so-called phase retrieval problem (which was motivated by the problem of recovering a function (up to global phase) from the magnitude of its Fourier transform). There are several results here (besides the general NP-hardness). I would like to point out some blog posts by Dustin Mixon, e.g.: - -Saving phase: Injectivity and stability for phase retrieval -Phase retrieval: Stability and recovery guarantees -Phase retrieval from very few measurements - -In these posts and the reference given there, Dustin discussion issues of solvabilty and indeed, the squared system is often solvable even if not quadratically overdetermined. There are also some algorithms around, e.g DOLPHin, PhaseMax or PhaseLift.<|endoftext|> -TITLE: Is this equivalent to RH - Riemann hypothesis? -QUESTION [15 upvotes]: $$\pi = 3\prod_{\zeta(1/2+it) = 0}\frac{9+4t^2}{1+4t^2}\iff\text{RH is true}.$$ - -REPLY [35 votes]: Yes, this is equivalent to RH (but not in any significant way). Recall the completed Riemann $\xi$-function -$$ -\xi(s) = s(s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s), -$$ -which, by Hadamard's factorization formula can be written as -$$ -e^{A+Bs} \prod_{\rho}\Big(1-\frac{s}{\rho}\Big) e^{s/\rho}, -$$ -where the product is over all non-trivial zeros $\rho$ of $\zeta(s)$. -Now one can check that $A=0$ (plug in $s=0$) and that $B= -\sum_{\rho}\text{Re }(1/\rho)$. It follows that -$$ -|\xi(s)| = \prod_{\rho: \text{Im}(\rho) >0} \Big| \frac{s-\rho}{\rho}\Big|^2, -$$ -by grouping complex conjugate zeros (and the product now converges). Now evaluate this at $s=2$: thus -$$ -\xi(2) =2 \times 1 \times \pi^{-1} \times \Gamma(1) \times \zeta(2) = \frac{\pi}{3} -$$ -equals -$$ -\prod_{\rho: \text{Im}(\rho) >0} \Big| \frac{2-\rho}{\rho}\Big|^2. -$$ -Split the product over zeros into two factors: the first one from zeros on the critical line, and the second one over zeros not on the critical line (if any). The first factor is simply -$$ -\prod_{\substack{\rho = 1/2 +i\gamma \\ \gamma >0}} \frac{(3/2)^2+\gamma^2}{(1/2)^2 +\gamma^2} = \prod_{\substack{\rho = 1/2 +i\gamma \\ \gamma >0}} \frac{9+4\gamma^2}{1+4\gamma^2}. -$$ -If RH is true then the second factor is $1$. If RH is false, then note that the contribution of the zeros $\beta+i\gamma$ and $1-\beta+i\gamma$ together to the second product is -$$ -\frac{(2-\beta)^2+\gamma^2}{\beta^2+\gamma^2} \frac{(1+\beta)^2+\gamma^2}{(1-\beta)^2 +\gamma^2} > 1; -$$ -(both factors are $\ge 1$ since $0\le \beta \le 1$, and at least one of them must be strictly larger than $1$). -There's a little bit more fun to be had with this problem. It can be used to show easily that if $\gamma_0$ is the first ordinate of a zero (not necessarily on the critical line) of $\zeta(s)$ then -$$ -\frac{\pi}{3} \ge \frac{9+4\gamma_0^2}{1+4\gamma_0^2}, -$$ -and since $\pi$ is so close to $3$, one can extract from this the fairly good bound that $\gamma_0 \ge 6.49\ldots$. This general idea is of course well known, but I thought this particular choice was pretty!<|endoftext|> -TITLE: Calculation of logarithmic capacity? -QUESTION [5 upvotes]: I am reading this paper about "Numerical approximation of the logarithmic capacity of domains", and there (on the third page) I found simple formulas for logarithmic capacity of simple figures like squares and equilateral triangles. -For a better understanding, I tried to recalculate those by myself several times, but I could not succeed. Is there a way that I can find those calculations or something similar that I can use as a guide? - -REPLY [9 votes]: In two dimensions, you have a powerful tool, the Riemann mapping. If you have a compact set in the plane whose complement is connected, knowing explicitly -the map of the complement onto the exterior of the unit disk gives you the log capacity. Now take a book on complex variables with many examples of conformal maps, and you can calculate capacities of many planar sets. I recommend the book -by -L. Volkovyskii, A collection of problems in complex analysis, there is an English Dover edition, which is a real encyclopedia of explicit conformal maps. -(If you can read some Russian, there is another book: M. A. Evgrafov, Problems in the theory of analytic functions). -All examples given in the paper that you cite can be obtained with this method. -For the disc, ellipse and the segment the conformal map is the Joukowski function, for the square and triangle use a modified Schwarz-Christoffel formula, and it leads to Gamma function. (All this is covered in the book of Volkovyskii cited above. -You can easily generalize the last two examples to regular $n$-gons with any $n$. The answer is in terms of Gamma function; it is written in Polya, Szego, Isoperimetric problems of mathematical Physics. -It is more difficult in dimension 3 and higher (Newtonian potential). A good collection of explicitly computed capacities is contained in the book N. S. Landkof, Foundations of modern potential theory (there is an English translation). -EDIT. Here are some detail for the regular $n$-gon. We want to map the exterior region onto the exterior of the unit disk so that $\infty\to\infty$ by a function $f$. -To do this, break the exterior region into $n$ congruent triangles with one vertex at infinity, -and other two vertices the adjacent vertices of the polygon. The interior angles -of a triangle are $2\pi/n$ at $\infty$ and $\pi/2+\pi/n$ at finite vertices. Now map one of these triangles onto the sector $\{ z: |\arg z|<\pi/n, |z|>1\}$. So that -$\infty\to\infty$ and other two vertices go to $e^{\pm i\pi/n}$. By the symmetry -principle this function will have an analytic continuation to the whole exterior -and map it onto the exterior of the unit disk, so it is our $f$. - The required map of the triangle onto a sector is performed -with the Schwarz-Christoffel formula, and the integral which is involved is reduced to the Gamma-function. Then $f(z)\sim az,\;z\to\infty$ and capacity is $1/|a|$.<|endoftext|> -TITLE: push-forward of linear algebraic group schemes -QUESTION [6 upvotes]: Suppose that $\pi: X \to S$ is a proper morphism of schemes, and $G \to X$ is a flat affine algebraic group scheme over $X$. Is the push-forward $\pi_* G$ also affine? -I'd be happy to assume more -- say that $\pi$ is smooth, and that $G$ is smooth over $X$, or even a torus... - -REPLY [3 votes]: If $\pi \colon X \to S$ is proper, flat and of finite presentation and $W$ is an affine $X$-scheme, then $\pi_*W \to S$ is affine and of finite presentation. Unless $W$ is etale over $X$, it is difficult to deduce much about the smoothness of $\pi_*W \to S$. -To see this, you can combine an affine representability result for modules, such as https://stacks.math.columbia.edu/tag/08JY or [EGAIII-2, 7.7.8], with the ideas from Proposition 2.5 of: -Lieblich, Max, Remarks on the stack of coherent algebras, Int. Math. Res. Not. 2006, No. 11, Article ID 75273, 12 p. (2006). ZBL1108.14003. -A precise reference is Theorem 2.3 of -Hall, Jack; Rydh, David, General Hilbert stacks and Quot schemes, Mich. Math. J. 64, No. 2, 335-347 (2015). ZBL1349.14013.1434731927. https://projecteuclid.org/euclid.mmj/1434731927<|endoftext|> -TITLE: Twin primes conjecture and extrapolation method -QUESTION [46 upvotes]: Let $(p_1, p_2)$ be a twin prime pair, where we include $(2, 3)$. If $p_1 \equiv 1$ mod $4$ then we let $t_{(p_1, p_2)} := p_1 ^ 2 / p_2 ^ 2$ otherwise, we let $t_{(p_1, p_2)} := p_2 ^ 2 / p_1 ^ 2$. -I conjecture that the product -$$ -\prod_{(p_1, p_2): \text{twin primes}}t_{(p_1, p_2)} -=\tfrac{3 ^ 2}{2 ^ 2} \cdot \tfrac{5 ^ 2}{3 ^ 2}\cdot -\tfrac{5 ^ 2}{7 ^ 2}\cdot\tfrac{13 ^ 2}{11 ^ 2} \cdot\tfrac{17 ^ 2}{19 ^ 2} \cdot\tfrac{29 ^ 2}{31 ^ 2} \cdot\tfrac{41 ^ 2}{43 ^ 2} \cdot \tfrac{61 ^ 2 }{59 ^ 2} \cdot \tfrac{73 ^ 2}{ 71 ^ 2}\cdot \tfrac{101 ^ 2 }{ 103 ^ 2}\cdots -$$ -is equal to $\pi$. (If this is true then twin prime numbers are infinity many.) -Some numerical values of partial products: -___________________________ $p_1 \equiv 3$ mod $4$ ___ $p_1 \equiv 1$ mod $4$ - -3.1887755102040816321 to $10^1$, ___________ 1 = ____________ 1 -3.2055606708805624550 to $10^2$, ___________ 4 = ____________ 4 -3.1290622219773513145 to $10^3$, __________ 16 < ___________ 19 -3.1364540609918890779 to $10^4$, _________ 100 < __________ 105 -3.1384537326021492746 to $10^5$, _________ 620 > __________ 604 -3.1417076006640026373 to $10^6$, ________ 4123 > _________ 4046 -3.1417823471756806475 to $10^7$, _______ 29498 > ________ 29482 -3.1415377533170544536 to $10^8$, ______ 219893 < _______ 220419 -3.1415215264211035597 to $10^9$, _____ 1711775 < ______ 1712731 -3.1415248453830039795 to $10^{10}$, ____13706087 < _____ 13706592 -3.1415126339547108140 to $10^{11}$, -3.1415144504088659201 to $10^{12}$, -3.1415142045284687040 to $10^{13}$, -3.1415144719058962626 to $10^{14}$, -3.1415384423175311229 to $10^{15}$ -Can we find a few more decimal places using the extrapolation method? - -REPLY [3 votes]: From the product -$$ -\prod_{(p_1, p_2): \text{twin primes}}t_{(p_1, p_2)} -=\tfrac{3 ^ 2}{2 ^ 2} \cdot \tfrac{5 ^ 2}{3 ^ 2}\cdot -\tfrac{5 ^ 2}{7 ^ 2}\cdot\tfrac{13 ^ 2}{11 ^ 2} \cdot\tfrac{17 ^ 2}{19 ^ 2} \cdot\tfrac{29 ^ 2}{31 ^ 2} \cdot\tfrac{41 ^ 2}{43 ^ 2} \cdot \tfrac{61 ^ 2 }{59 ^ 2} \cdot \tfrac{73 ^ 2}{ 71 ^ 2}\cdot \tfrac{101 ^ 2 }{ 103 ^ 2}\cdots -$$ -We keep the terms that are $p_1 \equiv 5$ mod $8$ or $p_1 \equiv 7$ mod $8$ -I conjecture that -$$ -\tfrac{5 ^ 2}{7 ^ 2}\cdot\tfrac{29 ^ 2}{31 ^ 2} \cdot\tfrac{73 ^ 2}{ 71 ^ 2}\cdot \tfrac{101 ^ 2 }{ 103 ^ 2}\cdot \tfrac{149 ^ 2 }{ 151 ^ 2}\cdot \tfrac{193 ^ 2 }{ 191 ^ 2}\cdot \tfrac{197 ^ 2 }{ 199 ^ 2}\cdot \tfrac{241 ^ 2 }{ 239 ^ 2}\cdots=5^.5/5 -$$ -or -$$ -\tfrac{5 ^ 4}{7 ^ 4}\cdot\tfrac{29 ^ 4}{31 ^ 4} \cdot\tfrac{73 ^ 4}{ 71 ^ 4}\cdot \tfrac{101 ^ 4}{ 103 ^ 4}\cdot \tfrac{149 ^ 4}{ 151 ^ 4}\cdot \tfrac{193 ^ 4 }{ 191 ^ 4}\cdot \tfrac{197 ^ 4 }{ 199 ^ 4}\cdot \tfrac{241 ^ 4 }{ 239 ^ 4}\cdots=0.2 -$$ -So, Pi (if the product converges to Pi) included in the remaining terms of product. -$$ -\tfrac{3 ^ 2}{2 ^ 2} \cdot \tfrac{5 ^ 2}{3 ^ 2}\cdot -\tfrac{13 ^ 2}{11 ^ 2} \cdot\tfrac{17 ^ 2}{19 ^ 2} \cdot\tfrac{41 ^ 2}{43 ^ 2} \cdot \tfrac{61 ^ 2 }{59 ^ 2}\cdot \tfrac{109 ^ 2 }{107 ^ 2}\cdots=\sqrt {5}*π -$$ -or -$$ -\tfrac{3 ^ 4}{2 ^ 4} \cdot \tfrac{5 ^ 4}{3 ^ 4}\cdot -\tfrac{13 ^ 4}{11 ^ 4} \cdot\tfrac{17 ^ 4}{19 ^ 4} \cdot\tfrac{41 ^ 4}{43 ^ 4} \cdot \tfrac{61 ^ 4 }{59 ^ 4}\cdot \tfrac{109 ^ 4 }{107 ^ 4}\cdots=5*π^2 -$$<|endoftext|> -TITLE: Is an open subset of a rigid space rigid? -QUESTION [10 upvotes]: Let $X$ be a locally compact Hausdorff space. Call $X$ rigid if its only autohomeomorphism is the identity, $\operatorname{Homeo}(X)=\{1\}$. -Questions: - -Let $X$ be rigid. Is it true that every open subset $Y\subset X$ is rigid, i.e., $\operatorname{Homeo}(Y)=\{1\}$? -Let $Y\subset X$ be open. Is it true that every autohomeomorphism of $Y$ extends (perhaps non-uniquely) to an autohomeomorphism of $X$? -What can we assume about $X$ additionally to make sure that the answer to question 1 is positive? Metrizability, separability? - -Discussion: -An affirmative answer to question 2 obviously implies an affirmative answer to question 1. Question 2 may be asking for too much, but there are simple cases where it is true. For instance, given a continuous bijective function $f:(0,1)\to(0,1)$, it can be continuously continued to a bijective function $F:\mathbb{R}\to\mathbb{R}$, hence every autohomeomorphism of $(0,1)$ can be continued to an autohomeomorphism of $\mathbb{R}$. A positive answer to question 1 seems more realistic to me. For instance, it is true for the Cook's curve, I believe. It is harder for me to check this condition for other examples of rigid spaces in the literature. -Thank you. - -REPLY [2 votes]: Here is a counterexample to the first question, a planar, non-separating one-dimensional locally connected continuum. -Let $A_1$ be the line segment in the $xy$-plane $\mathbb{R}^2$ with end points $(0,0)$ and $(1,0)$. Define set $A_2$ as the union of $A_1$ with a sequence of mutually disjoint trees (finite acyclic graphs) $T_i$, no two of them homeomorphic, each attached to $A_1$ at a single point other than the end point of $A_1$. The diameters of the trees should converge to zero, and the set of points of attachment should be dense in $A_1$ (see the first drawing). Also, $A_2\setminus A_1$ should be contained in the vertical open strip $0 -TITLE: rational points and a local perturbation of an elliptic curve -QUESTION [6 upvotes]: Let $E_{a,b}$ be an elliptic curve defined by the equation $y^{2}=x^3+ax+b$ where $a,b \in \mathbb{Q}$. -Suppose that for $a=a_{0}$ and $b=b_{0}$ the rank of $E_{a_{0},b_{0}}(\mathbb{Q})=1$. -question: -is there an $\epsilon> 0$ such that for any $(\alpha,\beta) \in \mathbb{Q}^{2}$ and $|\alpha-a_{0}|+|\beta -b_{0}|< \epsilon$ then the rank of $E_{\alpha,\beta}(\mathbb{Q})=1$. - -REPLY [9 votes]: A counter-example, showing that the answer is "no" for some $(a_0,b_0)$ can be constructed as follows. Take an elliptic surface over $\mathbb{Q}$ with rank $2$. By Silverman's specialisation theorem, the rank of each fibre is at least $2$ (but often larger) except for a finite number of fibres. At an exceptional fibre of rank $1$, you will get a counter-example. -For instance $a_0=2$, $b_0=1$ the curve is of rank $1$, but the family $E_t:y^2=x^3+(−t^2+t+2)x+1$ is of rank at least $2$ over $\mathbb{Q}(t)$ as it contains $(0,1)$ and $(t,t+1)$. For all $t$ close to zero the rank of $E_t(\mathbb{Q})$ will be $2$ or larger. -I believe a similar argument should work in general and show that the answer is always "no". -Arithmetic properties like the rank of the Mordell-Weil group are not continuous in the real or $p$-adic topology. For instance, the number of prime factors of numerator and denominator of $\Delta$ will have an influence on the rank. -In the above example as $t=1/n$ approaches $0$, the rank will jump around $2$ and $3$ and sometimes larger values rather randomly: -$$ -\begin{array}{c|cccc} -t=\tfrac{1}{n} & \tfrac{1}{2} & \tfrac{1}{3} & \tfrac{1}{4} &\tfrac{1}{5} -& \tfrac{1}{6} & \tfrac{1}{7} & \tfrac{1}{8} & \tfrac{1}{9} & \tfrac{1}{10} & -\tfrac{1}{11} & \tfrac{1}{12} & \tfrac{1}{13} & \tfrac{1}{14} &\tfrac{1}{15} & \tfrac{1}{16} & \tfrac{1}{16} & \tfrac{1}{17} & \tfrac{1}{18} & \tfrac{1}{19}\\ -\text{rank} & 1 & 3 & 2 & 2 & 3 & 2 & 3 & 3 & 3 & 3 & 4 & 2 & 2 & 3 & 3 & 3 & 3 & 3 &2 \\ -\end{array} -$$<|endoftext|> -TITLE: Property-like structure in a model category -QUESTION [11 upvotes]: In a model category, I have tools to show that mapping spaces are contractible. But if I want to show a mapping space is empty or contractible, is there anything I can do on general grounds? -The idea is this. Suppose I have a type of structure I'm interested in, and a collection of objects. To each object I assign a space parameterizing the choices required to equip that object with the structure in question. If all these parameter spaces are empty-or-contractible, that means the structure is "property-like": an object either has the structure or it doesn't; it can't carry different versions of the same structure. If the parameter spaces are all contractible, this means that in fact every object carries a unique version of the structure. Let me illustrate with an example of each. -A contractible space of choices: property-like structures that always exist. For example, in Joyal's model structure for quasicategories, suppose I want to show that the space of composites of two composable morphisms in a quasicategory $X$ is always contractible. This reduces to showing that the horn inclusion $\Lambda^1[2] \to \Delta[2]$ is an acyclic cofibration (since this implies that the fibers of $X^{\Delta[2]} \to X^{\Lambda^1[2]}$ are contractible). -Of course, this is immediate from typical descriptions of the model structure, but more generally I should consider myself lucky if all I have to do to answer a question is show that one explicit map is an acyclic cofibration, since I can work with explicit generators to make this a matter of combinatorics. (There's an asterisk in this case because in this model structure really I only know how to understand anodyne extensions combinatorially and not acyclic cofibrations in general, but in practice this understanding usually suffices.) -An empty-or-contractible space of choices: property-like structures that don't always exist. But now, again working in the Joyal model structure, suppose I want to show that the space of retracts of a given idempotent in a quasicategory is always either empty or contractible. This doesn't reduce to showing that the inclusion of the free idempotent into the free retract is an acyclic cofibration, because that would be too strong -- it would imply that every idempotent has a retract, which is not the case. -The closest thing I can think of is to perform a Bousfield localization to force every idempotent to have a retract, and then show that this map is an acyclic cofibration in the new model structure. But this only shows that if a quasicategory has all split idempotents (and I have to check that these are exactly the fibrant objects in the localized model structure), then the space of retracts for a given idempotent is contractible -- it doesn't tell me anything about quasicategories where some but not all idempotents split. -In this particular case, Lurie shows this fact using the theory of cofinality. But this is something specific to the example, not a general approach to the understanding property-like structures, even if we restrict our attention to quasicategories. -Question: Do model categories afford some general method for understanding property-like structures on their objects/morphisms, analogous to the method of showing that certain maps are acyclic cofibrations? Or can these things only be understood on a case-by-case basis? - -REPLY [4 votes]: Specifically for the case of quasi-categories (or any other model for $\infty$-categories) the following observation can be useful: suppose that $f: {\cal C} \to {\cal D}$ is a map of quasi-categories such that the induced map ${\rm Tw}(f):{\rm Tw}({\cal C}) \to {\rm Tw}({\cal D})$ on twisted arrow categories is coinitial, i.e., for every $e \in {\rm Tw}({\cal D})$ the comma category ${\rm Tw}({\cal C})_{/e}$ is weakly contractible (this is the dual property of cofinal, and implies in particular that reindexing diagrams along ${\rm Tw}(f)$ induces an equivalence on limits, rather than colimits). Then for any quasi-category ${\cal E}$ the induced map on functor categories ${\rm Fun}({\cal D},{\cal E}) \to {\rm Fun}({\cal D},{\cal E})$ is fully-faithful (and in particular, the induced map on functor spaces is an embedding). To see this, use the fact that if $g,h: {\cal D} \to {\cal E}$ are two functors then the space of natural transformations from $g$ to $h$ can be written as a limit on the twisted arrow category: -$$ {\rm Map}_{{\rm Fun}({\cal D},{\cal E})}(g,h) \simeq {\rm lim}_{x \to y \in {\rm Tw}({\cal D})} {\rm Map}_{\cal E}(g(x),h(y)) $$ -The coinitiality of ${\rm Tw}(f)$ now implies that ${\rm Map}_{{\rm Fun}({\cal D},{\cal E})}(g,h) \stackrel{\simeq}{\to} {\rm Map}_{{\rm Fun}({\cal C},{\cal E})}(g \circ f,h \circ f)$ is an equivalence. -Examples: -1) The map $\Delta^1 \to J$ from the walking arrow to the walking isomorphism induces a coinitial map on twisted arrow categories (this is because ${\rm Tw}(J) \simeq J \simeq \ast$ and ${\rm Tw}(\Delta^1)$ has a contractible geometric realization). This reflect the (simple, but important) fact that an arrow being an equivalence is a property. -2) Consider the map ${\rm Idem} \to {\rm Ret}$ from the free idempotent to the free retract (so that ${\rm Idem}$ has a single object equipped with an idempotent self map and ${\rm Ret}$ has two objects which sit in a retract diagram). Then one can show that the map ${\rm Tw}({\rm Idem}) \to {\rm Tw}({\rm Ret})$ is coinitial (this is a bit more tedious, but can still be done completely by hand: ${\rm Tw}({\rm Idem})$ has two objects, ${\rm Tw}({\rm Ret})$ has five). This reflects the fact you mentioned that a splitting of an idempotent is a property.<|endoftext|> -TITLE: Left adjoint pseudofunctor commutes with pseudocolimits -QUESTION [9 upvotes]: I'm looking for a reference for this seemingly basic fact: assume I have a 2-functor $G : {\cal X}\to {\cal Y}$ and assume I can define a left 2-adjoint $F$ for it, which is nevertheless only a pseudofunctor. - -Is it true that $F$ commutes with all pseudocolimits in $\cal Y$? - -REPLY [3 votes]: My idea would be to take the proof that left adjoint functors commute with colimits and then sprinkle "up to isomorphism" generously throughout the proof. - -The usual argument seems indeed to apply here, but there are a few details I am not convinced with, and I would like to understand the proof with a certain command. - -First of all the definition of pseudocolimit I'm using: if $W : {\cal A}° \to\bf Cat$ and $D : {\cal A}\to \cal B$ a pseudocolimit $W\boxtimes D$ of $D$ weighted by $W$ is an object of $\cal B$ such that there is an isomorphism of categories -$$ -{\cal B}(W\boxtimes D,B)\cong \text{Psd}({\cal A}°,{\bf Cat})(W, {\cal B}(D,X)) -$$ My first question is: shall I assume that this isomorphism is only an equivalence? -A rather nontrivial result now gives me that there is a (possibly ultra-complicated) weight $\bar W$ such that $W\boxtimes D$ is in fact an honest $\bar W$-weighted colimit. Is this true without further assumptions on the data? -If yes, the proof now boils down to the classical argument "sprinkled" with some canonical isos, since the hom-functor still commutes with ($\bar W$-)weighted colimits (I can assume $\cal B$ is co/tensored, this is not -a problem), and then -$$ -\begin{array}{c} -F(W\boxtimes D) \xrightarrow{\qquad \qquad} B\\\hline -W\boxtimes D \xrightarrow{\qquad \qquad} GB\\\hline -\{\!\!\{W,(D \xrightarrow{\qquad \qquad} GB)\}\!\!\}\\\hline -\{\!\!\{W,(FD \xrightarrow{\qquad \qquad} B)\}\!\!\}\\\hline -W\boxtimes FD \xrightarrow{\qquad \qquad} B -\end{array} -$$ (braces = weighted limit).<|endoftext|> -TITLE: A group, neither amenable, nor having a subgroup that looks like $F_2$ up to level $n$? -QUESTION [6 upvotes]: It is known that there are non-amenable groups not containing $F_2$, the free group on two generators. We can even have that every 2-generated subgroup is finite. -But is there a non-amenable group $G$ where for some $n$, $G$ is length-$n$ unfree in the following sense? - -Definition. $G$ is length-$n$ unfree if for all $a,b\in G$, there exist words $u\ne v$ of length $n$ over the alphabet $\{a,b\}$ such that $u=v$ in $G$. - -REPLY [12 votes]: Burnside groups of exponent $n$ are length $n$ unfree. If $n\ge 665$, odd, then the free Burnside group of exponent $n$ of rank 2 or more is not amenable (see Adian's book "The Burnside problem" or Olshanskii's book "Geometry of defining relations" or my book "Combinatorial algebra: syntax and semantics", Chapter 5).<|endoftext|> -TITLE: Relation between Hecke operators and coefficient of L-functions -QUESTION [5 upvotes]: This question has its seed in this one by Gory, which found an enlightening answer but one of the comments kept me wondering. I am beginning to discover Hecke operators, and there appears to be an ubiquitous relation between Hecke eigenvalues and coefficients of L-functions that I do not get at all. I will try to state everything in details. -Hecke operators. Let us fix a place $p$ and consider an unramified local component $\pi_p$ of an automorphic representation of $GL_2$ over $F$. Let $K_p$ denote $GL_2(\mathcal{O}_p)$. We define the Hecke operator $T_{p^i}$ as the convolution action of the characteristic function of -$$\bigcup_{\substack{a+b = i \\ a \geqslant b}} -K_p -\left( -\begin{array}{cc} -p^a & \\ - & p^b -\end{array} -\right) -K_p$$ -L-functions. The automorphic representation $\pi$ also has an attached $L$-function (built on the Satake parameters at unramified places and a specific completion defining the remaining factors) which can be written as (and this defines the $\lambda_\pi(n)$) -$$L(s, \pi) = \sum_{n \geqslant 1} \lambda_\pi(n) n^{-s}$$ -Coefficients as eigenvalues. With all those definitions in hand, if $\phi$ is a function in the (one-dimensional) subspace of $K_p$-invariant vectors of $\pi_p$, do we have that -$$T_{p^i} \star \phi = p^{1/2} \lambda_\pi(p^i) \phi \quad ?$$ -Questions. More precisely, I would like to ask both following (maybe elementary) questions: - -I know it for $i=1$ (for instance Gelbart), however does it remain for $i \geqslant 2$, and do you have a proof of that? -in the case where $\pi_p$ is ramified, those convolutions always give zero because there is no $K_p$ invariant vector in $\pi_p$ but the convolution creates such invariant vectors. In order to get the coefficient $\lambda_\pi(p^i)$ is this case, can I do exactly the same construction replacing $K_p$ by $K_1(p^f)$ where $f$ is the (additive) arithmetic conductor of $\pi_p$? (in that case the vector space of vectors fixed by it is one-dimensional) - -I would appreciate any details or good reference for this matters, thanks in advance! - -REPLY [6 votes]: A normalized version of your guess is right. First note that the $T_{p^n}$'s satisfy the relation -$$ T_{p^{n+1}} = T_p T_{p^n} - p T_{p^{n-1}} $$ -(e.g., Bump Prop 4.6.4). This gives you a recursion relation among Hecke eigenvalues of $\phi$. E.g., $T_{p^2} = T_p T_p - p T_1$ says if the eigenvalue for $T_p$ is $p^{1/2} a_p$, then the eigenvalue for $T_{p^2}$ is $p(a_p^2 - 1)$. -Now you want to compare with coefficients of the Dirichlet series. Say the Satake parameters of $\pi$ are $\alpha=\alpha_p$ and $\alpha^{-1}$, so $a_p = \alpha + \alpha^{-1}$ (e.g., Bump Prop 4.6.6--here I'm assuming trivial central character for simplicity). By the Euler product, you only need to look at the coefficients of the Dirichlet series for the factor at $p$, which is defined to be -$$ L_p(s,\pi) = \frac 1{1-\alpha p^{-s}} \frac 1{1-\alpha^{-1} p^{-s}} -= (\sum \alpha^i p^{-is} )( \sum \alpha^{-j} p^{-js} ) $$ -Here the coefficient of $p^{ns}$ is -$$ c_n := \lambda_\pi(p^n) = \sum_{i+j=n} \frac{\alpha^i}{\alpha^j}. $$ -Now it is not hard to see the $c_n$'s satisfy the relation -$$ c_{n+1} = c_1 c_n - c_{n-1}. $$ -For instance, for $n=2$, we have $c_2 = c_1 c_1 - c_0 = a_p^2 - 1$. -Comparing with the Hecke recurrence, one gets -$$ a_{p^n} = p^{\frac n2} c_n = p^{\frac n2} \lambda_\pi(p^n), $$ -where $a_{p^n}$ is the eigenvalue of $T_{p^n}$. -For ramified representations, yes you can define Hecke operators using appropriate congruence subgroups $K_0(p^n)$ or $K_1(p^n)$, but you should do this a little differently if you want to get coefficients of Dirichlet series of newforms as Hecke eigenvalues. That is, you shouldn't look at all double cosets in $GL_2(\mathbb Z_p$). See e.g. the book of Knightly and Li.<|endoftext|> -TITLE: Trace of non-commutable matrices -QUESTION [11 upvotes]: Let $M_1$ and $M_2$ be two symmetric $d\times d$ matrices. What is the relationship between -$tr(M_1M_2M_1M_2)$ and $tr(M_1^2 M_2^2 )$? -P.S. I tried a few examples and found -$$ -tr(M_1M_2M_1M_2) \le tr(M_1^2 M_2^2 ) -$$ -seems always true. Is there a theorem? - -REPLY [10 votes]: Not to take anything away from Suvrit's answer, but this is actually much simpler. First, we can assume $M_1$ is diagonal. Call it $diag(x_1, \dotsc, x_i).$ -Then the difference between the LHS and the RHS is -$$\sum_{i> j} a_{ij}^2 (x_i - x_j)^2,$$ -where the $a_{ij}$ are the entries of $M_2.$<|endoftext|> -TITLE: A modification of the Ljunggren-Nagell equation -QUESTION [5 upvotes]: [Thanks to Gerhard Paseman for helping me reformulate my original question.] -The equation -$$ -\frac{a^m-1}{a-1}=b^2 -$$ -was solved by Ljunggren, building on work of Nagell, who showed that if $a>1$, $b>1$, and $m>2$ are integers, there are exactly two (easy to find) solutions. Can the same methods explicitly give all solutions to the following equation? -$$ -\frac{a^m-1}{a-1}=2b^2 -$$ -Also, is this worked out anywhere in the literature? (If not, I'm thinking of giving it as a problem to an undergraduate researcher. But if there is some easy way to approach the problem I'd like to know.) - -REPLY [6 votes]: The equation -$$ -\frac{a^m-1}{a-1}=2b^2 -$$ -does not have solutions in positive integers for $m>2$ as shown below. -First, notice that $a$ must be odd. -Second, one can see that $m$ must be even. Indeed, if $a\equiv 3\pmod4$, then odd $m$ would produce an odd number in the left-hand side. On the other hand, if $a\equiv 1\pmod4$, then from Theorem 3 of the LTE it follows that -$$ -\nu_2(m)=\nu_2(\frac{a^m-1}{a-1})=\nu_2(2b^2)>0, -$$ -i.e. $m$ is even. -Let $m=2k$ where $k>1$. Then -$$ -\frac{a^m-1}{a-1}=\frac{a^k-1}{a-1}(a^k+1). -$$ -Since $\gcd(a^k-1,a^k+1)=2$, we have two cases to consider: $\frac{a^k-1}{a-1}=c^2$ or $a^k+1=c^2$ for some $c\mid b$. -In the former case: - -if $k=2$, we get $(a,c)=(t^2-1,t)$ for any $t$ satisfying $(t^2-1)^2+1=2d^2$ where $d=b/c$. However, it can be computationally verified (e.g., in Magma) that this elliptic curve has the only solution with $t=0$ (giving $a=-1$). -if $k>2$, we get Ljunggren equation giving $(k,a,c)=(5,3,11)$ and $(k,a,c)=(4,7,20)$, which however do not produce any solutions to the original equation. - -In the latter case, one can refer to Mihailescu's Theorem to obtain the only solution $(k,a,c)=(3,2,3)$, which again does not produce any solutions to the original equation. -Hence, the equation in question does not have any solutions for $m>2$.<|endoftext|> -TITLE: Practically calculating the domain of a power series for function of several complex variables -QUESTION [5 upvotes]: For simplicity, let us consider a function $f$ holomorphic on a domain $D \subseteq \mathbb{C}^2$. We may therefore write $f$ as a sum of power series $$f(z) = \sum_{\nu_1 \nu_2 =0}^{\infty} c_{\nu_1 \nu_2}(z_1 - w_1)^{\nu_1} (z_2 - w_2)^{\nu_2}.$$ -We should also note that the domain of convergence of a power series in $>1$ complex variable is not a polydisk as one would expect, but rather a logarithmically convex Reinhardt domain. -In the case of one complex variable, the standard way of determining the radius of convergence is by means of the ratio test. Is there an analogous "standard approach" for determining the domain of convergence of a power series of two complex variables? -For example, the power series $$\frac{1}{1 - z_1z_2} = \sum_{\nu=0}^{\infty} z_1^{\nu} z_2^{\nu}$$ converges for $\left| z_1 z_2 \right| <1$, while the series $$\frac{z_1}{(1-z_1)(1-z_2)} = \sum_{\left| k \right| =0}^{\infty} z_1^{k_1 + 1} z_2^{k_2}$$ converges in the bidisk $\{ \left| z_1 \right|< 1, \left| z_2 \right| < 1\}$, completed by the complex line $\{ z_1 =0 \}$. -[Reference: Shabat's Introduction to Complex analysis]. - -REPLY [3 votes]: The usual Cauchy-Hadamard formula has a generalization to several variables. -The numbers $r_1,\ldots,r_n$ are called conjugate radii of convergence if the -series converges in the open polydisk $B(r_1,\ldots,r_n)$ and does not converge on any open subset of -$\{ z:|z_j|>r_j, 1\leq j\leq n\}$. Then we have the formula -$$\limsup_{|k|\to\infty}\left(|c_{k_1\ldots,k_n}|r_1^{k_1}\ldots r_n^{k_n}\right)^{|k|}=1,$$ -where $|k|=k_1+\ldots+k_n.$ -Reference: B. A. Fuks, Theory of analytic functions of several variables, -vol. I, Chap. I, sect 3, Theorem 3.7.<|endoftext|> -TITLE: A problem involving the Error Function -QUESTION [11 upvotes]: I am looking at the following function on the domain $x\geq 0$: -$$F(x)=(x+a)e^{x^2}(1-\mathrm{erf}(x))-\frac{b}{\sqrt\pi},$$ -where $a>0$, $0z_a:=0\vee(-a)$, -where $0z_a$, we have -$$\rho'(x)>0 \iff a>1\text{ or } \big(0 < a \le 1\ \&\ x > x_a:=(1 - a^2)/a\big).$$ -Now we are ready to use l'Hospital-type rules for monotonicity. -Case 1: $a\le0$, so that $z_a=-a$. Then $\rho$ is (strictly) decreasing (on the entire interval $(z_a,\infty)$), -whence, by Proposition 4.1 in the mentioned paper, $r$ is decreasing, and so, equation (1) has at most one root (in $(z_a,\infty)$). In fact, in this case there is exactly one root, since $r(z_a+)=\infty$ and $r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$. -Case 2: $01$. Then $\rho$ is increasing (on the entire interval $(z_a,\infty)$), -whence, by the mentioned Proposition 4.1, $r$ is increasing. So, $r -TITLE: Is the following set convex or not? -QUESTION [7 upvotes]: Let $(E, \langle\cdot\;, \;\cdot\rangle)$ be a complex Hilbert space. Let $T\in\mathcal{L}(E)$ and $M\in \mathcal{L}(E)^+$. -Assume that $T(ker(M))\nsubseteq ker(M)$. We define the following subset: -\begin{eqnarray*} -S_M(T) -&=&\{\lambda\in \mathbb{C}\,;\;\; \exists\,(\alpha_n,\beta_n)\in ker(M)\times \overline{Im(M)}\,;\;\;\|M^{1/2}\beta_n\|=1, -\displaystyle\lim_{n\rightarrow+\infty}\langle MT \alpha_n\; |\;\beta_n\rangle+\langle MT \beta_n\; |\;\beta_n\rangle=\lambda,\\ -&&\phantom{+++++}\;\hbox{and}\;\displaystyle\lim_{n\rightarrow+\infty}\|M^{1/2} T(\alpha_n+\beta_n)\|<\infty\;\}. -\end{eqnarray*} -What do you think about the convexity of $S_M(T)$?? I try with an example of $M$ and $T$ such that $T(ker(M))\nsubseteq ker(M)$, I get $S_M(T)=\mathbb{C}$. -I claim that $S_M(T)=\mathbb{C}$. Do you think that my claim is true? -Thank you for your help!! - -REPLY [4 votes]: Suppose $M\geq 0$ and $T(\ker M) \nsubseteq \ker M$. So there exists an $\alpha \in \ker M$ such that $T\alpha \notin \ker M$ which gives that $ MT\alpha \neq 0$ and since $M$ is positive then $M^{1/2}MT\alpha\neq 0$ as well. -Let $\beta = \frac{MT\alpha}{\|M^{3/2}T\alpha\|} \in {\textrm Im}(M)$ then $\|M^{1/2}\beta\| = 1$ and $\|M^{1/2}T(\alpha + \beta)\| < \infty$. -If $\lambda \in \mathbb C$ then $(\lambda \alpha, \beta) \in \ker M \times {\mathrm Im}(M)$ and -\begin{align*} -\langle MT(\lambda\alpha)\ |\ \beta\rangle + \langle MT\beta\ |\ \beta\rangle &= \lambda\langle MT\alpha\ |\ \frac{MT\alpha}{\|M^{3/2}T\alpha\|}\rangle + \langle MT\beta\ |\ \beta\rangle -\\ & \lambda\frac{\|MT\alpha\|}{\|M^{3/2}T\alpha\|} + \langle MT\beta\ |\ \beta\rangle. -\end{align*} -Because $\alpha$ and $\beta$ are fixed and $\|MT\alpha\| \neq 0$ then this gives you any complex number for an appropriate choice of $\lambda$. Therefore, $S_M(T) = \mathbb C$.<|endoftext|> -TITLE: Is there a Hausdorff weakly Lindelof space which is not DCCC? -QUESTION [5 upvotes]: As we know, every regular weakly Lindelof space is DCCC. Here DCCC denotes discrete countable chain condition, a space $X$ has discrete countable chain -condition if every discrete family of non-empty open sets of $X$ is countable. -A space $X$ is said to be weakly Lindelof if every open -cover $\mathcal U$ of $X$ contains a countable subfamily $\mathcal V \subset \mathcal U$ such that $\bigcup \mathcal V$ is dense in $X$. - -Is there a Hausdorff weakly Lindelof space which is not DCCC? - -REPLY [4 votes]: The answer to this problem is negative because of the following -Theorem. If a topological space $X$ is weakly Lindelof, then each discrete (more generally, locally countable) family of open sets in $X$ is at most countable. -Proof. Let $\mathcal U$ be a locally countable family of open subset of $X$. -Then each point $x\in X$ has a neighborhood $O_x$ meeting at most countably many sets of the family $\mathcal U$. By the weak Lindelof property of $X$ the open cover $\{O_x:x\in X\}$ contains a countable subfamily $\{O_{x}:x\in C\}$ whose union $\bigcup_{x\in C}O_x$ is dense in $X$ and hence intersects each set $U\in\mathcal U$. -Assuming that $\mathcal U$ is uncountable and applying the Pigeonhole Principle, we can find a point $x$ of the countable set $C$ such that $O_x$ intersects uncountably many sets of $\mathcal U$. But this contradicts the choice of $O_x$.<|endoftext|> -TITLE: A path in the unit square that "doubles back" on itself in a nice way -QUESTION [6 upvotes]: Given a path $P$ in the unit square, and two points $p_{1},p_{2}$ -located on $P$, let $d_{P}(p_{1},p_{2})$ denote the distance from -$p_{1}$ to $p_{2}$ traversed along $P$. Given $a>1$, I am looking -for the shortest $P$ that satisfies the following property: -For any point $x$ in the unit square, there exist $p_{1},p_{2}$ -on $P$ such that $d_{P}(p_{1},p_{2})\geq a(\|p_{1}-x\|+\|p_{2}-x\|)$. -This is clearly a very difficult optimization problem, so I'm just -looking for some practical suggestions, not necessarily provably optimal (Spirals? Sawtooths?). -The motivation for this problem is as follows: imagine a car driving -along $P$ at unit velocity containing several people. We want it -to be possible for a person to leave the car at a certain point $p_{1}$, -walk at speed $1/a$ to $x$, and then rendezvous back with the -car at $p_{2}$ without requiring the car to stop. - -REPLY [6 votes]: This is just about the asymptotic behavior for $a\to+\infty$. I claim that the minimal length $\ell$ is about $\sqrt{2a}$ for large $a$. -The upper bound -Consider $2n$ horizontal lines splitting the square into strips of width $\frac 1{2n-1}$ Now travel along the odd-numbered lines (in the natural enumeration from the top) in the natural way: left to right on line 1, down to line 3, right to left on line 3, doun to line 5, left to right on line 5, and so on. Then return to line 2 in constant time and repeat the process going over even lines in the same fashion. Notice that every strip is bounded by one odd line and one even line and the time between going over those lines is about $n$ (with constant error). So, if $x$ is in the strip, we can drop off at the nearest point on the odd line and get picked up at the nearest point on the even line walking the total distance $\frac 1{2n-1}$. Thus if $\frac 1{2n-1}a -TITLE: When is countable direct-product of projective modules again projective ? -QUESTION [8 upvotes]: Let $R$ be a commutative ring with unity. The Bass-Papp theorem states that any countable direct sum of injective $R$-modules is injective iff $R$ is Noetherian . Chase's theorem states that any direct product of projective $R$-modules is projective iff $R$ is Artinian . My question is : Is any characterization for commutative rings (with unity) are known such that any countable direct-product of projective modules over the ring is again projective ? -Also asked on MSE : https://math.stackexchange.com/questions/2501945/when-is-countable-direct-product-of-projective-modules-again-projective - -REPLY [6 votes]: Under the assumption that a countable direct product of modules over $R$ means a direct product of countably many modules over $R$, I answer OP's question when $R$ is Noetherian. In addition, I outline a remark for $R$ an arbitrary countable ring with identity. - - -Claim 1. Let $R$ be a commutative ring with identity. If $R$ is Noetherian then the following are equivalent: - -$(i)$ Every direct product of countably many projective modules over $R$ is projective. -$(ii)$ $R$ is Artinian. - -Proof. The implication $(ii) \Rightarrow (i)$ is given by Stephen Chase's Theorem [1, Theorems 3.3 and 3.4]. Assume that $(i)$ holds. Then $M \Doteq R^{\aleph_0}$ is a projective module. If $R$ is moreover connected, then $M$ is free over $R$ by [2, Corollary 4.5]. $R$. By John O'Neill's result [3, Lemma 1.1], the ring $R$ is then Artinian. If $R$ is not connected, it is the direct product of finitely many non-trivial connected rings $R_i$. By hypothesis, the module $M_i \Doteq R_i^{\aleph_0}$ is projective over $R$, and hence over $R_i$, for every $i$. As $R$ is a product of finitely many Artinian rings, it is Artinian. - - -For an arbitrary countable ring, the following weaker result holds. - - -Claim 2. If $R$ is a countable ring with identity such that $R^{\aleph_0}$ is projective over $R$, then $R$ is left perfect. -Proof. Apply [1, Theorem 3.1] and Bass's characterization of left perfect rings. - - -Coming back to the commutative setting, we get - - -Corollary. - Let $R$ be an integral domain. If $R$ is countable then the following are equivalent: - -$(i)$ Every direct product of countably many projective modules over $R$ is projective. -$(ii)$ $R$ is a field. - -Proof. Any element of a perfect commutative ring is either a unit or a zero divisor. The result is then a direct consequence of Claim 2. - - - -[1] S. Chase, "Direct product of modules", 1960. -[2] H. Bass, "Big projective modules are free", 1962. -[3] J. O'Neill, "When a ring is an $F$-ring", 1993.<|endoftext|> -TITLE: Is the Golomb countable connected space topologically rigid? -QUESTION [28 upvotes]: The Golomb space $\mathbb G$ is the set of positive integers endowed with the topology generated by the base consisting of the arithmetic progressions $a+b\mathbb N_0$ with relatively prime $a,b$ and $\mathbb N_0=\{0\}\cup\mathbb N$. It is known that the space $\mathbb G$ is connected and Hausdorff. -It is also easy to check that the multiplication map $\cdot:\mathbb G\times \mathbb G\to\mathbb G$, $\cdot:(x,y)\mapsto xy$, is continuous, so $\mathbb G$ is a commutative topological semigroup. - -Problem. Is the Golomb space $\mathbb G$ topologically homogeneous? Or maybe rigid? - -We recall that a topological space $X$ is rigid if its homeomorphism group is trivial. -This problem was motivated by this question, which discusses the relation of the Golomb space to another countable connected Hausdorff space, called the rational projective space $\mathbb QP^\infty$. This space is easily seen to be topologically homogeneous. - -REPLY [21 votes]: [Edit, Dec 6, 2019] I have a pleasure to inform that this problem was finally resolved in affirmative by T.Banakh, D.Spirito and S.Turek who proved the following - -Theorem. The Golomb space is topologically rigid. - -Edit: now published: The Golomb space is topologically rigid. Comment. Math. Univ. Carolin. 62 (2021), no. 3, 347–360.<|endoftext|> -TITLE: Inclusion of lattices and fundamental domains -QUESTION [5 upvotes]: Let $G$ be a locally compact abelian group. A lattice in $G$ is a discrete subgroup $\Lambda$ such that the quotient $G / \Lambda$ is compact. A Borel fundamental domain of a lattice $\Lambda$ in $G$ is a Borel set $F \subseteq G$ that intersects each coset in $G / \Lambda$ in exactly one point, i.e., $F$ is a set of coset representatives. -My question is now the following: -Given two lattices $\Lambda_1, \Lambda_2 \subseteq G$ such that $\Lambda_1 \subseteq \Lambda 2$, is it always possible to take corresponding fundamental domains $F_i$ of $\Lambda_i$, where $i = 1,2$, such that $F_2 \subseteq F_1$? -By looking at special cases in, say, $G=\mathbb{R}$ or $G = \mathbb{Z}$, it seems that this is always possible. -Any comment or reference is highly appreciated. - -REPLY [8 votes]: The image of $\Lambda_2$ in $G/\Lambda_1$ is finite. Take a set of representatives $s_1, s_2, \ldots s_n$ in $\Lambda_2$. Now consider $\cup s_i F_2$. I claim this is a Borel fundamental domain for $F_1$. -It is Borel as it is a finite union of Borel sets. Let $g\Lambda_1$ be a coset of $\Lambda_1$. Then $g\Lambda_2$ intersects $F_2$ in a single point $x$. Now, $\Lambda_2 = \cup s_i \Lambda_1$, and so $x$ is in some $s_i$ translate of $F_2$, and hence in $F_1$. It is unique as otherwise we have $s_i \lambda_1 = s_j \lambda_1'$, which would mean our set of representatives was too big in the first place.<|endoftext|> -TITLE: Short embeddings for open manifolds and dimension reduction of sets -QUESTION [7 upvotes]: The question is maybe a bit technical, but I find the related construction very beautiful. -In the very famous work - "$C^1$-isometric imbeddings" by J.Nash (1954) the -author presented the fundamental theorem (which was especially recognized some years after) about isometric embeddings of Riemannian manifolds. -$\textbf{Theorem of J. Nash:}$ Any Riemannian $n$-manifold has $C^{1}$ an isometric imbedding in $E^{2n+1}$ (Euclidean $2n+1$-dimensional space). -There are also some similair other theorems of J.Nash and N.Kuiper which nowadays are formulated in a bit different form, especially in a view of famous Gromov's H-principle. -I have a question regarding the considerations of J.Nash in the aforementioned paper. I don't detail all the proof of this theorem, but I recall the main ingredient of the construction of such immersion (imbedding) - "short" immersion or imbedding. -$\textbf{Definition}:$ Immersion (imbedding) $z : (M,g) \rightarrow (E^k, h)$ is short if -$$ - z^*h \leq g \text{ in the sense of quadratic forms}, -$$ -where $g$ is a metric on $M$, $h$ is euclidean metric on $E_k$. -Having initial short immersion or imbedding J.Nash presents some sequence of immersions (imbeddings) $\{z_n\}_{n=1}^{\infty}$, where all $z_n$ are all short and monotonically increase induced metric $z_n^*h$. This sequence of $z_n$ converges in $C^1$ sense and also gives the desired metric tensor $g$ for induced metric. So we naturally come up to the following question: -$\textbf{Important question:}$ How to construct an initial short immersion (imbedding)? -1) For compact manifolds the answer is simple: use Whitney theorem and multiply the map by a small $\varepsilon > 0$ to decrease the induced metric. -2) For open (non-compact) manifolds it is a bit trickier. The construction is as follows: -Let $\{U_i\}_{i=1}^{\infty}$ locally finite open cover of $M$, $\{\phi_i\}_{i=1}^{\infty}$ is the partition of unity subordinated to this cover. Using Dawker's theorem (or Borsuk theorem) we choose $\{U_i\}$ such that no more than $n$ charts intersect other chart. So we say that the multiplicity of the cover $\{U_i\}$ is $s$. -We divide $\{U_i\}$ into $s$ separate classes: it is easy, we take the charts in order and give them any class that is not yet given to any other neighbor. Next we construct an imbedding in $s(n+2)$ dimensional space by defining the maps: -\begin{align} - &u_\sigma(x) = \begin{cases} - \varepsilon_i \phi_i(x), \text{ if $x$ belongs to vicinity of class } \sigma,\\ - 0, \text{ otherwise}, - \end{cases}\\ - &v_\sigma(x) = \begin{cases} - \varepsilon^2_i \phi_i(x), \text{ if $x$ belongs to vicinity of class } \sigma,\\ - 0, \text{ otherwise}, - \end{cases}, \\ - &w_{\sigma j}(x) = \begin{cases} - \varepsilon_i \phi_i(x) x_{ij}, \text{ if $x$ belongs to vicinity of class } \sigma,\\ - 0, \text{ otherwise}, - \end{cases}, -\end{align} -where $x_{ij}$ stands for the $j$-th coordinate in chart $U_i$. The numbers $\varepsilon_i$ is the sequence of positive constants monotonously decreasing to zero. -One can see that: if two points $x$ and $y$ lie in different charts $U_i, U_j$, then they are separated by maps $u_\sigma, \, v_\sigma$. If they belong to the same chart, then they are separated by $w_{\sigma i}$. All maps are $C^{\infty}$-smooth and, by the latter considerations, are injective. From the form of $w_{\sigma i}$ one can see that the differential is injective, so it is an immersion and as it is injective it is an imbedding. Well, injectivity of differential and global injectivity do not yet imply an embedding, but one can analyze the afforementioned maps and see that it is a homeomorphism onto its image. -Controllong the $\varepsilon_i$ we can make it short. So we've constructed short imbedding in $E^{s(n+2)}$. -Then J.Nash says that since the image of the manifold in $E^{s(n+2)}$ is $n$-dimensional set, "the classical process of generic linear projection can be applied and one can reduce the dimension of surrounding Euclidean space to $2n+1$ without introducing any singularities and self intersections". -$\textbf{Question:}$ What is this "classical process of generic linear projection?" -p.s. I found the latter construction for the case of open manifolds very beautiful. It is sad to miss the last point of such beauty! - -REPLY [2 votes]: It seems to be a general position argument. -You need to find a $2{\cdot}n+1$-dimensional subspace of $\mathbb{E}^{s\cdot(n+2)}$ such that (1) it contains no vector orthogonal to your submanifold and (2) no two points of your submanifold are projected to the same point. -All you need is to calculate dimensions correctly and apply Sard's theorem.<|endoftext|> -TITLE: The supremum value of $\int f(t) \log{\frac{1}{|t|}} \, dt$ for normalized Fourier pairs non-negative outside of $[-1,1]$ -QUESTION [10 upvotes]: Observe that for any Schwartz function $f \in \mathcal{S}(\mathbb{R})$ having -$$ -f(0) = \widehat{f}(0) = 1 -$$ -and -$$ - f, \widehat{f} \geq 0 \quad \textrm{outside of} \quad [-1,1], -$$ -the following ridiculous argument based on the prime number theorem yields the strict upper bound -$$ - \int f(t) \log{\frac{1}{|t|}} \, dt < 1 + \gamma = 1.57\ldots. -$$ -[Proof. Just take the $X \to \infty$ limit in the identity -$$ -\frac{1}{X}\sum_{k \in \mathbb{Z} \setminus \{0\}} f\Big( \frac{k}{X} \Big) \log{|k|} = \sum_{n \leq X} \Big( \frac{\widehat{f}(0)}{n} - \frac{f(0)}{X}\Big)\Lambda(n) + \sum_{n \leq X} \frac{\Lambda(n)}{n} \sum_{s \in \mathbb{Z} \setminus \{0\}} \widehat{f} \Big( \frac{sX}{n} \Big) \\ + \frac{1}{X} \sum_{n > X} \Lambda(n) \sum_{s \in \mathbb{Z} \setminus \{0\}} f\Big( \frac{sn}{X} \Big). -$$ -Our conditions on $f, \widehat{f}$ imply that the second and third sums on the right-hand side have only non-negative terms, while the first sum of the right-hand side is evaluated by the logarithmic form of the prime number theorem: $S(X) := \sum_{n \leq X} \Big( \frac{1}{n} - \frac{1}{X} \Big) \Lambda(n) = \log{X} - 1 - \gamma + o(1)$. ] -This begs the question of whether such an argument could conceivably be reversed. Any $f$ as above places an 'explicit' upper bound of $\log{X} - \int f(t) \log{\frac{1}{|t|}} \, dt + O_f((\log{X})/X)$ on the prime number sum $S(X)$, whence my -Question. Is $1+\gamma$ the supremum value of $\int f(t) \, \log{\frac{1}{|t|}} \, dt$ under the given conditions $f(0) = \widehat{f}(0) = 1$ and $f, \widehat{f} \geq 0$ outside of $[-1,1]$? Or may the $1+\gamma$ bound be improved? -Also, quite independently of such a motivation, I would be curious to see any alternative proofs of an upper bound on $\int_{\mathbb{R}} f(t) \, \log{\frac{1}{|t|}} \, dt$ by an absolute constant (under the above conditions on the Fourier pair $f, \widehat{f}$), even if these are weaker than the $1+\gamma$ bound here. -Of course, the normalization of the Fourier transform here is the following one: -$$ -\widehat{f}(y) = \int_{\mathbb{R}} f(x) e^{- 2\pi i xy} \, dx. -$$ - -An application. Here is an application added of the observed inequality. For $g : [0,1] \to \mathbb{R}^{\geq 0}$ any non-negative continuous function with $\int_0^1 g(t) \, dt = 1$ and $S := \int_0^1 g^2(t) \, dt$, we may apply the preceding to (an extension by zero and smoothing of) $f(t) :=S^{-1} \cdot (g * g)(t/S)$. Indeed, we have $f(0) = S^{-1} \cdot (g * g)(0) = 1$ and $\widehat{f}(0) = |\widehat{g}(0)|^2 = 1$, and both $f(t)$ and its Fourier transform $\widehat{f}(t) = |\widehat{g}(t)|^2$ are everywhere non-negative. The conclusion reads: -Special case. The inequality -$$ -\int_0^1 \int_0^1 g(x) g(y) \log{\frac{1}{|x-y|}} \, dx \, dy - \log \int_0^1 |g(t)|^2 \, dt < 1 + \gamma = 1.57\ldots -$$ -takes place for every continuous non-negative function $g \in C([0,1],\mathbb{R}^{\geq 0})$ on $[0,1]$ of unit integral: $\int_0^1 g(t) \, dt = 1$. -May one give a direct proof of the last elementary inequality, with any absolute constant bound in place of $1+\gamma$? And may one prove a strictly smaller bound? - -REPLY [8 votes]: One can take the continuum limit of your proof as $X \to \infty$, again using the prime number theorem, to obtain a proof that does not involve primes at all: -$$ \int f(t) \log \frac{1}{|t|}\ dt = \gamma - \sum_{\sigma = \pm 1} \int_0^\infty f(\sigma t) (\log t + \gamma)\ dt $$ -$$ = \gamma - \lim_{\varepsilon \to 0} \sum_{\sigma = \pm 1} \int_0^\infty f(\sigma t) (\log \frac{t}{\varepsilon} + \gamma)\ dt + \log \frac{1}{\varepsilon}$$ -$$ = \gamma - \lim_{\varepsilon \to 0} \sum_{\sigma = \pm 1} \int_0^\infty f(\sigma t) (\sum_{0 < s < t/\varepsilon} \frac{1}{s})\ dt + \log \frac{1}{\varepsilon}$$ -$$ = \gamma - \lim_{\varepsilon \to 0} \sum_{\sigma = \pm 1} \sum_{s>0} \frac{1}{s} \int_{\varepsilon s}^\infty f(\sigma t)\ dt + \log \frac{1}{\varepsilon}$$ -$$ = \gamma - \lim_{\varepsilon \to 0} \sum_{\sigma = \pm 1} \sum_{s>0} \int_{\varepsilon}^\infty f(\sigma s t)\ dt + \log \frac{1}{\varepsilon}$$ -$$ = \gamma - \lim_{\varepsilon \to 0} \int_{\varepsilon}^\infty \sum_{s \in \mathbb{Z} \backslash \{0\}} f(s t)\ dt + \log \frac{1}{\varepsilon}$$ -$$ = \gamma - A - \lim_{\varepsilon \to 0} \int_{\varepsilon}^1 \sum_{s \in \mathbb{Z} \backslash \{0\}} f(s t)\ dt + \log \frac{1}{\varepsilon}$$ -$$ = \gamma - A - \lim_{\varepsilon \to 0} \int_{\varepsilon}^1 (\sum_{s \in \mathbb{Z} \backslash \{0\}} f(s t) - \frac{1}{t})\ dt $$ -$$ = \gamma - A - \lim_{\varepsilon \to 0} \int_{\varepsilon}^1 (\frac{1}{t} \sum_{s \in \mathbb{Z} \backslash \{0\}} \hat f(s/t) - 1)\ dt $$ -$$ = \gamma + 1 - A - B$$ -where -$$ A := \int_1^\infty \sum_{s \in \mathbb{Z} \backslash \{0\}} f(st)\ dt$$ -$$ = \int_{|t| \geq 1} f(t) (\sum_{1 \leq s \leq |t|} \frac{1}{s})\ dt$$ -and -$$ B := \int_0^1 \sum_{s \in \mathbb{Z} \backslash \{0\}} \hat f(s/t)\ \frac{dt}{t}$$ -$$ = \int_1^\infty \sum_{s \in \mathbb{Z} \backslash \{0\}} \hat f(st)\ \frac{dt}{t}$$ -$$ = \int_{|t| \geq 1} \hat f(t) \frac{\lfloor |t| \rfloor}{t}\ dt.$$ -[In the language of distributions, what this identity is saying I think is that the distributional Fourier transform of $\lfloor |t| \rfloor/t - 1$ is $\gamma - \log \frac{1}{|t|} 1_{|t| \leq 1} - \sum_{1 \leq s \leq |t|} \frac{1}{s}$.] -Since $A,B$ are clearly non-negative, this gives your inequality. This also shows that one is within $o(1)$ of equality if and only if one simultaneously has -$$ \int_{|t| \geq 1} f(t) (1 + \log |t|)\ dt = o(1)$$ -and -$$ \int_{|t| \geq 1} \hat f(t)\ dt = o(1).$$ -By the Hahn-Banach theorem, these estimates are incompatible with the hypotheses $f(0)=\hat f(0)=1$, $f(t), \hat f(t) \geq 0$ for $|t| \geq 1$ if and only if there exist non-negative measurable functions $a(t), b(t)$ supported on $|t| \geq 1$ with $\sup_t \frac{a(t)}{1+\log |t|}, \sup_t b(t) < \infty$ and numbers $\alpha,\beta$ not summing to zero, such that -$$ \alpha f(0) + \beta \hat f(0) = \int_{\mathbb R} f(t) a(t)\ dt + \int_{\mathbb R} \hat f(t) b(t)\ dt $$ -for all Schwartz $f$, or equivalently that -$$ \alpha \delta + \beta = a + \check b$$ -in the sense of tempered distributions, where $\delta$ is the Dirac delta. But the right-hand side is continuous at the origin, so $\alpha$ must vanish; the Fourier transform of the right-hand side has a continuous antiderivative at the origin, so $\beta$ must vanish, contradiction. This shows that one can make $A$ and $B$ simultaneously $o(1)$, so $1+\gamma$ is in fact optimal. (But the invocation of the Hahn-Banach theorem makes it difficult to explicitly construct $f$ that come close to equality!) -One can solve this equation as follows. By Lemma 3 of -Amrein, W.O.; Berthier, A.M., On support properties of Lsup(p)-functions and their Fourier transforms, J. Funct. Anal. 24, 258-267 (1977). ZBL0355.42015, -one can find, for any $R>0$, a non-zero function $f \in L^2({\bf R})$ such that $f$ and $\hat f$ both vanish on $[-R,R]$ (this is basically because the compact operator $1_{[-R,R]} {\mathcal F} 1_{[-R,R]}$ is a strict contraction on $L^2$, which in turn follows from the uncertainty principle that a function and its Fourier transform cannot be simultaneously compactly supported), in fact Proposition 6 gives an infinite-dimensional space of such functions. By convolving $f$ by a suitable approximation to the identity, and then multiplying by the Fourier transform of a suitable approximation to the identity (and shrinking $R$ slightly), one can make $f$ Schwartz. -When one takes a second antiderivative of $f$, one obtains a new Schwartz function $f_1$ which is equal to a linear function $a+\beta x$ on $[-R,R]$, while the Fourier transform still vanishes on $[-R,R]$. If $a=0$ (which can be achieved due to the infinite dimensional nature of the space of $f$), one can divide by $x$ and obtain a further Schwartz function $f_2$ that is equal to a constant $\beta$ on $[-R,R]$, while the Fourier transform is equal to a constant $\alpha$ on $[-R,R]$. This gives the identity -$$ \alpha \delta + \beta = (\beta - f) + (\alpha - \hat f)^{\vee}$$ - I think one can work a little harder to ensure that $\alpha,\beta$ can be arbitrary real numbers while simultaneously keeping $a=0$, and in particular can have non-zero sum (otherwise by Hahn-Banach there would be a way to express some nontrivial combination of polynomials on a halfline and Fourier transforms of polynomials on a halfline as functions supported on $[-R,R]$ plus a function with Fourier transform supported on $[-R,R]$, which should be easy to rule out by the argument in strikethrough). This gives a constraint of the desired form (taking $R=1$). So some improvement to $1+\gamma$ is in fact possible. -Update: here are some details on the "working a little harder". For $f$ Schwartz with both $f$ and $\hat f$ vanishing on $[-R,R]$, one can write $\beta$ as the inner product of $f$ with $1_{(-\infty,0]}$ and $a$ as the inner product of $-1_{(-\infty,0]} x$. If $a$ vanishes, one can write $\alpha$ as the inner product of $f$ with a function $\phi$ whose Fourier transform is equal to a constant multiple of $1_{(-\infty,0]}(x) / x^2$ outside of $[-R,R]$ and is smooth in $[-R,R]$. So supposing for contradiction that there is a non-trivial constraint between $\alpha$ and $\beta$ when $a=0$, there must exist some non-trivial linear combination $g$ of $1_{(-\infty,0]}$, $1_{(-\infty,0]} x$, and $\phi$ such that all Schwartz functions $f$ with both $f$ and $\hat f$ vanishing on $[-R,R]$ are orthogonal to $g$. In particular, if $f \in L^2$ with $f$ and $\hat f$ vanishing on $[-2R,2R]$, and $\psi_1, \psi_2$ are suitable approximations to the identity (let's say real symmetric), then $(f \hat{\psi_1}) * \psi_2$ is orthogonal to $g$, or equivalently $f$ is orthogonal to $(g * \psi_2) \hat \psi_1$. Taking limits as $\psi_2$ approaches the Dirac delta, we conclude that $f$ is orthogonal to $g \hat \psi_1$. Taking duals, this means that we have a decomposition $g \hat \psi_1 = g_1 + \hat g_2$ where $g_1,g_2$ are $L^2$ function supported in $[-2R,2R]$. This implies in particular that $g \hat \psi_1$ extends to a holomorphic function on ${\bf C} \backslash [-2R,2R]$. Dividing by $\hat \psi_1$ (which one can choose to be non-zero at any given complex number), we conclude that $g$ extends to a holomorphic function on ${\bf C} \backslash [-2R,2R]$ (the extension is independent of $\psi_1$ by analytic continuation). -The function $x \phi''(x)$ has a test function for a Fourier transform with nonzero integral, so $\phi''(x)$ (as a distribution) is equal to a Schwartz function plus a non-zero multiple of $p.v. 1/x$, and extends to an entire function plus a non-zero multiple of $1/x$ away from the origin. This implies that $\phi$ is extends holomorphically to ${\bf C}$ with $[-2R,2R]$ and the negative imaginary axis (removed). On the other hand, by uniqueness of analytic continuation, any non-trivial multiple of $1_{(-\infty,0]}$ and $1_{(-\infty,0]} x$ cannot be extended to this region. These facts are only consistent if $g$ is a scalar multiple of $\phi$ alone. But $\phi$ has nontrivial monodromy around $[-2R,2R]$ (it behaves like the sum of an entire function and the multivalued function $z \log z$), while $g$ does not, giving the required a contradiction.<|endoftext|> -TITLE: Blowing-up an ideal generated by squares -QUESTION [5 upvotes]: Let $f_1,\dots,f_r$ be regular functions on a smooth projective variety $X$, and consider the ideals $I = (f_1^2,\dots,f_r^2)$ and $J = (f_1,\dots,f_r)$. Let $Y = Z(I)$ and $W = Z(J)$ be the subschemes of $X$ defined respectively by $I$ and $J$. Assume that $W$ is smooth. -Now, let $X_Y = Bl_YX$ and $X_W = Bl_WX$ be the blow-ups of $X$ respectively along $Y$ and $W$. -What is the relation between $X_Y$ and $X_W$? For instance, does there exist a morphism from one to the other? - -REPLY [2 votes]: If $\alpha \in \mathbb{N}^{r}$ satisfies $|\alpha| = r+1$ then $\exists i, \alpha_i \geq 2$ so that $f^{\alpha} \in f_i^2 J^{r-1}$. -Thus $J^{r+1} = I J^{r-1}$. In particular $I$ becomes invertible on $X_W$, so that $X_W \rightarrow X$ uniquely factors through $X_{Y}$.<|endoftext|> -TITLE: Thinnest covering of the plane by regular pentagons -QUESTION [9 upvotes]: Q. Is it known what is the thinnest covering of the infinite plane by regular pentagons? - -By covering I mean every point of the plane is covered. -By thinnest I mean the proportion of the plane covered more than once is minimal among all coverings. -This seems like it must be known, but I cannot find it, perhaps because I -don't know the correct terminology. -This is a natural attempt: - -          - - -If I've calculated correctly, this -covering doubly covers about $38\%$ of the plane: -$$\tfrac{1}{2} - \left(3-\sqrt{5}\right) \approx 0.382 \;.$$ -I am interested because the above covering can be achieved by "rolling" -a dodecahedron, and I'd like to know if there is a thinner cover -which might not be "rollable." - -REPLY [7 votes]: The thinnest known covering of the plane with congruent regular pentagons is shown in my answer to: Terrible tilers for covering the plane. What you see there is probably not "rollable". The covering is of the "double-lattice" type and is known to be the thinnest among double lattice coverings with regular pentagons (to be published). Also, it is conjectured to be the thinnest one among all coverings with congruent regular pentagons. The conjecture is still open. -By the way, Joe, your "natural attempt" is of a double-lattice type also, generated by a trapezoid contained in the pentagon. The trapezoid is a p-hexagon too, but not of maximum area, and the larger the p-hexagon, the thinner the covering generated by it.<|endoftext|> -TITLE: Fractional moments of Poisson distribution -QUESTION [5 upvotes]: I wonder if there is a formula to calculate the fractional (0,1) moments of a Poisson distribution? -Thanks in advance. - -REPLY [2 votes]: Fractional factorial moments $M_q$ and cumulants $C_q$ of the Poisson distribution are calculated in Fractional moments of distributions (1994)$^\ast$ -$$M_q=e^{-\lambda}\frac{\Phi(1,1-q;\lambda)}{\Gamma(1-q)},$$ -$$C_q=\frac{q\lambda}{\Gamma(2-q)},$$ -where $\Phi$ is the confluent hypergeometric function. For integer $q=n$ one recovers the usual results -$$M_n=\lambda^n,\;\;C_n=\lambda\delta_{n,1},$$ -in view of the limit -$$\lim_{q\rightarrow n}\frac{\Phi(1,1-q;\lambda)}{\Gamma(1-q)}=e^\lambda \lambda^n.$$ -In the interval $n -TITLE: Existence of certain cubes in finite fields -QUESTION [6 upvotes]: Consider $F := GF(q)$ where $q = p^e$ and $E := GF(q^2)$ where w is a primitive element of $E$. -Fix $\theta := w^{q - 2}$. -Starting point: can I always write $1 + \theta$ as a power of $\theta$? -If $Gcd (q^2 - 1, q - 2) = 1$, $\theta$ is a primitive element for $E$ and the answer is yes. -So we can restrict to the case $q \mod 3 = 2$ where all cubes of elements of $E$ are powers of $\theta$. -Is $1 + \theta$ a cube in $E$? No, when $e$ is odd; otherwise yes. -So in the remaining odd case we want to replace the role of $1 + \theta$ by $1 + \theta^j$ for some $j$, leading to the following conjecture. -Conjecture: -Consider $q = p^e$ where $p = 2 \mod 3$ and $e$ is odd. -There exists $j$ in $\{1, \dots, q^2 - 1\}$ such that -$Gcd (q^2 - 1, j) = 1$ and $1 + \theta^j$ is a cube in $E$. -Computational evidence in support of this conjecture is strong. -While it does not hold for $q = 5$ and $q = 8$, such $j$ exists for -all relevant $q$ in the range $[11\dots 10^8]$. Always $j$ is at most $q - 1$. -The context for the query is the study of presentations for the classical groups -$SU(3, q) \leq GL(d, q^2)$ where a generator has action described by $\theta$. -Pointers towards a proof would be much appreciated. - -REPLY [7 votes]: I think what you want is a solution of $1+x^3=y^3$ such that $x$ is a primitive element of $E$ by writing $x^3=\theta^j$. This is known to exist for $q$ large and is not very difficult to extract a bound. One uses the Weil bound to count the number of solutions of $1+x^3=y^3$ as well as the number of solutions of $1+x^{3d}=y^3$, for divisors $d$ of $q^2-1$ (and for large $d$ replace the Weil bound by a direct estimate), then apply inclusion exclusion to get the desired result. The technique goes back to the work of Bilharz (Math Ann 1937) on the function field analogue of Artin's conjecture.<|endoftext|> -TITLE: Products and Gale-Stewart games -QUESTION [6 upvotes]: For the purpose of this post, I will say that the Gale-Stewart game is the infinite two-player game of perfect information where players I and II alternate playing natural numbers, with I going first. A round of the game looks like: I plays $n_0$, II plays $n_1$, I plays $n_2$, II plays $n_3$, etc. The outcome of this round is the sequence $(n_i)_{i\in\omega}\in\omega^\omega$. -Given a set $A\subseteq\omega^\omega$, I will speak of a player having a strategy for playing into $A$, rather than using the usual winning/losing terminology. If $\sigma$ is a strategy for one of the players, I denote by $[\sigma]\subseteq\omega^\omega$ the set of all outcomes of the game when that player follows $\sigma$. -Question: Suppose we are given a set $C\subseteq\omega^\omega\times\omega^\omega$. Under which circumstances do there exists strategies $\sigma$ and $\tau$ for players (either one) in the Gale-Stewart game such that one of $[\sigma]\times[\tau]\subseteq C$ or $([\sigma]\times[\tau])\cap C=\emptyset$ holds? -Is it sufficient that $C$ is determined? To put that another way, is there a game which encodes this property? Or is there a (reasonably definable) counterexample? -The first thing that comes to mind is the game where the two players alternate, with I going first, playing pairs of natural numbers $(n_i,m_i)$, and whose outcome is the pair $((n_i)_{i\in\omega},(m_i)_{i\in\omega})\in\omega^\omega\times\omega^\omega$. If $C$ is determined, then either I has a strategy for playing into $C$ or II has a strategy for playing into its complement in this game. Suppose player I has a strategy for playing into $C$. From this, it is easy to construct two strategies $\sigma$ and $\sigma'$ in the Gale-Stewart game for player I with $[\sigma]\subseteq\pi_0(C)$ and $[\sigma']\subseteq\pi_1(C)$, where $\pi_0$ and $\pi_1$ are the first and second coordinate projections, respectively: read off the first (or second) coordinate of I's play in the game with pairs while II plays their move in the Gale-Stewart in the first (or second) coordinate, and an arbitrary number in the other coordinate. However, I have no reason to suspect that $[\sigma]\times[\sigma']\subseteq C$. One issue seems to be a lack of independence in the coordinates played according to a strategy; each of played coordinates can depend on either of the previously played coordinates. -Edit: Joel's answer below precludes the possibility that the strategies are from the same player in the Gale-Stewart game, but I want to also address the possibility that alternate players have such strategies, e.g., I has a strategy $\sigma$ and II has a strategy $\tau$ such that $[\sigma]\times[\tau]\subseteq C$ or $([\sigma]\times[\tau])\cap C=\emptyset$. - -REPLY [4 votes]: It is a very nice question. -I claim that it is not sufficient that $C$ is determined, and indeed, there -are counterexamples where $C$ is a game with only two moves. -Consider the two-dimensional game where player I plays $(x_0,y_0)$ -and player II responds with $(x_1,y_1)$. Player I wins, if -$y_1=x_0$. That is, player I wins, if player II copies on his second coordinate the first-coordinate move of player I. In your framework, the payoff set $C$ is the set of plays with -projections to $(x,y)$, where $y(1)=x(0)$. -Clearly, player II has a winning strategy in this game, which is -simply to make sure that $y_1$ is not the same as the -already-played $x_0$. -But I claim that there can be no strategies $\sigma$ and $\tau$ for -player I with $[\sigma]\times[\tau]\subset C$ or for player II, -with $[\sigma]\times[\tau]\subset\neg C$. -In the first case, for any one-dimensional strategies $\sigma$ and $\tau$ for -player I, we can devise a play that refutes them by having player -II actually play so as to violate the move-copying requirement. -In the second case, for any one-dimensional strategies $\sigma$ and $\tau$ for -player II, we can have player I first play $y_0$, in order to get -$\tau$'s response, and then play $x_0$ using that information. In -this way, player I can in effect look ahead in the second -coordinate to see how player II will play, and then using that -information complete the first move in the first coordinate by -playing $x_0$ in such a way that player II will in effect have -copied it. So this play will be in $C$, contrary to hypothesis. -So there are counterexamples with clopen games of very low -complexity.<|endoftext|> -TITLE: Can cobordisms of 3 or 4 manifolds be visualized by moves on kirby diagrams? -QUESTION [6 upvotes]: I'm mostly interested in the 4d case so I'll state the question in that form. Basically, it comes down to two parts: -1) What simple moves can be performed to a kirby diagram of a 4-manifold that guarantee that the resulting diagram is cobordant (via a 5d manifold) to the original manifold. -2) Can any cobordism be achieved through such moves? - -REPLY [7 votes]: I think that a complete set of moves for cobordisms are the following: Kirby moves (that preserve the 4-manifold), handle trading (dotted circles become 0-framed 2-handles), addition/deletion of pairs of trivial circles with framings $\pm 1$ (corresponding to connected sums with $CP^2 \,\#\, {\overline{CP^2}}$). This could be proved by using Kirby calculus for 3-manifolds (which is exact!) as in Kirby's original paper as follows. Take two cobordant 4-manifolds $M_1$ and $M_2$ with their Kirby diagrams (with one 0-handle and one 4-handle). Up to addition of complementary 2/3-handles, we can assume that they have the same number of 3-handles. Now, removing the 3- and 4-handles, you get 4D 2-handlebodies with the same boundary $\#_k\, (S^1 \times S^2)$ for some $k \geq 0$. Trading 1-handles completely remove them (without changing the boundary and the signature, because it is surgery along embedded circles). By Kirby's theorem, the resulting manifolds can be related by Kirby moves (now only on 2-handles) and addition/deletion of separated $\pm 1$-framed trivial knots. When the latter operation occurs, just compensate by adding another suitable $\mp 1$ trivial circle that will be not involved in subsequent moves. You will end up with the Kirby diagram for $M_2$ (maybe after 1-handle trading) plus some $\pm1$ trivial knots (the ones you added for compensation) and the $\pm 1$'s sum up to 0 since the signature must be the same.<|endoftext|> -TITLE: About $\omega_1^{CK}$ -QUESTION [20 upvotes]: Here we use $\omega_1^{CK}$ to denote the least nonrecursive ordinal. The following theorem is well known. - -$\mathbf{Theorem}$ $\omega_1^{CK}$ is an admissible ordinal. - -But its proof seems weird. The usual proof uses a nonstandard technique. I wonder whether there exists a pure standard proof. Or, to negate it, whether there is an $\omega$-model $M$ of $KP$ so that $$M\models \omega_1^{CK}\mbox{ exists but is not admissible ?}$$ -If there exists such a model, then it means that $KP$ is not enough to prove the theorem. So one has to use some assumptions beyond $KP$ and probably requires the existence of Kleene's $O$. So nonstandard techniques can be applied. - -It seems there is no such model. The ideal of the following proof is inspired by Philip. - -$\mathbf{Theorem}$: There is no $\omega$-model of $KP$ in which $\omega_1^{CK}$ exists, but is not admissible. - -Here is an outline of the proof. - -$\mathbf{Proof}$: Suppose not, fix such an model $M$. Through out the proof, all the notions are relative to $M$. So they may be nonstandard. -Since $\omega_1^{CK}$ exists in $M$, so does Kleene's $\mathscr{O}$. Let\begin{multline*}WO=\{e\mid R_e \mbox{ is a recursive linear order over }\omega - \\ \wedge \mbox{ there is no an infinite descending chain on }R_e\}\end{multline*} and $$WO^*=\{e\mid R_e \mbox{ is a recursive linear order over }\omega \mbox{ isomorphic to an ordinal} \}.$$ -Clearly $M\models WO^*\subseteq WO$. It is routine to prove that $M\models WO^*\leq_m \mathscr{O}$. Now for any $e\not\in WO^*$, there must be some $n_0$ so that $R_{e_{n_0}}=\{(m_0,m_1)\mid R_e(m_0,n_0)\wedge R_e(m_1,n_0)\wedge R_e(m_0,m_1)\}$ and $e_{n_0}\notin WO^*$. Repeat this, we may $\mathscr{O}$-recursively obtain an $R_e$-descending sequence $\{n_i\}_{i\in \omega}$. So $e\not\in WO$. In other words, $WO\subseteq WO^*$ and so $M\models WO=WO^*$. -Now it is clear that $WO$ is a $\Pi^1_1$-complete set in $M$. So is $\mathscr{O}$. Using this fact, Gandy basis holds in $M$.I.e. every nonempty $\Sigma^1_1$set contains a real $x\leq_T \mathscr{O}$ and $\omega_1^x=\omega_1^{CK}$, where $\omega_1^x$ is the least non-x-recursive ordinal. -Let $\mathscr{N}=\{N=(\omega,E)\mid N\mbox{ is an }\omega \mbox{-model of }BS+V=L \wedge \forall \alpha(\alpha<\omega_1^{CK}\rightarrow \alpha\in N)\},$ where $BS$ is the corrected Basic set theory as developed in Devlin book. In $M$, $\mathscr{N}$ is a $\Sigma^1_1$-set. Clearly $\mathscr{O}$ hyperarithmetically computes a copy of $(L_{\omega_1^{CK}})^M$. So by the existence of $\mathscr{O}$, $\mathscr{N}$ is nonempty in $M$. -Now by the Gandy basis, there is some $N\in \mathscr{N}$ so that $\omega_1^N=\omega_1^{CK}$. Let $N'=(L_{\omega_1^{CK}})^N$. Since $N$ is a theory of $BS$, we have that $N'=(L_{\omega_1^{CK}})^M$. Also note that $N'\in M$. -We claim that $N'\models KP$. Otherwise, let $f$ be a $\Sigma_1^{N'}$ total function from $\omega$ to $\omega_1^{CK}$. Then, by the totality, it is also $\Sigma_1^N$. In other words, we have that $\omega_1^{CK}$ is recursive in $N$, a contradiction. -This finishes proof. - - -So the question I asked may not be a right formalization to get rid of the nonstandard argument. - -REPLY [9 votes]: I claim there is no $\omega$-model $M\models KP$ such that -$$M\models \,\,"\omega_1^{ck} \mbox{ exists, but is not admissible ''.} -$$ -(In other words if $(\eta =\omega_1^{ck})^M$ then $(L_\eta \nvDash KP)^M$. Suppose $M$ were such a model; by truncating it, if need be, we may assume also that $\omega$ is the only admissible ordinal of $M$. The definition of Kleene's $\mathcal{O}$ is by way of a $\Sigma_1^{KP}$ recursion, and has the special property that for any $b\in \mathcal{O}$ we have $B=\{a|a<_{\mathcal{O}}b\}$ is r.e. Moreover at the stage that $b$ is put into $\mathcal{O}$ by the recursion, so have all elements of $B$. -For $M$ the recursion requires $On^M$ many steps to complete. But we do not need to establish the existence of $\mathcal{O}^M$ as a set. This would indeed go beyond $KP$, and is really the point of this answer. We only need, roughly speaking, the $\eta$'th iterate of this positive monotone recursion, and indeed by KP, the graph of any initial segment of the recursion is an element of $M$. We thus find eventually a $b\in \omega$ with $|b|_{\mathcal{O}^M}=\eta$, and the recursion up to $\eta$ is an element of $M$. -By the comments above if - $(B=\{a|a<_{\mathcal{O}}b\} )^M$ then $B$ is contained in the $\eta$'th iterate of the recursion in $M$ and so is a set in $M$, and is moreover r.e. in the sense of $M$. Letting $f:\omega \rightarrow B $ be a recursive bijection in $M$, we then define $n\prec m \leftrightarrow f(n) <_{\mathcal{O}^M} f(m)$, and then $((\omega, \prec) $ is recursive with $o.t.(\omega, \prec)= \eta \geq \omega_1^{ck} )^M$, a contradiction.<|endoftext|> -TITLE: How are constants/functions named after their discoverer? -QUESTION [10 upvotes]: In general, when a paper references an object discovered/defined in another paper by author X, it goes something along the lines of: -"Let $\tau$ be the constant defined by X in 1999 [1]$\ldots$", -or -"Let $f_{\mu}$ denote the function that generalizes the case $\ldots$ (X, [1])". -At what point does the literature start talking about "X's constant" or "The X function"? -Who/what determines that an object discovered by somebody deserves to take his name? - -REPLY [29 votes]: This is determined by informal consensus of researchers in an area. Anyone can propose a name for a mathematical object, just by using this name in a paper. Then the proposed name either sticks to the object or not. This depends on the opinions of other people working in the area. Eventually the name of the object becomes established. -Sometimes several names for the same object become established, and all of them popular. -For example Fatou set is the same as the set of normality. The name Fatou set -was proposed in two influential surveys in 1980s, and many people in the area use it. -The name does not have to be the name of the discoverer of the object. -For example, Drinfeld introduced an object which he named Yangian, -in honor of Chen-Ning Yang. -M. Lyubich and I gave the name "Baker's domain" to an object that we defined. -Unlike some other "personal names" we proposed in the same paper, this one is used by everyone who writes on the subject. Arnold once stated a principle that -"If the thing is named after someone, this indicates that the person had nothing to do with the thing". M. Berry remarked that "Arnold Principle applies to itself". -Sometimes a multitude of names reflects a priority dispute or nationalistic feelings (Young diagram vs Ferrers diagram, Schwarz inequality vs Cauchy or -Bouniakowski etc.) The name Casorati–Sokhotski–Weierstrass theorem reflects -some priority research, but the theorem in question was earlier stated by Briot and Bouquet. Same happened with "Gauss-Manin connection" which was in fact discovered by Legendre. -Sometimes the accepted name is changed ($\pi$ used to be "Archimedes number" and -$e$ used to be "Euler's number"). -A funny story happened with "Koebe constant". It turned out that it is equal to $1/4$, so one mathematician wrote "Now it cannot be called Koebe constant anymore, -because it turns out that it already has a name, namely $1/4$"). -Sometimes a name is based on a mistake. For example Abel's equation -is the accepted name for $f(x+1)=g(f(x))$. It is based on a manuscript of Abel, but the manuscript itself turns out to be just Abel's personal notes on -the work of Napier, where the equation was introduced. -People usually do not give their own names to mathematical objects. However, -in his influential book on functional analysis, Banach introduces -"spaces of type H", "spaces of type F" and "spaces of type "B". As his "spaces of type H and F" are nothing but Hilbert and Frechet spaces, people probably understood the hint:-) -G. Julia intensively lobbied (not publicly, in private correspondence, which is now published) that the "irregular set" be called Julia set. He succeeded. -Remark. I just found the paper, arXix:1204.4716v1 (Bernard Ycart, A case of mathematical eponimy: the Vandermonde determinant), which shows that there is a whole research area related to this question: it is called "mathematical eponimy". The paper cites several other publications on this subject.<|endoftext|> -TITLE: A trace formula for $\mathrm{GSp(4)}$ -QUESTION [5 upvotes]: The Arthur trace formula and its variations provide general results for reductive groups, however to the extent of my knowledge only few specific instances of the formula have been really worked out in details, i.e. with precise decomposition of the spectrum, as for $\mathrm{GL}(n)$. -Is there such a formula for $\mathrm{GSp(4)}$ or instances of use of the general trace formula in that setting? Or should I rather come back to the classification of representations of $\mathrm{GSp(4)}$ (perhaps as in Roberts and Schmidt)? -Thanks for any idea! - -REPLY [4 votes]: There are many kinds of trace formulas on a given group $G$, and different things you could mean by decomposition of the spectrum. As mentioned in the comments, there's Arthur's article in the Shalika volume, though I am not sure whether those results are unconditional as of yet. On the other hand, Arthur's work on the trace formula for classical groups including SO(5) $\simeq$ PGSp(4). -Precisely what Arthur is interested is functoriality (wrt the standard represenentation) to GL($N$). As a result you get a description of the discrete part of the $L^2$ spectrum for $G$. In the case of SO(5) (i.e., GSp(4) with trivial central character), Ralf Schmidt wrote down what this explicitly tells you about the automorphic representations appearing discretely in his Packet structure preprint. -If you are interested in the non-discrete spectrum, there are some things you should be able to say, but I don't know of an explicit description of this part of the spectrum.<|endoftext|> -TITLE: Realisation of maps between spheres by simplicial maps -QUESTION [14 upvotes]: Let $K^n_0 := \partial \Delta_{n+1}$ the simplicial set obtained by removing the $(n+1)$-simplex from the standard simplex. This gives a simplicial decomposition of the sphere $S^n$. More generally, let $K^n_m$ given by $m$-fold barycentric subdivision of $K^n_0$. -Given a map of simplicial sets $f: K^{n+k}_m \rightarrow K^n_0$, geometric realization induces a continuous map $F: S^{n+k} \rightarrow S^n$, and it is well-known that every such continuous map is homotopic to a simplicial map $f: K^{n+k}_m \rightarrow K^n_0$ for $m$ large enough. -For $k\geq 1$, the homotopy groups of sphere are finite, hence there exists a smallest $m$ such that all homotopy classes of maps $F: S^{n+k} \rightarrow S^n$ are represented by a simplicial map $f: K^{n+k}_m \rightarrow K^n_0$. -Question: Are there explicit bounds on this smallest $m$ in terms of $n$ and $k$? - -REPLY [12 votes]: First, $\pi_{4k-1}(S^{2k})$ has an infinite cyclic direct summand for every $k \geq 1$. As a simple example, you can think of $\pi_3(S^2) \cong \mathbb{Z}$, coming from the Hopf fibration. -We recently wrote a paper where we solved a problem, which is closely related to your question. Namely, we give an algorithm which for a given finite simply connected simplicial complex $Y$, and a given $d \geq 2$, under some technical conditions on $Y$ the algorithm outputs for every generator $\alpha \in \pi_d(Y)$, a simplicial map $f_\alpha : \Sigma^d_\alpha \rightarrow Y $, where $\Sigma^d_\alpha$ is a simplicial complex whose geometric realisation is homeomorphic to $S^d$. In particular this works for spheres. You can find the paper on archive (https://arxiv.org/pdf/1706.00380.pdf), and it will soon appear in the SODA 2018 proceedings. The complexity of the algorithm is singly exponential on the number of simplices of $Y$, and we show that this is optimal. The way we prove it is by constructing an example $X$ where the necessary size of the spheres $\Sigma^n_\alpha$ is exponential on the size of $X$. That shows that any general algorithm must necessarily have at least singly exponential complexity. -However, we do not work with barycentric subdivisions, but with a rather specific and non-canonical (if I might put it so) triangulations of the $d$-sphere. The construction of those was invented by Clemens Berger in his PhD dissertation (https://tel.archives-ouvertes.fr/tel-00339314/document - in French only). We believe that one could be able to fit those "non-canonical" models into an iterated barycentric subdivision of the boundary of a standard simplex. If this is true, it would mean that you need to subdivide $\partial \Delta[m+1]$ enough times to fit the exponentially large spheres produced by our algorithm, and you would be able to represent any homotopy element you wish. -On the other hand, even though our algorithm is exponential in the general case, it might happen that it runs in polynomial time for some very specific spaces. For example. It is not impossible that this turn out to be true for spheres, but I cannot say much more about it. -There is also another point of view. Spheres are compact Riemannian manifolds and you can rephrase your question in the language of Lipschitz constants. In this language, your question roughly translates to "what Lipschitz constants can one expect from representatives of homotopy groups of spheres". There is a connection between the Lipschitz constant of a Lipschitz function and the fineness of a triangulation required to homotope it to a simplicial map. This is somewhat expected because both notions measure some sort of geometric complexity of the function in mind. In particular, there is the following result by Gromov (http://www.ihes.fr/~gromov/PDF/4[19].pdf): -Theorem (Gromov): Let $X$ and $Y$ be compact simply connected Riemannian manifolds. Then -$$\# \{ [f] \in [X,Y] \, : \, Lip(f) \leq L \} = O(L^\alpha)$$, -where $\alpha$ depends only on the rational homotopy type of $X$ and $Y$. -You can check also this (http://www.pnas.org/content/110/48/19246.full) paper from Weinberger and Ferry. There is a long study of similar questions from that point of view. -I hope my comments are useful. In conclusion I would like to say that you question is very deep and very hard.<|endoftext|> -TITLE: Mean minimum distance for M and N uniformly random points on reals between 0 and 1 -QUESTION [6 upvotes]: Similar to Mean minimum distance for N random points on a one-dimensional line, but instead of only N random points, choose N and M random points and find the mean minimum distance between points of N and M, but NOT M and M or N and N. -So, given $x_1,x_2,...,x_N \in [0,1]$ and $y_1,y_2,...,y_M \in [0,1]$ (uniformly random), find the mean distance: $min(abs(x_i - y_j))$ with $i\in[1,N]$ and $j\in[1,M]$. -Ideally, the answer would include the probability of the points being a minimum of $d$ distance apart, like in the referenced question. - -REPLY [4 votes]: Let $D_1$ be the answer conditioned on the leftmost point being at 0 and the rightmost point at 1 (colors irrelevant). Let $0 = x_1 \leq \ldots \leq x_{n + m} = 1$ be the ordered coordinates of points. We can see that $s_i = x_{i + 1} - x_i$ are equidistributed subject to $s_1 + \ldots + s_{n + m - 1} = 1$, and coloring is independent of $s_i$. The probability that there are exactly $k$ adjacent color changes in the sequence is $$\frac{{n - 1 \choose \lceil \frac{k - 1}{2} \rceil}{m - 1 \choose \lfloor \frac{k - 1}{2} \rfloor} + {m - 1 \choose \lceil \frac{k - 1}{2} \rceil}{n - 1 \choose \lfloor \frac{k - 1}{2} \rfloor}}{n + m \choose n},$$ since the sequence either starts with a white group, and has $\lceil (k + 1) / 2 \rceil$ white groups and $\lfloor (k + 1) / 2 \rfloor$ black groups, or vice versa. -Conditioned on the colouring, the smallest distance is equal to the minimum of $k$ instances of $s_i$ (it is irrelevant which ones due to symmetry). For $0 \leq t \leq 1 / k$, the probability of $\min(s_1, \ldots, s_k) \leq t$ is $1 - (1 - tk)^{n + m - 2}$. Integrating with density, we have $$E\min(s_1, \ldots, s_k) = \int_0^{1 / k} k(n + m - 2)(1 - tk)^{n + m - 3} tdt =$$ $$(n + m - 2) \int_0^1 z^{n + m - 3}\frac{1 - z}{k} dz = \frac{1}{k(n + m - 1)}$$ (note that the answer is correct in the special case $k = n = m = 1$ where some transformations were illegal). Hence $$D_1 = \frac{1}{(n + m - 1){n + m \choose n}}\sum_{k = 1}^{\infty}\frac{{n - 1 \choose \lceil \frac{k - 1}{2} \rceil}{m - 1 \choose \lfloor \frac{k - 1}{2} \rfloor} + {m - 1 \choose \lceil \frac{k - 1}{2} \rceil}{n - 1 \choose \lfloor \frac{k - 1}{2} \rfloor}}{k}$$ -Finally, integrating by the span $s$ between extreme points we obtain the complete answer $$D = \int_0^1 (n + m)(n + m - 1)(1 - s)s^{n + m - 2} \cdot sD_1 ds = \frac{n + m - 1}{n + m + 1}D_1 =$$ $$\frac{1}{(n + m + 1){n + m \choose n}}\sum_{k = 1}^{\infty}\frac{{n - 1 \choose \lceil \frac{k - 1}{2} \rceil}{m - 1 \choose \lfloor \frac{k - 1}{2} \rfloor} + {m - 1 \choose \lceil \frac{k - 1}{2} \rceil}{n - 1 \choose \lfloor \frac{k - 1}{2} \rfloor}}{k}$$ -When $n = m = 2$, we have $D = \frac{1}{5 \cdot 6}(\frac{2}{1} + \frac{2}{2} + \frac{2}{3}) = \frac{1}{30} \cdot \frac{11}{3} = \frac{11}{90}$, which matches the computations in the comments. Not sure if the sum can be simplified further, still the answer is feasible to compute. -The probability of the smallest distance being at least $d$ can be found along the same lines by computing $$\frac{(n + m)(n + m - 1)}{n + m \choose n}\sum_{k = 1}^{\infty}\left({n - 1 \choose \lceil \frac{k - 1}{2} \rceil}{m - 1 \choose \lfloor \frac{k - 1}{2} \rfloor} + {m - 1 \choose \lceil \frac{k - 1}{2} \rceil}{n - 1 \choose \lfloor \frac{k - 1}{2} \rfloor}\right) \times \\ \int_0^1 (1 - s)s^{n + m - 2} \cdot \max(0, 1 - \frac{kd}{s})^{n + m - 2}ds$$<|endoftext|> -TITLE: Why do we care about Schur Positivity -QUESTION [13 upvotes]: Some of the most important open problems in Algebraic Combinatorics concern the Schur positivity of classes of symmetric functions. Why is this an important property to have? - -REPLY [20 votes]: In algebraic combinatorics, conjectures that certain numbers are integers or are positive or are unimodal are really implicit challenges to find the hidden structure. And finding hidden structure is a large part of what mathematics is all about. -As lambda said in the comments, in the specific case of combinatorially defined symmetric functions whose Schur-coefficients are integers but not obviously positive, a positivity conjecture is an implicit challenge to find the representation of $\mathfrak{S}_n$ or $GL(n)$. -Let me give an example from my own research. For the sake of simplicity, I will use slightly nonstandard notation. If $P$ is a finite partially ordered set on the set $\{1,2,\ldots,n\}$ then define a coloring of $P$ to be a map $\kappa:P\to \mathbb{N}$ such that every pre-image $\kappa^{-1}(i)$ is a totally ordered subset of $P$. Then, given variables $x_1, x_2, \ldots$ we can define -$$X_P := \sum_\kappa x_{\kappa(1)} x_{\kappa(2)}\cdots x_{\kappa(n)}$$ -where the sum is over all colorings $\kappa$. It is easy to check that $X_P$ is a symmetric function. What is less obvious is: - -Theorem (Haiman, Gasharov). If $P$ is a unit interval order then $X_P$ is Schur-positive. - -When a combinatorialist sees a statement like this, it is not the bare fact of positivity that is so important. Rather, long experience has shown that statements like this indicate the presence of some structure lurking beneath the surface. Gasharov's proof gave a combinatorial interpretation of the Schur-coefficients in terms of something called $P$-tableaux. I won't define $P$-tableaux here but the point is that the Schur-positivity led to the search for this combinatorial structure that was not apparent from the original definition. -This is not the end of the story, however, because one can ask if there is a natural representation of the symmetric group whose character is given by $X_P$. (Haiman's proof has a representation-theoretic flavor but does not construct such a representation directly.) For the answer to this question, we had to wait about twenty years, when Shareshian and Wachs conjectured (and Brosnan and myself—and independently, Guay-Paquet—proved) that a certain action called the dot action of the symmetric group on the cohomology of a regular semisimple Hessenberg variety associated to $P$ has character equal to $X_P$ (or actually $\omega X_P$ but never mind this technicality for now). The exact definition of all these terms is not important here; again, the point is that Schur positivity was the "canary in the coal mine" that signaled the presence of this hidden connection between colorings of partially ordered sets and the cohomology of an algebraic variety. -The biggest unanswered question in this subject is: - -Conjecture (Stanley–Stembridge). if $P$ is a unit interval order then $X_P$ is $e$-positive; i.e., it is a nonnegative combination of elementary symmetric functions. - -Once again, the conjecture challenges us to find the hidden structure that would explain this empirically observed fact.<|endoftext|> -TITLE: Inner automorphisms of Hopf algebras -QUESTION [7 upvotes]: Is there a reasonable notion of an inner automorphism of a Hopf algebra $H$ which in the case of a group ring $H=\mathbb KG$ for a group $G$ reduces to a conjugation by $g\in G$? - -REPLY [8 votes]: I am not sure if the following is the kind of answer you are expecting, but take the (left) adjoint action $(ad_l h)\triangleright k=\sum h_1 kS(h_2)$ of a hopf algebra $H$ on itself. -(It is known that the left adjoint action of $H$ on itself is always inner (in the sense of the def. 6.1.1, p. 87, in Montgomery's book "Hopf algebras and their action on rings") and turns $H$ into a $H$-module algebra). -If you set $H=kG$, then, the left adjoint action for $g\in G$ reduces to action by conjugation: $(ad_l g)\triangleright k=gkg^{-1}$. -More generally, for any inner action $\ \triangleright : H\otimes A\rightarrow A$, it is easy to show that any grouplike element $g\in G(H)$, acts as an inner automorphism of $A$. In case $H=A=kG$ and $\triangleright$ is the left adjoint action then this inner automorphism is the action by conjugation (of the corresponding group element). -Concluding, i would say that the notion of an "inner action" (p.674-675) may be the reasonable notion you are looking for.<|endoftext|> -TITLE: Undetermined games of "overdetermined" type -QUESTION [8 upvotes]: This is motivated by a previous question of mine, but I think it is ultimately more interesting (and hopefully easier to answer in the positive). In that question, a class of games (on $\omega$, of length $\omega$) is considered where the payoff set for such a game is determined by $\omega_1$-many, rather than the usual $2^\omega$-many, yes/no facts. Call such a class of games overdetermined; I'm interested in whether there are, provably in ZFC+$\neg$CH, undetermined overdetermined games (heh). -Formally, say that a game with payoff set $A$ is $\Gamma$-overdetermined (for $\Gamma$ a pointclass) if $A$ is $E$-invariant for some equivalence relation $E\in\Gamma$ with $\omega_1$-many classes. -Obviously this is a silly notion in full generality: take $E=\{(a, b): a\in A\iff b\in A\}$. Things become interesting, however, when we restrict attention to reasonably-definable $E$. For example, every copy/diagonalize game on ordinals (see the question linked above) is overdetermined with respect to a (fixed) $(\Sigma^1_1\vee\Pi^1_1)$-relation: changing notation slightly, a play $\pi$ yields a sequence of linear orderings $L_i^\pi$ $(i\in\omega)$, and two plays $\pi,\chi$ are guaranteed to be won by the same player if $L_i^\pi\approx L_i^\chi$ for each $i\in\omega$, where $A\approx B$ for linear orders $A, B$ if $A$ and $B$ are isomorphic or each is ill-founded. And this equivalence relation has only $\omega_1$-many classes. -My question is: - -Working in ZFC + $\neg$CH, can we prove the existence of an undetermined game which is overdetermined by some "tame" pointclass (e.g. projective, or even better $\Sigma^1_1\vee\Pi^1_1$)? - -I strongly suspect that the answer is "yes," even for $\Sigma^1_1\vee\Pi^1_1$, but I don't immediately see how to prove it. Note that I'm allowing the game itself to be as horrible as desired, as long as it respects some nice equivalence relation with few classes (incidentally, I'd also be happy to replace "$\omega_1$" with "$<2^\omega$") - the point is to restrict the usefulness of choice in terms of building an undetermined game by ensuring that any naive construction via transfinite recursion has to "end too early" to be guaranteed to work. -Beyond this question, I'm interested in any literature on games of arbitrary complexity but which respect some "tame" equivalence relation with few classes; I haven't managed to find any on my own. - -REPLY [4 votes]: The club game is overdetermined by a projective equivalence relation, so stationary co-stationary subsets of $\omega_1$ will give the counterexamples to overdeterminacy that you're looking for. The equivalence relation is actually $\Delta^1_2$, and if you're careful you can probably show it's low in the difference hierarchy on $\Pi^1_1$. -Recall that in the club game, two players alternate playing natural numbers $x(n)$ and $y(n)$. Thus Player I produces $x\in \mathbb R$ and Player II produces $y\in \mathbb R$. The outcome of the game depends on the reals $(x)_n$ and $(y)_n$ given by the canonical homeomorphism $\mathbb R\to \mathbb R^\omega$. Consider the sequence $(x)_0,(y)_0,(x)_1,(y)_1,\dots$. If some element of this sequence does not code a wellorder of $\omega$, we assign the play the value $\infty_\text{I}$ if Player $\text{I}$ is responsible for the first failure and $\infty_\text{II}$ otherwise. Let $|\cdot|$ be the rank function $\text{WO}\to \omega_1$. Assuming each $(x)_n,(y)_n$ codes a wellorder, if the sequence $|(x)_0|,|(y)_0|,|(x)_1|,|(y)_1|,\dots$ is not strictly increasing, we assign the play the value $\infty_\text{I}$ if Player $\text{I}$ is responsible for the first failure and $\infty_\text{II}$ otherwise. Otherwise all the rules have been followed, and we assign the play the value $\sup_{n<\omega} |(x)_n|,|(y)_n|$. Two plays are deemed equivalent if they have the same value. The equivalence relation is $\Delta^1_2$ since it is correctly computed by any transitive model of $\text{ZFC}^-$ that contains the play. -Given a subset $S\subseteq \omega_1$, let $A_S\subseteq \mathbb R$ be the collection of plays whose value is either $\infty_{\text{II}}$ or else lies in $S$. If Player $\text{I}$ wins $G(A_S)$ then $S$ contains a club, and if Player $\text{II}$ wins $G(A_S)$ then $\omega_1\setminus S$ contains a club. The argument is on pages 1972-1973 of the Handbook of Set Theory, Koellner-Woodin Theorem 2.12. Thus if $S$ is stationary co-stationary, $A_S$ is not determined.<|endoftext|> -TITLE: Distance to a geodesic ray -QUESTION [6 upvotes]: Given the Cayley graph of a group $G$ (for some fixed generating set $S$), consider the set $J_S$ of all the elements which lie on some infinite geodesic ray starting at the identity element of $G$, $e_G$. -Let $d$ denotes the distance in the Cayley graph -A simple example where $J_S \neq G$ is $G = \mathbb{Z} \times \mathbb{Z}_2$ with the generating set $S = \lbrace (\pm 1,0), (\pm 1 ,1) \rbrace$. In that case, $(0,1)$ does not lie on an infinite geodesic ray starting at $(0,0)$. -Question: Is there a choice of $G$, $S$ and $x \in G \setminus S$ so that $d(x,J_S) > \tfrac{1}{2} d(x,e_G)$. -[Edit: As David pointed out below, the interesting question would be: does $\limsup \frac{d(x_n,J_S)}{ d(x_n,e_G)} \leq \tfrac{1}{2}$ (where $x_n$ is some enumeration of the vertices)? - -REPLY [4 votes]: Let $G=\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}$. Let $S$ consist of $(1,1),(1,0),(0,1)$ and their inverses. Let $x=(0,3)$. -It seems clear that the neighbors of $x$ are contained in $\{-1,0,1\}\times\{2,3,4\}$, since each coordinate of a neighbor of $x$ will differ by $0,1$, or $-1$ from the corresponding coordinate of $x$. -If $(n,k)\in G$ and $n>7$, then it is clear that the word norm $|(n,k)|_S$ is equal to $n$ (representing $k$ by some element of $\{0,1,\ldots, 6\}$, we have that $a^kb^{n-k}$ represents $(n,k)$, where $a=(1,1)$ and $b=(1,0)$, and it is obvious that no shorter word can represent $(n,k)$). Hence, any geodesic word representing $(n,k)$ uses only the generators $(1,1)$ and $(1,0)$. Similarly, if $n<-7$, any geodesic word representing $(n,k)$ uses only the generators $(-1,-1)$ and $(-1,0)$. It follows that $J_S$ contains no neighbors of $x$, as -$J_S\cap (\{-1,0,1\}\times\mathbb{Z}/7\mathbb{Z})=\{(-1,-1),(-1,0),(0,0),(1,0),(1,1)\}$. -Thus, $d(x,J_S)\geq 2$. But $|x|_S=3$.<|endoftext|> -TITLE: Morphisms of formal group laws $\,F_a \rightarrow F_m\,$ and $\,F_m\to F_m$ -QUESTION [5 upvotes]: While studying cohomology theories on the stable homotopy setting, I have come up with the following basic question: -Consider the additive formal group law, $F_a$, and the multiplicative formal group law, $F_m$, both defined over a ring $R$. -If $R$ is a $\mathbb{Q}$-algebra, the exponential series defines a morphism of formal group laws -$$F_a \to F_m \quad , \quad x \mapsto e^x := \sum_{k=0}^\infty x^k / k! $$ and it is not difficult to check that any other morphism $\,f\colon F_a \to F_m\,$ is of the form $f (x) = e^{ax}$, for some $a\in R$. -I can extend the previous reasoning in case $R$ is an integral domain–of any characteristic–, and I am wondering how many morphisms of formal group laws $F_a \to F_m$ are there over a general ring. That is to say: -The question is to describe all the series $\,f \in R[[x]]\,$ such that -$$ f (s+t) = f(s)\cdot f(t) \qquad \mbox{ and } \qquad f(0) = 1 \ . -$$ - -On the other hand, I am also interested in endomorphisms of the multiplicative group law $F_m$. The obvious ones are rising to the $k^{th}$ power, for $k \in \mathbb{Z}$: -$$ F_m \to F_m \quad , \quad x \mapsto (1+x)^k \ , $$ and the analogous question would be: -Describe all the series $\,f \in R[[x]]\,$ such that -$$ f( x + y + xy) = f(x) \cdot f(y) \quad \mbox{ and } \quad f(0) = 1 \ . $$ -I think that there only exist these "rising to the $k^{th}$-power" maps, no matter the ring of coefficients. - -REPLY [3 votes]: For endomorphisms of $F_m$ one should consider the ring of numerical polynomials: -$$ P = \{f(a)\in \mathbb{Q}[a]: f(\mathbb{Z})\subseteq\mathbb{Z}\} $$ -The functions $b_i(a)=\left(\begin{array}{c}a\\ i\end{array}\right)$ give a basis for $P$ over $\mathbb{Z}$, with -$$ b_ib_j = \sum_{m=\max(i,j)}^{i+j} \frac{m!}{(m-i)!(m-j)!(i+j-m)!} b_m. $$ -We have a series $f(x)=\sum_{i>0}b_ix^i\in P[[x]]$ (which is morally "$(1+x)^a-1$") satisfying -$$ f((1+x)(1+y)-1)=(1+f(x))(1+f(y))-1. $$ -In other words, $f$ is an endomorphism of $F_m$. Any endomorphism of $F_m$ over any ring $R$ arises by applying some homomorphism $P\to R$ to the coefficients of $f(x)$. -Now let $R$ be an algebra over $\mathbb{Z}/p$. Consider the map $\mathbb{Z}\to R[[x]]$ given by $a\mapsto (1+x)^a-1$. This is continuous for the $p$-adic topology on $\mathbb{Z}$ and the $x$-adic topology on $R[[x]]$. This allows us to define $(1+x)^a-1$ for all $a\in\mathbb{Z}_p$, and this is still an endomorphism of $F_m$. -For a more complete story, we should consider $P/p$. One can identify this with the ring of continuous maps $\mathbb{Z}_p\to\mathbb{Z}/p$ (where $\mathbb{Z}_p$ has the $p$-adic topology, and $\mathbb{Z}/p$ has the discrete topology). To understand this, put $T=\{u\in\mathbb{Z}_p:u^p=u\}$. It is known that the reduction map $T\to\mathbb{Z}/p$ is bijective; the inverse map $\tau\colon\mathbb{Z}/p\to T$ is the Teichmuller lift map, given by $\tau(u)=\lim_ku_0^{p^k}$ for any lift $u_0$ of $u$. Given any element $a\in\mathbb{Z}_p$ there are unique elements $c_k(a)\in\mathbb{Z}/p$ such that $a=\sum_k\tau(c_k(a))p^k$. The functions $c_k$ are continuous, so they give elements of $P/p$. In fact, we find that -$$ P/p = \mathbb{Z}/p[c_0,c_1,c_2,\dotsb]/(c_k^p-c_k). $$ -Any homomorphism from this ring to $R$ will give an endomorphism of $F_m$ defined over $R$. -All of the above can be extracted from the literature on operations and cooperations in complex $K$-theory and its $p$-adic completion.<|endoftext|> -TITLE: Is it possible to compute André-Quillen cohomology by resolving the module variable? -QUESTION [7 upvotes]: Let $A\to B$ be a morphism of commutative rings. Let $\mathcal C$ be the category of commutative $A$-algebras augmented over $B$. Let $\mathcal M_B$ denote the category of $B$-modules. The cotangent complex can be defined like this (I will brush model-categorical considerations under the carpet). Consider the derived functor of the functor -$$s\mathcal C \to s\mathcal M_B$$ -obtained levelwise from the functor $\mathcal C\to \mathcal{M}_B$ given by $R\mapsto B\otimes_R \Omega_{R|A}$. Here $\Omega$ denotes Kähler differentials. -The image of $B\in \mathcal C$ under this derived functor is the cotangent complex $\mathbb L \Omega_{A|B}$. -If $M$ is a $B$-module, one defines the André-Quillen cohomology modules of $A\to B$ with coefficients in $M$ as -$$D^i(B|A,M)=H^q(\mathcal M_B(\mathbb L\Omega_{B|A},M)) \cong H^q \mathrm{Der}_A(P,M)$$ -where $P\to A$ is a projective resolution of $A$ in $s\mathcal C$. -So I guess this allows one to safely say that André-Quillen cohomology are some kind of "derived functor of the derivations". -I am wondering if it is (sometimes?) possible to resolve the module variable, i.e. to derive the functor $\mathrm{Der}_A(B,-):s\mathcal M_B \to s\mathcal M_B$, and get the same result. This feels somehow like asking whether $\mathrm Der_A(-,-): s\mathcal C \times \mathcal sM_B\to s\mathcal M_B$ is balanced. - -REPLY [9 votes]: The cotangent complex $\Bbb L_{B/A}$ is a (homologically) bounded-below complex of projective $B$-modules, and the bottom homology group is $H_0 (\Bbb L_{B/A}) = \Omega_{B|A}$. -If we apply $RHom_{B-mod}(-,M)$ to this, we get a Grothendieck spectral sequence for computing André-Quillen cohomology: -$$ -Ext^p_B(H_q \Bbb L_{B/A}, M) \Rightarrow D^{p+q} (B|A, M). -$$ -In particular, as Jason Starr points out, the derived functors of $Der_A(B,-) = Hom_B(\Omega_{B|A}, -)$ are the terms -$$ -Ext^p_B(H_0 \Bbb L_{B/A}, M) -$$ -on the edge of this spectral sequence, and so we get an equality precisely when the rest of the spectral sequence collapses to zero. The most common way that this could happen is if $H_q \Bbb L_{B/A} = 0$ for all $q > 0$ (smoothness, or certain local complete intersections), or maybe if the higher André-Quillen homology groups and $M$ are supported in different places, but is unlikely otherwise.<|endoftext|> -TITLE: Ramanujan's series $1+\sum_{n=1}^{\infty}(8n+1)\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}$ -QUESTION [18 upvotes]: This is a repost from MSE as I haven't got anything so far there. - -Ramanujan gave the following series evaluation $$1+9\left(\frac{1}{4}\right)^{4}+17\left(\frac{1\cdot 5}{4\cdot 8}\right)^{4}+25\left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{4}+\cdots=\dfrac{2\sqrt{2}}{\sqrt{\pi}\Gamma^{2}\left(\dfrac{3}{4}\right)}$$ in his first and famous letter to G H Hardy. The form of the series is similar to his famous series for $1/\pi$ and hence a similar approach might work to establish the above evaluation. Thus if $$f(x) =1+\sum_{n=1}^{\infty}\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}x^{n}$$ then Ramanujan's series is equal to $f(1)+8f'(1)$. Unfortunately the series for $f(x) $ does not appear to be directly related to elliptic integrals or amenable to Clausen's formula used in the proofs for his series for $1/\pi$. - -Is there any way to proceed with my approach? Any other approaches based on hypergeometric functions and their transformation are also welcome. Any reference which deals with this and similar series would also be helpful. - -REPLY [8 votes]: Ramanujan's result is a particular case of the Dougall's theorem -$${}_5F_4\left(\genfrac{}{}{0pt}{} -{\frac{n}{2}+1,n,-x,-y,-z} -{\frac{n}{2},x+n+1,y+n+1,z+n+1};1\right )=\frac{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+z+n+1)},$$ with $x=y=z=-n=-\frac{1}{4}$. See page 24 in the book B.C. Berndt, Ramanujan's Notebooks, Part II: http://www.springer.com/in/book/9780387967943 -A variant of the Dougall’s identity can be used to get many Ramanujan type series for $1/\pi$, see https://www.sciencedirect.com/science/article/pii/S0022247X1101184X (A summation formula and Ramanujan type series, by Zhi-Guo Liu).<|endoftext|> -TITLE: Growth rate of longest sequence of strings where no string is a subsequence of a later one -QUESTION [5 upvotes]: We define $STR(n)$ to be the longest sequence of strings with $n$ symbols such that the $k$th string has at most k symbols, the symbols of the string are taken from an alphabet consisting of $n$ characters, and no string is a subsequence of a later one. -For example, $STR(1)=2$, because the longest sequence is "A", "". $STR(2)=4$, because the longest sequence is "A", "BB", "B", "". I'm not sure how big $STR(3)$ is (although it is at least 11 due to the sequence "A","BB","BC","CB","B","CCCCC","CCCC","CCC","CC","C",""). -We do know, however, that $STR(n)$ is always finite, for the same reason that $TREE(n)$ and $SSCG(n)$ (see well-quasi-order). -My question, what is the growth rate of $STR(n)$? It is slower than $TREE(n)+1$, since strings can embed in trees. -(In particular, if we could find $n$ such that $STR(n)>TREE(3)$, that would be great.) - -REPLY [8 votes]: I suppose it's a good idea to turn my comment into an answer. -The function $STR$ is basically the function $F$ defined by Friedman in this paper (more precisely, it's easy to show $STR(k)=F(k-1)+1$). Friedman pinpoints the growth rate of this function quite precisely in theorem 5.19 of that paper. He uses the following variant of fast-growing hierarchy: $$H_1(x)=2x+1$$ $$H_{\beta+1}(x)=H_\beta^{x+1}(x)$$ $$H_\lambda(x)=H_{\lambda(x)}(x)$$ -(where $\lambda(x)$ denotes the $x$-th element of a standard fundamental sequence of $\lambda$ when $\lambda<\varepsilon_0$). -Theorem: The function $F$ eventually dominates all functions $H_\beta$ for $\beta<\omega^\omega$ and is itself eventually dominated by $H_{\omega^\omega+1}$. -Hence, in a very precise sense, $F$ has growth rate of approximately $H_{\omega^\omega}$. -On the other hand the function $TREE$ is growing very much faster. An often-quoted lower bound is $f_{\Gamma_0}$ in the fast-growing hierarchy, but I don't think an exact bound like above has been proven. -This way or another, regarding your last question - I haven't bothered to check the details, but I imagine that even with $TREE(3)$ we can construct sequences of trees of length exceeding things like $STR^{STR(5)}(5)$, so we are not going to easily find $n$ such that $STR(n)>TREE(3)$.<|endoftext|> -TITLE: Global orthogonal coordinates on the open unit ball -QUESTION [6 upvotes]: Is there any diffeomorphism $x:\mathbb{R}^n\to\text{Im }x=B_1\left(0\right)\subset\mathbb{R}^n$ such that - -$x$ is an orthogonal chart, i.e., the coordinate vector fields $X_i=\partial x/\partial u_i$ satisfy -$X_i\left(u\right)\cdot X_j\left(u\right)=0$ for all $1\leq i\neq j\leq n$ at every point $u=\left(u_1,\dots,u_n\right)\in\mathbb{R}^n$; -the Euclidean norm of the differential $dx$ satisfies $$\left\|dx\left(u\right)\right\|^2\mathrel{\mathop:}=\sum_{i=1}^n\left\|X_i\left(u\right)\right\|^2=\rho^2\left(x\left(u\right)\right)$$ -at every point $u\in\mathbb{R}^n$, where $\rho:B_1\left(0\right)\to\mathbb{R}$ is given by $\rho\left(x\right)=\frac{1-\left\|x\right\|^2}{2}$. In particular, $dx$ vanishes at infinity. - -The above assumptions lead to the following second-order PDE -$$\Delta x-\rho\left(x\right)x=\sum_{i=1}^n\frac{\partial}{\partial u_i}\ln\left(\left\|X_i\right\|^2\right)X_i,$$ -which might be useful somehow. -$\hspace{4pt}$ I believe that such a diffeomorphism does not exist. Actually, I can't even think of such an $x$ satisfying just condition 1. By Liouville's theorem we know that $x$ cannot be conformal. Moreover, radial diffeomorphisms typically used to show that $\mathbb{R}^n$ and $B_1\left(0\right)$ are diffeomorphic do not satisfy condition 1 nor 2. - -REPLY [12 votes]: This problem is just the classical problem of finding global Tchebychev coordinates on hyperbolic $n$-space. By Hilbert's Theorem, this is impossible when $n=2$. The problem remains open in higher dimensions, despite years of work. -Here is how one can see the reformulation: Let $J$ be the Jacobian of $x$ with respect to $u$, i.e., $\mathrm{d}x = J\,\mathrm{d}u$. Then condition 1 is that $J^TJ$ be diagonal, say, $J^TJ = \mathrm{diag}(\lambda_1^2,\ldots,\lambda_n^2)$, and condition 2 is that -$$ -\lambda_1^2+\cdots+\lambda_n^2 = (1-\|x\|^2)^2/4. -$$ -Writing $\lambda_i = \tfrac12(1{-}\|x\|^2) g_i$ yields -$$ -g_1^2+\cdots+g_n^2=1\tag1 -$$ -and the above equations become -$$ -(g_1\,\mathrm{d}u_1)^2 + \cdots +(g_n\,\mathrm{d}u_n)^2 -= \frac{4}{(1-\|x\|^2)^2}\bigl(\mathrm{d}x_1^2+\cdots+\mathrm{d}x_n^2\bigr) -\tag2 -$$ -The right hand side of (2) is the standard form of the conformal hyperbolic metric on the unit ball, so the question is asking for coordinates $u_i$ and functions $g_i>0$ on hyperbolic $n$-space satisfying equations (1) and (2). This is the classic problem of Tchebychev coordinates. -For example, when $n=2$, one can write $(g_1,g_2)=(\cos f,\sin f)$ for some function $f$ taking values in $(0,\tfrac12\pi)$, and then one is asking when -the metric -$$ -ds^2 = \cos^2f\,\mathrm{d}u_1^2 + \sin^2f\,\mathrm{d}u_2^2 -$$ -has Gauss curvature $-1$. It is well-known that this holds if and only if $f$ satisfies the Sine-Gordon equation -$$ -\frac{\partial^2f}{\partial u_1^2} - \frac{\partial^2f}{\partial u_2^2} -= \cos f\,\sin f.\tag3 -$$ -Hilbert proved that there do not exist global coordinates $u_i$ on the hyperbolic plane and a function $f(u_1,u_2)$ taking values in $(0,\tfrac12\pi)$ satisfying (3). -As I wrote, it is still not known whether global solutions (with $g_i>0$) exist in higher dimensions, but, as Cartan showed, there do exist many local solutions. Essentially, they depend on $n(n{-}1)$ functions of one variable. -Addendum: Of course, if one only wants global orthogonal coordinates, that is not hard: If you first apply inversion with respect to a point on the boundary of the ball $B_1(0)$ in $x$-space, you'll get an injective diffeomorphism $v:B_1(0)\to\mathbb{R}^n$ whose image is a half-space. Say $v\bigl(B_1(0)\bigr)$ is defined by $v_n>0$ (after some translation and rotation in $v$-space). Then define -$$ -(u_1,\ldots,u_{n-1},u_n) = \bigl(v_1,\ldots,v_{n-1},\log(v_n)\bigr). -$$ -You'll then have $u:B_1(0)\to\mathbb{R}^n$ being a bijective orthogonal diffeomorphism, and its inverse will be the diffeomorphism $x:\mathbb{R}^n\to B_1(0)$ that you desire. -When $n=2$, there are more options: Choose a conformal diffeomorphism $z:B_1(0)\to R\subset\mathbb{C}$ where $R = I_1\times I_2$ is a rectangle that is a product of two intervals $I_i\subset\mathbb{R}$, not both equal to $\mathbb{R}$, -and let $z = v_1 + i v_2$ and choose any bijective diffeomorphisms $h_i:I_i\to\mathbb{R}$. Then $u_1 = h_1(v_1)$ and $u_2 = h_2(v_2)$ will provide the desired functions. Now let $x:\mathbb{R}^2\to B_1(0)$ be the inverse of $u:B_1(0)\to\mathbb{R}^2$.<|endoftext|> -TITLE: Flatness and Cohen-Macaulay rings -QUESTION [9 upvotes]: Let $A$ be an Artin local ring and let $f:A \to B$ a local ring homomorphism to a Noetherian local one dimensional $A$-algebra $B$. - -If $B$ is Cohen--Macaulay and the localization of $B$ at any minimal prime is flat over $A$, then is $B$ flat over $A$? - -As far as I can see, this doesn't follow from any of the standard flatness criteria, but I could have missed something simple. -(I am mostly interested in the case that $B$ is the localization of a finite type $A$-algebra at some maximal ideal, but it seems likely that the answer to the question is the same with or without this extra assumption.) - -REPLY [5 votes]: That is not true. Let $k$ be a field. A module over the ring of dual numbers, $A=k[\epsilon]/\langle \epsilon^2 \rangle,$ is equivalent to a $k$-vector space $B$ with a square-zero $k$-linear self-map $L_\epsilon:B\to B.$ For every integer $q\geq 1$, the Tor module equals, $$\text{Tor}^A_{q\geq 1}(A/\langle \overline{\epsilon}\rangle,B) = H(L_\epsilon,B) = \text{Ker}(L_\epsilon)/\text{Image}(L_\epsilon).$$ Now let $B$ be the following one-dimensional, local, Noetherian $k$-algebra that is a local complete intersection, hence Cohen-Macaulay, $$B=k[x,y]_{\langle x,y \rangle}/ \langle y^2\rangle,$$ where $L_\epsilon$ is multiplication by $\overline{xy}.$ -The localization of $B$ at the unique minimal prime $\mathfrak{p}=\langle \overline{y}\rangle$ is $$B_\mathfrak{p} = k(x)[y]/\langle y^2\rangle.$$ The homology module for this localization is $$H(L_\epsilon,B_{\mathfrak{p}}) = \{0\}.$$ Thus, $B_{\mathfrak{p}}$ is $A$-flat. However, $B$ is not flat since $\text{Tor}_{q\geq 1}^A(B,A/\mathfrak{m})$ equals -$$ \text{Ker}(L_{\overline{xy}}:k[x,y]/\langle y^2\rangle \to k[x,y]/\langle y^2\rangle)/\text{Image}(L_{\overline{xy}}:k[x,y]/\langle y^2\rangle \to k[x,y]/\langle y^2\rangle) = $$ $$\langle \overline{y}\rangle / \langle \overline{xy} \rangle = k\cdot \overline{y},$$ which is one-dimensional.<|endoftext|> -TITLE: "Anti" fixed point property -QUESTION [19 upvotes]: Let $(X,\tau)$ be a topological space. If $f:X\to X$ is continuous, we say $x\in X$ is a fixed point if $f(x) = x$. -The space $(X,\tau)$ is said to have the anti fixed point property (AFPP) if the only continuous maps $f:X\to X$ with fixed points are the identity map $\text{id}_X:X\to X$, and the constant maps. -The only continuous self-maps of strongly rigid spaces are the identity and the constant maps, so they trivially have the AFPP. - -Q. Assume that $(X,\tau)$ is Hausdorff and it has the AFPP. Does this imply that $(X,\tau)$ is strongly rigid? - -REPLY [5 votes]: There is a ZFC example based on the so called Cook continuum -(this is a compact metric space with a very rigid structure with respect to selfmaps). -Let $C$ be a Cook continuum and $a, b$ two distinct points of $C$. Let $X$ be the quotient of $C \times \lbrace 0,1 \rbrace$, where the point $(a,0)$ is identified with $(b,1)$ and $(a,1)$ is identified with $(b,0)$. -Claim: Such a space $X$ has AFPP, but it is not strongly rigid. -There is a nonconstant homeomorphism $h$ which sends the points $(c,0)$ to $(c,1)$ and the points $(c,1)$ to $(c,0)$ for every $c$ from $C$. This $h$ is well defined on the quotient space $X$. -The proof that $X$ has AFPP is using the properties of Cook continuum... -For similar constructions using Cook continuum see the book of Pultr and Trnkova: Combinatorial, algebraic and topological representations of groups, semigroups and categories.<|endoftext|> -TITLE: What is an "integrable hierarchy"? (to a mathematician) -QUESTION [20 upvotes]: This is one of those "what is an $X$?" questions so let me apologize in advance. -By now I have already encountered the phrase "integrable hierarchy" in mathematical contexts (in particular the so called "Kdv hierarchy" which is apparently related to enumerative geometry of curves in some ways which are a total mystery to me) enough times to care about the meaning of these words. -However when i type these words in google most of the results are links to physics papers and I haven't been able to find anything clarifying about the meaning of this phrase, let alone a precise mathematical definition. Hence the question: - -What is an integrable hierarchy? Where do they come from? What are their applications? - -EDIT: Let me emphasize that this question is mainly about the "hierarchy" part. I understand (at least on a basic level) what is an integrable system and I know of several good mathematical references for this topic. The emphasis here is on what constitutes an integrable hierarchy as apposed to a plain old "integrable system". - -REPLY [2 votes]: In brief, an integrable hierarchy is an infinite (usually countable) set of integrable partial differential systems such that any two systems in this set are compatible. Such hierarchies are usually generated by recursion operators or master symmetries, cf. e.g. the introduction of this recent paper and the references given there regarding the recursion operators and this book and references therein regarding master symmetries.<|endoftext|> -TITLE: "Descent" of analytic functions along a finite morphism -QUESTION [6 upvotes]: Let $f : X \to Y$ be a finite surjective morphism of smooth affine algebraic varieties over the complex numbers. Is it true that a function on $Y$ whose pullback via $f$ is an analytic function on $X$, is itself analytic? -I ask because I am interested in knowing that, for a reductive complex algebraic group $G$, an analytic $W$-invariant function on a Cartan subalgebra $\mathfrak{h}$ lifts to an analytic $G$-invariant function on the Lie algebra $\mathfrak{g}$. So, if I am not mistaken, it is enough for me to know that an analytic function on $\mathfrak{h}$, invariant under $W$, gives rise to an analytic function on $W \backslash \mathfrak{h}$. -Thanks, -Sasha - -REPLY [9 votes]: Let $h:Y\to \mathbb{C}$ be such a function. If $U\subset Y$ is an open subset (for the complex topology) and $s:U\to X$ is an (analytic) local section of $f$ on $U$, then $h=h\circ f\circ s$ on $U$, hence $h_{\vert U}$ is analytic since $h\circ f$ is. -So, $h$ is analytic on $Y\smallsetminus B$ where $B\subset Y$ is the branch locus of $f$, which is a proper Zariski closed subset. Since $h$ is continuous on $Y$ by nfdc23's remark, it follows that it is analytic on $Y$. -[EDIT] It seems plausible that the argument works assuming $Y$ normal. Here is a counterexample where it isn't: take for $Y$ the cuspidal cubic in $\mathbb{C}^2$ (with equation $y^2=x^3$ and for $f:X=\mathbb{C}\to Y$ the normalization (mapping $t$ to $(t^2,t^3)$). Then $f$ is homeomorphism, so $t$ descends to a continuous function on $Y$, which is not analytic.<|endoftext|> -TITLE: What is the link between the Domino Tilings and the Ising Model? -QUESTION [7 upvotes]: Is there a link between the theory of Domino Tilings and the Ising Model? In the global qualitative sense that physicists use, the answer is "yes". The connections could go like this: - -The dimer model is the limit of the ising model (Onsager?) -The theory of domino tiling is dual to the theory of dimer tilings. (Kenyon?) - -However, when I put the two connections together, I wasn't able to say this Ising model configuration maps to this Domino Tiling. And I suspect, they are slightly different objects. First here's the section from Baxter's book on Exactly Solved Models giving us hope: - -He tells us, the Ising Model "freezes" into a Dimer problem, which is dual to a domino tiling problem. -From Fisher's original 1966 paper we find, the square Ising model maps to dimers model on "decorated" square lattice: - -The theories of exactly solved statistical mechanics and of integrable systems are not necessarily related. It's just that if I read one, then I inadvertently start reading the other. Here's a discussion of Korepin and Izergin. - -So now, there is also a link between 6-vertex model and domino tilings. Does that also have fine print? - -For all of the above questions, I cited papers from the 1960's and 1980's. I wonder what the modern perspective of integrable theories are. -I seem to have asked a question like this already. And there will be more clarifications yet to come. - -map from 6-vertex model to domino tiling - -REPLY [2 votes]: Since others have already commented on the link with domino tilings, let me point out a relation between the eight-vertex model and an Ising-type model on the square lattice, cf the comments. -These Ising-type models are a bit unusual: they feature both two- and four-point interactions. [The latter are like the interactions of an interaction-round-a-face (IRF) model on the dual lattice.] The weights can be made to match, and the relation is 2-to-1 at the level of configurations, so the partition functions match up to a factor of two. See e.g. eight-vertex model at Wikipedia and the references there. -(As you know I explained the relation between the six-vertex model and domino tilings of the Aztec diamond, cf the last figure in the OP, in my answer to your question that you mention.)<|endoftext|> -TITLE: Lower bounds on Kullback-Leibler divergence -QUESTION [8 upvotes]: This was originally a question on Cross Validated. -Are there any (nontrivial) lower bounds on the Kullback-Leibler divergence $KL(f\Vert g)$ between two measures / densities? -Informally, I am trying to study problems where $f$ is some target density, and I want to show that if $g$ is chosen "poorly", then $KL(f\Vert g)$ must be large. Examples of "poor" behaviour could include different means, moments, etc. -Example: If $f=\sum_ka_kf_k$ and $g=\sum_kb_kg_k$ are two mixture distributions, is there a lower bound on $KL(f\Vert g)$ in terms of $KL(f_k\Vert g_j)$ (and also the convex weights $a_k,b_j$)? Intuitively, we'd like to say that if $\inf_{k\ne j} KL(f_k\Vert g_j)$ is "big", then $KL(f\Vert g)$ cannot be small. -Anything along these lines (for mixtures or arbitrary measures) would be useful. Obviously, you can make assumptions about the quantities involved. Alternatively, references to any papers that study these kinds of problems (either directly or indirectly) would be helpful! - -REPLY [11 votes]: Pinsker's inequality states that -\begin{equation} - \text{KL}(f|g)\ge B_P:=\|f-g\|^2/2, -\end{equation} -where $\|f-g\|:=\int|f-g|$ is the total variation norm of the difference between the distributions with densities $f$ and $g$. -Another lower bound on $\text{KL}(f|g)$ can be given in terms of the Hellinger distance $d_H(f,g):=\frac1{\sqrt2}\|\sqrt f-\sqrt g\|_2$: -\begin{equation} - \text{KL}(f|g)\ge B_H:=2d_H(f,g)^2=\int(\sqrt f-\sqrt g)^2; -\end{equation} -see e.g. mathSE. -One may note note that either one of these two lower bounds, $B_P$ and $B_H$, may be better (that is, greater) than the other. E.g., if the densities $f$ and $g$ with respect to the counting measure on the set $\{1,2\}$ are given by the vectors $(1/2,1/2)$ and $(1/2-t,1/2+t)$ for $t\in(0,1/2)$, then $B_P>B_H$ for $t\in(0,t_*)$, and $B_P -TITLE: Example of a non-arithmetic Veech surface (other than regular polygon)? -QUESTION [5 upvotes]: I am reading this paper of Avila and Delecroix of the billiard flow on polygonal surfaces, but I have to get through some basic definitions first. What is a non-arithmetic Veech surface? - -A Veech surface is an “exceptionally symmetric” translation surface, in the sense that the Veech group is a (finite co-volume) lattice in $SL(2, \mathbb{R})$ (it is easily seen that the Veech group is never co-compact). Simple examples of Veech surfaces - are square-tiled surfaces, obtained by gluing finitely many copies of the unit square $[0, 1]^2$ along their sides: in this case the Veech group is commensurable with $SL(2, \mathbb{Z})$. - -These are the kinds of surface they want to rule out (because they will not lead to weak-mixing flows. The promise is, that as long as you're not scared of decimal error, these problems have an elementary flavor, can be simulated on a computer. The proofs might not be. - -Veech surfaces that can be derived from square-tiled ones by an affine - diffeomorphism are called arithmetic. Arithmetic Veech surfaces are branched covers - of flat tori, so their directional flows are never topologically weak mixing (they admit a continuous almost periodic factor). - -Their example of a non-arithmetic Veech surface is the regular polygon $P_n$ (with opposite sides identified) and or a copy of $P_n$ and its mirror $-P_n$ if $n = 2k+1$. $S_n = P_n$ or $S_n = P_n \cup (- P_n)$. Then they show two results. The first one is slightly more refined than the Veech dichotomy because it says these directions are weak-mixing. - -Thm 2 The geodesic flow in a non-arithmetic Veech surface is weakly mixing - in almost every direction. Indeed the Hausdorff dimension of the set of exceptional - directions is less than one. - -And we get rare examples of weak-mixing dynamical systems (that are not mixing). The elements of the Veech group are $2\times 2$ matrices living inside a number field $k/\mathbb{Q}$: - -Thm 3 Let $S$ be a Veech surface with a quadratic trace field (i.e., $r = [k: \mathbb{Q}] = 2$). - Then the set of directions for which the directional flow is not even topologically - weak mixing has positive Hausdorff dimension. - -E.g. the regular polygon has trace field $k = \mathbb{Q}[\cos \frac{\pi}{n} ]$ Certainly then, the regular polygons have mysteries yet to unfold, but also I am curious what these positive Hausdorff dimension fractal sets might be. Another question is if there are Veech surfaces other than than translation surfaces of regular polygons. - -REPLY [2 votes]: Some examples are given in the 2005 paper by Sergey Vasilyev.<|endoftext|> -TITLE: Sufficient cohesion and conservativity of "underlying stuff" -QUESTION [8 upvotes]: Consider a category $\mathsf C$ admitting a quadruple adjunction as below. -$$(\Pi_0 \dashv \text{disc} \dashv \Gamma \dashv \text{codisc}) : - \mathsf{C} - \stackrel{\stackrel{\longrightarrow}{\longleftarrow}}{\stackrel{{\longrightarrow}}{{\longleftarrow}}} - \mathsf{B} - \;$$ -A good example is the category of locally connected spaces with $\mathsf{B}=\mathsf{Set}$. In this case the connected-components functor $\Pi_0$ implies $\mathsf C$ is extensive. -Suppose now $\mathsf C$ is extensive and admits the exponentials below. In Axiomatic Cohesion, Lawvere defines an adjoint quadruple as above to be cohesive if: - -$\Pi_0$ preserves finite products and $\text{codisc}$ is fully faithful. -$\Pi_0$ preserves $\mathsf{B}$-parametrized powers in the sense of a natural isomorphism $$\Pi_0(X^{\text{disc}I})\cong \Pi_0(X)^{I}.$$ -The canonical arrow $\Gamma\Rightarrow \Pi_0$ is epimorphic. - -Lawvere later defines such an adjunction to be sufficiently cohesive if each object has a "contractible envelope": - -For every $X$ there exists a monic map $X\to Y$ with $Y$ contractible in the sense that $\Pi_0(Y^A)=\mathbf 1$ for all $A$. - -Then it is proved that a topos of cohesion ($\mathsf C$ is a topos) is sufficiently cohesive iff the truth-value object is connected, and also iff all injective objects are connected. Here, connectedness of $X$ means $\Pi_0(X)\cong \mathbf 1$. -For locally connected spaces it seems the injective objects are the inhabited indiscrete spaces, whose connectedness reflects their cohesion. This is opposed to case of sets where $\text{codisc}$ is the identity and sets are seldom singletons. - -Can a sufficiently cohesive category (quadruple adjunction) have a conservative $\Gamma$? - -I ask for the following reason: For locally connected spaces $\Gamma$ is the "points functor" and I think of the fact it is faithful but not conservative as reflecting that "discontinuous" arrows (i.e breaking cohesion) may be underlain by "isomorphisms on stuff"; I want to know if this is "correct" intuition or not. - -REPLY [2 votes]: Let ${\phi : \mathrm{disc} \rightarrow \mathrm{codisc}}$ be the canonical map. -The map ${\Gamma \phi }$ is an iso in the base so, if $\Gamma$ is conservative, $\phi$ is an iso. -In other words, if $\Gamma$ is conservative, the original string of adjoints is a quality type. So, if the original string is sufficiently cohesive then the whole thing is degenerate.<|endoftext|> -TITLE: Eigenvalue problem with two quadratic constraints -QUESTION [7 upvotes]: I would like to solve the following problem: -$$\begin{array}{ll} \text{minimize} & \mathbf{x}^T \mathbf{A} \mathbf{x}\\ \text{subject to} & \mathbf{x}^T\mathbf{B}\mathbf{x} = 0\\ & \mathbf{x}^T \mathbf{x} = 1\end{array}$$ -where $\bf x$ is a vector, $\bf A, \bf B$ are square matrices, and $\bf A$ is symmetric. - -Here is my thinking: -Use the Lagrange multiplier method, -\begin{equation} -\mathcal L (\bf x, \lambda, \mu) = \mathbf{x}^T \mathbf{A} \mathbf{x} - \lambda \mathbf{x}^T\mathbf{x} - \mu \mathbf{x}^T \mathbf{B} \mathbf{x}. -\end{equation} -Take the derivative with respect to $\bf x$, we get: -\begin{equation} -\bf{A x = \lambda x + \mu Bx} -\end{equation} -This is not exactly an eigenvalue problem or a generalized one. What's next? -I can apply the constraints and get $\lambda = \bf x^TAx$, $\mu = \bf x^TB^TAx/(x^TB^TBx)$. But I am looking for a method that can turn the problem to a linear problem, e.g. generalized eigenvalue problem, so that I can apply the standard numerical linear algorithms. -In principle, if I can solve $\det (A-\lambda I - \mu B) = 0$, I can -eliminate, say, $\mu$. But this is not feasible, numerically. A perturbative solution with $|\mu|\ll 1$ is acceptable. -Question: Are there any methods, ideally using standard numerical linear algorithm, to solve this problem? - -These problems are similar but not the same: -Linearly constrained eigenvalue problem -Thank you in advance. -Edit: In viewing of the comments, I removed the "full rank" condition and does not requires $\bf A$ to be "positively defined". Hopefully, the problem may have a solution? -The background of the problem is as follows: -$\bf A$ is a Hamiltonian. $\bf x$ is its eigenvector with lowest energy. $\bf x^T Bx = 0$ represents a constraint imposed by a symmetry. In practice, $\bf A$ is truncated, and $\bf x^T B x \ne 0$. -Now, I am trying to reformulate the problem to guarantee the symmetry constraint $\bf x^T B x = 0$. As a result, $\bf x$ may not be an eigenvector of $\bf A$, which is the price to pay. My hope is that as the symmetry violation is small enough, the problem may still have an efficient solution. Hope this helps. - -REPLY [2 votes]: Because no one has offered a solution meeting your ideal of using a standard numerical linear algorithm, I will offer an approach using the global numerical nonlinear optimizer BARON. -Here is a solution using BARON as the solver under YALMIP under MATLAB. I will use the B provided by @Federico Poloni in his comment above. I'm not sure what symmetric and positively defined is supposed to mean, so I chose a random A which is symmetric positive definite with all elements positive, which ought to comply with whatever it means. -n = 4; -B = [zeros(n/2) eye(n/2);eye(n/2) zeros(n/2)]; -A = rand(n); A = A*A'; % random instantiation of A -x = sdpvar(n,1); % declare x an an optimization vector -Constraints = [x'*B*x == 0,x'*x == 1] % the non-convex constraints -Objective = x'*A*x % objective function to be minimnized -% minimize the Objective, subject to the Constraints, using BARON -optimize(Constraints,Objective,sdpsettings('solver','baron')) - -For -A = - 1.716800970124081 0.998289669825227 1.266317282130762 0.970191833948101 - 0.998289669825227 1.486118602130391 1.165572239200317 0.702280553602394 - 1.266317282130762 1.165572239200317 1.679161019401491 0.884294705407438 - 0.970191833948101 0.702280553602394 0.884294705407438 0.729460526019744 - -The result is - optimal x = [-0.397502000000000 -0.061859500000000 -0.140779000000000 0.904625000000000]' - - optimal objective value = 0.116730782147915 - -The constraints are satisfied to within a tolerance of less than 1e-6, but a tighter tolerance could be used. -True, this will not scale in a friendly way as n increases.<|endoftext|> -TITLE: Is there a "universal" connected compact metric space? -QUESTION [35 upvotes]: Fact 1. The Cantor set $K$ is "universal" among nonempty compact metric spaces in the following sense: given any nonempty compact metric space $X$, there exists a continuous surjection $f\colon K \to X$. -Fact 2. The closed interval $I$ has a similar "universal" property among nonempty compact connected and locally connected metric spaces: given any such space $X$, there exists a continuous surjection $f \colon I \to X$. -This makes me wonder: is there a compact connected metric space $J$ such that for any nonempty compact connected metric space $X$, there exists a continuous surjection $f \colon J \to X$? -Such a space $J$, if it exists, would be 'intermediate' between $I$ and $K$: there would need to be continuous surjections -$$ K \to J \to I $$ -Fact 1 is sometimes called the Alexandroff–Hausdorff theorem, since appeared in the second edition of Felix Hausdorff’s Mengenlehre in 1927 and also in an article by Pavel Alexandroff published in Mathematischen Annalen in the same year. Fact 2 was proved by Hans Hahn in 1914 and reproved by him more nicely in 1928. For a nice history of these results, see: - -L. Koudela, The Hausdorff–Alexandroff theorem and its application in theory of curves, WDS'07 Proceedings of Contributed Papers, Part I, 2007, pp. 257–260. - -One may rightly complain that "universal" is the wrong word above, since we're not claiming there exists a unique continuous surjection, and indeed there's usually not. A better term is versal. There can be two non-homeomorphic spaces having the same versal property. For example, $I^2$ would work just as well as $I$ in Fact 2, thanks to the existence of space-filling curves. -Nonetheless we can create a category in which these versal properties become universal, by a cheap trick. Let $\mathrm{CompMet}$ be the collection of all homeomorphism classes of nonempty compact metric spaces, and put a partial order on this where $[X] \ge [Y]$ iff there exists a continuous surjection $f \colon X \to Y$. The homeomorphism class of the Cantor set is the top element of the poset $\mathrm{CompMet}$. My question asks if the subset of $\mathrm{CompMet}$ coming from connected compact metric spaces has a top element. -I'd also appreciate any interesting information on this poset $\mathrm{CompMet}$. -For example, I think that there's a map sending each element of $\mathrm{CompNet}$ to its number of connected components, and I think that this is an order-preserving map from $\mathrm{CompMet}$ to the cardinals less than or equal to the continuum. But there also seems to be an order-preserving map sending each element of $\mathrm{CompNet}$ to its number of path-connected components. Are there other interesting maps like this? - -REPLY [9 votes]: Here's a related positive result: -Alan Dow and KP Hart proved in 1998 that the (non-metrizable) continuum $\beta[0,1)\setminus [0,1)$ maps onto every continuum of weight $\leq \omega_1$, and therefore onto every metrizable continuum. See https://arxiv.org/pdf/math/9805008.pdf.<|endoftext|> -TITLE: Does $[0,1]\cap \mathbb{Q}$ have a connected $T_2$ quotient? -QUESTION [6 upvotes]: Is there an equivalence relation $R$ on $[0,1]\cap \mathbb{Q}$ such that $([0,1]\cap \mathbb{Q})/R$ is connected, Hausdorff, and has more than $1$ point? - -REPLY [11 votes]: Yes, there exists such a relation on $\mathbb Q$. -Just use the fact that the rational projective space $\mathbb QP^\infty$ from (the answer to) this question is a countable, Hausdorff, connected (and even topologically homogeneous). By definition, the space $\mathbb QP^\infty$ is a quotient (and even open) image of a countable metrizable space without isolated points. By the classical Sierpinski theorem such space is homeomorphic to $\mathbb Q$ (and to $\mathbb Q\cap[0,1]$, too). So, $\mathbb QP^\infty$ is a connected Hausdorff quotient (even open) image of $\mathbb Q$.<|endoftext|> -TITLE: Special $\Gamma$-categories and symmetric monoidal categories -QUESTION [8 upvotes]: Let $\Gamma^{op}$ be the category of finite pointed sets. A special $\Gamma$-category is a functor $Y:\Gamma^{op}\to Cat$ such that the canonical maps $Y[n]\to Y[1]^n$, called Segal maps, are equivalences of categories, for all $n\geq 0$. -There is a canonical way of associating to an unbiased symmetric monoidal category a special $\Gamma$-category (called a “homotopy monoidal category” by Tom Leinster in Higher Operads, Higher Categories). -I think that Moritz Groth, in his Example 3.4 of his course in infinity categories, is saying that this association is an equivalence. -However, Tom Leinster, pages 120-121 of the aforementioned book, says that “there is reasonable hope” that this is true. -I would like to know whether this result is true or not, and where do the delicate points lie. If it is true, I what are some references where the result is proven in detail? -EDIT: I would like to add some context. A commutative monoid in a cartesian monoidal category $\mathcal C$ is, equivalently, a functor $\Gamma^{op}\to \mathcal C$ that satisfies that the Segal maps are isomorphisms. -My point is that I would like to apply this to the concept of symmetric monoidal category itself. However, a (small) symmetric monoidal category is not monoid, but a symmetric pseudomonoid in the cartesian 2-category $Cat$. -So I guess I would like to know whether it is true that symmetric pseudomonoids in 2-categories $D$ are equivalently pseudofunctors $\Gamma^{op} \to D$ that satisfy that the canonical maps $Y[n]\to Y[1]^n$ are 1-equivalences. Note that this is not exactly what I wrote in the first line (here I'm letting $Y$ be a pseudofunctor). - -REPLY [3 votes]: Tom Leinster's book is very old. For higher category theory 2003 is like a previous epoch. In those times there were many competing definitions of higher category theory and higher algebra, for most of them it was unclear if they are equivalent or whether they even model all homotopy types. For some of them this was eventually proven false, e.g. the once-favoured derivators are unsufficient to properly work with higher algebra and the strict $\omega$-categories are obviously unsufficient for the homotopy hypothesis (homotopy types = $\infty$-groupoids). The main delicate points are in giving the proper definitions and handling all coherence conditions, so that the theory is flexible enough for all purposes but at the same time is still tractable. For this reason the aforementioned statement is stated not as a theorem (probably it is so in Leinster's foundations, but I don't have the print copy at hand) but rather as a guiding principle which should be true in any proper higher category theory. This is certainly true in the now-accepted foundations via quasicategories and higher Segal spaces. In fact, it is usually taken as the definition of a monoidal higher category, so the only question is whether it covers all classical example (it does). J. Lurie's "Higher Algebra" is nowadays the go-to reference for all questions about the foundations of higher algebra, and his "Higher Topos Theory" lays the used foundations of $(\infty, 1)$-category theory. -If you are working within classical 1-category theory, then the statement is very easy to prove explicitly. I am unaware of any specific reference for it. For $A_\infty$-monoids in homotopy theory I believe that the approach via $\Gamma$-spaces originates from the work of G. Segal. You can check the references in the relevant page on nLab. -Looking at the arXiv version of Leinster's paper, it also doesn't really discuss $\infty$-categories, nor does it even define an equivalence of $\omega$-categories (see par. 9.2), so the equivalence between monoidal categories and $\Gamma$-spaces was likely never really proved in that foundations.<|endoftext|> -TITLE: Homotopy domination of a wedge of two polyhedra -QUESTION [7 upvotes]: The topological space $A$ is called homotopy dominated by the space $X$ if there are maps $f:A\longrightarrow X$ and $g:X\longrightarrow A$ so that $g\circ f\simeq id_A$. -Question: Suppose that $X_1$ and $X_2$ are two polyhedra. If $A$ is homotopy dominated by $X_1\vee X_2$, then is $A$ of the form $A_1 \vee A_2$ (up to homotopy equivalent) where $A_i$ is homotopy dominated by $X_i$ for $i=1,2$? - -REPLY [6 votes]: The answer here is certainly yes under many sets of mild side hypotheses. -For example, once upon a time, I wrote a paper with Frank Adams (!) that seems of some relevance: [J.F.Adams and N.J.Kuhn, Atomic spaces and spectra, Proc. Edin. Math. Soc 32 (1989), 473-481]. We show that if $X$ is a space or spectrum that is $p$--complete and of finite type, then the monoid of homotopy classes of based self maps $[X,X]$ is a profinite monoid. In this case, one concludes that, if $X$ has no nontrivial retracts then every self map is either invertible or topologically nilpotent. -Let's apply this to the stated question, under the hypotheses $A$ has no nontrivial retracts and is complete of finite type. Consideration of homology shows that at least one of the two maps $A \rightarrow X \rightarrow X_i \rightarrow X \rightarrow A$ is not topologically nilpotent, and so must be an equivalence. Thus $A$ is a retract of either $X_1$ or $X_2$, i.e. the question has an affirmative answer in this case. -More generally, you are asking something closely related to a Krull-Schmidt type theorem for spaces: if a space $X$ is written as a wedge of `indecomposable' spaces in two different ways, must the pieces correspond? Issues here include: need a space be written in this way? and also: What is the difference between retracts and wedge summands? -If $X_1 \vee X_2$ is a suspension, and suitably complete, then certainly a Krull-Schmidt theorem holds and the answer is yes. In the more algebraic world of spectra, some of us were using these ideas all the time in the early 1980's in papers about stable splittings of classifying spaces.<|endoftext|> -TITLE: Separable bidual but nonseparable third dual -QUESTION [11 upvotes]: Does there exist a Banach space $X$ such that $X^{**}$ is separable but $X^{***}$ is non-separable? -More generally, for every natural $n$ can someone construct an example of Banach space $X$ such that $X^{n}$ is separable but not $X^{n+1}$? - -REPLY [10 votes]: Yes to both. Lindenstrauss extended James' construction to show that for any separable $X$ there is a separable $Y$ s.t. $Y^{**}/Y$ is isometrically isomorphic to $X$. Induct on that. Spaces built that way are called James-Lindenstrauss spaces. Another proof is contained in my "Factoring Weakly Compact Operator" paper with Davis, Figiel and Pelczynski.<|endoftext|> -TITLE: Can "ampleness" be detected inside the derived category? -QUESTION [12 upvotes]: Let $X$ be an algebraic variety (separated quasi-compact scheme of finite type) over a field $k$. -One of the possible definitions of an ample line bundle goes as follows: - -Def 1: A line bundle $\mathcal{L}$ on $X$ is said to be ample iff some tensor power of it $\mathcal{L}^{\otimes k}$ admits $n+1$-generating sections (for some $n$) s.t. the associated morphism $X \to \mathbb{P}^n$ is a closed embedding (then $\mathcal{L}^{\otimes k}$ is said to be very ample). - -I always found this definition rather subtle and mysterious. The classical story goes through showing that this definition is equivalent to the following one (which is manifestly much more useful in practice and much less easy to check): - -Def 2: A line bundle $\mathcal{L}$ on $X$ is said to be ample if for every coherent sheaf $\mathcal{F}$ there exists some $n>0$ (depending on $\mathcal{F}$) such that for all $m>n$ the sheaf $\mathcal{L}^{\otimes m} \otimes_{\mathcal{O}_X} \mathcal{F}$ is generated by global sections (i.e. is a quotient of a trivial vector bundle). - -The proof I know of this equivalence is subtle and goes through a reduction argument to the projective case and using serre vanishing (notice that may be why the relation between $k$ in the first definition and $n$('s) in the second is highly indirect). -Let $QCoh(X)$ denote the derived (stable $\infty$-)category of sheaves of quasi-coherent $\mathcal{O}_X$-modules with symmetric monoidal structure given by $\otimes_{\mathcal{O}_X}$. -Given this structure we can easily to detect (shifted) line bundles inside $QCoh(X)$ as those are given by the $\otimes$-invertible objects. Here's the question: - -Questions: Given a (shifted-)line bundle $\mathcal{L}$ in $QCoh(X)$ can we... (increasing level of difficulty). - -Detect whether $\mathcal{L}$ is ample (in the classical sense above) without "leaving" the derived category $QCoh(X)$? (using - $\otimes$-structure). -Detect whether $\mathcal{L}$ is ample by considering $QCoh(X)$ without the $\otimes$-structure, but remembering the action of - $Pic(X)$ (the $\infty$-picard groupoid of line bundles) on it. -Detect whether $\mathcal{L}$ is very ample without using the $\otimes$-structure at all?. ֿ - -REPLY [3 votes]: Yes, to be ample amounts to be a generator of the derived category taking into account $\otimes$-powers, specifically an invertible sheaf $\mathcal{L}$ is ample if and only if the family -$$ -\{\mathcal{L}^{\otimes t}[n]\,/\, n\in \mathbb{Z}, t \geq 0\} -$$ -generartes de derived category of quasi-coherent sheaves on the scheme. This is explained in greater detail in Neeman's JAMS 1996 "The Grothendieck Duality Theorem via Bousfield’s Techniques and Brown Representability" in the general setting of divisorial schemes. -I guess this answers question 1 at least.<|endoftext|> -TITLE: Orders of reductions of rational points on elliptic curves -QUESTION [10 upvotes]: I am looking for references where the following (or similar questions) have been studied: -Let $K$ be a number field or a function field in one variable over a finite field and let $E$ be an elliptic curve (or more generally, an abelian variety) over $K$. If $x \in E(K)$ is a point of infinite order then the order of its reduction modulo a good prime tends to infinity with the order of the residue field. - -Are there any results that are known about the prime factorisation of the order of the reduction of $x$? For example, is it known that there is an infinite sequence of rational primes $p_i$ and primes $P_i$ of (the ring of integers of) $K$ such that $p_i$ divides the order of the reduction of $x$ modulo $P_i$? - -I would also be interested in similar statements for the order of the group of rational points on the reduction of any elliptic curve $E$ modulo primes of $K$. -(I expect that much stronger results should be true, but don't know the literature in this area.) - -REPLY [4 votes]: For example, is it known that there is an infinite sequence of rational primes $p_i$ and primes $P_i$ of (the ring of integers of) $K$ such that $p_i$ divides the order of the reduction of $x$ modulo $P_i$? - -Yes, a weak and ineffective form of this at least follows from Siegel's integrality finiteness theorem. See the above comments for a reference. -A lot more can be said. Conditionally on the GRH for Dedekind zeta functions, Miri and Murty proved that $|E(\mathbb{F}_P)|$ has at most $16$ prime factors (counting multiplicities!) for $\gg_E X / (\log{X})^2$ of the primes $P$ of norm $N(P) \leq X$. This they did by adapting Chen's method for his almost twin primes theorem; note that the problem of getting infinitely many prime orders of $E(\mathbb{F}_P)$ could be regarded as an elliptic variant of the twin prime problem. See Theorem 5 with the original reference [25] in this paper of Cojocaru, as well as this paper of hers for an unconditional proof in the CM case that $|E(\mathbb{F}_P)|$ is essentially squarefree infinitely often (with the right density in fact).<|endoftext|> -TITLE: Pursuit solutions to the Rock-paper-scissors flow and delay differential equations -QUESTION [6 upvotes]: The Rock-paper-scissors flow is the following reaction-diffusion system -$$r_t = \Delta r + rs-rp,$$ -$$p_t = \Delta p + pr-ps,$$ -$$s_t = \Delta s + sp-sr.$$ -We can assume $r,p,s\geq 0$, $r+p+s$ is constant, and $1=\int_M (r+p+s) dV$. -Are there travelling wave solutions in one spacial variable of the form -$$r(x,t)=r(x-3,t)=p(x-2,t)=s(x-1,t)=u(x-vt)?$$ -Here $u(y)\geq0$ is some function and $v>0$ is some constant. Such a solution would represent distributions of rock-paper-scissors which pursue each other around a circle. -It suffices to find a non-trivial, three periodic solution to the following non-linear delay differential equation -$$u''(y)+vu'(y)+u(y)(u(y-2)-u(y-1))=0.$$ -Most of the references I've found on DDEs are very numerical in flavor or cover DDEs in rather specific forms. - -REPLY [4 votes]: Damn it! Such a nice equation with such an interesting behavior that I just started to understand after a few days of thinking, and the stupid positivity condition spoils all the fun. Why cannot we just trade this world for one in which burning a negative amount of fuel in a negative amount of oxidizer would produce a positive amount of heat? -Anyway, write $a(x)=u(x), b(x)=u(x-1), c(x)=u(x-2)$. Then $a''+va'=a(b-c)$, etc. If $a,b,c>0$, we can divide and write -$$ -\frac{a''}a+v\frac{a'}a=b-c\, -$$ -i.e. -$$ -(\log a)''+\frac{(a')^2}{a^2}+v(\log a)'=b-c\,, -$$ -etc. -Adding these three equations up and integrating over the period, we conclude that -$$ -\int\left(\frac{(a')^2}{a^2}+\frac{(b')^2}{b^2}+\frac{(c')^2}{c^2}\right)=0\,, -$$ -so $a,b,c=\operatorname{const}$. -Case closed to the total dissatisfaction of everyone involved. :-(<|endoftext|> -TITLE: Odd function on the 2-sphere whose integrals over all hemispheres is zero -QUESTION [5 upvotes]: Let $h:\mathbb{S}^2 \to \mathbb{R} $ be a smooth function satisfying: - -$h(-x)=-h(x)$ -For every hemisphere $A \subseteq \mathbb{S}^2$, $\int_{A}h\text{Vol}_{\mathbb{S}^2}=0$, where $\text{Vol}_\mathbb{S}^2$ is the standard volume form on the sphere. - -I am interested in finding an elementary proof that $h=0$. (Without relying on the invertibility of the Funk transform, see details below on the connection of this to the problem above). -Edit: As commented by Alex Degtyarev, if we assume $h$ is everywhere non zero, then it's trivial. (In that case $h$ has constant sign, so assumption $1$ alone immediately implies $h=0$). -Motivation: -Let $\omega$ be a 2-form on $\mathbb{S}^2$ with the property that the induced area of all the hemispheres is the same. -I want to find an elementary proof that $\omega$ is invariant under the antipodal map, i.e $f^*\omega=\omega$, where $f(x)=-x$. -Denote $$V=\{ \omega \in \Omega^2(\mathbb{S}^2) \, | \, \int_{A}\omega=\int_{A}f^*\omega \, \text{ for every hemisphere $A \subseteq \mathbb{S}^2$} \},$$ -$$W=\{ \omega \in \Omega^2(\mathbb{S}^2) \, | \, \omega=f^*\omega \}.$$ -We want to prove $V \subseteq W$. Let $\omega \in V$, and define $\tilde \omega:=\omega-f^*\omega$. Since $V$ is a vector space, closed under the operation $\omega \to f^*\omega$, we have $\tilde \omega \in V$. Note that $f^*\tilde \omega=-\tilde \omega$, and that we need to show $\tilde \omega=0$. -Thus, the problem is equivalent to the following: - -Let $\omega \in V$, satisfy $f^*\omega=-\omega$. Then $\omega=0$. - -The assumptions imply $\int_A \omega=0$ for every hemisphere $A$. Writing $\omega=h\text{Vol}_{\mathbb{S}^2}$, we obtain the formulation of the question as stated in the beginning. -Edit: If we assume $\omega$ is non-degenerate (i.e everywhere non-zero), then the question becomes trivial: In that case $h$ has a constant sign, hence must be zero due to the property $h(-x)=-h(x)$. -It turns out that using flows by Killing fields, one can reduce this problem to the invertibility of the Funk transform, but this is a non-elementary result which I prefer to avoid. -(Essentially the idea is that if $\int_A\omega=0$ on any hemisphere, then $\int_{A}L_X\omega=0$ for every Killing field $X$). For details see here and here. - -REPLY [4 votes]: This is Lemma 6.2 in -Gonzalez, Fulton B.; Kakehi, Tomoyuki, Dual Radon transforms on affine Grassmann manifolds, Trans. Am. Math. Soc. 356, No. 10, 4161-4180 (2004). ZBL1049.44001. -(actually, the lemma is for arbitrary dimension, so the special functionology might be simplified further for $\mathbb{S}^2$) -For convenience here is the Lemma (complete with proof):<|endoftext|> -TITLE: Why are coherent sheaves on $\Bbb P^1$ derived equivalent to representations of the Kronecker quiver? -QUESTION [17 upvotes]: I'm looking for an explanation or a reference to why there is this equivelence of triangulated categories: $${D}^b(\mathrm {Coh}(\Bbb P^1))\simeq {D}^b(\mathrm {Rep}(\bullet\rightrightarrows \bullet))$$ -It is my understanding that the only reason why $\Bbb P^1$ appears at all is because it is used to index the regular irreducible representations of the Kronecker quiver. I have also heard that this equivalence can be used to understand the geometry of $\Bbb P^1$. -I suppose more generally, adding arrows to the Kronecker quiver gives a similar result for $\Bbb P^n$. - -REPLY [8 votes]: I'd just like to say thank you for the answers I have gotten for this question. After doing some more reading I have come across a few results which I shall summarise here. - -Beilinson [1] gives the result that for projective space $\Bbb P^n$ the line bundles $\mathcal O_{\Bbb P^n}, \dots,\mathcal O_{\Bbb P^n}(n)$ freely generate $D^b(\mathrm{Coh}(\Bbb P^n))$. -Bondal [2] generalises this by observing that if line bundles $L_0,\dots,L_r$ freely generate $D^b(\mathrm {Coh} (X))$, then there is an equivilence of triangulated categories given by $${\operatorname{{\bf R}Hom}}_{\mathcal O_X}(\oplus_i L_i, -):D^b(\mathrm {Coh} (X))\longrightarrow D^b(\mathrm {mod}_A)$$ -where $\mathrm {mod}_A$ is the category of finitely generated right modules over the algebra $A = \mathrm{End}(\oplus_i L_i)$. -King [3,4] shows that Hirzebruch surfaces $\Bbb F_n $ and the smooth Fano toric surfaces have a collection of line bundles that generate $D^b(\mathrm{ Coh}(X))$ as above. To do this, King showed that each of these toric surfaces is a fine moduli space of $\theta$-stable representations of a quiver with relations $Q$ whose path algebra is the endomorphism algebra $A$. The tautological bundles of the moduli space are the line bundles that generate $D^b(\mathrm{Coh}(X))$. -Craw-Smith [5] generalise King's construction to projective toric varieties of arbitrary dimension. They show that $X$ is isomorphic to a component of the fine moduli space of $\theta$-stable representations of a quiver with relations. (I think this quiver is the Bondal quiver?) -Craw-Winn [6] further generalise this result to Mori-dream spaces. - -References: -[1] A. Beilinson. Coherent sheaves on $\Bbb P^n$ and problems in linear algebra. Funktsional. Anal. i Prilozhen., 12 (1978) 68-69. -[2] A. Bondal. Representations of associative algebras and coherent sheaves. Izv. Akad. Nauk SSSR Ser. Mat. 45 (1989) 25-44. -[3] A. King. Moduli of representations of finite-dimensional algebras. Quarterly J. Math. Oxford 45 (1994) 515-530. -[4] A. King. Tilting bundles on some rational surfaces. Unpublished article available from Alastair King's homepage http://people.bath.ac.uk/masadk/papers/ (near the bottom), (1997). -[5] A. Craw and G.G. Smith. Projective toric varieties as fine moduli spaces of quiver representations American Journal of Mathematics, 130 (2008) 1509-1534., arXiv:0608183 -[6] A. Craw and D. Winn. Journal of Pure and Applied Algebra 217 (2013) 172-189 arXiv:1104.2490 -Interestingly most of the authors names sound like mine.<|endoftext|> -TITLE: Enumerating all arrangements of intervals with given lengths -QUESTION [10 upvotes]: Suppose I am given a set of $n$ intervals, each having length $\ell_i$. Is there a bound on the number of possible orderings of their left and right endpoints? For example, if each interval is represented by $[x_i,y_i]$, with $y_i-x_i=\ell_i$, then one possible ordering would be $x_1\leq x_3\leq y_1\leq x_2 \leq y_2 \leq y_3$. If the lengths are not fixed, then it is clear that the number of orderings is $(2n)!/2^n$. Does fixing the lengths permit a significantly better upper bound? - -REPLY [7 votes]: Edit My criticism of Richard's post no longer applies after he edited it drastically, so I edited it out :-) -First, note that it is enough to bound the number of compatible pairs of permutations of $x$ and $y$ separately. Indeed, since we have at most ${2n\choose n}=e^{O(n)}$ insertions of those orders, we can that get the whole thing up to an exponential error, so if we are not after an exact asymptotics, we do not lose much this way. -To do it, we will just play with collision graphs. The idea is pretty simple: start moving intervals around without changing any of the two relevant orders until left ends or right ends of something nearly collide, after which move the tangled piece together until some more similar ends collide somewhere, and so on until we bring the configuration to a single piece. The graph is just the set of some disjoint red cliques corresponding to bunches of $x$-s that are all squashed nearly together and blue cliques corresponding to $y$'s. Note that the entire graph is connected (using both colors), so once we know the graph and the resolution of all cliques, we know the entire configuration. Suppose also that the lengths are generic (say, independent over $\mathbb Q$). If we are interested just in the strict orders, that can be done without any loss of generality. Our collision graph is then just a pair of disjoint clique graphs. Let us assume that the red graph has $k_j$ $j$-cliques and the blue graph $\ell_j$ $j$-cliques, so -$$ -\sum_j jk_j=\sum_j j\ell_j=n\,. -$$ -Now note that there are $\frac{n!}{(k_1!k_2!\dots)(1!)^{k_1}(2!)^{k_2}\dots}$ possible red graphs and, thereby, $\frac{n!}{k_1!k_2!\dots}$ resolved red graphs. Note also that such red graph has $\sum k_j$ components. Suppose that the red graph is given. Let us see where we can place the blue graph drawing its cliques in some natural order. Observe that when we do the first clique, we have to take all vertices from differentconnected components (otherwise we get a non-trivial relation between lengths) and once we draw it, we create a big stiff component that cannot have any more blue edges. Also, the number of components drops by the size of the clique minus one. The same argument (only the last sentence is relevant) applies to the second clique and so on. Thus, if the red and the blue graphs were compatible at all, we must have $n-\sum_j\ell_j\le \sum_j k_j$ (the component reduction should not exceed the number of components available from the beginning), i.e., $n\le\sum_j\ell_j+\sum_j k_j$ is a necessary compatibility condition. -Now we just forget anything except this last observation and write -$$ -\frac{n!}{k_1!k_2!\dots}\frac{n!}{\ell_1!\ell_2!\dots}\le \frac{n!^2}{n^{n}}\frac {n^{k_1+k_2+\dots}}{k_1!k_2!\dots}\frac {n^{\ell_1+\ell_2+\dots}}{\ell_1!\ell_2!\dots} -$$ -We need to sum such expressions over all $k$'s and $\ell$'s satisfying $\sum_j jk_j=\sum_j j\ell_j=n$. This amounts to the square of the $n$-the Taylor coefficient of $e^{nz}e^{nz^2}e^{nz^3}\dots=e^{\frac {nz}{1-z}}$. The Cauchy bound with the radius $1/2$ (yeah, one can optimize here, but we have already missed our chance for the exact asymptotics when deciding to separate the orders, so why to care now?) gives $(2e)^{2n}$ and the final estimate of -$$ -\frac{(2n)!}{n^n}(4e^2)^n=n!e^{O(n)} -$$ -The end :-)<|endoftext|> -TITLE: What is the "serious" name for the topograph (for a quadratic form) -QUESTION [20 upvotes]: One way to study (mixed signature) quadratic forms in two variables is to study the topograph. Looks like the signature doesn't matter: here is (1,1) and (1-,1). -The name is derived from τοποσ (Greek: "place") and γραφή ("writing"). I read that if you're really good at reading topographs you can extract information like the genus, class number, solve the Pell equation, and more. - -There are two resources I found for topographs: - -The Sensual Quadratic Form, John H Conway -Topology of Numbers, Allen Hatcher - -They are basically drawing the dual tree of the Farey Tesselation, which is a tiling of $\mathbb{H}$ or $\mathbb{D}$ by hyperbolic triangles. Is there a more serious name for putting trees on $\mathbb{H}$? -This question emerges, for example, trying to draw these things with a computer and I needed to decide a natural place to put the interior vertices, and I couldn't think of one. The "outer" vertices are indexed by $\text{P}\mathbb{Q}^1$ and the interior vertices could be in any reasonable place. -There could be a serious name for this structure, like the Bruhat-Tits building or maybe it's in Serre's book on Trees. Any guidance? - - -A figure similar to the topograph also appears in a discussion of the Bruhat-Tits tree for $\text{SL}(2, \mathbb{Q}_2)$. [notes] - -REPLY [4 votes]: There is a somewhat larger machinery - a generalization of the theory of dessins d'enfants, when drawing the topograph of a binary quadratic form. -The topograph is merely the quotient of the bipartite Farey tree by the translation action of the correct subgroup of the modular group--where correct means the automorphism group of the binary quadratic form in question. The quotient looks like a wheel so we call it a çark (pronounced as chark, the word has a common etymology with Indian chakra, Greek kyklos and English wheel). -Using the çark one can solve many problems around the corresponding binary quadratic form. Details can be found here and here. -There is also an android application here.<|endoftext|> -TITLE: Symplectic mapping class group and the "Lagrangian sphere complex" -QUESTION [13 upvotes]: For a genus $g$ surface $\Sigma_g$, the mapping class group $\mathrm{Mod}(\Sigma_g)$ acts on the curve complex $\mathcal C(\Sigma_g)$: vertices being isotopy classes of essential, nonseparating, simple, closed curves in $\Sigma_g$ with an edge connecting two vertices if the curves can be made disjoint. In fact, there are many cousins of $\mathcal C(\Sigma_g)$ that have been extensively studied. -In one dimension higher, we have the symplectic mapping class group $\mathrm{SMod}(X)$ of some symplectic manifold $X$. We can build a complex $\mathcal L(X)$ in a similar way: the vertices of $\mathcal L(X)$ are the isotopy classes of Lagrangian spheres in $X$ with an edge connecting two if they are disjoint. Is anything known about this complex and/or it's relatives? - -REPLY [2 votes]: I only just noticed this question, so maybe it's too late, but here's an -answer. -Note that some symplectic manifolds (like $\mathbf{CP}^2$) contain no Lagrangian spheres, so -this complex is then empty. Let's ignore that and focus on the other -examples. -For certain simple symplectic 4-manifolds (small blow-ups of the -projective plane or a quadric) this complex should be known explicitly -(I can tell you the vertices and tell you what you need to check to -get the edges). Let $X_k$ be the $k$-point blow-up of -$\mathbf{CP}^2$ with its monotone symplectic form. - -e.g. $S^2\times S^2$ with its monotone symplectic form has -precisely one Lagrangian sphere up to Hamiltonian isotopy (due to -Hind https://arxiv.org/abs/math/0311092), so your complex is just a -point. Same for $X_2$ (https://arxiv.org/abs/0902.0540). -e.g. For $X_3$ and $X_4$, Lagrangian spheres are uniquely -determined up to isotopy by their homology classes -(https://arxiv.org/abs/0902.0540 again), which gives: - - -$X_3$: the classes $L_{12}=E_1-E_2$, $L_{13}=E_1-E_3$, -$L_{23}=E_2-E_3$, $L_{123}=H-E_1-E_2-E_3$ (I'm counting the -two orientations of $L$ as the same sphere, so $[L]$ and -$-[L]$ give the same sphere). Just for homology reasons, -$L_{ij}$ must intersect $L_{jk}$ for $i\neq k$. The only -other thing to check is whether $L_{123}\cap L_{ij}$ can be made -empty (no homological obstruction). If so then your complex is a -$D_4$ Dynkin graph: otherwise it's a disconnected set of 4 -points (I didn't think about which case holds, sorry). -$X_4$: similarly we get $L_{ij}=E_i-E_j$ and -$L_{ijk}=H-E_i-E_j-E_k$. The possible edges are $L_{ij}$ to -$L_{k\ell}$ when $\{i,j\}\cap\{k,\ell\}=\emptyset$ and -$L_{ijk}$ to $L_{ij}$. If all these edges exist then your -complex is obtained from a tetrahedron by putting $L_{ijk}$ at -the four vertices and $L_{ij}$ on the edge between $L_{ijk}$ -and $L_{ij\ell}$, then connecting opposite edges. - -e.g. for $X_5$ you have similar homological data, but you can also -focus on a single homology class and get infinitely many knotted -spheres (as observed by Seidel -https://arxiv.org/abs/math/0309012). These are nonetheless -classified (Borman-Li-Wu, https://arxiv.org/abs/1211.5952): the -symplectic Torelli group acts transitively on them, and in this case -the symplectic Torelli group is the pure spherical 5-strand braid -group (https://arxiv.org/abs/0909.5622). I haven't thought about how -the Lagrangian "curve complex" would work in this example, but my -guess is it would be closely related to the curve complex on the -punctured sphere, in the following way. All the spheres are -vanishing cycles for algebraic degenerations of $X_5$ coming from -moving the 5 blow-up points. As three blow-up points become -collinear, the blow-up develops a holomorphic $-2$-curve (proper -transform of the line through these three points). If you take the -anticanonical model then you get a nodal variety (and the Lagrangian -sphere is the vanishing cycle of this node). To get 5 points on -$S^2$ from this, look at the conic through the 5 blow-up -points. You get a holomorphic sphere with 5 marked points. As the -node forms, two of these marked points collide, and you can -associate to the Lagrangian $S^2$ the simple closed curve which -encircles these two marked points. If you can form two nodes -simultaneously then these pairs of points collide simultaneously so -the simple closed curves will be disjoint (as will the vanishing -cycles). - -In higher dimensions, little is known. However, by focusing attention -on your favourite collection of spheres and computing Floer cohomology -you should be able to figure out subcomplexes. Perhaps a more -natural/manageable/useful construction here replaces vertices of the -complex with quasiequivalence classes of objects (maybe spherical objects if you want to specify that) in the Fukaya category, joined by an edge when their Floer -cohomology vanishes. This still admits an action of the symplectic -mapping class group (though factoring through the autoequivalence group of the Fukaya category) which is what you'd probably want to use this complex for anyway.<|endoftext|> -TITLE: How to compute cohomology using differentials of the second kind on a Fermat curve? -QUESTION [8 upvotes]: Differentials of the second kind -Gross and Rohrlich in the paper On the periods of abelian integrals and a formula of Chowla and Selberg state the claim below without citation (pg. 198), giving an explicit determination of the cohomology classes of the Fermat curve $X_d := \{x^d + y^d - z^d = 0\} \subset \mathbb{P}^2$ using differentials of the second kind. -Working on the affine chart $\{z=1\} \subset \mathbb{P}^2$ they define the following meramorphic 1-forms on $X_d$: -$$ - \eta_{r,s,t} = x^{r-1}y^{s-d} \mathrm{d} x. -$$ -These forms can be integrated over homology classes, giving the cohomology classes that we need. - -Claim: The middle cohomology of $X_d$ is generated by $\eta_{r,s,t}$ where $0 < r,s,t < d$ and $r+s+t \equiv 0 \,(\text{mod } d)$. - -Example: Let $d=3$. Then, $X_3$ is elliptic and we have $\mathrm{H}^1(X_3,\mathbb{C}) = \langle \frac{\mathrm{d}x}{y^2} , \frac{x\mathrm{d}x}{y}\rangle$. - -Main problem: How does one prove the claim above? - -Perhaps using Griffiths residues -We could try to prove this statement using the Griffiths' residue map: -$$ - \operatorname{res}: \bigoplus_{\ell \ge 1} \mathrm{H}^0(\mathbb{P}^2,\Omega_{\mathbb{P}^2}(\ell X_d)) \to \mathrm{H}^1(X_d,\mathbb{C}), -$$ -which implies that the residues of the following meramorphic forms generate the image: -$$ - \omega_{r,s,t} = \frac{x^{r-1}y^{s-1}z^{t-1}}{(x^d+y^d-z^d)^\ell }\cdot \Omega, -$$ -where $0 < r,s,t < d$, $r+s+t = \ell d$ and $\Omega = x \mathrm{d}y\mathrm{d}z - y \mathrm{d}x\mathrm{d}z + z\mathrm{d}x\mathrm{d}y$. -Therefore it remains to integrate the forms $\omega_{r,s,t}$ over a suitable tubular neighbourhood of $X_d$. Representation theoretic arguments imply that $\operatorname{res} \omega_{r,s,t} = c_{r,s,t} \eta_{r,s,t}$ for a scalar $c_{r,s,t} \in \mathbb{C}^*$, provided Gross and Rohrlich are correct. -However, an explicit computation reveals that the residue of $\omega_{r,s,t}$ is in fact a scalar multiple of $x^{r-1}y^{s-\ell d} \mathrm{d}x$ instead of $x^{r-1}y^{s-d} \mathrm{d}x$. Therefore we can answer Question 1 if we can answer the following. - -Equivalent problem: How can we lower the pole order of $x^{r-1}y^{s-\ell d} \mathrm{d}x$ modulo cohomological equivalence to get $x^{r-1}y^{s-d}\mathrm{d}x$? - -REPLY [2 votes]: If I recall correctly, you can find a proof of the claim in Lang's book "Introduction to Algebraic and Abelian Functions". He has a chapter on the Fermat curve and in fact (after looking at the google preview of the book) I think that the claim is essentially Theorem 2.2 in Chapter II of Lang's book.<|endoftext|> -TITLE: Stable Dold-Kan correspondence and symmetric group actions -QUESTION [5 upvotes]: There exists a Quillen equivalence between $HRModSpectra$ (model category of ring spectra over Eilenberg-MacLane spectra $EM(R)$, where $R$ is a commutative ring, with stable model structure) and $Ch$ (model category of unbounded chain complexes of $R$-modules). -I was wondering what the Quillen functors are that give the above Quillen equivalence. -One can start with an unbounded chain complex $X$ and apply the Dold-Kan functor $\Gamma$ to chain complex $X_{\geq 0}$ to get a simplicial abelian group $\Gamma(X_{\geq 0})$, and then consider $\Gamma(X[-n]_{\geq0})$ (shifting $X$ to the left by n places, truncating and then applying $\Gamma$). This way one gets an $\Omega$-spectrum $Y$ = {${Y_{0}, Y_{1},...}$}, with $Y_{n} = \Gamma(X[-n]_{\geq0})$. -How does then one proceed to prove that $Y$ is a symmetric spectrum? For that one needs an action of symmetric group $S_{n}$ on $Y_{n}$. Now each $Y_{i}$ is in fact as a simplicial set equivalent to $\prod K(\pi_{k}(Y_{n}), k)$ and $K(\pi_{n}(Y_{n}), n)$ has an $S_{n}$ action, though I'm not sure if this action will satisfy the compatibility conditions that are required of a symmetric spectrum. - -REPLY [2 votes]: I'm posting a CW answer so that this doesn't remain open and unanswered, even though the OP said in the comments he figured it out. I just posted an answer to an analogous question and in doing so found this question. The relevant paper is "Stable model categories are categories of modules" by Schwede and Shipley. The construction the OP describes (Dold-Kan plus shift) is on page 39, where it is also shown how to make this into an $HR$-module (via the Alexander-Whitney map). Remark B.1.10 discusses the symmetric group actions. This functor $\mathcal{H}$ takes a chain complexes to a naive $HR$-module, not a symmetric spectrum. It is not an extension of the usual Eilenberg-MacLane functor, $H$. The authors claim that there is no way to make $\mathcal{H}R$ into a symmetric spectrum that is level equivalent to $HR$. This explains why the Quillen equivalence has to zig-zag through naive $HR$-modules.<|endoftext|> -TITLE: Associative mean -QUESTION [9 upvotes]: Can there be a function $m(a,b)$ that is both associative and a mean, i.e., -$\min (a,b) \leq m(a,b) \leq \max (a,b)$? The obvious solutions are $m(a,b) = \max(a,b)$ or $\min(a,b)$, but are there more? - -REPLY [3 votes]: Let us, indeed, describe "monotone, continuous, associative means" on $[0,1]^2$. I'll just write $a*b$ instead of $m(a,b)$ to make long expressions easier to comprehend. Let $c=0*1,d=1*0$. Then, for every $x\le c\le y$, we have -$c\le c*y=(0*1)*y=0*(1*y)\le 0*y\le 0*1=c$ and, thereby, $c=0*y\le x*y\le c*y=c$, i.e., $x\le c\le y$ implies $x*y=c$. -Let now $a$ be any number with $q=0*aa$ such that $q'=0*a'>q$ at all, then, by continuity we can find such $a'$ with $q' -TITLE: Spherical and Wonderful varieties -QUESTION [6 upvotes]: A spherical variety is a normal variety $X$ together with an action of a connected reductive affine algebraic group $G$, a Borel subgroup $B\subset G$, and a base point $x_0\in X$ such that the $B$-orbit of $x_0$ in $X$ is a dense open subset of $X$. -A wonderful variety is a smooth complete variety $X$ with the action of a semisimple simply connected group $G$ such that there is a point $x_0\in X$ with open $G$ orbit and such that the complement $X\setminus G\cdot x_0$ is a union of prime divisors $E_1,\cdots, E_t$ having simple normal crossing, and such that the closures of the $G$-orbits in $X$ are the intersections $\bigcap_{i\in I}E_i$ where $I$ is a subset of $\{1,\dots, t\}$. -Now, fix a connected reductive affine algebraic group $G$ and a Borel subgroup $B\subset G$. Could there exist two non isomorphic smooth complete varieties that are spherical with respect to $(G,B)$ and wonderful with respect to $G$ ? - -REPLY [7 votes]: The only groups which act on only one wonderful variety are tori (with $X$ being a point). All other admit at least $X=G/B$ and $X=G/G$. -If one fixes the open $G$-orbit then there is at most one wonderful completion (Luna-Vust, Luna). -It is known that the number of wonderful varieties for $G$ is finite (work of Alexeev-Brion). By now, they are all classified (work of mostly Luna, Losev, and Bravi-Pezzini). The result is way too complicated to describe it here. Bravi and Luna have written a nice survey on the case where $G=F_4$.<|endoftext|> -TITLE: Rank-constrained least-squares solution of the Sylvester matrix equation -QUESTION [5 upvotes]: For the Sylvester matrix equation $AX+XB=C$, I want to find the least-squares solution $X$ via -$$\begin{array}{ll} \text{minimize} & \| AX + XB - C \|_{\text{F}}^2\\ \text{subject to} & \mbox{rank} (X) \leq k\end{array}$$ -I searched broadly online but couldn't find any literature on this. When dealing with a rank constraint, I think that we will use the Eckart-Young Theorem somewhere. However, if we solve the Sylvester equation without the rank constraint using the Kronecker product, it's hard to deal with the rank of $X$ when $X$ is in vectorized form. I also tried to do SVD on $A$ and $B$ but could not proceed either. -Does anyone have any idea about this problem? Or is there literature on this already? Thanks! - -REPLY [2 votes]: Check out the work of Beckermann and Townsend. -Beckermann, Bernhard; Townsend, Alex, On the singular values of matrices with displacement structure, ZBL06803120.<|endoftext|> -TITLE: Counting twin primes efficiently -QUESTION [5 upvotes]: This question, as well as its answers and comments, highlights a lot of unsettling numerical coincidences where certain sums over twin primes ostensibly converge to all kinds of weird values, however most of them look fishy in some aspects such as convergence rate (see comments by Terry Tao). To shed more light on this, it would be helpful to obtain further values of these (partial) sums, still linear computation of these quickly becomes prohibitive. -A simpler related problem is to count the number $f(N)$ of twin primes in a range $[1, N]$ (basically, summing $1$ instead of other stuff over primes $(p, p + 2) \leq N$). The question is: which algorithms would allow us to compute this number in $O(N^{\alpha})$ time for $\alpha < 1$? What if we want to approximate $f(N)$ with (arbitrary) relative tolerance $\delta$? -Just for the reference point, there are quite a few sublinear algorithms for counting primes in a range; good places to look are here, here and here. It is not clear how much of these results can carry over to twin prime counting though. - -REPLY [5 votes]: This is a well-studied problem - so well-studied that the well-known Pentium division bug was a by-product. See Nicely's web page.<|endoftext|> -TITLE: Intuition for torsion of a chain complex and application to lens spaces -QUESTION [7 upvotes]: I have read a bit about the torsion of an acyclic complex. One of my concrete hopes was that I could understand why $L(7,1)$ and $L(7,2)$ are not homeomorphic - I am under the impression that classifying lens spaces was I problem that motivated Reidemeister to introduce torsion. -All of the definitions of torsion that I have seen are totally opaque to me. How do people think of the torsion of a chain complex and how in trying to classify lens spaces could I have been led to defining/computing torsions? - -REPLY [12 votes]: Consider the special case of the simplest complex of real vector spaces $\newcommand{\pa}{\partial}$ -$$0\to U_0 \stackrel{\pa}{\to} U_1\to 0.$$ -(Ultimately everything can be reduced to this simple situation via some algebraic tricks.) -This complex is acyclic iff $\pa$ is an isomorphism. By chossing bases in $U_0$ and $U_1$ appropriately we can represent $\pa$ as the identity matrix. -Assume this complex is acyclic and set $n=\dim U_0=\dim U_1$. -The torsion arises when $U_0$ and $U_1$ have additional data attached to them. Assume that $L_0$ and $L_1$ are lattices in $U_0$ and respectively $U_1$ such that $\pa(L_0)\subset L_1$. (These are finitely generated Abelian subgroups that span their respective ambient spaces. $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bR}{\mathbb{R}}$ Think of the subgroup $\bZ^n$ of $\bR^n$.) -A $\bZ$-basis $\newcommand{\ue}{\underline{\mathbf{e}}}$ $\newcommand{\uf}{\underline{\mathbf{f}}}$ $\ue_i$ of $L_i$ is also an $\bR$-basis of $U_i$. By choosing $\bZ$-bases $\ue_0$, $\ue_1$ of $L_0$ and respectively $L_1$ we can represent $\pa$ as an $n\times n$ matrix $M(\pa,\ue_0,\ue_1)$. -Observe that if $\uf_0$ and $\uf_1$ are other $\bZ$-bases of $L_0$ resp $L_1$, then -$$|\det M(\pa,\ue_0,\ue_1)|=|\det M(\pa,\uf_0,\uf_1)|=|L_1/\pa L_0|, $$ -where $|S|$ denotes the cardinality of the set $S$. -We see that if the associated complex is obtained from a complex of Abelian groups we can associated an invariant, the above determinant, or the order of the quotient $L_1/\pa L_0$. The torsion of this complex is the defined (up to a sign) to be the number -$$\tau(\pa,L_0,L_1):=\pm \frac{1}{\det M(\pa,\ue_0,\ue_1)} =\pm \frac{1}{|L_1/\pa L_0|}. $$ -The chain complexes that appear in topology often come equipped with such lattices. Think of simplicial homology with local coefficients, or the chain complex associated to a $CW$-decomposition. You get One such invariant for every triangulation and every choice of local coefficients. For the Reidemeister torsion, the local coefficients are Abelian and correspond to group morphisms $\pi_1\to \mathbb{C}^*$. -It turns out that for smooth manifolds the resulting invariant is independent of the triangulations used to define the torsion. -This is only the beginning of the story and I have omitted many important details. It should help you navigate the first part of my book on torsion where you will find many other descriptions and applications.<|endoftext|> -TITLE: (Pre)orientation vs. formal completion -QUESTION [11 upvotes]: Let $\mathbb G$ be an abelian vatiety over an $\mathbb E_\infty$-ring $A$. That is to say, it consists of an abelian group object in the $\infty$-category of relative schemes $\mathbb G\to \operatorname{Spét} A$ which are flat (+ maybe more conditions). Denote its underlying classical abelian variety over $\pi_0A$ by $\mathbb G_0$. -Recall from Lurie's "Survey of Elliptic Cohomology": -Definition: A preorientation on $\mathbb G$ is a map of abelian topological groups -$ -\mathbf{CP}^\infty\to \mathbb G(A) -$ -or equivalently a map of formal spectral group schemes -$ -\operatorname{Spf}A^{\mathbf {CP}^\infty}\to\mathbb G, -$ -or equivalently yet an element of $\pi_2\mathbb G(A)$. -Viewed as a map $S^2\to\mathbb G(A)$, a preorientation induces through some adjunctions and restriction to $\pi_0$ a map $\beta :\omega\to \pi_2$, where $\omega$ denotes the invariant differentials of $\mathbb G_0$ over $\pi_0A$. -Question: In what way are the following two conditions related? -A) The preorientation map exhibits an equivalence $\operatorname{Spf}A^{\mathbf {CP}^\infty}\simeq \widehat{\mathbb G}$, where the RHS is the formal completion of $\mathbb G$ at the identity. -B) The preorientation is an orientation, in the sense that - -The map of underlying ordinary schemes $\mathbb G_0\to \operatorname{Spec} \pi_0A$ is smooth of relative dimnension $1$. -For all $n,$ the composition $$\pi_nA\otimes_{\pi_0A}\omega\xrightarrow{\operatorname{id}\otimes\beta}\pi_nA\otimes_{\pi_0A}\pi_2A\to\pi_{n+2}A,$$ -where the unlabeled arrow is the multiplication in the graded ring $\pi_*A$, is an isomorphism. - -Lurie seems to assert in "Survey" that they are equivalent, or at the very least that B) implies A). I would be very happy if somebody could explain why that is. -Remark: -A surely relevant fact is that for the classical formal group $\mathbb G_0 = \operatorname{Spf}\pi_0\left(A^{\mathbf {CP}^\infty}\right)=\operatorname{Spf}A^0(\mathbf {CP}^\infty)$, the $n$-th tensor power of its module of invariant differentials $\omega^n$ is isomorphic to $\pi_{2n}A$, as proved e.g. in Rezk's notes. But I don't see how this shows that B) $\Rightarrow$ A). -Namely, I feel like this should be saying something about the $\operatorname{Spf}A^{\mathbf{CP}^\infty}$ and its module (spectrum) of invariant differentials, but all I see is a statement about (tensor powers of) the module of invariant differentials of its underlying classical counterpart. -Any help will be warmly appreciated! - -REPLY [3 votes]: Okay, I think I may have figured out how B) $\Rightarrow$ A). Because nobody else has given an answer yet, I'm spelling it out (I hope that's not considered bad form). And if what I wrote doesn't make sense, please point it out. -Condition B) is equivalent to saying that: - -$A$ is even weakly periodic and -$\operatorname{Spf}A^0(\mathbf {CP}^\infty)\cong \widehat{\mathbb G}_0$, i.e. statement A) on the level of underlying classical formal groups. - -For both we use the fact that for formal group $\mathbb G_A:=\operatorname{Spf} A^0(\mathbf{CP}^\infty)$ invariant differentials are -$$ -\omega^n_{\mathbb G_A}\cong \pi_{2n}A. -$$ -The preorientation supplies, upon passage to underlying classical schemes and formal completion, a map of formal groups $\operatorname{Spf}A^0(\mathbf{CP}^\infty)\to \widehat {\mathbb G}_0$. This map is, thanks to assumption 1. of B), an isomorphism iff it induces an isomorphism on invariant differentials, which 2. of B), used for $n=1$, now implies. Then we get weak periodicity of $A$ directly from 2. of B) by using it for arbitrary $n$. -So the claim we're after, to show that B) implies A), is that a map of formal spectral schemes -$$\sigma:\operatorname{Spf}A^{\mathbf{CP}^\infty}\to \widehat{\mathbb G}$$ -is an equivalence iff it induces an isomorphism on the underlying formal groups. -This should I think follow from the flatness assumption on $\mathbb G$, since $A^{\mathbf {CP}^\infty}$ is also flat over $A$ by virtue of $A$ being even weakly periodic. -We already have a map of formal groups $\sigma$, so we just need to show that it is an equivalence. But since we are trying to prove this assertion for formal completions, it suffices to restrict on both sides, i.e. on $\operatorname{Spét}A^{\mathbf{ CP}^\infty}$ and on $\mathbb G$, to étale open affine neighbourhoods of the identity. Affine spectral schemes can always be exchanged for connective $\mathbb E_\infty$-rings by reversing the arrows, and since flatness is an étale-local property, we have reduced to showing that a certain map of flat $\mathbb E_\infty$-$A$-algebras $B\to B'$ is an equivalence. -Here we can use Lemma 7.2.2.17 from "Higher Algebra" which says that, in the presence of flatness, this will happen iff the induced map $\pi_0B\to \pi_0B'$ is an isomorphism. But repeating the arguments of the previous paragraph in reverse, we find this is equivalent to the condition that $\sigma$ induces an isomorphism on the underlying formal schemes $\mathbb G_A\to \widehat{\mathbb G}_0$, which we already know to be the case. Thus $\sigma$ is indeed an equivalence, proving A).<|endoftext|> -TITLE: Numbers of solutions equal on every finite commutative ring -QUESTION [10 upvotes]: Let $X,Y$ be two schemes finite type over $\Bbb Z$, assume $\#X(A)=\#Y(A)$ for every finite commutative ring $A$, then - -must these two schemes be isomorphic ? -What invariants of schemes coincide on $X,Y$ (dimension, cohomology and so on)? -If one is smooth, must the other one be? -Is the condition above equivalent to $\#X(A)=\#Y(A)$ for all but finitely many finite commutative ring $A$? - -The motivation for this problem is from Weil conjecture or Igusa zeta-function, which either count points over finite fields or quotient rings like $\Bbb Z/p^n\Bbb Z$ to get the information of underlying scheme. -Firstly, as every finite ring is Artinian we can only consider finite local rings. If $X$ is a equidimensional smooth scheme finite type over $\Bbb Z$, then by criterion of formally etaleness we know that $\#X(A)=\#X(A/I)(\#I)^{\text {dim} X}$ for every ring $A$ with square-zero ideal $I$ i.e $I^2=0$ if both sides are finite. -(Working locally, we can assume $X \overset{\text{etale}}\rightarrow \Bbb A_\Bbb Z ^{\text {dim} X} \rightarrow \text{Spec} \Bbb Z$, then $X(A)=X(A/I)\times_{\Bbb A_\Bbb Z ^{\text {dim} X}(A/I)} \Bbb A_\Bbb Z ^{\text {dim} X}(A)$ as sets and $\Bbb A_\Bbb Z ^{\text {dim} X}(A) \rightarrow \Bbb A_\Bbb Z ^{\text {dim} X}(A/I)$ is surjective with each fiber has $(\#I)^{\text {dim} X}$ elements). -Hence in above case we have $\#X(A)=\#X(A/m)\prod_{i=1}^{\infty}(\#m^i/m^{i+1})^{\text {dim}X}$ for any finite local ring $(A,m)$, so knowing information over finite fields is enough to decide information over every finite rings in the smooth case. So for equidimensional smooth proper schemes over $\text{Spec} \Bbb Z$ , the condition is equivalent to their Hasse-Weil function equal. -Secondly, two elliptic curves over $\Bbb F_q$ have same zeta-function if and only if they are isogenous. So two isogenous non-isomorphic elliptic curves over $Spec \Bbb Z$ may serve as a counter example, but there is no elliptic curves over $\Bbb Q$ with good reduction everywhere. (Probably the same holds for abelian varieties). In some sense, smooth proper schemes over $\text{Spec} \Bbb Z$ are rare (those connected etale ones are trivial by Minkowski's theorem), so I think we need to consider singular ones in general. -On the other hand, as in Classification of finite commutative rings shows finite rings are rich, we already have lots of finite rings like $\mathbb F_p[x,y] / \langle x^2, xy, y^2\rangle$. Furthermore, there is an estimate for the number of commutative rings of order $≤N$. It is -$exp[\frac{2}{27} \frac{log(N)^3}{(log 2)^2} \; +O(log(N)^{\frac {8}{3}})] \quad for N\to \infty$ -by Bjorn Poonen, and many related interesting theorems is in the article -So in the non-smooth case the condition over finite rings may not be totally determined by finite fields, there may be new phenomenons. This recent article explains some relationships of rational singularity and growth of $\#X(\Bbb Z/p^n\Bbb Z)$. -At last, are there some examples of computation of $\Bbb \#X(A)$ in the non-smooth case? I worked out some cases for $X=\text{Spec} \Bbb Z[x,y]/(y^2-x^3)$ but not completely. -Edit: I am mostly interested in the case $X$ is proper or $X \rightarrow \text{Spec} \Bbb Z$ is surjective, as there are counterexamples using isogenous elliptic curves for the first problem in below's comments. The title contains the words "commutative" because we can also define points over finite rings in general (like $M_n(\Bbb F_q)$) for $X$ finite type over $\Bbb Z$ and ask the same question, which I have no idea to deal with even in the smooth case at present. - -REPLY [6 votes]: No, that is not true. Let $B$ be a finite type scheme over $\text{Spec}\ \mathbb{Z},$ for instance, $B=\mathbb{P}^1_{\text{Spec}\ \mathbb{Z}}.$ Let $X,$ respectively $Y,$ be the relative Proj of the symmetric algebra of a locally free sheaf $E,$ resp. $F,$ on $B.$ If the rank of $E$ equals the rank of $F,$ then $X$ and $Y$ have the same number of points over every finite ring. Yet there is no reason that $X$ should equal $Y,$ e.g., for Hirzebruch surfaces the minimal self-intersection number of an irreducible curve on the surface is an invariant that distinguishes Hirzebruch surfaces.<|endoftext|> -TITLE: Is $\Omega J_{p^n-1}S^2$ commutative up to homotopy? -QUESTION [18 upvotes]: Fix a prime $p\geq 5$ and an integer $n>0$. All spaces in this question are implicitly $p$-localized. Consider the spaces $X=J_{p^n-1}S^2$ (the $p^n-1$'th stage in the James construction $JS^2\simeq\Omega S^3$) and $Y=\Omega X$. These appear naturally in a number of applications. The loop sum operation makes $Y$ into an $H$-space. Is it known whether this is commutative? (Here and elsewhere, "commutative" means "commutative up to homotopy".) -One basic idea is that the loop space of any $H$-space is a commutative $H$-space. However, it is standard that $H^*(X;\mathbb{Q})=\mathbb{Q}[x]/(x^{p^n})$, and it is easy to see that this does not admit any Hopf algebra structure, so $X$ is not an $H$-space. -On the other hand, there is a James-Hopf map $h\colon JS^2\to JS^{2p^n}$. A well-known calculation in cohomology shows that $X$ is the fibre of $h$, so $Y$ is the fibre of $\Omega h$. The domain of $\Omega h$ is $\Omega JS^2\simeq\Omega^2S^3\simeq\Omega^3\mathbb{H}P^\infty$. The codomain is $\Omega JS^{2p^n}\simeq\Omega^2S^{2p^n+1}$. Here $S^{2p^n+1}$ is not a loop space, but it is an old theorem that it admits a commutative product (as does any odd-dimensional $p$-local sphere). Thus, the domain and codomain of $\Omega h$ have some extra commutativity to spare. On the other hand, $h$ is not a loop map, so $\Omega h$ is not obviously a double loop map, so it may be that the extra structure on the (co)domain cannot be brought into play. -One can check that the map $H_*(Y;\mathbb{Z}/p)\to H_*(\Omega^2S^3;\mathbb{Z}/p)$ is injective, and $H_*(\Omega^2S^3;\mathbb{Z}/p)$ is commutative, so there is no obvious primary homological obstruction to commutativity of $Y$. -There is a canonical map $JS^2\to\mathbb{C}P^\infty$ which is a rational equivalence. This restricts to give a rational equivalence $X\to\mathbb{C}P^{p^n-1}$, which in turn gives a rational equivalence $Y\to\Omega\mathbb{C}P^{p^n-1}$. Using the fibration $S^{2p^n-1}\to\mathbb{C}P^{p^n-1}\to\mathbb{C}P^\infty$ one can check that $Y$ is rationally equivalent to $\Omega S^{2p^n-1}\times S^1$ and thus to $K(\mathbb{Q},2p^n-2)\times K(\mathbb{Q},1)$. This has an obvious commutative product, and I think it works out that this is the only possible product up to homotopy. We therefore deduce that $Y_{\mathbb{Q}}$ is commutative, but I do not think that this approach gives useful information integrally. - -REPLY [5 votes]: This was answered in the affirmative by Brayton Gray in his paper Homotopy Commutativity and the EHP Sequence. Specifically he shows that for all $n$ the space $\Omega J_{p^s-1} S^{2n}$ is homotopy commutative for $s\geq 1$ when localised at any prime $p\geq 3$. Moreover he claims to be able to show that $\Omega J_{jp^s-1}S^{2n}$ is homotopy commmutative for $s\geq 1$ and $j\leq p$ odd, although he does not give a full proof. -In the same paper he also obtains results on the homotopy commutativitivy of the classifying space $B_{2n-1,r}$ of the iterated suspension.<|endoftext|> -TITLE: On Siegel mass formula -QUESTION [17 upvotes]: I have asked this question exactly here. The question is as follows: -I am interested deeply in the following problem: -Let $f$ be a (fixed) positive definite quadratic form; and let $n$ be an arbitrary natural number; then find a closed formula for the number of solutions to the equation $f=n$. - -For special case $P_{-4}(x,y)=x^2+y^2$, here gives a closed formula for number of solutions. -Also, you can find another formula for the special cases $P_{-20}(x,y)=x^2+5y^2$ and $P_{-28}(x,y)=x^2+7y^2$ there. -You can finde a close formula here for $P_{-8}(x,y)=x^2+2y^2$, here. -You can finde a close formula here for $P_{-3}(x,y)=x^2+xy+y^2$, here. -Also, you can find the answer for (only finitely many) other forms there, -maybe this helps too. -You can finde a close formula here for $f(x,y,z,w)=x^2+y^2+z^2+w^2$, here. -By a more Intelligently search through the web; you can find similar formulas for the only finite limited number of positive definite quadratic forms. -[I think there exists such an explicit formula at most for $10000$ quadratic forms. Am I right?] - - - -As I have mentioned (I am not sure of it!) only for a finite number of quadratic forms we have such an explicit, closed, nice formula; and this way goes in the dead-end for arbitrary quadratic forms. -So Dirichlet tries to find the (weighted) sum of such representations by binary quadratic forms of the same discriminant. -That formula works very nice for our purpose if the genera contain exactly one form. In the Dirichlet formula, each binary quadratic forms appears by weight one in the (weighted) sum. -More precisely let $f_1, f_2, ..., f_h=f_{h(D)}$ be a complete set of representatives for reduced binary quadratic forms of discriminant $D < 0$; -then for every $n \in \mathbb{N}$, with $\gcd(n,D)=1$ we have: -$$ -\sum_{i=1}^{h(D)} -N(f_i,n) -= -\omega (D) -\sum_{d \mid n} -\left( \dfrac{D}{d}\right) -; -$$ -where $\omega (-3) =6$ and $\omega (-4) =4$ -and for every other (possible) value of $D<0$ we have $\omega (D) =2$. -Also by $N(f,n)$; -we meant number of integral representations of $n$ by $f$; i.e. : -$$ -N(f,n) -:= -N\big(f(x,y),n\big) -= -\# -\{(x,y) \in \mathbb{Z}^2 : f(x,y)=n \} -. -$$ - - -I have heard that there is a generalization of Dirichlet's theorem; for quadratic forms in more variables, due to Siegel. -I have searched through the web; -but I have found only this link : Smith–Minkowski–Siegel mass formula ; -also, I confess that I can't understand the whole of this wiki-article. - -Could anyone introduce me a simple reference in English; for Siegel mass formula? - - - -Also you can find better informations here, and may be here & here. - -REPLY [6 votes]: There are many, many references on quadratic forms. This is a huge area, depending on which way you want to go. One of the main approaches is to construct a theta series associated to your quadratic form whose Fourier coefficients give you the representation numbers you want. These are modular forms, and this is treated in many introductions to modular forms (in fact, I have some notes on this). -One specific reference which I think is nice is the book: - -Topics in Classical Automorphic Forms, by Henryk Iwaniec. - -He discusses the Siegel mass formula in a simple setting and explains how you can use can get formulas for representation numbers in special cases. In general, you can get "explicit formulas" for representation numbers in terms of Fourier coefficients of Hecke eigenforms. You can get nice asymptotics on representation numbers this way, though the precise arithmetic of Fourier coefficients of cuspidal eigenforms is mysterious.<|endoftext|> -TITLE: Given a filtration of a finitely generated module over a noetherian ring that "looks" split, is it split? -QUESTION [11 upvotes]: The following question came up while trying to determine whether the extension problems in a spectral sequence are trivial. -Given a noetherian ring $R$ and a finitely generated $R$-module $M$ with a filtration $M=F_0 \supset F_1 \supset \ldots \supset F_n \supset F_{n+1}=0$ such that $M \cong \bigoplus_{i=0}^n F_i/F_{i+1}$ (abstractly, but I do not want to assume that this isomorphism is induced by the maps in the filtration), is the filtration split in the sense that all the short exact sequences $0 \to F_{i+1} \to F_i \to F_i/F_{i+1} \to 0$ are split? -I can prove this for $R=\mathbb{Z}$, but my proof does not seem to generalize well. In that case, one can work $p$-locally and then show that a short exact sequence $0 \to A \to B \to C \to 0$ of $p$-groups, -$|\{x \in B|ord(x)\text{ divides }p^k\}|\geq |\{x \in A|ord(x)\text{ divides }p^k\}|\cdot|\{x \in C|ord(x)\text{ divides }p^k\}|$ -, with equality iff the sequence is split, and then use that by the assumed isomorphisms, the number of these elements can never strictly increase. -Steven Landsburgs answer in Do all exact sequences $0 \rightarrow A \rightarrow A \oplus B \rightarrow B \rightarrow 0$ split for finitely generated abelian groups? got me hoping that something like this might hold for finitely generated modules over noetherian rings, but a simple induction argument does not seem to work. -Does anyone know if this generalisation is true, and knows a proof? - -REPLY [7 votes]: Take the direct sum of the short exact sequences -$$0\to F_{i+1}\to F_i\to F_i/F_{i+1}\to0$$ -for $0\leq i\leq n$. -This has the form -$$0\to \bigoplus_{i=1}^n F_i\to \bigoplus_{i=0}^n F_i\to F_0\to 0$$ -and so splits by the linked answer of Steven Landsburg. -A direct summand of a split short exact sequence is split, so -$$0\to F_{i+1}\to F_i\to F_i/F_{i+1}\to0$$ -is split for each $i$.<|endoftext|> -TITLE: Consistency of "the sharp of every set exists" -QUESTION [10 upvotes]: If there exists a Jónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jónsson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals). -So the existence of a proper class of Jónsson cardinals is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound. -Is the exact consistency strength of "for every $x$, $x^\#$ exists" known? -Edit: as François G. Dorais pointed out in a comment, if $\kappa$ is Jónsson then $V_\kappa\models$ "for every $x$, $x^\#$ exists". (we aren't guarenteed that $V_\kappa\models ZFC$ but that doesn't matter consistency-strength-wise), so "there is a Jónsson cardinal" is an upper bound for the consistency strength of "for every $x$, $x^\#$ exists". - -REPLY [5 votes]: In one sense, closure under sharps is itself a standard point in the hierarchy of consistency strengths. Just like the exact consistency strength of "ZFC + measurable" is "ZFC + measurable", so is the case for closure under sharps. However, the value of the consistency strength hierarchy is in terms of the connections it offers, and here we can say more. -Closure under sharps is equivalent to the failure of the covering lemma (alternatively, weak covering lemma) for every $L[x]$. It also equivalent to $\mathbf{\Pi}^1_1$-determinacy in every generic extension of V. The nature of the assertion "for every $x$, $x^\#$ exists" is such that equiconsistent statements tend to be actually equivalent to it. -Jónsson cardinals are equiconsistent with Ramsey. A consistency-wise weaker assertion that is consistency-wise stronger than closure under sharps is existence of $ω_1$-Erdos cardinals. A still weaker assertion (that remains consistency-wise stronger than closure under sharps) is determinacy in level $ω^2$ of the difference hierarchy of analytic sets. Some other (somewhat technical) notions can be found in "On unfoldable cardinals, ω-closed cardinals, and the beginning of the inner model hierarchy" (P.D. Welch, 2004).<|endoftext|> -TITLE: Minimum probability that two Gaussian random variables are small -QUESTION [6 upvotes]: Let $X,Y$ be two centered Gaussian random variables each with variance at most $1$. Note that we do not assume independence. I would like to minimize -$$\mathbb{P}(|X|\leq 1, |Y|\leq 1).$$ -Is it true that the latter quantity is minimized when $X,Y$ are independent and both have variance $1$? - -REPLY [5 votes]: The fraction in the central square is not minimized with independent standard normals. -Start with that distribution. Take a pair of points $(x,y)$ and $(u,v)$ with $|x|<1, |y|<1, |u|>1, |v|>1,\ p(x,y)>p(u,v)$. Reduce the probabilities near those points by $p(u,v)$, and increase the probabilities near $(x,v)$ and $(u,y)$ by the same amount. -Repeating this will maintain the marginal distributions, while decreasing the part of the distribution in the central square from 46.6% to 36.5%. It may also be possible to reduce the percentage further. - -REPLY [5 votes]: The minimum value is simply $2\alpha-1 = 0.365379$ where $\alpha = \Phi(1)-\Phi(-1) = P(|X|<1)$ where $X \sim N(0,1)$. This can be achieved by translating the percentile of $X$ (considering the percentile $\mod 1$) to produce the percentile of $Y$. For example, let $T=\Phi(X)+\alpha \mod 1$ and then $Y=\Phi^{-1}(T)$. Other translations work, too. -This is optimal because $P(|X|<1)= \alpha = P(|Y|<1)$ and $1 \ge P(X \cup Y) = P(X) + P(Y)-P(X \cap Y) = 2 \alpha - P(X \cap Y)$ so $P(X\cap Y) \ge 2\alpha -1$.<|endoftext|> -TITLE: How do I see the equality $57 = 3 \times 19$ geometrically? -QUESTION [16 upvotes]: Consider the finite field ${\bf F}_p$ and its cubic extension ${\bf F}_{p^3}$. The multiplicative group ${\bf G}_m({\bf F}_{p^3})$ contains the multiplicative group ${\bf G}_m({\bf F}_p) \cong {\bf Z}/(p-1){\bf Z}$ as a subgroup. The quotient $A_p = {\bf G}_m({\bf F}_{p^3})/{\bf G}_m({\bf F}_p)$ is an abelian group. On the other hand, ${\bf F}_{p^3}$ can be considered simply as a $3$-dimensional vector space over ${\bf F}_p$, so this quotient is naturally a projective plane ${\bf P}^2({\bf F}_p)$. Its cardinality equals $p^2+p+1$, thus divisible by $3$ for $p=6k+1$. Therefore the group $A_p$ contains a subgroup of order $3$. Its cosets are represented by some triangles in ${\bf P}^2({\bf F}_p)$. Of course all the triangles in the projective plane go to each other under the projective transformations, but they should have some particular geometry with respect to all the markings on the plane coming from the identification of a $3$-dimensional vector space with the field ${\bf F}_{p^3}$. Say, can one speak of triangle centers for this case (at least if one fixes the line on the infinity)? It seems like the least possible case $p=7$, when ${\bf P}^2({\bf F}_p)$ contains $57$ points, is of interest. - -REPLY [3 votes]: I'm not sure if this does what you want, but the subgroup of order $3$ in the additive group $(\mathbb{Z}_{57},+)$ is $\{0,19,38\}.$ -To expand on that, one construction for the plane $\mathbb{P}_{7}$ of order $7$ is to take as points the elements of $\mathbb{Z}_{57}$ with lines $$\ell_k=[k,k+1,k+3,k+13,k+32,k+36,k+43,k+52]$$ for $k \in \mathbb{Z}_{57}.$ Of course the lines are unordered sets but keeping the order might make the cyclic collineation $\phi:x \mapsto x+1$ easier to follow. -The triangle $T=T_0=\{0,19,38\}$ (mentioned above) has $19$ cosets $T_i=\{i,i+19,i+38\}$. -The lines obtained by extending the sides are -$\ell_6=[6,\ \ 7\ ,9,\ \ {\large 19 },{\large 38},42,49,1]$ -$\ell_{25}=[25,26,34,{\large 38 },{\large 0},4,11,20]$ -$\ell_{44}=[44,45,47,{\large 0 },{\large 19},23,30,39]$ -If "the" center of a triangle $\{P,Q,R\}$ in some plane (like $\mathbb{R}^2$ or $\mathbb{P}_{7}$) is "the" solution $C$ to $C+C+C=P+Q+R$ then we see that in $\mathbb{P}_{7}$ (as constructed here) a triangle has three centers. $C,C+19,C+38$ for a unique $0 \leq C \leq 18.$ These are just the vertices of $T_C.$ -So the special triangles could be described those which are the triangle of centers for some triangle. OR as those triangles which coincide with their own triangle of centers.<|endoftext|> -TITLE: Constructive algebraic geometry -QUESTION [40 upvotes]: I was just watching Andrej Bauer's lecture Five Stages of Accepting Constructive Mathematics, and he mentioned that in the constructive setting we cannot guarantee that every ideal is contained in a maximal ideal---since that obviously requires Zorn's Lemma (or is equivalent to Zorn's Lemma, perhaps?)---so we need to approach algebraic geometry from a different point of view, such as locales. -I am curious how algebraic geometry looks from this constructive point of view, and if there are any good references on this subject? -I'm fairly new to constructive mathematics, though I have been lured in by Kock's synthetic differential geometry, and I'm starting to read the HoTT book. - -REPLY [50 votes]: Let me wrote a quick introduction to this idea: -1) Locales -I do not know if you are already familiar with the notion of locale that Andrej is referring to in his talk: They are a small variation on the idea of a topological space, where instead of defining a space by giving a set of points together with a collection of "open subsets" stable under arbitrary unions and finite intersections, one just gives an abstract ordered set of "open subspaces" which should have arbitrary supremums and finite infimums and such that binary infinimum distribute over arbitrary supremum. Such a poset is called a Frame (this is actually the same as a complete Heyting algebra), a morphism of frame is an ordered preserving map which commutes to finite infimums and arbitrary supremums. And locales are defined as just being "the opposite of the category of frames", i.e. a locale is just a frame, and morphisms of locales are morphisms of frame in the other direction. -It is rather easy to go from a topological space to a locales by just remembering the poset of open subspaces, and attaching to a continuous map its action on open subset by pre-image. -Conversely if one has a locale $X$, one can define a "point" of it as a morphism from the locale corresponding to the one point topological space to $X$, there is a topology on this set of points (generated by the element of the frame corresponding to $X$) and these two constructions form an adjunction between locales and topological spaces, which can be shown to be an equivalence between rather large subcategories. More precisely it induces an equivalence between "sober topological spaces" and "spatial locales", where spatial locales are the locales "having enough points" in a precise technical sense. -There are however some locales which have no points at all. At first you can regard them as pathological monster, but it in fact appears that at least some of them are extremely natural and interesting objects (things like the space of "generic real numbers" or "random real numbers" that Andrej mentioned in his talk, or the "space of bijection" between two infinite set of different cardinality whose non triviality is the key points for Cohen forcing to works). -Despite this, classically, locales and topological spaces are very similar (for example, if you are only interested in complete metric spaces or locally compact spaces you will not notice the difference) but the category of locales is overall (arguably) slightly better behaved than the category of topological spaces. -But constructively, it is a lot harder to construct "points" of locales and a lot more locale appears to be non-spatial. This make the gap between topological spaces and locales considerably larger. And in this weaker framework, locales become incredibly better behaved than topological spaces. For example without the law of excluded middle the correct definition of the locale of real number might be non-spatial but is always locally compact, while it is well known that constructively the topological space of real numbers can fail to be locally compact (this is one of the bad things that happen in constructive analysis that Andrej mentioned in his talk). In fact it is known that they agree if and only if the topological space of real number is locally compact. -Similarly, a lot of classical theorem which requiert the axiom of choices become fully constructive when formulated in terms of locales. This the case for example of the Tychonov theorem, the Hahn Banach theorem, or the Gelfand duality. -For example the idea for the Hahn Banach theorem is that instead of asking for the existence of certain linear form or extension of linear form, we construct a space (a locale) of linear form and show that this is not the empty space (even if it might not have points) and that the map between these spaces induced by restricting to a subspace is always a "surjection" in a good localic sense. And this roughly also what happen for the Gelfand duality or the case of ring spectrum that we will discuss below -I highly recommend to have a look to Peter Johnstone excellent non technical introduction paper to the subject "The point of pointless topology" which will expand on what I just said. and probably explains it more clearly. -2) The localic Zariski spectrum -So, the starting idea is that one can construct the Zariski spectrum of a ring, together with its structural sheaf directly as a locale without ever mentioning prime ideal or maximal ideal. For example, one can just says that the poset of open is given by the poset of radical ideal of the ring. One can also give a presentation by generator and relation of the corresponding frame which is more convenient to work with. This is done for example in section V.3 of P.T.Johnstone's book "Stone spaces". And while prime ideal are a little dangerous constructively as they might not exists, ideals are not problematic at all, you always have plenty of them. -This locale we are constructing is still morally the "space of prime ideal of the ring" but the question of whether the ring actually has any prime ideal or not become just the question of whether this space has points or not, which a rather unimportant question in locale theory. -In fact, if you are familiar with the notion of classifying topos, it is the "space of prime ideal" in the sense that it is the classifying space for prime ideal, more precesely for "prime ideal complement", i.e. subset $O$ of the ring such that ($xy \in O \Rightarrow x \in O \text{ and }y \in O$; $0 \notin O$; $1 \in O$ ; $x+y \in O \Rightarrow x \in O \text{ or } y \in O$) which classically are exactly the complements of prime ideal, and construcively are more important because they are the things you need to get a locale ring when localizing. -This let you go through with the definition of a scheme in a constructive setting. It appears than most classical results of algebraic geometry became constructive when formulated using locales instead of spaces (and replacing statement involving existence of prime or maximal ideal by statement about property of this "space of prime ideals" similarly to the Hahn banach theorem above). Unfortunately, except very basic things, these are mostly "folk theorem" and do not very often appears in literature (because let's face it, not many algebraic geometer are interested in constructivisme...). With the exception of the work of Ingo Blechschmidt that I mention below. -In fact (but that is a personal opinion) I've always found that, for someone already familiar with locale theory, the localic treatment of the basic property of the Zariski spectrum are notably simpler than their usual topological treatment. -3) Internal logic -The last step to this story relies on the use of "internal logic". The idea is that if you work internally in (the topos of sheaves over) the localic Zariski spectrum, then you actually have a "prime ideal of your ring" (more precisely, you have a "localizing system" which satisfies the properties to be the complement of a prime ideal mentioned above) and this prime ideal is in some sense the universal prime ideal of your ring (proving a property for this internal prime ideal proves property for all prime ideal of your ring) and a large number of proof that relies on the axiom of choice because they says at some point "let M be a maximal/prime ideal of your ring" (for example, if you prove that some element is nilpotent because it belongs to all prime ideal) become constructive if, instead of choosing a prime ideal, you move to this internal logic and use the universal one. -This produces a quite efficient and automatic method to make lots of results of algebra and algebraic geometry constructive. -This idea of exploiting internal logic in algebraic geometry tend to make everything constructive and has been pushed quite far by Ingo Blechschmidt (you can watch one of his talk here, or read his thesis work here) -To my knowledge the work of Ingo is the only place where you will find a non trivial treatment of algebraic geometry which use this picture. -For another example, another place, completely different from prime ideal, where you will use non-constructive things in algebra or algebraic geometry is when you want to talk about algebraic closure of a fields. But it appears that this point of view also provide a solution. In general you cannot construct a single algebraic closure of a given fields, but what you can instead do is to consider the "space of all algebraic closure of the field" in the sense of theory of classifying topsoes. This times this will not even be a locale, but a more general kind of topos, and it appears that this topos is something that have actually been studied a lot by algebraic geometers: indeed, if you start from some ring $R$ and study the "space of all algebraic closure of all residual fields of $R$" defined in a proper way what you get is nothing else than the "small étale topos of $R$". See the work of Gavin Wraith on the subject.<|endoftext|> -TITLE: When is it easier to work projectively? -QUESTION [8 upvotes]: There are many instances in which theory over $\mathbb{C}$ is cleaner than theory over $\mathbb{R}$. For example, continuously differentiable functions over $\mathbb{R}$ are not necessarily twice differentiable, whereas entire functions over $\mathbb{C}$ are infinitely differentiable with a convergent power series. Also, $\mathbb{C}$ is algebraically closed, and so systems of polynomial equations over $\mathbb{C}$ can by analyzed with Hilbert's Nullstellensatz, whereas the real case requires the more complicated Stengle's Positivstellensatz. -In my experience, there seems to be greater ease in working in projective geometry instead of affine geometry, much like this ease with $\mathbb{C}$ instead of $\mathbb{R}$. For example, in projective space, any two distinct lines intersect at exactly one point, whereas in affine space, it depends on whether the lines are parallel. -What are some examples in which theory over projective space is cleaner than theory over affine space? -I am interested in a wide spectrum of examples, i.e., elementary/deep examples from continuous/discrete spaces involving algrebra/geometry/combinatorics. Presumably, a big list of answers will illustrate the distinguishing characteristics of projective spaces that make them so nice to work with in so many areas. - -REPLY [4 votes]: From Timothy Chow: -Bézout's theorem comes to mind as an elementary example. - -REPLY [2 votes]: From Leo Alonso: -The fundamental theorem of projective geometry, any collineation comes from a semi-linear automorphism, may be stated for affine spaces but in a clumsier form. - -REPLY [2 votes]: From Pooter: -Over $\mathbf{C}$ say, projective space is compact, but affine space is not. There are numerous good properties that are essentially consequences of this: global holomorphic functions are constant, the image of closed subset under a holomorphic map is again closed, (fancier) cohomology groups of coherent sheaves are finite-dimensional vector spaces, and so on. - -REPLY [2 votes]: From Francois Ziegler: -More automorphisms $\rightarrow$ simpler classification of conics or quadrics.<|endoftext|> -TITLE: A short proof for simple connectedness of the projective line -QUESTION [17 upvotes]: The Riemann-Hurwitz formula implies that the projective line $\mathbb{P}^1_K$ over any algebraically closed field $K$ is simply connected (i.e., $\pi_1^{et}(\mathbb{P}^1_K) = 1$; equivalently, if $\phi\colon C\to \mathbb{P}^1_{K}$ is finite etale, then $\deg\phi=1$). -For $K=\mathbb{C}$, this follows from the connection with topology and from the fact that the complex plane $\mathbb{A}^1_{\mathbb{C}}$ is contractable. -In positive characteristic the affine line is not simply connected due to Artin-Schreier covers. -My question is whether there is a short proof for this fact in positive characteristic? - -REPLY [18 votes]: You can deduce this from the classification of vector bundles on $\mathbf{P}^1$. Say $f:C \to \mathbf{P}^1$ is a connected finite etale Galois cover of degree $n$. We must show $n=1$. -The sheaf $E := f_* \mathcal{O}_C$ is a rank $n$ vector bundle on $\mathbf{P}^1$, so we can write it as $E \simeq \oplus_{i=1}^n \mathcal{O}(a_i)$ for some integers $a_i$. As $C$ is connected, we have $h^0(\mathbf{P}^1,E) = 1$, so we must have $a_1 = 0$ and $a_i < 0$ for $i > 1$ (after rearrangement). It then follows that after pullback along any finite cover $g:D \to \mathbf{P}^1$ of smooth connected curves, we still have $h^0(D, g^* E) = 1$ as negative line bundles remain negative after pullback along finite maps, and thus cannot acquire sections. -But $f$ was a finite etale cover, so there is some $g:D \to \mathbf{P}^1$ as above with $g^* C \to D$ is just a disjoint union of $n$ copies of $D$ mapping down (in fact, one may take $g = f$ as $f$ was Galois). But then $h^0(D, g^*E) = h^0(D, \oplus_{i=1}^n \mathcal{O}_D) = n$. The only way this can happen is if $n=1$. - -REPLY [8 votes]: You can apply the following statement to $X = \mathbb{P}^1_K$ and $L = O(1)$ when $K$ is a separably closed field. - - -Let $L$ be a line bundle on a reduced connected scheme $X$ such that $H^{0}(X,\mathcal{O}_X)$ is a separably closed field. Assume that any vector bundle on $X$ is of the form $\oplus_{i=1}^{n} L^{\otimes \lambda_i}$ for some integers $\lambda_1 \geq \dots \geq \lambda_n$, and that $L^{\lambda}$ has no nonzero global section for $\lambda < 0$. Then $X$ is simply connected. - - -Indeed, let $\mathcal{A}$ be a finite etale $\mathcal{O}_X$-algebra. Then $\mathcal{A} = \oplus_{i=1}^{n} L^{\otimes \lambda_i}$ as an $\mathcal{O}_X$-module for some integers $\lambda_1 \geq \dots \geq \lambda_n$. Because $\mathcal{A}$ has a global section (the unit) one must have $\lambda_1 \geq 0$. -If $\lambda_1 > 0$ then the morphism $L^{2 \lambda_1} \rightarrow \mathcal{A} \otimes_{\mathcal{O}_X} \mathcal{A} \rightarrow \mathcal{A} \rightarrow L^{\lambda_i}$ must be zero for each $i$ since $2 \lambda_1 > \lambda_i$. The the multiplication of $\mathcal{A}$ is $0$ on the $L^{ \lambda_1}$-factor, contradicting the reducedness of $\mathcal{A}$. Thus $\lambda_1 = 0$. -Since $\mathcal{A}$ is a finite etale $\mathcal{O}_X$-algebra the discriminant morphism -$$ -\mathrm{det}(\mathcal{A})^{\otimes 2} \rightarrow \mathcal{O}_X -$$ -is an isomorphism, so that $\sum_i \lambda_i =0$. Since $\lambda_{i} \leq 0$ for all $i$ we get that $\lambda_i = 0$ and thus that $\mathcal{A}$ is a trivial vector bundle. In particular the morphism -$$ -H^0(X,\mathcal{A}) \otimes_{H^0(X,\mathcal{O}_X)} \mathcal{O}_X \rightarrow \mathcal{A} -$$ -is an isomorphism. In particular $H^0(X,\mathcal{A})$ is a finite etale $H^0(X,\mathcal{O}_X)$-algebra, hence a trivial one, so that $\mathcal{A}$ is a trivial finite etale $\mathcal{O}_X$-algebra.<|endoftext|> -TITLE: Kontsevich space in positive characteristic -QUESTION [7 upvotes]: What goes wrong if you try to define the Kontsevich space $\overline{\mathscr{M}_{0,n}}(\mathbb{P}^r,e)$ is positive characteristic? -It is a naive question, but I couldn't find much with a google search. I thought I found something saying they weren't DM stacks because you have inseparable maps, but I didn't clearly see the connection. - -REPLY [10 votes]: I am just writing my comments as an answer. The main computations have to do with the cotangent complex of a stable map. I will work with unpointed stable maps for simplicity (the associated cotangent complex is a bit more complicated in the pointed case). -Notation and Hypotheses. Let $k$ be an algebraically closed field. Let $C$ be a geometrically reduced, at-worst-nodal, proper $k$-curve $C$ (or even just a curve with local complete intersection singularities). Let $$u:C\to X$$ be a $k$-morphism. -Definition. For every irreducible component $C_i$ of $C,$ denote by $\{p_{i,j}\}_j$ the finite set of intersection points of $C_i$ with the union of the remaining irreducible components of $C_i.$ The component is contracted if the restriction of $u$ is constant. The component is inseparable if the restriction of $u$ is non-constant, yet the induced $k$-morphism, $C_i\to u(C_i),$ is inseparable. A component is non-hyperbolic if either the arithmetic genus of $C_i$ equals $1$ and $\{p_{i,j}\}$ is an empty set (i.e., $C$ equals $C_i$) or the arithmetic genus equals $0$ and the $k$-scheme $\{p_{i,j}\}$ has length $\leq 2.$ A component is unstable if it is contracted and non-hyperbolic, otherwise it is stable. A component is non-DM if it is inseparable and non-hyperbolic, otherwise it is DM. If every component of $C$ is stable, then $(C,u)$ is a stable map. If every component of $C$ is stable and DM, then $(C,U)$ is a DM stable map. -Proposition. Let $S$ be an excellent, Noetherian scheme. Let $X\to S$ be a locally finitely presented morphism. There is an algebraic $S$-stack, $\mathfrak{M}(X),$ whose objects over every $S$-scheme $T$ are pairs $$(\pi:C\to T,u:C\to X\times_S T)$$ of a proper, fppf morphism $\pi$ of relative dimension $1$ and a $T$-morphism $u$ and whose isomorphisms of objects, $$\phi: (\pi:C\to T,u:C\to X\times_S T) \to (\pi':C'\to T,u':C'\to X\times_S T),$$ are $T$-isomorphism $\phi:C\to C'$ such that $u'\circ \phi$ equals $u.$ There is an open substack $\overline{\mathcal{M}}(X)\subset \mathfrak{M}(X)$ parameterizing precisely those families where every geometric fiber $C_k$ is a connected, reduced, at-worst-nodal curve and $(C_k,u_k)$ is a stable map. The diagonal $\Delta:\overline{\mathcal{M}}(X) \to \overline{\mathcal{M}}(X) \times_S \overline{\mathcal{M}}(X)$ is finite. Finally, there is an open substack $\overline{\mathcal{M}}^{\text{DM}}(X)$ of $\overline{\mathcal{M}}(X)$ parameterizing precisely those families where every geometric fiber $(C_k,u_k)$ is a DM stable map. This is the maximal open substack of $\overline{\mathcal{M}}(X)$ that is a Deligne-Mumford stack. -Proof. The first result follows by Proposition 3.6 of the following. -MR2854858 -de Jong, A. J.; He, Xuhua; Starr, Jason Michael -Families of rationally simply connected varieties over surfaces and torsors for semisimple groups. -Publ. Math. Inst. Hautes Études Sci. No. 114 (2011), 1–85. -The assertion about $\overline{\mathcal{M}}(X)$ is implicit in the original article of Kontsevich. -MR1363062 (97d:14077) -Kontsevich, Maxim -Enumeration of rational curves via torus actions. The moduli space of curves (Texel Island, 1994), 335–368, -Progr. Math., 129, Birkhäuser Boston, Boston, MA, 1995. -For the final assertion regarding $\overline{\mathcal{M}}^{\text{DM}}(X),$ it is useful to use the cotangent complex. The cotangent complex of $u,$ $L_u,$ is a complex of $\mathcal{O}_C$-modules (already for the Zariski topology) concentrated in degrees $(-\infty,0]$ whose cohomology sheaves are coherent, and whose degree $0$ cohomology sheaf $h^0(L_u)$ is naturally isomorphic to the cokernel of the following homomorphism of $\mathcal{O}_C$-modules, $$u^*\Omega_{X/k}\xrightarrow{(du)^\dagger} \Omega_{C/k}.$$ -The Zariski tangent space of the automorphism group scheme of $\mathfrak{M}(X)$ at $[(C,u)]$, i.e., the Zariski tangent space of the fiber of $$\Delta:\mathfrak{M}(X_k)\to \mathfrak{M}(X_k)\times_{\text{Spec}\ k} \mathfrak{M}(X_k),$$ equals the hypercohomology group $$R^0\text{Hom}_{\mathcal{O}_C}(L_u,\mathcal{O}_C[0]).$$ Similarly, the space of first order deformations (in the sense of Rim-Schlessinger) is -$$R^1\text{Hom}_{\mathcal{O}_C}(L_u,\mathcal{O}_C[0]).$$ Finally, an obstruction space is given by $$R^2\text{Hom}_{\mathcal{O}_C}(L_u,\mathcal{O}_C[0]).$$ (I learned from Ragnar-Olaf Buchweitz not to call this group the obstruction group, because there are always many possible obstruction groups, and the one above might not equal the minimal obstruction group.) -Using one of the two $E_2$-stage spectral sequences of hypercohomology, namely the one computing hypercohomology from the Ext groups of the cohomology sheaves of $L_u$, we can write down some exact sequences relating the hypercohomology groups above to other natural cohomology groups. This is, roughly, the analysis in Fulton-Pandharipande where they prove that the stack of stable maps from genus $0$ curves to $\mathbb{P}^n$ is smooth in characteristic $0$. -Even without spectral sequences (and in any characteristic), we have, $$R^0\text{Hom}_{\mathcal{O}_C}(L_u,\mathcal{O}_C[0]) = \text{Hom}_{\mathcal{O}_C}(\text{Coker}(du)^\dagger,\mathcal{O}_C) \subset \text{Hom}_{\mathcal{O}_C}(\Omega_{C/k},\mathcal{O}_C).$$ -Since $C$ is geometrically reduced, the summand of $\text{Coker}(du)^\dagger$ coming from the zero-dimensional components of this sheaf has only the zero homomorphism to $\mathcal{O}_C.$ Thus, the trouble comes from irreducible components of $C$ that are contained in the support of this sheaf. -Now assume that $(C,u)$ is a stable map. For every irreducible component $C_i$ of $C,$ the restriction to $C_i$ of the $\mathcal{O}_C$-module $\Theta_C=\textit{Hom}_{\mathcal{O}_C}(\Omega_{C/k},\mathcal{O}_C)$ is an $\mathcal{O}_{C_i}$-submodule of $$\textit{Hom}_{\mathcal{O}_{C_i}}(\Omega_{C_i/k},\mathcal{O}_{C_i})\left(-\sum_j \underline{p}_{i,j}\right) = \Theta_{C_i/k}\left(-\sum_j \underline{p}_{i,j}\right).$$ For irreducible components $C_i$ of arithmetic genus $\geq 2,$ already $\Theta_{C_i/k}$ has no nonzero sections. If the arithmetic genus is $0$, then $\Theta_{C_i/k}$ has a one-dimensional vector space of sections, but there are no sections that vanish on the $\geq 1$ points $p_{i,j}.$ Finally, if $C_i$ has arithmetic genus $0$, i.e., if $C_i$ is isomorphisc to $\mathbb{P}^1,$ then $\Theta_{C_i/k}$ is $\mathcal{O}_{\mathbb{P}^1}(2)$, but twisting down by the $\geq 3$ points $p_{i,j}$ makes the degree negative. Thus, the hyperbolic components do not contribute to the global sections. In particular, since $u$ is stable, the contracted components of $u$ do not contribute to the Zariski tangent space of the automorphism group scheme. Therefore, the only nonzero contributions must come from non-hyperbolic components of $C$ that are contained in the support of $\text{Coker}(du)^\dagger.$ -For every non-contracted component of $C$ that maps separably to its image in $X$, the morphism $(du)^\dagger$ is generically surjective on that component, so the component is not in the support of $\text{Coker}(du)^\dagger.$ Thus, the only contributing components $C_i$ are the non-DM components. For every non-DM component, there will be nonzero global sections of $\Theta_{C_i/k}(-\sum_j \underline{p}_{i,j})$ that give nonzero elements of the Zariski tangent space of the automorphism group scheme. Thus, the Zariski tangent space of the automorphism group scheme is nonzero if and only if there exists a non-DM component. QED -If we specify also an ample invertible sheaf $\mathcal{O}_X(1)$ on $X,$ then the degree on every bad component $C_i$ of $u^*\mathcal{O}_X(1)$ is a positive integer that is divisible by $p.$ Thus, the total degree of $u^*\mathcal{O}_X(1)$ on all of $C$ is at least as positive as $p$. So for a finite type, polarized scheme $(X_R,\mathcal{O}_X(1))$ that is defined over a finitely generated $\mathbb{Z}$-algebra $\textbf{R}$ that is an integral domain with characteristic $0$ fraction field, for every fixed integer $e$, for the $R$-stack of stable maps with the degree of $u^*\mathcal{O}_X(1)$ bounded above by $e$, the stack is a Deligne-Mumford stack after base change to $R[1/e].$ So the non-Deligne-Mumford behavior only happens for "small" primes.<|endoftext|> -TITLE: Constructive analysis and synthetic differential geometry -QUESTION [9 upvotes]: I am curious if (any of) the various inequivalent constructions of the real line in constructive mathematics can be used to build a model of Kock and Lawvere's synthetic differential geometry? In other words, do any of the constructions of the real line (in say HoTT) satisfy the Kock-Lawvere axiom for a class of functions which deserve to be called smooth? If not, how can we "augment" the real line with nilpotent infinitesimals for this to be true? -I am a complete novice when it comes to constructive mathematics, but I'm reasonably comfortable with Anders Kock's synthetic differential geometry texts. Unfortunately, I haven't had a chance to read the Models of SDG text yet, so I apologize if this is covered there. - -REPLY [6 votes]: In the smooth-topos models of SDG, the situation is generally something like this. The internally-definable Cauchy real numbers $\mathbf{R}_c$ are the sheaf of locally constant $\mathbb{R}$-valued functions, while the internally-definable Dedekind real numbers $\mathbf{R}_d$ are the sheaf of continuous $\mathbb{R}$-valued functions. In between these there is the sheaf $\mathbf{R}_s$ of smooth $\mathbb{R}$-valued functions, $\mathbf{R}_c \to \mathbf{R}_s \to \mathbf{R}_d$, which in turn is a quotient of the "line object" $\mathbf{R}_l$ that contains infinitesimals. Despite much trying, I don't know of any internal construction that produces $\mathbf{R}_s$ or $\mathbf{R}_l$ in these toposes; it seems that they have to be considered extra structure with which the topos is equipped. However, they are related to other structure it has, such as differential cohesion. You may also be interested in smooth structures on a topos.<|endoftext|> -TITLE: Pic^0 and H^0(K,Pic^0) -QUESTION [9 upvotes]: Let $K$ be a field and $C$ a smooth and projective curve over $K$. Then the kernel $Pic^0(C)$ of the degree map injects into $H^0(K,Pic^0_C)$, where $Pic_C^0$ is the connected component of the Picard variety. -I am wondering if there are examples where this is not an isomorphism for $K$ a global field. I am especially interested if there are elements of order prime to $p$ in the cokernel if $K$ is a global field of characteristic $p$. - -REPLY [12 votes]: By the long exact sequence of low degree terms for the Leray spectral sequence computing $H^r_{\text{et}}(C,\mathbb{G}_m)$ via $H^p_{\text{et}}(\text{Spec}\ K,R^q f_*\mathbb{G}_m)$, the cokernel of the map $$\text{Pic}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K})$$ equals the kernel of the pullback map on Brauer groups, $$\text{Br}(f):\text{Br}(\text{Spec}\ K) \to \text{Br}(C).$$ Of course $\text{Coker}(\text{Pic}(f))$ surjects onto the cokernel of the degree map, $$\text{deg}(f):\text{Pic}(C) \to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}_{C/K}/\text{Pic}^0_{C/K}) =\mathbb{Z}.$$ By the Snake Lemma, the kernel of the map $$\text{Coker}(\text{Pic}(f))\to \text{Coker}(\text{deg}(f))$$ equals the cokernel of your homomorphism, $$\text{Pic}^0(f):\text{Pic}^0(C)\to H^0_{\text{et}}(\text{Spec}\ K,\text{Pic}^0_{C/K}).$$ The cokernel of $\text{deg}(f)$ is cyclic. Thus, if $\text{Ker}(\text{Br}(f))$ is not cyclic, then $\text{Coker}(\text{Pic}^0(f))$ is nonzero. More precisely, for a prime $\ell$, if $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell \mathbb{Z}$ has rank $r+1$ as a vector space over $\mathbb{Z}/\ell\mathbb{Z},$ then $\text{Ker}(\text{Br}(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$, possibly higher rank because of $\text{Tor}_1(\mathbb{Z}/\ell\mathbb{Z},\text{Coker}(\text{deg}(f))).$ -The Brauer group of a global field is described by class field theory. In particular, for any prime $\ell,$ the rank of the $\ell$-torsion subgroup $\text{Br}(K)[\ell]$ is infinite (infinitude of primes). Let $\alpha_0,\dots,\alpha_r$ be $\mathbb{Z}/\ell\mathbb{Z}$-linearly independent classes in $\text{Br}(K)[\ell].$ Let $P_0,\dots,P_r$ be associated Severi-Brauer $K$-schemes. Now let $C$ be a general complete intersection curve in the product variety $$P:=P_0\times_{\text{Spec}\ K}\dots \times_{\text{Spec}\ K} P_r.$$ The kernel of the pullback map $\text{Br}(K)\to \text{Br}(P)$ contains the classes $\alpha_0,\dots,\alpha_r.$ Thus, the $\mathbb{Z}/\ell\mathbb{Z}$-vector space $\text{Coker}(\text{Pic}^0(f))\otimes \mathbb{Z}/\ell\mathbb{Z}$ has rank at least $r$.<|endoftext|> -TITLE: What's the summation of formal series $\sum_{n\geq0}\binom{n\delta}{n}x^n$？ -QUESTION [6 upvotes]: $\delta$ is a positive number. Is this Taylor expansion of some function? - -REPLY [8 votes]: The Bürmann-Lagrange theorem gives that -$$\sum_{n\geq 0} {n\delta \choose n} t^n = \frac{1}{1-\delta t(1+z)^{\delta -1}}=\frac{1+z}{1+(1-\delta) z}$$ -where $z=z(t)=\sum_{n\geq 1} \frac{1}{n}{n\delta \choose n-1}t^n$ is the solution of $z=t(1+z)^\delta$ $\big($i.e. -$z(t)$ is the local inverse at $0$ of $z \mapsto \frac{z}{(1+z)^\delta}\big)$. -See e.g problem 216 in section 3 of Polya/Szegő, Problems and Theorems in Analysis I.<|endoftext|> -TITLE: Singular values of Stirling numbers matrix -QUESTION [8 upvotes]: Consider the Stirling numbers of the first kind $s(i, j)$, and form a matrix $S_1(n),$ where the $(i, j)$th entry is $s(i, j)$. (IMPORTANT NOTE the indices start at $0,$ so this matrix is $(n+1)\times (n+1).$) This matrix is the inverse of the corresponding matrix for the Stirling numbers of the second kind, and the question applies to both of them. -Now, for the question: I have computed the logs of the singular values of these things (you have to do this to high precision, since the entries grow very fast). Here is the plot of the singular values for $n=100:$ - -You will note two empirical facts: firstly, $1$ is a singular value, and secondly the singular values bigger than $1$ decay exponentially, as do the ones smaller than $1$, but with a different rate. Has this been observed? Can one prove it (or at least explain it)? -ADDITION In my comment on the answer, I speak of the shape of the curve. To underscore what I mean, here are two more interesting matrices: the binomial coefficient matrix (which should be, in principle, similar to the Stirling matrix, due to the recurrence relation), and the Hilbert matrix. Here are the singular value plots of both (in order), again logarithmically. You will note that all three curves are regular. While the Stirling curve is piecewise linear, the Binomial curve is convex (on the top segment) and the Hilbert curve is concave. This must mean something :) - -REPLY [3 votes]: I will make the first attempt, though my resulting bound can certainly be improved. -I will do the $N\times N$ Stirling matrix of the second kind. We have $S_2(n,k) = \left\{n\atop k\right\}$ for $0\leq n,k\leq N-1$, which satisfies the following recurrence relation: -$$\left\{n+1\atop k\right\} = k\left\{n\atop k\right\} + \left\{n\atop k-1\right\},\qquad n\geq 0, \quad k>0.$$ -Therefore, the matrix $S_2$ satisfies the Sylvester matrix equation: -$$\underbrace{\begin{bmatrix}0 & 1 \\ & & \ddots \\ & & & 1 \\ -1 \end{bmatrix}}_{=A} S_2 - S_2\underbrace{\begin{bmatrix} 0 & 1 \\ & 1 & \ddots \\ & & \ddots & 1 \\ & & & N-1 \end{bmatrix}}_{=B} = \begin{bmatrix}0 & \ldots & 0 & 0 \\\vdots & \ddots & \vdots & \vdots \\0 & \ldots & 0 & 0 \\ \times & \ldots & \times & \times \end{bmatrix},$$ -where the $\times$'s denote nonzero entries. We say that $S_2$ has an $(A,B)$-displacement rank of 1. -From now on assume that $N$ is an even integer. The same idea, with different details, works when $N$ is an odd integer. -Since the rhs of the above equation is of rank 1, the eigenvalues of $A$ lie at (shifted) roots-of-unity, and the eigenvalues of $B$ are in $\{0,\ldots,N-1\}$ we find that, see Corollary 2.2 of paper -$$\sigma_{2k+1}(S_2) \leq \|V\|_2\|V^{-1}\|_2 Z_{2k}(\{0,\ldots,N-1\},F_+\cup F_-)\|S_2\|_2,$$ -where $V$ is the eigenvector matrix for $B$, i.e., $B = V\Lambda_B V^{-1}$ and $\|\cdot\|_2$ is the spectral norm. Here, $Z_{k}(E,F)$ denotes a Zolotarev number (see Section 2 of paper) and -$$F_+ = \left\{e^{it}: t\in [\tfrac{\pi}{N},\pi-\tfrac{\pi}{N}\right\}, \quad F_- = \left\{e^{it}: t\in [-\pi + \tfrac{\pi}{N},-\tfrac{\pi}{N}\right\}.$$ -I do not have bounds on $Z_{2k}(\{0,\ldots,N-1\},F_+\cup F_-)$ at hand and would have to work them out. Instead, to use a previous result, I will use the fact that -$$Z_{2k}(\{0,\ldots,N-1\},F_+\cup F_-)\leq Z_{2k}(\mathbb{R},F_+\cup F_-)$$ -since $\{0,\ldots,N-1\}\subset \mathbb{R}$. -Therefore, when $N$ is an even integer, we obtain (see Lemma 5.1 of paper): -$$\sigma_{2k+1}(S_2) \leq 4\|V\|_2\|V^{-1}\|_2\left[\exp\left(\frac{\pi^2}{4\log(4N/\pi)}\right)\right]^{-k}\|S_2\|_2.$$ -This explains the rapid decay of the singular values of $S_2$, provided $\|V\|_2\|V^{-1}\|_2$ does not grow rapidly with $N$. With a little more work, I believe that one can show that $\|V\|_2\|V^{-1}\|_2<8$ for any $N$ (this appears to be true numerically for $1 -TITLE: Why aren't proceedings from ICM 2014 on mathscinet? -QUESTION [25 upvotes]: Articles from the Proceedings of the International Congress of Mathematicians, Seoul, 2014 don't appear to be on Mathscinet. Why is this? -(Someone pointed this out to me recently, and I was reminded of it today when I tried to cite a lecture.) - -REPLY [3 votes]: This is just a comment, in community-wiki format to make it easier to compactify links. -To expand on Ed Dunne's reply, it may be helpful to note that each ICM organizing committee makes its own arrangements about formal publication of proceedings. For the Seoul ICM see Seoul and for PDF files of their four volumes of proceedings see here. Presumably in due time all of this will be incorporated into the ICM site, where one can find older proceedings: ICM. -Nowadays most people post earlier manuscripts on the arXiv, but of course they are more troublesome to track down. In any case, the ICM databases of speakers and sessions are useful because published proceedings are usually organized into several volumes.<|endoftext|> -TITLE: Concepts in topology successfully transferred to graph theory and combinatorics with non-trivial applications? -QUESTION [17 upvotes]: What are some of the difficult concepts in topology that have been transferred to graph theory and combinatorics where a certain new application has been found. -A good example is Lovász's proof of Kneser's conjecture. - -REPLY [2 votes]: Here is a closed model defined in the category of undirected graphs, which maybe answer the question that you raised in your comment to my previous answer: I believe you want to define an homotopy theory which characterizes perfect matchings. The model here characterizes maximum matchings. -Homotopy theory in category theory have been defined by Quillen to generalize classical homotopy theory defined in the category of topological spaces to various settings. -In general, the notion of path does not exist in every category, but, it is possible to define a notion of weak equivalence, fibration, cofibration between two objects. It is on this purpose that Quillen has defined the notion of closed model: -${\it Definitions.}$ -Let $C$ be a category, and $W$ a subclass of the class of morphisms of $C$, we say that $W$ verifies the $2$-$3$ property if and only if for every morphisms $f:X\rightarrow Y$ and $g:Y\rightarrow Z$, if two morphisms of the triple $(f,g,g\circ f)$ are in $W$, so is the third. -Let $C$ be a category complete and cocomplete; we say that $C$ is endowed with a closed model if and only if there exist three classes of morphisms $(Fib,Cof,W)$ such that: - -$W$ satisfies the $2$-$3$ property, -Let $Fib'=W\cap Fib$, $(Cof,Fib')$ is a weak factorization system -Let $Cof' = W\cap Cof$, $(Cof',Fib)$ is a weak factorization system. - -See the reference for the definitions, the class $W$ is called the class of weak equivalences the class $Fib$ is called the class of fibrations, the class $Fib'$ the class of weak fibrations, the class $Cof$ the class of cofibrations, the class $Cof'$ the class of weak cofibration. -In th reference, I have described a method to generate closed models in a category: closed models defined by counting: -Let $C$ be a category complete and cocomplete whose initial object is denoted by $\phi$. - For every objects $X$ and $Y$ of $C$, we denote by $X+Y$ the sum of $X$ and $Y$. - Let $(X_i)_{i\in I}$ be a family of objects of -$C$ and $l_i: \phi\rightarrow X_i$ the canonical morphism. -There exist morphisms $j^i_1:X_i\rightarrow X_i+X_i$ and $j^i_2:X_i\rightarrow X_i+X_i$ such that for every -morphisms $f:X_i\rightarrow Z$ and $g:X_i\rightarrow Z$, there exists a unique morphism $m(f,g):X_i+X_i\rightarrow Z$ -such that $m(f,g)\circ j^i_1 = f$ and $m(f,g)\circ j^i_2 = g$. We set $m_i = m(Id_{X_i},Id_{X_i})$. -Such a morphism is often called a folding morphism. We suppose that the class of morphisms $l_i,m_i\in I$ admits the small element argument. -We denote by $W_I$ the class of morphisms which are right orthogonal to every morphisms $l_i$ and $m_i, i\in I$. -\medskip -{\it Proposition.} -There exists a closed model $(Hom(C),cell(I),W_I)$ defined on $C$ where $Hom(C)$ is the class of morphisms of $C$ and $cell(I)$ be the class of morphisms of $C$ which are retracts of transfinite compositions of pushouts of $l_i,m_i,i\in I$. -\medskip -Remark that a morphism $f:X\rightarrow Y $ of $C$ is an element of $W_I$ if and only if for every $i\in I$, the map $c^i_f:Hom_C(X_i,X)\rightarrow Hom_C(X_i,Y)$ -defined by $c^i_f(h) = f\circ h$ is bijective. -\medskip -This model can be apply to the category of undirected graphs: Let $C_U$ be the category which has two objects that we denote by $0$ and $1$. We suppose -that $Hom_{C_U}(0,1)$ contains two elements $s,t$, $Hom_{C_U}(0,0)$ contains one element, - $Hom_{C_U}(1,1)$ contains the identity and an involution $i$ such that $i\circ s =t$, and $Hom_{C_U}(1,0)$ is empty. An undirected graph is a presheaf on $C_U$. We denote by $UGph$ the category of undirected graphs, its a complete and cocomplete category since it is a Grothendieck topos. -Particular examples of undirected graphs are the undirected arc graph $A_U$ is the graph defined by $A_U(0)=\{u_1,u_2\}$, $A_U(1)=\{a_1,a_2\}$ such that $A_U(i)(a_1)=a_2$, $A_U(s)(a_1)=u_1$ -and $A_U(t)(a_1)=u_2$. -The graph $V_U$ is the graph defined by $V_U(0)=\{v_1,v_2,v_3\}$, $V_U(1)=\{b_1,b_2,c_1,c_2\}$ such that $V_U(s)(b_1)=V_U(s)(c_1)=v_1$, -$V_U(t)(b_1) = v_2, V_U(t)(c_1)=v_3$, $V_U(i)(b_1)=b_2$ and $V_U(i)(c_1) = c_2$. -\medskip -We consider the closed model $(Fib,Cof,W)$ defined by counting the object $V_U$ of $UGph$. -\medskip -{\bf Theorem.} -{\it Let $X$ and $Y$ be finite graphs, a weak equivalence $f:X\rightarrow Y$ for the closed model defined by counting $V_U$ induces a bijection between the maximum matchings of $X$ and $Y$. Conversely, if the cardinal of the set of arcs of $X$ and $Y$ are equal, then a morphism $f:X\rightarrow Y$ which induces a bijection between the maximal matchings of $X$ and $Y$ is a weak equivalence.} -\medskip -{\bf Proof.} -A maximum matching of $X$ can be defined by $p_Y:\sum_JA_U^j\rightarrow Y$ where $A_U^j$ is a graph isomorphic to $A_U$ such that: $p$ is injective on arcs, $p_Y(A_U^{j_1})\cap p_Y(A_U^{j_2})$ is empty and every arc of $Y$ shares a vertex with an arc of $p_Y(\sum_J A_U^j)$. -Let $p_X:\sum_JA_U^j\rightarrow X$ be a maximum matching let's show that $p_Y=f\circ p_X$ is a maximum matching. Consider $b$ an arc of $Y$, there exists an unique arc $a$ of $X$ such that $f(a)=b$ since $f$ induces a bijection on arcs. If $a$ is not in the image of $p_X$, $a$ shares a vertex with $p_X(A_U^j)$ for an element $j\in I$, this implies that $b=f(a)$ shares a vertex with the image of $f\circ p_X$. Suppose that $f(p_X(A_U^{j_1}))$ shares a vertex with $f(p_X(A_U^{j_2}))$, you can define a morphism $m:V_U\rightarrow Y$ whose image is the subgraph of $Y$ whose arcs are $f(p_X(A_U^{j_1}))$ and $f(p_X(A_U^{j_2}))$. Since $f$ is defined by counting $A_U$ and $V_U$, there exists a unique morphism $n:V_U\rightarrow X$ such that $m=f\circ n$, but the arcs of $n$ are $p_X(A_U^{j_1})$ and $p_X(A_U^{j_2})$, contradiction, since they do not share a vertex. -Now we show that $f$ induces a surjection between the sets of maximum matchings of $X$ and $Y$. Let $p_Y:\sum_JA_U^j\rightarrow Y$ be a perfect matching. Since $f$ is defined by counting $V_U$, it induces a bijection between the sets $Ar(X)$ and $Arc(Y)$ of arcs of $X$ and $Y$. We can define the morphism $p_X:\sum_J A_U^j\rightarrow X$ such that the restriction $p_X^j$ of $p_X$ to $A_U^j$ is the unique morphism $p^j_X:A_U^j\rightarrow X$ such that $p_Y^j=f\circ p_X^j$, $p_X$ is a maximum matching: -Consider an arc $a$ of $X$ which is not in the image of $p_X$, $f(a)$ is not in the image of $p_Y$, this implies that there exists $j\in J$ such that $f(A_U^j)$ shares a vertex with $f(a)$, we can define a morphism $g:V_U\rightarrow Y$ whose image is the subgraph of $Y$ which arcs are $f(a)$ and $p_Y(A_U^j)$. Since $f$ is defined by counting $A_U$ and $V_U$, there exists a morphism $h:V_U\rightarrow X$ such that $g=h\circ f$, the image of $h$ is $a$ and $p_X(A_U^j)$. -Remark that $p_X(A_U^{j_1})\cap p_X(A_U^{j_2})$ is empty since $p_Y(A_U^{j_1})\cap p_Y(A_U^{j_2})$ is empty. We deduce that $p_X$ is a maximum matching. We deduce that $f$ induces a surjective map on maximum matchings. -Suppose that $X$ and $Y$ are finite undirected graphs such that the set of arcs of $X$ and $Y$ have the same cardinal and $f$ is a weak equivalence: firstly we remark that $f$ induces a bijection between the sets of arcs of $X$ and $Y$. Let $b$ be an arc of $Y$, there exists a maximum matching $p\sum_JA_U^J\rightarrow Y$ whose image contains $b$, we can lift $p$ to $q:\sum_JA_U^j\rightarrow X$ such that $p=f\circ q$, this implies that there exists an arc $a$ of $X$ such that $f(a)=b$. Since $f$ is surjective on arcs and the cardinals of the set of arcs of $X$ and $Y$ are equal, we conclude that $f$ induces a bijection between the sets of arcs of $X$ and $Y$. Let $g:V_U\rightarrow Y$, there exists arcs $a,a'$ of $X$ such that the image of $(a,a')$ the subgraph of $X$ which is the union of $a$ and $a'$ by $f$ is $g(V_U)$, suppose that $a,a'$ do not have a common vertex, there exists a maximal matching $p\sum_JA_U^j\rightarrow X$ whose image contains $(a,a')$, but $f\circ p$ is not a matching: contradiction, since we have supposed that $f$ induces a map well defifined on maximal matchings. This implies that $g$ can be lifted to $X$ and $Hom(V_U,X)\rightarrow Hom(V_U,Y)$ induced by $f$ is surjective. This map is also injective since $f$ is induces a bijection on arcs. -\medskip -If one supposes that $X$ and $Y$ do not have isolated vertices, similar arguments show that a week equivalence between $X$ and $Y$ induces a bijection between their respective set of perfect matchings. -\medskip -This closed model is not trivial, consider the graph $X$ which is the undirected path of length $2$ (or $V_U$), it is obtained by identifying one vertex of two copies of $A_U$, and $Y$ which is the concatenation of $A_U$ and a circle. You can define $f:X\rightarrow Y$ obtained by identifying the middle vertex of $X$ with a vertex at its end. It is a weak equivalence. -Tsemo Aristide -Applications of closed models defined by counting to graph theory and topology -Algebra Letters, Vol 2017 (2017), Article ID 2<|endoftext|> -TITLE: One dimension random walk. Is hitting time Lipschitz with respect to target? -QUESTION [7 upvotes]: Consider a random walk $S_t = \sum_{i=1}^{t} X_i$, with $X_i$ i.i.d.. Assume that $X_i \in [0,1]$. Define $\tau(y) := \inf\{t: S_t\geq y\}$, i.e., $\tau(y)$ is the hitting time of $[y,\infty)$. Is this possible to show that $\mathbb{E}[\tau]$ is a Lipschitz function, under some "natural" condition? -One condition, I can imagine, is that the pdf of $X$, $f(x) \leq M$ for some large $M$. It makes $E[S_t]$ "smooth". But it is still unclear for me how to argue $\mathbb{E}[\tau]$ is Lipschitz rigorously. -Any hint? -Here my idea of proving it with a "$\log$" factor gap. Staring from Ori's idea, -$$ - \mathbb{E}[\tau(y)] = \sum_{t=1}^\infty \mathbb{P}[S_t < y]. -$$ -Then, -\begin{align} -\mathbb{E}[\tau(y) -\tau(y')] &= \sum_{t =1}^\infty \mathbb{P}[S_t\in (y,y')] \\ -&= \sum_{t =1}^T \mathbb{P}[S_t\in (y,y')] + \sum_{t =T+1}^\infty \mathbb{P}[S_t\in (y,y')]. -\end{align} -We choose $T$ such that the second term $\approx|y'-y|$. Assume $y$ is bounded, i.e. $y\in[0,K]$, so we can bound $\mathbb{P}[S_t\in(y,y')]$ uniformly by $\mathbb{P}[S_t \leq K]$, which, by Hoffding , $\lesssim \exp(-ct)$. So we need $T \approx \log(\frac{1}{|y-y'|})$, so that -$$ - \sum_{t =T+1}^\infty \mathbb{P}[S_t\in (y,y')] \approx |y-y'|. -$$ -With this $T$. -$$ - \sum_{t =1}^T \mathbb{P}[S_t\in (y,y')] \lesssim M|y-y'|\log(\frac{1}{|y-y'|}). -$$ - -REPLY [5 votes]: could you elaborate more on your high tech solution? -OK, but it is just a standard boring exercise giving one no intellectual pleasure whatsoever. Write $X=t+Y$ where $EY=0$. Note that $t\ge\frac 1{2M}$. Let $f=f_1$ be the pdf of $Y$. Then the pdf $f_n$ of $Y+\dots+Y$ ($n$ times) is $f*\dots*f$, so $\widehat {f_n}=(\widehat f)^n$. Now $\widehat f(z)$ is an analytic function that is bounded by $e^{-\delta}$ on $(\mathbb R\setminus[-\rho,\rho])\cup [-\rho-i\tau,-\rho+i\tau]\cup[\rho-i\tau,\rho+i\tau]$ and $|f(x+iy)|\le e^{-ax^2+by^2}$ whenever $|x|\le\rho, |y|\le\tau$ with some efficient $a,b,\rho,\tau,\delta>0$ depending on $M$ only. Now, for $n\ge 2$, we can write -$$ -f_n(u)=\frac{1}{2\pi}\int_C (\widehat f(z))^n e^{iuz}dz -$$ -where $C$ is the contour $-\infty\to -\rho\to -\rho+iy \to \rho+iy\to\rho\to +\infty$ with any $y\in[-\tau,\tau]$ we want such that $uy>0$. Bounding $|f(z)|^n$ by $e^{-\delta(n-2)}|f(z)|^2$ on $C\setminus [-\rho+iy,\rho+iy]$, we see that that part of the contour contributes at most $Ce^{-\delta n}$ (recall that $\int_{\mathbb R}|\widehat f|^2=\int_{\mathbb R}|f|^2$ is controlled and the integral over the remaining interval is at most $e^{(bny^2-uy)}\int_\mathbb R e^{-anx^2}\,dx$. We would like to take $bny=\frac u2$, i.e. $y=cu/n$ to get $n^{-1/2}e^{-\gamma u^2/n}$ from here. That is fine as long as $|u|c'n$, we only get $e^{-\tau|u|/2}$, but we have allowed ourselves an exponential error already, so the final bound -$$ -f_n(u)\le C(e^{-\delta n}+n^{-1/2}e^{-\gamma u^2/n}) -$$ -holds always. The pdf of the sum of $n$ independent copies of $X$ -is just $f_n(u-nt)$. I leave the estimation of the sum to you. -Edit. OK, here is a cute (in my taste) way to do it. Note that the statement can be reformulated as follows: For every $y>0,0<\varepsilon<\varepsilon_0$, the expected number of times the random walk $W_n=X_1+\dots+X_n$ hits $[y,y+\varepsilon]$ is at most $C\varepsilon$. Now, conditioning on $n$ such that $W_1,\dots,W_ny-1$, we see that it is enough to get the bound for $y<1$. Now the probability to hit our interval after $n$ steps is at most $P\{W_{n-1}<1+\varepsilon_0\}M\varepsilon$, so it will suffice to sum $P_n=P\{W_n<1\}$. But now we obviously have $P_n\le \frac{M^n}{n!}$, so the Lipschitz constant is at most $Me^M$ (of course, this bound can be greatly improved, but, when doing it, try not to raise the level of complexity :-) ). -Next edit: Now we shall remove the condition that $X<1$ (but will still keep the iid assumption; the next question will be if we need the second "i" in it). -Let $t$ be the median of $X$. Split the pdf of $X$ into $f_-=f\chi_{[0,t]}$ and $f_+=f\chi_{(t,+\infty)}$. Now consider $F=f+f*f+\dots+f^{*N}$. It is a bounded function ($N$ is finite), so let $Z$ be its maximum. Write -$$ -F\le f+F*f=f+F*f_++F*f_-\le M+\frac Z2+F*f_- -$$ -The last term has a clear meaning: it is the chance to hit the target from distance $t$ or less. Now, if our target is $[y,y+\varepsilon]$, as before, then we can condition on $n$ such that $W_1,\dots,W_{n-1}1$, $\mathcal F_n=\mathcal F_{n-1}*\mathcal F_1$, i.e. $\mathcal F_n$ consists of all possible convolutions of $n$ (possibly different) functions in $\mathcal F_1$. -The key high-tech (well, not too high, but still requiring some play with the Fourier transform) result we need is the bound $\|f\|_\infty\le cn^{-1/2}$ for all $f\in\mathcal F_n$ with some absolute constant $c>0$. -Our assumptions on the non-negative functions $f_k$ and $g_k$ supported on $[0,+\infty)$ are the following: -1) $f_k\in \mathcal F_n$ for all $k$. -2) $\|g_k\|_\infty\le Gn^{-1/2}$ with some absolute $G>0$ that can (and will) be explicitly computed in terms of $c$ from the high-tech result. -3) $f_k$ dominate $g_k$ in the sense that for every $t\ge 0$, we have $\int_t^\infty g_k\le\int_t^\infty f_k$. -It is worth mentioning here that we don't restrict $g_k$ in any other way. We do not even require that $\int_{-\infty}^\infty g_k=1$ (that it is $\le 1$ follows directly from the domination property), i.e., we allow a positive probability of a misfire. -The claim is that then our infinite sum is bounded by $Zn^{-1/2}$ where $Z$ is some constant to be determined in terms of $c$ and $G$. -Proof -Split $g_k$ into $g_k^-+g_k^+$ where $g_k^-$ is the part lying to the left from the median of $X_k$ (it depends on $k$ now; note also that $X_k$ is not a misprint: we, indeed, split the shot according to the median of the step) and $g_k^+$ is the part lying to the right of it. Split $f_k$ in the same way. It is easy to see that $g_k^+$ is dominated by $f_k^+$. -Our first relatively trivial (by now) observation is that the sum -$$ -g_1^-+f_1*g_2^-+f_1*f_2*g_3^-+\dots -$$ -(close range shooting) is bounded by $2Gn^{-1/2}$ regardless of the length of the walk for exactly the same reason as before: condition on the first time we are closer to the target than the median of the next step; the probability of the hit is then $Gn^{-1/2}\varepsilon$ but the chance that we shall still remain to the left of the target after the next step is only $1/2$ and if we do, the whole game starts over. -Another trivial observation is that if the length of the walk is $0$ (only the initial shot is fired), then we can bound the corresponding one-term sum by $Gn^{-1/2}$, so if $Z\ge G$, we have no trouble with walks of length $0$ for all $n$. -Now we shall run the induction on the walk length. Suppose that the bound for all shorter walks is already established and we want to investigate some walk of the currently suspicious length corresponding to some $n$. Chopping off the $g^-$ part, we see that our main task is to bound -$$ -Q=g_1^++f_1*g_2^++f_1*f_2*g_3^++f_1*f_2*f_3*g_4^++f_1*f_2*f_3*f_4*g_5^++f_1*f_2*f_3*f_4*f_5*g_6^++\dots -$$ -Write -$$ -\frac 32 Q= -\\ -(\frac {g_1^+}2+f_1*g_2^+)+(f_1*f_2)*(\frac{g_3^+}2+f_3*g_4^+)+(f_1*f_2)*(f_3*f_4)*(\frac{g_5^+}2+f_5*g_6^+)+\dots -\\ -+f_1*[(\frac {g_2^+}2+f_2*g_3^+)+(f_2*f_3)*(\frac{g_4^+}2+f_4*g_5^+)+(f_2*f_3)*(f_4*f_5)*(\frac{g_6^+}2+f_6*g_7^+)+\dots] -\\ -+\text{ a few ($\le 8$, say) individual endpoint terms due to the boundary effect and possibly wrong parity} -$$ -Estimate the individual terms by $Gn^{-1/2}$ each and note that each of the long lines represents a shooting walk of smaller length with steps $f_k*f_{k+1}\in \mathcal F_{2n}$ and the corresponding shots $\frac{g_k^+}2+f_k*g_{k+1}^+$ (the initial convolution with $f_1$ in the second long line is harmless for bounding the maximum and can be safely forgotten). -We would like to use the induction assumption now but we need to check the conditions first. That the new steps are in $\mathcal F_{2n}$ is a no-brainer. Next -$$ -\frac{g_k^+}2+f_k*g_{k+1}^+\le \frac{Gn^{-1/2}}2+\frac {cn^{-1/2}}2\le G(2n)^{-1/2} -$$ -provided that $\frac G2+\frac c2\le \frac{G}{\sqrt 2}$. Note that we played the uniform norm of $f_k$ and the $L^1$ norm of $g_{k+1}^+$, not the other way around, to bound the second term. Finally, $\frac {g_k^+}2$ is dominated by $\frac{f_k^+}2$, which, in turn, is dominated by $f_k^+*f_{k+1}^-$ (one half of any function is dominated by its convolution with any non-negative function supported on $[0,+\infty)$ and having integral $\frac 12$). Since $g_{k+1}^+$ is dominated by $f_{k+1}^+$, we get the whole shot dominated by $f_k^+*f_{k+1}^-+f_k*f_{k+1}^+\le f_k*f_{k+1}$, so the domination condition is satisfied as well. -Thus, by the induction assumption, the final bound on our entire suspicious shooting walk is -$$ -10Gn^{-1/2}+\frac 23[2Z(2n)^{-1/2}]=\left[10G+\frac{4}{3\sqrt 2}Z\right]n^{-1/2} -$$ -(note that it is $2n$ instead of $n$ that gives us the crucial $\sqrt 2$ improvement). -Since $\frac {4}{3\sqrt 2}<1$, we can choose $Z$ to be an appropriate positive multiple of $G$ to get the sum in brackets equal to $Z$, thus finishing the story. -I talked to a few probabilists about whether this is "well-known". The consensus seems to be that it is not but also that it is of no interest because nobody cares about random walks with positive steps anyway. Still, I found it rather amusing, so I'm posting it here to preserve it for posterity. Yuval Peres also noted that the sharp constant should be about $2$ rather than about $1000$ this proof yields, but, I guess, I'll leave the question of finding the sharp bound to somebody else. -The End.<|endoftext|> -TITLE: Largest ordered "field" in NBG without axiom of global choice -QUESTION [12 upvotes]: I know from Wikipedia that in NBG, the surreal numbers are the largest possible ordered field (if a proper class is allowed to be a field). But then, it is written: "in theories without the axiom of global choice [...] it is not necessarily true that the surreals are the largest ordered field". -How would such a field look like? Is it even possible to define it? The surreals can be defined in various ways, e. g. axiomatically or as a Hahn series over $\mathbb R$. Is this possible without global choice? -I am not very good in logic and set theory, it is thus very probable I have made a mistake. - -REPLY [12 votes]: There is no problem defining the surreal field without global choice. -One can define it in ZFC and considerably weaker theories, for -example with the hereditary birthday construction of left-sets and -right-sets, and also in other ways. -With global choice, the surreal field No is largest in the sense -of model-theoretic universality: all other ordered class fields are -isomorphic to a subfield of No. -One can prove this by means of the usual model-theoretic -back-and-forth argument (really only just the "forth" part), -combined with the fact that the surreal field No is set-saturated -and homogeneous. -That is, given any class field $F$, you use global choice to -well-order $F$, and then build up the embedding $j:F\to\text{No}$ -in stages. At any stage, you've embedded part of $F$ into the -surreals, and you consider the next point. By saturation, there is -a surreal number realizing the right type, and you can extend the -embedding one more step. -It is the same argument as showing that the rational line -$\mathbb{Q}$ is a universal countable linear order. -Without global choice, the argument seems to break down, and one -cannot seem to get started. Of course, what is left of the argument -is that even without global choice, the surreal field No remains -universal for all set-sized fields (can one handle well-orderable class fields?). But I don't think it is known whether -there is actually or can be a proper class field that does not -embed into No when global choice fails. -If universality fails, it might not mean that there is some other -"larger" field. Rather, what I would expect is simply that there -are some other ordered class fields that don't embed into No, -perhaps none of them maximal even. -I asked an analogous question about the surreals as a universal -class linear order: Is the universality of the surreal number line -a weak global choice -principle? -Basically, it seems that most of us expect that one really needs global choice for -the universality argument, but to my knowledge we don't yet have any proof of this. -I wonder how the universality of the surreals as a linear order relates to its universality as an ordered field? It has both universality properties under global choice, but can one separate these without global choice?<|endoftext|> -TITLE: Inner forms of $GL(2)$ -QUESTION [5 upvotes]: I know that inner forms of $GL(2)$ are quaternion algebras. However, I cannot find the proof myself. -First, since quaternion algebras are forms of $GL(2)$ by the usual embedding in matrices, they are automatically inner by Skolem-Noether theorem. -But how can I prove the converse, that is to say if I have a group "inner isomorphic" (i.e. by conjugation) to $GL(2)$ over an algebraic closure, then it is a quaternion algebra? -Thanks for any clue or reference! - -REPLY [9 votes]: If you have a group $G$ over $k$ such that $G_{\overline{k}}$ is isomorphic to $GL_2$ and, and each element of $\operatorname{Gal}(\overline{k}|k)$ acts on this $GL_2$ by the standard action times conjugation by some (unique up to scalars element), then we can use the same cocycle to form a twisting of the algebra $M_2(k)$, because $PGL_2$ acts by conjugation on $M_2$. -By Hilbert 90, the twisted form of $M_2(K)$ remains a $4$-dimensional vector space, with algebra structure. Any two-sided ideal must remain a two-sided ideal over $\overline{k}$, and similarly with central elements, so it is a (rank 4) central simple algebra, i.e. a quaternion algebra. -Because the group of units of this algebra is $GL_2$, and this restriction is compatible with the action of $PGL_2$ and Galois, so the multiplicative group of this algebra is a twisted form of $GL_2$.<|endoftext|> -TITLE: Branching rules for $SU(p,q)$ -QUESTION [6 upvotes]: What is the branching rule for the subgroup $SU(p,q-1)\subset SU(p,q)$, i.e., the structure of the restriction of irreducible, finite-dimensional representations of $SU(p,q)$ to $SU(p,q-1)$? I would appreciate any reference and comments. - -REPLY [5 votes]: As was indicated above your question is the equivalent to the branching rules for $sl_{n-1} \to sl_n$. This is a well-known branching rule and it is given by the following formula: If the representation of $sl_n$ is given by the highest weight $(\lambda_1 \geq \cdots \geq\lambda_n)$ then it decomposes upon restricting to $sl_{n-1}$ to the direct sum of irreducible representations with highest weights $(\lambda^{\prime}_1 \geq \cdots \geq\lambda^{\prime}_{n-1})$ satisfying $\lambda_{i+1} \leq \lambda^{\prime}_i\leq \lambda_{i}$. This is equivalent to removing some boxes from the corresponding Young diagramm.<|endoftext|> -TITLE: Generating symmetric groups with small cycles -QUESTION [7 upvotes]: This was asked but never answered at MSE. -Let $S_n$ denote the symmetric group and let $H$ be a subgroup which contains -an $n$-cycle. If $n$ is prime, and if $H$ also contains a 2-cycle, then necessarily $H = S_n$. However, this property does not hold in general which raises the question of how many additional cycles are required to restore the conclusion. -Define $f(n)$ to be the smallest integer such that whenever (in addition -to the $n$-cycle) $H$ contains cycles of length $2,3,4,\dotsc,f(n)$, then $H$ must be equal to the full symmetric group $S_n$. Note that only the cycle lengths are specified and not the entries. It is known that all cycle lengths do suffice to generate $S_n$. Therefore, $f(n)$ is well defined and takes values -in the range $2 \le f(n)\le n-1$. For example, $f(3)=2$ either directly or by noting that 3 is prime. However, $f(4) \neq 2$ because -the proper dihedral subgroup has cycles both of length $2$ and $4$. So we get -$f(4)=3$ instead (max possible value). -Next consider the even case $n = 2m$. $S_{2m}$ has a wreath product subgroup -$S_m\wr S_2$ of order $2(m!)^2$. This not only contains -all cycle lengths from 2 up through $m$ but also a full length $2m$-cycle. -All of which implies that $f(2m) \ge m+1$ and shows not only that $f(n)$ is unbounded but also that it can exhibit arbitrarily large jumps [e.g., $f(101)=2$ vs. $f(102)\ge52$]. -Questions: (1) For which $n$ besides $n=3$ and $n=4$ do we get the maximum value -$f(n) = n-1$? -(2) What is the set of values taken on by $f(n)$ as $n$ ranges over the natural -numbers? For large $n$, does $f(n)$ vary with number theoretic irregularity -or does it settle into predictable patterns? - -REPLY [3 votes]: Let $p$ be the smallest prime divisor of $n$. I believe that $f(n)=n/p+1$. The key points are as follows. -1) If $H \leq S_n$ is primitive and contains a $2$-cycle, then $H=S_n$. -2) If $1 -TITLE: Parahorics in nonsemisimple reductive algebraic groups -QUESTION [6 upvotes]: If $G$ is a semisimple algebraic group over a local field with finite residue field $K$ and $x$ a point in the Bruhat-Tits building $B(G, K)$ then the parahoric group scheme $P_x$ is a group scheme $P$ whose $O_K$ points are the connected component of the stabilizer of $x$. -If $G$ is not semisimple there is a construction of a group scheme $P$ associated to $x$ called a parahoric. But as the center of $G$ acts trivially on the building the $O_K$ points of $P$ aren't exactly the stabilizer of $x$. -Is it the case that $Z(G(K))P_{x}(O_K)$ is the connected component of the stabilizer of $x$? This seems believable from the construction. If $x$ is in the extended compartment instead will I just get the connected component of $P_{x}(O_K)$? Bruhat and Tits don't use the extended compartment, so I lack good references for that. - -REPLY [3 votes]: I don't know how Bruhat-Tits theory is used in representation theory, but I think there is a little confusion of notions in your question. -For a connected reductive algebraic group $G$, given a point $x$ in $B(G,K)$, Bruhat and Tits define $4$ integral models denoted $\mathfrak{G}_x^0$, $\mathfrak{G}_x$, $\hat{\mathfrak{G}}_x$ and $\mathfrak{G}_x^{\dagger}$. Here $\mathfrak{G}_x^0$ is the connected component of $\mathfrak{G}_x$, and in Bruhat-Tits terminology $\mathfrak{G}_x^0(\mathcal{O}_K)$ is called the "connected fixator" ("fixateur connexe" in French). I guess that's the group you are referring to as "the connected component of the stabiliser of $x$" (I put that in quote, because to me this doesn't mean anything: when compact, the stabiliser of $x$ in $G(K)$ is a profinite group, i.e. it has normal closed (for the strong topology) subgroups of finite index, and the index can be as big as you want). -Ok, now if we look at $GL_2$, take a point $x$ in $B(G,K)$ and an apartment $A$ containing it. Then throw in many basis as in Bruhat-Tits I, 10.2, so that x identifies with $0\in \mathbf{R} = A$. Let $S_{-1}$ be the stabiliser of $-1\in A$. By Bruhat-Tits I, $10.2.9$, $S_{-1}$ consists of matrices $g\in GL_2(K)$ such that the valuation of the entries are greater then $\begin{pmatrix} 0&1\\-1&0 \end{pmatrix}+\frac{\omega(\textrm{det}(g))}{2}$. For example, $\begin{pmatrix} 0&\pi \\\pi^{-1}&0 \end{pmatrix}$ belongs to $S_{-1}$. -On the other hand, by Bruhat-Tits II, $4.6.28$, matrices in $\mathfrak{G}_{-1}^0(\mathcal{O}_K)$ have determinant in $\mathcal{O}_K^{\times }$ and belongs to $S_{-1}$ as well. So we conclude that the element $g = \begin{pmatrix} 0&\pi^2\\1&0 \end{pmatrix}$ belongs to the stabiliser of $-1$, but not to $Z(G(K))\mathfrak{G}_{-1}^0(\mathcal{O}_K)$. -So this is a negative answer to your first question, even if we interpret "connected component of the stabiliser of x" to mean "stabilisers of $x$ acting type preservingly". -EDIT: Ok, as pointed by Watson Ladd, the above counterexample is just not one, sorry that was silly. At least there is still the counterexample if you take $x=\frac{1}{2}$. Then I believe $g = \begin{pmatrix} 0&1\\\pi&0 \end{pmatrix}$ stabilises $\frac{1}{2}$, but does not belong to $Z(G(K))\mathfrak{G}_{\frac{1}{2}}^0(\mathcal{O}_K)$. Here the action of $g$ is not type preserving, though, so this calls for the question whether $Z(G(K))\mathfrak{G}_{x}^0(\mathcal{O}_K)$ consist of stabilizers of $x$ acting type preservingly.<|endoftext|> -TITLE: Nontrivial lower bound on the sum of matrix norms -QUESTION [5 upvotes]: Let $X, V\in\mathbb{R}^{n\times r}$ such that $X^\top V$ is symmetric. The central quantity I care about is -\begin{equation} -\|XV^\top\|_{F}^2+\|X^\top V\|_{F}^2 +[\text{Tr}(X^\top V)]^2. -\end{equation} -An easy lower bound for this quantity is given by $2\sigma_{r}(X)^2\|V\|_{F}^2$, where $\sigma_{r}(X)$ is the smallest singular value of $X$. -I'm wondering whether there exists some absolute constant $c>0$ such that the following holds -\begin{equation} -\|XV^\top\|_{F}^2+\|X^\top V\|_{F}^2 +[\text{Tr}(X^\top V)]^2\geq c\|X\|_{F}^2\|V\|_{F}^2. -\end{equation} - -REPLY [3 votes]: Let function $f : \mathbb R^{m \times n} \to \mathbb R_0^+$ be defined as follows -$$f (\mathrm X) := \| \,\mathrm X \mathrm A^\top \|_\text{F}^2 + \| \,\mathrm X^\top \mathrm A \,\|_\text{F}^2 + \left( \langle \mathrm A, \mathrm X \rangle \right)^2$$ -where $\mathrm A \in \mathbb R^{m \times n}$ is given. Note that -$$\| \,\mathrm X \mathrm A^\top \|_\text{F}^2 = \| \,\mathrm A \mathrm X^\top \|_\text{F}^2 \geq \lambda_n ( \mathrm A^\top \mathrm A ) \, \| \mathrm X \|_\text{F}^2$$ -$$\| \mathrm X^\top \mathrm A \|_\text{F}^2 = \| \mathrm A^\top \mathrm X \|_\text{F}^2 \geq \lambda_m ( \,\mathrm A \mathrm A^\top ) \, \| \mathrm X \|_\text{F}^2$$ -and that $\left( \langle \mathrm A, \mathrm X \rangle \right)^2 \geq 0$. Hence, -$$f (\mathrm X) \geq \left( \lambda_n ( \mathrm A^\top \mathrm A ) + \lambda_m ( \,\mathrm A \mathrm A^\top ) \right) \| \mathrm X \|_\text{F}^2$$ -Suppose that $\rm A$ is tall (i.e., $m > n$) and has full column rank (i.e., $\mbox{rank} (\mathrm A) = n$). In this case, -$$\lambda_n ( \mathrm A^\top \mathrm A ) = \sigma_n^2 (\mathrm A) = \left( \frac{\| \mathrm A \|_2}{\kappa (\mathrm A)} \right)^2$$ -where $\kappa (\mathrm A)$ is the (finite) condition number of $\rm A$, and $\lambda_m ( \,\mathrm A \mathrm A^\top ) = 0$. Thus, -$$f (\mathrm X) \geq \left( \frac{1}{\kappa (\mathrm A)} \right)^2 \| \mathrm A \|_2^2 \, \| \mathrm X \|_\text{F}^2$$ -Since -$$\| \mathrm A \|_\text{F} \leq \sqrt{\mbox{rank} (\mathrm A)} \, \| \mathrm A \|_2 = \sqrt{n} \, \| \mathrm A \|_2$$ -we obtain -$$f (\mathrm X) \geq \underbrace{\frac 1n \left( \frac{1}{\kappa (\mathrm A)} \right)^2}_{=: c (\mathrm A)} \| \mathrm A \|_\text{F}^2 \, \| \mathrm X \|_\text{F}^2$$ -where $c$ is a function of matrix $\rm A$.<|endoftext|> -TITLE: Can a harmonic vector field possess a limit cycle? -QUESTION [8 upvotes]: Can a harmonic vector field $X$ on a Riemannian surface $(M,g)$ possess a limit cycle(An isolated periodic orbit)? -Note that the Laplacian of a vector field is defined via natural correspondence between the space of vector fields and the space of $1$-forms.(The natural correspondence arising from the Riemannian metric). If this correspondence is denoted by $i$ then the Laplacian of a vector field $X$ is defined as $\Delta X=i^{-1} \Delta (i(X))$. Where the latter Laplacian is the natural Laplacian on the space of differential forms. -In particular is there a quadratic vector field $$\begin{cases} x'= ax+by+\lambda(x^2-y^2)+txy \\ y'=cx+dy+\mu (x^2-y^2)+sxy\end{cases}$$ -which has at least one limit cycle? -The second question: Assume that $X$ is a vector field on a Riemannian surface - Assume that $\gamma$ is a periodic orbit of $X$. Is it true to say that there is a point $p\in \gamma$ such that $\Delta X$ is tangent to $\gamma$ at $p$? - -REPLY [4 votes]: Consider the infinite strip, $[0,1]\times\mathbb{R}$ with $[0,y]$ identified with $[1,y]$ for all $y\in\mathbb{R}$, and with the Riemannian metric $g=dxdx+dydy$. Consider $X=\partial_x +f(y)\partial_y$. The Laplacian is $\Delta X =f''(y) \partial_y$. -For the first question, let $f=-y$, so that the line $y=0$ is a limit cycle and $\Delta X=0$. For the second question, let $f=y^2$, so that $y=0$ is again a periodic orbit, but now, everywhere on the curve, $\Delta X=2\partial_y$ is orthogonal to the tangent $\partial_x$.<|endoftext|> -TITLE: The specificity of dimension $1+3$ for the real world -QUESTION [24 upvotes]: I have been asked sometimes, and I ask myself, to what extent the dimension $1+3$ is important for our real world, say compared to an hypothetical $(1+d)$-dimensional world. I have two answers in mind. - -The Huygens principle. If you switch off a point source of light, then a point situated at distance $L$ will be in dark after time $\delta t=L/c$ ($c$ the speed of light). This would be false in dimension $1+2$ for instance, even if the energy would be very low after $\delta t$. -Chemistry is a consequence of quantum mechanics. Mathematically it involves the linear representations of the rotation group. In $1+2$ dimensions, the group is $SO_2$, which is abelian and isomorphic to a circle ; its representations are one-dimensional, associated with linear characters. In our world, the groups $SO_3$ is not abelian and the situation is way richer. In particular, we have a notion of spin. - - -What are other manifestations of the dimension $1+3$ in our real world ? - -In order to limit this discussion to a reasonable extent, I assume that the Physics of a hypothetical world would be based on equations similar to those we already know. In particular, second-order differential operators would be at stake, because of their nice mathematical properties (maximum principle, ...) - -REPLY [2 votes]: One answer in the form of a paper is Tegmark’s “On the dimensionality of spacetime” at https://arxiv.org/abs/gr-qc/9702052<|endoftext|> -TITLE: Chebyshev polynomials of the first kind and primality testing -QUESTION [64 upvotes]: Can you provide a proof or a counterexample for the claim given below ? -Inspired by Agrawal's conjecture in this paper and by Theorem 4 in this paper I have formulated the following claim : - -Let $n$ be a natural number greater than two . Let $r$ be the smallest odd prime number such that $r \nmid n$ and $n^2 \not\equiv 1 \pmod r$ . Let $T_n(x)$ be Chebyshev polynomial of the first kind , then $n$ is a prime number if and only if $T_n(x) \equiv x^n \pmod {x^r-1,n}$ . - -You can run this test here . -I have tested this claim up to $5 \cdot 10^4$ and there were no counterexamples . -EDIT -Algorithm implementation in Sage without directly computing $T_n(x)$ . -Python script that implements this test can be found here. -The Android app that implements this test can be found on Google Play. -ADDED -I offer $100$ € for a proof of this claim. The proof must be published in one of the following journals: Journal of Number Theory , Algebra & Number Theory , Moscow Journal of Combinatorics and Number Theory . - -REPLY [7 votes]: I do not share excitement about this test and believe it admits false positives (i.e., pseudoprimes). Here are some supporting arguments. -Assuming that $n\equiv 2\text{ or }3\pmod{5}$, we will have $r=5$. A square-free composite integer $n$ will pass the test for $r=5$ if $T_n(x)\equiv x^n\pmod{x^5-1,p}$ for every prime $p\mid n$. At the same time, it can be seen that -\begin{split} -T_n(x) \equiv x^n\pmod{x^5-1,\ 3}\quad&\Longleftrightarrow\quad n\equiv 3,\ 27,\ 38,\text{ or } 137\pmod{205}\\ -T_n(x) \equiv x^n\pmod{x^5-1,\ 7}\quad&\Longleftrightarrow\quad n\equiv 7,\ 343,\ 858,\text{ or } 4797\pmod{6005}\\ -T_n(x) \equiv x^n\pmod{x^5-1,\ 13}\quad&\Longleftrightarrow\quad n\equiv 13,\ 2197,\ 14268,\text{ or } 54927\pmod{71405} -\end{split} -and so on. In general, for a prime $p\equiv 2,3\pmod5$ -$$T_n(x) \equiv x^n\pmod{x^5-1,\ p}\quad\Longleftrightarrow\quad n\equiv p,\ p^3,\ p^5,\text{ or } p^7\pmod{5q_p},$$ -where $q_p$ is the period of $T_n(x)\pmod{x^5-1,\ p}$. -(Similar congruences hold for $r=7$.) -It is not clear why certain $n$ cannot satisfy such congruences modulo every $p\mid n$. I do not say that it is easy to find such $n$, but its existence seems quite plausible. -P.S. Also, notice that in the AKS test the value of $r$ is taken satisfying $r>\log(n)^2$ (in fact, even the order $o_r(n)>\log(n)^2$), and this makes huge difference. Perhaps, the present test can be saved from pseudoprimes as well by requiring $r$ be of the magnitude of $\log(n)$ or so. -UPDATE (2021-10-02). Congruences above have been corrected. Here is a Sage code for computing $q_p$.<|endoftext|> -TITLE: recognition of symmetric groups in GAP -QUESTION [7 upvotes]: In GAP (https://www.gap-system.org), there is a function IsSymmetricGroup, which tells you whether a subgroup of $S_n$ generated by given permutations is all of the $S_n$. It looks like it takes virtually no time, even in large examples I tried ($n=50$). What is the method behind this function? Is it so easy to recognise symmetric groups? -On a related note, suppose that we know that $\sigma_1$, ..., $\sigma_k$ generate $S_n$. Are there some economic algorithms to write some standard generators of $S_n$, e.g. the adjacent transpositions, in terms of these sigmas? - -REPLY [12 votes]: The method (I assume) uses Jordan's theorem, which says that an primitive subgroup of $S_n$ with a cycle of prime order (at most $n-2,$ if memory serves) is either $A_n$ or $S_n.$ You rule out $A_n$ by looking at the generators, you show transitivity by randomly generating an $n$-cycle (of which there are a lot, so it does not take long to find one, there are other ways, too), and you show primitivity by finding a permutation which has a $p$ cycle for $n/2 < p < n-1$ (raising it to a power, you just get the $p$-cycle. There are lots of such, so basically generating a few thousand elements will do the trick. Notice that if, after generating the few thousand elements you DO NOT find the sorts of elements you want, the group is almost surely NOT the symmetric group (but the NO answer will be probabilistic, while the YES answer will be deterministic). -For more on this subject, check out my paper with Pemantle and Peres on invariable generation of symmetric groups: Pemantle, Robin; Peres, Yuval; Rivin, Igor, Four random permutations conjugated by an adversary generate (\mathcal{S}_{n}) with high probability, Random Struct. Algorithms 49, No. 3, 409-428 (2016). ZBL1349.05337.<|endoftext|> -TITLE: Can Homotopy Type Theory or algebraic geometry deal with homotopy fibers in terms of families? -QUESTION [11 upvotes]: ((In conclusion) -It was hard to choose which answer to accept. I decided for the one which addressed most of the various aspects of the question. -) -(Later addon) -I now decided to put a bounty on this, and want to put particular accent on one aspect of the question that puzzles me most. -As discussed in greater length below, at the first sight from the point of view of Homotopy Type Theory (HoTT) the situation seems next to trivial: if we view $f:X\to Y$ as a family $X_y:=f^{-1}(y)$ of spaces continuously varying over $Y$ then, if $Y$ is connected, HoTT squeezes water out of it and leaves the plain fact "all $y$ are equal to each other". This then implies that all $X_y$ are equal to each other. -But on the other hand, the above argument completely ignores the fact that the fibration associated to $f$ might be nontrivial, i.e. one might be unable to uniformly pick simultaneous equalities $X_y=X_{y'}$ for all pairs $y$, $y'$. So, on top of having this next to trivial description of homotopy fibers, seemingly HoTT must also possess techniques to detect nontriviality of the fibration, etc. What are they? -End(Later addon) -One of the most basic constructions in homotopy theory assigns to a map $f:X\to Y$ another map $\pi_f:P_f\to Y$ such that $P_f$ is equivalent to $X$ and (under connectedness assumption for $Y$) all $\pi_f^{-1}(y)$ are equivalent to each other. $P_f$ is the space of all pairs $(x,\gamma)$ where $x\in X$ and $\gamma$ is a path in $Y$ with $\gamma(0)=x$. -I would like to view this construction from a familial perspective. Namely, I want to view $f$ as a family $X_y:=f^{-1}(y)$ of spaces continuously varying over $Y$. Then, we are constructing another family $\check X_y$ from it, with$$ -\check X_y=\bigcup_{y'}\operatorname{paths}(y,y')\times X_{y'}. -$$That spaces in this family are pairwise equivalent is ensured by the fact that any path between $y_1$ and $y_2$ produces equivalences between $\operatorname{paths}(y_1,y')$ and $\operatorname{paths}(y_2,y')$ for all $y'$. -From this point of view, the equivalence of total spaces is clear from $$\bigcup_{y,y'}\operatorname{paths}(y,y')\times X_{y'}=\bigcup_{y'}\left(\bigcup_y\operatorname{paths}(y,y')\right)\times X_{y'},$$as the spaces in parentheses are contractible. -Now this way of formulating it suggests that there must be many other situations where a similar construction is present. -The most likely area is of course Homotopy Type Theory. I am pretty sure that something like this is used very frequently there. Could I have a specific reference - how is it called there and what does it mean conceptually from the type-theoretic/logical point of view? -It also looks like some sort of generalization for the associated bundle <-> principal bundle business. Is a generalization along these lines known? -I am especially interested in the context of moduli spaces in algebraic geometry. Some moduli spaces (like projective spaces or, more generally, Grassmanians) manifestly consist of equivalent objects but some others do not. Is there a way to modify moduli spaces in a way similar to the above, to arrive at some equivalent moduli spaces consisting of equivalent objects? -I have vague feeling that this construction might be related to switching from coarse moduli spaces to fine moduli stacks, does this make sense and can it be made rigorous? -If this is not too much, one more question. -A dual construction suggests itself,$$ -\hat X_y=\prod_{y'}X_{y'}^{\operatorname{paths}(y,y')}. -$$It is as easy to show that all $\hat X_y$ are equivalent to each other. However this time I think total space is not equivalent to the total space of the original family. Instead, the spaces of all global sections are equivalent. -Does this dual thing occur anywhere? This time I don't even know whether it is used in homotopy theory itself. - -REPLY [5 votes]: Yes, there is a certain sense in which your statements are true. As Mike Shulman and Qiaochu Yuan said, the strict fiber of a map cannot be defined in HoTT and doesn't make sense, but you can work from the other direction: instead of trying to associate a family of types to a map start with an arbitrary type family $P: X \to Type$. This includes any classical Serre fibrations: you associate to a point a fiber over it and to higher dimensional simplices some choice of their lifting, which is possible to do coherently by the path lifting property of fibrations. In the simplest case of a flat bundle you can make an explicit choice of lifts, which defines an equivalence between flat bundles and representations of the fundamental group of $X$. Homotopy-theoretically "every bundle is flat" in a sense that any two maps between fibers induced by a pair of equivalent paths are themselves (noncanonically) equivalent. The magic of HoTT and its models allows us to assume that all fibrations are given as representations of the path groupoid: path induction allows us to prove that any map of types is functorial in the sense that paths map to paths, composition to composition etc for all higher relations. This shows that a type family $P: X \to Type$ associates to any point $x: X$ a type $P(x): Type$, to any path $s: x=y$ a homotopy equivalence $ P(x) = P(y)$ etc (a map from paths in $Type$ to equivalences is automatic by path induction). In certain cases we can turn a coherent system of equivalences between fibers into a type family over $X$. The simplest case is when $X$ is defined as a higher inductive type, since higher inductive types are defined as representing objects of certain systems of equivalences between types. See chapter 6, in particular [6.12 The flattening lemma] in the HoTT book. -Given a type family $P: X \to Type$ we can construct its total space $Y := \sum_{x:X} P(x)$ with a canonical projection map $Y \xrightarrow{p} X$ defined by induction on $\Sigma$. Then we can consider the type family of homotopy fibers of $p$: $$x:X \vdash \mathrm{fib}_p(x) := \left( \sum_{y:Y} x = p(y) \right) : Type$$ -The resulting type family is equal to $P$ in the sense $\prod_{x:X} P(x) = \mathrm{fib}_p(x)$. By univalence we need to provide a homotopy equivalence between fibers. The map from left to right is the obvious inclusion of a strict fiber into a homotopy one. The map from right to left is constructed via induction on $\Sigma$ and paths. In the other direction, to any map $Y \xrightarrow{f} X$ we can associate its fiber type family $\mathrm{fib}_f$ and its total space $\sum_{x:X} \mathrm{fib}_f(x)$. It will be equivalent to $Y$ as objects over $X$. The relevant reference here is [4.8 The object classifier] in the HoTT book. -Essentially the big statement at work here is that from the PoV of higher category theory, the core groupoid for the overcategory $\mathbb{S}_{/X}$ is "representable" by a certain object $Type: \mathbb{S}$ (more formally, we consider a filtration of the overcategory by subcategories of bounded increasing cardinalities and require that each of these functors should be representable, thus getting an increasing sequence of objects $Type_{\kappa}$, see HTT [6.1.6. $\infty$-topoi and classifying objects]). In classical 1-topos theory and set theory this statement is limited to $(-1)$-truncated morphisms and type families and corresponds to the comprehension principle: any subset can be defined by a proposition which is true exactly for the elements of this subset, and vice versa. Type theory extends this principle to arbitrary constructions: any statement $P:X \to Type$ defines a type of "structures" on $X$ (the sigma type), and equivalently any map of types can be defined by some statement in a sense of a type family. -Regarding your "dual" construction: your type is defined as $\prod_{y^\prime} \prod_{y = y^\prime} X_{y^\prime}$. By path induction this type is equivalent to $X_y$, so your type family is equivalent to $X$. -Regarding the moduli spaces: I don't think this construction has really anything to do with them. Not that it's entirely irrelevant, but moduli spaces carry much more structure than homotopy types and you can't get around that unless you're willing to sacrifice most information. One possibility is to consider moduli spaces as stacks of groupoids on the Zariski (étale, flat, ...) site. Sheaves of groupoids naturally form a higher topos and should be a model of HoTT, however the information you get this way is mostly trivial: if your classified objects don't have any automorphisms, then the stack will be just a sheaf of sets, and if you have automorphisms, then you'll get the tautological description of locally isomorphic objects. Another possibility would be to enlarge your notion of equivalence, for example you could state that all fibers of the universal family are equivalent, e.g. by saying that an equivalence of $E$ and $E^\prime$ can be given as any fibration $X$ over $\mathbb{A}^1$ such that $X(0) \simeq E$ and $X(1) \simeq E^\prime$. This feels a lot like $\mathbb{A}^1$-homotopy theory, but I can't state anything more meaningful. In any case it feels like most (all?) of the algebraic information would be lost, even if one is very careful with definitions (I already see some problems in the one above). -Abstractly, the story of the moduli spaces is the story about classifying ringed topoi, while HoTT speak only about objects of a single higher topos (or topoi that arise as their overcategories). Sort of orthogonal beasts. -Regarding your "associated bundle --- principal bundle" comment. Indeed, one can view this construction for connected spaces in a similar way. Assume given a point $x:X$ and $X$ connected, then fibrations $P:X \to Type$ over $X$ are equivalent to homotopy representations of the ($A_\infty$-)group of loops $\Omega_x X$ in the automorphisms $Aut(P(x))$ of the fiber $P(x)$. This is certainly true in classical homotopy theory of spaces, but in HoTT there are complications. First there is a problem with the definition of $A_\infty$-groups and their representations, which we will ignore. The construction itself in one direction is obvious, we already know that each loop induces an automorphism of the fiber. In the other direction $y=x$ is a right torsor over $\Omega_x X$ so we can take the product over the loop group $(y = x) \times_{\Omega_x X} P(x)$. Classically this gives us back $P$, but in HoTT this requires clarification. -We define a $(\infty, 1)$-category $C$ as a type of objects $Ob_C: Type$ and a family of morphism types $a,b:C \vdash C(a,b): Type$, together with units and composition which we suppress in the following discussion. As usual, a groupoid is a category in which all morphisms are invertible, and for any category $C$ we have its core groupoid $C^\sim$ which is the largest subgroupoid in $C$. For any $X: Type$ its path groupoid $\Pi X: Cat$ has $Ob_{\Pi X} = X$ and $a, b:X \vdash \Pi X(a, b) := (a = b) : Type$. A category is called univalent if the natural functor $\Pi Ob_C \to C^\sim$ is an equivalence. Our definition of a category follows the Segal space approach, and univalent categories are precisely complete Segal spaces. The category of types $Type$ has $Ob_{Type} = Type$ and $X,Y : Type \vdash Type(X, Y) := (X \to Y)$. The classical axiom of univalence is precisely the statement that $Type$ is an univalent category. We also define the geometric realization functor $|C|$ as the left adjoint to the path groupoid functor, i.e. $Type(|C|, X) = Cat(C, \Pi X)$. For any ($A_\infty$-)group or monoid $G$ we define the delooping category $\mathbb B G$ as $Ob_{\mathbb B G} = 1$, $\mathbb B G(\ast, \ast) = G$. Presheaves of types on $\mathbb B G$ are the same as representations of $G$. -Now, the correspondence between type families and group representations stated above in these terms says that for any connected $X: Type$ with $x: X$ the categories $\mathbb B \Omega_x X$ and $\Pi X$ are Morita equivalent, i.e. their categories of type presheaves are equivalent. We have an obvious inclusion $\mathbb B \Omega_x X \to \Pi X$, however this is not an equivalence of categories, since an equivalence on object types would restrict to the statement $\forall y: X, x = y$ which means that $X$ is contractible. This is in stark contrast with the 1-categorical situation where any groupoid is equivalent to its full subgroupoid on one object. The classical statement relies on a choice principle to choose an equivalence for each nonempty $C(x, y)$, but constructively we can't have such strong choice. One can argue that this failure is due to non-univalence of $\mathbb B \Omega_x X$ and it's true in a sense, but I have two objections. The first is that the truth of this equivalence for univalent subcategories is itself a nontrivial theorem, essentially it is May's delooping theorem together with its version for path groupoids. Using such strong statements for a basic notion of equivalence doesn't seem like a good idea. The second is that the delooping category is an extremely natural object, and being unable to naturally talk about it would be a problem. I also find it troubling that the simple construction of delooping category for a monoid would suddenly blow up into a much more complex path category if the monoid is a group. Similarly, it is natural to select subcategories on a set of objects, or consider action groupoids for groups, and none of them would be categories under the univalence requirement (e.g. the action groupoid on a torsor would have "too few" morphisms to be univalent and to be equivalent to its geometric realization, which is a point). Even the basic 1-categorical definition of a category wouldn't be a category if we require all categories to be univalent. Sure, we could do the Rezk completion, but that extra step would be unnecessary in half of situations and very nontrivial in the other half. -Non-univalent categories will break the homotopy hypothesis, but I view it as a lesser evil in this case. Groupoids should be at least as complicated as homotopy types, but there is no reason why they couldn't contain more information. -That said, univalence of $Type$ is a very natural requirement. In particular, I don't know how to prove the Morita equivalence above without the univalence axiom, or even if it's true. With univalence it is easy to prove that for any groupoid its category of presheaves is equivalent to the category of type families over its geometric realization: -$$\begin {eqnarray} -C & \to & Type & \simeq\\ -C & \to & Type^\sim & \simeq \\ -C & \to & \Pi Type & \simeq \\ -|C| & \to & Type & \simeq \\ -\Pi |C| & \to & Type -\end{eqnarray} -$$ -Together with the delooping theorem $|\mathbb B \Omega_x X| = X$, $|\Pi X| = X$ this proves the Morita equivalence above. The part about path groupoids is easy to prove, but I'm not sure that the part about $\mathbb B$ can be proved without univalence. -In summary, assuming univalence, type families over connected inhabited types can be uniquely reconstructed from their fiber over some point together with an action of the loop group on it. At the moment there is no type-theoretic reference for the constructions that I performed above, but something like that should be in books on higher categories and topology.<|endoftext|> -TITLE: Diameter of a weighted Hamming cube -QUESTION [7 upvotes]: Let $C=(\mathbb{Z}/2\mathbb{Z})^N$ be the Hamming cube with its usual graph structure, and assume each edge $e=(x,x+\epsilon)$ (where $x\in C$ and $\epsilon$ has one $1$ and $N-1$ zeros) is given a length $\ell(x,\epsilon)$ satisfying the constraints -$$ \sum_\epsilon \ell(x,\epsilon) \le 1 \qquad \forall x\in C.$$ -Denote by $d_\ell$ the distance function induced by these lengths. - -Q1: What is the growth of the largest possible diameter $D(N)$ for such $(C,d_\ell)$? - -I think I can prove a bound $D(N) \lesssim log(N)$, but I wonder if a constant bound could be true. -In fact, this is not the real question I have, but it seemed a reasonable way to get to it. The real question is about functions $f:C\to \mathbb{R}$ such that -$$(1) \qquad \sum_\epsilon \lvert f(x+\epsilon) - f(x) \rvert \le 1 \qquad \forall x\in C$$ -and of vanishing average. - -Q2: What is the largest possible value $\lVert f\rVert_\infty$ for such a function? - -(My motivation is a toy example to test some concentration inequalities for Markov chains. In some cases I figured the semi-norm $\max_x \sum_\epsilon \lvert f(x+\epsilon)-f(x)\rvert$ should be efficient for the Glauber dynamic (aka lazy random walk) on $C$, but I need the above to get a good spectral gap in the corresponding functional space) - -REPLY [4 votes]: This was too long for a comment, Fedor Petrov gave a very nice answer with sharp constant (see also fedja's comment with a finite constant), and I just want to give the reference where the problem was solved back in 2000, for $f$ taking values in an arbitrary normed linear space $B$. -The OP is asking a particular case of what is now called Pisier's inequality for $p=\infty$ without $\log(n)$ factor. -Let $f:\{-1,1\}^{n} \to B,$ where $B$ is linear normed space. Define -$$ -D_{j}f(x) = \frac{f(x)-f(x^{j})}{2}, -$$ -where $x^{j}=(x_{1}, \ldots, -x_{j}, \ldots, x_{n})$, i.e., it puts negative sign in front of $j$-th coordinate. Then Pisier proved - -Theorem (Pisier, 1986). For all $1\leq p \leq \infty$, and any $f:\{-1,1\}^{n} \to B$ we have - $$ (\mathbb{E}\|f-\mathbb{E}f\|^{p})^{1p} \leq C - \log(n)\left(\mathbb{E}_{x} \mathbb{E}_{y}\|\sum_{j=1}^{n} - y_{j}D_{j}f(x) \|^{p}\right)^{1/p}. $$ - -Later Talagrand (1993) showed that in general for $p<\infty$, $\log(n)$ factor is sharp. However in case of $B=\mathbb{R}$, $\log(n)$ factor can be removed but with $C=C(p)$ (again $p<\infty$), and for the real-valued functions, the inequality is sometimes called Poincar\'e inequality, because the right hand side (by Khinchin's inequality) can be replaced by a more familiar discrete gradient: -$$ -\|\nabla f\|_{p}:= \left(\mathbb{E}\left(\sum_{j=1}^{n} |D_{j}f|^{2}\right)^{p/2}\right)^{1/p} -$$ -Finally, Wagner (2000) closed the gap by showing - -Theorem (Wagner, 2000). For $p=\infty$ Pisier's inequality holds without $\log(n)$ factor. - -This answers OP, because for $B=\mathbb{R}$, Pisier's inequality becomes (without $\log(n)$ factor) -$$ -\|f-Ef\|_{\infty} \leq C \| \sum_{j=1}^{n}|D_{j}f| \|_{\infty}. -$$ -Pisier, Gilles, Probabilistic methods in the geometry of Banach spaces, Probability and analysis, Lect. Sess. C.I.M.E., Varenna/Italy 1985, Lect. Notes Math. 1206, 167-241 (1986). ZBL0606.60008. -Wagner, R., Notes on an inequality by Pisier for functions on the discrete cube, Milman, V. D. (ed.) et al., Geometric aspects of functional analysis. Proceedings of the Israel seminar (GAFA) 1996-2000. Berlin: Springer. Lect. Notes Math. 1745, 263-268 (2000). ZBL0981.46021. -Talagrand, M., Isoperimetry, logarithmic Sobolev inequalities on the discrete cube, and Margulis’ graph connectivity theorem, Geom. Funct. Anal. 3, No. 3, 295-314 (1993). ZBL0806.46035.<|endoftext|> -TITLE: A criterion for second countability -QUESTION [8 upvotes]: Let $(X,\tau)$ be a topological space. -Assume for any arbitrary topological base $\mathcal{E}$ of $\tau$ we have that: the Borel sigma algebras coming form $\mathcal{E}$ and $\tau$ are the same. Can we conclude that $X$ is second countable ?! -This question is also asked when $X$ is a locally convex space. Please read the comments below. - -REPLY [4 votes]: A counterexample to this question (and its locally convex version) is any non-metrizable (locally convex) space $X$, which is hereditarily Lindelof. The hereditary Lindelofness of $X$ implies that any open set is a countable union of basic open sets and this implies that the $\sigma$-algebra generated by any base of the topology coincides with the Borel $\sigma$-algebra. -As for an example of non-metrizable hereditarily Lindelof spaces, take any metrizable separable locally convex space with a weaker non-metrizable topology. -Being a continuous image of a second-countable space, such space will have countable network and hence will be hereditarily Lindelof. -For example, you can take any infinite-dimensional separable Banach space endowed with the weak topology. It will be not metrizable but hereditarily Lindelof. -The function spaces $C_p(X)$ over cosmic spaces $X$ have countable network and hence are hereditarily Lindelof. The function spaces $C_k(X)$ with compact-open topology over $\aleph_0$-spaces $X$ have countable $k$-network and hence are hereditarily Lindelof. -The locally convex space $\mathbb R^\infty$, which is inductive limit of an increasing sequence of finite-dimensional spaces, has countable network and hence is hereditarily Lindelof and not metrizable. -So, there plenty of examples. But one can modify the question replacing the second countability by the hereditary Lindelofness: -Problem. Is a (locally convex) topological space hereditary Lindelof if the $\sigma$-algebra generated by any base of the topology coincides with the $\sigma$-algebra of Borel sets? -Remark. The example given by @Wille Liou in the comment to the original question is hereditarily Lindelof (even hereditarily compact), but not Hausdorff. -In fact, the hereditary Lindelofness admits the following characterization: -Theorem. A topological space $X$ is hereditary Lindelof if and only if for any subspace $Y\subset X$, the $\sigma$-algebra generated by any base of the topology of $Y$ coincides with the Borel $\sigma$-algebra of $Y$.<|endoftext|> -TITLE: Is the Mackey topology $\tau(l^{\infty},l^{1})$ strongly Lindelöf? -QUESTION [6 upvotes]: Let $l^{\infty}$ (respectively, $l^{1}$) be the space of bounded -(respectively, absolutely summable) real sequences. I need to find out if -$l^{\infty}$ equipped with the Mackey topology $\tau(l^{\infty},l^{1})$, i.e. -the finest locally convex topology that leads to the topological dual $l^{1}$, -is strongly/hereditarily Lindelöf. -This is a curious case because $l^{\infty}$ equipped with weak* topology is -strongly Lindelöf (as a countable union of second countable balls), while -it is not Lindelöf with respect to the norm-topology. The Mackey topology -is finer than the former and coarser than the latter. -Thanks in advance! - -REPLY [4 votes]: This is true and follows from the fact that in this case the Mackey topology agrees with the weak $\ast$ topology on balls.<|endoftext|> -TITLE: Is Bing's countable connected space topologically homogeneous? -QUESTION [14 upvotes]: In this paper R.H. Bing has constructed his famous example of a countable connected Hausdorff space. -The Bing space $\mathbb B$ is the rational half-plane $\{(x,y)\in\mathbb Q\times \mathbb Q:y\ge 0\}$ endowed with the topology consisting of the sets $U\subset \mathbb B$ such that for any $(a,b)\in U$ there exists $\varepsilon>0$ such that each point $(x,0)$ with $\min\{|x-(a-b/\sqrt{3})|,|x-(a+b/\sqrt{3})|\}<\varepsilon$ belongs to $U$. -It is easy to see that for every rational numbers $a>0$ and $b$ the affine map $f_{a,b}:\mathbb B\to\mathbb B$, $f_{a,b}:(x,y)\mapsto (ax+b,ay)$, is a homeomorphism of the Bing space $\mathbb B$. -This impliees that the action of the homeomorphism group on $\mathbb B$ has at most two orbits. -Problem. Is the Bing space $\mathbb B$ topologically homogeneous? -Remark. It seems that the first example of a topologically homogeneous countable connected Hausdorff space was constructed by Joseph Martin. A simple transparent example of such space is the rational projective space $\mathbb QP^\infty$ discussed in this MO post. It is interesting who first realized that the space $\mathbb QP^\infty$ is connected? - -REPLY [6 votes]: The Bing space is topologically homogeneous. -The proof of this fact can be found here. It is a bit long (to be reproduced here) and uses the standard back-and-forth argument.<|endoftext|> -TITLE: A weak version of the Whitehead Theorems -QUESTION [5 upvotes]: Let $f:X\longrightarrow Y$ be a map between CW-complexes $X$ and $Y$. By the Whitehead Theorems, if one of the conditions: -1- (homotopy version) $\pi_n (f):\pi_n (X)\longrightarrow \pi_n (Y)$ is an isomorphism for all $n\geq 1$, -or -2- (homology version) $\pi_1 (f):\pi_1 (X)\longrightarrow \pi_1 (Y)$ and $H_n (\tilde{f}):H_n (\tilde{X})\longrightarrow H_n (\tilde{Y})$ are isomorphisms for all $n\geq 2$, -hold, then there is a map $g:Y\longrightarrow X$ such that $g\circ f\simeq id_X$ and $f\circ g\simeq id_Y$. -Question: Is there any weaker condition (with respect to above conditions) under which there is a map $g:Y\longrightarrow X$ such that we have only $g\circ f\simeq id_X$? - -REPLY [5 votes]: Suppose you have a map $f\colon X\to Y$ of finite CW complexes such that $K(p,n)_*(f)$ is injective for all primes $p$ and integers $n\geq 0$ (where $K(p,n)$ is Morava $K$-theory). Then the Nilpotence Theorem of Hopkins, Devinatz and Smith implies that the map $\Sigma^kf^{(m)}\colon \Sigma^kX^{(m)}\to\Sigma^kY^{(m)}$ has a left inverse $g$ for sufficiently large $k$ and $m$. This may not seem very satisfactory but I think that it is the best that you can reasonably hope to do. The question of whether a map has a left inverse is just intrinsically much more subtle than the question of whether it is an equivalence. -One can also say that in the category of finite spectra, a map $f\colon X\to Y$ has a left inverse iff $\pi_*^S(f)\colon \pi_*^S(X)\to\pi_*^S(Y)$ is injective, provided that Freyd's Generating Hypothesis is correct. But the Generating Hypothesis has been any open question for fifty years now, which is another indication that the problem is intrinsically hard.<|endoftext|> -TITLE: Special values of adjoint $L$-functions of automorphic representations of $\mathrm{GSp}(4)$ as Petersson norms -QUESTION [10 upvotes]: Here I consider cuspidal automorphic representations $\pi$ over the similitude group $\mathrm{GSp}(4,\mathbb{A}_\mathbb{Q})$. Let $f$ be a non-zero vector in the representation $\pi$. I want to know if there is any reference/work on relating special values of the complex adjoint $L$-function $L(s,\pi,\mathrm{Ad})$ of $\pi$ to the Petersson norm $\langle f,f\rangle$. -I know there is an article of Atsushi Ichino ('On critical values of adjoint $L$-functions for $\mathrm{GSp}(4)$') on this topic. Yet this article assumes $\pi$ to be unramified over all the finite places and to be of a special type over the archimedean place. So can we remove or weaken these assumptions? Is there any work after this? -Any comment or suggestions would be welcome. - -REPLY [10 votes]: For $\mathrm{GL}_2$, the relationship between the Petersson norm of a newform $f$ and its adjoint $L$-function is roughly a statement of the form -\[\frac{|a_f(1)|^2}{\langle f, f\rangle} = \frac{c_f}{\Lambda(1, \operatorname{ad} f)},\] -where $a_f(n)$ denotes the first Fourier coefficient of $f$, $\Lambda(s,\pi)$ denotes the completed $L$-function (including the archimedean components, which are essentially products of gamma functions), and $c_f$ is an explicit constant that essentially only depends on the behaviour of $f$ at the ramified primes that one can make completely explicit. (However, I don't think this explicit formulation has been written down anywhere in the literature, though it certainly can be done by, say, combining Lemma 4.2 of this paper of mine with the results in Section 1 of this paper of Gelbart and Jacquet.) -For $\mathrm{GSp}_4$, the Fourier expansion is of the form -\[f(Z) = \sum_{S} a_f(S) e^{2\pi i \operatorname{Tr}(SZ)},\] -where the sum is over matrices $S = \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix}$ such that $a,b,c \in \mathbb{Z}$, $a > 0$, $b^2 - 4ac = D < 0$. -We should expect that $|a_f(1_2)|^2/\langle f, f\rangle$ is related to a special value of $L$-functions in some way. Unlike the $\mathrm{GL}_2$ case, this special value of $L$-functions is (conjectured to be) much more complicated for $\mathrm{GSp}_4$. This is a special case of Böcherer's conjecture, which in this setting states that -\[\frac{|a_f(1_2)|^2}{\langle f, f\rangle} = c_f \frac{\Lambda(\frac{1}{2}, f) \Lambda(\frac{1}{2}, f \otimes \chi_{-4})}{\Lambda(1, \operatorname{ad} f)}.\] -A particular case of this has recently been proved by Furusawa and Morimoto (Theorem 1). -Note that $a_f(S_1) = a_f(S_2)$ if $S_1 = A^t S_2 A$ for some $A \in \mathrm{SL}_2(\mathbb{Z})$. This means that for any imaginary quadratic field $K = \mathbb{Q}(\sqrt{D})$, noting that that there is a bijection between the class group $\mathrm{Cl}_K$ and equivalence classes of matrices $S$, we may study -\[R_f(K) = \sum_{S \in \mathrm{Cl}_K} a_f(S).\] -Then Böcherer's conjecture in this setting states that -\[\frac{|R_f(K)|^2}{\langle f, f\rangle} = c_{f,D} \frac{\Lambda(\frac{1}{2}, f) \Lambda(\frac{1}{2}, f \otimes \chi_D)}{\Lambda(1, \operatorname{ad} f)}.\] -Taking $D = -4$, so that $S \in \mathrm{Cl}_K$ implies that $S \sim I_2$, yields the previous formulation. One can additionally insert a class group character into the definition of $R_f(K)$, which leads to different $L$-functions in the conjecture. -Should one wish to make the constant $c_f$ explicit, one can look at the work of Dickson, Pitale, Saha, and Schmidt, where they have refined this conjecture to explicitly determine the constant $c_f$. They still require several assumptions; see Theorem 1.12 of their paper for the precise statement. -In general, these kinds of conjectures/statements are easy to state at unramified primes, but take a lot of work to understand at ramified primes; this can be seen, for example, in the work of many people (Nelson, Nelson-Pitale-Saha, Hu, Collins, etc.) of determining the explicit expression for the inner product of three $\mathrm{GL}_2$ newforms via the Watson-Ichino formula.<|endoftext|> -TITLE: Are there any useful conditions for a biclosed monoidal structure on presheaves to descend to a biclosed monoidal structure on sheaves? -QUESTION [5 upvotes]: Suppose $C$ is a small category with a monoidal structure. Then by the special case of the Day convolution theorem for presheaves, $\operatorname{Psh}(C)$ is equipped with a corresponding biclosed monoidal structure. If $C$ is equipped with a Grothendieck topology, is there any useful condition for when the biclosed monoidal structure on presheaves descends to a biclosed monoidal structure on the category of sheaves for that Grothendieck topology? - -REPLY [5 votes]: There is a more general form of Day's theorem that does pretty much that, at least for sub-canonical topologies: -Theorem (Day): Let $C$ be a complete and co-complete Category, and $D \subset C$ a full subcategory of $C$ endowed with monoidal structure which contains a full subcategory dense in $C$. Then if there exists functors: -$H',H: D^{op} \times C \rightarrow C$ and natural isomorphisms: -$Hom(S,H(S',X)) = Hom(S \otimes S',X) = Hom(S',H'(S,X))$ -for $S,S' \in D$ and $X \in C$. -Then there is a unique biclosed monoidal closed structure on C such that the inclusion $D \rightarrow C$ extend into a monoidal functor. -The theorem as stated above appears and is proved (in french) as proposition 6.3 of this paper (Ara, Maltsiniotis), a variant of it which seems to relax the fullness of $D$ in $C$ is stated without proof as proposition 9 in this paper (Street). In both case there are references to two papers of Day which I havn't look at yet. -You can applies it to your question as follow, take $D$ to the the subcategory of representable pre-sheaves and $C$ the category of sheaves. The condition for the existence of $H$ and $H'$ is then just that given a sheaf $X$ and a representable $S$ then the presheaves $S' \mapsto Hom(S \otimes S',X)$ and $S' \mapsto Hom(S' \otimes S,X)$ are sheaves, i.e. a the condition is that "tensoring a covering by a fixed element gives you a covering". I'm convinced that this extend well to non-sub-canonical topologies, but I'm lacking of time to check it today (if someone do it, I'll be interested to know the answer)<|endoftext|> -TITLE: A more dense analog of the Mills' constant -QUESTION [8 upvotes]: Is there a real number $A$ such that $$\left \lfloor n^{A} \right \rfloor$$ is a prime number (for all natural numbers $n$)? It is obvious that $A>1+\epsilon$ from the prime number theorem. - -REPLY [21 votes]: No, such an $A$ does not exist. First, $A$ cannot be an integer because then $\lfloor n^A\rfloor $ is never a prime for $n\geq 2$. So, assume that $A$ is not an integer. Then by Weyl's equidistribution theorem the fractional parts of $n^{A}/2$ are equidistributed modulo $1$. In particular $\{n^A/2\}\in [0,1/2)$ infinitely often. So $\lfloor n^{A}\rfloor$ is infinitely often an even number.<|endoftext|> -TITLE: For each $n$: show there is a genus $1$ curve over some field $k$ with no points of degree less than $n$, (simple argument / best reference)? -QUESTION [18 upvotes]: What is the simplest example (or perhaps best reference) for the fact that there are genus $1$ curves (over a field of your choice --- or if you wish, over $\mathbb{Q}$, to make it more exciting) with no points of degree less than $n$? Brian Conrad gave a slick answer here: https://math216.wordpress.com/2011/04/22/fourteenth-post/ but I wonder if there is something that can be explained more simply. My expectation is that this isn't possible, because describing genus one curves extrinsically is hard, but there may be a clever trick I am missing. - -REPLY [10 votes]: There are already two great answers, but I want to post an answer that works over all local fields, such as $\mathbb{Q}_p$, based on an alternative philosophy. Instead of starting with an elliptic curve $(E,0)$ and studying torsors $X$ for that curve that have large index, first we start with a "simpler" ambient scheme $Y$ that manifestly has large index, and then we try to find a genus $1$ curve $X$ in that scheme. -Open Problem. For every Severi-Brauer variety $Y$ over a field $K$, i.e., for every $K$-scheme $Y$ such that $Y\times_{\text{Spec}\ K}\text{Spec}\ \overline{K}$ is $\overline{K}$-isomorphic to $\mathbb{P}^{n-1}_{\overline{K}},$ does there exist a genus $1$ $K$-curve $X$ and a $K$-morphism $X\to Y?$ -This problem was explicitly stated as part of an open problem session for the conference, "Ramifications in Algebra and Geometry", cf. Problems 2 and 3 of the following: http://www.mathcs.emory.edu/RAGE/RAGE-open-problems.pdf -There are positive results for general Severi-Brauer varieties and general fields for small values of the integer $n$. One such result was reported by David Saltman at a seminar in Fall 2016 at Stony Brook University: -https://www.math.stonybrook.edu/deptcalendar/event.php?ID=3964&Date=2016-11-02 -For some reason, the link to the Stony Brook University calendar is demanding a password(!), so here is a link to a seminar announcement for a similar seminar by David Saltman at NYU. -https://math.nyu.edu/dynamic/calendars/seminars/algebraic-geometry-seminar/846/ -Further positive results are in an article of A. J. de Jong and Wei Ho. -MR3091612 -de Jong, Aise Johan; Ho, Wei -Genus one curves and Brauer-Severi varieties. -Math. Res. Lett. 19 (2012), no. 6, 1357–1359. -https://arxiv.org/abs/1207.4810 -The following proposition is an answer to Problem 2, and is also an answer to Problem 3 for cyclic algebras over a "large" field. -Proposition. For every Severi-Brauer $K$-variety $Y$ arising from a cyclic $K$-algebra of rank $n^2$, there exists a unique $\text{Aut}(Y)$-orbit of the Hilbert $K$-scheme of $Y$ parameterizing nodal, elliptic normal curves whose (geometric) irreducible components are lines, and this component has a $K$-point parameterizing such a curve $X_0.$ If $K$ is a "large" field in the sense of Florian Pop, e.g., the fraction field of a Henselian DVR, then there are also smooth elliptic normal curves $X$ in $Y$ obtained as deformations of $X_0.$ If the period of the cyclic algebra equals $n$, then every curve in $Y$ has index divisible by $n$. -Corollary. For every local field $K$, e.g., $\mathbb{Q}_p$ or $\mathbb{F}_p((t)),$ for every order-$n$ element in the Brauer group $\text{Br}(K)\cong \mathbb{Q}/\mathbb{Z},$ there exists a cyclic $K$-algebra $A$ of rank $n^2$ representing this class, and there exists a smooth, geometrically connected, genus-$1$ $K$-curve $X$ in $Y$ whose index is divisible by $n$. -Before giving the proof of the proposition, recall the definition of cyclic algebras, as in Roquette's beautiful history of class field theory. -MR2222818 (2006m:11160) -Roquette, Peter -The Brauer-Hasse-Noether theorem in historical perspective. -Schriften der Mathematisch-Naturwissenschaftlichen Klasse der Heidelberger Akademie der Wissenschaften, 15. -Springer-Verlag, Berlin, 2005. vi+92 pp. ISBN: 3-540-23005-X -https://www.mathi.uni-heidelberg.de/~roquette/brhano.pdf -Let $K$ be a field, and let $n$ be an integer $\geq 2.$ Let $L/K$ be a finite field extension of degree $n$ that is Galois with cyclic Galois group $\langle \sigma \rangle.$ Let $a\in K^\times$ be an element. The cyclic algebra over $K$ associated to $L$ and $a$ is the $L$-vector space of dimension $n$, $$A(L/K,\sigma,a) := L\cdot 1 \oplus L\cdot u \oplus \dots \oplus L\cdot u^{n-1}, $$ with the unique $K$-central algebra structure determined by the relations, $$1\cdot x = x = x\cdot 1, \ \ u^n = a\cdot 1, \ \ u\cdot b = \sigma(b) \cdot u,$$ for every $x\in A(L/K,\sigma,a)$ and for every $b\in L.$ This is a central simple $K$-algebra. -The Severi-Brauer variety of $A(L/K,\sigma,a)$ is the smooth, projective $K$-scheme $Y$ that represents the functor on $K$-schemes $T$ associating to $T$ the set of all left ideals in $A(L/K,\sigma,a)\otimes_K \mathcal{O}_T$ that are locally direct summands of (free) rank $n.$ There is an evident inclusion of $L$ in $A(L/K,\sigma,1)$ as $L\cdot 1.$ That inclusion induces a left ideal in $A(L/K,\sigma,a)\otimes_K L$ giving an $L$-point of $Y.$ The Galois orbit of this $L$-point, $\Gamma \subset Y$, is a smooth closed $K$-subscheme of dimension $0$ and length $n.$ -In particular, the index of $Y$, i.e., the least positive length of a $0$-dimensional closed $K$-subscheme of $Y$, equals $n.$ So for cyclic algebras of period $n,$ i.e., the order of the corresponding Brauer class in the Brauer group of the field, the index equals the period, irrespective of the field $K.$ The Merkurjev-Suslin Theorem, and the refinements by Merkurjev, imply that for every prime integer $n$, the $n$-torsion subgroup of the Brauer group is generated (as a group) by the classes of cyclic algebras of rank $n^2$ as above. -Even when every Brauer element of order $n$ is a linear combination of classes of cyclic algebras of rank $n^2$, a typical order-$n$ element of this group is not represented by a cyclic algebra. There are many examples of central simple algebras whose index is much larger than the period. (The period divides the index, and they have the same list of prime factors. The exponent is the smallest integer $e$ such that the index divides the $e^{\text{th}}$ power of the period.) It is an open problem to relate the symbol length of a Brauer class (i.e., the fewest number of cyclic $n$-algebras necessary to generate a specified $n$-torsion Brauer class) and the index of the class. Here is one example. -Open Problem. For a function field $K$ of a surface over an algebraically closed field, for which the period always equals the index by de Jong's Period-Index Theorem, is every Brauer class represented by a cyclic algebra? -Proof of the Proposition. For the Severi-Brauer variety $Y$ of a cyclic algebra, there is a unique $\text{Aut}(Y)$-orbit of the parameter space of nodal, elliptic normal curves whose geometric irreducible components are lines: the $n$-gon of lines is uniquely determined by the unordered $n$-tuple of nodes $\Gamma,$ and any two linearly nondegenerate ordered $n$-tuples of points in $\mathbb{P}^{n-1}$ are projectively equivalent (the set of such projective equivalences is naturally a torsor for a torus of rank $n-1$). The claim is that this orbit has a $K$-point parameterizing such a curve $X_0.$ -Here is the construction. For the Galois orbit $\Gamma\subset Y$ constructed above, the automorphism $\sigma$ restricts to an automorphism of $\Gamma.$ There is a unique minimal closed subscheme $X_0\subset Y$ that contains $\Gamma,$ whose geometric irreducible components are lines, and such that for every geometric point $p:\text{Spec}\kappa \to \Gamma$, there is a $\kappa$-irreducible component of $X_0$ that contains both $p$ and $\sigma(p).$ Concretely, after base change to $L$, the union of the $n$ lines $\Lambda_r=\text{span}(\sigma^r(p),\sigma^{r-1}(p))$, $r=0,\dots,n-1,$ is Galois-invariant, hence equals the base change of a $K$-curve $X_0\subset Y$. -The $K$-curve $X_0$ is geometrically connected and geometrically reduced. -The curve $X_0$ is nodal: the point $\sigma^r(p)$ is contained in two irreducible components $\Lambda_r$ and $\Lambda_{r+1}.$ The arithmetic genus of $X_0$ equals $1$. Geometrically, $X_0$ is an elliptic normal curve in $\mathbb{P}^{n-1},$ i.e., it is linearly nondegenerate and linearly normal (necessarily of degree $n$). In fact, any two such curves in $\mathbb{P}^{n-1}$ are conjugate under the group $\text{Aut}(\mathbb{P}^{n-1})$ of projective linear transformations. Thus, there is a unique $\text{Aut}(X)$-orbit of such curves $X_0$ in $Y.$ -Finally, an obstruction group for infinitesimal deformations of a curve $X$ in $Y$ with ideal sheaf $\mathcal{I}$ is $$O_{X,Y}=\text{Ext}^1_{\mathcal{O}_X}(\mathcal{I}/\mathcal{I}^2,\mathcal{O}_X).$$ This is compatible with base change from $K$ to $L$, where $Y_0$ equals a union of $n$ lines. Since the normal bundle of $\Lambda_r$ equals $\mathcal{O}(1)^{\oplus (n-2)}$, and since even after twisting down by $\sigma^r(p)$ and $\sigma^{r-1}(p)$, the twisted sheaf $\mathcal{O}(-1)$ on the line has vanishing $h^1$, it follows that the obstruction group is the zero group, and infinitesimal deformations smooth all nodes. Thus, the Hilbert $K$-scheme parameterizing elliptic normal curves in $Y$ is smooth at the point parameterizing $X_0,$ and the unique irreducible component of the Hilbert scheme containing this point has a dense open subscheme $U$ parameterizing smooth elliptic, normal curves. -If the field $K$ is "ample" or "large" in the sense of Florian Pop, then there are $K$-points of $U$. QED -Proof of the Corollary By Hensel's Lemma, the fraction field of every Henselian DVR is "large". For a local field $K$ such as $\mathbb{Q}_p$ or $\mathbb{F}_p((t)),$ every period-$n$ element in the Brauer group $\text{Br}(K)\cong \mathbb{Q}/\mathbb{Z}$ is represented by a cyclic $K$-algebra of rank $n^2$ by the Brauer-Hasse-Noether-(Albert) Theorem and Hasse's Structure Theorem. By the proposition, there exist $K$-points of $U$ parameterizing smooth elliptic normal curves $X$ in $Y$. QED<|endoftext|> -TITLE: Why are $\sigma$-algebras preferable to $\sigma$-rings? -QUESTION [6 upvotes]: The following is said without further explanation in Folland's Real Analysis: - -Some authors prefer to take the domains of measures to be $\sigma$-rings rather - than $\sigma$-algebras. The reason is that in dealing with "very large" - spaces one can avoid certain pathologies by not attempting to measure "very large" - sets. However, this point of view also has technical disadvantages, and it is no longer - much in favor. - - -Could anyone come up with some references where the authors "take the domains of measures to be $\sigma$-rings rather -than $\sigma$-algebras"? -What examples of "pathologies" may Folland refers to? -What "technical disadvantages" does "this point of view" have? - -REPLY [7 votes]: Reference: Measure Theory by Halmos. -An example of a pathology: If measures are not $\sigma$-finite, you have to be careful how to formulate Fubini's theorem. The old way of doing this was to have the measure defined on the $\sigma$-ring of those sets on which the measure is σ-finite. The current way of doing this is to assume that the measure is "locally determined". -An example of a technical disadvantage: Defining spaces such as $L^\infty$ is easier when the whole space is in the domain of the measure.<|endoftext|> -TITLE: How do I know if an irreducible representation is a permutation representation? -QUESTION [14 upvotes]: I have a vague question, a less vague question and a lot of vaguer questions about permutation representations of a finite group $G$. - -Vague question. Recall that if $G$ acts on a finite set $X$, we get a permutation representation $$G \to GL_{\lvert X \vert}(\mathbb C).$$ (Unless $X$ is very small,) this representation is never irreducible, for $\mathbb C (1, 1, \ldots, 1)^T$ splits off as a subrepresentation. What's left is the permutation representation $V_X$ associated to $G$. A classical fact is that $V_X$ is irreducible if and only if the original action was $2$-transitive. My question is: conversely, how do I know if some irreducible representation $V$ (whose character I know) is obtained by this construction? Clearly, a necessary condition is that $\chi_V$ should only take integer values $\geq -1$, but is it sufficient? If not, do we have another criterion? -Less vague question. (A special case of the first question). If I have not blundered, $\mathfrak S(6)$ has a degree-nine representation whose character only takes the values $-1$, $0$, $1$, $3$ and $9$. If you believe that $\mathfrak A(6) \simeq PSL_2(\mathbb F_9)$, you could get it by first considering the permutation representation $V_{\mathbb P^1(\mathbb F_9)}$ of $\mathfrak A(6)$, and inducing it to $\mathfrak S(6)$. This splits off as $$\mathrm{Ind}_{\mathfrak A(6)}^{\mathfrak S(6)} V_{\mathbb P^1(\mathbb F_9)} = V \oplus (V \otimes \epsilon),$$ where $\epsilon$ is the sign morphism. Is this $V$ a permutation representation? (Note that $\mathfrak S(6)$ is not isomorphic to $PGL_2(\mathbb F_9)$, so the action on the projective line does not extend to $\mathfrak S(6)$ in a trivial way.) I believe this representation is not permutation, but I have no proof and not much confidence in my intuition (a brute-force examination of all index-nine subgroups of $\mathfrak S(6)$ would work, but I'd rather avoid hard work and I don't know any conceptual arguments or how to use modern computer tools to figure this out). -Vaguer questions. The permutation representation machinery gives a morphism -$$\mathbf{Burn}(G) \to \mathbf{R}(G)$$ -from the Burnside ring of $G$ to its representation ring. Better still, it gives a morphism to all representation rings (whatever the field). What can we say about this morphism, beyond Wikipedia's example showing it can be noninjective and nonsurjective? Is there any textbook where this morphism is considered quite thoroughly? Do you know of any work on representation theory where the $\mathbf{Burn}(G)$-algebra structure on $\mathbf R(G)$ has been usefully exploited? - -I realize a lot of my questions are very imprecise and I apologise unreservedly for that. I would nonetheless appreciate any comment, answer or reference recommendation about this topic. -(Comment: I asked this question on MSE a few days ago, but got no answer, hence this question.) - -REPLY [11 votes]: The map from the Burnside ring to the representation ring is well studied. Andreas Dress wrote many papers on this in the 1970's. See also work of Tammo tom Dieck, who was applying things to equivariant topology. -There is a detecting character ring for each of these. Characters for elements of the Burnside ring are integer valued functions from conjugacy classes of subgroups of $G$, and an evident map between the two character rings. A classic character $\chi$ that is integer valued is in the image of this map exactly when $\chi(g) = \chi(h)$ if $g$ and $h$ generate conjugate cyclic groups. -In a 1987 paper in Comm. Math. Helv., Peter Webb gives an explicit basis for the kernel of the map from the Burnside ring to the Representation ring, after inverting the order of $G$. -This should get you going, if you want to learn about this.<|endoftext|> -TITLE: Ricci flow is not a gradient flow for $L^2$-space of metrics -QUESTION [6 upvotes]: I am reading Ben Andrews book about Ricci flow and at the start of the chapter about Perelman's gradient flow formulation for Ricci flow he says Robert Bryant exposed that there are no functionals defined on the $L^2$-space of Riemannian metrics that promotes Ricci flow as a gradient flow. -Does anyone know the name of this paper? The book does not include it on their references. - -REPLY [6 votes]: If there were such a functional $\mathcal{F}$, observe that - -Under Ricci flow the functional would have to decrease. That is, if $\partial_t g(t) = -2 Rc[g]$ then $\partial_t \mathcal{F}(g(t)) \leq 0$, with strict inequality of $-2 Rc[g] \neq 0$. -The functional would have to be invariant under diffeomorphisms, that is if $\Phi$ is a diffeomorphism of the manifold then $\mathcal{F}(\Phi^*g) = \mathcal{F}(g)$. This is because the Ricci flow is invariant under diffeomorphism. - -The point is that Robert Bryant found a metric on $\mathbb R^n$ which moves under Ricci flow only by diffeomorphisms (which is called a steady soliton). That is, there is a metric $g_S$ and a family of diffeomorphisms $\Phi_t$ such that $g(t) = \Phi^*_t g_S$ is the solution of Ricci flow starting from $g_S$. By point 2 above, $\mathcal{F}(g(t)) = \mathcal{F}(g_S)$, which contradicts point 1. -The details of the constructing the soliton are written by Robert Bryant here. Many other gradient steady solitons have been discovered since then. -Edit: As pointed out by Rbega, this argument only shows that such a functional would have to be infinite on the Bryant soliton. We can say that mean curvature flow is the gradient flow for area. The gradient of area is defined even for surface with infinite area, by consider compact perturbations. Therefore, we need a stronger argument. -If there was a compact steady soliton, then the argument above would work (assuming that $\mathcal{F}$ is always finite on compact manifolds). However, there is no compact steady soliton (except for Ricci flat manifolds). -The stronger argument comes from the existence of shrinking solitons on compact manifolds. The first known example is the Koiso-Cao soliton. A shrinking soliton is a pair of a metric $g$ and a vector field $X$ which solves -$$-2Rc = - g + \mathcal{L}_X g,$$ -equivalently -$$-2Rc_{ij} = - g _{ij}+ \nabla_i X_j + \nabla_j X_i.$$ -Under Ricci flow, a steady soliton shrinks and also move by diffeomorphisms integrating $X$. -If $-2Rc$ where the gradient of $\mathcal{F}$ then, -$$-Grad \mathcal{F} = - g + \mathcal{L}_X g$$ -I claim this can't be true unless $\mathcal{L}_X g = 0$. Note that the two parts of this tensor field are orthogonal in $L^2$, because -$$\int_{M} (g,\mathcal{L}_X g)_g dVol_g = \int_M tr(\mathcal{L}_X g) dVol_g = \int_M div_g(X) dVol_g = 0.$$ -Therefore we have -\begin{align} -|\mathcal{L}_X g|^2_{L^2} -&= (-Grad \mathcal{F}, \mathcal{L}_X g)_{L^2} \\ -&=-\partial_{\epsilon} \mathcal{F}(g + \epsilon \mathcal{L}_X g) \\ -&= -\partial_{\epsilon} \mathcal{F}(\Phi_{\epsilon}^* g) = 0 -\end{align} -The first equality is by the orthogonality of $-g$ and $\mathcal{L}_X g$, the second is the defining property of the gradient.<|endoftext|> -TITLE: A curious determinantal inequality I -QUESTION [10 upvotes]: Let $A, B$ be Hermitian matrices. Does the following hold? -$$\det(A^{2}+B^{2}+|AB+BA|)\leq \det(A^{2}+B^{2}+|AB|+|BA|)$$ -As usual, $|X|=(X^*X)^{1/2}$. Clearly, quantities on both sides are no less than $\det(A+B)^2$. - -REPLY [7 votes]: Consider the following matlab output: - - - -The square root function produces a very small imaginary error in the second case so I've just stripped that away in the final calculation. Therefore, the proposed determinantal inequality does not hold.<|endoftext|> -TITLE: Fraction of the sets receive each color -QUESTION [5 upvotes]: Let $S_1,S_2,\dots,S_k$ be subsets of the set $S=\{1,2,\dots,n\}$, not necessarily distinct. We will color each element of $S$ red, green, or blue. From this coloring, each set $S_i$ will receive one or more color according to the following rule: -Let $r_i,g_i,b_i$ denote the number of red, green, and blue elements of $S_i$, respectively, and let $m_i=\max(r_i,g_i,b_i)$. If $r_i\geq m_i-1$, we give the color red to $S_i$. Similarly for green and blue. -What is the maximum constant $d$ for which we can always color the elements of $S$ in such a way that for any color, at least a fraction $d$ of the sets $S_i$ receive that color? -An algorithm that starts with a two-coloring and change the color of one element at a time to the third color achieves $d=1/5$, while an example shows that $d=1/3$ is the best one can hope for. - -REPLY [3 votes]: The statement seems to follow from Sperner's lemma. We will give such a coloring that the first few elements are red, the middle ones are green and the last few are blue (for any ordering of the elements). -We can represent the colorings of the $n$ elements with a subdivision of a large triangle into $n^2$ smaller triangles in the standard way, where the side lengths of the small triangles are $1$, while the side lengths of the large triangle are $n$. Express the vertices of the small triangles using barycentric coordinates. All barycentric coordinates will be of the form $\frac{r_v}n,\frac{g_v}n,\frac{b_v}n$, where $r_v$, $g_v$ and $b_v$ are non-negative integers. To each vertex $v$, assign a coloring of the elements, $C(v)$, where $r_v$, $g_v$ and $b_v$ elements are red, green and blue, respectively. -Now we define the colors of the vertices (which will be used in Sperner's lemma). A vertex $v$ is colored red if in $C(v)$ for at least $k/3$ sets we have that $r_i\ge m_i$. (Attention, there is no $-1$!) A vertex is colored green if it is not red, and in $C(v)$ for at least $k/3$ sets we have that $g_i\ge m_i$. A vertex is colored blue if it is not red or green, and in $C(v)$ for at least $k/3$ sets we have that $b_i\ge m_i$. Note that every vertex gets colored and the coloring satisfies the conditions of Sperner's lemma. -Therefore, we get a $3$-colored small triangle. But it is easy to see that any of its vertices will give a coloring $C(v)$ that satisfies the required conditions. -Update: Oops, as pointed out by Dap, the "easy to see part" is incorrect (as usual ;). One possible fix is to use real coordinates, as described by fedja. Another would be to allow multiple colors for each vertex, i.e., the "if it is not red" type conditions should be ignored. For such a multicoloring, Sperner's lemma is still true, and we get a triangle whose vertices contain all 3 colors (some possibly multiple times). And now the "easy to see" part should work because the "danger sets", i.e., the ones for which we have, e.g., $m_i=r_i=g_i$, contribute to multiple color counts, like in this case to both $r_i$ and $g_i$. But I don't see how to finish this workaround without a cumbersome calculation.<|endoftext|> -TITLE: Deformation-Obstruction Theory of YM Instantons -QUESTION [8 upvotes]: In Donaldson-Kronhiemer Section 4.2.5. (local models of the moduli space of YM instantons) they first get local models of the moduli space $M$ inside the space of all connections modulo gauge $\mathcal{B}$ by taking $(F^+)^{-1}(0)/\Gamma_A$, where ($\Gamma_A$ is the isotropy group of the connection). -Now here comes my confusion. They then remark its differential at zero is $d^+_A$, but that $H^1$ of its deformation-obstruction complex $$\Omega^0(\mathfrak{g}_E) \xrightarrow{d_A} \Omega^1(\mathfrak{g}_E) \xrightarrow{d^+_A} \Omega^+(\mathfrak{g}_E)$$ -is ker $\delta_A$, where $\delta_A = d^+_A \oplus d_A^\ast : \Omega^0 \to \Omega^1 \oplus \Omega^+$ and this shows we have finite-dimensional local models of $M$ cut out by the zeros of a $\Gamma_A$-equivariant map $$f : \text{ker } \delta_A \to \text{coker }d^+_A.$$ -But why the extra operator $d^\ast_A$? Why should the space of infinitesimal deformations be ker $\delta_A$ rather than ker $d^+_A?$ After all, it is $d^+_A$ which is the differential of $\psi$, not $\delta_A$. Is the Zariski tangent space not defined as the kernel of the differential of the map $\psi$ which cuts out a local model? - -REPLY [4 votes]: $\newcommand{\A}{\mathscr{A}}$ $\newcommand{\G}{\mathscr{G}}$ Denote by $\A$ the space of connections, by $\A_-$ the space of ASD connections and by $\G$ the gauge group. For ssimplicity I will not keep track of various Sobolev decorations. The moduli space $\newcommand{\M}{\mathscr{M}}$ $\M$ is defined as a set by the equality -$$\M=\A_-/\G\cdot A. $$ -Thus the tangent space to $\M$ at an ASD connection ought to be -$$T_A\M=T_A\A_-/T_1\G\cdot A.$$ -Observe that $\newcommand{\gog}{\mathfrak{g}}$ -$$ T_A\A_-= \ker d_A^+,\;\; T_1\G\cdot A\cong\mathrm{Im}\;\big(d_A:\Omega^1(\gog_E)\to\Omega^1(\gog_E)\;\big). $$ -Now you need to invoke a basic fact in Hodge theory. If you have a cochain complex -$$ 0\to V_0\stackrel{d_0}{\to} V_1\stackrel{d_1}{\to} V_2 \stackrel{d_2}{\to}\cdots $$ -then, -$$ H^k(V_\bullet)\cong \ker\big(\; d_k\oplus d_{k-1}^*: V_k\to V_{k+1}\oplus V_{k-1}\;\big). $$ -This happens under certain assumptions, e.g. , if the above complex is elliptic. -When applied to the deformation complex $\mathbf{C}$ of the ASD equation which is elliptic we get -$$T_A\M=\ker d_A^+/ \mathrm{Im}\;(d_A)=H^1(\mathbf{C}) \cong \ker d_A^+ \oplus d_A^*. $$ -A very good exercise I recommend solving is to verify that the deformation complex $\mathbf{C}$ is indeed elliptic.<|endoftext|> -TITLE: A dense subset in $B(X)$ under the weak operator topology -QUESTION [5 upvotes]: Let $X$ be a Banach space and consider $B(X)$, the set of all bounded linear maps on $X$. By the W-topology on $B(X)$ we mean the topology induced by the semi-norms -$$B(X)\to [0,\infty): T\to |\langle Tx,x^*\rangle|$$ -where $x\in X$ and $x^*\in X^*$. -The algebraic tensor product $X\otimes X^*$ may be considered as a subspace of $B(X)$. Does the following hold? -Under the W-topology, $B(X)$ is the closure of $X\otimes X^*$. - -REPLY [4 votes]: I think yes, for the following general reason: If $E$ is a real vector space and $F$ a linear space of linear forms of $E$, then a linear subspace $V$ of $E$ is $\sigma(E,F)$-dense in $E$ if and only if no $f\in F\setminus\{0\}$ vanishes identically on $V$. (Equivalently, any proper $\sigma(E,F)$-closed linear subspace of $E$ is included in $\ker f$ for some $f\in F\setminus\{0\}$. This follows from the Hahn-Banach theorem, and from the fact that the $\sigma(E,F)$-continuous linear forms on $E$ are precisely the elements of $F$). -In your situation, $E:=B(X)$, $F:=X\otimes_{alg} X^*$, seen as a linear space of linear forms on $B(X)$ (that is, $f=\sum_{i=1}^m x_i\otimes x^*_i$ corresponds to the linear form $T\mapsto\langle T,f\rangle:= \sum_{i=1}^m \langle Tx_i, x^*_i\rangle$ on $B(X)$), and $V:=X\otimes_{alg} X^*$, seen as the subset of $B(X)$ of finite rank operators (that is, $T=\sum_{j=1}^m y_j\otimes y^*_j$ corresponds to the finite rank operator $u\mapsto Tu:=\sum_{j=1}^m y_j\langle u,y^*_j \rangle$ on $X$). -Thanks to the above general fact, to prove that $V$ is $\sigma(E,F)$-dense in $E$ it is sufficient to prove: for any $f\in F\setminus\{0\}$ there exists $T\in V$ such that $\langle T, f\rangle\neq 0$. -Indeed, let $f\in F$ be a non-zero linear form. Let $f=\sum_{i=1}^m x_i\otimes x^*_i$ be a representation of $f$ with minimum $m$. This implies that $x_1,\dots,x_m$ are linearly independent elements of $X$ and $x_1^*,\dots,x_m^*$ are linearly independent elements of $X^*$. Then there are $y_1^*,\dots,y_m^*$ in $X^*$ such that $\langle x_i,y^*_j\rangle=\delta_{ij}$, and there are $y_1,\dots,y_m$ in $X$ such that $\langle y_i,x^*_j\rangle=\delta_{ij}$. Then the operator $T=\sum_{j=1}^m y_j\otimes y^*_j$ is such that $\langle T,f\rangle:=\sum_{i=1}^m \langle Tx_i, x^*_i\rangle=\sum_{ij}\langle y_j,x^*_i \rangle\langle x_i,y^*_j \rangle=\sum_{ij}\delta_{ij}=m>0$, that is, no non-zero element of $F$ vanishes identically on $V$.<|endoftext|> -TITLE: A question concerning Tauberian theory -QUESTION [7 upvotes]: Let $D(s) = \sum_{n=1}^\infty a_n n^{-s}$ be a Dirichlet series with $a_n ≥ 0$ and abscissa of convergence $\sigma_a = 1$. Further, we assume that $D(s)$ is holomorphic in each point $\Re(s) = 1$ except a pole of order $k > 0$ in $s = 1$ and that $D(s)$ possesses a meromorphic continuation on some half-plane $\Re(s) > 1 - \epsilon_0$ with some $\epsilon_0 > 0$. -My question is, whether under the above assumptions we can already say that -$$ \sum_{n ≤ x} a_n = xP(\log x) + o(x)$$ -with some polynomial $P$ with degree $k-1$. Unfortunately I am not able to exclude other poles in the strip $1 > \Re(s) > 1 - \epsilon_0$. -In the simple case $k=1$ the question follows by the Wiener-Ikehara theorem, since then we have $\sum_{n ≤ x} a_n \sim Cx$ with $C = \mathrm{res}_{s=1}D(s)$. -Thank you! - -REPLY [5 votes]: The answer is no, even if we assume there are no other poles than $1$ in $\sigma > 1- \epsilon_0$. I give an example below with $\epsilon_0=1$. This is a variant of an example given by Karamata in $1952$. -Let $b_n = 1 + \cos(\log^2(n)) \geq 0$ and consider $B(s) = \sum_{n \geq 1} b_n n^{-s}$. Let us write -$$ -\sum_{n \leq x} b_n = \int_{1}^x (1+\cos(\log^2(t))) dt + R(x) -$$ -where $R(x) = O(\log^2(x))$. Then for $\sigma >1$ we have -$$ -B(s) = \int_{1}^{+ \infty}(1+ \cos(\log^2(t)) )t^{-s}dt + s \int_{1}^{+ \infty} R(t) t^{-s-1} dt. -$$ -The last integral extends holomorphically to $\sigma >0$. The first integral is equal to -$$ -\frac{1}{s-1} + \frac{1}{2} I_+(s) + \frac{1}{2} I_-(s) -$$ -where -$$ -I_{\pm}(s) = \int_{0}^{+ \infty} \exp{((1-s)u \pm i u^2)} dt. -$$ -Using contour integration one can move the integration line to $u \mapsto e^{\pm \frac{i \pi}{4}}u$, and this yields -$$ -I_{\pm}(s) = e^{\pm \frac{i \pi}{4}} \int_{0}^{+ \infty} \exp{((1-s)u e^{\pm \frac{i \pi}{4}} - u^2)} dt. -$$ -Thus $I_+$ and $I_-$ are entire functions. -Let us now choose $a_n = b_n \log(n) \geq 0$ so that $D(s) = - B'(s)$ is meromorphic on $\sigma >0$, with a unique pole at $s=1$ of order $2$. One checks that -$$ -\sum_{n \leq x} a_n = x \log(x) + x \left( \frac{1}{2} \sin(\log^2(x))-1 \right) +O(\frac{x}{\log x}) -$$ -Remark : The error term in Delange's theorem can be improved provided growth assumptions (on a strip) are made on $D(s)$.<|endoftext|> -TITLE: Statement of the pair correlation conjecture -QUESTION [10 upvotes]: In his paper "The pair correlation of zeros and the zeta function", -Montgomery defines a function -$$F(\alpha,T) = \left(\frac{T}{2 \pi} \log T\right)^{-1} \sum_{0 < \gamma, \gamma' < T} T^{i \alpha (\gamma'-\gamma)} w(\gamma'-\gamma)$$ -where $w(u)=\frac{4}{4+u^2}$, and the sum is over pairs of imaginary parts $\gamma, \gamma'$ of non-trivial zeros of the Riemann Zeta function. -Then he says that heuristic arguments suggest that "$F(\alpha)=1+o(1)$ for $\alpha \geq 1$, uniformly in bounded intervals". My question is: what does that mean? -Can someone please reformulate this conjecture with quantifiers in the right order? Thanks. - -REPLY [11 votes]: For any $\epsilon > 0$ and for any finite interval $I \subset [1, \infty)$, there is a $T_{0}$ such that for all $T > T_{0}$ and all $\alpha \in I$ we have $|F(\alpha,T) - 1| \leq \varepsilon$. The meaning of the conjecture is that the Fourier transform $F(\alpha, T)$ of the pair correlation of zeros up to height $T$, of the Riemann zeta-function converges to a limit $F(\alpha)$ as $T$ goes to infinity. You can invert the Fourier transform, and then read off the conjecture as saying that for any smooth Schwartz class function $f$, the limit -$$ -\lim_{T \rightarrow \infty} \frac{1}{N(T)}\sum_{0 < \gamma, \gamma' < T} f(\log T (\gamma - \gamma')) -$$ -exists, where $N(T)$ is the number of zeros up to $T$. Moreover the limit is an explicit linear functional of $f$. It coincides with the similar functional that one gets from considering the pair correlation of the eigenvalues of random GUE matrices. For better or for worse a lot of ink has been spilled on this last observation. -The importance of the conjecture is that it immediately implies that the normalized gaps between zeros of the Riemann zeta-function tend to get arbitrarily small. With a bit more work this then implies that there are no Siegel zeros. This was in fact Montgomery's original reason for considering the pair correlation of the zeros. -P.S: In your first display you should be dividing by $T \log T$ and not $T / \log T$, since $N(T) \sim (1/2\pi) T \log T$.<|endoftext|> -TITLE: Countably infinite connected Hausdorff space with the fixed point property -QUESTION [6 upvotes]: Is there an infinite, countable connected $T_2$-space $(X,\tau)$ such that $(X,\tau)$ has the fixed point property? (This means that for every continuous map $f:X\to X$ there is $x\in X$ such that $f(x) = x$.) - -REPLY [5 votes]: Yes, there is such an example. -This is nearly Problem 10705 in The American Mathematical Monthly, proposed by D. W. Brown in 106 #1 (January 1999), p. 67, where the problem asks for a countably infinite $T_{2}$ example. In an editorial comment following John Cobb's solution in 107 #4 (April 2000), pp. 375-376, it is stated that Prabir Roy's lattice space --- A countable connected Urysohn space with a dispersion point, Duke Mathematical Journal 33 #2 (1966), pp. 331-333 --- is an example that is a countable connected Urysohn space. For what it's worth, my notes on this problem say that this can be verified by making use of the analog in Roy's space of observations (a) and (b) at the beginning of the proof of the theorem at the bottom of p. 375.<|endoftext|> -TITLE: In constructive mathematics, why does the category of abelian groups fail to be abelian? -QUESTION [17 upvotes]: I was reading the paper Towards Constructive Homological Algebra in -Type Theory by Thierry Coquand and Arnaud Spiwack, and they state that constructively, the category of abelian groups fails to be abelian, because we cannot verify that every monic and epic map is normal. In particular, they say that given a monic map $u:A \rightarrowtail B$, we cannot prove constructively that it is the kernel of $B \to B/\mathrm{Im}(u)$. -I'm new to constructive mathematics, so it isn't obvious to me exactly where the issue is. At what stage do we use excluded middle or some form of choice when proving that monos and epis are normal? -Also, could we correct this lack of normality somehow with some additional assumptions on the groups (countably or finitely generated, for example)? - -REPLY [22 votes]: There are many different flavors of constructive mathematics. The theory that was used in this paper is weak, it lacks some useful constructions from the usual set theory such as quotient sets. Another problem is that it lacks function extensionality (that is, if you can prove that two functions are pointwise equal, this does not imply that they are equal). For these reasons we usually work with setoids (which authors of the paper call Bishop sets) in such weak theories. You can define the setoid of functions with the correct equivalence relation on it and you obviously can define quotient setoids. But this approach still has problems as the example of abelian group shows. -We can solve these problems by strengthening our theory (but still keeping it constructive and even predicative if we want). Homotopy type theory does precisely that. Note that if we don't care about higher types, we can define a theory which is between the ordinary type theory and HoTT. We just add axioms for function extensionality and propositional extensionlity (logically equivalent propositions are equal). This makes the theory into an internal language of a topos (assuming that the type of propositions is impredicative as in the paper). In this theory we can construct quotient types in the usual way as the type of equivalence classes. Even if we don't want to add an impredicative type of propositions, we still can add quotient types explicitly as a special case of higher inductive types. Then we can prove that the category of abelian groups is abelian without any problems.<|endoftext|> -TITLE: Inequality for functions on [0,1], continued -QUESTION [6 upvotes]: Let $02$ in arXiv:1708.07669. - -REPLY [6 votes]: OK, here goes. -We start with changing the notation ($z\to 20z^2$, $-z-3\to r$, $20rz\to y$ means that what was denoted by $z$ will be denoted by $20z^2$ from now on, $r$ is $-z-3$ with new $z$, so it is $-\sqrt{z/20}-3$ in terms of old $z$, and $y$ denotes $20rz$ with just (re)defined $r,z$). -So, execute the following sequence (the order matters!) -$$ -x\to e^{-x},\ a\to e^{-a},\ mkja\to T,\ ja\to a -$$ -Denote $\varphi(t)=-\log(1-e^{-t})$. For a function $f$ denote by $LRS(f,u,v; h)$ and $RRS(f,u,v;h)$ the left and the right Riemann sums of $f$ on the interval $[u,v]$ with step $h$ respectively (we will always assume that $(v-u)/h\in \mathbb N\cup\{\infty\}$). -Then the above substitutions induce the substitution -$$ --\log f_{mk}(a^jx)\to \frac{LRS(\varphi,x,\infty;a)+xT-RRS(\varphi,0,T;a)}a -$$ -The claim to prove reduces to the statement that this function decreases in $a$ as long as we run $a$ along admissible steps for $[0,T]$ with fixed $T$. -We have already seen that replacing Riemann sums by integrals results in the area $A\ge 0$ of the "excess triangle" formed by the vertical line through $x$, the horizontal line through $T$, and the graph of $\varphi$ in the coordinate system where $x$ runs over the horizontal axis and $T$ over the vertical one in the numerator. Thus, the function to investigate is -$$ -\frac Aa+\frac {\Phi(a)}a+\frac{\Psi(a)}a -$$ -where $\Phi(a)=LRS(\varphi,x,\infty;a)-\int_x^\infty \varphi(t)\,dt$ and $\Psi(a)=\int_0^T\varphi(t)\,dt-RRS(\varphi,0,T,a)$ -Note that we have a nice series expansion -$$ -\varphi(t)=\sum_{k\ge 1}\frac 1ke^{-kt} -$$ -The integration and the Riemann sum computation are linear operations with respect to the function, so we have -$$ -\Phi(a)=\sum_{k\ge 1}\frac 1k\Phi_k(a),\qquad \Psi(a)=\sum_{k\ge 1}\frac 1k\Psi_k(a) -$$ -where $\Phi_k$ and $\Psi_k$ are defined in the same way but using $e^{-kt}$ instead of $\psi(t)$. -Now, we (or, if you prefer, at least I) cannot integrate $\varphi$ or to sum it along an arithmetic progression. However everybody can do it for an exponential function. The direct computation yields -$$ -\Phi_k(a)=e^{-kx}\left[\frac a{1-e^{-ka}}-\frac 1k\right],\qquad \Psi_k(a)=(1-e^{-kT})\left[\frac 1k-\frac a{e^{ka}-1}\right] -$$ -Note that in the computation of $\Psi_k(a)$ we have used the fact that $T/a$ is an integer. However the resulting answer is formally defined for all $a>0$. So we will not use the "fitting condition" anywhere below. -Notice also that -$$ -\left[\frac a{1-e^{-ka}}-\frac 1k\right]+\left[\frac 1k-\frac a{e^{ka}-1}\right]=a -$$ -so -$$ -\frac{\Phi_k(a)+\Psi_k(a)}a=\frac{1-e^{-kx}-e^{-kT}}k\left[\frac 1a-\frac k{e^{ka}-1}\right]+ \operatorname{const} -$$ - Thus we need to show that -$$ -\frac Aa+\sum_{k\ge 1}\frac{1-e^{-kx}-e^{-kT}}{k^2}\left[\frac 1a-\frac k{e^{ka}-1}\right] -$$ -is decreasing or, equivalently, that -$$ -\frac A{a^2}+\sum_{k\ge 1}\frac{1-e^{-kx}-e^{-kT}}{k^2}\left[\frac 1{a^2}-\frac {k^2}{e^{ka}+e^{-ka}-2}\right]\ge 0\,. -$$ -When the excess triangle is absent or is above the graph, we have $e^{-x}+e^{-T}\le 1$ and the factor $k\ge 1$ can only improve this inequality, so each term is non-negative and the estimate is trivial, thus justifying my first remark. -The other case is slightly more interesting. -Let $t>0$ satisfy $e^{-t}+e^{-x}=1$. We have now $t>T$. Write -$$ -A=\int_T^{t}\varphi(s)\,ds-(t-T)\varphi(t)=\sum_{k\ge 1} -\frac 1k\left[\int_T^{t}e^{-ks}\,ds-(t-T)e^{-kt} -\right] -\\ -=\sum_{k\ge 1}\frac{e^{-kT}-e^{-kt}-k(t-T)e^{-kt}}{k^2} -$$ -We want to combine these terms multiplied by $\frac 1{a^2}$ with the corresponding terms in the main sum. It will be more convenient to diminish them first by multiplying them not by the full $\frac 1{a^2}$ but just by the expressions in the brackets in the main sum. Then the result can be rewritten as -$$ -\sum_{k\ge 1}\frac{1-e^{-kx}-e^{-kt}-kte^{-kt}}{k^2}\left[\frac 1{a^2}-\frac {k^2}{e^{ka}+e^{-ka}-2}\right]+ -\sum_{k\ge 1}\frac{Te^{-kt}}{k}\left[\frac 1{a^2}-\frac {k^2}{e^{ka}+e^{-ka}-2}\right] -$$ -The second sum is clearly non-negative. Let's look at the first one. -We have -$$ -\int_u^\infty\varphi(s)\,ds=\sum_{k\ge 1}\frac 1k\int_u^\infty e^{-ks}\,ds=\sum_{k\ge 1}\frac{e^{-ku}}{k^2}\,. -$$ -Using this with $u=0,x,t$ and flipping the geometric picture for the integral of $\varphi$ over $[t,\infty)$ to the vertical axis, as usual, we see that -$$ -\sum_{k\ge 1}\frac{1-e^{-kx}-e^{-kt}}{k^2}=tx -$$ -(the area of the remaining rectangle). -On the other hand, -$$ -\sum_{k\ge 1}\frac{kte^{-kt}}{k^2}=t\varphi(t)=tx -$$ -as well. So the first factors form a sequence whose numerators are (obviously, because $u\mapsto(1+u)e^{-u}$ is decreasing on $[0,+\infty)$) increasing and whose sum equals $0$. Thus, they break from $-$ to $+$ just once when $k$ goes up. However the second factors are increasing in $k$. Thus, the sum we are interested in is non-negative as well. -The End<|endoftext|> -TITLE: Orbits of $GL(n, \mathcal{O})$ on pairs of linear subspaces over non-Archimedean local fields -QUESTION [10 upvotes]: Let $F$ be a non-Archimedean local field. Let $\mathcal{O}$ be its ring of integers. Let $Gr_{i,n}$ denote the Grassmannian of $i$-dimensional linear subspaces in $F^n$. - -Can one describe explicitly the $GL(n,\mathcal{O})$-orbits on $Gr_{i,n}\times Gr_{i,n}$, i.e. on pairs of subspaces? - -Remark. $GL(n,\mathcal{O})$ acts transitively on $Gr_{i,n}$. - -REPLY [4 votes]: The $\text{GL}(n,\mathcal{O})$-orbits refine the $\text{GL}(n,F)$-orbits, i.e., the Bruhat cells. Up to replacing $i$ by $n-i$, assume that $2i\leq n.$ For every integer $m$ with $0\leq m\leq i,$ the Bruhat cell $U_m$ in $\text{Gr}_{i,n}\times \text{Gr}_{i,n}$ is the set of pairs $([V],[W])$ of $F$-vector subspaces of $F^{\oplus n},$ $V\cong F^{\oplus i}$ and $W\cong F^{\oplus i},$ such that $V\cap W$ is an $F$-vector subspace of codimension $m$ in both $V$ and in $W$. The Bruhat cell $U_m$ is an affine space over $F$ of dimenion $i(n-i)+m(n-m-i).$ -For every finite, free $\mathcal{O}$-module, say $\mathcal{O}^{\oplus n}$, with its associated $F$-vector space, $\mathcal{O}^{\oplus n} \otimes_{\mathcal{O}} F = F^{\oplus n},$ for every $F$-vector subspace $V\subset F^{\oplus n},$ denote by $V_\mathcal{O}$ the intersection $V\cap \mathcal{O}^{\oplus n}.$ -Lemma 1. The $\mathcal{O}$-submodule $V_{\mathcal{O}}$ of $\mathcal{O}^{\oplus n}$ is a finite, free $\mathcal{O}$-module that is a direct summand of $\mathcal{O}^{\oplus n}.$ -Proof. -Since $\mathcal{O}$ is a Noetherian ring, the $\mathcal{O}$-submodule $V_{\mathcal{O}}$ of $\mathcal{O}^{\oplus n}$ is a finitely generated $\mathcal{O}$-module. Since $V_\mathcal{O}$ is an $\mathcal{O}$-submodule of $\mathcal{O}^{\oplus n},$ it is also torsion-free. Since $\mathcal{O}$ is a DVR, also $V_{\mathcal{O}}$ is a finite, free $\mathcal{O}$-module. Finally, since $V_{\mathcal{O}}$ is a saturated $\mathcal{O}$-submodule of $\mathcal{O}^{\oplus n}$, also the cokernel $\mathcal{O}^{\oplus n}/V_{\mathcal{O}}$ is also a finitely generated and torsion-free $\mathcal{O}$-module. Thus, also $\mathcal{O}^{\oplus n}/V_{\mathcal{O}}$ is a finite, free $\mathcal{O}$-module. Altogether, this implies that $V_{\mathcal{O}}$ is a direct summand of $\mathcal{O}^{\oplus n}.$ QED -Denote by $\pi$ a uniformizing element of $\mathcal{O}.$ -Lemma 2. For every pair of $F$-linear subspaces of $F^{\oplus n},$ $V\cong F^{\oplus \ell}$ and $W\cong F^{\oplus m},$ giving a direct sum decomposition, there exists a free basis $(b_1,\dots,b_m)$ of $W_{\mathcal{O}},$ and an ordered sequence of integers $0\leq e_1\leq \dots \leq e_m$ such that for every $i=1,\dots,m,$ the image of $b_i$ in $\mathcal{O}^{\oplus n}/V_{\mathcal{O}}$ equals $\pi^{e_i}\cdot c_i$ for some $c_i,$ and $(c_1,\dots,c_m)$ is a free basis for $\mathcal{O}^{\oplus n}/V_{\mathcal{O}}.$ The ordered sequence of integers is independent of the choice of basis. -Proof. This is part of the theory of finitely generated modules over a Principal Ideal Domain. The sequence of powers $(\pi^{e_1},\dots,\pi^{e_m})$ are the elementary divisors. QED -Denote $i-m$ by $r$. By Lemma 1, the quotient $\mathcal{O}$-module $(V+W)_{\mathcal{O}}/(V\cap W)_{\mathcal{O}}$ is a finite, free $\mathcal{O}$-module with associated $F$-vector space $(V+W)/(V\cap W).$ The $F$-vector subspaces, $V/(V\cap W)\cong F^{\oplus m}$ and $W/(V\cap W)\cong F^{\oplus m}$ give a direct sum decomposition of $(V+W)/(V\cap W).$ For every sequence $\underline{e}=(e_1,\dots,e_m)$ of nonnegative integers, denote by $U_{\underline{e},m}\subset U_m$ the set of pairs such that the elementary divisors of the $\mathcal{O}$-module homomorphism, $$(W/(V\cap W))_{\mathcal{O}} \to (V+W)_{\mathcal{O}}/V_{\mathcal{O}},$$ equal $(\pi^{e_1},\dots,\pi^{e_m}).$ -Proposition. The group $\text{GL}(n,\mathcal{O})$ acts transitively on $U_{\underline{e},m}.$ -Proof. By Lemma 1, $(V\cap W)_{\mathcal{O}}$ is a finite, free direct summand in $V_{\mathcal{O}},$ in $W_{\mathcal{O}},$ and in $\mathcal{O}^{\oplus n}.$ Similarly, $(V+W)_{\mathcal{O}}$ is a finite, free direct summand in $\mathcal{O}^{\oplus n}.$ Thus, there exists a free basis $(d_1,\dots,d_r)$ for $(V\cap W)_{\mathcal{O}},$ and there exists an extension $(d_1,\dots,d_r,a_1,\dots,a_m)$ to a free basis of $V_{\mathcal{O}}.$ By Lemma 2, there exists an extension $(d_1,\dots,d_r,a_1,\dots,a_m,c_1,\dots,c_m)$ to a free basis of $(V+W)_{\mathcal{O}}$ and there exists a free basis $(d_1,\dots,d_r,b_1,\dots,b_m)$ of $W_{\mathcal{O}}$ such that for the images in $(V+W)_{\mathcal{O}}/V_{\mathcal{O}}$, each $\overline{b}_i$ equals $\pi^{e_i}\overline{c}_i.$ Finally, there exists an extension $(d_1,\dots,d_r,a_1,\dots,a_m,c_1,\dots,c_m,g_1,\dots,g_{n-(2m+r)})$ to a free basis for $\mathcal{O}^{\oplus n}.$ -For any other pair $(V',W')$ with associated bases $(d'_i),$ etc., there exists a unique $\mathcal{O}$-module isomorphism $V_{\mathcal{O}}\to V'_{\mathcal{O}}$ sending the free basis $(d_1,\dots,d_r,a_1,\dots,a_m)$ to $(d'_1,\dots,d'_r,a'_1,\dots,a'_m).$ There exists a unique $\mathcal{O}$-module isomorphism $(V+W)_{\mathcal{O}}/V_{\mathcal{O}}\to (V'+W')_{\mathcal{O}}/W'_{\mathcal{O}}$ sending $(\overline{c}_1,\dots,\overline{c}_m)$ to $(\overline{c}'_1,\dots,\overline{c}'_m).$ There is a unique lift of these moduli isomorphisms to an $F$-vector space isomorphism $V+W\to V'+W'$ that maps $(b_1,\dots,b_m)$ to $(b'_1,\dots,b'_m).$ Because the elementary divisors are equal, this $F$-vector space isomorphism maps the $\mathcal{O}$-module saturation $(V+W)_{\mathcal{O}}$ of $V_{\mathcal{O}}+W_{\mathcal{O}}$ to the $\mathcal{O}$-module saturation $(V'+W')_{\mathcal{O}}$ of $V'_{\mathcal{O}}+W'_{\mathcal{O}}.$ Finally, there is a unique extension to a $\mathcal{O}$-module isomorphism of $\mathcal{O}^{\oplus n}$ to $\mathcal{O}^{\oplus n}$ sending $(d_1,\dots,d_r,a_1,\dots,a_m,b_1,\dots,b_m,g_1,\dots,g_{n-(r+2m)})$ to $(d'_1,\dots,d'_r,a'_1,\dots,a'_m,b'_1,\dots,b'_m,g'_1,\dots,g'_{n-(r+2m)}).$ QED -Edit. As Uri Bader points out, the data of an integer $m$ satisfying $0\leq m \leq i$ together with the sequence of elementary divisors is equivalent to the data of the $\mathcal{O}$-module isomorphism class of the finitely generated $\mathcal{O}$-module, $$\mathcal{O}^{\oplus n}/(V_{\mathcal{O}}+W_{\mathcal{O}}).$$ So, as in Uri Bader's answer, a more intrinsic index for the $\text{GL}(n,\mathcal{O})$-orbits of $\text{Gr}_{i,n}\times \text{Gr}_{i,n}$ is an isomorphism class of an $\mathcal{O}$-module that is generated by $\leq n-i$ generators and whose free rank is at least $n-2i$. - -REPLY [3 votes]: I assume $2i\leq n$ (otherwise use duality and replace $\text{Gr}_{i,n}$ with $\text{Gr}_{n-i,n}$). -I claim that the set of orbits is parametrized by $i$-tuples $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_i)$, -where $\lambda_j\in \{0,1,2,\ldots,\infty\}$ and $\lambda_j\leq \lambda_{j+1}$ -(for $2i\geq n$ it will be $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_{n-i})$). -For a commutative ring $R$ I use here the term $\text{Gr}_{i,n}(R)$ to denote the space of direct summands of $R^n$ which are isomorphic to $R^i$. -Note that there is a natural (and $\text{GL}_n(\mathcal{O})$-equivariant) identification of $\text{Gr}_{i,n}(F)$ with $\text{Gr}_{i,n}(\mathcal{O})$, -thus we may forget about $F$ and focus on $\text{Gr}_{i,n}(\mathcal{O})$. -Let $p<\mathcal{O}$ be the maximal ideal. -Note that two pairs of modules in $\text{Gr}_{i,n}(\mathcal{O})$ are in the same $\text{GL}_n(\mathcal{O})$-orbit iff their reductions mod $p^k$ are in the same -$\text{GL}_n(\mathcal{O}/p^k)$-orbit for every natural $k$. -Fix $k$ and consider $\mathcal{O}/p^k$-modules. -You can check that two pairs $(x,y),(x',y') \in \text{Gr}_{i,n}(\mathcal{O}/p^k) \times \text{Gr}_{i,n}(\mathcal{O}/p^k)$ are in the same $\text{GL}_n(\mathcal{O}/p^k)$-orbit iff $x\cap y \simeq x' \cap y'$. -Up to an isomorphism, the intersection module $x\cap y$ could be an arbitrary $\mathcal{O}/p^k$-module of rank bounded by $i$ (here we use $2i\leq n$). The isomorphism types of such modules are parametrized by $i$-tuples $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_i)$, -where $\lambda_j\in \{0,1,2,\ldots, k\}$ and $\lambda_j\leq \lambda_{j+1}$, -where a module of type $\lambda$ is isomorphic to $\oplus_j \mathcal{O}/p^{\lambda_j}$. -A discussion and justifications are given in claim 2.2 here. -Now, for $(x,y)\in \text{Gr}_{i,n}(\mathcal{O})\times \text{Gr}_{i,n}(\mathcal{O})$, consider the limit of the tuples $\lambda$ associated with their reduction mod $p^k$ where $k$ tends to $\infty$.<|endoftext|> -TITLE: Tubular Neighborhood Theorem for $C^1$ Submanifold -QUESTION [14 upvotes]: Can anyone reference/disprove the theorem in the case where the embedded submanifold is merely $C^1$ instead of smooth? I have a compact $C^1$ embedded submanifold of $\mathbb{R}^n$ without boundary that I want to show there exists a tubular neighborhood of (with the radius of the tube not necessarily constant). Actually, I want to be able to create one piecewise using a finite covering. Much obliged. - -REPLY [13 votes]: The answer to your question depends on whether you are looking for a tubular neighborhood in the general differential topological sense or the more restrictive geometric sense. The answer in the topological sense is Yes, but in the geometric sense the answer in general is No. These two conceptions may coincide for $C^2$ submanifolds, but for $C^1$ submanifolds the construction is a bit more subtle, or non-canonical, as I will elaborate below. -A $C^1$ submanifold $M\subset R^n$ may not have any neighborhoods fibrated by normal vectors. So $C^1$ submanifolds do not in general have canonical tubular neighborhoods generated by the exponential map, and "radius" of the neighborhood does not really make sense; however, they still have a tubular neighborhood in the topological sense, i.e., a $C^1$ embedding of the normal bundle which extends that of $M$, as the zero section of the bundle. These may be constructed as follows. -The fastest way to construct a topological tubular neighborhood for $M$ is by noting that there exists a $C^1$ diffeomorphism $\phi\colon R^n\to R^n$ such that $\phi(M)$ is $C^\infty$ (e.g., see Thm. 3.6 in Chap. 2 of Hirsch's book). Then $\phi(M)$ is going to have a tubular neighborhood $U$ in the standard sense, obtained by applying the inverse function theorem to the exponential map (or taking the union of all sufficiently short normal vectors), and $\phi^{-1}(U)$ yields the desired tubular neighborhood of $M$ (which will be $C^1$-diffeomorphic to the normal bundle). -An example of a $C^1$ submanifold without a neighborhood fibrated by normals may be constructed as follows. Take an ellipse in $R^2$, which is not a circle, and let $M$ be the inner parallel curve of the ellipse which passes through the foci. Then $M$ will be $C^1$, but it will not have a tubular neighborhood in the geometric sense. Indeed, the nearest point projection map into $M$ will not be one-to-one in any neighborhood of $M$.<|endoftext|> -TITLE: Construction of combinatorial model categories with all objects fibrant -QUESTION [6 upvotes]: By abstract construction of a combinatorial model category, I mean starting from a locally presentable category satisfying some assumptions, e.g. equipped with a cylinder or a cocylinder satisfying some special hypothesis, and from these data build a model category structure. The question now is: - -What are the known abstract constructions of a combinatorial model - category with all objects fibrant and such that not all maps are - fibrations (to rule out the case of the discrete model structure) ? - -The only example I am aware of is the third section of Marc Olschok's PhD, Model structures from balls. - -REPLY [2 votes]: There are a ton of papers about what you are asking. Another is the thesis of Richard Williamson (arXiv:1304.0867v1). Also, Valery Isaev has a paper that produces a model structure with all objects fibrant, given some cylinder or path object information (https://arxiv.org/pdf/1312.4327.pdf). The thesis of Remy Tuyeres produces a model structure given some even more general category theoretic information. Maybe check the references of those three sources; I'll bet there are many others.<|endoftext|> -TITLE: Curvature of the boundary vs. normal derivative of the first eigenfunction -QUESTION [9 upvotes]: Disclaimer. I posted this question in Math.SE, but it haven't received enough attention. -Let $\varphi_1$ be the first eigenfunction of the zero Dirichlet Laplacian in a planar bounded domain $\Omega$. That is, -$$ -\left\{ -\begin{aligned} --\Delta \varphi_1 &= \lambda_1 \varphi_1 &&\text{in } \Omega \subset \mathbb{R}^2,\\ -\varphi_1 &= 0 &&\text{on } \partial \Omega. -\end{aligned} -\right. -$$ -If $\Omega$ is piece-wise smooth and has corners, and we look at the plot of $\varphi_1$, then we see that its normal derivative tends to zero near exterior (outward) corners, and tends to infinity near interior (inward) corners. -See the plot for the standard L-shape: - -This fact suggests that, in a smooth domain, there should be some connection between curvature of the boundary at a point and the normal derivative of $\varphi_1$ at this point. -That is, if the curvature is big positive, then the normal derivative is close to zero. And if the curvature is big negative, then the normal derivative is large. -However, I was not able to find corresponding inequalities in the literature. (Although I believe that such results should be well-known.) -I would appreciate some references to such facts and related results in this direction. -Thanks! - -REPLY [2 votes]: I do not there is a strong relation between the two notions in the general case. Curvature is obviously a local object. On the other hand, the behaviour of the first Dirichlet eigenfunction near a boundary point $p$ is highly non-local. -For example, if $\Omega$ is the union of two balls of different radii connected by a narrow channel, then the normal derivative at the boundary of the smaller ball will converge to zero as the width of the channel goes to zero, despite the fact that the curvature remains unchanged. -That said, curvature and the normal derivative should be related to each other for domains "without thin channels" — for example, for NTA domains. A possible approach would be via results similar to the boundary Harnack inequality. Unfortunately, I am not aware of any references in that direction.<|endoftext|> -TITLE: Sequence $(x_n)$ whose first $p$ terms is a complete residue system: value of $\lim\limits_{n\to\infty}\frac{x_n}{n}$? -QUESTION [10 upvotes]: Let $\{x_{n}\}$ be a sequence in $\mathbb{N}$ with $x_{1}=1$ such that for any prime $p$, the set $$A=\{x_{1},x_{2},\ldots,x_{p}\}$$ forms a complete residue system $\pmod{p}$. Now is it true that $\lim\limits_{n\to\infty}\frac{x_{n}}{n}$ exists? If yes what is it's value? - -REPLY [20 votes]: This problem is due to Imre Ruzsa who posed it at the 2015 Miklós Schweitzer Contest in Hungary: https://mathproblems123.wordpress.com/2015/10/31/miklos-schweitzer-2015-problems/ -Here is a solution (thanks also to YCor for his comments and encouragement). It is easy to see that $x_1=1$ and $x_2=2$. We claim that $\{x_1,\dots,x_p\}$ equals $\{1,\dots,p\}$ for every prime $p$. Let us assume that this holds for some prime $p$, and let us show that it also holds for the next prime $q$ in place of $p$. For this, it suffices to verify that $x_j\leq q$ holds for any $p -TITLE: partitions of Boolean algebras -QUESTION [11 upvotes]: A partition of a Boolean algebra is a collection of pairwise disjoint nonzero elements with supremum 1. For any infinite Boolean algebra $A$ let $a(A)$ be the least size of an infinite partition of $A$. Is $a(A+B)=\min\{a(A),a(B)\}$, where $A+B$ is the free product of $A$ and $B$? - -REPLY [6 votes]: It seems this is an open question. First note that (since the answer to the corresponding question for lottery sums is yes) the following are equivalent: - -$a(A+B)=\min\{a(A),a(B)\}$ for every $A,B$ -$a(A+A)=a(A)$ for every $A$. - -Regarding 2, one should read the article The minimal size of infinite maximal antichains in direct products of partial orders by Miloš Kurilić (Order, 2017). The article starts (and ends) with the open question: - -Is there a partial order $\mathbb{P}$ such that $a(\mathbb{P}) -TITLE: absolute Galois group of the function field of a curve over $\mathbb{F}_p^{alg}$ -QUESTION [7 upvotes]: In their 2008 paper "Torelli theorem for curves over finite fields" Bogomolov, Korotiaev and Tschinkel mention in the beginning of Section 9 that absolute Galois groups of curves over $\mathbb{F}_p^{alg}$ are free profinite on countably many generators. However, they do not give reference, and after talking to some colleagues I am at a loss with regard to the status of this statement. -How does one derive this fact (if it is true)? - -REPLY [8 votes]: The theorem is true. It seems to have been proved independently by Florian Pop and by David Harbater. In Pop's paper, it is the corollary on p. 556. -MR1334484 (96k:14011) -Pop, Florian -Étale Galois covers of affine smooth curves. The geometric case of a conjecture of Shafarevich. On Abhyankar's conjecture. -Invent. Math. 120 (1995), no. 3, 555–578. -http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002112868 -Harbater's paper is the following. -MR1352282 (97b:14035) -Harbater, David -Fundamental groups and embedding problems in characteristic p. (English summary) -Recent developments in the inverse Galois problem (Seattle, WA, 1993), 353–369, -Contemp. Math., 186, Amer. Math. Soc., Providence, RI, 1995.<|endoftext|> -TITLE: Is there an oracle that can compute something iff it is computable in every countable model that is equivalent to $(V, \in)$? -QUESTION [5 upvotes]: Let us work in Kelly-morse set theory, so we can talk about $V$. For some model $M=(\mathbb N, \in_M)$ that is elementary equivalent $(V, \in)$, we can have an oracle that corresponds to $(\mathbb N, \in_M)$'s truth predicate. We will say that $M$ is a true countable model of set theory, and we will call the oracle $O_M$. -My question is, is there an oracle that can compute a function iff every oracle $O_M$ for ever true countable model of set theory can compute the function? -Such an oracle would at least be able to determine if a first order sentence without free variables is true in $(V, \in)$. My conjecture is it does exist, and is defined by the previous statement. -Some notes: - -For any definable set, there is a Turing machine with oracle $O_M$ that outputs a number corresponding to that set. -Any $O_M$ (for true countable M) can calculate Rayo's function. (Another question to ask would be if a Turing machine equipped with an oracle for Rayo's function let's you determine truth of statements in $(V, \in)$ without free variables.) - -REPLY [6 votes]: Yes - this real is exactly the parameter-free theory of $V$ (or anything Turing-equivalent to it). -One direction is immediate: if $M$ is elementarily equivalent to $V$, then from the elementary diagram of $M$ we can compute the theory of $V$ (just look at the parameter-free sentences). It's the other direction that is interesting, and the key is that Henkinization is effective: from any complete consistent first-order theory $T$ we can uniformly compute a model of $T$ together with its elementary diagram. -Note that this applies to arbitrary structures, not just $V$: the reals computable from the elementary diagram of any structure $\mathcal{N}$ elementarily equivalent to $\mathcal{M}$ are exactly the reals computable from $Th(\mathcal{M})$. - -Incidentally, a very similar-sounding question has a very different answer in general. Given a structure $\mathcal{A}$, we can look at its copies - those structures $\mathcal{B}$ with domain $\mathbb{N}$ which are isomorphic to $\mathcal{A}$. We can ask what reals are computable from (the atomic diagram of) every copy of $\mathcal{A}$. Interestingly, even very complicated structures can have very little computing power in this sense; e.g. Linda Jean Richter showed in Degrees of Structures (JSTOR) that if $\mathcal{L}$ is a linear order, it has no computing power - for every noncomputable real $x$ there is some copy $\mathcal{J}$ of $\mathcal{L}$ whose atomic diagram doesn't compute $x$. If you're interested in this sort of question, Ash and Knight's book Computable Structures and the Hyperarithmetical Hierarchy is the standard reference. -It's also worth noting that for any computable theory $T$ and any noncomputable set $x$, there is a model of $T$ whose atomic diagram does not compute $x$; moreover, any such theory has a model with a low atomic (or even elementary) diagram - even a really complicated theory like ZFC.<|endoftext|> -TITLE: A singular value-eigenvalue inequality -QUESTION [9 upvotes]: Singular value or eigenvalue problems lie at the center of matrix analysis. One classical result is -$$\lambda_{j}(X^{*}X+Y^{*}Y)\geq 2\sigma_j(XY^*)$$ -for $j \in \{1, \ldots, n\}$, where $\lambda_j(\cdot)$ and $\sigma_j(\cdot)$ denote the $j$th largest eigenvalue and singular value, respectively, and $X$ and $Y$ are $n \times n$ complex matrices. I conjecture that the following does hold. -$$\lambda_{j}((I+X^{*}X)(I+Y^{*}Y))\geq\sigma_j^2(I+XY^*)$$ - -REPLY [4 votes]: The conjecture is true. -Lemma 1 : For every matrix $Z$ holds $\lambda_j(Z^* Z + Z^* + Z ) \leq \lambda_j(Z^* Z + 2 (Z^* Z)^{1/2})$ . -For a proof see the proof of $\lambda_j(Z^* + Z ) \leq \lambda_j(2 (Z^* Z)^{1/2})$ in Bhatia, Matrix Analysis, Proposition III.5.1 (Fan-Hoffman). -Lemma 2 : For $a \geq 0 , b > 0$ holds $(a+b)(1+b^{-1}) \geq a + 2 a^{1/2} + 1$ . -Proof left to the reader. -Proof of the conjecture : -Choose $Z = XY^*$, $A=Z^* Z$, $B=YY^*$ . -We may assume that $B$ is invertable. -Then we have to show : -$\lambda_j((I+B^{-1})^{1/2}(A+B)(I+B^{-1})^{1/2}) \geq \lambda_j(A + 2A^{1/2} + I)$ . -Let $V_j$ be the span of the eigenvectors to the j largest eigenvalues of $A$ and $W_j = (I+B^{-1})^{-1/2}(V_j)$ . -Then -$\lambda_j((I+B^{-1})^{1/2}(A+B)(I+B^{-1})^{1/2})$ -$ \geq min \{x^* ((I+B^{-1})^{1/2}(A+B)(I+B^{-1})^{1/2})x : x \in W_j$ and $x^* x = 1\}$ -$\geq min \{x^* ((I+B^{-1})^{1/2}(\lambda_j(A)I+B)(I+B^{-1})^{1/2})x : x \in W_j$ and $x^* x = 1\}$ -$\geq \lambda_j(A) + 2 \lambda_j(A)^{1/2} + 1 = \lambda_j(A + 2A^{1/2} + I)$ .<|endoftext|> -TITLE: Is there a $W^{2,2}$ isometric embedding of the flat torus into $\mathbb{R}^3$? -QUESTION [9 upvotes]: It is well known that there exists a $C^1$ isometric embedding of flat torus into $\mathbb{R}^3$, and that this embedding cannot be $C^2$. -Is there a $W^{2,2}$ isometric embedding? (i.e an isometric map $f \in W^{2,2}(\mathbb{T}^2,\mathbb{R}^3)$?) - -REPLY [3 votes]: This is just an expansion on my comment to @j.c. 's answer. -We want to use Pakzad's result cited in that answer in order to prove that there is no $f$ as in the original question. -My idea: -By Sobolev embedding theorems we have that $f\in C^{0,\alpha}$ for all $\alpha\in (0,1)$. In particular, the image $\Sigma=f(\mathbb{T}^2)$ is a compact set. As such, there is an $R_0>0$ so that -$$\Sigma\subset B_{R_0}(0)$$ -where here $B_{R_0}(0)$ is the open euclidean ball of radius $r_0$ centered at the origin. -Now let -$$ -R=\inf\{ \rho>0: \Sigma \subset B_\rho(0)\} -$$ -and observe that, as $\Sigma$ is compact $\Sigma\cap \partial B_{R}\neq \emptyset$. In particular, there is a $p'\in \Sigma\cap \partial B_{R}$ and so, there is a $p\in \mathbb{T}^2$ so $f(p)=p'$. -Now pick a small radius $r$ so that $\exp_p: B_r(0)\to \mathbb{T}^2$ is an isometry onto it's image (here we use that the metric on the torus is flat) and consider the map $\tilde{f}=f\circ \exp_p|_{B_r(0)}$. Observe, that $\tilde{f}(0)=p'$. By Pakzad's result, up to shrinking $r$ we know that either $\tilde{f}$ is affine or there is a line through $0$ on which $\tilde{f}$ is affine. As $f$ is isometric these affine functions are non-constant and so the image of $\tilde{f}$ contains a line segment through $p'$. This segment must leave $\bar{B}_R(0)$ which contradicts the choice of $R$.<|endoftext|> -TITLE: Am I allowed to say "first-order Vopěnka cardinal"? -QUESTION [6 upvotes]: For a cardinal $\kappa$ such that $V_{\kappa}$ satisfies Vopěnka's principle as a first-order axiom schema, am I allowed to say "first-order Vopěnka cardinal", or is there any kind of standard term for it? - -REPLY [11 votes]: In my paper, The Vopěnka principle is inequivalent to but conservative over the Vopěnka scheme, I distinguish between the (second-order) Vopěnka principle and the first-order version, which I call the Vopěnka scheme, and prove that these principles are not equivalent, although they are equiconsistent and moreover, the Vopěnka principle is conservative over the Vopěnka scheme for first-order assertions. -In the paper, I say that a Vopěnka cardinal is a cardinal $\kappa$ such that $V_\kappa$ satisfies the Vopěnka principle, and a Vopěnka scheme cardinal is a cardinal such that $V_\kappa$ satisfies the Vopěnka scheme. Although as I mentioned the Vopěnka principle and Vopěnka scheme are equiconsistent, nevertheless Corollary 10 in the paper shows in contrast that Vopěnka cardinals are strictly stronger in consistency strength than even a closed unbounded proper class of Vopěnka scheme cardinals. -So you are referring to what I call the Vopěnka scheme cardinals.<|endoftext|> -TITLE: Example of affine locally symmetric space -QUESTION [6 upvotes]: Who can give an example of an affine locally symmetric space that is not a Riemanian locally symmetric space? - -REPLY [5 votes]: Having had hard time visualizing the example from the answer by Robert Bryant, I decided to make some graphs for it. They turned out so beautiful that I decided to share them. -Geodesics passing through $(0,1)$: - -Geodesics passing through $(1,-0.3)$: - -And for the "dual" example, geodesics passing through $(0,-0.1)$: - -Geodesics passing through $(1,1)$:<|endoftext|> -TITLE: Pushforward maps for cohomology of coherent sheaves -QUESTION [5 upvotes]: Let $X$ be a smooth projective algebraic variety over a field $k$, of dimension $n$, and let $Z$ be a smooth closed subvariety of dimension $m$, with $i: Z \hookrightarrow X$ the inclusion map. -For any locally free coherent sheaf $\mathcal{F}$ on $X$, there is a pullback map -$$\imath^*: H^i(X, \mathcal{F}) \to H^i(Z, \iota^* \mathcal{F});$$ -and via Serre duality we have isomorphisms $H^i(X, \mathcal{F})^\vee = H^{n-i}(X, \mathcal{F}^\vee \otimes \omega_X)$ and $H^i(Z, \iota^* \mathcal{F})^\vee = H^{m-i}(Z, \iota^* \mathcal{F}^\vee \otimes \omega_Z)$, where $\omega_X$ and $\omega_Z$ are the dualising sheaves. Setting $j=m-i$ and $\mathcal{G} = \mathcal{F}^\vee$, we conclude that there is a pushforward map -$$\imath_*: H^j(Z, \iota^* \mathcal{G} \otimes \omega_Z) \to H^{j + - c}(X, \mathcal{G} \otimes \omega_X),$$ -for any $j$ and any locally free coherent sheaf $\mathcal{G}$ on $X$, where $c = n-m$ is the codimension of $Z$ in $X$. -Does this map have an intrinsic description (not using Serre duality)? Can it be defined without assuming that $X$ be projective, or that $\mathcal{G}$ be locally free? - -REPLY [5 votes]: The map is induced by the right adjoint $i^!$ of the pushforward functor and the adjunction morphism $i_*i^! \to \mathrm{id}$, in view of the formula $i^!(F) \cong i^*(F) \otimes \omega_{Z/X}[\dim Z/X]$. This works for any locally complete intersection closed embedding.<|endoftext|> -TITLE: $q$-analog of an integral from quantum field theory? -QUESTION [5 upvotes]: This question has been completely reformulated and a new property for the function $f_q$ has been added due to a series of helpful comments by fedja. -Consider the integral from quantum field theory due to F.A. Smirnov (see this MSE post): -\begin{align} -&\int_{-\infty}^{\infty}\prod_{j=1}^3\Gamma\Bigl(\frac 1 3 -\frac {\alpha-\beta_j}{2\pi i}\Bigr) -\Gamma\Bigl(\frac 1 3 +\frac {\alpha-\beta_j}{2\pi i}\Bigr)(3\alpha-\sum\beta_m)e^{-\frac{\alpha}2}d\alpha\nonumber\\ -&=\frac{(2\pi \Gamma(\frac 2 3))^2}{\Gamma(\frac 4 3)} -\prod_{k\neq j}\Gamma\Bigl(\frac 2 3 -\frac {\beta_k-\beta_j}{2\pi i}\Bigr) -e^{-\frac12\sum\beta_m}\sum e^{\beta_m},\quad |\text{Im}~\beta_j|<2\pi/3. -\end{align} -Some analytical and numerical calculations suggest that it has a $q$-analog of the form -\begin{align} -&\int_{-\infty}^\infty\prod_{j=1}^3\frac{\Gamma_q\left(\frac13+\frac{x-\beta_j}{2\pi i}\right)}{\Gamma_q\left(\frac23+\frac{x-\beta_j}{2\pi i}\right)}\frac{\sum q^{-\frac{\beta_m}{2 \pi i}}-\left(1+q^\frac13+q^{-\frac13}\right) q^{-\frac{x}{2 \pi i}}}{\prod_{m}\sin\left(\frac{\pi}3-\frac{x-\beta_m}{2 i}\right)}e^{-\frac{x}2}dx\\ -&=\frac{-2\pi i}{(q;q)_{\infty }^9}\frac{\Gamma_q^2\left(\frac23\right)}{\Gamma_q\left(\frac43\right)}\frac{q^{5/9}}{(1-q)^2}e^{-\frac{1}{2} (\beta_1+\beta_2+\beta_3)} q^{-\frac{\beta_1+\beta_2+\beta_3}{6 \pi i}}\prod_{k\neq j}\Gamma_q\left(\frac23+\frac{\beta_j-\beta_k}{2\pi i}\right)\\ -&\times\frac{\theta_q(q^{\frac{\beta_1-\beta_2}{2\pi i}})}{\sin\frac{\beta_1-\beta_2}{2 i}}\frac{\theta_q(q^{\frac{\beta_2-\beta_3}{2\pi i}})}{\sin\frac{\beta_2-\beta_3}{2 i}} \frac{\theta_q(q^{\frac{\beta_3-\beta_1}{2\pi i}})}{\sin\frac{\beta_3-\beta_1}{2 i}} \cdot f_{q_1}(\beta_1,\beta_2,\beta_3).\tag{5} -\end{align} -where $q_1=e^{-\frac{4 \pi ^2}{\ln(1/q)}}$, $\Gamma_q$ is the $q$-Gamma function and -$$ -\theta_q(z)=(q;q)_\infty(z;q)_\infty(q/z;q)_\infty. -$$ -The exact form of the function $f_q$ is unknown but there is evidence that it has a relatively simple closed form expression in terms of finite combination of theta functions. -Obviously $f_q(x,y,z)$ is symmetric in all three arguments $x,y,z$. It is known also that it has the following properties -\begin{align} -&f_q(x+c,y+c,z+c)=e^cf_q(x,y,z),\tag{1}\\ -\\ -&f_q\left(\frac{2\pi i}{3},0,z\right) -\begin{aligned}[t] - & =\left(e^z-1\right) \frac{\theta_q\left(e^{\frac{4 i \pi }{3}-z}\right)}{\theta_q\left(e^{-z}\right)}\\ - & =\left(e^z-1\right)e^{-\frac{\pi i}{3}} q_1^{-\frac{1}{9}+\frac{z}{6 \pi i}}\frac{\theta_{q_1}\left(q_1^{\frac13+\frac{z}{2\pi i}}\right)}{\theta_{q_1}\left(q_1^{\frac{z}{2\pi i}}\right)} - \end{aligned} \tag{2}\\ -\\ -&f_q(x,y,z)=e^x+e^y+e^z\\ -&+\frac32\left(e^x+e^y+e^z\right)\left(1+e^{x-y}+e^{y-x}+e^{y-z}+e^{z-y}+e^{z-x}+e^{x-z}\right)q\\ -&+i\frac{\sqrt{3}}{2}\left(e^{-x}+e^{-y}+e^{-z}\right)\left(e^{2x}+e^{2y}+e^{2z}\right)q+O(q^2) -\tag{3} -\\ -&f_{q}(x,y,z)= e^{-\frac{\pi i}{3}} q_1^{-\frac29} \small{\frac{\left(e^z-e^y\right) q_1^{\frac{-2 x+y+z}{6 \pi i}}+\left(e^x-e^z\right) q_1^{\frac{x-2 y+z}{6 \pi i}}+\left(e^y-e^x\right) q_1^{\frac{x+y-2 z}{6 \pi i}}}{\Big(1-q_1^{\frac{x-y}{2 \pi i}}\Big) \Big(1-q_1^{\frac{z-x}{2 \pi i}}\Big) \Big(1-q_1^{\frac{y-z}{2 \pi i}}\Big)}\left(1+O(q_1^{\frac13})\right)}\tag{4} -\\ -\end{align} -Eq. $(2)$ is a condition that the difference of LHS and RHS of $(5)$ is an entire function. $(3)$ has been extracted from numerical calculations. $(4)$ has been found from exact $q$-series representation of the integral assuming $\beta_j\in\mathbb{R}$. However numerical calculations show that asymptotics $(4)$ holds if all differences $|\text{Im}~(\beta_j-\beta_k)|\leq \frac{2\pi}{3}$, and one can check that $(4)$ is consistent with $(2)$. -Q: Can anybody make a good guess from these data $1-4$ what this function $f_q(x,y,z)$ might be? - -REPLY [5 votes]: The function -$$ -f_q(x,y,z)=\sum_{cyc}e^z\frac{\theta_q\left(e^{\frac{2 \pi i}{3}+x-z}\right) \theta_q\left(e^{\frac{2 \pi i}{3}+y-z}\right)}{\theta_q\left(e^{x-z}\right) \theta_q\left(e^{y-z}\right)} -$$ -satisfies all $4$ conditions and also has been confirmed numerically.<|endoftext|> -TITLE: A good reference to the general Chinese Remainder Theorem -QUESTION [13 upvotes]: I am writing a paper on the topology of the Golomb space and need a good (standard) reference to the following -General Chinese Remainder Theorem. For integer numbers $a_1,\dots,a_n$ and positive integers $b_1,\dots,b_n$ the intersection $\bigcap_{i=1}^n(a_i+b_i\mathbb Z)$ is not empty if and only if $a_i-a_j\in d_{i,j}\mathbb Z$ for any $1\le i -TITLE: Do prime ideals in polynomial ring generate prime ideals in the ring of holomorphic functions? -QUESTION [15 upvotes]: Suppose that $I \subset \mathbb C[z_1,\dots, z_n]$ is a prime ideal. Consider the ideal $I_{hol}$ in the ring of holomorphic functions $f: \mathbb C^n\to \mathbb C$ generated by polynomials from $I$. -Is $I_{hol}$ prime? - -REPLY [7 votes]: Edit. I added some lemmas to address the issue raised by David Speyer and the OP. The books by Grauert and Grauert-Remmert are wonderful sources. The proofs in those books are the "correct" arguments, using "sledgehammers" as little as possible. Even though it is a sledgehammer, Hironaka's Resolution of Singularities Theorem does quickly establish the result. Also, I will also use the smaller sledgehammer ("club hammer"?) of GAGA. -For every locally finite type $\mathbb{C}$-scheme $(X,\mathcal{O}_X)$, denote by $(i_X,i_X^\#):(X^\text{an},\mathcal{O}_X^{\text{an}})\to (X,\mathcal{O}_X)$ the associated complex analytic space. For every topological space $S$, denote by $C^0_S(\mathbb{C})$ the sheaf of continuous, $\mathbb{C}$-valued functions on $S$. For every complex analytic space $(S,\mathcal{O}_S)$, denote by $u_S:\mathcal{O}_S\to C^0_S(\mathbb{C})$ the natural homomorphism of sheaves of $\mathbb{C}$-algebras. -Question 0. For a finite type $\mathbb{C}$-scheme $(X,\mathcal{O}_X)$, what is the kernel of $u_{X^{\text{an}}}:\mathcal{O}_X^{\text{an}}\to C^0_{X^{\text{an}}}(\mathbb{C})$? -For the nilradical $\mathcal{N}\subset \mathcal{O}_X$, certainly $\mathcal{N}\cdot \mathcal{O}_X^{\text{an}}$ is contained in the kernel. Most of the comments for the question and for this answer focus on the problem of proving that the kernel equals $\mathcal{N}\cdot \mathcal{O}_X^{\text{an}},$ which it does. One approach uses "analytic reducedness" of the local rings $\mathcal{O}_{X,p}/\mathcal{N}_p$ at $\mathbb{C}$-points $p\in X^{\text{an}}$. Not all reduced local Noetherian rings are analytically reduced. However, excellent local rings that are reduced are analytically reduced, cf. the Wikipedia page for excellent local rings. -https://en.wikipedia.org/wiki/Excellent_ring#Properties -The approach here to the kernel of $u_{X^{\text{an}}}$ is different, using Hironaka's Resolution of Singularities. -Lemma 1. For every finite type $\mathbb{C}$-scheme, the associated complex analytic space is a complex manifold if and only if $(X,\mathcal{O}_X)$ is a smooth $\mathbb{C}$-scheme. In particular, if $(X,\mathcal{O}_X)$ is a smooth $\mathbb{C}$-scheme, then the homomorphism $u_{X^\text{an}}$ is injective. -Proof. This is local, so we may assume that $X$ is affine. Then the result follows from the Jacobian criterion for smoothness (on the algebraic side) and the complex analytic version of the Implicit Function Theorem (whose hypothesis is the rank condition on the Jacobian as in the Jacobian criterion). For a complex manifold, there are local biholomorphisms with polydisks, whose sheaf of holomorphic functions is manifestly a subsheaf of the sheaf of continuous functions. QED -Lemma 2. For every closed immersion of smooth $\mathbb{C}$-schemes, $(j,j^\#):(Y,\mathcal{O}_Y)\to (X,\mathcal{O}_X)$, with ideal sheaf $\mathcal{I}$, the analytic ideal sheaf $\mathcal{I}\cdot \mathcal{O}_X^{\text{an}}$ equals the subsheaf of $\mathcal{O}_X^{\text{an}}$ of germs of holomorphic functions that vanish identically on the underlying set of $Y^{\text{an}}$. -Proof. Since $\mathcal{O}_X^\text{an}$ is flat as a sheaf of $i_X^{-1}\mathcal{O}_X$-algebras, the following sequence is a short exact $$0\to \mathcal{I}\cdot \mathcal{O}_X^{\text{an}} \to \mathcal{O}_X^{\text{an}} \to j^\text{an}_*\mathcal{O}_Y^{\text{an}}\to 0.$$ By the previous lemma, $\mathcal{O}_Y^{\text{an}}$ is a subsheaf of $C^0_{Y^\text{an}}(\mathbb{C})$. QED -Lemma 3. For every closed immersion $(j,j^\#):(Y,\mathcal{O}_Y)\to (X,\mathcal{O}_X)$ of finite type $\mathbb{C}$-schemes, if $(X,\mathcal{O}_X)$ is smooth and quasi-projective, and if $(Y,\mathcal{O}_Y)$ is reduced, then the conclusion from Lemma 2 holds. -Proof. By Hironaka's Resolution of Singularities Theorem, there exists a projective, birational morphism, $$(\nu,\nu^\#):(\widetilde{X},\mathcal{O}_{\widetilde{X}})\to (X,\mathcal{O}_X),$$ and a smooth closed subscheme, $$(\widetilde{j},\widetilde{j}^\#):(\widetilde{Y},\mathcal{O}_{\widetilde{Y}})\to (\widetilde{X},\mathcal{O}_{\widetilde{X}}),$$ with ideal sheaf $\widetilde{\mathcal{I}}$ such that $\nu_*\widetilde{\mathcal{I}}\cdot \mathcal{O}_X$ equals $\mathcal{I}$. By Serre's GAGA, also $\nu^{\text{an}}_*\widetilde{\mathcal{I}}^{\text{an}}$ equals $(\nu_*\widetilde{\mathcal{I}})^{\text{an}}$, so that the same result holds for the associated analytic spaces. -Associated to the short exact sequence on $\widetilde{X}^{\text{an}}$, $$0 \to \widetilde{I}\cdot \mathcal{O}_{\widetilde{X}}^{\text{an}} \to -\mathcal{O}_{\widetilde{X}}^\text{an}\to \widetilde{j}_*^\text{an}\mathcal{O}_{\widetilde{Y}^{\text{an}}} \to 0,$$ there is an exact sequence on $X^{\text{an}}$, $$0 \to \nu^\text{an}_*\widetilde{I}\cdot \mathcal{O}_{\widetilde{X}}^{\text{an}} \xrightarrow{e} -\nu^\text{an}_*\mathcal{O}_{\widetilde{X}}^\text{an}\to -\nu^\text{an}_*\widetilde{j}_*^\text{an}\mathcal{O}_{\widetilde{Y}^{\text{an}}}.$$ Finally, by Lemma 1, also the natural map from the third term to the pushforward of $C^0_{\widetilde{Y}^\text{an}}(\mathbb{C})$ is injective. Thus, we also have injectivity of the induced homomorphism from the cokernel of $e$ to the pushforward of $C^0_{\widetilde{Y}^{\text{an}}}(\mathbb{C})$. -Since $X$ is already smooth, the natural homomorphism $\nu^\#:\mathcal{O}_X[0]\to R\nu_*\mathcal{O}_{\widetilde{X}}$ is a quasi-isomorphism compatible with arbitrary base change. In particular, $\mathcal{O}_X\to \nu_*\mathcal{O}_{\widetilde{X}}$ is an isomorphism. By GAGA, also $\mathcal{O}_X^{\text{an}} \to \nu_*^{\text{an}}\mathcal{O}^{\text{an}}_{\widetilde{X}}$ is an isomorphism. Together with the previous paragraph, it follows that the natural homomorphism from $\mathcal{O}_{X^\text{an}}/\mathcal{I}\cdot \mathcal{O}_{X^\text{an}}$ to the pushforward of $C^0_{\widetilde{Y}^\text{an}}(\mathbb{C})$ is injective. This homomorphism factors through the natural homomorphism, $$\mathcal{O}_{X^\text{an}}/\mathcal{I}\cdot \mathcal{O}_{X^\text{an}} \to j^{\text{an}}_*C^0_{Y^\text{an}}(\mathbb{C}).$$ Thus, also this second natural homomorphism is injective. QED -Lemma 4. For every finite type $\mathbb{C}$-scheme $(Y,\mathcal{O}_Y)$, the scheme is reduced if and only if the homomorphism $u_{Y^{\text{an}}}$ is injective. In particular, for the nilradical $\mathcal{N}\subset \mathcal{O}_Y$, the nilradical of $\mathcal{O}_{Y}^\text{an}$ equals $\mathcal{N}\cdot \mathcal{O}_Y^\text{an}$. -Proof. Since $\mathcal{O}_Y^\text{an}$ is flat over $i_Y^{-1}\mathcal{O}_Y$, the nilradical of $\mathcal{O}_Y^\text{an}$ contains $\mathcal{N}\cdot \mathcal{O}_Y^\text{an}$. Thus, if $(Y,\mathcal{O}_Y)$ is nonreduced, then $u_{Y^\text{an}}$ is not injective. -Conversely, assume that $(Y,\mathcal{O}_Y)$ is reduced. To prove that $u_{Y^{\text{an}}}$ is injective and that $\mathcal{N}\cdot \mathcal{O}_Y^{\text{an}}$ equals the entire nilradical, it suffices to work locally. Locally there are closed immersions of $(Y,\mathcal{O}_Y)$ into affine space. Thus, the result follows from the previous lemma. QED -Lemma 5. For every finite type, affine $\mathbb{C}$-scheme $(X,\mathcal{O}_X)$, for every surjection of coherent $\mathcal{O}_X$-sheaves, $\phi:\mathcal{F}\to \mathcal{G}$, the induced map $\phi^{\text{an}}(X^{\text{an}}):\mathcal{F}^{\text{an}}(X^{\text{an}})\to \mathcal{G}^{\text{an}}(X^{\text{an}})$ is surjective. -Proof. Probably there is a more direct proof, but this also follows from vanishing of the higher sheaf cohomology of coherent analytic sheaves on a Stein complex analytic space. Since $\mathcal{O}_{X^\text{an}}$ is flat over $i_X^{-1}\mathcal{O}_X$, the kernel of $\phi^{\text{an}}$ equals the coherent analytic sheaf associated to the kernel of $\phi$. Since $X$ is affine, the complex analytic space $X^{\text{an}}$ is Stein. By vanishing of the higher cohomology of coherent analytic sheaves on a Stein complex analytic space, the first sheaf cohomology of $\text{Ker}(\phi^{\text{an}})$ vanishes. Surjectivity of $\phi^{\text{an}}(X^{\text{an}})$ follows by the long exact sequence of sheaf cohomology. QED -Proposition. For a finite type $\mathbb{C}$-scheme $(X,\mathcal{O}_X)$ that is smooth and affine, for every closed subscheme $(Z,\mathcal{O}_Z)$ that is integral and with ideal sheaf $\mathcal{I}$, also the ideal $I^\text{hol}:=\mathcal{I}(X)\cdot \mathcal{O}_X^\text{an}(X^{\text{an}})$ is a prime ideal in $\mathcal{O}_X^\text{an}(X^\text{an})$. -Proof. If a product $g\cdot h$ of holomorphic functions is contained in $I^\text{hol}$ then it vanishes on $Z^\text{an}$, since each element of $I(X)$ vanishes on $Z^\text{an}$. Since $Z$ is irreducible, the smooth locus $Z_{\text{sm}} = Z\setminus Z_{\text{sing}}$ is a dense open subscheme that is an integral, smooth $\mathbb{C}$-scheme. The zero loci of $g$, and $h$ on $Z^\text{an}_{\text{sm}}$ are complex analytic subvarieties of a connected, complex manifold. If neither of these complex analytic subvarieties equals all of $Z^\text{an}_{\text{sm}}$, then they are each nowhere dense. In that case, also the union is nowhere dense, contradicting the hypothesis that $g\cdot h$ vanishes on $Z^\text{an}$. Thus, one of the factors, say $g$, vanishes identically on $Z^\text{an}_{\text{sm}}$. Since $Z^\text{an}_{\text{sm}}$ is dense in $Z^\text{an}$ for the analytic topology, also $g$ vanishes on $Z^\text{an}$. By Lemma 3, $g$ is a section of $(\mathcal{I}\cdot \mathcal{O}_X^\text{an})(X^\text{an})$. -Fix a finite generating set $f_1,\dots,f_r$ of the ideal $I = \mathcal{I}(X)$. By Lemma 5, the induced map, $$(\mathcal{O}_X^{\text{an}})(X^\text{an})^{\oplus r} \to (\mathcal{I}\cdot\mathcal{O}_X^{\text{an}})(X^\text{an}), \ \ (u_1,\dots,u_r)\mapsto u_1f_1 + \dots + u_rf_r,$$ is surjective. By construction, the image is in $I^{\text{hol}}$. Therefore $I^{\text{hol}}$ equals all of $(\mathcal{I}\cdot\mathcal{O}_X^{\text{an}})(X^\text{an})$. Thus, $g$ is an element of $I^{\text{hol}}$. QED<|endoftext|> -TITLE: Is there an explicit expression for Chebyshev polynomials modulo $x^r-1$? -QUESTION [19 upvotes]: This is an immediate successor of Chebyshev polynomials of the first kind and primality testing and does not have any other motivation - although original motivation seems to be huge since a positive answer (if not too complicated) would give a very efficient primality test (see the linked question for details). -Recall that the Chebyshev polynomials $T_n(x)$ are defined by $T_0(x)=1$, $T_1(x)=x$ and $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$, and there are several explicit expressions for their coefficients. Rather than writing them down (you can find them at the Wikipedia link anyway), let me just give a couple of examples: -$$ -T_{15}(x)=-15x(1-4\frac{7\cdot8}{2\cdot3}x^2+4^2\frac{6\cdot7\cdot8\cdot9}{2\cdot3\cdot4\cdot5}x^4-4^3\frac{8\cdot9\cdot10}{2\cdot3\cdot4}x^6+4^4\frac{10\cdot11}{2\cdot3}x^8-4^5\frac{12}{2}x^{10}+4^6x^{12})+4^7x^{15} -$$ -$$ -T_{17}(x)=17x(1-4\frac{8\cdot9}{2\cdot3}x^2+4^2\frac{7\cdot8\cdot9\cdot10}{2\cdot3\cdot4\cdot5}x^4-4^3\frac{8\cdot9\cdot10\cdot11}{2\cdot3\cdot4\cdot5}x^6+4^4\frac{10\cdot11\cdot12}{2\cdot3\cdot4}x^8-4^5\frac{12\cdot13}{2\cdot3}x^{10}+4^6\frac{14}{2}x^{12}-4^7x^{14})+4^8x^{17} -$$ -It seems that $n$ is a prime if and only if all the ratios in the parentheses are integers; this is most likely well known and easy to show. -The algorithm described in the above question requires determining whether, for an odd $n$, coefficients of the remainder from dividing $T_n(x)-x^n$ by $x^r-1$, for some fairly small prime $r$ (roughly $\sim\log n$) are all divisible by $n$. In other words, denoting by $a_j$, $j=0,1,2,...$ the coefficients of $T_n(x)-x^n$, we have to find out whether the sum $s_j:=a_j+a_{j+r}+a_{j+2r}+...$ is divisible by $n$ for each $j=0,1,...,r-1$. -The question then is: given $r$ and $n$ as above ($n$ odd, $r$ a prime much smaller than $n$), is there an efficient method to find these sums $s_j$ without calculating all $a_j$? I. e., can one compute $T_n(x)$ modulo $x^r-1$ (i. e. in a ring where $x^r=1$) essentially easier than first computing the whole $T_n(x)$ and then dividing by $x^r-1$ in the ring of polynomials? -(As already said, only the question of divisibility of the result by $n$ is required; also $r$ is explicitly given (it is the smallest prime with $n$ not $\pm1$ modulo $r$). This might be easier to answer than computing the whole polynomials mod $x^r-1$.) - -REPLY [5 votes]: The coefficient of $x^j$ in $(T_n(x)\bmod (x^r-1))$ equals the coefficient of $t^{n+r-j-1}$ in -$$\frac{(1+t^2)^{r-j}}{2^{r-j}} \frac{((1+t^2)^{r-1}t - 2^{r-1}t^{r-1})}{((1+t^2)^r - 2^rt^r)}.$$ -This coefficient can be explicitly computed as -$$\sum_{k\geq 0} 2^{rk-r+j} \left( \binom{r-1-j-rk}{\frac{n+r-j-2-rk}{2}} - 2^{r-1}\binom{-j-rk}{\frac{n-j-rk}{2}}\right).$$ -(here the binomial coefficients are zero whenever their lower indices are noninteger or negative)<|endoftext|> -TITLE: Internal Hom of Deligne' tensor product -QUESTION [5 upvotes]: I read the following statement (equation 22) in "Monoidal 2-structure of bimodule categories" by Justin Greenough: -Let $\mathcal{C}$ be a finite tensor category (abelian k-linear rigid monoidal category with simple unit and finite dimensional Hom spaces). Let $\mathcal{M}$ and $\mathcal{N}$ be exact left module categories over $\mathcal{C}$. -We introduce the left $\mathcal{C}$-module structure in $\mathcal{M} \boxtimes \mathcal{N}$ (the Deligne' tensor product of $\mathcal{M}$ and $\mathcal{N}$) by: -$$X \otimes (M \boxtimes N) = (X \otimes M) \boxtimes N,$$ -where $X \in \mathcal{C}$. -Then the equation 22 tells us that -$$\underline{Hom}_{\mathcal{M} \boxtimes \mathcal{N}}(M \boxtimes N, S \boxtimes T) = \underline{Hom}_{\mathcal{M}}(M, S) \otimes \underline{Hom}_{\mathcal{N}}(N,T),$$ -where $\underline{Hom}_{*}$ are internal hom for left $\mathcal{C}$ structure in -$\mathcal{M} \boxtimes \mathcal{N}$, $\mathcal{M}$ and $\mathcal{N}$. -Now, let us consider the simple case: -let $\mathcal{C}$ be a unitary fusion category and $\mathcal{M} = \mathcal{N} = \mathcal{C}$. Then by the definition of internal Hom and the equation above, we have -$$Hom_{\mathcal{C} \boxtimes \mathcal{C}}(1 \boxtimes 1, X \boxtimes X^*) \cong Hom_{\mathcal{C}}(1, \underline{Hom}_{\mathcal{C} \boxtimes \mathcal{C}}(1 \boxtimes 1, X \boxtimes X^*))\\ -\cong Hom_{\mathcal{C}}(1, \underline{Hom}_{\mathcal{C}}(1, X) \otimes \underline{Hom}_{\mathcal{C}}(1,X^*)), $$ -where $1$ is the unit of $\mathcal{C}$ and $X$ is a simple object in $\mathcal{C}$ such that $X \ncong 1$ and $X^*$ is the left (or right) adjont of $X$. -Since $\underline{Hom}_{\mathcal{C}}(1, X) = X$ and $\underline{Hom}_{\mathcal{C}}(1, X^*) = X^*$, we have -$$\{0\} = Hom_{\mathcal{C}}(1,X) \otimes Hom_{\mathcal{C}}(1, X^*) \cong -Hom_{\mathcal{C} \boxtimes \mathcal{C}}(1 \boxtimes 1, X \boxtimes X^*) \cong -Hom_{\mathcal{C}}(1, X \otimes X^*) \neq \{0\}.$$ -So it seems that we have a contradiction here. Can anyone tell me if I made a mistake somewhere? Does the equation (22) in "Monoidal 2-structure of bimodule categories" hold? Thank you in advance! - -REPLY [4 votes]: That equation is not correct. You should be suspicious because the definition of the $\mathcal{C}$-module category structure on $\mathcal{M} \boxtimes \mathcal{N}$ doesn't use the $\mathcal{C}$-module category structure on $\mathcal{N}$, only the one on $\mathcal{M}$. So how could the $\mathcal{C}$-enriched hom of $\mathcal{N}$ show up? -The correct equation is: -$$ \underline{Hom}_{\mathcal{M} \boxtimes \mathcal{N}}(M \boxtimes N, S \boxtimes T) \cong \underline{Hom}_{\mathcal{M}}(M , S) \otimes {Hom}_{\mathcal{N}}( N, T)$$ -Here the second factor is just the usual vector space valued hom in $\mathcal{N}$ (and we are using that $\mathcal{C}$ is naturally a $Vect$-module category). -With this corrected equation you no longer get a contradiction because it is $Hom_{\mathcal{C}}(1, X^*) = 0$ which shows up instead of $\underline{Hom}_{\mathcal{C}}(1, X^*) = X^*$.<|endoftext|> -TITLE: How does $\zeta^{\mathfrak{m}}(2)$ and relate to $\zeta(2)$? -QUESTION [5 upvotes]: EDIT There appears to be a numerical zeta function $\zeta(2)$ as well as at least two different "motivic" zeta function realizations (Betti and de Rham) $\zeta^{\mathfrak{m}}(2)$. The "period map" of equally mysterious properties relates the two objects and proves the Hoffman conjecture. -Implicit in the comment is that $\zeta(2)$ is a period, which got lifted to a motive... Possibly $\zeta(2)$ has become a homological object? The comments did not explain anything. - -I tried to read a paper of Francis Brown Mixed Tate Motives over $\mathbb{Z}$ and I was confused about the status of the conjecture he states: - -Conjecture Every multiple zeta value $\zeta(n_1, \dots, n_r)$ is a $\mathbb{Q}$-linear combination of multiple zeta values at $n \in \{ 2, 3\}$: - $$ \big\{ \zeta(n_1, \dots, n_r): \text{ where } n_i \in \{ 2, 3\} \big\} $$ - -He proves a verion of this using motivic zeta functions, which do not seem to be numerical at all. Brown proves: - -Theorem The set of elements $ \{ \zeta^{\mathfrak{m}}(n_1, \dots, n_r): \text{ where } n_i \in \{ 2, 3\} \} $ are a basis in the space of motivic multiple zeta values. - -In fact, I haven't any idea what a motive is. Despite it's elementary appearance, it seems to be related to work of Goncharov and Deligne [1]. -Wikipedia's example for the affine line and projectiv line could potentially make sense: -$$ -Z(\mathbb{A}^n, t) = \frac{1}{1 - \mathbb{L}^n t} \hspace{0.25in}\text{ and }\hspace{0.25in} -Z(\mathbb{P}^n, t) = \prod_{i=0}^n \frac{1}{1 - \mathbb{L}^i t}$$ -Wikipedia offers another exampls with the Hilbert scheme of points. None of that seems relevant. -Confusingly Brown, notes that in his conjecture $\zeta^{\mathfrak{m}}(2) \neq 0$ unlike in Goncharov's work. -How does $\zeta^{\mathfrak{m}}(2)$ and relate to $\zeta(2)$ ? Does Brown's Theorem prove the Conjecture (of Hoffman)? - -Right now Brown's discussion does not mean very much to me because it rests on some rather difficult concepts: - -Tannakian Categories (and it's Galois Group) -Mixed Motives -Motivic Iterated Integrals -Betti and de Rham realizations of motives -Why are Lyndon words relevant? -Do the motivic relations he finds descend to relations of plain zeta values? -Which "period map" is being used? - -My impression is that the Shuffle Relations (which one might find in Kaneko-Ihara-Zagier) get lifted to something very complicated (which could be the Tannakian Category). He has two notes on Feynman Amplitudes [1, 2] which are broader than multiple zeta functions. - -REPLY [7 votes]: I am not sure what exactly your question is, but the question in the title ``How does $\zeta^{\mathfrak{m}}(2)$ relate to $\zeta(2)$?'' can be answered without really knowing what a motivic multiple zeta values is. -First, it follows essentially from the representation of multiple zeta values as iterated integrals on $\mathbb P^1 \setminus \{0,1,\infty\}$ that the $\mathbb{Q}$-algebra $\mathcal{Z}$ of multiple zeta values is a quotient -$$ -\mathbb{Q}\langle e_0,e_1\rangle/I -$$ -of the shuffle $\mathbb{Q}$-algebra $\mathbb{Q}\langle e_0,e_1\rangle$ by some ideal $I$. Here, the multiple zeta value $\zeta(n_1,\ldots,n_r)$ corresponds to the class modulo $I$ of the word $e_1e_0^{n_1-1}\ldots e_1e_0^{n_r-1}$. -On the other hand, in Mixed Tate motives over $\mathbb Z$, Brown defines the $\mathbb Q$-algebra $\mathcal{Z}^{\mathfrak m}$ of motivic multiple zeta values to be the quotient -$$ -\mathbb{Q}\langle e_0,e_1\rangle/J, -$$ -where $J \subset I$ is now the ideal of motivic relations (whatever those are), and the motivic multiple zeta value $\zeta^{\mathfrak m}(n_1,\ldots,n_r)$ is defined to be the class of $e_1e_0^{n_1-1}\ldots e_1e_0^{n_r-1}$. Now since $J \subset I$, basic algebra implies that the association $\zeta^{\mathfrak m}(n_1,\ldots,n_r) \mapsto \zeta(n_1,\ldots,n_r)$ induces a well-defined surjection of $\mathbb Q$-algebras -$$ -\mathcal{Z}^{\mathfrak m} \cong \mathbb{Q}\langle e_0,e_1\rangle/J \rightarrow \mathbb{Q}\langle e_0,e_1\rangle/I \cong \mathcal{Z}. -$$ -As already mentioned in the comments, this is precisely what Brown calls the period map (it should be noted that there is also another definition of motivic multiple zeta values and the period map which adapts better to other situations, see for example Brown's ICM 2014 talk). - -As you can see, there is no difficulty at all in relating motivic multiple zeta values and multiple zeta values; the hard part is the definition of motivic multiple zeta values themselves, and more precisely the definition of the ideal $J \subset I$ of motivic relations (this is exactly where Tannakian categories, Betti and de Rham realization functors, etc. come into play). Besides Brown's articles on the subject, which you seem to be well aware of, there is also the excellent book Multiple zeta values: From numbers to motives, by Burgos and Fresán. -Finally, it should be said that one conjectures that $J=I$, i.e. that every relation between multiple zeta values is motivic. In particular, the algebras of motivic and ordinary multiple zeta values would be isomorphic, $\mathcal{Z}^{\mathfrak{m}} \cong \mathcal{Z}$. In order to appreciate the extraordinary depth of this conjecture, note that the odd motivic zeta values $\zeta^{\mathfrak{m}}(2k+1)$ are all known to be transcendental, while the analogous statement is not known for a single odd zeta value $\zeta(2k+1)$. Even worse, the only odd zeta value we know is irrational is $\zeta(3)$, thanks to Apéry.<|endoftext|> -TITLE: Proof of the conjecture of Lehmer: a Dobrowolski type minoration -QUESTION [20 upvotes]: A few weeks ago Jean-Louis Verger-Gaugry announced a proof of Lehmer's conjecture, see https://arxiv.org/pdf/1709.03771.pdf. The key result (Theorem 5.28, p. 122) is a Dobrowolski type minoration of the Mahler Measure $M(\beta)$, namely -\begin{align}\label{eq:1} M(\beta) \geq \Lambda_r\mu_r-\frac{\Lambda_r\mu_r \arcsin(\kappa/2)}{\pi}\frac{1}{\log(n)},\end{align} -where $\Lambda_r\mu_r = 1.15411\ldots$ and $\kappa=0.171573\ldots$ are some constants and $n=\mathrm{dyg}(\beta)$ is some function of $\beta$ assumed to be at least 260. -This result should follow directly from the asymptotic expansion in Theorem 5.27 given by -$$\log M_r(\beta) = \log \Lambda_r \mu_r + \frac{\mathcal{R}}{\log(n)} + O\left(\left(\frac{\log\log n}{\log n}\right)^2\right),$$ -where $M(\beta)\geq M_r(\beta)$ and $\mathcal{R}$ depends on $\beta$ and $n$ satisfying -$$|\mathcal{R}| < \frac{\arcsin(\kappa/2)}{\pi}.$$ -Indeed, if we would now that the error term is positive we obtain the desired lower bound for $M(\beta)$ by taking the exponential of $\log M_r(\beta)$ and using the series expansion of the exponential. However, it is not shown that this error is positive and I don't see how one can show this fact (if it indeed turns out to be true). So, my question is: -What happened with the error term occurring in the expansion of $\log M_r(\beta)$, but no longer occurring in the lower bound for $M(\beta)$? - -REPLY [7 votes]: There are other strange assertions in the paper. For instance p. 131. The author wants to prove (Theorem 7.3) that that a certain meromorphic function $P/f$ has no pole, where the holomorphic function $f$ has (at least) a simple zero at $\omega\in \mathbb C$, and $P$ is a polynomial. The proof is very odd:First of all, it is never proved that $U=P/f$ has no pole. Secondly, $U$ is treated... like it was already proved that it has... no pole. More precisely, the author considers $U$ as a formal series in $X$ (like $1/X= 1/z$ is a Laurent series). Then he derives $P=Uf$ and gets $P'=U'f +Uf'$. But now he writes "specializing the formal variable $X$ to the complex variable $z$" and gets $$P'(z) = U'(z)f(z) +U(z)f'(z),$$ forgetting that this formal hocus-pocus does not make $U(z)$ (and $U'(z)$) a defined complex number! He then makes $z=\omega$ and obtains $P'(\omega) = U(\omega) f'(\omega)$, forgetting that $U$ is still a formal series. In fact this argument holds for $P=1$ and $f=z$. It would then be proved that $U(z) =1/z$... is holomorphic at $0$! -Added: there are (at least) two other places where the same argument is used : p110 and p125. The 'fracturability' defined p. 4 is this kind of strange factorization of polynomials, like $1$ is factorized as $1=(1/z)z$, and $1/z$ becomes a holomorphic function...<|endoftext|> -TITLE: When is $A\otimes R$ a free $R$-module? -QUESTION [13 upvotes]: Let $R$ be a commutative ring. If I am not mistaken, there is the following fact: - -For a finitely generated abelian group $A$, the $R$-module $A\otimes R$ is free if and only if we can write the torsion part of $A$ as a direct sum of cyclic groups of the form $\mathbb{Z}/k$, where $k$ is invertible or zero in $R$ - -While elementary, I found this surprisingly tricky to prove and my proof takes about half a page. But I assume, this fact should be known. As I need it for a paper of mine, I want to ask whether someone knows a reference for it. In absence of a reference, I would also be happy with a slick 5-line proof! - -REPLY [19 votes]: Here is a 7-line proof of your statement. - - -Claim. Let $R$ be a commutative ring with identity $1_R$. Let $A \simeq \mathbb{Z}/d_1\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/d_k\mathbb{Z}$ be a finitely generated Abelian group given with its invariant factor decomposition, i.e., $d_i \ge 0, d_i \neq 1$ and $d_i$ divides $d_{i + 1}$ for every $1 \le i \le k - 1$ (note that $d_i = 0$ is allowed). -Then $A \otimes_{\mathbb{Z}} R$ is free over $R$ if and only if either $d_i 1_R$ is zero or a unit of $R$ for every $i$. -Proof (7 lines). Let $M \Doteq A \otimes_{\mathbb{Z}} R \simeq -R/d_1 R \oplus \cdots \oplus R/d_k R$. If $d_i 1_R$ is zero or a unit of $R$ for every $i$, then $M$ is clearly free over $R$. Assume now that $M$ is free over $R$. We reason by contradiction and consider the smallest $j \ge 1$ such that $d_j 1_R$ is neither zero nor a unit. Then we have $M \simeq R/d_j R \oplus \cdots \oplus R/d_k R$. Considering $M/\mathfrak{m}M$ for a maximal ideal $\mathfrak{m}$ which contains $d_jR$, we see that the minimal number of generators of $M$ over $R$ is $s \Doteq k -j + 1$. Therefore $M \simeq R^s$ and consequently $d_jM$ cannot be generated by less than $s$ elements. As $d_jM \simeq -(d_jR/d_{j + 1} R) \oplus \cdots \oplus (d_jR/d_k R)$, we get a contradiction. - - -You may turn the above proof into a one-liner if you refer to this more general result of Irving Kaplansky [1, Theorem 9.3]. - - -Kaplansky's Theorem. - Let $R$ be a ring in which every one-sided ideal is two-sided. In particular, $R$ may be any commutative ring. Suppose the $R$-module $M$ is isomorphic to the direct sum of cyclic modules $R/S_1, \dots, R/S_m$, and also to the direct sum of $R/T_1, \dots, R/T_n$, where $S_i,\, T_¡$ are ideals each containing its successor, $S_1, T_1\neq R$. Then $m = n$ and $S_i = T_i$ for all $i$. - - - -Addendum. I found the discussion of Qiaochu Yuan and YCor about projective modules (see comments above) particularly enlightening. This is why I would like to assemble all its fragments here. - - -Qiaochu Yuan's Claim on projectives. Let $R$ be a commutative ring with identity and let $A$ be a direct sum of cyclic groups $\mathbb{Z}/k_i\mathbb{Z}$ where $k_i \in \mathbb{Z}$ and $i$ ranges in some arbitrary set. - Then $A \otimes_{\mathbb{Z}} R$ is a projective module over $R$ if and only if $k_iR = k_i^2R$ for every $i$. -Proof. Let $M \Doteq A \otimes_{\mathbb{Z}} R \simeq -\bigoplus_i R/k_i R$. Then $M$ is projective over $R$ if and only if $R/k_iR$ is projective for every $k$. This holds if and only if the natural map $R \twoheadrightarrow R/k_iR$ splits for every $i$, which is in turn equivalent to the fact that $k_iR$ is an idempotent ideal. As observed by YCor, $k_iR$ is generated by an idempotent in this case, see e.g, [2, Exercise 2.1], a classical application of Nakayama's lemma. - - -If $\mathbb{Z}1_R \simeq \mathbb{Z}/n\mathbb{Z}$ for some $n > 1$, then -the condition $\text{gdc}(k, n) = \text{gdc}(k^2, n)$, or equivalently $\text{gcd}(k, \frac{n}{\text{gcd}(n, k)}) = 1$, implies $kR = k^2R$. This is certainly a necessary condition when $R = \mathbb{Z}/n\mathbb{Z}$. -YCor outlined the fact that every module over $\mathbb{Z}/6\mathbb{Z}$ is projective. He further extends this remark by mentioning below that every module over a commutative ring $R$ with identity is projective if and only if $R$ is the direct product of finitely many fields, see this MO post for references and a more general statement. - - -Remark on projective modules over $\mathbb{Z}/n\mathbb{Z}$. Let $n$ be a positive integer. Then the following are equivalent: - - -The integer $n$ has no square factor. -Every module over $\mathbb{Z}/n\mathbb{Z}$ is projective. - - -Proof. By the first Prüfer Theorem, a module over $R = \mathbb{Z}/n\mathbb{Z}$ is a direct sum of cyclic factors $\mathbb{Z}/k\mathbb{Z}$ where $k$ divides $n$. If $n$ is square-free, any such factor is projective over $R$ by Qiaochu Yuan's Claim. Hence any module over $R$ is projective. If $n$ has a square factor $k^2$ with $k > 1$, then $R/kR$ is not projective over $R$ by Qiaochu Yuan's Claim. - - - -[1] I. Kaplansky, "Elementary divisors and modules", 1949. -[2] H. Matsumura, "Commutative Ring Theory", 1986.<|endoftext|> -TITLE: Minkowski's theorem for non-0-symmetric sets -QUESTION [7 upvotes]: Let $\Lambda \subseteq \mathbb{R}^n$ be a full-rank lattice, i.e. $\Lambda = A \mathbb{Z}^n$ for some $A \in \mathrm{GL}_n (\mathbb{R})$, and let $C \subseteq \mathbb{R}^n$ be a $0$-symmetric convex body. Then Minkowski's theorem asserts that -$$ -\# | \Lambda \cap C | \geq \frac{|C|}{2^n | \mathbb{R}^n / \Lambda|}. -$$ -I was curious if there is a version of Minkowski's theorem that holds for convex bodies not necessarily symmetric around $0$. Clearly, the body should contain $0$, otherwise we might have $\# | \Lambda \cap C | = 0$. Are there other conditions on $C$ than symmetry around $0$ that still allow for a lower bound on the number of lattice points in $C$? -Any comment or reference is highly appreciated. - -REPLY [3 votes]: I am not sure whether this will necessarily be of interest to you, but Athreya and Margulis jointly proved a probabilistic version of the Minkowski Theorem; here is the arXiv version: https://arxiv.org/pdf/0811.2806.pdf . The random Minkowski theorem is Theorem 2.2 on the third page. Str\"ombergsson then showed that the bound that they obtain is sharp: see https://arxiv.org/pdf/1008.3805.pdf . In a recent preprint, I then managed to generalize the random Minkowski theorem of Athreya-Margulis to higher "probabilistic successive minima" (so to speak); see https://arxiv.org/pdf/1909.05205.pdf . (I should probably change "Lebesgue measurable" to "Borel measurable," though.)<|endoftext|> -TITLE: What is the essential image of $AbVar$ in $p-div$? -QUESTION [9 upvotes]: Given an abelian variety $A$ over a base scheme $\text{Spec } \mathcal{O}_{K_p}$, we define the functor $P$ as taking $A \mapsto \text{colim}_n A[p^n]$, its associated $p$-divisible group. What is the essential image of $P$? -Further, take the functor $\circ$, which takes a $p$-divisible group $B[p^\infty]$ to its associated formal group $B[p]^\circ$ (the connected component of the $p$-divisible group, under Tate's categorical equivalence of connected $p$-divisible groups and formal Lie groups). What is the essential image of $\circ$? -In other words, how much of $p-Div$ and $FG$ can we touch via the study of abelian varieties? - -REPLY [2 votes]: In short, I still don't know over a general ring, but for $\mathbb{F}_q$, we have an answer: The idempotent completion of the essential image of $AbVar_{/\mathbb{F}_q}$ in $FG_{/\mathbb{F}_q}$ is categorically equivalent to $FG_{/\mathbb{F}_q}$. -In Manin's paper, The Theory of Formal Commutative Groups over Fields of Finite Characteristic, he calls formal groups "algebroid" if they are the completion of algebraic varieties over any field $k$. He proves that every formal group over $\mathbb{F}_q$ is isogenous to a subgroup of an algebroid formal group over $\mathbb{F}_q$. The proof outline is: - -Every formal group is isomorphic to $\oplus_i G_{n_i,m_i}$ up to isogeny. -Every such $G_{n,m}$ (pick any coprime m, n) can be realized in the formal group decomposition associated to an abelian variety. - -Notation: $G_{n,m}$ is the formal group whose Dieudonné $W(\mathbb{F}_p)$-module is isomorphic to $W(\mathbb{F}_p)/(F^n - V^m)$. -For step 2, he uses the Jacobian variety $J_a$ associated to the family of curves $y^p - y = x^{p^a-1}$ over $\mathbb{F}_{p^a}$, where $a = 1, 2, 3, ...$. The decomposition of the formal group $\hat{J}_a$ contains all $G_{n, m}$ such that $m+n \mid a(p-1)$. So, for any pair of coprime $(n, m)$ and any $p$, $\hat{J}_{n + m}$ contains a piece isogenous to $G_{n,m}$. -(There is the subtlety that we are working in $FG$ up to isogeny rather than isomorphism here, so this answer only tells us about the category $FG$ up to isogeny.) -A note on the difference in base field from my original question: I asked about $\mathcal{O}_{K_p}$ where $K_p$ is a finite extenstion of $\mathbb{Q}_p$, whereas Manin considers $\mathbb{F}_q$. It seems that since the varieties $A$ are defined over finite extensions of $\mathbb{Q}_q$, if they have good reduction we can extend them (via the Néron model?) to $\mathbb{Z}_q$, and take their special fiber to get to varieties defined over $\mathbb{F}_q$.<|endoftext|> -TITLE: Strict and non-strict orderings -QUESTION [7 upvotes]: Consider a set $A$ equipped with two binary relations $\le$ and $<$, related in the appropriate ways for the strict and non-strict version of an ordering. One might make different choices about exactly what axioms to impose, but some reasonable ones to choose from are: - -$\le$ is a preorder (or maybe a partial order). -$<$ is irreflexive. -$<$ is transitive. -$x\le y$ and $y -TITLE: Most natural connection on Lie group: comparison of different pictures -QUESTION [5 upvotes]: Let $G$ be a Lie group (not necessarily compact). One can equip $G$ with the left invariant metric (or -right invariant but in general there is no biinvariant metric in the noncompact case). Once the metric on -$G$ is chosen we can consider the Levi Civita connection for this metric i.e. the unique metric connection -which has no torsion. If our metric was chosen to be (left)invariant we arrive at the connection which -is in some sense compatible with the group structure. However there are several notions of compatibility -with the group structure: -1. If $X,Y$ are left invariant vector fields then we can require $\nabla_X Y$ to be left invariant as well. -2. We may require that the geodesic defined by our connection coincide with the one parameter subgroups. -3. Finally we may require that the parallel transport defined by our connection coincide with the -(differential of) group translation. -Of course we have also the previous mentioned -4. Condition of being metric connection, with respect to some chosen (say) left invariant metric - -I would like to understand what are the relations between conditions 1,2,3 and 4? - -REPLY [2 votes]: From the formula for the Levi-Civita connection you can check that the LC connection for a left invariant metric will satisfy 1, but will not satisfy 2 unless the metric is bi-invariant. -For 3, you have to specify what parallel translation is taken along. If it is along geodesics then both the geodesics and parallel translation are given by the group action on a symmetric space.<|endoftext|> -TITLE: Classifying functions up to suitable pre-composition and/or post-composition -QUESTION [5 upvotes]: What's a name for a general technique I've seen used many times? -Given any family $\mathcal{F}$ of functions such that $f:X\to Y$ for all $f\in \mathcal{F}$ when one wishes to study in general for an arbitrary $y\in Y$ what (if any) $x\in X$ satisfy $f(x)=y$ it seems often the approach taken is to first find some permutation group $P\subseteq \text{Sym}(Y)$ such that: -$$\forall \sigma\in P\left(f\in \mathcal{F}\implies f\circ \sigma\in \mathcal{F}\right)$$ -Then from here viewing this as a right group action of $P$ on $\mathcal{F}$ we form an equivalence relation $\sim$ over $\mathcal{F}$ with each equivalence class an orbit e.g. $\text{Orb}(f)=\{f\circ \sigma\in \mathcal{F}:\sigma\in P\}$ so that we get: -$$f\sim g\iff \text{Orb}(f)=\text{Orb}(g)\iff \exists \sigma\in P:f=g\circ \sigma$$ -At this point for every $f\in \mathcal{F}$ given the equation $f(x)=y$ we find a unique "nice" $C_{f}\in \text{Orb}(f)$ (nice in its easier to study what $x\in X$ satisfy $C_f(x)=y$) and let this be a canonical form for our equivalence class $\text{Orb}(f)=[f]_{\sim}$ then our original problem reduces to just studying the family of functions $\mathcal{G}\subseteq \mathcal{F}$ where $\mathcal{G}=\{C_f\in \mathcal{F}:f\in \mathcal{F}\}$. To make this idea more explict here are three examples where I've seen this sort of idea used: - -A simple first example is the notion of "reduced echelon form" in introductory linear algebra courses where here $\mathcal{F}=\mathcal{M}_{m\times n}(\mathbb{F})$ would be the set of all matrices over some field $\mathbb{F}$ and $P=\text{GL}_n(\mathbb{F})$ would be a general linear group over $\mathbb{F}$. https://en.wikipedia.org/wiki/Row_echelon_form - -A second example is many techniques used when studying univariate polynomials over again just to make things simple lets say those polynomials with coefficients in $\mathbb{F}$ so we get $\mathcal{F}\subseteq\mathbb{F}[X]$. From here $P$ would take the form of scaling/shifting for example reducing each polynomial to a monic one, or shifting along its input to eliminate a monomial in the resulting composition. E.g. if we have: -$$f(x)=x^n+ax^{n-1}+bx^{n-2}+\cdots+c$$ -Then if $\sigma:x\to x-\frac{a}{n}$ expanding $(f\circ \sigma)(x)$ the coefficient on the monomial $x^{n-1}$ vanishes. Where in particular if $1 -TITLE: How is Borger's approach to $\mathbb{F_{1}}$ related to previous approaches (e.g. Deitmar's)? -QUESTION [24 upvotes]: The "traditional" approach to the so-called "field with one element" $\mathbb{F}_{1}$ is by using monoids, or, to put it in another way, by forgetting the additive structure of rings. In Deitmar's approach to the subject, $\mathbb{F}_{1}$ is declared to be the trivial monoid $\{1\}$. Furthermore, $\mathbb{F}_{1}$-modules (or vector spaces, since we are calling $\mathbb{F}_{1}$ a "field" after all, although it's not really a field) are pointed sets, and $\mathbb{F}_{1}$-algebras are commutative monoids. The motivations for these ideas may predate Deitmar's work, and these ideas may also be found in this unpublished work by Kapranov and Smirnov. -Deitmar's approach is kind of a "template" for later approaches to the field with one element, such as the one by Toen and Vaquie. However, one approach that is different is the one by Borger. In this approach, $\mathbb{F}_{1}$-algebras are lambda rings, which are rings together with Frobenius lifts. The idea is that the extra structure provided by the Frobenius lifts serves as descent data to $\mathbb{F}_{1}$, while the forgetful functor that forgets that extra structure is the base change to $\mathbb{Z}$, i.e. it is the functor $-\otimes_{\mathbb{F}_{1}}\mathbb{Z}$. -In other words, the $\mathbb{F}_{1}$-algebras have less structure than rings ($\mathbb{Z}$-algebras) in Deitmar's approach, while they have more structure in Borger's approach. But it is often said that the two are related; in fact the nLab article on the field with one element claims that Borger's approach subsumes many aspects of previous approaches. For instance, in Lieven Le Bruyn's paper, Absolute Geometry and the Habiro Topology, it is shown that in Borger's approach one can arrive at the idea of Soule (also apparently going back to Kapranov and Smirnov) that -$\mathbb{F}_{1^{n}}\otimes_{\mathbb{F}_{1}}\mathbb{Z}=\mathbb{Z}[\mu_{n}]$. -In what other ways is Borger's approach related to these other approaches? With ideas similar to lambda rings apparently finding their way into algebraic topology and homotopy theory (e.g. Adams Operations in Cohomotopy by Guillot and Power Operations and Absolute Geometry by Morava and Santhanam), do we have an analogue of Toen and Vaquie's "schemes over $\mathbb{S}$" and related homotopy-theoretic constructions? What about Connes and Consani's "arithmetic site?" Can we obtain analogous constructions but with lambda rings instead of commutative monoids playing the role of $\mathbb{F}_{1}$-algebras? - -REPLY [23 votes]: Given a monoid $M$, the ring $\mathbb{Z}.M$ of $\mathbb{Z}$-linear combinations of elements of $M$ has a natural $\lambda$-ring structure given by setting the elements $m \in M$ to have rank $1$ (i.e. $\lambda^k(m)=0$ for all $k>1$). We then have functors -$$ -\text{Monoids} \xrightarrow{\mathbb{Z}.-} \lambda\text{-rings} \xrightarrow{\text{forget}} \text{Rings} -$$ -(monoid and ring structures assumed commutative) with right adjoints -$$ -\text{Monoids} \xleftarrow{\text{Rank } 1 \text{ elements}} \lambda\text{-rings} \xleftarrow{W} \text{Rings}, -$$ -where $W$ denotes the big Witt vectors. -Descent data from rings $A$ to monoids take the form of ring homomorphisms $A \to \mathbb{Z}.A$ for which the two compositions $A \to \mathbb{Z}.A \to \mathbb{Z}.(\mathbb{Z}.A)$ agree. On the other hand, a $\lambda$-ring structure is a ring homomorphism $A \to WA$ for which the two compositions $A \to WA \to W(WA)$ agree. As for instance in Cuntz-Deninger arXiv:1410.5249, WA is often a natural completion of $\mathbb{Z}.A$. -If we think of $\mathbb{F}_1$-algebras as monoids, then this suggests we should think of $\lambda$-rings as rings with formal descent data to $\mathbb{F}_1$, or as affine schemes over an analogue $\mathrm{dR}(\mathbb{Z}/\mathbb{F}_1)$ of Carlos Simpson's de Rham stack. -$\lambda$-rings don't easily fit into the Toën-Vaquié setting, since they don't come from an obvious monoidal category. However, to build schemes (resp. DM stacks, resp. Artin stacks) from affine objects, you only really need a good notion of open immersions (resp. étale maps, resp. smooth maps). Things probably work satisfactorily if you just require the underlying morphism of affine schemes to be an open immersion. There are good notions of modules $M$ over $\lambda$-rings $A$ (just put operations $\lambda^k$ on $M$ such that $A \oplus M$ becomes a $\lambda$-ring, which amounts to saying something like $\lambda^k$ is $\Psi^k$-semilinear), giving you a theory of quasi-coherent sheaves as well.<|endoftext|> -TITLE: Is a Morita equivalent functor an exact functor(Module protective direct sum) ? -QUESTION [5 upvotes]: We say that two finite dimensional algebras $A$ and -$B$ are stably equivalent if there is an equivalence $F:\underline{mod} A\longrightarrow \underline{mod} B$ -between the associated module categories modulo projective modules, where $mod A$ and $mod B$ are finitely generated modules categories over $A$ and $B$, respectively. -Question: -For any exact sequence $$0 \longrightarrow X_{1}\longrightarrow X_{2}\longrightarrow X_{3}\longrightarrow 0$$ -in $mod A$, can we obtain an exact sequence -$$0 \longrightarrow F(X_{1})\oplus P_{1}\longrightarrow F(X_{2})\oplus P_{2}\longrightarrow F(X_{3})\oplus P_{3}\longrightarrow 0$$ in $mod B$, -where $P_{i}$ is projective module in $mod B$ for $i=1,2,3$ -? - -REPLY [7 votes]: Let $A=kQ$ be the path algebra of Dynkin type $\mathcal{A_2}$ and $B=K[x]/(x^2)$. Both algebras have a unique simple non-projective module and all other indecomposable modules are projective. Thus their stable module categories consists of just one indecomposable objects and are isomorphic. -Let $S$ be the simple non-projective $A$-module and M be the simple $B$-module. -Then $S$ has projective dimension one and there is a short exact sequence: -$0 \rightarrow P_1 \rightarrow P_0 \rightarrow S \rightarrow 0$. -This sequence is necessarily mapped to a sequence of the form: -$0 \rightarrow Q_1 \rightarrow Q_0 \rightarrow M \oplus Q_2 \rightarrow 0$, with $Q_1$, $Q_2$ and $Q_0$ projective. But this sequence can not be exact or else $M$ would have finite projective dimension. But the algebra $B$ is selfinjective and not semi-simple and thus every non-projetive $B$-module has infinite projective dimension.<|endoftext|> -TITLE: A duality result for Coxeter groups -QUESTION [9 upvotes]: Short version: if $G$ is a Coxeter group and $H \subset G$ is a parabolic subgroup, both acting on a space $V$, is it true that the invariant-coinvariant algebra $(S(V)_G)^H$ has a natural bilinear form induced by taking coefficients in the top component? -Now the long, detailed version. Let $G$ be a Coxeter group acting on a real vector space $V$ with a fixed generating set of reflections $S$. We can form the symmetric algebra $S(V)$, the invariant subalgebra $S(V)^G \subset S(V)$, and the ideal $I_G \subset S(V)$ generated by elements of $S(V)^G$ of positive degree. The coinvariant algebra is defined as $S(V)_G = S(V)/I_G$. By the Cehvalley-Shephard-Todd theorem it is isomorphic to the regular representation of $G$ as a $G$-module. If $H \subset G$ is a parabolic subgroup then the $H$-invariants $(S(V)_G)^H$ form a subalgebra of $S(V)_G$, which is finite dimensional and whose top nonzero component is $1$-dimensional over $\mathbb R$, say spanned by $s$. Then $(S(V)_G)^H$ has a nice bilinear pairing given by taking $(f,g)$ to be the coefficient of $s$ in the product $fg$ for all $f, g \in (S(V)_G)^H$. -If $G$ is a Weyl group, the algebra $(S(V)_G)^H$ is isomorphic to the cohomology ring of a generalized flag variety by a famous theorem of Borel. Plus the bilinear form described above is the pullback of the Poincaré pairing in said cohomology, and in particular it is nondegenerate. Questions: - -Can the non-degeneracy of the bilinear form be proved by purely combinatorial methods, without appealing to Borel's theorem? -Can it be extended to all pairs $(G,H)$ where $G$ is just a finite Coxeter group? - -REPLY [10 votes]: Yes. By Chevalley-Shepard-Todd, $S(V)^G$ and $S(V)^H$ are polynomial rings. Let $S(V)^G=\mathbb{R}[g_1, \ldots, g_n]$ and $S(V)^H = \mathbb{R}[h_1,\ldots, h_n]$ where the $g_i$ and $h_i$ are homogenous. Then -$$S(V)_G^H = \mathbb{R}[h_1,\ldots,h_n]/\langle g_1,\ldots, g_n \rangle.$$ -Here the denominator is the ideal of $\mathbb{R}[h_1,\ldots,h_n]$ generated by the $g_i$, and $g_i \in \mathbb{R}[h_1,\ldots,h_n]$ because $S(V)^G \subseteq S(V)^H$. -So $S(V)^H_G$ is a complete intersection, and therefore Gorenstein. Saying $R$ is a finite dimensional Gorenstein $k$ algebra (for $k$ a field) exactly means that there is a linear functional $\int: R \to k$ such that $\langle f,g \rangle = \int fg$ is a perfect pairing. If $R$ is graded, then this functional is "take the top degree piece". See Eisenbud, Commutative Algebra with a view toward Algebraic Geometry Chapter 21.2 for a good overview of finite dimensional Gorenstein algebras.<|endoftext|> -TITLE: $K_0$-equivalence of varieties -QUESTION [10 upvotes]: Let $k$ be an algebraically closed field of characteristic zero. -Let $A$ be the $\mathbf{Z}$-subalgebra of the Grothendieck ring of $k$-varieties $K_0(\text{Var}_k)$ generated by classes of semi-abelian varieties. -Question 1. What can we say about the ring homomorphism $A\to K_0(\text{Var}_k)$? -It is obviously very far from being surjective. Mainly, is it integral/finite/flat/essentially of finite type? -Question 2. The class of smooth projective varieties over $k$ generates $K_0(\text{Var}_k)$ in the obvious sense. -Is there a more manageable generating class of $k$-varieties, possibly containing semi-abelian varieties at the very least? -Question 3. In light of Question 2, does the class $\mathcal{P}$ of varieties over $k$, constructed below, have any hope to be a generating class for $K_0(\text{Var}_k)$, and if not, upon calling $B$ the $\mathbf{Z}$-subalgebra of $K_0(\text{Var}_k)$ it generates, same question as in Question 1, for the ring homomorphism $B\to K_0(\text{Var}_k)$. -We call $\mathcal{P}_0$ the class of $k$-varieties whose objects are: - -abelian varieties -semi-abelian varieties (extension of abelian varieties by a torus) -smooth hypersurfaces in large projective spaces over $k$ -smooth projective varieties of dimension $\le 3$ -smooth projective toric varieties -Calabi-Yau varieties -products of the above - -Let $\mathcal{P}_1$ be the class of $k$-varieties whose objects are obtained by blowing up an element of $\mathcal{P}_0$ along a closed $k$-subvariety that is again an element of $\mathcal{P}_0$. -Finally, let $\mathcal{P}$ be the class of all those $k$-varieties such that their class in the Grothendieck ring of varieties $K_0(\text{Var}_k)$ belongs to the subring $B\subset K_0(\text{Var}_k)$ generated by the classes of the elements of $\mathcal{P}_1$. -Question 4. Does there at least exist a morphism $f: Y\to X$ over $k$, with $Y$ in $\mathcal{P}$, and such that $f$ is surjective/flat/such that $f^*\omega_X\simeq\omega_Y$? -(This last question is expected to have negative answer for $f$ required to be surjective, and a good strategy to see this is suggested in the comments.) - -REPLY [6 votes]: (Expanding my comments into an answer for more visibility.) -By Larsen-Lunts $K_0(\operatorname{Var}_k)/[\mathbb{A}^1]$ is the free abelian group on stable birational equivalence classes. It thus suffices to find a variety $X$ that is not stably birational to any variety in $\mathcal{P}_0$. -Take $X$ to be a smooth general type hypersurface in $\operatorname{Gr}(2,n),$ $n$ large. If two varieties are stably birational, then the bases of their MRC (maximally rationally connected) fibrations must also be birational. Now note that the operation of taking base of MRC fibration preserves the property of having a birational representative in $\mathcal{P}_0$. As $X$ is general type and thus its own MRC base, this reduces the problem to showing $X$ is not birational to any variety in $\mathcal{P}$. -Next we reduce birational equivalence to actual isomorphism. The operation of taking canonical model preserves the class of general type varieties in $\mathcal{P}_0$, and $X$ is its own canonical model. Thus we just need to rule out $X$ actually lying in $\mathcal{P}_0$. -As $\operatorname{Pic}X\cong\mathbb{Z}$ (by Lefschetz), $X$ can't be a product of two varieties. It remains to show $X$ is not a hypersurface in a projective space. Another application of Lefschetz shows that $H^4(X)\cong\mathbb{Z}\oplus\mathbb{Z},$ while a (high dimensional) hypersurface has $H^4\cong\mathbb{Z}$, giving us the final contradiction. QED. -Note that in the definition of $\mathcal{P}_1$, one can actually allow all (smooth) blow-ups! This is because blow-ups only change the class in $K_0(\operatorname{Var}_k)$ by a multiple of $[\mathbb{A}^1]$. This demonstrates that the Grothendieck ring of varieties is much more rigid than true motives - it really remembers the birational geometry of your variety. -Also, here is a sketch of an argument that for a very general high degree hypersurface $X$ in $\mathbb{P}^n\times\mathbb{P}^n$ so that it admits no dominant rational morphisms from varieties in $\mathcal{P}.$ The above arguments reduce this to showing $X$ admits no dominant rational morphisms from varieties in $\mathcal{P}_0$. My understanding is that Schoen proves in "Varieties dominated by product varieties" that a very general sufficiently ample hypersurface in any variety does not admit a dominant map from a product of varieties of smaller dimension, so it suffices to show that $X$ cannot be mapped to rationally by a smooth hypersurface in projective space. -Take $X$ to contain no rational curves. (To see this is possible, embed $\mathbb{P}^n\times\mathbb{P}^n$ in a projective space and intersect with a very general high degree hypersurface. It is known that this hypersurface will not contain any rational curves; I believe this statement was first proven by Clemens, in "Curves on generic hypersurfaces".) Under this assumption, any rational morphism to $X$ must be an actual morphism (see Kollar-Mori, Corollary 1.3.) -Now $X$ has an effective divisor class whose square is zero. But this stops us from being able to map a hypersurface in projective space dominantly to $X$, as the pullback of this divisor must also be effective of square zero, impossible on a high dimensional hypersurface (again, by Lefschetz).<|endoftext|> -TITLE: Proof of Witt's result about quaternion extensions -QUESTION [8 upvotes]: I'm searching for a proof of Witt's result that a biquadratic extension $K(\sqrt{a},\sqrt{b})/K$ extends to a Galois extension $L/K$ with quaternion group $Q_8$ iff the quadratic forms $, <1,1,1>$ are equivalent iff $(a,b)(a,a)(b,b) = 0 \in Br(K)$. -I know there is a proof in his original paper "Konstruktion von galoisschen Körpern der Charakteristik p zu vorgegebener Gruppe der Ordnung pf". However, I am searching for a proof in English. I would also like to find proofs from a more modern perspective as well as generalizations of the result. -Thank you. - -REPLY [7 votes]: A complete proof can be found in the first few pages of -https://mathscinet.ams.org/mathscinet-getitem?mr=977759 -Jensen, Christian U.(DK-CPNH); Yui, Noriko(3-TRNT) -Quaternion extensions. Algebraic geometry and commutative algebra, Vol. I, 155–182, Kinokuniya, Tokyo, 1988. -I will scan the relevant pages and post them here as soon as electricity is restored in my office. -Addendum: Here is the paper : -https://drive.google.com/open?id=0B8EHtI8F9qdIODNpZTQ1c1JjQTlIUUdsWm1nZmxRWVNTM1Bj<|endoftext|> -TITLE: Birational automorphisms of varieties of Picard number one -QUESTION [14 upvotes]: Let $X$ be a smooth projective variety of Picard number one, and let $f:X\dashrightarrow X$ be a birational automorphism which is not an automorphism. -Must $f$ necessarily contract a divisor? - -REPLY [3 votes]: In contrast to an earlier answer on this question, I claim the answer here is yes. -In the accepted answer to this question: -Pseudo-automorphisms on Fano varieties -abx explains that for any smooth variety $X$ of Picard number 1, any pseudo-automorphism of $X$ (i.e. a birational automorphism which is an isomorphism in codimension 1) must in fact be an automorphism. -So it remains to argue that if $f$ does not contract any divisor, then it is a pseudo-automorphism. The only thing to check is that $f^{-1}$ does not contract a divisor either. You can do this by looking at a resolution of $f$ : -$$ X \leftarrow^p \widetilde{X} \rightarrow^q X $$ -where $\widetilde{X}$ is smooth: the numbers of $p$-exceptional prime divisors and $q$-exceptional prime divisors must be equal, but the hypothesis that $f$ doesn't contract a divisor says that every $q$-exceptional divisor is $p$-exceptional. Hence the two sets are the same, and so $f^{-1}$ doesn't contract a divisor either.<|endoftext|> -TITLE: Constructive homological algebra in HoTT -QUESTION [14 upvotes]: I'm curious how much of homological algebra carries over to a constructive setting, like say HoTT (or some other variety of intensional type theory) without AC or excluded middle. There doesn't seem to be a lot of literature on this topic (or at least it's difficult for an outsider like me to find). -The category of abelian groups is fortunately still abelian in HoTT since we have set-quotients, so epis and monos are all normal. So I assume that a good amount of the general theory of homological algebra in abelian categories still applies (such as the snake lemma, 5 lemma, etc.) -But unfortunately it seems like the classical approach to derived functors suffers a setback because we cannot guarantee that we have enough projectives. I'm not sure if the existence of enough injectives is in jeopardy though, since abelian sheaves have enough injectives, but the comments in this related question suggest that indeed the existence of enough injectives is questionable. -I would guess that the derived category and Kan extension definition of derived functors (as given in Emily Riehl's homotopy theory textbook, for example) is better behaved in constructive mathematics, and it looks like Mike Shulman advocated this approach here. But from my naive point of view there's still the issue of showing existence of the Kan extension via functorial deformations of homotopical categories, and I'm also not sure if the usual localization procedure for constructing the derived category is complicated by the lack of AC. -Or perhaps something like $\infty$-topos theory in the style of Lurie, Riehl, Verity etc. (of which I know almost nothing) is the best approach to constructive homological algebra in HoTT, since the logic natively supports higher groupoids. Has any work been done on this since the advent of HoTT? -I'm pretty new to constructive mathematics (still making my way piece-wise through the HoTT volume), and I just have a "working mathematician's" knowledge of homological algebra, but I'm excited about the future of the univalent foundations project. - -REPLY [18 votes]: As regards HoTT, my own current opinion is that the best way to do "homological algebra" therein is by working directly with spectra. -With only a working mathematician's knowledge of homological algebra you may not know what a spectrum is. If you know the Dold-Kan theorem that a nonnegatively-graded chain complex is equivalent to a simplicial abelian group, hence in particular yields a simplicial set, i.e. a "space", then you can think of a spectrum as a space-like thing that corresponds similarly to a chain complex graded on the integers. -The definition of spectrum in HoTT is easy: it's a sequence of pointed types $E:\mathbb{N}\to Type_\ast$ together with equivalences $E_n \simeq \Omega E_{n+1}$. (If you're not familiar with spectra, the idea is that $E_n$ is like the part of a $\mathbb{Z}$-graded chain complex in degrees $\ge -n$, regraded to start from 0, with $\Omega$ corresponding to dropping the 0-chains and shift down by one.) See for instance this blog post and this paper. Similarly, you can define maps of spectra, homotopies between them, homotopy classes of maps between them, smash products of spectra, and so on. You can also define spectrification of a prespectrum, suspension prespectra which spectrify to suspension spectra, and the Eilenberg-MacLane spectrum $H M$ associated to an abelian group $M$. -Now you can define various homological notions by working with Eilenberg-MacLane spectra and then taking homotopy groups. For instance, $\mathrm{Ext}_S^n(M,N) = \pi_n(\mathrm{Hom}_{S}(H M, H N))$ (or maybe $\pi_{-n}$? I forget) and $\mathrm{Tor}^S_n(M,N) = \pi_n(H M \wedge_S H N)$. -These are not (even classically) the usual notions of Ext and Tor (over the integers), but rather "Ext and Tor over the sphere spectrum". This is roughly because the functor from simplicial abelian groups to simplicial sets forgets stuff. (It doesn't forget as much as you might think, though, because the higher homotopy groups are abelian.) To remember that stuff, we can work with modules over the ring spectrum $H\mathbb{Z}$, which classically coincide (as $(\infty,1)$-categories) with $\mathbb{Z}$-graded chain complexes. (We call the general version "over the sphere spectrum" because the sphere spectrum $S$ is the unit object of the monoidal structure on spectra, so that all spectra are $S$-modules in the same way that all abelian groups are $\mathbb{Z}$-modules.) Classically, this sort of Ext and Tor over $H\mathbb{Z}$ do coincide with the traditional versions defined using resolutions and chain complexes, so constructively I believe they should be the "correct" definitions. -Unfortunately, we don't know how to talk about ring spectra or modules over them in ordinary HoTT, because both require infinitely many coherence data. This could be solvable with two-level type theory, but I don't think anyone has pursued such a direction yet.<|endoftext|> -TITLE: Connected $T_2$-space with $\text{Cont}(X,X)$ not dense in $X^X$ -QUESTION [5 upvotes]: Disclaimer: Feel free to downvote or vote to close, if this is again trivial (I seem to have a bad day today; I promise that if this is again a bummer question, I will wait $\geq 1$ day before asking new questions). -For any space $(X,\tau)$, let $\text{Cont}(X,X)$ denote the set of continuous self-maps of $X$ and let $X^X$ denote the set of all self-maps of $X$, endowed with the product topology. -What is an example of a connected $T_2$-space $(X,\tau)$ such that $\text{Cont}(X,X)$ is not dense in $X^X$? - -REPLY [2 votes]: What you called strongly rigid spaces on this question are extreme examples where $C(X,X)$ is actually closed in $X^X$. Note that strongly rigid spaces are necessarily connected.<|endoftext|> -TITLE: Dealing with unwanted co-authorship requests -QUESTION [36 upvotes]: This post is a sequel to: Collaboration or acknowledgment? -The following has come to my attention. A senior mathematician (let us call him or her Alice) suggested a problem to a young mathematician (Bobby) who proceeded to solve it on her own and wrote up the result. Bobby agreed (*) to let Alice be listed as a coauthor, but Alice also insisted to include her PhD student (Charlotte) as a coauthor because they were thinking about the same problem, despite the fact that Alice and Charlotte did not even have partial results. -(*) Bobby had no problem with Alice joining her as a coauthor for the reasons mentioned by Igor Rivin below (I include you as a coauthor, you write me a good recommendation). Thus, the credit was unfairly diluted by including Charlotte who had not contributed. -Question: Is there a way for Bobby to manage such a situation without creating conflict? -Full disclosure: I am Bobby's PhD advisor. I can not interfere directly because Alice is a powerful person in the field known for aggressive backing of her PhD students and I do not want to inadvertently hurt Bobby's career. -Update: Following the advice of Joel David Hamkins, Bobby will be the sole author. There is nothing in the paper that Alice and Charlotte could point to as an idea they already had in mind. Looking back, what bothered me the most was not that Bobby's credit would be diluted but that someone who did not deserve it would be rewarded. -I will award bounty points to JDH for his uplifting Thanksgiving Day Answer, but, of course, any new comments are welcome. - -REPLY [15 votes]: Unwelcome requests should be denied. Period.<|endoftext|> -TITLE: When is the inverse image of the structure sheaf the structure sheaf? -QUESTION [14 upvotes]: For which morphisms of schemes $f : X\rightarrow Y$ do we have $f^{-1}\mathcal{O}_Y = \mathcal{O}_X$? (note that I don't mean $f^*$, I really just mean the basic inverse image sheaf as a sheaf of rings/abelian groups/sets) -This is obviously true if $f$ is an open immersion, and intuitively I feel like this is the only time when it is true. Is this correct? (If not, is it possible to completely characterize the morphisms $f$ for which this is true?) -I was led to think about this when I read here that for a morphism of stacks $f : \mathcal{X}\rightarrow\mathcal{Y}$ (say over $(\textbf{Sch}/S)_{et}$, there is a canonical identification $f^{-1}\mathcal{O}_\mathcal{Y} = \mathcal{O}_\mathcal{X}$, which I feel is somewhat strange. I suppose this is just a peculiarity of working with big sites? - -REPLY [11 votes]: Here is a trivial lemma. - -Lemma. Let $f \colon X \to Y$ be a morphism of schemes. Then the natural map $f^{-1}\mathcal O_Y \to \mathcal O_X$ is an isomorphism if and only if for each $x \in X$ the natural map $\mathcal O_{Y,f(x)} \to \mathcal O_{X,x}$ is an isomorphism. - -Proof. A morphism of sheaves on $X$ is an isomorphism if and only if it is so at the stalk of every $x \in X$. The stalk of a sheaf $\mathscr F$ at $\iota \colon \{x\} \to X$ is given by $\iota^{-1}\mathscr F$, so $(f^{-1}\mathcal O_Y)_x = \mathcal O_{Y,f(x)}$ since $\iota^{-1}f^{-1} = (f \circ \iota)^{-1}$. $\square$ -Under reasonable finiteness assumptions, this gives a local isomorphism as Simon Henry suggested: - -Lemma. Let $f \colon X \to Y$ be a morphism of schemes that is locally of finite presentation. Then $f^{-1}\mathcal O_Y = \mathcal O_X$ is and only if for every point $x \in X$, there exists an open neighbourhood $U \subseteq X$ of $x$ and an open neighbourhood $V \subseteq Y$ of $f(x)$ such that $f$ induces an isomorphism $U \stackrel\sim\to V$. - -Proof. Suppose $f^{-1}\mathcal O_Y = \mathcal O_X$, and let $x \in X$. By the lemma above, we get $\mathcal O_{Y,f(x)} \stackrel\sim\to \mathcal O_{X,x}$. Let $V \subseteq Y$ be an affine open neighbourhood of $f(x)$ and $U \subseteq f^{-1}(V)$ an affine open neighbourhood of $x$. If $V = \operatorname{Spec} A$ and $U = \operatorname{Spec} B$, then the map $g \colon A \to B$ is of finite presentation (Tag 01TQ). -Letting $\mathfrak p \subseteq A$ and $\mathfrak q \subseteq B$ be the primes corresponding to $f(x) \in V$ and $x \in U$ respectively, we conclude that $g^{-1}(\mathfrak q) = \mathfrak p$, and the map $A_\mathfrak p \to B_\mathfrak q$ is an isomorphism. By Tag 00QS, this implies that there exist $a \in A\setminus \mathfrak p$ and $b \in B\setminus \mathfrak q$ such that $A_a \cong B_b$. -Replacing $U$ by $D(b)$ and $V$ by $D(a)$ gives open neighbourhoods $U \subseteq X$ of $x$ and $V \subseteq Y$ of $f(x)$ such that $f$ induces an isomorphism $U \stackrel \sim \to V$. The converse is trivial. $\square$ -Examples include open immersions, disjoint unions, but also the map from a line with double origin to the line with a single origin. -However, if we drop the finite presentation assumption, we get all sorts of examples: -Example. Let $Y$ be any scheme, and consider the map -$$X = \coprod_{y \in Y} \operatorname{Spec} \mathcal O_{Y,y} \to Y.$$ -For a point $x \in X$ in the component $\operatorname{Spec} \mathcal O_{Y,y}$, we get an isomorphism $(\mathcal O_{Y,y})_x \to \mathcal O_{Y,f(x)}$, as can be seen easily by passing to an affine open neighbourhood of $y$ in $Y$ and using standard properties of localisation of rings.<|endoftext|> -TITLE: What are the compact Lagrangian submanifolds of a twisted cotangent bundle? -QUESTION [9 upvotes]: In Hamiltonian dynamics and symplectic geometry a twisted cotangent bundle is the cotangent space $T^*N$ of a closed (compact without boundary) $n$-manifold $N$ equipped with a twisted symplectic structure: $T^*N$ carries the canonical symplectic structure $\omega=d\lambda$, where $\lambda$ is the Liouville 1-form. One can "twist" $\omega$ by adding a closed two-form $\sigma$ on $N$ as follows: -$$ -\omega_{\sigma}:=\omega + \pi^*\sigma. -$$ -Here $\pi:T^*N \to N$ denotes the footpoint map. It is easy to check that $\omega_{\sigma}$ is symplectic. Twisted cotangent bundles play an important role in Hamiltonian dynamics, but I am here interested in their symplectic topology. Many classical questions in symplectic topology concern the closed Lagrangian submanifolds of $(T^*N,\omega)$. But what about closed Lagrangian submanifolds in $(T^*N,\omega_{\sigma})$? Does anyone know a non-trivial example (meaning $\sigma$ is not exact) where $(T^*N,\omega_{\sigma})$ contains closed Lagrangian submanifolds with "good properties" (say weakly exact, monotone etc.)? Are any general statements known? Any examples, ideas, references or proofs will be highly appreciated! -It is easy to find non-compact Lagrangians in $(T^*N, \omega_{\sigma})$: If $X\subset N$ is a submanifold such that $\sigma|_{X}=0$ then its conormal space -$$ -\nu^*(X):=\{ p\in T^*N\ |\ p|_{TX}\equiv 0 \}\subset T^*N -$$ -is a non-compact Lagrangian submanifold. My questions therefore concerns closed Lagrangian submanifolds! It is easy to find closed Lagrangians when $\sigma$ is exact. Hence, my interest is really in the case when $\sigma$ is not exact. -Thanks in advance! - -REPLY [4 votes]: Let's begin by pointing out the following: you will not find monotone examples for the simple reason that a nontrivial such deformations creates a class of nonzero symplectic area, while the Chern class is always vanishing. The best you could hope for is Calabi-Yau, and such examples indeed exist. However, I know of no examples where a closed Lagrangian has a nonvanishing Floer homology. -Now a general observation: In the case when $T^*N \setminus N$ has no second cohomology with $\mathbb{R}$-coefficients, e.g. if $N=S^2,$ then for small closed forms $\sigma$ one can use Moser's trick to show that any compact Lagrangian submanifold of $T^*N \setminus N$ is preserved (up to smooth isotopy) after turning on a sufficiently small magnetic potential. -A more concrete example: taking $\sigma$ to be the area form on $S^2,$ we obtain the total space of the line bundle $\mathcal{O}(-2)$ on $\mathbb{C}P^1$ with its standard Kähler form. (The first reference coming to my mind is 2.4A in [Y. Eliashberg and L. Polterovich; Unknottedness of Lagrangian surfaces in symplectic 4-manifolds] but maybe there is something more to the point). Unlike $T^*S^2$, the latter symplectic manifold is an open toric Calabi-Yau manifold. Unfortunately, according to Theorem 5 in [Ritter; Floer theory for negative line bundles via Gromov-Witten invariants], its symplectic homology vanishes: this twisted cotangent bundle therefore contains no Lagrangians with interesting Floer homology. See [Ritter-Smith; The monotone wrapped Fukaya category and the open-closed string map] where a closed-open map is constructed in this setting. -A side note: if you compactify a subset of the total space of $\mathcal{O}(-2)$ to the Hirzebruch surface $F_2(\alpha)$ as studied in [Fukaya-Ohta-Ono-Oh; Toric degeneration and non-displaceable Lagrangian tori in $S^2\times S^2$] then Fukaya category actually becomes nontrivial.<|endoftext|> -TITLE: BGG resolution for characters of reductive groups -QUESTION [5 upvotes]: Take $G$ a split reductive group (over a field of char 0) with Borel $B$, opposite Borel $\overline{B}$ and maximal split torus $T\subset B$. We write $X = G/\overline{B}$. Let $O \subset X$ be the big cell of $X$ for the Bruhat decomposition, it is a a dense open subset of $X$ which is isomorphic to $U$ the unipotent radical of $B$. -Let $\chi$ be a character of $T$ which is dominant with respect to $B$ and let $\mathcal{L}_\chi$ be the locally free sheaf on $X$ associated to $\chi$. We have that $\Gamma(X,\mathcal{L}_\chi) = Ind_{\overline{B}}^G\chi$ and $\Gamma(O,\mathcal{L}_\chi) = Ind_{T}^B \chi$. Restriction of sections gives us the inclusion : -$$ -0 \to Ind_{\overline{B}}^G\chi \to Ind_{T}^B \chi -$$ -This is the beginning of the BGG resolution. I would like to understand what the third term is. The only reference I know for the BGG resolution is Humphreys book on the BGG category but he works with lie algebras and the category O and I have not been able to translate his result in the language above. Looking at his book i'm guessing we must find something along the lines of $\oplus_{\alpha \in \Delta} Ind_{T}^B (s_{\alpha} \bullet \chi)$ where $\Delta$ is the basis of the root system associated to $\overline{B}$ and $\bullet$ means the dot action. But i'm not sure so if someone could help that would be great. -In fact I'm only interested in the weights of the third term of the sequence of the BGG resolution. -I think the paper "The Grothendieck-Cousin complex of an induced representation" of Kempf, which gives a geometric approach to the BGG resolution, might be what I'm looking for but again I wasn't able to extract a simple formula. You can find the paper here the important results for me I think are Theorem 12.5, lemma 12.6 and lemma 12.8. - -REPLY [3 votes]: Modulo the messy bookkeeping sometimes encountered when passing to dual modules, what Kempf does in his paper is to exploit the geometric setting of Cousin complexes in order to find an independent approach to the BGG resolution. Though this is mostly limited to working over a sufficiently large field of characteristic 0, the result is (essentially) the dual of the usual BGG resolution but translated into the language of algebraic groups. In particular, the guess by bob about the next term of the sequence is correct, though with the usual Verma modules here replaced by their duals in the BGG category. -Note that the type of algebraic group which is of primary interest here is semisimple, though the reductive case (with nontrivial semisimple derived group) is then easily treated. But characteristic 0 is above all essential in order to start with a simple module, one affording an irreducible representation. Here the induction process, or equivalently the functor of global sections of line bundles on a suitable flag variety, yields such a simple module for the group or its Lie algebra: this is essentially the Borel-Weil Theorem. Of course, these ideas originated in differential geometry relative to Lie groups, but the Chevalley classification of semisimple algebraic groups shows that this is the same result as for semisimple algebraic groups (and in fact almost the same in any characteristic). So one can shift without extra difficulty into algebraic geometry, at least in characteristic 0. -Which approach you prefer depends mostly on the language with which you are most comfortable and the further steps you have in mind. Kempf himself thought mostly in geometric terms, even when proving his important 1976 vanishing theorem for higher cohomology groups of dominant line bundles in prime characteristic (analogue of Kodaira vanishing); but soon afterward both Andersen and Haboush found a simpler algebraic proof (incorporated in Jantzen's book Representations of Algebraic Groups, Chapter II.4, Academic Press, 1987; Amer. Math. Soc., 2nd ed., 2003). To understand what is going on, it's probably easiest in any case to start with rank 1, where the ideas in Kempf's version of the BGG resolution (or else the original Lie algebra version) are transparent. -One other remark is that there is of course a reasonable generalization of Kempf's set-up to the parabolic framework (as rafaelm indicates in a comment): here a Borel subgroup is replaced by an arbitrary parabolic subgroup. This dualizes the way Alvany Rocha (in her Rutgers thesis work with Wallach) generalized the classical BGG resolution to the parabolic case for semisimple Lie algebras.<|endoftext|> -TITLE: Are cofibrant commutative S-algebras flat? -QUESTION [9 upvotes]: Let $R$ be a cofibrant commutative $S$-algebra (in the sense of Elmendorf-Kriz-Mandell-May; they call them "$q$-cofibrant") and $A$ be a cofibrant commutative $R$-algebra. - -Does $A\wedge_R-:RMod→RMod$ preserve all weak equivalences? - -This could be rephrased as "$A$ is a flat $R$-module". It is true that cofibrant $R$-modules are flat (EKMM III.3.8), however, the underlying $R$-modules of cofibrant commutative $R$-algebras are not generally cofibrant. Another positive result is that in the above hypotheses, $A\wedge_R-$ preserves weak equivalences of cofibrant commutative $R$-algebras (EKMM VII.7.4). - -REPLY [3 votes]: I'm a bit out of practice with these types of questions, and I've never really worked with $S$-modules, but I think the answer is yes. Here's why I think so. First, it's sufficient to check it for the case when A is a (co)domain of a generating cofibration in CAlg(R), by a standard cellular induction (for details, see Theorem A.2 in Hovey's paper on Smith ideals). Thus, we can assume A has the form $Sym(B) \wedge R$ where $B$ is a (co)domain of a generating cofibration of $S$-algebras. Thus, $A\wedge_R -$ is a weak equivalence of $R$-modules if and only if $Sym(B) \wedge -$ is a weak equivalence (in the underlying category of $S$-modules). Since (co)domains of the generating cofibrations of $S$-algebras are cofibrant (see MMSS), $B$ is cofibrant. Thus, $Sym(B)$ is cofibrant as a commutative $S$-algebra. Next, Lemma 3.7 of my paper on commutative monoids (accepted to JPAA), proves that $Sym(B)$ is cofibrant as an $S$-algebra (so the EKMM result you cite finishes the proof). I hasten to note that it's not recorded in print anywhere that the category of $S$-algebras satisfies the strong commutative monoid axiom, but for the crucial place in the proof of 3.7 where this is needed, you can use Proposition 4.2 of Shipley's A convenient model category for commutative ring spectra instead. Basically, the idea is that it's good enough to work with positive cofibrations to get the result you want. The last section of chapter VII of EKMM is also relevant, and could perhaps avoid the shift into positive cofibrations.<|endoftext|> -TITLE: Resolution of Gorenstein rational singularities on a surface -QUESTION [6 upvotes]: I am reading Artin's notes "Lipman's Proof of Resolution of Singularities for Surfaces" from the book "Arithmetic Geometry". I am very confused by the proof of Lemma $6.5.$ (I am formulating it below in a little bit different way than it appears in the text) - -Lemma 6.5: Let $(A,\mathfrak m, k)$ be a normal complete excellent ring of dimension $2$ that defines a rational double point (rational Gorenstein singularity). Denote by $X$ the blow-up of the unique closed point in Spec $A$. Assume that the exceptional divisor $E$ is equal to $2C$, where $C$ is a line in $\mathbb P^2_k$. And let $X' \to X$ be a sequence of blow-ups in closed points $p_1, \dots, p_n$, s.t. $X'$ is regular at evert point of the strict transform $C'$, then $\Sigma_{i} [k(p_i):k]=3$. - -The key step is to compute $\deg_C \mathcal O_X(-C)|_C$ (Note that $\mathcal O_X(-C)$ isn't locally free since $C$ isn't a Cartier divisor, but it is always reflexive, in particular torsion-free. Hence, $\mathcal O_C(-C)$ is always an invertible sheaf). Artin claims that it is equal to $-1$, but I don't understand his argument. - -In our case, since $2C$ is isomorphic to a double line in $\mathbb P^2$, the degree is the same as for such a line, i.e., $[-C,C]=-1$. - -How could one put this into a rigorous argument? It is not clear how to relate $\deg_C \mathcal O_C(-C)$ with this immersion since $\mathcal O_C(-C)^{\otimes 2} \neq \mathcal O_C(-E)$. -P.S. By a rational singularity I mean that for any normal modification $f:X \to Spec A$ we have $H^1(X,\mathcal O_X)=0$. If $A$ is also Gorenstein, it is called rational double point. The latter condition is equivalent to $\dim_k \mathfrak m/\mathfrak m^2 \leq 3$. -P.S.2. In the formulation of Lemma $E$ should be equal to a double line $2C$ with respect to the natural immersion $X \to \mathbb P^2_A$ defined by the sheaf $\mathcal O_X(-E)$. -UPD: Jason Starr mentioned in the comments that if $A$ is defined over a field $k$, then $A\cong k[[x,y,z]]/(F(x,y,z)-G(x,y,z))$, where $F$ is homogeneous quadratic polynomial and $G$ is of degree at least $3$. We can do almost the same without assuming that $A$ is defined over a field. Namely, since $A$ is a rational double point $\dim_k \mathfrak m^n/\mathfrak m^{n+1}=2n+1$. Then we have $3$ generators for $\mathfrak m$ and there is precisely one relation in degree $2$ between them in $gr_{\mathfrak m} A$. Let this relation be $F(x,y,z)=G(x,y,z)$, where $F,G\in k[T_1,T_2,T_3]$ are polynomials of degree $2$ and $3$ respectively ($F$ is also homogeneous). Since $E\cong Proj(gr_{\mathfrak m} A)$ we conclude that $E\cong V(F) \subset \mathbb P^2_k$. Taking into account that $E=2C$ we can actually choose (after a suitable linear change of coordinates) $F(T_1,T_2,T_3)=T_1^2$. -But I still don't understand what is the connection between $\deg \mathcal O_C(-C)$ and the intersection number $[-C,C]$ inside $\mathbb P^2_k$. - -REPLY [4 votes]: The explanation given in the text indeed seems to be too terse. What follows is taken from an insert I have in my copy of that book (since the argument I came up with when I read the article many years ago did not fit in the margin as other clarifications did). I hope it is helpful, and that I haven't made some blunder. -We'll use the notation in Artin's article (which I won't explain here). The idea is to give a description of $O_C(-C)$ that is intrinsic to the infinitesimal thickening $C \hookrightarrow Z$, without direct reference to the ambient $X_1$. That will allow us to make a switch to transfer the degree calculation to a more convenient choice of $X_1$! -By definition the $O_C$-module $O_C(-C)$ on $C \simeq \mathbf{P}^1_k$ is -$$(O_{X_1}(-C) \otimes_{O_{X_1}} O_C)/(O_C\mbox{-}{\rm{torsion}}) = (O_{X_1}(-C) \otimes_{O_{X_1}} (O_{X_1}/O_{X_1}(-C)))/(O_C\mbox{-}{\rm{torsion}})$$ -that is torsion-free with generic rank 1 by design, so it is invertible. Moreover, we claim that the canonical surjection of $O_{X_1}$-modules -$$q:O_{X_1}(-C) \twoheadrightarrow O_C(-C)$$ -has kernel exactly $O_{X_1}(-2C)$. The quotient $O_{X_1}(-C)/O_{X_1}(-2C)$ is easily checked to be torsion-free as an $O_C$-module and has generic rank 1 as such, so it is invertible as such. Thus, since a surjection between invertible sheaves is an isomorphism, to identify $\ker q$ as claimed it suffices to show that the $O_{X_1}$-linear restriction $q: O_{X_1}(-2C) \to O_C(-C)$ vanishes. But the target -of this restriction is a torsion-free $O_C$-module, so it suffices to check the vanishing near the generic point of $C$, hence over the regular locus $X_1^{\rm{reg}}$, where it is clear. The upshot is that we have -$$O_C(-C) \simeq O_{X_1}(-C)/O_{X_1}(-2C)$$ -as $O_C$-modules. -Let $I$ denote the coherent ideal sheaf of $C$ in $Z$, so this is a square-zero ideal by definition of $Z$ (as the ideal sheaf $O_{X_1}(-C)$ inside $O_{X_1}$ has square contained in $O_{X_1}(-2C)$) and hence $I$ is naturally an $O_C$-module. As such, we have -$$I := {\rm{image}}(O_{X_1}(-C) \to O_Z = O_{X_1}/O_{X_1}(-2C)) = -O_{X_1}(-C)/O_{X_1}(-2C) \simeq O_C(-C),$$ -the final isomorphism being what we established above. -Thus, we have an intrinsic description of the $O_C$-module $O_C(-C)$ as the coherent ideal sheaf of $C$ inside the square-zero thickening $Z$ of $C$. The conclusion is that $O_C(-C)$ as an $O_C$-module depends -only in the data of the infinitesimal closed immersion $C \hookrightarrow Z$ and not on the ambient $X_1$. -But we can identify $Z$ scheme-theoretically as a doubled-line in $\mathbf{P}^2_k$ with $C = Z_{\rm{red}}$ a straight line in this projective plane. Hence, to compute the degree of the $O_C$-module $O_C(-C)$ it suffices to do the calculation in an ambient $\mathbf{P}^2_k$. Now we can run the calculations in reverse, with $C$ inside $\mathbf{P}^2_k$ having ideal sheaf $O_{\mathbf{P}^2}(-1)$: -$$O_C(-C) \simeq O_{\mathbf{P}^2}(-C)/O_{\mathbf{P}^2}(-2C) \simeq -O_{\mathbf{P}^2}(-1)/O_{\mathbf{P}^2}(-2).$$ -Since $C = \mathbf{P}^1_k$ has structure sheaf $O_C$ with Euler characteristic equal to 1, so ${\rm{deg}}_C(L) = \chi(L) - 1$ for any line bundle $L$ on $C$, to show ${\rm{deg}}_C(O_C(-C)) = -1$ is the same as showing $\chi(O_C(-C))=0$. From the above displayed expression for $O_C(-C)$ in terms of $O_{\mathbf{P}^2}(-r)$'s, it is the same to show -that $\chi(O_{\mathbf{P}^2}(-1))=\chi(O_{\mathbf{P}^2}(-2))$. Both of -these latter Euler characteristics are equal to 0.<|endoftext|> -TITLE: Where to find the correct result in Higher Algebra, incorrect reference -QUESTION [11 upvotes]: I'm looking at the proof of Higher Algebra Proposition 6.1.6.27, and in the very first sentence of the proof, Lurie states: - -The functor $(F\delta)_{\Sigma_n}$ is n-homogeneous by Proposition 6.1.5.4. - -Checking back to 6.1.5.4, I see the statement - -Let $C$ be a small $\infty$-category which admits finite colimits and let $D$ be an $\infty$-category which admits finite limits and small filtered colimits. Assume that filtered colimits in $D$ are left exact. Then composition with the Yoneda embedding $j : C \to Ind(C)$ induces a fully faithful - functor $$\theta : Exc^n_c (Ind(C),D)) \to Fun(C,D)$$ - whose essential image is the full subcategory $Exc^n(C,D) \subset Fun(C,D)$ spanned by the n-excisive functors. - -which seems totally irrelevant. -Does anyone know what the correct reference is, or if the reference is correct, how to apply the proposition to achieve the result? - -REPLY [18 votes]: The correct reference is 6.1.4.14. (And the hypothesis of 6.1.6.27 should refer to countable limits and colimits, rather than finite limits and colimits.)<|endoftext|> -TITLE: In what respect are univalent foundations "better" than set theory? -QUESTION [56 upvotes]: It was an ambitious project of Vladimir Voevodsky's to provide new foundations for mathematics with univalent foundations (UF) to eventually replace set theory (ST). -Part of what makes ST so appealing is its incredible conciseness: the only undefined symbol it uses is the element membership $\in$. UF with its type theory and parts of higher category theory seems to be a vastly bigger body to build the foundation of mathematics. To draw from a (certainly very imperfect) analogy from programming: ST is like the C programming language (about which Brian Kernighan wrote: "C is not a big language, and it is not well served by a big book"), but UF seem more like the vast language of Java with all its object-oriented ballast. -Questions. Why should mathematicians study UF, and in what respect could UF be superior to ST as a foundation of mathematics? - -REPLY [21 votes]: The two main things I find compelling about type theory as a foundation are: - -A more direct way to deal with $\infty$-groupoids. -A closer match to how mathematicians actually talk about mathematics. - -Timothy Chow's answer touches on point 1, so let me just add that personally I had a lot of trouble understanding and dealing with set theoretic definitions of $\infty$-groupoids, and find the HoTT definition natural and easy to work with. -The second one I think is pretty important. If you ask an actual mathematician whether 5 is an element of 7 or what the intersection of the Monster group and the real numbers is, they'll tell you "that question doesn't make sense!" But in set theoretic foundations those kinds of questions do make sense and have answers! In type theory the answer is "that doesn't type check," which is the slightly formal version of "that doesn't even make sense." -This sort of "type checking" is really important practically! It's like dimensional analysis and often quickly tells you when a formula is wrong or gives you a good guess as to what could possibly be true. -Even for really simple things like the set theoretic definitions of particular natural numbers, of ordered pairs, and of functions, the answer that set theory gives is not at all like the typical mathematicians intuition. But the type theoretic definitions do closely match my intuition. An ordered pair is a new type of thing and you're allowed to make an ordered pair by telling me the first entry and the second entry. A function is a new type of thing and if you tell me something in the source it tells you something in the target.<|endoftext|> -TITLE: Pointwise evaluation of the Beck-Chevalley map in $\infty$-categories -QUESTION [5 upvotes]: In Higher Algebra Lemma 6.1.6.3, most of the proof is pretty straightforward, but after thinking I understood it all correctly, I realized I had a gap in my understanding. -Suppose we have a homotopy cartesian square of $\infty$-categories (the Lemma has these as Kan complexes, but I think this is irrelevant to my question here) $$\begin{matrix}X'&\xrightarrow{f'}&Y'\\ \downarrow^{g_X}&&\downarrow^{g_Y}\\X&\xrightarrow{f}&Y -\end{matrix}$$ -and for simplicity, $C$ is an $\infty$-category with enough limits for $f^*$ and $f'^*$ to admit right adjoints. -In the proposition (which has more conditions, though they are irrelevant to this question), we want to see if the Beck-Chevalley transformation $$g^*_Yf_* \to f'_*f'^*g_Y^*f_* \simeq f'_*g_X^*f^*f_*\to f'_*g_X^*$$ -is an is an equivalence. -Lurie suggests we prove this by pointwise evaluation on a functor $F:X\to C$ and an object $y$ of $Y'$ of the Beck-Chevalley transformation as the induced map $$\operatorname{lim}(F|X\times_Y Y_{g_Y(y)/}) \to \operatorname{lim}(F|X'\times_Y' Y'_{y/}).$$ -The trouble is, it's not clear to me why this map, induced by the diagram -$$\begin{matrix} X' & \to & Y'& \leftarrow & Y'_{y/}\\ -\downarrow &&\downarrow&&\downarrow\\ -X&\to&Y&\leftarrow&Y_{g_Y(y)/} -\end{matrix}$$ -is homotopic to the component of the Beck-Chevalley map at $F$ and $y$. -I tried evaluating the intermediate terms, but they are huge and messy, and I can't find a reference showing that they are indeed homotopic. -Is the proof easy? Does anybody have a reference? -Edit: I think this is related to the proof of 4.3.3 in Ambidexterity paper of Lurie and Hopkins. It looks like he uses cartesianness of the square there to identify (the adjuncts of) these maps (in the dual case of left Kan extensions), so I have added a cartesianness condition. - -REPLY [2 votes]: The Beck-Chevalley transformation $g^*_Y f_* \rightarrow f'_* g^*_X$ from a square of $\infty$-groupoids as above is an equivalence iff it's an equivalence when evaluated at every point of $Y'$, i.e. iff the transformation $p^*g^*_Y f_* \rightarrow p^*f'_* g^*_X$ is an equivalence for all maps $p : * \rightarrow Y'$. Consider the homotopy pullback square -$$\begin{matrix} -F & \xrightarrow{r} & *\\ -\downarrow^{q} & &\downarrow^{p}\\ -X' & \xrightarrow{f'} & Y'. -\end{matrix}$$ -The corresponding Beck-Chevalley transformation $p^*f'_* \rightarrow r_{*}q^*$ is clearly an equivalence - this just says that $f'_*$ evaluated at $p$ is the limit over the homotopy fibre $F$. Indeed, the Beck-Chevalley transformation for a square of this form (with $*$ in the upper right corner) is an equivalence (for all targets) iff the square is Cartesian. By the naturality of Beck-Chevalley transformations, the composite $(g_Y p)^* f_* \simeq p^*g^*_Y f_* \rightarrow p^*f'_* g^*_X \rightarrow r_{*}q^*g^*_X \simeq r_*(g_X q)^*$ is equivalent to the Beck-Chevalley transformation for the composite square -$$\begin{matrix} -F & \xrightarrow{r} & *\\ -\downarrow^{g_X q} & &\downarrow^{g_Y p}\\ -X & \xrightarrow{f} & Y -\end{matrix}$$ -(this follows from the adjunction identities). This transformation is again an equivalence if this square is Cartesian. So by 2-of-3 the original transformation is an equivalence at every $p$ if the square -$$\begin{matrix} -X' & \xrightarrow{f'} & Y'\\ -\downarrow^{g_X} & &\downarrow^{g_Y}\\ -X & \xrightarrow{f} & Y -\end{matrix}$$ -is Cartesian. (And conversely, this square is Cartesian if the Beck-Chevalley transformation is always invertible (or if it is invertible for functors to spaces).) -For $\infty$-categories you have to replace the homotopy fibre square with the lax pullback square -$$\begin{matrix} -X'_{p/} & \to & *\\ -\downarrow & \Rightarrow &\downarrow\\ -X' & \xrightarrow{f'} & Y' -\end{matrix}$$ -(which commutes up to a natural transformation in a direction I'm too lazy to work out) - the Beck-Chevalley transformation for this gives the limit formula for a right Kan extension. I'm not aware of a nice criterion for a square of ($\infty$-)categories to have an invertible Beck-Chevalley transformation, but the same argument shows that it can be checked pointwise.<|endoftext|> -TITLE: Well definedness of square roots of separating Dehn Twists -QUESTION [8 upvotes]: Let $c_1,c_2,c_3,c_4,c_5$ be a five chain of circles on a genus 2 surface (i.e $i(c_k,c_{k+1})=1$ and zero otherwise). Then $(T_{c_1} T_{c_2})^6 = (T_{c_4} T_{c_5})^6 = T_c$ where $c$ is a separating curve that bounds a neighborhood of $c_1$ and $c_2$ (and also bounds a neighborhood of $c_4$ and $c_5$). This follows from the two chain relation. Then we can think of $(T_{c_1} T_{c_2})^3$ and $(T_{c_4} T_{c_5})^3$ as square roots of $T_c$. Are these two equal as mapping classes? - -REPLY [8 votes]: They are different. In fact, they act differently on $H_1(\Sigma_2;\mathbb{Z})$. Let $V \subset H_1(\Sigma_2;\mathbb{Z})$ be the span of the homology classes of $c_1$ and $c_2$, and let $W \subset H_1(\Sigma_2;\mathbb{Z})$ be the span of the homology classes of $c_4$ and $c_5$. You can then calculate the $(T_{c_1} T_{c_2})^3$ acts by $-1$ on $V$ while fixing $W$, but $(T_{c_4} T_{c_5})^3$ acts by $-1$ on $W$ while fixing $V$. -There is by now a large literature on what kinds of roots a Dehn twist has, starting with -D. Margalit, S. Schleimer -Dehn twists have roots. -Geom. Topol. 13 (2009), no. 3, 1495-1497. -and continuing with -D. McCullough, Darryl, K. Rajeevsarathy, Kashyap -Roots of Dehn twists. -Geom. Dedicata 151 (2011), 397-409. -and -K. Rajeevsarathy -Roots of Dehn twists about separating curves. -J. Aust. Math. Soc. 95 (2013), no. 2, 266–288. -and -N. Monden -On roots of Dehn twists. -Rocky Mountain J. Math. 44 (2014), no. 3, 987–1001.<|endoftext|> -TITLE: Are pseudo-Anosov foliations dense? -QUESTION [9 upvotes]: A pseudo-Anosov foliation of a compact orientable surface $F$ is a one whose class in the space $\mathcal{PMF}(F)$ of projective measured foliations is preserved by some pseudo-Anosov homeomorphism of $F$. I saw it casually mentioned that pseudo-Anosov foliations are dense in $\mathcal{PMF}(F)$. What is a proper reference for that result? - -REPLY [6 votes]: The pseudo-Anosov foliations form a subset of $\mathcal{PMF}(F)$ which is invariant under the action of the mapping class group $MCG(F)$, because if $\Lambda_+(\phi) \in \mathcal{PMF}(F)$ is the stable lamination of a pseudo-Anosov $\phi \in MCG(F)$ then $\psi(\Lambda_+(\phi)) = \Lambda_+(\psi\phi\psi^{-1})$ is the stable lamination of the pseudo-Anosov $\psi\phi\psi^{-1}$. -So the fact that the pseudo-Anosov foliations are dense is an immediate corollary of Theorem 6.1 from "Thurston's Work on Surfaces" (originally published in 1979 as "Travaux de Thurston sur les Surfaces"): - -The action of $MCG(F)$ on $\mathcal{PMF}(F)$ is minimal, meaning that every orbit is dense.<|endoftext|> -TITLE: Ordered Nim game -QUESTION [9 upvotes]: Consider the following variant of Nim: -There are two players and $n$ piles of stones, with sizes $a_1,\dots,a_n$, such that $a_i\leq a_j$ for any $i -TITLE: Do there exist two points seeing one another? -QUESTION [11 upvotes]: Let $n\ge1$ be an integer number. Let $n$ nonoverlapping closed line segments and -$n+2$ distinct points which do not belong to those line segments be given in the plane -$\mathbb{R}^2$. -Can two points among the $n+2$ given points be chosen such that the ones - see one another i.e. the closed line segment connecting these points does not intersect - any of the given $n$ line segments? In the case $n=1$ the affirmative answer is simple. -However, I don't know the answer even for $n=2$. -Addition. Here is a modification of Joseph picture which demonstrates the difficulties when realizing the idea by Per - -If the points $A(\frac {38} {20},\frac {38} {20}) $ and $B(\frac {32} {10},\frac {32} {10} ) $ are assumed to belong to $\{(x,y):y \ge x,y\ge 3-x\} $, - then the proof suggested by Per fails. -Addition 2. -I'd like to demonsrate another difficulty which appears in the Per's approach. Let us consider -three segments $S_1:=[(-1,-1),(1,-1)],\,S_2:=[(1,1),(1,2)],\,S_3:=[(-1,0),(2,1)]$. The points -are not of importance here. After the first step the plane is divided into two regions $R_1$ and $R_2$. - -After the second step we obtain -the regions $R_1$ and $R_3$, where the latter is not convex. - -It is unclear for me whether after the final step we obtain the convex regions only (the larger the value of $n$, the more complicated the situation). -One more problem consists in the definitions of the borders of the regions. In order -to apply the pigeonhole principle the regions considered with their borders must not intersect. -My advice to Per is to read the book I. Lakatos. Proofs and Refutations. Cambridge: Cambridge University Press.(1976). -https://books.google.com.ua/books/about/Proofs_and_Refutations.html?id=1n6SFdXCOBQC&redir_esc=y - -REPLY [7 votes]: This very problem was proposed to St. Petersburg olympiad (selection round) in 2007 by Konstantin Kokhas (problem 6 for 10-th grade in the linked pdf). In the same year it was proposed - independently, I guess, - to the journal Matematicheskoe Prosveschenie (problem 5 in the linked pdf) by Maxim Kontsevich himself. -Here is a solution which is hopefully self-contained and complete. -Assume the contrary. At first, we slightly enlarge the segments so that new segments still do not overlap. After that each segment $AB$ between our $n+2$ points intersects one of $n$ segments in an interior point. This property is preserved under small perturbation of the $n$ segments. Such perturbation allows to get $n$ segments such that no three lines containing these segments have a common point, and no three out of $2n$ their endpoints lie on a line (for example, you may choose new endpoints one by one so that each time you fix an endpoint it does not lie on a line between two already fixed endpoints; and each time you fix both endpoints of a segment, the line containing it does not pass through a common point of other two lines. This is clearly possible, since already finitely many lines are forbidden and a small disc is allowed.) -Now we enlarge the segments one by one, each time either to infinity or to the intersection with another segment. We get a plane partitioned onto several convex regions. Let's prove that there are exactly $n+1$ regions and they are convex. Draw a large square and remove the parts of rays outside the square. Then we get a planar graph with all degrees equal to 3, and the number of vertices equals $2n$ (2 vertices at endpoints of each of extended segments). Each region is a convex polygon (since going along the boundary of each of the regions we see that all angles are less than $\pi$). Thus the number of edges equals $3n$, and the number of faces equals $n+2$ by Euler formula. One face is external, so $n+1$ inner faces as desired.<|endoftext|> -TITLE: "Formally unramified iff trivial Kähler differentials" using only universal properties? -QUESTION [19 upvotes]: For convenience we work with commutative rings instead of commutative algebras. - -Fix a commutative ring $R$. Consider the functor $\mathsf{Mod}\longrightarrow \mathsf{CRing}$ defined by taking an $R$-module $M$ to $R\ltimes M$ (with dual number multiplication). Following the nlab, the module of Kähler differentials of $R$ is defined as the universal arrow from $R$ to this functor. -There's also the equivalence between the category of split zero-square extensions and the category of modules. These result in the natural isomorphisms ($\operatorname{dom}$ takes an arrow to its domain). -$$\begin{aligned}\mathsf{CRing}(R,\operatorname{dom}(A^{\prime}\twoheadrightarrow A)) & \cong{\substack{\text{split square-zero}\\ -\text{extensions} -} -}(R\ltimes\Omega_{R}\twoheadrightarrow R,A^{\prime}\twoheadrightarrow A)\\ - & \cong \mathsf{Mod}(\Omega_{R},\operatorname{Ker}(A^{\prime}\twoheadrightarrow A)). -\end{aligned}$$ -So $\Omega _R\cong \bf 0$ iff $R$ has exactly one arrow to the domain of every -split square-zero extension. - -On the other hand, we say $R$ is unramified if given a zero-square extension $A^\prime \twoheadrightarrow A$, the induced post-composition $$\mathsf{CRing}(R,A^\prime)\longrightarrow \mathsf{CRing}(R,A)$$ is injective. The diagrams of interest are below. -$$\require{AMScd} \begin{CD} A^\prime /I @<<< R\\ @AAA \\ A^\prime & \end{CD} -\; \; \; \begin{CD} A^\prime /I @<<< R\\ @AAA @VVV\\ A^\prime @<<< I \end{CD}$$ - -Now, if $\Omega _R\cong \bf 0$ then there exists and is unique a diagonal arrow $R\to A^\prime $, which seems to easily give "unramified w.r.t split square-zero extensions". However, I see no way to deduce anything about the non-split ones. - -The thing is, I feel a proof is very close: suppose $g,h$ are diagonal fillers $R\to A^\prime$ on the left below. Viewing them as module arrows this is equivalent to saying $g-h$ lands in $I$, i.e factors through $I\vartriangleleft A^\prime $. Finally $g=h$ iff this factorization $R\to I$ of $g-h$ must be zero. If only this could be given by the universal property... -Perhaps the correct way to proceed is using some universal property of the diagonal? (Not that it realizes the Kähler differentials.) - -REPLY [2 votes]: Your question can be rephrased as follows: I know something about split square-zero extensions. How can I (categorically) conclude something about all square-zero extensions? The key observation is that square-zero extensions are the torsors for the split square-zero extensions. I first learned about this general idea from section 7.4.1 of Lurie's Higher algebra and I formulated this idea more precisely here. -Let $\varphi: B\to A$ be a square-zero extension. Then $\ker(\varphi)$ is an $A$-module and $A\oplus \ker(\varphi)\to A$ is an abelian group object in $\text{CRing}/A$. We have a natural group action $\tau: (A\oplus \ker(\varphi))\times_A B\to B$ which is a torsor in the sense that $$(\tau, \pi_B): (A\oplus \ker(\varphi))\times_A B\to B\times_A B$$ is an isomorphism. -Now we'll relate this to your question. Let $C$ be such that $\Omega_C = 0$. Let $\varphi: B\to A$ be a square-zero extension. We want to show that the canonical map $\text{Hom}(C, B)\to\text{Hom}(C, A)$ is injective. Let $\alpha, \beta\in\text{Hom}(C, B) $ be such that $\varphi\circ\alpha = \varphi\circ\beta$. Thus $$(\alpha, \beta)\in \text{Hom}(C, B)\times_{\text{Hom}(C,A)} \text{Hom}(C, B).$$ As you already showed, $\Omega_C = 0$ implies $\text{Hom}(C, A\oplus\ker(\varphi))\approx\text{Hom}(C, A)$. Thus $$\begin{split} \text{Hom}(C, B)\times_{\text{Hom}(C,A)} \text{Hom}(C, B) &\approx \text{Hom}(C, B\times_A B) \\ &\approx \text{Hom}(C, (A\oplus \ker(\varphi))\times_A B) \\ &\approx \text{Hom}(C, A\oplus \ker(\varphi))\times_{\text{Hom}(C, A)} \text{Hom}(C, B)\\&\approx \text{Hom}(C, A)\times_{\text{Hom}(C, A)} \text{Hom}(C, B)\\&\approx \text{Hom}(C, A\times_A B)\\&\approx \text{Hom}(C, B)\end{split}$$ and this final bijection maps $(\alpha, \beta)$ to $\beta$. This establishes that there can be at most one lift for a general square-zero extension. -This same argument works in any category with a notion of Kähler differentials, as I show here.<|endoftext|> -TITLE: Let $X$ be the class of all classical Laver tables. Is $HS(X)=S(X)$? -QUESTION [6 upvotes]: Let $A_{n}=(\{1,\ldots,2^{n}\},*_{n})$ be the algebra defined by -$x*_{n}1=x+1\mod 2^{n}$ and $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$ for all $x,y,z\in\{1,\ldots,2^{n}\}$. Suppose that $X$ is a subalgebra of some $A_{n}$ and $\simeq$ is a congruence on $X$. Then does there necessarily exist some $m$ along with some embedding $i:(X/\simeq)\rightarrow A_{m}$? -If so, then for all $n$, let $t_{n}$ be the least natural number $m$ such that whenever $X$ is a subalgebra of $A_{n}$ and $\simeq$ is a congruence on $X$, then there is some embedding $i:X/\simeq\rightarrow A_{m}$. Then what are some bounds on $t_{n}$? Or what is the exact value of $t_{n}$? - -REPLY [3 votes]: It is known that if $X$ is a cyclic subalgebra of $A_n$ and $\simeq$ is a congruence on $X$, then $X/\simeq$ is isomorphic to $A_m$ for some $m\leq n$. See Propositions 1.6 and 1.21 of Chapter X of Braids and Self-Distributivity by Dehornoy for the proof. -I don't think it is known whether the result is true for arbitrary subalgebras $X\leq A_n$. I experimented with $A_3$ and $A_4$ using UACALC, and could not find any counterexamples.<|endoftext|> -TITLE: On a finitary version of mixing -QUESTION [5 upvotes]: Let $(X_1,X_2,\ldots)$ be a stationary, mixing sequence of real random variables. Then it holds (for example) for any event $A$ that is measurable in $\sigma(X_1,X_2,\ldots)$ and any $S \subseteq \mathbb{R}$ that -$$ -\lim_{i \to \infty} \big|\mathbb{P}[A,X_i\in S] - \mathbb{P}[A] \cdot \mathbb{P}[X_i \in S]\big|=0. -$$ -I am looking for a finitary version of this statement. That is, something of this sort: for every $\varepsilon>0$ there is an $n$ large enough such that, if $A$ is measurable in $\sigma(X_1,\ldots,X_n)$, and if $i$ is chosen uniformly in $\{1,\ldots,n\}$, then with probability $1-\varepsilon$ -$$ -\big|\mathbb{P}[A,X_i\in S] - \mathbb{P}[A] \cdot \mathbb{P}[X_i \in S]\big| < \varepsilon. -$$ - -REPLY [2 votes]: Let $J_n$ be a random variable independent of $X:=(X_1,X_2,\ldots)$ and uniformly distributed in the set $\{1,\ldots,n\}$. For each $i\in\{1,\ldots,n\}$, let -\begin{equation} - p(i):=P(A,X_i\in S). -\end{equation} -We need to show that -\begin{equation} - p(J_n)\to P(A)P(X_1\in S) -\end{equation} -in probability uniformly in all $A\in\sigma(X_1,\ldots,X_n)$; the convergence here everywhere is as $n\to\infty$. -Let us show a bit more -- that this convergence is uniform over all $A$ in the underlying sigma-algebra (say $\Sigma$). Suppose then that the weak mixing condition holds (which of course will hold if the strong mixing holds). Then the sequence $X$ is ergodic; see e.g. the Remark on page 13 in Ergodic Theory. -Now the ergodic theorem implies that -\begin{equation} - d_n(S):=\frac1n\,\sum_{i=1}^n I\{X_i\in S\}-P(X_1\in S)\to0 -\end{equation} -almost surely and hence in probability, where $I\{\cdot\}$ denotes the indicator function. So, for every $\varepsilon>0$ there is a natural $n_\varepsilon$ such that for all natural $n>n_\varepsilon$ we have $P(|d_n(S)|>\varepsilon)<\varepsilon$; therefore and because $|d_n(S)|\le1$, -\begin{equation} - E|d_n(S)I\{A\}|\le E|d_n(S)| \le\varepsilon P(|d_n(S)|\le\varepsilon)+P(|d_n(S)|>\varepsilon)\le2\varepsilon. -\end{equation} -So, $d_n(S)I\{A\}\to0$ in $L^1$ uniformly in all $A$. So, -\begin{equation} - Ep(J_n)= - \frac1n\,\sum_{i=1}^n P(A,X_i\in S) - =Ed_n(S)I\{A\}+P(A)P(X_1\in S)\to P(A)P(X_1\in S) -\end{equation} -uniformly in all $A$. -Similarly, letting $(\tilde A,\tilde X)$ denote an independent copy of $(A,X)$, we have -\begin{multline*} - Ep(J_n)^2= - \frac1n\,\sum_{i=1}^n P(A,X_i\in S)^2 - =\frac1n\,\sum_{i=1}^n P(A,\tilde A,X_i\in S,\tilde X_i\in S) \\ - \to P(A,\tilde A)P(X_1\in S,\tilde X_1\in S)=P(A)^2P(X_1\in S)^2 -\end{multline*} -uniformly in all $A$; -here we use the fact that the weak mixing property is preserved under the direct product (see e.g. page 5 in http://arxiv.org/abs/math/0603575v1 ) and hence the sequence of pairs $((X_1,\tilde X_1),(X_2,\tilde X_2),\dots)$ is ergodic. -Thus, $Ep(J_n)\to P(A)P(X_1\in S)$ and $Var\,p(J_n)\to0$ uniformly in all $A$. So, by Chebyshev's inequality, indeed $p(J_n)\to P(A)P(X_1\in S)$ in probability uniformly in all $A$.<|endoftext|> -TITLE: Rationality of trace of endomorphism of Iwasawa-thing -QUESTION [5 upvotes]: Let $n$ be a positive integer, and $p$ a prime number. Let $K_i$ be the cyclotomic field containing exactly the $np^i$th roots of unity. Let $H$ be the inverse limit of $p$-power torsion of the class groups of the $K_i$. Let $V$ be the $\mathbb{Q}_p$ - vector space $H \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$. It is a result of Iwasawa theory that $V$ is finite-dimensional. -For any integer $q$ that is relatively prime to $np$, we have an automorphism $f_q$ of $\bigcup_{i = 1}^{\infty} K_i$ which, for each root of unity $\omega$, brings $\omega^q$ to $\omega$. This induces an automorphism of $H$ and thus an automorphism of $V$, say $f^*_q$. -What I'm interested in is whether $\mathrm{Tr}(f^*_q) \in \mathbb{Q}_p$ is rational, and related questions. Are there situations where we know this is the case, besides obvious cases like $q = 1$ or $V$ is trivial etc.? Are there situations where we know this isn't the case? Are there related operators on $V$ that nontrivially have rational trace? Can we ever say something weaker like $\mathrm{Tr}(f_q)$ is algebraic over $\mathbb{Q}$? -The reason I ask: Given a smooth projective curve $C$ over a finite field $K$ with algebraic closure $\overline{K}$, the $q$th power Frobenius morphism on $C$ induces the $\overline{K}$-linear Frobenius morphism on $\overline{C} = C \otimes_{K} \overline{K}$, which induces an endomorphism of the $\ell$-adic cohomology vector space $H^1(\overline{C}, \mathbb{Q}_{\ell})$. The traces of powers of this morphism yield information about the zeta function of the curve, but the only reason this works is that these traces are rational numbers (more specifically, integers) so they can be seen as both $\ell$-adic numbers or as real numbers, the latter being what we need for applications to the zeta function. -To relate this to the original question, I don't think these $f_q$ morphisms are analogous to powers of the $\overline{K}$-linear Frobenius, but rather a closely related morphism, which could be briefly described as the "inverse Frobenius on coefficients", and whose powers still induce endomorphisms of the $\ell$-adic cohomology vector spaces which have rational traces which give us essentially the same information as the $\overline{K}$-linear Frobenius. So basically I want to know if a similar thing holds in the number field setting, with $V$ playing the role of $H^1(\overline{C}, \mathbb{Q}_{\ell})$, in hopes that it might yield interesting information pertaining to non-p-adic L-functions (although I don't expect anything nearly as straightforward as in the function field case). - -REPLY [2 votes]: This is not really an answer, just a (very!) long comment. Everything I write is obvious for people working in Iwasawa theory, and I apologize for the trivialities. -Let me start by your final paragraph, where you discuss the analogy with curves over finite fields and the action of Frobenius: I guess you are aware that this was one of the motivations for Iwasawa theory, and this is discussed a bit in Washington's book, Chapter 13 (both in the introductory paragraph and in 13.6). But my feeling is that $V$ is somehow analogous to a $p$-adic theory (like crystalline, or overconvergent), rather than étale $l$-adic: in the analogy, we consider $K_\infty/K_0$ as the constant field extension, which has only one Frobenius. What changes in the rationality business are the coeffients: here, your coefficients are always the same, and $p$-adic, and you change "Frobenius". -Passing to your question, anyhow, let me stick for notational ease to the case $n=p$ and write $K_\infty=\varinjlim \mathbb{Q}(\zeta_{p^{k+1}})$. The Galois group $\operatorname{Gal}(K_\infty/K_0)$ is procyclic, isomorphic to $1+p\mathbb{Z}_p$ through the cyclotomic character $\kappa\colon\operatorname{Gal}(K_\infty/K)\to 1+p\mathbb{Z}_p$ (or rather its inverse, to be consistent with your choice): let me fix a topological generator $\gamma_0$ of this Galois group which, for notational ease, I suppose to be sent via $\kappa$ to the element -$(1+p)\in(1+p\mathbb{Z}_p)$. -Introduce the logarithm $\mathcal{L}$ "in base $1+p$" -$$ -\mathcal{L}(u)=\frac{\log_p(u/\omega(u))}{\log_p(1+p)}%\quad\text{ and }\quad (\mathbb{Z}/p)^\times\ni\ell\colon x\mapsto 1\in\mathbb{Z}/(p-1) -$$ -where $\log_p$ is Iwasawa's $p$-adic logarithm, $\omega$ is the$\mod{p}$ Teichmüller character, and $u\in\mathbb{Z}_p^\times$. Then, for every integer $q\not\equiv 0\pmod{p}$ we have $f_q=\gamma_0^{%-\ell(\omega(q))+ -\mathcal{L}(q)}$. -Now, if the action of $\gamma_0$ is semi-simple, then the matrix of $\gamma_0$ acting on $V\otimes \overline{\mathbb{Q}}_p$ is -$$ -M_{\gamma_0}=\begin{pmatrix}\alpha_1&&\\&\ddots&\\ &&\alpha_{\lambda}\end{pmatrix}, -$$ -where the $\alpha_i$'s are the roots of the characteristic polynomial, and then $\operatorname{Tr}(\gamma_0)=\alpha_1+\dots+\alpha_\lambda$. By the above remark, $\operatorname{Tr}(f_q^\ast)=\alpha_1^{%\omega^{-1}(q) -\mathcal{L}(q)}+\dots+\alpha_\lambda^{%\omega^{-1}(q) -\mathcal{L}(q)}$ and to address your question it seems natural to start studying these roots $\alpha_i$'s themselves. The first thing we can observe is that $\alpha_i\equiv 1\pmod{\mathfrak{m}_{\mathbb{C}_p}}$ because $\gamma_0$ is topologically unipotent and thus $\lim_{n}\alpha_i^{p^n}=1$. Moreover, by definition, $$\alpha_i^{\mathcal{L}(q)}=\exp_p(\log_p(\alpha_i)\mathcal{L}(q))=\exp_p\Bigl(\log_p(\alpha_i)\frac{\log_p(q/\omega(q))}{\log_p(1+p)}\Bigr); -$$ -I am not an expert in $p$-adic transcendence, but granted that $q/\omega(q),(1+p)$ and $\alpha_i$ are algebraically independent (for non-pathological $\alpha_i$), I would be surprised if the above expression were algebraic. -To go further, observe that $V$ is actually a direct sum $V^+\oplus V^-$ of two subrepresentations, which are the $\pm$-eigenspaces for the action of the complex-conjugation $c\in\operatorname{Gal}(K_\infty/K_0)$. A well-known conjecture by Greenberg predicts that $V^+=0$ (this has been checked numerically in many cases), so it is natural to restrict only to the subspace $V^-$. Greenberg himself has proven in his paper On a certain $l$-adic representation (Invent. Math., 1973) that the action of $\gamma_0$ on $V^-$ is semi-simple and that its minimal polynomial is $f_{\gamma_0}^-(T)=(T-1)^sg(T)$ where $s$ is the number of primes above $p$ in $K_0^+$ which split in $K_0$: so, in our case $n=p$ we have $s=0$ and $\alpha_i\neq 1$ for all $i$. The roots of $f_{\gamma_0}^-(T)$ (again, conjecturally $f_{\gamma_0}^-(T)=f_{\gamma_0}(T)$ because $V^+$ should be $0$) are connected, via the Main Conjecture of Iwasawa Theory (now a theorem), to the zeros of the $p$-adic $L$-function $L_p(s,\chi)$ of Kubota--Leopoldt, where $\chi$ runs through all even characters of $\operatorname{Gal}(K_0/\mathbb{Q})$. This goes as follows: the representation $V^-$ can be further decomposed as $V^-=\oplus_{\chi}V(\chi^{-1}\omega)$ where $\chi$ are the even characters of $\operatorname{Gal}(K_0/\mathbb{Q})$ and $V(\chi^{-1}\omega)$ is the subspace on which the action of $\operatorname{Gal}(K_0/\mathbb{Q})$ is given by $\chi^{-1}\omega$: accordingly, $f_{\gamma_0}^-(T)=\prod_\chi f(T,\chi)$. Then Mazur--Wiles and Rubin have proven that $L_p(s,\chi)=f((1+p)^s,\chi)$ and thus if $\alpha_i$ is a root of $f_{\gamma_0}^-(T)$, then $f(\alpha_i,\chi)=0$ for some $\chi$ and $L_p(\mathcal{L}(\alpha_i),\chi)=0$; and conversely, if $L_p(s_0,\chi)=0$ then $(1+p)^{s_0}$ is a root of $f(T,\chi)$ and hence of $f_{\gamma_0}(T)$. -Now comes the sad side of the story, namely that (in Washington's words, see the Remark following Theorem 5.11 in his book Introduction to Cyclotomic Fields) "the nature of the zeros of the $p$-adic $L$-function is not yet understood". Computations by Ernvall and Metsänkylä show that, in general, the $\alpha_i$ are not rational numbers, because the $p$-adic valuation of $\alpha-1$ is non-integral (see Propositions 4 and 5 in their paper Computation of the zeros of $p$-adic $L$-functions, Math. Comp., 58, 1992). It is worth quoting the final sentence of the paper: - -On the basis of the numerical data computed so far it seems natural - to expect that the zeros $T_j$ (corresponding to my $\alpha_j-1$) - and $s_i$ (corresponding to my $\mathcal{L}(\alpha_i)$) are distributed - randomly as regards their $p$-adic value (within the prescribed - limits) and their inclusion in various extensions of $\mathbb{Q}_p$. - Also the $p$-adic expansions of the zeros fail to show any regularity. - -Remark I have chosen a highly non-standard normalization, so if you look at the litterature, pay attention: in particular, what is called $f(T)$ is normally the characteristic polynomial of $\gamma_0-1$ rather than of $\gamma_0$.<|endoftext|> -TITLE: Coming out as transgender in the mathematical community -QUESTION [90 upvotes]: I don't know if MO is the right place to ask such a question, but anyway it's my only hope to get an answer, and it's very important for me (not to say 'vital'); so let's try. -I'm at this time a Ph.D. student, and I plan to defend in the spring of 2018. I'm currently looking for a postdoc position for next year. I am, at this time, known as a man in the mathematical community, but I'm actually a trans woman, beginning my gender transition. I have two problems. Firstly, I will have to come out as transgender, in at most a few years, in the mathematical community, and I'm quite fearful about the consequences (for example, for my career). Secondly, I have to ensure before applying for a postdoc that in the country where I apply, I will be able to pursue my transition, I will be accepted as I am at the university, and that there won't be any major threat to my security (because of the policy of the country regarding trans people, for instance). For this reason, having contacts in these countries who are reaserchers in maths and are familiar with transidentity questions would be very helpful for me, as I have no other means to get the info I need. -So my first question is: are there, here, trans mathematicians who would be willing to talk with me, in private, about how they came out (if they had to) in the mathematical community, how it was accepted, what has been the consequences for their career, and more generally what was their experience as trans mathematicians? (I also have other specific questions like, for instance, how to deal with a change of your first name when you already have published under your former name?) Even if you're not trans, if you have information about all of this (if you know a trans mathematician for example), I would be interested. -My second question is: in the countries where I am interested in applying for a postdoc, that is Spain, the Czech Republic, Canada, the US, and Brazil, do you have any contacts, in the academic world, who are familiar with LGBT questions, and who could give me an idea about the situation of trans people in their country, and especially at the university? (In order for me to know if it's safe to apply there or not?) -If some of you are yourselves why I don't ask these questions directly to researchers of the universities where I want to apply: that's simply because it's not safe. Trans people have to face a lot of discrimination and you never know if speaking about your transidentity with someone you don't know is safe or not - that's the reason why I choosed to ask it anonymously here, first. -(You can contact me in private at rdm.v[at]yahoo[dot]com.) - -REPLY [38 votes]: If you will be at the Joint Math Meetings in San Diego, there will be a Spectra-sponsored MAA panel on Out in Mathematics: Professional Issues Facing LGBTQ Mathematicians, Thursday, January 11th, 1pm - 2:20pm, Room 1A, Upper Level, San Diego Convention Center. It will be moderated by Lily Khadjavi (Loyola Marymount University), with panelists Shelly Bouchat (Indiana University of Pennsylvania), Juliette Bruce (University of Wisconsin Madison), Ron Buckmire (NSF), Frank Farris (Santa Clara University), and Emily Riehl (Johns Hopkins University). -Spectra is hosting an informal reception later that day, Thursday, January 11th, 2018, 6pm-8pm, Catalina Room, Marriott Marquis, which is a great way to meet people. This event has been held at every JMM since 1995. Here is the story of how it started, and shows how our professional societies took a courageous stand even during the anti-gay political storms of in the 1990s. It continues to have relevance today. -In November, 1992, Colorado residents passed an amendment to their Constitution targeting three local anti-discrimination ordinances, nullifying them and making any further attempts unconstitutional. The 1995 Joint Math Meetings were scheduled for Denver. Two mathematicians wrote to the AMS and MAA boards in late 1992 requesting that the JMM be moved, citing the blatant anti-gay animus of the amendment (so blatant that in 1995 the US Supreme Court ruled it unconstitutional). The boards met in a joint session at the 1993 JMM and passed a resolution recommending this move, arguing that the societies have a duty to protect participants from possible discrimination. Shortly afterwards, the Joint Meetings Committee moved the 1995 JMM to San Francisco. The Denver cancellation resulted in a lawsuit from the conference hotels, ultimately settled for $35,000, split evenly between the AMS and MAA. Increased attendance at the San Francisco meeting likely covered most of this cost. -Despite short notice and communication difficulties (this was pre-Facebook!), an informal event at the San Francisco JMM held at a restaurant near the conference hotel drew nearly 100 people. A group of concerned mathematicians felt it would be very useful to have a more visible presence at future JMMs, and formed a steering committee, together with setting up an email group. This group eventually was formalized as Spectra. -Spectra is holding a brainstorming session on Friday, January 12th, 2018, 3pm-5pm, Torrey Pines 2, Marriott Marquis and everyone is welcome to attend and contribute ideas.<|endoftext|> -TITLE: A question about the paper "The Condition Number of a Randomly Perturbed Matrix" -QUESTION [5 upvotes]: My question pertains to this paper by Terence Tao and Van Vu, https://arxiv.org/abs/math/0703307 -Both my questions pertain to the argument presented in this paper in its section 6 (page 5). We are looking at a $n-$dimensional square random matrix $M_n$ satisfying the conditions stated through Definitions 2.15, Definition 2.17 and Theorem 2.18 on page 3. -Now we are saying that lets assume the existence of a unit vector $v \in \mathbb{R}^n$ such that for some $B >10$ we have, $\Vert M_n v\Vert < n^{-B}$. The vector $\tilde{v}$ is created by truncating each coordinate of $v$ to the nearest multiple of $n^{-B-2}$. So if I understand this correctly then we have that for each coordinate $i$, $\vert v_i - \tilde{v}_i\vert \leq \frac{1}{2n^{B+2}}$. This is now supposed to imply a number of things which arent clear to me, - -The paper claims, $0.9 \leq \Vert \tilde{v}\Vert \leq 1.1$. Why? -If I just think directly then we have, $\Vert \tilde{v}\Vert \leq \Vert v + (\tilde{v}-v)\Vert \leq \Vert v\Vert + \sqrt{n} \frac{1}{2n^{B+2}} = 1 + \frac{n^{-B-\frac{3}{2}}}{2}$.For this to match the claim we need, $\frac{n^{-B-\frac{3}{2}}}{2} = 0.1$. But this is now incompatible with the initial statement that we need $B>10$. What am I missing? -The paper claims that the following is also true that, $\Vert M_n \tilde{v}\Vert \leq 2n^{-B}$. Why? -Just as above if I again think just directly then we have, that $\Vert M_n \tilde{v}\Vert = \Vert M_n(v+(\tilde{v}-v))\Vert \leq \Vert M_n v \Vert + \Vert M_n (\tilde{v}-v)\Vert \leq n^{-B} + \Vert M_n \Vert \frac{n^{-B-\frac{3}{2}}}{2}$. One way this can be compatible with the claim is if we have, $\Vert M_n \Vert \leq 2n^{1.5}$. -One might now go back to the ``boundedness" part of Definition 2.15 and the first bullet point of Theorem 2.18 on page 3 to see that with probability $1$ the entries of the matrix $M_n$ are integers bounded as $n^C$ for some constant $C>0$. This gives by Frobenius norms, $\Vert M_n \Vert \leq \Vert M_n \Vert_F \leq n^{1+C}$. So for compatibility with the previous bound we need, $n^{1+C} \leq 2n^{1.5}$. But then such an equation is now an upperbound on the constant $C$ and that is not something that Theorem 2.18 enforced. What am I missing? - -REPLY [9 votes]: One does not need to have $n^{-B-3/2}/2$ to be equal to $0.1$, it is enough for it to be less than or equal to $0.1$, which is certainly the case for $n$ large enough. -Thanks for pointing out this typo (or more precisely, set of typos) in this paper. As you point out, the exponents here are adapted to the case of small $C$ (in particular $C=0$, which is the most frequent case in applications) and require some adjustment for large $C$. [We do remark several times in the paper that the precise exponents are not to be taken too seriously.] Basically, if one replaces all occurrences of $n^2$ in this argument with, say, $n^{2(1+C)}$ (and makes some similar adjustments to some other similar factors such as $n^{-4}$) then this issue will be avoided. - -It is also worth mentioning that the results in this paper have been superseded by subsequent stronger results, e.g. -Tao, Terence; Vu, Van, Smooth analysis of the condition number and the least singular value, Math. Comput. 79, No. 272, 2333-2352 (2010). ZBL1253.65067.<|endoftext|> -TITLE: Evaluation of irreducible representations of the hyperoctahedral group at bipartition $(\lambda,\mu)=([n],\emptyset)$ -QUESTION [7 upvotes]: There is a very simple formulation for the character of irreducible representations of $S_n$ evaluated on an n-cycle, i.e. that it is 0 on all non-hook partitions, and $(-1)^m$ on hooks. Is there an analogous computation for irreducible characters of $B_n$, the hyperoctahedral group, evaluated on signed 2n-cycles? That is, on the conjugacy class indexed by the bipartition $([n],\emptyset)$? - -REPLY [4 votes]: In general, if $(\lambda,\mu)$ is a bipartition of $n$, then - $$ \prod_i(p_{\lambda_i}(x)+p_{\lambda_i}(y))\cdot\prod_j - (p_{\mu_j}(x)-p_{\mu_j}(y)) = \sum_{(\alpha,\beta)} - \chi^{\alpha,\beta}(\lambda,\mu)s_\alpha(x)s_\beta(y), $$ -where $(\alpha,\beta)$ ranges over all bipartitions of $n$ and $\chi^{\alpha,\beta}$ is the irreducible character of $B_n$ indexed by $(\alpha,\beta)$. Setting $(\lambda,\mu)= -(n,\emptyset)$ gives - $$ p_n(x)+p_n(y) = \sum_{(\alpha,\beta)} - \chi^{\alpha,\beta}(n,\emptyset)s_\alpha(x)s_\beta(y). $$ -But (as alluded to in the question) - $$ p_n = \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}, $$ -so - $$ p_n(x)+p_n(y)= \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(x) + - \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(y). $$<|endoftext|> -TITLE: A conjecture regarding prime numbers -QUESTION [59 upvotes]: For $n,m \geq 3$, define $ P_n = \{ p : p$ is a prime such that $ p\leq n$ and $ p \nmid n \}$ . -For example : -$P_3= \{ 2 \}$ -$P_4= \{ 3 \}$ -$P_5= \{ 2, 3 \}$, -$P_6= \{ 5 \}$ and so on. -Claim: $P_n \neq P_m$ for $m\neq n$. -While working on prime numbers I formulated this problem and it has eluded me for a while so I decided to post it here. I am not sure if this is an open problem or solved one. I couldn't find anything that looks like it. -My attempts haven't come to fruition though I have been trying to prove it for a while. If $m$ and $n$ are different primes then it's clear. If $m \geq 2n$, I think we can find a prime in between so that case is also taken care of. My opinion is that it eventually boils down to proving this statement for integers that share the same prime factors. My coding is kind of rusty so would appreciate anybody checking if there is a counterexample to this claim. Any ideas if this might be true or false? Thanks. -PS: I asked this question on mathstackexchage and somebody recommended I post it here as well. Here is the link to the original post -https://math.stackexchange.com/questions/2536176/a-conjecture-regarding-prime-numbers - -REPLY [9 votes]: I am working on a related project involving Grimm's conjecture. The hope is to show that every interval of consecutive composite numbers below $10^{12}$ contains an injective divisor map, see On comparing two almost injective divisor maps for more detail. The upshot is that there are about 700 opportunities for your event to happen (because the map $L(m)$ being largest prime factor of m is often injective, and in your case it won't be) below 2.5 times $10^{10}$, and that your event won't happen because the numbers involved are too close. (Specifically, $L(m)=L(n)=p$, and $m-n =kp$ where $L(k)$ is less than $p$ and usually less than 3, and in those cases $m/p$ and $n/p$ have sufficiently different sets of prime factors.). If I can achieve my aims while offloading data regarding your claim (e.g. a data file of the estimated 3000 $L$ pairs below $10^{12}$), I will do so and report back. If you have several months of computer cycles to spare, I can provide a program so that you can join in the fun, AND get some data on your problem. -Gerhard "Another Opportunity For Communal Computing" Paseman, 2017.11.26.<|endoftext|> -TITLE: Frankl's conjecture restricted to finite topological spaces -QUESTION [12 upvotes]: A finite topological space is a finite family of finite sets that is closed under both union and intersection. -Frankl's conjecture states that for any finite union-closed family of finite sets, other than the family consisting only of the empty set, there exists an element that belongs to at least half of the sets in the family. -Is Frankl's conjecture known to be true when restricted to finite topological spaces? - -REPLY [21 votes]: Consider the smallest nonempty set $S$ in our family $\mathcal F$ and pick any $s\in S$. Let $\mathcal F_0$ be the subfamily of sets not containing $s$ (including $\varnothing$) and $\mathcal F_1$ the subfamily of sets containing $s$. -If $s\not\in A\in\mathcal F$, then $S\cap A$ is a smaller element of $\mathcal F$, so it must be empty. Hence the map $\mathcal F_0\to\mathcal F_1,A\mapsto S\cup A$ is an injection (with inverse $B\mapsto B\setminus S$). Hence $|\mathcal F_1|\geq|\mathcal F_0|,|\mathcal F_1|\geq|\mathcal F|/2$.<|endoftext|> -TITLE: Finding the subgraph with the largest diameter -QUESTION [8 upvotes]: Given an undirected connected graph $G=(V, E)$. Find the induced subgraph $G[W]$ of $G$ with the largest diameter $d$, where the diameter is the largest distance between any pair of vertices. -The diameter of the graph below is 2, because we can get from every node to every other node over at most 2 edges. However, when removing a node, for example C (and it's adjacent edges), the diameter increases to 3, because we need 3 edges from A to E. - -One possible solution would be to generate all subgraphs, calculate the diameter and select the largest. However, the number of subgraphs rises exponentially with the number of vertices, so this is infeasible. - -REPLY [6 votes]: Equivalently, we want to find the longest induced path. According to Wikipedia, it is NP-hard to find it: -It is NP-complete to determine, for a graph G and parameter k, whether the graph has an induced path of length at least k. Garey & Johnson (1979) credit this result to an unpublished communication of Mihalis Yannakakis. -http://en.wikipedia.org/wiki/Induced_path<|endoftext|> -TITLE: Geometric Interpretation of Multiplication in Pure Cubic Number Fields and Beyond -QUESTION [8 upvotes]: I got interested in the question of possible geometric interpretations of the multiplication in algebraic number fields of degree $>2$ (with application to multiplication of units in the ring of algebraic integers of that field), inspired by a talk of Franz Lemmermeyer in honor of Peter Roquette's 90th anniversary some weeks ago. -First, let's consider a pure cubic field $\mathbb{Q}(\sqrt[3]{d})$, an element $(x,y,z)^{\top} = x + d^{1/3}y + d^{2/3}z$ with norm $N((x,y,z)^{\top}) = x^3 + dy^3 + d^2z^3 - 3dxyz$ and multiplication -$$ \left( \begin{array}{c} x_1 \\ y_1 \\ z_1 \end{array} \right) \cdot \left( \begin{array}{c} x_2 \\ y_2 \\ z_2 \end{array} \right) = \left( \begin{array}{c}x_1x_2 + d(y_1z_2+z_1y_2)\\ x_1y_2 + y_1x_2 + dz_1z_2 \\ x_1z_2 + y_1y_2 + z_1x_2 \end{array} \right).$$ -Due to -$$ -\begin{aligned} -& x^3 + dy^3 + d^2z^3 - 3dxyz = \\ -& (x + d^{1/3}y + d^{2/3}z)(x^2 + d^{2/3}y^2 + d^{4/3}z^2 - d^{1/3}xy - d^{2/3}xz - dyz) = 1 -\end{aligned} -$$ -one realizes that the norm-1-surface is funnel-shaped with an asymptotic plane $x + d^{1/3}y + d^{2/3}z = 0$ and the opening of the funnel showing in direction $(d^{2/3}, d^{1/3}, 1)^{\top}$. The angle between the normal of the plane and the funnel direction is $\arccos(3d^{2/3}/(1+d^{2/3}+d^{4/3})$. Cutting the surface with planes parallel to the asymptotic plane yields ellipses. -See figure - -[Norm-1-surface for $d=2$ shown with $(1,1,1)$ (red) and $(1,1,1) \cdot (1,1,1) = (5,4,3)$ (green). The green line indicates the normal of the asymptotic plane, the (nearly not visible) blue line the funnel direction.] -For the number field irrelevant, geometrically yet reasonable case $d=1$ the norm-1-surface gives a rotationally symmetric funnel (cf. - -[Norm-1-surface for $d=1$ shown with powers (multiples) of a point, spiraling around the funnel]) and the group multiplication is just multiplication of the height above the asymptotic plane and complex multiplication (i.e. addition of angles) in the plane perpendicular (parallel to asymptotic plane). -Interestingly this is also valid for the case of general $d$: -Applying the linear transformation $$ -\left( \begin{array}{c} x' \\ y' \\ z' \end{array} \right) = -\left( -\begin{array}{ccc} - 1 & d^{1/3} & d^{2/3} \\ - 1 & -\frac{1}{2}d^{1/3} & -\frac{1}{2}{d^{2/3}} \\ - 0 & \frac{1}{2} \sqrt{3} d^{1/3} & -\frac{1}{2} \sqrt{3} d^{2/3} \\ -\end{array} -\right) -\left( \begin{array}{c} x \\ y \\ z \end{array} \right) -$$ -gives the group multiplication law in the new coordinates as -$$ \left( \begin{array}{c} x'_1 \\ y'_1 \\ z'_1 \end{array} \right) \cdot \left( \begin{array}{c} x'_2 \\ y'_2 \\ z'_2 \end{array} \right) = \left( \begin{array}{c} x'_1x'_2 \\ y'_1y'_2 - z'_1z'_2 \\ y'_2z'_1 + y'_1z'_2 \end{array} \right),$$ -i.e. a simple multiplication in $x'$-direction und complex multiplication (i.e. addition of angles and multiplication of modulus) in the $y'-z'$-plane. -I don't know but would assume that this has been remarked before (I'm not at all an expert in algebraic number theory and don't have an overview of the literature.) I found it pretty interesting that by a pure linear transformation one can bring the group multiplication in such a simple form and one thus has a kind of geometric interpretation of units in the corresponding ring. -My question is whether this observation could be generalized to higher degree ($>3$) algebraic fields, e.g. quartic or quintic fields. -For the cubic field I found the linear transformation above rather easily by hand and a bit of Mathematica calculation. -But the more proper (and for higher degrees, thus higher dimensions necessary) way to proceed would be some sort of normal form theory of 3-tensors, which I'm not familiar with and which I could not find easily. -Here my thoughts: If the multiplication of the algebraic numbers in vector components (generalization of the multiplication given above) is given by a 3-tensor $M$, i.e. $(x_1 \cdot x_2)_k = \sum_{i,j=1}^{\text{degree of the field}} M_{ijk}x_{1i} x_{2j}$, then by applying a linear transformation $x_i = \sum B_{ii'} x'_{i'}$ one gets $M'_{i'j'k'} = \sum_{ijk}B^{-1}_{k'k}M_{ijk}B_{ii'}B_{jj'}$, and we are looking for $B$ such that $M'$ gets as simple as possible (ideally diagonal or with $2\times2$-blocks, or even $4\times 4$-blocks and quaternion multiplication(?); for degree three we find one one-dimensional and one two-dimensional block, see above the multiplication law in the prime coordinates). -As I have it on my hand, here a 3d slice of the norm-1-surface for a pure cubic field of degree 4: - -Any comments or hints are welcome. -Edit 03.12.17 -I have made some progress in finding a method that gives the linear transformation for which the multiplication law has the simplest possible form. For the degree-3-case the result above is reproduced. -For the degree-4-case: Until now I could only handle the special (algebraically irrelevant) case $d=1$, for which the multiplication is given by -$$ \left( \begin{array}{c} x_1 \\ y_1 \\ z_1 \\ w_1 \end{array} \right) \cdot \left( \begin{array}{c} x_2 \\ y_2 \\ z_2 \\w_2 \end{array} \right) = \left( \begin{array}{c}x_1x_2 + (y_1w_2+ z_1z_2 + w_1y_2)\\ x_1y_2 + y_1x_2 + (z_1w_2 +w_1z_2) \\ x_1z_2 + y_1y_2 + z_1x_2 + (w_1w_2) \\ x_1w_2 + y_1z_2+z_1y_2+w_1x_2 \end{array} \right).$$ -(The brakets indicate where for general $d$ a factor $d$ occurs.) -Applying the linear transformation $$ -\left( \begin{array}{c} x' \\ y' \\ z' \\w' \end{array} \right) = -\left( -\begin{array}{ccc} - 1 & 1 & 1 & 1 \\ - 1 & -1 & 1 & -1 \\ -1 & 0 & -1 & 0 \\ -0 & 1 & 0 & -1 \\ -\end{array} -\right) -\left( \begin{array}{c} x \\ y \\ z \\w \end{array} \right) -$$ -gives the group multiplication law in the new coordinates as -$$ \left( \begin{array}{c} x'_1 \\ y'_1 \\ z'_1 \\w'_1 \end{array} \right) \cdot \left( \begin{array}{c} x'_2 \\ y'_2 \\ z'_2 \\ w'_2 \end{array} \right) = \left( \begin{array}{c} x'_1x'_2 \\ y'_1y'_2 \\ z'_1z'_2 - w'_1w'_2 \\ z'_1w'_2 + z'_2w'_1 \end{array} \right),$$ -i.e. a simple multiplication in $x'$- and $y'$-directions und complex multiplication (i.e. addition of angles and multiplication of modulus) in the $z'-w'$-plane. -I assume/hope that my method also works for general $d$, but could not yet verify it due to very cumbersome expressions. I will try for special values of $d$ numerically. -Edit 05.12.17 -Following the hint given by Lee Mosher, I have had a look into Stewart/Tall, Algebraic Number Theory and Fermat's Last Theorem, 3rd edition. Chapter 8 contains more or less what I was looking for. -(Lesson learnt for me: Obviously the goal of transforming the multiplication law in the most simple form is far easier achievable by means of algebraic number theory than with my approach of (multi)linear algebra (sort of generalized eigenwert equation for 3-tensors), but it was good to see it worked, at least for low degrees.) - -REPLY [4 votes]: My rep. is not high enough to leave a comment so this has to go here. I comment that I have done a detailed study of matrices similar to those you mention for cubic fields only. A small part of that will be in: A cubic generalization of Brahmagupta's identity, J. Ramanujan Math. Soc., 32, (2017) no. 4, 327-337. Also, see : http://mathoverflow.net/questions/61859/, which is related. bf Edit 29/11/17: I can give a partial answer to the question in the following: Let $$N_{\mathcal{C}}^{(\alpha )} = \left( -\begin{array}{ccc} - u & -a d y & -a d x-b d y \\ - x & u-b x-c y & -c x-d y \\ - y & a x & u-c y \\ -\end{array} -\right) ,$$ where $\mathcal{C}(x, y) = (a, b, c, d)$ is an index form for the cubic field $K = \mathbb{Q}(\delta )$, $\delta \in \mathbb{R} $ satisfies $\mathcal{C}(\delta , 1) = 0$, and $u,x,y$ are coefficients of Belabas' integral basis $\{ 1, \theta , \phi \}$, $\theta = a \delta $, $\phi = a \delta^2 + b \delta $,and $\alpha = u + x \theta + y \phi $. With $\alpha \in K$ or the ring of integers, the matrices commute with one another, their determinant is the norm, their sum can be used to produce the sum $\alpha_1 + \alpha_2$, the trace of the matrix is the trace of $\alpha $, the inverse of the matrix can be used to produce the inverse of the $\alpha \in K$. One extracts the first column and builds $u + x \theta + y \phi \in K$ from it. Now the proofs of all of this should work if instead of letting $u,x,y \in \mathbb{Q}$, we let $u, x, y \in F$, another field for which we know the associated matrices, e.g. the quadratic field $\mathbb{Q}(\sqrt{D})$. From that it should be possible to find matrices that do the same job as these do for the appropriate extension field, e.g. the normal closure of $K$.<|endoftext|> -TITLE: Calculating limits progressively -QUESTION [9 upvotes]: Consider the problem of finding the limit of the following diagram: -$$ \require{AMScd} \begin{CD} -& & & & E -\\ & & & & @VVV -\\ && C @>>> D -\\ & & @VVV -\\A @>>> B -\end{CD} $$ -The abstract definition of the limit involves an adjunction related to collapsing the entire index category to a point. However, one could break this operation into two stages: first collapsing the upper three objects to a point reduces it to -$$ \require{AMScd} \begin{CD} - & & C \times_D E -\\ & & @VVV -\\A @>>> B -\end{CD} $$ -and then we finish computing the limit as $A \times_B (C \times_D E)$. -This is a particularly convenient thing, since it implies a way to work locally with more complicated diagrams where you ultimately want a limit — i.e. take limits or perform other modifications to smaller pieces of the diagram while leaving the rest unchanged. - -However, not every variation works out so nicely. If we try the same thing but instead collapse the middle three objects to a point, the intermediate diagram becomes -$$ \require{AMScd} \begin{CD} - & & C \times_D E -\\ & & @VVV -\\A \times_B C@>>> C -\end{CD} $$ -So, trying to perform this operation isn't local at all; it modifies the value of the diagram at the other two vertices. -To clarify what I mean, this diagram together with the appropriate "cone" is (I believe) universal among all diagrams with "cones" of the form -$$ \require{AMScd} \begin{CD} - & & \bullet &\to& E -\\ & & @VVV @VVV -\\\bullet @>>> \bullet & \to & D -\\ \downarrow & & \downarrow & \searrow @AAA -\\ A @>>> B @<<< C -\end{CD} $$ - -It seems clear what the the abstract theory behind this sort of calculation should be; just factor the usual adjunction into a sequence of adjunctions. -But my interest in such things is very much not in the abstract — these are the sorts of operations one would like to have as a practical calculus of diagrams. -So my question is if such a calculus is known? Is there worked out how to predict and recognize which sorts of operations really should be local? Or for those operations that are not local, to easily work out how the rest of the diagram gets modified? -(and the bonus question: how much of this carries over to homotopy limits?) - -REPLY [5 votes]: This sort of calculus is central to the abstract study of homotopy limits via derivators. See this paper and this one for some examples of "detection lemmas" that decompose limits using Kan extensions in various ways; they all follow from the calculus of homotopy exact squares.<|endoftext|> -TITLE: Sums of four coprime squares -QUESTION [6 upvotes]: The Four Squares Theorem says that every natural number is the sum of four squares in $\mathbb Z$. What is known about coprime representations? Here we call a presentation $n=a^2+b^2+c^2+d^2$ coprime if the g.c.d. of the four numbers $a,b,c,d$ is 1. Does every natural number have a coprime presentation? If not, is there a simple criterion characterising the numbers that have coprime presentations? What is known about the number of different coprime presentations of a given $n$? - -REPLY [12 votes]: Let $R(n)$ denote the number of ways of writing $n$ as a sum of $4$ squares, and $r(n)$ the number of ways where gcd of $(a,b,c,d) =1$. Then grouping representations of $n$ as a sum of $4$ squares according to the gcd of the variables, clearly we have -$$ -R(n) = \sum_{k^2 | n} r(n/k^2), -$$ -and so by Mobius inversion -$$ -r(n) = \sum_{k^2| n} \mu(k) R(n/k^2). -$$ -Now by Jacobi's four square theorem, $R(n)$ is given explicitly as $8$ times a multiplicative function $F(n)$ defined on prime powers by -$$ -F(2^k) = 3 \text{ for all } k\ge 1, -$$ -and, for odd primes $p$, -$$ -F(p^k) = p^k + p^{k-1} + \ldots + 1. -$$ -So $r(n)$ is $8$ times a multiplicative function $f(n)$ which is defined on prime powers by -$$ -f(2)= 3; \ \ f(4) = 2;\ \ f(2^k)=0 \text{ for } k \ge 3, -$$ -and for odd primes $p$ and $k\ge 1$ -$$ -f(p^k) = p^k + p^{k-1}. -$$<|endoftext|> -TITLE: Why do we need model categories? -QUESTION [37 upvotes]: I cannot give a good answer to this question. And -2) Why this definition of model category is the right way to give a philosophy of homotopy theory? Why didn't we use any other definition? -3) Has model category been used substantially in any area not related to algebraic topology? - -REPLY [6 votes]: Extensive answers have already been given in this thread. Just a few remarks here and there. - -I think the question “why we need” assumes something about “we”, and in some extent, about “need”. There are people who study those objects for their own sake, there are those who study applied PDE and have no need of model categories. Even those who enter the domain which can be covered by model categories often work in a specialised situation where more adapted techniques exist (homological algebra and triangulated categories for algebraic geometry being an example). I guess the meta-reason is that at the current historic moment (or maybe that will remain so indefinitely) any description of entities ‘’up to homotopy’’ from scratch cannot be done without having this ‘’up to homotopy’’ notion already defined. One has to break the vicious circle and pull things down from their ‘’platonic’’ world. Which brings me to the answer for -Since categorical philosophy has proven itself to be reasonably natural, introducing categories with weak equivalences is both a natural step and something reflected in many examples. As we now know, this principle of modelling homotopy phenomena is, formally speaking, just as good as any other higher-categorical approach. Working with an arbitrary category with weak equivalences $(C,W)$ can be impossible in practice, but there are many ways to (homotopically) embed it into a model category. One can consider the category of simplicial presheaves on $C$ for instance, with a suitable Bousfield localisation of the projective model structure. (The examples usually present themselves with better, often canonical, embeddings.) One can thus work with the objects of $C$ by performing operations on the bigger model category and verifying that the answer is sensible for $C$. This adds to the explanation as why the notion is quite ubiquitous. -Algebraic geometry (both the field and the community) has different tradition from algebraic topology, yet model categories have found their way here as well. For example, Kontsevich defines a noncommutative space as a suitable DG-category, and considers them up to Morita equivalence. There are a few model structures on DG-categories, which cover both the usual DG- and Morita equivalences, and they have been used to get various results, such as the theorem representing DG-functors between quasicoherent sheaf categories as bimodules. Another issue is the structure of noncommutative cohomological invariants, Hochschild cohomology, Deligne conjecture, and related matters. Many of the results (including laying a foundation of derived algebraic geometry) were obtained by people outside of the core algebraic topology community, with the use of model categories.<|endoftext|> -TITLE: Minimal polynomial of cos(π/n) -QUESTION [25 upvotes]: I know that $\cos(\pi/n)$ is a root of the Chebyshev polynomial $(T_n + 1)$, in fact it is the largest root of that polynomial, but often that polynomial factors. For example, if $n = 2 k$ then $\cos(\pi/n)$ is the largest root of $T_k$, which is a polynomial of lower degree, and if $n = 3$ then $\cos(\pi/n)$ is a root of $2 x - 1$, again lower degree than $T_3 + 1$. -How can I compute, for a given $n$, a polynomial in $\mathbb{Q}[x]$ of minimal degree that $\cos(\pi/n)$ is a root of? - -REPLY [3 votes]: There is a Wikipedia entry dedicated to this, which contains an alternative method to compute the minimal polynomial of $2\cos(\pi/n)$, which is essentially the same as for $\cos(π/n)$. In fact, denoting the minimal polynomial of $2\cos(π/n)$ by $\Psi_n(x)$, we have by the quoted AMM article in Vladimir Dotsenko's answer for odd $n=2k+1$ $$\prod_{d \mid n}\Psi_d(x) = 2\Big(T_{k + 1}(\frac x2) - T_k(\frac x2)\Big)$$ and for even $n=2k$ $$\prod_{d \mid n}\Psi_d(x) = 2\Big(T_{k + 1}(\frac x2) - T_{k-1}(\frac x2)\Big).$$ -But there is in fact no need to calculate the Chebyshev polynomials. Defining for odd $n=2k+1$ -$$\chi_{n}(x):= \sum_{k=0}^m (-1)^{\lfloor k/2\rfloor}\binom {m-\lfloor (k+1)/2\rfloor}{\lfloor k/2\rfloor} x^{m-k},$$ we have directly -$$\prod _{d\mid n}\Psi_{d}( x)=(x-2)\chi_{n}(x),$$ noting that $\Psi_{1}( x)=x-2$. -For even $n=2k$, we just need to sum up two of those polynomials: $$\prod _{d\mid n}\Psi_{d}( x)=(x-2)\Big(\chi_{n+1}(x)+\chi_{n-1}(x)\Big).$$<|endoftext|> -TITLE: Is there a geometric interpretation of skew Schur functions? -QUESTION [11 upvotes]: Consider the cohomology ring of the Grassmannian of k-planes in complex n-space. It has a standard presentation as a quotient of the ring of symmetric functions. In this presentation, the Schur functions are mapped to the Schubert classes, thus have a nice geometric interpretation. -One can generalise the Schur functions to skew-Schur functions. Do these also have a nice geometric interpretation? - -REPLY [9 votes]: This is discussed in Stanley's paper Some combinatorial aspects of the Schubert calculus. Corollary 3.7 says that under the natural isomorphism given by the Borel presentation of $H^*(G/P)$ which sends an ordinary Schur function $s_{\lambda}$ to the class of the Schubert variety $X_{\lambda}$, a skew Schur function $s_{\lambda / \mu}$ is sent to the class of the Richardson variety $X_{\lambda}^{\mu}$. -Note that Stanley calls Richardson varieties skew Schubert varieties in this paper. Unfortunately, the version of the paper that is online at Stanley's website has some printing defects that make some of the pages illegible.<|endoftext|> -TITLE: Algorithm to determine whether there is an injective homomorphism between two Lie algebras -QUESTION [5 upvotes]: Let $\mathfrak{g}$ be a $n$-dimensional Lie algebra and $\mathfrak{h}$ be a $k$-dimensional Lie algebra ($k < n$). The multiplication tables for these Lie algebras are known. Is there a way to show that $\mathfrak{h}$ is isomorphic to a subalgebra of $\mathfrak{g}$? Or there is no "algorithm"? -To be more specific, I'd like to know the answer to this question if $\mathfrak{g}$ is the 8-dimensional real Lie algebra $\mathfrak{g}$ with multiplication table - -and $\mathfrak{h}$ is some 3-dimensional real Lie algebra from the Bianchi classification, e.g. -$$[E_1,E_2] = E_2, [E_2,E_3] = 0, [E_3,E_1] = -E_3.$$ - -REPLY [5 votes]: The new question amounts to classifying, up to isomorphism, 3-dimensional subalgebras of the real Lie algebra $\mathfrak{su}(2,1)$. -If I'm correct, the answer is: - -simple split subalgebras (isomorphic to $\mathfrak{sl}_2$). -simple non-split subalgebras (isomorphic to $\mathfrak{so}_3$). -Heisenberg subalgebras -subalgebras with basis $(T,X,Y)$ with $[T,X]=X$, $[T,Y]=2Y$, $[X,Y]=0$. - -In particular, this discards the Lie algebra with basis $(T,X,Y)$, $[T,X]=X$, $[T,Y]=\lambda Y$, $[X,Y]=0$ whenever $\lambda\notin\{2,\frac12\}$ (the one written as an example by the OP is $\lambda=1$). -To prove this: the first part is to realize each of these algebras. The first two appear as $\mathfrak{su}(1,1)$ and $\mathfrak{su}(2)$. The last two will pop up in the proof. -Consider the action of a 3-dimensional subalgebra $\mathfrak{h}$ on $\mathbf{C}^3$. If it is irreducible, $\mathfrak{h}$ is semisimple and this falls in one of the first two cases. -So it preserves a line or plane, and preserves its orthogonal too, so preserves a line. If the line is non-isotropic, its orthogonal is a stable supplement, and thus $\mathfrak{h}$ falls into $\mathfrak{s}(\mathfrak{u}(1)\times\mathfrak{u}(2))$ or $\mathfrak{s}(\mathfrak{u}(1)\times\mathfrak{u}(1,1))$, which are 4-dimensional and isomorphic to a product of an abelian 1-dimensional Lie algebra and a simple 3-dimensional Lie algebra. This forces $\mathfrak{h}$ to be simple, and this case is done (of course, if we classify up to conjugation, this is a distinct case! but we don't.) -The last case is when it preserves an isotropic line. In this case let us say that the hermitian form is given by the matrix $\begin{pmatrix} 0 &0&1\\0&1&0\\1&0&0\end{pmatrix}$. Then $\mathfrak{u}(2,1)$ consists of the matrices $$\begin{pmatrix} x&y&it\\z&iu&-\bar{y}\\iv&-\bar{z}&-\bar{x}\end{pmatrix}\quad x,y,z\in\mathbf{C},\;t,u,v\in\mathbf{R},$$ -that is, which are skew-hermitian with respect to the antidiagonal; $\mathfrak{u}(2,1)$ consists of those trace zero matrices therein, i.e., for which $iu=-x+\bar{x}$. Since all isotropic vectors are in the same orbit, it is enough to consider the case when $\mathfrak{h}$ preserves the line $(\mathbf{C},0,0)$, that is, is contained in the 5-dimensional subalgebra $\mathfrak{w}$ consisting of those matrices -$$\begin{pmatrix} x&y&it\\0&-x+\bar{x}&-\bar{y}\\0&0&-\bar{x}\end{pmatrix}\quad x,y\in\mathbf{C},\;t\in\mathbf{R}.$$ -So we have to classify those 3-dimensional subalgebras of $\mathfrak{w}$ up to isomorphism. Let $\mathfrak{h}$ be solvable and 3-dimensional. We discuss on the dimension $D\in\{0,1,2\}$ of $[\mathfrak{h},\mathfrak{h}]$. -The following facts hold for an arbitrary solvable 3-dimensional Lie algebra over a field of characteristic zero (and certainly also all characteristic except maybe 2,3): -$D=0$ iff $\mathfrak{h}$ is abelian, $D=1$ iff $\mathfrak{h}$ is Heisenberg or product of a 1-dimensional abelian with the non-abelian 2-dimensional Lie algebra (according to whether the 1-dimensional derived subalgebra is central or not). So we have to discard the abelian case, and to treat the cases when $D=2$. -For the abelian case, observe that for a matrix as above, if $x\neq 0$ it is diagonalizable with distinct eigenvalues; then its centralizer is diagonalizable but is constrained to have trace zero, so has dimension $\le 2$. The for $x=0$ what remains precisely consists of a Heisenberg subalgebra. -Let us discard the non-abelian product case; it has 1-dimensional center. Consider a matrix as above: as in the abelian case, if $x\neq 0$, it has abelian centralizer, which is now excluded. So $x=0$. If $x=0$ and $y\neq 0$, this is conjugate to a Jordan nilpotent matrix and again has abelian centralizer. So the center is the line generated by $E_{13}$. So $\mathfrak{h}$ is contained in the centralizer in $\mathfrak{w}$ of $E_{13}$ which consists of those matrices as above with $x$ purely imaginary, namely those -$$\begin{pmatrix} is&y&it\\0&-2is&-\bar{y}\\0&0&is\end{pmatrix}\quad y\in\mathbf{C},\;t,s\in\mathbf{R}.$$ -This 4-dimensional Lie algebra has a basis $T,X,Y,Z$ with $Z$ central and $[T,X]=Y$, $[T,Y]=-X$, $[X,Y]=Z$. It is easy to check that its only 3-dimensional subalgebra is the Heisenberg one with basis $(X,Y,Z)$ (here we use that we work with reals). So this case is excluded. -Finally let us treat the $D=2$ case. Then $[\mathfrak{h},\mathfrak{h}]$ is contained in $[\mathfrak{w},\mathfrak{w}]$ (which is Heisenberg), and is abelian. In a Heisenberg Lie algebra, the 2-dimensional subalgebras consists of those planes containing the center. Hence $[\mathfrak{h},\mathfrak{h}]$ consists, for some nonzero complex number $y_0$, of those -$$\begin{pmatrix} 0&\lambda y_0&it\\0&0&-\lambda\bar{y_0}\\0&0&0\end{pmatrix}\quad t,\lambda\in\mathbf{R}.$$ -Then $\mathfrak{h}$ is contained in the normalizer of this 2-dimensional subalgebra in $\mathfrak{w}$, which consists of those matrices as above with $x$ real, namely -$$\begin{pmatrix} x&y&it\\0&0&-\bar{y}\\0&0&-x\end{pmatrix}\quad y\in\mathbf{C},\;x,t\in\mathbf{R}.$$ -So $\mathfrak{h}$ has a basis $(T,X,Y)$, with -$$T=\begin{pmatrix} 1&y&it\\0&0&-\bar{y}\\0&0&-1\end{pmatrix},X=\begin{pmatrix} 0&y_0&0\\0&0&-\bar{y_0}\\0&0&0\end{pmatrix},Y=\begin{pmatrix} 0&0&i\\0&0&0\\0&0&0\end{pmatrix},$$ -which indeed satisfy $[T,X]=X,[T,Y]=2Y$, $[X,Y]=0$.<|endoftext|> -TITLE: For which functions is the (generalized) Riemann hypothesis known? -QUESTION [6 upvotes]: In [1], Lin Weng shows that the Riemann hypothesis (RH) holds for certain linear combinations of shifted completed Riemann zetas. Further, Deligne's result on the Riemann hypothesis for function fields gives a RH for the latter. I would like to know whether these are all examples for which the RH is known. -To be precise, we say that a meromorphic function on $\mathbb C$ -satisfies the generalized Riemann hyporthesis (genRH), if all of its poles and zeroes lie in the union of a finite number of vertical lines $c+i\mathbb R$ and the real line. Are there any known examples, other than those in [1], or those derived from Deligne's work, for which the genRH has been proven? -Of particular interest would be a function which can be written as a Dirichlet series for $\Re(s)>>0$. -[1] Symmetries and the Riemann hypothesis. Algebraic and arithmetic structures of moduli spaces (Sapporo 2007), 173–223, -Adv. Stud. Pure Math., 58, Math. Soc. Japan, Tokyo, 2010. - -REPLY [2 votes]: The answer is that there's trivially many, even if you want it to be defined by a Dirichlet series. Dirichlet polynomials (such as $1+2^{-s}$) will work. But that's cheap because these are too small (bounded on vertical lines). Starting with any Dirichlet series $f(s)=\sum a_n n^{-s}$ that analytically continues to an entire function, you can take the function $\exp f(s)$, which can also be defined as a Dirichlet series and is nowhere zero. But that's cheap because these are too big (not order 1 functions) and hence cannot be defined as a product over their zeros. -If you wanted functions that are neither too big nor too small then conjecturally all L-functions satisfy this, but there isn't a single function that you can prove this for, and for all we know no such function exists. There is an earlier mathoverflow question that defines this precisely and asks for a function satisfying the weaker property of having a "zero-free region", which is provided in the answer.<|endoftext|> -TITLE: Counterexamples to gluing complexes of sheaves -QUESTION [13 upvotes]: Note: I asked the question below last week on MathSE but received no answer. -Background: -I have read the claim that perverse sheaves behave more like sheaves than like complexes of sheaves. This refers to the fact that they can be glued. -For instance, suppose that $X$ is a complex analytic space and $P_1^{\bullet}, P_2^{\bullet}$ are perverse sheaves defined on the open sets $U_1, U_2$ respectively. Then if we are given an isomorphism $\alpha_{12}: P_1^{\bullet}|_{U_1 \cap U_2} = P_2^{\bullet}|_{U_1 \cap U_2},$ then there exists a unique (up to canonical isomorphism) perverse sheaf $P^{\bullet}$ defined on $U_1 \cup U_2$ with the property that there are isomorphisms $\beta_i:P^{\bullet}|_{U_i} \to P_i^{\bullet}$ with ${\beta_2}|_{U_1 \cap U_2}={(\alpha_{12} \circ \beta_1) }|_{U_1 \cap U_2}.$ -More generally, if one has an open cover $\{U_i\}$ of $X$ and perverse sheaves $P_i^{\bullet}$ of with isomorphisms on the overlaps $U_i \cap U_j$ satisfying the co-cycle condition, then this data glue in the usual way. -My question: -What goes wrong if one tries to glue ordinary complexes of sheaves? Are there counterexamples showing that the gluing property cannot hold in $C^{\bullet}(Sh(X))$ (the category of sheaves on $X$) or in $D^b(X)$ (the bounded derived category of sheaves on $X$)? - -REPLY [3 votes]: I'm not sure if you are still interested in this question. Actually for an open cover $\{U_i\}$ and complexes of sheaves on each $U_i$, we could give the "higher" descent data and "higher" cocycle conditions in terms of twisted complexes, which was first introduced by O’Brian, Toledo and Tong in late 1970's and 1980's. I have written a paper to describe twisted complexes as a dg-enhancement of the derived category of sheaves on the space (sorry for the self-citation). Later in a joint paper with Block and Holstein we proved that twisted complexes indeed give the homotopy limit of cosimplicial diagrams of dg-categories coming from geometry (we can consider homotopy limit as a more formal way to say descent data). -I would like to mention that in a recent preprint, the authors generalized the main result in the paper of Block, Holstein and me to arbitrary cosimplicial diagrams of dg-categories.<|endoftext|> -TITLE: The elliptic regularity theorem for differential operators with variable coefficients -QUESTION [6 upvotes]: I'm following the book "Introduction to the theory of distributions" by Friedlander and Joshi. There is the following result p. 109 -Theorem (8.6.1). Let $X \subset \mathbb{R}^n$ be an open set, and let $P$ be an elliptic operator with constant coefficients. Then -$$\mathrm{singsupp}(u)=\mathrm{singsupp}(Pu)$$ -As an observation after the demonstration says: -"This principle, applied to Schwartz kernels and backed by an appropriate construction, gives the elliptic regularity theorem for differential operators with variable coefficients." -Would you give me references for this more general case? Is the theory of pseudo-differential operators necessary? - -REPLY [4 votes]: I was able to find the right reference. This construction is present in the book "Introduction to pseudo-differential and Fourier Integral volume 1" by J.F. Treves. More specifically the sections are as follows - -Parametrices of Elliptic Equations -Definition and Continuity of the "Standard" Pseudodifferential Operators -in an Open Subset of Euclidean Space. Pseudodifferential Operators Are -Pseudolocal - -up to page 12, there is the following result: -Lemma (2.2). Let $P$ denote a differential operator with variable coefficients in $\Omega$. Suppose that there is a very $K: \mathcal{E}'(\Omega) \longrightarrow \mathcal{D}'(\Omega)$ regular operator such that $KP-I$ is regularizing (which is sometimes expressed by saying that $K$ is a left parametrix of $P$). Then $P$ is hypoelliptic, i.e., it has the following property: Given any open set $U$ of $\Omega$, then every distribution $u$ in $U$ such that $Pu \in C^{\infty}(U)$ is a $C^\infty$ function in $U$. -In other worlds this lemma says that -$$\mathrm{singsupp}(u)=\mathrm{singsupp}(Pu)$$ -The theory of pseudo-differential operators is not necessary, in fact this construction is a particular introduction to it. -PS. To understand this construction, it is necessary to study Chapter 6 (Schwartz kernels and kernel theorem) of the book "Introduction to the theory of distributions" by Friedlander and Joshi.<|endoftext|> -TITLE: How do computer algebra packages like Sagemath implement rank of a matrix -QUESTION [11 upvotes]: I am not sure if this is the right place to ask this question, but I believe there will be people here who do computations on computer algebra packages like Sage in their work. -I have been using Sagemath to perform some matrix rank computations. It turns up a few bizarre results occasionally. -For example, I had to find the rank of a matrix ($100 \times 150$) with large integer entries (entries of magnitude in the range of $1$ to $10^{15}$). When I wrote the code with the matrix M declared as matrix(ZZ, R, C), or as matrix(QQ, R, C), it returns a rank of around 90 (which I believe is correct), whereas if I declare the matrix over the reals as matrix(RR, R, C), it returns a rank of around 50, which I believe is too low based on some conjectures I have. -So, overall I am curious, what are the standard way(s) to implement rank computation (and does it differ based on reals, or rationals) and where can I read more about these? If these issues arise due to precision errors, how can I get around them? And more importantly, how do I know beforehand that my computation is susceptible to precision errors? -(I tried looking up the source code of Sagemath a couple of times, but I was quickly lost, so I hope someone can point me to the precise documentation/source code) - -REPLY [8 votes]: I don't know what algorithm Sage actually uses, but computing rank over the integers is fun and easy: Complexity of computing matrix rank over integers . It is NOT so easy if you want good running time, and for that there are a number of papers of Arne Storjohann, which show that one can do it asymptotically as fast as for real matrices (a surprising result, in view of coefficient blow-up). Storjohann has actually implemented his algorithms, and I am sure Sage uses this or something like it. -Chen, Zhuliang; Storjohann, Arne, A BLAS based C library for exact linear algebra on integer matrices, Kauers, Manuel (ed.), Proceedings of the 2005 international symposium on symbolic and algebraic computation, ISSAC’05, Beijing, China, July 24--27, 2005. New York, NY: ACM Press (ISBN 1-59593-095-7). 92-99 (2005). ZBL1360.65086. -As for the reals, the fastest way to compute rank is to compute the singular value decomposition, and throw away singular values below some cutoff (probably around $10^{-6}$). As pointed out by many in the comments, rank is very unstable, so unless you want the "pseudo-rank" (as above, defined by the size of the singular values), don't go there.<|endoftext|> -TITLE: Propositional vs Definitional extentionality in type theory -QUESTION [5 upvotes]: There are essentially two ways to impose extentionality on a type theory (I know, it is not very fashionable to impose extentionality these days, but please, bear with me) you can either have a "propositional extentionality axiom" like UIP (uniqueness of identity proof) which says that every two inhabitant of Id(x,y) are propositionally equals, or a "reflexion rules" which says that if you have an inhabitant $a:Id(x,y)$ then $x=y$ and $a=r(x)$. (and if you do it for all types you might not even it to ask that $a=r(x)$) -One expect that the two are equivalent in some precise sens: clearly the reflexion principle implies UIP, so if you can prove something using UIP you can also prove it using the reflection principle, but the converse is a bit harder and one only expect that if we are able to proves that x=y using the reflection principle then one can prove that there is an inhabitant of Id(x,y) using UIP. -I assume it can be formalized by somehow quotienting types & terms using the equivalence relation defined by $\exists v \in Id(x,y)$ but I would be interested to see it done concretely. -So I would be interested by any references making some cases of this into rigorous theorem (for various kind of type theory). If it exists I'm also interested in situation where you only impose UIP/Reflection on some types and not on all types. -(and I don't really care about the specific form of the axioms used to impose extentionality other than the distinction between definitional and propositional) - -REPLY [3 votes]: Martin Hofmann proves in his thesis (theorem 3.2.5) that whenever we have a type $\Gamma \vdash A$ in intensional type theory (ITT) and a term $|\Gamma| \vdash a : |A|$ in ETT there is a term $\Gamma \vdash a' : A$ such that $|\Gamma| \vdash |a'| \equiv a : |A|$. In particular, this implies that whenever some statement is provable in ETT it is also provable in ITT. -This is related to the question that I'm working on, so let me discuss it if you don't mind. Hofmann's theorem implies that ITT and ETT are equivalent in some precise sense. We can define a notion of weak equivalences between type theories (of course, we need to define a category of type theories first) either as a map with some syntactic properties similar to the one described above or as a map that induces (Quillen) equivalence of categories of models (these definitions are equivalent). There are many natural examples of weak equivalences between type theories. This is a work in progress and will be a part of my thesis.<|endoftext|> -TITLE: Using a known result without a specific reference -QUESTION [20 upvotes]: This is a question of mathematical writing. Let me know if it would be better suited to academia.SE. -I am writing a paper in invariant theory. It uses some slightly heavy commutative algebra. There are a few points where I use facts of which I am convinced, and I believe they are widely known, but I am not sure how to look for a print reference. For example: -1) "flat of relative dimension $n$" is preserved under arbitrary base change -2) the functor of invariants commutes with flat base change -I learned (1) from Ravi Vakil's algebraic geometry notes (exercise 24.5.L). I learned (2) from the thesis of my coauthor (the proof is easy). There are other results like this I'm not thinking of right now that I probably learned from the Stacks Project. -I guess a thesis can be cited in print if needed, but it seems inappropriate to cite either Vakil's notes or the Stacks Project in a print article since they have not undergone formal peer-review, as authoritative as they are. I imagine I might be able to find one or both of these things in EGA, but then again, I might not, and I would spend a lot of time looking. As a young scholar, I do not yet feel I have a beat on what is regarded as common knowledge. My question is: - -What guidelines does one use to decide if results such as these require a reference in an article or can be used as common knowledge? - -REPLY [12 votes]: I agree with RvDdB's answer, but want to add some thoughts. To quote from another SE answer of mine: -First, my general philosophy is that one should try to make papers reasonably accessible to young people who have not spent months or years working on this specific problem. I generally feel that most math papers should do a better job of providing references than they do. -There I also espouse the view that writing (and reading) papers and getting feedback (via reviews, discussions, questions during/after talks) provides a process whereby you learn what is more common knowledge to other people in your field and what is less common. When you are young (and sometimes even when you're not), you probably don't have a good sense of this, so it's good to err on the safe side of including a citation if you're not sure, and often a referee will help you out by suggesting a citation is needed (or less often, not needed). -But one rule of thumb is: think about what you knew before you started working on this specific problem. You may need to modify this depending on your situation (e.g., if you started working on the problem before you knew anything about the field), but perhaps thinking about this sentiment is still helpful. -Anyway, the main thing I wanted to add to the already existing answers is: think about your intended audience. I would give this advice for writing the paper in general, and it's also applicable to your particular question. For instance, if I'm writing a paper targeted at people who do local representation theory I might not bother to reference some basic properties of the local Jacquet-Langlands correspondence known 30-40 years ago (e.g., the correspondence for 1-dimensionals of division algebras), but if I think my paper should be of interest to, say, people in classical modular forms who aren't all experts in local representation theory, then I definitely will. (I'm not saying that the amount of time passed -since a result was known should be the only factor here--the older less well known facts I would also cite.) -Do you think your paper will/should be of interest to people who aren't intimately familiar with properties of base change? If so, try to provide a citation.<|endoftext|> -TITLE: Could we always find a line to intersect transversally with a given compact manifold? -QUESTION [7 upvotes]: This problem may be an embarrassing one, but I could not prove it even for the $1$ dimensional case. Here is the problem: - -Question 1. $M$ is a compact $n$-dimensional smooth manifold in $R^{n+1}$. Take a point $p\notin M$. Prove there is always a line $l_p$ pass $p$ and $l_p\cap M\neq \emptyset$, and $l_p$ intersect transversally with $M$. - -You can naturally generalise it to: - -Question 2. $M$ is a compact $n$-dimensional smooth manifold in $R^{n+m}$. Take a point $p\notin M$. Prove $\forall 1\leq k\leq m$, there is always a hyperplane $P_p, dim(P_p)=k$ pass $p$ and $P_p\cap M\neq \emptyset$, and $P_p$ intersect transversally with $M$. - -Thanking for Piotr pointing out, assume "transverse" means "the tangent spaces intersect only at 0". -We focus on question 1 for simplicity. -Even in $1$ dimension it is not easy at least for me, warning: a line $l$ pass $p$ may be intersect $M$ at several points combine a set $A_l$, $A_l$ could be finite, countable or even it is not countable (consider $M$ is induced by a smooth function for which the zeros set is Cantor set.)... And if there is one point $a\in A_l$, $l$ is tangent with the tangent line of $M$ at $a$, then $l$ is not intersect transversally with $M$. -My attempt: -I could use a dimensional argument and Sard's theorem to establish a similar result but instead of a fix point $p$, we prove for generic point in $R^{n+1}$ which is not in $M$ we can choose such a line. -So it seems reasonable to develop the dimensional technique to attach the question 1, in 1 dimensional, it will relate to investigate the ordinary differential equation: -$$\frac{f(x)-b}{x-a}=f'(x)$$ -Where $p=(a,b)$, $M$ have a parameterization $M=\{x,f(x)\}$. If there is a counterexample for the question 1, then there is another solution which satisfied the ODE in the sense: -At least for every line $l$ there is a intersection point $a_l\in l\cap M$, $f$ satisfied ODE at $a_l$. -This is just like the uniqueness of the solution of such a ODE is destroyed at some subspace of a line which has some special linear structure, I do not know if this point of view will be helpful. -I will appreciate for any useful answers and comments. - -REPLY [5 votes]: For the codimension 1 case. -Using Thom transversality theorem. -Consider the maps $f_s:\mathbb{R} \to \mathbb{R}^n$ parametrized by $s \in S^{n-1}$ and given by $f_s(t) = p + t \cdot s$. The map $F(s,t) = f_s(t)$, $F:S^{n-1} \times \mathbb{R} \to \mathbb{R}^n$ is clearly transverse to $M$, thus Thom's transversality says that $f_s$ is transverse to $M$ for almost all $s$. Now it suffices to prove that for an open set in $S^{n-1}$, the line given by $f_s$ intersects $M$. Proven below. -Using Sard's theorem directly. -Thom's transversality is usually proven using Sard's theorem. Here is the idea. -Consider the projection $\Pi:\mathbb{R}^n \setminus \{p\}\to S^{n-1}_p$ onto a sphere centered at $p$. A line $l_p$ through $p$ intersects $M$ transversally if the two points $l_p \cap S^{n-1}_p$ are regular values of $\Pi$ (indeed, the critical points of $\Pi$ are exactly the points $x \in M$ at which the normal $\vec n_x$ is perpendicular to the radial direction (with respect to $p$)). By Sard's theorem, the set of regular values is dense in $S^{n-1}_p$. -We need to choose any point $s$ on the sphere for which both $s$ and $-s$ are regular values, and the line $f_s$ through $p$ and $s$ actually intersects $M$. It suffices to prove that the set of points $s$ for which this line intersects $M$ contains an open set. We could now use the Jordan-Brouwer Separation Theorem and we would be done, but we can do it more directly (and in a way that seems to generalize). -The set of points $s \in S$ for which $f_s$ intersects $s$ has nonempty interior. -For each point $q \notin M$ the projection $\Pi:M \to S_{q,\varepsilon_q}^{n-1}$ onto the sphere centered at $q$, of radius $\varepsilon_q$ small enough so that the sphere does not intersect $M$, has some (topological) degree $d_q$. It is easy to check that if one takes any point $x \in M$ and considers the points $x \pm \delta \vec n_x$ for small $\delta$, the degrees of the corresponding maps differ by $1$. It follows that we can find a point $q$ for which $d_q \neq d_p$, which guarantees that for every point $q'$ in a small open ball $B$ around $q$ (all these points have same degree $d_q$), the line joining $p$ and $q'$ intersects $M$. Projection of $B$ on $S_p^{n-1}$ is an open set which we sought. -For the general case (partial solution). -I think a similar reasoning should work, however, notice that for $k < m$ we cannot make $P_p$ intersect transversally with $M$ because of dimensional reasons: the dimensions of $M$ and $P_p$ don't add up to at least $n+m$. Recall that transversality implies Thus, either (1) you want to consider $k \geq m$, or (2) define "transversal intersection" for such manifolds saying that the tangent spaces have to intersect at an empty set. -Also, for $k>n$ we can just take any plane $P_p$ which works for $k=m$ and just extend it to a $k$-dimensional plane. -Assuming $k = m$. -A similar reasoning should work for $f_s:\mathbb{R}^m \to \mathbb{R}^{n+m}$ with $s = (s_1, \ldots, s_m)$ going over all families of pairwise perpendicular unit vectors, and $f_s(t_1,\ldots,t_m) = p+\sum_{j=1}^m t_i \cdot s_i$. Thom's transversality says that for almost all choices of $s$, the plane $f_s$ is transverse to $M$. -The nonempty interior issue. -The only thing left is to prove that the set of $s$ for which the intersection is nonempty has nonempty interior. Last time we proved that there is a zero-dimensional sphere containing $p$, namely $\{p,q\}$, which has nonzero linking number with $M$, and by deforming if to spheres $\{p,q'\}$ and taking lines through pairs $p,q'$, we got an open set of parameters for which the line intersects $M$. -Here should be able to do a similar trick by finding a $m-1$-dimensional sphere with nonzero linking number with $M$. The ball that bounds that sphere has to intersect $M$, thus the plane $P$ containing the sphere has to intersect $M$. By perturbing the sphere we get spheres with the same linking numbers, and get all the planes that lie in a neighbourhood of $P$; in particular, we get an open set of parameters $s$ for which $f_s$ intersects $M$. -Well, we don't actually need a round sphere, but we do need a smooth sphere that lies in a $m$-dimensional plane. There's some trickery needed to do this, but I am sure something like this can be done. -Maybe somebody else can do it better? -For $k -TITLE: Acyclic aspherical spaces with acyclic fundamental groups -QUESTION [10 upvotes]: A space $X$ (by which I mean a CW complex) is acyclic if its reduced singular homology $\tilde H_\ast(X;\Bbb Z)$ is trivial in all degrees. -A discrete group $\pi$ is said to be acyclic if its classifying space $B\pi$ is acyclic. -A space $X$ is aspherical if its universal cover is contractible. A space $X$ is non-aspherical if its universal cover isn't contractible. -Question: Does there exist a non-aspherical, acyclic space $X$ whose fundamental group $\pi$ is also acyclic? - -REPLY [16 votes]: Yes, such things exist. -Take any finitely presented infinite acyclic group $G$, for example, Higman's group. -It is a theorem by Kervaire (''Smooth homology spheres and their fundamental groups'') that for each $n \geq 5$, there is an integral homology sphere $M^n$ with fundamental group $G$. Consider $X=M-\ast$. The integral homology of $X$ is trivial, the fundamental group of $X$ is $G$. Suppose that $X$ were aspherical. Observe that $M\simeq X \cup_f D^n$ is obtained by attaching an $n$-cell. Since $\pi_{n-1} (X)=0$, $f$ is nullhomotopic, and hence $M \simeq S^n \vee X$. It follows that the universal cover of $M$ has the homotopy type of $\tilde{X}$ with infinitely many $S^n$'s attached (since $G$ is infinite). So $H_n (\tilde{M};\mathbb{Z}) \neq 0$, a contradiction, because $\tilde{M}$ is a noncompact $n$-manifold.<|endoftext|> -TITLE: Semi group of polynomials which all roots lie on the unit circle -QUESTION [9 upvotes]: Let $X=\{f\in \mathbb{C}[z]\mid |z| \neq 1 \implies f(z) \neq 0\} $. -The motivation for consideration of such an $X$ is the the concept of Lee-Yang polynomials. -With the standard multiplication, $X$ is an Abelian semigroup with cancellation property. -Let $G$ be the Grothendieck group associated with $X$. -Is there a well known group which is isomorphic to $G$? In other words, is there an alternative formulation of $G$ in terms of some well known group? Is there a natural topology on $G$ which makes it a locally compact topological group? - -REPLY [6 votes]: Every locally compact group topology on $G$ makes the constants $\mathbf{C}^*$ an open subgroup. -Indeed, as observed in Fetisov's answer, $G$ is a direct product $\mathbf{C}^*\times A$ with $A$ free abelian. So $\mathbf{C}^*$ is the intersection of all kernels of homomorphisms $G\to\mathbf{Z}$. -R. Alperin (1980) proved that every homomorphism from a locally compact group into $\mathbf{Z}$ is continuous. It follows that $\mathbf{C}^*$ is closed in $G$. -So, working in the quotient, it is enough to prove: the only locally compact group topology $T$ on a free abelian group $A$ is the discrete one. Indeed, every subgroup of $A$ is free abelian. So $(A,T)$ cannot have a non-trivial compact subgroup (e.g., using again Alperin's result). It follows (by Hilbert's fifth problem, but which was previously known, probably due to Pontryagin) that the zero component $(A,T)^\circ$ is isomorphic to $\mathbf{R}^k$, and again this forces $k=0$, so $A$ is discrete. -Now $\mathbf{C}^*$ admits plenty of exotic locally compact group topologies, but any reasonable assumption (e.g., $\sigma$-compact + evaluation at some point is continuous) will force the topology to be the canonical one.<|endoftext|> -TITLE: Finiteness aspects of Deligne cohomology -QUESTION [8 upvotes]: Say $X$ is a smooth projective variety over $\mathbf{C}$, and $\mathcal{X} = X^{\rm an}$ its $\mathbf{C}$-analytic space. -For what integers $i,d$ is the Deligne cohomology $H^i_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(d))$ of $\mathcal{X}$, an extension of a finitely generated abelian group by a compact group? - -REPLY [4 votes]: The following ought to answer your question. -Prop. Let $\mathcal{X}$ be a smooth proper complex analytic space. We have a long exact sequence of abelian groups: -$$\cdots \to H^i_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(j))\to H^i(\mathcal{X},\mathbf{Z})\to H^i(\mathcal{X},\mathbf{C})/F^jH^i(\mathcal{X},\mathbf{C})\to H^{i+1}_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(j))\to\cdots$$ -where $F^{\bullet}H^i(\mathcal{X},\mathbf{C})$ is the Hodge filtration on $H^i(\mathcal{X},\mathbf{C})$. -Sketch of Proof. -By design of the Deligne complex $\mathbf{Z}(j)_{\mathcal{D}}$, we have a triangle in $D(\text{Ab})$: -$$\Omega_{\mathcal{X}/\mathbf{C}}^{\le j-1}[1]\to \mathbf{Z}(j)_{\mathcal{D}}\to\mathbf{Z}.$$ -Form hypercohomology and prove $\mathbb{H}^i(\mathcal{X},\Omega_{\mathcal{X}/\mathbf{C}}^{\le j-1}) = H^i(\mathcal{X},\mathbf{C})/F^jH^i(\mathcal{X},\mathbf{C})$. -To see this, use the triangle: -$$\Omega_{\mathcal{X}/\mathbf{C}}^{\ge j}\to \Omega_{\mathcal{X}/\mathbf{C}}^{\bullet}\to \Omega_{\mathcal{X}/\mathbf{C}}^{\le j-1}$$ -and reduce to check $\mathbb{H}^i(\mathcal{X},\Omega_{\mathcal{X}/\mathbf{C}}^{\ge j})=F^jH^i(\mathcal{X},\mathbf{C})$, which follows almost by definition of the Hodge filtration. QED -In particular, for $i=2j$, we get: -$$0\to J^j(\mathcal{X})\to H^{2j}_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(j))\to \text{Hdg}^{j}(\mathcal{X},\mathbf{Z})\to 0$$ -upon being careful about grading conventions (eg. I have seen $J^n$ often denoted $J^{2n}$, etc.) -Rem. Note that $H^i(\mathcal{X},\mathbf{Z})$ is always finitely generated, hence your question really is about when the image of $H^i(\mathcal{X},\mathbf{Z})$ in $H^i(\mathcal{X},\mathbf{C})/F^jH^i(\mathcal{X},\mathbf{C})$ under the above edge map, is a full lattice. If $i$ is odd this is true.<|endoftext|> -TITLE: Fanos with $\chi_{top} = 0$ -QUESTION [8 upvotes]: Question 1:Do there exist smooth projective Fano varieties over $\mathbb{C}$ with topological Euler characteristic $0$? -Question 2:If so what is the lowest dimension in which such examples occur? - -REPLY [10 votes]: Every odd $n$-dimensional smooth (2,2)-complete intersection has Euler characteristic 0. I'll leave this website as evidence (http://pbelmans.ncag.info/cohomology-tables/), although this result should also follow from a careful Chern class calculation of $c_n(T_X)$. -When $n\ge 3$ these are Fano varieties. By Remy's argument $n=3$ is the smallest dimension where this can happen. -The reason for this, I believe, is that the intermediate Jacobian of these Fano threefolds is the Jacobian of a hyperrelliptic curve of genus (n+1)/2 (this should be contained in Miles Reid's thesis http://homepages.warwick.ac.uk/~masda/3folds/qu.pdf). Thus the middle cohomology has dimension n+1. The sum of the dimensions of the even cohomology groups is also n+1, which easily follows from the Lefschetz theorems.<|endoftext|> -TITLE: Pro-representability -QUESTION [7 upvotes]: In his descent II Bourbaki paper, Grothendieck lays out some criteria for a functor $F:C\rightarrow\mathrm{Set}$ to be strictly pro-representable; i.e. isomorphic to an inductive limit of representable functors $Hom(X_i,-)$ where X_i form a projective system with transition morphisms which are epimorphims. Assume $C$ has finite projective limits. -Then Grothendieck forms the comma category $CC=(1\downarrow F)$ and makes the following definitions. Object $A$ of $CC$ dominates object $B$ of $CC$ if there is an arrow $A\rightarrow B$ in $CC$, An object $A$ of $CC$ is minimal if whenever it is dominated by some $B$, such that the morphism in $CC$ is induced by a strict monomorphism $f$ in $C$, $f$ is an isomorphism. -He then makes the following claims for $A,B$ in $CC$. -1) If $A$ is minimal and dominates $B$ by an arrow induced by an arrow $v$ in $C$, and $F$ is left exact (commutes with finite projective limits) then $v$ is unique. -2) if $B$ is minimal and dominated by $A$ then $v$ is epi. -3) $F$ is strictly pro-representable iff it is left exact and every object in $CC$ is dominated by a minimal object. -I don't see these things basically because I don't know how to cook up strict monomorphisms to use the minimality hypothesis. For (3) I'm aware of how the proof that any functor is a colimit of representable functors uses the same comma category. But still I need pointers… - -REPLY [4 votes]: If I try to prove 1: -Take objects $A,B$ given by a morphism $f:x\rightarrow y$ in $C$ and objects $u\in F(x)$, $v\in F(y)$ such that $v=F(f)(u)$. Consider another morphism $g:x\rightarrow y$ such that $v=F(g)(u)$. Then take $h:z\rightarrow x$ to be the equalizer of $f,g$. If I understand correctly $h$ is a typical example of a strict monomorphism. And since $F$ is left exact, $F(h):F(z)\rightarrow F(x)$ is the equalizer of the two maps $F(f), F(g): F(x) \rightarrow F(y)$. Of course $u\in F(z)$ since $F(f)(u) = F(g)(u)$. Then the minimality axiom implies that $h$ is an isomorphism, to that $f=g$. -I have no proof for 2 for the moment but this should not be more difficult. -To prove 3: -It's just the usual proof that $F$ is canonically a colimit of representable functors, but using only minimal objects. Then by 2 the transition morphisms will automatically be epimorphism, and you construct a strict pro-object. The usual proof goes as follows: there is a canonical map -$$\operatorname{colim}_{(x,u)\in 1\downarrow F}\quad \hom(x,-) \longrightarrow F $$ -basically because by Yoneda lemma a morphism $\hom(x,-)\rightarrow F$ is defined uniquely by an object of $F(x)$, corresponding to the image of $1_x \in \hom(x,x)$. This canonical map is always a injective (since you take the colimit over all these couples). In the classical case where you consider all of the comma category it is also surjective for more or less tautological reasons. Here the hypothesis tells you can restrict to minimal couples in the comma category and this will again be surjective.<|endoftext|> -TITLE: When we count the same set, must the number always be the same? -QUESTION [12 upvotes]: Return to Frege's question, What justifies arithmetic? And consider the ur-proposition that counting a finite set always produces the same number, and ask whether this has a logical justification, that is a justification from assumptions which are necessarily true. Note that few of us actually first believed this assertion because of a logical proof. We accepted it say by authority - we were told it was so - or by experience - we experimented and discovered that counting the same set in two different ways always produced the same. Neither of these methods of justification, of course, necessarily ensure truth. -OK you say, but there is a proof from bedrock principles using induction which is entirely from logical principles. Model a counting of a finite set $A$ as an injective function from an initial segment of the natural numbers (beginning with 1) onto $A$, that is a sequence without repetitions. For the base case suppose a counting of a set produces the number 1. Then there is one thing, and the only way to count the set would produce the number 1. So check for the base case. -Now suppose $m$ is the successor of $n$ and the induction hypothesis holds for $n$ (that is, if we count any set to have the number $n$, then any other counting of the set must also produce the number $n$). Consider a set $A$ where one counting produces the number $m$, and another counting produces the number $v$. It does seem an acceptable (and logical) principle that if $m$ is the successor of $n$, and we have counted to $m$, then we need to have counted first to $n$, and then counted one more. That is, if $\boldsymbol f$ counts $A$ as $m$, then there exists $\boldsymbol f',B,b$ such that $b \in A \land B = A $\ $\{b\} \land\boldsymbol f'$ counts $B$ as $n$. Induction proves that $v \neq 1$ has a predecessor $u$. Letting $\boldsymbol g$ count $A$ as $v$, then there exists $\boldsymbol g',C,c$ such that $c \in A \land C = A$ \ $ \{c\} \land \boldsymbol g'$ counts $C$ as $u$. And - we can't apply the induction hypotheses immediately because while $B$ is of size $n$, and $C$ is of size $u$, $B$ and $C$ need to be the same set to apply the hypothesis, but they may be different since $b$ and $c$ may be different. -The way the proof must continue is that (assuming that $b$ and $c$ are different, since if they are the same there is no problem) we assert the existence of a sequence $\boldsymbol g''$, where we replace $b$ by $c$, that is where $\boldsymbol g'(i) = b$, assign instead $\boldsymbol g''(i) = c$ and leave everything else the same. Then by simple logic $\boldsymbol g''$ counts $B$ and indeed counts it as as $u$. Then one can apply the induction hypothesis, and the ur-proposition follows. -Only the assumption that there exists a sequence with the one replacement - that is, given a sequence $abcd...x...$, there must exist a sequence $abcd...y...$, for any $x$ and $y$ - is, or does not appear to be, a logical assumption. The new sequence is not a part of the older one, and its existence is not implicit in the older's existence. The assumption simply does not fall out of the meaning of "sequence." It is in fact very useful - I claim that with it and some other, only necessarily true, assumptions, one can develop arithmetic -, and so it can be given the justification that it leads to results which are useful in the sciences, and therefore it is desirable to make it. But this is different from being logical, and the question must be raised if in some cases reality is better represented without this assumption or perhaps making a different one. Modern physics is full of non-intuitive explanations; perhaps it can be better presented by using a non-intuitive mathematics. -So is the uniqueness of counting necessarily true? That one proof depends on a non-logical assumption, is not conclusive, since after all perhaps there is another, this time logical, proof where the replacement assumption is not needed. I have tried a bit and failed from assumptions which satisfy me, so I conjecture that it is essential in the framework of these assumptions. Anyway, my question is, is it? I formalize the question in the following way. -Consider FOL with equality, with one constant 1, with one two-place predicate $\sigma$ (successoring), and with two types (standard and bold). Standard types are natural numbers, and bold types are finite sequences of natural numbers. -Define: -$\boldsymbol f^D(x)$ abbreviates $\exists y \space \boldsymbol f(x) = y$ -This defines $x$ being in the domain of $\boldsymbol f$. -$\boldsymbol f^H(x)$ abbreviates $\boldsymbol f^D(x) \land \forall y (\sigma(x,y) \Rightarrow \neg \space \boldsymbol f^D(y))$ -This defines $x$ as the top or high number in the initial segment defining the sequence $\boldsymbol f$. -$\boldsymbol f =_k \boldsymbol g$ abbreviates $\forall x \neq k \forall y(\boldsymbol f(x) = y \iff \boldsymbol g(x) = y)$ -This says $\boldsymbol f$ and $\boldsymbol g$ are defined and have the same values for the same $x$, except perhaps at $k$. -Consider the following axioms: -1/ $\forall x \forall y \forall z (\sigma x,y \land \sigma x,z \Rightarrow y = z)$ -2/ Induction schema, i.e. $\phi (1) \land \forall n \forall m (\sigma n,m \land \phi (n) \Rightarrow \phi (m)) \Rightarrow \forall n \space\phi (n)$ -3/ $\forall\boldsymbol f \space \boldsymbol f^D (1)$ -4/ $\forall\boldsymbol f \exists n \space \boldsymbol f^H (n)$ -5/ $\forall\boldsymbol f \forall n \forall m \space (\boldsymbol f^H(m) \land \sigma n,m \Rightarrow \exists \boldsymbol f' ((\boldsymbol f')^H(n) \land \boldsymbol f' =_m \boldsymbol f)) $ -6/ $\forall\boldsymbol f \forall n \forall m \space (\boldsymbol f^D(m) \land \sigma n,m \Rightarrow \boldsymbol f^D(n)) $ -7/ $\forall \boldsymbol f \forall i \forall c (\boldsymbol f^D(i) \Rightarrow \exists \boldsymbol f' (\boldsymbol f'(i) = c \land \boldsymbol f' =_i \boldsymbol f))$ -Now 1/ through 7/ are sufficient to prove: -(UNIQUENESS) $\forall k \forall n \forall \boldsymbol f \forall \boldsymbol g (Img(\boldsymbol f) = Img(\boldsymbol g) \land Injection(\boldsymbol f) \land Injection(\boldsymbol g) \land \boldsymbol f^H(k) \land \boldsymbol g^H(n)\Rightarrow k = n)$, -where -$ Img(\boldsymbol f) = Img(\boldsymbol g) $ abbreviates $\forall y(\exists x \space \boldsymbol f(x,y) \iff \exists x \space \boldsymbol g(x,y))$ and -$Injection(f)$ abbreviates $\forall x \forall y (\boldsymbol f(x) = \boldsymbol f(y) \Rightarrow x = y)$ -(Indeed I would claim 1/ through 7/ are sufficient to develop a satisfactory arithmetic.) -My question is: Is there a model in which 1/ through 6/ (i.e. eliminate 7/) are true but in which UNIQUENESS is false? -NOTE. Let $\phi (k)$ abbreviate -$\forall n \forall \boldsymbol f \forall \boldsymbol g (Img(\boldsymbol f) = Img(\boldsymbol g) \land Injection(\boldsymbol f) \land Injection(\boldsymbol g) \land \boldsymbol f^H(k) \land \boldsymbol g^H(n)\Rightarrow k = n)$ -Then using only 1/ through 6/ I can prove $\phi(1)$ and, for any $k$, $\forall x_1,x_2,...,x_k (\sigma 1,x_1 \land \sigma x_1,x_2 \land \space ... \space \land \sigma x_{k-1},x_k \Rightarrow \phi (x_k)) $ -UNIQUENESS however of course is $\forall k \phi (k)$, and this I cannot prove, hence my question. - -REPLY [17 votes]: Such a model can be constructed using known independence results in bounded arithmetic as follows. -Let $I\Delta_0(f)$ be a theory in the usual language of arithmetic ($0,S,+,\cdot,<$) augmented by a new function symbol $f$, axiomatized by Robinson’s arithmetic plus induction for all bounded formulas in the expanded language. By a result of Ajtai [1,2], $I\Delta_0(f)$ does not prove the bijective pigeonhole principle $\mathrm{PHP}_n(f)$: this means that there exists a model $(\mathcal M,f)\models I\Delta_0(f)$, and a (necessarily nonstandard) element $n\in M$ such that $f$ restricted to the interval $[0,n]_\mathcal M$ defines a bijection $[0,n]\to[0,n-1]$. -Now, let us define a model $\mathcal N$ of the theory from the question: - -the number sort of $\mathcal N$ consists of $[0,n]$, with $\sigma$ being the graph of $S$ restricted to $[0,n-1]$ -the sequence sort consists of all sequences of the form $f\restriction_{[0,m]}$ and $\mathrm{id}\restriction_{[0,m]}$ for $m\le n$ - -It is straightforward to see that $\mathcal N$ satisfies axioms 1 and 3–6, and the functions $\mathrm{id}\restriction_{[0,n-1]}$ and $f\restriction_{[0,n]}$ witness that UNIQUENESS fails. -In order to see that $\mathcal N$ satisfies axiom schema 2 (induction) as well, notice that under the obvious interpretation of $\mathcal N$ in $\mathcal M$, number quantifiers $\exists x$, $\forall x$ translate to bounded quantifiers $\exists x\le n$, $\forall x\le n$, and sequence quantifiers can be simulated by bounded quantifiers as well: for example, we may code $\mathrm{id}\restriction_{[0,m]}$ as $2m$, and $f\restriction_{[0,m]}$ as $2m+1$. Thus, $\exists\mathbf g$, $\forall\mathbf g$ become $\exists z_\mathbb g\le2n+1$, $\forall z_\mathbb g\le2n+1$, and under this encoding, $\mathbb g(x)=y$ is defined by the bounded formula -$$\exists m\le n\,\bigl((z_\mathbb g=2m\land y=x\land x\le m)\lor(z_\mathbb g=2m+1\land y=f(x)\land x\le m)\bigr).$$ -Thus, any formula $\phi$ translates to a $\Delta_0(f)$ formula in $\mathcal M$, and consequently satisfies induction. -References: -[1] Miklós Ajtai: The complexity of the pigeonhole principle, in: Proceedings of the 29th Annual Symposium on Foundations of Computer Science, IEEE, 1988, pp. 346–355, doi: 10.1109/SFCS.1988.21951. -[2] Miklós Ajtai: The complexity of the pigeonhole principle, Combinatorica 14 (1994), no. 4, pp. 417–433, doi: 10.1007/BF01302964.<|endoftext|> -TITLE: Complexity of a Fibonacci numbers discrete log variation -QUESTION [17 upvotes]: In my work I encountered the following -FIBMOD PROBLEM: -Given $k,m$ in binary, decide if there exists $n$ such that -$\, F_n = k \,$ (mod $m$). Here $F_n$ is a Fibonacci number. -This is a variation on the discrete log problem, but in a larger -field. For example, let $m=p$ be a large prime. Then the problem is -asking if there exists $n$ such that $\alpha^n + \beta^n = k$ (mod $p$), -where $\alpha$ and $\beta$ are the roots of $\, x^2 - x - 1$. -Note, however, that discrete log asks also to find $n$ which is potentially harder. -Questions: -0) Are there any references on this problem? -1) Is this problem in NP $\cap$ co-NP? -2) Is this problem in BQP? -3) Is there a reason to believe that FIBMOD is hard? For example is there a way to show that FIBMOD is DISCRETE-LOG - hard? -Note: Fibonacci numbers mod $m$ are periodic with period $\le 6 m$, as explained in this Pisano period Wikipedia article. Recall that Fibonacci numbers can be computed by taking powers of the matrix: -$$\begin{pmatrix} -0 & 1 \\ 1 & 1 -\end{pmatrix}^n \begin{pmatrix} -0 \\ 1 -\end{pmatrix} = \begin{pmatrix} -F_n \\ F_{n+1} -\end{pmatrix} -$$ - Using the Chinese Remainder Theorem this implies that FIBMOD is in NP. - -REPLY [3 votes]: The Binet formula for Fibonacci numbers is -$$F_n = \frac{\phi^n - (-\phi)^{-n}}{\phi - (-\phi)^{-1}},\qquad\text{where}\ \phi:=\frac{1+\sqrt{5}}2.$$ -Then the congruence $F_n\equiv k\pmod{m}$ reduces to a pair of quadratic equations (indexed by the parity of $n$): -$$z^2 - k(\phi - (-\phi)^{-1})z - (-1)^n \equiv 0\pmod{m}$$ -with respect to $z:=\phi^n$. So, if we can solve these quadratic equations (which is easy if $m$ is prime), then the problem reduces to the classic discrete log problem base $\phi$ modulo $m$. - -As for references: -Lucas sequences have been previously adopted for the needs of cryptography, providing an alternative to modular exponentiation, which most notably resulted in the LUC cryptosystem. It was however shown that many of the supposed security advantages of LUC over cryptosystems based on modular exponentiation are either not present, or not as substantial as claimed. You can find corresponding references in Wikipedia. -For a general theory of linear recurrences over rings, see - -V. L. Kurakin, A. S. Kuzmin, A. V. Mikhalev, and A. A. Nechaev. Linear recurring sequences over rings and modules. Journal of Mathematical Sciences 76:6 (1995), 2793-2915.<|endoftext|> -TITLE: Beilinson-Bernstein localization, equivariant modules -QUESTION [5 upvotes]: I have a question regarding the equivariance in the Beilinson-Bernstein localization. Let $G$ be an simply connected algebraic group over a discrete valuation ring $R$ of mixed characteristic $(0,p)$ and $K$ a closed subgroup of $G$ with corresponding lie algebras $\mathfrak{g}, \mathfrak{k}$ and $X$ be the flag variety. One can define then the notions of $K$-equivariant $U(\mathfrak{g})$ modules and $K$-equivariant $D$-modules. -The Beilinson-Bernstein localization states that there is an equivalence between $K$-equivariant finitely generated $U(\mathfrak{g})$ modules with trivial central character and $K$-equivariant $D_X$ modules. This true when working over fields of characteristic 0 (One proof can be found in HTT theorem 11.5.3). The proof uses the crucial fact that $X$ is $D_X$- affine. -However, when working over a discrete valuation ring the flag variety is no longer $D_X$-affine. Can we still prove that the localization functor sends $K$-equivariant finitely generated $U(\mathfrak{g})$ modules with trivial central character to $K$-equivariant $D_X$ modules? My problem is that without $D$-affine condition I can not pass to the global sections. I am looking for any hints/references. -Remark: This is very much in the spirit of the question I asked here and got no response. https://math.stackexchange.com/questions/2498085/beilinson-bernstein-localization-equivariant-modules - -REPLY [2 votes]: I think the answer to your question might be partly contained in the paper "On irreducible representations of compact p-adic analytic groups" by Ardakov and Wadsley. You can find a link to a pdf on Simon Wadsley's webpage. -If you look at Proposition 5.15 there (putting $n=0$), you get that the cohomology groups, and so in particular the global sections, of every coherent $D_X$-module are finitely generated over $U(\mathfrak{g})$. By using Belinson-Bernstein over the fraction field of $R$, it should be easy to deduce that even though $X$ may not be $D$-affine, it is "up to bounded torsion" as far as coherent modules are concerned, meaning that the adjunction morphisms have kernels and cokernels which are killed by a fixed power of $\pi$ (where $\pi$ is the uniformizer of your ring $R$). -This paper does not work with $K$-equivariant modules but I don't think it would change much of their arguments.<|endoftext|> -TITLE: Image of boundary circle under map from punctured elliptic curve to ℂ -QUESTION [8 upvotes]: Let $E=\mathbb C/\Lambda$ be an elliptic curve, -and let $D\subset E$ be a very small disc. -($D$ is round for the usual flat metric on $E$) -By the main result of [1], there exists a holomorphic immersion $f:E\setminus D \to \mathbb C$. The image $f(\partial D)$ is a closed curve in $\mathbb C$ that self-intersects a bunch of times. - -What is the shape of $f(\partial D)$? - -I understand that such a curve is not unique. -What I want is a qualitative description of an example of such a curve. A drawing would be great. -Example: -When $D\subset E$ is a "rather big" disk, the curve ⌘ is the image of $\partial D$ under an immersion $E\setminus D \to \mathbb C\mathbb P^1$ (points in the central square have two preimages; points in the four lobes have zero preimages). But that only works when $D$ is a rather big compared to the size of $E$, and I don't know how to modify this curve as the size of $D$ tends to zero. - -[1]: Gunning, R. C., Narasimhan, R., Immersion of open Riemann surfaces. Math. Ann. 174, 103–108 (1967). - -REPLY [2 votes]: Based on David Speyer's answer, I got what I want, namely some intuition about what these closed paths look like. -It's a bit hard to draw them, so I include an inline description. - -Let $X:=3i$, $Y:=0$, $Z:=-3i$. -Given $T\in[0,\infty)$, I'll define a closed path in $\mathbb C$, which I'll call $\gamma_T$. -Let $R:=5e^T$ and $r:=e^{-T}$. -Let $C_X$ be the circle of radius $r$ centred around $X$. -Let $C_{Y}$ be the circle of radius $r$ centred around $Y$. -Let $C_{Z}$ be the circle of radius $r$ centred around $Z$. -Let $C_\infty$ be the circle of radius $R$ centred around $0$. -The path $\gamma_T$ is as follows: - -• Start at $Re^{-iT}$ - • straight line to $X+re^{-iT}$ - • run along $C_X$ for an angle of $4\pi+2T$, until the point $X+re^{iT}$ - • straight line to $Re^{iT}$ - • run along $C_\infty$ for an angle of $-2T$, until the point $Re^{-iT}$ - • straight line to $Y+re^{-iT}$ - • run along $C_Y$ for an angle of $4\pi+2T$, until the point $Y+re^{iT}$ - • straight line to $Re^{iT}$ - • run along $C_\infty$ for an angle of $-2T$, until the point $Re^{-iT}$ - • straight line to $Z+re^{-iT}$ - • run along $C_Z$ for an angle of $4\pi+2T$, until the point $Z+re^{iT}$ - • straight line to $Re^{iT}$ - • run along $C_\infty$ for an angle of $-4\pi-2T$, until the point $Re^{-iT}$ - -Remark: The path $\gamma_0$ is easy to visualise. -After deforming it a tiny bit, it becomes isotopic to the image of the curve ⌘ under the map $z\mapsto (z-a)^{-1}$, where $a$ is the middle point of one of the lobes. - -I believe that there exists an elliptic curve $E=\mathbb C/\Lambda$ and an immersion $$f:E\setminus\{0\}\to\mathbb C$$ such that $\forall T\in [0,\infty)$, the path $\gamma_T$ is the image under $f$ of a small counterclockwise loop (not quite round, but of bounded distortion) of radius $\simeq T^{-1/3}$ around the puncture.<|endoftext|> -TITLE: Automorphisms of genus 6 surfaces -QUESTION [14 upvotes]: I have found in Wikipedia that the number of automorphisms of a Riemann surface of genus 6 does not exceed 150. The page offers neither a proof nor a reference. Can someone help me? - -REPLY [3 votes]: The question is purely combinatorial - the Hurwitz bound comes from the observation that the quotient of a surface by its automorphism group is a hyperbolic orbifold, and the hyperbolic orbifold of smallest area is the $(2, 3, 7)$ triangle orbifold, Dividing the area of your surface by the area of the orbifold gives you the Hurwitz bound. However, for some genera, there is no triangulation where the vertices have the right degrees, then you have to look for bigger triangle groups. For the surface of genus $6,$ you get the $(2, 3, 10)$ triangle group, but in any case, this is just a combinatorial exercise (for any fixed genus).<|endoftext|> -TITLE: A conjecture in which both "if" and "only if" are near misses -QUESTION [16 upvotes]: [Migrated from Math Stack Exchange] -More than a year ago, I posted the following on the Math Stack Exchange. - -Consider $2^n-1$. Based on checking a few small numbers for $n$ (in - fact, the first ten natural numbers) we might "conclude" that "$n$ is - prime if and only if $2^n-1$ is prime." However, further investigation - shows that for $n=11$, $2^n-1$ is not prime. Thus our claim that "if - $n$ is prime then $2^n-1$ is prime" is a near miss. However, we can - prove its reverse, that "if $2^n-1$ is prime then $n$ is prime". I - am looking for some elementary number theoretic examples that both - directions of our eventually fake biconditional are near misses. - -To my great surprise, I haven't received even a near miss comment, let alone an answer. I know that the notion of "near miss" is disputable. Thus you might want to ignore my example above, and give an example of a biconditional statement (number-theoretic or not, historical or not, made-up or not) that seems to be true for some reasons (e.g., observing a number of cases, or intuition, etc) in both directions, but then it turns out that it is, in fact, false in both directions. - -REPLY [25 votes]: False claim: A Hausdorff topological space is compact if and only if it is sequentially compact. -It's believable if your intuition of Hausdorff spaces comes entirely from metric spaces (where the claim is, in fact, true). However, both directions are false for different reasons: -Counterexample I: the Alexandroff line is sequentially compact but not compact. -Counterexample II: the (Tychonoff) product of $|\mathbb R| $-many copies of the unit interval is compact but not sequentially compact.<|endoftext|> -TITLE: Erich Stiemke biography -QUESTION [12 upvotes]: Can someone provide some biographical details, especially the dates of birth and death, about Erich Stiemke? According to the Mathematics Genealogy Project, he obtained his Dr. phil at Universität Berlin in 1914 from Frobenius and Schottky. - -REPLY [3 votes]: Two small additions. -(1) Ordinal number of Stiemke's doctorate since the start of the Berlin mathematics faculty in the 1810s. According to p.361 of [K.-R. Biermann, Die Mathematik und ihre Dozenten an der Berliner Universität 1810-1933. Akademie-Verlag Berlin 1988. ISBN 3-05-500402-7], a book which is out of print, Stiemke was the 161st to obtain a doctorate in mathematics from University of Berlin (there was only one such before WW2). (The first was one Samuel Ferdinand Lubbe in 1818.) Notably, Biermann lists July 1, 1925 as the date of Stiemke's doctorate, while he lists 16 July 1914 for the date of the oral exam, with an exclamation mark (presumably to draw attention to the delay between exam and doctorate). Again, the ordinal number 161 depends on Biermann's decision to take the publication date of Stiemke's dissertation as the date of the doctorate. Whether this was more than an arbitrary decision by Biermann is not apparent from the book (I checked the index of loc. cit.; at the few places where Stiemke appears in the text, nothing more is said than what is already in this thread.) In particular, nothing is said about who was the driving force behind a posthumous doctorate eleven years after the `fact'; presumably it was Emmy Noether, but that is my guess only. -(2) Stiemke's bibliography. Stiemke has exactly three published articles: -(2.1913) Stiemke, E. Sur les modules dénombrables. (French) C. R. 157, 273-274 (1913). (gallica) -(2.1915) Stiemke, E. Über positive Lösungen homogener linearer Gleichungen. Math. Ann. 76, 340-342 (1915). (eudml, DOI: 10.1007/BF01458147) -(2.1926) Stiemke, E. Über unendliche algebraische Zahlkörper. Mathematische Zeitschrif 25, 9-39 (1926). (eudml, DOI: 10.1007/BF01283824) -By the way, (2.1915) contains an interesting little result giving a sufficient condition for the existence of a solution to a homogeneous system of linear equations such that each component of the solution is positive; the result was new to me (I bet it is a special case of something more modern, but I didn't manage to make such a connection): - -Theorem (Stiemke 1915). For any $(n,m)\in\mathbb{N}^2$ and $A\in\mathbb{R}^{n\times m}$, if for every $v\in\mathbb{R}^n$ the vector $v^{\mathrm{t}}A$ is either the zero vector or has at least one negative component, then there exists $x\in\mathbb{R}^m$ having all components strictly positive such that $Ax=0$.<|endoftext|> -TITLE: A non-commutative analog of: two polynomials are algebraically dependent iff their Jacobian is zero -QUESTION [5 upvotes]: Let $f,g \in \mathbb{C}[x,y]$. -There is a well-known result, that can be found for example -here, pages 19-20, that says the following: - -$f,g$ are algebraically dependent over $\mathbb{C}$ if and only if their Jacobian $Jac(f,g):=f_xg_y-f_yg_x$ is zero. - -Actually, this result is valid for $f_1,\ldots,f_n \in \mathbb{C}[x_1,\ldots,x_n]$, any $n \in \mathbb{N}$. -Now, let $f,g \in A_1(\mathbb{C})$, where $A_1(\mathbb{C})$ -is the first Weyl algebra over $\mathbb{C}$, namely, the $\mathbb{C}$-algebra -generated by $x,y$ such that $[y,x]=yx-xy=1$. -I wonder if there exists an analog result to the above in $A_1(\mathbb{C})$, -namely: - -$f,g \in A_1(\mathbb{C})$ are 'algebraically dependent' over $\mathbb{C}$ if and only if $[f,g]=0$. - -One has to be careful because: -(1) One has to define algebraic dependence over $\mathbb{C}$ of two non-commuting elements $f$ and $g$. -Should it be $\sum \lambda_{ij}f^ig^j=0$, with $\lambda_{ij} \in \mathbb{C}$ not all zero, -or $\sum \lambda_{ij}f^ig^j + \sum \mu_{ij}g^if^j=0$, with $\lambda_{ij}, -\mu_{ij} \in \mathbb{C}$ not all zero?. -(Perhaps the first definition should be called 'one-sided algebraic dependence', -while the second definition should be called 'two-sided algebraic dependence'). -(2) Perhaps this question is relevant. The example there (of Dixmier) is of $U,V \in A_1$, $[U,V]=0$, and $U^3-V^2+1=0$, so it does not contradict my plausible analog result, since those $U$ and $V$ are algebraically dependent. -((3) I am not sure if this is relevant, but in the above mentioned reference, on page 11, the Gelfand-Kirillov dimension is mentioned with connection to transcendence degree; for $\mathbb{C}[x,y]$ those notions coincide. Is the fact that the Gelfand-Kirillov dimension of $A_1(\mathbb{C})$ is two relevant to my question?). -Thank you very much for any help! - -REPLY [7 votes]: The reverse implication is true in a considerably more general setting (Burchnall-Chaundy theory). Namely, for any pair $(U,V)$ of commuting meromorphic coefficient differential operators in one variable of order at least one, there is a two-variable polynomial $P(z,w)$ such that $P(U,V)=0$ (the polynomial evaluation is unambiguous because $U$ and $V$ commute). There is an enormous body of literature on this topic related to integrable systems. -The proposed forward implication does not appear interesting or meaningful to me. (However, see a very interesting formulation proposed by David Speyer in the comments.) For example, the defining relation $yx-xy-1=0$ between $x$ and $y$ may be viewed as a form of "noncommutative algebraic dependence", yet of course $x$ and $y$ do not commute. And the proposed "one-sided algebraic dependence" need not have good formal properties, such as symmetry or transitivity. At best, one can attempt to extract such properties from commutativity.<|endoftext|> -TITLE: What's there to do in category theory? -QUESTION [63 upvotes]: Disclaimer: I posted this question on MSE only a few days ago; and received very few comments. I know that the etiquette is to wait a bit more than that before moving a post from MSE to MO, but I figured that posting it on MO would be an actual improvement because there would be some actual researchers in category theory on this site, willing to give details about what it is they do, what's interesting about it, etc, whereas there may be less of them on MSE. If this isn't appropriate, I'll remove this post, and if it's the case I'm sorry for the disturbance. - -I'm sure anyone who's heard of categories has also heard the classical "Well obviously there aren't any real theorems in category theory, it's much too general", or something in the likes of it. -Now obviously this argument is invalid (although its conclusion may be correct) because the same could be said of set theory, but there are clearly many really important theorems and results in set theory (I guess I don't have to justify that's it a huge field of research). -Now these theorems come from the fact that, when we do set theory, we don't just look at $\in$, we look at "derived stuff", like transitive sets, well-ordered sets, models of certain things, filters, etc. (I'm just giving a few examples to explain what I mean, I perfectly know that there's much much more to set theory than just those). -So the same thing should apply to category theory : of course we're not going to prove of we just stand there with our arrows and objects; you have to consider interesting ones, with more properties etc. -My question is about these (sorry for the lengthy intrduction). I know that a big part of category theory (although I don't really know in what proportion) is devoted to studying topoi(/ses ?) and for instance cartesian closed categories. -But I'm also guessing that there's much more than that to category theory; and my problem is that I don't know much about what is currently studied, what the major subfields of category theory are, or for that matter what subfields there are; so that when I want to refute the argument given at the very beginning I'm a bit stuck because I feel like I'm reducing category theory to topos theory and abelian categories. -Here's the actual question (after the too wordy introduction) : could you give some examples of subfields of research (if possible, currently, or previously very active fields) in category theory, paradigmatic questions or theorems in those subfields; how they're interesting in themselves and for some, how they can be interesting for other areas in maths (more than just giving a common language) ? - -REPLY [18 votes]: I apologize in advance for this very long answer. I am pretty sure that many people could write a better version of it. Unfortunately, they are not doing it. So, here we are. -The very beginning of the question asks for some even classical results in category theory that are relevant outside category theory. My knowledge is very limited thus I will list just few examples. A good research on google could produce the same, even better, result. Few example are better exposed because they are closer to my understanding. - -The theory of algebraic theories, known also with the name of Lawevere theories. -The theory of monads. -Abelian categories. -Between 70's and 80's there was a very interesting fashion, that eventually died and I do not understand why. -Consider a category, say Hausdorff spaces, can we find a full embedding into a category of algebras? -This is a very natural question, that we can rephrase in the following way: given a category, how effective can algebra be when studying it? -Pultr and Kucera gave a partial answer to this question in The category of compact Hausdorff spaces is not algebraic if there are too many measurable cardinals. In the same fashion Freyd proved that the homotopy category of topological spaces is not concrete. Thus there is no way of building a beautiful fundamental group which classifies homotopical algebra, somehow higher algebra is required by the complexity of the category. One could also relate this article to this subject. -The theory of locally presentable categories. Given a category $\mathcal{K}$ there is a natural notion of size for its objects called presentability rank. This was introduced by Makkai and Paré. Presentability rank coincides with cardinality (up to some subtleties) in category of models of a theory. It coincides with density of a space in metric spaces, with cardinality of an Hilbert basis in the case of Hilbert spaces. Do you think that such a unifying gadget should be studied? This is just a feature of this theory, the easiest to formulate. - - -As people told you the community seems to be working on HoTT and higher categories but there are also other projects around. - -There is an open connection between accessible categories and abstract elementary classes that lately attracted the interest of Grossberg, Boney and Vasey. -Still some people are working on categorical logic à la Makkai, extending some completeness results to infinitary logic. This is the case of C. Espindola. -Still some people are working on topos theory, this is the case of Nate Ackerman. O. Caramello is trying to apply this framework to algebraic geometry. The same is trying to do Ingo Blechschmidt. -W. Kubis gave a shorter proof of the uniqueness of the Gurarii space with a strongly category theoretic approach. This is linked to its presentation of Fraissé theory. - - -As a final remark, I will list three question that I asked on this platform. Obviously this is not interest of any research but they witness how naturally questions in category theory are related to other fields of mathematics and are not just a rewriting of known results. - -How to compute cocontinuity of a functor. -Free objects in first order theories. -Model existence theorem in elementary topoi. - - -Now that I have produced some mathematical argumentation in defense of my position, I would like to say that MO is an online community that everybody reads, especially young students. -I do believe that category theory has proven to be useful in many fields and it is time to stop answering to this kind of questions. One could say that this is just a genuine attempt to ask for the activity of a field but this is just rhetoric. It is impossible for people outside a subject to get into its research just asking "what are you guys doing?" Imagine to do it with algebraic geometry. You simply wouldn't understand the answer. -This kind of questions cannot be well answered because of the ignorance of the asker. Unfortunately this produces a misperception in the average reader, that will pretend there is no technical research in the subject.<|endoftext|> -TITLE: What is known about this cohomology operation? -QUESTION [10 upvotes]: Let $X$ be some space, $C^*(X,R)$ its cochain complex. Then there is a multiplication -$$ -\mu : C^*(X,R) \otimes C^*(X,R) \rightarrow C^*(X,R) -$$ -inducing the cup product, and a homotopy -$$H : C^*(X,R) \otimes C^*(X,R) \rightarrow C^{*-1}(X,R)$$ -"witnessing" the graded commutativity of the cup product, i.e. such that $dH(x,y) - Hd(x,y) = \mu(x,y) \pm \mu(y,x)$. -I'm interested in the cohomology operation $x \mapsto H(x,x)$. Note that this goes $H^p(X,R) \rightarrow H^{2p-1}(X,R)$. For $R = \mathbb{Z}/2$, I think this is the Steenrod square $Sq_1 = Sq^{p-1}$ (notation from chapter 2 this book), where $p = |x|$. As a first step, I'd like to understand it when $R = \mathbb{Z}/4$. This is some lift of $Sq_1$, but what more is known about it? Slightly more concretely, what methods are there for computing this kind of cohomology operation? -(I have reason to believe that this operation is NOT a ``Steenrod square'' for $\mathbb{Z}/4$, i.e. does not appear in the cohomology of the spectrum $H \mathbb{Z}/4$. But I also have reason to wish that what I just said were not true.) - -REPLY [11 votes]: I'm going to refer to $H$ as the cup-$1$ product. -So when $p = |x|$ is odd and $x$ is a cycle, the boundary formula says -$$d(x \cup_1 x) = 2x^2$$ -which is not necessarily zero. In this case, we don't get a cohomology operation unless (e.g.) when $2 = 0$ in $R$, and then the cohomology operation is "mostly like $Sq^{p-1}$". -When $p$ is even, then $d(x \cup_1 x) = 0$ and so we do get a cohomology operation (once we verify that it's well-defined, of course). However, in this case the next product (the cup-2 product) has an identity -$$ -d(x \cup_2 x) = 2 x\cup_1 x. -$$ -Here are some consequences. - -If 2 is invertible in the ring $R$, then $x \cup_1 x$ is automatically zero in cohomology. -If $R = \Bbb Z$, then $x \cup_2 x$ is a cochain which reduces, mod 2, to $Sq^{p-2} r(x)$ (the Steenrod square on the mod-2 reduction $r(x)$). However, the integral Bockstein $\beta$ of $Sq^{p-2} r(x)$ is calculated by taking a cocycle representative over $\Bbb Z/2$, lifting it to a cochain representative over $\Bbb Z$, taking the coboundary, and dividing by $2$. This means that $x \cup_1 x$ is a representative for $\beta Sq^{p-2} r(x)$. (Note that if we then mod-2 reduce, we see it become $Sq^1 Sq^{p-2} r(x) = Sq^{p-1} r(x)$ because $p$ is even.) -If $R = \Bbb Z/4$, I believe that a similar calculation shows that $x \cup_1 x$ is a representative for $\beta' Sq^{p-2} r(x)$ where $r$ is reduction mod $2$ and $\beta'$ is the Bockstein associated to the exact sequence -$$0 \to \Bbb Z/4 \to \Bbb Z/8 \to \Bbb Z/2 \to 0.$$ -However, in my time available I can only make this work if $x$ lifts to mod-$8$ cohomology, and so there's possibly a correction factor. EDIT: This computation works, but you need to use a cup-3 product to show that there are no extra correction factors. - -More information on the cup-$i$ products is available in the McClure-Smith's paper (http://arxiv.org/abs/math/0106024) referenced in Henrik Rüping's answer, and in Mosher and Tangora's book.<|endoftext|> -TITLE: bad reduction of a rational surface -QUESTION [6 upvotes]: Let $X$ be the blowing-up of $\mathbb{P}^2_{\mathbb{Q}}$ at four rational points on a line. Can one show that $X$ has bad reduction at 2? Or does $X$ secretly has good reduction at 2? -What one can be sure is that the closures in $\mathbb{P}^2_{\mathbb{Z}}$ of any such four points cannot be disjoint over $(2)\in{\rm Spec\ }\mathbb{Z}$, since each rational line in $\mathbb{P}^2_{\mathbb{F}_2}$ contains only 3 rational points. If (one of) the cross ratio of the four points is $a/b$ with ${\rm gcd}(a,b)=1$, the same consideration suggests that $X$ has bad reduction at $p$ iff $p\mid a$ or $p\mid b$ or $p\mid(a-b)$. - -REPLY [5 votes]: You are completely right that for any 4 rational points on the projective line, two have equal reductions mod two. However as ulrich points out, this does not create a singularity of the underlying surface. -The reason is that if we blow up $[1,0,0], [0,1,0],[1,1,0]$ in $\mathbb P^2_{\mathbb Z}$, say, we can then blow up, not the inverse image of the copy of $\operatorname{Spec} \mathbb Z$ defined by $[1,-1,0]$, but instead its strict transform. The strict transform can be defined as the closure in the blown-up surface of the relevant rational point. From this it is easy to see that the strict transform is isomorphic to $\operatorname{Spec} \mathbb Z$ and that the surface is smooth at it, so there is no problem blowing it up to obtain a smooth surface. -Explicitly, we can see that the strict transform of $[1,-1,0]$ intersects $[1,1,0]$ in the direction given by $\frac{ [1,-1,0]- [1,1,0]}{2} = [0,-1,0]$ - i.e. at the intersection of the exceptional divisor and the strict transform of the line that both points lie on.<|endoftext|> -TITLE: Tate twists and cohomology of $\mathbf{P}^1$ -QUESTION [12 upvotes]: I was wondering if anyone could give me some intuition as to why, for a smooth projective variety $X$ over $\mathbf{C}$ of complex dimension $d$, the Tate twist on $H^n(X(\mathbf{C}),\mathbf{Z})$ to be incorporated to have a pairing with $H_{d-n}(X(\mathbf{C}),\mathbf{Z})$ into $\mathbf{Z}(-d)$, is related to (tensor powers of) the cohomology group $H^2(\mathbf{P}^1,\mathbf{Z})\simeq H^1(\mathbf{G}_m,\mathbf{Z})$. -In other words, why does one often define $\mathbf{Z}(-1) := H^2(\mathbf{P}^1,\mathbf{Z})$? Is there some Kunneth formula lurking behind the scene? Should I think about $X(\mathbf{C})$ as a family $X(\mathbf{C})\times \mathbf{P}^1\to \mathbf{P}^1$? Is there a map $H^2((\mathbf{P}^1)^d,\mathbf{Z})\to H^d(X(\mathbf{C}),\mathbf{Z})$? Where does it come from? -I can understand the notation comes from $H_0(\mathbf{P}^1,\mathbf{Z})\simeq H_1(\mathbf{G}_m,\mathbf{Z})\simeq\pi_1^{\rm ab}(\mathbf{G}_m) =\pi_1(\mathbf{G}_m) := \mathbf{Z}(1)$, the reason I'm asking the question: even if there is $H_0(X(\mathbf{C}),\mathbf{Z})\to H_0(\mathbf{P}^1,\mathbf{Z})$, I see no map $X(\mathbf{C})\to\mathbf{P}^1$ in general. Unless somehow $X(\mathbf{C})$ can always be realized as the fiber over some point of $\mathbf{P}^1$ of some map $\mathcal{X}\to\mathbf{P}^1$, with $X$ homotopy equivalent to $\mathcal{X}$. Can one just choose $\mathcal{X} = X(\mathbf{C})\times\mathbf{P}^1$? (can it possibly be so simple? In algebraic geometry, this would amount to taking the trivial deformation of $X\to *$ to $\mathcal{X} \to\mathbf{P}^1$ along a fixed point $*\to\mathbf{P}^1$, and trivial deformations strike me as usually not so interesting). I'll appreciate any insight on the matter a lot -Thanks - -REPLY [13 votes]: The Tate twist is what we need to express Poincaré duality without making any choice. Such a choice appears in the choice of an orientation of the affine line minus the origin, and have shadows in the description of the Thom isomorphism, hence, in the description/construction of Gysin maps, trace maps, and so forth. A natural way to make this transparent goes through the theory of Chern classes. In what follows, when I write the symbol $=$, I mean that there is a canonical isomorphism, the construction of which does not involve any choice: it always comes by functoriality. -If we define $\mathbf Z(1)$ as the first homology group of the affine line minus the origin, then it is a free abelian group of rank one, and one defines $\mathbf Z(n)$ as a tensor product of $n$ copies of $\mathbf Z(1)$ for $n\geq 0$, and as the dual of $\mathbf Z(-n)$ for $n<0$. Given an abelian group $A$, one defined $A(n)=A\otimes\mathbf Z(n)$. Since, by definition, cohomology is the dual of homology, and since we are dealing with free groups, we have a canonical identification -$$H^1(\mathbf A^1-\{0\},\mathbf Z)=\mathbf Z(-1)\, .$$ -By an elementary Mayer-Vietoris argument, one can deduce that $H^2(\mathbf P^1,\mathbf Z(1))=\mathbf Z$, but that is just a computation, not an explanation. -What precedes means that the classifying space of the topological abelian group $\mathbf C^\times=\mathbf G_m(\mathbf C)$ is not really a $K(\mathbf Z,2)$, but rather a $K(\mathbf Z(1),2)$. This is why we have Chern classes of line bundles -$$c_1:Pic(X)=[X,B\mathbf G_m]\to[X(\mathbf C),B\mathbf C^\times]=H^2(X,\mathbf Z(1))$$ -(where $[A,B]$ means the homotopy classes of maps from $A$ to $B$ in the appropriate sense). This is used to prove the projective bundle formula. For a vector bundle $E$ on $X$ of rank $r$, with associated projective bundle $\mathbf P(E)$, if $t$ denotes the first Chern class of the tautological line bundle on $\mathbf P(E)$, then the map $(x_0,\ldots,x_{r-1})\mapsto \sum^{r-1}_{i=0}t^ix_i$ is an isomorphism: -$$\bigoplus_{i=0}^{r-1}H^{n-2i}(X,\mathbf Z(-i))=H^n(\mathbf P(E),\mathbf Z)$$ -Now, given a closed immersion $i:Z\to X$ between smooth schemes over $\mathbf C$, one defines $H^n(X,Z)=H^n(Z,i^!(\mathbf Z))$. In other words, this is the $n$th cohomology group of the homotopy fibre of the restriction map $R\Gamma(X,\mathbf Z)\to R\Gamma(X-Z,\mathbf Z)$. The Thom isomorphism is a canonical identification: -$$H^n(X,Z)=H^{n-2c}(Z,\mathbf Z(-c))$$ -where $c$ is the codimension of $Z$ in $X$. (When $i=s$ is the zero section of a vector bundle $E$ the Thom isomorphism is obtained from the projective bundle formula applied to the direct sum of $E$ and of the trivial line bundle. The general case follows from a deformation to the normal cone argument.) -For $X=\mathbf P^1$ and $Z=\{\infty\}$, this means that -$$H^2(\mathbf P^1,\{\infty\},\mathbf Z)=H^0(\{\infty\},\mathbf Z(-1))=\mathbf Z(-1)\, .$$ -The meaning of all this is that $\mathbf Z(-1)$ is not the second cohomology group of the projective line. It is rather the second cohomology group of the projective line pointed at infinity. Of course, by homotopy invariance, there is a canonical isomorphism $H^2(\mathbf P^1,\{\infty\},\mathbf Z)=H^2(\mathbf P^1,\mathbf Z)$, but, in some sense, that is misleading: the canonical identification of $H^2(\mathbf P^1,\{\infty\},\mathbf Z)$ with the dual of $H_1(\mathbf A^1-\{0\},\mathbf Z)$ is through a (baby version of the) Thom isomorphism, which is yet another expression of the theory of Chern classes. -When we write the motive of $\mathbf P^1$ as a direct sum of the constant motive with a Tate motive, by definition, the Tate motive is the motive of the -projective line pointed at infinity. Therefore, its Betti realisation goes to $H^2(\mathbf P^1,\{\infty\},\mathbf Z)$.<|endoftext|> -TITLE: Why study unirational and rational varieties? -QUESTION [8 upvotes]: I am new to the study of unirational and rational varieties, but I want to know the motivation for why mathematicians started to study these conditions. The reasons that I could list to study unirational and rational varieties are the following. - -The study of unirational varieties can be used to answer questions about whether a scheme has a $ k $-rational point. -In the minimal model program, rational varieties are in some sense the ones "most like $ \mathbb{P}^{n} $". Unirationality is a weakening of the rationality condition, but if we can answer questions about unirationality we may be able to a) strengthen the condition to deal with rationality and/or b) generalize the minimal model program by focusing on dominant, generically finite, rational maps from a variety $ X $ to a variety $ Y $ instead of looking at birational maps. Therefore, in order to answer questions about the minimal model program, or some kind of generalization, we would like to understand these conditions. - -However, I wanted to pose the question to people who have studied these questions in detail, as the answers I gave to my own question seem too loose and I suspect that there is something that I am missing. -Thanks, -Schemer - -REPLY [3 votes]: For a given variety it is important to know whether it is rational or unirational, because if this is the case, then the variety admits a nice parameterization. I believe this is the historical reason for studying these notions.<|endoftext|> -TITLE: A kind of supercompactness -QUESTION [11 upvotes]: Is there a notion of supercompactness of a cardinal $\kappa$ that implies the following? -For every sequence $\langle \alpha_i : i < \kappa \rangle \subseteq \kappa$ and every $i_0 < \kappa$, there is an elementary embedding $j : V \to M$ such that if $\lambda = j(\langle \alpha_i : i < \kappa \rangle)(\kappa + i_0)$, then $M^\lambda \subseteq M$. -This works with almost-hugeness, but I’m wondering if there is an intermediate notion (which already has a definition, or a simpler one that just stating the above). - -REPLY [13 votes]: Your notion is the same as Shelah-for-supercompact. The part with -$i_0$ can be eliminated by suitable translation with a new function. Indeed, the property is equivalent to handle all $i<\kappa$ at once, and even much more than this, as in statement 4. -Theorem. The following are equivalent for any cardinal -$\kappa$. - -Your definition. For every $f:\kappa\to\kappa$ and every -$i_0<\kappa$, there is an elementary embedding $j:V\to M$ into a -transitive class $M$ with critical point $\kappa$ and -$M^{j(f)(\kappa+i_0)}\subset M$. -Shelah-for-supercompact. For every $f:\kappa\to\kappa$, there is -$j:V\to M$ with critical point $\kappa$ and -$M^{j(f)(\kappa)}\subset M$. -Uniform version of your definition. For every -$f:\kappa\to\kappa$, there is $j:V\to M$ with critical point -$\kappa$ such that $M^{j(f)(\kappa+i)}\subset M$ for every -$i<\kappa$. -Souped-up uniform version. For every $f:\kappa\to\kappa$ and every $h:\kappa\to\kappa$, there is $j:V\to M$ with critical point $\kappa$ and $M^{j(f)(\kappa+i)}\subset M$ for all $i -TITLE: Functional equation and contragredient -QUESTION [7 upvotes]: I am often annoyed by the presence of the contragredient $\tilde{\pi}$ of the representation we consider, when we write the functional equation of its L-function: -$$L(s, \pi) = \varepsilon(s, \pi) L(1-s, \tilde{\pi})$$ -I am interested in the $GL(2)$, and maybe $GL(n)$ case: when do we have $\pi = \tilde{\pi}$? Are there explicit examples where this is not the case and where we can compute $\tilde{\pi}$? - -REPLY [7 votes]: [Expanding my earlier comment into an answer]: -Self-dual cuspidal automorphic representations of $GL(n)$ have been classified by Jim Arthur, as one of the main results of his monograph on automorphic forms for classical groups. -The general shape of the classification is this. If $\pi$ is self-dual, then the Rankin--Selberg $L$-function $L(\pi \times \pi, s)$ has a simple pole at $s = 1$. On the other hand, this $L$-function factors as $L(Sym^2 \pi, s) \times L(\wedge^2 \pi, s)$, and neither can vanish at $s = 1$. So if $\pi$ is self-dual, exactly one of these functions has a pole. We say a selfdual $\pi$ is of orthogonal type if $Sym^2 \pi$ has a pole, and of symplectic type if $\wedge^2 \pi$ has a pole. -If $\pi$ has a Galois representation attached, then you can easily convince yourself that orthogonal / symplectic type is precisely saying that the Galois representation lands in an orthogonal, resp. symplectic, group; and this leads one to expect that $\pi$ should be a functorial transfer from the Langlands duals of these groups. This, if I understand correctly, is exactly what Arthur proves. More precisely: - -if $\pi$ is symplectic, then $n = 2m$ has to be even, and $\pi$ should be a functorial transfer from the dual group of $Sp(2m)$, which is $SO(2m+1)$. -if $\pi$ is orthogonal and $n = 2m+1$ is odd, then, up to a quadratic twist coming from the quotient $O(2m+1)/SO(2m+1)$, $\pi$ must be a functorial transfer from the dual group of $SO(2m+1)$, which is $Sp(2m)$. -if $\pi$ is orthogonal and $n = 2m$ is even, then either the Langlands parameter of $\pi$ factors through $SO(2m)$, in which case $\pi$ comes from the Langlands dual of $SO(2m)$ which is $SO(2m)$ again; or there is a non-trivial quadratic character intervening, in which case $\pi$ comes from a twisted form of $SO(2m)$. - -Let me make this explicit in some small cases. - -If $n = 2$ then we have $\tilde\pi = \pi \otimes \omega_{\pi}^{-1}$, as @PeterHumphries has pointed out. Hence there are two ways $\pi$ can be selfdual: we may have $\omega_{\pi} = 1$, in which case $\pi$ is of symplectic type and is a functorial lift from $PGL_2 \cong SO(3)$; or we may have $\omega_\pi \ne 1$ but still $\pi \cong \pi \otimes \omega_\pi^{-1}$, so $\omega_{\pi}$ has to be quadratic and $\pi$ is a dihedral representation induced from the corresponding quadratic field -- which is precisely saying that $\pi$ is of orthogonal type and comes from a twisted form of $SO(2)$. -If $n = 3$, then this recovers Ramakrishnan's theorem that $\pi$ has to come from $SL(2)$ via the adjoint square lift, up to twisting by a quadratic character. -If $n = 4$, then there are two possibilities again: either $\pi$ is symplectic, so it comes from $SO(5) \cong PGSp(4)$; or $\pi$ is orthogonal, in which case it comes from a (possibly) twisted form of $SO(4)$. The split form of SO(4) is a quotient of $SL(2) \times SL(2)$, so we get $\pi$'s that are transfers from $GL_2 \times GL_2$ via the Rankin--Selberg lifting, and the non-split forms correspond to Asai transfers from $GL_2$ over a quadratic field extension.<|endoftext|> -TITLE: Holomorphic vector fields tangent to a hypersuface singularity -QUESTION [7 upvotes]: Let $(V,0)\subset (\mathbb{C}^n,0)$, $n\geq 3$ a (germ of) hypersurface given by $V =\{f=0\}$, $f$ a germ of holomorphic function. A (germ of) holomorphic vector field $X$ on $(\mathbb{C}^n,0)$ is "tangent to $V$" if $X(f) = {\rm d}f(X)$ belongs to the ideal generated by $f$ in $\mathcal{O}_{\mathbb{C}^n,0}$. -My question is: For any hypersurface germ $V$, does there exist a germ of vector field with an isolated singularity that is tangent to $V$? -If $V$ is smooth or has an isolated singularity, this is fairly simple. However, for germs of non-isolated hypersurface singularities I could neither prove this nor find a counterexample. - -REPLY [2 votes]: I think that the answer to this question is no for $\mathbb C^3$, for -$$f=zy(z-y)(z-xy).$$ -I'll assume that $v$ is holomorphic and tangent to $f=0$ in a small neighbourhood of $(0,0,0)$. -Proof. Suppose by contradiction that $v$ is a holomorphic vector field tangent to $f=0$ near $(0,0,0)$ with an isolated zero at $(0,0,0)$. Note that the surface $f=0$ has a singularity along the $x$-axes. So $v$ should be tangent to the $x$-axes and also should be non-zero on it by our assumptions. I claim that this is not possible. -Indeed, note that the tangent cone of the surface $f=0$ at a point $(x_0,0,0)$ is the union of $4$ planes, $z=0$, $y=0$, $z-y=0$ and $z-x_0y=0$. These planes have double ratio $x_0$ (to define such a double ratio intersect these $4$ planes with any plane passing through $(x_0,0,0)$ and take the double ratio of $4$ lines in the intersection). On the other hand if $v$ were non-zero on the $x$-axis, its flow would send points of the $x$-axes to different points and so it would not preserve the double ratio. This is a contradiction, since a byholomorphism must preserve such a ratio. QED.<|endoftext|> -TITLE: The cofibration/fibration $\leftrightarrow$ epi/mono confusion -QUESTION [11 upvotes]: A recent question Why do we need model categories? reminded me of this long-standing confusion of mine -- I mentioned it in an answer there, and then decided to ask a separate question about it. I even dare not to use the soft-question tag. -What confuses me is this. There is an obvious resemblance between model categories and factorization systems. Yet, one thing goes totally wrong: in factorization systems, "left halves" tend to be "like epimorphisms" and "right halves" - "like monomorphisms". At the same time, in model categories cofibrations have distinct flavor of monos to them, and fibrations - of epis. -You see, I cannot even formulate this rigorously, yet I hope you agree that what I now described contains undeniable truth. -So what is this truth? Does this phenomenon have any explanation? Living with this puzzle for many years, the only consideration that I've been able to come up with is this: take one step from sets to categories. In sets, monos are just inclusions. In categories, the "correct" notion of mono starts to involve some amount of "epiness", since "good" monos in categories are full and faithful functors, and fullness involves some surjectivity condition. I don't even know whether there is some sort of such "first-step-consideration" from the opposite end, relating "betterness" of epis with some sort of additional requirements which have to do with monomorphy. -I repeat - this is a strange question: although it is full of most vague handwaving, I hope you agree that it touches on something very rigorous, which I just fail to capture. - -REPLY [9 votes]: The (epi,mono) factorization system in Sets is part of a model structure on Sets whose weak equivalences are the epis, fibrations are monos and cofibrations are everything. This is a model for the homotopy theory of (-1)-truncated sets. One can also define a similar model structure on simplicial sets, where the weak equivalences are the maps which are surjective on pi_0 (i.e, homotopy epis) and whose fibrations are the Kan fibrations whose homotopy fibers are all (-1)-truncated (i.e., homotopy monos). This is a also a model for (-1)-truncated spaces/sets. A similar thing can be done in others contexts as well, for example, if you can replace simplicial sets by a model category of simplicial presheaves which presents a certain $\infty$-topos, then you can often left Bousfield localize $M$ such that the new weak equivalences are homotopy epis and the fibrations (at least between old fibrant objects) are the old fibrations which are also homotopy monos. The new model category will present the homotopy theory of (-1)-truncated objects in the original $\infty$-topos. I will go on a limb here and suggest that the (epi,mono) type factorization systems, which appear very typical from a 1-categorical point of view, are, from a model categorical point of view, essentially the particular case of (certain kinds of) (-1)-truncated homotopy theories.<|endoftext|> -TITLE: Upper bound on the number of permutations in a set during an algorithm -QUESTION [6 upvotes]: Fix $n\geq 2$ and let $S_n$ be the symmetric group on $n$ letters with identity $e$. We consider elements of $S_n$ to be bijections $[n]\to [n]$ as well as sequences (one line notation). For $1\leq i -TITLE: Do convex closed semialgebraic hyperplane cross-sections imply semi-algebraicity? -QUESTION [7 upvotes]: Let $S\subset\mathbb{R}^n$, with $n\geq 3$, such that for any hyperplane $L$ one has $L\cap S$ closed, semialgebraic, and convex. Is it true that $S$ itself is semialgebraic? -A colleague explained to me that closedness is necessary, as one can take an open ball in $\mathbb{R}^n$ and paste a sufficiently nasty curve into its boundary. -It is probably possible to weaken the convexity assumption to something like "no isolated points", and still have a meaningful question. -PS. the is basically a question recently asked by Lev Birbrair and Aris Daniilidis. -PPS. The if $n=2$ then for any convex set $S$ the intersection with a line is semialgebraic, so the answer is obvious "no" in this case. - -One way, suggested by Birbrair, to make question more interesting is to assume in addition that $S$ is defined in an $o$-minimal structure. This would preclude examples like in the my own answer below. - -REPLY [3 votes]: Here I record an answer based on very kind comments above (and offline). -The construction starts from a closed ball $B\subset\mathbb{R}^3$ and an infinite sequence $\mathcal{L}:=\{L^+_k\}$ of open half-spaces, satisfying the condition -$$B\cap L^+_i\cap L^+_j=\emptyset\quad\text{ if and only if $i\neq j$.}\qquad\qquad (*)$$ Denote $L^-_k:=\mathbb{R}^3\setminus L_k^+$, and set $$B^-:=B\cap L^-_1\cap L^-_2\cap\dots\cap L^-_k\cap\dots$$ -Zariski closure of the boundary $\partial B^-$ of $B^-$ contains infinitely many planes $\partial L_k^-$, and thus it cannot be a surface. -Theorem 3.20 in lecture notes by Michel Coste says that the Zariski closure of the boundary $\partial S$ of a semialgebraic set $S$ is algebraic of the same dimension as $\partial S$. Applying it to the $B^-$, we obtain that $\partial B^-$ is not semialgebraic, threfore $B^-$ is not semialgebraic, as well. -It remains to construct an appropriate $\mathcal{L}$, so that any plane $P$ intersects only finitely many $B\cap L_k$. Each $B\cap L_k$ is a spherical cap of radius $r_k$ and centre $C_k\in\partial B$. Let $\pi$ be the stereographic projection of $\partial B$ from the north pole onto the tangent plane $\Pi$ at the south pole $O\in\Pi$. Let $M\gg 0$, $c_k:=((k+M)^{-1},(k+M)^{-2})\in\Pi$, for $k\geq 1$; set $C_k=\pi^{-1}(c_k)$ and $r_k:=e^{-k-M}$, thus specifying a particular $\mathcal{L}$. Then $(*)$ holds. As $O$ is the limit point of $\{C_k\}$, any plane $\Omega$, to have a chance to intersect infinitely many $B\cap L_k$, and thus provide non-semialgebraic $B^-\cap\Omega$, must pass through $O$. -To establish that such an $\Omega$ does not exist, it suffices to show that any line $\ell$ in $\Pi$ intersects only finitely many disks $D_k\subset\Pi$ with centre at $c_k$ of radius $r_k$. As the points $c_k$ lie on a parabola $y=x^2$, -$\ell$ won't hit inifinitely many $D_k$, unless it is tangent to it at $O$. So it remains to consider the case of $\ell$ being the line $y=0$. As $r_k$ decreases exponentially, but the $y$-coordinate $(k+M)^{-2}$ of $c_k$ only quadratically, there will be $K$ so that for any $k>K$ the -intersection of $\ell$ and $D_k$ is empty. Thus any plane intersects $B^-$ in a semialgebraic set, and we are done.<|endoftext|> -TITLE: Mazur's Question on Mod $N$ Galois representations -QUESTION [8 upvotes]: In Rational Isogenies of Prime Degree, Mazur poses: -"the problem of determining all elliptic curves $E'/\mathbb{Q}$ with symplectic $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ isomorphisms $E'[N]\cong V$. -He then claims: -"This can be reduced to (or rephrased as) the problem of determining all $\mathbb{Q}$-rational points of a certain twisted form $X(V)$ of the modular curve $X(N)$." -My question is, how does one define $X(V)$ and how can one see that it is a twist of $X(N)$. Over which field is this twist defined? - -REPLY [3 votes]: In addition to the previous comments, let me also say that this question has been considered by Kraus in joint works with Freitas, Halberstadt, and Oesterlé. -The properties of the modular curve associated to the problem you mention are given in Proposition 1 of the paper by Kraus and Oesterlé, Sur une question de B. Mazur, Math. Ann. 293 (1992). The proof of this proposition is given in the reference [6] therein. Email me if you want an electronic copy of it.<|endoftext|> -TITLE: Why do sheaves embed in presheaves? -QUESTION [7 upvotes]: Presheaves on a category $C$ form the free co-completion, in the sense that every functor from $C$ to a cocomplete category extends in a unique way to the presheaf category. -If $C$ is equipped with a Grothendieck topology, then a similar statement holds if we ask that our cocontinuous extension also send some diagrams formed out of coverings to colimits. -A rough but pleasant analogy is that the presheaf category is like the free abelian monoid on a set with colimits replacing sums, and the sheaf category is what you get by quotienting by some linear relations. -From this point of view it is obvious that there must be a cocontinuous functor from presheaves to sheaves, which of course is sheafification. However from this perspective it seems somewhat surprising to me that the category of sheaves embeds as a full subcategory of presheaves. -Is there some reason to expect this to be the case from thinking only about cocompletions and related abstract nonsense? - -REPLY [5 votes]: One small additional remark to Qiaochu Yuan response and David Roberts comment to show that it is really the existence of an adjoint that is the important point here. (and that was really too long for a comment) -We can look at a more general situation, where the existence of an adjoint might fail: If $C$ is a non-small category, the "free co-completion of $C$" still exists: it is the category of presheaf over $C$ that are small colimits of representable (the presheaf of small co-finality). I will call them the "small presheaves" they always form a locally small category. -If I have a topology on $C$ then I can also try to construct a category of "small sheaves" that has the universal property you mentioned: I take the category of sheaves over $C$ which are the sheafification of small presheaves. -Depending on the topologies, several things might go wrong: the sheafification can be undefined, or one can get a non locally small category. -But for a well chosen topology it can happen that sheaficiation of small presheaves exists and that you get a locally small category this way. In this case this would be the solution to your universal problem. -In this situation the category of "small sheaves" that you constructed in still included in the category of all presheaves but is not necessarily included in the free co-completion. So you do have a sheafication functor from from the "free co-completion" to the "free co-completion with relation" as expected, but you no longer get this surprising functor in the other direction. -Let me give an explicit example where all of this happens: -Take $C$ to be the category of all ordinals (with morphisms given by the order relation) with the atomic topology (every non-empty sieve is a covering, in this case this is the same as saying that every map is a covering). -Then one can check the following: -1) A presheaf is an ordinal indexed collection of set with transition map $F(x) \leftarrow F(y)$ for $x -TITLE: Formalizations of The Matchstick Diagram Representation of Ordinals -QUESTION [8 upvotes]: The matchstick diagram is a really interesting and intuitive method of representing countable ordinals. However, because of how difficult it is to graphically represent ordinals with it, I started wondering about how large the ordinals can be. -The "$\omega_1$ of matchstick diagrams", or $\omega_1^{md}$, is what I came up with. - -Let $\mathbb{Q}_{2}:=\{\frac{1}{2^n}:n\in\mathbb{N}\}$. -For a countable limit ordinal $\alpha$, let a matchstick diagram of $\alpha$ be a function $f:\alpha\rightarrow\mathbb{Q}_{2}$ such that: - -$f(0)=1$ -$f(\beta+1)=\frac{f(\beta)}{2}$ for any $\beta<\alpha$. -$\sum_{\beta<\alpha}f(\beta+1)$ converges to a rational number. - -Let a matchstick diagram $f$ of $\alpha$ have length $z$ for some $z\in\mathbb{Q}$ iff $\sum_{\beta<\alpha}f(\beta+1)=z$. -$\omega_1^{md}$ is the smallest limit ordinal with no matchstick diagram. Equivalently, $\omega_1^{md}$ is the supremum of all ordinals with matchstick diagrams. - -Notably, if $\alpha$ has a matchstick diagram at all, then for every $z>1$, there is a $y0$ can be written as the sum of a sequence $\sum_{i<\omega} \beta_i$ where each $\beta_i < \alpha$ is itself a power of $\omega$. -Proof. If $\gamma = \gamma'+1$ is successor, write it as the sum of $\omega$ terms all equal to $\omega^{\gamma'}$. Otherwise, write $\gamma$ as limit of an increasing sequence $(\gamma_i)_{i<\omega}$, then $\omega^\gamma = \sum_{i<\omega} \omega^{\gamma_i}$ (because in fact $\sum_{i\leq n} \omega^{\gamma_i} = \omega^{\gamma_n}$), as announced. -(In my JavaScript program, this lemma is implemented by the function summingSequence, essentially with the proof I just gave.) -Now inductively define a matchstick ordinal representation inside $[0,1)$ for any $\alpha < \omega_1$ which is a power of $\omega$ as follows: if $\alpha = 1$ just use a single matchstick at position $0$; otherwise, write $\alpha = \sum_{i<\omega} \beta_i$ as given by the lemma, divide the interval $[0,1)$ into countably many consecutive subintervals, each one $p$ times the length of the previous one (for what you ask, take $p=1/2$, whereas my program uses $p=3/10$), and place a scaled copy of the matchstick representation of each $\beta_i$ inside each consecutive interval. -You asked for the interval between the two first matchsticks to be $1$, but this is immediately solved by scaling the above construction (simply note that if $p=1/2$, the interval it provides is of the form $1/2^k$ for some $k$). The case of ordinals which aren't pure powers of $\omega$ is easily handled by the Cantor normal form and I won't describe it. -Anyway, my answer is essentially the same as Joel David Hamkins's, but I wanted to emphasize that it can be more or less implemented (up to whatever ordinal you are willing to code an ordinal notation system for). -Of course, the disappointing fact is that such representations are not at all enlightening. When I first saw matchstick representations, I thought, oh, this is wonderful, if I could draw one for some difficult-to-imagine ordinal like $\varepsilon_0$ it would help me visualize $\varepsilon_0$, but of course this is not at all true, as the interactive JavaScript page shows: with the construction described above, all large ordinals look boringly the same, they are visually indistinguishable from the obvious self-similar set consisting of countably many copies of itself laid end to end (each $p$ times the size of the previous one).<|endoftext|> -TITLE: $p$-groups in which all normal abelian subgroups are cyclic -QUESTION [6 upvotes]: It is well-known that any finite $p$-group in which all its abelian subgroups are cyclic is either a cyclic group or a generalized quaternion group. -What can be said about $p$-groups in which every normal abelian subgroup is cyclic? - -REPLY [6 votes]: See Gorenstein, Finite Groups, Chapter 5, Theorem 4.10. Such groups are as follows: - -if $p$ is odd, then $G$ is cyclic; -if $p=2$, then $G$ is either cyclic, or generalised quaternion of order $2^l$ for some $l\geq 3$, or dihedral of order $2^m$ for some $m\geq 4$ (note that $D_8$ does not satisfy the hypothesis), or semidihedral of order $2^n$ for some $n\geq 4$.<|endoftext|> -TITLE: Quasi-isometric groups without common virtual geometric model -QUESTION [11 upvotes]: Are there known examples of finitely generated groups $G$ and $H$ that are quasi-isometric but do not admit finite-index subgroups $G' -TITLE: References for computation of 2-primary stable 64-stem ${_2\pi_{64}^s}$? -QUESTION [6 upvotes]: I want to learn about the $2$-primary component of the stable homotopy groups of spheres in dimension $64$. Since the triviality of $61$-st stem has been proved just recently, I thought that either I am unable to find about ${_2\pi_{64}^s}$ or still there is something unknown about this. I did not find anything in Ravenel's Green book about his, and not sure if there is anything in Toda's book? I will be very grateful for any advise on this!?! - -REPLY [5 votes]: This is the result from Kochman and Mahowald: -   -Which you can get from "On the Computation of Stable Stems", Contemporary Mathematics Volume 181, 1995.<|endoftext|> -TITLE: Fontaine, J.-M.; Illusie, L. p-adic periods------Does any one have the following article? -QUESTION [8 upvotes]: Fontaine, J.-M.; Illusie, L. -p-adic periods: A survey. (English) -Ramanan, S. (ed.) et al., Proceedings of the Indo-French conference on geometry held in Bombay, India, 1989. -If anyone has the above paper can you please share it? Thank you. - -REPLY [4 votes]: The desired paper is freely available on Fontaine's webpage! Here is the first part, and here is the second.<|endoftext|> -TITLE: Singularities at worst like a hyperplane arrangement -QUESTION [6 upvotes]: Is there a standard name for the type of singularities a codimension-$1$ subvariety of a smooth algebraic variety has when it looks locally (possibly analytically) like an arrangement of hyperplanes? «Normal crossings» refers to the special situation in which locally it looks like a subarrangement of the Boolean arrangement, that is, some coordinate hyperplanes. - -REPLY [5 votes]: This paper uses the term "arrangement of smooth, complex algebraic hypersurfaces", or simply, "arrangement of smooth hypersurfaces". To quote: "Our goal here is to further generalize these results to a much wider class of arrangements of hypersurfaces, by which we mean a collection of smooth, irreducible, codimension 1 subvarieties which are embedded in a smooth, connected, complex projective algebraic variety, and which intersect locally like hyperplanes." -In another paper on the subject, Clément Dupont simply uses the term "hypersurface arrangement". To quote: "We develop a model for the cohomology of the complement of a hypersurface arrangement inside a smooth projective complex variety. This generalizes the case of normal crossing divisors, discovered by P. Deligne in the context of the mixed Hodge theory of smooth complex varieties."<|endoftext|> -TITLE: Reference request for Functional Analysis -QUESTION [8 upvotes]: Does anyone know a book that motivates the beginning of functional analysis in a clear way? -By "clear," I mean that it shows why one would want to define Hilbert spaces and why the theorems are motivated. I know of Dieudonne's History of Functional Analysis, but I am looking for something that also explains the theory. - -REPLY [6 votes]: In addition to books on history (J. Dieudonne, History of functional analysis (review) is very good and easy reading), I recommend some -books by the founding fathers: - -John von Neumann, Mathematische Grundlagen der quantum Mechanik, (there are English and Russian translations). Functional analysis (in Hilbert space) is related to quantum mechanics in the same way as Calculus to classical mechanics. -F. Riesz and B. Szőkefalvi-Nagy, Functional analysis (multiple editions in French, German, Russian, English). -P. Levy, Problèmes concrets d'analyse fonctionnelle. Avec un complément sur les fonctionnelles analytiques par F. Pellegrino. 2d ed. Gauthier-Villars, Paris, 1951. (There is a Russian translation). -L. Schwartz, Mathematics for the physical sciences. Hermann, Paris; Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont. 1966.<|endoftext|> -TITLE: "Urban paths" in a Riemannian manifold -QUESTION [6 upvotes]: Fix a compact Riemannian manifold $M$, a base point $p\in M$, a basis $B$ for the tangent space $T_p$, and a destination $q$. -The metric on $M$ determines its Levi-Civita connection. -My goal consists of finding a short path from $p$ to $q$ of a special form: such a path must leave $p$ traveling along a geodesic lying in a direction contained in $B$ and, carrying $B$ along according to the connection, arrive at new point $p_1$; if $p_1\not=q$, then at $p_1$ the path must turn in a new direction belonging the the transported basis. The steps then repeat until finally we arrive at $q$ after a journey possessing a finite number of legs. Call such a path urban (because we must stay on the "city streets" determined by $B$) unless the literature already has a name for this. -Clearly, with a bound on the number of legs, some urban path attains the -infinimum length of all urban paths with no more than that many legs. -My questions: without a bound on the number of legs, must some urban path still attain the infinimum? If yes, can we get a bound, based solely on the dimension of $M$, for the number of legs in this ``urban geodesic''? - -REPLY [3 votes]: Let us modify the problem a bit. -Consider all paths from $p$ to $q$. -For each path $\gamma$, transport the frame along $\gamma$ and measure the $\ell^1$-length $\gamma$ --- the length of $\gamma$ in the $\ell^1$-norm provided by the frame. -Note that for urban paths the $\ell^1$-length coincide with the usual length -Now consider a path $\gamma_0$ from $p$ to $q$ which minimizes $\ell^1$-length. -Note that arbitrarily close to $\gamma_0$ there is an urban path with arbitrarily close length. -So if $\gamma_0$ is a strict minimum and it is not an urban path then there's no chance of getting a minimum among the urban paths. -An example can be constructed the following way: -Take Euclidean plane and multiply its metric by conformal factor $1+(x-y)^2$; -the path $\gamma_0(t)=(t,t)$ satisfies the condition above.<|endoftext|> -TITLE: Non-invertible version of unitary intertwiners between correspondences of $C^\ast$-algebras -QUESTION [6 upvotes]: There is a version of the Eilenberg-Watts theorem for $C^\ast$-algebras, where functors between appropriate module categories correspond to what are called 'correspondences' of $C^\ast$-algebras. These are one-sided Hilbert modules for one of the $C^\ast$-categories together with an action on this module by the other $C^\ast$-algebra. In the literature one finds a (weak) (2,1)-category where the objects are $C^\ast$-algebras, 1-arrows are correspondences, and 2-arrows are unitary intertwiners. However, clearly if one thinks of the categories of representations, the appropriate functors between such categories, and the natural transformations between these functors, then one could have non-invertible 2-arrows. - -What intertwiners play the analogue of the non-invertible natural transformations? - -REPLY [3 votes]: As you note in your comment, a lot depends on which category of modules you choose to consider. -If you work with the category of Hilbert $C^*$-modules, with adjointable operators as morphisms, then the appropriate kind of maps between correspondences would be the adjointable bimodule maps: for $C^*$-algebras $A$ and $B$, the category of (nondegenerate) $C^*$-correspondences from $A$ to $B$, with adjointable bimodule maps as morphisms, is equivalent to the category of strongly continuous $*$-functors from $C^*$-modules over $A$ to $C^*$-modules over $B$, with natural transformations as morphisms. This is due to Blecher (Mathematische Annalen, 1997). -If, on the other hand, you consider operator modules, as in the paper of Blecher cited in the comments, then the appropriate class of morphisms would be completely bounded bimodule maps---and, one could argue, the appropriate class of bimodules is that of operator bimodules. In the paper you cited (Math. Scand. 2001), Blecher proves the remarkable fact that every invertible operator bimodule between $C^*$-algebras is a $C^*$-correspondence.<|endoftext|> -TITLE: Explicit forms for the roots of Eulerian polynomials -QUESTION [12 upvotes]: Let $E_n(z)$ be the Eulerian polynomial -$$E_n(z) = \sum_{\tau \in \mathfrak{S}_n} z^{\operatorname{des}(\tau)}$$ -where $\mathfrak{S}_n$ denotes the set of all permutations of $\{1,\ldots,n\}$ and $\operatorname{des}(\tau)$ is the number of descents $\tau(i) > \tau(i+1)$ in the permutation $\tau$. -These are studied in great detail, see its OEIS wiki page for other definitions and several properties. -In particular, it is known the Eulerian polynomial has only negative and simple roots, -$$E_n(z) = \prod_{i=1}^n (z+q^{(n)}_i)$$ -for different positive numbers $q^{(n)}_i$. -My question now is - -What is known about the $q^{(n)}_i$'s? Do they have an explicit description? - -(It is known that the roots of $E_n$ separate the roots of $E_{n+1}$. -This is, -$$q_1^{(n+1)} < q_1^{(n)} < \cdots < q_{n}^{(n+1)}< q_n^{(n)} < q_{n+1}^{(n+1)}$$ -when they come in sorted order. That's not the type of property I am looking for, but only properties towards their explicit values.) -Here are the first two examples and the type of property I would like to have answered from a description I search for: -$$ -\begin{align*} -E_2(z) &= z^2 + 4z + 1 - = (z+2-\sqrt{3})(z+2+\sqrt{3}) \\ -E_3(z) &= z^3 + 11z^2 + 11z + 1 - = (z+5+2\sqrt{6})(z+5-2\sqrt{6})(z+1) -\end{align*} -$$ -(The roots become more complicated than $a\pm b\sqrt{c}$ for bigger $n$'s.) -It is known that the mean value of the discrete distribution given by $E_n$ is $n/2$ and the variance is $(n+2)/12$. -This can be used to show that -$$ -\begin{align*} - \sum_i \frac{1}{1+q^{(n)}_i} &= \frac{n}{2} \\ - \sum_i \frac{q^{(n)}_i}{\big(1+q^{(n)}_i\big)^2} &= \frac{n+2}{12} -\end{align*} -$$ -Doing this computation in the first example yields -$$ - \frac{1}{1+2+\sqrt{3}} + \frac{1}{1+2-\sqrt{3}} = 1 -$$ -and -$$ - \frac{2+\sqrt{3}}{(1+2+\sqrt{3})^2} + \frac{2-\sqrt{3}}{(1+2-\sqrt{3})^2} = \frac{1}{3} -$$ - -Is there any known immediate property of the roots that can be used to get these identities? - -REPLY [7 votes]: Here is some partial information. If $E_n(z)=\sum A_n(i)x^i$, then -$A_n(i)$ grows roughly like $(i+1)^n$ for fixed $i$. Now $A_n(i)$ is -(up to sign) the $i$th elementary symmetric function of the -roots. Hence $|q_1^{(n)}|$ is about $2^n$. Then $|q_2^{(n)}|$ is about -$(3/2)^n$, $|q_3^{(n)}|$ is about $(4/3)^n$, etc. It is not hard to -show in fact that for fixed $i$, - $$ \lim_{n\to\infty} |q_i^{(n)}|^{1/n} = - \frac{i+1}{i}. $$ -Note also that there is an explicit formula for $A_n(i)$ as a sum with -$i+2$ terms (https://en.wikipedia.org/wiki/Eulerian_number). Thus any -rational symmetric function of the roots can be written in terms of -these sums.<|endoftext|> -TITLE: An equivalent condition for separability of $X^*$ -QUESTION [7 upvotes]: Let $X$ be a Banach space. By the weak operator topology on $B(X)$, we mean the locally convex topology implemented by the following semi-norms: -$$B(X)\to[0,\infty) : T\to|\langle Tx,x^*\rangle|$$ -where $x\in X$ and $x^*\in X^*$. -Q. True or false: Assume that with respect to the weak operator topology, the unit ball of $B(X)$ is second countable. Then the Banach space $X^*$ is separable. -I had two abortive attempts: -1st. Yes it is true: The wot-separability of $B(X)_{\|.\|\leq1}$ guarantees existence two sequences $\{x_n\}\subseteq X$ and $\{x^*_m\}\subseteq X^*$ such that $N_{n,m}$ forms a sub-basic nbhds (at 0) for the weak operator topology where -$$N_{n,m}=\{T\in B(X)_{\|.\|\leq1}: |\langle Tx_n,x^*_m\rangle|<1\}$$ -Let us consider $A$=conv$_r(\{x^*_m\})$, the rational convex hull of $\{x^*_m\}$, which is a countable set. If $A$ is not dense in $X^*$ then one may find $\phi\in X^{**}$ with $\phi(x^*_m)=0$ for all $m\geq1$. It implies that all bounded linear maps of the form of $f\otimes\phi\in B(X,X^{**})$, given by $x\to f(x)\phi$, are contained in the intersection $\bigcap N_{n,m}$. Note that there is no nonzero bounded linear map $T\in B(X)$ contained in the intersection $\bigcap N_{n,m}$but likely there are some operators in $B(X,X^{**})$ contained in the intersection $\bigcap N_{n,m}$! -2ed. To find a counterexample, let us put $X=\ell^1$. We have in general $B(X,Y^{*})\simeq (X\hat{\otimes}Y)^*$. Therefore $B(\ell^1)=(\ell^1\hat{\otimes}c_0)^*$. One may check easily that the inclusion $\iota: B(\ell^1)\to(\ell^1\hat{\otimes}c_0)^*$ is wot-weak star continues but not homeomorphism (to see this, it is enough to consider shifts $T_n(e_k)=e_{n+k}$). Note that the unit ball of $(\ell^1\hat{\otimes}c_0)^*$ is $w^*$-compact and metrizable. Since the inclusion is not hoemorphism, one may not conclude the unit ball of $B(\ell^1)$ is wot-compact and metrizable but probably it remains second countable at least! -Finally note if $X^*$ is separable then the unit ball of $B(X)$ is wot-second countable metrizable space. To see this assertion, note that the projective tensor product $X\hat{\otimes}X^*$ is also separable. Therefore the unit bal of dual space $(X\hat{\otimes}X^*)^*$ is weak-star compact and metrizable. Since the inclusion from ($B(X)_{\|.\|\leq1}$,wot) into the unit ball of $(X\hat{\otimes}X^*)^*$ is a homeomorphism, one may conclude that the unit ball of $B(X)$ is wot-second countable metrizable space. - -REPLY [2 votes]: I think this works. Suppose $B(X)$ with WOT is second countable, so there is a countable base to the topology $(U_n)$. This means that if $(T_i)$ is a net in $B(X)$, then $T_i\rightarrow 0$ WOT if and only if, for each $n$ with $0\in U_n$, there is $i_n$ so that $i\geq i_n \implies T_i\in U_n$. -Each WOT open set is of the form -$$ \mathcal U(T_0,(x_j),(x_j^*)) := \{ T\in B(X) : |\langle (T-T_0)(x_j), x_j^* \rangle|\leq 1 \ (j=1,\cdots,n)\} $$ -where $T_0\in B(X), (x_j)_{j=1}^n\subseteq X, (x_j^*)_{j=1}^n\subseteq X^*$. -Let $Q\subseteq X^*$ be the collection of all $x_j^*$ which occur for some $U_n$. Thus $Q$ is countable, and so the rational linear combinations of $Q$ is also a countable set. -Towards a contradiction, suppose that $X^*$ is not separable. Hence the closed linear span on $Q$ is not all of $X^*$. Thus there is $f_0\in X^*$ and a (bounded) net $(x_i)$ in $X$ such that $f_0(x_i)=1$ for each $i$ but $x^*(x_i)\rightarrow 0$ for each $x^*\in Q$. -Choose some $f_1\in X^*$ and let $T_i$ be the rank-one operator $x\mapsto f_1(x) x_i$. Let $U_n$ have the form $\mathcal U(T_0,(x_j),(x_j^*))$. As each $x_j^*\in Q$ we have that -$$ \langle T_i(x_j), x_j^* \rangle = f_1(x_j) x_j^*(x_i) \rightarrow 0 $$ -and so as $0\in U_n$, if $i$ is sufficiently large also $T_i\in U_n$. -This is a contradiction, as if $f_1(x_1)=1$ say, then $\langle T_i(x_1), f_0 \rangle = f_0(x_i) = 1$ for all $i$, and so $T_i\not\rightarrow 0$ WOT. -Thus $X^*$ is separable.<|endoftext|> -TITLE: $p$-adic exponentials for $p$-adic Lie groups -QUESTION [8 upvotes]: Let $G$ be a $p$-adic Lie group, $\text{Lie}(G)$ its Lie algebra. -Is there any reasonable notion of exponential map $\text{exp} : \text{Lie}(G)\to G$? - -REPLY [5 votes]: Besides the books already mentioned, I highly recommend Michel Lazard's Groupes analytiques p-adiques, which is the original source for a lot of the material in both Dixon-DuSautoy-Mann-Segal's Analytic pro-p groups and Schneider's p-adic Lie groups. Lazard's text was most probably written in close collaboration with Serre and is freely available online: http://www.numdam.org/item?id=PMIHES_1965__26__5_0 -The exponential map and the Hausdorff formula are treated in particular (with a look towards $\mathbb{Z}_p$-integrality) in section 3.2.<|endoftext|> -TITLE: Is it possible to find a copy of the Cartan seminar (1954–1955) on homotopy and Eilenberg–Mac Lane spaces? -QUESTION [8 upvotes]: I am trying to find a comprehensive reference on Cartan's construction, the comparison theorem of Moore and related subjects. I have the following references - -Cartan, H., Algebres d’Eilenberg-Mac Lane, Seminaire Cartan, ENS, 1954-55, expos´es 2 to 11. -(128, 180, 194, 197, 224, 232, 468) -Moore, J.C., Cartan’s constructions, the homology of $K(\pi,n)$’s, and some later developments, -Colloque “Analyse et Topologie” en l’Honneur de Henri Cartan (Orsay, 1974), pp. 173–212. -Asterisque, No. 32-33, Soc. Math. France, Paris, 1976. (194) - -Does anybody know where one might obtain a copy of these, or know about other -references about the topics in those notes and articles? - -REPLY [9 votes]: The Cartan Seminar is at: http://www.numdam.org/actas/SHC -Actually most French journals and seminars have been digitized and can be fount at: http://www.numdam.org/ -Unfortunately, Astérisque has not been digitized completely and is not in the list; perhaps the only ones that are digitized are the volumes corresponding to Bourbaki seminars published in Astérisque.<|endoftext|> -TITLE: Motivic $\mathbf{Z}(1)$ -QUESTION [7 upvotes]: I know that the Bloch higher Chow complex, $\mathbf{Z}(i)_{\mathcal{M}}$, on a smooth scheme over a field $k$, reads, in degree $1$: -$$\mathbf{Z}(1)_{\mathcal{M}}\simeq\mathbf{G}_m[-1].$$ -How to see this? I did not find it anywhere clearly stated and proved in the literature. -Is there a motivic exponential sequence of some kind? (not that I know) so as to say the complex $0\to \mathbf{Z}(1)\to\mathbf{G}_a\to 0$, which ought to be $\mathbf{Z}(1)_{\mathcal{M}}$, should be quasi-isomorphic to $\mathbf{G}_m[-1]$ via some exponential? (but what is $\mathbf{Z}(1)$, let alone the exponential?!) -This is what happens for Deligne cohomology, for instance. - -REPLY [2 votes]: This statement is proven in the article "The Bloch Complex in Codimension One and Arithmetic Duality". The proof is very elementary. -Also it can be found in the "Lecture Notes on Motivic Cohomology" by -Carlo Mazza,Vladimir Voevodsky and Charles Weibel, Part 1 lecture 4.<|endoftext|> -TITLE: Asphericity of 2-complexes -QUESTION [6 upvotes]: Is it decidable whether a finite group presentation is diagrammatically aspherical (that is there is no reduced spherical diagram over this presentation)? Probably - not, but I cannot find a reference. - -REPLY [9 votes]: The answer is no. -This follows from a theorem of Collins and Miller, who constructed a recursive sequence of presentations $P_n$ such that the set of $n$ for which $P_n$ presents the trivial group is recursively enumerable but not recursive, and $P_n$ is aspherical if and only if it presents a non-trivial group. I'll add a precise citation later today.<|endoftext|> -TITLE: The two ways Feynman diagrams appear in mathematics -QUESTION [45 upvotes]: I've heard about two ways mathematicians describe Feynman diagrams: - -They can be seen as "string diagrams" describing various type of arrows (and/or compositions operations on them) in a monoidal closed category. -They are combinatorial tools that allow one to give formulas for the asymptotic expansion of integrals of the form: - -$$ \int_{\mathbb{R}^n} g(x) e^{-S(x)/\hbar} $$ -when $\hbar \rightarrow 0$ in terms of asymptotic expansions for $g$ and $S$ around $0$ (with $S$ having a unique minimum at $0$ and increasing quickly enough at $\infty$ and often with a very simple $g$, like a product of linear forms), as well as some variation of this idea, or for the slightly more subtle ``oscilating integral'' version of it, with $e^{-i S(x)/\hbar}$. -My question is: is there a relation between the two ? -I guess what I would like to see is a "high level" proof of the kind of formula we get in the second point in terms of monoidal categories which explains the link between the terms appearing in the expansion and arrows in a monoidal category... But maybe there is another way to understand it... - -REPLY [8 votes]: Before interpreting them in more advanced language like "string diagrams" or "monoidal closed categories" it might be good to stress that Feynman diagrams are very elementary combinatorial objects which encode contractions of tensors. -By tensor I mean an array of numbers like $A=(A_{a,b,c})$ with indices $a,b,c$ running over some finite sets which need to be specified. -If you have such objects say $B_{abcd}$ and $C_{ab}$ you can construct new ones like -$$ -T_{a,b,c,d}:=\sum_{e,f,\ldots,l} C_{ae}C_{bg} B_{eghf} C_{fi}C_{hk}B_{iklj}C_{jc}C_{ld} -$$ -Obviously one can produce tons of similar examples of increasing complexity and it is desirable to have a way of encoding precisely such constructions. A natural way of doing that is basically to use pictures or graphs which is what Feynman diagrams are. -Linear algebra "done wrong", i.e., matrix algebra is the particular case where tensors have one (vectors) or two indices (matrices) only. Although, the $n$-dimensional determinant introduces a higher (Levi-Civita) tensor -$\epsilon_{i_1\ldots i_n}$ given by the sign of the permutation $i_1\ldots i_n$ (and zero if indices are repeated). -A significant part of functional analysis is about studying what happens when discrete summation -indices $a,b,\ldots$ become continuous variables that are integrated over. Then, matrices $C_{a,b}$ become kernels $C(x,y)$ which one can make sense of, e.g, as distributions, by invoking the Schwartz Kernel Theorem. -Feynman diagrams, as they are used in quantum field theory, typically correspond -to this infinite-dimensional generalization. For instance, the diagram for the expression $T_{abcd}$ above becomes the main second order contribution to the four-point function of the $\phi^4$ model in $d$ dimensions if one decides that the tensor $C_{ab}$ -now becomes the kernel $C(x,y)$ of the operator $-\Delta+m^2 I$ in $\mathbb{R}^d$ -and one lets the tensor $B_{abcd}$ become the kernel $B(x,y,z,u)$ of the distribution in $S'(\mathbb{R}^{4d})$ -given by the action -$$ -f\longmapsto\ \int_{\mathbb{R}^d} f(x,x,x,x)\ d^dx -$$ -on test functions $f\in S(\mathbb{R}^{4d})$. -As for why this should have to do with the Laplace/stationary phase method, the reason is because Gaussian integration is an "algebraic" operation. Namely, it can be expressed as a differential operator (albeit of infinite order). -For example if $\mu$ is the centered Gaussian measure on $\mathbb{R}^n$ with covariance $C_{a,b}$, then for any polynomial -$P\in \mathbb{R}[x_1,\ldots,x_n]$, -$$ -\int_{\mathbb{R}^n} P(x)\ d\mu(x)=\left.\exp\left(\frac{1}{2}\sum_{a,b=1}^n C_{a,b} \frac{\partial^2}{\partial x_a\partial x_b}\right)\ P(x)\ \right|_{x=0}\ . -$$ -Note that Haar integration on $SU(n)$ can also be expressed as an infinite order differential operator (see my two answers to How to constructively/combinatorially prove Schur-Weyl duality? ). -So Feynman diagrams also appear in invariant theory/representation theory (see my answer to Who invented diagrammatic algebra? for some examples and pictures by Kempe in the case of the invariants of the binary quintic that are given explicitly in nondiagrammatic fashion in my answer to - Explicit formulas for invariants of binary quintic forms ).<|endoftext|> -TITLE: If $f,g \in D[x,y]$ are algebraically dependent over $D$, then $f,g \in D[h]$ for some $h\in D[x,y]$? -QUESTION [6 upvotes]: This question asks: If $f,g \in k[x,y]$ are two algebraically dependent polynomials over an arbitrary field $k$, is it true that there exists a polynomial $h \in k[x,y]$ such that $f,g \in k[h]$, -namely, $f=u(h)$ and $g=v(h)$ for some $u(t),v(t) \in k[t]$; -the answer is positive. - -Is it possible to replace the field $k$ by an integral domain $D$? - Namely: If $f,g \in D[x,y]$ are two algebraically dependent polynomials over an arbitrary integral domain $D$, is it true that there exists a polynomial $h \in D[x,y]$ such that $f,g \in D[h]$? - -Denote the field of fractions of $D$ by $Q(D)$. -It is clear that if $f,g \in D[x,y] \subset Q(D)[x,y]$ are two algebraically dependent polynomials over $D$, then from the above question there exists a polynomial $h \in Q(D)[x,y]$ such that $f,g \in Q(D)[h]$, -namely, $f=u(h)$ and $g=v(h)$ for some $u(t),v(t) \in Q(D)[t]$. -I do not see why, for example, $D[x][y] \ni f=u(h)=u_mh^m+\cdots+u_1h+u_0$ should imply that $h \in D[x,y]$ and $u_j \in D$ -(changing variables does not seem to help, namely if the leading term is $cy^l$, -with $c \in Q(D)$). -Any comments are welcome. - -REPLY [9 votes]: No. Choose a field $k$ and $D=k[u^2,u^3,v^2,v^3,uv]\subset k[u,v]$, so $D$ is a noetherian domain. In $D[x,y]$, choose $f=(ux+vy)^2$ and $g=(ux+vy)^3$; they are clearly algebraically dependent (but $ux+vy\notin D[x,y]$). Write $K=k(u,v)=\mathrm{Frac}(D)$. - -Claim: there is no $P\in D[x,y]$ such that $f,g\in D[P]$. - -By contradiction, let $P\in D[x,y]$ such that $f,g\in D[P]$. So $f=r_1(P)$, $g=r_2(P)$ with $r_1,r_2\in D[t]$. Since $f^3=g^2$ and $K[t]$ is a UFD, there exists $r\in K[t]$ such that $r_1=r^2$ and $r_2=r^3$. So $r(P)^2=f$. So $r(P)=\pm (ux+vy)$; up to change $(u,v)$ to $(-u,-v)$, let us suppose $r(P)=ux+vy$. This implies that $r\in K[t]$ and $P\in K[x,y]$ have degree 1. Write $r=at+b$ and $P=cx+dy+e$; then $r(P)=acx+ady+ae+b=ux+vy$. So $ac=u$, $ad=v$, $ae+b=0$. -So $c,d$ are nonzero; then $c/d=u/v$, hence $cv=du$. Since $c,d\in k[u,v]$, we can write $c=uq$ and $d=vq$ with $q\in k[u,v]$. We have $aq=1$. Since $r^2\in D[t]$, we have $a^2\in D\subset k[u,v]$. So $a^2$ is invertible in $k[u,v]$, and hence $a\in k^*$. So $u=a^{-1}c\in D$, a contradiction.<|endoftext|> -TITLE: Finding a PA cut in a nonstandard model of PA -QUESTION [7 upvotes]: For a certain project I am currently working on, I need to be able to find PA cuts in nonstandard models of PA, in desirable intervals. For example, I wonder if the following is true, where $\newcommand\PA{\text{PA}}\PA_k$ refers to the theory with only $\Sigma_k$ induction. -Question. If $M$ is a model of $\PA$ in which $\PA_{k-1}$ is consistent, but $\PA_k$ is not (so $k$ is nonstandard), then is there a $\PA$ cut in $M$ above $k$ in which $\PA_k$ is consistent? -That is, I want to cut $M$ below the first proof of a contradiction in $\PA_k$, but above $k$, and have $\PA+\text{Con}(\PA_k)$. -Alternatively, is there some other $\Sigma_1$ property of $k$, other than $\neg\text{Con}(\PA_k)$, such that I can always find a $\PA$ cut in $M$ between $k$ and the witness of that property? Kameryn Williams suggested that the Paris-Harrington result may provide this, since it is designed to ensure $\PA$ cuts below the corresponding PH-Ramsey number. But I would need, however, that one can always end-extend the model so as to make the $\Sigma_1$ property true. Does the PH construction have both these features? -With the consistency statements, for example, for any nonstandard $k$ in any model $M$ of $\PA$, there is always an end-extension of $M$ to a model of $\PA$ with $\neg\text{Con}(\PA_k)$. - -REPLY [5 votes]: Q1 -Theorem1. Let $M$ be a countable nonstandard model of $PA$, let $r,c,a\in M\setminus\mathbb{N}$, $M\models r,c\leq a$, and suppose $M\models Con_{{\bf I}\Sigma_r}$. Then there exists a model $K$ of $PA$ such that $a\in K$ and - -$M|_a=K|_a$, -$M|_{2^a}\subseteq K$, -$K\models\exists d<2^{a^c}Pr_{{\bf I}\Sigma_r}(d,\ulcorner\bot\urcorner)$, -$K\models Con_{{\bf I}\Sigma_{r-1}}$. - -Proof. See this paper. - -The proof of theorem 1 uses finite Godel's theorem and some other provability results: -Theorem2. Let $T\supseteq {\bf I}\Delta_0+\mathsf{EXP}$ be a consistent theory with provability predicate $Pr_T(x,y)$ such that $Pr_T(x,y)\in\mathsf{P}$, then there exists a $\epsilon >0$ such that the length of the shortest $T$-proof of $Con_T(\bar{n})$ defined by $\forall x(|x|\leq \bar{n}\to \neg Pr_T(x,\ulcorner \bot \urcorner))$ is at least $n^\epsilon$. -Proof. See this paper. - -Let $M$ be a countable nonstandard model of $PA+Con_{{\bf I}\Sigma_{r}}$ for some nonstandard $r\in M$, then by theorem 1, there exists a countable nonstandard model $K\models PA$ such that: - -$M|_{2^{r+1}}\subseteq K$, -$K\models\exists d<2^{(r+1)^r}Pr_{{\bf I}\Sigma_r}(d,\ulcorner\bot\urcorner)$, -$K\models Con_{{\bf I}\Sigma_{r-1}}$. - -This implies that for any $PA$-cut $K'$ in $K$ above $r$, $K|_{2^{(r+1)^r}}\subseteq K'$, hence $K'\models \exists d<2^{(r+1)^r}Pr_{{\bf I}\Sigma_r}(d,\ulcorner\bot\urcorner)$, therfore the answer of the question is no. - -Q2 -Let $Y(x,y)$ be the $PA$ indicator defined in theorem 3.23 of metamathematics of first-order arithmetic. Suppose $(Y(x,y)=z) \equiv \exists w \psi(x,y,z,w)$ for some $\psi\in\Delta_0$. Define $\phi(x)$ by $\exists y \left(\psi(x,(y)_0,(y)_1,(y)_2)\land x< (y)_0\land x< (y)_1\right)$. Let $M$ be a nonstandard model of $PA$. Suppose for some nonstandard element $k\in M$, $M\models \phi(k)$. This implies that there exists a least element $c\in M$ such that $M\models \psi(k,(c)_0,(c)_1,(c)_2)\land k< (c)_0\land k< (c)_1$. This implies: - -$Y(k,(c)_0)=(c)_1$, -$(c)_1$ is nonstandard, -$k < (c)_0$, - -therefore by definition of indicator there exists a cut $I$ in $M$ such that: - -$I\models PA$, -$k\in I$, -$(c)_0\not \in I$, and hence $c\not\in I$.<|endoftext|> -TITLE: Map between homology of spectra -QUESTION [5 upvotes]: Let $X$ be a suspension spectra whose $BP$-homology is infinitely generated -($BP_*(X) = \Sigma^d BP_*/I$, where $I$ has the form $I=(v_0^{i_0}, \dots , v_n^{i_n})$ such that the homology is a $BP_*(BP)$ coalgebra). -Let $C_nX$ the fiber of the map $ X \to L_nX$ and let $\Sigma C_n X$ be its cofiber. -What can be said about the natural map $$BP_*(\Sigma C_n X) \to BP_*(\Sigma C_{n-1} X) ? $$ -My guess would be that it's always injective, but i'm not entirely sure about it. -Thanks - -REPLY [9 votes]: For any $m$, there is a Kunneth spectral sequence -$$ -Tor_{BP_*} (K(m)_*, BP_*(X)) \Rightarrow K(m)_* X. -$$ -For your $X$, $BP_*(X)$ is acted on nilpotently by $v_m$ for $0 \leq m \leq n$, and so this spectral sequence starts with zero. -Thus $K(m)_* X = 0$ for $0 \leq m \leq n$, and so for any $d \leq n$ we have -$$ -0 = L_{K(0) \vee K(1) \vee \dots \vee K(d)} X = L_d X, -$$ -by a result of Ravenel's. -Thus the fiber $C_d(X)$ of the map $X \to L_d(X)$ is equivalent to $X$, and $C_{n} X \to C_{n-1} X \to X$ are all equivalences. In particular, we get an isomorphism on $BP_*$.<|endoftext|> -TITLE: Infinite projective plane with small edges -QUESTION [7 upvotes]: Let $\kappa$ be an infinite cardinal. We say $E\subseteq {\cal P}(\kappa)$ is an infinite projective plane on $\kappa$ if - -$e_1\neq e_2\in E$ implies $|e_1\cap e_2| = 1$, and -whenever $n\neq m\in \kappa$, there is $e\in E$ with $\{n,m\}\subseteq e$. - -Is it possible to find an infinite projective plane $E$ on an infinite cardinal $\kappa$ such that for all members of $E$ we have $|e|<\kappa$? - -REPLY [3 votes]: Encouraged by (Martin?) Goldstern, I resurrect some half-remembered arguments from Emil Artin's Geometric Algebra, although there likely are more recent sources and perhaps cleaner presentations than below. -Dominic's setup and condition 2 implies there is at least one line. If there is exactly one line, then it is the whole space. Moving on to the case of there being at least two lines with none of them singleton sets, given a line (warning: notational conflict with question ahead) $E$ there is then a point $p$ which is not on $E$. The number of lines through $p$ is in 1-1 correspondence with the number $\lambda$ of points on $E$; condition 1 shows there are at most $\lambda$ many lines through $p$, while condition 2 gives at least that many lines. However, this is true for every point $p$ off of $E$. If there is another line $F$ different from $E$ not containing $p$ (and there is since we have enough points and lines, otherwise $p$ is the only point off of $E$), the same argument shows $F$ has as many points as does $E$. This further extends to all lines through $p$: each of those has $\lambda$ many points. -Finally we have $\lambda \leq \kappa \leq \lambda \cdot \lambda$. The left inequality follows because $E$ is a proper subset of the space. The right hand follows because all the lines through $p$ cover the space. If we are in a set theory with nice cardinal arithmetic, then the left hand side equals the right hand side, and the answer to the question becomes no. The case involving singleton sets is left to Joel. -Gerhard "More Comfortable With Finitary Logic" Paseman, 2017.12.29.<|endoftext|> -TITLE: Frobenius automorphisms of cohomology of a variety -QUESTION [6 upvotes]: Suppose $X$ is a smooth variety defined over $\mathbb{Q}$. There are (at least) two automorphisms of cohomology groups of $X$ that are called "Frobenius", and I would like to understand how they are related. - -If $p$ is a prime of good reduction for $X$, then $H^n_{dR}(X)\otimes\mathbb{Q}_p$ depends functorially on the special fiber of an integral model of $X$. The Frobenius endomorphism of the special fiber induces an automorphism $F_p\in GL\big(H^n_{dR}(X)\otimes\mathbb{Q}_p\big)$. -Fix a prime $\ell$ and an algebraic closure $\bar{\mathbb{Q}}$ of $\mathbb{Q}$. Then $H^n_{et}(X_{\bar{\mathbb{Q}}},\mathbb{Q}_{\ell})$ comes with an action of $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ that is unramified for all but finitely many primes $p$. For unramified $p$, there is a well-defined conjugacy class $\Phi_p\subset GL\big(H^n_{et}(X_{\bar{\mathbb{Q}}},\mathbb{Q}_{\ell})\big)$ coming from the conjugacy class of the Frobenius at $p$ in $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. - - -What is the relationship between $F_p$ and $\Phi_p$? - -REPLY [5 votes]: I think you are intending to take everything with smooth proper varieties (or else there is no good notion of a prime of good reduction). -In this case, the two Frobenius elements have the same characteristic polynomial. -In the smooth projective case, this is due to Katz-Messing. Translating this into your language requires using the isomorphism between crystalline cohomology of the special fiber and de Rham cohomology of the lift, and the isomorphism between etale cohomology of the special fiber and the generic fiber. -In the general case it is sufficient to know that both cohomology theories are pure and satisfy the Grothendieck-Lefschetz fixed point formula. Then the characteristic polynomial of Frobenius in each case will be a term of the Weil zeta function, which is independent of the cohomology, and hence will be equal in both cases. For crystalline cohomology the purity should follow from Katz-Messing by an alteration argument, as in Chiarellotto, or from Kedlaya's $p$-adic proof of the Weil conjectures.<|endoftext|> -TITLE: Criterion for homotopy pullback square of simplicial categories -QUESTION [10 upvotes]: Assume given a pullback square of simplicial categories -$$\begin{array}[c]{ccc} -A&{\rightarrow}&B\\ -\downarrow&&\downarrow\\ -C&{\rightarrow}&D. -\end{array}$$ -Suppose further that one of the induced arrows $Ho (B) \to Ho(D)$ or $Ho(C) \to Ho(D)$ is an isofibration, and for each couple of objects $x,y \in A$, the induced pullback square of simplicial mapping spaces (I abuse the notation by writing $x,y$ instead of their images in $B,C,D$) -$$\begin{array}[c]{ccc} -A(x,y)&{\rightarrow}&B(x,y)\\ -\downarrow&&\downarrow\\ -C(x,y)&{\rightarrow}&D(x,y) -\end{array}$$ -is a homotopy pullback. -Does this imply that the original square is in fact a homotopy pullback of simplicial categories? - -REPLY [5 votes]: Yes. In fact such a square can be replaced with a weakly equivalent Reedy fibrant pullback square without changing the object set of any of the simplicial categories. For a proof see, e.g., Lemma 3.1.11 in this paper.<|endoftext|> -TITLE: Where was $I_x/I_x^2$ first introduced? (DG or AG) -QUESTION [15 upvotes]: Cotangent space appears in both differential geometry (DG) and algebraic geometry (AG). -In DG, given a smooth manifold $M$ and $x\in M$ one has an isomorphism $I_x/I_x^2 \cong T^*_xM$, where $I_x$ is an ideal of functions vanishing at $x$ in ring $C^\infty(M)$ (or $C_x^\infty(M)$). One can prove this isomorphism algebraically, or using Hadamard's lemma like argument. -In AG, given an affine variety $V$ and $x\in V$ one defines Zariski cotangent space as $\mathfrak{m}_x/\mathfrak{m}_x^2$, where $\mathfrak{m}_x$ is defined as an ideal of functions vanishing at $x$ in ring $O_{V,x}$. -Clearly AG introduced its version of tangent and cotangent spaces based on DG. However, It seems to me that in DG the description of cotangent space as $I_x/I_x^2$ is rarely used. - -Question. Was the description of the cotangent space $T^*_xM$ as $I_x/I_x^2$ been known before cotangent Zariski space was introduced? - -REPLY [23 votes]: Zariski formulated the criterion for smoothness at a point in terms of the dimension of $I_x/I_x^2$ (to use your notation) as Theorem 3.2 in the paper https://www.jstor.org/stable/pdf/2371499.pdf from 1939 (see p. 260). See the paragraph and theorem preceding that also. He singles out Theorem 3.2 in the first paragraph of the paper, with no reference to earlier work by anyone else. In his later paper http://www.ams.org/journals/tran/1947-062-01/S0002-9947-1947-0021694-1/S0002-9947-1947-0021694-1.pdf from 1947, the discussion on pages 2 and 3 compares two ways of describing simple points: what Zariski calls a "novel and intrinsic" definition A from his earlier work (which involves $I_x/I_x^2$) and a "time-honored and classical" definition B (which involves a Jacobian rank). He writes on page 3 "There -is ample evidence in the present paper, as well as in previous -papers of ours, in support of the thesis that it is the more general -concept of a simple point, as defined in A, that constitutes the natural -generalization of the classical concept of simple point." -In summary, Zariski is the person who realized the importance of $I_x/I_x^2$ in geometry, so it first arose in algebraic geometry rather than differential geometry.<|endoftext|> -TITLE: About taking an expectation over orthogonal matrices -QUESTION [6 upvotes]: Say $Q$ is a random variable which is sampling orthogonal matrices in $m$ dimensions using the Haar measure on $O(m)$. Let $A$ and $B$ be some (fixed) subset of rows and columns of $Q$ such that $\vert A \vert = \vert B \vert = k$. -Now is such an identity true? (If yes then could you kindly give the proof or a reference for it!) -$\mathbb{E}_{Q \sim O(m)}[det^2(Q_{A,B})] = \frac {1}{\binom {m}{k} }$ -where by $Q_{A,B}$ we mean the submatrix of $Q$ corresponding to the rows from $A$ and columns from $B$. - -If it seems necessary for the proof feel free to assume that $Q \sim SO(m)$ - -REPLY [8 votes]: Let's denote $S=\{1,2,\dots m\}$ and $A\subset S$ with $|A|=k$. Then from orthogonality of $Q$ we have -$$Q_{A,S}Q_{A,S}^{T}=I_k$$ therefore Cauchy Binet tells us that -$$\sum_{B\in \binom{S}{k}}\det(Q_{A,B})^2=1.$$ -If you apply expectation on both sides and notice that $E(\det(Q_{A,B})^2)$ doesn't depend on $B$ by invariance of the Haar measure you get your identity.<|endoftext|> -TITLE: Levi-Civita connections from metrics on the orthogonal frame bundle -QUESTION [10 upvotes]: Following Kobayashi and Nomizu, a connection on a manifold is given by a establishing a notion of horizontal vector in the tangent space of a frame bundle. (Alternative approaches make covariant differentiation foundational.) -An important step in developing Riemannian geometry consists of isolating the Levi-Civita connection as that connection with zero torsion that preserves the metric. -Could an alternative approach to defining the Levi-Civita connection go like this: Given a manifold $M$ with Riemannian metric, construct some natural (family of?) Riemannian metrics on the orthogonal frame bundle of $M$. Then simply define "horizontal" to mean orthogonal (in the sense of the constructed metric) to vertical? -Pedagogically, this might offer a bypass around defining and studying torsion. -About my "family of" hedge. There may be no canonical way to compare the scale of vertical vectors, essentially elements of the Lie algebra of the orthogonal group, with more general vectors. -If this is worked out anywhere, I'd appreciate a reference. If there's some obstruction to this approach, I'd appreciate an explanation. -======== -I do see that this is related to A geometric interpretation of the Levi-Civita connection? -and -Intuition for Levi-Civita connection? . -I have also asked this on Mathematics Stack Exchange without feeling satisfied by the responses there, as you can see from the comments: -https://math.stackexchange.com/questions/2529479/levi-civita-connections-from-metrics-on-the-orthogonal-frame-bundle/2530478?noredirect=1#comment5228620_2530478 - -REPLY [3 votes]: Sure. Take the structure equations of Cartan, with soldering forms $\omega^i$ and connection forms $\omega^i_j$, and then use them as the orthonormal basis of 1-forms for a metric: $ds^2=\sum (\omega^i)^2 + \sum (\omega^i_j)^2$. This metric appears in some work in physics, I think. The same idea works even for a Lorentzian metric down on the manifold: you get a Riemannian metric on the frame bundle.<|endoftext|> -TITLE: The integral cohomology of real projective space -QUESTION [9 upvotes]: I've run across a way of combining the integral cohomology of the real projective space $RP^\infty$ with its cohomology with twisted coefficients, that seems very simple and natural, but which I don't recall every seeing before, so my question is: Has this been noticed before and, if so, where is it published? -Let me describe the result in a relatively simple way, then comment on how I actually came to it. We know that $H^n(RP^\infty;\mathbb{Z})$ is $\mathbb Z$ if $n=0$, $\mathbb Z/2$ if $n>0$ is even, and 0 otherwise. If $\mathbb Z_-$ denotes the local coefficient system on $RP^\infty$ on which the nontrivial loop acts as $-1$ on $\mathbb Z$, we know that $H^n(RP^\infty;\mathbb Z_-)$ is $\mathbb Z/2$ if $n>0$ is odd, and 0 otherwise. -Let $R = \mathbb{Z}\times\mathbb{Z}/2$ and write elements of $R$ as $n + \epsilon\gamma$, where $n\in\mathbb Z$ and $\epsilon\in\mathbb Z/2$. Define a group graded on $R$ by -$$ - H^{n+\epsilon\gamma}(RP^\infty) = - \begin{cases} - H^n(RP^\infty;\mathbb Z) & \text{if $\epsilon=0$} \\ - H^n(RP^\infty;\mathbb Z_-) & \text{if $\epsilon=1$.} - \end{cases} -$$ -This is a graded ring, using the pairing of coefficient systems $\mathbb Z_- \otimes \mathbb Z_- \cong \mathbb Z$, etc. With this definition, -$$ - H^*(RP^\infty) \cong \mathbb Z[w]/\langle 2w \rangle -$$ -where $w\in H^{1+\gamma}(RP^\infty) = H^1(RP^\infty;\mathbb Z_-)$. (The class $w$ is the Euler class of the canonical line bundle over $RP^\infty$.) This could also be viewed as a description of the cohomology of the group $\mathbb Z/2$, of course. Similar simple descriptions can be given for the cohomologies of the truncated spaces $RP^k$. -Where this actually came from: Stefan Waner and I recently published a Springer Lecture Notes volume in which we describe ordinary equivariant (co)homology graded on "representations of the fundamental groupoid" of a space. This can be applied in the nonequivariant case as well — the group of representations of the fundamental groupoid of $RP^\infty$ is exactly the $R$ above, and the general theory leads to the calculation of $H^*(RP^\infty)$ above with this grading. - -REPLY [9 votes]: This description is given in Lemma 1 of - -M. Cadek. The cohomology of $BO(n)$ with twisted integer coefficients. J. Math. Kyoto Univ. 39-2 (1999), 277-286. - -(and Cadek goes on to give similar descriptions of the cohomology of infinite real Grassmannians)<|endoftext|> -TITLE: Which 3-regular graphs are Schreier coset graphs? -QUESTION [5 upvotes]: Given a group $G$ and a subgroup $H$ the Schreier coset graph (w.r.t. some set $S$ of $G$) is the directed (and labelled) graph whose vertices are the cosets of $H$ (i.e. the set $G/H$) and $x \sim y$ if there is a $s \in S$ so that $x = sy$. -When $S$ is symmetric (i.e. $s \in S \implies s^{-1} \in S$), then one can associate to this graph an undirected graph (by replacing the two edges with opposite directions and labels $a$ and $a^{-1}$ by an undirected edge). Note that it could happen that $a = a^{-1}$ which is unproblematic unless the edge makes a loop. So in the undirected graph one needs to distinguish the degree 2 loops (which come from $a$ and $a^{-1}$ acting trivially on the coset $x$) and the degree 1 loops (which come from $a = a^{-1}$ acting trivially on the coset $x$). -Using this construction (and this convention for loops), a Schreier graph is always a regular graph (of degree $|S|$). -Given an even degree regular graph (without "degree 1" loops), then it is possible to find a group $G$ (a free group actually), subgroup $H -TITLE: Strong approximation for principal ideal domains -QUESTION [11 upvotes]: A well known consequence of the strong approximation theorem for semisimple simply connected algebraic groups over a number field is that certain reduction maps are surjective, for example, the canonical projection $Sp_n({\mathbb Z}) \to Sp_n({\mathbb Z}/m{\mathbb Z})$ is surjective for any modulus $m$. This latter fact has been proven by Newman and Smart (Acta Arithmetica 9, 1964) in a rather elementary way, and their proof appears to carry over to an arbitrary principal ideal domain (with an obvious modification in the proof of Lemma 1: Replace $E^+,E^-$ by the upper triangular part of the matrix $E$ (with zeros below the diagonal) and its transpose). -I couldn't locate an explicit reference for this statement or similar results on surjectivity of reduction maps for other groups over a general principal ideal domain $R$ (or maybe other substitutes for the strong approximation theorem), except the review MR0865878 (88b:20072) of an article of Zhang and You in Dongbei Shida Xuebao (found via "Citations from reviews" in MathSciNet), which mentions in passing that the reduction map modulo an arbitrary $q$ for $Sp_n(R)$ is onto. -The question then is: Does anybody know more references for this problem? - -REPLY [8 votes]: For $G(R) = SL_n(R)$ (see, e.g., this MO post) or $G(R) = Sp_{2n}(R)$, there is a common line of reasoning in order to prove that the reduction modulo $\mathfrak{a}$, say $\varphi_{\mathfrak{a}}: G(R) \rightarrow G(R/\mathfrak{a})$, is surjective. It consists in showing that $G(R/\mathfrak{a})$ is generated by elementary matrices any of which has a lift in $G(R)$. -For $SL_n(R)$, these elementary matrices are $E_{ij}(r) = 1_n + re_{ij}$ with $1 \le i \neq j \le n$ and where $1_n$ is the $n$-by-$n$ identity matrix and $e_{ij}$ differs from the zero matrix only by its $(i, j)$ entry, which is $r \in R$. -For $Sp_{2n}(R)$, these elementary matrices are -$ -SE_{ij}(r) = \left\{ -\begin{array}{ccc} -1_{2n} + re_{ij} & \text{ if } i = j',\\ -1_{2n} + re_{ij} - (-1)^{i + j}re_{j'i'} & \text{otherwise}. -\end{array}\right. -$ -with $1 \le i \neq j \le n$ and -where $i \mapsto i'$ is the permutation of $\mathbb{N}$ defined by $(2i)' = 2i - 1$ and $(2i - 1)' = 2i$; alternatively, this is the infinite product of transpositions $(12)(34)(56) \cdots$. -If $SL_n(R)$ is generated by the matrices $E_{ij}(r)$ with $r \in R$, the ring $R$ is said to be $GE_n$-ring in P. M. Cohn's terminology [1]. There are many examples of rings which are, or which are not $GE_n$-rings, see [1, 3] or this MO post. I don't know about the terminology for $Sp_{2n}(R)$, but here is a claim which addresses both groups. - - -Claim. Let $R$ be a Noetherian one-dimensional domain, e.g., $R$ is a principal ideal domain. Let $G(R) = SL_n(R)$ or $G(R) = Sp_{2n}(R)$. Then $\varphi_{\mathfrak{a}}$ is surjective for any ideal $\mathfrak{a}$ of $R$. -Proof. We can assume that $\mathfrak{a} \neq 0$. Since $R/\mathfrak{a}$ is zero-dimensional, its Bass'stable rank is $1$. Therefore $G(R/\mathfrak{a})$ is generated by elementary matrices, see [1] and [2, Theorem 7.3.b]. Because any of these matrices has a lift in $G(R)$, the map $\varphi_{\mathfrak{a}}$ is surjective. - - -As explained by Marty, the above should apply more generally to simply connected Chevalley groups. Let $\Phi$ be a reduced irreducible root system, $R$ be a commutative ring, and let $G = G(\Phi, R)$ be the simply connected Chevalley group of type $\Phi$ over $R$. Let $E(\Phi, R)$ be the elementary subgroup of $G$. - - -Claim to be checked. Let $\Phi$ be a reduced irreducible root system and let $R$ be a Noetherian one-dimensional domain. Then $\varphi_{\mathfrak{a}}$ is surjective for any ideal $\mathfrak{a}$ of $R$. -Proof. The condition on $R$ should be sufficient to ensure that - $G(\Phi, R/\mathfrak{a}) = E(\Phi, R/\mathfrak{a})$ for any non-zero ideal $\mathfrak{a}$. - - - -[1] P. Cohn, "On the structure of the $GL_2$ of a ring", 1966. -[2] L. Vaserstein, A. Suslin, "Serre's problem on projective modules over polynomial rings, and algebraic $K$-theory", 1976. -[3] F. Grunewald, J. Mennicke and L. Vaserstein, "On the groups $SL_2(\mathbb{Z}[x])$ and $SL_2(k[x,y])$", 1994.<|endoftext|> -TITLE: Are irreducible subgroups Zariski-dense? -QUESTION [6 upvotes]: A subgroup $H$ of an algebraic group $G$ is said to be Zariski-dense if its Zariski closure is all of $G$ (or alternatively, if every polynomial which vanishes on all elements of $H$ vanishes identically). -My question: Is any irreducible subgroup of $SL(2,\mathbb{C})$ Zariski-dense? (It's easy to see that the converse is true). -Remark: The original motivation for this question is that I have seen stated in two papers that a subgroup of $SL(2, \mathbb{C})$ is Zariski dense if and only if its natural action on $\mathbb{P}^1$ has no fixed points (which is equivalent to this subgroup being irreducible). However, as explained in the answers and comments below, this statement is in general false. - -REPLY [5 votes]: To summarize the discussion. Let $G$ be a subgroup of $\mathrm{SL}_2(\mathbf{C})$, $H$ its Zariski closure. - -Proposition. Equivalent statements: -(i) $G$ acts irreducibly on $\mathbf{C}^2$; -(ii) $H$ acts irreducibly on $\mathbf{C}^2$; -(iii) $G$ fixes no point on $\mathbb{P}^1_\mathbf{C}$; -(iv) $H$ fixes no point on $\mathbb{P}^1_\mathbf{C}$; -(v) One of the following holds: -$\quad$(a) $H$ (and hence $G$) is finite and non-abelian; -$\quad$(b) $H$ is conjugate to the monomial group, made up of diagonal matrices and anti-diagonal matrices with determinant 1; -$\quad$(c) $H=\mathrm{SL}_2(\mathbf{C})$ (i.e., $G$ is Zariski-dense). - -Proof. The equivalence between (i),(ii),(iii),(iv) is trivial (although that iii/iv implies i/ii is specific to dimension 2). -For a finite group in dimension 2 in characteristic zero, non-irreducibility implies that the action is diagonalizable; hence (a) implies (i). The only fixed points by the group of diagonal matrices with determinant 1 are the two coordinate axes; then are switched by the monomial group, hence (b) implies (iv), and (c) implies (iv) follows (and is clear anyway). -Conversely, suppose that none of (a),(b),(c) holds. If $G$ is finite, this means that $G$ is abelian, its irreducible representations have dimension 1 and the negation of (i) follows. Otherwise, we discuss on the Lie algebra of $H$. If it is conjugate to the upper unipotent or upper triangular subalgebra, then the corresponding connected group fixes a unique point, which is then fixed by $H$ and we obtain the negation of (ii). Since (c) does not hold, it is then conjugate to the subalgebra of diagonal matrices. Hence the Zariski connected component of $H$ consists of the group $D$ diagonal matrices with determinant 1; it has index two in its normalizer (monomial matrices). Since (b) does not hold, we deduce that $H=D$, and again is abelian and does not act irreducibly. $\Box$<|endoftext|> -TITLE: Naturally occuring counting process with a 1/log asymptotics? -QUESTION [7 upvotes]: Besides prime numbers, is there another physically realizable counting process that exhibits a 1/log density ? The reason I am posting this question is that we are measuring the response of a quantum tunneling based counting device and it is showing 1/log asymptotics which we have matched up to 50 million primes. The measurement results can be found here -https://arxiv.org/abs/1711.11032 -In other words, is the 1/log density unique to prime number generation ? -Any suggestions would be helpful before we set up an experiment to precisely measure the electron transport process. - -REPLY [13 votes]: No. One can create many sets of integers whose density decays like 1/log. An easy variant of the set of primes is the set of integers $n$ not divisible by any prime less than $n^\alpha$, for some fixed $0<\alpha<\frac12$. -Here's another example: the set of all integers $n$ such that $n$ is divisible by the number of digits of $n$. (So all one digit numbers, all even two-digit numbers, all three-digit multiples of 3, ..., all ten-digit numbers ending in 0, etc.) The number of digits of $n$ is $\lfloor \log_{10}(n) \rfloor + 1$, so the 1/log behavior emerges naturally from the definition. -A stranger possibility is to take numbers $n$ such that both $n$ and $n+1$ can be written as the sum of two squares, $n=a^2+b^2$ and $n+1=c^2+d^2$. The set of numbers that can be written as the sum of two squares has density decaying like 1/sqrt(log), and this "twin" construction essentially multiplies those two "probabilities" together (up to the leading constant). -As a general rule: when people see that the set of primes has a particular property, they tend to overestimate how unique the set of primes is with respect to that property (or how characteristic that property is of the set of primes).<|endoftext|> -TITLE: Has there been further work on Bender-Brody-Müller approach to RH? -QUESTION [12 upvotes]: Earlier this year (April 4, 2017), a seemingly tantalizing approach of the Riemann Hypothesis based on ideas dating back to Hilbert and Pólya by Bender, Brody and Müller was made publicly available. I remember having read a criticism thereof later, but remain ignorant of what happened since then. A quick googling only brings out blog articles that don't seem to give real news on its status. So, has the heuristics presented in the considered paper been comforted by some further work or not ? - -REPLY [3 votes]: A new paper in Physical Review Letters on related approaches:https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.127.241602 -Also, a summary appears in Physics :https://physics.aps.org/articles/v14/s157<|endoftext|> -TITLE: Generating function in graph theory -QUESTION [19 upvotes]: I am looking for a simple illustration of generating functions in graph theory. -So far, the matching polynomial seems to be the best. But I want something bit richer; at least a derivative should show up. -Please help. - -REPLY [3 votes]: Generating functions are ubiquitous in the enumerative theory of maps, that is, graphs on surfaces. For planar graphs this theory was first developed in the seminal work of Tutte in the late 60s through early 80s. But in recent years there has also been a lot of work devoted to the higher genus maps as well. It turns out that the generating functions for these objects satisfy sophisticated differential equations (coming from the "KP hierary"); see for instance Goulden and Jackson, "The KP hierarchy, branched covers, and triangulations" (https://arxiv.org/abs/0803.3980) or Carrell and Chaupy, "A simple recurrence formula for the number of rooted maps on surfaces by edges and genus" (https://dmtcs.episciences.org/2424/)<|endoftext|> -TITLE: Existence of second order potential for PDE -QUESTION [5 upvotes]: There is a statement in the literature (see the paragraph between equations (18) and (19) in http://aip.scitation.org/doi/10.1063/1.523863), which I would like to generalise, but I don't have a nice proof of the original claim. The statement is the following: - -Given any symmetric tensor field $T_{ab}$ on a 3-dimensional hyperboloid $H$ with - \begin{equation} -D_{[a}T_{b]c}=0, \tag{1} -\end{equation} - there exists a scalar field $T$ on $H$ satisfying - \begin{equation} -T_{ab} = D_a D_b T + T h_{ab}, \tag{2} -\end{equation} - where $h_{ab}$ is the metric on the hyperboloid and $D$ is the covariant derivative on the hyperboloid. The Riemann tensor on $H$ satisfies $R_{abcd} = h_{ac}h_{bd} - h_{ad} h_{bc}$. - -(I am also happy to assume that $T_{ab}$ is traceless, which together with (1) implies that $T_{ab}$ is divergence-less.) -Physically this is saying the the tensor $T_{ab}$ admits a (second order) potential $T$. I would like to generalise it to other dimensions, possibly to other manifolds, etc. But the only proof of this statement which I know, relies on explicit use of spherical harmonics on $H$. -Are there any general methods to determine if given tensor satisfying certain PDE can be written as some differential operator (possibly involving geometric data) acting on some other tensor? -After some search I found Characterizing Hessians among symmetric bilinear tensors, which I can mimic to show that given $T$ then $T_{ab}$ constructed in (2) indeed solves (1). But how can I show that all solutions to (1) are given by (2)? - -REPLY [7 votes]: First of all, you have a sign wrong in your formula for the curvature. The curvature tensor you gave has positive constant sectional curvature +1 while you claim that you want negative sectional curvature (i.e., hyperbolic space), which would flip the sign of $R$. Second, when you speak of spherical harmonics, I believe you must be copying the formula for the unit $n$-sphere, $S^n$, not hyperbolic space $H^n$. This also causes an error in your potential formula (2), which would be correct for the $n$-sphere, but should be $T_{ab} = D_aD_bT - T h_{ab}$ for hyperbolic space. -I'll give the answer for hyperbolic space, since that is what you want, but be aware that you'll need to flip signs to get the same answer for the $n$-sphere. -The result you are seeking follows immediately from the Frobenius theorem, using the techniques mentioned in the MO question you cite at the end. The idea is simply this: Take the tensor $T$ as given. Let $\omega_i$ be any $h$-orthonormal frame field (which can be chosen globally on $H^n$ since it is contractible) and let $\theta_{ij}=-\theta_{ji}$ be the unique $1$-forms satisfying $\mathrm{d}\omega_i = -\theta_{ij}\wedge\omega_j$. (Here and below, I am using the 'Einstein' summation convention.) By the assumption that the sectional curvature is identically $-1$, we have $\mathrm{d}\theta_{ij} = -\theta_{ik}\wedge\theta_{kj} - \omega_i \wedge\omega_j$. -Now, on $X = H^n\times\mathbb{R}\times\mathbb{R}^n$, with projections $u:X\to\mathbb{R}$ and $(u_i):X\to\mathbb{R}^n$ onto the second and third factors, consider the Pfaffian system $\mathcal{I}$ generated by the $(n{+}1)$ linearly independent $1$-forms -$$ -\xi = \mathrm{d}u - u_i\ \omega_i -\quad\text{and}\quad -\xi_i = \mathrm{d}u_i +\theta_{ij}\ u_j - (T_{ij}+u\,\delta_{ij})\ \omega_j\,. -$$ -By the hypotheses on $T = T_{ij}\omega_i\omega_j$ and the curvature of $h$, this system is Frobenius, i.e., it satisfies $\mathrm{d}\xi \equiv \mathrm{d}\xi_i\equiv 0 \mod \mathcal{I}$. -Thus, $X$ is foliated by the leaves of $\mathcal{I}$, which are transverse to the fibers of the projection $\pi:X\to H^n$ onto the first factor. Since the system is affine linear in $(u,u_i)$, it follows that each leaf $L\subset X$ of $\mathcal{I}$ becomes a covering space of $H^n$ under the projection $\pi:L\to H^n$. Since $H^n$ is connected and simply-connected, such a projection is a diffeomorphism of the leaf $L$ with $H^n$ and hence has an inverse, which can be written in the form $\sigma:H^n\to L\subset X$ of the form $\sigma(p) = \bigl(p,f(p),f_i(p)\bigr)$. By construction, the function $f = u\circ\sigma:H^n\to\mathbb{R}$ and $(f_i) = (u_i)\circ\sigma:H^n\to\mathbb{R}^n$ must satisfy -$$ -df = f_i\ \omega_i -\qquad\text{and}\qquad -df_i = -\theta_{ij}\ f_j + (T_{ij}+f\delta_{ij})\ \omega_j\,. -$$ -The function $f$ is the potential that you seek.<|endoftext|> -TITLE: About the validity of a new conjecture about a diophantine equation -QUESTION [32 upvotes]: Let us consider the following conjecture: -Conjecture: There are no integer solutions of the equation $$x^{y-z}z^{x-y}=y^{x-z}$$ with $x,y,z$ distinct positive integers greater than or equal to $2$. -I came across this result when studying some diophantine equations. Several attempts were made to find a solution, but without any success. By this question I want to see if someone can give me a conterexample to this conjecture. - -REPLY [86 votes]: The conjecture is true, in fact the equation has no solution in distinct positive real numbers. To see this, let us write the equation in the more symmetric form -$$ x^y y^z z^x = x^z y^x z^y. \tag{$\ast$}$$ -We get the same equation after interchanging $x$ and $y$, or $y$ and $z$, i.e., after permuting the variables arbitrarily. Hence we can assume without loss of generality that $x>y>z>0$. Then, with the notation $a:=x-y$ and $b:=y-z$, the original equation becomes -$$ (y+a)^b (y-b)^a = y^{a+b}, $$ -where each factor and each exponent is positive. Equivalently, -$$ (1+a/y)^b (1-b/y)^a = 1, $$ -where each factor and each exponent is positive. However, this is impossible, since -$$ (1+a/y)^b (1-b/y)^a < (e^{a/y})^b (e^{-b/y})^a = 1.$$ -Added on 22 January 2021. Recently I posted the equation $(\ast)$ to a non-professional discussion board, and to my surprise two entirely new solutions arose. They are not mine, but I sketch them here as they are really nice and instructive. I will assume that $x,y,z>0$ are distinct and $(\ast)$ holds. I will derive a contradiction in two new ways. -First new proof (sketch). By assumption, $u:=y/x$ and $v:=z/x$ satisfy $u^{v-1}=v^{u-1}$. This contradicts (after some thought) the fact that the function $t\mapsto\frac{\ln t}{t-1}$ is strictly decreasing on the positive axis (the function is not defined at $t=1$, but it extends analytically there). -Second new proof (sketch). By assumption, the determinant -$$\begin{vmatrix}1&x&\ln x\\1&y&\ln y\\1&z&\ln z\\\end{vmatrix}$$ -vanishes, hence its rows are linearly dependent. This contradicts (after some thought) the fact that the function $t\mapsto\ln t$ is strictly concave on the positive axis.<|endoftext|> -TITLE: Absolute continuity of measures on infinite binary sequences -QUESTION [5 upvotes]: Suppose $P$ and $Q$ are two probability measures on the space $\Omega = \{0,1\}^{\mathbb N}$ of infinite binary sequences equipped with the product $\sigma$-algebra generated by its cylinder sets, with -$$ p_k = P(\omega_k = 1) \in (0,1), \quad q_k = Q(\omega_k = 1) \in [0,1].$$ -Question: Why does $Q\ll P$ hold if and only if -$$\sum_k \left( 1 - \sqrt{p_k q_k} - \sqrt{(1-p_k)(1-q_k)} \right) < \infty?$$ -A special case has already appeared on MO. I found this question through social media, where people were sharing a test from the Indian Statistical Institute that featured Donald Trump. A scan of the test is here. The question I'm asking is 4(a), and part (b) follows from it after some algebra. I was able to solve the other questions on the exam but could not make much progress on this. - -REPLY [10 votes]: Ok, Loïc Teyssier, Gerry Myerson, Nik Weaver, Stefan Waldmann, coudy, and a few other people will scold me badly for answering a question about a relatively simple exercise in undergraduate probability instead of crying loudly that "We do research here!" and closing, but here goes. -Note first of all that all examiners have an irritating habit of write the conditions in a slick form that hides what is going on as much as possible. So your first task on any exam is to trade beauty for clarity. -Since $2\sqrt(xy)=x+y-(\sqrt x-\sqrt y)^2$ and $(\sqrt x-\sqrt y)^2\asymp\frac{(x-y)^2}{x+y}$, the condition really is -$$ -\sum_k (p_k-q_k)^2\left[\frac 1{(p_k+q_k)}+\frac{1}{(1-p_k)+(1-q_k)}\right]<+\infty -$$ -Let $\Delta_k=q_k-p_k$. -Now the formal density of $Q$ with respect to $P$ is -$$ -\prod_k \left(1+\frac{\Delta_k}{p_k}\right)^{\omega_k}\left(1-\frac{\Delta_k}{1-p_k}\right)^{1-\omega_k} -$$ -We want to show that the partial products $D_n=\prod_{k=1}^n(\dots)$ converge in $L^1(P)$. The trivial sufficient condition would be the uniform $L^2(P)$ bound and we almost have it. Indeed, -$$ -\mathcal E_P D_n^2=\prod_{k=1}^n\left[1+\Delta_k^2\left(\frac 1{p_k}+\frac{1}{1-p_k}\right)\right] -$$ -So everything would work if we had the series without $q$'s in the denominator convergent. Unfortunately, we do not. Note, however, that if $\frac{\Delta_k^2}{p_k}>10\frac{\Delta_k^2}{p_k+q_k}$, then the quantity in the right is at least $\frac 12q_k$, so the sum over such $k$ of $q_k$ converges and, thereby, with $Q$-probability $1$ only finitely many of such $\omega_k$ are $1$. Similarly for $1-$ fractions. Thus, if we just condition upon the corresponding finite sets of bad indices (and finite subsets of integers are countably many, so we just partition the interesting part of the probability space into countably many chunks), we can use the $L^2$ criterion in each chunk separately. -This is sufficiency. Now, since it is an undergraduate exercise, do the other part yourself (or ask the people mentioned above to help you). I'll stop here.<|endoftext|> -TITLE: Jacquet Module of an Essentially Square Integrable Representation -QUESTION [5 upvotes]: Let $F$ be a $p$-adic field. Let $G$ be a connected reductive group and $\rho$ an irreducible admissible representation of $G(F)$. Let $P$ be a parabolic subgroup of $G$ and suppose further that $\rho$ is essentially square integrable. It seems like the Jacquet module $\mathrm{Jac}_P(\rho)$ is essentially square integrable or at least a sum of essentially square integrable representations. Is this true/is there a reference for this fact? - -REPLY [3 votes]: Yes, your claim is true. -It follows from proposition 43 of the Bernstein notes on representation of $p$-adic groups. -That proposition gives a characterization of essentially square integrable representations (called square integrable modulo center, in this source) in terms of certain exponents of all the (dual) Levi subalgebras. -Your claim follows then by the transitivity of the Jacquet functor.<|endoftext|> -TITLE: For a stable matrix $B$ and anti-symmetric $T$, such that $B(I+T)$ is symmetric, show that $\mbox{tr}(TB)\leq0$ -QUESTION [14 upvotes]: Let stable matrix (i.e., its eigenvalues have negative real parts) $B \in \mathbb R^{n \times n}$ and anti-symmetric matrix $T \in \mathbb R^{n \times n}$ satisfy -$$B^\top - T B^\top = B + B T$$ - -Prove that $\mbox{tr}(TB) \leq 0$. -What are necessary and sufficient conditions on $B$ such that $\mbox{tr}(TB) = 0$. (e.g. it is sufficient for $B$ to be symmetric. Is that also a necessary condition?) - - -Note that for each $B$, there is a unique $T$ satisfying this condition. -Motivation: -There is a claim in this paper that nonorthogonality of the eigenvectors of linear stability operator of a stochastic dynamical system amplifies the effect of noise, which is proven for the $2\times2$ case, and stated for the general case. This claim can be reduced to the above statement. Here is how it goes: -Consider a linear stochastic dynamics described by $$d x = Ax\, dt+\sigma dW,$$ where $\sigma>0$, $t\in\mathbb R^+$, $x(t)\in\mathbb R^n$, $A\in\mathbb R^{n\times n}$ with eigenvalues in left half plane, and $W$ is the $n$-dimensional Wiener process. This is an $n$-dimensional Ornstein-Uhlenbeck process. -If $A$ is symmetric, the distribution of $x$ at long time approaches a multivariate normal distribution with its covariance given by $A^{-1}$, and -$$\left\langle ||x||^2 \right\rangle = -\frac12\sigma^2\mbox{tr}(A^{-1}).$$ -When $A$ is not symmetric, the covariance can be written as the inverse of a symmetric matrix $GA$, where $$\frac12(G^{-1}+(G^{-1})^\top) = I_{n\times n}.$$ This relationship along with the symmetry of $GA$ uniquely defines $G$. In this case -$$\left\langle ||x||^2 \right\rangle = -\frac12\sigma^2\mbox{tr}(G^{-1}A^{-1}).$$ -The ratio of mean squared norm of $x$ to its value for a symmetric matrix with the same eigenvalues is what is called the amplification $$\mathcal H=\frac{\mbox{tr}(G^{-1}A^{-1})}{\mbox{tr}(A^{-1})}.$$ The claim is that $\mathcal H\geq 1$. -Let $B = A^{-1}$, and $T$ be the anti-symmetric part of $G^{-1}$. Now, $GA$ is symmetric iff $B^\top-TB^\top=B+BT$, and $\mathcal H\geq 1$ iff $\mbox{tr}(TB)\leq 0$: -$$ -\begin{cases} -T = \frac12 (G^{-1}-(G^{-1})^\top)\\ -I_{n\times n} = \frac12 (G^{-1}+(G^{-1})^\top) -\end{cases}\implies G^{-1} = I_{n\times n}+T\\ -(GA)^\top=GA\Leftrightarrow (G^{-1})^\top(A^{-1})^\top=A^{-1}G^{-1}\Leftrightarrow -B^\top-TB^\top=B+BT.$$ -$$\mathcal H=\frac{\mbox{tr}(G^{-1}A^{-1})}{\mbox{tr}(A^{-1})} = \frac{\mbox{tr}(B+TB)}{\mbox{tr}(B)}= 1+\frac{\mbox{tr}(TB)}{\mbox{tr}(B)}\geq 1\Leftrightarrow \mbox{tr}(TB)\leq 0.$$ -I am having difficulty understanding how the equation $B^\top - T B^\top = B + B T$, puts a restriction on the $\mbox{tr}(TB)$, since taking the trace of both side of this equation gives no new information. - -REPLY [8 votes]: First let us check that $T$ exists and is unique. Let $\mathrm{Sym}_n$ be the space of symmetric matrices (with real coefficients), $\mathrm{M}_n$ the space of all matrices and $\mathrm{Alt}_n$ the space of antisymmetric matrices. -Claim: the map $\mathrm{Alt}_n \rightarrow \mathrm{M}_n / \mathrm{Sym}_n, T \mapsto BT$ is injective. Let $T$ be in the kernel, i.e. $BT$ is symmetric. Up to conjugating by an orthogonal matrix, we can assume that $D = BT$ is diagonal, with exactly the first $r$ coefficients non-zero, for a well-defined $0 \leq r \leq n$. Let $C = B^{-1}$. The fact that $CD$ is antisymmetric implies that $C_{i,j} = 0$ for $i>r$ and $j \leq r$, i.e. $C$ is block upper diagonal. If $r >0$, the matrix $C' = (C_{i,j})_{1 \leq i,j \leq r}$ also has the properties that all of its eigenvalues have negative real part. But the fact that $CD$ is antisymmetric implies that $C'$ has vanishing diagonal, contradiction. -Comparing dimensions, the above linear map is an isomorphism, and the preimage of (the class of) $-B$ gives the unique $T \in \mathrm{Alt}_n$ such that $S := B(1+T)$ is symmetric. -The set of $B$ satisfying the assumption is open in $\mathrm{M}_n$ and path-connected ($t \mapsto t-1 + t B$ connects $-1$ to $B$), and the construction of $T$ is continuous, so the invertible symmetric matrix $S$ stays negative definite. -Now write -$$ BT = S(1+T)^{-1} T = S T(1-T)(1-T^2)^{-1} = ST(1-T^2)^{-1} - ST^2(1-T^2)^{-1}. $$ -The matrix $T(1-T^2)^{-1}$ is antisymmetric (it is the commuting product of an antisymmetric and a symmetric matrix) so by symmetry of $S$ we have $\operatorname{tr}(ST(1-T^2)^{-1}) = 0$, and $\operatorname{tr}(BT) = \operatorname{tr}(-ST^2(1-T^2)^{-1})$. Now $-S$ is positive definite and $T^2(1-T^2)^{-1}$ is negative semi-definite, so this trace is $\leq 0$ (write $-S$ as the square of a positive definite matrix). We also get that we have equality iff $T = 0$, i.e. iff $B$ is symmetric.<|endoftext|> -TITLE: Limit of decomposable bundles -QUESTION [10 upvotes]: Let $(E_b)_{b\in B}$ be a family of vector bundles on a smooth projective variety $X$, parameterized by a smooth curve $B$. Let $\mathrm{o}\in B$. Assume that $E_b$ is decomposable (= direct sum of two lower rank bundles) for $b\neq \mathrm{o}$. Can we conclude that $E_{\mathrm{o}}$ is decomposable? If not, what would be a counter-example? -Background : For $X=\mathbb{P}^n$, $n\geq 2$, I believe this is an open problem (see Mohan's answer to this MO question). I am primarily interested in the case of rank 2 vector bundles on a curve. - -REPLY [3 votes]: Though Jason Starr has given an excellent answer above, let me just describe another path which might be of interest. -Constructing deformations as required by abx on the projective space $P$ is equivalent to finding a vector bundle $G$ on $P$ with a nilpotent endomorphism $t$ and a map $\phi:E\to G$ such that $\phi(E)+t(G)=G$. Then you can write down a family on $P\times A$ where $A=\mathbb{A}^1=\mathrm{Spec} k[t]$ (same $t$ intentionally) by treating $G$ as a sheaf over $X\times A$ supported on $t^n=0$ for some $n$. Then, you have a surjection on $P\times A$, $E\to G$, using $\phi$ and the kernel is a vector bundle whose generic fibers over $A$ are $E$. In general, if $G$ is not a direct sum of line bundles and $E$ is, you get a family whose general member is decomposable and special member is not. -On $P=\mathbb{P}^2$, this is easy to do and for higher dimensional case (at least in characteristic zero) this is an open question. Here is one of the simplest such. Take the usual resolution of the ideal sheaf $I$ of a point and take the push-out $\mathcal{O}(-2)\to \mathcal{O}$, by a quadric not passing through the point to get $0\to \mathcal{O}\to G\to I\to 0$, a rank two vector bundle with a nilpotent endomorphism, got by sending $G\to I\to \mathcal{O}\to G$ and $\phi:\mathcal{O}(-1)^2\to G$. -This bundle gives you a family of rank two vector bundles on $P$ with general member $\mathcal{O}(-1)^2$ and special member indecomposable and thus unstable. -Finally, if $M$ is a rank 2 unstable non-split bundle on $P$, one may assume after twisting, that we have an exact sequence $0\to \mathcal{O}(k)\to M\to J\to 0$, $J$ defining a zero dimensional non-empty scheme. If we restrict this to a curve not passing through $V(J)$, $M$ splits if and only if it splits as $\mathcal{O}(k)\oplus\mathcal{O}$. If we take the curve $C$ to be of sufficiently large degree, one can make sure the map $H^0(M)\to H^0(M|C)$ to be an isomorphism and then $M|C$ can not be split.<|endoftext|> -TITLE: Riemann surfaces with an atlas all of whose open sets are biholomorphic to $\mathbb{C}$? -QUESTION [10 upvotes]: Is there a compact Riemann surface other than the sphere with an atlas consisting of open subsets biholomorphic to $\mathbb{C}$? Is there a compact Riemann surface other than the sphere which possesses an open subset biholomorphic to $\mathbb{C}$? - -REPLY [19 votes]: Using the uniformization theorem, we can prove that the only compact connected Riemann surface admitting an open set biholomorphic to $\mathbb{C}$ is the Riemann sphere $\mathbb{P}^1(\mathbb{C})$. This answers negatively both questions of the OP. -Indeed, let $X$ be a compact connected Riemann surface with an open set $U$ biholomorphic to $\mathbb{C}$. Choose a base point in $U$, and let $\tilde{X}$ be the universal cover of $X$. By the uniformization theorem, $\tilde{X}$ is isomorphic either to $\mathbb{P}^1(\mathbb{C})$, or to $\mathbb{C}$, or to the open unit disk $D$. In the first case we have necessarily $X \cong \mathbb{P}^1(\mathbb{C})$, so let us assume we are not in this case. Since $\mathbb{C}$ is simply connected, the inclusion map $f : \mathbb{C} \to X$ lifts to a map $\tilde{f} : \mathbb{C} \to \tilde{X}$ which is necessarily injective. If $\tilde{X}=D$ then we get a nonconstant bounded entire function, which is impossible. Finally if $\tilde{X}=\mathbb{C}$ then $\tilde{f}$ is a nonconstant entire function, so by Picard's theorem misses at most one value. But $X \cong \mathbb{C}/\Lambda$ and $\tilde{X} \to X$ is far from being injective, which gives a contradiction. (Alternatively, we can use the fact that any injective entire function $\tilde{f}$ is of the form $\tilde{f}(z)=az+b$ with $a \neq 0$, see this question at MSE.) -EDIT. Here is a more elementary argument avoiding the use of the uniformization theorem. The idea was suggested by an attempt of a student of mine. -Let $X$ be a compact connected Riemann surface with an open set $U$ biholomorphic to $\mathbb{C}$. Let $\omega \in \Omega^1(X)$ be a holomorphic $1$-form. The $2$-form $\sigma = i \cdot \omega \wedge \overline{\omega}$ is everywhere non-negative, and is integrable on $X$ since $X$ is compact. It follows that $\sigma$ is integrable on $U$. Let us write $\omega |_U = f(z) dz$ with $f(z) = \sum_{n \geq 0} a_n z^n$ an entire function. Then -\begin{equation*} -\int_U \sigma = \int_{\mathbb{C}} |f(z)|^2 i dz \wedge d\overline{z} = \int_0^\infty \int_0^{2\pi} |f(re^{i \theta})|^2 2r dr d\theta. -\end{equation*} -Now we have -\begin{equation*} -\int_0^{2\pi} |f(re^{i\theta})|^2 d\theta = \int_0^{2\pi} \sum_{m,n \geq 0} a_m \overline{a_n} r^{m+n} e^{i(m-n)\theta} d\theta = 2\pi \sum_{n \geq 0} |a_n|^2 r^{2n}. -\end{equation*} -Assume that $f$ is nonzero, so that there exists an integer $k \geq 0$ such that $a_k \neq 0$. Then -\begin{equation*} -\int_U \sigma \geq 4\pi |a_k|^2 \int_0^\infty r^{2k+1} dr -\end{equation*} -and the last integral is infinite, contradicting the integrability of $\sigma$ on $U$. It follows that $f=0$ and thus $\Omega^1(X)=\{0\}$, so that $X$ has genus $0$ and $X \cong \mathbb{P}^1(\mathbb{C})$. - -REPLY [11 votes]: Compact Riemann surfaces of genus at least $2$ are uniformized by the unit disk, hence they -do not admit any non-constant holomorphic map from $\mathbb{C}$.<|endoftext|> -TITLE: Is there an infinite set $X$ such that for every isotone $f\colon[X]^\omega\to[X]^\omega$ there is a free decreasing sequence? -QUESTION [17 upvotes]: (Pierre Gillibert asked me this question and I post it with his permission.) -Let $X$ be an infinite set, and $f\colon[X]^\omega\to[X]^\omega$. We say that $\{x_n\mid n<\omega\}\subseteq X$ is a free decreasing sequence (for $f$) if for all $n$, $x_n\notin f(\{x_k\mid k>n\})$. - -Is there an infinite set $X$ such that for every isotone $f\colon[X]^\omega\to[X]^\omega$ there exists a free decreasing sequence? Is it at least consistent from assumptions such as $V=L$, large cardinals or strong forcing axioms? - -Some observations: - -It is clear that $X$ is uncountable, because otherwise $f(A)=X$ for all $A\in[X]^\omega$ would pose a counterexample. -If there is no such set of size $\kappa$, then there is no such example of size $\kappa^+$. -If $\kappa$ has uncountable cofinality, and there is no such set of size $<\kappa$, then there is no example of size $\kappa$, since we can "glue" counterexamples and use the fact that every countable set is bounded (this is in effect the same proof for the previous observation). - -REPLY [5 votes]: Here's a long, possibly unhelpful comment making use of a presumably excessive large cardinal assumption. -Suppose that $\lambda$ is an uncountable cardinal and that there is a nontrivial elementary embedding -$j \colon L(V_{\lambda + 1}) \to L(V_{\lambda + 1})$ with critical point less than $\lambda$. So we are assuming that the large cardinal axiom I0 holds (see https://en.wikipedia.org/wiki/Rank-into-rank), with $\lambda$ here as the $\lambda$ there. -Note that (1) $j(\lambda) = \lambda$, (2), $j[\lambda]$ is in $L(V_{\lambda + 1})$ and (3) $[\lambda]^{\omega} \subseteq L(V_{\lambda + 1})$. -It seems that $L(V_{\lambda + 1})$ thinks that $\lambda$ is an $X$ as desired. -Claim : In $L(V_{\lambda + 1})$, for any $f : [\lambda]^{\omega} \to [\lambda]^{\omega}$ (not necessarily isotone) there exist an $\alpha < \lambda$ and an $X \subseteq \lambda$ of cardinality $\lambda$ such that $\alpha$ is not in $\bigcup f[[X]^{\omega}]$. -Applying the claim iteratively ought to let us build an independent family for any given $f$ in $L(V_{\lambda + 1})$. -The output of our iterative construction is then in $L(V_{\lambda + 1})$, showing that $L(V_{\lambda + 1})$ thinks that $\lambda$ is as desired. Note that $L(V_{\lambda + 1})$ is not a model of Choice, so maybe this doesn't address the question. -Proof of claim. Let $f : [\lambda]^{\omega} \to [\lambda]^{\omega}$ in $L(V_{\lambda + 1})$ be given. Let $Z$ denote $j[\lambda]$. Let $\alpha$ be an element of $\lambda \setminus Z$ (for instance, the critical point of $j$). If $x$ is a countable subset of $Z$, then $x = j(y)$ for some $y \in [\lambda]^{\omega}$ (the pointwise $j$-preimage of $x$), so $j(f)(x) = j(f(y))$, which is contained in $Z$, so $j(f)(x)$ does not have $\alpha$ as a member. Then, in $L(V_{\lambda + 1})$, $Z$ is a subset of $\lambda$ of cardinality $\lambda$, and $\alpha$ is not in $\bigcup j(f)[[Z]^{\omega}]$. By the elementarity of $j$, then, we have the conclusion of the claim : there exist an $\alpha < \lambda$ and an $X \subseteq \lambda$ of cardinality $\lambda$ such that $\alpha$ is not in $\bigcup f[[X]^{\omega}]$. -Since I'm not using the isotone condition, then at least one of the following should hold : (1) I'm making a mistake (very likely), (2) the first sentence of the December 7 comment above uses more Choice than holds in $L(V_{\lambda + 1})$ or (3) I0 is inconsistent.<|endoftext|> -TITLE: History of "natural transformations" -QUESTION [22 upvotes]: (Edit #1 after Carlo's response) -It is often claimed that the notion of natural transformations existed in mathematical vocabulary long before it had a definition. In fact, I quoted the statement in italic from [1, p. 2]. As another example, in [2, p. 70] Ralf Kromer says: The claim is that there was, at the time when [3] was written, a current informal parlance consisting in calling certain transformations natural and that Mac Lane and Eilenberg tried (and succeeded) to grasp this informal parlance mathematically. -Three years after [3], Eilenberg and Maclane discovered the fact that this notion could be mathematically defined in [4]. -However, Ralf Kromer casts doubt on the above mentioned claim, due to lack of evidence (see: [2, p. 70]). -(Edit #2 after Eric's comment) My question is, can you supply an evidence of use of the phrase "natural transformations" or its variants in mathematical literature prior to [3]? I must add that in [2] Kromer gives a number of examples of use of phrases such as "natural homomorphism" or "natural projection" in the literature prior to or around the same time as [3], but in each case they turn out to have different meanings. So I am asking for example(s) of use of the phrase "natural transformations", which are really natural transformations. -References: - -Peter Freyd: Abelian Categories (1964). -Ralf Kromer: Tool and Object: a history and philosophy of category theory (2007). -Samuel Eilenberg and Saunders Maclane: Group extensions and homology, Annals Math. (2) 43, p.p. 757–831 (1942). -Samuel Eilenberg and Saunders Maclane: General theory of natural transformations, Trans. AMS, 58, p.p.: 231-294 (1945). - -REPLY [2 votes]: The words "natural homomorphism" and "natural isomorphism" are also used (mainly in the context related to the First Isomorphism Theorem) in Pontryagin's "Topological groups" (Russian edition 1938, English translation 1939). Google books confirms my memory of this here.<|endoftext|> -TITLE: Relation between commutator length and stable commutator length in free groups -QUESTION [6 upvotes]: In Bardakov, Algebra and Logic, Vol. 39, No. 4, 2000 I have found the following (page 225, see https://link.springer.com/article/10.1007/BF02681648) - -We pronounce tile validity of the following: -Conjecture. For every element z in the derived subgroup of a free non-Abelian group F and for any natural m, - $$ -\mathrm{cl}(z^m) \geq (m+1/2)\mathrm{cl}(z) -$$ - -Where cl denotes the commutator length of an element (ie. the minimal number to express it as a product of commutators). -This inequality is not true, and $$z = [a, b]$$ may be a counterexample. However, I belive that there may be a typo, so it should rather be -$$ -\mathrm{cl}(z^m) \geq (m+1)/2 \cdot \mathrm{cl}(z) -$$ -Unvortunatelly, I could not find it in any other paper/book (including Calegari's "scl"). And the proof in Bardakov is unclear to me. -Do you know any paper, with a proof of the above inequality? Or maybe some counterexample? Or maybe has anybody have any clue why Bardakov did not prove this inequality? - -REPLY [2 votes]: The misconception is due to bad translation. The original text was in Russian and can be found here: http://www.mathnet.ru/links/f40b0cf29e7b19a9b2ab1a95ef70baba/al284.pdf. -It reads: "Можно высказать предположение о том" which means "we can phrase a guess". -I e-mailed Valeiry Bardakov about that conjecture. And it occured to be false. In fact there exists a sequence $z_n$ of elements of a free group such that $cl(z_n) \to \infty$, but $cl(z_n^2) < const$. It can be seen using the result by O.Kharlampovich and A.Myasnikov [1,Section 6 and 7]. -According to Theorem 3 of [1], there exist solutions to -$$x_1^2 x_2^2 x_3^2 x_4^2 = 1 \quad (*)$$ -such that $cl(x_1x_2x_3x_4)$ is arbitrarily big. But we can observe, that $(x_1x_2x_3x_4)^2 = x_1x_2x_3x_4 x_1x_2x_3x_4$ can be expressed as $x_1^2 x_2^2 x_3^2 x_4^2$ times a bounded number of commutators. And now we see that (*) implies that $cl((x_1x_2x_3x_4)^2)$ is bounded by a fixed constant, that we can easily calculate. -[1]O.Kharlampovich and A.Myasnikov, "Implicit Function Theorem over Free -Groups and Genus Problem", in: Knots, braids, and mapping class -groups—papers dedicated to Joan S. -Birman (New York, 1998), Amer. Math. Soc., Providence, RI, 2001, 77–83<|endoftext|> -TITLE: Can Woodin's fast function forcing kill Shelah cardinals? -QUESTION [5 upvotes]: Definition 1. An uncountable cardinal $\kappa$ is Shelah if for every function $f:\kappa\rightarrow \kappa$ there exists a transitive class $M$ and a non-trivial elementary embedding $j:V\rightarrow M$ such that $^\kappa M\subseteq M$, $crit(j)=\kappa$ and $V_{j(f)(\kappa)}\subseteq M$. -Definition 2. Woodin's fast function forcing on $\kappa$ consists of partial functions $p$ from $\kappa$ to $\kappa$ ordered by inclusion such that: - -The domain of $p$ consists of inaccessible cardinals $\lambda <\kappa$ which are closed under $p$, namely for every $\lambda, \theta\in dom(p)$ if $\theta<\lambda$ then $p(\theta)<\lambda$. -For every $\lambda\in dom(p)$ we have $|dom(p)\cap \lambda|<\lambda$ - -Remark. Following Joel's comment below, it is worth mentioning that the above definition is NOT the only variant of fast function forcing. There are other versions with slightly different properties. For a more complete argument along these lines see Joel's answer in this MO post. -On one hand, preserving Shelah cardinals through lifting arguments often needs dominating the corresponding functions $f:\kappa\rightarrow \kappa$ by functions in the ground model. -On the other hand, we know that the fast function forcing $\mathbb{P}_{\kappa}$ adds a very fast (and so non-dominatable) function of this type into the universe (i.e. the fast function) and simultaneously fails to satisfy $\kappa$-cc property which is a usual condition for providing dominating functions in the ground model. - -Thus it is somehow natural to expect that (all variants of) Woodin's fast function forcing can kill Shelah cardinals. Is it true? Any concrete example? - -REPLY [2 votes]: Assume it preserve the Shelahness of $\kappa$, where $\kappa$ is th eleast Shelah cardinal. - Let $f: \kappa \to \kappa$ be the fast function added by the forcing $\mathbb{P}$. Let $G$ be the generic filter. -By our assumption, there exists $j: V[G] \to M$ such that $M$ is closed under $\kappa$-sequences -(in $V[G]$), $crit(j)=\kappa$ and $V[G]_{j(f)(\kappa)} \subset M.$ -Let us force below $(\kappa, wt^V(\kappa)+\omega) \in j(\mathbb{P}),$ where $wt^V(\kappa)$ -is the witnessing number of $\kappa$ in $V$. This implies $V[G]_{wt^V(\kappa)+\omega} \subset M,$ in particular, - the results of An Easton like theorem in the presence of Shelah cardinals -show that $\kappa$ is not the least Shelah cardinal in $V[G]$. But the forcing creates no new Shelah cardinals, - so $\kappa$ is not the least Shelah cardinal in $V$ as well, a contradiction.<|endoftext|> -TITLE: Why is $K_{\upsilon}|K$ separable for a global field $K$? -QUESTION [6 upvotes]: I asked this question on math.stackexchange but maybe it fits here better. If not, I apologize in advance and will remove the question. -Let $K$ be a global field and $\upsilon$ a prime of $K$. Then in a paper, which I read, it is used that $K_{\upsilon}|K$ is separable, where $K_{\upsilon}|K$ is the completion. This is obviously true if $charK=0$. -Why is $K_{\upsilon}|K$ separable if $charK=p>0$? -Things I tried but could not get to work (maybe I missed something and they would work): - -Use that $\hat A \cap Frac(A)^{alg} \subset A^h$ for an excellent ring $A$ (this is mentioned here henselization and completion) -Use Exercise 1 of section II.4 of Serre's Local Fields: If $A$ is a DVR for which every finite purely inseparable extension $Frac(A)$ has integral closure finite over $A$ then the completion of $Frac(A)$ is separable over $Frac(A)$ -Try to use that $K_{\upsilon}=\mathbb F_{p^n}((T))$ - -The statement seems to me like it should be in a book somewhere but I just can't find it. -Thanks in advance. - -REPLY [5 votes]: Since there is already an answer, I want to give my slightly different answer in a special case: Say $K = k(t)$ where $k$ is a finite field and $K_v = k((t))$. We only need to check that -$$ -K_v \otimes_K K^{1/p} -$$ -is reduced, see Tag 030W. Since $K^{1/p} = k(t^{1/p})$ this is true because the tensor product is visibly equal to $k((t^{1/p}))$. I claim this argument works in general too, but since we have a fully written out proof above, I won't insist. Enjoy!<|endoftext|> -TITLE: Higher arcsin wanted -QUESTION [15 upvotes]: There is a known proof of the identity $$\sum_{k=0}^\infty \frac1{(2k+1)^2}=\frac{\pi^2}8\,\,\,(*)$$ (equivalent to $\sum \frac1{n^2}=\frac{\pi^2}6$) by expanding $\arcsin x$ as a power series $$\arcsin x=\sum_{k=0}^\infty \frac{(2k-1)!!}{(2k)!!(2k+1)}x^{2k+1},$$ -substituting $x=\sin t$ and integrating for $t$ from 0 to $\pi/2$. The double factorials cancel by some magic and we get $(*)$. -I wonder whether the higher values of zeta at even integer points may be obtained on this way for other functions similar to $\arcsin$ or other clever integrals. - -REPLY [11 votes]: In $[1]$ Borwein and Chamberland show that for $|x|\le 2$ -\begin{align*} -\arcsin^{2N+1}(x/2) &= (2N+1)!\sum_{k=0}^\infty \frac{G_N(k){2k\choose k}}{2(2k+1)4^{2k}} x^{2k+1}, - & N = 0,1,\ldots \tag{1} \\ -\arcsin^{2N}(x/2) &= (2N)!\sum_{k=1}^\infty \frac{H_N(k)}{{2k\choose k}k^2} x^{2k} - & N = 1,2,\ldots \tag{2} -\end{align*} -where -\begin{align*} -G_0(k) &= 1 \\ -G_N(k) &= -\sum_{n_1=0}^{k-1}\frac{1}{(2n_1+1)^2} -\sum_{n_2=0}^{n_1-1}\frac{1}{(2n_2+1)^2} -\cdots -\sum_{n_{N}=0}^{n_{N-1}-1}\frac{1}{(2n_{N}+1)^2} -\end{align*} -and -\begin{align*} -H_1(k) &= 1/4 \\ -H_N(k) &= \frac{1}{4} -\sum_{n_1=1}^{k-1}\frac{1}{(2n_1)^2} -\sum_{n_2=1}^{n_1-1}\frac{1}{(2n_2)^2} -\cdots -\sum_{n_{N-1}=1}^{n_{N-2}-1}\frac{1}{(2n_{N-1})^2}. -\end{align*} -Letting $x=2\sin t$ in $(1)$ and $(2)$ and integrating from $0$ to $\pi/2$ we find the same type of cancellation described in the question statement, with the result -\begin{align} -\sum_{k=0}^\infty \frac{G_N(k)}{(2k+1)^2} &= \frac{1}{(2N+2)!}\left(\frac{\pi}{2}\right)^{2N+2} \tag{3} \\ -\sum_{k=1}^\infty \frac{H_N(k)}{k^2} &= \frac{1}{(2N+1)!}\left(\frac{\pi}{2}\right)^{2N}. \tag{4} -\end{align} -Equations $(3)$ and $(4)$ are natural generalizations of $(*)$. -They give the value of $\zeta(2)$ and relate the values of $\zeta(n)$ for $n=4,6,\ldots$ to other interesting sums. -Equation $(3)$ implies for $N=0,1$ that -\begin{align*} -\sum_{k=0}^\infty \frac{1}{(2k+1)^2} &= \frac{\pi^2}{8} \\ -\sum_{k=0}^\infty \frac{\psi^{(1)}(k+1/2)}{(2k+1)^2} &= \frac{5\pi^4}{96}, -\end{align*} -where $\psi^{(n)}(k)$ is the polygamma function. -The last result can be reexpressed as -$$\sum_{k=1}^\infty \frac{1}{k^4} -= \frac{\pi^4}{72} - \frac{4}{15}\sum_{k=0}^\infty\frac{\psi^{(1)}(k+3/2)}{(2k+1)^2}.$$ -(Given $\zeta(4) = \pi^4/90$ this implies -$\sum_{k=0}^\infty{\psi^{(1)}(k+3/2)}/{(2k+1)^2} = {\pi^4}/{96}.$) -Equation $(4)$ implies for $N=1,2$ that -\begin{align*} -\sum_{k=1}^\infty \frac{1}{k^2} &= \frac{\pi^2}{6} \\ -\sum_{k=1}^\infty \frac{H_{k-1}^{(2)}}{k^2} &= \frac{\pi^4}{120}, -\end{align*} -where $H_k^{(n)}$ are the generalized harmonic numbers of order $n$. -The last result can be reexpressed as -$$\sum_{k=1}^\infty \frac{1}{k^4} = \sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2} - \frac{\pi^4}{120}.$$ -(Given $\zeta(4)$ this implies -$\sum_{k=1}^\infty{H_k^{(2)}}/{k^2} = {7\pi^4}/{360}.$) -$[1]$ Borwein, J. M. and Chamberland, M. "Integer Powers of Arcsin." Int. J. Math. Math. Sci., Art. 19381, 1-10, 2007.<|endoftext|> -TITLE: What is the cofinality of $([\kappa]^\omega, \subseteq)$? -QUESTION [10 upvotes]: For an uncountable cardinal $\kappa$, we are interested in the least size of a cofinal subset of the partial order $([\kappa]^\omega, \subseteq)$. It is obvious that this cofinality is at least $\kappa$ and a rather simple induction shows that $cof([\aleph_n]^\omega, \subseteq)=\aleph_n$ for each natural number $n \geq 1$. -What is known in $ZFC$ for bigger cardinals? -Also I am intrigued by the following statement made by Alan Dow in Efimov spaces and the splitting number, Topology Proc. (2005): - -It is a "large cardinal" hypothesis to assume that there is a cardinal $\kappa$ with uncountable cofinality such that this cofinality is greater than $\kappa$. - -What exactly does he mean by that, and what is a good reference to read about it? - -REPLY [10 votes]: I think he means the consistency of "there exists a cardinal $\kappa$ with uncountable cofinality so that $cf([\kappa]^\omega, \subseteq) > \kappa$" requires large cardinal axioms. -To motivate this, note that for example if $0^\sharp$ does not exist, then for any cardinal $\kappa > 2^{\aleph_0}$ of uncountable cofinality we have $cf([\kappa]^\omega, \subseteq) = \kappa$. -In fact one can say more. A related argument is given in Theorem 3.1 of Singular cofinality conjecture and a question of Gorelic.<|endoftext|> -TITLE: Does the rate of decay of an entire function dictate the global growth rate? -QUESTION [5 upvotes]: If $f$ is entire and we assume a decay rate on $\mathbb{R}^+$, does this dictate a growth rate on $\mathbb{C}$? -For example, is there a function that decays as $e^{-x^2}$ on $\mathbb{R}^+$ and is of exponential type? -What I am actually interested in is whether there exists an entire function such that $|f(z)|\le Me^{-a z}$ for all $z>0$ and $ |f(z)| \le Me^{b |z|} $ for all $z$, with $b -TITLE: The defining characteristic representations of Lie type groups -QUESTION [5 upvotes]: Let $\mathbf{G}$ be a connected reductive group over $\mathbb{F}_q$, let $G=\mathbf{G}(\mathbb{F}_q)$, and let $k$ be a large enough finite extension of $\mathbb{F}_q$. -If $\mathbf{G}=\mathrm{SL}_2$, then in page-109 of Bonnafe's book it is said that "It turns out that the simple $kG$-modules are the restrictions of simple "rational representations" of the algebraic group $\mathbf{G}_{\overline{k}}$." -Question 1: Where can I find a proof of this statement? -Question 2: Is this true for a general $\mathbf{G}$? -Any help would be appreciated. - -REPLY [6 votes]: Here is an extended comment on Jay Taylor's answer to both questions (in community-wiki style), with some other references added. Note however that the results for simple (or semisimple) algebraic groups and their relevant finite subgroups are well-known by now. Also, the results for the rank 1 groups treated by Bonnafe go back further to Steinberg's adviser Richard Brauer (in Toronto), so the history here is rather long. -1) In Steinberg's seminal 1963 paper linked by Jay (in a journal issue dedicated to Brauer), the approach relies on "lifting" irreducible representations of a (simple) Chevalley group to projective representations of the ambient algebraic group. This is then modified suitably for some twisted groups. At that time Chevalley's construction focused on adjoint algebraic groups, so there also needed to be some study of central extensions and covering groups (which Steinberg did basic work on). -Later on Steinberg, in his 1967-68 Yale lectures on "Chevalley groups" and the contemporary AMS Memoir cited by Jay, provided a more unified formulation of his earlier results. All of this depends on knowing how to characterize the irreducible representations of the simple (or semisimple) algebraic groups in the "highest weight" formulation of the 1956-58 Chevalley seminar, as well as Steinberg's generalization of the twisted tensor product theorem (worked out by Brauer-Nesbitt in rank 1) to arbitrary ranks. The main point is to reduce the study of irreducibles for finite groups of Lie type in the defining characteristic to the study of "restricted" highest weight representations for the algebraic groups. This is far from complete, however, in spite of many results involving Lusztig's ideas based on a characteristic $p$ analogue of Kazhdan-Lusztig theory. -Meanwhile, Jantzen's 1987 approach based partly on Kempf's ideas here were later exposed in my 2006 book here: see Chapter 2 along with the posted revisions. Jantzen's own 2003 AMS expanded edition of his 1987 text Representations of Algebraic Groups remains the best source for a thorough treatment of the overall theory for semisimple algebraic groups. But recent work by Achar, Riche, Williamson. and others shows the subtleties of the unsolved problems about weight space dimensions for all $p$. This is all relatively easy in rank 1, but not in general. -2) As Jay indicates, the other question here about reductive groups is addressed by the more recent work of Brunat-Lubeck. Recall that Borel and Tits explored their notion of "reductive" in arbitrary characteristic (in 1965) and showed that a connected reductive algebraic group is just the almost-direct product of a semisimple group and a torus. While the representation theory of semisimple groups reduces soon to the study of simple algebraic groups, the (algebraic) representations of a torus all come from dimension 1. So the main result of Brunat-Lubeck mainly involves assembling the pieces carefully, with attention to the fine points of the group structure.<|endoftext|> -TITLE: On the limit of partial sum of infinite doubly stochastic matrix -QUESTION [8 upvotes]: Let $A=(a_{ij})$ be an infinite doubly stochastic matrix. Does there necessarily exist a subsequence $\{n_k\}_{k=1}^\infty$ such that -$$ \lim_{k\to\infty}\frac{1}{n_k}\sum_{i=1}^{n_k}\sum_{j=1}^{n_k}a_{ij} >0?$$ -In a previous post A question on the partial sum of infinite doubly stochastic matrix, Iosif Pinelis constructed a counterexample for the whole sequence. - -REPLY [11 votes]: No. Enumerate all positive integers which are not powers of $2$: $3=n_1 -TITLE: Homology of a limit of spectra + Cofiber -QUESTION [5 upvotes]: I have a countable sequence of finite suspension spectra $X_i$, whose $BP$-homology is a $BP_*(BP)$-comodule. Let's assume $BP_*(X_i) = \Sigma^{d_i} BP_* / (v_0^{k_0}, \dots v_i^{k_i}),$ for some $d_n$ big enough that makes them being suspension spectra. -Then, if we let $X = \bigvee_i X_i$, -$$BP_*( X)= \bigoplus_i BP_* / (v_0^{k_0}, \dots v_i^{k_i}).$$ -Is that right? If yes/no, why? -Now, let's assume i have a spectrum $Y = \prod_i X_i$, which means -$$BP_* (Y) = \prod_i BP_* / (v_0^{k_0}, \dots v_i^{k_i})$$ -and the obvious map $$i: X \to Y$$ -How would the cofiber $ C_i$ of this map look like? More in detail, i would like $C_i$ to have non torsion elements. Is this the case? -Thank you! - -REPLY [5 votes]: So far as I can tell, it depends. -The map $\oplus BP_*(X_i) \to \prod BP_*(X_i)$ is injective and the cokernel $M$ is $BP_*$ of the cofiber. In particular, for any $N$ we have an isomorphism -$$ -M \cong \prod_{i \geq 0} BP_*(X_i) \Big/ \bigoplus_{i \geq 0} BP_*(X_i) \cong \prod_{i \geq N} BP_*(X_i) \Big/ \bigoplus_{i \geq N} BP_*(X_i) -$$ -As a result, properties of the modules $BP_*(X_i)$ that hold for all but finitely many exceptions tend to be inherited by the module $M$. -The first case is if the natural numbers $d_n$ go to $\infty$ as $n$ goes to $\infty$. In the identities $BP_*(X) = \oplus BP_* (X_i)$ and $BP_*(Y) = \prod BP_*(X_i)$, the direct sum and product are taken in the category of graded abelian groups. If the $d_n$ go to infinity, then for any fixed $k$ there are only finitely many $X_i$ with $BP_k(X_i) \neq 0$. As a result the group $M$ is zero in grading $k$. (The same is true for homotopy groups, and so the cofiber of the map $X \to Y$ is contractible). If you're asking for the $X_i$ to be suspension spectra, then I suspect that you're forced to have $d_n \to \infty$ (but I don't have a proof). -The second case is when the integers $d_n$ do not go to $\infty$. In this case, multiplication by $v_j^{k_j}$ acts by zero on all but finitely many $BP_*(X_i)$, and so the same is true for $M$. Thus means that $M$ consists entirely of things that are torsion. - -What you might have had in mind was an "ultrafilter" type argument that did the exact opposite. If, instead, you have -$$ -BP_*(X_i) \cong \Sigma^{d_i} BP_* / (v_0^{k_{0,i}}, v_1^{k_{1,i}}, \dots, v_i^{k_{i,i}}) -$$ -such that, for any fixed $n$, $k_{n,i} \to \infty$, then you have the opposite case: for all but finitely many $i$, $BP_m(X_i)$ has no $v_i^k$-torsion. As a result, the module $M$ has no torsion. - -(You probably already know this, but I feel obligated to mention that the identity $BP_*(\prod X_i) \cong \prod BP_*(X_i)$ isn't always true, and relies on the $X_i$ being connective and $BP$ having ($p$-local) finite type.)<|endoftext|> -TITLE: Find eigenvalues with given multiplicity in presence of errors -QUESTION [5 upvotes]: Let $A^0$ a $3 \times 3$ real symmetric matrix with eigenvalues $\lambda_1^0 = \lambda_2^0 \neq \lambda_3^0$. Hence, $\lambda_1^0$ has multiplicity $2$. -In real-world applications, I will have a perturbed matrix $A=A^0 + \Delta A $. -Hence, I will find 3 different eigenvalues, but, if the errors are "small", two of them will be similar: $\lambda_1 \simeq \lambda_2 \neq \lambda_3$. -Is it possible to find the "best" approximation of the "original" eigenvalue $\lambda_1^0$? -The first idea I came up was to use the average of $\lambda_1$ and $\lambda_2$, but I would like to have a sort of optimality criterion. - -REPLY [2 votes]: Let $M$ be the set of symmetric $3\times3$ matrices that admit a multiple eigenvalue. It is a closed subset, a codimension-$2$ submanifold (with a singular line along the homotheties). A reasonnable approach is to project $A$ over $M$, in the euclidian structure defined by the Frobenius norm. -Let $S\in M$ be generic (not a homothety). It has a double eigenvalue, corresponding to an eigenplane $P$. The tangent space to $M$ at $S$ is made of those symmetric $T$ such that the restriction of $T$ to $P$ (in the sense of quadratic forms) is proportional to the identity. That is $v^TTv\equiv\alpha|v|^2$ over $P$, for some $\alpha$. -Projecting $A$ on $M$ consists therefore in finding a plane $P$ on which the quadratic form $q_A(v)=v^TAv$ coincides with $\lambda|v|^2$ ; and this $\lambda$ is the eigenvalue you are searching for. In practice, you consider a level set of $q_A$ and look for a plane that cuts it along a circle. -Edit. There exists orthonormal coordinates in which $q_A$ writes $ax^2+by^2+cz^2$, where $a,b,c$ are the eigenvalues of $A$. A rather simple calculation shows that if a plane cuts a level set of $q_A$ along a circle, then its equation has to be of the form $z=\alpha x$, up to a permutation of the coordinates (i.e. the plane contains a coordinate axis). Then $q_A(v)\equiv b|v|^2$ on this plane and $\alpha$ is given by $(b-c)\alpha^2=(a-b)$. This is possible only if $b$ lies between $a$ and $c$. -In conclusion, we are led to choose one eigenvalue of $A$, precisely the intermediate one $\lambda_2$ (with the order $\lambda_1\le\lambda_2\le\lambda_3$). This is the best approximation of $\lambda_1^0=\lambda_2^0$ in the sense of the Frobenius-projection onto the set of symmetric matrices having a double eigenvalue.<|endoftext|> -TITLE: Is the inverse of surreal numbers actually well-defined? -QUESTION [7 upvotes]: J.Conway wrote in his book "On numbers and games" (1st edition, 1976) on p. 66 - -It seems to us, however, that mathematics has now reached the stage where formalization within some particular axiomatic set theory is irrelevant, even for foundational studies. - -I happen to disagree. In fact, I'm having some fun formalizing mathematics in the Mizar system, which has just one article on Conway games yet. I want to find out if further formalization would be fruitful, i.e. if the operations defined in the book can be formalized at all. The cited article only defines $-x$ for a game $x$, but gives a pretty good idea how to deal with the highly inductive nature of games and I'm certain I could formalize addition and multiplication. What bothers me is the definition of $y=\frac{1}{x}$, given by $$y=\left\{\left.0,\frac{1+(x^R-x)y^L}{x^R},\frac{1+(x^L-x)y^R}{x^L}\right|\frac{1+(x^L-x)y^L}{x^L},\frac{1+(x^R-x)y^R}{x^R}\right\}$$ -for $x$ positive and only positive $x^L$ considered, which is needed for defining division. Conway, after giving this definition on p.21, writes himself - -Note that expressions involving $y^L$ and $y^R$ appear in the definition of $y$. It is this that requires us to "explain" the definition. The explanation is that we regard these parts of the definition as defining new options for $y$ in terms of old ones. - -In a footnote, the rather trivial example of $\frac{1}{3}=\{0,\frac{1}{4},\frac{5}{16},\ldots|\frac{1}{2},\frac{3}{8},\ldots\}$ is given to show "how the definition works". -I can't see why $y$ is well defined in general. For example, given two uncountable cardinals $\alpha<\beta$ I'm having a hard time seeing how $\frac{1}{\beta-\alpha}$ should be computed. The emphasis here lies on "uncountable". -Claus Tøndering gave a seemingly equivalent definition of the inverse here on p.44 (if the definition should not be equivalent, please point out why). He defines $y$ through $y^L$ and $y^R$ as such: -$$0\in y^L$$ -$$z\in y^L \Rightarrow \tfrac{1+(x^R-x)z}{x^R}\in y^L, \tfrac{1+(x^L-x)z}{x^L}\in y^R$$ -$$z\in y^R \Rightarrow \tfrac{1+(x^L-x)z}{x^L}\in y^L, \tfrac{1+(x^R-x)z}{x^R}\in y^R$$ -This is still "too" recursive to be formalized. One of my problems is that I can't comprehend the cardinality of $y^L$ and $y^R$. I mean, I could define $y^L_0 = \{0\}, y^R_0 = \{\}$ and for $n\in\mathbb{N}, n>0$ change Tønderings definitions to $z\in y^L_{n-1} \Rightarrow \ldots\in y^L_n$ and so on (or better: $$y^L_n = \left\{\left.\tfrac{1+(x^R-x)z}{x^R}\right|z\in y^L_{n-1}\right\}\cup\left\{\left.\tfrac{1+(x^L-x)z}{x^L}\in y^L \right|z\in y^R_{n-1}\right\}$$ and $y^R$ analogue) and conjecture $$y^L = \{0\}\cup\bigcup_{n\in\mathbb{N}} y^L_n,\quad y^R = \bigcup_{n\in\mathbb{N}} y^R_n$$ -Here is my problem: Could the left or right options of $y$ accidentally be a proper class? I really doubt I could prove that's wrong. First off, I'm having trouble believing the equality holds, that I could miss something by merely having a countable union. Secondly, $y^L$ and $y^R$ are required to be sets and I can image how they accidentally could become classes this way with some $x$ nefarious enough (maybe $x=\beta-\alpha$ is enough already?), because maybe the set generation process never stops at a certain day. I get couldn't information about this topic at all. In papers about surreal numbers either they are just given like here without further doubt or not explicitly given at all. Some papers, like these from Philip Ehrlich, go deeper into cardinality or other theories above my understanding, so if the issue would be resolved there, I wouldn't have noticed. -On the matter of the $y^L_n$ and $y^R_n$ being sets, Conway writes - -Theorem 10. We have (i) $xy^L<1 -TITLE: Maximum zero converges to $\sqrt{2}$ -QUESTION [9 upvotes]: In my research I came upon a recursively defined sequence, and I'm pretty sure it converges to $\sqrt{2}$ though I can't prove it easily. I don't think it is a difficult question but I'm not sure. -Consider the following sequence of functions over $\mathbb{R}$, where it makes sense: -$f_0(x)=0$, -$\displaystyle f_{n+1}(x)=\frac{1}{2(x-f_n(x))}\hspace{1cm}$ ($n\geq 0)$. -Now let us define the sequence $(x_n)$ by $\displaystyle x_n:=\max\left\{y\in[0,\sqrt{2}[\,:\quad y=\frac{1}{2y}+f_n(y)\right\}$. - -Question: is is true that $x_n\rightarrow\sqrt{2}$ when $n\rightarrow +\infty$? - -Numerical evidence strongly suggest that, and it completely makes sense with the problem it originated from. The issue is that the functions $f_n$ have more and more poles as $n$ grows, and there is no function it converges to. It looks like the set of the poles of $f_n$ tends to be dense in $[-\sqrt{2},\sqrt{2}]$ when $n\rightarrow +\infty$, and $f_n$ is always decreasing outside of the poles. The poles seem to accumulate more around $\pm\sqrt{2}$ than around $0$. -For visual reference, one can see a graph of $f_{10}$ here -In advance, thank you for your interest/time. -Edit: I added the fact that I'm only interested in the $y\in[0,\sqrt{2}[$. I don't care what happens outside the interval since then it's trivial. - -REPLY [3 votes]: No need for any analysis: the roots are roots of shifted Tchebyshev -polynomials, all of the form $\sqrt{2}\cos(\pi/(2k))$ for suitable $k$ -in arithmetic progression.<|endoftext|> -TITLE: Is every square root of an integer a linear combination of cosines of $\pi$-rational angles? -QUESTION [6 upvotes]: For example, $\sqrt 2 = 2 \cos (\pi/4)$, $\sqrt 3 = 2 \cos(\pi/6)$, and $\sqrt 5 = 4 \cos(\pi/5) + 1$. Is it true that any integer's square root can be expressed as a (rational) linear combinations of the cosines of rational multiples of $\pi$? -Products of linear combinations of cosines of rational multiples of $\pi$ are themselves such linear combinations, so it only needs to be true of primes. But I do not know, for example, a representation of $\sqrt 7$ in this form. - -REPLY [23 votes]: Someone should actually record the formula. If $p$ is a prime $\equiv 1 \bmod 4$, then -$$\sqrt{p} = \sum_{k=1}^{p-1} \left( \frac{k}{p} \right) \cos \frac{2 k \pi}{p}$$ -where $\left( \tfrac{k}{p} \right)$ is the quadratic residue symbol. Note that $\left( \tfrac{k}{p} \right) = \left( \tfrac{p-k}{p} \right)$, so every term appears twice. -Similarly, if $p \equiv 3 \bmod 4$, then -$$\sqrt{p} = \sum_{k=1}^{p-1} \left( \frac{k}{p} \right) \sin \frac{2 k \pi}{p}.$$ -Again, $k$ and $p-k$ make the same contribution. -These are usually both written together as -$$\sqrt{(-1)^{(p-1)/2} p} = \sum_{k=1}^{p-1} \left( \frac{k}{p} \right) \exp \frac{2 k \pi i}{p}.$$ -This is a formula of Gauss. - -REPLY [11 votes]: Yes, that is true. The general case is the Kronecker-Weber theorem as Lucia mentioned in the comments. For square roots one can be more explicit and prove $\mathbb{Q}(\zeta_p) \cap \mathbb{R} = \mathbb{Q}[\sqrt{ (-1)^{(p-1)/2} p }]$ for odd prime numbers $p$ using properties of the Legendre symbol. Therefore you can write $\sqrt{p}$ either as $\mathfrak{Re}(\sqrt{p})$ or $\mathfrak{Im}(\sqrt{-p})$ depending on $p\mod 4$ so that you can write it as rational (in fact: integral) linear combination of real- or imaginary parts of powers of $\zeta_p$, i.e. cosine or sine values.<|endoftext|> -TITLE: Nielsen-Schreier theorem for monoids -QUESTION [7 upvotes]: Let $S$ be a finitely generated free abelian semigroup (or monoid), and let $T \subset S$ be a sub-semigroup (sub-monoid). Does the Nielsen-Schreier theorem hold in this case, that is, will $S$ still be free? Moreover, will $S$ be finitely generated as in the groups case? - -REPLY [3 votes]: Here is (partially) what is known about commutative monoids and their submonoids. - -The elementary theory of every finitely generated commutative semigroup (monoid) is decidable (Taiclin). -The isomorphism problem for finitely generated commutative semigroups (monoids) is decidable (Taiclin). -Every finitely generated commutative monoid is representable by matrices over $\mathbb Z$, hence is residually finite and Hopfian (Malcev). -Every finitely generated commutative monoid is finitely presented (Redei) moreover it admits a finite Church-Rosser presentation and for every finite presentation, the Knuth-Bendix algorithm terminates (all these essentially follow from the Hilbert Finite Basis theorem). -The lattice of subsemigroups of the infinite cyclic semigroup does not satisfy any non-trivial identity (Repnitskii, Katsman). In contrast, the lattice of subgroups of any commutative group is modular.<|endoftext|> -TITLE: What is the cohomological dimension of the commutator subgroup of the pure braid group? -QUESTION [22 upvotes]: I'm interested in computing the cohomological dimension of the commutator subgroup $[P_n,P_n]$ of the pure braid group $P_n$. I wasn't able to find a reference in the literature. -Because $[P_n,P_n]$ has an abelian subgroup of rank $\lfloor(n-1)/2\rfloor$ we have$$(n-2)/2\le\text{cd}([P_n,P_n])\le n-2.$$ My guess is that in fact $\text{cd}([P_n,P_n])=n-2$. -There is a right split short exact sequence $$1\to[F_n,F_n]\to[P_{n+1},P_{n+1}]\to[P_n,P_n]\to1,$$ which implies that $[P_n,P_n]$ is an iterated semidirect product of infinitely generated free groups. This suggests an inductive spectral sequence argument but I'm having problems understanding the cohomology groups $H^{n-2}\left([P_n,P_n];H^1([F_n,F_n])\right)$. -EDIT: The commutator subgroup $[B_n,B_n]$ of the full braid group has been studied by Gorin and Lin in "Algebraic equations with continuous coefficients and some problems of the algebraic theory of braids" (1969) Math. USSR Sb. 7 569-596. In particular, it follows from their results that $\text{cd}([B_3,B_3])=1$ and $\text{cd}([B_4,B_4])=2$. However, this is not immediately helpful because $[P_n,P_n]$ is not a finite index subgroup of $[B_n,B_n]$. - -REPLY [18 votes]: Here is the answer: for $n\geq 2$ we have $\mathrm{cd}([P_n,P_n])=n-2$. -https://arxiv.org/abs/1905.05099<|endoftext|> -TITLE: Does Laver Forcing add an infinitely often equal real? -QUESTION [7 upvotes]: Given a model $V$ of set theory and an inner model $W \subseteq V$, a real $x \in V \cap \omega^\omega$ is infinitely often equal over $W$ if for each real $y \in \omega^\omega$ there are infinitely $n$ with $x(n) = y(n)$. Cohen reals have this property and it was an open question for a while as to whether every forcing adding an infinitely often equal real also adds a Cohen real. In fact this is false. -What I want to know is whether it is ever the case that Laver forcing adds a real which is infinitely often equal over the ground model. I expect the answer to be no given the aforementioned question about finding infinitely often equal reals that are not Cohen. However, I'm struggling to prove it. -For reference, recall that Laver forcing, $\mathbb L$, is defined to be the set of trees on $\omega$ with a distinguished stem and infinitely branching at every node above the stem. A Laver real is defined in the generic extension to be the union of the stems of the trees in the generic. It's not hard to see that such a real is always dominating over the ground model. Moreover it is well known (and proved in Jech) that $\mathbb L$ never adds Cohen reals over the ground model, even when iterated. - -REPLY [9 votes]: The answer is no. -This follows easily from the Laver property. Every new real which is bounded by a ground model real (say: by the identity function) is contained in a small slalom from the ground model, hence will be eventually different from lots of ground model reals.<|endoftext|> -TITLE: Annihilation operators in a vertex algebra -QUESTION [9 upvotes]: Let $V=\bigoplus_{d\in\mathbb N}V(d)$ be a Möbius-covariant vertex algebra with $V(0)=\mathbb C$. -Recall that a vector $v\in V$ is called quasi-primary if $L_1v=0$. -For $v\in V(d)$, we write $Y(v,z)=\sum_{n\in\mathbb Z} z^{-n-d}v_{(n)}$. -With that convention, $v_{(n)}$ is an operator $V(k)\to V(k-n)$. - -Let $d>n$. - Is it true that for any quasi-primary $v\in V(d)$ and any vector $w\in V(n)$, we have $v_{(n)}w=0$? - -If the above relation does not always hold, are there reasonable extra assumptions that one can impose on $V$ that imply it? -Do the above relations hold when $v$ is required to be primary instead of quasi-primary? (Add the assumption that $V$ is a VOA so that the notion of a primary vector make sense) - -REPLY [4 votes]: The answer seems to be yes for quasi-primary $v$ if $V$ has a suitable invariant bilinear form. Then one can identify $v_{(n)} w$ with its pairing with the vacuum, and obtains it as the appropriate coefficient of -$$(\mathbf{1}, Y(v, x)w) = (-x^{-2})^d (Y(v, x^{-1})\mathbf{1}, w) = (-x^{-2})^d (e^{x^{-1} L(-1)} v, w) = 0$$ -since the weight of $v$ is greater than that of $w$. -Of course, this argument does not work if $v$ is not quasi-primary since then $v$ must be replaced with $e^{x L(1)} v$ after the first equality in the calculation. -Note that it isn’t necessary to assume the bilinear form is nondegenerate, just nondegenerate on the one-dimensional vacuum space.<|endoftext|> -TITLE: An extremal problem related either to an uncertainty principle on the circle, or else to the prime number theorem -QUESTION [10 upvotes]: Consider for $X = 1,2, \ldots$ the set $\mathcal{S}_X$ of trigonometric polynomials $f(t) := \sum_{|k| \leq X} c_k e^{2\pi i kt}$ on the circle $\mathbb{T} := \mathbb{R}/\mathbb{Z}$ of degree $\leq X$ (Fourier transform supported on $\{-X,\ldots, X\}$), such that $f(0) = 1$ and $c_0 = 0$ (the last assumption is probably inessential). Let -$$ -M_X(f) := \sup_{\mathbb{T} \setminus [-\frac{1}{X},\frac{1}{X}] }|f|. -$$ -and -$$ -B_X := \inf_{f \in \mathcal{S}_X} M_X(f). -$$ -Is the $X \to \infty$ limit of $B_X$ strictly positive, or zero? -The motivation is the same as in this question that I have asked previously. Observe that, for all $n \leq X$, the average of $f(q)$ over $\mu_{n} \setminus \{1\}$, $q := e^{2\pi i t}$, does not exceed $M_X(f)$ in magnitude. As a result, Chebyshev's bound, the orthogonality relation in $\mathbb{Z} / n$ and the Dirichlet convolution identity (definition of the Mangoldt function) $\log = 1 * \Lambda$ yield for any $f \in \mathcal{S}_X$ the estimate -$$ -\sum_{n \leq X} \frac{\Lambda(n)}{n} = \sum_{k \neq 0} c_k \log{|k|} + O(M_X(f)), -$$ -where the implied coefficient is absolute and explicit. This would be an asymptotic formula if the $O(\cdot)$ term could be made to approach zero as $X \to \infty$, suggesting that $M_X(f)$ should perhaps be bounded away from zero. If not, then of course the next question would be to ask for the asymptotic computation of the extremal sequence $f_X$ and its decay rate $M_X(f_X) = B_X$. -It occurred to me that the linked question may have possibly been about functions on the circle $\mathbb{T} \leftrightarrow \mathbb{Z}$ rather than on the real line $\mathbb{R} \leftrightarrow \mathbb{R}$. For (I could be wrong about this) it seems to be a rather special situation to have $c_k \sim \frac{1}{X}\varphi(k/X)$ with $\varphi \in \mathcal{S}(\mathbb{R})$ a fixed Schwartz function supported on $[-1,1]$ and with $\varphi(0) = 0$ and $\widehat{\varphi}(0) = 1$. It does follow from the same argument as in Terry Tao's solution of the linked problem that there is an absolute $\epsilon_0 > 0$ such that $\lim_{X \to \infty} M_X(\sum_{k} \frac{1}{X} \varphi(k/X) e^{2\pi i kt}) \geq \epsilon_0$ for all such $\varphi$; for this unpacks to stating that $\sup_{\mathbb{R} \setminus [-1,1]} |\widehat{\varphi}| \geq \epsilon_0 > 0$ whenever $\mathbb{supp}(\varphi) \subset [-1,1]$ and $\varphi(0) = 0, \widehat{\varphi}(0) = 1$: a version of the uncertainty principle on the real line. But it isn't clear to me whether the sequence of solutions to our extremal problem should have such a limiting distribution $\varphi$. Also it would be nice to know of an argument that is directly about trigonometric polynomials. -Is there a version of the uncertainty principle on the circle that would yield the $M_X(f) \geq \epsilon_0 > 0$ answer in the present question too? -[Note: The conditions $c_0 = 0$ and $\varphi(0) = 0$ are probably irrelevant to the discussion; but they are convenient, so let me impose them for concreteness' sake. Other natural choices would be to take $c_0 = 1/X$ (corresponding to $\varphi(0) = 1$), or to drop them altogether. ] - -REPLY [11 votes]: A compactness argument shows that for sufficiently large $X$ one has the bound -$$ \sup_{x \in {\mathbb T} \backslash [-1/X,1/X]} |f(x)| \gg \sup_{x \in [-1/X,1/X]} |f(x)|$$ -whenever $f$ is a trigonometric polynomial of degree at most $X$; this would imply that $B_X \gg X$. (Perhaps there is a normalising factor of $1/X$ missing in your question?) -Proof: Suppose the claim failed, then one could (after normalising) find a sequence $X_n \to \infty$ and a sequence $f_n$ of trigonometric polynomials of degree at most $X_n$ such that -$$ \sup_{x \in [-1/X_n,1/X_n]} |f_n(x)| = 1$$ -and -$$ \sup_{x \in {\mathbb T} \backslash [-1/X_n,1/X_n]} |f_n(x)| = o(1).$$ -We can find $x_n \in [-1/X_n,1/X_n]$ such that $|f_n(x_n)|=1$. Writing $F_n: {\mathbf R} \to {\mathbf C}$ for the $X_n$-periodic function -$$ F_n(x) := f_n( \frac{x}{X_n} - x_n \hbox{ mod } 1)$$ -we see that $|F_n(0)|=\|F_n\|_{L^\infty}=1$, that $\sup_{2 \leq |x| \leq X_n/2} |F_n(x)| = o(1)$, and that $F_n$ is band-limited to $[-1,1]$ (i.e., its Fourier transform is supported in $[-1,1]$). In particular, if $\varphi$ is a Schwartz function whose Fourier transform equals $1$ on $[-1,1]$, then $|\langle F_n, \varphi \rangle| = |F_n(0)| = 1$. -By passing to a subsequence one can assume that $F_n$ converges weakly to another function $F$ in the unit ball of $L^\infty$, which is then non-zero by testing against $\varphi$. On the other hand, $F_n$ vanishes outside of $[-2,2]$ and its distributional Fourier transform is supported on $[-1,1]$, which is a contradiction as the Fourier transform is also analytic. -The compactness argument does not give an effective bound for the implied constant, but presumably one can do so if one uses a suitable effective version of the uncertainty principle, such as the one in this blog post of mine.<|endoftext|> -TITLE: nef vs. 1-nef vector bundles -QUESTION [7 upvotes]: Let $X$ be a compact, connected, Kähler manifold, of dimension $d$, with Hermitian metric $\omega$; let $E$ be a vector bundle on $X$ of rank $r\geq2$. -By [1] definition 3.1.2: - -A line bundle $L$ over $X$ is said numerically effective (nef, for short) if for any $\epsilon>0$ there exists a smooth Hermitian metric $h_{\epsilon}$ on $L$ such that - \begin{equation*} -\Omega_{h_{\epsilon}}(L)\geq-\epsilon\omega; -\end{equation*} - that is the curvature form $\Omega_{h_{\epsilon}}(L)$ of the Chern connection on $L$ (with respect to $h_{\epsilon}$) can have an arbitrary negative part. $E$ is nef if the tautological bundle $\mathcal{O}_{\mathbb{P}(E)}(1)$ is nef. - -By [1] definition 3.1.3: - -$E$ is $1$-nef if for any $\epsilon>0$ there exists a Hermitian metric $h_{\epsilon}$ on $E$ such that $\Omega_{h_{\epsilon}}(E)\geq-\epsilon\omega$. - -By [1] proposition 3.2.4, the $1$-nef bundles $E$ over $X$ are nef; but the inverse is unknown in general; excepted for: - -$d=1$ (i.e. algebraic curves), see [1] theorem 3.3.1; -on toric and Abelian varieties $E\otimes\det E$ is $1$-nef; -tangent bundle $TX$ of $X$, where it is nef and $d\in\{2,3\}$; - -(2) is justified in [1] at page 113, (3) follows by [2] theorems 6.1, 7.1 and [1] proposition 3.2.4. -Question: Are there other examples of manifolds over which the nef bundles are $1$-nef? Or is there an example of nef not $1$-nef bundle over some manifold $X$? - -[1] M. A. A. De Cataldo - Singular Hermitian metrics on vector bundles, J. reine ang. Math. 502 (1998) 93-122 -[2] J.-P. Demailly, T. Peternell, M. Schneider - Compact complex manifolds with numerically effective tangent bundles, J. Algebraic Geom. 3 (1994) 295-345 - -REPLY [2 votes]: Let $(X,\omega)$ be a compact Kähler manifold of dimension $d\geq2$; let $E$ be a nef vector bundle of rank $r\geq2$ such that $\displaystyle\deg(E)=\int_Xc_1(E)\wedge\omega^{d-1}=0$. $E$ is a $1$-nflat vector bundle. - -Proof. By [DPS, propositions 1.14.i and 1.15.i] $\det(E)$ is a ($1$-)nef line bundle; by [BGO2, remark 3.2.ii and lemma 3.13] $\det(E)$ is a ($1$-)nflat line bundle; because -\begin{equation*} -E^{\vee}\cong\bigwedge^{r-1}E\otimes(\det(E))^{\vee} -\end{equation*} -$E^{\vee}$ is a nef vector bundle. By definition $E$ is a nflat vector bundle. -By [DPS, theorem 3.2] $E$ has a filtration whose quotients are locally free and flat, by [BGO2, theorem 3.16] this is equivalent to say that $E$ is $1$-nflat. $\Box$ -Remark. Using the notion of Numerically Flat Higgs bundle (H-nflat, for short) introduced in [BGO1], and $1$-Numerically Flat Higgs bundle ($1$-H-nflat, for short) introduced in [BGO2], and applying [BC, theorems 3.2 and 5.2] indeed of [DPS, theorem 3.2], one proves the following (partial) generalization to Higgs bundles case. - -Let $(X,H)$ be a smooth complex projetive variety of dimension $d\geq2$, with a polarization $H$; let $\mathfrak{E}=(E,\phi)$ be a H-nef Higgs bundle of rank $r\geq2$ such that $\displaystyle\deg(E)=c_1(E)\cdot H^{d-1}=0$ and $c_2(E)\cdot H^{d-2}=0$. $\mathfrak{E}$ is a $1$-H-nflat Higgs bundle. - - -[BC] U. Bruzzo, A. Capasso, Filtrations of numerically flat Higgs bundles and curve semistable Higgs bundles on Calabi-Yau manifolds, arXiv:1904.10069 [math.AG] -[BGO1] U. Bruzzo and B. Graña Otero, Numerically Flat Higgs bundles, Commun. Contemp. Math. 9 (2007) 437-446 -[BGO2] U. Bruzzo and B. Graña Otero, Metrics on semistable and numerically effective Higgs bundles, J. reine ang. Math. 612 (2007) 59-79. -[DPS] J.-P. Demailly, T. Peternell, M. Schneider - Compact complex manifolds with numerically effective tangent bundles, J. Algebraic Geom. 3 (1994) 295-345<|endoftext|> -TITLE: Automated search for bijective proofs -QUESTION [14 upvotes]: In enumerative combinatorics, a bijective proof that $|A_n| = |B_n|$ (where $A_n$ and $B_n$ are finite sets of combinatorial objects of size $n$) is a proof that constructs an explicit bijection between $A_n$ and $B_n$. -Bijective proofs are often prized because of their beauty and because of the insight that they often provide. Even if a combinatorial identity has already been proved (e.g., using generating functions), there is often interest in finding a bijective proof. -In spite of the importance of bijective proofs, the process of discovering or constructing a bijective proof seems to be an area that has been relatively untouched by computers. Of course, computers are often enlisted to generate all small examples of $A_n$ and $B_n$, but then the process of searching for a bijection between $A_n$ and $B_n$ is usually done the "old-fashioned" way, by playing around with pencil and paper and using human insight. -It seems to me that the time may be ripe for computers to search directly for bijections. To clarify, I do not (yet) envisage computers autonomously producing full-fledged bijective proofs. What I want computers to do is to search empirically for a combinatorial rule—that says something like, take an element of $A_n$ and do $X$, $Y$, and $Z$ to produce an element of $B_n$—that appears to yield a bijection for small values of $n$. -One reason that such a project has not already been carried out may be that the sheer diversity of combinatorial objects and combinatorial rules may seem daunting. How do we even describe the search space to the computer? -It occurs to me that, now that proof assistants have "come of age," people may have already had to face, and solve (at least partially), the problem of systematically encoding combinatorial objects and rules. This brings me to my question: - -Does there exist a robust framework for encoding combinatorial objects and combinatorial rules in a way that would allow a computer to empirically search for bijections? If not, is there something at least close, that could be adapted to this end with a modest amount of effort? - -In my opinion, Catalan numbers furnish a good test case. There are many different types of combinatorial objects that are enumerated by the Catalan numbers. As a first "challenge problem," a computer program should be able to discover bijections between different kinds of "Catalan objects" on its own. If this can be done, then there is no shortage of more difficult problems to sink one's teeth into. - -REPLY [9 votes]: As mentioned in the comments, the FindStat project is aiming at what you want. Concerning the size: it contains currently about 1000 'combinatorial statistics', that is maps $s:\mathcal C_n\to \mathbb Z$ on some (graded) set of 'combinatorial' objects $\mathcal C_n$ and about 150 'combinatorial maps' between two collections. What makes FindStat powerful is the (trivial) ability to compose maps. For example, for Catalan objects we obtain about 1.500.000 a priori different statistics. -Let me point out some possible ways of using it, in the spirit of the question. - -'automatically producing a bijection mapping one statistic to another' is demonstrated in Two statistics on the permutation group and Combinatorics problem related to Motzkin numbers with prize money I. -'automatically producing conjectures' is achieved by clicking on 'search for distribution' on any of the statistics in the statistics database. The result is a list of statistics that are conjecturally equidistributed with the given statistic, but where a map transforming the first into the second might not be known. A classic is http://findstat.org/St000012. -it is easy to write a script that iterates 2. to find a 'partner' for a given pair of equidistributed statistics. An example is given in the comments to https://math.stackexchange.com/questions/2511943/leaf-labelled-unordered-rooted-binary-trees-and-perfect-matchings. Note that this meanwhile has a proof, also (essentially) discovered by FindStat. -a different kind of conjectures is provided by the list of 'experimental identities' found when selecting any of the maps at http://findstat.org/MapsDatabase. I am guessing that not all identities at http://findstat.org/Mp00101 are immediately obvious. -I am also working on a new package that checks whether a statistic satisfying given constraints can possibly exist. But that's for later...<|endoftext|> -TITLE: What are the matrices preserving the $\ell^1$-norm? -QUESTION [17 upvotes]: So I am inspired by unitary matrices which preserve the $\ell^2$-norm of all vectors, so in particular the unit norm vectors. But then I saw that the $\ell^1$-norm of probability vectors is preserved by matrices whose columns are probability vectors. And this got me thinking: But what are the matrices preserving the $\ell^1$-norm of arbitrary real unit $\ell^1$-norm vectors? So basically we extend a probability vector to also allow a sign, but ignoring the signs, this should still be a probability vector; and then we ask for the corresponding structure-preserving matrices. -It is already clear that the columns of such a matrix should be this 'extended' kind of probability vector, because we can multiply the matrix with a standard basis vector which has $\ell^1$-norm 1. But not all of such matrices preserve this, take for example -$$ M = \frac{1}{2} \left(\begin{matrix} 1 & 1\\ 1 & -1 \end{matrix}\right) $$ -and -$$ x = \left( \begin{matrix} 0.3 \\ -0.7 \end{matrix} \right) $$ -Then we have -$$ Mx = \left(\begin{matrix} -0.2 \\ 0.5 \end{matrix}\right) $$ -which fails the test. - -REPLY [4 votes]: Here is yet another (sketch of the) proof that the matrices preserving the $p$-norm for $p\neq 2$ are generated by permutation matrices, and the diagonal ones with diagonal elements of absolute value $1$. -$\let\eps\varepsilon$I address the real case; but the complex one should be similar. -If $2 -TITLE: Haar Measure Integral -QUESTION [6 upvotes]: I am a physicist and I am wondering whether the following integral over Haar measure （edit: say $U$ is unitary, orthogonal or symplectic matrix) -\begin{align} -\int dU \: \exp\left( \mathrm{tr}(UX) + \mathrm{tr}(X^\dagger U^\dagger) \right) -\end{align} -have an explicit expression in terms of the matrix X. For example, if the group is $U(1)$, then the result would be the modified Bessel function $I_0(2|X|)$. For the general case, I guess one can at least expand the exponent and use the Weingarten functions, and then perform a re-summation. But I know too little about the properties of the Weingarten functions to organize the re-summation into any simple, explicit form. Does someone know how to do this, or perhaps where formulae like this can be found? - -REPLY [6 votes]: Depending on what you mean by "explicit", in the unitary case this can be read off -from a generalization of the Harish--Chandra-Itzykson-Zuber formula. To see that, note that your integral can be rewritten as -$$J=\int_{U_N}\int_{U_N} \exp(\Re (\mbox{tr} V YU)) dU dV,$$ -where $Y$ is a diagonal real matrix whose entries are the singular values of $X$. -Now, for fixed diagonal $A,B$ consider the integral -$$J(A,B)=\int_{U_N} \int_{U_N} \exp(\Re (\mbox{tr} A V BU)) dU dV.$$ -Then $J=J(I,Y)$. For $A,B$ with distinct entries, $J(A,B)$ has an explicit formula involving determinants in Bessel functions, see for example formula (3.6) in the review paper of Zinn-Justin and Zuber, https://arxiv.org/pdf/math-ph/0209019.pdf (they attribute the result to Balantekin and to Guhr--Wetting, although I guess one can trace it all the way back to Harish-Chandra). Now in your case $A=I$ and in particular the entries of $A$ are not distinct, but resolving this involves a straight-forward limit, replacing $I$ by $A=I+\epsilon \Delta$ where $\Delta$ has distinct real entries, and taking $\epsilon \to 0$. -I suspect that the case of $A=I$ has an even simpler formula, but I don't see it. Maybe somebody else, more versed in representations than me, can -comment on that.<|endoftext|> -TITLE: Integral $\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}=?$ -QUESTION [25 upvotes]: How to evaluate this integral: -$$\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}=?$$ -I'm making use of the integral identity: -$$\int_{0}^{+\infty }e^{-t(x_{1}+x_{2}\cdots +x_{n})}dt=\frac{1}{x_{1}+x_{2}\cdots +x_{n}}$$ and then reversing the order of integration with respect to time and space variables. -But for $n=1$, then such that, $$\int_{0}^{\infty }dt\int_{0}^{1}x^{2}e^{-tx}dx=\int_{0}^{\infty }\frac{2 - e^{-t}(2 + 2t+t^2)}{t^3}dt=\int_{0}^{1}x\,dx=\frac{1}{2},$$ and $$\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}\\=n\int_{0}^{+\infty }\frac{2 - e^{-t}\left ( 2 + 2t+t^2 \right )}{t^3}\left ( \frac{1-e^{-t}}{t} \right )^{n-1}dt.$$ - -REPLY [12 votes]: Here is another approach, which also gives the rational term. -(I) To see how it works let $n\geq 2$ and consider first -the simpler case -\begin{align*} -\mathbb{E}\bigg(\frac{1}{X_1+\ldots+X_n}\bigg)=\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt -\end{align*} -Using $\frac{1}{t^n}=\int_0^\infty \frac{z^{n-1}}{(n-1)!} e^{-zt}\,dz$ we find -\begin{align*} -\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt&=\int_0^\infty\int_0^\infty \frac{z^{n-1}}{(n-1)!} e^{-zt}(1-e^{-t})^n\, dz\,dt\\ -&=\int_0^\infty\int_0^\infty \frac{z^{n-1}}{(n-1)!} e^{-zt}(1-e^{-t})^n\, dt\,dz\\ -&=\int_0^\infty \frac{z^{n-1}}{(n-1)!}\, \mathrm{Beta}(z,n+1)\,dz\\ -&=n\,\int_0^\infty \frac{z^{n-1}}{z(z+1)\cdots(z+n)}\,dz -\end{align*} -The following observation will be the key: -Lemma: let $x_0,\ldots,x_n$ be distinct positive numbers and $k\leq n-1$. Then -$$I_{k,n}(x_0,\ldots,x_n):=\int_0^\infty \frac{z^{k}}{(z+x_0)(z+x_1)\cdots(z+x_n)}\,dz=(-1)^{n+k+1}\Delta^n(x^k\log(x);x_0,\ldots,x_n)$$ -where (for a real function $f$) $\Delta^n(f;x_0,\ldots,x_n)$ -denotes the divided difference of $f$ corresponding to $x_0,\ldots,x_n$. -Recall that (Newton-interpolation) -(1) the divided differences are for $f$ and mutually distinct $x_0,\ldots,x_n$ are defined recursively by -$\Delta^0(f;x_0)=f(x_0)$, -$\Delta^n(f;x_0,\ldots,x_n)=\frac{\Delta^{n-1}(f; x_1,\ldots,x_n)-\Delta^{n-1}(f; x_0,\ldots,x_{n-1})}{x_n-x_0}$ -(2) they are explicitly -given by $$\Delta^n(f;x_0,\ldots,x_n)=\sum_{i=0}^n \frac{f(x_i)}{\prod_{j\neq i} (x_i-x_j)}\;\;(**)$$ -(3) $$\Delta^n(f;x,x+1,\ldots,x+n)=\frac{1}{n!} \sum_{i=0}^n {n \choose i} (-1)^{n-i} f(x+i)$$ -Proof of the lemma: -For $k=0, n=1$ we have $$\int_0^\infty \frac{1}{(z+x_0)(z+x_1)}=\frac{\log(x_1)-\log(x_0)}{x_1-x_0}=\Delta^1(\log(x);,x_0,x_1)$$ -For $k=0,n>1$ the repeated use of $\frac{1}{(z+a)(z+b)}=\frac{-1}{b-a}\left(\frac{1}{z+b}-\frac{1}{z+a}\right)$ shows that $(-1)^{n+1}I_{0,n}(x)=\Delta^n(\log(x),x)$. -The validity for $k>0,n=k+1$ follows from $\frac{z}{z+b}=1-\frac{b}{z+b}$ and $(**)$. End of proof. -The lemma and (3) now give that -\begin{align*} - \int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt &=n \Delta^{n-1}(x^{n-2}\log(x);x+1,x+2,\ldots,x+n)\\ - &=\frac{n}{(n-1)!}\sum_{i=0}^{n-1}{ n-1 \choose i} (-1)^{n-1-i} (i+1)^{n-2}\log(i+1) -\end{align*} -(II) Now let $n\geq 1$ and consider -$$ Q_{n+1}:=\mathbb{E}\bigg(\frac{X_1^2+\ldots + X_{n+1}^2}{X_1+\ldots+X_{n+1}}\bigg)$$ -Write -$$Q_{n+1}=(n+1)\int_0^1 u^2 \bigg(\int_0^\infty e^{-ut}\bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt\bigg)\,du$$ - Proceeding as above shows that for $u>0$ -\begin{align*} - \int_0^\infty e^{-ut}\bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt -&=\frac{n}{n!} \sum_{i=0}^n {n \choose i} (-1)^{n-i} (u+i)^{n-1}\log(u+i) -\end{align*} -Thus -$$q_{n+1}:=\frac{Q_{n+1}}{n(n+1)}=\frac{1}{n!} \sum_{i=0}^n {n \choose i} (-1)^{n-i} \int_0^1 u^2(u+i)^{n-1}\log(u+i)\,du$$ -Now expand $u^2=(u+i-i)^2$ and integrate partial to find -that $q_{n+1}=L-R$ -where $L=\frac{1}{n!}\sum_{i=0}^n{n\choose i}(-1)^{n-i} \ell(i),\;R=\frac{1}{n!}\sum_{i=0}^n {n \choose i}(-1)^{n-i} r(i)$ with -\begin{align*} - \ell(i)=&\frac{1}{n+2}\left((i+1)^{n+2}\log(i+1)-i^{n+2}\log(i)\right)\\ - &-\frac{2i}{n+1}\left((i+1)^{n+1}\log(i+1)-i^{n+1}\log(i)\right)\\ - &+\frac{i^2}{n}\left((i+1)^{n}\log(i+1)-i^n\log(i)\right)\\[0.2cm] - r(i)=&\frac{1}{(n+2)^2}\left((i+1)^{n+2}-i^{n+2}\right)\\ - &-\frac{2i}{(n+1)^2}\left((i+1)^{n+1}-i^{n+1}\right)\\ - &+\frac{i^2}{n^2}\left((i+1)^{n}-i^n\right) -\end{align*} -Collecting terms in the logarithmic part $L$ shows that the coefficient of $\log(i)$ in $q_{n+1}$ is $c_{i,n+1} =(-1)^{n-i} \frac{i^n}{n}{ n \choose i-1}\left(\frac{n^2+3n+2-2i}{(n+2)!}\right)$, matching Fedor Petrov's answer. -The rational part $R$ can readily be summed (note that only the contributions of the powers $n$ and $n+1$ (with coefficients $-\frac{1}{n(n+1)(n+2)}$ and $\frac{2}{n(n+1)(n+2)}$) need to be considered) to -give $$R=\frac{1}{n+2}-\frac{1}{n(n+1)(n+2)}\;\;,$$ -so that the rational term of $Q_{n+1}$ is $-\frac{n(n+1)-1}{n+2}=-\frac{(n+2)(n-1)+1}{n+2}$, confirming Sylvain JULIEN's guess.<|endoftext|> -TITLE: Binary weight of shifted integers -QUESTION [6 upvotes]: Suppose $n_2$ denotes the binary representation of the integer number $n$. Let $X_2(n)=[1_22_2\ldots n_2]$,$n\geq3$, be a binary vector which is obtained by concatenating of binary representation of the numbers from $1$ to $n$. Also, let $X_2^m(n)$,$0\leq m\leq n-1$,denotes the cyclically $m$ shift of the entries of the vector $X_2(n)$. For example, we have -$$X_2^0(n)=X_2(n)$$ -$$X_2^1(n)=[n_21_22_2\ldots(n-1)_2]$$ -$$X_2^2(n)=[(n-1)_2n_21_22_2\ldots(n-2)_2]$$ -and so on. -For two binary vectors $X$ and $Y$ (with same length), suppose $|X\cap Y|$ denotes the number of ones common to both $X$ and $Y$. -The conjecture is: -Let $n\geq 3$ be a natural number. For all $m$ and $k$, we have $|X_2^m(n)\cap X_2^k(n)|\cong 0 \mod 2$ if and only if $n=2^s-1$, for some integer number $s$. -Note: I use $\lfloor \log_2n\rfloor +1$ bits for binary representation of each integer number from $1$ to $n$. So, fedja's example is as follows: -$X_2(3)=[011011]$, $X_2^1(3)=[110110]$ and $X_2^2(3)=[101101]$. We can see the claim is true. -The conjecture is tested for many integer numbers. I appreciate any helpful comments and answers. -$\textbf{Added later}(19/09/2017)-9:03:$ The other easy way to see the problem is as follows: -For the natural number $n\geq 3$, let $G$ be a graph with $n$ vertices such that the vertices of $G$ are labeled by numbers $1$ to $n$. Then we represent each vertex label in uniform binary form and we connect the vertex $i$ to the vertex $j$ with the edge weight equal to the number of ones common to both $i_2$ and $j_2$. Note that the resulting graph is complete and each edge has a weight between $0$ to $\lfloor \log_2n\rfloor$. See the below example: - -By this representation, we have an equivalent graph theory problem. In this context we must show that the summation of the weights of edges of some special subgraphs are even or odd. - -REPLY [2 votes]: This is not a full answer, just the easy implication that if $n=2^{s}-1$ with $s\ge 2$, then, indeed, all overlaps are even. -We will think of our binary representations of length $s$ as written on the circle. We'll look at each digit sequence separately, so we get (starting with the right bit) -$$ -1010101\dots101, -\\ -01100110011\dots0011, -\\ -000111100001111\dots00001111 -$$ -etc. -Let's do elementary Fourier analysis on $\mathbb Z_n$. -The $r$-th sequence $(r=0,\dots,s-1)$ has the Fourier coefficients -(up to conjugation, which I'm too lazy to write at 11:15PM) -$$ -F_r(z)=\frac 1n z^{2^r-1}(1+z+z^2+\dots+z^{2^r-1})(1+z^{2^{r+1}}+z^{2\times 2^{r+1}}+\dots +z^{n+1-2^{r+1}}) -\\ -=\frac 1nz^{2^r-1}\frac{z^{2^r}-1}{z-1}\frac{z^{n+1}-1}{z^{2^{r+1}}-1} -=\frac 1nz^{2^r-1}\frac 1{z^{2^r}+1} -$$ -for $z^n=1, z\ne 1$. $F(1)=\frac{n+1}{2n}$ regardless of $r$. -The cardinality of the intersection of the $r$-th sequence with the shift of itself by $m$ can be written as -$$ -n\sum_{z:z^n=1}|F_r(z)|^2z^m=n\sum_{z:z^n=1}\frac{z^m}{|z^{2^r}+1|^2} -$$ -(Plancherel) -This looks pretty useless for determining the individual parities but it shows immediately ($z\mapsto z^{2^r}$ is just a rearrangement of the $n$th roots of unity) that the number of overlaps in the $r$-th position for $m$ is the same as the number of overlap in the $0$-th position for $m/2^r$ where the division is understood in the sense of $\mathbb Z_n$. Thus, instead of asking what happens for the individual $m\ne 0$ in all positions, we can ask what happens in the $0$-th position for $m,m/2,\dots, m/2^{s-1}$ ($m=0$ is a trivial case because the self-overlap in each position is $\frac{n+1}2=2^{s-1}$, which is even when $s>1$ and $n=2^1-1=1$ is, indeed, problematic). -Now we forget all high-tech and just keep this conclusion in mind. Of course, once we know it, we can show it in an elementary way too. -If $m\in\{1,\dots,n-1\}$, then we need to count the pairs of odd numbers in $\{1,2,\dots,n\}$ at distance either $m$ or $n-m$. But that is easy: if $m$ is even, they are $\frac{n+1-m}2=2^{s-1}-\frac m2\equiv \frac m2\mod 2$. Otherwise they are $\frac{m+1}2\equiv_2 \frac{m+1}2+2^{s-1}=\frac{m+n}2+1$. -Thus, if we consider the full cycle $m_r=m/2^r$ ($r=0,\dots,r-1$) in $\mathbb Z_n$ using the representatives $m_r\in\{1,\dots,n-1\}$ (some numbers may repeat, that's OK), then the parity of the number of overlaps in the $0$-th position for the shift by $m_r$ is the parity of $m_{r+1}=\frac{m_r}2$ if $m_r$ is even and of $1+m_{r+1}=1+\frac{m_r+n}2$ if $m_r$ is odd. This means that to get the parity of the full sum over the cycle, we should add the number of odd members of the cycle and the number of $+1$ corrections. But these two numbers are the same because the corrections are generated exactly by the odd members of the cycle, so the final result is always even. -This is quite simple and relatively clean. It is the other implication that seems to be a headache. Any bright ideas?<|endoftext|> -TITLE: (Co)limits of locally cartesian closed categories -QUESTION [6 upvotes]: Do locally cartesian closed $\infty$-categories form a presentable $\infty$-category? It seems like they should, and that the inclusion $\text{LCC}\rightarrow\text{Cat}$ preserves limits and maybe even colimits. -Here is a possible strategy for proof. There is a functor $\text{Cat}\rightarrow\text{Fun}(\Delta^1,\text{Cat})_\text{cart}$ taking an $\infty$-category $\mathcal{C}$ to the cartesian fibration $\text{Fun}(\Delta^1,\mathcal{C})\rightarrow\mathcal{C}$; this fibration classifies the functor $\mathcal{C}\rightarrow\text{Cat}$ which sends $X$ to $\mathcal{C}_{/X}$. The condition that $\mathcal{C}$ is locally cartesian closed is precisely that for all $X\rightarrow Y$, $\mathcal{C}_{/X}\rightarrow\mathcal{C}_{/Y}$ has a chain of two right adjoints (its right adjoint has a right adjoint). This should correspond to some condition on the cartesian fibration, so that $\text{LCC}$ is a pullback of $\text{Cat}\rightarrow\text{Cat}^\text{cart}$ along some subcategory of $\text{Cat}^\text{cart}$. A pullback of presentable categories should be presentable. -But I'm not sure how to handle the "chain of two right adjoints", and I wonder if I am missing an easier argument. - -REPLY [5 votes]: First, a clarification: what are the morphisms of locally cartesian closed categories meant to be? -Now, a word on the "two adjoints" thing. -For every category $C$, there is a cocartesian fibration $\mathrm{cod}: C^{\Delta[1]} \to C$ with fiber $C/c$ over $c \in C$; reindexing is given by composition. $C$ has pullbacks if and only if $\mathrm{cod}$ is simultaneously a cartesian fibration (i.e. it is a bifibration); reindexing is given by pullback. -If $C$ has pullbacks, then the cartesian fibration $\mathrm{cod}$ classifies a functor $C^\mathrm{op} \to \mathsf{Cat}$. This functor is also classified by a cocartesian fibration $\overline{\mathrm{cod}}: E \to C^\mathrm{op}$ (different from $\mathrm{cod}$). In the "total category" $E$, an object over $c \in C^\mathrm{op}$ is a map $f: d \to c$ in $C$, while a morphism from $f$ to $f'$ over $\gamma: c \leftarrow c'$ is a map $c \times_d c' \to d'$. Now $C$ is locally cartesian closed if and only if the cocartesian fibration $\overline{\mathrm{cod}}$ is simultatneously a cartesian fibration. Pavlovic calls such a functor a "trifibration"; his paper is referenced in the nlab page on bifibrations linked to above. -The process of straightening a cartesian fibration and then unstraightening it to get a cocartesian fibration may sound unwieldy in the $\infty$-categorical setting, but it has been studied by Barwick, Glasman, and Nardin, and I think they give a more direct construction. -Upshot: -This allows one to check that (small) locally cartesian closed categories form a presentable category, since they lie in a pullback diagram of accessible right adjoints between such: -$\require{AMScd} -\begin{CD} -Cat^{lcc} @>>> Cat^{cart} \\ -@VVV @VVV\\ -Cat^{pb} @>>> Cat^{\Delta[1]} -\end{CD}$<|endoftext|> -TITLE: Relationships between homology maps of cobordant manifolds -QUESTION [16 upvotes]: Let $W$ be a cobordism between $n$-manifolds $M$ and $N$, and $f\colon W \mapsto X$ be a map to some manifold $X$. -Does anybody know of any nice examples of general relationships between the images of the maps $g_*\colon H_*(M)\mapsto H_*(X)$ and $h_*\colon H_*(N)\mapsto H_*(X)$ induced by the restrictions $g=f\mathrel{|M}$ and $h=f\mathrel{|N}$, based on some given data about about these manifolds? -For Example, if $M=N$ and $W=[0,1]\times M$, then $f$ is a homotopy between $g$ and $h$, so $g_*=h_*$. Also, if our manifolds are oriented and compact, $X$ is a connected $n$-manifold, and $f$ is smooth, then the degree of $f\mathrel{\mathrel{|}\partial W}=f\mathrel{\mathrel{|}M\coprod N}$ is zero, so the degrees of $g$ and $h$ are equal up to sign, as are the images of $g_*$ and $h_*$ in degree $n$ homology. - -REPLY [10 votes]: If we consider cohomology with $\mathbb{Z}/2$-coefficients, if $M$ is a manifold $H^*(M;\mathbb{Z}/2)$ it satisfies Poincaré duality, and is an unstable algebra over the Steenrod algebra. Brown and Peterson in their paper "Algebraic bordism groups" (Annals of maths, 1964) had the wonderful idea to give an algebraic model of geometric bordisms in terms of algebraic bordisms of unstable algebras over the Steenrod algebra. -They showed that this algebraic bordism group $N^{alg}_*$ is isomorphic to Thom's bordism group of smooth manifolds. For any unstable algebra algebra $A$ they define an algebraic bordism group $N^{alg}_*(A)$. In fact, they proved that for any space $K$, Conner, Floyd's bordism group $N_*(K)$ of non-oriented smooth manifold over $K$ is isomorphic to $N^{alg}_*(H^*(K;\mathbb{Z}/2))$. The isomorphism is given by a natural transformation: -$$\phi_K:N_*(K)\rightarrow N^{alg}_*(H^*(K;\mathbb{Z}/2))$$ -that sends a bordism class $f:M\rightarrow K$ to the algebraic bordism class of the induced map $f^*:H^*(K;\mathbb{Z}/2)\rightarrow H^*(M;\mathbb{Z}/2)$. -Not only thanks to Wu formulas (that use Poincaré duality and Steenrod squares action on $H^*(M;\mathbb{Z}/2)$), you can recover Stiefel-Whitney numbers, and you can determine if two manifolds are cobordant but using the unstable algebra structure of $H^*(K;\mathbb{Z}/2)$ you can recover the geometric bordism group of $K$.<|endoftext|> -TITLE: Eigenvalues of Laplace operator -QUESTION [23 upvotes]: Assume that $(M,g)$ is a Riemannian manifold. -Is there any relation between the sequence of eigenvalues of Laplace operator acting on the space of smooth functions and the sequence of eigenvalues of Laplace operator acting on the space of smooth $k$-forms for $k\neq 0$? - -REPLY [7 votes]: It sounds to me like you have heard, and are trying to remember the statement of, the following fact about the Hodge Laplacian $d^*d+dd^*$ on forms. -Slogan: eigenforms come in pairs. -Sketch proof: If $(d^*d+dd^*)\eta=-\lambda\eta$, then $(d^*d+dd^*)(d\eta)=dd^*d\eta=d(d^*d+dd^*)\eta=-\lambda d\eta$. -Statement: Recall the Hodge decomposition $\Gamma(M,\Lambda^kM)=A_k\oplus B_k\oplus H_k=\operatorname{im}(d^*_{k+1})\oplus\operatorname{im}(d_{k-1})\oplus (\operatorname{ker}(d_k^*)\cap\operatorname{ker}(d_k))$, such that for all $k$, $d_k:A_k\to B_{k+1}$ is an isomorphism. Then for each $\lambda$, $d_k$ sends the $\lambda$-eigenspace of $A_k$ to the $\lambda$-eigenspace of $B_{k+1}$. -Special case: Each nonzero eigenvalue of the Laplace operator on functions is also an eigenvalue of the Laplace operator on 1-forms. -Also, if you take any sensible choice whatsoever of Laplacian on $\Gamma(M,\Lambda^kM)$, then the asymptotic distribution (à la Weyl) of the eigenvalues will be the same as the asymptotic distribution of the eigenvalues of the Laplacian on scalars (up to multiplication by the rank of $\Lambda^kM$). See, eg, Theorem 2.41 of Berline-Getzler-Vergne.<|endoftext|> -TITLE: NP-hardness of finding 0-1 vector to maximize rows of {-1, +1} matrix -QUESTION [16 upvotes]: Consider the following discrete optimization problem: given a collection of $m$-dimensional vectors $\{ v_1, \dots, v_n \}$ with entries in $\{-1, +1\}$, find an $m$-dimensional vector $x$ with entries in $\{0,1\}$ that maximizes the number of vectors $v_i$ having positive dot product with $x$. -For example, for the collection of vectors given by the rows of the matrix -$$ \begin{bmatrix} -1 & -1 & -1 & -1 \\ +1 & +1 & -1 & -1 \\ +1 & -1 & +1 & -1 \\ -1 & +1 & +1 & -1 \\ -1 & -1 & -1 & -1 \end{bmatrix} $$ -the optimal choice of $x$ is $[1,\ 1,\ 1,\ 0]^T$, which has positive dot products with the middle three rows. -Is this problem known to be NP-hard? If so, are any polynomial-time approximation algorithms available? -EDIT: Cross-posted to cstheory stackexchange here: https://cstheory.stackexchange.com/questions/39735/np-hardness-of-finding-0-1-vector-to-maximize-rows-of-1-1-matrix - -REPLY [8 votes]: Here is a simple embedding of 3-SAT into the current setup (the question is just if we can get all vectors good). -Call the first column special with $1$'s. -Split the other variables into pairs $(a,b)\in\{(0,0),(1,1),(1,0),(0,1)\}$. -Our first task will be to eliminate any $(1,0)$ or $(0,1)$ options. For that, -use the Hadamard matrix without the identically $1$ column, interpreting $1$ as $(-1,1)$ and $-1$ as $(1,-1)$ and put $1$ in the special column in these rows. Then the sum of dot products without special column is $0$ and if we have a single pair of bad type, some dot product is not $0$, so we have $\le -1$ somewhere forfeiting our chance. Also we forfeit it if we use $0$ for the special variable. Using $1$ for the special variable and having only $(0,0)$ and $(1,1)$ in the pairs is still OK. -So, put $-1$ in the remaining rows in the special column. -Now we have 3 options for other "pair entries" in the matrix: $(-1,-1),(1,-1),(1,1)$, which effectively work as $-2,0,2$ against $(0,0)$ and $(1,1)$ interpreted as $0$ and $1$ respectively, so we'll switch to this new representation. -Use the first 3 (new) variables as controls. We will use only $0$ and $2$ for them in the matrix, so, obviously, the controls should be all set to $1$. -Also, since everything is even now, we can forget about the cutoff at $+1$ -(forced by the special column) and come back to the $0$ cutoff with the matrix entries $-1,0,1$. -Now if we have Boolean $a,b,c$ -and a 3-disjunction with them, we will create the corresponding row where we put $0$ everywhere except the corresponding variables and controls. The remaining 6 entries are as follows: -$a\vee b\vee c$ - $0,0,0$ controls, $1$ at $a,b,c$; -$\bar a\vee b\vee c$ - $1,0,0$ controls, $-1$ at $a$, $1$ at $b,c$; -$\bar a\vee \bar b\vee c$ - $1,1,0$ controls, $-1$ at $a,b$, $1$ at $c$. -$\bar a\vee \bar b\vee \bar c$ - $1,1,1$ controls, $-1$ at $a,b,c$. -So the exact solution is, indeed, NP. As to approximations up to a constant factor, I don't know yet.<|endoftext|> -TITLE: Proof there is no algorithm to compute the intersection of a line and sinusoidal wave? -QUESTION [12 upvotes]: There is obviously a set of situations where one lack an algorithm to compute the exact solution of an equation via symbolic manipulation only, for example x = sin(x). -One has to resort to numerical analysis and iteration methods (for example) to estimate a value resolving the above. -My question is: is there an accepted mathematical proof or set of evidence demonstrating that it is impossible to resolve this equation via symbolic manipulation or is there a possibility that one comes with a solution through some clever trick in the future? -If such a proof exist, which mathematical concepts do I need to Google for further investigation? - -REPLY [15 votes]: Your question is addressed in my paper What is a closed-form number? -The first step is to decide which "symbols" or functions you accept as furnishing a "symbolic solution." In my paper I focus on perhaps the most restrictive set, namely exp and log and the arithmetic operations. -The second issue you have to address is whether you're asking for a symbolic expression for a function or for a symbolic expression for a number (namely, the root of some equation that you're trying to solve). This distinction seems not to have been emphasized in the literature prior to my paper. There is quite a bit of literature about closed-form functions, but knowing, for example, that the smallest positive root of the equation $kx = \sin x$, thought of as a function of $k$, has no closed-form expression, doesn't answer the question of whether there is some ad hoc expression for specific values of $k$. -In my paper I show that Schanuel's conjecture implies that there is no closed-form expression for the roots of equations such as $x + e^x = 0$. I think that this is pretty much the best answer we can currently give to this type of question.<|endoftext|> -TITLE: Curious Catalan convolutions -QUESTION [10 upvotes]: Question. Do these identities involving even-index Catalan numbers have a known combinatorial interpretation? They look as though they should. I haven’t seen one in the literature. -$$\sum_{a+b=n}C_{2a}C_{2b}=4^nC_n$$ -$$\sum_{a+b=n}C_{2a}{4b\choose 2b}=4^n{2n\choose n}$$ - -Added. Thanks to Richard Stanley’s helpful answer, I now know that the first of these identities is known in the literature as Shapiro’s Catalan convolution, and has a bijective proof due to Hajnal and Nagy (2014). -Added further. I hadn’t realised quite how straightforwardly each of these identities implies the other. For example, if the first holds then: $$\begin{align} -2\sum_{a+b=n}C_{2a}{4b\choose 2b}&=2\sum_{a+b=n}(2b+1)C_{2a}C_{2b}\\ -&=\Bigl(\sum_{a+b=n}(2a+1)C_{2a}C_{2b}\Bigr)+\Bigl(\sum_{a+b=n}(2b+1)C_{2a}C_{2b}\Bigr)\\ -&=(2n+2)\sum_{a+b=n}C_{2a}C_{2b}\\ -&=(2n+2)4^nC_n\\ -&=2\cdot 4^n {2n\choose n} -\end{align}$$ -In light of that, a proof of one is essentially also a proof of the other. - -Notes. -I came across these identities while thinking about an unanswered question of Mike Spivey from 2012. I can define a bijection for the first one – I imagine the second can be done in a similar way – but it isn’t obviously a very nice bijection, and it uses Garsia-Milne involution. Here is a description of it, in the language of species. -If $D$ is the species of Dyck paths with matching up/down pairs labelled (or binary plane trees with internal nodes labelled, etc.) then $D=1+XD^2$, where $1$ is the species of the empty set (sometimes denoted $E_0$) and $X$ is the species of singletons ($E_1$). Split $D=D'+D''$, where $D'$ represents paths of odd semilength and $D''$ even, so we have the mutually recursive isomorphisms $$\begin{align}\tag{1}D''&=1+2XD'D''\\\tag{2}D'&=X(D'^2 + D''^2)\end{align}$$ (using the $=$ sign to denote isomorphism). The aim is to define an isomorphism between $D''^2$ and $1 + 4X^2D''^4$. We shall do this by defining an isomorphism $$D''^2+Y=1 + 4X^2D''^4+Y$$ for the species $Y=(2XD'D'')^2$, and appealing to the Garsia-Milne principle. Here is the isomorphism, from right to left: -$$\begin{align} -1 + 4X^2D''^4+Y&=1 + 4X^2D''^2(D'^2+D''^2)\\ -&=1 + 4XD'D''^2\tag{using 2}\\ -&=1 + 4XD'D''(1+2XD'D'')\tag{using 1}\\ -&=1 + 4XD'D'' + 8(XD'D'')^2\\ -&=D''^2 + 4(XD'D'')^2\tag{using 1}\\ -&=D''^2 + Y -\end{align}$$ - -REPLY [11 votes]: For the first identity, see additional problem A33 in my book Catalan Numbers. References are given to bijective proofs by Andrews and Nagy.<|endoftext|> -TITLE: Simplicial complex construction from given Betti numbers? -QUESTION [5 upvotes]: Is it possible given a set of Betti numbers to construct a (possibly set of) simplicial complex with the given Betti-described topology? I understand there can be an infinity of simplicial complexes with the same Betti numbers but does a "minimal" simplicial complex construction algorithm exist? -Such an algorithm will greatly help reduce the solution space of constructing simplicial complexes of specified topology. - -REPLY [5 votes]: One way to make things "minimal" (given the lack of any further information) is to construct a simplicial complex whose cup products are all trivial, so the (co)homology generators don't interact with each other in exciting ways. That is to say, just build a simplicial wedge of spheres. All you need to do is have the ability to construct hollow $n$-simplices for all $n>1$ that is smaller than the top non-trivial betti dimension. -More explicitly: if the betti numbers are $(b_0, b_1, \cdots, b_k)$, introduce $b_0$ vertices. Note that $b_0 \geq 1$ if your original simplicial complex is non-empty, so you have at least one vertex, let's call it $a$. Now for each non-zero $b_i$ for $i > 1$, throw in $b_i$ hollow $(i+1)$-simplices which have $a$ as one of their vertices and no other vertices in common.<|endoftext|> -TITLE: Galois invariant Picard group elements -QUESTION [9 upvotes]: Let $X$ be a smooth variety over a perfect field $k$ with $X(k) \neq \emptyset$. Then is the natural map - \begin{equation} - \mathrm{Pic}(X) \to (\mathrm{Pic}(X_{\bar{k}}))^{\mathrm{Gal}(\bar{k}/k)} \qquad (1) -\end{equation} - surjective? - -Remarks: - -If $X$ is a proper, then the map (1) is in fact an isomorphism. This is usually proved using the Hochshild--Serre spectral sequence. -The map (1) need not be injective in general. Take $X$ to be the complement of a closed point of degree two in $\mathbb{P}^1_k$. Then $\mathrm{Pic}(X) = \mathbb{Z}/2\mathbb{Z}$ but $\mathrm{Pic}(X_{\bar{k}}) = 0$. - -REPLY [6 votes]: Updated. The example by @Lucifer is completely correct. Thanks to @Count Dracula who explained the example proposed by @Lucifer. -That example is fine. I am keeping the counterexamples below, since they arise in a different way: as Severi-Brauer schemes over multiplicative group schemes over a field (and schemes mapping to such Severi-Brauer schemes). The counterexamples are given after a general statement that can (sometimes) be used to prove surjectivity of the morphism (1). -Setup. Denote by $S$ the scheme $\text{Spec}\ k$ (to save typing). As in Borovoi-- Colliot-Thélène -- Skorobogatov, denote by $\mathfrak{g}$ the Galois group $\text{Gal}(\overline{k}/k)$. Let $\pi:X\to S$ be a smooth, separated, quasi-compact morphism with geometrically irreducible fiber. Let $\sigma:S\to X$ be a section of $\pi$. Denote by $G_X$, resp. $G_{X,\sigma}$, the group $S$-scheme that represents the functor sending every smooth $S$-scheme $T$ to $\mathbb{G}_{m,X}(X\times_S T)$, resp. to the kernel of $$\sigma^*(T):\mathbb{G}_{m,X}(X\times_S T) \to \mathbb{G}_{m,S}(T).$$ Pullback by $\sigma$ defines a group homomorphism, $$\sigma^*:\text{Br}(X)\to \text{Br}(S).$$ Denote the kernel by $\text{Br}(X)_\sigma$. Pullback by $\pi$ defines a group homomorphism, $$\pi^*:H^2_{\mathfrak{g}}(G_{X,\sigma}(\overline{k})) \to \text{Br}(X)_\sigma.$$ -Lemma. The cokernel of the natural restriction morphism (1) equals the kernel of $H^2_{\mathfrak{g}}(G_{X,\sigma}(\overline{k})) \to \text{Br}(X)_\sigma$. The group $G_{X,\sigma}(\overline{k})$ is a finitely generated, torsion-free Abelian group. In particular, the cokernel of the natural restriction morphism is a torsion group. -Proof. The relevant terms in the Hochschild-Serre spectral sequence give the following long exact sequence, $$0 \to H^1_{\mathfrak{g}}(G_X(\overline{k})) \to \text{Pic}(X)\xrightarrow{\text{res}} \text{Pic}(X_{\overline{k}})^{\mathfrak{g}} \xrightarrow{\delta} H^2_{\mathfrak{g}}(G_X(\overline{k})) \to \text{Br}(X).$$ By Hilbert's Theorem 90, the first term is $H^1_{\mathfrak{g}}(G_{X,\sigma}(\overline{k}))$. Pullback by the section $\sigma$ defines a splitting of the map $$H^2_{\mathfrak{g}}(\mathbb{G}_{m,S}(\overline{k}))\to \text{Br}(X).$$ Thus, the cokernel of $\text{res}$ equals the kernel of $\pi^*$. -The group $G_{X,\sigma}(\overline{k})$ is finitely generated by Rosenlicht. Every torsion-element satisfies an integral equation over $\overline{k}$. Since $X_{\overline{k}}$ is integral, $\overline{k}$ is integrally closed in $H^0(X_{\overline{k}},\mathcal{O}_X)$. Thus, $G_{X,\sigma}(\overline{k})$ is a finitely generated, torsion-free Abelian group. The profinite Galois cohomology group $H^2_{\mathfrak{g}}(G_\sigma(\overline{k}))$ is torsion. Thus, $H^2_{\mathfrak{g}}(G_\sigma(\overline{k}))$ is a torsion group. QED -The contravariant functor $G$ that sends a pointed $S$-scheme, $(X,\sigma)$, to the group scheme $G_{X,\sigma}$ has an adjoint. For every smooth group $S$-scheme $M$ with $M\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ isomorphic to $\mathbb{G}_{m,\overline{k}}^\rho$, the group scheme $G_{M,e}$ is the Cartier dual group scheme $M^D$. For every morphism of group $S$-schemes, $$\phi:M^D\to G_{X,\sigma},$$ there is a unique $S$-morphism, $$f:X\to M,$$ sending $\sigma$ to the identity section $e$ and such that the pullback map on unit group $S$-schemes is $\phi$. -Proposition. The Galois cohomology group $H^2_{\mathfrak{g}}(M^D)$ is canonically isomorphic to the kernel, $\text{Br}_1(M)$, of the pullback map $$\text{pr}_M^*:\text{Br}(M)\to \text{Br}(M\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}).$$ -Proof. Since $\text{Pic}(\mathbb{G}_{m,\overline{k}}^\rho)$ is the trivial group, this follows from the previous lemma. QED -The Cartier dual group $S$-scheme to $G_\sigma$ is a smooth group $S$-scheme $M$ of multiplicative type. There is a unique $S$-morphism $$f:X\to M,$$ mapping $\sigma$ to the identity section $e$ and such that the induced homomorphism of groups of units is an isomorphism of $S$-group schemes. -Proposition. The cokernel of $\text{res}$ is identified with the kernel of the group homomorphism $$f^*:\text{Br}_1(M) \to \text{Br}(X).$$ -In particular, if there exists a degree $1$ zero-cycle in the generic fiber of $f$, then $\text{res}$ is surjective. -Proof. This follows from the previous proposition and functoriality of the Hochschild-Serre spectral sequence with respect to $f$. If the fiber of $f$ over $\eta=\text{Spec}\ k(M)$ has a degree $1$ zero-cycle, then the pullback map $\text{Br}(k(M))\to \text{Br}(X_\eta)$ is injective by the lemma. Since $M$ is smooth, it follows that also $f^*$ is injective. QED -Counterexamples. This suggests how to construct a counterexample, namely as a Severi-Brauer scheme over a multiplicative group scheme $M$ over $k$. -Let $k$ be a finite field $\mathbb{F}_q$. Let $M$ be $\mathbb{G}_{m,k}= \text{Spec}\ k[t,t^{-1}]$. The Cartier dual group scheme $M^D$ has $M^D(k) = M^D(\overline{k}) \cong \mathbb{Z}$, generated by the element $t$. -For every integer $\ell$, let $k_\ell=\mathbb{F}_{q^\ell}$ be a degree $\ell$ extension of $k$ with cyclic Galois group generated by a Frobenius map. This extension together with the generator of the Galois group is unique up to unique isomorphism. Associated to the cyclic extension $k_\ell/k$ and the element $t\in M^D(k)$, there is a cyclic algebra $A_{\ell,t}$ that is a locally free $\mathcal{O}_M$-module of rank $\ell^2$ and that is an Azumaya algebra over $\mathcal{O}_M$. The Brauer class of $A_{\ell,t}$ in $\text{Br}(M)$ is an element of order $\ell$ generating the full $\ell$-torsion subgroup of $\text{Br}(M)$. By Tsen's Theorem, the Brauer group of $M\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ is trivial. Thus, the class of $A_{\ell,t}$ generates a cyclic subgroup of order $\ell$ in $\text{Br}(M)_k$. -Now let $f:X\to M$ be the Severi-Brauer scheme parameterizing left ideals in $A_{\ell,t}$ of rank $\ell$. Since the Brauer group of $k$ is trivial (by Wedderburn or Chevalley-Warning), the restriction of $f$ over the $k$-rational point $e\in M(k)$ is a trivial Severi-Brauer scheme. Thus, there exists a section $$\sigma:\text{Spec}\ k \to X,$$ with image in the fiber of $f$ over $e$. The Picard group of $X\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ is $\mathbb{Z}$, and the generator restricts on each projective space (geometric) fiber of $f$ to a generator $\mathcal{O}(1)$ of the projective space. However, there cannot be such an invertible sheaf on $X$, or else $\mathcal{A}_{\ell,t}$ would have zero Brauer class.<|endoftext|> -TITLE: Cohomology of sheaf of Schwartz distributions with support in a submanifold -QUESTION [5 upvotes]: Let $M$ be a smooth manifold. Let $Z\subset M$ be a smooth submanifold which is a closed subset. Let $F$ denote the sheaf of generalized functions (equivalently, Schwartz distributions) on $M$, namely for any open subset $U\subset M$, $F(U)$ is the space of generalized functions on $U$. - -Is it true that $H^i_Z(M,F)=0$ for any $i>0$? - -Remark. I think that if one considers the sheaf of smooth functions instead of generalized, then the anologous statement is not true in general even when $M =\mathbb{R}$ and $Z$ is a point. - -REPLY [3 votes]: The long exact sequence cotains $H^0(M,F)\to H^0(M\smallsetminus Z,F|_{M\smallsetminus Z})\to H_Z^1(M,F)\to H^1(X,F)$. -The last term is zero, as $F$ is a fine sheaf. But the first arrow is not surjective in general, so the first cohomology group is not trivial. The same argument shows that for $i>1$ the cohomologies are indeed trivial.<|endoftext|> -TITLE: Left adjoint of $I\colon \mathrm{Kan}\hookrightarrow\mathrm{WeakKan}$? -QUESTION [5 upvotes]: The inclusion $I\colon \mathbf{Grpd}\hookrightarrow\mathbf{Cat}$ of groupoids into categories has both a left and a right adjoint $L,R\colon \mathbf{Cat}\to \mathbf{Grpd}$, with $R(C)$ being largest groupoid contained in $C$ and $L(C) = C[C^{-1}]$ being $C$ with all morphisms brutally inverted. Going into $\infty$-category theory, we may replace $\mathbf{Cat}$ by weak Kan complexes (AKA quasi-categories) and $\mathbf{Grpd}$ by Kan complexes ($\infty$-groupoids). The inclusion $I\colon \mathbf{Kan}\hookrightarrow\mathbf{WKan}$ still has a right adjoint, given by taking largest contained Kan complex, as documented in Corollary 1.5 in Joyal, A., Quasi-categories and Kan complexes, J. Pure Appl. Algebra 175, No.1-3, 207-222 (2002). ZBL1015.18008. -However, the question now is: Does it also have a left adjoint (possibly in some higher sense)? This would have to be something like taking a weak Kan complex and brutally inverting all $1$-morphisms. Does this make sense in some way, possibly non-canonically? -If no, does there at least exist some $\infty$-groupoid analogue (“$\Delta^n_{\text{$\infty$-grpd}}$”) of the standard simplex $\Delta^n\in\mathbf{SSet}$, behaving a bit like $L(\Delta^n)$, something like free Kan complex on $n+1$ objects? And in this case, why does this construction not solve my first problem by putting $$L(K) = \varinjlim_{\Delta^n\to K} \Delta^n_{\text{$\infty$-grpd}}? $$ - -REPLY [2 votes]: As Valery Isaev has pointed out, $\mathbf{Kan} \hookrightarrow \mathbf{WKan}$ does not have a left adjoint. However, instead of considering $\mathbf{Kan} \hookrightarrow \mathbf{WKan}$, you can consider the forgetful functor $\mathbf{sKan} \hookrightarrow \mathbf{sWKan}$ of (weak) Kan complexes with chosen Kan fillers and morphisms preserving these choices. (Don't google this, I've just made up the notation.) The two categories are locally presentable (they can be axiomatized as essentially algebraic theories), and every forgetful functor between such categories is a right adjoint. Because the corresponding left adjoint is given by repeated pushouts along the inclusion $\Lambda^k_n \rightarrow \Delta^n$, it follows that the unit of the adjunction is a fibrant replacement wrt. the model structure on simplicial sets that has the Kan complexes as fibrant objects.<|endoftext|> -TITLE: Pull-back of knots in branched covers and the Alexander polynomial -QUESTION [5 upvotes]: Given a knot $K \subset S^3$ one can form its double branched cover $\Sigma_2(K)$ and consider the pull-back knot $\widetilde{K} \subset \Sigma_2(K)$ of $K$ to $\Sigma_2(K)$ (the locus fixed by the involution of $\Sigma_2(K)$). Here a few questions I would like to ask - -How do I compute the Alexander polynomial of $\widetilde{K}$? -Can I compute $\Delta_{\widetilde{K}}(t)$ from $\Delta_K(t)$ plus some torsion invariants of $\Sigma_2(K)$? -Is there a formula for $\Delta_{\widetilde{K}}(t)$ in the case when $\widetilde{K}$ comes from a torus knot $K \subset S^3$? - -REPLY [3 votes]: Corollary 4.2 from the paper "Metabelian representations, twisted Alexander polynomials, knot slicing, and mutation" by Herald, Kirk and Livingston gives the followwing recipe to compute Alexander polynomial of the lift of $K$ to the $n$-fold cover $\Sigma_n(K)$: -$$\Delta_{\widetilde{K}_n}(t) = \prod_{i=0}^{n-1} \Delta_{K}(\xi_n^i t^{1/n}),$$ -where $\xi_n = \exp(\frac{2 \pi i}{n})$.<|endoftext|> -TITLE: Sullivan conjecture for compact Lie groups -QUESTION [9 upvotes]: Let $G$ be a topological group, and $M$ a connected compact smooth manifold. I'm studying -$$ \pi_0 (map (BG,M)). $$ -For $G$ a finite group, we know that this is just a point by the Sullivan conjecture on maps from classifying spaces which was proven by Miller. (This does not require smoothness of $M$.) -On the other hand, if $G$ is an infinite discrete groups, this $\pi_0$ can be larger (take $G=\mathbb{Z}$ and $M=S^1$). -Question What happens if $G$ is a compact Lie group? Are there examples where this $\pi_0$ is more than a point? - -REPLY [20 votes]: You were right to single out Lie groups as potentially interesting. In [Topology 5 (1966), 241-243], Brayton Gray showed that the homotopy group of maps $[BS^1, S^3]$ was uncountable. Indeed, he showed that the subgroup of phantom maps -- maps null on every finite subcomplex - was uncountable. Then Alex Zabrodsky, in [Isreal J. Math. 58 (1987)], has a theorem that refines this: all maps in this case are phantom, and the group is isomorphic to $\hat Z/Z$. -More generally, there was a decade of work, after Miller's theorem, exploiting the Sullivan conjecture, and much of it was focused on understanding maps out of $BG$ for $G$ compact Lie. In particular, there are a number of papers identifying mapping spaces of the form $Map(BG,BH)$, for well chosen pairs of compact Lie groups. Now note that $\Omega Map(BG,BH) \simeq Map(BG,H)$ which is of the form you were asking about. Look up papers of Dwyer, Wilkerson, McClure, Oliver, and Lannes to get going in the literature.<|endoftext|> -TITLE: Relations for the algebra of differential operators on a smooth affine variety -QUESTION [9 upvotes]: Over a ground field of characteristic zero, the algebra of differential operators $\mathcal{Diff}(X)$ on a smooth affine variety $X$ is generated by the functions $O_X$ and the derivations $Der(O_X)$, both viewed as subspaces of the endomorphism algebra $End(O_X)$ of the vector space $O_X$. This is a theorem of Grothendieck that one can find proved in detail in a few places, likethe book Noncommutative Noetherian Rings J. C. McConnell and J. C. Robson or, if I recall correctly, J. E. Bjork's Rings of differential operators (but maybe Bjork does only the analytic case? I don't have the book at hand) -Now there are some obvious relations one knows that hold in $\mathcal{Diff}(X)$ in that case: functions commute, the product of a function and a derivation is given by the module structure of the derivations, the commutator of two derivations is, well, their commutator, which is also a derivation, and the commutator of a derivation and a function is the value of the former on the latter. - -Are these relations enough to give a presentation of the algebra? - -One could phrase this in the language of Lie-Rinehart pairs: is $\mathcal{Diff}(X)$ the universal enveloping algebra of the Lie—Rinehart pair $(O_X,Der(O_X))$? -The answer to this question is yes, and it is part of what one needs to prove in order to show that the associated graded algebra of $\mathcal{Diff}(X)$ with respect to the filtration on differential order is the coordinate ring of the cotangent bundle on $X$, for example. - -Has a detailed proof of this been written down? - -REPLY [4 votes]: A paper by Vladimir Bavula addresses this question. Indeed the relations in question are sufficient to determine $Diff(X)$. Moreover, Bavula gives an explicit finite set of generators of $Diff(X)$ (functions and vector fields) and presents defining relations between the generators.<|endoftext|> -TITLE: (Variation of an old question) Are these functions polynomials? -QUESTION [5 upvotes]: This is a followup to the question here: How to show that the following function isn't a polynomial over Q?. -As before, let $b_1,b_2,\dots$ be an enumeration of $\mathbb Q$. The question might be sensitive to the enumeration (but probably not). -Question 1: Suppose I define $$f_3(x) = (x-b_1)^3 + (x-b_1)^3(x-b_2)^3 + \dots,$$ -how do I show that this function is not a polynomial? I do not believe the answers to the old question extend to this case since they all seem to use the positivity of squares. -Can we show that $f(x)$ is not a polynomial? -There is a even more general version (and the one I was interested in from the beginning) that I was interested in, however I forgot to omit the trivial cases/there was some ambiguity in the second question, so I asked it again here: -Is the following function a polynomial? - -REPLY [4 votes]: $f_3(x)$ is not a polynomial function, which can be seen as follows. Call $b_n$ a champion if it exceeds all the earlier terms $b_1,\dots,b_{n-1}$. Clearly, there are infinitely many champions, and they tend to infinity. Let $m$ be fixed, and consider the champions $b_n$ with $n>m$. We get, as $b_n$ tends to infinity, -$$ f_3(b_n)>\prod_{k=1}^m (b_n-b_k)^3 = (1+o(1))\ b_n^{3m}. $$ -This shows that $f_3$ has degree at least $3m$. As $m$ is arbitrary, the claim follows.<|endoftext|> -TITLE: Simultaneous "orthonormalization" in $\mathbb{C}^4$ -QUESTION [20 upvotes]: Let $A$ be a positive, invertible $4 \times 4$ hermitian complex matrix. -So we have a positive sesquilinear form $\langle Av,w\rangle$. Say that a pair $(v,w)$ of vectors in $\mathbb{C}^4$ is good for $A$ if they are orthogonal and have the same norm relative to the hermitian form given by $A$, i.e., $\langle Av,w\rangle = 0$ and $\langle Av,v\rangle = \langle Aw,w\rangle$. - -Let $A$, $B$, and $C$ be positive, invertible $4 \times 4$ complex matrices. - Can we always find a pair of nonzero vectors which is simultaneously good for $A$, $B$, and $C$? - - -It was suggested that the answer should be yes for generic $A$, $B$, and $C$. Note that this implies it is true for all $A$, $B$, and $C$, as the set of triples for which it holds is closed. (If $(A_n, B_n, C_n) \to (A,B,C)$ then let $(v_n,w_n)$ be a good pair for $A_n$, $B_n$, and $C_n$ with $\langle A_nv_n,\rangle = \langle A_nw_n,w_n\rangle = 1$ and let $(v,w)$ be a cluster point of $(v_n,w_n)$.) - -REPLY [5 votes]: I can now prove the existence of a good pair if, after rescaling so $A = I_4$, some nonzero Hermitian matrix in the span of $B$ and $C$ has a repeated eigenvalue. (But as I learned from Robert Bryant here, that generally will not be the case.) -But in this special case, since $(v,w)$ is good for $A$, $B$, and $C$ iff it is good for every Hermitian matrix in their span, wlog we can assume $B$ has an (at least) double eigenvalue. Subtracting a scalar multiple of $A = I_4$, we can assume this double eigenvalue is $0$. -If $0$ is a triple eigenvalue, then it is easy to find a pair of vectors in this three-dimensional eigenspace which is good for $I_4$ and $C$, and that solves the problem. The solution is also easy if the two nonzero eigenvalues of $B$ have opposite sign: say $B = {\rm diag}(a,-b,0,0)$ with $a,b > 0$, wlog with $\frac{1}{a^2} + \frac{1}{b^2} = 1$. Then let $W = \left[\matrix{\frac{1}{a}&0&0\cr\frac{1}{b}&0&0\cr 0&1&0\cr 0&0&1}\right]$, so -that $W^*BW = 0$. Then find a pair of vectors $v_0,w_0 \in \mathbb{C}^3$ which is good for $I_3$ and $W^*CW$ (easy) and set $v = Wv_0$, $w = Ww_0$. -The hard case is the one where both nonzero eigenvalues have the same sign. The case where they are equal is the one I treated in an earlier answer, which I'm retaining below. If they are not equal, wlog say $B = {\rm diag}(1,a,0,0)$ with $a > 1$. Similarly to the case where $a = 1$ presented below, it will suffice to find $0 \leq \lambda \leq 1$ and a $2\times 2$ unitary $U$ such that $$sC_1s + cUC_2^*s + sC_2U^*c + cUC_3U^*c$$ is a scalar multiple of $I_2$, where $C = \left[\matrix{C_1&C_2\cr C_2^*&C_3}\right]$ and $s = {\rm diag}(\sqrt{\lambda},\sqrt{\lambda/a})$, $c = {\rm diag}(\sqrt{1-\lambda},\sqrt{1 - \lambda/a})$. This is done as in Robert Bryant's solution when $a = 1$ where again, when $\lambda = 0$ the expression is just $UC_3U^*$, which becomes the Hopf map when you pass to the $S^3$-$S^2$ picture (which we can do if there is no solution to the problem). But it's a little harder here because the $\lambda = 1$ extreme no longer reduces to a constant map. However, not so much harder because some computation shows that the image of $S^3$ under the $\lambda=1$ map misses a point on $S^2$, and is therefore null homotopic, leading to the same contradiction. - -Previous answer: -It follows from Robert Bryant's brilliant answer to this question that the answer is yes in an important special case, when $A = I_4$ (as pointed out in the comments, there is no essential loss of generality in assuming this) and $B$ is a rank $2$ projection. -Namely, work in an orthonormal basis that diagonalizes $B$ and write $C = \left[\matrix{C_1&C_2\cr C_2^*&C_3}\right]$ where $C_1$, $C_2$, and $C_3$ are $2\times 2$ matrices and $C_1$ and $C_3$ are Hermitian. We seek a rank $2$ projection $P$ with the property that $PBP$ and $PCP$ are both scalar multiples of $P$; if so, then any orthonormal vectors $v$ and $w$ in the range of $P$ will be good for $A$, $B$, and $C$. -If $C_3$ is a scalar multiple of $I_2$ then $P = I_4 - B$ is the desired projection. Otherwise, according to the answer cited above we can find $a \geq 0$ and a $2\times 2$ unitary $U$ such that $$C_1 + a(C_2U + U^*C_2^*) + a^2U^*C_3U$$ is a scalar multiple of $I_2$. A short computation then shows that $P = \frac{1}{1 + a^2}\left[\matrix{I_2& aU^*\cr aU&a^2I_2}\right]$ has the desired properties.<|endoftext|> -TITLE: Computing an explicit homotopy inverse for $B(*,H,*) \hookrightarrow B(*,G,G/H)$ -QUESTION [6 upvotes]: Suppose that $G$ is a finite group, $M$ is a right $G$-set and $N$ is a left $G$-set. Then we have a simplicial set $B(M,G,N)$ whose $n$-simplicies are $M \times G^n \times N$. -Now suppose that $H \subseteq G$ is a subgroup. There is a natural inclusion $B(*,H,*) \hookrightarrow B(*,G,G/H)$. Both of these spaces have fundamental group $H$ and no higher homotopy groups, and the inclusion induces an isomorphism on fundamental groups, so it is a weak equivalence. - - - Question: Is there a way to write down an explicit inverse weak equivalence $B(*,G,G/H) \to B(*,H,*)$? In practice, I would be happy with just the $B_{\leq 2}(*,G,G/H) \to B_{\leq 2}(*,H,*)$ part of such a map. - - -My instincts tell me that the answer is no: simplicial sets are not sufficiently squishy enough to do this, but I want to be sure before I start trying harder stuff. - -REPLY [14 votes]: Yes, there is an explicit algorithm for doing this. Pick a set of representatives $a_i \in G$ for the left cosets of $G/H$. Then the inverse map is as follows. -Given any element $(g_n,\dots,g_1, g_0H)$ in $G^n \times G/H$, let $a_i$ be the chosen representative of the left coset $g_i g_{i-1} \dots g_0 H$. Then $g_{i+1} a_i H = a_{i+1} H$, and so we get an element in $H^n$ given by -$$ -(a_n^{-1} g_n a_{n-1}, \dots, a_2^{-1} g_2 a_1, a_1^{-1} g_1 a_0). -$$ -One can check (and should!) that this is compatible with the face and degeneracy maps. -But really, what is going on here? -The simplicial set $B(*,H,*)$ is the nerve of a groupoid with one object and $H$ as its set of self-maps. The simplicial set $B(*,G,G/H)$ is the nerve of a translation groupoid with $G/H$ as its set of objects and maps $g: aH \to gaH$ for any $g \in G$. The inclusion of simplicial sets comes from an equivalence of groupoids. -To construct the inverse to this equivalence of categories, we choose for each object (a left coset $aH$) an isomorphism from $eH$ to $aH$: these are our chosen coset representatives. Any map $g: aH \to gaH = bH$ is then replaced, under these isomorphisms, with the self-map $b^{-1} g a: eH \to eH$ (provided that $a$ and $b$ were our chosen coset representatives). -The fact that these are groupoids is at the core of this argument, because it means that the associated simplicial sets are "fibrant": they have enough flexibility to construct most desired maps into them.<|endoftext|> -TITLE: Elements of absolute value 1 in cyclotomic extension of dyadic rationals -QUESTION [7 upvotes]: Let $\zeta$ be a primitive $2^k$th root of unity, and consider the ring $ \mathbb{Z}[\zeta, \frac{1}{2}] \subset \mathbb{C}$. Are there any elements of absolute value 1 other than the powers of $\zeta$? -Note that if $\frac{1}{2}$ were replaced by $\frac{1}{5}$, then the number $\frac{4}{5} + \frac{3}{5} i$ would be an example. -I can prove that the answer is 'no' for $k \leq 3$. In particular, for $k = 3$ we want to find integers $a, b, c, d$ and positive integer $n$ such that: -$$ (a + b \zeta + c \zeta^2 + d \zeta^3)(a + b \zeta^{-1} + c \zeta^{-2} + d \zeta^{-3}) = 4^n $$ -Expanding this yields: -$$ a^2 + b^2 + c^2 + d^2 + (ab + bc + cd - da) \sqrt{2} = 4^n $$ -By Jacobi's four-square theorem, either all but one of $a,b,c,d$ is zero (and we get a cyclotomic unit) or $a^2 = b^2 = c^2 = d^2 = 4^{n-1}$. In this case, $|ab| = |bc| = |cd| = |da|$ and an even number of them are negative. It follows that the coefficient of $\sqrt{2}$ is nonzero, whence we derive a contradiction. -Does this result hold for all $k$? - -REPLY [3 votes]: The answer is yes. Let $K=\mathbb{Q}(\zeta)$ and suppose that there is an element $u\in\mathbb{Z}[\frac{1}{2},\zeta]$ such that for some embedding $\sigma_0\colon K\hookrightarrow \mathbb{C}$ the absolute value of $\sigma_0(u)$ is $1$. Then I claim that $u\in\mathbb{Z}[\frac{1}{2},\zeta]^\times$ and that for all embeddings $\sigma\colon K\hookrightarrow\mathbb{C}$ we have $\Vert\sigma(u)\Vert=1$. Indeed, $K$ is a $CM$ field, so there exists an automorphism $c\colon K\to K$ such that for all embeddings $\sigma$, the composition $\sigma\circ c$ equals $\overline{\sigma}$, or in other words $\sigma\bigl(c(x)\bigr)=\overline{\sigma(x)}$ for all $x\in K$. Thus the equation $\Vert\sigma_0(u)\Vert=1$ can be written $\sigma_0(u)\sigma_0(cu)=1$ and as $\sigma_0$ is injective this forces $cu=u^{-1}$. This proves my two claims at once, using that $c\bigl(\mathbb{Z}[\frac{1}{2},\zeta]\bigr)=\mathbb{Z}[\frac{1}{2},\zeta]$. -Now, set $\pi=(\zeta-1)$ which generates the prime ideal of $\mathbb{Z}[\zeta]$ above $2$ (in the sense that $2\mathbb{Z}[\zeta]$ is the principal ideal generated by $(\pi)^d$ with $d=\dim_{\mathbb{Q}}(K)$). This $\pi$ verifies -$$ -c\pi=\frac{1}{\zeta}-1=\frac{1-\zeta}{\zeta}=\xi\pi -$$ -where $\xi=-\zeta^{-1}$ is again a root of unity. Our element $u$ can be written as $u=\alpha/\pi^n$ for some $\alpha\in\mathbb{Z}[\zeta]$ and $n\in\mathbb{N}$: we can also assume that $n$ is minimal, which amounts to saying that $\alpha\notin(\pi)$. Then -$$ -1=u\cdot cu=\frac{\alpha}{\pi^n}\frac{c\alpha}{\pi^n\xi^n}=\frac{\alpha\cdot c\alpha\xi^{-1}}{\pi^{2n}} -$$ -and this forces $n=0$ since the ideals $(\pi)$ and $(c\pi)$ coincide and hence $\alpha\cdot c\alpha\notin\pi$. But then $u\in\mathbb{Z}[\zeta]$, hence it is an algebraic integer all of whose conjugate lie on the unit circle. It is therefore a roots of unity, by Kronecker's theorem.<|endoftext|> -TITLE: The projective covers of Artinian module -QUESTION [9 upvotes]: The injective hull for a module always exists, however over certain rings modules may not have projective covers. I have a question. - If $A$ is an Artinian module on a Noetherian local ring $R$ then $A$ has projective cover? If not, please give a counter example. - -REPLY [10 votes]: No: as soon as the (local noetherian) ring itself is not artinian, there exists an artinian module with no projective cover. -First recall (over any ring) that a projective cover of $M$ is $N$ projective and an epimorphism $N\to M$ that is non-surjective in restriction to any proper submodule of $N$. -In the current context ($R$ local noetherian), for $k$ the residual field, say that $M$ is aperiodic if $\mathrm{Hom}(M,k)=0$. If $M$ is aperiodic and nonzero then it has no projective cover $L\to M$: indeed, since $N\neq 0$, $M\neq 0$; also every projective $R$-module is free. Hence $L/P\simeq k$ for some submodule $P$; then since $M$ is aperiodic, the image of $P$ in $M$ should be all of $M$, contradicting that $L\to M$ is a projective cover. -Then it suffices to check the existence of a nonzero artinian aperiodic module. Let $I$ be the injective hull of $k$. It is artinian (by an old result of Vámos, see 19.1 in Lam's book "Lectures on modules and rings"). -The Matlis dual of $I$ (see remainder below) is the free module of rank one over the completion $\hat{R}$ of $R$. Since $R$ is non-artinian, so is $\hat{R}$. Hence the latter has a quotient that is a non-artinian domain $D$ (e.g., of Krull dimension 1). Hence $\mathrm{Hom}(k,D)=0$. By Matlis duality, $D$ corresponds to a submodule $J$ of $I$ with $\mathrm{Hom}(J,k)=0$. Since $D$ has infinite length, so does $J$. Hence $J$ is an aperiodic module as requested. - -(Added) Remainder on Matlis duality. This is a very useful tool, which mostly reduces the understanding of artinian modules to standard knowledge about finitely generated modules. -First observe that any artinian $R$-module is canonically a module over the completion $\hat{R}$ of $R$, and $R$-module homomorphisms between artinian $R$-modules are $\hat{R}$-module homomorphisms. The Matlis dual of an artinian $R$-module $M$ is $T(M)=\mathrm{Hom}_{R\mathrm{-mod}}(M,I)$. The first important result in Matlis duality is that $T(M)$ is a finitely generated $\hat{R}$-module. If $N$ is a finitely generated $\hat{R}$-module, write $T'(N)=\mathrm{Hom}_{\hat{R}\mathrm{-mod}}(N,I)$; this is an artinian $\hat{R}$-module and we can view it as an $R$-module. One can view $T'(T(M)$ as the bidual of $M$; Matlis duality says that the canonical $R$-module homomorphism $M\to T'(T(M))$ is an isomorphism for every artinian $R$-module $M$. Thus $T$ is a contravariant equivalence of categories between artinian $R$-modules and finitely generated $\hat{R}$-modules. (We have $T(k)=k$ and it restricts to a contravariant involutive self-equivalence of categories of the category of $R$-modules of finite length.) A standard reference for Matlis duality is the book by Bruns and Herzog. - -REPLY [9 votes]: Take $R=\mathbb{Z}_p$, the $p$-adic integers, and $A=\mathbb{Q}_p/\mathbb{Z}_p$. Then $A$ is an Artinian $R$-module, but doesn't have a projective cover. -[Since $R$ is local, projectives are free. If $\varphi=\pmatrix{\varphi_1\\\varphi_2}:\mathbb{Z}_p\oplus\mathbb{Z}_p\to A$ is a homomorphism then $\varphi_1$ factors through $\varphi_2$ (or vice versa), say $\varphi_1=\varphi_2\theta$. Then $\left\{\left(x,-\theta(x)\right)\vert x\in\mathbb{Z}_p\right\}$ is a direct summand of $\mathbb{Z}_p\oplus\mathbb{Z}_p$ in the kernel of $\varphi$. So the kernel of any map to $A$ from a free module of rank greater than one has a non-zero direct summand in its kernel.]<|endoftext|> -TITLE: Gauss map of general K3 surface -QUESTION [7 upvotes]: Let $S\subset\mathbb{P}^g$ be a polarised smooth projective K3 surface of genus $g$ (and degree $d=2g-2$) over $\mathbb{C}$. Denote by $\phi: S\to G(3,g+1)$ the Gauss map, taking a point $s\in S$ to its tangent 2-plane $\mathbb{T}_{S,s}\cong\mathbb{P}^2$ in $\mathbb{P}^g$. This is known to always be finite and birational, i.e. it is the normalisation onto its image. - -Q: If $g$ is large enough and $S$ is general, is it true that $\phi$ is a closed embedding? - -If $g=3$ then the Gauss map $\phi$ of a general quartic K3 $S\subset\mathbb{P}^3$ is ramified along the Hessian curve in $\mathcal{O}_S(8)$, on which it has degree 2, so in particular $\phi$ is not a closed embedding here. - -REPLY [4 votes]: I think that a sufficient condition for the Gauss map to be a closed embedding is that each tangent plane $\mathbb{T}_{S,s}$ intersects the surface $S$ only at the point $s$ and with the expected multiplicity $3$. -In turn, a sufficient condition to guarantee this is that the polarization is $3$-very ample: a paper of Knutsen "On kth-order embeddings of K3 surfaces and Enriques surfaces" characterizes completely the polarizations with this property, and in particular one gets that this happens for a general K3 as soon as $g\geq 7$.<|endoftext|> -TITLE: Lavrentiev phenomenon between $C^1$ and Lipschitz -QUESTION [8 upvotes]: Does there exist a (onedimensional) integral functional of calculus of variations (with $f$ finite everywhere) -$$ -F(y)=\int_a^b f(t,y(t),y'(t))\,dt
 -$$ -such that -$$ -\inf_{y\in Lip([a,b])}F(y)<\inf_{y\in C^1([a,b])}F(y) -$$ -that is, it shows the Lavrentiev phenomenon between $C^1$ and Lipschitz. - -REPLY [7 votes]: Then trivially yes. Just take $F(y)=1+\sum_{q\in \mathbb Q}\frac{a_q}{|q-y|^2}$ with $a_q>0$ such that the series converges a.e. Then change all $+\infty$ values of $F$ to $1$. Now take $[a,b]=[-1,1]$ and define -$$ -f(t,y,\xi) = -\begin{cases} -0 &\text{if }y=|t|\\ -F(y) &\text{otherwise} -\end {cases}. -$$ -A Lipschitz function can just stay on the safe path $y=|t|$ and pay $0$ price for the trip, but a $C^1$ function will have to deviate from this path and either have $y\ne |t|$, $y'\ne 0$ at at least one point, which is enough to blow the integral up to $+\infty$, or stay constant, which gives the cost of at least $2$.<|endoftext|> -TITLE: General existence theorem for cup products -QUESTION [5 upvotes]: I'm curious if it is possible to formulate cup products and prove that they exist in a general way which would subsume a lot of examples: e.g. group cohomology, sheaf cohomology for sheaves on topological spaces, quasicoherent sheaf cohomology, things like etale cohomology where we look at sheaves on more general sites, etc. -In the sources I've looked at (e.g. Cassels-Fröhlich, Lang's Topics in Cohomology of Groups), the existence of cup products is proven in terms of cochains, Čech cohomology, etc., even when more abstract definitions and uniqueness theorems are given. -What I'm looking for is something like this: Let $\mathscr{A}$ is an abelian category with a symmetric monoidal structure $\otimes$ such that $(\mathscr{A}, \otimes)$ satisfies certain conditions (e.g. enough injectives, existence of $\mathrm{Hom}$-objects, whatever other features are common in practice) and $\{H^i\}$ is a universal $\delta$ functor from $\mathscr{A}$ to another abelian symmetric monoidal category $\mathscr{B}$ (assuming whatever we want for $\mathscr{B}$, even $\mathscr{B} = \mathbf{Ab}$, the category of abelian groups). If $\phi \colon H^0(M) \otimes H^0(N) \rightarrow H^0(M \otimes N)$ is an additive bi-functor, then there is a unique sequence of additive bi-functors $\Phi^{p,q} \colon H^p(M) \otimes H^q(N) \rightarrow H^{p+q}(M \otimes N)$ such that: - -$$\Phi^{0,0} = \phi$$ -$\Phi$ is a "map of $\delta$-functors separately in $M$ and $N$": if \begin{equation}\tag{1} \label{s1} 0 \rightarrow{A'} \rightarrow A \rightarrow A'' \rightarrow 0 -\end{equation} -is an exact sequence in $\mathscr{A}$ with \begin{equation} -\tag{2}\label{s2} 0 \rightarrow A' \otimes B \rightarrow A \otimes B \rightarrow A'' \otimes B \rightarrow 0 -\end{equation} -still exact, then $\Phi^{p +1 ,q} \circ (\delta_1 \otimes H^0(\mathrm{id}_B)) = \delta_2 \circ \Phi^{p+1, q}$. Here, $\delta_1 \colon H^p(A'') \rightarrow H^{p+1}(A')$ and $\delta_2 \colon H^p(A'' \otimes B) \rightarrow H^{p+1}(A' \otimes B)$ are the maps provided by the $\delta$-functor structure on $H$ via the sequences (\ref{s1}), (\ref{s2}). Similarly, if we swap the roles of $A$ and $B$, we require that $\Phi^{p +1 ,q} \circ (\delta_1 \otimes H^0(\mathrm{id}_B)) = (-1)^{p} (\delta_2 \circ \Phi^{p+1, q})$. - -The answers to this question shed some light on this matter: Suppose we are in a setting where $H^0(M) = \mathrm{Hom}(O, M)$ for some object $O$ of $\mathscr{A}$ (e.g. group cohomology where we can take $O = \mathbf{Z}$, sheaf cohomology where we can take $O = \mathscr{O}_X$, etc.) Then $H^p(M) = \mathrm{Ext}^p(O, M)$, so we should get a pairing $H^p(O) \otimes H^p(O) \rightarrow H^{p+q}(O)$ induced by the 'composition' mapping $\mathrm{Hom}(O, O) \otimes \mathrm{Hom}(O, O) \rightarrow \mathrm{Hom}(O,O)$. I'm not sure exactly how to prove this part in general either, but I've at least seen it discussed in terms of classes of extensions of modules (I'm not sure how generally the result that $\mathrm{Ext}$ describes extension classes holds). This also doesn't allow general group objects, and I'm not sure how to do the extension. -The above question also discusses a more homotopical/$\infty$-categorical way to think about cup products, but I'm not familiar enough in that language to really get what's going on: I'd much prefer an argument working in ordinary abelian categories. - -REPLY [2 votes]: For -$\mathcal A=$ category of sheaves of abelian groups on a space X -$\mathcal B=$ category of abelian groups -see Theorems 6.2 and 7.1 in Chapter II of Glen Bredon's Sheaf theory (Spring GTM 170, 2nd edition, 1997).<|endoftext|> -TITLE: Axiomatic characterization of virtual fundamental classes? -QUESTION [15 upvotes]: There are several objects called "virtual fundamental classes." For example, certain Deligne-Mumford stacks, quasi-smooth derived schemes, etc. will admit a "perfect obstruction theory" as defined by Behrend-Fantechi and then the mojo of intrinsic normal cones produces a "virtual fundamental class" $[X] \in A_d(X)$ of virtual dimension $d$ in the Chow ring of the stack (or whatever) $X$. -Already here, there could be different perfect obstruction theories e.g. different morphisms $\varphi : E^\cdot \to \mathbb{L}_X$, $\varphi' : E'^\cdot \to \mathbb{L}_X$. How can we compare the virtual classes $[X]_{\varphi}, [X]_{\varphi'}$ coming from the two different perfect obstruction theories? Are they always the same? -Further, there are many approaches to defining virtual classes for moduli problems in symplectic geometry. There, one does not use perfect obstruction theories. Rather, one looks for "Polyfold structures" or "Kuranishi structures" which can be used to produce a homology class which is then called "the virtual fundamental class." I suppose one would like to compare the classes gotten from Polyfolds with the ones gotten from Kuranishi structure. And there are many appraoches to Kuranishi structure, so I suppose one would like to compare the virtual classes gotten from those approaches. -I apologise if there is a big well-known theorem stating asserting they are all equal. But even in that event, I think comparing the virtual classes in algebraic geometry with those in symplectic geometry is interesting. -And so, I ask: What makes these classes "virtual fundamental." Is there an axiomatic list of properties which either determines the thing or allows us to say general "model independent" things? - -REPLY [2 votes]: Joyce has some work in this direction: in particular, he defines a functor from Deligne-Mumford $\mathbb{C}$-stacks with perfect obstruction theories to what he names 'd-orbifolds'. These should be thought of, essentially, as 'derived' $\mathbb{R}$-stacks. Since we are in characteristic 0, this means these spaces come with a homotopy sheaf of simplicial $\mathbb{R}$-algebras, or equivalently, commutative differential graded algebras over $\mathbb{R}$. I strongly suspect that from this perspective, one can define an actual quasi-smooth derived geometric stack from which one can obtain a virtual fundamental cycle. Indeed, one would need to work over the appropriate site, and the Joyce's work gives some indications on what one should do.<|endoftext|> -TITLE: What did Ramanujan get wrong? -QUESTION [45 upvotes]: Quoting his Wikipedia page (current revision): - -compiled nearly 3,900 results -Nearly all his claims have now been proven correct - -Which of his claims have been disproven, can any insight be gained from the mistakes of this genius? - -REPLY [44 votes]: Hardy wrote some things about this, as I learned when writing this blog post. Here is a mistake which was even featured in the Ramanujan movie: in his letters to Hardy, Ramanujan claimed to have found an exact formula for the prime counting function $\pi(n)$, but (in Hardy's words) - -Ramanujan’s theory of primes was vitiated by his ignorance of the theory of functions of a complex variable. It was (so to say) what the theory might be if the Zeta-function had no complex zeros. His method depended upon a wholesale use of divergent series… That his proofs should have been invalid was only to be expected. But the mistakes went deeper than that, and many of the actual results were false. He had obtained the dominant terms of the classical formulae, although by invalid methods; but none of them are such close approximations as he supposed. - -Based on the second sentence in particular it sounds like what happened, although I haven't checked, was that Ramanujan's formula was the explicit formula but missing the contribution from the complex zeroes of the zeta function.<|endoftext|> -TITLE: Invariants of exterior powers -QUESTION [9 upvotes]: Let $\mathfrak{g}$ be the Lie algebra of $GL(n,\mathbb{R})$. Let $\theta(X) = - X^T$ be the Cartan involution on $\mathfrak{g}$; it induces decomposition as $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ of $+1$-eigenspace $\mathfrak{k}$ and $-1$ eigenspace $\mathfrak{p}$. Then $\mathfrak{k}$ is the Lie algebra of $K = O(n)$. I want to compute the dimensions of $K$-invariants $\textrm{Hom}_K(\wedge^q \mathfrak{p}, \mathbb{C})$, which, I suppose, is equal to $(W/ \mathfrak{k} W)^*$ where $W = \wedge^q \mathfrak{p}$, where $\mathfrak{p}$ is viewed as $\mathfrak{k}$-module by adjoint-representation. Could you someone point a way further? - -REPLY [9 votes]: To offer a slightly more geometric viewpoint on the same, the space $\bigoplus_q \mathop{\mathrm{Hom}}_K(\Lambda^q(\mathfrak{p}),\mathbb{C})$, which is the direct sum of all spaces you are considering, is the cohomology of the Lie algebra $\mathfrak{gl}_n$ relative to its subalgebra $\mathfrak{o}_n$, which, by a standard argument, is the same as $H^*(U(n)/O(n))$. It is well known (it really is; if you need some reference, see, for example, page 28 of lecture notes of Haefliger - http://www1.mat.uniroma1.it/people/piazza/HaeDC.pdf) to be the exterior algebra generated by elements $\xi_{4k-3}$, for all $k$ such that $2k-1\le n$. In fact, $\xi_{d}$ is the anti-symmetrisation of the trace of product of $d$ matrices. This confirms the very nice answer given by Abdelmalek Abdesselam.<|endoftext|> -TITLE: Vanishing line on Conway's game of life -QUESTION [30 upvotes]: If the initial state of Conway's game of life is a line of $n \in [0,100]$ alive cells, then it vanishes completely after some steps iff $n \in \{0,1,2,6,14,15,18,19,23,24 \}$. See below for $n=24$. - -Question: Is such a line non-vanishing for any $n \in [25,\infty)$? -Edit -Definition: a finite pattern $p$ has a weak period $wp$ if for any cell $c$ in the grid, there is $k>0$ such that the set of cells which are neighbours of neighbours of neighbours... ($k$ times) of $c$, is periodic of period $wp$ after sufficiently many generations, from the initial state $p$. -The sequence A061342 gives the weak period $wp_n$ of a line of $n$ alive cells. By combining the checking above with the fact that $wp_n \ge 2$ for $n \in [84,1000]$, we deduce that the pattern is non-vanishing for $n \in [25,1000]$. We observe that for $n=500$, four gliders are produced on the boundaries after $435$ steps, but $435<500$, so this must happens $\forall n \ge 500$. Assuming that these gliders (or others) are perpetual (as stated implicitly by Nathaniel Johnston in A061342, although without reference, while the proof could be non-trivial, as pointed out by Will Sawin in the comments), the answer to the above question would be yes. -Definition: a finite pattern $p$ is weakly-vanishing if any cell $c$ in the grid becomes perpetually dead after sufficiently many generations (depending on $c$), from the initial state $p$. -Improved question: Is there a weakly-vanishing line of $n$ alive cells with $n \in [25,\infty)$? -Stronger question: Is $wp_n \ge 2$ for any $[84,\infty)$ ? - -Tobias Fritz pointed out in the comments that there is a one-cell thick pattern with infinite growth (see this page), but it is disconnected. Bonus question: Can that happen in the connected case? - -REPLY [6 votes]: Just to help develop intuition - here is a fragment of a typical evolution for large $n$ (made with Golly)<|endoftext|> -TITLE: Unirationality of Fermat varieties in characteristic $p$ -QUESTION [7 upvotes]: In Shioda's famous paper "An Example of Unirational Surfaces in Characteristic $p$" (MSN), the author proved that the Fermat surface over the characteristic-$p$ field $k$ -$$ -x_1^n +x_2^n + x_3^n+x_4^n =0 -$$ -is unirational if $q \equiv -1 \pmod n$ for some $p$-power $q$. -At end of this paper, he mentioned that he proved it in the higher-dimension case with similar method, i.e., -$$ x_1^n +x_2^n + \cdots +x_{m+1}^n =0$$ -is unirational if $m \geq 3$ is odd and $q \equiv -1 \pmod n$ for some $p$-power $q$. But I can't prove it with his method in the surface case. How does his method generalize? - -REPLY [7 votes]: Let $k$ be a field of characteristic $p > 2$. -Fix a $\nu \in \mathbf{Z}_{\geq 1}$, and set $q := p^\nu$. -Let $r \in \mathbf{Z}_{\geq 1}$ and let -$$ -X^{2r+1}_{q+1}\colon \{x_0^{q+1} + \cdots + x_{2r+1}^{q+1} = 0\} \subset \mathbf{P}^{2r+1}. -$$ -be the Fermat hypersurface of degree $q + 1$ in $\mathbf{P}^{2r+1}$. Our aim is to show the - -Proposition. $X^{2r+1}_{q+1}$ is unirational. - -As Shioda does in his article, the general statement that $X^{2r+1}_n$ is unirational whenever $q \equiv -1 \pmod{n}$ is reduced to this case by mapping $X_{q+1}^{2r+1}$ surjectively onto $X_n^{2r+1}$ via $x_i \mapsto x_i^{(q+1)/n}$. -Proof. We basically follow Shioda's method verbatim. -Rearrange the defining equation and change coordinates by a root of $-1$ to obtain -$$ -(x_0^{q + 1} - x_{2r+1}^{q + 1}) + (x_{2r - 1}^{q + 1} - x_{2r}^{q+1}) - = \sum_{i = 1}^{r - 1} (x_{2i-1}^{q + 1} - x_{2i}^{q+1}). -$$ -Making the change of variables -$$ -\begin{align*} - y_0 & = x_{2r+1} + x_0, & y_{2i-1} & = x_{2i-1} + x_{2i}, \\ - y_{2r+1} & = x_{2r+1} - x_0, & y_{2i} & = x_{2i-1} + x_{2i}, -\end{align*} -$$ -the index $i$ ranging from $1$ to $r$, the equation above can be written as -$$ - y_0y_{2r+1}(y_0^{q - 1} + y_{2r+1}^{q-1}) + y_{2r - 1}y_{2r}(y_{2r - 1}^{q - 1} + y_{2r}^{q - 1}) - = \sum_{i = 1}^{r - 1} y_{2i-1}y_{2i}(y_{2i-1}^{q - 1} + y_{2i}^{q - 1}). -$$ -Now pass to the affine chart with $y_0 \neq 0$, i.e. set $y_0 = 1$ in the equation above so that the function field $K = K(X_{q+1}^{2r+1})$ of $X_{q+1}^{2r+1}$ is $k(y_1,\ldots,y_{2r+1})$ in which the variables $y_i$ are related by the equation above. -Consider the field extension $L$ of $K$ obtained by adjoining a $q$th root $t$ of $y_{2r}$. -The game now is to show $L$ is purely transcendental. -To do this, consider the change of variables, in which a strangely convenient new variable $u$ is pervasive: -$$ -\begin{align*} - y_{2r-1} & = uy_{2r}, \\ - y_{2r+1} & = uv, \\ - y_i & = uv_i\;\;\text{for}\;i = 1,\ldots,2r - 2. -\end{align*} -$$ -With these new variables, $L$ is the field $k(t,u,v,v_1,\ldots,v_{2r-2})$ subject to the single relation -$$ -v(1 + u^{q - 1}v^{q - 1}) + t^{q(q + 1)}(u^{q-1} + 1) - = u \sum_{i = 1}^{r - 1} v_{2i-1}v_{2i}(v_{2i-1}^{q - 1} + v_{2i}^{q - 1}) -$$ -where I have already cancelled a common factor of $u$. -Now to isolate $v$. -Throwing the term $t^{q(q+1)}$ to the right, -$$ -v + u^{q - 1}(v + t^{q+1})^q - = -t^{q(q+1)} + \sum_{i = 1}^{r - 1} v_{2i-1}v_{2i}(v_{2i-1}^{q - 1} + v_{2i}^{q - 1}). -$$ -Make the change of variables $s = u(v + t^{q+1})$, so that $L \cong k(s,t,v,v_1,\ldots,v_{2r-2})$ subject to the relation -$$ -v + s^{q-1}(v + t^{q+1}) = -t^{q(q+1)} + \sum_{i = 1}^{r - 1} v_{2i-1}v_{2i}(v_{2i-1}^{q - 1} + v_{2i}^{q - 1}). -$$ -Rearranging one more time, we solve for $v$ in terms of the other $2r$ variables -$$ -v = \frac{-t^{q+1}(s^{q-1} + t^{q^2 - 1}) + \sum_{i = 1}^{r - 1}v_{2i-1}v_{2i}(v_{2i-1}^{q-1} + v_{2i}^{q-1})}{1 + s^{q-1}} -$$ -thereby eliminating the single relation imposed on the variables of our field. -Thus $L \cong k(s,t,v_1,v_2,\ldots,v_{2r-2})$, as desired. $\square$ - -REPLY [6 votes]: Let $F^r$ be the Fermat hypersurface of fixed degree $n$ in $\mathbb{P}^{r+1}$, so that $\dim(F^r)=r$ (in any characteristic). There is a rational dominant map $\varphi :F^r\times F^s \cdots\!\!\twoheadrightarrow F^{r+s}$: if you write the equation of $F^r$ in affine coordinates $x_1^n+\ldots +x^n_{r+1}=-1$, and that of $F^s$ $\ y_1^n+\ldots +y^n_{s+1}=-1$, put $\varphi (x,y)= (x_1,\ldots ,x_{r+1},\lambda y_1,\ldots ,\lambda y_{s+1})$ in $\mathbb{P}^{r+s+1}$, with $\lambda ^n=-1$. Thus once you know that $F^2$ is unirational, the same holds for $F^r$ for any $r$ even. I don't know if this is what Shioda had in mind, but he certainly used a lot that trick in his subsequent work.<|endoftext|> -TITLE: $K$-theory backwards -QUESTION [12 upvotes]: Let $R$ be a ring. The $K$-theory of $(Mod(R)^{f.g.proj},\oplus)$ is obtained by first throwing out non-isomorphisms and then group completing. What happens if these steps are reversed? -That is, consider the symmetric monoidal $\infty$-category $C = - (Mod(R)^{f.g.proj},\oplus)$ of finitely-generated projective $R$-modules under direct sum (of course, this is in fact an ordinary category). To compute $K(R)$, we first pass to the core $\iota(C)$ to get an $\infty$-groupoid (which in this case is a 1-groupoid) and then we group complete to get $K(R) = \iota(C)[C^{-1}]$. -I'm thinking of group completion as an $\infty$-categorical process: for a symmetric monoidal $\infty$-category $C$, the group completion $C[C^{-1}]$ is the free symmetric monoidal $\infty$-category on $C$ where every object has a monoidal inverse; this agrees with the usual notion for symmetric monoidal $\infty$-groupoids. So it makes sense to reverse these steps and consider $\iota(C[C^{-1}])$. So my question is: -Question: - -If $C$ is a symmetric monoidal $\infty$-category, then do we have $\iota(C)[C^{-1}] = \iota(C[C^{-1}])$? -If not, does this at least hold in the case $C = Mod(R)^{f.g.proj}$, yielding an alternate description of $K(R)$? - -Motivation: My motivation here is basically to explore how robust the definition of $K$-theory is to perturbation. For example, an alternate description of $K(R)$ that definitely does work is the following: $K(R) = \iota_\oplus(\iota(Mod(R))[R^{-1}])$: that is, take the groupoid of all $R$-modules and rather than group completing, monoidally invert the single object $R$ itself viewed as an $R$-module; finally, throw away all objects which are not monoidally invertible (note in particular that since we're not monoidally inverting everything, we've sidestepped the Eilenberg swindle). I like this description because you don't have to put the "finitely-generated projective" condition in by hand -- it falls out naturally just from knowing that that the symmetric monoidal category $(Mod(R),\oplus)$ has a distinguished object $R$. If the answer to my question is "yes", then one could simplify this further to $K(R) \overset{?}{=} \iota_\oplus(\iota(Mod(R)[R^{-1}]))$; then even the choice of the $\oplus$ monoidal structure becomes more "canonical" in the sense that it has a universal property in $Mod(R)$, unlike in $\iota(Mod(R))$. Other variants start to suggest themselves, too. - -REPLY [3 votes]: The only way I know to speak of `group-completion' of a symmetric monoidal $\infty$-category involves not only adjoining inverses of all objects but also inverses of all morphisms. In that sense, group-completion corresponds to first taking the classifying space (rather than the core) and then group-completing. For example, if $\mathcal{C}$ has an initial or terminal object, its classifying space is trivial, so its group completion is trivial. -Of course there are plenty of decent symmetric monoidal categories where every object has an inverse but which are not groupoids (for example, the full subcategory of $\text{Mod}_R$ on invertible modules). But maybe this does not deserve to be called `group-complete' because the inverse is not functorial (at least, not covariantly). -If there is a functorial inverse $-\text{id}:\mathcal{C}\rightarrow\mathcal{C}$ such that $-\text{id}\otimes\text{id}$ is the zero functor, then the category is a groupoid. Specifically, each morphism $f:X\rightarrow Y$ has an inverse (up to equivalence) $X\otimes Y\otimes(-f):Y\rightarrow X$. -EDIT: Why would we want the inverse to be covariantly functorial? Let's say for example that $\mathcal{R}$ is a semiring $\infty$-category which has additive inverses. Then multiplication by any element (in particular -1) is functorial $\mathcal{R}\rightarrow\mathcal{R}$, so $\mathcal{R}$ is an $\infty$-groupoid as above. -Thus group-complete symmetric monoidal $\infty$-categories are exactly grouplike symmetric monoidal $\infty$-groupoids, also known as grouplike $\mathbb{E}_\infty$-spaces or connective spectra. The universal example of one of these is the sphere spectrum $\mathbb{S}$. Modules over $\mathbb{S}$ (in the sense of modules over a semiring $\infty$-category) are precisely the connective spectra, and the group-completion operation you want is $$-\otimes\mathbb{S}:\text{SymMon}\rightarrow\text{Mod}_\mathbb{S}\cong\text{Sp}_{\geq 0},$$ analogous to $$-\otimes\mathbb{Z}:\text{ComMon}\rightarrow\text{Mod}_\mathbb{Z}\cong\text{Ab}.$$ I discuss all of this in section 4 of my paper https://arxiv.org/abs/1606.05606<|endoftext|> -TITLE: Characterizing SP-DAGs by Forbidden Minors? -QUESTION [6 upvotes]: So it's well-known that an alternative way to define a series-parallel (undirected graph) is by the forbidden minor $K_4$. Is there a known analog of this definition for directed graphs — specifically, for DAGs? - -REPLY [4 votes]: You might want to have a look at the paper -Jacobo Valdes, Robert E. Tarjan, Eugene L. Lawler; The recognition of Series Parallel digraphs; STOC 1979; doi:10.1145/800135.804393 -Section 4 is called "Forbidden subgraph characterizations", and it contains the following two statements: - -A digraph with a single source and a single sink is a two-terminal series-parallel digraph if and only if it does not contain a subdigraph homeomorphic to the digraph with nodes 1,2,3,4 and arcs (1,2), (1,3), (2,4), (3,4) and (2,3). -A digraph is a general series-parallel digraph if and only if its transitive closure does not contain the digraph with nodes 1,2,3,4 and arcs (1,3), (1,4) and (2,3) as an induced subdigraph.<|endoftext|> -TITLE: Conjecture on Odd Perfect Numbers -QUESTION [6 upvotes]: I'm a grad student in mathematics and I've been working with a very gifted high school student (likely the smartest high school student I've ever met) on problems he's brought up and some competition math problems. This student has developed an interest in perfect numbers and the question regarding existence of odd perfect numbers. He has come up with a conjecture about odd perfect numbers, but I have not studied number theory and hence am not necessarily aware of well-known results of the field. So, here we are. -His idea: -Suppose $N \in \mathbb{N}$, with prime decomposition $N = p_1^{q_1}\cdots p_n^{q_n}$ Define $\tilde{N} = p_1\cdots p_n$. -Conjecture: If $N$ is an odd perfect number, then the sum of reciprocals of all factors of $\tilde{N}$ (excluding 1, including $\tilde{N}$) is less than 1. - -Q. Does this conjecture appear to be equivalent to something that has already been established? If this conjecture is true, does it appear to have any obvious implications? - -REPLY [3 votes]: The earliest result appears to be -$$\frac{1}{2}<\sum_{p\mid N}\frac{1}{p}<2 \log \left (\frac{\pi}{2} \right ),$$ -by Perisastri. Other authors have sharpen this result. For example, for a perfect number $N\equiv 0 \mod 3$, the sum has -$$\sum_{p\mid N}\frac{1}{p}=\frac{1}{3}+ \cdots+\frac{1}{p},$$ -see Ribenboim, New Book of Prime Records, p. 101, for a discussion and other references. -Perisastri, M. On the non-existence of odd perfect numbers of a certain form. Math. Student 28 1960 85–86 (1962). MR0142494.<|endoftext|> -TITLE: Idea behind Grothendieck's proof that formally smooth implies flat? -QUESTION [11 upvotes]: From this answer I learned that Grothendieck proved the following result. -Theorem. Every formally smooth morphism between locally noetherian schemes is flat. -The book Smoothness, Regularity, and Complete Intersection by Majadas and Rodicio cites the following result. -Theorem. Let $(A,\mathfrak m,K)\to (B,\mathfrak n, L)$ be a local homomorphism of noetherian local rings. Then TFAE. - -$B$ is a formally smooth $A$ algebra for the $\mathfrak n$-adic topology; -$B$ is a flat $A$-module and the $K$ algebra $B\otimes_AK$ is geometrically regular. - -The authors then write: - -This result is due to Grothendieck [EGA 0$_{\rm{IV}}$ , (19.7.1)]. His proof is - long, though it provides a lot of additional information. He uses this - result in proving Cohen’s theorems on the structure of complete noetherian local rings. An alternative proof of (I) was given by M. André [An1], - based on André –Quillen homology theory; it thus uses simplicial methods, that are not necessarily familiar to all commutative algebraists. A - third proof was given by N. Radu [Ra2], making use of Cohen’s theorems - on complete noetherian local rings. - -Questions: - -Are there any English references for the proof of Grothendieck or of André? -What is the conceptual outline of Grothendieck's proof? - -André's Homologie des Algèbres Commutatives does not look very geometric to a novice like me and I was hoping perhaps Grothendieck's path was more geometric. I would also like to at least glimpse the big picture of the proof. - -REPLY [4 votes]: You have a very short proof by Jesús Conde in the paper -A short proof of smooth implies flat. -Comm. Algebra 45 (2017), no. 2, 774–775. -https://doi.org/10.1080/00927872.2016.1175461 -see, alternatively, https://arxiv.org/abs/1504.05709<|endoftext|> -TITLE: Brief history of cardinal characteristics of the continuum -QUESTION [15 upvotes]: Cardinal characteristics of the continuum (CCC) are cardinals which are associated with naturally arising combinatorial properties of "the continuum". -The reason that "the continuum" is qualified, is that sometimes it is better to think about it as the Cantor space, other times as the Baire space, and sometimes as $\Bbb R$ itself, or as $[0,1]$. While different, they are all Polish spaces, and therefore Borel isomorphic. -For example, we can ask what is the smallest size family of functions $\cal F\subseteq\omega^\omega$ such that for every $g\in\omega^\omega$ there is some $f\in\cal F$ such that $g\leq f$ everywhere, or at least $g\leq^f$, namely there is some $m$, such that for all $n>m$, $g(n)\leq f(n)$. -One can show that this cardinal, also known as the dominating number and usually denoted by $\frak d$, is uncountable, and of course bounded by $2^{\aleph_0}$. We can prove, for example, that its cofinality is uncountable. And it is consistently taking many different values. -Some are more topological or measure-theoretic in nature, e.g. what is the smallest cardinality of a non-meager set, or how many null sets are needed to cover the whole space. -There are many more CCCs, some more famous (e.g. $\frak p$ and $\frak t$ which made headlines over the last summer), and some are less famous (e.g. $\frak h$ the shattering number). - -Where can I find a brief history of the research into CCCs? - -REPLY [9 votes]: Juris Steprans's article in the Handbook of the History of Logic: "History of the Continuum in the 20th century". -http://www.math.yorku.ca/~steprans/Research/PDFSOfArticles/hoc2INDEXED.pdf<|endoftext|> -TITLE: Pulling Back Cohen-Macaulay Sheaves -QUESTION [9 upvotes]: Suppose $f:X\to Y$ is a finite morphism of varieties and $\mathcal{F}$ is a Cohen-Macaulay sheaf on $Y$. Under what conditions on $f$ is $f^*\mathcal{F}$ Cohen-Macaulay? - -REPLY [8 votes]: As far as being true for the most general situation, you need: $X,Y$ to be Cohen-Macaulay, $\dim \mathcal F_x =\dim Y_x$ ($\mathcal F_x$ is maximal Cohen-Macaulay) for all $x$ in the support of $\mathcal F$, and $f$ to have finite flat dimension. (more is true with a little extra technical assumption, you need the ``Cohen-Macaulay defect" to be constant along $f$, see the last paragraph). -Then what you need follows from the two local statements about f.g modules over a local ring $R$. -1) If M is maximal CM, and N has finite projective dimension, then $Tor_i(M,N)=0$ for all $i>0$ (M, N are Tor-independent). -See: http://www.math.lsa.umich.edu/~hochster/711F06/L11.20.pdf -2) If $M,N$ are Tor-independent and $pd_RN<\infty$, we also have the depth formula: -$depth(M\otimes N) + depth(R) = depth(M)+depth(N)$ -See: https://www.math.unl.edu/~siyengar2/Papers/Abform.pdf -In particular, if N is CM then $M\otimes N$ is also CM of same depth. (here we use that $R$ is CM, so $depth(R) =depth(M)$). -If you drop any condition, it is not hard to find examples to show that the statement is no longer true. On the other hand, for special $X,Y,\mathcal F$, sometimes one can say a bit more. For example, if $R$ is a complete intersection, Tor-independence forces the depth formula to hold, without knowing that $N$ has finite flat dimension. -Added in response to OP's request: By looking at the depth formula in 2), the following more technical, but general statement is true: suppose $f: R\to S$ is a finite, local map of finite flat dimension, and $dim(R)-depth(R)=dim(S)-depth(S)$. Then for a maximal CM module $M$ over $R$, $M\otimes_R S$ is (maximal) CM over $S$. This cover both cases when $R,S$ are CM ( both sides of the equality is $0$), or if $f$ is flat (both dim and depth are preserved).<|endoftext|> -TITLE: A question on Voevodsky´s categories -QUESTION [6 upvotes]: I want to try to understand the Voevodsky´s big triangulated categories of motives $DM$ and $DM^{eff}$. Unfortunately, I am being not able to find answers to the following, too vague, questions: - -1.- What is the idea behind his construction and what is one possible motivation for this? -2.- What are the basic features of this categories, besides of being triangulated? -3.- Could you suggest me a good reference where I can find a detailed discussion of the construction of this categories? - -REPLY [11 votes]: One could say that the story begins with Beilinson's conjectures on the existence of a theory of motivic cohomology. In accordance with the insights of the Grothendieck school that cohomology theories in nature always come with a corresponding "category of coefficients", and that it is very often profitable to work at the categorical level instead of at the level of cohomology groups, Beilinson also conjectured that there should exist triangulated categories $DM(S, \mathbf{Z})$, where the conjectural motivic cohomology groups can be computed as Ext-groups. By a category of coefficients I mean a system of (triangulated) categories, for each scheme $S$, equipped with the formalism of six operations (the objects are thought of as "coefficients" for cohomology). -When Voevodsky set out to construct these categories $DM(S,\mathbf{Z})$, he was guided strongly by the principle that motivic cohomology should play the role in algebraic geometry of singular cohomology of smooth manifolds, whose category of coefficients is the derived category of chain complexes of abelian groups. In fact, singular cohomology is only one of many "generalized cohomology theories" on smooth manifolds (other examples include topological K-theory and cobordism), and unlike singular cohomology, most cohomology theories cannot be computed as the cohomology groups of a chain complex, but rather must be defined as the homotopy groups of a space or spectrum. Thus, inspired by the Eilenberg-Steenrod axioms, let's define a cohomology theory on smooth manifolds to be a presheaf $\mathcal{F}$ of spaces on the category of smooth manifolds satisfying two conditions. First we require that it is homotopy invariant: we have $H^i(X \times \mathbf{R}, \mathcal{F}) \simeq H^i(X, \mathcal{F})$ for all smooth manifolds $X$, where $H^{-*}(X, \mathcal{F})$ denote the homotopy groups of the space $\mathcal{F}(X)$. Secondly we require the existence of Mayer-Vietoris long exact sequences (or more precisely that $\mathcal{F}$ satisfies a homotopical version of the sheaf condition). Then one can prove that the category of cohomology theories in this sense is equivalent to the homotopy category of spaces (to make precise sense of the "category of cohomology theories" one should use homotopical or higher categorical language). If we take presheaves valued in chain complexes instead, we get a category equivalent to the derived category of abelian groups (= the category of coefficients for singular cohomology). -Voevodsky's first construction of DM, which goes back to his 1992 thesis, was directly inspired by the above. Thus the idea was to take presheaves of chain complexes on schemes (over some fixed base $S$), and impose homotopy invariance with respect to the affine line $\mathbf{A}^1$ instead of the real line. An important question is then which Grothendieck topology to take for the sheaf condition. Instead of the Zariski or Nisnevich topologies, Voevodsky used a topology he called the "h-topology", which is essentially what you get when you take the Zariski topology and you add in proper surjections as coverings (the name came from his expectation that this topology seemed to be "suitable for the developing of the homotopy theory of schemes"). With this definition he was able to construct Gysin sequences and prove projective bundle and blow-up formulas in motivic cohomology. -Let's come back to the setting of smooth manifolds for a moment. The Dold-Thom theorem says that the (reduced) singular homology groups of a smooth manifold $X$ can be computed as the homotopy groups of the infinite symmetric power $S^\infty(X)$. Inspired by this, Suslin constructed a singular homology theory for schemes by taking an algebraic analogue of infinite symmetric powers. In 1996, Suslin and Voevodsky were able to prove a comparison of Suslin's singular homology construction with étale cohomology with torsion coefficients. After this work, I think Voevodsky realized that what was really important was perhaps not the h-topology itself, but just the fact that h-sheaves automatically admit transfers. Indeed, the existence of Gysin sequences and projective bundle formula only actually require the Nisnevich topology. In view of this he then defined (the effective version of) DM in the form we use today, which is more or less rigged so that hom groups in the category compute Suslin homology (at least over a field). The idea is that instead of using the h-topology, we just use the Nisnevich topology but then force our presheaves to have the additional structure of transfers in a different way: we take presheaves on the category whose objects are smooth $S$-schemes, but whose morphisms are Voevodsky's finite correspondences. -Above I have been discussing the effective categories $DM^{eff}(S,\mathbf{Z})$. The non-effective version $DM(S,\mathbf{Z})$ is just obtained by forcing the Tate twist operation $X \mapsto X(1) := X \otimes \mathbf{Z}(1)$ to become invertible. This is analogous to the relationship between the bounded-below derived category $D^+(\mathbf{Z})$ and the unbounded derived category $D(\mathbf{Z})$, where the latter can be thought of as taking $D^+(\mathbf{Z})$ and forcing the suspension functor $[1]$ to become invertible. The main motivation is to be able to express duality phenomena: for example, one can prove that for a smooth proper $S$-scheme $X$ of relative dimension $n$, the dual of the motive $M(X) \in DM(S,\mathbf{Z})$ is given by $M(X)(-n)[-2n]$. -For a reference on triangulated categories of motives, I would recommend the book on the subject by Cisinski and Déglise. A shorter reference that could be easier to start with is the paper "Finite correspondences and transfers over a regular base" by Déglise.<|endoftext|> -TITLE: Dodecahedral rolling distance -QUESTION [11 upvotes]: Let a dodecahedron sit on the plane, -with one face's vertices on an origin-centered unit circle. -Fix the orientation so that the edge whose indices are $(1,2)$ is horizontal. -For any $p \in \mathbb{R}^2$, define the dodecahedral distance $dd(p)$ from $o=(0,0)$ -to $p$ to be the fewest number of edge-rolls that will result in -a face of the dodecahedron landing on top of $p$. -Equivalently, imagine reflecting a regular pentagon over edges, -as illustrated below: It takes $4$ rolls/reflections to cover $p=(5,\pi)$: - -          - - -          - -$p=(5,\pi)$, $dd(p)=4$, $s=(3,1,4,2)$. - - -My main question is: - -Q. Given $p$, how can one calculate $dd(p)$? - -Greedily choosing, at each step, the roll that is best aligned with the vector $p-o$ -does not always succeed. -Could one characterize the sequences of roll indices $s$, where rolling over -edge $(i,i+1)$ of the pentagon is index $i\,$? These sequences -feel analogous to continued-fraction approximations. -What do all the points $p$ of $\mathbb{R}^2$ with $dd(p)=k$ look like, -i.e., what is the shape of a $dd$-circle $C_k$? - -          - - -          - -$p=(18.3,-1.4)$, $dd(p) \le 12$, $s=(2, 4, 1, 3, 5, 2, 4, 1, 3, 5, 2, 5)$. - - -That the dodecahedral distance is well-defined follows, e.g., -from "Thinnest covering of the plane by regular pentagons." - -REPLY [5 votes]: Here are a few pictures and speculations to add to the the great ones from @j.c. -Here are the pentagons which arise from starting with the central green one and doing the reflections which do not overlap any previous ones. - -These are just the outer frontiers of the $C_i.$ (Specifically, this shows the frontiers for $i \le 7$ and half the frontier for $i=8$.) It seems clear (though I do not claim a proof) that for a target point inside one of the pentagons one does best to roll out to it using just these pentagons. For a target point in one of the rhombi one should roll out to a neighboring pentagon and at the last step reflect over a previously unused (for reflections) edge. One can use one of the bordering pentagons closest to the center for the closer half of the rhombus and a small part of the further half. - -Here is a central portion of the same thing with lines connecting the centers of pentagons which share an edge. The resulting hexagons tile the plane with a $5$-fold rotational symmetry. - - -Finally, here is a picture of just these hexagons going out further. For each of the $5$ orientations, the hexagons of that orientation fall into $2$ connected components. One orientation is highlighted in blue.<|endoftext|> -TITLE: Inequality on permutation polytope -QUESTION [6 upvotes]: Let $a = (a_1, \cdots, a_n), b = (b_1, \cdots, b_n), c = (c_1, \cdots, c_n) \in \mathbb{R}^n$ with $a_1 \geq \cdots \geq a_n, b_1 \geq \cdots \geq b_n, 0 < c_1 \leq \cdots \leq c_n$. -In addition, assume that $\sum_{i=1}^k b_i \leq \sum_{i=1}^k a_i$ for all $k \in \{1, \cdots, n-1\}$ and $\sum_{i=1}^n b_i = \sum_{i=1}^n a_i$. -Let $A := \{(a_{\sigma(1)}, \cdots, a_{\sigma(n)}) \,|\, \sigma \in S_n\}$ and $K_A$ be the unique minimal convex set containing $A$. -For $b \in K_A$, does $$\sum_{i=1}^n c_i b_i \geq \sum_{i=1}^n c_i a_i$$ hold? - -REPLY [3 votes]: Write $c_k=p_1+\cdots+p_k$ with $p_k=c_k-c_{k-1}\ge0$. Then -$$\sum_ic_ib_i-\sum_ic_ia_i=\sum_kp_k(s_k(b)-s_k(a))$$ -with $s_k(a)=a_k+\cdots+a_n$. By assumption, $s_k(b)-s_k(a)\ge0$ and you are gone. - -This proof is just Abel's resummation, that is discrete integration by parts.<|endoftext|> -TITLE: Cutting a Gaussian in two pieces that are maximally separated in the Wasserstein metric -QUESTION [10 upvotes]: Denote the standard Gaussian probability measure on $\mathbb R^n$ by $\gamma$. We partition $\mathbb R^n$ into two sets $A$ and $A^c$ such that $\gamma(A) = \gamma(A^c) = 1/2$. -Denote by $\gamma_{A}$ to Gaussian measure restricted to $A$, and normalized so that it is a probability measure. Similarly, define $\gamma_{A^c}$ to be the Gaussian measure restricted to $A^c$ and normalized. -My question is the following: -What is the optimal $A$ such that $\gamma_A$ and $\gamma_{A^c}$ are the farthest apart; i.e., solving -$$\arg\max_{A} W_2(\gamma_A, \gamma_{A^c}),$$ -where $W_2$ is the 2-Wasserstein distance? -Possible generalization: Instead of constructing $\gamma_A$ and $\gamma_{A^c}$ as above, we could start with any two probability measures $\gamma_1$ and $\gamma_2$ such that $\gamma = \frac{\gamma_1 + \gamma_2}{2}$ and find $\arg \max_{\gamma_1, \gamma_2} W_2(\gamma_1, \gamma_2)$. -Finding upper bounds on the $W_2$ distance is also of interest. A natural conjecture, inspired by the Gaussian isoperimetric inequality, would be that $A$ should be a half-space. Counterexamples to this are also welcome! - -REPLY [2 votes]: Here are calculations for the two possibilities posed in the comments in the case $n=2$. -The element of probability for $\gamma$ is $(1/\pi)\exp(-x^2-y^2)\,dx\, dy = (1/\pi)\exp(-r^2)\,r\,dr\, d\theta$. -So the dividing lines can be either a line or a circle, $|$ or $\circ$, at $x=0$ or $r=\sqrt{\ln(2)}$. -In the circular case the probability between the circle and $r$ is the same as the probability between the circle and $f(r)=\sqrt{-\ln(1-\exp(-r^2))}$. These give -\begin{align} -W_2^{^|} &= \int_{x=0}^{\infty}\int_{y=-\infty}^{\infty} 2x\frac{2}{\pi}\exp(-x^2-y^2)\,dx\, dy \\ -&= \frac{2}{\sqrt{\pi}}\\ -&\simeq 1.13\\ \\ -W_2^{^{\Large\circ}} &= \int_{r=0}^{\sqrt{\ln(2)}}\int_{\theta=0}^{2\pi} -\left(f(r)-r\right) -\frac{2}{\pi}\exp(-r^2)\,r\,dr\, d\theta \\ -&= \sqrt{\pi}(1-2\text{erf}(\sqrt{\ln 2}))+2\sqrt{\ln 2}\\ -&\simeq 0.74 -\end{align} -So of these two options, the half-plane gives the larger Wasserstein distance.<|endoftext|> -TITLE: Abandoned LCAs on Cantor's Attic : Grand Reflection cardinals, universe cardinals, weak universe cardinals -QUESTION [8 upvotes]: Cantor's Attic is a really great website for the various descriptions of large finite numbers, large countable ordinals, and large cardinal axioms. -However, after looking through the archives of the website, I have found that originally, the following cardinals were included and never given a definition: - -Grand reflection cardinals -Universe cardinals -Weak universe cardinals - -The universe cardinals and weak universe cardinals were replaced by the worldly cardinals in the same spot, so it makes sense that the term "universe" was renamed to worldly. However, that doesn't explain what "weak universe" cardinals are. -The grand reflection cardinals were created and never replaced. They still remain on the upper attic today, although hidden by code. You can see the link there, but it links to nothing. -So what are these cardinals? Does anybody know? The best person I could think of to answer this would be @JoelDavidHamkins himself, who was the one to put these on Cantor's Attic. - -REPLY [8 votes]: Thanks for the question. I had briefly used the universe/weak-universe terminology in 2010 in my answer to a question on MathOverflow, What interesting/nontrivial results in Algebraic geometry require the existence of universes? -My thinking at that time was that $\kappa$ should be a universe cardinal, if $V_\kappa$ was an (uncountable) Grothendieck universe, which means that this is the same as an inaccessible cardinal. -What we now call the worldly cardinals can be seen as a weak version of this, where one drops the requirement that $\kappa$ is regular. So $\kappa$ is worldly (or a weak universe cardinal in that MO answer), if $V_\kappa\models\text{ZFC}$. -That answer also mentions a very weak universe concept, which is simply a transitive model of $\text{ZFC}$. -I think I had created the Cantor's Attic entries at first at about the same time I had posted that answer, which was just a little before I had started using the worldly cardinal terminology, which I think is better and which I am happy to see has become comparatively established. So I think I may have created the early entries, and then switched over to the worldly terminology later. -The Grand Reflection cardinals, in contrast, were introduced by Philip Welch. This is a cardinal $\kappa$ relating truth in $V_\kappa$ for subsets of $V_\kappa$ to truth in $V$ for proper classes. -Incidentally, Cantor's Attic is a community-run affair, and knowledgeable people are welcome to help with adding to or improving entries.<|endoftext|> -TITLE: Chow Groups of varieties over number fields -QUESTION [12 upvotes]: I believe that there is a conjecture that for any smooth projective variety $X$ over a number field $K$, its Chow groups $CH^i(X)$ (or at least $CH^i(X)\otimes_{\mathbf Z} \mathbf Q$) are finitely generated. -What is the standard name for this conjecture? In private communication people referred to it as Beilinson's conjecture. I assume that it should have been formulated before Beilinson. What is the best paper to cite for this conjecture? Is it known in any non-trivial case? - -REPLY [4 votes]: Just adding to Lucifer's answer: This is Conjecture 5 in Beilinson's paper -https://mathscinet.ams.org/mathscinet-getitem?mr=902590 -For a smooth projective variety over a number field and a fixed codimension, Tate's conjecture predicts that the rank of algebraic cycles modulo homological equivalence is given by the order of the pole of an appropriate L-function; Beilinson's conjecture predicts that the rank of homologically equivalent cycles (modulo rational equivalence) is the order of the zero of an appropriate L-function; special cases were stated by Swinnerton-Dyer, Tate, Bloch before. See 6.2 and 6.5 of Nekovar's article below for the precise statements. -Nice introductions to Beilinson's conjectures are -1) Nekovar's article -http://math.stanford.edu/~conrad/BSDseminar/refs/BeilinsonintroII.pdf -2) Ramakrishnan's article -https://mathscinet.ams.org/mathscinet-getitem?mr=991982 -3) The Book: available at Professor Schneider's webpage -https://ivv5hpp.uni-muenster.de/u/pschnei/publ/beilinson-volume/<|endoftext|> -TITLE: Historical question about the $\aleph_2$-Souslin hypothesis -QUESTION [7 upvotes]: For an uncountable regular cardinals $\kappa,$ let $\kappa$-Souslin hypothesis, denoted $SH(\kappa)$ be the assertion that there are no $\kappa$-Souslin trees. -By a result of Jensen, $GCH+SH(\aleph_1)$ is consistent. -However, the problem of the consistency of $GCH+SH(\aleph_2)$ is still open. - -Question. $(a)$ Where the above question first appeared? -$(b)$ Who first asked the question (possibly not in a published paper or book) and when? - -Any references are welcome. - -REPLY [13 votes]: First, in case your question suggests that you managed to prove the consistency of $GCH+SH(\omega_2)$, then let me congratulate you wholeheartedly! -Second, to put things in context, let us recall that an $\omega_2$-Souslin tree is a non-special $\omega_2$-Aronszajn tree which is (obviously) an $\omega_2$-Aronszajn tree. -Back to your question. By the time that Jensen proved the consistency of $GCH+SH(\omega_1)$, all of the following were already known: -(1) If $CH$ holds, then there exists a special $\omega_2$-Aronszajn tree (Specker, 1949). -(2) If $V=L$, then $GCH$ holds and there is an $\omega_2$-Souslin tree (Jensen, 1972). -(3) It is equiconsistent with the existence of a weakly compact cardinal that there are no $\omega_2$-Aronszajn trees (Mitchell [and Silver], 1972). -Anyone who was aware of the last three results was basically led to the question of whether $GCH$ (or even just $CH$) is consistent with $SH(\omega_2)$. -I do not think this ``automatic'' question is attributed to anyone. -As far as I can tell, the first paper to address this problem was Gregory's (1976). -Recall that Jensen's theorem concerning $V=L$ actually gave two sufficient conditions for the existence of an $\omega_2$-Souslin tree: (1) $CH+\diamondsuit(S^2_1)$, and (2) $\square_{\omega_1}+\diamondsuit(S^2_0)$. -In the above-mentioned paper, Gregory proved that $\diamondsuit^+(S^2_0)$ follows from $GCH$ (this should be compared with a 1980 theorem of Shelah who proved that $\diamondsuit(S^2_1)$ does not follow from $GCH$). -In the same paper, Gregory furthermore proved that $GCH$ and the existence of a non-reflecting stationary subset of $S^2_0$ (a consequence of $\square_{\omega_1}$) already yields such a tree. -Gregory's result and the surrounding questions were soon after echoed in a survey paper by Kanamori and Magidor (1978). -Around the same time, Laver proved that the consistency of a measurable cardinal yields the consistency of $CH+SH(\omega_2)$. -I recently learned that in a visit to Berkeley, Laver said that he is looking to reduce the hypothesis into a weakly compact cardinal, since ``this would gave an equiconsistency''. -In a phone conversation I had with Shelah this very week, he told me the following. -When Laver proved his consistency result, Mati Rubin (who got his Ph.D. with Shelah) was in Colorado. Rubin informed Shelah of the proof, -and Shelah managed to reduce the measurable cardinal hypothesis into a weakly compact. -After a wait period, Shelah airmailed his proof to Laver, who wrote down a joint paper (1981). That is, the paper was physically written by Laver. -The last paragraph of the Laver-Shelah paper (1981) reiterates the fact it is open whether $GCH+SH(\omega_2)$ is consistent. -This paragraph refers to Gregory's 1976 paper that shows that $GCH+SH(\omega_2)$ implies the consistency of a Mahlo cardinal, -and to an upcoming paper of Shelah and Stanely (that appeared in 1982) showing that $CH+SH(\omega_2)$ implies the consistency of an inaccessible cardinal. -A proof (or just a claim) that the Laver-Shelah theorem gives an equiconsistency result never appeared in print. -Shelah also told me that for many years Gregory was claiming that he knows how to get the consistency of $GCH+SH(\omega_2)$. -Again, such a paper never appeared. -The last page of the Laver-Shelah paper points out that the same technique of their proof shows that the consistency of a weakly compact cardinal yields the consistency of $CH$ together with all $\omega_2$-Aronszajn trees are special. -This should be compared with a theorem of Shelah-Stanley (1988) and Todorcevic (1989) that if $\omega_2$ is not weakly compact in $L$, then there is a non-special $\omega_2$-Aronszajn tree. -Moving forward to 1992, Kojman and Shelah published a paper that improves Gregory's 1976 theorem to show that if $GCH$ holds and there exists a non-reflecting stationary subset of $S^2_0$, -then there exists an $\omega_2$-Souslin tree which is moreover countably complete. Their proof replaced Jensen's $\diamondsuit(S^2_1)$ hypothesis with some weak form of club-guessing at $S^2_1$. -In their paper, Kojman and Shelah mention that the referee of the paper claimed that Gregory knew how to get a countably complete $\omega_2$-Souslin tree from $GCH+\square_{\omega_1}$. -Personally, I doubt that Gregory anticipated club-guessing, -and have no idea how else could his proof go. In any case, Gregory's proof was never published. -Coming back to your question: the fact that $GCH+SH(\omega_2)$ is open appears in the opening paragraph of the Kojman-Shelah paper, -and the reader is referred to Gregory (1976), Laver-Shelah (1981), and Shelah-Stanley (1982). The paper is concluded by reiterating the problem of whether $GCH+SH(\omega_2)$ requires the consistency of a weakly compact. -According to Shelah (and to the literature), these questions remained unanswered. -In a paper from 2007, Konig, Larson, and Yoshinobu introduced a generalized club-guessing principle at $S^2_1$ and proved that the conjunction of their principle with $GCH$ yields a (countably closed) $\omega_2$-Souslin tree. In response, in a paper from 2011, I proved that their principle does not follow from $GCH$, and hence their approach cannot shed any light on the problem of $GCH+SH(\omega_2)$. -In a paper from from 1974, Erdos and Hajnal asked whether a poset of size $\aleph_2$ that does not contain a copy of $\omega_2$ or of $-\omega_2$, -must contain a suborder of size $\aleph_1$ that does not contain a copy of $\omega_1$ or of $-\omega_1$. -In paper from 1981, Todorcevic proved that the consistency of a weakly compact cardinal yields the consistency of $GCH$ together with the assertion that any $\omega_2$-Aronszajn tree contains an $\omega_1$-Aronszajn subtree. -In paper from 2017, I proved that if $GCH$ holds and $\omega_2$ is not weakly compact in $L$, -then there exists an $\omega_2$-Souslin tree with no $\omega_1$-Aronszajn subtrees. -The proof is motivated by the ``Ostaszewski square'' principle that I introduced in a paper from 2014. -The latter asserts the existence of a $\square_{\omega_1}$-sequence with some built-in (strong form of) club-guessing. -Roughly speaking, the new ingredient of the 2017 paper was to show that $\square_{\omega_1}$ may be relaxed to $\square(\omega_2)$. -In the same paper, I also pushed further the Kojman-Shelah argument and showed that $GCH+\square(\omega_2)$ entails the existence of an $\omega_2$-Souslin tree which is countably complete. -There are further exciting news about $\omega_2$-Aronszajn trees obtained recently by Krueger, and by Lambie-Hanson and Lucke, -but these do not directly address the problem of $GCH+SH(\omega_2)$.<|endoftext|> -TITLE: Is there a mathematical and information theoretic explanation for this cube packing phenomenon? -QUESTION [58 upvotes]: I saw this unintuitive result on dice packing: - -A jumble of thousands of cubic dice, agitated by an oscillating -rotation, can rapidly become completely ordered, a result that is hard -to produce with more conventional shaking. - -The paper is not very rigorous. - -Is there a mathematical explanation of this result? - -Is such phenomenon studied? - - - -Can the presence of centripetal force in oscillating rotation help explain the phenomenon? - - -Update: Why does entropy seemingly reduce? Is there information theoretic explanation? - - - -Is there a minimum and maximum volume below and beyond respectively of both the cylinder and the dice which the phenomenon fails? - - -How about radius of the cylinder having an effect and the mass of individual dice? - -Note the complexity remains the same. One is simply explainable as random and the other has total order explainable succinctly. But crucially they go through an intermediate transition phase of very high complexity. But unlike the example in https://www.scottaaronson.com/blog/?p=762 the entropy reduces and s something non-intuitive is happening. I do not think there is any elementary explanation. The example in the link of mixing coffee produces high entropy coffee no matter how you stir (slow or fast with or without rotation but there might be a totally non-intuitive way to mix coffee and lower entropy). - -REPLY [5 votes]: This should be seen as a comment, but it is too long for a comment: -One aspect, that has not been addressed in the answers, is the influence of the container geometry. In the experiment, a cylindrical container has been subjected to two different kinds of motions: a coaxial rotation with alternating directions and tapping to the side of side of the cylinder, which can also be interpreted as a horizontal motion that keeps the axis upright. -Assume for a moment, that the experiment were carried out without the influence of gravity and the cubes at least a distance of $\epsilon$ apart. then, - -Both kinds of motions would result in horizontal motions of the cube's centers of gravity. -in the case of rotational motion the primary source of impact on the cubes would be friction between the cylinder and the cubes touching it, resulting in a pure rotation of those cubes around an axis through their center of gravity that is parallel to the one of the container. The "wave of motion" will proceed outside-in as the rotating cubes start to collide and the axes of rotation will start to deviate from the cylinder axis; there will also be forces on the cubes' centers that have a vertical component. All in all, the effects will be fairly moderate. -in the case of tapping motion cubes will collide with cylinder and are roughly reflected like rays of light on a mirror, which means that there will be a caustic of the horizontical motion vectors or, put differently, the shockwaves will be bundled and result in a much higher motion energy acting on certain cubes. A secondary effect will be, that after a while cubes will be reflected from the opposite "side" of the cylinder and then collide with other cubes, that are still on their way to that side; the effect will be approximately the same as if two vertically aligned balls are dropped to the ground, namely that the reflection of the second cube will be accelerated towards the cylinder axis again. - -So, in a nutshell the difference can be explained by the different magnitude of shockwaves that are generated with the different kinds of motions; that can easily be checked by replacing the cubes with water and measuring the deviation of the surface from equilibrium. -Further experimental investigations could check the effect of container geometry (elliptical or prismatic instead of cylindrical) or rotation aroung an axis off the center of symmetry. -But I am neither a physicist, nor do I have any background in statistical mechanics, so forgive me if my answer is nonsense.<|endoftext|> -TITLE: Density of a saturated random packing of congruent circles -QUESTION [9 upvotes]: The problem of the expected density of a saturated random packing of unit circles in the plane can be described as follows. -In a circular region $C$ of a large radius pick a point at random and draw a unit circle centered at the chosen point. Then pick another point in $C$ at random and draw again a unit circle, centered at the chosen point. If this circle overlaps with the previously drawn one, discard it and pick another point in $C$. Continue this process until the probability of finding one more point in $C$ for which the unit circle drawn about it does not overlap with any of the previously drawn circles is zero. The density of such packing of unit circles in $C$ is computed as the sum of the areas of the unit circles divided by the area of $C$. Let $d(C)$ denote the expected value of the density, and let $d$ denote the limit of $d(C)$ as the radius of $C$ tends to infinity. -Numerical experiments indicate that $d$ should be around $0.82$ (see for example https://www.sciencedirect.com/science/article/pii/0021979771903389), but, to the best of my knowledge, the exact value of $d$ is unknown. Are there any rigorous results obtained in this direction? Any reasonably good estimates, perhaps? Any references will be appreciated. - -REPLY [3 votes]: Following user @j.c.'s lead, here is another paper on RSA (Random Sequential Adsorption), which concludes with a density of $0.77$. From the abstract: - - - - -Hinrichsen, Einar L., Jens Feder, and Torstein Jøssang. "Random packing of disks in two dimensions." Physical Review A 41.8 (1990): 4199. - (Journal link.) - - -          - - -          - -Fig.2: The first iteration step.<|endoftext|> -TITLE: Generators of an ideal with small degree -QUESTION [5 upvotes]: Let $P_1,\ldots, P_d, Q_1, \ldots, Q_k \in \mathbb{C}[x_0,\ldots, x_n]$ be homogenous polynomials of degree at most $r$. -Assume that $P_1 \cdot P_2 \cdots P_{d-1} \cdot P_d \in \langle Q_1, \ldots, Q_k \rangle$. -Here $\langle h_1, \ldots, h_s \rangle$ is the ideal with the generators $h_1, \ldots, h_s$. -Is it true that for some $\{ i_1, \ldots, i_f \} \subseteq \{1, \ldots, d\}$ the polynomial $P_{i_1} \cdots P_{i_f} \in \langle Q_1, \ldots, Q_k \rangle$, where $f= f(k,r)$? - -REPLY [4 votes]: Yes. The point is that many invariants of the ideal $I=(Q_1,\dots,Q_k)$ can be bounded depending only on $k$ and $r$. It is pure luck that the following paper has collected many of them in a very convenient Proposition 4.6. -http://www-personal.umich.edu/~asnowden/papers/genstillman-071517.pdf -In particular, you can find a primary decomposition of $I = I_1\cap\dots \cap I_l$ such that each of the $I_i$ is $\mathfrak p_i$-primary and the number of generators of $I_i$ as well as degrees and $l$ itself are bounded by some function of $k$ and $r$ (using part 6 of the Prop. cited above). -So the problem reduces to the case when $I$ is $\mathfrak p$-primary. But there is a $B$ such that $\mathfrak p^B\subseteq I$, and this number can also be bounded on the degrees and number of generators of $I$ (part 10 of loc. cit.). Such $B$ works.<|endoftext|> -TITLE: What are compact objects in the category of topological spaces? -QUESTION [9 upvotes]: Let $\mathscr C$ be a locally small category that has filtered colimits. Then an object $X$ in $\mathscr C$ is compact if $\operatorname{Hom}(X,-)$ commutes with filtered colimits. -On the other hand, the category of topological spaces has a competing notion of compactness. Not every compact topological space is a compact object in $\operatorname{\underline{Top}}$, as is explained here. Todd Trimble asked (in the $n$-category café) if the situation is any better if $X$ is assumed compact Hausdorff. -More generally, is there some sort of classification of compact objects in $\operatorname{\underline{Top}}$? - -REPLY [12 votes]: Proposition. Let $X$ be a topological space. Then $X$ is a compact object if and only if $X$ is a finite discrete space. - - -Before giving the proof, we state an easy lemma. -Lemma. Suppose $X$ is a compact object in $\operatorname{\underline{Top}}$. Then $X$ is finite. -Proof. Let $Y$ be the indiscrete topological space with underlying set $X$; i.e. $\mathcal T_Y = \{\varnothing,Y\}$. It is the union of its finite subsets, and this gives it the colimit topology because a subset $U \subseteq Y$ is open if and only if its intersection with every finite subset is. Indeed, if $U$ were neither $\varnothing$ nor $Y$, then there exist $y_1, y_2 \in Y$ such that $y_1 \in U$ and $y_2 \not\in U$. But then $U \cap \{y_1,y_2\}$ is not open, because $\{y_1,y_2\}$ inherits the indiscrete topology from $Y$. -Since $X$ is compact, the identity map $X \to Y$ factors through a finite subset of $Y$. This forces $X$ to be finite. $\square$ -For the remainder of the proof, we will use the auxiliary construction of a specific colimit given here. We will recall the notation. For reasons that become clear later, we have swapped the roles of $0$ and $1$. -Definition. For all $n \in \mathbb N$, let $X_n$ be the topological space $\mathbb N_{\geq n} \times \{0,1\}$, where the nonempty open sets are given by $U_{n,m} = \mathbb N_{\geq m} \times \{0\} \cup \mathbb N_{\geq n} \times \{1\}$ for $m \geq n$. They form a topology since -\begin{align*} -U_{n,m_1} \cap U_{n,m_2} &= U_{n, \max(m_1,m_2)}, \\ -\bigcup_i U_{n,m_i} &= U_{n,\min\{m_i\}}. -\end{align*} -Define the map $f_n \colon X_n \to X_{n+1}$ by -$$(x,\varepsilon) \mapsto \left\{\begin{array}{ll} (x,\varepsilon), & x > n, \\ (n+1,\varepsilon), & x = n. \end{array}\right.$$ -This is continuous since $f_n^{-1}(U_{n+1,m})$ equals $U_{n,m}$ if $m > n+1$ and $U_{n,n}$ if $m = n+1$. Let $X_\infty$ be the colimit of this diagram. -Since the elements $(x,\varepsilon), (y,\varepsilon) \in X_n$ map to the same element in $X_{\max(x,y)}$, we conclude that $X_\infty$ is the two-point space $\{0,1\}$, where the map $X_n \to X_\infty = \{0,1\}$ is the second coordinate projection. Moreover, the colimit topology on $\{0,1\}$ is the indiscrete topology. Indeed, neither $\mathbb N_{\geq n} \times \{0\} \subseteq X_n$ nor $\mathbb N_{\geq n} \times \{1\} \subseteq X_n$ are open. -Proof of Proposition. It's easy to check that finite discrete spaces are compact: any map out of them is continuous, and finite sets are compact in $\operatorname{\underline{Set}}$. -Conversely, if $X$ is compact, then $X$ is finite by the Lemma. Let $U \subseteq X$ be any subset, and let $f \colon X \to X_\infty = \{0,1\}$ be indicator function $\mathbb 1_U$; this is continuous because $X_\infty$ has the indiscrete topology. Since $X$ is a compact object, there exists $n \in \mathbb N$ such that $f$ comes from a map $g \colon X \to X_n$. Let $h \colon X \to X_n \to \mathbb N_{\geq n}$ be the first coordinate projection, i.e. -$$g(x) = \left\{\begin{array}{ll} (h(x),1), & x \in U, \\ (h(x),0), & x \not \in U. \end{array}\right.$$ -Let $m \in \mathbb N_{\geq n}$ be a number larger than $h(x)$ for all $x \in X\setminus U$ (we can do this because $X$ is finite). Then $g^{-1}(U_{n,m}) = U$, hence $U$ is open. Since $U$ was arbitrary, we conclude that $X$ is discrete. $\square$ - -Remark. We have only used the surjectivity part of the natural map -\begin{equation} -\operatorname{colim}_i \operatorname{Hom}(X,X_i) \to \operatorname{Hom}(X,\operatorname{colim}_i X_i).\label{1}\tag{1} -\end{equation} -In particular, we see that in $\operatorname{\underline{Top}}$, surjectivity for all systems $X_i$ implies injectivity for all systems. This is not completely formal; for example for the system $X_n$ above, the maps -$$f,g \colon \mathbb N^{\operatorname{disc}} \to X_0$$ -given by $f(x) = (x,0)$ and $g(x) = (x+1,0)$ give the same morphism $\mathbb N^{\operatorname{disc}} \to X_\infty$, but they don't give the same morphism $\mathbb N^{\operatorname{disc}} \to X_n$ for any $n$. Thus, the map in (\ref{1}) is not injective in general. I don't know in what generality surjectivity implies injectivity.<|endoftext|> -TITLE: Interpretation of the cohomology of compact lie groups and their classifying spaces in DAG? -QUESTION [12 upvotes]: I'll be using homological grading throughout this question. -Let $G$ be a compact connected lie group. The following isomorphisms are classical and can be proven using several methods: -$$H^{\bullet}(BG;\mathbb{R}) \cong Sym^{\bullet}(\mathfrak{g}^*[2])^G$$ -$$H^{\bullet}(G; \mathbb{R}) \cong Sym^{\bullet}(\mathfrak{g}^*[1] -)^G$$ -Where on the RHS we take invariants w.r.t. the adjoint action of $G$ on the lie algebra. Replacing $G$ with its complexification (which is the complex points $\mathbb{G} (\mathbb{C})$ of a connected semi-simple complex algebraic group) doesn't change the cohomology so let us do this. Next let's interpret the graded rings above as (derived if you like) prestacks (functors $CommAlg_{\mathbb{C}} \to \infty$-groupoids). Translating to this language we get the following (which is just a different way to notate the above): -$$SpecH^{\bullet}(B\mathbb{G}(\mathbb{C});\mathbb{C}) \cong \mathfrak{g}_{\mathbb{C}}[-2] \text{ //}G \cong \mathbb{T}_{B\mathbb{G}}[-1] \text{ //} \mathbb{G}$$ -$$Spec H^{\bullet}(\mathbb{G}(\mathbb{C}); \mathbb{C}) \cong \mathfrak{g}_{\mathbb{C}}[-1] \text{ //}G \cong \mathbb{T}_{B\mathbb{G}} \text{ //} \mathbb{G}$$ -I find it weird that both sides of the equation naturally take as input a connected semisimple complex algebraic group $\mathbb{G}$ but the isomorphism itself is of a transcendental nature (by passing through the real points of the compact real form as a topological group). This mysterious relationship does not manifest itself at the level of the proofs of these statement (at least in the proofs I know). It would be nice to know whether these interpretations can be turned to proofs (at least giving one of the results given the other one): - -Question 1: Is their a higher mechanism at work here? Phrased differently: Are these isomorphisms a formal consequences of the formalism of DAG together with some basic facts about groups? - -REPLY [9 votes]: I don't know of a DAG mechanism that implies these statements, any more formally than the standard proofs -- i.e., we reduce to a maximal torus, where the statement follows from the shape of the cohomology of the circle. -There are higher mechanisms though of which this statement is a hint. On one side you have topology (of BG), on the other side (derived) algebraic geometry -- suggesting that rather than something formal this is a shadow of mirror symmetry. Indeed the statement is [essentially equivalent to] the mirror symmetry between the G-equivariant A-model on a point and the B-model into a shift of the Cartan mod Weyl group. That is probably a natural setting for this question. -Or if you prefer, the statement can be seen (at the risk of tremendous overkill) as a feature of geometric Langlands (roughly the case of ${\mathbf P}^1$), specifically of the wonderful derived Geometric Satake theorem of Bezrukavnikov-Finkelberg https://arxiv.org/abs/0707.3799 (though of course the basic statements above are used in the proof!) -Namely there's an equivalence of monoidal dg categories -$$D_{LG_+}(LG/LG_+) \simeq D_{coh}({\mathfrak g}^{\vee,\ast}[2]/G^\vee)$$ -between equivariant sheaves on the affine Grassmannian for $G$ and equivariant coherent sheaves on the coadjoint representation of the Langlands dual. Your statement is an "affinization" of this -- describing modules for the endomorphisms of the unit (or natural coefficient rings) on both sides -- on the left there's the equivariant category -$$D_{LG_+}(pt)=D_G(pt)=D(BG)$$ -and on the right there's the affinization of the coadjoint quotient, -$${\frak g}^{\vee,\ast}[2]//G^\vee \simeq {\frak h}^{\vee,\ast}[2]//W -\simeq {\frak h}[2]//W\simeq {\frak g}[2]//G.$$<|endoftext|> -TITLE: indecomposable module over a local ring -QUESTION [7 upvotes]: I ask this in mathematics for some days.it doesn't have an answer up to now. https://math.stackexchange.com/questions/2565828/indecomposable-module-over-a-local-ring -As we all know, for an arbitrary ring, any finite length module is indecomposable iff the endomorphism ring is local. -I have a question: - -If $R$ is local ring, $M\in R$-$\mathrm{Mod}$, is $M$ indecomposable iff $\mathrm{End}(M)$ is local? - -Of course, if $\mathrm{End}(M)$ is local, then we have $M$ is indecomposable. For the converse is there a counterexample？Especially in commutative Noetherian local rings. -any help will be appreciated! - -REPLY [7 votes]: If $M$ is allowed to be infinitely generated, then there are counterexamples even for finite dimensional local algebras. -Let $R=\mathbb{C}[x,y]/(x,y)^2$, a three-dimensional local $\mathbb{C}$-algebra. -Let $M=\mathbb{C}[t]\oplus\mathbb{C}[t]$ as a vector space, with $R$-module structure given by $\left(p(t),q(t)\right)x=\left(0,p(t)\right)$ and $\left(p(t),q(t)\right)y=\left(0,tp(t)\right)$. -Then an easy calculation shows that the $R$-module endomorphisms of $M$ are the maps of the form -$$\left(p(t),q(t)\right)\mapsto \left(f(t)p(t),f(t)q(t)+\varphi\left(p(t)\right)\right),$$ -where $f(t)\in\mathbb{C}[t]$ and $\varphi:\mathbb{C}[t]\to\mathbb{C}[t]$ is an arbitrary linear map. -There are no non-trivial idempotent endomorphisms, so $M$ is indecomposable, but restricting $f(t)$ to any maximal ideal of $\mathbb{C}[t]$ gives a maximal right (and left) ideal of $\text{End}_R(M)$, so $\text{End}_R(M)$ is not local. -Of course, this example is not finitely generated. Swan showed that a finitely generated indecomposable module for a complete local ring has local endomorphism ring, but without completeness there are counterexamples (for local Noetherian rings) to Krull-Schmidt (which must also give examples without local endomorphism ring) in -Evans, E.Graham jun., Krull-Schmidt and cancellation over local rings, Pac. J. Math. 46, 115-121 (1973). ZBL0272.13006, -the first of which is also credited to Swan.<|endoftext|> -TITLE: Mathematical research interrupted by a war -QUESTION [43 upvotes]: I am not sure that this is appropriate at MO, so if not, please, delete this. -This is inspired by David Hansen's question where he asks about mathematics done during the WWII. I would like to ask the opposite question: - -what are some examples of mathematical research interrupted by a war? - -Everyone is aware of the terrible damage inflicted by the war on the Polish mathematical school. The dramatic destinies of Stefan Banach (who lived in very difficult conditions during the WWII and died soon after it), Juliusz Schauder (killed by Gestapo), Józef Marcinkiewicz (killed by NKVD) and of many others have much influence on the conscience of mathematicians in Central Europe (including Russia, and I believe, not only here). -When I was a student an idea was popular in Soviet Union that war moves science. I must confess, I am a partisan of the opposite one: war kills science. I would be grateful to people here who would share their knowledge and give illustrations. -P.S. By war I mean any war, not necessarily WWII. - -REPLY [6 votes]: Eugenio Elia Levi's death during WWI had a profound impact on Italian mathematics. Though still quite young, Levi had signalled himself as a mathematician willing to recognize the relevance of the then emerging -Lie theory, following the work of his mentor Bianchi. When he was just 22 he published the paper in which what will thereafter be called Levi factors were introduced. -Moved by very strong patriot feelings, despite the fact that he could have been exempted from military duties, he voluntereed and was killed by a stray bullet after Caporetto's defeat. -Lie theory in Italy remained largely unexplored, with Luigi Bianchi being one of the few trying to convince his students of its relevance. -A "missed opportunity" for the Italian mathematics.<|endoftext|> -TITLE: Infinite, residually-finite, finitely generated, groups with very limited composition factors of their finite images. -QUESTION [5 upvotes]: Let $S$ be a finite collection of finite non-abelian simple groups, for example $S=\{ A_5 \}$. I am looking for infinite residually-finite finitely-generated groups $G$ such that the composition factors of each finite image of $G$ are in $S$. Are there such examples? - -REPLY [6 votes]: Theorem 1.27 of https://arxiv.org/pdf/math/0510294.pdf gives 63-generated, residually finite, just infinite groups with the property you want for any set of simple nonabelian groups. The construction is due to Dan Segal. See Dan Segal, The finite images of finitely generated groups, Proc. London Math. Soc. (3) 82 (2001), no. 3, 597–613. -Original Answer. Peter Neumann constructed finitely generated just infinite branch groups all of whose finite quotients are iterated permutational wreath products of $A_6$ and hence have only $A_6$ as a composition factor. See example 1.6.6 of https://arxiv.org/pdf/math/0510294.pdf.<|endoftext|> -TITLE: Question about denoting/designating of algebraic structures -QUESTION [5 upvotes]: I saw this image on Wikipedia (Template:Group-like structures, current revision): - -Since there are five "properties" that we can have (in this context), namely: totality, associativity, identity, invertibility and commutativity, there are $\binom {5}{1}+ \binom{5}{2}+ \binom {5} {3} + \binom {5}{4} + \binom {5}{5}=5+10+10+5+1=31$ possible structures that we can have and there are only $10$ of them listed here, which are well known. -It could be that among other $21$ structures not listed here some are also studied and have a name but as the number of operations on a structure increases or/and as a number of properties of operations increases, then it would be very hard to remember the names of all structures, so I would like to propose the following: -If we denote totality by $T$, associativity by $A$, identity by $I$, invertibility by $In$ and commutativity by $C$, then we could denote a group as a $(T,A,I,In)$ structure. where it is immediately seen that group has properties of totality, associativity, has an identity, and every element has an inverse. -Another example, loop would be $(T,I,In)$ structure. -All of this can be made more precise but it is just some start, so maybe more rigour is needed. -This could be more generalized to structures over which there are more than one operation defined, but I will not do that here, I just want to ask: - -Is this a well-known approach to designation of algebraic structures? Has anything similar been done? Would this lead to simplifications of some sort? What are your thoughts about this approach? - -REPLY [3 votes]: As Gerhard Paseman mentions, you should definitely look up Universal Algebra. Also worth a read is the wikipedia page Outline of algebraic structures. You will encounter a link to Jipsen's list of algebraic structures -- currently at 328, but that cheats a little because it includes a lot of structures with many symbols (and even many sorts). My old question on this topic similarly didn't get any takers. -If you continue with this line of thinking, you will eventually ask the same question of other structures -- in particular, of logics. Because, well, logics can also be seen as theories, in a meta-logic. And there you'll find John Halleck's list invaluable. -But these are 'just lists', and do not have kind of systematic exploration. As far as being systematic, the only attempt that I am aware of is ⟨warning, self-promotion ahead⟩ my work on the MathScheme Library, as part of the MathScheme project. -Long story short: at the 'bottom' of the hierarchy, almost every theory has been given a name. Every time I think I've found the 'smallest' one which is new, I subsequently find that it has been looked at before. The only ones that seem to be nameless are ones with quite odd. For example, theories involving $\left(x*y\right)*x = x*y$ as an (apparently nameless?) axiom.<|endoftext|> -TITLE: Does birational imply D-equivalent? -QUESTION [5 upvotes]: It is well-known that there are Calabi-Yau's who are not birational but are derived equivalent. However I am interested in seeing D-equivalence as a weakening of birationality. - -Q. If $X$ and $Y$ are smooth projective varieties and $X$ is birational to $Y$, does it follow that $D(X)\simeq D(Y)$? - -My guess would be that no such result exists so far, but I cannot see a good obstruction to it happening. I know a lot of work is done to show that flips, flops and blowups are derived equivalent so it would mean that if the result above were true, the later would be easier to prove. - -REPLY [8 votes]: The answer to the question as literally asked is "no". If $X$ is an surface and $Y$ is $X$ blown up at a point, then the rank of the Grothendieck $K_0$-group of $Y$ is one larger than that of $X$. Since $K_0$ is recoverable from the derived category, this shows that $\mathcal{D}(X) \not \cong \mathcal{D}(Y)$. -This is why people ask not only that $X$ and $Y$ be birational, but impose conditions on the canonical class, such as that $X$ and $Y$ are Calabi-Yau, or the notion of $K$-equivalence in Kawamata. - -REPLY [6 votes]: This is conjectured to hold for varieties with trivial canonical bundle (Bondal-Orlov), and is known to be true for Calabi-Yau 3-folds (Bridgeland established this in his seminal paper https://arxiv.org/abs/math/0009053). In higher dimensions this is one of the basic problems in the field, and, to the best of my knowledge, very little progress has been made towards its solution. -Let me relate it to another question. A folklore conjecture is that derived invariant varieties have the same Hodge numbers (this is proven in dimension at most 3 by Popa and Schnell, an independent proof was given by Abuaf). However, birational varieties need not have the same Hodge numbers; the easiest counterexample is $X=\mathbf{P}^2$ and $Y=\mathbf{P}^1\times\mathbf{P}^1$. These are birational but cannot be derived equivalent (by Bondal-Orlov they would have to be isomorphic!).<|endoftext|> -TITLE: periods of higher weight modular forms -QUESTION [10 upvotes]: For now let $f\in S_2(\Gamma_0(N))$ and define $I(\alpha,\beta)=\int_\alpha^\beta f(z)dz$. If $M\in\Gamma_0(N)$ it is immediate to prove that $I(\alpha,M(\alpha))$ is independent of $\alpha$: this uses (1) modularity which shows that $I(M(\alpha),M(\beta))=I(\alpha,\beta)$, and (2) the fact that $f$ is holomorphic, so that the evident contour integral vanishes. -My question is what happens in higher weight $f\in S_k(\Gamma_0(N))$, where $f(z)dz$ is no more an invariant differential. It is natural to consider the period polynomial $I(\alpha,\beta)=\int_\alpha^\beta(X-z)^{k-2}f(z)dz$, and modularity again shows $I(M(\alpha),M(\beta))=I(\alpha,\beta)|_{2-k}M^{-1}$. -But I cannot seem to use holomorphy to deduce something analogous to the weight $2$ case. Am I missing something ? - -REPLY [3 votes]: In the case $k >2$, the integral $I(\alpha,M(\alpha))$ does indeed depend on $\alpha$. -I think the right point of view is using group cohomology as follows. Let $f \in S_k(\Gamma_0(N))$ and let $\alpha \in \mathcal{H}$. Using your notations, the map $M \in \Gamma_0(N) \mapsto I(\alpha,M(\alpha))$ is a 1-cocycle on $\Gamma_0(N)$ with values in the space of polynomials $V_k = \mathbf{C}[X]_{\leq k-2}$. If you change $\alpha$ then the cocycle changes by a coboundary since by Stokes' formula -\begin{equation*} -I(\beta,M(\beta))-I(\alpha,M(\alpha)) = I(M(\alpha),M(\beta)) - I(\alpha,\beta) = I(\alpha,\beta) | (M^{-1}-1) -\end{equation*} -as you indicated in your question. -So there is a well-defined cohomology class $\phi_f \in H^1(\Gamma_0(N),V_k)$. If I recall correctly, the original version of the Eichler-Shimura isomorphism (maybe only for $N=1$?) says that $S_k(\Gamma_0(N))^2$ is isomorphic to $H^1(\Gamma_0(N),V_k)$.<|endoftext|> -TITLE: Does $V = \textit{Ultimate }L$ imply GCH? -QUESTION [19 upvotes]: In his Midrasha Mathematicae lectures ("In Search of Ultimate $L$", BSL 23 [2017]: 1–109), Woodin notes that $V = \textit{Ultimate }L$ implies $\textrm{CH}$ (Theorem 7.26, p.103). Is it known whether $V = \textit{Ultimate }L$ implies $\textrm{GCH}$? - -REPLY [8 votes]: During this year's conferene on inner model theory in Münster, Gabriel Goldberg proved that the so-called Ultrapower Axiom implies that $\mathrm{GCH}$ holds above a supercompact cardinal (and since then lowered the bound to a strongly compact cardinal). It seems very likely (it might even be known) that $\mathrm{Ultimate } \ L$ satisfies this requirement. Hence, given enough large cardinals, it will satisfy $\mathrm{GCH}$ at least on a tail end. -For more information, see G. Goldberg. Strong Compactness and the Ultrapower Axiom.<|endoftext|> -TITLE: Characterizing freely adjoining K-filtered colimits as K-continuous presheaves -QUESTION [6 upvotes]: Note: This question has a 1-categorical and an $\infty$-categorical versions. I am interested in the $\infty$-categorical one so this is the version that I write below, but an answer for the 1-categorical version would be interesting too. -Given an $\infty$-category $\mathcal{C}$ and a collection of simplciial sets $\mathcal{I}$, there is a notion of "freely adjoining to $\mathcal{C}$ all colimits of shape $\mathcal{I}$". This is the full subcategory $P_{\mathcal{I}}(\mathcal{C})$ of all presheaves on $\mathcal{C}$ generated from the image of the Yoneda by $\mathcal{I}$-colimts (HTT 5.3.6). Now, assume that $\mathcal{I}$ is not any collection, but the collection of all $I$, such that $I$-colimits commute in spaces with all $K$-limits for all $K$ in some collection $\mathcal{K}$. e.g. -1) $\mathcal{K}$ is finite simplicial sets and $\mathcal{I}$ is filtered simplicial sets. -2) $\mathcal{K}$ is finite discrete simplicial sets and $\mathcal{I}$ is sifted simplicial sets. -Assuming $\mathcal{C}$ admits all $K^{op}$-colimits for all $K\in \mathcal{K}$, in these two examples it is known that $P_{\mathcal{I}}(\mathcal{C})$ can be described as the full subcategory $P^{\mathcal{K}}(\mathcal{C})$ of the $\infty$-category of presheaves spanned by the ones which preserve $\mathcal{K}$-limits (for (1) HTT 5.3.5.4 and for (2) HTT 5.5.8.15). The proofs of these two facts seem to be quite different, so I am wondering -Question: Under the above conditions on $\mathcal{I}$, $\mathcal{K}$ and $\mathcal{C}$, does $P_{\mathcal{I}}(\mathcal{C})$ always coincide with $P^{\mathcal{K}}(\mathcal{C})$? -It is easy to see that we have an inclusion of $P_{\mathcal{I}}(\mathcal{C})$ into $P^{\mathcal{K}}(\mathcal{C})$, but the converse seems more complicated. I also have a creepy feeling that the words "sound doctrine" may come up... - -REPLY [3 votes]: For the 1-categorical case, it seems to be indeed a question of soundness. More precisely, the condition that $P_{\cal I}({\cal C}) = P^{{\cal K}}({\cal C})$ is equivalent to the condition that every ${\cal K}$-limit preserving presheaf $F:{\cal C}^{op} \to {\rm Set}$ is an ${\cal I}$-colimit of representables, or, in this case, a ${\cal K}$-filtered colimit of representables. Looking in this paper we see that this indeed holds whenever ${\cal K}$ is sound (theorem 2.4), and that furthermore, this being the case appears to be the main motivation for the definition of soundness. For example, it seems to me that $P_{\cal I}({\cal C})$ doesn't contain the terminal presheaf in the non-sound case described in Example 2.3 (vi) of loc.cit. -I would say that it's quite likely that the $\infty$-categorical case will behave similarly, once one defines soundness appropriately (maybe replace "connected" with "weakly contractible" in Definition 2.2 of loc.cit)?<|endoftext|> -TITLE: Which topological manifolds do not correspond to strongly Hausdorff locales? -QUESTION [9 upvotes]: I'm toying with the idea of using locales as a way to define topological manifolds without beginning with points, largely for philosophical reasons. -In this context I think I want to redefine a topological manifold as a locale that is paracompact and strongly Hausdorff in the sense of Johnstone Stone Spaces p.82. -My understanding is that this may make my version of the theory strictly smaller than the traditional one. That might be OK for my purposes, depending on what kinds of objects are "missing". -My question: is there any way to characterise those paracompact topological spaces that are Hausdorff but whose corresponding locales are not strongly Hausdorff? -I'm just beginning to learn about locales so would also appreciate answers that explain why my question (or whole project) is completely misconceived... - -REPLY [10 votes]: Let me expand a bit my comment as this is a rather subtle property. -As I said any locally compact Hausdorff topological space is a strongly hausdroff locally compact locales. (and under the axiom of choice the two notion are completely equivalent) So this does not exaclty answer the precise question you ask, but as it applies to all topological manifold, it should be sufficient for your purpose. -The result you need for this is lemma C4.1.8 of sketches of an elephant which says that: -Lemma : (C4.1.8 in Sketches) The product of a locally compact spatial locale with a spatial locale is spatial. -The reason that it gives you what you want is because the only reason why Hausdroff and strongly Hausdorff are differents is because in general the product $X \times X$ in the category of locales can be non-spatial even if $X$ is spatial, hence asking that the diagonal map $X \rightarrow X \times^{top} X$ in the category of topological space is closed only say that the map from $X$ to the largest spatial sublocale of $X \times X$ is closed. But this largest spatial sublocale can be itself not closed and so it does not implies that the locale is strongly Hausdorff. But because of the lemma above, if $X$ is locally compact then $X \times X$ is spatial and so the product in the category of topological space and the product in the category of locales are the same. -This is enough for the treatment of manifolds. -I wouldn't be surprise if results of this kind also holds under only paracompactness assumption. But having only studied the constructive theory of locales I know very little of paracompactness in this framework so I leave to someone else to comment or answer about this. -Side note : I would like to add a last remark regarding your project: classically (i.e. assuming the axiom of choice and the law of excluded middle) there is a complete equivalence between locally compact strongly Hausdorff locales and locally compact Hausdorff topological spaces, so you will have no problem formulating differential geometry in this framework and it will not say anything new. -In constructive settings this is no longer true. In fact if you look at section D4.7 of Sketches of an elephant you will see a very interesting feature: -In constructive mathematics one can define a ``locale of real numbers'' which is always locally compact and Hausdorff but which is in general not spatial, i.e. not the same as the topological spaces of real numbers. And constructively The topological space of real numbers can even fail to be locally compact ! In fact it will be isomorphic to the locale of real number if and only if it is locally compact. -So clearly, if one wants to do differential geometry in constructive mathematics one HAS to use locales instead of topological space (or we need to say goodbye to local compactness... which would basically the end of everything in differential geometry) and I have never seen anyone develop that point of view. -(Also note that this is not related to AC, but really to LEM: the law of excluded middle alone is already enough to prove that the topological space of real number is locally compact/the locale of real number is spatial)<|endoftext|> -TITLE: Local endomorphism rings and indecomposable modules -QUESTION [7 upvotes]: Given a finite dimensional algebra $A$. -For which such algebras is it true that a (not necessarily finitely generated) module $M$ is indecomposable iff its endomorphism ring is local? This is for example true for representation-finite algebras. Are there other examples? - -REPLY [8 votes]: Probably there are no other examples. -The paper -Brenner, Sheila; Ringel, Claus Michael, Pathological modules over tame rings, J. Lond. Math. Soc., II. Ser. 14, 207-215 (1976). ZBL0356.16010. -doesn't explicitly discuss indecomposable modules with non-local endomorphism ring, but does discuss pathological phenomena such as having indecomposable modules $M$, $N_1$ and $N_2$ with $M\oplus N_1\cong M\oplus N_2$, but $N_1\not\cong N_2$, which couldn't happen if $M$, $N_1$ and $N_2$ all had local endomorphism rings. -They remark that such pathologies were known to occur for $k\langle x,y\rangle$-modules and prove that they do for $k[t]$-modules (which also proves it for $k\langle x,y\rangle$-modules, of course). -Therefore, if $A$ is a finite dimensional algebra with a representation embedding of $\text{Mod-}k\langle x,y\rangle$ or $\text{Mod-}k[t]$ into $\text{Mod-}A$ (i.e., an exact functor preserving non-isomorphism and indecomposability), then $A$ must have indecomposable modules with non-local endomorphism rings. -For any wild algebra $A$ there is a representation embedding of $\text{Mod-}k\langle x,y\rangle$ into $\text{Mod-}A$. -In the same paper they also remark that for every known tame algebra $A$ there is a representation embedding of $\text{Mod-}k[t]$ into $\text{Mod-}A$. But that was 1976, and maybe more is known now.<|endoftext|> -TITLE: Union of pairwise almost disjoint sets -QUESTION [5 upvotes]: I asked this question on stackexchange : -https://math.stackexchange.com/questions/2570199/union-of-pairwise-almost-disjoint-sets -but I got no answer after 24 hours, so I ask it here. -Let $r$ and $n$ be two natural numbers, with $n \geq 2$. What is known about the least possible cardinality of the union of $r$ sets with cardinality $n$, such that any two of these sets have at most one element in common ? -Let $f(r, n)$ denote this least possible cardinality. -If $n = 2$, the condition that two of the sets have at most one common element amounts to say that these sets are distinct, thus $f(r, 2)$ is the least natural number $k$ such that ${k \choose 2} \geq r$. -In the general case, each of the $r$ sets contains $n \choose 2$ pairs and two sets never contain a same pair, thus the union of the $r$ sets contains at least $r {n \choose 2} $ pairs, thus -(1) $f(r, n) \geq k(r, n)$, where $k(r, n)$ denotes the least $k$ such that ${k \choose 2} \geq r {n \choose 2}$. -This is not optimal, in the sense that $f(r, n) > k(r, n)$ happens. For example, $f(2, 3) = 5$ and $k(2, 3) = 4$. -I have two questions : -1° do you know a better minoration of $f(r, n)$ than (1) ? -2° (1) gives $f(30, 4) \geq 20$; can it be proved that $f(30, 4) \geq 21$ ? -Thanks in advance. - -REPLY [5 votes]: Your question is a special case of the set packing problem. My reference for this stuff (which I know nothing about) is A. E. Brouwer, Packing and covering of $\binom kt$ sets, in: A. Schrijver (ed.), Packing and Covering in Combinatorics, Mathematical Centre Tracts 106 (1979), 89–97. (This is obviously not the last word on the subject, but for some reason I happen to have a copy.) Paraphrasing the definition from Brouwer: - -Let $0\le t\le k\le v,$ and define - $$D(t,k,v)=\max\{|\mathcal B|:\mathcal B\subset\mathcal P_k(v)\text{ and no two elements of }\mathcal B\text{ have }t\text{ points in common}\}$$ - where $\mathcal P_k(v)$ is the collection of $k$-subsets of a fixed $v$-set. - -In this notation, your $f(r,n)$ is the least $v$ such that $D(2,n,v)\ge r.$ -According to Brouwer's 1979 survey, the exact values of $D(2,k,v)$ are known for $k=3$ and $k=4.$ For $k=4$ the results are attributed to A. E. Brouwer, Optimal packings of $K_4$'s into a $K_n,$ Math. Centre Report no. zw 92/97, Math. Centre, Amsterdam, 1977; J. Combinatorial Theory (A) 26 (1979), 278–297, and are as follows (paraphrased): - -For $v\ne8,9,10,11,17,19,$ we have - $$D(2,4,v)=\left\lfloor\frac v4\left\lfloor\frac{v-1}3\right\rfloor\right\rfloor-\varepsilon,$$ - where $\varepsilon=1$ for $v\equiv7$ or $10\pmod{12}$ and $\varepsilon=0$ otherwise; -and for $v=8,9,10,11,17,19$ we have $D(2,3,v)=2,3,5,6,20,25.$ - -In particular, $D(2,4,19)=25$ and $D(2,4,20)=30,$ so $f(30,4)=20.$<|endoftext|> -TITLE: Murnaghan-Nakayama rule when all cycles have same size -QUESTION [5 upvotes]: Let $\lambda \vdash nk$. Let $n^k$ denote the partiton with $k$ parts of size $n$. We can compute $\chi^\lambda(n^k)$ by using the Murnaghan-Nakayama rule, as a signed sum over border-strip tableaux, (shape $\lambda$ and border strips of size $n$). -Now, by checking several cases on the computer, it seems like each term in the (Murnaghan-Nakayama) sum has the same sign. Combinatorially, it reduces to proving the following: -Define the height of a border-strip to be the maximal row index that it touches, minus the minimal row index. The height of a border-strip decomposition of the diagram $\lambda$ is the sum of the heights of the border-strips. Show that the parity of this sum is independent of the actual decomposition (it only depends on $\lambda$ and $n$). -I am looking for a reference of this (I do think I can in fact prove a stronger statement, but I have not worked out the details yet). - -REPLY [3 votes]: This is corollary 10 in "A bijection proving orthogonality of the characters of $S_n$", by Dennis E. White.<|endoftext|> -TITLE: Decoupling, efficient congruencing and Vinogradov's main theorem -QUESTION [6 upvotes]: It seems to be word in the generic corridor that decoupling (as in Bourgain-Demeter-Guth) and efficient congruencing (Wooley) are deeply related, and even that they are deep down the same thing - with efficient congruencing being the p-adic approach and decoupling the real one. -Has this remained at a pure "moral" level - an impressionistic statement based on many similarities? Or is it an insight that has given rise to an actual dictionary between the two approaches? What has been actually carried out? - -REPLY [5 votes]: Shaoming Guo, Zane Kun Li, Po-Lam Yung, and I found a bilinear proof of the decoupling inequality for the moment curve that uses the same inductive procedure as Wooley's efficient congruencing proof (Zane was the graduate student mentioned in Terry's answer from 2017). -The overall structure of the (Bourgain-Demeter-Guth) multilinear and the bilinear proofs of decoupling turned out to be quite similar, and both can be written concisely using some definitions introduced in a blog post by Tao. -However, as opposed to the multilinear argument, we do not use any Kakeya-type inequalities.<|endoftext|> -TITLE: Bohr compactification as a topological compactification -QUESTION [6 upvotes]: Let $G$ be a locally compact Hausdorff group. Denote its Bohr compactification by $bG$. -Despite group structure, $G$ has several (Hausdorff) compactifications that, in a sense, the smallest one is the one-point compactification, and the largest one is the Stone-Čech compactification. -Is $bG$ is isomorphic to one of the (topological) compactifications of $G$ as a topological space? In case of a positive answer, to which one? I don't know the answer even in the case of $G=\mathbb{R}$. - -REPLY [3 votes]: As Francois Ziegler answered, for a locally compact group $G$, the Bohr compactification of $G$ is a compactification in the usual sense iff $G$ is compact. This is true with no further restrictions. -Recall that the forgetful functor from compact groups to topological groups has a left adjoint functor, denoted here $b$. For any topological group $G$, the identity in $\text{Hom}(bG,bG)$ corresponds to an element of $\text{Hom}(G,bG)$ called the unit, $u:G\to bG$. -It is easy to check that $u(G)$ is dense in $bG$ (this is usually seen directly from the construction of $b$, but also follows from the abstract nonsense). -The homomorphism $u:G\to bG$ is known as the -Bohr compactification of $G$. -This terminology is a bit unfortunate, as $u$ is not a compactification in the sense usually used in topology, unless $G$ was compact to begin with -(in fact, $u$ is rarely injective - groups for which $u$ is injective are called "maximally almost periodic"). -By a "compactification in the usual sense" of a locally compact space $X$ one means a continuous map $f:X\to Y$ where $f(X)$ is dense in $Y$ and $f:X\to f(X)$ is a homeomorphism. In that case $f(X)$ is necessarily open in $Y$ (this an easy exercise). -In case $X$ and $Y$ were groups and $f$ a group homomorphism, $f(X)$ was also closed, as any open subgroup is closed, thus $f(X)=Y$ by density. So $X$, being homeomorphic to $Y$, must have been compact to begin with. -More generally, let me note that any locally compact subgroup of any topological group is necessarily closed (and if the ambient group is locally compact, a subgroup is locally compact iff it is closed).<|endoftext|> -TITLE: Constructive proofs of existence in analysis using locales -QUESTION [13 upvotes]: There are several basic theorems in analysis asserting the existence of a point in some space such as the following results: - -The intermediate value theorem: for every continuous function $f : [0,1] \to \mathbb{R}$ such that $f(0) \leq 0$ and $f(1) \geq 0$, there exists a point in the space $f^{-1}(0)$. -Brouwer's fixed-point theorem: for every function $f$ on (say) the closed disk $D$, there exists a point in the space $\{ x \in D\ |\ f(x) = x \}$. -The fundamental theorem of algebra: for every nonconstant polynomial $p$ with complex coefficients, there exists a point in the space $p^{-1}(0)$. - -These results are not provable constructively as stated. Sometimes such theorems can be reformulated so that they become provable. These modifications are all ad hoc, but there is one general idea that we can apply in all of these cases. First, we need to replace spaces with corresponding locales. Moreover, since the existence of points of locales is often not provable constructively, we should replace the condition "has a point" with "is nontrivial". - -Are these theorems provable constructively when formulated in this way? - -Note that if this is true, then this implies classical results since assuming LEM these locales are spatial. - -REPLY [5 votes]: I claim that the following result have constructive* proof: -1) Let $f : [0,1] \rightarrow \mathbb{R}$ be a uniformly continuous function such that $f(0)\leqslant 0$ and $f(1) \geqslant 0$ then (as a locale) $\{x, |f(x)=0 \}$ is not empty. -2) Let $f :D \rightarrow D$ be a uniformly continuous function then the locale $\{x | f(x)=x\}$ is not empty. -3) let $p$ be a complex polynomial, with at least one coeficient $|a_i|>0$ other than the constant coeficient, then the locale $\{x |p(x)=0 \}$ is not empty. -each time " $F$ not empty" really just mean that $F = \emptyset \Rightarrow False$ -and the "*" on constructive refer to the fact that these are proof relying on Barr's theorem, cf below. -All of them follow the same scheme which I will explain below (and I will explain precesely what I mean by 'constructive' as there is a small subtleties here) -A first remark: I added "uniformly continuous" because, asking that a function from $[0,1]$ or $D$ to $\mathbb{R}$ extend to a function between the corresponding locales is the same as saying that the function is uniformly continuous. So if every time you interpret those as functions on the corresponding locale, then you can remove the uniform continuity hypothesis, and I only put it to remind of this. -Now let's move to the proof of those claim. They point is to prove them internally in a Grothendieck $T$ using Barr's covering theorem which assert that every topos $T$ admit a surjection from a topos $E \twoheadrightarrow T$ such that the internal logic of $E$ satisfies LEM and AC. (surjection mean that $f^*$ is conservative) -More precisely: - -Every Grothendieck topos admit a surjection $E \twoheadrightarrow T$ where $E$ is a boolean locale. (hence its logic satisfies LEM) -Assuming AC (in the topos of set) the internal logic of every boolean locale satisifes AC. - -The proof strategy for each points is the following: -You start with the problem in a topos $T$, you pull it back to a problem in $E$ where you now how to prove it because you have LEM and AC in $E$. And then you are able to deduce the result in $T$ using that $E \twoheadrightarrow T$ is a surjection. I will explain in more detail at the very end of my answer how this is done. -This gives a non constructive proof that these theorems hold in every Grothendieck topos. If you have a proof of the claims $1,2$ or $3$ that uses LEM but not AC, then you actually obtain a constructive but non predicative proof that it holds in every Grothendieck topos, and in particular in the topos of sets. -Moreover in each case you can construct a "classyfing topos for the assumption", for exemple for $(1)$ one can construct the classifying locales of all function $f:[0,1] \rightarrow \mathbb{R}$ such that $f(0) \leqslant 0$ and $f(1) \geqslant 0$ simply because $[0,1]$ is exponentiable in the category of locale and similarly. You can then apply the argument above to these classyfing topos for the assumption to deduce that the theorem holds in these specific toposes (locales). As classyfing topos have a "syntactical" description in terms of geometric logic, this actually implies that each of the claim $1$, $2$ and $3$ have a proof purely in the language of geometric logic. this puts us is the following situation: -We have given a non-constructive proof that these claim have constructive and even predicative proofs. -Whether you accept that as a constructive proof depends on your philosophical stance: - -If you are interested in constructivity only because of topos theoretic interpretation (or other models of constructive mathematics) then you should accept that as a constructive proof. -If you are interested in constructivity for philosophical reason, this shouldn't be a satisfying proof. -If you are interested in "constructive proof <-> algotrithm" correspondence, then this is only partially satisfying: it says that "there exists an algorithm" that will do the corresponding task, but it does not say what the algorithm is. - -And in all case, it does mean that a constructive proof do exists, so we should be able to find it. (and if those results are important to you or you plan to publish somehting that rely on them I would suggest to try to find actual construct proof for them instead of this argument, it is a lot easier to find a proof once one knows that it exists !) -Finally, let me clarify the proof that the validity of such statement in $E$ implies their Validity in $T$. I will do that for point $2$ in full detail, but this is exactly the same for all of them. -First, if we have $p:E \rightarrow T$, and interally in $T$ some function $f:D \rightarrow D$ then this can be pulled back as: -$p^{\sharp} f : p^{\sharp} D \rightarrow p^{\sharp} D$ in $E$, where $p^{\sharp}$ is how I denote the pullback of locales along a geometric morphism. And (for the locales of real numbers) $p^{\sharp} D $ is just $D$ so we have the exact same situation in $E$. -The sublocale of $D$, $\{ x \in D |f(x)=x \}$ is defined as a certain pullback so it is preserved by pullback as well: -$p^{\sharp} (\{ x \in D | x= f(x) \}) = \{ x \in p^{\sharp} D | x = p^{\sharp}(f)(x) \}$ -The key point is the following: if $K$ is a locale in some topos, one can consider the proposition $``K=\emptyset"$ (which, as any proposition is a subterminal object of our topos. I claim that when $K$ is a compact locale (internally in $T$) then: -$p^*(``K=\emptyset") = `` p^{\sharp} K = \emptyset" $ -intuitively, this is becasue (say when $T$ is a locale), a compact locale $K$ in $T$ is a proper geometric morphism $K \rightarrow T$, in particular, it is a closed map, hence its image is closed and the open complement of its image is the proposition in $T$, $K= \emptyset$ and this is stable under any pullback (but it only works because $K \rightarrow T$ is stably closed). -For a more precise proof, it is essentially the Beck-Chevalley condition for proper map of toposes (C3.2.6 in sketches of an elephant). -This implies that if $K$ is some locale in $T$ such that for $p : E \twoheadrightarrow T$ a surjection, if $ p^\sharp K = \emptyset \Rightarrow False $ in $E$ than $K= \emptyset \Rightarrow False$ in $T$. -Indeed $K= \emptyset \Rightarrow False$ can be written as the fact that $\emptyset \hookrightarrow `` K = \emptyset''$ is an isomorphism, and th $\emptyset \hookrightarrow `` p^{\sharp}K = \emptyset''$ is exactly the pullback of this map by $p^*$, so the conservativity of $p^*$ implies the result.<|endoftext|> -TITLE: Cryptographic Secret Santa -QUESTION [10 upvotes]: Is there a protocol for conducting a Secret Santa without a central authority? Precisely, we want to sample uniformly a permutation that has no one-cycles and reveal to each member his or her successor without revealing any other part of the permutation. Alternatively, we can uniformly sample an $n$-cycle, where $n$ is the number of participants in the Secret Santa. - -REPLY [13 votes]: Cryptographic Protocols with Everyday Objects (section 5) - -Typical cryptographic Secret Santa protocols require a fully - homomorphic encryption system. These protocols are thus suitable for - those with some planning, some expertise and access to computational - assistance, but not to a group of friends without access to computers. - Here we present an alternative solution, which does not lower the - security requirements, and does not require anything beyond card, - envelopes and pens.<|endoftext|> -TITLE: A problem involving the inverse Collatz map -QUESTION [7 upvotes]: Let $C$ be the Collatz map on the natural numbers, defined by: -$$C(n) := -\begin{cases} -n/2 & \text{if} \;n \;\text{even} \\ - (3n+1)/2 & \text{if} \;n \;\text{odd} -\end{cases}$$ - -The inverse of $C$ is: -$$C^{-1}(\{n\}) = -\begin{cases} -\{2n\} & \text{if} \;n \not \equiv 2 \pmod 3 \\ - \{2n, (2n-1)/3\} & \text{if} \;n \equiv 2 \pmod 3 -\end{cases}$$ -Let $n$ be a natural number with $n \equiv 2 \pmod 3$. Let $C^{-k}$ be $C^{-1} \circ \cdots \circ C^{-1}$ ($k$ times). -Consider the cardinals $c_1(n,k):= \vert C^{-k}(\{2n\}) \vert $ and $c_2(n,k):= \vert C^{-k}(\{(2n-1)/3\}) \vert$. -Question: Is it true that $\forall n \equiv 2 \pmod 3$, $\exists k>0$ such that $c_1(n,k) \neq c_2(n,k)$? -Remark: It is checked for $n \le 10^8$. -Assuming the answer is no, let $n$ be a counter-example. -Motivation-Question: Is $n$ also a counter-example of the Collatz conjecture? -Assuming the answer is yes, let $\alpha$ be the map defined by: -$$\alpha (n) := -\begin{cases} -1 & \text{if} \;n \not \equiv 2 \pmod 3 \\ - 1+\min\{k>0 \ \vert \ c_1(n,k) \neq c_2(n,k) \} & \text{if} \;n \equiv 2 \pmod 3 -\end{cases}$$ -By looking to the graph below, we get for example that $\alpha(8)=2$ and $\alpha(20)>3$. -A natural number $N$ is called a champion if $\forall n -TITLE: For which values of $k$ is it known that there are infinitely many $n$, such that $2^{n+k}\equiv 1\pmod{n}$? -QUESTION [9 upvotes]: I know that there are no solutions to $2^n\equiv 1\pmod{n}$ for $n>1$ and I can prove that there are infinitely many $n$ such that $2^{n+1}\equiv1\pmod{n}$. -My question is: - -Do we know other fixed values $k\in \mathbb{N}$ such that - $2^{n+k}\equiv 1\pmod{n}$ holds infinitely often? - -REPLY [10 votes]: For any $k\geq 1$, there are infinitely many solutions of the congruence $2^{n+k}\equiv 1\pmod{n}$. To see this, observe first that there is always a solution $n\geq 1$ satisfying $n+k\geq 7$. Indeed, for $k\geq 6$ this is verified by the trivial solution $n=1$, while for $1\leq k\leq 5$ it is verified by the pairs -$$ (k,n) \ = \ (1,15),\ \ (2,7),\ \ (3,5),\ \ (4,31),\ \ (5,3).$$ -Now it suffices to show that, for any fixed $k$ and for any solution $n\geq 1$ satisfying $n+k\geq 7$, there is a bigger solution $n'>n$ for the same $k$. Indeed, let $p$ be a primitive prime divisor of $2^{n+k}-1$, which exists by Zsigmondy's theorem. Then, the order of $2$ modulo $p$ equals $n+k$, whence $n+k\mid p-1$. This implies that $pn+k=(p-1)n+(n+k)$ is divisible by $n+k$, hence -$$p,n\mid 2^{n+k}-1\mid 2^{pn+k}-1.$$ -However, $p>n+k>n$, so $p$ and $n$ are coprime, and the above implies that $pn\mid 2^{pn+k}-1$. That is, $n'=pn$ is a solution bigger than $n$.<|endoftext|> -TITLE: Lower bounds for the top rational cohomology of arithmetic groups -QUESTION [9 upvotes]: I would like to know what estimates exist for the dimension of $H^d({\rm GL}_2(\mathcal{O}_{K,S}),\mathbb{Q})$ where $\mathcal{O}_{K,S}$ is a ring of $S$-integers in a number field $K$ and $d$ is the virtual cohomological dimension of the group ${\rm GL}_2(\mathcal{O}_{K,S})$. As the title says, I am mostly interested in lower bounds or asymptotic statements measuring the growth of the top cohomology in terms of arithmetic information (like genus, discriminant of $K$ or some such number , and the number of places in $S$). Moreover, information on the possible methods to obtain such estimates (from Euler characteristic computations, via study of Hecke operators, representation theory or so) would be very welcome. -An example of a formula in the spirit of the question is the estimate of the cuspidal cohomology of Bianchi groups in - -J. Rohlfs. On the cuspidal cohomology of the Bianchi modular groups. Math. Z. 188 (1985), no. 2, 253–269. - -I'm wondering if there are similar statements for general rings of $S$-integers. I haven't even been able to find asymptotics for the rational cohomology of ${\rm GL}_2(\mathbb{Z}[1/n])$ for $n\to \infty$ in the literature. -Maybe one more point describing my motivation: in the case of a function field of a curve $C$ over a finite field, the situation is easier to understand. The top rational cohomology can be computed in terms of Gauss-Bonnet formulas and is then related to the zeta-value $\zeta_C(-1)$ for the curve. Estimates for the rational cohomology then follow from the Weil conjectures (so the growth is roughly $q^{2g+s-3}$ with $g$ the genus of the curve, $s$ the number of places, and $q$ the size of the ground field). Essentially I want to know if a similar picture exists (possibly conjecturally?) on the number field side. - -REPLY [3 votes]: Let $\mathcal{O}$ be the ring of integers in an algebraic number field $k$ and let $\text{cl}(\mathcal{O})$ be the class number of $\mathcal{O}$. In my paper "Integrality in the Steinberg module and the top-dimensional cohomology of $\text{GL}_n(\mathcal{O}_k)$" (joint with Church and Farb), I prove the following two theorems: - -In its vcd, the dimension of the rational cohomology group of $\text{SL}_n(\mathcal{O}_k)$ is at least $(\text{cl}(\mathcal{O})-1)^n$. In particular, it is nonzero if $\text{cl}(\mathcal{O}) > 1$. -Conversely, assume that $\text{cl}(\mathcal{O})=1$. Furthermore, assume either that $k$ has a real embedding or that $\mathcal{O}$ is Euclidean. Then in their vcd's the rational cohomology of $\text{SL}_n(\mathcal{O}_k)$ and $\text{GL}_n(\mathcal{O}_k)$ vanishes. There is a similar vanishing result for the homology with certain twisted coefficients. - -These are Theorems C and B of that paper (we remark that Theorem C also claims a similar bound for GL, but there is a subtle error in the proof for GL. Our techniques give some lower bound, but we haven't pinned down precisely what it is. That should be corrected soon.). -I remark that in the Euclidean case the vanishing result in 2 above is an old theorem of Lee and Szczarba from -R. Lee and R. H. Szczarba, On the homology and cohomology of congruence subgroups, Invent. Math. 33 (1976), no. 1, 15-53. -For $n=2$, the lower bound in 1 above is pretty classical, but the vanishing result in 2 seems to be new (at least in the non-Euclidean case). I'm not sure what happens for $S$-arithmetic groups. -The paper can be downloaded from my webpage here.<|endoftext|> -TITLE: Determinant of a sub-matrix of the classical adjoint -QUESTION [7 upvotes]: Let $A$ be a square matrix of order $n$, say with complex coefficients, and let $M$ be the plain matrix of minors of $A$ of order $n-1$ (no transpose, no sing changes). Let $I$ and $J$ be $r$-subsets of the index set $[n]$. Then, apparently -$$\det M_{I\times J}=\det A_{([n]\setminus I)\times ([n]\setminus J)}\det(A)^{r-1}.$$ -(One can also write the identity for the classical adjoint of $A$ instead of $M$, or also assume $A$ invertible, and express the identity in terms of $A^{-1}$). -So for $r=1$ this is just the definition of $M_{ij}$. For small values of $n$, it is possible to check the identity, keeping track of the terms of the expansion of the determinant. But is there a more synthetic proof, and an interpretation of it? - -REPLY [8 votes]: For me, a standard book to look for such things is Prasolov's linear algebra. This is Theorem 1.2.6.1. Here is the Russian edition, but the proof is essentially a formula. - -REPLY [4 votes]: You might try using the very interesting combinatorial approach of Doron Zeilberger in A Combinatorial Approach to Matrix Algebra, Discrete Mathematics 56 (1985), 61-72. There he gives short proofs of various matrix results such as the Cayley-Hamilton Theorem and the Matrix-Tree Theorem as combinatorial identities, which for me is the "right" way to understand these, rather than appealing to, for example, properties of complex matrices and eigenvalues.<|endoftext|> -TITLE: Functorial multiplication on commutative rings -QUESTION [13 upvotes]: Let $C$ be the category of associative commutative rings with 1 and let $F:C\to C$ be a functor which commutes with the forgetful functor to abelian groups (i.e. $F$ is a functorial way to define another multiplication on every associative commutative ring with 1). Assume also that on $\mathbb Z$ the new multiplication is equal to the old one. Is it true that $F$ is the identity functor? - -REPLY [15 votes]: Yes, $F$ is the identity functor. -The multiplication on $R$ would come in the form of a natural transformation $R \times R \to R$ which is bilinear in each variable. In commutative rings, the functor $R \mapsto R \times R$ is represented by the polynomial algebra $\Bbb Z[x,y]$, and so by the Yoneda lemma natural transformations $R \times R \to R$ are represented by polynomials $f(x,y) \in \Bbb Z[x,y]$. In short, your new multiplication on $F(R)$ must be of the form $x * y = f(x,y)$ for $f$ some polynomial with integer coefficients. -In order for $x * y$ to be distributive, we need $f(x + x',y) \equiv f(x,y) + f(x',y)$, so that $f$ is linear in each variable. This forces $x * y = axy$ for some $a \in \Bbb Z$. However, the only choices of $a$ that ensure $F(R)$ is always unital are $a = \pm 1$, and the only one that preserves the unit of $\Bbb Z$ is $a=1$.<|endoftext|> -TITLE: $P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-...}}}}$ -QUESTION [25 upvotes]: Vassilev-Missana - A note on prime zeta function and Riemann zeta function¹ claims the following remarkable identity: -$$ -P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-...}}}}, -$$ -for integer $s>1$, where $\zeta(s)$ is the Riemann zeta function defined as $\sum_{n=1}^\infty \frac{1}{n^s}$ and $P(s)$ denotes the prime zeta function $\sum_{p\in\mathbb{P}}^\infty \frac{1}{p^s}$, and $\mathbb{P}$ is the set of prime numbers. -I am not a number theoretician, but the paper looks like it might contain some errors. I have already raised the question here, and it looks like I may be right. However, after performing some calculations using Wolfram Alpha, it looks like the identity might have some truth to it. -If we denote by $\epsilon(s)$ the error term $P(s)-\left(1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-...}}\right)$, we get the following results for successive finite approximations to the nested radical, in which the $k^{\rm th}$ approximation is $\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{...-\sqrt{\frac{2}{\zeta(2^{k-1}s)}}}}}$, and in this case by taking $s=2$: -\begin{array}{|c|c|} -\hline -k & \epsilon(2)\\ \hline -3 & 0.19732\ldots \\ \hline -4 & -0.13198\ldots \\ \hline -6 & 0.04839\ldots \\ \hline -5 & -0.035665\ldots \\ \hline -7 & 0.007512\ldots \\ \hline -8 & -0.013753\ldots \\ \hline -9 & -0.00304\ldots \\ \hline -\end{array} -so indeed it looks like $\epsilon(2)$ is converging to $0$ (note the expression "looks like", as I have not run any calculations for really large $k$). -Additionally, by taking $s=3$ the identity above would remarkably imply that: -$$ -\zeta(3) = 2\left((1-P(3))^2+\sqrt{\frac{2}{\zeta(6)}-\sqrt{\frac{2}{\zeta(12)}-\sqrt{\frac{2}{\zeta(24)}-...}}}\right)^{-1}, -$$ -and indeed taking the finite nested radical approximation for $k=9$, the error term is already $\epsilon=-0.002516\ldots$ and seems to approach $0$ as $k\to\infty$ (note again the expression "seems to"). -Is it possible that the identity still holds, nonwithstanding the issues in the paper, and if so, how may one prove it? Or do the approximations for large $k$ actually not produce $\epsilon\to0$? If so, how can the apparent convergence be explained? - -¹ Note that the journal in question is a reputed one, with Paul Erdős being at some point the editor (between '95 and '96), not a second-rate journal. - -REPLY [14 votes]: This so-called "functional equation" -$$(1-P(s))^{2}+1-\frac{2}{\zeta(s)}=P(2s)$$ -which was published in a 2016 paper has been used in a 2019 paper claiming to prove the generalized Riemann hypothesis. As all here have pointed out, that equation is false. A disproof has been recently published in 2021 and can be found here. It interested me to discover this, because in 2015 I had derived this same false equation. You can find my posts about it here and here. I wanted to take it upon myself to show a generalization that highlights the error. -There is a formula that relates the partial euler prouct to the partial prime zeta function. -$$\prod_{k=1}^n\left(1-\frac{1}{{p_k}^s}\right)=\sum_{r=1}^n\sum_{\{r\}}\prod_{j=1}^l\frac{(-1)^{i_j}}{i_j!}\left(\frac{{P_n(s\cdot r_j)}}{r_j}\right)^{i_j}$$ -where the sum is over partitions of $r$ and $P_n(s)=\sum_{k=1}^n\frac{1}{{p_k}^s}$, the partial prime zeta function. $p_*\in\text{Primes}$ (Note: this formula holds if we use any set of real numbers, not just the primes). For more details see Grunberg's paper On asymptotics, Stirling numbers, Gamma function, and polylogs -An equivalent representation is -$$ -\prod_{k=1}^n\left(1-\frac{1}{{p_k}^s}\right)=\sum_{k=0}^n{\frac{B_k(-0!P_n(s),-1!P_n(2s),\dots,-(k-1)!P_n(ks))}{k!}} -$$ -where $B_n(x_1,\dots,x_n)$ is the complete exponential Bell polynomial. -On a separate note, this also is equal to the analytic form, $\exp{\left(-\sum_{k=1}^\infty{\frac{P_n(ks)} -{k}}\right)}$ -This means we can write -$$ -\begin{align} -\left(1-\frac{1}{2^{s}}\right) -&=1-P_1\left(s\right) -\\ -&=\color{red}{1-P_1\left(s\right)}\underbrace{\color{red}{-\frac{P_1\left(2s\right)}{2}+\frac{P_1\left(s\right)^{2}}{2}}}_0 -\\ -&=1-P_1\left(s\right)\underbrace{-\frac{P_1\left(2s\right)}{2}+\frac{P_1\left(s\right)^{2}}{2}}_0\space\underbrace{-\space\frac{P_1\left(3s\right)}{3}+\frac{P_1\left(2s\right)P_1\left(s\right)}{2}-\frac{P_1\left(s\right)^{3}}{6}}_0 -\\ -&=\cdots -\\ -\left(1-\frac{1}{2^{s}}\right)\left(1-\frac{1}{3^{s}}\right) -&=\color{red}{1-P_2\left(s\right)-\frac{P_2\left(2s\right)}{2}+\frac{P_2\left(s\right)^{2}}{2}} -\\ -&=1-P_2\left(s\right)-\frac{P_2\left(2s\right)}{2}+\frac{P_2\left(s\right)^{2}}{2}\underbrace{-\frac{P_2\left(3s\right)}{3}+\frac{P_2\left(2s\right)P_2\left(s\right)}{2}-\frac{P_2\left(s\right)^{3}}{6}}_0 -\\ -&=\cdots -\\ -\left(1-\frac{1}{2^{s}}\right)\left(1-\frac{1}{3^{s}}\right)\left(1-\frac{1}{5^{s}}\right) -&=1-P_3\left(s\right)-\frac{P_3\left(2s\right)}{2}+\frac{P_3\left(s\right)^{2}}{2}-\frac{P_3\left(3s\right)}{3}+\frac{P_3\left(2s\right)P_3\left(s\right)}{2}-\frac{P_3\left(s\right)^{3}}{6} -\\ -&=\cdots -\end{align} -$$ -and so on. -In my case, I derived those lines with the red RHS in a different way and falsely assumed that the number of terms in the product on the LHS, as well as the number of summands in the partial prime zeta functions on the RHS could be continued beyond two terms. This is where the Riemann zeta function and the so-called "functional equation" came from. -These equations are true for any set of real numbers not just powers of primes starting from $2$. That's why as Lucia notes that the 2016 “identity” is true only if the Riemann zeta (as an Euler product) and the prime zeta functions involve a single prime term or two prime terms. -Thus a true generalized identity analogous to the false one is -$$\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)=1-\left(\frac{1}{a}+\frac{1}{b}\right)-\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)}{2}+\frac{\left(\frac{1}{a}+\frac{1}{b}\right)^{2}}{2}$$ -from which we can derive the nested radical identity -$$1-\frac{1}{a}-\frac{1}{b}=\\ -\lim\limits_{n\to\infty}\sqrt{2\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)-\sqrt{2\left(1-\frac{1}{a^{2}}\right)\left(1-\frac{1}{b^{2}}\right)-\cdots-\sqrt{2\left(1-\frac{1}{a^{2^n}}\right)\left(1-\frac{1}{b^{2^n}}\right)-1}}}$$<|endoftext|> -TITLE: Persistent homology over the integers -QUESTION [9 upvotes]: Is it likely that in the future, there will be interest in computing persistent homology over the integers (or other PIDs)? -Currently, persistent homology is usually done over a field (like $\mathbb{Z}/2$), as the algorithms for producing the barcode only work for a field. -However, working over a field loses a lot of information. Also, there are algorithms that can compute the persistent groups over a PID (http://geometry.stanford.edu/papers/zc-cph-05/zc-cph-05.pdf Section 5). -Thanks for any help. References will be very welcome. - -REPLY [10 votes]: As mentioned in Carlsson and Zomorodian's paper (to which you have linked), the problem of computing persistence barcodes with coefficients in a ring $R$ relies essentially on classifying graded modules over the polynomial ring $R[t]$. If (and only if!) $R$ is a field, $R[t]$ is a principal ideal domain and isomorphism classes of modules over it are easy to describe. When $R = \mathbb{Z}$, the modules are well-known to be hideous. Similar problems have plagued multi-dimensional persistence, since $R[s,t]$-modules are also pretty hard to classify (even when $R$ is a field). -But if you're willing to invest a bit in the Grothendieck construction, there might be hope: https://arxiv.org/abs/1601.03107<|endoftext|> -TITLE: Does there exist a notion of discrete riemannian metric on graph? -QUESTION [9 upvotes]: I would like to know if there is any notion of a discrete Riemannian metric on graphs. C. Mercat has worked on discrete Riemann Surfaces, but that's not exactly what I am working on. -To be more precise, when one defines a weighted Laplacian $\Delta : \mathbb{R}^V \mapsto \mathbb{R}^V $ on a graph $G=(V,E)$ by -$$ -\forall v \in V, \quad \delta (f)(v)=\sum_{w \sim v} a_{\{v,w\}} \big( f(v)-f(w) \big),$$ -the family $ \big( a_{e} \big)_{e \in E}$ of reals indexed by the set of edges can be viewed as a metric on the graph. But the analogy with the case of riemannian metric on manifolds is not so obvious. For instance, what would mean that two such metrics $ \big( a_{e} \big)_{e \in E}$ and $ \big( b_{e} \big)_{e \in E}$ are conformal ? -Thanks in advance, -LC. - -REPLY [6 votes]: This circle of questions is studied in this old paper by D. Jakobson and I. Rivin.<|endoftext|> -TITLE: Crafting Suspension Spectra -QUESTION [7 upvotes]: There is a theorem by Hopkins and Smith which states that for every $n > 0$ there is an ideal $I_n = (v_0^{k_0}, \dots, v_n^{k_n})$ such that there exist a spectrum $X_n$ with the following homology: -$$ BP_*(X_n)=BP_*/I_n.$$ -I was wondering if it was possible to have a similar but more detailed result. -Is there, for every $n$, an ideal $I_n = (v_0^{k_{0,n}}, \dots, v_n^{k_{n,n}})$ such that: -1) For every $i$, the limit of the $k_{i,n}$ goes to infinity when $n$ goes to infinity, -2) There exist a suspension spectrum $X_n$ with $ BP_*(X_n)=\Sigma^{d_n}BP_*/I_n$, for a natural number $d_n$ -3) The $d_n$ don't go to infinity when $n$ goes to infinity. -if it's not possible to satisfy 1,2,3, what about just 1,2? -Thanks!!! - -REPLY [2 votes]: For large $i_0,\cdots,i_n$ we can realize $BP_\ast/(v_0^{i_0},\cdots,v_n^{i_n})$ as the $BP$-homology of a finite spectrum $S/(v_0^{i_0},\cdots,v_n^{i_n})$. This follows from Devinatz-Hopkins-Smith. Suspending a finite spectrum enough times gives a suspension spectrum. This means that there is some space $X_n$ such that $\Sigma^{k_n} S/(v_0^{i_0},\cdots,v_n^{i_n}) \simeq \Sigma^\infty X_n$ for $k_n\gg 0$. I'm not sure how to address the question about the limit of the $k_n$'s.<|endoftext|> -TITLE: A regular first countable space of cellularity at most $2^\omega$ -QUESTION [5 upvotes]: Let $X$ be a regular first countable space of cellularity at most $2^\omega$. -Is it true that the cardinality of $X$ is at most $2^\omega$? - -A cellular family is a family of pairwise disjoint non-empty open sets. - The cellularity of a space $X$ (denoted by $c(X)$) is defined as the supremum of the cardinalities of the cellular families in $X$. - -REPLY [7 votes]: First of all let me note that a first-countable space of density $2^\omega$ has cardinality $\leq 2^\omega$, so a counterexample $X$ to your question must satisfy $c(X) \leq 2^\omega < d(X)$. That leads us naturally to higher Suslin Lines. -A continuous linear order $X$ (endowed with the order topology) is called a $\kappa$-Suslin line if $c(X) \leq \kappa < d(X)$. -The existence of a $\kappa$-Suslin Line is equivalent to the existence of a $\kappa$-Suslin tree and since higher Suslin trees are consistent with ZFC (Jensen proved they exist under $V=L$ for example), the existence of higher Suslin lines is also consistent with ZFC. A higher Suslin line is not necessarily first-countable but we can get something similar to a Suslin Line which is first-countable using an idea of this paper of Juhasz, Soukup and Szentmiklossy: -https://arxiv.org/pdf/math/0703728.pdf -Assume the existence of a $\mathfrak{c}$-Suslin Line $X$. Let $Y$ be the set of all points of countable cofinality in $X$. The continuity of the linear order implies that $Y$ is dense in $X$ and hence we can fix a dense subset $Z \subset Y$ of cardinality $\mathfrak{c}^+$. On $Z$ consider the topology $\tau$ generated by intervals of the form $(x,y]$, for $x < y \in Z$. Note that $(Z,\tau)$ is a first-countable space of cardinality $\mathfrak{c}^+$ and $c((Z, \tau))\leq \mathfrak{c}$, so that provides a consistent counterexample to your question. -Let me finish by noting that although it's not true that the continuum is a bound on the cardinality of first-countable spaces of cellularity $\leq 2^\omega$, the Hajnal-Juhasz inequality $|X| \leq 2^{\chi(X) \cdot c(X)}$ implies that $2^{2^\omega}$ is a bound on the cardinality of such a space (I'm assuming this is what originally motivated your question).<|endoftext|> -TITLE: Can the relative degree and ramification index can be read off the characteristic polynomial? -QUESTION [5 upvotes]: Let $L/K$ be a Galois extension of global fields with Galois group $G$. Assume that for a prime $p$ of $K$ we are given the datum $(L_{p}(s,\rho))_{\rho}$, where $\rho$ varies over the irreducible representations of $G$ and $L_{p}(s,\rho)$ is the Euler factor in the sense of Artin; i.e. it is the reciprocal of the characteristic polynomial of the Frobenius $\sigma_p$ after we restrict $\rho$ to the decomposition group $D$ and then we replace $\rho$ with the representation of $D/I$ on $V^I$, where $I$ is the inertia group. So in formulas $$L_{p}(s,\rho) = \det ({\rm id}_{V^I}- X\cdot \rho(\sigma_p)|_{V^I} )^{-1},$$ where $X=N(p)^{-s}$. (Here $V=V_{\rho}$ is the underlying space of the representation.) -My question: -Does this datum encode the relative degree and ramification index of the prime $p$? - -REPLY [4 votes]: Yes. The regular representation of $G$ decomposes as the sum of $\rho$'s with multiplicities $\dim\rho$, hence the product of $L(s,\rho)^{\dim\rho}$ over the various $\rho$'s equals the Dedekind zeta function $\zeta_L(s)$. That is, if $e$ (resp. $f$) is the ramification (resp. inertia) degree in $L/K$ of a given prime $p$ of $K$, then the Euler factors at $p$ reveal that -$$ \prod_\rho L_p(s,\rho)^{\dim\rho}=(1-N(p)^{-fs})^{n/(ef)},$$ -where $n=(L:K)$. So your datum determines $f$ and $n/(ef)$, hence also $e$, because it clearly determines $n$. - -REPLY [2 votes]: I will try to answer my question, at least on how to recover $e$, following a discussion with Chantal David on this. -First put $n=[L:K]$, $e_p$ the ramification index of $p$, $f_p$ the relative degree (aka, inertia degree, aka residue degree), and $g_p$ the number of primes of $L$ lying above $p$. Then -$$ -\frac{\zeta_L(s)}{\zeta_K(s)} = \prod_{p} (1-N(p)^{-f_p s})^{-g_p}(1-N(p)^{-s}). -$$ -(Here $N(p) = |O_K/p|$.) -On the other hand, Artin proved that -$$ -\frac{\zeta_L(s)}{\zeta_K(s)} = \prod_{\rho\neq \rho_0}\prod_p L_p(s,\rho)^{\dim \rho}, -$$ -where $\rho_0$ is the trivial irreducible representation. -So comparing these locally (at our $p$) and introducing a new variable $u=N(p)^{-s}$, we get that -$$ -(1-u^{f_p})^{-g_p}(1-u) = \prod_{\rho\neq \rho_0} L_p(s,\rho)^{\dim \rho}. -$$ -Comparing degrees in $u$ gives that -$$ -1-g_pf_p = \sum_\rho \dim\rho \deg_u L_p(s,\rho) -$$ -so that we recovered $g_pf_p$ and thus also $e_p=\frac{n}{g_pf_p}$.<|endoftext|> -TITLE: Independent/Easy fraction of sentences over PA -QUESTION [7 upvotes]: Let $S(n)$ be the set of all sentences over PA of length at most $n$ (counting the quantifier symbols, boolean connectives, arithmetic operations and constants, and counting each variable as length $1$). -Let $I(n) = \{ φ : φ \in S(n) \land \text{$φ$ is independent of PA} \}$. -Let $E(n) = \{ φ : φ \in S(n) \land \text{PA$^-$ proves or disproves $φ$} \}$.   [PA$^-$ is defined here.] -$ -\def\lfrac#1#2{{\large\frac{#1}{#2}}} -$ - -I think $\lfrac{\#(I(n))}{\#(S(n))} \to 0$ as $n \to \infty$, contrary to the conjecture at the end of this paper. - -My intuition is that it is relatively easy for a random sentence to be provable or disprovable just because of some provable example or counter-example. But I am not sure how to go about proving this. Is there any simple trick I am missing? - -Also, I suspect $\lfrac{\#(I(n))+\#(E(n))}{\#(S(n))} \to 1$ as $n \to \infty$, but I am unsure. - -This is actually an attempt to capture the idea that most statements that are not decided by PA$^-$ cannot be decided by PA. In intuitive terms I am trying to say that most statements are either easy to prove or disprove or independent of PA. Is any such thing true? -I posted this question on Math SE about a year ago but did not get any response, so I hope someone here can help. I would be quite surprised if my first conjecture is false! But neither conjecture seem to yield to structural induction or padding tricks. - -REPLY [2 votes]: The question of limiting fractions is obviously sensitive to the presentation and ordering of sentences. Meanwhile we can get some results for sentences of the form $$Q\vec{x}\ p(\vec{x})=0$$ where $Q\vec{x}$ is a sequence of quantifiers over variables in $\mathbf{N}$, and $p$ is a polynomial with coefficients in $\mathbf{Z}$, in which every variable in $\vec{x}$ appears non-trivially. We can regard these as abbreviations of sentences of the form $Q\vec{x}\ p(\vec{x})=q(\vec{x})$, where $p$ and $q$ have coefficients in $\mathbf{N}$. -Since every sentence of first-order arithmetic is equivalent under $PA^-$ to such a sentence, this is a reasonable set of sentences to consider. We can order the sentences to put the low-coefficient, low-degree, low-variable polynomials first. We find the following. - -When $Q$ is $\emptyset$: $PA^-$ settles all sentences of the form $p(1)=0$. -When $Q$ is $\forall x$: $PA^-$ settles all sentences of the form $\forall x \ p(x)=0$, and proves them false. -When $Q$ is $\exists x$: $PA^-$ settles all sentences of the form $\exists x \ p(x)=0$. Given $p$, we can find a bound on the roots, prove in $PA^-$ that $p$ is positive for $x$ above that bound, and test in $PA^-$ whether any there are any roots with $x$ below that bound. -When $Q$ is $\forall x \forall y$: $PA^-$ settles all sentences of the form $\forall x\forall y \ p(x,y)=0$, and proves them false. -When $Q$ is $\exists x \forall y$: $PA^-$ settles all sentences of the form $\exists x\forall y \ p(x,y)=0$, mostly proving them false. Either we can instantiate an $x$, and $PA^-$ proves the claim $\forall y$; or we can choose a number of $y$'s depending on the degree of $p$ and $PA^-$ proves that $p(x,0),\ p(x,1),\ldots p(x,d)$ have no common root. - -After this things get more interesting: - -When $Q$ is $\forall x \exists y$: $PA^-$ cannot settle all of these, but $PA$ can. For instance, $PA$ proves $\forall x \exists y\ (x-2y)(x-2y-1)=0$. However, this is false in the model of $PA^-$ whose domain is the eventually non-negative polynomials in $\mathbf{Z}[t]$, so it is independent of $PA^-$. -When $Q$ is $\exists x \exists y$: $PA^-$ cannot settle all of these, and it's open whether $PA$ can. For instance, $PA$ disproves $\exists x \exists y\ (x+1)^2-2y^2=0$. However, this is true in the model of $PA^-$ whose domain is the eventually non-negative polynomials in $\mathbf{Z}[\sqrt{2}][t]$ with integer constant terms, so it is independent of $PA^-$. An effective version of Falting's theorem would presumably show that $PA$ settles all of these sentences. - -It seems to me that as the degrees and quantifiers increase, a positive fraction of these sentences will be settled by $PA$ but not by $PA^-$, which is contrary to the second conjecture. In any case, this seems to be one interesting way to focus the question.<|endoftext|> -TITLE: How many positions of a tiling polygon can occur simultaneousy? -QUESTION [10 upvotes]: Let $T$ be a polygon which tiles the plane. For an instance of $T$ (mirrored or not), call the set of its translates a position of $T$. -My question: - - -How many different positions can occur in a tiling $\mathcal T$ of the plane? - - -Call this number $p(\mathcal T)$ and define for a tile $p(T):=\max_{\mathcal T} p(\mathcal T)$. -For example, the Voderberg tile has $p(T)\ge60$ and supposedly $p(T)=60$. -But is it even clear that $p(T)$ is always finite? Theoretically a tile containing irrational angles might occur in infinitely many positions, or is there an easy combinatorial argument against that? -A related probably easier question arises if we require $T$ to be convex. - - -For convex tiles $T$, is $p(T)=12$ best possible? - - -The picture shows a tiling of a certain pentagon with $p(\mathcal T)=12$, where the $12$ marked tiles form a fundamental domain. -It comes from the bottom right tiling of this picture quoted here. -. -Are there similar constructions, possibly spiral tilings, with bigger $p(\mathcal T)$? -Note that the Voderberg tilings cannot be reduced to tilings with isoceles 12-degree triangles. But in fact, talking about triangles: by subdividing the regular hexagon paving, we see that the triangle $T$ with angles $30°,60°,90°$ has already $p(T)=12$, so there is no need for "sophisticated" tilings as the above to illustrate the (supposedly) extremal case. But I'll keep the picture because of the funny symmetries (with two privileged "directions") which that particular tiling exhibits. - - -And of course we can ask similar questions for tilings in higher dimensions. - -REPLY [15 votes]: Infinitely many, even for a triangle. -Let $T$ be the "Pythagorean" triangle with sides of length $3,4,5$. First, tile the $3\times 4$ rectangle $R$ by two copies of $T$. Tile the $60\times 60$ square $Q_1$ with $R$. Then, attach a $12$-times dilated copy of $T$ to each side of the square $Q_1$, forming a square $Q'_2$. - -Tile a square $Q_2$ with $25$ copies of $Q'_2$. Repeat these two steps (attaching dilated copies of $T$ to $Q_i$ and multiplying the resulting square $5$ times in each direction). - -In each iteration we get one new rotation of $T$, assuming we tile only with rotated copies of $T$, since each square $Q_{i+1}$ is rotated by $\mathrm{arcsin}(3/5)$ with respect to $Q_i$, and this angle is irrational (https://arxiv.org/abs/1006.2938).<|endoftext|> -TITLE: Automorphism of $\mathbb{P}_A^n$ -QUESTION [13 upvotes]: Let $k$ be a field. -The proof that $Aut(\mathbb{P}_k^n)=PGL(n+1:k)$ relies on the fact for any automorphism $\alpha$ of $\mathbb{P}_k^n$, $\alpha^*(\mathcal{O}_{\mathbb{P}_k^n}(1)) = \mathcal{O}_{\mathbb{P}_k^n}(1)$. -It is not necessarily true that $\pi^*O_{\mathbb{P}_A}(1)\simeq O_{\mathbb{P}_A}(1)$ if $\pi:\mathbb{P}_A^{n}\to\mathbb{P}_A^{n}$ is an automorphism. -See counterexample: https://math.stackexchange.com/questions/979958/automorphism-of-the-projective-space-mathbbp-an -In particular, we can't use the same type of proof to show that $Aut(\mathbb{P}_A^n)=PGL(n+1:A)$. Furthermore, I don't think such an equality will exist for an arbitrary ring $A$. -What is the intuitive reason as to why we might not have such equality for arbitrary rings? What are the conditions on A for which this might be true? (other than the obvious one such as A a UFD or a field) - -REPLY [12 votes]: Let me summarize the comments of R. van Dobben de Bruyn. The $A$-automorphisms of $\mathbb{P}^n_A$ correspond in a one-to-one way to couples $(u,L)$, where $L$ is an element of $\operatorname{Pic}(A) $ and $u:A^{n+1}\rightarrow L^{n+1}$ an isomorphism. In other words, there is an exact sequence -$$1\rightarrow \operatorname{GL}(n+1,A)/A^*\rightarrow \operatorname{Aut}(\mathbb{P}^n_A)\xrightarrow{\ \lambda \ } \operatorname{Pic}(A)\, , $$where the image of $\lambda $ is the subgroup of elements $L\in\operatorname{Pic}(A) $ such that $L^{n+1}\cong A^{n+1}$ (note that this implies in particular $L^{{\scriptscriptstyle\otimes }(n+1)}\cong A$).<|endoftext|> -TITLE: Essential maps of spectra which are null when localized at any prime -QUESTION [9 upvotes]: There are maps of spaces which are not null-homotopic, but when localized at any prime become null. I don't know explicit constructions of any, but an example is given in Section 6 of Chapter 25 of the Handbook of Algebraic Topology (this chapter was written by C.A. McGibbon), of a phantom map $\Omega^2S^5\to \mathbb{H}P^\infty$ which has this property. -My question is really simple to state: are there maps of spectra which are globally non-trivial, but become null when localized at any prime? Or, can anyone give a proof that this is impossible? -It's well known that there's an "arithmetic square" for reconstructing spectra from their rationalizations and their $p$-completions (cf. these notes), and I believe this might have some bearing on this situation? - -REPLY [10 votes]: Such maps exist, even when the target is the sphere $S^{0}$. -A map $X \rightarrow S^{0}$ is $p$-locally trivial for each $p$ if it vanishes when composed with each of the $p$-localizations $S^{0} \rightarrow S^{0}_{(p)}$. This is the same as being in the kernel of the map $[X, S^{0}] \rightarrow [X, \prod S^{0}_{(p)}]$ induced by the diagonal $S^{0} \rightarrow \prod S^{0}_{(p)}$. -Assume, by contradiction, that all such maps are zero. This would mean that $S^{0} \rightarrow \prod S^{0}_{(p)}$ is a monomorphism in the stable homotopy category. Since the latter is triangulated, a monic necessarily splits, but $S^{0} \rightarrow S^{0}_{(p)}$ is not split, as that would in particular mean that we have a splitting on $\pi_{0}$, even though $\mathbb{Z} \hookrightarrow \prod \mathbb{Z}_{(p)}$ is not split. We deduce that $S^{0} \rightarrow \prod S^{0}_{(p)}$ is not split and hence not a monomorphism. -The proof of the fact that in a triangulated category any monic splits (see the answer to this question) allows one to construct an explicit example of a map that vanishes $p$-locally at each prime. Namely, the inclusion of the fibre $F \rightarrow S^{0}$ of the map $S^{0} \rightarrow \prod S^{0}_{(p)}$ is non-zero, as otherwise the latter would be split, although the composition $F \rightarrow S^{0} \rightarrow \prod S^{0}_{(p)}$ does necessarily vanish.<|endoftext|> -TITLE: The Euclidean norm and $k$ largest elements -QUESTION [6 upvotes]: This is not a homework problem, although I fear it may turn out to be at that level. For any nonnegative $x\in\mathbb{R}^n$, let $f_k(x)$ be the sum of the $k$ largest values in $x$, and define $$f(x)=\max_{k} \frac{f_k(x)}{\sqrt{k}}$$ over all $k\in\{1,\dots,n\}$. It is easy to see that $f(x)\leq \|x\|_2$ for all nonnegative $x$. Is the ratio $\|x\|_2/f(x)$ bounded from above? - -REPLY [5 votes]: I assume that your question concerns only nonnegative elements of $\mathbb{R}^n$, since for instance $f$ vanishes on $(-1,0)\in \mathbb{R}^2$. -Now on $\mathbb{R}_+^n$, $f$ coincides with $f\circ g$ where $g:\mathbb{R}^n \rightarrow \mathbb{R}_+^n$ turns every coordinate to its absolute value. -$$f\circ g(x)=\operatorname{max}_{S\subset\left\lbrace 1, \ldots , n\right\rbrace} \dfrac{\sum_{i\in S}|x_i|}{\sqrt{|S|}}\text{ if $x\neq\underline{0}, $ } 0 \text{ otherwise}$$ -One easily checks that $f\circ g$ is a norm on $\mathbb{R}^n$, so that the answer is yes since any two norms are equivalent on a finite-dimensional vector space. -However, there is no uniform bound on $n$ as shown by the following example. -Let $x=(x_1,\ldots, x_n)\in \mathbb{R}^n$ be defined by $x_i=\sqrt{i\phantom{.}}-\sqrt{i-1}$. Clearly, $x_i>x_{i+1}$ for all $i$, so that -\begin{eqnarray} -f(x)&=&\max_{k\in \left\lbrace 1, \ldots , n\right\rbrace} \dfrac{1}{\sqrt{k\phantom{.}}}\sum_{i=1}^k x_i \\&=&1 -\end{eqnarray} -However, $$\parallel\, x\!\parallel_2^2 = \sum_{i=1}^n\left( \sqrt{i\phantom{.}}-\sqrt{i-1}\right) ^2$$ tends to infinity with $n$. -This is an interesting problem to solve, since the extreme cases (e.g. $(1,\ldots, 1)$ or $(1, 0, \ldots, 0)$) do not provide counterexamples. The coordinates need to be different but not too far apart.<|endoftext|> -TITLE: "Equivalence" is to "group" as "adjoint" is to ....? -QUESTION [8 upvotes]: The collection of all self-equivalences of a category $C$ constitutes a $2$-group, which is a categorification of the notion of a group. My question is about what happens when one replaces equivalences by general adjoint functors. -The n-lab page on adjunctions says: - -a morphism in an adjunction need not be invertible, but it has in some sense a left inverse from below and a right inverse from above. - -I would like to know if this can be formalised in an interesting way. -In order to ask a specific question, I want to take a single adjoint pair of functors $(L,R):C\to C$, and let $M$ be the monoid of endofunctors of $C$ generated by $L$ and $R$ (together with some suitably chosen family of natural transformations). - -Q. What kind of ($2$-)algebraic structure is $M$? - -I apologise that this is a somewhat vague question. An ideal answer would have the form "$M$ is a (lax/colax/...) $2$-blah", where "blah" is an interesting algebraic structure that arises in other contexts not a priori having anything to do with adjoint functors. If it makes more sense to change the setting a little---e.g., to consider instead of $M$ the collection of all functors on $C$ admitting a two-sided adjoint---then please go ahead and do so. - -REPLY [3 votes]: The free adjunction, i.e. the 2-category generated by an adjunction $C^\to_\leftarrow D$ (with $C$ and $D$ distinct) is described by Schanuel and Street. In particular, if you restrict to the full sub-2-category on $C$, you get the free monad, and the free full sub-2-category on $D$ is the free comonad. The free monad itself is an interesting 2-category; it's the delooping of the strict mononoidal category $\Delta_a$ of augmented simplices, otherwise known as finite linear orders. Since a monad is just a monoid in the endomorphism category, this is equivalent to the fact that $\Delta_a$ is the free strict monoidal category on a monoid (see the last link). -The free adjunction with $C=D$ will be more complicated. I suspect it's not that much more complicated, but I don't know of a reference where it's written down.<|endoftext|> -TITLE: Closed vector subspaces of large powers of R -QUESTION [10 upvotes]: By a large power of $\mathbb R$ is meant a topological vector space which is the product of infinitely many copies of the real line. -Is every closed subspace of such a TVS linearly homeomorphic to a power of $\mathbb R$? - -REPLY [8 votes]: Yes, that's true. This is called a space of minimal type (see. H.H.Schaefer p.191): - -A locally convex space $X$ over $\mathbb R$ is isomorphic to some ${\mathbb R}^I$ ($I$ being a cardinal number) if and only if $X$ has no weaker (Hausdorff) locally convex topology or, what is equivalent, if $X$ is weakly complete. - -As a corollary, each closed subspace (and each Hausdorff quotient space) of ${\mathbb R}^I$ is isomorphic to some ${\mathbb R}^J$.<|endoftext|> -TITLE: "Standard arguments" in Mahowald's eta_j paper -QUESTION [9 upvotes]: In “A new infinite family in $_{2}\pi^S_*$" (1976), Mark Mahowald constructs elements $\eta_j \in \pi_{2^j}(S^0)$ for $j \neq 2$ which come from permanent cycles in the Adams Spectral Sequence that are generated by $h_1h_j \in Ext_A^{2, 2^j}(\mathbb{Z}_2, \mathbb{Z}_2)$. Let $H^*$ denote reduced mod-2 cohomology and for $Y$ a CW-complex let $Y_\ell$ denote the $\ell$-skeleton. Mahowald actually constructs a certain map from a stable complex $f_j: X_j \to S^0$ where $X_j$ has dimension $2^j-1$, as well as a map $g_j: S^{2^j} \to X_j$, so that $X_j/(X_j)_{2^j-2} \simeq S^{2^j-1}$, the composition of $g_j$ with the quotient $X_j \to X_j/(X_j)_{2^j-2}$ is the Hopf map, $H^{< 2^j - 2^{j-3}}(X_j) = 0$, and $Sq^{2^j}$ is nonzero in the mapping cone of $f_j$. Then he defines $\eta_j$ be the composition $f_j \circ g_j$ and concludes that by ``standard arguments'', $h_1h_j$ is a permanent cycle, etc. -What are these standard arguments? I know that a good way to show that maps are nonzero in the $\pi^S_*$ is to show that they are detected by a primary or secondary stable cohomology operation. I know that secondary cohomology operations come from relations in the Steenrod algebra; I know that relations in the Steenrod algebra give rise to the second column in the Adams Spectral Sequence. Unfortunately, I can't quite put things together to see why, for example $f_j \circ g_j$ ``represents $h_1h_j$'' (as Mahowald says)! In particular, I have no idea why the product on the ASS should be related to composition of maps of complexes. -(Why I am asking this question: I am not much of a homotopy theorist, but for some reason I had to read a later paper of Mahowald's that was based on observations of this one in which he shows that certain Eilenberg-Maclane spectra are Thom spectra. This paper seemed interesting.) - -REPLY [10 votes]: Since the paper actually refers to a secondary operation associated with the Adem relation -$$Sq^{2^i+1}Sq^1+Sq^2Sq^{2^i}+Sq^4Sq^{2^i-2}+Sq^{2^i}Sq^2=0$$ -then a standard argument to show that the composition $S^{2^i}\stackrel{g_i}\to X_i\stackrel{f_i}\to S^0$ is essential is to proceed and prove by contradiction. If the composition is null then one may look at the cohomology of the double mapping cone $S^0\cup_{\overline{f_i}}C(X_i\cup e^{2^i+1})$ where existence of a map $\overline{f_i}:C_{g_i}\to S^0$ follows from the assumption that $f_i\circ g_i$ is null. Now, evaluation of the above Adem relation on the $0$-dimensional class of in the cohomology of the double mapping cone gives the desired contradiction. This is a standard argument prior to the use of Adams Spectral Sequence arguments.<|endoftext|> -TITLE: Conjugacy of Regular Semisimple Subalgebras -QUESTION [5 upvotes]: Let $\mathfrak{g}$ be a complex semisimple Lie algebra. In his paper Semisimple Subalgebras of Semisimple Lie Algebras, Dynkin states that by a result of Weyl, -1) Two regular semisimple subalgebras are conjugate if and only if their simple roots are conjugate under the Weyl group. -From this Dynkin concludes that for the classical Lie algebras, -2) Any regular semisimple subalgebra of a given type (e.g. $A_2\oplus B_2$ in $B_5$) is unique up to conjugacy, with some exceptions if $\mathfrak{g}$ is of type $D_n$. The same is true of the exceptional Lie algebras, except $E_7$ and $E_8$ have several exceptions. -Does anyone know where I can find a proof of Statements 1 and 2? The first statement supposedly follows from a result in Theorie der Darstellung kontinuierlicher halb-einfacher Gruppen durch lineare Transformationen by Weyl, but I am unable to find an English translation. I am more interested in Statement 2, however, but Dynkin does not cite any source, nor does he hint at a proof. - -REPLY [5 votes]: It would help to recall Dynkin's definition of "regular semisimple subalgebra" of a complex semisimple Lie algebra (say $\mathfrak{g}$), which has not become standard in later literature. This just means a semisimple subalgebra of $\mathfrak{g}$ which contains a Cartan subalgebra (sometimes called a "maximal toral subalgebra" in this special case, any two being conjugate under the adjoint group). Since the subalgebra is semisimple, it involves a symmetric set of roots of $\mathfrak{g}$. The tricky step is to determine all such finite sets of roots and their conjugacy under the Weyl group. There are of course obvious examples, determined by subsets of a fixed set of simple roots: these subalgebras are now usually called "Levi subalgebras" of parabolic subalgebras. Other examples are more subtle, such as type $A_2$ inside $\mathfrak{g}$ of type $G_2$. -The paper by Dynkin which you mention appeared first in 1952 in Russian here, but an English translation has apparently only been published in an AMS translation volume (with four other papers) in 1957. By now this article is somewhat old-fashioned, of course, but along with Dynkin's other early work it has had a lot of influence on later work. -An early (independent) approach by Borel and de Siebenthal was announced in 1948, with full details provided in a longer paper in the Swiss journal Comment. Math. Helv. 23 (1949), 200-222. This paper is in French and relies on the extended Coxeter-Dynkin diagram attached to a compact semisimple Lie group. Here one has to know that such Lie groups correspond precisely to complex semisimple Lie groups or their Lie algebras. (It was shown in the late 1950s by Chevalley that such Lie groups over $\mathbb{C}$ correspond naturally to semisimple algebraic groups, which opens up the algebraic approaches further.) -Dynkin himself returned to Lie theory only sporadically, working out another approach with a collaborator here. Besides this treatment, there are by now a lot of differing accounts of the classical semisimple theory in textbooks and articles, but for example the details you ask for in 2) may be hard to locate in precisely that form. In any event, I'd urge you to get familiar with some of the later literature, which may be easier to read than the translation of Dynkin's paper.<|endoftext|> -TITLE: Examples of analytic functions to motivate a first course in complex variables -QUESTION [7 upvotes]: [Changed title as a plea to re-open the question.] -If one is to motivate a course in complex variables, what specific analytic (holomorphic/meromorphic) function of one variable would you cite as an example that comes up in other areas of mathematics? Particularly if one can comment on its natural domain (either a subset of $\mathbb C$ or an extension, i.e. a Riemann surface)—perhaps with special mention of any pole structure, conjectural or not—without getting into too much details of that particular field. -Number theory of course offers a whole slew of examples: $L$-functions, modular forms (not to forget elliptic functions), where they encapsulate a lot of very deep mathematics. Would be great to be more specific. Of course the Riemann zeta function would be high on everyone's list. -What about "special functions" that solve differential equations (on the complex domain)? I'd also love to see a layman's example of a Painlevé transcendent. -I seem to recall Weierstrass's nowhere-differentiable function was discovered as the boundary value of an analytic function. Anyone know what it was? What about the smooth but nowhere-analytic function? -ADDED later: Let me explain a bit why I thought of asking this here––and I do it with some reservation. I was trying to write up an explanatory note on analytic continuation, and found that one could get across the idea (if not the actual theorem) without having to develop the basic theory of complex variables (along the line of Cauchy). In particular, I managed to give a list of "classes" of functions with increasingly sophisticated natural domains. - -polynomials and entire functions -rational functions and meromorphic (on $\mathbb C$) -algebraic functions like $\sqrt z$, and functions like $\log z$ -modular forms, etc., on the upper half plane that can't be extended at all - -For each I wanted to give some nontrivial examples, so the students would see a wide variety of functions before going into the general theory. ("To see" may be taken literally: https://en.wikipedia.org/wiki/Domain_coloring.) I'm afraid too often in the standard complex variable course, when we are speaking of holomorphic functions, the students could only think of polynomials or rational functions when it comes to counterexamples or "checking" theorems. One of the books that tried to remedy that is - -Stalker, Complex Analysis: Fundamentals of the Classical Theory of Functions - -which denotes the first half on "special functions" before going into the general theory. (I'm not sure how it would work in an actual course.) -So, the purpose for this question is solicit help in expanding and/or enriching the above list with more examples, especially ones that "open up" a whole subject (similar to the MO question Fundamental Examples), in a way that would be accessible and beneficial to students. Not knowing much myself, I have the impression that many new examples have come up since the "classical" special functions. -I do agree with the comment that there are way too many examples of analytic functions useful in other areas, which makes it even more startling since an analytic function is completely determined by its restriction on a small neighborhood, or by a sequence of numbers (be it Taylor series coefficients, or Dirichlet, or Fourier)––a point that perhaps is obvious but may seem surprising in the standard introductory course. My impression is that we don't stress this point enough. - -REPLY [3 votes]: These are the Mittag-Leffler function, entire functions of completely regular growth, and the Blaschke product.<|endoftext|> -TITLE: What's a (infinity-) semi-stack? -QUESTION [7 upvotes]: A stack is an object that mixes the notions of (algebraic) space and group. The key insight of stack theory is that most things you would want to do with spaces you can do with stacks: namely, you have a symmetric monoidal category of quasicoherent sheaves, you have forms and differentials, you have pullbacks, (several) notions of locality, an étale theory, etc. -There is a generalization to infinity-stacks, which mix the notions of algebraic space and infinity-groupoid (equivalently, $(\infty, 0)$-category, or topological space). -I want a generalization of this to a notion of "semi-stack", which mixes the concepts of algebraic space and semigroup -- and more generally, in the derived context, a notion of $(\infty, 1)$-stacks which interpolates between spaces and $(\infty, 1)$-categories. Probably because I'm not familiar with higher algebra, I haven't been able to find a definition. Here are some properties I would like from the category of "derived semistacks", $Stck_{(\infty,1)}$, roughly in decreasing order of importance (non-derived versions to be supplied by the reader): -(1) Given any derived semistack $X\in Stck_{(\infty, 1)}$ I want defined a symmetric monoidal dg category (at least in characteristic zero, with some infinity-categorical modifications in characteristic p) of (quasi-)coherent sheaves $QCoh(X)$. -(2) It should be true that any suitably finite $(\infty, 1)$-category $c$ has a corresponding object $c\times \text{pt}\in Stck_{(\infty,1)}.$ The category $QCoh(c\times \text{pt})$ is the symmetric monoidal representation category $Rep_k(c)$. (Tensor product of representation of categories is defined object-wise, like for quivers.) -(3) The category $Stck_{(\infty, 1)}$ is fibered over the category of small categories (i.e. $Hom(X, Y)$ is a small, "discrete" category for X, Y derived semistacks, with composition given by functors in a functorial way). -(4) The category $Stck_{(\infty)}$ (viewed as fibered in $(\infty)$-groupoids) is fully faithfully contained in $Stck_{(\infty,1)}$. Moreover, I want a "stackification" functor $X\mapsto X^\times$ from $Stck_{(\infty,1)}\to Stck_{\infty}$ which satisfies the following adjunction property: -$Hom(X, Y)^\times = Hom(X, Y^\times)$, for $X\in Stck_\infty$ and $Y\in Stck_{\infty, 1}$. Here $Hom(X, Y)^\times$ is the $(\infty, 0)$-category underlying $Hom(X, Y)$ (consisting of the same objects and all invertible morphisms). -(5) Let $\overline{G}_m$ be the semigroup scheme given by $\mathbb{A}^1$ with product given by $x\times y:\, = xy$, extending the group scheme $G_m = \mathbb{A}^1\setminus \{0\}$. I want an object $\text{pt}/\overline{G}_m$ (one point with "endomorphisms $\overline{G}_m$") -such that $(\text{pt}/\overline{G}_m)^\times = BG_m$ is the classifying space of the multiplicative group, and such that for an ordinary scheme (or stack) $X$, we have $$Hom(X, \text{pt}/\overline{G}_m)$$ given by the category with objects line bundles $L$ over $X$ and morphisms $$\pi_0Hom(L, M) \cong Hom_{Coh}(L, M)$$ -("morphisms between line bundles which are allowed to go to zero".) - -Does anyone know of such a construction? Would it be sufficient to consider infinity-categories (in Lurie's simplicial set formulation) over, say, the fppf site on a point? -It would be nice if this category (perhaps under some additional conditions, like smoothness of some sort) were to exhibit some of the good behavior of stacks, viz. good notions of locality, an étale homology theory, other motivic invariants, etc... - -REPLY [8 votes]: If we remove the $\infty$ from your question, the idea you want already has the name "stack" or "stack of categories", and it's been considered for about 50 years. See for example Giraud's Non-abelian cohomology book (from 1971). Stacks are just fibered categories that satisfy good descent properties with respect to some chosen topology. Standard examples fibered over schemes include the stack $QCoh$ of all quasi-coherent sheaves and the stack $Ell^{isog}$ of elliptic curves, where morphisms are pullback squares composed with isogenies. Your example of line bundles, where morphisms are allowed to be non-invertible, is a substack of $QCoh$. -I don't think adding infinity really changes anything essential, but I'm out of practice with the infinity-stuff. -Notational suggestion: As David Ben-Zvi mentioned in the comments, some people are introducing new terms to distinguish between stacks in categories and stacks in groupoids. In fact, many people have now done so, and these new terms generally don't agree with each other. It might be less confusing to add numerical prefixes describing the type of fiber, following the practice of higher categories (and the part of this question that doesn't use "semi-stack"). That is, a $(1,0)$-stack is a stack in groupoids, a $(0,0)$-stack is a sheaf of sets (or setoids), a $(1,1)$-stack is a stack in categories, an $(\infty,0)$-stack is what is commonly called an infinity-stack, and an $(\infty,1)$-stack is what the question seems to be seeking.<|endoftext|> -TITLE: Example of an unstable map between finite complexes which is the identity on homotopy but not homotopic to the identity? -QUESTION [17 upvotes]: Stably, phantom maps (nonzero maps which are zero on homotopy) exist, but it's not known if they exist between finite complexes (Freyd's Generating Hypothesis). Unstably, it's easy to find maps which are the same on homotopy but not homotopic (even between finite complexes), however I don't know an example of a map which is the identity on homotopy but not homotopic to the identity. -I presume I could take $\Omega^\infty(1+f)$ where $f$ is a known stable phantom map -- and I would be interested to see the details worked out -- but known stable phantom maps seem to hinge crucially on $\varprojlim^1$ issues, and so are "inherently infinitary". I'd like to see an unstable example which is not "inherently infinitary"; concretely it would be nice to see an example on a finite complex, but I'm not wedded to this interpretation of "not inherently infinitary". To sum up: -Question: What is an example of a self-map $f: X \to X$ of a (pointed, connected) CW complex which is not homotopic to the identity but such that $\pi_\ast(f): \pi_\ast(X) \to \pi_\ast(X)$ is the identity? Bonus points if $X$ is a finite complex. -An equivalent question is: what's an example of a space $X$ such that $Aut(X)$ doesn't act faithfully on $\pi_\ast(X)$, where $Aut(X)$ is the group of homotopy classes of self-homotopy-equivalences of $X$? -Another way of putting this is: Whitehead's theorem tells us that the functor $\pi_\ast$ reflects isomorphisms, but does it reflect identities, i.e. is it faithful with respect to isomorphisms? -For that matter, I'm having trouble coming up with a space $X$ such that $Aut(X)$ doesn't act faithfully on its homology, either. - -REPLY [9 votes]: George Cooke gave an example in Trans, AMS 237 (1978) 391-406. Define a map $h\colon S^n \times S^n \vee S^{2n} \to S^n \times S^n \vee S^{2n}$ by -$$ -\begin{cases} -h|_{S^n \vee S^n \vee S^{2n}} = \mathrm{id} \\ -h|_{2n-cell} \ \mathrm{wraps\ non-trivially\ around}\ S^{2n}. -\end{cases} -$$ -This induces the identity on homotopy, but is non-trivial on homology. This in fact gives a homotopy action of $\mathbb{Z}$ on the space and Cooke showed how this could be replaced by a true action on a homotopically equivalent space. -As far as the question about $\mathrm{Aut}(X)$, the paper by Pak and me in -Topology and its Applications 52 (1993) 11-22 might be relevant. We constructed examples where the representation from components of the homeomorphism group to $\mathrm{Aut}(\pi_*(X))$ is not faithful. This involved interactions between the notions of simple space, the Gottlieb group, principal bundles over tori and the $h$-rank of a space. It all went back to the old question of Gottlieb: is there a finite simple complex with non-trivial fundamental group, but trivial Gottlieb group.<|endoftext|> -TITLE: The difference between $q$-deformations and $h$-deformations -QUESTION [6 upvotes]: What is the difference between $q$-deformations and $h$-deformations of universal enveloping algebras? -In chapter XVI of Quantum groups by Kassel, a very precise definition of a quantum enveloping algebra is given. Such an algebra is called an $h$-deformation. In chapter XVII the Drinfeld-Jimbo algebras are introduced and it is proved that they are $h$-deformations. Chapter XVIII gives some uniqueness results of these $h$-deformations (the rigidity theorems). -Along the way, $q$-deformations are also introduced (although there is not really a proper definition). One result states that you can view a $q$-deformation as a Hopf-subalgebra of an $h$-deformation. One major difference is the quasi-triangular structure. By definition $h$-deformations have a quasi-triangular structure, but $q$-deformations don't. By Kassel's definition, this implies that $q$-deformations are not quantum enveloping algebras. -An obvious question arises: Are there still rigidity results for $q$-deformations? -Further I'd like to point out that other good books such as 'Quantum groups and their represenations' of Klimyk and Schmüdgen introduce Drinfeld-Jimbo algebras using the parameter $q$. However, Drinfeld introduced $h$-deformations first. Whenever people talk about $q$-deformations, they seem to be interested only in the representation theory and seem to forget what a deformation should be. -Other than the above question, I would very much appreciate it if anyone could shed some light on these issues. In particular I am confused why some many people accept the $q$-deformations without rigidity results (or I can't find these). For sake of completeness I would like to state that I also asked this question today, which is related but focuses on something different. I am not trying to spam the site with a bunch of questions. -EDIT: I would like to mention that the Hall algebra approach to quantum groups seems to recover $q$-deformations, but it looks very unlikely to find $h$-deformations in this way (how on earth would you get $e^h$ from counting extensions?) - -REPLY [5 votes]: While $\hbar$-quantum groups are really to be understood as deformations, $q$-quantum groups are somewhat different. -A quantum group in the $\hbar$-setting is a $\mathbb C[[\hbar]]$-Hopf algebra $\mathcal A_\hbar$. As such, the deformation parameter can be given a specific value only at $\hbar=0$, by letting $\mathcal A_0=\mathcal A_\hbar/(\hbar\mathcal A_\hbar)$. On the contrary a $q$--quantum group is a $\mathbb C(q)$-Hopf algebra $A_q$. After fixing a $\mathbb C[q,q^{-1}]$-rational form (and depending upon this choice) the parameter $q$ can be given any value in $\mathbb C\setminus 0$. In a sense the $\mathcal A_\hbar$ is a local deformation, while $A_q$ is a global deformation. -The difference is indeed made clear by rigidity results. Rigidity results for $\mathcal A_\hbar$ means that in an infinitesimal neigbourhood of $0$ any deformation of $\mathcal A_0$ is isomorphic to $\mathcal A_0$ itself (usually what one proves is rigidity of the associative algebra structure). On the other hand it is not true that in a neigbourhood of $q=1$ in the complex plane any specialized $q$-algebra is isomorphic to $A_1$. In fact, any neigbourhood of $q=1$ contains some roots of unity and therefore contains some special algebras which are not isomorphic as associative algebras to $A_1$. It is true, however, that if you restrict $q$ to be real inside $]0,1[$ you can prove some rigidity results. -As for the relation between the two, I haven't clear the embedding result that you mention. It is however possible that having fixed $q\in\mathbb R$ one can restrict $\mathcal A_\hbar$ to those power series that are convergent when $e^\hbar=q$. -ADDED -To answer your additional question: -$q$-deformations, as I said, are those that allow you to put the deformation parameter to a complex number therefore: - -They allow phenomena like the roots of unity case, which is the part of the theory that connects to braid invariants and was, historically, one of the main sources of interest in qg-theory. -They allow to put a $C^*$--algebra structure on the quantum group thus connecting to operator algebraic approaches. You do not have anything like a $C^*$-algebra on $\mathbb C[[\hbar]]$-deformations. - - -$C^*$-algebraic approach is essential to try to connect quantum group theory to NC geometry à la Connes. - -Through representation theory, $q$--deformations give rise to $q$-special functions and their properties like summation formulas, thus generalinzing the representation theoretic approach to classical special function theory. - -This are but few (personal, idiosincratic) of the reasons of interests in $q$- deformations. Of course reconciling the two approaches is very interesting. For example: -- on the $\mathbb C[[\hbar]]$-side much is known about NC index theorems. Understanding completely how this relates to the $q$-side of the story is what would bring in an understanding of NC index theory in terms of global geometry of the underlying Poisson group. -As for $q$ rigidity results I think you should read this: -Reference request quantum SU(3)<|endoftext|> -TITLE: Boundary terms of formal adjoints of differential operators -QUESTION [7 upvotes]: Let $M$ be a compact manifold with boundary. If we have two vector bundles $E, F \to M$ with inner products and a differential operator $D: C^{\infty}(E) \to C^{\infty}(F)$ then $D$ admits a formal adjoint $D^*: C^{\infty}(F) \to C^{\infty}(E)$. -This satisfies, for smooth sections with compact support in the interior $\phi \in C^{\infty}_c(E)$, $\psi \in C^{\infty}_c(F)$ the identity -$$\int_M \langle D\phi, \psi \rangle_F \text{dvol} = \int_M \langle \phi, D^*\psi \rangle_E \text{dvol}.$$ -My question is, when can we understand the value of -$$\int_M (\langle D\phi, \psi \rangle_F - \langle \phi, D^*\psi \rangle_E)\text{dvol}$$ -when $\phi, \psi$ are not compactly supported in the interior? There should be some boundary term, but I can't figure out exactly what it should be. -My motivating example is the following "integration by parts" for the spin Dirac operator on the positive spinor bundle over a spin$^c$ manifold with boundary: -$$\int_M \langle D_A^+\phi, \psi \rangle = \int_M \langle \phi, D_A^-\psi \rangle - \int_{\partial M}\langle \phi, \rho(\nu)\psi \rangle.$$ -Here $\rho(\nu)$ is Clifford multiplication by the unit normal vector to the boundary. -There is a similar identity for the Levi-Civita connection, etc. - -REPLY [12 votes]: Yes, there is a generalization of such a "boundary term" for an arbitrary linear differential operator (smooth coefficients assumed, of course). My favorite way to define the formal adjoint $D^*$ of a differential operator $D$ doesn't involve any integrals. Given $D$, $D^*$ is defined by the requirement to satisfy an identity of the form -$$ - \langle D[\phi], \psi \rangle \mathrm{dvol} - - \langle \phi, D^*[\psi] \rangle \mathrm{dvol} - = dW[\phi,\psi] , -$$ -where $d$ is the exterior derivative, for some differential operator $W[\phi,\psi]$ that is bilinear in its arguments and is valued in $(\dim M - 1)$-forms. Such an identity is sometimes called of Green or Liouville type. If such a $D^*$ exists, then it automatically unique and (by Stokes' theorem) satisfies the adjoint formula involving integration over $M$, which you gave in the question. It is possible to go in the other direction (start from the adjoint formula with the integral and end up with the identity that I wrote), but that requires some slightly sophisticated machinery from the calculus of variations on jet bundles. -The point of the above definition is that the $W$ form is what gives you the "boundary term" that you are looking for. Namely, without the restriction of compact support (but still assuming that $M$ is compact as a manifold with boundary), you get -$$ - \int (\langle D[\phi], \psi \rangle - \langle \phi, D^*[\psi] \rangle)\mathrm{dvol} - = \int_{\partial M} W[\phi,\psi] . -$$ -Note that, by itself, $W$ is not unique, since it can equally well be replaced by $W[\phi,\psi] + dU[\phi,\psi]$. But the $dU$ term will cancel from the boundary integral, since $\partial M$ now itself has no boundary. -To actually find $W$, it is easiest to work in local coordinates, as suggested by Deane Yang. In local coordinates $(x^1,\ldots, x^n)$, -$$ - dW = d (W^a dx^1 \cdots dx^{a-1} \wedge dx^{a+1} \cdots dx^n) - = \partial_a W^a d^nx -$$ -and the key identity that you need is of course -$$ - (\partial_b f) g d^nx - f (-\partial_b g) d^nx - = \partial_a (\delta^a_b f g) d^nx . -$$ -Heuristically, as you use integration by parts to get $D^*$ from $D$, instead of throwing away all the boundary terms, you progressively collect them in $W$.<|endoftext|> -TITLE: "Monoid objects" without points -QUESTION [7 upvotes]: Let $\mathbf{C}$ be a category with binary Cartesian products. Let's say that a map $f : X \to Y$ is constant if, for every pair of maps $g,h : A \to X$, maps $f \circ g$ and $f \circ h$ are equal. Now, we can define a "monoid object without points" in $\mathbf{C}$ as follows: it is an object $X$ together with a map $* : X \times X \to X$ and a constant map $e : X \to X$ such that $*$ is associative in the usual sense and both compositions $X \xrightarrow{\langle id, e \rangle} X \times X \overset{*}\to X$ and $X \xrightarrow{\langle e, id \rangle} X \times X \overset{*}\to X$ are equal to the identity morphism. We also can define a "group object without points" in a similar way. -If $\mathbf{C}$ has a terminal object and there is a map $x : 1 \to X$, then $X$ is a monoid object since we can define the neutral element as $1 \overset{x}\to X \overset{e}\to X$. Moreover, this defines a bijection between the set of monoid structures and "monoid without points" structures. In particular, the only "monoid objects without points" in $\mathbf{Set}$ are usual monoids and the empty set, but there might be more examples in an arbitrary category. - -Is there a name for such "monoid" and "group" objects? Are there any interesting examples of them? Were they considered in the literature? - -The reason I'm asking is that (I believe) I have an example of a locale which might not have points but has a structure of a group in the sense I described. - -REPLY [8 votes]: I don't believe there are non-trivial examples of this concept. Assume that $e: X \to X$ is a constant map, then $e$ is idempotent. Any category can be embedded fully faithfully into a Cauchy complete category, so we can assume that $e$ splits as $X \xrightarrow{p} T \xrightarrow{i} X$ with $p\circ i=1_T$. I claim that $T$ is a terminal subterminal object. Indeed, let $f,g: A \to T$, then -$$\begin{eqnarray} -A \xrightarrow f T & = & -A \xrightarrow f \left( T \xrightarrow i X \xrightarrow p T\right) \xrightarrow 1 \left(T \xrightarrow i X \xrightarrow p T \right) \\ -& = & -A \xrightarrow f T \xrightarrow i \left(X \xrightarrow p T \xrightarrow 1 T \xrightarrow i X \right) \xrightarrow p T \\ -& = & -A \xrightarrow g T \xrightarrow i \left(X \xrightarrow p T \xrightarrow 1 T \xrightarrow i X \right) \xrightarrow p T \\ -& = & A \xrightarrow g T -\end{eqnarray} -$$ -Regarding your supposed example of a locale, it is most likely that what you have is either a groupoid in locales or a localic group over some non-trivial base.<|endoftext|> -TITLE: representations of an algebraic group and extension of scalars -QUESTION [5 upvotes]: Let $G$ be an algebraic group over an algebraically closed field of characteristic zero $K$ and let $L$ be another algebraically closed field, together with an embedding $K \hookrightarrow L$. -Why is it true that the extension of scalars is an equivalence of categories from finite dimensional $K$-representations of $G$ to finite-dimensional $L$ representations of $G \times_K L$? -Are all the assumptions on the fields (algebraically closed, characteristic zero) needed? - -REPLY [6 votes]: First of all, even setting aside the issue with scalar automorphisms noted in comments, at the level of objects there is a "problem": the functor is not essentially surjective for unipotent groups (by a consideration of Ext$^1$-groups with their natural structure of vector space over the ground field). Probably nobody cares, so I won't get into the details (but if someone does care then hopefully someone else will summon the energy to write out an actual argument, since I don't feel like going down that road here). -Now let's focus on positive statements that someone might care about. -Let $K'/K$ be an extension of fields (any characteristic, but characteristic 0 is especially nice when $G^0$ is reductive because of the Remark below), and $G$ a smooth affine group over $K$. Assume $K$ is algebraically closed (but $K'$ is not assumed to be algebraically closed). Then we claim: -Theorem: For any semisimple linear representation $\rho':G_{K'} \to {\rm{GL}}(V')$, there exists a semisimple linear representation $\rho:G \to {\rm{GL}}(V)$ unique up to isomorphism such that $\rho' \simeq \rho_{K'}$. -Remark: In characteristic 0 the semisimplicity hypothesis on the representations automatically holds if $G$ has reductive identity component. There is a version of the Theorem in characteristic 0 that doesn't require $K$ to be algebraically closed if $G$ is split connected reductive, but that involves an entirely different ingredient than is used below, so I'll pass over it in silence. -To prove the Theorem, we begin with: -Lemma The coefficients of the characteristic polynomial of $\rho'$ come from $K[G] \subset K'[G_{K'}]$. -Proof. An element $f' \in K'[G]$ comes from $K[G]$ if and only if its restriction to a Zariski-dense subset of $G(K)$ belongs to $K$. -(Indeed, if we apply "spread out and specialize" to $f’$ then we get some $f \in K[G]$ such that $f_{K'}$ agrees with $f'$ on a Zariski-dense -subset of $G(K)$, but such a subset is also Zariski-dense in $G_{K'}$ (exercise!), so $f' = f_{K'}$ as desired.) -Let $B$ be a Borel $K$-subgroup of $G$, so the $G(K)$-conjugates of $B(K)$ cover $G(K)$ (since $K$ is algebraically closed). Hence, it suffices to show that the coefficients on all such $G(K)$-conjugates have values in $K$. Let $T$ be a maximal $K$-torus in $B$, so $B = T \ltimes U$ for $U := \mathscr{R}_u(B)$. Applying Lie-Kolchin to $B_{K'}$ acting on $V'$, those coefficients on any $b \in B(K)$ only depend on the $T$-component of $b$. Thus, since the characteristic polynomial is conjugation-invariant, we're reduced to studying these coefficients on points of $T(K)$. All weights of $T_{K'}$ are "$K$-rational" (since $T$ is a split $K$-torus, as $K$ is algebraically closed), so we win. -QED Lemma -Now if we apply "spreading out and specialization" followed by semisimplification, from $\rho'$ we get a semisimple -$$\rho: G \to {\rm{GL}}(V)$$ -such that $\rho_{K'}$ has the same characteristic polynomial as $\rho'$ due to the Lemma. But $\rho_{K'}$ is semisimple because $\rho$ is semisimple with $K$ algebraically closed (i.e., $V$ is a semisimple representation of the abstract group $G(K)$ with $K$ algebraically closed, so it is "absolutely semisimple" and hence — by consideration of the endomorphism algebra — $V_{K'}$ is semisimple over $K'$ as a representation of the abstract group $G(K)$ and thus as a representation of the algebraic group $G_{K'}$ by Zariski-density of $G(K)$ in $G_{K'}$). Hence, $\rho_{K'}$ and $\rho'$ are semisimple representations of $G(K')$ with the same characteristic polynomial, so by Brauer-Nesbitt (which applies to semisimple representations of finite dimension for any abstract group at all) these representations are isomorphic. That isomorphism amounts to a single conjugation in the language of matrices, so it says that as algebraic representations they’re $K'$-isomorphic. This gives that $\rho'$ descends to a semisimple representations of $G$ over $K$ as desired. -QED Theorem<|endoftext|> -TITLE: (Non-)formality for ADE preprojective algebras -QUESTION [6 upvotes]: Given a quiver $Q$, I can associate to $Q$ a certain 2-Calabi-Yau (dg-)algebra $\Gamma_Q$ by a 2-dimensional version of the "Ginzburg dga" construction: i.e., start by doubling $Q$, and then impose the "preprojective" relation $\sum_{x\in Q_1}[x,x^*]=0.$ (See, e.g., this article of Van Den Bergh for more details.) Then $H^0(\Gamma_Q)$ is what is normally referred to as the "preprojective algebra" associated to $Q$. -I am interested the formality of $\Gamma_Q.$ According to this article, we have the following result, assuming the quiver $Q$ is acyclic: -Theorem [Hermes]: The algebra $\Gamma_Q$ is formal if and only if the quiver $Q$ is not of ADE type. -This result is quite surprising to me. First, the heuristic I have always had in mind for ADE quivers is that they are strictly better-behaved than non-ADE quivers, whereas in this case it is the ADE type quivers which seem more complicated. Second, the algebras $\Gamma_Q$ show up in geometrical situations where there is a natural bigrading, and there doesn't seem to be any natural difference in the ADE & non-ADE geometries, so I might hope that I might be able to make a formality argument there which works uniformly across quiver types--but clearly such an argument is bound to fail in type ADE. -The proof in loc. cit. is a calculation, which I don't find particularly enlightening or at least in which I can't find answers to the above questions which satisfy me. So my question is a request for another explanation of this phenomenon. In other words: - -Question: What is special about preprojective algebras in type ADE? Why should I have expected a formality argument to fail in this case where it succeeded for all other quivers? - -REPLY [2 votes]: This is a case in which various heuristics about the 'niceness' of ADE quivers can interfere with each other. One result in this area that might conform to your expectations is that the algebra $\mathrm{H}^0(\Gamma_Q)$ is finite-dimensional if and only if $Q$ is a Dynkin quiver. (If $Q$ is extended Dynkin, then it is infinite dimensional but Noetherian, and for other quivers even Noetherianity fails.) From a certain point of view, this makes the Dynkin case better behaved. -However, it also removes any possibility of $\Gamma_Q$ being formal. As you point out, $\Gamma_Q$ is a $2$-Calabi–Yau dg-algebra, so when it is formal, the ordinary algebra $\mathrm{H}^0(\Gamma_Q)$ is also $2$-Calabi–Yau. But finite-dimensional algebras (at least ungraded ones) cannot have this property: see this answer. -I am not sure which geometrical situations you have in mind, but sometimes one should replace a Dynkin diagram by its affine counterpart when passing from geometry to representation theory. For example, if $R$ is a Kleinian singularity corresponding to a Dynkin diagram $\Delta$, its category of Cohen–Macaulay modules is equivalent to projective modules over $\mathrm{H}^0(\Gamma_Q)$ for $Q$ any orientation of the extended Dynkin diagram $\tilde{\Delta}$ (and this preprojective algebra is a non-commutative crepant resolution of $R$.)<|endoftext|> -TITLE: How to compute (enriched) Cauchy completions? -QUESTION [8 upvotes]: Lawvere famously explained that the following three constructions are all secretly "the same" construction: - -Completing an ordinary category by including splittings of all idempotents. -Completing a linear category by including all direct sums and splittings of idempotents. -Completing a metric space by including limits of all Cauchy sequences. - -Indeed, for any reasonable type of enriched category, there is an "enriched Cauchy completion". The above examples correspond to categories enriched in $\mathrm{Set}$, in $\mathrm{AbGp}$, and in $\mathbb R_{\geq 0}$. -See Karoubi envelope and Cauchy completion in the nLab. -Suppose I hand you some interesting world in which to enrich categories. What techniques are there to work out the meaning of "Cauchy completion" in that world? By "work out the meaning of," I mean for example the statement that an $\mathrm{AbGp}$-enriched category is (enriched) Cauchy complete iff it contains direct sums and splittings of idempotents. -In my case, I have some known (enriched) absolute limits, and I'm mostly trying to prove that my list is complete — that if I have limits for every entry on my list, then I have all absolute limits. - -REPLY [4 votes]: Another approach is to use the fact that absolute weights are small-projective in the enriched functor category, i.e. mapping out of them preserves all colimits. Now take a weight $W$, assumed to be absolute and hence small-projective, and express it as a canonical weighted colimit of representables $W = \mathrm{colim}^W Y$, with $Y$ the Yoneda embedding. Mapping out of $W$ preserves this colimit, so $$\hom(W,W) \cong \hom(W,\mathrm{colim}^W Y) \cong \mathrm{colim}^W \hom(W,Y-).$$ -In particular, the identity map of $W$ corresponds to a map $I \to \mathrm{colim}^W \hom(W,Y-)$. If you have an explicit construction of colimits in your enriching category, you can work out what this means, and then use some naturality to express $W$ in terms of your basic colimits. -For instance, in the case $V=\mathrm{Set}$, colimits are quotients of disjoint unions, so this map gives us a particular object $x$ and a map $W \to Y x$, which naturality implies is a section of the coprojection $Y x \to W$. Hence $W$ is a retract of a representable, so all absolute $\mathrm{Set}$-colimits are generated by splitting idempotents. Similarly, in the case $V=\mathrm{Ab}$, colimits are quotients of direct sums, so this map gives us a finite number of objects $x_1,\dots,x_n$ and a map $W \to Y x_1 \oplus \cdots\oplus Y x_n$, exhibiting $W$ as a retract of this direct sum. Hence all absolute $\mathrm{Ab}$-colimits are generated by finite direct sums and splitting idempotents. And so on.<|endoftext|> -TITLE: Who first defined locally convex topological vector spaces? -QUESTION [12 upvotes]: Who first defined the class of locally convex topological vector spaces? - -REPLY [6 votes]: The reference to Kolmogorov is Studia Math, 5 (1934), 29-33. In my opinion, the most reliable source of references to classical papers is the book by Dunford-Schwartz, Linear Operators. They mention also the paper of von Neumann. -Kolmogorov writes about convexity in his note.<|endoftext|> -TITLE: A property of 47 with respect to partitions into five parts -QUESTION [9 upvotes]: Is 47 the largest number which has a unique partition into five parts (15, 10, 10, 6, 6), no two of which are relatively prime? - -REPLY [6 votes]: Here is a quick demonstration that it is effectively solvable for any number of parts. -A sufficiently large number that has exactly one partition with the property must be prime. Otherwise we can write it as $n = a \cdot b$ with $1 < a \leq b$ and partition $b$ into $k$ parts in two different ways, then multiply them by $a$ to get two different partitions of $n$ with the property. -Now suppose $n = 6 \cdot m + 35$. A sufficiently large $m$ has two different partitions into $k - 1$ parts such that each part of each partition is divisible by either $5$ or $7$. Multiplying by $6$ and appending $35$ then gives two different partitions of $n$ with the property. -On the other hand, suppose $n = 6 \cdot m + 55 = 6 \cdot m + 5 \cdot 11$. The same logic above applies, and that covers all the cases. Filling in some more details results in a simple formula for an upper bound in terms of $k$: the first part says if $n$ is composite then $n \lt (k+2)^2$, and the second part chases down the primes relatively more quickly placing the largest one in $6 \cdot k + O(1)$. -More generally still, we can require that the pairwise $\text{gcd}$ of the parts be greater than or equal to some $d \ge 2$ (so the question of $47$ corresponds to the case $d = 2$ and $k = 5$). An effective upper bound on unique solutions can still be obtained: we just have to carry out the above argument for all congruence classes mod $d$, in each case representing the remainder as a semiprime with prime factors larger than $d$. -EDIT: That led me to the following even simpler version, working mod $2$ instead of mod $6$: if $m$ has two different partitions into $5$ parts, then $n = 2 \cdot m$ has two different partitions into $5$ parts, no two parts of which are relatively prime. So $n < 2 \cdot (5+2) = 14$ if $n$ is even. Otherwise, suppose $n = 2 \cdot m + 15$, and observe that -$18 = 3+3+6+6 = 3+3+3+9$, -$19 = 3+5+5+6 = 3 + 3 + 3 + 10$, and -$20 = 3+3+5+9 = 3+5+6+6$ -to conclude $m \le 17$ and $n \le 49$.<|endoftext|> -TITLE: On the homotopy type of $\mathbb{QP}^\infty$ -QUESTION [6 upvotes]: It can be shown that the infinite-dimensional rational projective space $\mathbb{QP}^\infty$ is a connected, Hausdorff topological space. What can be said about its homotopy type (is it simply connected, is there any hope to compute its cohomology algebra)? - -REPLY [3 votes]: It is nice that you asked a question about the space $\mathbb Q P^\infty$. I have thought about this space for a long time and came to the conclusion that $\mathbb Q P^\infty$ is the most "regular" space among countable connected Hausdorff spaces. -It seems that $\mathbb Q P^\infty$ is a unique space among countable connected Hausdorff spaces that admits a simle topological characterization: -Theorem. A topological space $X$ is homeomorphic to $\mathbb QP^\infty$ if and only if $X$ is countable, Hausdorff, and has a countable base $\mathcal B$ of the topology such that for any $n\ge 2$ and basic open sets $U_1,\dots,U_n\in\mathcal B$ the intersection $\bar U_1\cap\dots\cap \bar U_n$ is connected, non-empty, and has zero-dimensional complement $X\setminus (\bar U_1\cap\dots\cap \bar U_n)$. -The proof can be done by a (more-or-less) standard back-and-forth argument.<|endoftext|> -TITLE: Is every non-negative test function the limit of a sequence of sums of squares of test functions? -QUESTION [11 upvotes]: Let $0\leq f\in\mathscr{D}(\mathbb{R}^n)$. As shown e.g. by J.-M. Bony, F. Broglia, F. Colombini and L. Pernazza, Nonnegative functions as squares or sums of squares, J. Funct. Anal. 232 (2006) 137-147 (see also this MO question and this math.SE question), not every such an $f$ is of the form $f=g^2$ for $g\in\mathscr{D}(\mathbb{R}^n)$ or even a finite sum of such. On the other hand, as mentioned in the paper above, C. Fefferman and D. H. Phong sketched a proof (On positivity of pseudo-differential operators, Proc. Natl. Acad. Sci. U.S.A. 75 (1978) 4673-4674) of the fact that any $0\leq f\in\mathscr{C}^\infty(\mathbb{R}^n)$ can be written as a sum of squares $$f=\sum^k_{j=1}g_j^2$$ with $g_j\in\mathscr{C}^{1,1}(\mathbb{R}^n)$ (i.e. $g_j$ is a differentiable function whose derivatives are locally Lipschitz) for all $1\leq j\leq k$ for some $k\in\mathbb{N}$. This fact was a key ingredient of the proof of the important inequality for scalar pseudodifferential operators with non-negative symbols that bears their name. For a modern, more detailed proof of the above formula, see N. Lerner, Some Facts About the Wick Calculus, in L. Rodino, M.W. Wong (eds.), Pseudodifferential Operators: Quantization and Signals, Lectures given at the C.I.M.E. Summer School held in Cetraro, Italy June 19–24, 2006 (Springer Lecture Notes in Mathematics 1949, 2008), pp. 135-174, particularly Theorem 5.2, pp. 167-172, and the discussion right after Theorem 1.1 in the paper by Bony et alii above on page 139. It immediately follows by multiplication by squares of smooth bump functions that the $g_j$'s may be chosen to be compactly supported if $f\in\mathscr{D}(\mathbb{R}^n)$. Bony et alii showed above that this regularity for $n\geq 4$ is sharp. -All this leads naturally to the following - -Question: Is every $0\leq f\in\mathscr{D}(\mathbb{R}^n)$ the limit of a - sequence of sums of squares in $\mathscr{D}(\mathbb{R}^n)$ in the latter's topology? - -In other words, is the cone of non-negative elements of $\mathscr{D}(\mathbb{R}^n)$ the (sequential) closure of the cone of sums of squares in $\mathscr{D}(\mathbb{R}^n)$? -I am particularly interested in arguments that do not rely on the result by Fefferman and Phong. - -EDIT - Follow-up subquestion: (suggested by André Henriques) Is every $0\leq f\in\mathscr{D}(\mathbb{R}^n)$ the limit of an increasing sequence of sums of squares in $\mathscr{D}(\mathbb{R}^n)$ in the latter's topology? Particularly, can $f$ be written as $$f=\sum^\infty_{j=1}g_j^2$$ with $g_j\in\mathscr{D}(\mathbb{R}^n)$ for all $j$ and convergence in $\mathscr{D}(\mathbb{R}^n)$? - -REPLY [3 votes]: This is a comment on Pietro Majer's answer, with graphics. The first figure shows $x^2$ (red) and the first five functions of Pietro's method approaching it (blue). The second figure shows the double derivatives thereof. As you can see, the functions are converging but their second derivatives are not.<|endoftext|> -TITLE: Is the connected sum of a triangulable manifold with a non-triangulable manifold a non-triangulable manifold? -QUESTION [5 upvotes]: Let $M,N$ be topological manifolds such that $M$ does not admit a $PL$ structure and $N$ does. Is $M\#N$ still a triangulable manifold? - -REPLY [12 votes]: In high dimensions Galewski and Stern (and independently Matsumoto) proved that a manifold $M$ is triangulable iff $\beta \Delta(M)=0$. Here $\Delta \in H^4(M;\Bbb Z/2)$, and $\beta$ is the Bockstein corresponding to the short exact sequence $\ker \mu \to \Theta_{\Bbb Z} \to \Bbb Z/2$. The middle term is the homology cobordism group of 3-manifolds and the last map is the Rokhlin invariant of 3-manifolds. -These invariants are additive under connected sum. Therefore if $\beta \Delta(N)$ is nonzero, so is $\beta \Delta(M \# N)$ as long as the dimension is at least 6, and so $M \# N$ remains non-triangulable. -The only leftover mystery cases are dimensions 4 and 5. In dimension 4 triangulable manifolds are automatically PL and therefore smooth. So we see that $M \# N$ is smoothable only if $M$ and $N$ are not both definite manifolds of the same signature, but otherwise we are out of luck. In dimension 5, $M \# N$ is not triangulable if $M$ is but $N$ is not: $$\beta \Delta(M \#N) = \beta \Delta(M) + \beta \Delta(N) = \beta \Delta(N) \neq 0.$$ But eg $M \# M$ is always triangulable.<|endoftext|> -TITLE: Splitting of exact triangles in derived category -QUESTION [6 upvotes]: Let $A\to B\to C\to A[1]$ be a distinguished triangle in a (bounded below) derived category of an abelian category. - -Is there a necessary and sufficient condition that it splits, namely $B\simeq A\oplus C$ in a way compatible with the morphisms in the triangle? - -Remark. A necessary condition is that the image of the identity morphism in $id\in Hom(C,C)\to Hom(C,A[1])$ vanishes. -This condition is sufficient provided all the objects $A,B,C$ belong to the given abelian category (rather than derived one) and form a short exact sequence. I do not know if this condition is sufficient in general. - -REPLY [13 votes]: In any triangulated category, the necessary and sufficient condition for a distinguished triangle $A\to B\to C\to A[1]$ to split is that the morphism $C\to A[1]$ in this distinguished triangle vanishes. This morphism is the same thing as "the image of the identity morphism under $Hom(C,C) \to Hom(C,A[1])$" that you mention in your Remark. -Indeed, this condition is necessary, because, given a splitting $C\to B$ of the morphism $B\to C$, i.e., assuming that the composition $C\to B\to C$ is the identity morphism, our morphism $C\to A[1]$ decomposes as $C\to B\to C\to A[1]$, which is zero since the composition $B\to C\to A[1]$ in any distinguished triangle vanishes. -This condition is sufficient, because, for any distinguished triangle $A\to B\to C\to A[1]$ with a zero morphism $C\to A[1]$, applying $Hom(X,-)$ for any object $X$ in your triangulated category produces an exact sequence $0\to Hom(X,A) \to Hom(X,B) \to Hom(X,C) \to 0$. -In any additive category (and a triangulated category is additive by definition; in particular, any derived category is additive), exactness of such sequences for all objects $X$ implies a splitting $B\simeq A\oplus C$ compatible with the morphisms $A\to B$ and $B\to C$. -Indeed, taking $X=C$ you can see that there is a morphism $C\to B$ such that the composition $C\to B\to C$ is the identity morphism. Denoting by $f\colon B\to B$ the result of subtracting the composition $B\to C\to B$ from $id_B$ and taking $X=B$, you can see that there is a morphism $B\to A$ such that $f$ is equal to the composition $B\to A\to B$. So $id_B$ is the sum of the two compositions $B\to A\to B$ and $B\to C\to B$. -Finally, the composition $A\to B\to A\to B$ is equal to $A\to B$, since the composition $A\to B\to C\to B$ vanishes. As the morphism $Hom(A,A)\to Hom(A,B)$ is injective (take $X=A$), it follows that the composition $A\to B\to A$ is equal to $id_A$. The two morphisms $A\to B$ and $B\to C$ that we had to begin with, together with the two morphisms $C\to B$ and $B\to A$ that we have found, provide the desired isomorphism $B\simeq A\oplus C$.<|endoftext|> -TITLE: Does Lackenby's polynomial bound on knot moves imply polynomial mixing in "Quantum Money From Knots?" -QUESTION [14 upvotes]: In the 2010 paper Quantum Money from Knots Farhi, Gosset, Hassidim, Lutomirski, and Shor give a doubly stochastic Markov chain acting on grid diagrams. Transitions in the Markov chain are permutations of the configuration space of grid diagrams, given by random Cromwell moves. They imply that the security of their quantum money system is dependent on the rapid polynomial-time mixing of their Markov chain. They comment that, at least at the time it was written, even for the unknot it wasn't known if there were two equivalent grid diagrams requiring a superpolynomial number of moves to go from one to the other. -In the 2013 paper A polynomial upper bound on Reidemeister moves Lackenby shows that the number of Reidemeister moves needed to untangle a diagram of the unknot with $c$ crossings is, at most, $(236c)^{11}$. From other comments, it appears that Lackenby has extended the above to show that arbitrary knots can be converted to one another with a polynomial number of Reidemeister moves. - -Is Lackenby's result strong enough to show, or lend credence to, Farhi, Gosset, Hassidim, Lutomirski, and Shor's conjectured polynomial-time mixing? - -I envision Lackenby's result as putting a polynomial upper bound on the diameter/God's number of the graph of Reidemeister moves - similar to the graph of Cromwell moves that can be randomly walked with Farhi, Gosset, Hassidim, Lutomirski, and Shor's Markov chain. - -EDIT January 6, 2018 -To give more detail, states in the version Farhi, Gosset, Hassidim, Lutomirski, and Shor's Markov chain ("the Markov chain") that I am picturing consist of grid diagrams $G$ reachable from a starting diagram $\tilde{G}$ of dimension $\bar{D}$, that do not increase the dimension beyond some maximum $\alpha$. (In their paper, $\alpha$ was set to $2\bar{D}$; however, for the purposes of the question I think we must require $\alpha$ to be polynomial in $\bar{D})$. -The Markov chain as described in the paper further includes a weighting parameter $i$; the limiting distribution of the Markov chain is uniform over pairs $(G,i)$. However, for the purposes of the question, I think $i$ is not needed, and the limiting distribution need merely be uniform over all grid diagrams $G$ reachable from $\tilde{G}$ that don't increase the dimension by more than $\alpha$. Of course, most grid diagrams $G$ will likely be of dimension $\alpha$, because as shown below the Markov chain is more likely to randomly tie a knot than to untie it. -In more detail, transitions may be decided from the following parameters. - -Let $j$ be a integer drawn uniformly from $2$ to $8$ - this is used to decide which Cromwell move to apply, Cromwell moves being one of left/right/up/down cyclic permutations if $2\le j\le 5$, horizontal/vertical transpositions if $6\le j\le 7$, or stabilizations/destabilizations if $j=8$. -Let $x$, $y$ be integers drawn uniformly from $1$ to $\alpha-1$ - these are used when $6\le j\le 8$ to decide which row/column in the grid diagram to apply transpositions or stabilizations/destablizations, if possible. If the vertex does not allow such a move, then no transition occurs, and the Markov chain simply does a self-loop. Stabilizations increase the grid dimension and require a marker at $(x,y)$. Destabilizations reduce the grid dimension and require an absence of a marker at $(x,y)$ and a presence of a marker at $(x+1,y+1)$, $(x+1,y)$, and $(x,y+1)$. Thus, following the Markov chain, it's easier to randomly tie a knot than to untie the knot. -Let $k$ be an integer drawn uniformly from $0$ to $3$. This is used to rotate the grid diagram (by $90k$ degrees clockwise) prior to applying the stabilization/destabilization, and rotate it again (by $90k$ degrees counterclockwise) after the stabilization/destabilization. - -The authors argue that because each transition can be a permutation of the grid space, the Markov chain is doubly stochastic, hence the limiting distribution will be uniformly distributed. -Thus, given an initial grid diagram $\tilde{G}$ of dimension $\bar{D}$, repeatedly choose random tuples $(j,x,y,k)$ a polynomial number of times, applying the above transitions. Knowing the Lackenby bound, will the distribution be uniform over all grid diagrams equivalent to $\tilde{G}$ of dimension $\le \alpha$, or could there still be some bottleneck? How about if $\tilde{G}$ were known to be of a particular knot type, say the unknot? - -REPLY [8 votes]: Thanks to HJRW2 for the flattering invitation here, and I will give an answer, but it might be not all that deep. In fact I haven't been on MO much lately; maybe I should visit it more. -I don't see any basis to say that Lackenby's result proves the mixing property of the quantum mixing proposal. There are many graphs where the diameter is better understood than the mixing time. Lackenby's methods are based on Dynnikov monotonicity for grid diagrams, and for that reason and others there isn't any statement about getting from here to there with random moves. -The picture is even worse for other knot types, since the exponent in Lackenby's polynomial bound depends on the knot type. -Nonetheless you can fairly say that it lends some credence to the mixing time question, at least for the unknot. Lackenby's bounds are spectacular, but no one thinks that they are sharp. Moreover no one knows how to make examples of the unknotting problem that look hard, i.e., nothing like the RSA factoring challenges with money at stake. An algorithm with a rigorous bound on running time is hard, but that's not the same thing. The graph isomorphism problem was in the same state of deadlock before Babai's algorithm. -On the other hand the quantum money from knots protocol would need more credibility than just the mixing time proposal for people to trust it. It's perfectly respectable as a provocative proposal, but I don't know that it's convicing, at least not yet. There is another quantum money proposal due to Aaronson and Christiano that could be more convincing; on the other hand it's less glamorous in that it uses linear algebra over finite fields rather than knot theory.<|endoftext|> -TITLE: Classifying space as the geometric realization of the nerve of $G$ viewed as a small category -QUESTION [9 upvotes]: Let $C$ be a small category: we define its nerve $(N(C)_k)_k$ as the following simplicial set: $N(C)_0=Ob(C)$ (the set of objects), $N(C)_1=Mor(C)$ (the set of all morphisms) and $N(C)_k$ to be a set of all $k$-tuples of compasable morhpisms $(f_1,...,f_k)$. This is equipped with the face maps $d_i$ defined by $d_i(f_1,...,f_k):=(f_1,...,f_{i-1},f_i \circ f_{i+1},f_{i+2},...,f_k)$ except of $d_0$ and $d_k$ which omit the first and the last morphism. Degeneracies act as follows $s_j(f_1,...,f_k):=(f_1,...,f_{j-1},id,f_j,...,f_k)$. -When $G$ is a group then $G$ can be viewed as a small category with only one object and the set of morphisms being $G$ itself. Thus everything is composable and $N(G)_k=G^k$. -One thus arrives at the well defined simplicial set. Therefore one can speak about its geometric realization $|N(G)|$: there are in fact two possible versions, the standard one, sometimes called ,,thin realization'' and another version, which is called ,,thick'' realization which is denoted by $\| N(G) \|$ and which is defined using only relations involving the face maps. See also this discussion. -I was told that this geometric realization is in fact (possible version of) the classifying space $BG$. I wonder why it is true. This question is motivated by this discussion. Since my original motivation was the statement that the group cohomology coincides with the (singular for example) cohomology of the classifying space and with the help of the cited discussion I was able to understand why is it true directly (not invoking the fact that $BG$ is geometric realization of $N(G)$) I'm posting this question as another topic. So my question is: - -Why the geometric realization of $N(G)$ is (homotopy equivalent to) $BG$? - -Some ideas can be found in Hatcher's book ,,Algebraic Topology'' however, as I understood correctly, the face maps $d_i$ from this book would act rather as $(g_1,...,g_k) \mapsto (g_1,...,g_{i-1},g_{i+1},...,g_k)$ which is not the same as for $N(G)$. -EDIT: The classifying space is not unique up to homeomorphism, but it is unique up to homotopy: the definition which is most familiar for me is the abstrack definition as the quotient of contractible principial $G$-bundle $EG$. There is also specyfic construction due to Milnor using infinite join which is also familiar to me. -EDIT 2: As I understood correctly the answer below: each element from $\mathcal{E}G_n$ being of the form $(g_0,g_1,...,g_n)$ may be presented in the form $(g_0',g_0'g_1',...,g_0'g_1'...g_n')$. Solving in $g_0',...,g_n'$ gives $g_0'=g_0,g_1'=g_0^{-1}g_1,...,g_n'=g_{n-1}^{-1}g_n$. Thus we obtain a bijective map $\alpha:\mathcal{E}G_n \to \mathcal{E}G_n$ given by $\alpha(g_0,...,g_n)=(g_0,g_0^{-1}g_1,...,g_{n-1}^{-1}g_n)$. Let $d_i,s_j$ be the face maps and degeneracies which you described (omitting $i$-th vertex and repeating) and $d_i',s_j'$ be faces and degeneracies as in the definition of the nerve of the category. Then $\alpha s_j=s_j' \alpha$ and $\alpha d_i=d_i' \alpha$ for all $j$ and for all $i$ except $i=0$ where we obtain the difference on the zeroth coordinate $(\alpha d_0)(g_0,...,g_n)=(g_1,...)$ and $(d'_0 \alpha)(g_0,...,g_n)=(g_0^{-1}g_1,...)$ (the remaining entries are the same). Therefore if we pass to the quotient we get an equality. This suggest that there is something correct in this procedure: however I'm little bit confused with the following: $\mathcal{E}G$ is a simplicial set. Its geometric realization is just topological space: the bar notation which you are using refers to $EG/G$ which is also topological space. However you described face and degeneracies maps for $EG/G$ which looks like the relevant maps for $NG$ but the former is just topological space while the latter is (abstract) simplicial set. - -REPLY [7 votes]: You should ignore simplicial objects at first, and just consider groupoids. In the following, you can let $G$ be a topological group such that $e\hookrightarrow G$ is a closed cofibration. All groupoids will be by default internal to $\mathbf{Top}$, even if not stated. -Consider three groupoids: $G$ itself considered as a groupoid with only identity arrows; $\mathbf{E}G$ with objects $G$ and a unique arrow between any two objects (thus the arrows are $G\times G$ with projections as source and target); $\mathbf{B}G$ with one object and $G$ the automorphisms of that object. -There are functors $G \to \mathbf{E}G \to \mathbf{B}G$. The first is the identity on objects, the second sends the arrow $(g,h)$ to $gh^{-1}$. Both $G$ and $\mathbf{E}G$ are (strict) group objects in $\mathbf{Gpd}$, and there is a free action $G\times \mathbf{E}G \to \mathbf{E}G$ of $G$ on $\mathbf{E}G$ by right multiplication after inclusion. The quotient by this action is exactly $\mathbf{B}G$, by the given functor. This can be seen directly. -Further, $\mathbf{E}G$ is canonically equivalent to the trivial groupoid $\ast$, in the sense that $\mathbf{E}G \to \ast$ has a canonical section and $\mathbf{E}G \to \ast \to \mathbf{E}G$ is canonically isomorphic (by $\alpha$, say) to the identity. So you can view $\mathbf{E}G \to \mathbf{B}G$ as being a principal $G$-bundle object with contractible total space. -Now the nerve functor preserves products, so there is a free action of the constant simplicial object on $G$ on $N\mathbf{E}G$, and moreover the quotient by this action is exactly $N\mathbf{B}G$, by the simplicial map $N\mathbf{E}G \to N\mathbf{B}G$. The natural isomorphism $\alpha$ gives a simplicial homotopy $N\alpha$, so that $\mathbf{E}G$ is a contractible simplicial object. So we have a $G$-bundle in simplicial objects with contractible total space. -Now do geometric realisation (not the thick geometric realisation), which preserves products, so that you get a free $G$-action on $EG := |N\mathbf{E}G|$ whose quotient is $BG :=|N\mathbf{B}G|$. The simplicial homotopy gives a contracting homotopy for $EG$. Local triviality of $EG \to BG$ is a bit harder, but possible. The open cover over which one trivialises has a subordinate partition of unity, by construction. -So $BG$ constructed by the geometric realisation of the nerve of $\mathbf{B}G$ is the base of a (numerable, even) principal $G$-bundle with contractible total space.<|endoftext|> -TITLE: Allowing $G$-CW complexes to have more general cells -QUESTION [18 upvotes]: Let $G$ a finite group. I've seen three options discussed for making $G$-cell complexes: in increasing generality, one might allow $X_n$ to be constructed from $X_{n-1}$ by attaching cells of the form - -$G/H\times D^n$, where $D^n$ has trivial $G$-action, or -$G/H\times D(V)$, where $D(V)$ is the unit disk of a $G$-representation $V$, or -$G\times_H D(V)$, where $D(V)$ is the unit disk of an $H$-representation $V$ - -(The subgroup $H$ and representation $V$ is allowed to vary for different cells. Cells of type 1 are what are used in the standard definition of $G$-CW complex.) - -Megan Shulman, on p.47 of her thesis (link goes directly there), says regarding standard $G$-CW complexes that - -If we are interested in putting CW structures on spaces found in nature, this is a very restrictive definition... - -However, I seem to remember that even for a $G$-cell complex $X$ constructed with cells of type 3, we can always rearrange (triangulate?) somehow to give $X$ a standard $G$-CW complex structure. Is that true or false? Which papers look into this? - -Ferland and Lewis, on p.23 of their paper (link goes directly there), say that cells of type 3 - -... are of interest because they arise naturally from equivariant Morse theory (see, for example, [21]). Further, if $G$ is a finite abelian - group, then the usual Schubert cell structure on Grassmannian manifolds generalizes in an obvious way to a generalized $G$-cell structure on the Grassmannian manifold $G(V, k)$ of $k$-planes in some $G$-representation $V$ (see Chapter 7). -[21] A. G. Wasserman, Equivariant differential topology, Topology 8 (1969), 127–150. - -However, skimming over at that section of Wasserman's paper (link goes directly there), I must admit I don't see where anything like cells of type 3 show up, much less demonstrate their usefulness. Could anyone explain their relationship to equivariant Morse theory? -Also, in their Chapter 7 (link goes directly there; look at the bottom of p.83 & top of p.84 in particular), I'm not sure I see where cells of type 3 are necessary, because it seems like $W$ is inherently a $G$-representation, and hence they are only obtaining cells of type 2. Could anyone clarify this? - -So to summarize, I'd greatly appreciate any of the following: - -simple / natural examples that illustrate the additional flexibility of each generalized cell: - -spaces which can easily be seen to have a cell structure when cells of type 3 are allowed, but not when only cells of type 2 are allowed -spaces which can easily be seen to have a cell structure when cells of type 2 are allowed, but not when only cells of type 1 are allowed - -any other reasons why cells of type 2 and 3 are reasonable -a citation / proof / counterexample for this half-remembered "fact" that cells of type 1 suffice - -REPLY [3 votes]: Just a quick note adding to Steve's answer, with a focus on the stable setting. In the Hill-Hopkins-Ravenel paper on the Kervaire problem, they introduce the "positive complete model structure" on G-spectra, and the "complete" part gives you cells of type 3. This is on page 161 of the last arxiv version (which I think is super close to the published annals version) https://arxiv.org/pdf/0908.3724v4.pdf -They needed these cells in many places in the paper, but most notably to prove that, for every cofibrant spectrum $X$, the natural map $(E\Sigma_n)_+ \wedge X^{\wedge n} \to X^{\wedge n}/\Sigma_n$ is a weak equivalence (where $\Sigma_n$ denotes the symmetric group on $n$ letters). They use this property to set up a good homotopy theory of commutative monoids. This property was claimed without proof in earlier sources setting up the homotopy theory of equivariant spectra, but the proof is really non-trivial. It follows for the positive and positive flat model structures, but I only know how to prove it using the positive complete model structure. I believe Doug Ravenel has continued to investigate this model structure. I guess my point is that you should not restrict yourself to looking at classical sources, because the decision about which cells to use was still in flux till very recently. Ravenel gave a great talk a few times in 2015 where he argued that this positive complete model structure is the definitive right answer (at least, for equivariant stable homotopy theory).<|endoftext|> -TITLE: Negative curves on surface of general type -QUESTION [8 upvotes]: Let $X$ be a complex projective surface of general type, $K_X$ be the very ample canonical divisor (which is automatically minimal). Is there an example for such a $X$ that there exist a sequence of negative curves $\{C_i\}_{i=1}^{\infty}$ with $C_i^2$ being fixed, but the geometric genus $g(C_i)$ going to infinity? - -REPLY [9 votes]: Yes. Start with a rational surface $S$ containing infinitely many $(-1)$-curves $E_n$ (for instance $\mathbb{P}^2$ blown up along 9 general points). Choose a very ample divisor $H$ on $S$, a smooth curve $B$ in the linear system $|2H|$, and consider the double covering $\pi : X\rightarrow S$ branched along $B$. Then $K_X\cdot \pi ^*E_n=2(K_S+H)\cdot E_n=-2+2H\cdot E_n\ $ goes to infinity with $n$, and so does $g(\pi ^*E_n)$. -Note that if $\pi ^*E_n$ is not irreducible, it is the union of two smooth rational curves, and $X$ contains only finitely many such curves (Lu-Miyaoka); so $\pi ^*E_n$ is irreducible for all but finitely many $n$.<|endoftext|> -TITLE: Difference of Beilinson conjecture and equivariant Tamagawa number conjecture -QUESTION [6 upvotes]: As stated in the title, I am wondering the main difference between Beilinson conjecture and eTNC. If I read correctly, I can see that there are many literature treating both conjectures in the same line - special values of L-functions. If my impression is completely wrong, I am sorry for this stupid question. But, if they have some commonness, it would be greatly helpful if anyone can clarify their difference. - -REPLY [8 votes]: The key difference between these conjectures is the coefficient ring that is involved. You also are leaving out an important "middle" conjecture -- the (non-equivariant) Tamagawa number conjecture, as formulated in Bloch and Kato's article in the Grothendieck Festschrift -- and knowing what this conjecture says might clarify the relation between the other two conjectures a bit. -In Beilinson's conjecture, the coefficient ring is $\mathbf{Q}$: the conjecture predicts the leading terms of $L$-functions up to rational factors, in terms of the determinant of a regulator matrix defined using a $\mathbf{Q}$-basis of some motivic cohomology space. -In the Bloch--Kato Tamagawa number conjecture (TNC), the coefficient ring is $\mathbf{Z}_\ell$ for some prime $\ell$, and the conjecture predicts the leading terms of $L$-functions up to $\ell$-adic units. The extra integral structure comes from comparing motivic cohomology with Galois cohomology, which (unlike motivic cohomology) works well with $\mathbf{Z}_\ell$ coefficients. -In the equivariant Tamagawa number conjecture (ETNC), one introduces further extra structure by considering a Galois extension $K / k$ and studying the arithmetic of a motive over $k$ base-extended to $K$, which means that all the cohomology groups are modules over the group ring $\mathbf{Z}_\ell[G]$ where $G = \operatorname{Gal}(K/k)$ is some finite group. Then the conjecture relates the leading terms of equivariant (i.e. group-ring-valued) $L$-functions to the isomorphism classes of cohomology groups as $\mathbf{Z}_\ell[G]$-modules. -(You could also formulate an "equivariant Beilinson conjecture" involving $\mathbf{Q}[G]$-modules, but it would be fairly trivally equivalent to the original Beilinson conjecture for each of the motives $M \otimes \rho$ where $\rho$ varies over representations of $G$. For the same reason, ETNC reduces to TNC when $\ell \nmid \#G$. The interesting cases are when $G$ is an $\ell$-group, in which case there are non-trivial congruences between $\mathbf{Z}_\ell$-representations of $G$ which the ETNC detects.) -So, to sum up, the triple of conjectures (Beilinson conj, TNC, ETNC) all seek to relate $L$-values to arithmetically-interesting cohomology modules, but the cohomology modules are over the rings $(\mathbf{Q},\ \mathbf{Z}_\ell,\ \mathbf{Z}_\ell[G])$ respectively.<|endoftext|> -TITLE: Does the equation $x^2+x=a$ have an integer solution? -QUESTION [7 upvotes]: I am writing a paper on the topological structure of the Golomb space (defined here) and arrived to the following question: -Question 1. Is it true that for a number $a\in\mathbb N$ the equation $x^2+x=a$ has an integer solution $x$ if and only if for any number $b\in\mathbb N$, coprime with $a$, the equation $x^2+x=a \mod b$ has a solution $x$ (i.e., a solution in the ring $\mathbb Z/b\mathbb Z$). -In fact, I need a more general fact. -Question $2^n$. Is it true that for any number $n\ge 0$ and any number $a\in\mathbb N$ the equation $(x^2+x)^{2^n}=a$ has an integer solution $x$ if and only if for any number $p\in\mathbb N$, coprime with $a$, the equation $(x^2+x)^{2^n}=a \mod p$ has a solution $x$ (i.e., a solution in the ring $\mathbb Z/p\mathbb Z$). -We can also ask a more general -Problem. For which monic polynomials $f\in\mathbb Z[x]$ the following local-to-global principle holds: -$(*)$: for every $b\in\mathbb N$ the equation $f(x)=a$ has a solution in $\mathbb Z$ if and only if for any $b\in\mathbb N$, relatively prime with $a$ the equation $f(x)=a \mod b$ has a solution in the ring $\mathbb Z/b\mathbb Z$? -Remark. I have edited the second question a little bit. - -REPLY [9 votes]: Let me try to answer the more general problem. -First, there will probably not be any useful statement if the polynomial is not monic. The reason is that it is almost impossible to distinguish a rational root from an integer root, as is shown by examples like $(2x-1) (x^2+1)$ where the first factor has roots modulo every prime but $2$, and the second factor has roots mod $2$, but neither factor has integer roots. -Second, there are issues with certain special Galois groups. One example is $(x^2-2)(x^2-3)(x^2-6)$. The point is that if $a$ and $b$ are nonsquares in $\mathbb F_p$, then $ab$ is always a square. A lower-degree example is $(x^3+x+1)(x^2+ 31)=0$. If the cubic has no roots mod a prime, its discriminant must be a perfect square. -The way to avoid this is to put some condition on the Galois group of the polynomial that forces the implication "every element of the Galois group fixes some point $\implies$ some point is fixed by every element of the Galois group" to hold. This holds fine for subgroups of $S_2,S_3,S_4$ and thus the implication on roots holds for monic polynomials of degree at most $4$.<|endoftext|> -TITLE: Consistency of a non-measurable set of reals when the continuum cannot be well-ordered -QUESTION [6 upvotes]: Can it be shown, on the assumption that $ZF$ is consistent, that there is a model of $ZF$ in which the reals cannot be well-ordered but there does exist a set of reals which is not Lebesgue measurable? - -REPLY [13 votes]: Shelah proved that assuming $\sf ZF+DC$, if every set of reals is Lebesgue measurable, then $\omega_1$ is inaccessible to reals. -This alone should hint you that it is easy to arrange models where $2^{\aleph_0}$ cannot be well-ordered, $\sf DC$ holds, and there are non-measurable sets. -More specifically, any free ultrafilter on $\omega$ generates a non-measurable set of reals. So working in any model where $\omega$ has free ultrafilters is enough. -(As Joel remarked, assuming $\sf DC$ is sort of essential, since otherwise the usual notion of measure could break apart, e.g. the reals can be a countable union of null sets, or even countable sets.)<|endoftext|> -TITLE: Diophantine equation $x^p+ax=y^p+by$ -QUESTION [5 upvotes]: Problem. Is there a prime number $p$ (desirably $p\le 3$) and an infinite set $A\subset\mathbb N$ such that for any distinct numbers $a,b\in A$ the Diophantine equation $x^p+ax=y^p+by$ has no positive integer solutions? - -REPLY [3 votes]: My suspicion is that every prime except $2$ has this property. I gave up on $p=2$ after realizing that (except for a=1,2,4) $x^2+ax=y^2$ always has solutions. -Here is some circumstantial evidence for $p=3,$ which should be harder than larger $p.$ -If my calculations are correct, the following set $$A=\{0,40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 62, 64, 65, 66,$$ $$ 67, 68, 69, 70, 72, 73, 74, 75, 76, 78, 80, 81, 82, 84, 85, 88, 92, 93, 94, 96, 100, 102\}$$ of size $44$ is such that there are no solutions of $x^3+ax=y^3+by$ in positive integers $y \gt x$ with $a \gt b$ and $a,b\in A.$ - -Here are some details: -It is pretty fast to check if $x^3+ax=y^3+by$ has integer solutions. We may assume $a \gt b \geq 0.$ If any solutions occur, they must have $ \frac{a-b-3}3 \gt x:$ -To have $x^3+ax=y^3+by$ requires that $y \geq x+1.$ Thus there can only be solutions provided that $$x^3+ax \geq (x+1)^3+b(x+1) \gt x^3+3x^2+3x+bx$$ So $$(a-b-3)x>3x^2.$$ -Actually, I don't think that $x$ can be anywhere near that large. Here are all the cases $[a,[x,y]]$ with $0 \lt x \lt y\ $ and $x^3+ax=y^3$ such that $ a \leq 104:$ -$ [7, [1, 2]], [26, [1, 3]], [28, [2, 4]], \mathbf{[38, [4, 6]], [61, [8, 10]]}, [63, [1, 4], [3, 6]], [104, [2, 6]]$ -So the other $97$ values of $a$ with $1 \leq a \leq 103$ are all compatible with being in $A$ if $0 \in A.$ But some pairs are not compatible with each other. Reducing to the set $A$ listed above makes all pairs compatible. -It is a fairly ad-hoc set selected by making a graph with the $97$ values as vertices and all $323$ bad pairs $(a,b)$ as edges, deleting some vertices of highest degree and repeating. - -Note that whenever $[a,[x,y]]$ makes $x^3+ax=y^3$, so also does $[d^2a,[dx,dy]].$ Also $[a,[x,y]]=[k^3-1,[1,k]]$ makes $x^3+ax=y^3.$ These two facts together account for five of the seven pairs above with $a \lt 104,$ the other two are in bold. -Up to $a=5000$ there are only $140$ values of $a$ incompatible with $0 \in A.$ $114$ of them have only one bad pair $x,y$ with $x^3+ax=y^3$ and the other $26$ have two bad pairs. About half of the bad values of $a$ are of the form $a=d^2(k^3-1)$ and explained by the two facts above. -Of these $140$ there are $46,28,25,23$ and $18$ respectively in the intervals $[1,1000],[1001,2000],[2001,3000],[3001,4000]$ and $[4001,500].$<|endoftext|> -TITLE: Complexes in stable categories -QUESTION [6 upvotes]: Generalizing from 1-category theory, there's a simple definition of a "naive complex" in a stable $\infty$-category. Considering bounded positive graded chain complexes, they are a sequence of maps -$$ A_n \xrightarrow{d_n} A_{n-1} \xrightarrow{d_{n-1}} \ldots \xrightarrow{d_1} A_0 $$ -with the property that $d_k d_{k+1} \simeq 0$. If I've not made any errors, then at first blush they seem to have a reasonable theory, along with a nice "realization" given by iteratively taking homotopy cofibers: -$$ |A| = \mathrm{cofib}(\mathrm{cofib}(\ldots) \to A_0) $$ -that parallels taking the total complex of a bicomplex. This realization can also be given by the colimit of the diagram -$$ A_n \rightrightarrows^{d_n}_0 A_{n-1} \rightrightarrows \ldots \rightrightarrows^{d_1}_0 A_0$$ - -However, naive complexes don't seem to be studied. Instead, in Higher Algebra, Lurie proves a Dold-Kan correspondence stable $\infty$-category asserting equivalences between the categories of: - -Simplicial objects -Filtered objects (i.e. arbitrary $\mathbb{Z}_{\geq 0}$-indexed diagrams) -upper-triangular array with zeroes along the diagonal, in which every square is a pushout - -and it is this last sort of thing that Lurie calls a complex. (Specifically, a $\mathbb{Z}_{\geq 0}$-complex) -It is only when looking at the homotopy category does Lurie say anything about a correspondence between simplicial objects and naive complexes. -So my question is whether this is an oversight; i.e. - -Is the $\infty$-category of simplicial objects in a stable $\infty$-category equivalent to the $\infty$-category of (unbounded, positively graded) naive complexes? - -and if the answer is no, what goes wrong? - -REPLY [5 votes]: Here is the problem with the notion of naive complex. Suppose we have a naive complex -$$ \require{AMScd} \begin{CD} -A @>f>> B @>g>> C @>h>> D \end{CD} $$ -If we propose to compute the realization iteratively, the first step would be to produce the sequence -$$ \require{AMScd} \begin{CD} -\mathrm{cofib}(f) @>>> C @>h>> D \end{CD} $$ -However, this need not be a naive complex! So everything falls apart. -A more elaborate explanation is to look at the big diagram of pushout squares one can construct from the complex: -$$ \require{AMScd} \begin{CD} -A @>f>> B @>>> 0 -\\ @VVV @VVV @VVV -\\ 0 @>>> \mathrm{cofib}(f) @>>> \Sigma A @>>> 0 -\\ & & @VVV @VVV @VVV -\\ & & C @>>> \mathrm{cofib}(g) @>>> \mathrm{cofib}(\mathrm{cofib}(f) \to C) -\\ & & & & @VVV -\\ & & & & D -\end{CD} $$ -There is no direct way to produce the expected map $\mathrm{cofib}(\mathrm{cofib}(f) \to C) \to D$. In fact, by the rightmost square, one can only possibly exist if $(\Sigma A \to D) \simeq 0$. -The map $\Sigma A \to D$ is the Toda bracket mentioned in the comments. (wikipedia nlab) - -We can give an explicit counterexample. In the $\infty$-category of chain complexes, suppose our naive complex is given by: -$$ \require{AMScd} \begin{CD} -0 @>>> 0 @>>> \mathbb{Z}/4 @>1>> \mathbb{Z}/4 -\\ @VVV @VVV @VV4V @VVV -\\ \mathbb{Z} @>2>> \mathbb{Z}/4 @>2>> \mathbb{Z}/8 @>>> 0 -\end{CD} $$ -That this is a naive complex can be checked on the homology groups. This isn't a bicomplex as is, and the point is that it cannot be improved to become one! One can show that after replacing $A \xrightarrow{f} B$ with $\mathrm{cofib}(f)$, the diagram becomes -$$ \require{AMScd} \begin{CD} -\mathbb{Z} @>1 + 2n>> \mathbb{Z}/4 @>1>> \mathbb{Z}/4 -\\ @VV2V @VV4V @VVV -\\ \mathbb{Z}/4 @>2>> \mathbb{Z}/8 @>>> 0 -\end{CD} $$ -where $n$ encodes the choice of chain homotopy and on homology groups, the top row does not compose to zero for any choice of $n$. - -The fix is to redefine "naive complex" to include the parallel zeroes: i.e. so that the index 1-category is a sequence of parallel pairs of arrows such that each arrow coequalizes the pair before it. -(the second arrow in each pair is still required to be zero!) -I interpret Pavlov's comment above as saying that this redefinition of "naive complex" results in a category of complexes that is equivalent to simplicial objects, although I have not proved that myself.<|endoftext|> -TITLE: Homotopy equivalence of diffeomorphism groups -QUESTION [23 upvotes]: Let $M$ be a closed connected smooth manifold and let ${\rm Diff}^r(M)$ be the group of $C^r$-diffeomorphisms equipped with the compact-open $C^r$-topology. I am looking for a reference to the fact that the natural inclusion ${\rm Diff}^r(M)\to {\rm Diff}^1(M)$ is a homotopy equivalence for $1\leq r\leq \infty$. - -REPLY [10 votes]: I am not aware of a detailed reference but here is a sketch. - -$\mathrm{Diff}^r(M)$ is a Hilbert manifold (i.e., it is locally homeomorphic to a separable Hilbert space). This can be found e.g., in section 10 of Michor's "Manifolds of differentiable mappings". EDIT: Instead of referring to Michor who indeed focuses on the $C^\infty$ case let me give a direct proof. Let $\exp$ be the exponential map of some smooth Riemannian metric on $M$. Then any self-diffeomorphism $\phi$ of $M$ defines a vector field $X$ given by $\phi(p)=\exp_p X(p)$. Conversely, any vector field on $M$ can be exponentiated to a self-map of $M$ so that the vector field is close to zero if and only if the corresponding diffeomorphism is close to the identity. Also $\phi$ is $C^r$ if and only if $X$ is $C^r$. This defines a homeomorphism of a neighborhood of the identity in $\mathrm{Diff}^r(M)$ onto a neighborhood of the origin in the separable Frechet space of $C^r$ vector fields on $M$. Since $\mathrm{Diff}^r(M)$ is topologically homogeneous we get a chart in every point (not just at the identity). Finally any separable Frechet space is homeomorphic to $\ell^2$, so $\mathrm{Diff}^r(M)$ is a Hilbert manifold. -Any Hilbert manifold is an ANR (because $\ell^2$ is an ANR, and any local ANR is an ANR). Also any ANR it is homotopy equivalent to a CW complex. In particular, there is Whitehead's theorem and to show that the inclusion $\mathrm{Diff}^r(M)\to\mathrm{Diff}^1(M)$ is a homotopy equivalence it is enough to prove that it induces an isomorphism on homotopy groups. -Build a continuous smoothing operator that takes as an input a $C^r$ self-map of $M$, and instantly makes it $C^\infty$. This is a good elementary exercise in differential topology. (Sketch: $C^\infty$ embed the compatible $C^\infty$ structure on $M$ into some $\mathbb R^n$, use methods of Chapter 2 section 2 of Hirsch's "Differential topology" to continuously approximate $C^r$ maps $M\to\mathbb R^n$ by $C^\infty$ maps, and then use tubular neighborhood projection to push the maps to $M$). -The smoothing operator instantly pushes a continuous map from a disk to $\mathrm{Diff}^1(M)$ into $C^\infty(M, M)$, and hence into $\mathrm{Diff}^\infty(M)$ because by the inverse function theorem diffeomorphisms form an open subset in smooth maps. Thus any singular sphere in $\mathrm{Diff}^1(M)$ can be deformed to $\mathrm{Diff}^r(M)$, and if a singular sphere $\mathrm{Diff}^r(M)$ contracts in $\mathrm{Diff}^1(M)$, then the null-homotopy can be pushed into $\mathrm{Diff}^r(M)$. - -In summary, -the inclusion $\mathrm{Diff}^r(M)\to \mathrm{Diff}^1(M)$ induces an isomorphism on homotopy groups, and hence a homotopy equivalence.<|endoftext|> -TITLE: Which constants are ambivalent and why? -QUESTION [6 upvotes]: This question is possibly a bit more philosophical $-$ compatible with the Christmas season, which is an appropriate moment to look at the world from a more universal angle... My last question with a similar flavor was a long time ago. -Some mathematical constants come in pairs with their reciprocal. One of the first examples that comes to mind is the golden ratio, sometimes defined as $\phi=\frac{\sqrt{5}+1}{2}$, sometimes as $\phi=\frac{\sqrt{5}-1}{2}$, with the role of both being perfectly ambivalent. This happens of course because, up to sign, $\phi$ and $1/\phi$ are Galois conjugates. Similarly for $i$ and $-i$ (and in fact any pair of conjugate complex numbers of norm $1$, with a special mention of the 3rd roots of unity), and one could argue that $\phi $ and $i$ and $e^{2\pi i/3}$ are essentially just the simplest non trivial cases of solutions to $x^2+px\pm1=0$. -But before you say "this is trivial, and after all, how should one even define a constant?", let me state that for $e$ and $e^{-1}$ the situation is very similar. While exponential growth is of course quite different from exponential decay, the same principle is at work in both. We may reduce this to a simple sign change in the defining formulae $\lim\limits_{n\to\infty}\left(1\pm\frac1n\right)^n$, or take just the sign change between $e^x$ and $e^{-x}$. - -Is that already all there is to say about "ambivalent" constants? - -On the other side, a striking example of a non ambivalent one is $\pi$, which is a period while $1/\pi$ probably isn't, likewise e.g. for $\ln2$. - -REPLY [2 votes]: I think we get insight here by asking a more specific question: - -Geometrically, why is $e$ vs $1/e$ ambivalent when $\pi$ vs $1/\pi$ is not? - -It’s easier to talk about $\pi$ first. Consider a square $S$ inscribed in a circle $C$, and their areas. - -Then are these ratios ambivalent, or is one more distinguished: -$$\frac{area(C)}{area(S)} \ \text{ or } \ \frac{area(S)}{area(C)}\ ?$$ -Maybe surprisingly, the answer is clear: the first ratio is more distinguished, because we can approximate it by approximating the area of the circle with the area of smaller squares. More generally, it’s easier to approximate curved areas with polygonal areas than vice versa. (Archimedes would have been ridiculous to calculate $\pi$ by filling a square with small circles.) And therefore $\pi/2$ is more distinguished than $2/\pi$. -We get the same distinction with the circle inscribed in the square, or using perimeters instead of areas. -Talking about $e$ geometrically requires a little more work. Let $h$ be a hyperbola with asymptotes $x$ and $y$. Let $HX$ and $H'X'$ be line segments from $h$ to $x$ which are parallel to $y$. - -Suppose the area of the curved region $HH'X'X$ equals the area of the parallelogram between $H$ and the asymptotes. Then are these ratios ambivalent, or is one more distinguished: -$$\frac{HX}{H'X'} \ \text{ or }\ -\ \frac{H'X'}{HX}\ ?$$ -Asked this way, the answer is again clear: these two ratios are ambivalent, because they are both ratios of simple straight line segments. And therefore $1/e$ and $e$ are ambivalent.<|endoftext|> -TITLE: Structure of the Cantor part of the derivative of a BV function -QUESTION [12 upvotes]: It is well known that an integrable function $u \colon \mathbb R^d \to \mathbb R$ is said to be of bounded variation iff the distributional gradient $Du$ is (representable by) a finite Radon measure, still denoted by $Du$. -Then it is also well known that the measure $Du$ can be decomposed into three terms, $D^{a}u, D^{j}u, D^c u$ resp. absolutely continuous part, jump part and Cantor part. Quite a plethora of fine results are present in the literature (see, e.g. the monography by Ambrosio-Fusco-Pallara also for the notation below, which is however quite standard). For instance, $D^au= \nabla u \mathscr L^d$ (being $\nabla u$ the approximate differential) and $D^j u = (u^+-u^-) \otimes \nu_{J_u} \mathscr H^{d-1}$. -On $D^c u$ little is present: it is always generically said that, as a measure, is something intermediate between the jump part $\mathscr H^{d-1}$ and the a.c. part $\mathscr L^d$. As an example, it is always considered the Cantor-Vitali staircase (the devil's function), whose derivative has only Cantor part and is exactly $\mathscr H^{\alpha}$, for $\alpha = \log_3 2$ which is - incidentally - the Hausdorff dimension of the standard Cantor set. -So here is my question: - -Is it always true that $D^c u$ is (absolutely continuous w.r.t.) $\mathscr H^\alpha$ for suitable $\alpha \in (d-1, d)$? In other words, is $D^c u$ always an Hausdorff measure (up to a density) restricted to some Cantor-like set of certain dimension $\alpha \in (d-1,d)$? - -That should have something to do with densities and Besicovitch Theorem... but I do not know. - -REPLY [8 votes]: Certainly not with a single $\alpha$, but it is tempting to decompose $D^cu$ into an integral $\int_{d-1}^d\mu_\alpha\ d\nu(\alpha)$ with $\mu_\alpha$ having a density with respect to $\mathcal H^\alpha$ on an $\alpha$-dimensional set. I don't know if it is always possible.<|endoftext|> -TITLE: The "derived drift" is pretty unsatisfying and dangerous to category theory (or at least, to me) -QUESTION [49 upvotes]: I'm currently a young, not-so-young mathematician, finishing its second postdoc. I developed an interest for rather different topics in the last few years but constantly, slowly converged towards something that has to do with (but at this point I'm quite unsure is) category theory and its applications. What motivated me in the study of mathematics back in the days was the desire to understand the mechanisms ruling algebraic topology; then the word "functor" came in, and I fell into the rabbit-hole. -At this point most of you would expect I'm not unsatisfied with the shape category theory has nowadays: isn't a blurry mixture of homotopy theory and category theory precisely what I'm tackling? -I'm instead profoundly disappointed by the drift that categorical thinking has taken in the last ten years or so. And this is because the more I dwell into what "higher category theory" and "formal homotopy theory" became, the less I like both of them (I will somewhat refer as both with the portmanteau term "HTT"; I hereby stress that this acronym has no particular meaning whatsoever): - -It is still absolutely unclear what good is HTT for category theorists. To my eye, it is certainly a masterpiece of applied mathematics (in the sense that its tasks rest on the use of conceptualization as a tool, not as a target), but it doesn't seem to add a single grain of sand to the sea of category theory; instead, it re-does all the things you need to know to behave "as if" your homotopy-things were things, or to compactly bookkeep an infinite amount of data into a finite amount of space. These are honest practical motivations, addressed in a way I'm unable to judge; what I am able to judge, is the impact this impressive amount of material is having on category theory intended not as a part of mathematics, but as a way to look at mathematics from the outside. I feel this impact is near to zero. Not to mention that to my eye you do category theory only the australian way; everyone else is applying category theory towards the solution of a specific mathematical problem. And yet, I couldn't think of two more distant languages than Australian CT and HTT; what's wrong with me? What's wrong with the community? Sure there have been attempts to circumvent this; I feel this is a beginning, and somehow the first example of HTT done by truly categorical means. But in the end, you open and read these papers, only to find that you still need to know simplicial sets and homotopy theory and the lingo of topologists. This is not what I'm after. -When you use HTT, you are not providing a foundation for (higher) category theory; instead, you are relying quite heavily on the structure of a single category (simplicial sets), and on its quite complicated combinatorics. I am perplexed by the naivety of people that believe HTT can serve as a foundation for higher category theory; I am frightened by the fact that these people seems to be satisfied by what they have. So should I? Or shall I look for more? And where? Struggling with the books I had, I haven't been able to find a single convincing word about neither of these terms (foundation, category, theory). Again, it seems that HTT is a framework to perform computations (be them in stable homotopy theory or intersection theory or something else), instead than a language explaining the profound reason why you already know what things intimately are (this is what category theory does, to me). It is also quite schizophrenic that HTT exhibits the double nature of a device taking (almost all) category theory for granted, and at the same time it wants to rebuild it from scratch. Do I have to already know this stuff, to learn this stuff? -There's a rather deep asymmetry between category theory and homotopy theory: these two fields, although intimately linked, live different planets when it comes to outreach and learning. By its very nature, categorical thinking is trivial; there are few things to prove, and all of them are done with the same toolset, and instead there's an extreme effort in carving deep definitions that can turn into milestones of thought (I take "elementary topos" as an example of such a definition). On the contrary, homotopy theory is a scattered set of results, fragmented in a cloud of subfields, speaking different dialects; every proof is technically a mess, uses ad-hoc ideas, complicated constructions, forces to re-learn things from scratch... in a few words, there is no Bourbaki for algebraic topology [edit: now I know there is one, but it's evidently insufficient]. - -This double nature entails that there's no way to learn HTT if you (like me) are not so acquainted with the use of concrete and painful arguments; in a few words, if you are not a good enough mathematician. The complexity of techniques you are requested to master is daunting and leaves outside some beginners, as well as some people caught at the wrong time in their formation process. Sure, the situation is changing; but it's doing it slowly, too slow to perceive a real change in the pace, or in the sensibility, or in the sense of priority of the community. -Until now, every single attempt I made to enter the field failed in the most painful way. I feel there's no way I can understand fragmented, uncanny arguments like those. The few I can follow, I'd be absolutely unable to repeat, or reshape to prove something I need: they simply lie outside the language I'm comfortable with. Every time I have to check whether something is true, I have absolutely no clue how to operate, apart from pretending that what I do happens in/for a 1-category. And this disability is not conceptual, it is utterly practical, and seemingly unsolvable. -Learning HTT requires to abandon categorical thinking from time to time; you are forced to show that something is true in a specific model, using a rather specific and particular technique, without relying on completely formal arguments. It is an unsatisfying, poor language from the point of view of a category theorist and people seem to avoid tackling foundations to do geometry and topology. Which is fine, but not my cup of tea. -It is at this point extremely likely that, by lack of ability, or simply because I can't recognize myself in (the absence of) their philosophy, I won't be part of the crew of people that will be remembered for their contributions to higher category theory. What shall I do then? The echo-chamber where I live in seems to suggest a "love or leave it" approach, without any space for people that couldn't care less about chromatic homotopy theory, algebraic geometry, differential geometry, deformation theory... -So, what shall I do? I can list a few answers, all equally frightening: - -Settle down, learn my lesson, and fake to be a real mathematician, even though I know barely anything about the above mentioned homotopy theory, algebraic geometry, differential geometry, deformation theory? To a certain extent, it is working: my thesis received surprisingly positive reports, I happen to be able to maintain a position, even though scattered and temporary. But I'm also full of discomfort; I fear that my nature is preventing me from becoming a good mathematician; I am unsatisfied and I feel I'm denying my true self. What's worse: I feel I have to deny it, posting this rant with a throwaway account, because the ideas I proposed here are unpopular and could cost my academic life. -Shall I quit mathematics, since at this point there's no time to learn something new (I have to employ my time writing to avoid death)? I have to do mathematics with what I have; I feel what I have, what I know at the deep level I want, is barely nothing. And I can't use things I don't know, that's the rule. -Shall I face the fact that I've been defeated in my deepest desire, becoming exactly the kind of mathematician (and human being) I've always hated, the one who uses a theorem like a black box and makes guesses about things he ignores the true meaning of? But mathematics works this way: there is no point in knowing that something is true, until you ignore why it's true. Following a quite common idea among category theorists, I would like to go further, knowing why something is trivial. I don't want to know a definition, I want to know why that definition is the only possible way to speak about the definiendum. And if it's not, I want to be aware of the totality of such ways: does this totality carry a structure? The presence/absence of it have a meaning? Is there a totality of totalities, and how it behaves? When I first approached HTT I thought that answering these very questions was its main task. You can see how deeply I'm disappointed. And you can see the source of my sense of defeat: I feel stupid, way more limited, distracted from learning technicalities, way more than people that do not tackle this search for an absolute meaning. Younger than me, many colleagues began studying HTT, rapidly reaching a certain command of the basic words and subsequently began producing mathematics out of this command. To them, category theory is just another piece of mathematics, not different from another (maybe more beautiful); you do your exercises, learn to prove theorems, that's it. To me, category theory is the only satisfying way to think. Am I burdened by this belief to the point that it's preventing me from being a good mathematician? -The questions I raised at point 3 do not pertain mathematics; I should do something else. In fact, the only reason why I tried to become a mathematician was that I felt that mathematics is the only correct meaning of the word "philosophy", and the only correct way to pursue it. But turning to philosophy would be, if possible, even a more unfortunate choice: philosophers tend to be silly, ignorant people who claim to be able to explain ethics (=a complicated and elusive task) ignoring linear algebra (=something that shall be the common core of knowledge of every learned person). - - - -One of the answers below advises me to "give HTT another try". - -This is what to do. I've no clue about how, and this is why I'm looking for mathematical help. I can't find a way out of this cul-de-sac: doing new, unpolished mathematics is a social event, but I've lived the years of my PhD isolated and without a precise guidance aside from myself. - -REPLY [68 votes]: I'm coming a bit late to this party, but I'll put in my two cents anyway because they are rather different from everything else I've heard so far. In a nutshell, my response is: - -Yes, I agree that this is a problem (though I do think you would have done better to post only the question and not the rant), and -What you can do is be part of the solution. - -For a long time I resisted "homotopical higher category theory" too, for reasons that I think are not unrelated to yours. I even wrote a somewhat whiny blog post about it. What eventually "brought me on board" was not the applications to algebraic geometry or what-have-you (which is not, of course, to denigrate those applications), but the truly category-theoretic conceptual insights arising from what you call HTT. Examples include: - -Colimits in a 1-category cannot be as well-behaved as we would like them to be, and the reason is because a 1-category doesn't have enough "room"; an $(\infty,1)$-category fixes this. For instance, Giraud's axioms for a 1-topos assert "descent" only for coproducts and quotients of equivalence relations; the analogous axioms for an $(\infty,1)$-topos assert descent for all colimits. -Passing "all the way to $\infty$" has a "stabilizing" effect that enables $(\infty,1)$-categories and $(\infty,1)$-category theory to "describe itself" in ways that 1-category theory can only approximate. For instance, the 1-category of 1-categories does not include enough information to characterize the "correct" notion of "sameness" for 1-categories, namely equivalence (at least, not unless you hack it with something like a Quillen model structure); for that you need the 2-category of 1-categories, or at least the $(2,1)$-category of 1-categories. But the $(\infty,1)$-category of $(\infty,1)$-categories does characterize them up to the correct notion of equivalence. (Although for many purposes one still needs the $(\infty,2)$-category of $(\infty,1)$-categories, pointing towards the still largely-unexplored territory of $(\infty,\infty)$-categories.) Similarly, a 1-topos can only have a subobject classifier, classifying those objects that are "internally $(-1)$-categories", i.e. truth values; but an $(\infty,1)$-topos can have an object classifier that classifies all objects (up to size limitations). -Various mysterious phenomena in 1-topos theory are explained as shadows of $(\infty,1)$-topos-theoretic phenomena. For instance, the analogy between open geometric morphisms and locally connected ones is explained by seeing them as the steps $k=-1$ and $k=0$ of a ladder of locally $k$-connected $(\infty,1)$-geometric morphisms, and similarly for proper and tidy geometric morphisms. Moreover, various apparently ad hoc notions of the "homotopy theory of toposes", such as cohomology, fundamental groups, shape theory, and so on, are explained as manifestations of the $(\infty,1)$-topos-theoretic "shape", which is characterized by a simple universal property. -Perhaps most importantly, the fundamental idea that the basic objects of mathematics are not just sets, but $\infty$-groupoids. Thus, for instance, the really good notion of "ring" should be an $\infty$-groupoid with a coherent multiplication and addition structure (i.e. a ring spectrum), including the set-based notion of "ring" as simply a special case. And so on. - -Note that none of these ideas depends on any concrete model for $(\infty,1)$-categories, and most of them have nothing to do with homotopy theory; they are purely category-theoretic ideas. So I think even a category theorist who cares nothing about homotopy theory ought to be interested in a kind of "category theory" where these are true. -That said, I think a good category theorist should care at least somewhat about homotopy theory, if for no other reason then for the same reason that a good category theorist should care about other applications of category theory. Like all fields of mathematics, category theory is supported and invigorated by its connections to other fields of mathematics, and the close tie between higher category theory and homotopy theory has great potential to stimulate both subjects. That this potential has been realized more fully on the homotopy-theoretic side is, I think, largely an accident of history and personality. -Why is $(\infty,1)$-category theory not usually done "Australian-style"? I believe it is just because people doing $(\infty,1)$-category theory don't know, or at least don't appreciate, Australian-style 1- and 2-category theory, while many Australian-style category theorists don't know or appreciate $(\infty,1)$-category theory. This creates a tremendous opportunity for anyone who is willing to put in the effort to be a bridge, teaching category theorists how to think about $(\infty,1)$-categories "category-theoretically" and teaching $(\infty,1)$-category theorists the benefits of "really thinking like a category theorist". -One way to be such a bridge is to learn the simplicial technology that's currently used for $(\infty,1)$-category theory and "do them Australian-style". For instance, as far as I know there is still no $(\infty,2)$-monad theory with the power and flexibility of 2-monad theory; someone should do it. Enriched $(\infty,1)$-categories are only starting to be investigated. The $(\infty,2)$-category of $(\infty,1)$-profunctors has been used for some applications, but its category-theoretic potential is largely unexplored. As far as I know, no one has even defined $\infty$-double-categories yet. (Edit: They've been defined, but apparently not systematically studied; see comments.) What about generalized $\infty$-multicategories? Etc. etc. -While a worthy endeavor, I suspect that this is not what you want to do. In particular, it sounds like you don't feel able to spend the time to really understand simplicial technology. I can sympathize with that; it's difficult enough for me, and I was already exposed to lots of simplicial stuff as a graduate student since my advisor was an algebraic topologist. So I generally avoid using simplicial technology as much as possible. One way to do this, which I have pursued myself, is to study $(\infty,1)$-categories using 1- and 2-categorical machinery, including Quillen model categories (which, by the way, have an algebraic version that is rather more pleasing to a category theorist's heart) but also homotopy-level structures such as derivators, homotopy 2-categories, and homotopy proarrow equipments. -This works quite well for surprisingly many things, and doesn't require you to learn any simplicial technology. However, it does often depend on the fact that someone has proven something using simplicial technology in order to "get into the world" where you're working. Moreover, you've also expressed some skepticism about the very idea of simplicial technology and concrete models. I think it'd be good if you can get over this to a degree — mathematics has to move forward with what we have, even if it's not perfect, and later on someone can make it better — but I do also sympathize with it, because for instance of the last conceptual insight I mentioned above: - -The basic objects of mathematics are not just sets, but $\infty$-groupoids. - -How can this be, if an $\infty$-groupoid is defined in terms of sets (e.g. as a Kan complex)? -Well... there is now a way to study $\infty$-groupoids directly, without defining them in terms of sets: it's called homotopy type theory (HoTT). HoTT is (among other things) a foundational theory, on roughly the same ontological level as ZFC, whose basic objects can be regarded as $\infty$-groupoids; I wrote a philosophical introduction to it from this perspective. (There's also work on an analogous theory whose basic objects are $(\infty,1)$-categories.) Thus, HoTT offers the promise of an approach to homotopy theory and higher category theory that's almost completely free of simplicial technology, and incorporates the conceptual insights of $(\infty,1)$-category theory "from the ground up", allowing us to build intuition for, and work directly with, higher-categorical and higher-homotopical structures without having to construct them explicitly out of sets. When I read or write a proof in $(\infty,1)$-topos-theoretic language, I'm never quite sure whether I've dotted enough "i"s to make all the coherence come out right; but when I instead write it in HoTT then I am, not only because with HoTT I understand the profound reason why you already know what things intimately are (as you put it), but because a HoTT proof can be formalized and verified with a computer proof assistant. There are already some graduate students who have "grown up" with HoTT and can "think in it" in ways that surpass those of us who "came to it late". -Now, this "promise" of HoTT is not yet fully realized. Many coherent higher-categorical structures can be represented simply and conceptually in HoTT; but many others we don't know how to deal with yet. So here's another way you can be part of the solution: improve the ability of HoTT to represent higher category theory, so that eventually it becomes powerful enough that even the "applied" $(\infty,1)$-category theorists can do away with simplices. This is, in large part, what I am now working on myself.<|endoftext|> -TITLE: Seeking references for finding primes infinitely often -QUESTION [12 upvotes]: I've been pondering this weakened version of the finding primes problem for a while: - -Is there an algorithm which given $k$ outputs a prime $p > 2^k$ in time $F(\log_2(p))$? - -This differs from the ordinary finding primes problem, which I'll state as: - -Is there an algorithm which given $k$ outputs a prime $p > 2^k$ in time $F(k)$? - -Equivalently, the weak version is that there is a deterministic algorithm to find a prime $p > 2^k$ which runs in time $F(k)$ on infinitely many inputs $k$. Another formulation of the weak problem is to determine the values of $\displaystyle\liminf_{p\rightarrow\infty}{\frac{L(p)}{\log_2(\log_2(p))}}$ and $\displaystyle\liminf_{p\rightarrow\infty}{\frac{L(p)}{\log_2(p)}}$ where $p$ is prime and $L$ is the Levin complexity. -I consider both problems to be defined with respect to the class of models of computation possessing a time-translation to a single-tape Turing machine satisfying $T' \in T^{1+o(1)} \cdot S^{O(1)}$, where $T$ is the time used in the other model and $S$ is the space. This class includes the various RAM models and multi-tape Turing machines. However, the equivalence is sharper than a polynomial time-translation, and we can witness a specific value of the time exponent with an algorithm satisfying $S \in T^{o(1)}$ in any model in the class. For example, searching an interval $[2^n, 2^n + 2^{\epsilon \cdot n})$ for a prime can be accomplished in time $2^{\epsilon \cdot n + o(n)}$ in all of these models because the AKS test is simultaneously in subexponential time and subexponential space. But we won't be able to place the AKS test itself in a particular time class like $n^{6+o(1)}$ unless we can adapt it to simultaneously use space $n^{o(1)}$, until then the best we can say is that its runtime is $n^{O(1)}$ in every model. If either finding primes problem can be solved in polynomial time, assuming model-invariance won't make it any harder to prove that, and (suspending disbelief) it can only help to prove a lower bound on the exponent of exponential-time algorithms. It's plausible that it would present an obstruction to improving the exponential upper bound — for example, an algorithm that finds a prime $p > 2^k$ using $p^{\frac{1}{3}}$ time and $p^{\frac{1}{3}}$ space on some particular machine wouldn't qualify as an improvement under model-invariance. That's because the time exponent doesn't translate when the space is exponential — $\frac{1}{3}$ becomes $\frac{b+1}{3}$ after a $T' = T \cdot S^b$ time-translation. The exponential-time algorithms I refer to below are all in subexponential space anyway so this issue doesn't seem to actually come into play. -For the ordinary version, all we know is $F(k) \in 2^{0.525 \cdot k + o(k)}$, and I haven't been able to find any better bound for the weak version. -My understanding from the polymath page is that we don't know if having factoring for free can help us find primes. Can it help us find primes infinitely often? -An attractive aspect of the finding primes problem is that there are many conjectures which imply improvements. For example, the Riemann hypothesis puts $F(k) \in 2^{\frac{k}{2} + o(k)}$, Cramér's conjecture implies $F(k) \in k^{O(1)}$, and $\text{P}=\text{NP}$ implies $F(k) \in k^{O(1)}$. For my version of the problem, we additionally have that infinitely many Mersenne or Fermat primes gives $F(k) \in k^{O(1)}$, infinitely many $n^2+1$ primes implies $F(k) \in 2^{\frac{k}{2}+o(k)}$, and Bunyakovsky's conjecture implies $F(k) \in 2^{o(k)}$. Some other conjectures have similar implications. A world where we can't find primes infinitely often would be a very weird place, even weirder than one where we just can't find primes! -That is my main motivation behind studying this problem, to try and understand what that world would be like. -I'm looking for more information about finding primes infinitely often. In particular, is there any better time bound than what is known for the ordinary version? I haven't found it discussed in the polymath threads or elsewhere, and I haven't identified any open problems that any improvement implies, despite all the open problems that imply improvements. So for all I know, there's a simple and provably fast algorithm that I just can't think of myself. - -REPLY [2 votes]: You might want to check out Shanks' nice paper: -Shanks, D., On maximal gaps between successive primes, Math. Comput. 18, 464-651 (1964). ZBL0128.04203.<|endoftext|> -TITLE: The graph of Rule 110 and vertices degree -QUESTION [8 upvotes]: Consider the elementary cellular automaton called Rule 110 (famous for being Turing complete): - -It induces a map $R: \mathbb{N} \to \mathbb{N}$ such that the binary representation of $R(n)$ is that of $n$ after the application of Rule 110. For examples, $R(0)=0$, $R(1)=3$, $R(2)=6$ and $R(3)=7$. - See below the illustration for $R(571)=1647$: - -There is an OEIS sequence computing $R(n)$ for $n<2^{10}$: A161903. -Definition: Let $\mathcal{G}$ be the graph with $\mathbb{N}$ as set of vertices and $\{\{n,R(n)\}, n \in \mathbb{N} \}$ as set of edges. -The graph $\mathcal{G}$ has infinitely many connected components and its vertices degree is not bounded above. -The proof follows from Lemmas D and E, below. -Question: Is the vertices degree bounded above on each connected component of $\mathcal{G}$? - -Lemma A: For $n,r>0$, we have that $2^{r-1}n0$, $R^r(n) = 2kn$ iff $n=0$. -Proof: If $n>0$ then there is $s \ge 0$ such that $n=2^sm$ and $m$ odd. But $R^r(2^sm) = 2^s R^r(m)$, so $R^r(m)=2km$, it follows by Lemma B that $m$ is even, contradiction. $\square$ -Lemma D: For $n,r>0$, $n$ and $2^rn$ are not in the same connected component of $\mathcal{G}$. -Proof: If $n$ and $2^rn$ are in the same connected component then there are $r_1,r_2>0$ such that $R^{r_1}(n) = R^{r_2}(2^rn) = 2^r R^{r_2}(n)$, so by Lemma A, $r_1 = r_2 +r$. Then, $R^{r}(m)=2^rm$ with $m=R^{r_2}(n)$, thus $m=0$ by Lemma C, contradiction. $\square$ -Lemma E: For any $n>0$, there is a vertex of $\mathcal{G}$ of (finite) degree $\ge n$. -Proof: The pattern of $r \ge 2$ successive black cells (corresponding to the vertex $2^r-1$) is the image of any pattern of $r-1$ cells without successive white cells, without three successive black cells and whose first and last cells are black, see for example below with $r=27$: - -Clearly, for $r$ large enough, the vertex $2^r-1$ has degree $\ge n$. $\square$ -Exercise: Show that $|R^{-1} (\{ 2^r-1 \}) | = \sum_{2a+b = r} {a \choose b}$, known as the Padovan sequence. - -Bonus question: What is the sequence $\mathcal{S}$ of minimal numbers of the connected components of $\mathcal{G}$? -Note that $\mathcal{S} \subset \mathbb{N} \setminus R(\mathbb{N}^*) = \{0, 1, 2, 4, 5, 8, 9, 10, 11, 16, 17, 18, 19, \dots \}$, so: - -$\{0,1,2,4,8\} \subset \mathcal{S}$, by Lemmas B and D. -$5 \in \mathcal{S}$ iff $\forall r_1, r_2>0$ then $R^{r_1}(1) \neq R^{r_2}(5)$. -$11 \not \in \mathcal{S}$ because $R^4(1) = R(11)=31$. -I don't know for the other numbers appearing above. - -A search for "$0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18$" gives one result only, the Fibbinary numbers $\mathcal{F}$, i.e. the integers whose binary representation contains no consecutive ones (the number of such numbers with $n$ bits is the $n$th Fibonacci number). It is related to Zeckendorf's theorem. -Is it true that $\mathcal{F} \subseteq \mathcal{S}$, or even $\mathcal{F} = \mathcal{S}$? -Note that $\mathcal{F} \subset \mathbb{N} \setminus R(\mathbb{N}^*)$ because a portion $(0,0,1,\epsilon)$ gives $(1,1)$ by applying Rule 110. -I don't know whether $19 \in \mathcal{S}$, but $19 \not \in \mathcal{F}$. - -REPLY [4 votes]: I guess I'll indulge in my guilty pleasure a bit. The connected component of the number $1$ has unbounded degree. -Let $X = \{x \in \{0,1\}^{\omega} \;|\; \sum x < \infty\}$, the finite support configurations (which you identified with the naturals) and let $f : X \to X$ be rule $110$ with left and right flipped. After a lot of trial and error I managed to figure out this is called rule 124 in Wolfram's (very convenient and intuitive) naming scheme. This is just so natural numbers go to the right. -We will show that the connected component of the configuration $x = 1000...$ satisfies that for all $k \in \mathbb{N}$ there are infinitely many $n$ such that $f^n(x)$ has at least $k$ preimages in $X$. Let $y \in \{0,1\}^{\omega \times \omega}$ be the spacetime diagram of $x$, so $y_{(i,0)} = x_i$ and $y_{(i,j+1)} = f((y_{(h,j)})_h)_i$. The key to proving it is the following: - -The set $\{(m,n) \;|\; y_{(m,n)} = 1\}$ is semilinear. - -Semilinear subsets of $\mathbb{Z}^d$ are finite unions of linear sets, which in turn are sets of the form $a + V$ where $V = \langle v_1,v_2,...,v_n \rangle$ where $\langle \cdots \rangle$ denotes the smallest submonoid containing the $v_i$. Equivalently it means definability in Presburger arithmetic. -I won't actually give a precise proof, but this is seen from a simulation of the first ~3000 steps of evolution, and unless I've missed some very subtle period-breaking phenomenon -- which I doubt because eyes usually jump at those --, this should not take long to prove rigorously by pen-and-paper, or by computer. (You could (at least in theory) do it in Walnut, since semilinear implies $k$-automatic, presumably there are also direct implementations of Presburger arithmetic.) -Now, all we need to observe is that in some of the periodic growing areas of $y$ (there are two) you can locate a diamond, i.e. a word $w$ such that the preimage used above it is $u$, and there is another word $v$ with the same image $w$ and the same two left- and rightmost symbols. Then more and more instances of $u$ appear on the line above and you can change any subset of them to $v$. - -For all $k$, $y$ has a row with at least $k$ disjoint occurrences of $01011101100 \mapsto 111011111$ -any subset of which can be changed to $01100101100 \mapsto 111011111$. - -(Those are in terms of the flipped rule. In terms of $110$, these would be $00110111010 \mapsto 111110111$ vs $00110100110 \mapsto 111110111$. Also, I write $u \mapsto w$ for the usual application of the rule to words, so it shortens them by one on both sides.) -Description of spacetime -I will describe what is seen in a non-rigorous fashion, but of course the finite steps are completely proved by just seeing them in the simulator, and given these initial steps, as I've stated above, the final conclusion is something that can be proved by pen and paper (disclaimer: I did not actually write a proof, as it is still a lot of work). -I recommend drawing a spacetime diagram yourself, I used Golly and you can find Golly compatible rule files at the end of this post. -So, the support will grow steadily as a triangle, the leftmost border stays put (as you observed) and the right border moves at speed $1$. After around 100 steps, the roughly 70 rightmost cells of the spacetime diagram form a complicated glider, then there is a slightly denser area, then there is another glider, and the rest is again this denser stuff. (The leftmost "glider" is not really a glider, because it's not moving periodically, but in this informal description I use this term anyway.) We can already observe that the rightmost glider will never change its behavior as it is evolving with a period of 16 and moving at the speed of light. But that is not enough to solve your question. - -The pattern looks nice for a few hundred steps, but around 600 steps the glider on the left suddenly blows up! - -Fast forward to step 1200 (lots of things happen on the way) and you have two gliders moving to the right. Now the leftmost one suddenly turns into a left-going glider (marked with a green 1 in the figure). Then the central glider suddenly disappears at step 2086 (marked with a green 2), and right after that the leftmost glider hits the left boundary at step 2160 and turns into another glider going to the right. - -It takes a few hits from glider gunfire, but then seems to become resistant to it. - -Fast-forward a few thousand steps and it is still going strong. -In fact... -there no more surprises! -You can now prove (by induction) that this pattern persists: the glider on the left keeps moving at speed $(18, 213)$, and shoots additional gliders to the left which hit the left boundary every $140$ steps, and the rightmost glider sends additional gliders periodically, which hit the left glider but don't destroy it. The phases match up after some time, which is really the reason why nothing strange happens. -Alternatively, you can prove as I suggested above that $y$ is semilinear: just write the formulas, and check that this semilinear set is closed under rule $110$: it's decidable whether a given semilinear set is the spacetime diagram of a given CA (for example by the Presburger characterization). I'm sure $y$ is semilinear, because all the things that are happening are is periodic, different types of things happen in areas bounded by rational inequalities (or rational lines), and there are only finitely many phenomena (like, four) going on. And then there's about a million bits of garbage at the beginning, which you just code in. -Ok, so now we need to find occurrences of $01011101100 \mapsto 111011111$. Well, that's one of the steps of one of the gliders that are being gunned periodically from right to left. -Golly compatible rules: -Here's rule 110, call it (e.g.) W110.rule and put it in the Rules folder. Then open it in Golly. -@RULE W110 -@TABLE -n_states:2 -neighborhood:Moore -symmetries:none - -var a={0,1} -var b={0,1} -var c={0,1} -var d={0,1} -var e={0,1} -var f={0,1} -var g={0,1} - -# C,N,NE,E,SE,S,SW,W,NW,C' -0,0,0,a,b,c,d,e,0,0 -0,0,1,a,b,c,d,e,0,1 -0,1,0,a,b,c,d,e,0,1 -0,1,1,a,b,c,d,e,0,1 -0,0,0,a,b,c,d,e,1,0 -0,0,1,a,b,c,d,e,1,1 -0,1,0,a,b,c,d,e,1,1 -0,1,1,a,b,c,d,e,1,0 - -And here's rule 124: -@RULE W124 -@TABLE -n_states:2 -neighborhood:Moore -symmetries:none - -var a={0,1} -var b={0,1} -var c={0,1} -var d={0,1} -var e={0,1} -var f={0,1} -var g={0,1} - -# C,N,NE,E,SE,S,SW,W,NW,C' -0,0,0,a,b,c,d,e,0,0 -0,0,1,a,b,c,d,e,0,0 -0,1,0,a,b,c,d,e,0,1 -0,1,1,a,b,c,d,e,0,1 -0,0,0,a,b,c,d,e,1,1 -0,0,1,a,b,c,d,e,1,1 -0,1,0,a,b,c,d,e,1,1 -0,1,1,a,b,c,d,e,1,0<|endoftext|> -TITLE: Newman's Lemma or Diamond Lemma -QUESTION [8 upvotes]: I'd like to learn the Newman's Lemma or Diamond Lemma (the one used in abstract rewriting system), can someone recommend me some books where I can read it? I'd appreciate self-contained books with examples. Thank you. - -REPLY [16 votes]: Franz Baader and Tobias Nipkow, Term Rewriting and All That is a book fully devoted to term rewriting; much of it is about applying the diamond lemma. -Vincent van Oostrom, Newman's proof of Newman's lemma proves the lemma itself on just 1 page. That's not the shortest proof. -Gert Smolka, Confluence and Normalization in Reduction Systems proves the diamond lemma in §15, after showing all sorts of related things. These notes are clearly written for computer scientists (the notations resemble Coq), but from what I've seen are fairly readable. -Kimmo Eriksson, Strong convergence and the polygon property of 1-player games, Discrete Mathematics, Volume 153, Issues 1--3, 1 June 1996, Pages 105--122 explores the "Polygon property theorem", which is similar to the diamond lemma. (It requires the play sequences to be of the same length, which makes it somewhat more restrictive, but its conclusion is accordingly stronger.) -George M. Bergman, The diamond lemma for ring theory, Advances in Mathematics, Volume 29, Issue 2, February 1978, Pages 178--218 (errata) applies the diamond lemma to the study of algebras. -As @MarkSapir pointed out, Mark Sapir, Combinatorial algebra: syntax and -semantics, Springer 2014 has a section (§1.7) on confluence. -Marc Bezem, Thierry Coquand, Newman's Lemma -- a Case Study in Proof Automation and Geometric Logic proves a "bi-colored" analogue of Newman's lemma. -Lars Hellström, The Diamond Lemma for Power Series Algebras gives an analogue of the ring-theoretical diamond lemma for infinite-but-sort-of-convergent rewriting sequences (I am being deliberately vague here, since I've read only the motivating sections of this). - -Self-containedness isn't really an issue in this subject; the proof of the lemma is usually less than a page long, and there are zero prerequisites. I once gave an expository talk about confluence that includes a proof of the diamond lemma.<|endoftext|> -TITLE: intuition for hochschild homology -QUESTION [21 upvotes]: According to this post Intuition for group homology, I wonder what is the intuition for Hochschild homology. - -The Hochschild homology is defined as the homology of this complex - chain. - Given a ring $A$ and a bimodule $M$. - Define an $A$-module by setting \begin{equation} \displaystyle -\notag C_{n}(A,M)=M\otimes_{k}A^{\otimes n} \end{equation} - and for each $n$ the maps $d_{i}:C_{n}(A,M)\rightarrow C_{n-1}(A,M)$ - as follows. Define \begin{equation} \displaystyle d_{0}(m\otimes -a_{1}\otimes\cdots\otimes a_{n})=ma_{1}\otimes -a_{2}\otimes\cdots\otimes a_{n} \end{equation} - and \begin{equation} \displaystyle d_{i}(m\otimes -a_{1}\otimes\cdots\otimes a_{n})=m\otimes a_{1}\otimes\cdots\otimes -a_{i}a_{i+1}\otimes\cdots\otimes a_{n},\quad 1\leq i\leq n-1 \end{equation} - and \begin{equation} \displaystyle d_{n}(m\otimes a_{1}\otimes \cdots -\otimes a_{n})=a_{m}m\otimes a_{1}\otimes\cdots\otimes a_{n-1}. \end{equation} - When $i -TITLE: The square of a ccc topological group -QUESTION [7 upvotes]: Jensen proved that under $\Diamond$ there is a homogeneous Suslin continuum, so the square of a ccc homogeneous space can fail to be ccc. What about ccc topological groups? - - -Is there a ccc topological group whose square is not ccc? - - -The obvious thing to try would be the free topological group over a Suslin continuum, but that doesn't work because it's a $\sigma$-compact group and, by a result of Tkachenko, $\sigma$-compact groups are ccc (and the square of a $\sigma$-compact group is clearly a $\sigma$-compact group). - -REPLY [4 votes]: Yes. There are such groups after adding a Cohen real (see Theorem $4$ in "Nonpreservation of properties of topological groups on taking their square", Malykhin, 1987) or under RVM (see Theorem $0^c$ in "Some applications of S and L combinatorics", Todorcevic, 1993).<|endoftext|> -TITLE: Convex sets in Alexandrov spaces -QUESTION [6 upvotes]: Let $X$ be a compact Alexandrov space with $curv\geq 1$ (and without boundary). Does $X$ always have a nontrivial compact convex subset without boundary? -Definition of a convex subset: $A\subseteq X$ is called convex if for every two points $p ,q\in A$, there exists a minimizing geodesic between them which is completely in $A$. - -REPLY [3 votes]: This is extremally rare, even if $X$ is a Riemannian manifold. -If the convex set $A$ has interior points, then any boundary point of the subset $A$ in $X$ lies on the boundary of Alexandrov space $A$. -So if $X\ne A$ then $\dim A<\dim X$. -Note that $A$ has to be totally geodesic, otherwise an end of geodesic would be a boundary point of $A$. -Generic Riemannian manifold do not have totally geodesic submanifolds of dimension at least 2. -So, we are left with the case $\dim A=1$. -In other words, $A$ is a closed geodesic in $X$. -Since $A$ is convex, it has to be length minimizing on each half. -This is also extremally rare thing --- generic Riemannian manifold does not have such geodesics.<|endoftext|> -TITLE: A curious determinant of quadratic forms -QUESTION [6 upvotes]: In a work about the Wave Equation, I encountered the following symmetric matrix of size $1+n$, whose entries are quadratic forms. The arguments are a scalar $a$ and a vector $X\in k^n$. -$$S(a,X)=\begin{pmatrix} -\frac12(a^2+|X|^2) & -aX^T \\ -aX & X\otimes X+\frac12(a^2-|X|^2)I_n -\end{pmatrix}.$$ -It turns out that the determinant factorizes at a high degree: -$$\det S=\left(\frac{a^2-|X|^2}2\right)^{1+n}.$$ - -Is there any conceptual explanation ? Is this an example of some theory, or of a more general family ? - -For the sake of completeness, the solutions of the WE $\partial_t^2u=c^2\Delta u$ satisfy the additional conservation laws ${\rm Div}\,S(\partial_t u,c\nabla u)=0$ (read the divergence row-wise). The first line of this vector-valued equation is the conservation of energy. The determinant is a power of $(\partial_tu)^2-c^2|\nabla u|^2$, which is a null-form, in Klainerman's terminology. - -REPLY [6 votes]: Let's put $x:=(0,X)\in k^{n+1}$ and $y:=-ae_1+x\in k^{n+1}$. Then $S$ writes as a symmetric rank-$2$ perturbation of a multiple of the identity, $S=S_0+ \lambda I_{n+1},$ with -$$S_0:= -\big(a^2-|x|^2\big)\ e_1\!\otimes e_1+y \otimes y,$$ -and $$\lambda:={1\over2}\big(a^2-|x|^2\big).$$ -This gives $\lambda$ as eigenvalue of $S$ of multiplicity $n-1$, with eigenspace $\ker S_0 =e_1^{\perp}\cap u^{\perp}$. On the orthogonal of the latter, $(\ker S_0)^\perp=\mathrm{span}(e_1,u)$, writing the operator $S$ in the orthogonal basis $(e_1,x/|x|)$ -one finds the eigenvalues $|x|^2+\lambda\pm a|x|$, and -$$\det(S_0+\lambda I_{n+1})=\lambda^{n-1}\big((\lambda+|x|^2)^2-a^2|x|^2 \big)$$ This of course holds for any value of $\lambda$, but $\lambda:={1\over2}\big(a^2-|x|^2\big)$ is special in that it makes the determinant equal to $\lambda^{n+1}$.<|endoftext|> -TITLE: What is the chromatic number of the Erdős–Rényi graph G(n,d/n) when d < 1? -QUESTION [5 upvotes]: What is the chromatic number of the ER graph $G(n,d/n)$, when $d < 1$ (there exist expressions for $d > 1$, but what if the graph is super sparse?). Here $n$ is the number of vertices and $d/n$ is the edge generation probability. - -REPLY [12 votes]: Consider subgraphs consisting of two cycles with an edge in common (i.e. a theta-graph or something more complex). The number of such labelled graphs with $t$ vertices is at most $n^t$, and the probability of each is at most $(d/n)^{t+1}$ since they have at least $t+1$ edges. Summing over $t$ shows that the expected number of such subgraphs goes to 0 for $d\le 1$, implying that the probability of any such graphs appearing also goes to 0. So a random graph in this range has only cycles that can be coloured independently, plus tree-like stuff. The chromatic number is 3 if any cycle is odd (which has constant probability $c(d)$ I think) and 2 or less otherwise. -ADDED: The expected number of odd cycles is asymptotically -$$E(d) = \sum_{i=1}^\infty \frac {d^{2i+1}}{4i+2} = -\frac d2 + \frac14\ln\frac{1+d}{1-d}.$$ -Since the distribution of the number of odd cycles is asymptotically Poisson, the probability that there are no odd cycles is asymptotically -$$P(d) = \exp(-E(d)) = e^{d/2}\biggl( \frac{1-d}{1+d}\biggr)^{1/4}.$$ -Since the probability of having no edges at all is infinitesimal, asymptotically the probability that $\chi(G)=2$ is $P(d)$ and the probability that $\chi(G)=3$ is $1-P(d)$. I suggest you plot these functions for $0\le d\le 1$ to see what they look like.<|endoftext|> -TITLE: Representing mathematical statements as SAT instances -QUESTION [15 upvotes]: The following problem (call it THEOREMS) belongs to class NP. - -Input: Mathematical statement $S$ (written in some formal system such as ZFC) and positive integer $n$ written in unary. -Output: "Yes" if $S$ has a formal proof of length at most $n$. "No" otherwise. - -It is known that there is an algorithm which, in a number of steps polynomial in $n$ (and in the length of $S$, which is usually negligible if $n$ is large), converts every instance of THEOREMS to an instance of (say) SAT. -My questions are the following: - -Has this algorithm been implemented in full generality? That is, I could download it, input my $S$ and $n$, press a button, wait for some time, and get an instance of SAT? -If not, is there at least theoretical analysis what could be the length of the resulting SAT in terms of $n$, $n^{100}$ or $n \log n$? -Erdos Discrepancy Problem, at least for discrepancy 2, has been reduced to SAT. Are you aware about similar conversion of any other famous problem, like Riemann Hypothesis? - -Note: I am not discussing how feasible would be to actually solve the SAT instance corresponding to "proving Riemann Hypothesis in a million of steps". My question is just about GENERATING this and similar SAT instances. - -REPLY [4 votes]: Improving on Brumleve's answer, I have a method for encoding a length-$n$ proof with a quasilinear $\tilde {O} (n)$-bit 3SAT instance. -In most formal systems, proofs and objects that appear in proofs (propositions, formulas, etc.) have a tree-like structure in that each such object can be built from other such objects in an enumerated list of ways. For example the proposition $P \to Q$ is built out of the two propositions $P$ and $Q$, and a proof of $Q$ can be made with modus ponens out of a proof of $P \to Q$ and a proof of $P$. Such a proof-tree can be encoded with a pointer architecture, as some sequence $x_0, \dots, x_{n-1}$ where each $x_i$ encodes a formula or a proposition or a proof of a proposition, possibly encoding including references to earlier objects $x_j$, $j < i$. For example, if $x_k$ encodes the proposition $P \to Q$, which I will denote with the notation $k \mapsto (P \to Q)$, then $x_k$ might have information of the form $(\to, i, j)$ where $i \mapsto P$ and $j \mapsto Q$. Similarly, denoting $k \vdash Q$ for $x_k$ encoding a proof of $Q$, a proof $k \vdash Q$ by modus ponens might be given in the form $(\mathtt {ModusPonens}, i_0, i_1, i_2, i_3, i_4)$ where $i_0 \mapsto P$, $i_1 \mapsto Q$, $i_2 \mapsto (P \to Q)$, $i_3 \vdash P \to Q$, and $i_4 \vdash P$. -In addition, if $x_j$ includes references $i_0, \dots, i_{k-1}$ to earlier objects, define $y_j = (x_{i_0}, \dots, x_{i_{k-1}})$ as a description of all the objects referenced by $x_i$. -To check if the objects $(x_0, \dots, x_{n-1})$ with the supplementary information $(y_0, \dots, y_{n-1})$ encodes a proof of the Riemann Hypothesis, you must check the following conditions: - -(Local validity) Each $x_i$ represents a valid construction of a proof object out of earlier proof objects. For example if $x_i = (\to, j, k)$ then it is necessary to check that $x_j$ and $x_k$ encode propositions and not some other kind of object. For steps such as modus ponens where it is necessary to check the syntactic equality of two subexpression, check for pointer equality of the corresponding objects instead. This change does not affect which statements we can prove in a given number of steps since it is always possible to deduplicate so that each syntactic expression appears at most once in the list $(x_0, \dots, x_{n-1})$, and this can only shorten the proof. If $x_i$ is constructed with enough references then checking the local validity of $x_i$ only requires $O (\log n)$ bit-operations on the inputs $(x_i, y_i)$, for a total of $O (n \log n)$ . -(Non-cyclicness) If $x_j$ contains references $i_0, \dots, i_{k-1}$ then we have $i_0, \dots, i_{k-1} < j$. -(Right conclusion) $x_{n-1}$ encodes a proof of the Riemann Hypothesis. One way to check this is to fix $x_0, \dots, x_{k-1}$ to hardwired constant values for some $k = O (1)$ so that $x_{k-1}$ formulates a statement of the Riemann Hypothesis, and ask that $x_{n-1}$ encodes a proof of $x_{k-1}$. -(Correctness of $(y_j)$) The values $y_k$ correctly dereference the references given in $x_k$. The rest of my answer explains how to check this. - -We've reduced the problem of checking proof validity to the following reference-checking problem: Given an array $x_0, \dots, x_{n-1}$ of $\ell$-bit values and a list $(k_0, y_0), (k_1, y_1), \dots, (k_{m-1}, y_{m-1})$ where each $k_j$ has $\lceil \log_2 n \rceil$ bits and each $y_j$ has $\ell$ bits, check that $x_{k_j} = y_j$ for all $j < m$. I claim that this can be done with $\tilde {O} ((n + m) \ell)$ additional variables and constraints. -First of all, we may append $(0, x_0), (1, x_1), \dots, (n-1, x_{n-1})$ to the list $((k_j, y_j))$. Therefore we may assume without loss of generality that $m \geq n$ and $(k_i, y_i) = (i, x_i)$ for $i < n$. Moreover, it is sufficient to check that $((k_i, y_i))$ is self-consistent: That if $k_i = k_j$ then $y_i = y_j$. This is checked by sorting $((k_i, y_i))$: Let $(\tilde {k}_0, \tilde {y}_0), \dots, (\tilde {k}_{m-1}, \tilde {y}_{m-1})$ be a permutation of $(k_0, y_0), \dots, (k_{m-1}, y_{m-1})$ with $\tilde {k}_0 \leq \tilde {k}_1 \leq \dots \leq \tilde {k}_{m-1}$ (this is checkable in $\tilde {O} (m \ell)$ variables and constraints). Then it suffices to check the that if $\tilde {k}_{i+1} = \tilde {k}_i$ then $\tilde {y}_{i+1} = \tilde {y}_i$. This requires $O (m (\ell + \log n))$ bits and constraints.<|endoftext|> -TITLE: Silver forcing and Cichoń's diagram -QUESTION [6 upvotes]: Recall that the Silver forcing $\mathbb{P}$ is defined as the set of all partial functions $p\in 2^{\le\omega}$ such that $\omega\setminus dom(p)$ is infinite. As usual, $p\le_\mathbb{P}q$ if $p$ extends $q$. The Silver model is the model $V$ obtained by the countable support iteration of length $\omega_2$ of the Silver forcing over a model of ZFC+CH. -I'm interested in values of the cardinal characteristics of the continuum in the van Douwen and Cichoń diagrams in $V$. Halbeisen in his brilliant book "Combintorial Set Theory" proves that: -$V\models \omega_1=\mathfrak{d}<\mathfrak{r}=\mathfrak{c}=\omega_2$. -It follows immediately that $V\models\mathfrak{u}=\mathfrak{i}=\omega_2$ and that all characteristics of the van Douwen diagram below $\mathfrak{d}$ are equal to $\omega_1$ in $V$. The only unknown in the diagram is the almost-disjointness number $\mathfrak{a}$. -Question 1: What is the value of $\mathfrak{a}$ in $V$? -Let $\mathcal{M}$ and $\mathcal{N}$ denote the ideal of meager subsets and the ideal of Lebesgue null subsets of $\mathbb{R}$, respectively. It can be shown that the ground model reals are non-meager in $V$, so $non(\mathcal{M})=\omega_1$ in $V$, and hence the left-hand half of the Cichoń diagram is also equal to $\omega_1$. Since $\mathfrak{d}=\omega_1$ in $V$, $cov(\mathcal{M})=\omega_1$ in $V$ as well. The rest of the Cichoń diagram is unfortunately unknown to me. -Question 2: What are the values of $cof(\mathcal{M})$, $non(\mathcal{N})$ and $cof(\mathcal{N})$ in $V$? - -REPLY [4 votes]: Answer to question 2: $cof(\mathcal N)=\aleph_1$, so all cardinals in Cichoń's diagram stay small (and their smallness is witnessed by the set of reals from the ground model). -Proof sketch: For $q\le p$ in Silver forcing, write $q\le_n p$ if $\omega\setminus dom(p)$ and $\omega\setminus dom(q)$ agree on their first $n$ elements. This is an axiom A structure with the following "pure extension" property: Whenever $\dot \alpha$ is a name of an ordinal, $p$ a condition, $n\in \omega$, then there is a set $A$ of size $2^n$ and a condition $q\le_n p$ such that $q\Vdash \dot \alpha\in A$. -This shows that for every new function (in the single Silver extension), $f:\omega\to\omega$ there is an old slalom $(S_n:n\in\omega)$, $|S_n|=2^n$, $\forall n: f(n)\in S_n$, in other words: the single Silver extension has the Sacks property. -The Sacks property is preserved in countable support iterations of proper forcings. (Shelah, Proper and Improper Forcing, chapter XVIII.) But for an iteration of Silver forcing, or any sufficiently nice Axiom A forcing, this can alternatively be proved "directly", with the usual $\le_{F,n}$-fusion argument. (E.g. in Baumgartner's classical paper "Iterated Forcing", if I remember correctly.) -Hence also the model after adding $\omega_2$ Silver reals has the Sacks property over the ground model. -By theorem 2.3.12 in Bartoszyński-Judah (first proved by Bartoszyński in 1984, I think: "Additivity of measure implies additivity of category", TAMS), the Sacks property is equivalent to the following property: Every new measure zero set is covered by an old measure zero set.<|endoftext|> -TITLE: $2 \times 2$ matrix question -QUESTION [10 upvotes]: Let $A$, $B$, and $C$ be $2\times 2$ complex matrices, with $A$ and $C$ rank $1$ Hermitian. Can we find a real number $a$ and a $2\times 2$ unitary $U$ such that -$$A + BV + V^*B^* + V^*CV$$ -is a scalar multiple of the identity, where $V = aU$? -The motivation is that this would answer this question in a large special case. - -REPLY [15 votes]: Yes, one can always do this. In fact, one can assume that $\det(U)=1$, as this follows by a homotopy argument, using the fact that $\pi_3(S^2)\simeq \mathbb{Z}$. -Here are the details: Assume that $\det(U)=1$ so that $V$ has a non-negative real determinant. Then one can write $V$ uniquely in the form -$$ -V(x) = \begin{pmatrix} x_0+i\,x_1 & -x_2 + i\,x_3\\ - x_2-i\,x_3 & \phantom{-}x_0-i\,x_1 - \end{pmatrix} -$$ -for some $x = (x_0,x_1,x_2,x_3)\in\mathbb{R}^4$. -Now consider the map $F$ from $\mathbb{R}^4$ into $\mathbb{R}^3$ (regarded as the traceless Hermitian $2$-by-$2$ matrices) defined by -$$ -F(x) = \left[A +BV(x) + V(x)^*B^* + V(x)^*CV(x)\right]_0\,, -$$ -where, for a Hermitian $2$-by-$2$ matrix, $H$, we write $H - \tfrac12\mathrm{tr}(H)\,I_2 = H_0$. For use below, define the norm on traceless Hermitian $2$-by-$2$-matrices $M$ by the rule $|M|^2 = \tfrac12 \mathrm{tr}(M^2)$. -Our task is to show that there exists an $x\in\mathbb{R}^4$ such that $F(x)=0$. -Note that $F(x)$ is a quadratic polynomial in $x$ taking values in $\mathbb{R}^3$. -When $|x|$ very large, $F(x)/|x|^2$ is close to $G(x) = (V(x)/|x|)^*C_0(V(x)/|x|)$, a vector whose norm is $|C_0|$. Since $C$ has rank 1, $C_0$ is not zero. Thus, when $|x|>>0$, $\bigl|F(x)\bigr|/|x|^2\approx |C_0|>0$. Moreover, when restricted to the $3$-sphere of radius $R>0$ in $\mathbb{R}^4$, the homogeneous map $G: S^3(R)\to S^2(|C_0|)$ is simply the Hopf map $S^3\to S^2$, appropriately scaled, which is known not to be homotopically trivial. (In fact, it is a generator of $\pi_3(S^2)\simeq\mathbb{Z}$.) -If there were not a $y\in\mathbb{R}^4$ that satisfied $F(y)=0$, then, for $R>>0$ the map $H(x) = F(x)/|F(x)|$ would map the $4$-ball of radius $R$ to $S^2$ in such a way that the map on its boundary $3$-sphere would be homotopic to the Hopf map, which is impossible. -Thus, there is a $y\in\mathbb{R}^4$ such that $F(y)=0$, i.e., -$$ -A +BV(y) + V(y)^*B^* + V(y)^*CV(y) = \mu\,I_2 -$$ -for some number $\mu$, as desired.<|endoftext|> -TITLE: Extending a weak*-converging sequence onto a superspace -QUESTION [6 upvotes]: Let $X$ be a real Banach space and $Y\subset X$ be a (closed) subspace of $X$. Assume that a sequence $y_n^*\in S_{Y^*}$ weak*-converges to some $y^*\in S_{Y*}$. (Here $S_{Y^*}$ stands for the dual unit sphere.) -Let $x_0 \in X \setminus Y$ and define $Z:= \textrm{span} (Y\cup \{x_0\})$. -Is the following true? -There exist a sequence $z_n^*\in S_{Z^*}$ with $z_n^*\restriction_Y = y_n^*$, and $z^*\in S_{Z^*}$ with $z^*\restriction_Y = y^*$, such that $z_n^*$ converges weak* to $z^*$. -Remarks: - -It is easy to obtain that a SUBSEQUENCE of $z_n^*$ converges weak* to $z^*$. -I'd like to consider general Banach spaces without additional assumptions (e.g. reflexivity, separability). -I wouldn't mind using an additional set-theoretic axiom consistent with ZFC. - -REPLY [6 votes]: The norm preserving condition, is as Mikhail points out, essential to prove that the such an extension does not exist. The question is stated for Banach spaces, but notice that the same question for normed spaces is equivalent to the question for Banach spaces because a bounded sequence of functionals converges weak$^*$ if it converges pointwise on a dense set. -So assume that the question has an affirmative answer in the category of normed spaces. We claim that a stronger thing is true; namely, in the set up of the OP, the superspace $Z$ of $Y$ can be taken to be $X$ itself. Indeed, the union of a nested family of extensions that are weak$^*$ convergent gives extension to the union of the domains of the extending functionals, so one can Zornicate just as in the standard proof of the Hahn-Banach theorem. More explicitly, consider all $(Z, (z_n^*)_n)$ such that $Y \subset Z $, $Z$ is a subspace of $X$, and $(z_n^*)_n$ are linear functionals in the unit ball of $Z^*$ such that for all $n$, $z_n^*$ extends $y_n^*$, and $(z_n^*)_n$ converges pointwise on $Z$. Partially order by $(Z, (z_n^*)_n) \le (W, (w_n^*)_n)$ provided $Z\subset W$ and for all $n$, $z_n^* \subset w_n^*$, and observe that the hypothesis in Zorn's Lemma applies. -To see that the OP's question has a negative answer, let $X=\ell_\infty$, $Y=c$ (the space of convergent sequences under the supremum norm), and let $y_n^*$ be the unit vector basis for $\ell_1 \subset c^*$. There are not uniformly bounded weak$^*$ convergent extensions to $\ell_\infty$ because weak$^*$ converging sequences in $\ell_\infty^*$ are weakly convergent. -EDIT 12/30/17: If you want an explicit example in the separable setting to the OP's original formulation, again let $X=\ell_\infty$, $Y=c$ (the space of convergent sequences under the supremum norm), let $y_n^*$ be the unit vector basis for $\ell_1 \subset c^*$, and let $z_0 $ be the element of $\ell_\infty$ whose odd coordinates are all $1$ and whose even coordinates are all $-1$.<|endoftext|> -TITLE: Depth under localization over a Cohen-Macaulay ring -QUESTION [7 upvotes]: Let $A$ be a Cohen-Macaulay local ring, and let $M$ be an $A$-module that is (S$_2$) and has depth $\ge n$ for some fixed $n$. Let $\mathfrak{p} \subset A$ be a prime of height $\ge n$. Is it true that $\mathrm{depth}_{A_{\mathfrak{p}}} M_{\mathfrak{p}} \ge n$? -I would be happy with a counterexample or a proof even in the special case $n = 3$. - -REPLY [9 votes]: Here is a counter-example when $A$ is regular local of dimension $4$ and $n=3$, the first non-trivial case. Let $A = k[[x,y,z,t]]$, and $P$ be a prime ideal of height $3$, say $P=(x,y,z)$. Let $M=\Omega P$, the syzygy of $P$. Obviously, $M$ is also $\Omega^2 (A/P)$. -The depth of $M$ at various localizations can be computed easily by tracking depth along short exact sequences. In particular, $M$ is $(S_2)$ since any second syzygy module over $A$ would be. Since $depth(A/P)=1$, it follows that $depth(M)=1+2=3$. On the other hand, when you localize at $P$, $M_P$ is a second syzygy of the residue field in a $3$-dimensional regular local ring, so $depth(M_P)=2$. -PS: as can be seen in the comments below the question, there are different definitions of $(S_n)$ in the literature, see What is Serre's condition (S_n) for sheaves? However, the notion I assumed for this answer is the most restrictive one ((1) in the one cited), so it applies to others as well.<|endoftext|> -TITLE: Bounds on the number of zeros of real analytic functions -QUESTION [6 upvotes]: Let $F(A)$ be a class of real-analytic function on an interval $A \subset \mathbb{R}$ minus the zero function. -We have the following theorem for $F(A)$. - -If $f \in F(A)$ then $f$ has at most finitely many zeros $A$. - -Proof Suppose $f\in F(A)$ has infinitely many zeros on a bounded interval. Then by Bolzano-Weierstrass the set of zeros has a convergent subsequence in $A$. Therefore, by identity theorem, $f$ must be zero on all of $A$. -However, this contradicts our assumption that $f$ is non-zero. Q.E.D. -My question: Are there ways of sharpening the bound on the number of zeros? -Let $N(f)$ be the number of zeros of $f$. Clear, there is no uniform bound on $N(f)$ for all $f\in F$. -However, there a subset of $F$ for which we do have good upper bounds like a set of polynomials of degree $n$ in which case $N\le n$. -My second question (or refined first question) is: For a give $f$ which is analytic on $A$, but not a polynomial, are there ways of finding an upper bound on the number zeros of $f$? - -REPLY [7 votes]: As noted by the comments, you must require that your interval $A$ is compact (otherwise, $\sin(1/x)$ has infinitely many zeroes on $\mathopen]0;1\mathclose[$. -Moreover, you cannot have a bound valid for every class $F(A)$, even if it only consists of polynomials — there are nonzero polynomials with as many zeroes as you wish on your interval $A$. -However, such polynomials will have unbounded degree. -This is a typical theme in o-minimality: if you bound the complexity of your class of functions, then there is a bound on the number of zeroes. -In some cases, this bound can be effective. For exemple, if $F(A)$ consists -of exponential polynomials with at most $m$ terms of the form $x^\alpha e^{\beta x}$, then the number of zeroes is bounded from above by something like $m-1$ (without guarantee...). You can find such results in Khovanskii's book Fewnomials.<|endoftext|> -TITLE: Who first proved the generalization of Bertrand's postulate to (2n,3n) and (3n,4n)? -QUESTION [19 upvotes]: In Wikipedia's page for Bertrand's postulate, it is said that its (2n,3n) version was proved by El Bachraoui in 2006. Seems likely that it was first proved way before than that! Can anyone point to the first source, or at least to a previous one? -Analogously, Wikipedia said until recently that the (3n,4n) version was due to Andy Loo in 2011. I'm aware of a proof by Denis Hanson in 1973, so I have updated the page with that info, but I don't know if his proof is the first one. Do you know of previous proofs? - -REPLY [24 votes]: I have finally found the following papers and results, which predate Nagura's paper of 1952. I cite them from newest to oldest: - -(Molsen, 1941): - - -For $n\geq 118$ there are primes in $(n,\frac43n)$ congruent to 1,5,7,11 modulo 12. -For $n\geq 199$ there are primes in $(n,\frac87n)$ congruent to 1,2 modulo 3. This result implies that of Nagura. - - -K. Molsen, Zur Verallgemeinerung des Bertrandschen Postulates, Deutsche Math. 6 (1941), 248-256. MR0017770 - -(Breusch, 1932): - - -For $n\geq 7$ there are primes in $(n,2n)$ congruent to 1,2 modulo 3 and to 1,3 modulo 4. -For $n\geq 48$ there is a prime in $(n,\frac98n)$. This result implies those of Nagura and Molsen (but not the congruences part). - - -R. Breusch, Zur Verallgemeinerung des Bertrandschen Postulates, dass zwischen x und 2x stets Primzahlen liegen, Math. Z. 34 (1932), 505–526. MR1545270 - -(Schur, 1929, according to Breusch in the previous paper): - - -For $n\geq 24$ there is a prime in $(n,\frac54n)$. This result already implies those of Hanson and El Bachraoui. - - -I. Schur, Einige Sätze über Primzahlen mit Anwendungen auf Irreduzibilitätsfragen I, Sitzungsberichte der preuss. Akad. d. Wissensch., phys.-math. Klasse 1929, S.128.<|endoftext|> -TITLE: positive sum of sines -QUESTION [5 upvotes]: This was asked but never answered at MSE. -Let $f(x) = \sin(a_1x) + \sin(a_2x) + \cdots + \sin(a_nx)$, where the $a_i$'s -represent distinct positive integers. Suppose also that $f(x)$ satisfies the inequality $f(x) \geq 0$ on the open interval $0 < x < \pi.$ -In the case n=1 of a single summand it is obvious that only $f(x) = \sin x$ satisfies the condition. For two summands, a short argument shows that only -$f(x) = \sin x + \sin(3x)$ works. Following up on this, we define $g_n(x) = \sin x -+ \sin(3x) + \sin(5x) + \cdots + \sin((2n-1)x)$. Computing the sum explicitly yields $g_n(x) = \frac{\sin^2(nx)}{\sin x}$ which makes it clear that $g_n(x)$ is -nonnegative on $(0,\pi).$ -Questions: (1) Are there any other examples of $f(x)$ as above besides $g_n(x)$? -If so, can one classify them all? -(2) The special case $f(x) = g_1(x) = \sin x$ satisfies the stronger condition of being strictly positive over $0 < x < \pi$. Is $\sin x$ the unique -such instance of $f(x) > 0$ on $(0,\pi)$? -Thanks - -REPLY [7 votes]: Let me begin by restating your conjecture. Consider the sum -$$S(x):=\sum_{j=1}^nb_j\sin jx,\quad b_n=1,\quad b_j\in\{0,1\}.$$ -Then $S(x)\geq 0$ on $[0,\pi]$ if and only if $b_j=1$ for all odd $j$ and $b_j=0$ for all even $j$ (in particular, $n$ is odd). -First of all, notice that $S(x)\geq 0$ on $[0,\pi]$ if and only if $S^*(x)\geq 0$ on $[-\pi,\pi]$, where -$$S^*(x):=2S(x)\sin x=\sum_{j=1}^nb_j\left(\cos(j-1)x-\cos(j+1)x\right)\geq 0.$$ -Second, we must have $b_1=1$, by the well-known theorem that a trigonometric polynomial without a constant term must change sign on $[-\pi,\pi]$. -(This is because its integral over $[-\pi,\pi]$ equals zero). -Now transform $S^*$ as follows: -$$S^*(z)=1-\cos(n+1)x+A(x)+B(x),$$ -where -$$A(x)=b_2\cos x+(b_3-1)\cos 2x+(1-b_{n-2})\cos(n-1)x-b_{n-1}\cos nx,$$ -and -$$B(x)=\sum_{j=3}^{n-2}(b_{j+1}-b_{j-1})\cos jx.$$ -Let us set $z=\exp(ix),\;|z|=1$ and consider the first two summands in $S^*$: -$$1-\cos(n+1)x=(2-z^{n+1}-z^{-n-1})/2.$$ -This is non-negative and has double zeros at the roots $z_k$ of degree $n+1$ of unity. -So for $S^*$ to be non-negative, it is necessary that $A+B$ be non-negative at -the points $x_k=2\pi k/(n+1)$ corresponding to $z_k$. -On the other hand, we have -$$\sum_{k=0}^n \left(A(x_k)+B(x_k)\right)=0, \quad x_k=2\pi k/(n+1),$$ -by the well-known "orthogonality relations", -$$\sum_{k=0}^n z_k^m=0,\quad\mbox{when}\quad |m|\leq n.$$ -Therefore $A+B$ must be zero at all $n+1$-st roots of unity, moreover, -all these zeros must be multiple (if not, $S^*(x)$ will change sign near some $x_k$), from which we conclude that $A+B\equiv 0$, because a non-zero trigonometric polynomial of degree $n$ cannot have $n+1$ multiple zeros. This means that -$$b_2=0,\; b_3=1,\; b_{n-2}=1,\; b_{n-1}=0,$$ -and $b_{j+1}=b_{j-1}$ for all $j\in[3,n-2]$. This proves your statement.<|endoftext|> -TITLE: Norming subspaces of duals of quotient spaces -QUESTION [5 upvotes]: In Davis, William J., and William B. Johnson. "Basic sequences and norming subspaces in non-quasi-reflexive Banach spaces." Israel Journal of Mathematics 14.4 (1973): 353-367., the authors discuss the following problem -"If $M$ is a norming subspace of $X^*$ and $Y$ is an $M$-closed subspace of $X$, then is $M \cap Y^\perp$ a norming subspace of $(X/Y)^*$ (where $Y^\perp$ is identified with $(X/Y)^*$ in the canonical way)?" -In the paper, the authors conclude that if $X$ is not quasi-reflexive, then there exists a norming subspace $M$ such that the conclusion fails. -What I am interested in is whether the result can be salvaged in some way by adding some reasonable conditions on $M$, and I would like to know if there is anything in the literature that has addressed this. The papers citing this paper do not discuss this particular question any further. -Thank you for your time. - -REPLY [2 votes]: Some comments on the problem (I print them as an answer, because they are too lengthy for a comment): -(1) As I understand, the condition that $Y$ is $M$-closed is -equivalent to the condition that $M\cap Y^\perp$ is total over the -quotient $X/Y$. -(2) So, if $X/Y$ is quasireflexive, then "$M\cap Y^\perp$ is -total" implies "$M\cap Y^\perp$ is norming" (we do not need -quasireflexivity of $X$ itself). -(3) You would like to get a condition on $M$ under which for -any subspace $Y$ in $X$ the condition "$M\cap Y^\perp$ is total" -implies "$M\cap Y^\perp$ is norming". -(4) There is a trivial condition of this type: $M$ is of finite -codimension in $X^*$ (but I doubt that it is satisfactory for -you). -(5) It is tempting to claim that any $M$ of infinite codimension -in $X^*$ fails to satisfy this condition for some $Y$, but this is -not true. If $X$ itself is a dual space and $M$ is its predual, -then $M$ is of infinite codimension in $X^*$ for -non-quasireflexive $X$, on the other hand, the condition from (3) -is satisfied. In fact, if $Y$ is $M$-closed, then $X/Y$ is the -dual of $Y^\perp\cap M$, and so "$Y^\perp\cap M$ is norming over -$X/Y$" -(6) Of course, one can investigate this matter further. My survey Weak* sequential closures in Banach space theory and their applications -could be helpful for this. The survey was written in 1999, but I -did not notice any further work in this direction except for my -note Weak$^*$ closures and derived sets in dual Banach spaces. Possibly I missed something because I do not work in this -direction now.<|endoftext|> -TITLE: Are finite spaces a model for finite CW-complexes? -QUESTION [17 upvotes]: Are finite topological spaces (i.e. topological spaces whose underlying set is finite) a model for the homotopy theory of finite simplicial sets (= homotopy theory of finite CW-complexes) ? -Namely, is there a reasonable way to: -(1) given a finite topological space $X$, construct a finite simplicial set $nX$. -(2) given two finite topological spaces $X$ and $Y$, construct a simplicial set $Map(X,Y)$ whose geometric realisation is homotopy equivalent to the mapping space $Map(|nX|,|nY|)$ between the geometric realizations of $nX$ and $nY$. -(3) etc. (higher coherences) -Note: One can, of course, define $Map(X,Y)$ to be the derived mapping space between $nX$ and $nY$. But I'm wondering whether there's something more along the lines of "take the (finite) set of all continuous maps $X\to Y$ and then ... " - -REPLY [10 votes]: Andre, the best answer to your very first question is given by Emily Clader, who -proved that every finite simplicial complex is weak homotopy equivalent to an inverse limit of finite spaces. -A small mistake is corrected and much further work is done in Matthew Thibault's unpublished 2013 University of Chicago thesis. -The answer to your question (1) is classical, going back to McCord as in Nardin's answer. I don't know a really good answer to (2). -I should apologize that my book referred to by Quaochu Yuan is still unfinished. It will be some day. It uses the finite space of continuous maps between finite spaces to discuss homotopies in Section 2.2, but of course that is too small to realize properly. The generalization of this to A-spaces (T_0 Alexandroff spaces) is subtle and is studied by Kukiela, -but he does not address your question (2). -In answer to a question raised in Nardin's answer, the category of A-spaces is isomorphic to the category of posets. It was implicit in Thomason's model structure on the category of small categories that there is a similar model structure on the category of posets, and that was made explicit by Raptis. It is Quillen equivalent to the standard model structure on simplicial sets. That was generalized to posets with action by a discrete group G by Stephan, Zakharevich and myself. In passing, that paper somewhat streamlines the nonequivariant proof.<|endoftext|> -TITLE: Semantics of derivations as derivatives -QUESTION [7 upvotes]: My understanding of how derivations on commutative rings are like derivatives is that a derivation on $R$ is differentiation with respect to a vector field on $\text{Spec}(R)$. But derivations are supposed to be thought of as like derivatives in a wider context than commutative rings, and I don't really understand how. -Take anti-derivations on the exterior algebra of differential forms on a manifold, for instance. The exterior derivative and Lie derivatives both give you information about infinitesimal change in a differential form, but the interior derivative is defined pointwise, as an anti-derivation on the exterior algebra of the tangent space at each point, which ruins any attempt to think of anti-derivations on differential forms as capturing some information about infinitesimal change. So how can you think about interior differentiation as being like a derivative in any more concrete sense than that it obeys similar syntactic rules? More generally, how can you think about anti-derivations on exterior algebras (or more generally still, on anti-commutative graded rings) as being like a derivative? -There's also derivations on non-commutative rings. The adjoint action of an element of a ring $\text{ad}_x(y):=xy-yx$ is a derivation, but I don't see the significance of this. For example, the Pincherle derivative seems to act like a sort of "differentiation with respect to $d/dx$" insofar as it sends $d/dx$ to $1$, and the fact that it is a derivation forces certain other facts that this heutristic naively suggests to be true (for instance, that the shift operator $S_1=e^{d/dx}$ is its own Pincherle derivative). Is there some more precise way to describe the Pincherle derivative as differentiation with respect to $d/dx$? What about a way to characterize arbitrary derivations on non-commutative rings? -How about derivations on Lie algebras? The Jacobi identity can be interpreted as saying that adjoint actions are derivations, but as in the analogous fact I mentioned for derivations on non-commutative rings, I'm curious about what the significance of this is. And about how to think of arbitrary derivations on Lie algebras. - -REPLY [11 votes]: In all of these contexts, derivations are infinitesimal automorphisms, in the sense that $D$ is a derivation on $A$ (an algebra, a Lie algebra, etc.) iff $\exp(Dt)$ is an automorphism of $A \otimes k[t]/t^2$. On a commutative ring automorphisms correspond to automorphisms of the spectrum so derivations correspond to infinitesimal automorphisms of the spectrum, or vector fields. -On a noncommutative ring or a Lie algebra the commutator $[x, -]$ exponentiates to conjugation by $\exp(xt)$, which is an inner automorphism; that's why these derivations are called inner derivations. This observation leads to the sometimes useful identity -$$\exp(Xt) Y \exp(-Xt) = \sum_{n \ge 0} \frac{\text{ad}_X^n(Y)}{n!}.$$ -Said another way, derivations always form a Lie algebra, and the "Lie group" that they're the Lie algebra of is, heuristically speaking, the automorphism group.<|endoftext|> -TITLE: Derivatives of Gaussian processes -QUESTION [6 upvotes]: A Gaussian process $X$ on Euclidean space $\mathbb R^d$ has a radial basis kernel if for any $u,w\in\mathbb R^d$, we have -$$ \mathrm{Cov}(X_u, X_w) = \sigma^2 \exp\left ( -\frac{\left\lVert u-w \right \rVert^2}2\right)$$ -Draws from Gaussian processes with zero mean and radial basis covariance kernels are smooth almost surely. Is there any work on the distribution of the derivative of a draw from such a GP? - -REPLY [5 votes]: In the following, I assume a zero mean function, for simplicity. -The realizations of a stationary Gaussian process are (a.s.) $n$-times differentiable if the covariance function $k(t_1-t_2):=K(t_1,t_2)$ is $2n$-times differentiable at 0. Hence, in the following, I assume the Gaussian radial basis function -$$k(t) = \exp\left(-\frac{1}{2} t^2\right)$$ -as covariance function, which is smooth at zero. -How do the realizations of such a Gaussian process look like? The covariance function $k$ is strictly positive, hence any function value is positively correlated with any other function value. However, this correlation is almost zero for long time differences. Hence, such realizations tend to go back of the mean after a short time. In short: what goes up must come down (a.s.). -Now let us have a look at the covariance of the derivative Gaussian process. By the formulas provided by cknoll, its covariance function is -$$k_\partial(t) :=-\frac{\operatorname{d}^2}{\operatorname{d}\!t^2} k(t) = (1-t^2) \exp\left(-\frac{1}{2} t^2\right) = (1-t^2)k(t)$$ -and it is easy to see that such processes are again smooth. -So how do the realizations of this derivative Gaussian process look like? -For simplicity, I want to distinguish three regions of this covariance function $k_\partial(t)$. - -Close to zero, the covariance function is positive and "looks like" $k(t)$. Hence, the derivative Gaussian process locally looks like the original Gaussian process. -$k(\pm1)=0$ and it is negative for $|t|>1$. Hence, derivatives have a negative correlation for time difference $>1$. The minimum of $k_\partial(t)$ is $\sqrt3$. Hence, the original Gaussian process will probably come down again after $1$ time steps and the derivative Gaussian process will probably have switched sign after $\sqrt3$ time steps. -For $|t|$ big, the covariance function is almost zero again, hence we would expect it to go back to the mean with a high variance. - -In summary, the derivatives of the realizations are again smooth (a.s.), look like the original Gaussian process over very short and very long time scales, but have the tendency to regularly switch their sign over medium time scales.<|endoftext|> -TITLE: A curious identity involving a covariant of binary cubic forms -QUESTION [10 upvotes]: Let $F(x,y) = a_3 x^3 + a_2x^2 y + a_1 xy^2 + a_0 y^3$ be a binary cubic form, say with real coefficients. Put $H(x,y) = H_F(x,y)$ for the Hessian covariant of $F$, defined by -$$\displaystyle H_F(x,y) = \frac{1}{4} \begin{vmatrix} F_{xx} & F_{xy} \\ F_{xy} & F_{yy} \end{vmatrix},$$ -and put -$$\displaystyle G(x,y) = G_F(x,y) = \begin{vmatrix} F_x & F_y \\ H_x & H_y \end{vmatrix}.$$ -$H,G$ are both covariants of $F$ under the substitution action of $\operatorname{GL}_2$, and in particular, $H, G, F$ and the invariant $\Delta(F)$, the discriminant of $F$, generate the ring of polynomial covariants. They are connected by a single syzygy, given by -$$\displaystyle 4H(x,y)^3 + G(x,y)^2 = -27 \Delta(F) F(x,y)^2.$$ -As can be verified by immediate calculation, we have -$$\displaystyle \Delta(G) = 729 \Delta(F)^3,$$ -which is a perfect cube. -My question is, suppose that $G$ is a binary cubic form with integer coefficients satisfying $\Delta(G) = 729 n^3$ for some non-zero integer $n$. Does it follow that $G$ is the $G_F$-covariant for some binary cubic form $F$ with integer (rational) coefficients? If not, what is a counter-example, and what would be a stricter condition that guarantees $G$ is such a covariant? -On a related note, if $G'$ is the non-trivial cubic covariant of $G_F$, then $G' = -729 \Delta(F)^2 F(x,y)$. - -REPLY [6 votes]: As Abdelmalek notes, if $G$ is to be the cubic covariant $G_F$ of another form $F$, and has a nonzero discriminant of the expected shape $\Delta(G)=3^6n^3$, then $F$ must divide $G'=G_G$ - indeed the only candidate (comparing discriminants and exploiting homogeneity) is $F=-G_G/(3^6n^2)$, which will have rational coefficients when $n$ is rational, and integral coefficients when (and only when) those of $G_G$ are integral multiples of $3^6n^2$. -And when this is the case, this $F$ does indeed work: -$$G_F = -G_{G_G}/(3^6n^2)^3 = 3^6\Delta(G)^2G/(3^{18}n^6) = 3^6(3^6n^3)^2G/(3^{18}n^6) = G\,.$$<|endoftext|> -TITLE: Birational Calabi-Yau varieties with non-isomorphic cohomological invariants -QUESTION [6 upvotes]: We know from the work of Kontsevich, for example, that birational Calabi-Yau complex varieties have the same Hodge numbers. I want to understand to what extent the equivalence of cohomological invariants fails for two such varieties. Precisely, I am looking for examples of birational Calabi-Yau varieties over any field that have some non-isomorphic cohomological invariants (Chow groups, Hodge structures, K-groups, cohomology groups, etc.) - -REPLY [2 votes]: Take a look at arXiv:math/0703315. -It gives an explicit pair of birational Calabi-Yau threefolds which are cohomologically non-isomorphic.<|endoftext|> -TITLE: Maximal Albanese variety -QUESTION [9 upvotes]: I am looking for a proof for the following statement - -A smooth projective variety $X$ has maximal Albanese dimension if and - only if the cotangent bundle of $X$ is generically generated by its - global sections, that is,$$ H^0(X,\Omega_X^1)\otimes\mathcal O_X\to - \mathcal \Omega_X^1$$ is surjective at the generic point of $X$. - -REPLY [9 votes]: In characteristic $0$ that follows from Sard's theorem / generic smoothness. In characteristic $p$, I believe that this is false (I still need to check how double point singularities, $\text{Zero}(x_1^2 + \dots + x_{n-1}^2+x_n^2 + x_{n+1}^p)$, affect the cotangent sheaf). -Characteristic 0.In characteristic $0$, the map of sheaves above is the pullback map, $$d(\text{alb}_X)^\dagger: \text{alb}_X^*\Omega_{\text{Alb}_X/k} \to \Omega_{X/k}.$$ Since the Albanese morphism from $X$ to its image, $\text{alb}_X(X)$, is generically smooth, the dimension of the image equals the rank of $d\text{alb}_X^\dagger$ at a generic point of $X$. -Positive characteristic. However, this seems to be false in positive characteristic. Here is an e-mail that I wrote a few years ago about positive characteristic counterexamples to a similar question. I added some TeX formatting to the e-mail. -"Dear Chris, -I saw Davesh today, and we talked more about the issue we were discussing on Friday regarding purely inseparable covers of Abelian varieties. I now believe that there are serious problems whenever the dimension is $> 1.$ Let $k$ be an algebraically closed field of characteristic $p > 0$. Let $A$ be an Abelian variety of dimension $g > 1$. For a "generic" element $f$ in $k(A)$, the corresponding rational transformation $f : A \dashrightarrow \mathbb{P}^1$ "regularizes" on a blowing up, $u : A' \to A$, such that $A'$ is smooth over $k$. Now consider the following Cartesian diagram, $$ \begin{array}{rrl} X & \xrightarrow{i} & A' \\ g~\downarrow & & \downarrow~f \\ \mathbb{P}^1 &\xrightarrow{F}& \mathbb{P}^1\end{array}$$ where the bottom horizontal $k$-morphism $\mathbb{P}^1 \to \mathbb{P}^1$ is the arises from the Frobenius morphism . The purely inseparable morphism to consider is $i$. -The issue is this: for $f$ generic, the "critical" points of $f$ are just finitely many ordinary double points (I am assuming that $p$ is different from $2$). If $\text{dim}(A) = 1$, then the points of $X$ above such critical points are non-normal. When you normalize, then $X$ is smooth, and your previous argument is valid, i.e., $X$ is an elliptic curve. However, if $\text{dim}(A) > 1$, then the points of $X$ above these finitely many critical points are, in fact, already normal. Thus, $X$ is normal yet singular. So $X$ cannot be an Abelian variety. -Best regards, -Jason" -By construction, the morphism $i$ is finite. Thus, $X$ has maximal Albanese dimension. By construction $H^0(A',\Omega_{A'/k})$ generically generates $\Omega_f$, and thus the $i$-pullback also generically generates $\Omega_g$. If the dimension of the Albanese of $X$ were strictly greater than $\text{dim}(A)$, then there would be more etale $\ell$-sheeted covers of $X$ than there are of $A$, for $\ell$ prime to the characteristic. Yet, since $i$ is purely inseparable, the pullback map on etale fundamental groups is an isomorphism. If the Albanese morphism of $X$ were different from $i$, it would be a generically finite morphism that factors $i$. Since $i$ has prime degree $p$, that is impossible. Thus, $i$ is the Albanese morphism. -Since $i$ is purely inseparable, the rank of $d(\text{alb}_X)^\dagger$ at a general point is at most $\text{dim}(A)-1$. Finally, since $\Omega_i$ is the pullback of $\Omega_F$, i.e., the pullback of $\mathcal{O}_{\mathbb{P}^1}(-2)$, there are no nonzero global sections of $\Omega_i$. Thus, from the usual exact sequence, $$0 \to \text{Image}(d(\text{alb}_X)^\dagger) \to \Omega_{X/k} \to \Omega_i \to 0,$$ it follows that the rank of the following sheaf homomorphism at a general point of $X$ equals $\text{dim}(A)-1$, $$H^0(X,\Omega_{X/k})\otimes_k \mathcal{O}_X \to \Omega_{X/k}.$$<|endoftext|> -TITLE: What is the symmetric monoidal structure on the $(\infty,1)$-category of spectra? -QUESTION [11 upvotes]: The $(\infty, 1)$ category $Sp$ of spectra as defined by Lurie in Higher Algebra has the structure of a symmetric monoidal category. Although I know the definition of symmetric monoidal category in the $(\infty, 1)$ setting and can reasonably follow Lurie's arguments in Higher Algebra as to why $Sp$ has such a structure, I don't understand it well enough to think about it intuitively. -So my question is, what does the symmetric monoidal structure on $Sp$ look like? How is this related to the symmetric monoidal structure on symmetric or orthogonal spectra in the ordinary categorical setting? How may I picture ring spectra and other such objects arising from the monoidal structure? - -REPLY [6 votes]: I thought it might be worth mentioning, as an addendum to the other nice answers, that there is also an $\infty$-categorical approach to the smash product that looks more like the traditional constructions: namely, you can just redo the most classical construction of the smash product on the stable homotopy category (as in Adams's book) using $\infty$-categories. (I suppose this might count as a "folk theorem", and unfortunately I don't think it's been written down in detail anywhere; I learned the idea from Justin Noel.) -Start with the $\infty$-category $\mathrm{PSp}$ of "prespectra", meaning sequences of pointed spaces $X_n$ with maps $\Sigma X_n \to X_{n+1}$. Using the smash product of pointed spaces you can enhance this to an $\infty$-operad: a multimorphism $(X, Y) \to Z$ is given by maps $X_n \wedge Y_m \to Z_{n+m}$ compatible with the structure maps in $X,Y,Z$ (and similarly for multimorphisms with more inputs). -Now consider the subcategory $\mathrm{Sp}$ of "spectra" (or "$\Omega$-spectra"), meaning prespectra $X$ where the adjoint maps $X_n \to \Omega X_{n+1}$ are equivalences. If you restrict the operad structure to this subcategory, I claim you get a symmetric monoidal $\infty$-category: for spectra $X$, $Y$, $Z$, multimorphisms $(X,Y) \to Z$ are represented by maps $X \wedge Y \to Z$ for a spectrum $X \wedge Y$. This spectrum you can define by choosing sequences $i_n$, $j_n$ with $i_n+j_n = n$ (with both unbounded) and setting $(X \wedge Y)_n := X_{i_n} \wedge Y_{j_n}$ - what Adams calls a "handicrafted smash product", if I recall correctly. -Actually working this out precisely would take a fair bit of (probably quite annoying) work, however, and in terms of applications I doubt it would have any advantages over the constructions Dylan and Denis describe.<|endoftext|> -TITLE: This group is "dual" to the Mathieu group $M_{23}$. Is it known? -QUESTION [10 upvotes]: Inspired by this question, in particular by the indeed elegant description of the Mathieu group $M_{23}$ it starts with, I am wondering about the following: -Instead of $C$, defined as the multiplicative subgroup of order $23$ in the field $F=\mathbb F_{2^{11}}$ with $2^{11}=23\cdot89+1$ elements, let us take the "complementary" subgroup $D$ of order $89$. Knowing that $M_{23}$ is the group of additive maps of $F$ to itself which permute the set $C$, what about the group $G$ of additive maps of $F$ to itself which permute the set $D$ instead? Naively, I would expect this group $G$ to be also a simple group by "duality", but it cannot be a sporadic one because of the divisor $89$. - -What can be said about $G$? - -Of course, any composite pernicious Mersenne number which is a product of two primes defines two "dual" groups in a similar way. Is there any interesting relationship between the groups of such a pair? Disclaimer: I am not a group theorist. - -REPLY [10 votes]: This group is the semidirect product $H=C_{89}\rtimes C_{11}$. Note that $H\le G$, where $C_{89}$ is multiplication by elements of order $89$ (and $1$), and $C_{11}$ is generated by the Frobenius automorphism $x\mapsto x^2$ of $\mathbb F_{2^{11}}$. -As $89$ is prime, $G$ is a primitive group. Primitive groups of such small degrees (actually up to degree $4000$, if I remember right, possibly with the exception of some affine cases) have been classified. In this case $G$ is either $A_{89}$, $S_{89}$, or the Sylow $89$-subgroups are normal. The first two cases cannot hold, because $\lvert\text{GL}_{11}(\mathbb F_2)\rvert$ is way too small (or because $13$ doesn't divide the order $2^{55}(2-1)(2^2-1)\dots(2^{11}-1)$ of $\text{GL}_{11}(\mathbb F_2)$). -Thus $G$ normalizes $C_{89}$. Let $g$ be an additive bijection of $\mathbb F_{2^{11}}$ which fixes $1$. Write $g$ as an additive polynomial map, i.e. $g(x)=\sum_{i=0}^{10}a_ix^{2^i}$. Then there is an integer $m$ with $1\le m\le 88$ such that $\sum_{i=0}^{10}a_i\zeta^{2^i}=\zeta^m$ for each element $\zeta$ of order $89$ (or $1$). Let $X$ be a variable, and $r(X)$ be the remainder of the division of $\sum_{i=0}^{10}a_iX^{2^i}$ by $X^{89}-1$. So $r(X)-X^m$ has at least $89$ roots, but degree $\le 88$, so $r(X)=X^m$. However, the remainders of $2^i$ modulo $89$, $0\le i\le 10$, are pairwise distinct. This forces that precisely one of the $a_i$ is $1$, the others are $0$, and $m$ is a power of $2$ modulo $89$.<|endoftext|> -TITLE: Where is my mistake in calculating duals? -QUESTION [7 upvotes]: I'm confused and probably have a thinking error. Exercise 6 on page 420 of Lam's Lectures on Modules and Rings says essentially: - -Let $R$ be a ring and $C$ a cyclic right $R$-module: $C=R/A$ with some right ideal $A$ in R. Let $(-)^{*}$ denote the functor $\operatorname{Hom}_R(-,R)$. - Then $C^{*} \cong \operatorname{ann}_l(A)$ as right $R$-modules. - -I would think that this is correct. -Specialising to a $A=J$ being the Jacobson radical this gives: -$C^{*} \cong \operatorname{ann}_l(J) \cong \operatorname{soc}(R)$ as the left annihilator of the Jacobson radical is the socle of the algebra; see Lemma 8.3 of Landrock's Finite Group Algebras and Their Modules. So Exercise 6 tells us that taking duals of simple modules gives again simple modules. In particular, the double dual of a simple module should again be the direct sum of simple modules?! -Now take $R$ to be a local non-selfinjective algebra and $C=R/J$ (which is the unique right simple $R$-module). Let $S$ be the simple left $R$-module. -Then we get $$C^{*} \cong \operatorname{soc}(R) \cong \bigoplus_{k=1}^{n}{S}$$ for some $n$. Then we get doing the same again, $C^{**} \cong {\oplus}_{k=1}^{m}{C}$ for some $m$. But taking for example the algebra $$R=K\langle x,y\rangle / (x^2,y^2,xy),$$ my computer tells me that $C^{**}$ is not semisimple and has an indecomposable direct summand of dimension 2. -I do not see the mistake at the moment, so maybe Lemma 8.3. in the book of Landrock might be wrong? - -REPLY [13 votes]: If $C$ is a right module then $C^*$ is a left module. -In Landrock, $\text{soc}(R)$ is the right socle. As he proves, it is a two-sided ideal, but it may not be semisimple as a left module (it is not necessarily equal to the left socle). Your example illustrates this: the right socle is spanned by $x$ and $yx$, but the left socle is spanned by $y$ and $yx$.<|endoftext|> -TITLE: Characterization of a sphere: every "sub-sphere" has two centers -QUESTION [15 upvotes]: Let me ask this question without too much formalization: -Suppose a smooth surface $M$ has the property that for all spheres $S(p,R)$ (i.e. the set of all points which lie a distance $R\geq 0$ from $p \in M$, with distance as measured inside the surface), there is always a different point $q(p,R) \in M$ and a distance $D(p,R)\geq 0$ so that $$S(p,R)=S(q(p,R),D(p,R))$$ -In words: both $p$ and $q(p,R) \neq p$ are a center of the sphere $S(p,R)$. -Q: Is $M$ (a part of) a sphere? More formally: is $M$ isometric to (a subset of) a sphere? -Remark: Spheres clearly possess the mentioned property, with every sub-sphere having one point $p$ ánd its antipodal $\pi(p)$ at its center. This property is maintained when we remove a finite number of pairs of antipodal points from such a sphere or when we take a few disconnected spheres. -This question has a counterpart on stackexchange. - -REPLY [15 votes]: Blaschke's conjecture might be relevant. In particular, if the surface is $S^2$ with a $C^3$ metric, then it follows from Blaschke's conjecture, proved by Green. This states that a metric on the sphere in which every point has a unique conjugate point must be the round metric on the 2-sphere. In your condition, one may observe that $q(p,R)$ does not depend on $R$, essentially by the Gauss lemma. The point is that the sphere of radius $r$ about $B(p,R)$ (the ball of radius $R$ about $p$) is $S(p,r+R)$. Hence $q(p,R)=q(p,r+R)=q(p)$. Then $q(p)$ is the unique point conjugate to $p$. -For the general case, it might follow by taking the completion of the metric, and showing that it must be a (topological) sphere.<|endoftext|> -TITLE: preliminary reading recommendation before embarking on Connes non commutative geometry book? -QUESTION [6 upvotes]: I want to try to understand non commutative geometry by reading Connes's book -..and I am discovering it is a hard book to read :-) as I miss a lot of background specially in operator algebra and homology theory ( my field is nonlinear PDE so I know a bit of functional analysis already- at least the one used in my field). -So my question: what reading could be recommended in order to prepare a non expert mathematician to read Connes's book? -For example, there are so many book on operator algebra or homology theory ...and my personal pick will be random, so I am seeking for expert recommendations: -Thanks in advance -JF - -REPLY [8 votes]: To understand everything in Connes' book you would need expertise in many different fields. My advice would be to browse it and see if anything attracts your interest. Then you can read up on the relevant background for that topic.<|endoftext|> -TITLE: Application of simple Lie algebras over finite fields -QUESTION [5 upvotes]: I am now interested in simple Lie algebras over finite fields. In Lie algebras over the complex numbers, there are several applications and some related topics. -Is there any potential application for simple Lie algebras over finite fields, or anything related? Perhaps, in coding theory or graph theory? - -REPLY [6 votes]: Fourier transform of invariant functions on finite reductive Lie algebras have been developed in here: http://www.springer.com/de/book/9783540240204 -This can be considered the Lie algebra analogue of character tables of finite groups -The MacWilliams identities in error correcting codes https://www.encyclopediaofmath.org/index.php/MacWilliams_identities -have a Fourier transform interpretation for example here: -https://www.jstor.org/stable/pdf/25098937.pdf -This can considered Fourier transform on abelian Lie algebras. -Finally Fourier transform on finite Lie algebras was used in the proof of a conjecture of Kac in the representation theory of quivers -https://arxiv.org/abs/1204.2375 - -REPLY [4 votes]: I can add one more example from geometry, topology and physics, where simple modular Lie algebras arise. In John Milnor's article On Fundamental Groups of Complete Affinely Flat Manifolds there was an open question for Lie algebras $L$, namely which Lie algebras admit a bijective $1$-cocycle in $Z^1(L,M)$ with $\dim(L)=\dim (M)$. For the adjoint module, this asks for a non-singular derivation. Jacobson proved that in characteristic zero such a Lie algebra must be nilpotent. -The question was formulated there also for Lie groups, namely which connected Lie groups admit a left-invariant affine structure. It can also be formulated in terms of Yang-Baxter groups, or left braces, see the article Counterexample to a conjecture about braces by D. Bachiller. -From the first Whitehead Lemma it follows that a simple Lie algebra of characteristic zero does not admit such a $1$-cocycle. In prime characteristic, this is no longer true. For modular simple Lie algebras there are such examples, see First Cohomology Groups for Classical Lie Algebras by Jens-Carsten Jantzen. - This has then interesting applications to $p$-groups, Yang-Baxter groups, left braces, pre-Lie algebras, post-Lie algebras, and other things. -Benkart, Kostrikin and Kuznezov classified Finite-Dimensional Modular Simple Lie algebras with a Nonsingular Derivation.<|endoftext|> -TITLE: Cotangent complex of perfect algebra over a perfect field -QUESTION [11 upvotes]: Let $A$ be a perfect $\kappa$-algebra over a perfect field $\kappa$ of positive characteristic $p$. Then the algebraic (= classical) cotangent complex $L_{A/\kappa}^{\operatorname{alg}}$ is known to vanish, due to the Frobenious automorphism having simultaneously to induce on the cotangent complex an automorphism and multiplication by $p$. -But we can also view $A$ and $\kappa$ as discrete $\mathbb E_\infty$-rings. The cotangent complex $L_{A/\kappa}$, which we obtain that way, is generally different from $L^{\operatorname{alg}}_{A/\kappa}$, since their homotopy groups give topological Andre-Quillen homology and (ordinary) Andre-Quillen homology respectively. -Q: Can we still say something about $L_{A/\kappa}$? -For instance: - -Does it perhaps vanish? -Are there at least any finiteness results (e.g. when $A$ is a field, is $\dim_A \pi_n L_{A/\kappa} < \infty$)? - -Perhaps a bit more broad afterquestion: what is in general the relationship between $L_{B/A}$ and $L^{\operatorname{alg}}_{B/A}$ for a discrete commutative $A$-algebra $B$? Other than that they coincide over the rationals and that $\pi_0$ of both is the module of Kähler differentials $\Omega_{\pi_0B/\pi_0A}$, of course. -Thanks in advance! - -REPLY [7 votes]: If $f: A \rightarrow B$ is a morphism of simplicial commutative rings (for example, a morphism of ordinary commutative rings), then the "topological" cotangent complex comes with additional structure: the action -of $B$ on $L_{B/A}$ can be promoted to a (left) action of a certain -associative ring spectrum $B^{+}$. There are canonical maps of ring spectra -$B \rightarrow B^{+} \leftarrow \mathbf{Z}$ which induce (via the multiplication on $B^{+}$) an equivalence between $B^{+}$ and the smash product of $B$ with $\mathbf{Z}$ (beware that $B^{+}$ is not commutative and the order of the multiplication matters). -There's also a canonical map of ring spectra $B^{+} \rightarrow B$, -and the "algebraic" cotangent complex can be recovered as the tensor product $B \otimes_{ B^{+} } L_{B/A}$. -You can use this description (and the fact that $B^{+}$ is not too different from $B$) to answer a lot of the sorts of questions that you're asking. For example, $L_{B/A}$ is zero, or almost perfect, or connected through some range, if and only if the algebraic cotangent complex $L^{\mathrm{alg}}_{B/A}$ has the same property.<|endoftext|> -TITLE: What is an "Instanton" in classical gauge theory? (to a mathematician) -QUESTION [15 upvotes]: There's already a question about the same topic but I think its aim is different. -Classical (non-quantum) gauge theory is a completely rigorous mathematical theory. It can be phrased in completely differential-geometric terms (where the main players are bundle with connections on a manifold). -I think I have a basic understanding of what gauge theory is about and what various words mean in this context (yang mils, potential, energy, etc...). However I have still not managed to figure out what "Instanton" means in this context. - -What is an Instanton? - -Is it something special to Yang-Mils theory? Is it something special to Quantum Gauge theory? Are there any mathematical interpretations/applications for Instantons? - -REPLY [6 votes]: Generally speaking, you could say they are a special type of solution to the field equations of gauge theories. More specifically, an instanton is a classical solution in a classical Euclidean field theory with finite non-zero action. -The name is due to the fact that they happen for an 'instant' (a point) of Euclidean time and so they are important in the path-integral formulation of a theory which uses Euclidean signature and as critical points. As other commentators have mentioned, in QFT we are generally talking about the Yang-Mills instanton (where a Yang-Mills theory is a QFT with a non-abelian gauge group). For a mathematical interpretation, the instanton solution of the Euclidean Yang-Mills equation leads an $SU(2)$ fibre bundle over $S^{4}$ but it can also be proved that any finite action solution of the Euclidean Yang-Mills equations leads to a fibre bundle over the four-sphere (see Uhlenbeck, 1979). -If we consider the best-known example, we take a pure Yang-Mills theory with symmetry group $SU(2)$ in $R^{4}$ with Euclidean signature. The equations of motion (ie. the Yang-Mills equations) are $D*F=0$ and $DF=0$. When we introduce the condition called the anti-self-duality equation, these equations reduce to ODEs for the gauge potential $A$. If we make an ansatz for the solution to the anti-self-duality equation which only differs from a pure gauge by a function of $r$ at infinite radius, we guarantee that our solutions have finite, non-zero action. Our ansatz for the gauge field is such that it becomes a pure gauge as $r$ tends to infinity and the associated field strength disappears, meaning that the action is finite. -If we choose the appropriate gauge transformation and use this to evaluate the field strength, we then obtain an equation whose full-solution is the 'instanton potential' which is regular on all of $R^{4}$. The action is equal to $-8 - \pi^{2}/g^{2}$, so it is obviously finite. See Gockeler and Schucker (1987) for more on the fibre bundle interpretation: this looks at other related fibre bundle structures such as the Dirac monopole. The fibre bundle structure of the Dirac monopole is extremely similar (basically the same) as the Yang-Mills instanton fibre bundle. (I only add this as seems like you were after a more interesting mathematical interpretation). -Following this logic we can attempt to construct a 'gravitational instanton' ie. we look for a metric with Euclidean signature described locally by an orthonormal frame and solve the Einstein field equations without matter. We end up with a solution such that $g$ and $f$ tend to $1$ as $r$ tends to infinity: this is similar to the Yang-Mills case where the Yang-Mills instanton potential becomes a pure gauge as $r$ tends to infinity. This is similar to the way that gravity ends up being 'analogous' to other gauge theories, rather than directly comparable, since gravity does not quantize well.<|endoftext|> -TITLE: The concept of convex foliation -QUESTION [6 upvotes]: A $n-1$ dimensional submanifold $N\subset \mathbb{R}^n$ is called a convex submanifold if for every $x\in N$ ,ther is a neighborhood $W$ of $x$ in $N$ such that $W$ entirly lies at one side of $T_x N$. A (local) diffeomorphism $\phi$ on $\mathbb{R}^n$ is called a convex diffeomorphism if $\phi$ and its inverse preserves the convexity of all codimension $1$ submanifolds. For example every affine linear isomorphism is a convex diffeomorphism but the diffeomorphism $\phi(x,y)=(x,y-x^2+x^3)$ is not a convex diffeomorphism because it maps the convex curve $y=x^2$ to non convex curve $y=x^3$. - -1.In the above definition, is the word "its inverse", redundant? - -Is there a well known description of the group of all smooth convex diffeomorphisms of $\mathbb{R}^n$? - -3.Does every manifold admit an atlas whose all transition maps are convex diffeomorphism? - -If the answer to the later question is affirmative, then we can define the concept of convexity for any codimension $1$ submanifold of an arbitrary manifold $M$. In particular we can speak of "convex foliation" of a manifold $M$, a codimension $1$ foliation of $M$ whose all leaves are convex submanifold. -In this case the next question would be the following: - -4.Is there a manifold which admit a codimension $1$ foliation but does not admit any convex foliation?In particular does $S^3$ admit a convex foliation? - -REPLY [7 votes]: If $n>1$, and a smooth diffeomorphism $f:U\to V$ (where $U$ and $V$ are, say, convex, open subsets of $\mathbb{R}^n$) carries convex sets to convex sets, then it is easy to show that it must carry each intersection $U\cap H$, where $H\subset\mathbb{R}^n$ is a hyperplane, to an intersection $V\cap H'$, where $H'\subset\mathbb{R}^n$ is another hyperplane. Hence $f$ must carry intersections of hyperplanes to intersections of hyperplanes, and hence line segments to line segements. In particular $f$ must be the restriction of a projective transformation of $\mathbb{RP}^n$ to $U\subset\mathbb{R^n}\subset\mathbb{RP}^n$. Conversely any such restriction carries convex sets to convex sets, as does its inverse. This answers Questions 1 and 2. -For Question 3, you are asking whether every smooth $n$-manifold $M$ admits a flat projective structure. The answer is no, for then the associated developing map on the simply connected cover would have a smooth immersion into $S^n$. However, already this is impossible for $n=4$ if one starts with a simply-connected compact $4$-manifold other than the $4$-sphere. -However, note that $n=2$ does work, since, by uniformization, every two-dimensional surface carries a metric of constant Gauss curvature, and the underlying projective structure associated to such a metric is projectively flat. -Finally, $S^3$, which does admit a codimension $1$ foliation and does admit a flat projective structure (as the double cover of $\mathbb{RP}^3$), does not admit a convex foliation. The reason is that the flat projective structure is unique (since $S^3$ is simply-connected) and at least one of the leaves of the foliation would have to be a compact torus because every codimension $1$ foliation of the $3$-sphere must have a Reeb component. However, it is not possible to immerse the standard torus in the standard $S^3$ smoothly as a convex surface (with respect to the projective structure): Endow $S^3$ with its standard 'round' metric of constant sectional curvature $+1$; its underlying projective structure is the flat one. For any torus $T^2\subset S^3$, the Gauss curvature of the induced metric must vanish somewhere (since the integral is zero), and, at such a point, the product of the principal curvatures must be $-1$, by the Gauss equation. In particular, the two principal curvatures are nonzero, with opposite signs, and the torus is not convex on a neighborhood of such a point. Thus, this gives an example of the kind asked for in Question 4.<|endoftext|> -TITLE: Underlying structure behind the infamous IMO 1988 Problem 6 -QUESTION [25 upvotes]: This is the infamous Problem 6 from the 1988 IMO which has recently been popularised by the YouTube channel Numberphile: - -Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^{2} + b^{2}$. Show that $\frac {a^{2} + b^{2}}{ab + 1}$ is the square of an integer. - -The elementary proof is well known and based on infinite descent using Vieta jumping. -What are non-elementary ways of solving it? Which mathematical structure is useful in creating the context for this problem? Is there an insight of the type where, once the problem is put into the right algebraic context, the solution is obvious? -What I have in mind is: notice, for example, how most of the steps in Euler's proof of the Fermat's prime sum of squares theorem become trivial once re-interpreted in the ring $\mathbb Z [i] $, the infinite descent itself being replaced by Euclidean division. Is something similar applicable here? - -REPLY [17 votes]: Indefinite binary quadratic forms, integer coefficients and discriminant not a square, possess an automorphism group; taking the Hessian matrix $H,$ an automorphism element is an integer matrix $P$ such that $P^T H P = H.$ The oriented part ($P$ has positive determinant) is infinite cyclic. -For the form $A x^2 + B xy + C y^2,$ with discriminant $\Delta= B^2 - 4 AC$ positive but not a square, Hessian -$$ -H = -\left( -\begin{array}{cc} -2A & B \\ -B & 2 C -\end{array} -\right), -$$ -all positive determinant $P$ are given by integer solutions to $\tau^2 - \Delta \sigma^2 = 4,$ with -$$ -P = -\left( -\begin{array}{cc} -\frac{\tau - B \sigma}{2} & - C \sigma \\ -A \sigma & \frac{\tau + B \sigma}{2} -\end{array} -\right) -$$ -For the case of $x^2 + Bxy+y^2$ with $B^2 > 4,$ the automorphisms take on the familiar Vieta appearance, and the negative determinant ones can be taken to be interchanging the variables. With a fixed target $T,$ any expression $x^2 + B xy + y^2 = T$ can be transported, by automorphisms, to a region satisfying desired inequalities. These desired solutions can be thought of as representative points in a group orbit. Siegel's description of counting solutions comes down to counting the representative solutions, that is the number of orbits, as the number of literal representations of a fixed target number is infinite. -Probably worth pointing out that finding a generator for the (oriented part) of the automorphism group requires solving $\tau^2 - \Delta \sigma^2 = 4,$ where $\Delta > 0$ is not a square. This would be a bit much. However, for the special case $x^2 + B xy + y^2,$ the discriminant is $\Delta = B^2 - 4,$ so that $B^2 - \Delta 1^2 = 4.$ There is also the single-variable "jumping" argument, done extremely well in Hurwitz 1907 -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -LEMMA 1 -Given integers $$ M \geq m > 0, $$ along with positive integers $x,y$ with -$$ x^2 - Mxy + y^2 = m. $$ -Then $m$ is a square. -PROOF. -First note that we cannot have integers $xy < 0$ with $ x^2 - Mxy + y^2 = m, $ since then - $ x^2 - Mxy + y^2 \geq 1 + M + 1 = M + 2 > m.$ If we have a solution with $x > 0$ and $xy \leq 0,$ it follows that $y=0.$ -This is the Vieta jumping part, with some extra care about inequalities. We begin with -$$ y > x $$ and -$$ y < Mx. $$ We have -$$ x^2 - Mxy + y^2 > 0, $$ -$$ x^2 > Mxy - y^2 = y(Mx - y) > x(Mx-y), $$ -$$ x > Mx - y > 0. $$ -That is, the jump -$$ (x,y) \mapsto (Mx - y,x) $$ -takes us from one ordered solution to another ordered solution while strictly decreasing $x+y.$ -Within a finite number of such jumps we violate the conditions we were preserving; we reach a solution $(x,y)$ with $y \geq Mx,$ that is -$x > 0$ but $Mx-y \leq 0.$ Since $(Mx - y,x) $ is another solution we know that $Mx-y = 0.$ Therefore $x^2 = m$ and $m$ is a square. -...... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -This one takes a little more work. -LEMMA 2 -Given integers $$ m > 0, \; \; M \geq m+3, $$ there are no integers $x,y$ with -$$ x^2 - Mxy + y^2 = -m. $$ -The contrapositive of lemma 2 is that when there are solutions, $M \leq m+2.$ The bound is sharp, achieved at $x=y=1 \; .$ As a side note, if $M \leq 2,$ then $x^2 - M xy + y^2$ is positive or positive semi-definite. So, the contrapositive gives the other example at the wikipedia article, that if $xy$ divides $x^2 + y^2 + 1$ and $x,y >0,$ then actually $x^2 + y^2 + 1 = 3xy.$ -Compare the contrapositive with Lemma 3 below... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -This one took a whole bunch more work. I also needed some help from Gerry Myerson. -LEMMA 3 -Given integers $$ m > 0, \; \; M > 0, $$ such that there are integers $x,y$ with -$$ x^2 - Mxy + y^2 = -Mm, $$ then -$$ M \leq (m+1)^2 + 1 \; . $$ -The bound is sharp, achieved with $x=1, \; y=m+1 \; .$<|endoftext|> -TITLE: For a computable binary tree, is having no computable branches the same as having no probabilistic algorithm for producing branches? -QUESTION [18 upvotes]: It is a classical result of computability theory that there is a -computable infinite binary tree $T\subset 2^{<\omega}$ with no -computable infinite branch. -One way to construct such a tree is to fix a pair $A$, $B$ of -computably inseparable c.e. sets, and to consider the tree of -attempts to find a separation of them. Thus, a finite binary -sequence of length $n$ is in the tree, if the set it describes -forms a separation of the numbers that have been enumerated into -$A$ and $B$ by stage $n$, that is, containing all such numbers from the stage $n$ approximation to $A$ and none from the stage $n$ approximation to $B$. This tree is computable, but any computable -branch would provide a computable separation, contrary to -hypothesis. -My question is whether a probabilistic Turing machine can have a -non-zero chance to find a branch through $T$. Let us consider a -probabilistic model of Turing machines where the Turing machine -program transitions are given by computable probabilities on the -outcome of each step, rather than single-value determistic -outcomes. -Question 1. If a computable infinite binary tree has a probabilistic -algorithm for producing a branch with non-zero probability, then -must it have a computable infinite branch? -In other words, if a computable tree has no computable branch, then -must it also admit no probabilistic algorithm to find a branch -with non-zero probability? -One can imagine following the greedy algorithm, say, and -deterministically following the most likely outcome at each step. -But it is easy to see that this won't always work, since we can -easily design a tree for which such a move leads to a dead part of -the tree on the first move. To turn the probabilistic algorithm -into an actual computable branch, we have to imagine that the -probabilistic algorithm will often be producing different branches. -One can also imagine trying to use the probabilistic algorithm by -running it far ahead until one sees that a certain accumulation of -probability, based on a comparison to the fixed probability of -success, means that certain choices are good choices. But I haven't yet been able to make this idea work. -The question above is asking whether the property of having no computable -branches is the same as having no probabilistic algorithm for -producing branches with non-zero probability. But a weaker question -would be: -Question 2. Is there a computable infinite binary tree with no -probabilistic algorithm for computing an infinite branch with non-zero -probability? -That is, can we strengthen the classical result from -no-computable-branches to produce a computable tree with no -probabilistic algorithm for producing branches? I expect that we -can. -These question arose from the discussion thread on a post of John -Baez's -concerning infinite chess. - -REPLY [16 votes]: For question 2, you're asking if there's a $\Pi^0_1$-class which is Medvedev-below no set of positive measure. -For an example, take the tree of completions of PA (or any other tree where every path has PA degree). No difference random computes a PA degree, so the unions of upper cones of paths contains no difference random, and thus is measure 0. So no positive measure set is Muchnik-above the paths through the tree, let alone Medvedev-above. -Edit: -One class of reals is Medvedev above a second class if there is a Turing functional that sends any element of the first class to an element of the second class. One class is Muchnik above a second class if every element of the first class computes an element of the second class; in other words, it's like Medvedev reducibility, but different reals can use different functionals. -Your notion of probabilistic machines can be instead thought of as a deterministic machine that consults a random real given as an oracle to determine which transition to follow. At each step, it consults the real in a way that's independent of the way it checked at previous steps. So a positive probability of getting a path turns into a positive set of real oracles which give a path, and the machine witnesses a Medvedev reduction from this set to your set of paths.<|endoftext|> -TITLE: Elliptic curves: for $P = aG$ for some $a$, what is $Q = a^{-1}G$? -QUESTION [5 upvotes]: Given an elliptic curve group with a generator $G$ where $G$ has a prime order, p. -Given a point $P=aG$ for some unknown $a$. Is it possible to efficiently calculate $Q=a^{-1}G$ without a discrete log operation? -With a discrete log, the problem is simple: first calculate $a$, then $a^{-1} = a^{p-1} $ mod $p$. -But I can't reduce a diffie-hellman problem to this to break it. Nor do I have the background to prove it directly (I have a background in NP-complete problems). -I see that the possibility of this operations would break a tiny subset of shared secrets but this should be negligible. So unless I'm wrong the existence of this algorithm isn't inconsistent with the original proof. - -REPLY [2 votes]: The name of the problem is `the Inverse Diffie-Hellman problem'. It is as hard as solving the computational Diffie-Hellman problem. A proof can be found in chapter 21, p.448-449 of Mathematics of Public Key Cryptography by Steven Galbraith (2012). -Source: this stackoverflow question.<|endoftext|> -TITLE: Mathematical objects whose name is a single letter -QUESTION [5 upvotes]: (Not research-level, but perhaps not easily answered elsewhere — you decide if MO can afford the innocent fun. If so, it should likely be “community-wiki” i.e. one object per answer.) - -I am seeking stories of mathematical objects that, in context, eat out namespace because their (most usual) name literally is a letter (e.g., in calculus, $e$). - -Per discussion in the comments, please rather exclude letters that are frozen out by being merely common notation ($e$ in group theory, $g$ in Riemannian geometry, whole alphabets in semisimple Lie theory), and not really the name of any single object. But include: how such (poor?) practice came about; what did or didn’t help reclaim letters (new names, new typography,...); or any good story. -Wikipedia’s disambiguation pages can suggest many ($c$, $e$, $i$, $j$, $k$, $o$, $q$, $t$, $F$, $G$, $J$, $K$, $L$, $O$, $P$, $W$, $Y$, $\mathcal O$, $\wp$, $\delta$, $\zeta$, $\eta$, $\vartheta$, $\varkappa$, $\lambda$, $\xi$, $\pi$, $\sigma$, $\tau$, $\chi$, $\mathrm B$, $\Gamma$, $\mathrm H$, $\Xi$, $\Omega$,...), but I am sure that is not all. - -REPLY [5 votes]: Imagine how different would be our mathematical discourse if students struggled -with $\alpha$-$\beta$ proofs rather than $\epsilon$-$\delta$ proofs! -From an earlier MO question: "Why do we use $\epsilon$ and $\delta$?," -an answer quoting Judith Grabiner: - -$\epsilon$ corresponds to the initial letter in the word "erreur" (or "error"), and Cauchy in fact used $\epsilon$ for "error" in some of his work on probability." - -As @RyanBudney summarizes, -"So it's $\epsilon$ for error in the answer, and presumably $\delta$ is in reference to difference in the input variables." -Or, as @SimonRose says, "Or possibly that $\delta$ is just the next letter over..." - -REPLY [2 votes]: Trivial example, but not to be omitted, the Greek and Latin numerals. E.g.:<|endoftext|> -TITLE: Group schemes over ring of Witt vectors and their representing algebras -QUESTION [6 upvotes]: Let $G$ be an affine groups scheme over $\mathbb Z$. As such it has an associated Hopf algebra, $A=\mathbb Z[G]$ such that $G(R)$ is naturally identified with the set $\hom_{Rng}(A,R)$ of ring homomorphisms, where the group operations (multiplication, inverse, unit) are given on this set from the co-operations of the algebra $A$. -Fix $m\in\mathbb{N}$ and $p$ a prime number, and let $W_m$ be the functor of $p$-typical Witt vectors of length $m+1$. The functor $W_m$ is represenatble as well, with representing algebra $\mathbb{Z}[x_0,\ldots,x_m]$. -I am interested in the structure of the group scheme $R\mapsto G(W_m(R))$. Greenberg's results imply that this functor is a an affine group scheme as well, and hence representable. Is there any known construction for the associated Hopf-algebra of $G\circ W_m$? - -Some obvious facts- -In the case where the prime $p$ is invertible in $R$, the Witt ring $W_m(R)$ is isomorphic to the product ring $\prod_{i=0}^m R$. Since the group scheme $R\mapsto G(\prod_{i=0}^{m} R)=\prod_{i=0}^{m}G( R)$ is represented by the Hopf algebra $A^{\otimes m+1}=\underbrace{A\otimes\cdots\otimes A}_{m+1\text{-fold}}$, I somehow expect there to exist a map between the representing algebra of $G\circ W_m$ and $A^{\otimes m+1}$, which becomes an isomorphism under localization by $p$. -On the other hand, in the complementary case, the group $G(W_m(R))$ is usually nothing like $G(R)\times G(R)$. For example, in the case $G= GL_n$, $R=\mathbb{F}_p$ and $m=2$, we have an exact sequence -$$1\to M_n(R)\to GL_n(W_2(R))\to GL_n(R)\to 1.$$ -In particular, $|GL_n(W_2(R))|\ne |GL_n(R)\times GL_n(R)|$, and one cannot expect any sort of bijection to exist between the two. - -I'm new to the subject, and my foundations on algebraic groups or algebraic geometry are not incredibly solid, so I apologize if anything I wrote above does not make complete sense, or is obviously false. -I would very much appreciate any clue or reference to the construction of the Hopf algebra of $G\circ W_m$, or any other interesting facts regarding the structure of $G\circ W_m$. -Thank you very much! -Shai - -REPLY [4 votes]: The Hopf algebra of $G\circ W_m$ does have an explicit construction. It's appeared in a few of my papers and zillions of Buium's papers, and no doubt many others which aren't coming to mind right now. Some people call $G\circ W_m$ the `order $m$ arithmetic jet space of $G$' and denote it $J^mG$. The reason is that the usual jet space is defined in the same way but where you replace $W_m$ by the functor $R\mapsto R[t]/(t^{m+1})$. But one key difference is that any explicit description of the function algebra of the arithmetic jet space has to be fundamentally nonlinear because the addition law on Witt vectors is nonlinear. So you won't be able to write it down by combining multilinear constructions in a nice way. -So let $A_m$ denote the function algebra $\mathscr{O}(J^mG)$. You want to describe $A_m$ explicitly. It can be done in a few different ways, which roughly correspond to how you think about the Witt vectors. This is similar to how you can construct tensor products in different ways. In fact this is not an accident, as both are left adoints---in other words, constructions of objects which are free in a certain sense. -Let me start with the most concrete first. So fix a presentation $\mathbb{Z}[x_i]/(f_j)$ of the function algebra $A=\mathscr{O}(G)$. - -There's the way with the usual Witt components/coordinates. $A_m$ is (canonically) $\mathbb{Z}[x_i^{[k]}]/I^{[m]}$, where the notation is the following: $x_i^{[k]}$ is an indeterminate, $k$ runs from $0$ to $m$, and $i$ runs over the indexing set of the original generators; $I^{[m]}$ is the ideal generated by all elements $f_j^{[k]}$, as $j$ runs over the indexing set for the relations (and $k$ from $0$ to $m$) and where $f_j^{[k]}$ comes in a certain purely syntactic way from the $k$-th Witt component of the expansion of $f_j$. It's easier (for me) to explain it with examples: if $f=x_1+x_2$, then $f^{[0]}$ will be $x_1^{[0]}+x_2^{[0]}$, and $f^{[1]}$ will be $x_1^{[1]}+x_2^{[1]}+\sum_{0 -TITLE: intersection cohomology and nearby cycles -QUESTION [7 upvotes]: This seems like a really basic question, but I somehow don't know and haven't been able to find the answer. -I suspect that (at least under suitable assumptions) there should be a relation between the following two constructions, but I'm looking for a precise theorem. -If I have a family $X$ over a disk, which I'm thinking of as a degeneration of smooth fibers $X_t$ to a singular fiber $X_0$, then there are two natural perverse sheaves I can cook up on $X_0$: one is the intersection cohomology of $X_0$, and the other is the nearby cycles of the intersection cohomology (which is constant) on $X-X_0$. What is the relation between these two? -To make things a little more precise: let $\eta \rightarrow S \leftarrow s$ be a henselian trait, $X \rightarrow S$ a map with $X_s$ and $X_{\eta}$ the special and generic fibers. I'm imagining that $X_{\eta} \rightarrow \eta$ is smooth, but perhaps this is unnecessary. How can I describe $IC(X_s, \mathbb{Q}_{\ell}))$? in terms of $\Psi(IC(X_{\eta}, \mathbb{Q}_{\ell}))$? (I expect an answer to involve the weight filtration and monodromy operator on nearby cycles, in addition to the "bare sheaf".) - -REPLY [7 votes]: Let me give a quick and lazy answer. Throughout all statements are up to shift in the derived category. -Let us assume that $X$ and $f : X \to \eta$ is smooth. Then the nearby cycles sheaf $\Psi_f \mathbb{Q}_X$ is generically a constant sheaf (generically the "nearby cycles" are points, in other words generically the vanishing cycles sheaf is trivial). Now Gabber's theorem identifies the weight filtration on $\Psi_f \mathbb{Q}_X$ with the monodromy filtration. It follows that $\Psi_f \mathbb{Q}_X$ has a filtration such that one subquotient of this (i.e. $W_0/W_{-1}$) is (geometrically) semi-simple and contains $IC(X_0)$ as a summand. Thus, in a certain sense the intersection cohomology is "right in the middle" of the nearby cycles. -This has various concrete consequences. For example, one has a spectral sequence with $E_1$ page the successive subquotients of the weight filtration, which converges to the cohomology of the nearby cycles = cohomology of a smooth fibre. In general it is tricky to work out what the other subquotients will be (metaphorically this is akin to calculating which summands occur in the Decomposition Theorem). However in certain cases (e.g. semi-stable reduction) one can be precise. For more on these topics look at the Rapoport-Zink and Clemens-Schmid spectral sequences.<|endoftext|> -TITLE: supersymmetry and the de Rham complex -QUESTION [11 upvotes]: In Alvarez-Gaume's paper "Supersymmetry and the index theorem" there is -given a certain supersymmetric Lagrangian whose quantization, apparently, leads to the de Rham Laplacian on the exterior algebra of a manifold. For what it's worth, the Lagrangian in question is -$$L = \frac{1}{2} g_{ij}(\phi) \dot{\phi}_i \dot{\phi}_j + -\frac{\sqrt{-1}}{2} g_{ij}(\phi)\overline{\psi}^i \gamma^0 \frac{D}{dt} \psi^j + \frac{1}{12} R_{ijkl} \overline{\psi}^i \psi^j \overline{\psi}^k \psi^l$$ -where $\phi$ is a map from $\mathbb{R}$ into a target manifold $M$, $g_{ij}$ is a metric on $M$, and $\psi^i$ is a spinor field on $M$ (I guess). -The same Lagrangian appears in several places, e.g. in Witten's paper -on "Supersymmetry and Morse theory" (although seemingly with $1/12$ replaced by $1/8$). Witten says ``How canonical quantization of [this Lagrangian] leads to the exterior algebra was discussed in [21].'' -But I cannot find anything of the sort in [21]="Dynamical breaking of supersymmetry", nor in any other source that I know about. -Can anyone give an explanation, or a reference, for $L$ above is related to the de Rham Laplacian? I am happy with a physics level of rigor but I am hoping to see some details spelled out. -Many thanks for any help. - -REPLY [10 votes]: I think there is a typo in the references to "Supersymmetry and Morse theory", [21] should be replaced by [22]="Constraints on supersymmetry breaking". The quantization of non-linear sigma models and its relation with the de Rham complex is discussed in Section 10 of this paper. -In addition to the big "Mirror symmetry" book already mentionned in the comments, a useful reference is in the book "Quantum Fields and Strings: A Course for Mathematicians" (http://bookstore.ams.org/qft-1-2-s/ ), Part 1, Supersolutions, Chapter 3 (p261-276), where one can find a detailed derivation of the supersymmetry of the Lagrangian, and Chapter 4 (p279-289), where a more conceptual point of view on this Lagrangian is given via some superspace formalism (in particular, in superspace formalism, the expression of the Lagrangian looks as simple as a single quadratic term, and one can recovers all the complicated terms such as the Riemann curvature-4-fermions terms by a simple systematic expansion in components). -Here is a brief sketch of the relation with the de Rham complex. Fields of the theory are $\phi^i$, $\psi^i$, $\overline{\psi}^i$. Upon canonical quantization, the Hilbert space of the quantum theory is the $L^2$-space of differential forms on $M$, quantum supercharges are $d$ and $d^{*}$ operators and so the quantum Hamiltonian is the Laplacian $\Delta=\{d,d^{*}\})$. -The relation between the explicit form of the Lagrangian and the Laplacian=quantum Hamiltonian is quite subtle because of the general phenomenon of ordering ambiguities in quantization. It is a good exercise to write explicitely the classical Hamiltonian from $L$ but it will not tell you immediately what the quantum Hamiltonian is the Laplacian because there are ordering ambiguities. The point of the supersymmetric story is that in this case the Hamiltonian is constructed from the simplest objects which are the supercharges, and that there is no ordering ambiguity to construct the quantum supercharges.<|endoftext|> -TITLE: Can any sequence of consecutive integers be realized as winding numbers? -QUESTION [8 upvotes]: For a closed plane curve $C$, define its sequence of winding numbers to -be the sorted list of the winding numbers of each of the distinct regions -of the plane demarcated by $C$. -For example, this curve (if I've calculated correctly) has sequence -$001111223 = 0^2 1^4 2^2 3$. - -          - - -A winding number sequence must include $0$ for the unbounded region of the plane. -I am wondering if there are any other restrictions: - -Q. Can any winding sequence of consecutive integers - that includes $0$ be realized by some curve $C$? - -REPLY [21 votes]: Isn't this easy by induction? Delete one of the largest numbers, say $m$, from your sequence, realize the remaining numbers, then in the realization pick any region with winding number $m-1$, and make an extra loop somewhere close to its boundary.<|endoftext|> -TITLE: Infinitely many solutions to $a^4+b^4+c^4=18$ over $\mathbb{Z}[i]$ -QUESTION [9 upvotes]: We got infinitely many solutions to $a^4+b^4+c^4=18$ over -$\mathbb{Z}[i],i^2=-1$. Probably we can get infinitely many solutions -to $a^5+b^5+c^5=N$ over $\mathbb{Z}[\alpha]$ for algebraic $\alpha$. - -Are these results known and/or trivial? - -There is some chance an improvement of the method to solve similar -equations over the integers. - -REPLY [2 votes]: Let $K$ be a number field. Then the Bombieri–Lang–Vojta–... conjecture asserts that for $d \geq 5$, all but finitely many of the $K$-points of the smooth projective surface of general type -$$X = V(x_0^d+x_1^d+x_2^d-Nx_3^d) \subseteq \mathbb P^3$$ -are contained in some strict subvariety $Z \subsetneq X$. A stronger version asserts that we can choose $Z$ independent of $K$, and that we may take it to be the union of all rational and elliptic curves in $X$. This is a good thing to try. -Over $\mathbb C$ and if $d$ is odd, we have a map -\begin{align*} -\mathbb P^1 &\to X\\ -[x:y] &\mapsto [x:-x:y\sqrt[d]{N}:y]. -\end{align*} -There are obvious modifications one can make for $d$ even. This exhibits a specific $\mathbb P^1$ on $X$, which is actually defined over $\mathbb Q[\sqrt[d]{N}]$. -This already gives infinitely many $\mathbb Q[\sqrt[d]{N}]$-points on $X$. Setting $y = 1$ and varying $x$ over $\mathbb Z[\sqrt[d]{N}]$, we actually get infinitely many integral points. Explicitly, I have exhibited for you the points -$$(a,b,c) = (x,-x,\sqrt[d]{N}),$$ -where $x \in \mathbb Z[\sqrt[d]{N}]$ is arbitrary. -One could probably have found these without referring to fancy conjectures. On the other hand, the conjecture can serve as a guide: conjecturally, this type of construction is the only way in which we can produce infinitely many rational points. In general, however, trying to produce maps $\mathbb P^1 \to X$ is probably not your winning strategy for producing rational or integral points...<|endoftext|> -TITLE: Ext in symmetric algebras and group algebras -QUESTION [6 upvotes]: Let $A$ be a selfinjective algebra and for an indecomposable module $M$ define $\psi_M:= \inf \{ i \geq 1 | Ext_A^i(M,M) \neq 0 \}$. -Questions: - -In case $A$ is symmetric, do we have $\psi_M \leq max \{ \psi_S | S $ is simple $\}$ for each indecomposable non-projective module $M$? -This should be true in case $A$ is representation-finite. -In case $A=kG$ is a group algebra over a field of characteristic $p$. Do we have even $\psi_M \leq \psi_K$ when $K$ is the trivial module and each indecomposable non-projective $M$ ? I can prove this for $p$-groups and in case $p$ does not divide the dimension of $M$. - -REPLY [5 votes]: I think this example answers both questions. -Let $k$ have characteristic $3$, and let $G=C_3\times S_3$. -Then $kG$ has two simple modules, both one-dimensional, and for each simple module $S$, $\text{Ext}^1(S,S)$ is one-dimensional. -But if $M=kC_3$, with $S_3$ acting trivially, then $\text{Ext}^i(M,M)=0$ for $i=1,2$.<|endoftext|> -TITLE: Existence of double eigenvalue -QUESTION [15 upvotes]: Let $A$ and $B$ be complex $4\times 4$ matrices. Assume both are Hermitian, and that they are linearly independent. -Must there exist a nonzero real linear combination $aA + bB$ which has a repeated eigenvalue? - -REPLY [23 votes]: The answer is 'no'. The generic pair $A$ and $B$ of $4$-by-$4$ Hermitian symmetric matrices will not have any nonzero real linear combination that has a double eigenvalue. -For a specific example, take -$$ -A = \begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&-2&0\\0&0&0&2\end{pmatrix} -\quad \text{and}\quad -B = \begin{pmatrix}0&i&0&0\\-i&0&0&0\\0&0&0&2i\\0&0&-2i&0\end{pmatrix}. -$$ -Then -$$ -\det(aA+bB - tI_4) = (t^2-a^2-b^2)(t^2-4a^2-4b^2), -$$ -and the roots of this polynomial in $t$ are distinct unless $a=b=0$. (Recall that we are assuming that $a$ and $b$ are real, which, of course, implies that $t$ is real.) -Added Remark: To see the claim that this property holds for a generic linearly independent pair of Hermitian symmetric $4$-by-$4$ matrices $A$ and $B$, it is only necessary to observe the following: The question is whether, for a generic such pair $A$ and $B$ in the $16$-dimensional real vector space $\mathcal{H}_4$ consisting of $4$-by-$4$ Hermitian symmetric matrices, the (real) span of $A$, $B$, and $I_4$ contains a nonzero element of rank at most $2$. Now, it is not difficult to show that the cone $C_2$ of elements in $\mathcal{H}_4$ that have rank at most $2$ is a closed algebraic cone of dimension $12$ (one that is singular along the $7$-dimensional locus of elements of rank at most $1$). Hence the generic $3$-dimensional subspace of $\mathcal{H}_4$ will meet this cone only at the zero matrix. It is also an open (though not dense) condition on a $3$-dimensional subspace that it contain a positive definite element. Since the standard $\mathrm{GL}(4,\mathbb{C})$-action on $\mathcal{H}_4$ acts transitively on the space of positive definite elements, it follows that the generic pair $A$, $B$, together with $I_4$ will span a $3$-plane that meets $C_2$ only at the origin, which is what was to be shown.<|endoftext|> -TITLE: Lalouvère's activities as censor -QUESTION [12 upvotes]: Fermat had a friend at Toulouse named Lalouvère. Lalouvère was censor, jesuit, and mathematician (in alphabetical order). -Antonella Romano writes on page 512 of her book La Contre-Réforme Mathématique. Constitution et diffusion d'une culture mathématique jésuite à la Renaissance (1540-1640): - -Un premier constat regarde les jésuites confrontés à la censure : sur tous les cas du Fondo Gesuitico qui concernent la France, deux professeurs seulement appartiennent à la liste établie dans le cadre de cet ouvrage, B. Labarthe et V. Léotaud. Si d'autres mathématiciens y apparaissent, c'est tout aussi exceptionnellement, et au titre de censeur, comme Antoine Lalouvère 114. - -Footnote 114 there reads: - -C'est lui qui porte un jugement négatif sur l'ouvrage de B. Labarthe. - -On on page 515 Romano writes: - -Antoine Lalouvère, le mathématicien toulousain qui s'est illustré dans le débat sur la cycloïde128, émet un jugement négatif sur le Hermetis Mathematici praeludium de B. Labarthe. - -Thus, Lalouvère in his capacity as a censor sank at least one mathematical book, namely that by his fellow jesuit Labarthe. -Questions 1. Why did Lalouvère report negatively on Labarthe's book? -Question 2. Were there other books reviewed negatively by him? -Question 3. If anybody is familiar with Labarthe's work on Aristotelianism I would appreciate a summary. -It would be of particular interest to determine whether or not the grounds for Lalouvère's negative report on Labarthe's book had to do with possible use of indivisibles or the infinitely small. See a related 2018 publication in Foundations of Science. - -REPLY [4 votes]: [Community Wiki to follow up on Matt's comment] - -Screenshot from De Backer's bibliography of Jesuit authors.(There is also an entry on Lalouvère.) Item B is a public lecture given by Labarthe when he was professor of philosophy at Clermont-Ferrand university, on a "more clear exposition" of Aristotelian philosophy. I have not been able to locate this document. (The bibliography entry says it should be in the library at Clermont, but their catalogue does not list it.)<|endoftext|> -TITLE: Discrete central subgroup of a connected Lie group is finitely generated -QUESTION [6 upvotes]: Every discrete central subgroup of a connected Lie group is finitely generated. - -This result was alluded to without comment in a book I was reading (Lie Group Actions in Complex Analysis by D. Akhiezer Proposition on page 38). -Assuming it was a trivial result, I posted to math.stackexchange where YCor was kind enough to give the not so trivial reference: Corollary 8.A.23 here. -That Corollary uses a handful of results + definitions that are not so familiar to me and would take up some time to understand. -So before I begin going through the reference, I wanted to see if someone knew of a more concise/elementary proof of this result. -Note: I am aware that the case where the center of the Lie Group has finitely many components is exercise level. -Thanks. - -REPLY [4 votes]: An old result of Iwasawa is that in any connected Lie group $G$, every compact subgroup is contained in a maximal compact subgroup, and all maximal compact subgroups are conjugate. -Let $G$ be a connected Lie group and let $Z$ be a discrete central subgroup. Then $Z$ has an infinite torsion quotient $Z'=Z/B$ (lemma below). Write $H=G/B$. Then $Z'$ is a discrete central subgroup of $G$ and is infinite torsion. Let $K$ be a maximal compact subgroup of $H$. Every finite $F$ subgroup of $Z'$ is contained in a maximal compact subgroup $K'_F$ of $H$; since $F$ is central and $K$ is conjugate to $K'_F$, we deduce that $F\subset K$. Hence $Z'\subset K$, a contradiction. -This would be fine with a proof of Iwasawa's result; the only problem is that I don't know if it relies on the result of finite generation of discrete central subgroups! - -Lemma: if $Z$ is an infinitely generated abelian group, then $Z$ has an infinite torsion quotient. -Proof: let $B$ be a maximal free subgroup. Then $Z/B$ is torsion. If $B$ is finitely generated then $Z/B$ is infinite. Otherwise, choose an infinite subset $(e_n)_{n\ge 1}$ in $B$ and replace $e_n$ by $ne_n$ for all $n$, to obtain a smaller free subgroup $B'$, with $B/B'$ infinite torsion; then $Z/B'$ is infinite torsion. - -Actually, instead of all the power of Iwasawa's result, it's enough to prove the result using the weaker result: ($*$) for any connected Lie group $G$ -and any increasing sequence $(K_n)$ of compact subgroups of $G$, we have $\overline{\bigcup K_n}$ compact. -Possibly this can done by hand (say, without using such things as Levi factors, just more basic Lie theory), I'll think twice. - -Edit: here's a proof of the OP's question relying on little (only on the semisimple case, where the center is discrete and finitely generated): -Let $G$ be a counterexample of minimal dimension. By the previous lemma, we can suppose (up to mod out by a discrete normal subgroup) that $G$ has an infinite, discrete torsion central subgroup $Z$. Taking the semisimple case for granted, $G$ is not semisimple. Let $V$ be the closure of the last nontrivial term of the derived series of its radical. Let $p$ be the projection $G\to G/V$. Since $\dim(G/V)<\dim(G)$, we have $\overline{p(Z)}$ compact. Hence its unit component has finite index, and hence some finite index subgroup of $Z$ is contained in the inverse image $H$ in $G$ of $\overline{p(Z)}^\circ$. Hence $H$ is a counterexample; by minimality, we deduce $\dim(H)=\dim(G)$ and hence $H=G$. That is, $\overline{VZ}$ is dense. We have $[V,Z]\subset V\cap Z$, which is a torsion discrete subgroup of the connected abelian Lie group $V$, and hence is finite. Hence $[G,G]$ is contained in $[V,Z]$, which is finite; by connectedness of $G$, we deduce that $G$ is abelian, and in turn this implies $Z$ finitely generated, and a contradiction. - -As regards the semisimple case, if by contradiction $G$ is semisimple and $Z$ is an infinitely generated central subgroup, then the quotient $H$ of $G$ by its center is semisimple and not compactly presented. Then one way to get a contradiction is to use that $G$ is quasi-isometric to its symmetric space, which is non-positively curved and hence large-scale simply connected, and for $G$ this means compactly presented, a contradiction. Of course this latter proof relies on some Riemannian and metric material.<|endoftext|> -TITLE: Vanishing of the cotangent complex for non-perfect algebras -QUESTION [5 upvotes]: Let $k$ be a perfect field of characteristic $p>0$. If $R$ is a perfect $k$-algebra, then the cotangent complex $L_{R/k}$ vanishes (the Frobenius is zero and induces an isomorphism on $L_{R/k}$ since $R$ is perfect). Is the converse true? Namely, if $L_{R/k}$ vanishes, then is $R$ perfect over $k$? I suspect not, but I can't find a counterexample. What happens if $k$ is a field of characteristic $0$? - -REPLY [3 votes]: see Bhargav Bhatt's example (suggested by Gabber) of an imperfect ring with trivial cotangent complex -http://www-personal.umich.edu/~bhattb/papers.html<|endoftext|> -TITLE: Is the order on repeated exponentiation the Dyck order? -QUESTION [21 upvotes]: The Catalan numbers $C_n$ count both - -the Dyck paths of length $2n$, and -the ways to associate $n$ repeated applications of a binary operation. - -We call the latter magma expressions; we will explain below. -Dyck paths, and their lattice structure -A Dyck path of length $2n$ is a sequence of $n$ up-and-right strokes and $n$ down-and-right strokes, all having equal length, such that the sequence begins and ends on the same horizontal line and never passes below it. A picture of the five length-6 Dyck paths is shown here: -A: B: C: D: E: - /\ - / \ /\/\ /\ /\ - / \ / \ / \/\ /\/ \ /\/\/\ - -There is an order relation on the set of length-$2n$ Dyck paths: $P\leq Q$ if $P$ fits completely under $Q$; I'll call it the height order, though in the title of the post, I called it "Dyck order". I've been told it should be called the Stanley lattice order. For $n=3$ it gives the following lattice: - A - | - B - / \ - C D - \ / - E - -For any $n$, one obtains a poset structure on the set of length-$2n$ Dyck paths using height order, and in fact this poset is always a Heyting algebra (it represents the subobject classifier for the topos of presheaves on the twisted arrow category of $\mathbb{N}$, the free monoid on one generator; see this mathoverflow question). -Magma expressions and the "exponential evaluation order" -A set with a binary operation, say •, is called a magma. By a magma expression of length $n$, we mean a way to associate $n$ repeated applications of the operation. Here are the five magma expressions of length 3: -A: B: C: D: E: - a•(b•(c•d)) a•((b•c)•d) (a•b)•(c•d) (a•(b•c))•d ((a•b)•c)•d - -It is well-known that the set of length-$n$ magma expressions has the same cardinality as the set of length-$2n$ Dyck paths: they are representations of the $n$th Catalan number. -An ordered magma is a magma whose underlying set is equipped with a partial order, and whose operation preserves the order in both variables. Given an ordered magma $(A,$•$,\leq)$, and magma expressions $E(a_1,\ldots,a_n)$ and $F(a_1,\ldots,a_n)$, write $E\leq F$ if the inequality holds for every choice of $a_1,\ldots,a_n\in A$. Call this the evaluation order. -Let $P=\mathbb{N}_{\geq 2}$ be the set of natural numbers with cardinality at least 2, the logarithmically positive natural numbers. Equipped with the operation given by exponentiation, $c$•$d\:=c^d$, we obtain an ordered magma, using the usual $\leq$-order. Indeed, if $2\leq a\leq b$ and $2\leq c\leq d$ then $a^c\leq b^d$. -Question: Is the exponential evaluation order on length-$n$ expressions in the ordered magma $(P,$^$,\leq)$ isomorphic to the height order on length-$2n$ Dyck paths? -I know of no a priori reason to think the answer to the above question should be affirmative. A categorical approach might be to think of the elements of $P$ as sets with two special elements, and use them to define injective functions between Hom-sets, e.g. a map -$$\mathsf{Hom}(c,\mathsf{Hom}(b,a))\to\mathsf{Hom}(\mathsf{Hom}(c,b),a).$$ -However, while I can define the above map, I'm not sure how to generalize it. And the converse, that being comparable in the exponential evaluation order means that one can define a single injective map between hom-sets, is not obvious to me at all. -However, despite the fact that I don't know where to look for a proof, I do have evidence to present in favor of an affirmative answer to the above question. -Evidence that the orders agree -It is easy to check that for $n=3$, these two orders do agree: - a^(b^(c^d)) A := A(a,b,c,d) - | | - a^((b^c)^d) B - / \ / \ - (a^b)^(c^d) (a^(b^c))^d C D - \ / \ / - ((a^b)^c)^d E - -This can be seen by taking logs of each expression. (To see that C and D are incomparable: use a=b=c=2 and d=large to obtain C>D; and use a=b=d=2 and c=large to obtain D>C.) Thus the evaluation order on length-3 expressions in $(P,$^$,\leq)$ agrees with the height order on length $6$ Dyck paths. -(Note that the answer to the question would be negative if we were to use $\mathbb{N}$ or $\mathbb{N}_{\geq 1}$ rather than $P=\mathbb{N}_{\geq2}$ as in the stated question. Indeed, with $a=c=d=2$ and $b=1$, we would have $A(a,b,c,d)=2\leq 16=E(a,b,c,d)$.) -It is even easier to see that the orders agree in the case of $n=0,1$, each of which has only one element, and the case of $n=2$, where the order $(a^b)^c\leq a^{(b^c)}$ not-too-surprisingly matches that of length-4 Dyck paths: - /\ - /\/\ ≤ / \ - -Indeed, the order-isomorphism for $n=2$ is not too surprising because there are only two possible partial orders on a set with two elements. However, according to the OEIS, there are 1338193159771 different partial orders on a set with $C_4=14$ elements. So it would certainly be surprising if the evaluation order for length-4 expressions in $(P,$^$,\leq)$ were to match the height order for length-8 Dyck paths. But after some tedious calculations, I have convinced myself that these two orders in fact do agree for $n=4$! Of course, this could just be a coincidence, but it is certainly a striking one. -Thoughts? - -REPLY [3 votes]: (This is what I have written just before my wife killed the internet connection 12 hours ago before she went to bed. I only show that $D\leq E \Rightarrow A\leq B$ where $D$ and $E$ are Dyck paths and $A$ and $B$ the corresponding binary trees. I didn't look at Timothy's answer yet, but I am guessing it's the same.) -Indeed, the bijection between (ordered, full) binary trees (with leaves labelled $a,b,c,\dots$ from left to right) and Dyck paths (traversing the binary tree starting at the root, first traversing the right subtree, and writing an up step for a right branch and a down step for a left branch) induces an order preserving map between the Stanley lattice and the exponential evaluation order. -A path $D$ is covered by a path $E$ in the Stanley lattice, if and only if a peak in $D$ is converted to a valley in $E$, all other steps remaining the same. -In terms of binary trees, a peak in the Dyck path corresponds to a pair of siblings where the right sibling $x$ does not have further children and there is a left branch somewhere after $x$, in the order the tree is traversed. -To see what the covering relation in the Stanley lattice corresponds to, we first do an easy special case: -Suppose that, in the binary tree $B$ corresponding to the Dyck path $E$, the parent $y$ of $x$ is a right child. -Let $L_1$ be the subtree rooted at the sibling of $x$, and let $L_2$ be the subtree rooted at the sibling of $y$. The magma expression corresponding to the subtree rooted at the parent of $y$ is $L_2 (L_1 x)$. -Then the binary tree $A$ corresponding to $D$ is obtained from $B$ by replacing the subtree rooted at the parent of $y$ with the binary tree corresponding to the magma expression $(L_2 L_1) x$, which is smaller than $L_2 (L_1 x)$. -The general case is only superficially more complicated: -Suppose that, in the binary tree $B$ corresponding to the Dyck path $E$, there is a (maximal) path of $k$ left branches from a node $y$ to the parent of $x$, with (right) siblings having subtrees $D_1,D_2,\dots,D_k$. Let $L_1$ be the subtree rooted at the (left) sibling of $x$ and $L_2$ be the subtree rooted at the (left) sibling of $y$. The magma expression corresponding to the subtree rooted at the parent of $y$ is -$$L_2(\cdots((L_1 x)R_1)\cdots R_k).$$ -Then the binary tree $A$ corresponding to $D$ is obtained from $B$ by replacing the subtree rooted at the parent of $y$ with the binary tree corresponding to the magma expression -$$(L_2 L_1)(x (R_1(\cdots R_k))).$$ -Setting $R=R_1\cdots R_k$, it remains to check that $L_2^{(L_1^{xR})} \geq (L_2^{L_1})^{(x^R)}$.<|endoftext|> -TITLE: Defining $SU(n)$ in HoTT -QUESTION [34 upvotes]: From a recent answer by Mike Shulman, I read: - -"HoTT is (among other things) a foundational theory, on roughly the same ontological level as ZFC, whose basic objects can be regarded as $\infty$-groupoids" - -Now, $\infty$-groupoids are the same thing as spaces, and I have a couple of spaces which I like. -Here's one which I like a lot: the Lie group $SU(n)$. - -Question: How do I say "let $X:=SU(n)$" in HoTT? - Here, $n$ can be either a variable, or a definite number (say $3$). - -Disclaimer: I (unfortunately) don't know much about HoTT. So please be gentle. - -REPLY [8 votes]: As time passes I've become increasingly unhappy with my original answer, so I want to give a new one. -One of the most interesting things about HoTT is that it lets you ask questions of a logical flavor about homotopy theory. For example, I was on Robert Rose's defense committee where the thesis problem was suppose you look at a version of HoTT where you don't assume you have the natural numbers, but you do have the circle, what can you still say? Can you still show that you have infinite sets? In what sense of the word infinite? How much number theory do you still get? These are interesting questions about the logic of homotopy theory that you can't even ask in the usual space-based foundations (classically how are you going to even define the circle without having access to infinite sets?). -So what I'd say now is that the question of whether (and how) you can define BSU(n) in book HoTT is an extremely interesting "logical" question about homotopy theory. As Mike explains you can certainly do this if you give yourself a stronger type theory than book HoTT, so if your goal is to just give yourself these tools you can do so, but it's still interesting to ask whether you can do this with the weaker book HoTT. -In particular, as you well know, classically there's a homotopy coherent action of SO(n) on n-loops via the geometric action on $E_n$-algebras. But this definition is highly geometric, which makes it difficult in practice to understand what this action actually looks like. For example, you might want to understand the action of $SO(4)$ on 4-loops in an $\infty$-groupoid well enough to be able to write down the generator of $\pi_7(S^4)$ algebraically. Furthermore, you might want to know whether you can keep doing this in general. The question of whether you can define the group $SO(n)$ in book HoTT and define its action on $\Omega^n S^n$ in book HoTT gives you a logical framework for asking this question! So the fact that you can't easily define the group $SO(n)$ in book HoTT is a feature, not a bug, because it lets you precisely phrase this very interesting mathematical question. (Of course, since it's an interesting difficult question, simply stating the question doesn't get you very far towards solving it.)<|endoftext|> -TITLE: Lefschetz fixed-point theorem for the Frobenius map -QUESTION [7 upvotes]: Where can one find a proof of Lefschetz fixed-point theorem for the Frobenius map on elliptic curves over algebraic closures of $F_{p}$ ? - -This could immediately follow if their coholomogies (for the sheaf of regular functions) were Weil cohomologies. But the proof of this is also hard to find. -Yet, there are references to this fact in connection with the use of Picard-Fuchs equation and counting rational points on such curves. - -REPLY [4 votes]: I am confused by the reference to sheaves of regular functions, and suspect the OP is confused. If $E$ is an elliptic curve over $\mathbb{F}_p$, then the cohomology groups of the sheaf of regular functions on $E$ is an $\mathbb{F}_p$ vector space, and Weil cohomology theories are normally required to take values in a characteristic $0$ vector space. If you take a cohomology theory with values in a vector space of characteristic $p$, then you can only count fixed points modulo $p$. It is true that, for $X$ a projective variety over $\mathbb{F}_p$, we have -$$\# X(\mathbb{F}_p) \equiv \sum_j (-1)^j\ \mathrm{Tr}(\mathrm{Frob}^{\ast} : H^j(X, \mathcal{O}) \to H^j(X, \mathcal{O})) \bmod p.$$ -A reference for this is W. Fulton, A fixed point formula for varieties over finite fields. But I suspect that isn't what the OP wants. -If you want to count points, not just count modulo $p$, then you need some definition of $H^j$ which takes values in modules over a characteristic zero ring. For an elliptic curve, you can use the Tate module: $T_{\ell}(E) := \lim_{\infty \leftarrow n} E[\ell^n]$ where $\ell$ is a prime not equal to $p$ and the maps in the inverse limit are multiplication by $\ell$. -You can make an ad hoc definition $H^1(E) = T_{\ell}(E)$ and $H^2(E) = \bigwedge^2 H^1(E)$. With this definition, as anon says, Lefschetz trace goes back to the 1930's. (See Silverman III.7.1 for $T_{\ell}(E) \cong \mathbb{Z}_{\ell}^2$, so $\bigwedge^2 T_{\ell}(E) \cong \mathbb{Z}_{\ell}$. Silverman III.8.3 implies that $\mathrm{Frob}$ acts on $\bigwedge^2 T_{\ell}(E)$ by $p$. The computations in Silverman V.1 and V.2 make it clear that $\# E(\mathbb{F}_p) = 1-\mathrm{Tr}\left( \mathrm{Frob}^{\ast} T_{\ell}(E) \to T_{\ell}(E)\right) + p$.) -If you want to show this matches some other definition of $H^j$, you have to say what definition you are using.<|endoftext|> -TITLE: Can a knot be cables of two different knots? -QUESTION [5 upvotes]: I wonder if there is an example of a knot $K$ in the 3-sphere which can be realized as cables of two distinct (up to isotopy) knots $K_1 \neq K_2$. -It is known that if a knot $K$ is the $(p,q)$-cable of another knot $K'$, then there is a unique annulus in its exterior $X_K$ with slope $pq$ on $\partial X_K$. This is the only restriction on possible cabling configurations of $K$ that I am aware of. Can we somehow argue that $K'$ is uniquely determined by $K$? - -REPLY [8 votes]: Yes, $K'$ is uniquely determined by $K$ and so, no, a knot cannot be a cable in two different ways. This follows from the Gordon-Luecke theorem and from a result of Feustel and Whitten. See Lemma 5 in this paper of Malyutin.<|endoftext|> -TITLE: Bilinear product of two summable families -QUESTION [11 upvotes]: Consider the following statement, which I suspect is false as written: - -Let $E,F,G$ be (Hausdorff) topological vector spaces (over $\mathbb{R}$), let $\varphi\colon E\times F\to G$ be continuous and bilinear, and let $(x_i)_{i\in I}$ and $(y_j)_{j\in J}$ be summable families in $E$ and $F$ respectively with $\sum_{i\in I} x_i = x$ and $\sum_{j\in J} y_j = y$. Then $(\varphi(x_i,y_j))_{(i,j)\in I\times J}$ is summable with $\sum_{(i,j)\in I\times J} \varphi(x_i, y_j) = \varphi(x,y)$. - -(Lest there be any doubt, “$(z_k)_{k\in K}$ summable with $\sum_{k\in K} z_k = z$” means that for every neighborhood $V$ of $z$ there is $K_0\subseteq K$ finite such that for any $K_0 \subseteq K_1 \subseteq K$ finite we have $\sum_{k\in K_1} z_k \in V$”.) -I am interested both in “nice” counterexamples to the statement above and in strengthenings of the hypotheses which would make it true—or basically any information regarding variations of this statement (I know essentially nothing except for the pretty much trivial fact that when $E,F,G$ are finite-dimensional it is correct). Since the rules of MO are to ask one specific question, and since I am mostly interested in counterexamples, let me ask: -Question: Is there a counterexample to the above statement with “nice” spaces $E,F,G$ (e.g., locally convex, complete, metrizable… or even Banach spaces)? -—but again, any information concerning it is welcome. -Comments (added 2018-01-04): - -It is clear that, under the hypotheses of the statement above, $\sum_{(i,j)\in I_1\times J_1} \varphi(x_i, y_j)$ converges to $\varphi(x,y)$ where $I_1$, $J_1$ range over the finite subsets of $I$ and $J$ respectively; what is to be proven is that $\sum_{(i,j)\in K_1} \varphi(x_i, y_j)$ converges to $\varphi(x,y)$ where $K_1$ ranges over the finite subsets of $I\times J$. The subtlety, of course, is that $K_1$ can fail to be a rectangle. -The following result is found in Seth Warner's book Topological Rings (1993), theorem 10.15: if $E,F,G$ be are Hausdorff commutative topological groups, and $\varphi\colon E\times F\to G$ is continuous and $\mathbb{Z}$-bilinear, and $(x_i)_{i\in I}$ and $(y_j)_{j\in J}$ summable families in $E$ and $F$ respectively with $\sum_{i\in I} x_i = x$ and $\sum_{j\in J} y_j = y$, then provided $(\varphi(x_i,y_j))_{(i,j)\in I\times J}$ is summable, it sum is $\sum_{(i,j)\in I\times J} \varphi(x_i, y_j) = \varphi(x,y)$. So the crucial question is the summability of $(\varphi(x_i,y_j))$, not the equality with $\varphi(x,y)$. Even with the very weak hypothesis that $E,F,G$ are commutative topological groups, I still don't have a counterexample! -The following possibly related result is found in Kamal Kant Jha's 1972 paper “Analysis of Bounded Sets in Topological Tensor Products” (corollary 3.3): If $(x_i)$ is a totally summable family in a locally convex space $E$ [meaning that there exists $L\subseteq E$ closed, absolutely convex and bounded, such that $\{x_i\}\subseteq L$ and $\sum_i p_L(x_i) < +\infty$ for $p_L$ the gauge of $L$] and ditto for $(y_j)$ in $F$, then $(x_i \otimes y_j)$ is totally summable in $E \mathbin{\otimes_\varepsilon} F$. - -REPLY [5 votes]: I found an answer to my question in Raymond Ryan's book Introduction to Tensor Products of Banach Spaces (2002), example 4.30, which I reproduce here with only minor modifications. -Specifically, this is with $E = F = \ell^2$ and $G = \mathbb{R}$, and furthermore $(y_j) = (x_i)$, i.e., we construct a bilinear form $\varphi$ on $\ell^2$ and an unconditionally convergent ($\Leftrightarrow$ summable) series $\sum_{i=1}^{+\infty} x_i = x \in \ell^2$ such that $\sum_{i,j} \varphi(x_i,x_j)$, while convergent in $\mathbb{R}$ for a certain ordering of the pairs $(i,j)$, is not absolutely convergent (so, not summable in $\mathbb{R}$). -Let $(e_i)$ (numbered from $1$, say) be the standard Hilbert (orthonormal) basis of $\ell^2$. Let $I_n$ be the integer interval between $2^n$ and $2^{n+1}-1$ inclusive (so $\#I_n = 2^n$), let $u_n := 2^{-n/2} \sum_{i\in I_n} e_i$ so that $(u_n)$ is an orthonormal sequence, and let $x := \sum_{n=1}^{+\infty} \frac{1}{n} u_n$. So we have $x = \sum_{i=1}^{+\infty} x_i e_i$ where $x_i = \frac{1}{n\,2^{n/2}}$ when $i \in I_n$ (and $x_1 = 0$ but no matter): this sum is unconditionally convergent. -Now let $A_n$ be the $2^n \times 2^n$ real matrix defined inductively by $A_0 = (1)$ and -$$A_{n+1} = \begin{pmatrix}A_n&A_n\\A_n&-A_n\end{pmatrix}$$ -[Ryan puts a minus sign on the lower left entry instead of the lower right, I don't think this changes anything but the above sign convention will make $\varphi$ symmetric]. The (Euclidean operator) norm $\|A_n\|$ of $A_n$ is $2^{n/2}$ (because $2^{-n/2} A_n$ is orthogonal). -Construct the bilinear form on $\ell^2 \times \ell^2$ which is block diagonal with blocks $2^{-n/2} A_n$ in the obvious sense, i.e., $\varphi(\sum_{i=1}^{+\infty} z_i e_i, \sum_{j=0}^{+\infty} z'_j e_j)$ is defined as $\sum_{n=0}^{+\infty} 2^{-n/2} \sum_{(i,j)\in {I_n}^2} (A_n)_{i,j} z_i z'_j$ (where $A_n$ is considered as indexed by ${I_n}^2$). Then $\varphi$ is continuous with $\|\varphi\| \leq 1$ (in the sense that $|\varphi(z,z')|^2 \leq \|z\|^2\,\|z'\|^2$) because $\|2^{-n/2} A_n\| \leq 1$. -In particular, $\varphi(x,x)$ is well-defined (with value $\sum_{n=1}^{+\infty} \frac{1}{n^2\,2^{n/2}}$, but no matter), whereas $\varphi(x_i e_i, x_j e_j) = \frac{\pm 1}{n^2\,2^{3n/2}}$ gives $\sum_{(i,j)\in {I_n}^2} |\varphi(x_i e_i, x_j e_j)| = \frac{2^{n/2}}{n^2}$ which does not converge (so $\sum_{i,j} \varphi(x_i e_i,x_j e_j)$ is not absolutely convergent).<|endoftext|> -TITLE: Chow group and base change -QUESTION [5 upvotes]: Let $k$ be a field with algebraic closure $\overline{k}$. Let $f\colon X\to k$ be a smooth projective variety(geometrically connected) over $k$. -Is the base change map $$\phi_i\colon \mathrm{CH}^i(X)\to\mathrm{CH}^{i}(X_{\overline{k}})$$ always injective? -(If $i=1$, $\mathrm{CH}^1(X)=\mathrm{Pic}(X)$, the Hochschild-Serre spectral sequence gives $$0\to H^1(\mathrm{Gal}(\overline{k}/k),f_{\overline{k},*}\mathbb{G}_m)\to\mathrm{Pic}(X)\to\mathrm{Pic}_{X_\overline{k}/\overline{k}}(k)\to H^2(\mathrm{Gal}(\overline{k}/k),f_{\overline{k},*}\mathbb{G}_m)$$ -The condition implies $f_{\overline{k},*}\mathbb{G}_m=\mathbb{G}_m$, so $H^1(\mathrm{Gal}(\overline{k}/k),f_{\overline{k},*}\mathbb{G}_m)=0$, and $\mathrm{Pic}_{X_{\overline{k}}/\overline{k}}(k)\to\mathrm{Pic}_{X_{\overline{k}}/\overline{k}}(\overline{k})$ is injective, we know $\phi_1$ is injective. -If $X$ is quasi-projective, then $\phi_1$ is not injective, for example $X=\mathbb{P}^1_{\mathbb{R}}-\{\pm i\}$, then $\mathcal{O}(1)$ is a nontrivial element in $\mathrm{ker}(\phi_1)$.) - -REPLY [5 votes]: No, this is not true in general. -A counterexample occurs already for Severi-Brauer varieties. Since the Chow group $\text{CH}(\mathbf{P}^n)$ is torsion free, it's enough to show there are Severi-Brauer varieties with torsion in their Chow groups. This was (I think) first observed in: - -Merkurjev, A. S. Certain K-cohomology groups of Severi-Brauer varieties. K-theory and algebraic geometry: connections with quadratic forms and division algebras (Santa Barbara, CA, 1992), 319–331, Proc. Sympos. Pure Math., 58, Part 2, Amer. Math. Soc., Providence, RI, 1995. - -Merkurjev uses the BGQ spectral sequence to show that certain differentials that compute the Chow groups are nonzero. Since it's also known from the Grothendieck-Riemann-Roch without denominators that the image of these differentials is contained in the torsion subgroup, this proves the claim. -More precise results in this direction appeared in: - -Karpenko, Nikita A. Codimension 2 cycles on Severi-Brauer varieties. K-Theory 13 (1998), no. 4, 305–330. (Available at the authors website https://sites.ualberta.ca/~karpenko/publ/ch2.pdf). - -Karpenko uses the gamma filtration on the Grothendieck ring of Severi-Brauer varieties to explicitly construct torsion elements in $\text{gr}_\gamma^2K(-)$ for certain Severi-Brauer varieties. He then shows this is equal to $\text{CH}^2(-)$ for Severi-Brauer varieties of certain indecomposable algebras. -More recently, Karpenko has provided a complete description of the Chow group of certain Severi-Brauer varieties in: - -Karpenko, Nikita A.(3-AB-MS) - Chow ring of generically twisted varieties of complete flags. (English summary) - Adv. Math. 306 (2017), 789–806. - -A number of these have a lot of torsion in their Chow groups (see Examples 3.17-3.22).<|endoftext|> -TITLE: Where do the (Akizuki)-Nakano Identities First Appear -QUESTION [6 upvotes]: The answers to this M.O. question give a history of the Kaehler identities. The identities can be extended to the vector bundle-valued setting, and play a central role in the proof of the Kodaira vanishing theorem. Where do these identities first appear? Moreover, is it more common to refer to these identities as the Nakano identities or the Akizuki-Nakano identities? -A consequence of these identities is the Bochner-Kodaira-Nakano identity, or as Demailly refers to it, the Bochner-Calabi-Kodaira-Nakano identity. Where does this first appear, and how are the contributions of Bochner, Calabi, Kodaira, and Nakano related? - -REPLY [4 votes]: Curvature and Betti -Numbers, Salomon Bochner (1948). -On a differential-geometric -method in the theory of analytic stacks, Kunihiko Kodaira (1953). -On complex analytic vector bundles, Shigeo Nakano (1955). -Note -on Kodaira-Spencer's Proof of Lefschetz Theorems, Yasuo Akizuki -and Shigeo Nakano (1954). - -The Bochner-Kodaira-Nakano identity expresses the antiholomorphic Laplace operator in terms of its conjugate operator plus extra terms involving the curvature of the manifold and the torsion of the metric. If the manifold is Kähler the torsion vanishes and one recovers the Akizuki-Nakano identity. -Calabi was Bochner's Ph.D. student and related results appear in his 1953 thesis, which is probably why Demailly refers to the Bochner-Calabi-Kodaira-Nakano identity.<|endoftext|> -TITLE: Do actions of BS(1,n) on finite sets factor through abelian quotients? -QUESTION [8 upvotes]: Suppose $BS(1,n)$ is the Baumslag-Solitar group and $S_m$ is the -symmetric group. If $\Phi: BS(1,n) \to S_m$ is a homomorphism, must the -image of $\Phi$ be abelian? - -REPLY [7 votes]: Another way to see the answer is no: the dihedral group $D_{2k}$ (sometimes written $D_k$) of order $2k$ is generated by $a$ ('rotation through $2\pi/k$') and $t$ (a reflection) and satisfies the relation $tat^{-1}=a^{-1}=a^{k-1}$. This is clearly a finite non-abelian group provided $k>2$, and taking $k=n+1$, we see that $D_{2k}$ is an image of $\mathrm{BS}(1,n)$ under the obvious map. -Finally, since $D_{2k}$ acts on the vertices of a regular $k$-gon, $D_{2k}$ embeds in $S_m$ for any $m\geq k$. This gives a negative answer to the question provided $n\geq 2$ and $m\geq 3$.<|endoftext|> -TITLE: Happy new semiprime after prime year! -QUESTION [12 upvotes]: After the change of the year I realized, as everyone did, that $2018=2\times1009$ and of course $1009$ is a prime number. $2017$ is also a prime number. Furthermore $2019=3\times 673$ and $673$ is also a prime number. So here is the conjecture. -For every number $n\in \mathbb{N}$ there exists a prime number $p$ such that -$p+1=2\times p_1$, -$p+2=3\times p_2$, -$p+3=4\times p_3$, -$p+4=5\times p_4$, -. . . -$p+n=(n+1)\times p_n$ -where $p_1,p_2,p_3,.....,p_n$ -are some prime numbers. -One can notice that this is an optimal situation in the sense that in every 2 numbers at least one should be divided by $2$ in every 3 numbers at least one should be divided by $3$ and in every n numbers at least one should be divided by $n$. -Of course the conjecture is too hard in the sense that it implies the open conjecture that there are infinitely many primes such that $2p-1$ is also prime, or you can ask the same question for the opposite direction meaning: -For every number $n\in \mathbb{N}$ there exists a prime number $p$ such that -$p-1=2\times p_1$, -$p-2=3\times p_2$, -. . . -$p-n=(n+1)\times p_n$ -where $p_1,p_2,p_3,.....,p_n$ -are some prime numbers that implies that the Sophie Germain Primes are infinite, -or both directions simultaneously,etc. -I don't know if this is a known question, I tried to find related ones but I didn't made it. -GENERALIZATION AND THOUGHTS: -Thinking to this direction one can make much more general conjectures that implies these ones. Par example: -For any number $n\in \mathbb{N}$ and for every prime $p' -TITLE: Prikry forcing and Cohen generic -QUESTION [5 upvotes]: Let $\kappa$ be a measurable cardinal and let $\mathcal{U}$ be a normal measure on $\kappa$. Let $\mathbb{P}$ be the standard Prikry forcing using $\mathcal{U}$. Let $\mathbb{Q} = \text{Add}(\kappa, 1)$ be Cohen forcing for adding a new subset to $\kappa$ using partial functions from $\kappa$ to $2$ of size ${<}\kappa$. -Question: Is there a projection from $\mathbb{P}$ to $\mathbb{Q}$? - -REPLY [4 votes]: The following result of Tom Benhamou and Gitik might be related: -Theorem. Suppose $V$ satisfies $GCH$ and $\kappa$ is a measurable cardinal. Then in a cofinality preserving generic extension, there exists a $\kappa$-complete ultrafilter $U$ on $\kappa$ such that Prikry forcing with $U$ adds a Cohen subset of $\kappa$ over $V$. -See page 69 of the paper Sets in Prikry and Magidor Generic Extensions.<|endoftext|> -TITLE: The 1-step vanishing polyplets on Conway's game of life -QUESTION [22 upvotes]: A $n$-polyplet is a collection of $n$ cells on a grid which are orthogonally or diagonally connected. -The number of $n$-polyplets is given by the OEIS sequence A030222: $1, 2, 5, 22, 94, 524, 3031, \dots$ -See below the five $3$-polyplets: - -A polyplet will be called $1$-step vanishing on Conway's game of life, if every cell dies after one step. -We observed that for $n\le 4$, a $n$-polyplet is $1$-step vanishing iff $n \le 2$. -We found $1$-step vanishing polyplets with $n=9, 12$, see below: - -Question: Is there an other $1$-step vanishing $n$-polyplet with $n \le 12$? -If yes, what are they? (note that there are exactly $37963911$ $n$-polyplets with $n \le 12$). -We can build infinitely many $1$-step vanishing polyplets with such pattern, see below with $n=142$: - -Bonus question: Are there $1$-step vanishing polyplets of an other kind? - -REPLY [5 votes]: Theorem: The set of positive integers $n$ such that there is a $1$-step vanishing polyplet with $n$ cells is $$ \{1,2,9,10,12,14 \} \cup \mathbb{N}_{\ge 15}$$ It can be represented by $1$-step vanishing polyominoes without hole and a non-trivial symmetry group. -Proof: The finite part is given by: - -For the infinite part, we will do three steps. -Firstly, the family beginning as follows: - -provides the integers $6a+b$ with $b \in \{ 3 , \dots , a+2 \}$. Now, $$6a + (a+2) +1 \le 6(a+1)+3$$ if and only if $a \ge 6$. But $6 \times 6+3 = 39$, so it is ok for $\mathbb{N}_{\ge 39}$. -Next, the gaps of the previous family for $14 -TITLE: The von Neumann algebra generated by a non-closable operator -QUESTION [7 upvotes]: Let $H$ be a separable Hilbert space and let $M$ be a densely defined operator $\mathcal{D}(M) \subset H \to H$. It is closable iff its adjoint $M^{\star}$ is densely defined, and then its closure $\overline{M}$ is $M^{\star \star}$. Let $\mathcal{M}$ be the smallest von Neumann algebra that $\overline{M}$ is affiliated with; it is called the von Neumann algebra generated by $M$. -Question 1: Is there a bounded operator $X \in B(H)$ such that $W^{\star}(X) = \mathcal{M}$? -In other words: Can a von Neumann algebra generated by a densely defined closable operator, be also generated by a bounded operator? -Question 2: Is there a way to generalize the generation of a von Neumann algebra to any densely defined operator (i.e. non necessarily closable)? -If an answer to Question 1 gives a process defining $X$ from $M$ and if this process works for any densely defined operator, that would also answer Question 2. - -Some investigations for answering Question 2 -Let first suppose that the operator $M$ is associated to an integer map, i.e. $H = \ell^2(\mathbb{N}^*)$ and there is a map $m: \mathbb{N}^* \to \mathbb{N}^*$ such that $Me_n = e_{m(n)}$, with $\mathcal{D}(M) = c_{00}(\mathbb{N}^*)$, dense in $H$. -Assume that $M$ is non-closable iff $\exists n \in \mathbb{N}^*$ with $m^{-1}(\{ n\})$ infinite. -First of all, there is the following natural way to associate a bounded operator to $m$: $$Ye_n = \frac{1}{n}e_{m(n)}.$$ Unfortunately, if $M$ is closable, it does not generate the same von Neumann algebra than $Y$ in general, because if $m=id$, then $M=I$ and $Y = diag(1/n \ | \ n \in \mathbb{N}^*)$, so $W^{\star}(M) = \mathbb{C}$ and $W^{\star}(Y) = \ell^{\infty}(\mathbb{N}^*)$. -But, there is a way to avoid this problem, by using the operator defined as follows: - $$\tilde{M}e_n = \begin{cases} e_{m(n)} & \text{if} \;m^{-1}(\{m(n)\}) \;\text{finite} \\ \frac{1}{n}e_{m(n)} & \text{if} \;m^{-1}(\{m(n)\}) \;\text{infinite} \end{cases}$$ Then $\tilde{M}$ is densely-defined and closable, even if $M$ is non-closable. Moreover, if $M$ is still closable then $\tilde{M} = M$ by construction, so they generate the same von Neumann algebra. -Example: If $m(n)=1$ $\forall n$, then $Me_n = e_1$ and $M$ is non-closable, whereas $\tilde{M}e_n = \frac{1}{n}e_1$ defines a bounded operator. Note that $\tilde{M}$ is not normal because $\tilde{M}^{\star}\tilde{M}e_1 = \sum_n \frac{e_n}{n}$ and $\tilde{M}\tilde{M}^{\star}e_1 = \frac{\pi^2}{6}e_1$. Then, $\mathcal{M}:=W^{\star}(\tilde{M})$ is non-abelian. Now, $\tilde{M}^2 = \tilde{M}$ and $\tilde{M}\tilde{M}^{\star}\tilde{M} = \frac{\pi^2}{6}\tilde{M}$, so $\dim(\mathcal{M}) = 4$. It follows that $\mathcal{M} \simeq M_2(\mathbb{C})$. -Invariance: Let $\sigma \in S(\mathbb{N}^*)$ be a permutation, $m_{\sigma}$ the deformation $\sigma \circ m \circ \sigma^{-1}$. Let ${M}_{\sigma}$ and $\tilde{M}_{\sigma}$ be the corresponding operators. Do $\tilde{M}$ and $\tilde{M}_{\sigma}$ generate isomorphic von Neumann algebras? -Hard example: consider Conway's game of life and let $\mathcal{S}$ be the set of states of the grid with only finitely many alive cells. It is countable infinite, so there is a bijection $b: \mathcal{S} \to \mathbb{N}^*$. Conway's rule (B3/S23) produces a map $r:\mathcal{S} \to \mathcal{S}$. Let $m$ be the integer map $b \circ r \circ b^{-1} : \mathbb{N}^* \to \mathbb{N}^*$. We can then define $\tilde{M}$ as above. Let $\mathcal{M}$ be the von Neumann algebra generated by $\tilde{M}$. Assuming that the above invariance is true, $\mathcal{M}$ is independant of the choice of $b$. Thus $\mathcal{M}$ can be called the von Neumann algebra generated by Conway's game of life. Bonus question: What is $\mathcal{M}$? -We can do the same with any other cellular automaton. -Conclusion: the use of $\tilde{M}$ is a way to generalize the generation of a von Neumann algebra to any densely defined operator associated to an integer map. Can we extend it to any densely defined operator (i.e. non necessarily associated to an integer map)? -A naive attempt of generalization: let $H$ be a separable infinite dimensional Hilbert space and let $M$ be any densely defined operator, with maximal domain $\mathcal{D}$. Let $\mathcal{B}$ be a countable basis and $b: \mathcal{B} \to \mathbb{N}^*$ a bijection. Take $e_n:= b^{-1}(n)$ and let $K$ be the compact operator defined by $Ke_n = \frac{1}{n}e_n$. Let $H_{\infty}$ be the closure of the subspace of vectors $v \in \mathcal{D}$ such that there is a countable orthonormal basis $\mathcal{B}_v$ of $\overline{M^{-1}(\mathbb{C}Mv)}$ with $\mathcal{B}_v \subset \mathcal{D}$ and $$\sum_{b \in \mathcal{B}_v} |\langle Mb,Mv \rangle|^2 = \infty.$$ Let $P$ be the orthogonal projection on $H_{\infty}$. Consider the following operator $$\tilde{M}:=MKP + M(I-P).$$ -Is $\tilde{M}$ well-defined? densely-defined? closable? Is the von Neumann algebra generated by $\tilde{M}$ independent of the choice of $\mathcal{B}$ and $b$? If so (...!), this construction would answer Question 2. -Invariance: Would such a von Neumann algebra be invariant replacing $K$ by any positive compact operator $L$ with all eigenvalues distinct and $L^{1+\epsilon}$ trace-class $\forall \epsilon >0$? - -REPLY [2 votes]: Only a comment on Question 2: It is already not easy to define the von Neumann algebra generated by a family of closable operators, see e.g. https://projecteuclid.org/euclid.cmp/1103899047 Def. 2.5 and Rem. 2.7. -Closability in this context appears naturally for the following reason: Assume that you have some (possibly non-closed) unbounded operator $M$ on some dense domain $\mathcal D$. To get towards defining a corresponding von Neumann algebra $\mathcal M$, I presume you would first try to compute a formal adjoint $M^*$ and I think you will agree that $M^*$ should also be densely defined. But then you already obtain that $M$ is closable.<|endoftext|> -TITLE: Large cardinals and reflection properties -QUESTION [7 upvotes]: Strong and supercompact cardinals are $\varSigma_2$-reflecting; extendible cardinals are $\varSigma_3$-reflecting. It is of course possible to build larger degrees of correctness into the definitions of large cardinals; if I understand correctly, this is essentially what Joan Bagaria does in his paper "$C^{(n)}$-Cardinals" (Arch. Math. Logic 51 (2012): 213–40; doi: 10.1007/s00153-011-0261-8). -But is there any 'standard' large cardinal notion—basically, one not specifically formulated using the notion of $\varSigma_n$-correctness or something equivalent—such that cardinals of that type are $\varSigma_n$-reflecting for $n > 3$? - -REPLY [2 votes]: There is actually a type of cardinal that satisfies this, called the "stationarily superhuge" cardinals. A cardinal $\kappa$ is called stationarily superhuge if the $\{\lambda|\kappa\text{ is huge with target }\lambda\}$ is stationary. If $\kappa$ is stationarily superhuge, $V_\kappa\prec V$, and moreover $L_\kappa\prec L$. The reason for this is that $\{\lambda|V_\lambda\prec V\}$ is club, assuming the existence of a stationarily superhuge cardinal. Then $V_\kappa\vDash\phi$ if and only if $M\vDash(V_\lambda\vDash\phi)$ if and only if $V_\lambda\vDash\phi$. Setting $\lambda$ is correct, we have $V_\lambda$ reflects $\phi$. -A similar argument works for $L_\kappa$, and for $H_\kappa$ (Or more simply, as $\kappa$ is superhuge and so inaccessible, $H_\kappa=V_\kappa$). Moreover, the set of such cardinals form a normal measure beneath $\kappa$. Let $D=\{X\subseteq\kappa|\kappa\in j(X)\}$. Then $M\vDash(j(\kappa)\text{ is reflecting})$, and so $U\in D$, where $U=\{\lambda<\kappa|\lambda\text{ is reflecting}\}$. -Reference: - -Julius B. Barbanel, Carlos A. Diprisco and It Beng Tan: Many-Times Huge and Superhuge Cardinals, -The Journal of Symbolic Logic, Vol. 49, No. 1 (Mar., 1984), pp. 112-122. https://doi.org/10.2307/2274094 https://www.jstor.org/stable/2274094<|endoftext|> -TITLE: Is the homotopy category of an abelian model category abelian? -QUESTION [9 upvotes]: A model structure on an abelian category $A$ is called an abelian model structure if the cofibrations are precisely the monomorphisms with cofibrant cokernel, and if the fibrations are precisely the epimorphisms with fibrant kernel. This terminology was introduced by Mark Hovey in Cotorsion pairs, model category structures, and representation theory; see also the survey article Cotorsion pairs and model categories. -In his book, Hovey introduced monoidal model categories and proved that their homotopy categories are also monoidal. Similarly, a model category is stable if and only if it's homotopy category is triangulated. I always assumed from the terminology that the condition about (co)fibrations was there to guarantee that the homotopy category of an abelian model category would be an abelian category, but now I can't find a reference for this, and I'm not even sure if it's true. I'm trying to branch out a bit into homological algebra and representation theory, and abelian model categories are new to me. Any help would be much appreciated! - -REPLY [20 votes]: No. The projective model structure on chain complexes of modules over a ring is an abelian model category, and the homotopy category is the derived category, which is never abelian unless the ring is semisimple.<|endoftext|> -TITLE: Which of these sums appear most often? -QUESTION [11 upvotes]: Let $N=\{1,2,3,\ldots, n\}$. -We sum all the elements of every nonempty subset of $N$. -Which sum(s) appears most often? (Let's call this sum a champion). -Using a simple pigeonhole argument a champion must appear at least $\frac{2^n-1}{T_n}$ times. ($T_n$ denotes the $n$-th triangular number). -It seems that the champion should be somewhere around $T_n/2$ but I cannot prove it. -Am I missing something obvious here? - -REPLY [16 votes]: In fact this question was already asked at MO, although in disguise: see here. Richard Stanley answered it wonderfully. The champions are the nearest integers to $n(n+1)/4$. -For a quick proof, see Lemma 6.13 on Page 93 (and the preliminaries on Page 92) in Stanley's Topics in algebraic combinatorics.<|endoftext|> -TITLE: Colimits, limits, and mapping spaces -QUESTION [11 upvotes]: It is true that in the category of topological spaces -$ \mathrm{Map}(\underset{i\in I}{\mathrm{colim}}\, X_i, Y)\cong -\underset{i\in I}{\mathrm{lim}}\,\mathrm{Map}(X_i,Y)$ ? Here mapping spaces are endowed with the compact-open topology. One has a bijective map from the left-hand side to the right-hand side, but is this map a homeomorphism? For example, if $X_1\subset X_2$, is the inclusion $\mathrm{Map}(X_2/X_1,Y)\subset \mathrm{Map}(X_2,Y)$ a homeomorphism on its image? If not, what if this inclusion is a cofibration? -(Compare with my previous question Mapping space from a quotient space.) - -REPLY [11 votes]: This is true if, instead of topological spaces, you work in a convenient category of topological spaces, in the sense of Steenrod. These are the place you want to do homotopy theory in (assuming you want to do it using topological spaces and not, for example, Kan complexes). In this case, the functor: -$$\mathrm{Map}(-,Y):C^{op}\to C$$ -is right adjoint to the functor -$$\mathrm{Map}(-Y):C\to C^{op}$$ -(careful: adjunctions with the opposite category are confusing :)). -Indeed, if we denote by $C(X,Y)$ the set of continuous maps from $X$ to $Y$ -$$C(Z, \mathrm{Map}(X,Y))=C(Z\times X,Y)=C(X,\mathrm{Map}(Z,Y))$$ -(in this case $\mathrm{Map}(X,Y)$ will in general be different from the compact open topology, for example if we work in compactly generated spaces it will have the Kelleyfication of the compact open topology).<|endoftext|> -TITLE: Is a symplectic submanifold of a Kähler manifold Kähler? -QUESTION [8 upvotes]: Is a symplectic submanifold of a Kähler manifold Kähler? -That is, if $X$ is a Kähler manifold with symplectic form $\omega$ and $i:Y\hookrightarrow X$ is an embedded submanifold such that $i^*\omega$ is symplectic, is $Y$ a Kähler submanifold of $X$? - -REPLY [18 votes]: No. In $\mathbf C^2$ with standard 2-form and complex structure, the real span of $U=\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)$ and $V=\left(\begin{smallmatrix}i\\1\end{smallmatrix}\right)$ is a symplectic but not complex (hence not Kähler) subspace, since -$$ -\omega(U,V)=1, -\qquad -JU=\left(\begin{smallmatrix}i\\0\end{smallmatrix}\right)\notin\operatorname{span}_{\mathbf R}(U,V). -$$<|endoftext|> -TITLE: How many triangulations of a regular octahedron are there, without introducing new vertices? -QUESTION [6 upvotes]: It is easy to find three triangulations, each consisting of four tetrahedra. Are there more? - -REPLY [15 votes]: No, these are all. The edge graph of the octahedron has no $K_4$ subgraph, so you have to add a new edge to make a triangulation. The only possible places for a new edge are connecting opposite vertices. You can only add one such edge, as any two meet in their interior. So every triangulation of the octahedron (without new vertices) adds exactly one of the three edges $e$ between opposite vertices. The only $K_4$'s in that graph are the four tetrahedra arranged around $e$, so any traingulation using $e$ must use a subset of those tetrahedra. We need all of them to fill in the octahedron.<|endoftext|> -TITLE: Schauder basis $L^p(\mathbb{R})$ -QUESTION [6 upvotes]: Let $\{e_{n}(x)\}_{n=0}^{\infty}$ be orthnormal basis of Hilbert space $L^2(\mathbb{R})$. If $\{e_{n}(x)\}_{n=0}^{\infty} \subset L^p(\mathbb{R})$ for some $p\geq 1$, is the $\{e_{n}(x)\}_{n=0}^{\infty}$ Schauder basis for $L^p(\mathbb{R})$? -Any reference? - -REPLY [2 votes]: One can derive the answer "Not in general" working out suitable variations of the following lemma. -Lemma. Let $X$ be a separable Banach space which is not isomorphic to a Hilbert space, but is continuously and injectively embedded into a Hilbert space $H$. Then $X$ contains a sequence $\{x_i\}_{i=1}^\infty$, which is linearly independent, has dense span in $X$, is an orthonormal sequence in $H$, but is not a Schauder basis in $X$ because the linear spans of finite subsets $\{x_i\}_{i=1}^n$ have indefinitely increasing projection constants in $X$. -Proof: By the Lindenstrauss-Tzafriri theorem [Israel J. Math. 9 (1971), 263–269] we can find a sequence $\{y_i\}_{i=1}^\infty$ in $X$ with the dense linear span satisfying the last condition. Applying the Gram-Schmidt ortonormalization, we get the desired sequence $\{x_i\}_{i=1}^\infty$.<|endoftext|> -TITLE: Two approaches to periodic cyclic cohomology -QUESTION [5 upvotes]: Cyclic cohomology may be defined in several ways: the easiest way to define it is via a subcomplex $C^*_{\lambda}$of Hochschild complex consisting from cyclic cochains. There are also other definitions for example using the cyclic bicomplex or $(b,B)$ (mixed) complex. Relevant definition can be found here. One can define the so called periodicity operator $S:HC^{n}(A) \to HC^{n+2}(A)$ and using this operator one can define the periodic cyclic cohomology as the inductive limit of $HC^{2n}(A)$ or $HC^{2n+1}$ (where the arrows of inductive system are just iterations of $S$). - -Why is periodic cyclic cohomology of $A$ isomorphic with the cohomology of the totalization of $(b,B)$-bicomplex? - -Several remarks: -1. To define periodic theory one uses the bicomplex which is no longer concentrated in the first quadrant (see again my previous question linked above). -2. I would be happy if I could see the map yielding quasi-isomorphism explicitly. -3. There are still another variants of cyclic theory such as negative theory for instance. If there is a way to prove that the two approaches: using $(b,B)$ bicomplex and ordinary cyclic complex and $S$ operator are equivalent, without invoking the negative theory, I would be happy to see it. - -REPLY [2 votes]: The original reference is Theorem 40, c) of http://www.alainconnes.org/docs/noncommutative_differential_geometry.pdf -A maybe more pedagogical version is Theorem 29 c) of http://www.alainconnes.org/docs/book94bigpdf.pdf<|endoftext|> -TITLE: Motivic vs Deligne cohomology -QUESTION [12 upvotes]: Where can I find the construction of the cycle class map from motivic cohomology to Deligne cohomology of smooth projective varieties over the complex numbers? -It should be a construction by Bloch using his cycle complex, but it seems I can't find on the web any paper spelling it out. - -REPLY [7 votes]: In addition to the papers mentioned, I would like to add Kerr, Lewis, Müller-Stach, The Abel–Jacobi map for higher Chow groups, Compositio (2006). They give an explicit construction (if you like that sort of thing -- I do), using currents, mapping a cubical Bloch complex to a complex computing Deligne cohomology. There is also a follow up paper by the first two authors (with obvious title) in Inventiones.<|endoftext|> -TITLE: Cauchy real numbers with and without modulus -QUESTION [16 upvotes]: In constructive mathematics there are many possible inequivalent definitions of real numbers. The greatest variety seems to be in Dedekind-style approaches: in addition to "the" Dedekind real numbers which satisfy locatedness (for all $qn. |x_p-x_q|<2^{-k}$. But almost without exception, constructive mathematicians define Cauchy sequences by "building a Skolem function" into the definition, assuming that there is a "modulus of Cauchy-ness" $N:\mathbb{N}^{\mathbb{N}}$ such that $\forall k \forall p,q > N_k. |x_p-x_q|<2^{-k}$. And once one has such a modulus, one can massage the sequence $(x_n)$ to make the modulus coincide with a standard one such as $N_k = k$. -My question is: what is the difference in constructive mathematics between Cauchy real numbers having a modulus of Cauchy-ness and without it? It seems to me that the no-modulus Cauchy reals sit in between the with-modulus Cauchy reals and the Dedekind reals; we shoudn't need a modulus to get locatedness, all we need is the existence of some point of the sequence that's closer than $\frac{|q-r|}{2}$ to the limit so we can compare it to $q$ and $r$. -I understand that someone who cares about algorithms and computation will want a modulus to compute with, but if I am a topologist and am happy with Dedekind reals, leaving out a modulus in the Cauchy reals is not a priori a bad thing to do. Are the no-modulus Cauchy reals any less well-behaved intrinsically than the with-modulus ones? Is their algebraic and order structure any different? And are there (hopefully topological) models (necessarily failing countable choice) that distinguish the no-modulus Cauchy reals from both the with-modulus ones and from the Dedekind reals? - -REPLY [6 votes]: I don't see an algebraic difference. Even in constructive math without countable choice, we can define the total functions $+, -, \times, \min, \max, \sqrt[3]{}$ and the partial functions $1/x, \sqrt{x}$ on the no-modulus Cauchy reals. -However, there is a difference in that Cauchy completeness does not hold for the no-modulus reals. Consider the situation in the recursive topos or in recursive mathematics: - -Let $R$ be the no-modulus Cauchy reals, considered as -functions $r : N \rightarrow Q$. -Let $S$ be the set of with-modulus Cauchy sequences of no-modulus -reals, considered as functions $s : N \rightarrow R$, such that -$\forall m, m', |s(m)-s(m')| < 1/m + 1/m'.$ -Suppose (for contradiction) that $\forall s \in S\ \exists r \in R\ \lim s=r$. -Then there would be a recursive function $f:S\rightarrow R \ \lim s=f(s)$. -Now consider $s=0^S$, i.e. the sequence such that $\forall n\ s(n)=0^R$, i.e. the sequence such that $\forall n\forall m\ s(n)(m)=0^Q$. The calculation of $(f(s))(1)$ uses at most $k_1$ terms of $s$, the calculation of $(f(s))(2)$ uses at most $k_2$ terms of $s$, etc. -Let $t(n)(m)=0$ if $n+m < k_{n+m}$, and 1 otherwise. Then $t$ is in $S$. -Then $f(s)=f(t)$, but $\lim s = 0$ and $\lim t=1$, which is a contradiction. - -If we tried to replace $R$ and $S$ by the with-modulus Cauchy reals and sequences thereof, we would find that $t(1)$ might not be a with-modulus real, $t$ might not be in the new $S$, and the argument would not go through. -So in the recursive world: the with-modulus reals lead to all Cauchy sequences converging; the no-modulus reals lead to Cauchy sequences that may not converge. - -REPLY [3 votes]: Let $(a_n)$ be a Specker sequence - a computable, bounded, increasing sequence of rationals so that the limit is not a computable real. -First, assume for simplicity that we work in any system of second-order arithmetic (classical or not) that has REC as a model. This is the model with the standard natural numbers and the computable sets of natural numbers. In this model, the Specker sequence is a non-modulus Cauchy sequence (because it is a Cauchy sequence in the standard model, and being Cauchy is defined arithmetically). But there is no computable modulus of convergence for a Specker sequence, so the model does not believe that the sequence is a modulus Cauchy sequence. -Thus our system cannot prove that "every non-modulus Cauchy sequence has a modulus". Systems affected by this include many systems of constructive second-order arithmetic as well as the classical system $\mathsf{RCA}_0$. -In fact, every modulus of a Specker sequence computes $\emptyset'$. So the previous argument also goes through for any system of second-order arithmetic (classical or not) that has a $\omega$-model without any Turing complete reals. This applies to $\mathsf{WKL}_0$ and to constructive systems with various forms of compactness. -Similar arguments will work for systems that are not fragments of second-order arithmetic, as long as they have $\omega$-models that do not contain any Turing complete reals.<|endoftext|> -TITLE: Can we cut and rotate a particular region of a hyperbolic 3-manifold to get another (non-homeomorphic) hyperbolic 3-manifold? -QUESTION [7 upvotes]: I'm trying to learn more about hyperbolic 3-manifolds, in particular the geometric implications of doing hyperbolic Dehn surgery to suitable knot complements. -Following this paper by Christian Millichap ( https://arxiv.org/abs/1209.1042 ), we can construct a hyperbolic Montesinos knot $K$ with several tangle regions (see p. 8). We then mutate these tangle regions along a suitable Conway sphere, before doing hyperbolic Dehn surgery to fill up the knot complements. Because of how we constructed the hyperbolic Montesinos knot $K$ (in particular, the relevant pair of $S^0$'s are unlinked on $K$, see What does it mean exactly for a pair of $S^0$'s to be unlinked on a knot $K$?), it can be shown that $K$ and all its mutations yield a hyperbolic 3-manifold with the same volume. Further, by an argument in this paper I don't fully understand (for details, see the proof of Theorem 3 on pg. 11), we can pick appropriate coefficients for the Dehn surgery so that the hyperbolic manifolds resulting from $K$ and its mutations are all non-homeomorphic. -I want to have a more precise (and concrete) understanding of how "different" these hyperbolic 3-manifolds are, beyond the fact that they are non-homeomorphic. In particular, suppose we have a hyperbolic knot $K$ and its mutant $K'$ and we know that they yield non-homeomorphic hyperbolic 3-manifolds (call them $M_{K}$ and $M_{K'}$) with the same finite volume via Dehn surgery. -My question is: since mutations of knots involve "cutting up" a certain region of a knot and rotating it appropriately to get a different knot, is it possible for us to cut out an analogous region of $M_{K}$ and (via some isometry in $H^3$) "rotate" the region to obtain $M_{K'}$? -In particular, I don't understand the geometric aspects of hyperbolic Dehn surgery well enough to know if filling in the knot complement with a solid torus presents an obstruction to this. All I know is that hyperbolic Dehn surgery involves picking a suitable pair of integers $(p,q)$ and choosing a basis $(m,l)$ for the fundamental group of the torus boundary of the knot complement (where m and l corresponds to the meridional and longitudinal side of the torus respectively). We may interpret the knot complement as a hyperbolic 3-manifold with a cusp corresponding to the torus boundary; via hyperbolic Dehn surgery, we can cut off the cusp and glue in a solid torus by mapping the boundary of the meridian disc of the torus to $s=pm+ql$ where $s$ is a slope on the torus boundary of the knot complement. (I am following p. 7 of the Millichap paper here). From a geometric standpoint, all I take from this is that there is a very specific way of gluing the solid torus along the torus boundary of the knot complement, and this might give us problems if we were to "cut and paste" the region in order to get a new hyperbolic 3-manifold. Am I right about this? - -If it helps to have something more visual, consider the following two knots, which are the prime Kinoshita-Teraska knot (LH) and the prime Conway knot (RH). This diagram shows how the two are related by a mutation along the Conway sphere (marked in red) by rotation about the y-axis. - -As Ian Agol mentions below, it can be verified that this example of knot mutation satisfies the unlinking condition mentioned above, hence Dehn surgery on the knot complements of these two knots yield hyperbolic 3-manifolds with equal volume. Now, suppose we have filled the knot complements out via Dehn surgery and gotten two non-homeomorphic 3-manifolds with the same volume - can we then cut and rotate whatever is analogous to the red region in the hyperbolic 3-manifold (derived from the LH knot) to get the hyperbolic 3-manifold derived from the RH knot? - -REPLY [8 votes]: Look at the proof of Theorem 5.5 in Ruberman's paper. The way that he proves that the Dehn fillings on mutants have the same volume is to show that there is a closed genus $\leq 2$ surface on which one may perform mutation, which preserves volume. This surface is obtained from the 4-punctured sphere by tubing along annuli which are tubular neighborhoods of the arcs of the knot lying to one side and whose endpoints are flipped by the mutation. The mutation on the 4-punctured sphere is compatible with mutation on the tubed genus 2 surface, in the sense that mutation along the 4-punctured sphere gives the same manifold as mutation along the genus 2 surface. Hence after Dehn filling, there will still be a mutation along a genus 2 surface which relates the two Dehn fillings. -The examples in your picture do satisfy the unlinking condition (Definition 5.4) which is needed in the hypothesis of Theorem 5.5. -Here's a picture of the tubed genus 2 surface:<|endoftext|> -TITLE: Simple question about polynomials -QUESTION [10 upvotes]: Starting from a problem in combinatorics, I ended up with a very simple problem about polynomials, which, unfortunately, I am not able to solve. -Say we work over $\mathbb C$. Fix $d>1$. -Is it possible to find 3 homogenous polynomials $F_0,F_1$ and $F_2$ of degree $d$ in $\mathbb C[x,y,z]$ without any non-trivial zeroes in common and such that -$$F_1\cdot F_2=F_0^2- x^{d+1}\cdot G$$ -for some homogeneous polynomial $G$ of degree $d-1$? -My guess is that the answer is no, but I couldn't rephrase it to a more classical algebraic geometry problem. -It is easy to show that the answer is negative if we take $G=0$ or $G=x^{d-1}$ or if $d=2$. - -REPLY [9 votes]: This equation has no solutions when $d$ is odd. (EDIT: See below for the general case.) -Actually, for $d$ odd, there are no triples $(F_0,F_1,F_2)$ with $F_1\cdot F_2-F_0^2$ a multiple of $x$, let alone $x^{d+1}$. One can see this in a completely hands-on way by setting $x=0$ and looking at the resulting triple of polynomials in $y$ and $z$, which can be seen to have a common root. Here is a more geometric argument. -The problem is equivalent to asking for a map (necessarily finite) $\mathbb{P}^2\rightarrow\mathbb{P}^2$ such that the pullback of $\mathcal{O}(1)$ is $\mathcal{O}(d)$ and the pullback of the conic defined by $xy=z^2$ contains the line $x=0$ with multiplicity $d+1$; the strengthening I gave above says that for $d$ odd, this divisor cannot contain the line at all. Indeed, if it does, the pushforward of the line must be some multiple of the conic. However, the pushforward of the line is of degree $d$, and so this can't happen. -EDIT: Here is an attempt to settle the problem for even $d$ as well. This argument uses a surprising amount of machinery , considering how elementary the problem is to state. This makes me a little suspicious that this argument is either wrong or more complicated than necessary, but here goes. -As above, we have a map $f:\mathbb{P}^2\rightarrow\mathbb{P}^2$ defined by $(F_0,F_1,F_2)$ satisfying $f^*\mathcal{O}(1)\cong\mathcal{O}(d).$ Let $X$ be the $d$th formal neighborhood of the line $L$ defined by $x=0$, or in other words, $X$ is the subscheme defined by $x^{d+1}$. Our hypothesis implies that the restriction of $f$ to $X$ lands (scheme-theoretically) in the conic $yz=x^2.$ Treating this conic as an abstract $\mathbb{P}^1$, we get a map $g:X\rightarrow\mathbb{P}^1.$ -Now as $\mathcal{O}_{\mathbb{P}^2}(1)$ restricts to $\mathcal{O}_{\mathbb{P}^1}(2)$ under the embedding of this conic, we know that $g^*\mathcal{O}_{\mathbb{P}^1}(2)\cong\mathcal{O}_X(d)$. We would like to know the stronger statement that $g^*\mathcal{O}_{\mathbb{P}^1}(1)\cong\mathcal{O}_X(d/2),$ which would follow if we knew that the Picard group of $X$ is torsion free. -To see that $\operatorname{Pic}(X)$ is torsion-free, note that we have a short exact sequence of sheaves $0\rightarrow I\rightarrow \mathcal{O}_X^*\rightarrow\mathcal{O}_{\mathbb{P}^1}^*\rightarrow 0.$ This in turn gives us the exact sequence $0\rightarrow H^1(I)\rightarrow H^1(\mathcal{O}_X^*)\rightarrow H^1(\mathcal{O}_{\mathbb{P}^1}^*)\rightarrow H^2(I).$ But $I$ is in fact a sheaf of $\mathbb{C}$-vector spaces, so its cohomology groups are also $\mathbb{C}$-vector spaces. (One can actually compute all of the cohomology groups of $I$, but we will not need this here.) As $H^1(\mathcal{O}_{\mathbb{P}^1}^*)\cong\operatorname{Pic}(\mathbb{P}^1)\cong\mathbb{Z},$ we see immediately that $\operatorname{Pic}(X)\cong H^1(\mathcal{O}_X^*)$ is torsion-free. -Now that we know $g^*\mathcal{O}_{\mathbb{P}^1}(1)\cong\mathcal{O}_X(d/2),$ this tells us that there are sections $s_1,s_2$ of $\mathcal{O}_X(d/2)$ and a unit $u$ of $\mathcal{O}_X$ such that we have $(F_0,F_1,F_2)=(us_1s_2,us_1^2,us_2^2).$ A quick cohomology calculation shows that all sections of $\mathcal{O}_X(d/2)$ are restrictions of sections of $\mathcal{O}_{\mathbb{P}^2}(d/2)$ and that all units of $\mathcal{O}_X$ are constants. So we can take $u=1$ and identify $s_1,s_2$ with two homogeneous degree $d/2$ polynomials (which we will again call $s_1,s_2.$) -All in all, we have $(F_0,F_1,F_2)\equiv (s_1s_2,s_1^2,s_2^2)\pmod{x^{d+1}}.$ But this implies in fact that $(F_0,F_1,F_2)=(s_1s_2,s_1^2,s_2^2),$ and so that $F_1F_2-F_0^2=0,$ aka that $f$ in fact comes from a morphism $\mathbb{P}^2\rightarrow\mathbb{P}^1.$ But no non-constant such morphisms exist, a contradiction. -Just for fun: The above argument breaks down in characteristic $2$. In fact, the original problem has an affirmative answer in characteristic $2$, even for $d$ as small as $2$. Take $F_0=xy+xz+yz, F_1=x^2+y^2, F_2=x^2+z^2.$ - -REPLY [2 votes]: It is a comment which is too long for a comment. You probably tried this but here it is anyway. You can eliminate $x$ and the condition that $F_i$ are homogeneous by substitution $y'=y/x, z'=z/x$. So you want to prove that there are no polynomials in two variables $F_i(y',z'), i=1,2,3$ of degree $d$ without zeroes in common such that $F_1F_2-F_3^2$ is a polynomial $G$ of degree $d-1$. If you take the highest degree homogeneous components $H_1, H_2, H_3$ of $F_1,F_2, F_3$, then you have $H_1H_2=H_3^2$. So the question is: -Given three homogeneous polynomials in 2 variables of the same degree $d$, satisfying $H_1H_2=H_3^2$ can you deform $H_1, H_2, H_3$ by adding polynomials in the same two variables of degree $d-1$ so that the resulting polynomials do not have common roots?<|endoftext|> -TITLE: Self-adjointness and choosing appropriate function spaces -QUESTION [6 upvotes]: Consider the following operator on some (yet undecided) space $S$ of functions over $[0\:\:1]$ -$$L(u)=\sin(x)u-x\dfrac{\partial u}{\partial x}$$ -Now, its formal adjoint is -$L^*(v)=\sin(x)v+\dfrac{\partial{(xv)}}{\partial x}$. -My questions are: -1) How do we determine $S$, the space of functions that will make operator $M=LL^*$ self-adjoint? -E.g. if I naively choose $L^2([0 \:\ 1])$ as domain of $L$, it doesn't result in self-adjointness of $M$ (at least this is what I found numerically). -2) Related to 1), does the boundary conditions affect this situation ? E.g., if I take the domain of $L$ to be $L^2([0\:\:1])$ with $u(0)=u(1)=0$, does it change the self-adjointness properties of $M$? - -REPLY [12 votes]: The question of self-adjointness is quite often all about the boundary conditions. In order to get the domains of the operator and its adjoint to match, boundary conditions need to be 'distributed' equally between the two. -In this particular case, the first step is to choose a domain $S$ such that the operator $M$ is symmetric, i.e. $(M(u),v)=(u,M(v))$ for all $u,v\in S$. -We have -$$M(u)=\big[\sin^2(x)+\sin(x)-x\cos(x)\big]u-x(xu)'',$$ -and when we integrate by parts, we obtain -$$(M(u),v)=(u,M(v))+v'(1)u(1)-v(1)u'(1).$$ -These boundary terms have to vanish for $M$ to be a symmetric operator. -Here you already witness the problem of distributing boundary conditions on $u$ and $v$. If we pick $S$ such that $u(1)=u'(1)=0$ for $u\in S$, we only need that $v(1)$ and $v'(1)$ are finite. It seems that choosing $u'(1)=\alpha u(1)$ and $v'(1)=\alpha v(1)$ for some $\alpha\in\mathbb{R}$ is the way to go. -How do we make this rigorous? We typically choose an initial domain $S$ that is a dense subspace of a Hilbert space. In this example, a natural choice for the Hilbert space is $L^2([0,1])$ and a dense subspace on which $M$ can be defined and is symmetric is the subspace -$$S=\big\{u\in W^{2,2}([0,1])\;|\;u(1)=u'(1)=0, \; |u(0)|<\infty \text{, and } |u'(0)|<\infty \big\}$$ -of the Sobolev space $W^{2,2}([0,1])$. -With this choice, the domain of $M^{\ast}$ is -$$\mathcal{D}(M^{\ast})=\big\{v\in W^{2,2}([0,1])\; | \; \max\{|u(0)|,|u'(0)|,|v(1)|,|v'(1)|\}<\infty\big\}.$$ -The next step is to find the deficiency subspaces $\mathcal{K}_{\pm}\doteq \ker(i\mp M^{\ast})$ and deficiency indices $n_{\pm}=\dim \mathcal{K}_{\pm}$ and then apply theorem $X.2$ (due to von Neumann I believe) on p.140 of Reed, M., Simon, B.: Methods of Modern Mathematical Physics. II. Fourier -Analysis, Self-Adjointness. Academic Press Inc., New York (1975), which I quote here for convenience: - -Let $A$ be a closed symmetric operator. The closed symmetric extensions of $A$ are in one-to-one correspondence with the set of partial isometries (in the usual inner product) of $\mathcal{K}_+$ into $\mathcal{K}_-$.* If $U$ is such an isometry with initial space $I(U)\subset\mathcal{K}_+$, then the corresponding closed symmetric extension $A_U$ has domain - $$\mathcal{D}(A_U)=\big\{\phi+\phi_++U\phi_+\big|\phi\in\mathcal{D}(A),\phi_+\in I(U)\big\}$$ - and - $$A_U(\phi+\phi_++U\phi_+)=A\phi+i\phi_+-iU\phi_+.$$ - If $\dim I(U)<\infty$, the deficiency indices of $A_U$ are - $$n_{\pm}(A_U)=n_{\pm}(A)-\dim I(U).$$ - *$\mathcal{K}_{\pm}\doteq \ker(i\mp A^{\ast})$ - -In your case, most likely you will find $n_+=n_-> 0$ and the above theorem will give you all the possible ways to extend the operator $M$ (i.e. its domain) such that it becomes self-adjoint. I expect (but haven't checked) that you find a one-parameter family of operators corresponding to the boundary condition $u'(1)=\alpha u(1)$. -There are some examples after the quoted theorem in the above-quoted book, and maybe this is already sufficient to get you going. Otherwise, please ask and I will try to give further details.<|endoftext|> -TITLE: Riemannian vs Non-Riemannian curvature -QUESTION [12 upvotes]: If you neither know the metric nor the holonomy group, how do you recognize a curvature tensor is Riemannian? -I assume a curvature, by definition, satisfies Bianchi identities. I know it is Riemannian if there exists a symmetric non degenerate tensor $g_{ab}$ such that these satisfy the condition $$g_{ea}R^e{}_{bcd}+g_{eb}R^e{}_{acd}=0 \, ,$$ but its solutions are not unique for the metric (a homogeneous equation). -In $n$ dimensions, these $\frac{n(n+1)}{2}\frac{n(n-1)}{2}$ conditions bring us down from the $\frac{n^2(n+1)(n-1)}{3}$ components of the curvature to $\frac{n^2(n+1)(n-1)}{3\cdot 2 \cdot 2}$ of the Riemann Tensor. Yet it seems to me these equations don't tell me what the metric should be. - -REPLY [13 votes]: NB: In what follows, to save typing, I will be working on a manifold $M$, but I will write $T$, $T^*$, etc. to denote the bundles $TM$, $T^*M$, etc. and let $M$ be understood. -It seems that the OP wants to be able to test whether a $(1,3)$-tensor $R$ with the symmetries of a Riemann curvature tensor is actually the curvature of a Riemannian metric. By the symmetries of the Riemann curvature tensor, we mean that $R$ is a section of $T\otimes T^*\otimes \Lambda^2T^*$ that is in the kernel $B$ of the natural skew-symmetrization mapping -$$ -T\otimes T^*\otimes \Lambda^2T^* \longrightarrow T\otimes \Lambda^3T^*. -$$ -There are local and global aspects of this problem, even in dimension $2$, and there are degeneracy problems that show up when the tensor is allowed to vanish at some places. -For example, the OP's response to Ben's comment about the vanishing of the curvature at a point doesn't quite make sense because there is no connection specified in the problem, just the tensor $R$, so there is no parallel transport available to transport either $g$ or $R$ to other places. -Of course, the first algebraic condition is that there must be a nondegenerate section $g$ of $S^2T^*$, for which the natural 'index-lowering' pairing $\langle g, R\rangle$ (which, a priori takes values in $T^*\otimes T^*\otimes \Lambda^2 T^*$) takes values in $\Lambda^2T^*\otimes\Lambda^2T^*$. As the OP remarks, this condition does not determine a $g$ uniquely, since, for example, one can always replace such a $g$ by a multiple of $g$. -The existence of any such nondegenerate $g$ does put algebraic conditions on $R$, though, since, for example, it implies that $R$ must take values in $(T\otimes T^*)_0\otimes \Lambda^2 T^*$, where $(T\otimes T^*)_0\subset T\otimes T^*$ is the space of endomorphisms of trace $0$. Call the corresponding subbundle $B_0\subset B$. Of course, there are further conditions, in order that there be a nondegenerate $g$ for which $\langle g, R\rangle$ takes values in $\Lambda^2T^*\otimes\Lambda^2T^*$, even in dimension $n=2$. -For example, in dimension $2$, if $R$ is a nonzero section of $B_0$ (which has rank $3$) and $R$ happens to take values in the subset $N\otimes \Lambda^2T^*$, where $N\subset (T\otimes T^*)_0$ is the set of (nonzero) nilpotent endomorphisms, then there is no nondegenerate $g$ satisfying the given condition. On the other hand, if $R$ is a nonzero section of $B_0$ and nowhere takes values in $N\otimes \Lambda^2 T^*$, then, locally, there always exists a $g$ whose Riemann curvature tensor is $R$. In fact, the local solutions depend on $2$ functions of one variable. However, there still might not be any global solutions. For example, if $g_0$ is the standard metric on the $2$-sphere with Gauss curvature $+1$ and $R_0$ is its $(1,3)$-curvature tensor, then $-R_0$ cannot be the curvature tensor of any nondegenerate $g$ globally defined on $S^2$. -However, the real problems start when $n=3$. If $R$ is a section of $B_0$, then we can form its space of curvature endomorphisms $R(X,Y)$ where $X$ and $Y$ are vector fields. These, of course, take values in $(T\otimes T^*)_0$, and, at each point $p\in M$, they span a subspace $E(R_p)\subset (T_p\otimes T^*_p)_0$ of dimension at most $3$. Suppose that $R$ satisfies the generic condition that this span has dimension $3$ at every point $p$. Then, at each point $p$, the subspace $E(R_p)$ must span a simple Lie algebra of dimension $3$ in $(T\otimes T^*)_0$, either isomorphic to $\mathfrak{so}(3)$ or $\mathfrak{so}(2,1)$. This is a further pointwise algebraic condition on $R$. Moreover, in either of these cases, -the space of 'compatible' nondegenerate quadratic forms $g$ that satisfy the condition that $\langle g, R\rangle$ take values in $\Lambda^2T^*\otimes\Lambda^2T^*$ are all multiples of a single nondegenerate quadratic form $g_0$ (of signature $(3,0)$ or $(2,1)$. Thus, the problem you are trying to solve is whether there is a nonzero function $u$ so that the nondegenerate quadratic form $g = u g_0$ has $R$ as its curvature. This is a wildly overdetermined problem, with at most a finite-dimensional space of solutions even locally. Thus, even with the algebraic necessary condition on $E(R)$, you cannot expect to solve this problem, even locally, although determining the existence of a solution is essentially reduced to an ODE problem that can be solved algorithmically. I can supply details if you are interested. -There are more subtle cases, when the rank of $E(R)$ at each point is $1$ or $2$, -but, again, since the explicit analysis is long, let me save that unless there is interest. -For $n>3$, there are, again, many further algebraic conditions and then differential conditions involved in the test for when a $(1,3)$-tensor satisfying the symmetries of a Riemann curvature tensor is actually a Riemann curvature tensor, but the calculations rapidly become unmanageable, though they are theoretically doable.<|endoftext|> -TITLE: Why is the motivic category defined over the site of smooth schemes only? -QUESTION [30 upvotes]: Fix a base scheme $S$. Stable and unstable motivic categories over $S$ are defined as certain categories of higher stacks on the Nisnevich site $Sm_S$ of smooth schemes over $S$. Why smooth? -As a casual observer, I see two broad classes of possible reasons: - -Technical reasons: Perhaps the Nisnevich topology doesn't make sense over non-smooth schemes, or maybe $\mathbb{A}^1$ or $\mathbb{G}_m$-locality become more complicated over non-smooth schemes. -Goals of the construction: Perhaps the stable and unstable categories make perfect sense over non-smooth schemes, but maybe motivic cohomology or algebraic K-theory simply don't extend to stacks over the resulting category. - -In either case, I wonder: is it clearly impossible, by some modification of the construction of these categories, to get things to work over non-smooth schemes? Would this be desireable? - -REPLY [25 votes]: It's worth noting first that smooth schemes are essentially the smallest possible category from which one can define the motivic homotopy category: to make sense of $\mathbb A^1$-homotopy and of the Nisnevich topology you need étale extensions of affine spaces in your category, and every smooth $S$-scheme is (Zariski) locally such. -That said, I can think of two fundamental places in the theory where smoothness is crucial, both of which also showcase the relevance of the Nisnevich topology and of $\mathbb A^1$-homotopy. -(1) The first is the localization property already addressed in Adeel's answer, which is itself crucial for many things, such as the proper base change theorem. -A characterizing property of henselian local schemes $S$ (which are the points of the Nisnevich topology) is that for $f:X\to S$ étale every section of $f$ over the closed point $S_0\subset S$ lifts uniquely to a section of $f$ over $S$ ("Hensel's lemma"); this is what makes localization work for sheaves on the small Nisnevich or étale sites. If $f: X \to S$ is smooth, it is still the case that every section $s_0:S_0 \hookrightarrow X$ lifts to $S$ (this uses smoothness in an essential way), but not uniquely. However, once a lift $s:S\hookrightarrow X$ has been chosen, then $X$ can be presented as an étale neighborhood of $S$ in the normal bundle of $s$, and it follows that the Nisnevich sheafification of the "space of lifts" of $s_0$ to $S$ is $\mathbb A^1$-contractible. Thus, in some precise sense, lifts are still unique up to $\mathbb A^1$-homotopy. This is the proof of the Morel-Voevodsky localization theorem in a nutshell. -Another nice consequence of the localization property is that fields form a "conservative family of points" in motivic homotopy theory (at least for the $S^1$-stable theory, though one can also say something unstably), something that could not be achieved using a Grothendieck topology alone. -(2) The second is Cousin/Gersten complexes. This is now specific to the case of a base field $k$, in which localization is useless. The key input here is a geometric presentation lemma of Gabber, a statement of which can be found as Lemma 15 in the introduction to Morel's book (http://www.mathematik.uni-muenchen.de/~morel/Prepublications/A1TopologyLNM.pdf). A consequence of this lemma is that every $\mathbb A^1$-invariant Nisnevich sheaf $F$ (of spaces or spectra, say) is "effaceable" on smooth $k$-schemes, in the sense of Colliot-Thélène-Hoobler-Kahn (https://webusers.imj-prg.fr/~bruno.kahn/preprints/bo.dvi). This implies that the coniveau spectral sequence degenerates at $E_2$ and hence that the Cousin complex -$$ 0 \to \pi_nF(X) \to \bigoplus_{x\in X^{(0)}} \pi_{n}F_x(X_x) \to \bigoplus_{x\in X^{(1)}} \pi_{n-1}F_x(X_x) \to \dots $$ -is exact when $X$ is smooth local. Here, $X^{(n)}$ is the set of points of codimension $n$, $X_x=\operatorname{Spec}(\mathcal O_{X,x})$, and $F_x(X_x)$ is the homotopy fiber of the restriction map $F(X_x) \to F(X_x-x)$. -These Cousin complexes are the basis for many computations in motivic homotopy theory, for example in the above-mentioned work of Morel. They can perhaps be viewed as a replacement for cell decompositions in topology. -So, in summary, the smooth/Nisnevich/$\mathbb A^1$ combo simultaneously allows us to (1) reduce questions over general base schemes to the case of base fields via localization, and (2) perform interesting computations in the latter case.<|endoftext|> -TITLE: Infinite cuboids In space -QUESTION [8 upvotes]: Given an arbitrary $X \subseteq \mathbb{R}^3$, can we always find infinite sets $A, B, C \subseteq \mathbb{R}$ such that either $A \times B \times C \subseteq X$ or $A \times B \times C \subseteq (\mathbb{R}^3 \setminus X)$? Here $\mathbb{R}$ denotes the real line. -I believe that the answer should be well known (and positive?), but I don't see it. - -REPLY [13 votes]: This is Problem 28 in the paper Unsolved problems in set theory of Erdos and Hajnal https://old.renyi.hu/~p_erdos/1971-28.pdf (at least if CH holds). In an 1982 paper https://old.renyi.hu/~p_erdos/1982-24.pdf Erdos reports that Mills and Prikry solved it negatively, but I could not locate their paper.<|endoftext|> -TITLE: If the homotopy category is well-generated, must the $\infty$-category be presentable? -QUESTION [8 upvotes]: Suppose $\mathcal{C}$ is a stable $\infty$-category whose homotopy category is a well-generated triangulated category in the sense of Neeman's book. Must $\mathcal{C}$ be a presentable $\infty$-category? -Rosicky proved that if $\mathcal{M}$ is a combinatorial stable model category then $Ho(\mathcal{M})$ is well-generated, so I'm asking for a kind of converse, and I'm happy for an answer either about model categories being combinatorial or about $\infty$-categories being presentable. -The starting point is probably trying to figure out whether or not the isomorphisms of $Ho(\mathcal{M})$ form an accessible subcategory of the arrow category, which seems like something you'd need (since in order for $\mathcal{M}$ or $\mathcal{C}$ to be presentable you need to know the weak equivalences are an accessible subcategory of the arrow category, see A.2.6.6 of Lurie's HTT). It feels like the sort of thing that should be true, but I also don't know if it's enough to then deduce $\mathcal{C}$ is presentable (though Lurie's A.2.6.9 is surely relevant). - -REPLY [5 votes]: I don't think this is actually fully answered by HA 1.4.4.2, which says that a stable $\infty$-category is presentable if and only if it's got coproducts (hence is cocomplete) and a $\kappa$-compact generator together with a locally small homotopy category (so that it's locally small.) That last condition sounds as if it would mean the homotopy category was $\kappa$-well generated, but there's no clear implication between being $\kappa$-compact in a stable $\infty$-category and being $\kappa$-compact in its underlying triangulated category. That's because being $\kappa$-compact in the homotopy category means being in the largest class of $\kappa$-small objects that is "$\kappa$-perfect" in the sense of Neeman, and it's only $\kappa$-smallness that has any clear connection to $\kappa$-filtered colimits. -However, a nearby claim is known, and was proved via a different model by Heider in his paper Two results from Morita theory of stable module categories. Heider shows that the homotopy category of a spectral model category is well generated if and only if it's triangulated equivalent to the localization of the derived category of a small spectral category at a set of objects. The derived category of a small spectral category models an arbitrary finitely presentable stable $\infty$-category, and its localizations at small sets model arbitrary accessible localizations, thus arbitrary $\alpha$-presentable stable $\infty$-categories. -Now, I'm pretty sure it's unknown precisely which stable $\infty$-categories underlie spectral model categories, so this doesn't quite answer the original question. I think it would be possible to imitate Heider's proof in general, though. Given an $\alpha$-compact generator $\mathcal G$ of the homotopy category of a stable $\infty$-category $\mathcal T$, seen as a full additive subcategory, take a small spectral category $\mathcal E$ with homotopy coherent nerve equivalent to the full subcategory of $\mathcal T$ spanned by the objects of $\mathcal G$. Then as far as I can tell, Heider's argument goes through without change to exhibit $\mathcal T$ as (the homotopy coherent nerve of) an $\alpha$-accessible localization of $D(\mathcal E^{\mathrm{op}})$.<|endoftext|> -TITLE: Is Deligne cohomology the motivic cohomology of analytic spaces? -QUESTION [12 upvotes]: Let $X$ be a smooth projective complex analytic space. -We can cook up a complex analytic version of Bloch's cycle complex by declaring -$z^n(X^{\rm an}, m)$ -is the free abelian group on all codimension $m$ analytic cycles on $X\times\Delta^n$ ($\Delta^n$ being the usual standard $n$ simplex in complex analytic spaces, ie. the spectrum of $\mathbf{C}\{u_0,\ldots, u_n\}/(u_0+\ldots+u_n -1)$) in good position (ie. intersecting every face in the appropriate codimension, as in Bloch's paper). The differential $d_m$ is the same as in Bloch's original definition, turning $(z^n(X^{\rm an}, m), d_m)$ into a complex of abelian groups. -Call $$\mathbf{Z}(n)_{\mathcal{M}} := (z^n(X^{\rm an}, m), d_m)[2m]$$ -and its hypercohomology "motivic cohomology of $X$". -Here's the question. Is motivic cohomology of $X$ at all related to the Deligne cohomology of $X$? More optimistically, does there exist a quasi-isomorphism -$$\mathbf{Z}(n)_{\mathcal{M}}\to\mathbf{Z}(n)_{\mathcal{D}} ?$$ -How should one think about Deligne cohomology, in other words? (if not as "the motivic cohomology of complex analytic spaces?) -Remarks -I can imagine a regulator map $\text{reg} : \mathbf{Z}(n)_{\mathcal{M}}\to\mathbf{Z}(n)_{\mathcal{D}}$ can be defined using currents, as done for the classical regulator. -This is for sure going to be a (rather uninteresting) quasi isomorphism, since $X$ is smooth, when $n = 0$. -For $n = 1$ this is likely going to be a quasi-isomorphism too (if one doesn't screw the definition of $\text{reg}$): both sides are just $\mathbf{G}_m[-1]$. - -REPLY [7 votes]: The answer is no. -So far, you have checked only very special cases, such as weight $n$ and degree $2n$, for $n = 0,1$. -Consider the analytic hypercohomology of your $\mathbf{Z}(n)_{\mathcal{M}}$ in degree $2n$, denoted $H^{2n}_{\rm an}(X,\mathbf{Z}(n))$. The same (elementary) argument as in several references (eg. Bloch's "Algebraic Cycles and Higher $K$-Theory", 1986) shows -$$H^{2n}_{\rm an}(X,\mathbf{Z}(n)) \simeq\text{CH}^n(X^{\rm an})$$ -the Chow group of analytic cycles on $X^{\rm an}$. By GAGA, there is a canonical isomorphism of abelian groups $\text{CH}^n(X)\simeq\text{CH}^n(X^{\rm an})$. -Upon running the same construction of the cycle map to Deligne cohomology as in Kerr-Lewis, as you suggest, you must end up with a cycle map -$$\text{CH}^n(X^{\rm an})\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$$ -that, under the above isomorphism from GAGA, should better agree with the usual cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$. -Rem. If your conjecture were true, then, in particular, the cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$ should be surjective, this latter group being surjecting onto the finitely generated abelian group of Hodge cycles $\text{Hdg}^{n,n}(X^{\rm an})$, the composition being the cycle map to Betti cohomology. As a consequence, the Hodge conjecture is implied by your contention. -There are two problems. I am not aware of any expectation along the lines of surjectivity of the cycle map onto Deligne(-Beilinson) cohomology. Rather, the only claim being usually made is surjectivity onto the abelian group of Hodge classes. -On the other hand, if your contention were correct, then the cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$ should be injective too, which rarely happens, as this puts strong constraints on the Hodge numbers of $X^{\rm an}$. -A paper of Esnault and Levine proves that if such cycle map is injective, then it is also surjective, and implies strong conditions on the shape of the Hodge diamond of $X^{\rm an}$. There are many examples of an $X$ that does not meet such conditions.<|endoftext|> -TITLE: The number of representations of an integer as the inner product of integral lattice points -QUESTION [13 upvotes]: I was looking through some old notes of mine and I came across a couple lemmas/identities I wrote down in regards to a question I asked about four years ago. In particular I wrote that for an arbitrary fixed integer $k>1$ we have the following asymptotic expansion: -$$\Psi_k(n)=\left|\{(\mathbf{u},\mathbf{v})\in \mathbb{N}^k\times \mathbb{N}^k:\mathbf{u}\cdot\mathbf{v}=n\}\right|\sim \frac{\sigma_{k-1}(n)\text{log}(n)^{k}}{\zeta(k)(k-1)!}$$ -With $\sigma_{k-1}(n)=\sum_{d\mid n}d^{k-1}$ and $\zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^k}$ the Riemann zeta function. Now alternatively if one expands the inner product of $\mathbf{u}\cdot\mathbf{v}$ we see $\Psi_k$ can be expressed in a number of other ways: -$$\Psi_k(n)=\sum_{\substack{\mathbf{u}\cdot\mathbf{v}=n\\(\mathbf{u},\mathbf{v})\in \mathbb{N}^k\times \mathbb{N}^k}}1=\left|\{(u_1,v_1,\ldots u_k,v_k)\in \mathbb{N}^{2k}:n=u_1v_1+\cdots +u_kv_k\}\right|\\=\sum_{\substack{m_1+m_2+m_3+\cdots +m_k=n\\(m_1,m_2,m_3,\ldots ,m_k)\in \mathbb{N}^k}}d(m_1)d(m_2)d(m_3)\cdots d(m_k)$$ -Where $d(m)=\sum_{d\mid m}1=\sigma_{0}(m)$ counts the divisors of any natural number $m$. Which unearths a number of $q$-series esque representation for the ordinary generating function of $\Psi_k$, for example: -$$\sum_{n=1}^{\infty}\Psi_k(n)q^n=\left(\sum_{n=1}^{\infty}q^{n^2}\frac{1+q^n}{1-q^n}\right)^k=\frac{\text{log}(1-q)^k}{\text{log}(q)^k}\sum_{j=0}^k\binom{k}{j}\left(\frac{\psi_{q}(1)}{\text{log}(1-q)}\right)^j$$ -Where $\psi_{q}(z)=\frac{1}{\Gamma_{q}(z)}\frac{d}{dz}\Gamma_{q}(z)$ is the $q$-analog of the digamma function defined analogously in terms of the $q$-gamma function, expressible as $\Gamma_{q}(z)=(1-q)^{1-z}\prod_{n=0}^{\infty}\frac{1-q^{n+1}}{1-q^{n+z}}$ for $|q|<1$. - -Now working with just some of these alternate representations, as well as fiddling with the order of the summands involved I was able to prove by induction on the integer $k>1$ that we have both: -$$\sum_{n\leq N}\Psi_k(n)=\frac{N^k\text{log}(N)^{k}}{k!}+\mathcal{O}(N^k\text{log}(N)^{k-1})$$ -$$\sum_{n\leq N}\frac{\sigma_{k-1}(n)\text{log}(n)^{k}}{\zeta(k)(k-1)!}=\frac{N^k\text{log}(N)^{k}}{k!}+\mathcal{O}(N^k\text{log}(N)^{k-1})\\$$ -So using the same heuristics as before it seems reasonable that $\Psi_k(n)\sim \frac{\sigma_{k-1}(n)\text{log}(n)^{k}}{\zeta(k)(k-1)!}$ which at least for the case at $k=2$ would coincide with the answer to my previous question. However I'm unable to find a concrete proof of this result and would therefore appreciate any help in the matter. - -REPLY [12 votes]: The asymptotic formula is true for even dimensions $k\geq 2$. We can prove this by induction on $k$, inspired by Rodrigo's observation on Eisenstein series. -The case $k=2$ is classical and addressed in the OP's previous post that he linked. Now it suffices to show that if the formula is true for even dimensions $k,\ell\geq 2$, then it is also true for dimension $k+\ell$. Moreover, we can assume in this implication that either $k,\ell\geq 4$ or $k+\ell\in\{4,6\}$. Let us start from the obvious identity -$$ \Psi_{k+\ell}(n)=\sum_{r+s=n}\Psi_k(r)\Psi_\ell(s).$$ -Using the fact that the logarithm is a slowly changing function, we are left with proving that -$$ \frac{\sigma_{k+\ell-1}(n)}{\zeta(k+\ell)\Gamma(k+\ell)}\sim\sum_{r+s=n}\frac{\sigma_{k-1}(r)}{\zeta(k)\Gamma(k)}\cdot\frac{\sigma_{\ell-1}(r)}{\zeta(\ell)\Gamma(\ell)}.$$ -If $k,\ell\geq 4$, then the left hand side is $(2\pi i)^{-k-\ell}$ times the $n$-th Fourier coefficient of the standard holomorphic Eisenstein series $E_{k+\ell}$ of weight $k+\ell$ and full level, while the right hand side is $(2\pi i)^{-k-\ell}$ times the $n$-th Fourier coefficient of $E_kE_\ell$. As $E_{k+\ell}-E_kE_\ell$ is a cusp form, the result follows from standard (or even weaker) upper bounds for the Fourier coefficients of cusp forms. -If $k+\ell\in\{4,6\}$, then the result follows from the explicit identities -$$5\sigma_3(n)+(1-6n)\sigma_1(n)=12\sum_{r+s=n}\sigma_1(r)\sigma_1(s),$$ -$$21\sigma_5(n)+(10-30n)\sigma_3(n)-\sigma_1(n)=240\sum_{r+s=n}\sigma_1(r)\sigma_3(s).$$ -These identities can also be proved with the help of Eisenstein series, or by elementary means, see e.g. (3.10) and (3.12) in the paper of Huard et al. -The proof is complete.<|endoftext|> -TITLE: Examples of integer sequences coincidences -QUESTION [33 upvotes]: For the time being, the OEIS website contains almost $300000$ sequences. Each of these sequences is the mark of a specific mathematical concept. Sometimes two (or more) distinct concepts have the same mark, which suggests a connection between a priori independent mathematical areas. The most famous example like that is perhaps the Catalan numbers sequence: A000108. -Question: What are the examples of pair of integer sequences coinciding on all the known terms, but for which the coincidence for all the terms is unknown? -Cheating is not allowed. By cheating I mean artificial examples like: - $u_n = v_n =n$ for $n \neq 10$, and if RH is true then $u_{10} = v_{10} = 10$, else $u_{10}+1 = v_{10} = 1$. -The existence of an OEIS entry could act as safety. -EDIT: I would like to point out that all the answers below are about pair of integer sequences which were already conjectured to be the same, and of course they are on-topic (and some of them are very nice). Note that such examples can be found by searching something like "conjectured to be identical" on OEIS, as I did for some of my own examples below... -Now, a more surprising kind of answer would be a (non-cheating) pair of integer sequences which are the same on the known entries, but for which there is no evidence a priori that they are the same for all the entries or that they are related (i.e. the precise meaning of a coincidence). Such examples, also on-topic, could reveal some unexpected connections in mathematics, but could be harder to find... - -REPLY [3 votes]: Another example I know is A191363, i.e., numbers $n$ such that $\sigma(n) = 2n - 2$. According to OEIS all known terms are $(a_k-1)a_k/2$ where $a_k$ is the sequence of Fermat primes (A019434).<|endoftext|> -TITLE: Any simple concrete proof of Faltings theorem? -QUESTION [9 upvotes]: Are there simple proofs of some concrete special cases of Faltings's theorem? Any help would be appreciated. - -REPLY [3 votes]: Recently Lawrence and Venkatesh gave a proof that uses Faltings' setup but looks at variation of p-adic Galois representations in a family of algebraic varieties. Neither simpler nor a concrete special case, though.<|endoftext|> -TITLE: Toda Hierarchy and Quantum Cohomology of $\mathbb{P}^1$ Frobenius manifolds -QUESTION [6 upvotes]: People usually say that the quantum cohomology of $\mathbb{P}^1$ Frobenius manifold $QH^*(\mathbb{P}^1)$, corresponds to dispersionless extended Toda hierarchy (e.g. page 6 of https://arxiv.org/pdf/math/0308152.pdf). I'm trying to understand this correspondence in more detail but I'm some having troubles. More specifically how do I recover the $\tau$-function? Should I be able to get the Frobenius potential $F$ of $QH^*(\mathbb{P}^1)$ from the $\tau$-function by -\begin{equation} -F = \frac{1}{2}(t^1)^2t^2 + e^{t^2} = \lim_{\epsilon \rightarrow 0}\epsilon^2\log \tau |_{t^{\alpha,p > 0} = 0, t^\alpha := t^{\alpha,0}} ? -\end{equation} -(Where $\epsilon \rightarrow 0$ is the dispersionless limit. I'm sort of getting this from Theorem 6.1 of https://arxiv.org/pdf/hep-th/9407018.pdf, but the result is for KdV). If I didn't set $t^{\alpha,p > 0} = 0$, should I get the Gromov-Witten potential of $\mathbb{P}^1$ with descendences from $\lim_{\epsilon \rightarrow 0}\epsilon^2\log \tau$? -Here is what I have gathered so far (mainly from https://arxiv.org/pdf/nlin/0306060.pdf), please correct me if I misunderstood something. The extended Toda hierarchy is given by the bihamiltonian structure on the loop space $\mathcal{L}(\mathbb{R}^2)$: -\begin{align*} -\{u(x),u(y)\}_1 = \{v(x), v(y)\}_1 &= 0, \\ -\{u(x), v(y)\}_1 &= \frac{1}{\epsilon}(e^{\epsilon \partial_x} - 1)\delta(x - y),\\ -\{u(x), u(y)\}_2 &= \frac{1}{\epsilon}(e^{\epsilon \partial_x}e^{v(x)} - e^{v(x)}e^{-\epsilon\partial_x})\delta(x-y)\\ -\{u(x),v(y)\}_2 &= \frac{1}{\epsilon}u(x)(e^{\epsilon\partial_x} - 1)\delta(x-y)\\ -\{v(x),v(y)\}_2 &= \frac{1}{\epsilon}(e^{\epsilon\partial_x} - e^{-\epsilon\partial_x})\delta(x-y). -\end{align*} -From this, I'm guessing some sort of Miura transformation can be used to put $\{.,.\}_2 - \lambda\{.,.\}_1$ into the normal form and from that we get a set of Hamiltonian densitites $\{h_{\alpha,p} = h_{\alpha,p}(x,u,v,u_x,v_x,...)\}$. The Hamiltonians are given by $H_{\alpha, p} := \int_0^{2\pi}h_{\alpha,p}dx$. The $t^{\alpha,p}$-evolution is -\begin{equation} -\frac{\partial w^\alpha}{\partial t^{\beta,p}} = \{w^\alpha,H_{\beta,p}\}_1, \qquad w = u,v;\ \alpha,\beta = 1,2;\ p \geq 0. -\end{equation} -All of these came from Theorem 3.1 of https://arxiv.org/pdf/nlin/0306060.pdf. From here we can use the standard result of biharmiltonian theory that $\partial h_{\alpha,p-1}/\partial t^{\beta,q} = \partial h_{\beta,q-1}/\partial t^{\alpha,p} =: \frac{1}{\epsilon}(e^{\epsilon\partial_x} - 1)\Omega_{\alpha,p;\beta,q}$, therefore there is a function $\log \tau$ such that -\begin{equation} -\Omega_{\alpha,p;\beta,q} = \epsilon^2\frac{\partial^2\log \tau}{\partial t^{\alpha,p}\partial t^{\beta,q}}. -\end{equation} -Ideally, I would like to calculate $\log\tau$ from here and directly and answer my question above. However, the explicit form of $h_{\alpha,p}$ as given in Theorem 3.1 is complicated and I'm not sure how to proceed. Also, I have never seen anyone written down the equation $F = \lim_{\epsilon \rightarrow 0}\epsilon^2\log \tau |_{t^{\alpha,p > 0} = 0, t^\alpha := t^{\alpha,0}}$ explicitely anywhere, so maybe it could be wrong. Could anyone give me some advices or correct my understand or provide me with a suitable reference? - -REPLY [3 votes]: I would say that basically everything you wrote is correct, and in particular the equation $F=\lim_{\epsilon\to 0} \epsilon^2 \log \tau|_{t^{\alpha,p>0}=0,t^{\alpha,0}=t^\alpha}$. -It is true that, more in general, $\epsilon^2 \log \tau$ gives you the total descendant potential $\mathcal F(t^{*,*};\epsilon)$ at all genera. -Notice that you don't even need to specify which coordinates you are using (the Miura transformation you write about), because no dynamical variable appears here, only the time variables $t^{\alpha,p}$. -However the correct equation for the tau structure $h_{\alpha,p}$ is -$$\frac{\partial h_{\alpha,p-1}}{\partial t^{\beta,q}} = \frac{\partial h_{\beta,q-1}}{\partial t^{\alpha,p}} = \partial_x \Omega_{\alpha,p;\beta,q}$$ -(the dispersive Poisson structure only enters this equation through the time derivatives $\partial_{t^{\alpha,p}}$). This equation relates to the tau function / total descendant potential via -$$\left. \frac{\partial h_{\alpha,p-1}}{\partial t^{\beta,q}}\right|_{w^\gamma_k=(w^{\mathrm{top}})^\gamma_k(x=0,t^{*,*};\epsilon)} = \frac{\partial^3 \mathcal F}{\partial t^{1,0} \partial t^{\alpha,p} \partial t^{\beta,q}},$$ -where $w^\mathrm{top}$ is the topological solution, with initial datum $(w^\mathrm{top})^\gamma(x,t^{*,*}=0;\epsilon) = x \delta^\gamma_1$ (here $w^\gamma$ are the normal coordinates $w^\gamma=\eta^{\gamma \mu}h_{\mu,-1}$ of the tau structure). -About your question on the explicit computation of the tau function starting from the hierarchy and its tau structure, you are right to say it is not a trivial exercise. The best way I can think of is to use the explicit relation between the Dubrovin-Zhang tau structure and the Lax representation of the extended Toda system. You can find this information scattered in the literature, including the paper of Carlet, Dubrovin and Zhang you already cited. -However we also gathered this information in section 6.1 of our paper with A. Buryak, https://arxiv.org/abs/1411.6797. It is about the double ramification hierarchy, but since it also deals with its relation to the Dubrovin-Zhang construction, you find the change of coordinates between each of the tree representations: Lax representation, DR hierarchy and DZ hierarchy. A word of warning: the Dubrovin-Zhang hierarchy comes in two version, with ancestors or descendants psi-classes. You need to use Givental's $S$-matrices as explained in the paper to pass from one to the other. -Finally, I gathered this example and several others (including ILW and Gelfand-Dickey) at the end of my review https://arxiv.org/abs/1703.00232. That too is about the double ramification hierarchy, but again you will find the change of coordinates between Lax representation, DR hierarchy and DZ hierarchy. All those formulae come from our papers with Buryak, Dubrovin and Guéré, so please look into the references therein to find out more.<|endoftext|> -TITLE: Do $\mathbb{A}^1-S$ and $\mathbb{A}^1-\{0,1\}$ have a finite etale cover in common? -QUESTION [15 upvotes]: We work over the field of complex numbers. (But remarks in characteristic $p$ are very welcome.) -Let $S$ be a finite set of points in $\mathbb{A}^1$ containing $0$ and $1$. [Edit: Assume $S$ contains only algebraic numbers.] - - -Do $\mathbb{A}^1-S$ and $\mathbb{A}^1-\{0,1\}$ have a finite etale cover in common? - - -That is, does there exist a curve $X$, a finite etale morphism $X\to \mathbb{A}^1-S$ and a finite etale morphism $X\to \mathbb{A}^1-\{0,1\}$? -For which $S$ is the answer positive? - -REPLY [14 votes]: The answer is positive if and only if $\mathbb{A}^1\setminus S$ is an arithmetic curve, i.e., $\pi_1(\mathbb{A}^1\setminus S)\subset \mathrm{Aut}(\mathbb{H}) = PSL_2(\mathbb{R})$ is an arithmetic subgroup. -This however does not happen "very often". Let me be more precise. -Note that the Euler characteristic of $\mathbb{A}^1\setminus S$ equals $1-\# S$. -For any fixed integer $e$, there are only finitely many isomorphism classes of arithmetic curves $X$ with Euler characteristic $e$ by Takeuchi's theorem; see Theorem 2.1 in https://projecteuclid.org/download/pdf_1/euclid.jmsj/1230396454 -Thus, if you fix an integer $n$, there are only finitely many $\mathbb{A}^1\setminus S$ with $\# S = n$ which share a common finite etale cover with $\mathbb{A}^1\setminus \{0,1\}$. -In characteristic $p>0$, the answer is positive (over any field $k$). Indeed, let $k$ be a field of characteristic $p>0$. Then, by Prop. 5.2 in Achinger's paper Wild ramification and $K(\pi,1)$-spaces, $\mathbb{A}^1_k\setminus S$ is a finite etale cover of $\mathbb{A}^1_k$. Let $f:\mathbb{A}^1_k\setminus S \to \mathbb{A}^1_k$ and $g:\mathbb{A}^1_k\setminus\{0,1\} \to \mathbb{A}^1_k$ be finite etale maps. Now define $X$ to be (a connected component of) the fibre product $\mathbb{A}^1_k\setminus S\times_{f,\mathbb{A}^1_k,g} \mathbb{A}^1_k\setminus\{0,1\}$. This comes with finite etale maps to $\mathbb{A}^1_k\setminus S$ and $\mathbb{A}^1_k\setminus \{0,1\}$.<|endoftext|> -TITLE: To what extent does a torsor determine a group -QUESTION [12 upvotes]: Let $k$ be a field, and suppose $G$ is a group-scheme over $k$ (I am happy to assume that $k=\mathbb{Q}$ and that $G$ is affine). A $G$-torsor over $k$ is a non-empty $k$-scheme $T$ equipped with an action $a:G\times T\to T$, such that $(a,\pi_2):G\times T\to T\times T$ is an isomorphism. There is an induced morphism -$$ -\begin{align*} -T\times T\times T&\to T,\\ -(t_1,t_2,t_3)&\mapsto a\bigg(\pi_2\big((a,\pi_2)^{-1}(t_1,t_2)\big),t_3\bigg)=``t_1t_2^{-1}t_3". -\end{align*} -$$ -I would like to know to what extent the map $T\times T\times T\to T$ determines $G$. I mean the following: - -For which pairs of groups $G$ and $G'$ does there exist a $G$-torsor $T$, a $G'$-torsor $T'$, and an isomorphism $T\cong T'$ compatible with the maps $T\times T\times T\to T$ and $T'\times T'\times T'\to T'$? - -REPLY [11 votes]: Under fairly general exactness assumptions on the ambient category, one can perform the construction of a group from $(T,m)$, if $T$ is an object equipped with a heap structure $m:T^3\to T$, i. e. an associative Maltsev operation. That is, $m$ must just satisfy (diagrammatic versions of) $m(x,x,y)=m(y,x,x)=y$ and $m(m(x,y,z),t,u)=m(x,y,m(z,t,u))$. -This is done exactly as one does construct a vector space from an affine space: -one gets a group as a quotient $T\times T\twoheadrightarrow G$ (which becomes the division map of the torsor $T$ over $G$, $(x,y)\mapsto x/y$ where $x/y$ is the unique element of $G$ with $(x/y)y=x$). -The quotient is by identifying $(x,y)$ with $(m(x,y,z),z)$ for all $x,y,z$. That is, one forms the coequalizer $T^3\rightrightarrows T^2\twoheadrightarrow G$ of appropriate morphisms. -The action of $G$ on $T$ is determined by $(x/y)\cdot z=m(x,y,z)$, and gives a $G$-torsor. It is an easy exercise to do this "without elements". Moreover any other group over which $T$ is a torsor and such that $m(x,y,z)$ is $(x/y)\cdot z$ for that structure too, is isomorphic to $G$, by an isomorphism compatible with the actions on $T$. -In fact switching sides one also has another group $G'$ such that $T$ is a $G$-$G'$-bitorsor. -This even works without global support assumption: defining support $S$ of $T$ as the coequalizer $T\times T\rightrightarrows T\twoheadrightarrow S$ of the projections $T\times T\rightrightarrows T$, one gets a group and a torsor structure in the slice category over $S$. -There are way too many references to choose from, spread across decades, so I think all this must be considered folklore.<|endoftext|> -TITLE: Complexity for calculating number of Perfect Matchings in k-regular hypergraph -QUESTION [6 upvotes]: Let $G(V,E)$ be a unweighted, k-regular hypergraph, with vertices $V=(v_1, ... v_n)$ and edges $E=(e_1, ... e_m)$. The k-regularity leads to $|e_i|=k$ (i.e. every edge contains exactly $k$ vertices). -A perfect matching (PM) is a subset of $E$, such that every vertex $v_i$ is contained exactly one time. -A 2-regular hypergraph is a usual graph. To get the number of PMs, one calculates the Hafnian of the corresponding $n \times n$ adjacency matrix A. Calculating the Hafnian is in the complexity class $\#P$. For bipartite 2-regular graphs, one can calculate the permanent of the adjacency matrix via the Ryser formular. $perm(A)$ can be evaluated in $O(2^{n-1}n^2)$ steps. -For $k>2$, the hypergraphs can be written as incidence matrix or high-dimensional $n \times n \times ... \times n$ adjacency matrix A. -My three questions: - - -With which algorithm can the number of perfect matchings be calculated -for unweighted k-regular graphs, and what is its runtime? - -With which algorithm can the number of perfect matchings be calculated -for a unweighted hypergraph with $|e_i|\leq k$, and what is its runtime? - -With which algorithm can the number of perfect matchings be calculated -for k-regular graphs with complex weights, and what is its runtime? - - - -Any answer or link to literature would be highly appreciated. - -REPLY [4 votes]: In the literature this problem also goes by "set packing" which can help find references. The set-up in this language is given a universe $V$ of $n$ elements and family of subsets $E$ a packing is a collection of mutually disjoint sets from $E$ (i.e. a matching). A $t$-packing is a packing consisting of $t$ sets. -The problem of counting $t$-packings in the $k$-uniform case where each element of $E$ contains exactly $k$ elements is considered in - Björklund, Andreas; Husfeldt, Thore; Kaski, Petteri; Koivisto, Mikko. - Counting paths and packings in halves. Algorithms—ESA 2009, 578--586, - Lecture Notes in Comput. Sci., 5757, Springer, Berlin, 2009. MR2557785 - -where it is shown that such $t$-packings can be counted in $O^*(\binom{n}{tk/2})$ time. Here $O^*$ suppresses a factor polynomial is the mentioned parameters. For perfect matching we need $tk = n$ so we end up with $O^*(\binom{n}{n/2})$. Methods in this approach use a disjoint sum problem, inclusion-exclusion, and dynamic programming. So, this provides some answer to question 1. It was the fastest algorithm I found, but I did not attempt to perform an exhaustive search of the literature. Perhaps this article, references within, and the term "set packing" can help to up further relevant information. -Also, again in the $k$-uniform case, parameterized by $t$ the problem of counting $t$-packing is shown to by $W[1]\#$-hard in -Liu, Yunlong; Wang, Jianxin. On counting parameterized matching and -packing. Frontiers in algorithmics, 125--134, Lecture Notes in Comput. -Sci., 9711, Springer, [Cham], 2016. MR3571845 - -since there is a reduction with parameterized matching counting in graphs.<|endoftext|> -TITLE: Fastest deterministic factoring algorithm in subexponential space? -QUESTION [5 upvotes]: Strassen's factoring algorithm shows that $\text{FACTORING} \in \text{DTIME}(N^{\frac{1}{4}+o(1)})$, but if I'm not mistaken in my analysis it also uses a similar amount of space. By making a trade-off I think it is possible to show $\text{FACTORING} \in \text{DTISP}(N^{k+o(1)}, N^{\frac{1}{2}-k+o(1)})$ for $\frac{1}{4} \leq k \leq \frac{1}{2}$. -On the other hand, trial division up to $\sqrt{N}$ demonstrates $\text{FACTORING} \in \text{DTISP}(N^{\frac{1}{2}+o(1)}, \log(N)^{O(1)})$. The only extra space we need is to keep a counter and perform the divisibility test. -Is there any deterministic factoring algorithm known to be in $\text{DTISP}(N^{k + o(1)}, N^{o(1)})$ for $k \lt \frac{1}{2}$? -I know that there is an $N^{\frac{1}{3}+o(1)}$-time algorithm due to Michael Rubinstein but I can't tell what the space usage would be. This would qualify as an example if the space can be made subexponential. -Otherwise, is trial division the best we can do in $N^{o(1)}$ space? -We won't be able to prove $\text{FACTORING} \notin \text{DTISP}(\log(N)^{O(1)}, O(\log(N))) = \text{L}$, since that would imply $\text{NP} \neq \text{L}$, an open problem. - -REPLY [5 votes]: Lehman's algorithm uses $O(N^{\frac{1}{3}})$ time $O(\log N)$ space. The algoritm is the following. -0) Check that $n$ is odd and $n > 8$. -1) Check that every $a = 2, \ldots, [n^{\frac{1}{3}}]$ is not a divisor of $n$. -2) For every $k=2, \ldots, [n^{\frac{1}{3}}]$ and for every $d= 0,1 \ldots, [n^{\frac{1}{6}} / (4 \sqrt{k})]$ check is the number -$$([\sqrt{4kn}]+d)^2- 4kn $$ a square of an integer. If this is the case then consider $A:=[\sqrt{4kn}]+d$ and $B:= \sqrt{A^2 - 4kn}$. Note that -$$A^2 \equiv B^2 (\text{mod } n ).$$ -Check: $1 < \text{gcd}(A \pm B, N) < N$. -If we have this inequality then of course we can factorize $N$. Otherwise $N$ is prime by a theorem in the paper.<|endoftext|> -TITLE: Possibly new solution to equal-mass three-body problem; refinement required -QUESTION [6 upvotes]: (This is a repost of this question from 18 months ago on the main Mathematics SE site, as the response there has been underwhelming, and I thought here would be a better authority. As you can probably tell, I am rather inexperienced on the professional scene.) -While messing around in this Wolfram Demonstrations applet, I found a suspicious pattern, in which I could see physical similarities between the path of the first object, the path of the second object at a later time, and even the path of the third object at an even later time. Through gradually tweaking that set of initial conditions, I was able to find an apparently new solution to the equal-mass three-body problem that wasn't listed here. -Here are the initial conditions, with my restrictions in parentheses: - -$x_1(0)=0.7812$ -$y_1(0)=-0.2465\;(\text{holding $x_1(0)^2+y_1(0)^2$ constant to avoid scaling})$ -$x_2(0)=-0.2465\;(=y_1(0))$ -$y_2(0)=0.7812\;(=x_1(0))$ -$x_3(0)=-0.5347\;(=-x_1(0)-x_2(0))$ -$y_3(0)=-0.5347\;(=x_3(0))$ -$x_1'(0)=-0.6087$ -$y_1'(0)=-0.286$ -$x_2'(0)=0.286\;(=-y_1'(0))$ -$y_2'(0)=0.6087\;(=-x_1'(0))$ -$x_3'(0)=0.3227\;(=-x_1'(0)-x_2'(0))$ -$y_3'(0)=-0.3227\;(=-x_3'(0))$ - -(I would estimate the uncertainty for all parameters to be roughly $0.01$ in either direction. It would be smaller, but a slight change in the position parameter can be practically canceled out by a slight change in one or both of the velocity parameters, apparently resulting in a relatively 2D structure in 3D parameter space.) This gives an approximate choreography with period $p\approx 17.0874$, shaped somewhat like a rosette: - -Going further in time, the entire system as presented apparently rotates, but doesn't break apart: - -At this point, what algorithms or heuristics should I use to increase the precision of the parameters (for example, trying to make the paths match over increasing lengths of time)? Is this system even new? (It seems way too simple to be new, but this relationship doesn't always hold true.) Is this question even relevant on the research scene? - -REPLY [4 votes]: This is known, as Šuvakov redirected me to "Three on a Celtic Knot" at http://www.maths.manchester.ac.uk/~jm/Choreographies/ after I emailed him about the subject.<|endoftext|> -TITLE: Realizing $\mathcal{A}(2)//\mathcal{A}(1)$ by a finite spectrum -QUESTION [8 upvotes]: Let $\cal A$ denote the mod 2 Steenrod algebra. Can the $\mathcal{A}(2)$-module structure on $\mathcal{A}(2)//\mathcal{A}(1)$ be enriched to an $\cal A$-module structure? If so, is there a finite spectrum $X$ such that $H^*(X) = \mathcal{A}(2)//\mathcal{A}(1)$? - -REPLY [19 votes]: The $A(2)$-module structure on $A(2)//A(1)$ does not extend to an $A$-module structure. In particular, there is no spectrum $X$ with $H^*(X; F_2) = A(2)//A(1)$ as an $A(2)$-module. -Additively, $A(2)//A(1)$ is generated by classes $g_i$ in degree $i$ -for $i = 0, 4, 6, 7, 10, 11, 13$ and $17$. The Adem relation $Sq^4 Sq^6 = Sq^{10} + Sq^8 Sq^2$ implies $Sq^{10}(g_0) = g_{10}$. The Adem relation $Sq^2 Sq^8 = Sq^{10} + Sq^9 Sq^1$ implies $Sq^{10}(g_0) = 0$. This contradicts the existence of any $A$-module structure extending the given $A(2)$-module structure. -PS: Bruner's ext code (http://www.math.wayne.edu/~rrb/papers/) contains a script (newconsistency) that lets you verify that a purported presentation really defines an $A(2)$-module, and tells you what is needed to extend the presentation to an $A$-module structure. It could be handy if you want to realize other $A(2)$-modules by spectra.<|endoftext|> -TITLE: Martingale version of Bernstein-type inequality for (slightly) heavy-tailed distributions? -QUESTION [7 upvotes]: It is known that for sub-exponentially distributed martingale difference sequence, the following Bernstein-type inequality holds: -$$ -ℙ\left(\left| - \sum_{i=1}^N a_i X_i -\right| \ge t \right) - \le -2\exp\left( -−c \min\left( -\frac{t^2}{K^2 \|a\|_2^2}, \frac{t}{K \|a\|_\infty} -\right) -\right) -$$ -where $K = \max \|X_1\|_{\psi_1}$ (in the conditional sense) and $a\in ℝ^d$. -Does similar concentration inequalities hold for heavy-tailed random variables where $X_i$ satisfies $ℙ(X_i > t) \le C\exp(ct^{-\alpha})$ for $\alpha \in (0,1)$? -In the independent increment (random-walk) case, Theorem 6.21 of Ledoux and Talagrand (Probability in Banach Spaces) gave a positive answer. What about a general martingale version (even just in one dimensional)? - -REPLY [5 votes]: This is worked out in some detail in the paper of Fan, Grama and Liu, -J. Math Anal. Appl. 448 (2017), 538-566 (see in particular Theorem 2.1 there, and the references). Unfortunately I do not have an open access to an electronic copy, and it doesn't seem to exist on arXiv.<|endoftext|> -TITLE: Which rational cohomology classes on a product of elliptic curves come from subschemes? -QUESTION [12 upvotes]: Let $X=E_1\times\cdots\times E_n$, where $E_i$ is the elliptic curve $E_i=\mathbb{C}/(\mathbb{Z}+\mathbb{Z}\alpha_i)$. In Grothendieck's "The Hodge Conjecture is False for Trivial Reasons," $X$ is stated to fail the original formulation of the Generalized Hodge Conjecture, for appropriate choices of $\alpha_i$. Namely, there exist classes in $F^1H^n(X,\mathbb{C})\cap H^n(X,\mathbb{Q})$ that are not generated by images of Gysin maps $H^{n-2q}(Y,\mathbb{Q})\to H^{n}(X,\mathbb{Q})$, where $Y$ is a resolution of a codimension $q\ge1$ subscheme of $X$. (Here $F^1H^n(X,\mathbb{C})=H^{1,n-1}(X)\oplus\cdots\oplus H^{n,0}(X)$ is the first term in the Hodge filtration for $X$.) -He states without proof that the dimension of the subspace of $H^n(X,\mathbb{Q})$ generated by images of Gysin maps is $2^n-N$, where $N$ is the dimension of the $\mathbb{Q}$-vector space generated by distinct products of $\alpha_i$. I am confused by this, because it seems that when the $\alpha_i$ are generic, this number is 0. In particular, there are no non-zero Hodge classes in $H^2(E_1\times E_2,\mathbb{Q})$, but don't $E_1\times 0$ and $0\times E_2$ define nonzero Hodge classes? -Would appreciate it if someone could clear up my confusion and also provide a hint for how to prove the claim in the above paragraph. - -REPLY [6 votes]: This definitely looks like an error in Grothendieck's paper! I think I have figured out what he meant. To simplify notation, I will concentrate on the case where Grothendieck builds his example: $H^3$ of a product of $3$ elliptic curves. I prefer to think in homology for this purpose, so I'll be talking about $H_3(E_1 \times E_2 \times E_3)$. -Put $X = E_1 \times E_2 \times E_3$ where $E_i = \mathbb{C}/(\mathbb{Z}+\mathbb{Z} \tau_i)$. I'll write $z_i$ for the coordinate on the universal cover of $E_i$, so $dz_i$ is the nonvanishing holomorphic $(1,0)$-form on $E_i$. -Let $V$ be the subspace of $H_3(X, \mathbb{Q})$ spanned by the images of all $H_3(Y, \mathbb{Q})$, as $Y$ ranges over $2$-dimensional subvarieties of $X$. Hodge's "general conjecture" is an attempted description of $V$. (The famous Hodge conjecture deals with $H_{2 \dim Y}(Y)$, so we are talking about the span of the fundamental classes; this less famous one allows the subscript on homology to be independent of the dimension of $Y$.) -Specifically, Hodge conjectures that $\alpha \in H_3(X, \mathbb{Q})$ lies in $V$ if and only if $\int_{\alpha} \omega=0$ for any $\omega \in H^{(3,0)}(X)$. In fact, $H^{(3,0)}(X)$ is one dimensional, spanned by $d z_1 \wedge d z_2 \wedge dz_3$, so this criterion says that $\int_{\alpha} d z_1 \wedge d z_2 \wedge dz_3=0$. Let's denote the space of $\alpha$ obeying this condition by $V'$. It is easy to see that $V \subseteq V'$, since $ d z_1 \wedge d z_2 \wedge dz_3|_Y=0$ for any algebraic surface $Y$. -Grothendieck observes that, for "trivial reasons", $V$ is even dimensional. However, he shows that $V'$ need not be. -By Kunneth, we have -$$H_3(X, \mathbb{Q}) = \bigoplus_{i_1+i_2+i_3=3} H_{i_1}(E_1, \mathbb{Q}) \otimes H_{i_2}(E_2, \mathbb{Q}) \otimes H_{i_3}(E_3, \mathbb{Q}).$$ -Every summand other than $H_1 \otimes H_1 \otimes H_1$ is easily seen to be in $V$. So it is natural to consider the inclusion $V \cap (H_1\otimes H_1 \otimes H_1) \subseteq V' \cap (H_1\otimes H_1 \otimes H_1)$, which will be equality if and only if $V \subseteq V'$ is equality. I believe that Grothendieck, without bothering to say so, has switched to working with these subspaces in his dimension computation. Each of the other summands is two-dimensional, so this doesn't effect the parity of the dimension. -So, let's compute the dimension of $V' \cap (H_1\otimes H_1 \otimes H_1)$. Let $\gamma^0_i$ be the $1$-cycle $\mathbb{R}/\mathbb{Z}$ in $E_i$, and let $\gamma^1_i$ be the $1$-cycle $\mathbb{R} \tau_i/\mathbb{Z} \tau_i$. So $\gamma^0_i$, $\gamma^1_i$ is a basis of $H_1(E, \mathbb{Q})$. So a basis for $H_1 \otimes H_1 \otimes H_1$ is the $8$ products $\gamma_1^{j_1} \times \gamma_2^{j_2} \times \gamma_3^{j_3}$, where each $j_i$ is $0$ or $1$. We have $\int_{\gamma_i^j} dz_i = \tau_i^j$, so -$$\int_{\gamma_1^{j_1} \times \gamma_2^{j_2} \times \gamma_3^{j_3}} d z_1 \wedge d z_2 \wedge dz_3 = \tau_1^{j_1} \tau_2^{j_2} \tau_3^{j_3}.$$ -Putting -$$\alpha = \sum_{j_1, j_2, j_3 =0}^1 c(j_1, j_2, j_3)\ [\gamma_1^{j_1} \times \gamma_2^{j_2} \times \gamma_3^{j_3}] \in H_3(X, \mathbb{Q}),$$ -we see that $\alpha \in V'$ if and only if -$$\sum_{j_1, j_2, j_3 =0}^1 c(j_1, j_2, j_3) \ \tau_1^{j_1} \tau_2^{j_2} \tau_3^{j_3}=0.$$ -Thus, $V' \cap (H_1\otimes H_1 \otimes H_1)$ is identified with the $\mathbb{Q}$-vector space -$$\left\{ c \in \mathbb{Q}^{\{0,1\}^3} : \sum_{j_1, j_2, j_3 =0}^1 c(j_1, j_2, j_3) \ \tau_1^{j_1} \tau_2^{j_2} \tau_3^{j_3}=0 \right\}.$$ -We see that the dimension of $V' \cap (H_1\otimes H_1 \otimes H_1)$ is $2^3-N$ where $N$ is the dimension of the $\mathbb{Q}$-vector space spanned by the monomials $\tau_1^{j_1} \tau_2^{j_2} \tau_3^{j_3}$. -As Grothendieck points out, if $\tau_1=\tau_2=\tau_3$ is a cubic irrational, these monomials span a $3$-dimensional vector space. As you say, for generic $\tau_i$, the monomomials span an $8$ dimensional vector space, so $V' \cap (H_1\otimes H_1 \otimes H_1)=0$. This reflects that the obvious classes, coming from $H_3(E_i \times E_j)$ with $1 \leq i < j \leq 3$, are all in the other summands of $H_3(X)$.<|endoftext|> -TITLE: Is it possible to define cardinals that are distinct from either the $\aleph$ numbers or $\beth$ numbers? -QUESTION [9 upvotes]: I am wondering if there are ways of defining "structure" on infinite sets that generate sequences of cardinals that cannot be proved to have the same cardinality as either the $\aleph$ or $\beth$ numbers? -Let me explain: -Start by defining $\beth_0=|\mathbb N|$, which then is also the cardinality of any other countably infinite set. Next use the power set axiom to define a sequence of sets: P($\mathbb N$), P(P($\mathbb N$)), P(P(P($\mathbb N$))), ... -So the first kind of "structure" we introduce is that of subsets. -By Cantor's Theorem, $|A|<|$P($A$)$|$. So define the $\beth$ numbers as the cardinality of these successive power sets: $\beth_1=|$P($\mathbb N$)$|$, $\beth_2=|$P(P($\mathbb N$))$|$, ... -Other kinds of structure on infinite sets would generate the same sequence of $\beth$s. For example, if we define F($A$) to be the set of all functions $A\rightarrow$ {0,1}. Then we will have $|$F($\mathbb N$)$|=|$P($\mathbb N$)$|=\beth_1$ and further $|$F(F($\mathbb N$))$|=\beth_2$, ... -A second kind of structure uses ideas of well ordered sets to build up ordinals. Define $\aleph_0=|\omega_0|$ and $\aleph_{\gamma^+}=$ {$\alpha:\alpha$ is an ordinal and $|\alpha|\le\aleph_{\gamma}$}. This will get us the sequence of $\aleph$s. -Again other kinds of structure (e.g. building ordinals using sets with a maximum element instead of using wellorderd sets) would generate the same sequence of $\aleph$s. -The independence of the (generalized) continuum hypothesis means that after $\aleph_0=\beth_0$ there is not much (other then some weak restrictions) we can prove about the relative size of $\aleph$s and $\beth$s in ZFC. And to me it seems like this is because once we get to uncountable sets there is just not enough "structure" avalaible to construct one-to-one functions and nail down their relative sizes. -So my question is are there other ways of defining "structure" on infinite sets to generate another sequence of cardinals (say $\gimel_0,\gimel_1, \gimel_3$ ...), which, at least within ZFC, then can't be shown to have the same cardinality as either some $\aleph$s or some $\beth$s? -Or alternatively, is there a proof that within ZFC that defining cardinals using either the method of $\aleph$s or $\beth$s somehow exhausts possible ways of cardinal definition? -Clarification: I want be clear that I understood comments that in ZFC all cardinals are $\aleph$s, since any set can be well-ordered. But it is undecidable which $\aleph$s the $\beth$s correspond to, right? Even though we know there exists a well-ordered set with cardinality $\beth_1$, we cannot say which $\aleph$ that is without deciding the continuum hypothesis. So the question is are there other ways of defining "structure" to generate cardinals that are similarly undecidable in ZFC? Or are the $\aleph$s or $\beth$s all there is? - -REPLY [2 votes]: I think the answer to your question lies in Scott's cardinals, the cardinality of a set x is defined as the equivalence class of all sets bijective to x that belongs to the minimal possible rank that a set bijective to x appears as a subset of. And along the same way ordinals also can be defined. If you work in ZF alone, then there are models of ZF in which some Scott's cardinals are not comparable to any $\aleph$ or $\beth$ number at all, like for example the Scott cardinals of Tarski Infinite Dedekind finite sets. Also I do think that there are many ways to define 'structure' even in $ZFC$ that can possess indecidability of cardinal comparisons similar to those between the $\aleph$s and the $\beth$s, like for example the cardinality of the set of all Hereditarily subnumerous sets to a set, you can build up stages of those in a similar manner to how you do with powers, call those as the $\daleth_i$ numbers (I don't know if this is used in other contexts), so there will be no clear comparisons between for example $\aleph_i$ and $\daleth_i$ numbers, nor even between the $\beth_i$s and the $\daleth_i$s.<|endoftext|> -TITLE: Convergence in Lebesgue measure -QUESTION [6 upvotes]: It is well known that if $K_n$ are compact sets in $\mathbb{R}^n$ converging in Hausdorff distance to $K$ compact as well, then it does not follow that their Lebesgue measures converge (even if the Lebesgue measure of the $K_n$ is constant). -Given this, I would like to ask the following question: If we assume that $K_n$ does not only converge in HD to $K$ but also in measure $\lambda(K_n) \rightarrow \lambda(K)$ and let $F$ be a Lipschitz function, does it then follow that $\lambda(F(K_n)) \rightarrow \lambda(F(K))$? -This question is related to the previous since Lipschitz continuity still preserves metric convergence of $F(K_n)$ to $F(K)$, but now we have also the information that $\lambda(K_n) \rightarrow \lambda(K)$. Does this allow us to deduce the claim? - -REPLY [2 votes]: Note the following lemma: under the given assumptions, we have $\lambda(K_n \triangle K) \to 0$. Fix $\epsilon > 0$ and choose a $\delta$ so small that the $\delta$-thickening $K^\delta$ of $K$ has measure $\lambda(K^\delta) < \lambda(K) + \epsilon$. (This is possible since $K$ is compact and hence $\bigcap_i K^{1/i} = K$.) Now if $n$ is so large that $K_n \subset K^\delta$ (by Haudsdoff convergence) then $$\lambda(K_n \setminus K) \le \lambda(K^\delta) - \lambda(K) < \epsilon$$ -and if we further take $n$ large enough that $\lambda(K_n) > \lambda(K) - \epsilon$, then we also get -$$\lambda(K \setminus K_n) \le \lambda(K^\delta) - \lambda(K_n) < 2 \epsilon$$ -proving the lemma. -Now note that $F(K_n) \triangle F(K) \subseteq F(K_n \triangle K)$. As such, we have -$$\lambda(F(K_n) \triangle F(K)) \le \lambda(F(K_n \triangle K)) \le C_F \lambda(K_n \triangle K) \to 0$$ -where $C_F$ is the Lipschitz constant of $F$. In particular $\lambda(F(K_n)) \to \lambda(F(K))$. -(Thanks to Iosif Pinelis for suggesting this argument which is much simpler than what was here before.)<|endoftext|> -TITLE: Equivalence of the term "divisor" -QUESTION [6 upvotes]: In the theory of Riemann surfaces, for a meromorphic function $f$ on a Riemann surface $X$, we define the divisor of $f$ to be the function $D$ which maps $$f \longmapsto \sum_{p \in X} \text{ord}_p \cdot p,$$ where $\text{ord}_p$ denotes the order of the root or pole of $f$ at $p$. -In the context of Stein manifolds however, we define a divisor to be an element $d \in H^0(X, \mathscr{D})$, where $\mathscr{D}$ is the quotient sheaf of $\mathscr{M}^{\times}$ by $\mathscr{O}_X^{\times}$, where $\mathscr{M}^{\times}$ is the sheaf of germs of invertible meromorphic functions and $\mathscr{O}_X^{\times}$ is the sheaf of germs of invertible holomorphic functions. -Are these definitions equivalent? If they are, I can't see why. -Note that this does seem like a question that should be posted on Math.StackExchange, but I have found little to no luck asking questions of anything north of the elementary theory of analytic functions of several complex variables on that site. - -REPLY [11 votes]: These two definitions are not the same and if you think about it, it is not at all surprising. In the first one you are looking at the definition of a divisor of a meromorphic function while in the second one the more general definition of a divisor. -So, perhaps you are thinking of comparing the definition of a divisor as an integral linear combination of points (or more generally of codimension 1 subspaces). This is matching your first description, but it allows divisors that do not come from meromorphic functions. And this is essentially what's in the second definition. It defines a divisor to be locally what you have in your first definition, but allows for the possibility that it does not come from a single meromorphic function. -More generally, the first definition gives you a Weil divisor, the second gives you a Cartier divisor. They are not equivalent in general, but they are on manifolds. More precisely, every Cartier divisor (i.e., one that is locally defined by a single equation) gives you a Weil divisor (i.e., an integral linear combination of codimension 1 subspaces) by looking at its zero set with multiplicities (essentially as in your first definition). A Weil divisor gives you a Cartier divisor if every codimension 1 subspace is defined locally by a single equation. This is true on manifolds, so you have nothing to worry about. Also, when you are looking at divisors of meromorphic functions, for those the two notions are automatically the same.<|endoftext|> -TITLE: How hard is it to guess Kuperberg's certificate of knottedness? -QUESTION [11 upvotes]: Kuperberg's Knottedness is in $\mathsf{NP}$, modulo GRH provides a certificate that a knot $K$ given by a knot diagram on $c$ crossings is not trivial. The certificate is a prime $p$, along with a solution in $\mathbb{Z}/p$ to a system of $m$ polynomial equations on $n$ variables describing a representation of the knot group $\pi_1(S^3\setminus K)$ onto $\mathrm{SU(2)}$. Assuming the Generalized Riemann Hypothesis, the bit complexity $\log_2 p$ is polynomial in $m,n$, which are polynomial in $c$. -Letting each polynomial $f_i$ be of degree at most $d$, the problem is to find a solution to: -$$f_1[x_1,x_2,\cdots,x_n]=f_2[x_1,x_2,\cdots,x_n]=\cdots=f_m[x_1,x_2,\cdots,x_n]=0$$ -in $\mathbb{Z}/p$. Thought of as a $\mathsf{SAT}_p$ problem, for each $f_i$ we are asked to find a solution $f_i=0\mod p$. -However, we don't a-priori know $p$- the prime modulo which a solution is guaranteed to exist. We only know that there is at least one solution for at least one $p$ not bigger than some number that I'll call $B$. - -Accordingly, is there any advantage to repeatedly guessing $n$ integers $\le B$, plugging these numbers in to the $m$ polynomials, and calculating the $\gcd$ of the resulting $f_i$, stopping when the $\gcd\ne1$? - -If the $\gcd\ne 1$, we can think of it as a product of primes $p_1\times p_2\times\cdots$. Thus, there will be a solution in $\mathbb{Z}/p_1,\mathbb{Z}/p_2,\cdots$, and we know Kuperberg's certificate exists (although we might not know what the prime is unless we can factor the $\gcd$.) -There's an old adage that we can shoot an arrow at a wall and aim for a bulls-eye, but perhaps it's more efficient to shoot the arrow at the wall and draw a bulls-eye around wherever the arrow lands. But perhaps there are too few primes with solutions, or too few solutions modulo any given prime, to make such guessing worthwhile. - -REPLY [3 votes]: This question is probably very open if we want to be rigorous. With the roughest heuristics, the chance $p$ divides a polynomial value $f_i$ is $1/p$, and so the chance of recovering $p$ is $1/p^n$ where $n$ is the number of polynomials. Now if we sum over all primes and $n>2$ the expected number of solutions is less than $1$. While we know there is a solution, we won't find it by random search by this heuristic argument.<|endoftext|> -TITLE: Maximal subgroups of a finite simple group which have some minimal subgroups in common -QUESTION [5 upvotes]: Let $G$ be a finite simple group and $M$ and $M'$ be two maximal subgroups of $G$. Also let $m_{M}$ be the set of minimal subgroups of $M$ and similarly $m_{M'}$ be the set of minimal subgroups of $M'$. Is it possible that $m_{M}\subseteq ‎m_{M'}$. - -REPLY [6 votes]: Yes. Let $G = {\rm PSL}(3,4)$, and let $M$ be a maximal subgroup with the structure $3^2:Q_8$. -The minimal subgroups (i.e. the subgroups of prime order) of $M$ have order $2$ or $3$ and generate a subgroup of $M$ with structure $3^2:2$. This subgroup is also contained in a maximal subgroup $M'$ of $G$ with $M' \cong A_6$. -Another example is $G = M_{22}$ with $M = M_{10} = A_6 \cdot 2$ and $M' = 2^4:A_6$. -Yet another is $G = {\rm PSL}(3,7)$ again with $M = 3^2:Q_8$ and $M' = (3 \times A_4):2$. -It seems likely that there will be infinitely many examples of type $G = {\rm PSL}(3,q)$ and $M = 3^2:Q_8$, , and this should not be hard to prove, since the maximal subgroups of ${\rm PSL}(3,q)$ are well understood.<|endoftext|> -TITLE: How are the left and the right group of a bitorsor related? -QUESTION [7 upvotes]: This question arose from my answer to To what extent does a torsor determine a group: it turns out that I do not know one thing about it. -Let $G$, $G'$ be groups in some nice enough category (you may assume a topos, if you feel like that). Can one find a nice intrinsic simplification of the condition "there exists a $G$-$G'$-bitorsor"? -It is clear that a necessary condition for the existence of a bitorsor is that $G$ and $G'$ are locally isomorphic, i. e. there is an object $B$ with global support such that the groups $B\times G\to B$ and $B\times G'\to B$ over $B$ are isomorphic over $B$. Is this also sufficient? -Can one do better? By this I mean not quantifying over objects but rather concocting some condition out of $G$ and $G'$ alone? - -REPLY [2 votes]: [I don't have enough 'points' to comment; below is really just a comment.] -If you consider instead your 'nice enough' category to be locales, then having a bi-torsor between two open localic groups G and G' implies that their toposes of sheaves are equivalent. Quite different localic groups can define the same topos, so my instinct is that there is no easy or natural way to determine (Morita) equivalence by inspecting the groups other than to tautologically give the definition. (Refer to Remark C5.2.14(d) for a reference to a concrete example, for the groupoid case at least.) -I was intrigued by the question because you seemed to be interested in working on things to do with torsors in more general categorical contexts. Since I have done some work on torsors (effectively Hilsum-Skandalis maps) in a cartesian category, which I think is the most general possible context, I hope you don't mind my providing a link to that work: -http://www.christophertownsend.org/Documents/HilsumSkandalisFrobenius.pdf -The general gist of the question seems to be: what can we say about how group(oids) are related/constructed given information about how their categories of equivariant sheaves are related; I find this to be an interesting avenue and only know partial answers.<|endoftext|> -TITLE: Is $\mathbb{R}\cong\text{Cont}(X,Y)$ for some non-trivial spaces $X,Y$? -QUESTION [31 upvotes]: For topological spaces $X,Y$ let $\text{Cont}(X,Y)$ be the collection of continuous functions $f:X\to Y.$ We endow $\text{Cont}(X,Y)$ with the topology inherited from the product topology on $Y^X.$ -Are there spaces $X,Y$ such that $X$ has more than one point and $Y\not\cong\mathbb{R}$ such that $\mathbb{R}\cong\text{Cont}(X,Y)$? - -REPLY [23 votes]: Following the answer of Uri Bader, we can show that $Y$ is a retract of the real line, so can be identified with a closed convex subset of $\mathbb R$. Without loss of generality we can assume that $0,1\in Y$ and hence $[0,1]\subset Y$. It follows that the function $C(X,Y)$ is a convex subset of $Y^X\subset \mathbb R^X$ and $C(X,[0,1])\subset C(X,Y)$. The assumption $Y\not\cong\mathbb R\cong C(X,Y)$ implies that $C(X,Y)$ contains a non-constant function $f$. Consider the constant functions $\mathbf 0:X\to\{0\}\subset Y$ and $\mathbf 1:X\to\{1\}\subset Y$ and observe that the set $T=\{\mathbf 0,\mathbf 1,f\}\subset C(X,Y)\subset Y^X\subset\mathbb R^X$ is affinely independent and its convex hull $conv(T)\subset C(X,Y)$ is homeomorphic to the 2-dimensional symplex, which cannot be contained in the real line $\mathbb R\cong C(X,Y)$. This contradiction completes the proof. -The negative answer can also be deduced from the following theorem. -We recall that a topological space $X$ is functionally Hausdorff if for any distinct points $x,y\in X$ there exists a continuous function -$f:X\to\mathbb R$ such that $f(x)\ne f(y)$. -Theorem. If for non-empty topological spaces $X,Y$ the function space $C(X,Y)$ is functionally Hausdorff and path-connected, then either $C(X,Y)$ is homeomorphic to $Y^n$ for some $n\in\mathbb N$ or $C(X,Y)$ contains a topological copy of the Hilbert cube. -Proof. The space $Y\cong C(\{x\},Y)$ is functionally Hausdorff and path-connected, being a retract of the functionally Hausdorff path-connected space $C(X,Y)$. If $Y$ is a singleton, then $C(X,Y)\cong Y^1$ is a singleton, too. So, we assume that $Y$ contains more than one point. In this case $Y$ contains a subspace $I$, homeomorphic to the closed interval $[0,1]$. -Consider the canonical map $\delta:X\to Y^{C(X,Y)}$, $\delta:x\mapsto (f(x))_{f\in C(X,Y)}$. If the image $\delta(X)$ is finite of cardinality $n$, then $C(X,Y)$ is homeomorphic to $Y^n$ (since each function $f\in C(X,Y)$ is constant on each set $\delta^{-1}(y)$, $y\in \delta(X)$). -So, we assume that the set $\delta(X)$ is infinite. Taking into account that the space $Y^{C(X,Y)}$ is functionally Hausdorff, we can construct a continuous map $g:Y^{C(x,Y)}\to I$ such that the image $Z=g(\delta(X))$ is infinite. The surjective continuous map $p:=g\circ\delta:X\to Z$ induces a continuous injective map $p^*:C(Z,I)\to C(X,I)$, $p^*:f\mapsto f\circ p$. It is easy to see that the function space $C(Z,I)\subset I^Z$ contains a topological copy $Q$ of the Hilbert cube $I^\omega$. Then $p^*(Q)$ is a topological copy of the Hilbert cube in $C(X,I)\subset C(X,Y)$.<|endoftext|> -TITLE: Iterated forcing and the super tree property at $\omega_2$ -QUESTION [8 upvotes]: It is a theorem of Baumgartner and Laver that iterating Sacks forcings of weakly compact length gives rise to the tree property at $\omega_2$. Natural questions (at least for me) are: do we get stronger tree properties at $\omega_2$ if we start with larger cardinals? In particular, if the length is strongly compact, do we get the strong tree property at $\omega_2$? If the length is supercompact, do we get the super tree property at $\omega_2$? I suspect it is known so I appreciate pointers to references. - -For uncountable cardinals $\kappa\leq \lambda$ with $\kappa$ being regular, we say $F\subset \bigcup_{u\in [\lambda]^{<\kappa}} 2^u$ is a $(\kappa,\lambda)$-tree if it is closed under taking restrictions, and for each $u\in [\lambda]^{<\kappa}$, $|F_u:=\{f\in F: dom(f)=u\}|<\kappa$. A function $d: \lambda\to 2$ is a cofinal branch if for all $u\in [\lambda]^{<\kappa}$, $d\restriction u\in F_u$. Given $\bar{f} = \langle f_u\in F: dom(f_u)=u\rangle$ (called a level sequence), we say $d: \lambda\to 2$ is an $\bar{f}$-ineffable branch if $\{u\in [\lambda]^{<\kappa}: d\restriction u = f_u\}$ is stationary in $[\lambda]^{<\kappa}$. -We say $\kappa$ has the strong tree property if for any $\lambda\geq \kappa$, any $(\kappa, \lambda)$-tree has a cofinal branch. -We say $\kappa$ has the super tree property if for any $\lambda\geq \kappa$, any $(\kappa, \lambda)$-tree $F$ and any level sequence $\bar{f}$, there exists an $\bar{f}$-ineffable branch. - -Edit: I just realized the answer in the case where the length is strongly compact is yes: the countable support iteration of Sacks forcings of strongly compact length forces the Semistationary Reflection Principle (equivalent to the statement that all stationary preserving forcings are semiproper and in fact stronger statement, i.e. the Baire version of Rado's conjecture, holds), along with $\neg CH$, implies the strong tree property holds at $\omega_2$ by Torres-Perez and Wu (https://www.researchgate.net/publication/306394279_Strong_Chang's_Conjecture_Semi-Stationary_Reflection_the_Strong_Tree_Property_and_two-cardinal_square_principles). The question remains for the case where the length is supercompact. - -REPLY [5 votes]: The answer to the supercompact case is yes. More specifically, in the forcing extension obtained by iterating Sacks forcing of supercompact length, the super tree property at $\omega_2$ holds. This follows from the following: -1) Countable support iteration of Sacks forcing satisfies $\omega_1$-approximation property. This was essentially shown in Baumgartner and Laver's classical paper about iterated Sacks forcing -2) Theorem 5.4 in Weiss' paper: Combinatorial Essence of Supercompactness (https://www.sciencedirect.com/science/article/pii/S0168007211001904).<|endoftext|> -TITLE: Limit associated with complementary sequences -QUESTION [7 upvotes]: Define $A=(a_n)$ and $B=(b_n)$ as follows: $a_0=1$, $a_1=2$, $b_0=3$, $b_1=4$, and $$a_n=a_0b_{n-1}+a_1b_{n-2}$$ for $n \geq 2$, where $A$ and $B$ are increasing and every positive integer occurs exactly once in $A$ or $B.$ Can someone prove that $\lim_{n \to \infty} a_n/n = 4$? -Here are first terms and some evidence regarding the limit: -$$A = (1,2,10,13,16,19,22,25,29,34,38,43,47,52, \dots)$$ -$$B = (3,4,5,6,7,8,9,11,12,14,15,17,18,20,21,23, \dots)$$ -I've checked that $-9 -TITLE: Low-dimensional irreducible 2-modular representations of the symmetric group -QUESTION [10 upvotes]: I apologize if this question is a little too basic for MathOverflow, but it's somewhat outside of my background and I'm frustrated that the answer doesn't seem to be explicit in the literature even though I suspect it's easily known by the experts. -What are the lowest-dimensional irreducible representations over $\mathbb{F}_2$ of the full symmetric group $S_n$ and in particular whether there exist any of dimension $\leq n - 2$ (resp. $\leq n - 3$) if $n$ is odd (resp. even)? -I know that for each $n \geq 3$ we have the $(n - 1)$-dimensional standard representation over any field. Moreover, if $n$ is even, we get an $(n - 2$)-dimensional representation as a quotient of this over characteristic $2$ (because then the vector with all $1$'s lies in the standard representation but is fixed by all of $S_n$). It's easy to see that these representations are irreducible. What I would specifically like to know is whether or not they are the lowest-dimensional irreducible representations of $S_n$ over $\mathbb{F}_{2}$. -(I'm curious about slightly more general questions as well, like whether or not the analogous representations to the ones I described above are the lowest-dimensional over any given positive characteristic.) - -REPLY [10 votes]: An earlier reference for the minimality of these degrees over any field is L.E.Dickson, Representations of the general symmetric group as linear groups in finite and infinite fields, Trans. Amer. Math. Soc. 9 (1908), 121-148. This can be found in Dickson's Collected Works.<|endoftext|> -TITLE: Why is persistent cohomology so much faster than persistent homology -QUESTION [34 upvotes]: I refer to this paper: de Silva, Vin; Morozov, Dmitriy; Vejdemo-Johansson, Mikael. Dualities in persistent (co)homology. Inverse Problems 27 (2011), no. 12, 124003, 17 pp. (Journal link, arXiv link). -According to the results in the paper, especially the experiments in page 15 it shows that persistent cohomology is faster than persistent homology by a factor of around 30 to 50. -That seems quite amazing to me, considering that over fields, homology and cohomology are dual. The paper does explain the reason why, but I don't really get the key idea how it accounts for a 3000% to 5000% improvement over persistent homology. -The paper's explanation (also on pg 15) is based on the difference between row operations and column operations. Apparently, row operation is supposed to be the better one, and since persistent cohomology can use the row operation (while it is difficult for homology), it results in better results. Also, the column algorithm (the worse algorithm) has to store all dead cycles, while for the row algorithm we can delete those cycles that died. -Also, theoretically, I am curious if such wonderful optimizations can be done for persistent cohomology, why can't the same be done for the dual persistent homology? Is there any theoretical reason for the impediment in persistent homology algorithms? Ideally I would imagine that the "best persistent cohomology algorithm" would perform as well as the "best persistent homology algorithm". -Thanks for any enlightenment. - -REPLY [30 votes]: There are several factors contributing to the improved performance of the algorithm reported in the paper; the use of cohomology is one, but there is also a computational shortcut involved, and the top Betti number (more precisely, the dth Betti number of the (d+1)-skeleton) also plays a role. -The computation of persistence barcodes is based on a variant of Gaussian elimination, applied to the columns of the boundary matrix of a filtered chain complex. Since the filtration order is important, no columns are exchanged, and eliminations are only done using left-to-right column additions. Gaussian elimination does not make use of the fact that the matrix is a boundary matrix. But this fact can be exploited to take a significant shortcut in the matrix reduction. -This shortcut, called the “clearing” optimization, allows one to clear out entire columns at once. The shortcut has been found independently by Chao and Kerber as well as by Morozov et al. (its use is implicit in the description of the cohomology algorithm of the mentioned paper). Clearing applies to both the homological and cohomological versions of barcode computation. In the mentioned paper, the cohomology algorithm uses clearing, while the homology algorithm does not. -The experimental results of the paper thus tell us that persistent cohomology with clearing is faster than persistent homology without clearing. As it turns out, however, cohomology with clearing is also faster than homology with clearing on typical examples, clarifying the role of cohomology for computational purposes. -Applying clearing requires performing the column operations in an appropriate order. For homology, clearing the column of a q-simplex requires reducing the column of a (q+1)-simplex first. In cohomology, in contrast, clearing the column of a q-simplex requires reducing the column of a (q-1)-simplex first. -If the goal is to compute persistence barcodes only up to a certain homological degree d, the computation starts with reducing columns of (d+1)-simplices. For those columns, clearing is not available. The number of columns that will be reduced to a zero column is exactly the -(d+1)st Betti number of the (d+1)-skeleton. -Now especially when computing persistence for Vietoris–Rips filtrations, the degree d is typically chosen small, and there are a lot more (d+1)-simplices than simplices of lower dimension, and also the (d+1)st Betti number will be large. When computing persistent homology, clearing only zeroes out columns of lower-dimensional simplices, and so the speedup obtained might be quite small. On the other hand, for cohomology, clearing is unavailable only for 0-simplices, but the number of 0-simplices is small, and persistence in dimension 0 can be computed very quickly using a union-find data structure instead. -In summary: clearing speeds up the computation of persistence, and cohomology typically allows for more clearing than homology, especially in the case of Vietoris–Rips filtrations.<|endoftext|> -TITLE: When is the Thom spectrum of a virtual vector bundle effective? -QUESTION [10 upvotes]: Remark: My question is valid in the classic setting of the stable homotopy category of spectra of CW-complexes. An answer on that setting will also be valid. - -Denote as $SH(X)$ Voevodsky's stable homotopy category over a scheme $X$. Denote $SH(X)^{\mathrm{eff}}$ its effective variant, that is to say, the smalles triangulated subcategory of $SH(X)$ which is closed under direct sums and contains suspension spectra of spaces but not their $\mathbb{P}^1$- desuspensions -(cf section 2). -Let $V\to X$ be a vector bundle. Denote as $\mathrm{Th}(V)$ its Thom space, and denote as well its infinite supension, which belongs to $SH(X)$. More concretely, it belongs to $SH(X)^{\mathrm{eff}}$. -Let $\xi$ be a virtual vector bundle over $X$ of rank $r\in \mathbb{Z}$ and denote $\mathrm{Th} (\xi)$ its associated Thom spectrum (cf. section 4.1 of this paper). My question is: - -Is it true that $$ \mathrm{Th}(\xi) \in SH(X) ^{\mathrm{eff}}\Leftrightarrow r\geq 0 \hspace{.5cm} ?$$ - -REPLY [7 votes]: Yes. -A bit more generally, if $\xi$ is a perfect complex of rank $\geq 0$, then $Th(\xi)$ is effective (even very effective): the question is Nisnevich-local on $X$ and $\xi$ is locally a complex of trivial vector bundles. -Conversely, suppose $Th(\xi)$ is effective. Since pullback preserves effective spectra, we can assume $X$ the spectrum of a field, so $Th(\xi)=(\mathbb P^1)^{\wedge r}$ and $r$ must be $\geq 0$.<|endoftext|> -TITLE: History of publication of von Neumann's characterization of orthogonally invariant matrix norms -QUESTION [11 upvotes]: Von Neumann has a result (rather well-known in convex analysis circles) which states that every orthogonally invariant matrix norm (meaning $\| P M Q\| = \| M \|,$ for any orthogonal $P, Q$) is a symmetric gauge of the singular values. This is a very nice and very useful result, for which the reference is (see also bibtex below): -J. von Neumann, Some matrix inequalities and metrization of matric space. -Mitt. Forschungsinst. Math. Mech. Kujbyschew-Univ. Tomsk 1, 286-300 (1937). -The question is really one of history: - -how did von Neumann (who was already in Princeton at the time) decide to publish in a Tomsk University proceedings? - -Tomsk is very far from anywhere in Russia, and I am not aware of anything else that appeared there. -Bibtex reference: -@Misc{zbMATH02519847, - Author = {J. {von Neumann}}, - Title = {{Some matrix inequalities and metrization of matric space.}}, - Year = {1937}, - Language = {English}, - HowPublished = {{Mitt. Forschungsinst. Math. Mech. Kujbyschew-Univ. Tomsk 1, 286-300 (1937).}}, - Zbl = {63.0037.03} -} - -REPLY [12 votes]: Probably there is no way to know it for certain, but it's a safe bet that this was related to Fritz Noether https://en.wikipedia.org/wiki/Fritz_Noether . He was a brother of Emmy Noether and also a mathematician (although not nearly as famous as his elder sister). Apparently, -he is best known for his work in relativity (Herglotz-Noether theorem). -As is well known, Emmy Noether moved to US after Nazi government dismissed Jews. Her brother, instead, moved to USSR. (Which, in hindsight, was not a very good idea.) He was appointed there to a professorship at the University of Tomsk. (Look at his publications in 1935-1937 - http://www-history.mcs.st-and.ac.uk/Extras/Fritz_Noether_publications.html.) -Of course, Von Neumann should have being familiar with Fritz Noether, so it is not a miracle if he sent him a paper or two. -Then there was the Great Purge. In 1937 (the same year the mentioned paper was published) Fritz Noether was arrested at his home in Tomsk. He was shot in Orel on September 10, 1941, and his burial place is unknown.<|endoftext|> -TITLE: Gambler's ruin: The fair game is the longest -QUESTION [13 upvotes]: Consider a gambler who, in every trial of a game, wins or loses a dollar with -probability $p\in\left( 0,1\right) $ and $q=1-p$, respectively. Let his -initial capital be $z>0$ and let him play against an adversary with the same -capital $z>0$. The game continues until one of the players is ruined. -In the language of random variables, this amounts to consider a sequence -$X_{1}^{\left( p\right) },X_{2}^{\left( p\right) },...\ $of random -variables (on a sample space $\Omega$) taking on two values $+1$ and $-1$ with -probabilities $\Pr\left[ X_{n}^{\left( p\right) }=+1\right] =p$ and -$\Pr\left[ X_{n}^{\left( p\right) }=-1\right] =q$. In particular, -$X_{n}^{\left( p\right) }$ describes the gambler's gain on the $n$th trial -and -$$ -S_{n}^{\left( p\right) }\left( \omega\right) =X_{1}^{\left( p\right) -}\left( \omega\right) +\cdots+X_{n}^{\left( p\right) }\left( -\omega\right) ,\quad S_{0}^{\left( p\right) }\left( \omega\right) -\equiv0\qquad\forall \,\omega\in\Omega -$$ -describes his net cumulated gain after $n$ trials. The duration of the -game is the random number of trials before he is either ruined or wins the -game (and his adversary wins or is ruined) -$$ -D^{\left( p\right) }\left( \omega\right) =\min\left\{ t\in\mathbb{N}% -:\left\vert S_{t}^{\left( p\right) }\left( \omega\right) \right\vert -=z\right\} \qquad\forall \,\omega\in\Omega. -$$ - -Q. I need a reference or a simple proof of the folk result according to which, - for each $p\in\left( 0,1\right) $, - $$ -\Pr\left[ D^{\left( p\right) }\leq s\,\right] \geq\Pr\left[ D^{\left( -1/2\right) }\leq s\,\right] \qquad\forall\, s=1,2,... -$$ - -REPLY [11 votes]: Let us show a bit more, that $F_p(s):=P(D^{(p)}\le s)$ is nondecreasing in $|p-1/2|$, for any real $s$. That is, take any $p$ and $p_1$ in $(0,1)$ such that $|p_1-1/2|>|p-1/2|$, and let, for brevity, $F:=F_p$ and $G:=F_{p_1}$. We shall show that then $F\le G$. -Indeed, by the formula (10), referenced by Carlo Beenakker, -\begin{equation*} - f_p(n):=P(D^{(p)}=n)=a_{n,z} b_{p,z}(pq)^{n/2} -\end{equation*} -for natural $n$, where $a_{n,z}$ depends only on $n$ and $z$; $b_{p,z}:=(p/q)^{z/2}+(q/p)^{z/2}$; and $q:=1-p$. Since $pq$ decreases in $|p-1/2|$, the family $(f_p)$ of pmf's has the monotone likelihood ratio (MLR) property, meaning that -\begin{equation*} - r(n):=\frac{f(n)}{g(n)} -\end{equation*} -increases in $n\in N$, where $N:=N_z:=\{n\colon a_{n,z}\ne0\}$, $f:=f_p$, $g:=f_{p_1}$, -and $p$ and $p_1$ are as before. -Therefore, for any given real $s$, -\begin{align*} - F(s)-G(s)&=F(s)[1-G(s)]-G(s)[1-F(s)] \\ - &=\sum_{m,n\colon\, N\ni m\le s -TITLE: Regularity of Fourier transforms of $L^p$ functions for $2n(\frac{1}{2}-\frac{1}{p})$, and $H_{-s}$ consists of distributions of order $k$ for $s\leq k\in\mathbb{N}$. Conversely, Theorem 7.6.6, pp 209-210 of the same book shows that if the Fourier transform of $L^p(\mathbb{R}^n)$ consists of distributions of order $k$, then $k\geq n(\frac{1}{2}-\frac{1}{p})$. In other words, there certainly are distributions of order greater but not smaller (ahem) than $n(\frac{1}{2}-\frac{1}{p})$ which are Fourier transforms of elements of $L^p(\mathbb{R}^n)$ if $p>2$.<|endoftext|> -TITLE: Naturally occurring, non-amenable Zappa-Szep products of discrete amenable groups? -QUESTION [10 upvotes]: We say $G$ is the Zappa-Szep product of two subgroups $K$ and $P$ if $K\cap P = \{e\}$ and the function $K\times P \to G$, $(k,p)\mapsto kp$, is bijective. -The Iwasawa decomposition shows that we can have amenable $K$ and $P$ such that $G=KP$ is non-amenable. However in such examples $K$ is usually not amenable when considered as a discrete group. -Where is a good place for me to look for examples of "naturally occurring" Zappa-Szep products where the two "factors" are amenable as discrete groups, yet the larger group itself is non-amenable? Can this ever happen for e.g. lattices in semisimple Lie groups? - -REPLY [6 votes]: Here is an observation of Yair Glasner that I previously missed. -Let $k$ be the field of real algebraic numbers and consider $G=\text{PSL}_2(k)$. -Let $P=B(\mathcal{k})$ be the standard Borel subgroup and $K=\text{PSO}_2(k)$. -Then $K$ and $P$ are solvable, $K\cap P=\{e\}$ and $G=KP$, but $G$ is not amenable (as it contains free groups). -All countable examples I am aware of are variants of this example. -In particular, I am not aware of any finitely generated example.<|endoftext|> -TITLE: A "dense" extension of the set of primitive recursive functions -QUESTION [6 upvotes]: Let $\mathcal{PR}$ be the set of primitive recursive functions. Let $\mathcal{PR}(f)$ be $\mathcal{PR}$ which we have amplified by adding (a recursive) $f$ the in the set of initial functions. To make this a true extension, assume $f$ outgrows all primitive recursive functions. -Now, $\mathcal{PR}$ is "dense" in the sense that it has the Ritchie-Cobham property, i.e. every function computable in primitive recursive time is primitive recursive and vice versa. - -Does $\mathcal{PR}(f)$ have the Ritchie-Cobham property? - -To me it seems natural the extension preserves the property and I would guess there is some "direct" argument for this, which I just fail to see. - -REPLY [5 votes]: The answer is yes. -Suppose a function $g$ is computable by a procedure $p$ whose -computation running time is bounded by a function -$h\in\newcommand\PR{\text{PR}}\PR(f)$. I claim that $g\in\PR(f)$. -To see this, let's first argue that $\PR(f)$ is closed under the -bounded search operator $(x,z)\mapsto\mu y -TITLE: Example of a smooth family of projective surfaces with non-vanishing integrals of Todd classes -QUESTION [7 upvotes]: Motivation: -Let $\pi\colon S \rightarrow B$ be smooth projective morphism of relative dimension 2 over a smooth projective scheme $B$. If the stucture sheaves of the fibres do not have higher cohomology, we have by GRR: -$$ \pi_* td(S/B) = ch(\mathcal{O}_B) = 1$$ -so push forwards of higher Todd classes are 0. -For trivial families of surfaces, we have obviously $td(S/B)_i = 0$ for $i>2$. -Therefore, the following question is not entirely trivial: -Question: -For arbitrarily $i>0$, what are examples of smooth families of projective surfaces $\pi\colon S\rightarrow B$ with non-zero $\pi_* td(S/B)_i$? - -REPLY [7 votes]: Edit. The odd Bernoulli numbers are zero, of course! So the first attempt below is wrong. The revised examples use complete subvarietes of moduli spaces of curves. These revised examples use nonvanishing of the even Bernoulli numbers. -Revised examples from complete subvarieties of moduli spaces of curves. Let $g\geq 2$ be an integer. Denote by $\mathcal{M}_g$ the Deligne-Mumford stack over $\text{Spec}\ \mathbb{C}$ parameterizing families of smooth, projective, geometrically connected curves of genus $g$. Denote the universal curve as $$\pi:\mathcal{C}_g \to \mathcal{M}_g.$$ Denote by $\omega_{\pi}$ the relative dualizing sheaf on $\mathcal{C}_g$. -Definition. The Hodge bundle is the pushforward of the relative dualizing sheaf to $\mathcal{M}_g$, $$\mathbf{E}_g := \pi_* \omega_{\pi}.$$ For every integer $r\geq 0$, the kappa class of degree $r$ is the (cohomological) cycle class $$\kappa_r = \pi_*(c_1(\omega_{\pi})^{r+1})\in \text{CH}^{\ r}(\mathcal{M}_g).$$ -The Hodge bundle is a rank $g$, locally free $\mathcal{O}_{\mathcal{M}_g}$-module (for the étale site with the usual structure sheaf). The kappa classes are elements in the Chow groups of the stack; the tensor product of these Chow groups with $\mathbb{Q}$ equal the associated $\mathbb{Q}$-Chow groups of the quasi-projective coarse moduli space $M_g$ by work of Angelo Vistoli. Using the Grothendieck-Riemann-Roch formula, Mumford computed the Chern character of the Hodge bundle on the compactified moduli space $\overline{\mathcal{M}}_g$. -MR0717614 (85j:14046) -Mumford, David -Towards an enumerative geometry of the moduli space of curves. -Arithmetic and geometry, Vol. II, 271–328, -Progr. Math., 36, Birkhäuser Boston, Boston, MA, 1983. -http://www.dam.brown.edu/people/mumford/alg_geom/papers/1983b--EnumGeomModuli-NC.pdf -Restricting to the open substack $\mathcal{M}_g$ of $\overline{\mathcal{M}_g}$ simplifies Mumford's formula. -Theorem [Mumford, Formula (5.2), p. 304, Towards an enumerative geometry ...] For every integer $s\geq 1$, $\text{ch}_{2s}(\mathbf{E}_g)$ is torsion, and $$\text{ch}_{2s-1}(\mathbf{E}_g) =(-1)^{s+1}\frac{|B_{2s}|}{(2s)!}\kappa_{2s-1}\in \text{CH}^{2s-1}(\mathcal{M}_g)\otimes \mathbb{Q},$$ where $B_{2s}$ is the Bernoulli number (nonzero since $2s$ is even). -Corollary. The $\mathbb{Q}$-cycle class of the Chern character of $R\pi_*\mathcal{O}_{\mathcal{C}_g}$ equals $$(1-g) + \sum_{s\geq 1}(-1)^{s+1} \frac{|B_{2s}|}{(2s)!} \kappa_{2s-1}.$$ -Proof. By relative duality, the K-theory class of $R\pi_*\mathcal{O}_{\mathcal{C}_g}$ equals $1-[\mathbf{E}_g^\vee]$. Thus, the Chern character of $R\pi_*\mathcal{O}_{\mathcal{C}_g}$ equals $$(1-g) + \sum_{r\geq 1}(-1)^{r+1}\text{ch}_r(\mathbf{E}_g).$$ QED -Let $f:V\to \mathcal{M}_g$ be a $1$-morphism that is generically finite to its image from an integral, projective $\mathbb{C}$-scheme of dimension $r$. The following theorem is discussed in the book of Harris and Morrison (one version of the theorem is attributed to Arakelov-Raynaud-Szpiro, but I have not been able to confirm this). -MR1631825 (99g:14031) -Harris, Joe; Morrison, Ian -Moduli of curves. -Graduate Texts in Mathematics, 187. Springer-Verlag, New York, 1998. xiv+366 pp. -Theorem. [cf. Harris-Morrison, Theorem 6.33, Moduli of Curves] For every irreducible, proper substack of $\mathcal{M}_g$, the restriction of $\omega_\pi$ to the inverse image under $\pi$ of the proper substack is "ample". In particular, the pullback to $V$ of $\pi_*(c_1(\omega_\pi)^{r+1})$ is a zero-cycle of positive (rational) degree. -As always for Deligne-Mumford stacks, an invertible sheaf is "ample" if every sufficiently positive and divisible tensor power of the invertible sheaf is isomorphic to the pullback from the coarse moduli space of an ample invertible sheaf (in the usual sense). -Corollary. If $r$ is odd, $r=2s-1$, then the pullback to $V$ of $\text{ch}_{2s-1}(R\pi_*\mathcal{O}_{\mathcal{C}_g})$ is a zero-cycle of nonzero degree. -Kodaira constructed complete curves in $M_g$. This was generalized to higher-dimensional complete subvarieties of $M_g$ by E. Y. Miller in the 1980s. I found a very readable reference for higher-dimensional complete subvarieties written by Zaal. -Christiaan Zaal -Complete subvarieties of moduli spaces of algebraic curves. -Thesis, Universiteit von Amsterdam -https://pure.uva.nl/ws/files/3915897/36235_Thesis.pdf -Theorem. [Kodaira, Miller, Zaal] For every integer $n\geq 0$, there exists an integer $g(n)\geq 2$ and a $1$-morphism $f:V\to \mathcal{M}_{g(n)}$ that is generically finite to its image from an integral, projective, $\mathbb{C}$-scheme $V$ of dimension $n$. -(In the opposite direction, I believe that the theorem of Steven Diaz remains the best upper bound on the maximal dimension $n(g)$ of a complete subvariety of $M_g$.) -Denote by $\pi_f:\mathcal{C}_f\to V$ the pullback by $f$ of the universal family of curves $\mathcal{C}_g$. Then for every integer $s\geq 1$ with $2s-1\leq n$, $\text{ch}_{2s-1}(R(\pi_f)_*\mathcal{O}_{\mathcal{C}_f})$ is nonzero, and the even degree graded pieces of the Chern character (of degree $\geq 2$) are zero. -In particular, there exists an integer $g(1)\geq 2$ and there exists a generically finite $1$-morphism, $$e:W\to \mathcal{M}_{g(1)},$$ from a smooth, projective $\mathbb{C}$-curve $W$. For $V$ as above of dimension $n$, denote the $(n+1)$-dimensional product $V\times W$ by $B$. Denote the product $\mathcal{C}_f\times \mathcal{C}_e$ by $S$. Denote by $\rho:S\to B$ the morphism $\pi_f\times \pi_e$. As a fiber product of smooth, projective morphisms of relative dimension $1$, the morphism $\rho$ is smooth and projective of relative dimension $2$. -Corollary. For every integer $r$ with $1\leq r \leq n+1$, the cycle class $\text{ch}_r(\rho_*\mathcal{O}_S)$ has nonnegative degree on some irreducible, closed subvariety of $B$ of dimension $r$. -Proof. By Künneth's formula, $$\rho_*\mathcal{O}_S = \text{pr}_V^*((\pi_f)_*\mathcal{O}_{\mathcal{C}_f})\otimes \text{pr}_W^*((\pi_e)_*\mathcal{O}_{\mathcal{C}_e}).$$ By the multiplicative property of the Chern character, for every integer $s\geq 1$, $$\text{ch}_{2s-1}(\rho_*\mathcal{O}_S) = (1-g(1))\text{pr}_V^*\text{ch}_{2s-1}((\pi_f)_*\mathcal{O}_{\mathcal{C}_f}),$$ and this has nonzero degree on irreducible closed subvarieties of the form $X\times\{w_0\}$ of dimension $2s-1$. Also, $$ \text{ch}_{2s}(\rho_*\mathcal{O}_S) = \text{pr}_V^*\text{ch}_{2s-1}((\pi_f)_*\mathcal{O}_{\mathcal{C}_f})\cup \text{pr}_W^*\text{ch}_1((\pi_e)_*\mathcal{O}_{\mathcal{C}_e}).$$ This has nonzero degree on irreducible closed subvarieties of the form $X\times W$ of dimension $2s=(2s-1)+1$. QED -First attempt with ridiculous error because the odd Bernoulli numbers equal zero. -You can create many such examples with isotrivial families. The simplest isotrivial morphism is a smooth, projective morphism of relative dimension $1$ whose geometric fibers are isomorphic to the projective line, $$\rho:C\to X.$$ The pushforward of $T_\rho =\textit{Hom}_{\mathcal{O}_C}(\Omega_\rho,\mathcal{O}_C)$ is a locally free $\mathcal{O}_X$-module of rank $3$ whose total Chern class equals $1-a$ for a cycle class $a$ of degree $2$. For a locally free $\mathcal{O}_X$-module $E$ of rank $2$, if $C/X$ represents the functor on $X$-schemes of invertible quotients of the pullback of $E$, then the class $a$ equals $$a = c_1(E)^2 - 4c_2(E).$$ Up to scaling and torsion, this is the unique polynomial expression in Chern classes of $E$ of cohomological degree $2$ that is compatible with arbitrary pullback and that is preserved by the operation of tensoring $E$ with invertible sheaves. -The first Chern class $c = c_1(T_\rho)$ has square equal to a pullback, $$c^2 = \rho^* a.$$ Thus, the relative Todd class of $\rho$ equals $$\text{Td}(\rho) = [C]\cap\left(\sum_{m\geq 0} \frac{B_{2m}}{(2m)!}\rho^*(a^m) - \sum_{m\geq 0}\frac{B_{2m+1}}{(2m+1)!}c\cup \rho^*(a^m) \right). $$ Thus, the pushforward of the Todd class equals, $$\rho_*\text{Td}(\rho) = -2[X]\cap \sum_{m\geq 0} \frac{B_{2m+1}}{(2m+1)!} a^m.$$ -In particular, for a second smooth, projective morphism of relative dimension $1$, $$\sigma:X\to B,$$ for $S$ defined to be $C$, and for the composition $\pi=\sigma\circ \rho$, the pushforward of the relative Todd class equals $$\pi_*\text{Td}(\pi) = \sigma_*\left(-2\text{Td}(\sigma)\cap \sum_{m\geq 0}\frac{B_{2m+1}}{(2m+1)!} a^m \right).$$ In the very simplest case that $X$ equals $\mathbb{P}^1_B$, $\sigma$ is the natural projection, and for any choice of constant cross-section $s:B\to \mathbb{P}^1_B$ (the $\infty$-section, for instance), this gives $$\pi_*\text{Td}(\pi) = -[B]\cap \left(\sum_{m\geq 0} \frac{B_{2m+1}}{(2m+1)!} (s^*a)^m +2\sum_{m\geq 0} \frac{B_{2m+1}}{(2m+1)!}\sigma_*(a^m) \right).$$ To be very explict, if $E$ is the locally free sheaf $\sigma^*\mathcal{L}\oplus \mathcal{O}_{\mathbb{P}^1}(1)$ for some choice of invertible $\mathcal{O}_B$-module $\mathcal{L}$ with first Chern class $c_1(\mathcal{L})=\lambda$, this evaluates to $$\pi_*\text{Td}(\pi) = -[B]\cap \left( \sum_{m\geq 0} \frac{B_{2m+1}}{(2m+1)!}\lambda^{2m} + 4\sum_{m\geq 1} \frac{B_{2m+1}}{(2m+1)!} m\lambda^{2m-1}\right).$$ If $\mathcal{L}$ is ample, then $\lambda^r$ is nonzero for every integer $r=0,\dots,\text{dim}(B).$<|endoftext|> -TITLE: 3-manifolds with all geodesics closed -QUESTION [6 upvotes]: A theorem of Bott states that if a manifold admits a metric with all geodesics closed, then its homology is isomorphic to the homology of one of the manifolds from the list: $S^n, \mathbb{RP}^n, \mathbb{CP}^n, \mathbb{HP}^n$ or $\mathbb{C}a\mathbb{P}^2$. -The problem of constructing such metrics is known to be extremely hard. -So my particular interest is: what else is known about this problem in the case of dimension $3$? -For example, are all homology spheres known to admit such metrics? If a homology sphere (or a homology $\mathbb{RP}^3$ --- by the way, are there any examples which are not $\mathbb{RP}^3$?) has finite $\pi_1$, its universal cover is $S^3$, and I believe that all geodesics will be closed in the metric induced by the standard metric on $S^3$. -But I've just learned that there are examples of homology spheres with infinite fundamental group. Their Thurston geometry is modeled on the universal cover of $\operatorname{SL}(2, \mathbb{R})$. It would be very interesting to know if such homology spheres admit metrics with all geodesics closed. - -REPLY [6 votes]: There are integral homology spheres in the following Thurston geometries: $S^3$, $\mathrm{PSL}(2,\mathbb{R})$, and $H^3$. No manifold of the latter two types (when equipped with any metric) can have all of its geodesics being closed. This follows from observing that the fundamental group contains a non-trivial free group, and then a Gromov-Hausdorff limiting argument.<|endoftext|> -TITLE: Integers $h$ such that $xy(x+y) = h$ has many integer solutions -QUESTION [14 upvotes]: In the following paper (http://www.ams.org/journals/jams/1991-04-04/S0894-0347-1991-1119199-X/), C.L. Stewart showed that there exist infinitely many integers $h$ such that the equation -(1) -\begin{equation} xy(x+y) = h -\end{equation} -has at least 18 solutions in co-prime integers $x$ and $y$. In the same paper, he also conjectured (his conjecture covers far more) that there exists positive numbers $r,c$ such that for all integers $h$ with $|h| \geq r$, the equation (1) has at most $c$ solutions in co-prime integers $x$ and $y$. -My question is, is it reasonable to conjecture that one may take $c = 18$? That is, does there exist infinitely many $h$ for which (1) has at least 19 solutions in co-prime integers $x$ and $y$? - -REPLY [13 votes]: There seem to be lots of $h$ which have $20$ solutions. Among squarefree $47$-smooth numbers I found -$$101623830, 363993630, 455885430, 1418488890, 15427730010, 31983962010, 322640788470, 1087394017710, 11406069164490, 304250263527210$$ with $20$ solutions and $903210$ and $23730036330$ with $24$. -EDIT: Oops: my program was counting only solutions with $x > 0$ (but $y$ of any sign). Those numbers $101623830, \ldots, 304250263527210$ actually have $30$ solutions each, and $903210$ and $23730036330$ have $36$.<|endoftext|> -TITLE: Who first used the word "Simplex"? -QUESTION [17 upvotes]: Who first used the word "Simplex" to describe the considered geometric figure? - -REPLY [19 votes]: According to Jeff Miller's Earliest Known Uses of -Some of the Words of Mathematics, the first known occurrence is in Schoute’s Mehrdimensionale Geometrie of 1902. - -REPLY [8 votes]: The earliest references found by Google date back to 1909. W.H. Bussey mentions it in an article from that year, On the tactical problem of Steiner published in the Bulletin of the AMS. -Precisely, what he defines to be a simplex is the boundary of what is called a simplex today: - -The $l+1$ points of such a set, if taken $l$ at a time, determine a -number of $l-1$-spaces whose points constitute a set that may -conveniently be called a simplex$^*$ of order $l$. The $l + 1$ points are called vertices. -$^*$*The word is used in geometry of n-dimensions to denote the configuration analogous to the triangle in the plane or the tetrahedron in 3-space. - -http://www.ams.org/journals/bull/1909-16-01/S0002-9904-1909-01845-2/S0002-9904-1909-01845-2.pdf<|endoftext|> -TITLE: Does a Kloosterman sum composed with a rational function exhibit square root cancellation? -QUESTION [5 upvotes]: Denote the classical Kloosterman and Salié sums, respectively, as $KL(a,b) = \sum_{r \in F_*} e(ar+\frac{b}{r})$ and $SL(a,b) =\sum_{r \in F_*} \chi(r) e(ar+\frac{b}{r})$, where $\chi(\cdot)$ is the quadratic character. It follows from Weil's proof of the Riemann hypothesis for curves (or more elementary considerations in the case of $SL(a,b)$) that for $a,b \neq 0$ that $|K(a,b)|, |SL(a,b)| \lesssim |F|^{1/2}$. -Informally, I'd like to know if we can prove square root cancellation in the composition of a Kloosterman (respectively, Salié) sum with a rational function. In particular: - - -Let $R(t)$ be a rational function in $t$. For $b \neq 0$ can one prove $$ |\sum_{t \in F}'KL(R(t),b)|\lesssim |F|$$ and/or $$|\sum_{t \in F}'SL(R(t),b) \lesssim |F|$$ - where $\sum'$ denotes the exclusion of the poles of $R(t)$. Of course the implicit constant should depend on $R$ but be independent of the field? - - -I'm particularly interested in the special case when the numerator and denominator of $R$ have degree $2$, but would be very interested in the general case as well. -There's been a lot of work by Bombeiri, Deligne, Hooley, Katz, Sperber, and others on deducing multivariate exponential sum estimates from Deligne's theorems, but many of these theorems are hard to parse for someone who isn't an expert in algebraic geometry. I'd also be interested in the above sums twisted by a multiplicative character in $t$. - -REPLY [2 votes]: It's fine for Kloosterman sums, as long as R is non-constant. The simplest argument (formally) being to use Deligne's general form of the Riemann Hypothesis. Briefly, KL(R(t),b) is the trace function of the sheaf R^*Kl, where Kl is the Kloosterman sheaf of rank 2 defined by Deligne. Katz has shown that the geometric monodromy group of that is SL_2, so R^*Kl has the same geometric monodromy group (a priori it could become a finite-index subgroup, but SL_2 has no finite-index algebraic subgroup). Then the Grothendieck-Lefschetz trace formula and the Riemann Hypothesis apply to get what you want. This kind of things is sketched in Section 6 of the Polymath8 paper (see, e.g., Prop. 6.11), or in "A study in sums of products" by Fouvry-Kowalski-Michel, among other places, and the same argument applies with a twist by a multiplicative character<|endoftext|> -TITLE: Can more polynomial time compensate for less polynomial memory? -QUESTION [12 upvotes]: I'm wondering what is known about the relation between time and memory for polynomial-time algorithms (which are necessarily also polynomial-space). In particular, I would like to learn what is known about less time vs. less memory, in the following sense: -Is it known whether (say) any language $L$ recognized by a polynomial-time algorithm $A$ which is polynomial-space $O(n^s)$ ($s>2$ say), can also be recognized by a polynomial-time algorithm $B$ which is polynomial-space $O(n^{s/2})$? Presumably any $B$ with less memory would need significantly more time, but can we restrict this extra time to polynomial-time? -So, recapping: is there for poly-time algorithms, some general way to compensate for less memory, in polynomial time? Answers and references greatly appreciated. -update: Dan Brumleve and Timothy Chow have given valuable insights and references in the comments and in this chat. It seems that time-space trade-offs like the one in this question are still a bridge too far. - -REPLY [3 votes]: I suspect that what you're hoping for—i.e., that for all $a$ and $b$ there exists $c$ such that $\mathrm{DTISP}(n^a, n^b) \subseteq \mathrm{DTISP}(n^c, n^{b/2})$—is false, but for sure nobody has proved this. In fact I suspect it may not even be true that $\mathrm{DTISP}(n^a,n^b) \subseteq\mathrm{DSPACE}(n^{b/2})$ (i.e., I suspect you cannot compensate for the loss of space with any amount of time). Certainly, if we consider all problems solvable in space $O(n^b)$ (with no restrictions on time), then -the space hierarchy theorem tells us that we will be able to solve certain problems that are simply unsolvable in space $O(n^{b/2})$, no matter how much time we give ourselves. But settling your question will probably require proving a timespace hierarchy theorem, and very little is known unconditionally about such things (see this question on CS Theory Stackexchange for some information). As another illustration of our ignorance about time-space tradeoffs, note that it is not known whether a problem that can be solved in polynomial time, and that can also be solved (possibly by a different algorithm) in polylogarithmic space, is in SC (i.e., is solvable is polynomial time and polylogarithmic space simultaneously).<|endoftext|> -TITLE: Direct bijections for $s,t$-Fibonomial identities -QUESTION [6 upvotes]: Sagan and Savage gave a combinatorial interpretation of a polynomial generalization of Fibonomial coefficients. Their proof uses the recurrence relation for the Lucas polynomials that generalize the Fibonacci numbers, namely $\{0\}=0$, $\{1\}=1$, $\{n\}=s\{n-1\}+t\{n-2\}$ for $n\ge 2$. Define $s$ as a weight of a monomino, $t$ is a weight of a domino, and extend the weight function multiplicatively as usual. Then the $s,t$-Fibonacci number $\{n+1\}$ is the sum of all possible weights of an $n\times 1$ strip of cells. Subsequently, $\{n\}!=\{n\}\{n-1\}\dots\{1\}$, $\{0\}!=1$, and the $s,t$-Fibonomial coefficient $$ -{m+n \brace m}=\frac{\{m+n\}!}{\{m\}!\{n\}!} -$$ -is the sum of weights of fillings of an $m\times n$ rectangle by pairs of complementary partitions $(\lambda,\lambda^*)$ so that rows of $\lambda$ and columns of $\lambda^*$ are filled by monominos and dominos, and every column of $\lambda^*$ starts with a domino (see Figure 2 in the paper). -The proof that ${m+n \brace n}$ is a combinatorial interpretation of the $s,t$-Fibonomial coefficients is recursive, using the identity -$$ -{m+n \brace m}=\{n+1\}{m+n-1 \brace m-1} + t\{m-1\}{m+n-1 \brace m}. -$$ -However, I am wondering if there have since been any direct bijections, using this combinatorial interpretation of $s,t$-Fibonomials, corresponding to the following identities: -$$ -\begin{split} -{m+n \brace m}&={m+n \brace n}\\ -{m+n+1 \brace m+1}\{m+1\}&={m+n+1 \brace m}\{n+1\}={m+n \brace m}\{m+n+1\}\\ -{m+n \brace m}\{m\}!&=\{m+n\}\{m+n-1\}\dots\{m+1\} -\end{split} -$$ - -REPLY [10 votes]: I'm glad to see you're interested in this question. A number of people have thought about it including Art Benjamin and myself. The person who seems to have gotten the furthest is Curtis Bennett who came up with a new way to view the rectangular tilings in terms of tilings of a triangle with a lattice path through it which explains those special dominoes in an intuitive way. Last I heard he and a student were trying to prove identities like the ones you state using their interpretation. -If you are looking for other open problems involving the Fibonomials, see my paper with Xi Chen, The fractal nature of the Fibonomial triangle, Integers, 14 (2014), A3, 12 pp., as well as the article Generalized Fibonacci polynomials and Fibonomial coefficients (with Tewodros Amdeberhan, Xi Chen and Victor H. Moll), Ann. Combin., 18 (2014), 541-562. -Finally, one can play the same game (replacing n by F_n or an appropriate polynomial) with Catalan numbers and even with Catalan analogues for Coxeter groups. The resulting fractions always give integers (or polynomials in the weighted case) and nobody really understands why. See the slides for my talk Open Problems for Catalan Number Analogues at -http://users.math.msu.edu/users/sagan/Slides/OpcH.pdf<|endoftext|> -TITLE: Duistermaat and Kolk's lost chapters on Lie groups -QUESTION [15 upvotes]: In Duistermaat and Kolk's book Lie Groups, it is written in the preface that "the text contains references to chapters belonging to a future volume". I could not find this second volume anywhere. Has it been published? Is it available somewhere? -For example, on page 51, they refer to Chapter 14, while the book only has 4 chapters. I would be very much interested to read these additional chapters. - -REPLY [8 votes]: I was the last PhD student of Duistermaat, and I'm pretty sure there is no second volume. I also don't remember Hans Duistermaat speaking of working on a second volume. -The last textbook that Duistermaat and Kolk completed was Distributions, published in 2010.<|endoftext|> -TITLE: Systems of imprimitivity for unitary representations - reference request -QUESTION [5 upvotes]: Let $G$ be a finite subgroup of the group $U_d(\mathbb{C})$ of unitary transformations of $\mathbb{C}^d$. Suppose that $G$ acts irreducibly but is imprimitive, meaning that there is a nontrivial direct sum decomposition $\mathbb{C}^d = \bigoplus_{i = 1}^r V_i$ such that each $g \in G$ permutes the $V_i$. -Then it seems that the $V_i$ are necessarily orthogonal: I wrote up a proof here. However I'm happy to admit that it took me quite some time to find this proof, and I still don't know of a reference. This must surely be well-known, and was probably known to Frobenius. Can anyone supply me with a reference? - -REPLY [6 votes]: This is not an answer, but here is another proof in the same spirit as yours. -Write $(-\vert -)$ for the canonical scalar product of ${\mathbb C}^d$. -Since the $G$-representation ${\mathbb C}^d$ is irreducible, we have -$$ -{\mathbb C}^d = \bigoplus_{g\in G/G_1} g.V_1 -$$ -where $G_1$ is the stabilizer of $V_1$ in $G$. Write $(-\vert -)_1$ for the restriction of $(-\vert -)$ to $V_1\times V_1$; it is $G_1$-invariant. For all $g\in G$ define a scalar product on $g.V_1$ by -$$ -(gv_1 \vert gw_1 )_{gV_1} = (v_1 \vert w_1 ) -$$ -It does not depend on the choice of $g$. Let $<-\vert ->$ be the scalar product on ${\mathbb C}^d$ given by the orthogonal sum of the $(-\vert -)_{gV_1}$, $g\in G/G_1$. It is clearly $G$-invariant. Now since the representation ${\mathbb C}^d$ is irreducible the scalar products $(-\vert -)$ and $<-\vert ->$ are proportional and we are done. N.B. Of course Schur's lemma is hidden in my proof.<|endoftext|> -TITLE: On measurable cardinals -QUESTION [8 upvotes]: Let $\kappa$ be an uncountable cardinal and $(P(\kappa),\cap,\cup, ^c,\kappa,\emptyset)$ the Boolean Algebra of all subsets of $\kappa$. -Fact: If there exists a countably complete non-principal ultrafilter on $P(\kappa)$, then $\kappa$ is larger than or equal to a measurable. -My question: Is there a Boolean Algebra $\mathcal{A}$ of subsets of $\kappa$, $\mathcal{A}$ of size strictly less than $2^\kappa$, and the existence of a countably complete non-principal ultrafilter on $\mathcal{A}$ would imply that $\kappa$ is (added: larger than or equal to) a measurable/inaccessible/strong limit? -I am interested even in special cases, e.g. $\kappa$ is regular. - -REPLY [9 votes]: Your question is very interesting in the case $\kappa^+<2^\kappa$. -For this case, recall that a cardinal $\kappa$ is weakly measurable if every family of $\kappa^+$ many subsets of $\kappa$ admits a $\kappa$-complete nonprincipal filter measuring them. This notion was introduced and studied by my student Jason Schanker in his dissertation. There are a variety of equivalent characterizations. -Jason proved that if $2^\kappa$ is larger than $\kappa^+$, then this concept is not necessarily equivalent to measurability, for he constructed models of ZFC with a weakly measurable cardinal, but no measurable cardinal. -In such a model, therefore, where $2^\kappa=\kappa^{++}$, we have measures for all the families of sets of size smaller than $2^\kappa$, but $\kappa$ is not measurable. So this provides a negative answer to your question. -Meanwhile, weakly measurable cardinals are necessarily weakly compact, inaccessible and much more, so there is a positive answer to the version of your question at the end where you ask for less than measurability. -Indeed, the existence of $\kappa$-complete nonprincipal filters measuring families of $\kappa$ many subsets of $\kappa$ is equivalent to $\kappa$ being weakly compact. In this sense, weak measurability is a generalization of weak compactness from $\kappa$ to $\kappa^+$. -Thus, if $2^\kappa=\kappa^+$, the answer to your question is no, since the existence of such filters for families of at most $\kappa$ many subsets of $\kappa$ is equivalent to $\kappa$ being weakly compact, which is a strictly weaker large cardinal concept than measurability.<|endoftext|> -TITLE: What's the motivation of entropy as a combinatorical tool? What problems is it able to solve? -QUESTION [32 upvotes]: I am interested in using Shannon's entropy in combinatorics. It is often presented with a motivation of how much information can be passed, but assume I am not interested in that, I want to understand why it is useful as a tool to bound objects (with examples now). -Several examples that demonstrate that very nontrivial results follow from relatively entropy arguments are: - -Bregman's inequality about permanents. -Whiteny's inequality about volume and its projections. -Shearer's lemma and numerous consequences; bounds on the number of independent sets in bipartite d-regular graph, number of homomorphisms to graph... -The largest size of an intersection unique family (see https://link.springer.com/article/10.1007/BF02579460 ). - -If you could help me with the following I'd be very happy; - -How would someone just interested in counting and bounding combinatorics arrive at entropy? -A common theme (noted by my proffessor when teaching about it) to some of the results are that original proofs used convexity arguments, how healthy is it to be think of entropy in combinatorics as a clean way to state convexity arguments and deriving nontrivial results by assigning weights to objects and bounding sums? -When can proofs be translated to entropy proofs? For instance I find it strange there is the $k-intersection$ proof with entropy, while I'm not aware of an entropy proof for sperner's theorem about antichains. -When are problems suspectible to entropy methods? - -Cheers. - -REPLY [3 votes]: I wanted to share another answer, that I feel comfortable with after trying to look for examples and intuition. -We start with the following riddle: -Find the least $k(n)$ so that there are $k$ subsets $D_1,...,D_k$ of $[n]$ s.t $|D_1\cap A|,...,|D_k \cap A|$ uniquely determine $A$ for each $A\subset[n]$. -This is an interesting question (coming from- there are $n$ bits, at each turn you pick a subset and are told how many of them are turned on, but you ask the questions nonadaptively, and need to determine the bits). -An easy initial observation is this; -Each intersections has at most $n+1$ options, and so $(n+1)^k\geq 2^n$, so $k\geq n/log_2(n)$. -Another way to phrase this, is that we can encode $n$ bits using $k\cdot log_2(n+1)$. Entropy for me, is all about encoding things. Now here will come the twist of entropy, it is not about deterministically encoding things, but rather, encoding most of the things, that is, we want to be able to encode\decode a random message of the possible inputs (with the given distribution on it). This is formalized, and this is where the expression for entropy most naturally arises for me, in the theorem of Shannon: -https://en.wikipedia.org/wiki/Shannon%27s_source_coding_theorem -I won't repeat it here, but the point is that we have $X_1,..,X_n$ iid in a probability space $W$. Given $w$ in our world, we define the random variable $Y(w)$ to be $P(w' s.t X_1(w')=X_1(w),X_2(w')=X_2(w)..)$. We can calculate the expectation, but we don't have any theorem to handle concentration of product. So we do the usual, we take $log$. From this and the large law of numbers we easily obtain the theorem. -It is nice because some easy corollaries are: $H(X,Y)\leq H(X)+H(y)$, $H(uniform)$ is maximizes given size of range. -Back to the riddle, the point is that if we take a bunch copies of random $A$, if we insisted to deterministically encode each one, we'd get the same bound. -But there are $2$ excellent twists: - -For uniform distribution, even though we now allow to encode only with h.p, it is still hard and demands many bits. -For other distributions, this "tensorization" made a difference, a formula we can calculate. - -Indeed, we have $n=H(uniform A)=H(|D_1\cap A|,...,|D_k \cap A|)\leq \sum H(|D_i \cap A|)$. So we want to bound $H(|D_i \cap A|)$. -Now if we're brute, and just insist on deterministically encoding all possible values of $|D_i \cap A|$, we don't get any improvement, but the whole point is that we allow us to only do so with h.p. -Since we expect this thing to be concentrated around $|D_i|/2$ with an interval of around $~\sqrt(|D_i|)$ around, we need only ~$log(\sqrt|D_i|$ bits, which is about half our original! (i.e we got a bound of about $2n/log(n)$). -To summarize, the trick is to tensorize (maybe there is a slight connection to the proof of the crossing inequality) by allow a small degree of freemdom, which traps uniform the uniform distribution.<|endoftext|> -TITLE: SO(3) monopole Floer homology -QUESTION [13 upvotes]: From what I understand about work on the Witten conjecture relating Donaldson and Seiberg-Witten invariants, the main strategy has been to relate them with the use of the "SO(3) monopole" theory developed by Feehan and Leness (see their paper here). -Both the Donaldson invariants and the Seiberg-Witten invariants have closely related Floer homologies for $3$-manifolds. -Is it possible to define such a Floer homology for $SO(3)$ monopoles? Are there clear reasons that this may or may not be possible? -As a first step, I am curious about whether there is a natural Chern-Simons-like functional $\mathcal{L}$ such that $SO(3)$ monopoles on a cylinder $\mathbb{R} \times Y$ correspond to negative gradient flowlines of $\mathcal{L}$. - -REPLY [5 votes]: It seems that not much work has been done for $SO(3)$ monopoles on three-manifolds. As far as I know, nobody has yet defined a Floer homology for that. However, Feehan, Leness and Kutluhan have started to investigate that, see section 3 of -https://www.math.ias.edu/files/feehanreport.pdf . -They don't seem to be interested in constructing an $SO(3)$ monopole homology, but they have worked on some issues crucial to developing such a theory (finding appropriate perturbations, solving compactness issues and studying the structure of the flow lines). -They say that $SO(3)$ monopoles on $\mathbb{R} \times Y$ correspond to the gradient flow lines of the $U(2)$ Chern-Simons-Dirac functional. If I am not mistaken, it is just the $U(2)$ Chern-Simons functional plus the usual spinor contribution (pairing between $\Phi$ and $D_B \Phi$).<|endoftext|> -TITLE: For which number of pairs is it an advantage to start in memory -QUESTION [10 upvotes]: Players A and B play memory starting with $n$ pairs of cards. We assume that they can remember all cards which have been turned. At his turn a player will first recall if two cards already turned match. If so he will choose such a pair, turn them and get a pair. If not he will turn uniformly at random a card which has not yet been turned. Then if the corresponding match has turned already he will turn that card and get a pair. If not he will again turn uniformly at random a card which has not yet been turned. As usual if a player gets a pair he can continue. Player A will start. For which $n$ does Player A have an advantage by being able to start and for which $n$ is it actually a disadvantage? -Here is what I considered so far: If $E_{k,l}$ denotes the expected number of pairs the starting player will get when $k$ cards are known and $l\geq k$ cards are unknown one can derive the following recursion formula -$$ -E_{k,l}=\frac{k}{l} \left( E_{k-1,l-1}+1 \right)\\ -+\frac{(l-k)}{l} \frac{1}{(l-1)} \left( E_{k,l-2}+1 \right)\\ -+\frac{(l-k)}{l} \frac{k}{(l-1)} \left( \frac{k+l}{2}-1-E_{k,l-2} \right)\\ -+\frac{(l-k)}{l} \frac{(l-k-1)}{(l-1)} \left( \frac{k+l}{2}-E_{k+2,l-2}\right). -$$ -Furthermore we have $E_{k,k}=k$. The expected number of pairs players $A$ resp. $B$ will get are $E_{0,2n}$ resp. $n-E_{0,2n}$. The following list show the expected number of pairs for A and B for $n\leq 24$. - -It seems that for larger $n$ the expected number of pairs for $A$ and $B$ are closer together. Can this be proven rigorously? For $n=18$ the expected number of pairs of $A$ and $B$ are equal and the game is fair in some sense. Do there exist other numbers $n$ with that property? Is it possible to give a characterization of the numbers for which $A$ has an advantage? -A has an advantage for $n\in \{ 1,4,7,8,10,11,14,17,18,20,21,23,24,\dots\}$. -B has an advantage for $n\in \{ 2,3,5,6,12,13,15,16,19,22,\dots\}$ -Octave code: - -n=50; A=diag(0:n-1);A(1,3)=1; -for s=4:2:n-1 - for l=s/2+1:s - k=s-l; - if k>0 - A(k+1,l+1)=k/l*(1+A(k,l)); - end - if l>k - A(k+1,l+1)=A(k+1,l+1)-(l-k)/l*k/(l-1)A(k+2,l); - end - if l>k+1 - A(k+1,l+1)=A(k+1,l+1)-(l-k)/l(l-k-2)/(l-1)A(k+3,l-1); - end - if l>k && l>2 - A(k+1,l+1)=A(k+1,l+1)+(l-k)/l/(l-1)(1+A(k+1,l-1)); - end - end end B=zeros(n/2-1,3); for i=1:n/2-1 B(i,:)=[i (i+A(1,2*i+1))/2 (i-A(1,2*i+1))/2]; disp([num2str(i) ':' num2str(A(1,2*i+1)) ' ' - num2str((i+A(1,2*i+1))/2)]); end -B(:,1) rats(B(:,2)) rats(B(:,3)) -Awins=''; Bwins=''; for i=1:n/2-1 if A(1,2*i+1)>0 Awins=[Awins ',' - num2str(i)]; else Bwins=[Bwins ',' num2str(i)]; end end Awins Bwins - -REPLY [4 votes]: This problem has been studied (and solved for many values) in this MSc thesis: -Erik Alfthan: Optimal strategy in the childrens game Memory. - He defines four strategies: - -Bad: open a known card, then an unknown that may match, -Safe: open an unknown card, then a known card, matching if possible, -Risky: open an unknown card, then match if possible, i.e. an known matching card - if possible, otherwise an unknown card that may match, -Passive: open two, known, unmatching cards. - -He proves that you should never play 'Bad' and there are situation when 'Passive' would be needed (easy example: 100 pairs and 99 unmatched cards are known, i.e., there's only 1 unknown pair), so the game would never terminate, so he disallows this strategy, i.e., players need to open with an unknown card. -I copy here my favorite table, followed by open problems. - -n = number of pairs on board -j = number of unknown pairs -Chosen strategy, 1 : risky, 0 : safe, # : no choice -  - j:0 1 2 3 4 5 6 7 8 9 ... - n - 2 # 1 # - 3 # 1 0 # - 4 # 1 0 1 # - 5 # 1 0 1 0 # - 6 # 1 0 1 0 0 # - 7 # 1 0 1 0 0 1 # - 8 # 1 0 1 0 0 1 0 # - 9 # 1 0 1 0 0 0 0 0 # -10 # 1 0 1 0 1 0 1 0 1 # -11 # 1 0 1 0 1 0 1 0 1 0 # -12 # 1 0 1 0 1 0 0 0 1 0 1 # -13 # 1 0 1 0 1 0 0 0 1 0 1 0 # -14 # 1 0 1 0 1 0 0 0 1 0 1 0 1 # -15 # 1 0 1 0 1 0 0 0 1 0 1 0 1 0 # -16 # 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 # -17 # 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 # -18 # 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 # -19 # 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 # -20 # 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 # -The pattern of the table continues, computation with as many as 200 -pairs, renders that, except for very small boards, the strategy should be Safe -when $j$ is even and Risky when $j$ is odd. -Conjecture 1. [Alfthan] Except for smaller boards, $n < 16$, the optimal strategy -is to use the risky strategy when $j$ is odd and safe strategy when $j$ is even. -Conjecture 2. [Alfthan] Except for smaller $j$, $j < 18$, the expected number of gained cards -is monotone in $n$ for fixed $j$, growing if $j$ is even and declining if $j$ is odd. -Update: I've just discovered this nice blogpost about the same game: -https://possiblywrong.wordpress.com/2011/10/25/analysis-of-the-memory-game/ -This has a 5. possible strategy that Alfthan seems to have missed: - -NON: pick up a face-down card and, if its match is known, a non-matching face-up card -(leaving a known pair on the table!), otherwise another face-down card. - -This might be advantageous in some cases, so if we allow this move, the above table gets changed! See the blogpost for details. - -REPLY [2 votes]: There is also an older paper which the above thesis doesn't refer to: U. Zwick & M. S. Paterson, "The memory game", Theoretical Computer Science 110 (1993), 169-196. I havent looked at the details.<|endoftext|> -TITLE: When does the following congruence identity hold? -QUESTION [5 upvotes]: Let $m$,$l$ be coprime integers where $m,l\geq 2$. For any integer $a$ and positive base $b \ (b\geq 2)$, let -$ -[a]_b -$ denote the element of $\{0,\ldots, b-1\}$ that satisfies the equivalence -$[a]_b \equiv a \bmod b$. -For any integer $n$, one can write -$$ -nl[l^{-1}]_m - nm[(-m)^{-1}]_l = n, -$$ as Bézout's Identity yields -$$ -l[l^{-1}]_m - m[(-m)^{-1}]_l = 1 -$$ (the existence of inverses is assured by the coprimality of $m$ and $l$). -Question: Under what conditions on $n$ does the equality -$$ -l[nl^{-1}]_m - m[n(-m)^{-1}]_l = n -$$ hold? - -REPLY [3 votes]: Each integer number $n$ can be uniquely expressed in the form $n=lx+my$, where $0\le x\le m-1$ and $y\in \mathbb{Z}.$ This artificial numeral system is more suitable for the riven problem because for $n=lx+my$ -$$l[nl^{-1}]_m - m[n(-m)^{-1}]_l = lx -m[-y]_l. $$ -The last expression is equal to $n$ iff $-l -TITLE: $p$-adic Hodge Theory for rigid spaces, after P. Scholze -QUESTION [32 upvotes]: I was going over P. Scholze's paper on $p$-adic Hodge Theory for rigid analytic varieties. -This question is around the "Poincaré Lemma" in the paper. -Throughout, let $X$ be a proper smooth rigid analytic variety over $\mathbf{Q}_p$, of pure dimension $d$. - -Corollary 6.13 says that the "$B_{\rm dR}$-version" of the Poincaré Lemma for de Rham-étale cohomology follows directly from Proposition 6.10, for smooth and proper rigid analytic varieties. - -Prop. 6.10 is a description of the sheaf $(\mathcal{O}\mathbb{B}_{\rm dR}^+)_X$ pro-étale locally on $X$ as $\mathbb{B}_{dR}^+[\![t_1,\ldots,t_d]\!]$ for local sections $t_1,\ldots,t_d$ of $\mathcal{O}\mathbb{B}_{\rm dR}^+$. -Question 1: I don't quite get the implication Prop. 6.10 $\Rightarrow$ Cor. 6.13. -Could anyone who's gone over the paper please clarify it for me? -It should be something trivial along the lines: " pro-étale locally on $X$ the de Rham complex looks like this, hence the augmentation from $\mathbb{B}_{dR}^+[0]$ is a quasi-isomorphism". - -Remark 1. Let's make an example, and call $x := (x_1,\ldots, x_d)$, $x^{\pm\infty} := (x_1^{\pm 1/p^{\infty}},\ldots, x_d^{\pm 1/p^{\infty}})$, and "restrict" the complex $\mathcal{O}\mathbb{B}^+_{\rm dR}\otimes_{\mathcal{O}_X}\Omega^{\bullet}_X$ to a $V\to X\widehat{\otimes}_{\mathbf{Q}_p}\mathbf{C}_p$ pro-étale, with $V$ small enough to admit a finite étale map to a "perfected" torus $\text{Spa}\ \mathbf{C}_p\{x^{\pm\infty}\}$, from which we pull back coordinates $x^{\pm\infty}$. Briefly, pro-étale locally on $X\widehat{\otimes}_{\mathbf{Q}_p}\mathbf{C}_p$, the complex $\mathcal{O}\mathbb{B}^+_{\rm dR}\otimes_{\mathcal{O}_X}\Omega^{\bullet}_X$ is: -$$\mathbb{B}_{\rm dR}^+[\![t_1,\ldots,t_d]\!]\otimes_{\mathbf{C_p}\{x^{\infty}\}}DR^{\infty}_{\mathbf{C}_p}$$ -where the $\mathbf{C}_p\{x^{\infty}\}$-algebra structure on $\mathbb{B}_{\rm dR}^+[\![t_1,\ldots,t_d]\!]$ should be spelled out in Lemmata 6.11, 6.12, -$$DR^{\infty}_{\mathbf{C}_p}: 0\to\mathbf{C}_{p}\{x^{\infty}\}\to\bigoplus_{a=1}^d\mathbf{C}_{p}\{x^{\infty}\}\text{d}x_a\to \bigoplus_{a -TITLE: Which metrics on exterior power are induced from metrics on the base? -QUESTION [6 upvotes]: $\newcommand{\id}{\text{id}}$ -$\newcommand{\Hom}{\text{Hom}}$ -This is a cross-post. -Let $V$ be a $d$-dimensional real vector space, and let $2 \le k \le d-1$. Every inner product on $V$ induces an inner product on $\Lambda^k V$: -$$ \langle v_1 \wedge \dots \wedge v_k , w_1 \wedge \dots \wedge w_k \rangle:=\det (\langle v_i ,w_j \rangle). $$ - -What are necessary and sufficient conditions on a product on $\Lambda^k V$ to to be induced from a product on $V$? - -For $k=d-1$ the answer is that every product on $\Lambda^{d-1} V$ is induced from a product on $V$. -Edit 1: -If there exist an inducing product at the base, this product is unique (details are provided under the "edit" here). Perhaps we can construct an "inverse map" which is defined on the space of products on $\Lambda^k V$, and see when the result is an honest inner product on $V$ (and not just a bilinear form). -Edit 2: -Here is an equivalent formulation of the question: -A choice of a product on $V$ is equivalent to a choise of a linear isomorphism $ g:V \to V^*$ that satisfies -$$ g(v)(w)=g(w)(v) \, \, \text{and}\, \,g(v)(v) \ge 0 \, \, \text{with equality only when } \, v=0. \tag{1}$$ -The equivalence is via $g(v)(w):= \langle v,w \rangle$. Using this perspective, the metric on $\Lambda^{k} V$ induced by $g$ is $\Lambda^kg:\Lambda^{k} V \to \Lambda^{k} (V^*) \cong (\Lambda^{k} V)^*$. -So, the question becomes the following: - -For which maps $h:\Lambda^{k} V \to (\Lambda^{k} V)^*$ which are symmetric and positive in the sense of $(1)$, there exist a symmetric and positive $g$ such that $h=\Lambda^kg$? -(The symmetry and positivity requirements on $g$ are in fact redundant-if there exist a "root" $g$ such that $h=\Lambda^kg$, then $g$ is symmetric and can be taken to be positive definite). - -Peter Michor suggests using Plucker relations as a necessary condition. These relations give an equivalent conditions for an element $h\in \Lambda^k (V^* \otimes V^*)$ to be decomposable, i.e. $h=g_1\wedge g_2\wedge\dots\wedge g_k$, where $g_i \in V^* \otimes V^*$. -However, in the formulation above, $ g:V \to V^*$, and $\Lambda^kg:\Lambda^{k} V \to \Lambda^{k} (V^*) \cong (\Lambda^{k} V)^*$ is the induced map on exterior powers, that is an element of $\Lambda^{k} V^* \otimes \Lambda^{k}V^*$. Thus, in order to use the Plucker relations, we need a way to consider it as an element in $\Lambda^{k}(V^* \otimes V^*)$. Moreover, we need that this identification will map "power-elements" to "decomposable elements". -I am not sure such a map exists. Here is a precise question on this. - -Yet another equivalent formulation of the question...: -Given $(\binom {d}k)^2$ numbers, indexed by ordered pairs $\big((i_1,\dots,i_k),(j_1,\dots,j_k)\big)$ where $1 \le i_1 0$. - -REPLY [3 votes]: A necessary condition are the Pluecker relations. -Namely, let $W= V^*\otimes V^*\ni g$, then $h=\Lambda^k g$ is decomposable. In detail, $h\in \Lambda^k W$ is decomposable, i.e., -$h=g_1\wedge g_2\wedge\dots\wedge g_k$, if and only $i_{\Phi}h\wedge h = 0$ for all $\Phi\in\Lambda^{k-1}W^* = \Lambda^{k-1}(V\otimes V)$. -It then remains to ensure that all $g_1$ are the same and are positive definite. -See here for various equivalent versions of the Pluecker relations.<|endoftext|> -TITLE: Shtukas for $\mathrm{Spec}\,\mathbf{Z}$ -QUESTION [8 upvotes]: This is a very soft and speculative question. Please feel free to downvote, close or delete it. -Studying the cohomology of moduli spaces of shtukas, Drinfeld proved the Langlands program for $\mathrm{GL}_2$ for global function fields, which was later extended by L. Lafforgue to $\mathrm{GL}_n$. -Peter Scholze has a conjectural section Shtukas for $\mathrm{Spec}\,\mathbf{Z}$ in his 2018 ICM report https://arxiv.org/abs/1712.03708 on arXiv. - -Might this be used to attack the Langlands conjectures for $\mathbf{Q}$ (and how)? -What could this be used for? - -REPLY [6 votes]: Scholze's "mixed characteristic shtukas" are already the basis for his attack on the local Langlands conjectures, so in some sense they have already found application to ``Langlands over $\mathbb{Q}$''. The basic strategy here is to transpose the ideas of Vincent Lafforgue's work on the global Langlands correspondence to shtukas over the Fargues-Fontaine curve. -As for the global problem, it should be noted that Drinfeld's moduli spaces of shtukas are themselves function field analogues of objects that you have (sometimes) in the number field case, namely Shimura varieties. The application to Langlands comes through the fact that the cohomology of moduli spaces of shtukas "realize" the Langlands correspondence. The analogous hope for Shimura varieties is false; for instance we have Shimura varieties for $GL_2$ (i.e. modular curves) but their cohomology cannot capture all automorphic representations, e.g. those attached to Maass forms. -The "universal cohomology theory over $\mathbb{Z}$" doesn't necessarily seem to solve the problem of being able to find all the expected automorphic/Galois representations in geometry, which is the fundamental premise for the work of Drinfeld/Lafforgue, but maybe Scholze has a grand plan for this. - -REPLY [4 votes]: After their work on integral $p$-adic Hodge Theory, Bhatt, Morrow and Scholze realized a certain $q$-de Rham cohomology theory was underlying their $A_{\rm inf}$ cohomology in their paper. This seemed to have the nice feature of "interpolating" among all sorts of cohomology theories: $\ell$-adic étale, crystalline, singular. -He's just speculating about how to make this completely precise, and possibly show such cohomology theory, once constructed one day, is "universal". It is unlikely that this will have applications towards global Langlands over number fields, let alone conjectures on algebraic cycles. -It simply seems to me Scholze just wants to understand "what's going on". -Merlin's comments and the answer by user84144 say it all<|endoftext|> -TITLE: Is every category equivalent to the fundamental category of a directed space? -QUESTION [8 upvotes]: I was wondering if there was some work in directed algebraic topology to define a classifying directed space of a category? By that I mean the following. -In (undirected) algebraic topology, we define the classifying space of a category as the geometric realization of its nerve. This space has the particularity property that its fundamental groupoid is equivalent to the free groupoid generated by this category. I guess (I actually do not know, I have never seen a complete proof of this statement, and I would be interested in a reference for that), this can be proved in two steps: - -Proving that the fundamental category of the nerve is isomorphic to the initial category. I think this is proved in Joyal and Tierney's notes. -Proving that the fundamental groupoid of a simplicial set is equivalent to the fundamental groupoid of its geometric realization. I think I see how to construct the equivalence, but proving it actually works needs to prove something like ``a homotopy between paths in the geometric realization can be approximate by a simplicial homotopy''. I would be grateful if someone knows an elementary reference for that. - -Anyway, my real question comes now. I think we can use these steps to prove a similar result in directed algebraic topology. - -This step is still relevant. -Here we need a directed geometric realization of a simplicial set such that its fundamental category is equivalent to the fundamental category of the simplicial set. I think I can provide such a realization by equipping standard geometric simplexes with a structure of $d$-space. For example, I can prove that the fundamental category of the standard geometric $n$-simplex with this d-space structure is equivalent to the poset $\{0,1,...,n\}$ with the natural order, as expected. But to wrap up, I get the same issues as in the undirected case. - -This question seems very basic, so I guess someone should have already addressed this problem. Do you know of a work in that direction ? - -Edit: I am adding the d-space structure on the standard geometric simplexes that I am considering. -If $\Delta_n = \{(t_0, t_1, \ldots, t_n) \in [0, 1]^{n+1} \mid \sum_i t_i = 1\}$, for $i \in \{0, \ldots, n\}$, define -$$D_i = \{(t_0, \ldots, t_n) \in \Delta_n \mid \forall j < i, t_j < t_i \wedge \forall j > i, t_j \leq t_i\}.$$ - -We will say that a continuous map $\gamma: [0,1] \longrightarrow \Delta_n$ is a dipath of $\Delta_n$ if there exist $k \geq 1$, $0 \leq i_1 < \ldots < i_k \leq n$ integers and $0 < t_1 < \ldots < t_{k-1} < t_k = 1$ real numbers with: - -$\forall t \in [0,t_1]$, $\gamma(t) \in D_{i_1}$, -$\forall j \in \{2, \ldots, k\}$, $\forall t \in ]t_{j-1},t_j]$, $\gamma(t) \in D_{i_j}$ - -REPLY [2 votes]: The obvious subdivision to try is the ordinal one (see paper by Phil Ehlers and myself). I don't guarantee that it will do the job but .... -You might look at things on quasi-categories as well. -There is also the idea of exit paths and Lurie has something on stratified spaces and this has been explored further by -D. Ayala, J. Francis and N. Rozenblyum, 2015, A stratified homotopy hypothesis, arXiv preprint arXiv:1502.01713. I hope this helps. (Do contact me by e-mail if you want to discuss those sources in more depth.)<|endoftext|> -TITLE: Examples of residually-finite groups -QUESTION [18 upvotes]: One of the main reasons I only supervised one PhD student is that I find it hard to find an appropriate topic for a PhD project. A good approach, in my view, is to have on the one hand a list of interesting questions and on the other hand rich enough family of examples so that the student would be able to find answers to some of these questions for some of the examples. I have many questions that I would like to know the answer for interesting examples of residually-finite groups, in particular, for finitely presented ones. Unfortunately, I am very far from being an expert on this topic. -Is there anywhere I can find a list of examples of families of residually-finite groups, preferably, including some of their interesting properties? - -REPLY [2 votes]: Here's a few examples in line with classical combinatorial group theory. - -Though small cancellation groups as a whole have already been mentioned, one important subclass of these are the one-relator groups with torsion $\langle A \mid w^n = 1 \rangle$, which were shown to be linear rather recently. Baumslag conjectured in the 60s that they would be residually finite, and Allenby-Tang resolved a good number of special cases of this quite some time before the recent work of Agol-Wise-Haglund settled it completely. -Polycyclic groups (due to Hirsch) and f.g. abelian-by-nilpotent groups (due to P. Hall). -Not all one-relator groups are residually finite, as the example non-Hopfian Baumslag-Solitar group $\langle a, b \mid b^{-1} a^2 b = a^3 \rangle$ shows. However, an important class is the class of cyclically pinched one-relator groups, i.e. those which admit a presentation of the form $\langle A \cup B \mid U = V\rangle$, where $U$ is a word over the generators $A$ and their inverses, and $V$ is one over $B$ and their inverses (and $A \cap B = \varnothing$). Baumslag, again, showed that these are residually finite in 1969. Some familiar examples, which have already appeared in other answers, are the fundamental groups of compact surfaces. -Baumslag showed in 1963, in an exceptionally short and beautiful paper (the proof is one paragraph long, and the remainder of the paper is a couple of paragraphs of applications) that the automorphism group of a finitely generated residually finite group is again residually finite. In particular, the automorphism group of the automorphism group of a free group is residually finite -- I do not know whether there is any other proof of this fact available using only the known presentations for these groups. -Finally, a non-example, but perhaps of some relevance/interest: it was once conjectured (by G. Lallement in 1974, in a paper on semigroup theory!) that positive one-relator groups are residually finite. Here a positive one-relator group is one $\langle A \mid w = 1 \rangle$ where $w \in A^\ast$ is a word not including any inverse symbols. However, as noted by Perrin-Schupp, the non-Hopfian Baumslag-Solitar group $BS(2, -3)$ admits the positive presentation $\langle a, b \mid (ab)^2(ba)^3 = 1 \rangle$.<|endoftext|> -TITLE: Is it consistent with ZF that $V \to V^{\ast \ast}$ is always an isomorphism? -QUESTION [35 upvotes]: Let $k$ be a field and $V$ a $k$-vector space. Then there is a map $V \to V^{\ast \ast}$, where $V^{\ast}$ is the dual vector space. If we are in ZFC and $\dim V$ is infinite, then this map is not surjective. As we learned in this question, there are models of $ZF$ where $V \to V^{\ast \ast}$ is an isomorphism when $V$ has a countable basis. I think the same argument shows that it is consistent with ZF that this is an isomorphism whenever $V$ has a basis. - -Is it consistent with $ZF$ that $V \to V^{\ast \ast}$ is an isomorphism for all vector spaces $V$? - -I ask because I'm teaching a rigorous undergrad analysis class. My students keep asking me whether they have to believe that $V \to V^{\ast \ast}$ can fail to be an isomorphism. Of course, I'm trying to change their intuition to point out why most mathematicians find the failure of isomorphism plausible and point out that there are more subtle ways to salvage the claim, such as Hilbert spaces, but I'd also love to be able to give them a choice free proof that there is some vector space where this issue comes up. - -REPLY [46 votes]: No, it’s not consistent. -Let $V=k^{(\omega)}$ be the vector space of finite sequences of elements of $k$. Then $V^*$ can be identified with the vector space $k^\omega$ of all sequences, and elements of the image of the natural map $V\to V^{**}$, considered as maps $k^\omega\to k$, are determined by their restriction to $k^{(\omega)}$. -So if $V\to V^{**}$ is an isomorphism, then, taking $W=k^\omega/k^{(\omega)}$, there are no nonzero linear maps $W\to k$, and hence $W^{**}=0$. But $W$ is nonzero. -So the map to the double dual must fail to be an isomorphism either for $V$ or for $W$. - -REPLY [6 votes]: Not an answer to your question but a variant which is perhaps more à propos given that you said you are teaching a rigorous undergrad analysis class. -Let $V=\oplus_{\mathbb{N}}\mathbb{R}$ be the space with countable basis which motivated your question. Another way to salvage the isomorphism $V\simeq V^{\ast\ast}$ (without giving up any cherished axiom) is to think of $V$ as equipped with the finest locally convex topology and consider duals as topological duals always given the strong topology. It is easy to see that $V^{\ast}\simeq \prod_{\mathbb{N}}\mathbb{R}$ with the product topology. Although perhaps counterintuitive, when one takes the proper (i.e., strong topological) dual of $\prod_{\mathbb{N}}\mathbb{R}$, then one gets back to $\oplus_{\mathbb{N}}\mathbb{R}$.<|endoftext|> -TITLE: (Binary) Theta functions and Atkin's U-operator -QUESTION [7 upvotes]: Let $D<0$ be a fundamental discriminant and consider the theta series $$\vartheta_Q(\tau)=\sum_{v\in\mathbb{Z}^2} q^{Q(v)}$$ -associated to a quadratic form $Q$ of discriminant $D$. It appears to be true that $\vartheta_Q=\vartheta_Q|U(|D|)$, where $U$ is the usual Atkin U-operator, defined by its action on $q$-series -$$\big(\sum_n a(n)q^n\big)|U(m):=\sum_n a(mn)q^n,$$ -but I can't seem to find a proof of this. I strongly suspect that this is well-known (probably in much greater generality), but I'm not sure where to look. If anyone has a reference or of course a proof, that would be much appreciated! - -REPLY [7 votes]: I think you want to show that in each ideal class the number of ideals of norm a is equal to the number of ideals of norm $a \cdot |D|$. This should follow directly from the following 2 facts. -(1). All primes $p$ dividing $|D|$ ramify in the ring of integers of $\mathbb{Q}(\sqrt{D})$. -(2). The unique ideal of norm $|D|$ (uniqueness and existence follow from (1), taking a little care about the prime $2$) is principal, generated by $\sqrt{D}$. -In view of (1) and (2) the map $J \rightarrow \sqrt{D} \cdot J$ gives the desired bijection.<|endoftext|> -TITLE: Distribution of primitive roots, as p varies -QUESTION [8 upvotes]: For a prime number $p$, let $\Phi(p)$ be the subset of $\{ 1, 2, \ldots, p-1 \}$ consisting of primitive roots modulo $p$. (Thus $\# \Phi(p) = \phi(p-1)$, where $\phi$ denotes the totient.) -I am curious about the distribution of $\Phi(p)$ as $p$ varies. I imagine that this is classic analytic number theory, but I'm finding it surprisingly hard to locate a precise statement/reference. -To make my question precise, consider the following: For any continuous function $f \colon [0,1] \rightarrow {\mathbb R}$, define -$$D_p(f) = \frac{1}{\phi(p-1)} \cdot \sum_{x \in \Phi(p)} f \left( \frac{x}{p-1} \right).$$ -Does this sequence of distributions $D_p$ converge (e.g., weakly in measure, or in some stronger sense?) to the uniform measure or something else on $[0,1]$? What's the best current result along these lines? - -REPLY [3 votes]: A more general result (for general finite fields) is proved in. -Perel’muter, G.I.; Shparlinskij, I.E., The distribution of primitive roots in finite fields, Russ. Math. Surv. 45, No.1, 223-224 (1990); translation from Usp. Mat. Nauk 45, No.1(271), 185-186 (1990). ZBL0705.11078.<|endoftext|> -TITLE: Good books on the divisor sum function $\sigma(n)$? -QUESTION [5 upvotes]: I would like gain detailed knowledge about properties of the divisor sum function $\sigma(n)$, special equation that have been studied (e.g. $\sigma(n) = 2n$ perfect numbers, ...) and progress that was made in the past centuries. -I am interested in arithmetic functions in general so the books mentioned in this question Good books on Arithmetic Functions ? is already a good start. -However, it would be great if you could recommend a book or a comprehensive paper that introduces/summarizes knowledge about the $\sigma(n)$ function. - -REPLY [3 votes]: Sándor, Jó.; Mitrinović, D. S. & Crstici, B. Handbook of number theory. I Springer, 2006 -This book describes a lot of results concerning almost all arithmetical functions.<|endoftext|> -TITLE: Is it consistent with ZF that $V\to V^{\ast \ast}$ is always surjective? -QUESTION [23 upvotes]: In a comment to a recent question, Jeremy Rickard asked whether it is consistent with ZF that the map $V \to V^{**}$ from a vector space to its double dual is always surjective. We know that "always injective" is consistent (since that's what happens in ZFC) and Jeremy Rickard's argument shows that "always an isomorphism" is not consistent. But what about "always surjective"? - -Gro-Tsen points out that Harry West has already showed this is impossible elsewhere on MO. I am missing one step in West's answer. I thought I'd write up the issue here, and someone can explain to me what I am missing. -First of all, for any field $F$ and any set $X$, we can form the free vector space with basis $X$, call it $FX$, and the vector space of functions from $X$ to $F$, call it $F^X$. It is easy to see that $(FX)^{\ast} \cong F^X$ so, if $V \to V^{\ast \ast}$ is always surjective, then the obvious injection $FX \to (F^X)^{\ast}$ must always be an isomorphism. So we may and do assume: -Key Consequence For every set $X$, the obvious injection $FX \to (F^X)^{\ast}$ is an isomorphism. -Now, suppose that $\alpha$ is an ordinal with cofinality $>\omega$. -Lemma 1 Let $X \subset \alpha$ have the property that $X \cap \beta$ is finite for every $\beta < \alpha$. Then $X$ is finite. -Proof If not, then $X$ is an infinite well-ordered set (by restricting the order from $\alpha$) so it contains a copy of $\omega$. By the hypothesis on $\alpha$, there is some $\beta_0 < \alpha$ containing this copy of $\omega$. But then $X \cap \beta_0$ is infinite. $\square$ -Let $V$ be the subspace of $F^{\alpha}$ consisting of functions which are supported on $F^{\beta}$ for some $\beta< \alpha$. -Lemma 2 $V^{\ast} = F \alpha$. -Proof: Let $\phi \in V^{\ast}$. We can restrict $\phi$ to $F^{\beta}$ for each $\beta < \alpha$ and, by the definition of $V$, the functional $\phi$ is determined by the list of these restrictions. On each $F^{\beta}$, by the Key Consequence, $\phi|_{F_{\beta}}$ coincides with some unique vector from $F \beta$. Let the support of that vector be $X_{\beta}$ and let $X = \bigcup_{\beta} X_{\beta}$. Then $X \cap \beta= X_{\beta}$ for each $\beta < \alpha$ so, by Lemma 1, $X$ is finite. The functional $\phi$ is then induced by a functional in $FX \subset F \alpha$. $\square$. -But then $V^{\ast \ast} = (F \alpha)^{\ast} = F^{\alpha}$, whereas $V$ is a proper subspace of $F^{\alpha}$. QED - -My only issue is, in ZF, are we sure that there are ordinals of cofinality $>\omega$? In ZFC, one simply takes the first uncountable ordinal, $\omega_1$. If we had a cofinal sequence $0=x_0$, $x_1$, $x_2$, \dots, in $\omega_1$, then $\omega_1$ would be the union of the countable intervals $[x_i, x_{i+1})$, and would hence be countable. -But in ZF, a countable union of countable sets doesn't have to be countable. I tried some tricks to get around this and failed; please let me know what I missed. - -REPLY [15 votes]: $\def\fin{\text{finite}}\def\count{\text{countable}}$This is a CW answer to write up the proof which Harry West gave on another answer, as pointed out by Gro-Tsen. Thanks to Prof. West for the clever solution and Gro-Tsen for pointing it out. I'm just rewriting it to record details which took me a while to follow. The answer is "no", there is always some vector space with $V^{\ast \ast}$ not spanned by the image of $V$. -Let $F$ be a field. For any set $X$, the vector space $F^X$ is the vector space of functions $X \to F$. It contains the subspace $F^X_{\fin}$ of finitely supported functions, which is isomorphic to the free vector space with basis $X$. It is straightforward that $(F^X_{\fin})^{\ast} = F^X$ so, if $(F^X)^{\ast} \supsetneq F^X_{\fin}$, then we are done already. Thus, we may make the key assumption that $(F^X)^{\ast} = F^X_{\fin}$ for every set $X$. -Now, let $X$ be a well-ordered uncountable set, such as the first uncountable ordinal $\omega_1$. The only place that we will use the well-order on $X$ is to be sure that every infinite subset of $X$ contains a copy of $\omega$ (in other words, infinite subsets of $X$ can't be Dedekind finite). -Let $F^X_{\count}$ be the subspace of $F^X$ consisting of countably supported functions. -Claim: $(F^X_{\count})^{\ast} = F^X_{\fin}$. -Proof: Let $\phi$ be any linear function $\phi : F^X_{\count} \longrightarrow F$. For any countable subset $Y$ of $X$, we have $F^Y \subset F^X_{\count}$. So the restriction of $\phi$ to $F^Y$ is a vector in $(F^Y)^{\ast}$ and, by the key assumption, $(F^Y)^{\ast} = F^Y_{\fin}$. So, for each $Y$, there is some unique finitely supported vector $\phi_Y$ in $F^Y$ such that $\phi(\psi)$ is the dot product $\psi \cdot \phi_Y$ whenever $\psi$ is supported on $Y$. By the uniqueness of $\phi_Y$, if $Y_1 \subseteq Y_2 \subset X$ are finite sets, then $\phi_{Y_1}$ is the restriction of $\phi_{Y_2}$ to $Y_1$. -Let $S = \bigcup_{X \subset Y,\ \count} \text{Support}(\phi_Y)$. For every countable $Y$, we know that $S \cap Y$ is finite. We claim that this forces $S$ to be finite; if $S$ is infinite then (using the well order on $X$) there is a countably infinite subset $Z$ of $S$ and $S \cap Z= Z$ is infinite, contradicting that $S \cap Y$ is finite for every countable $Y$. -So $S$ is finite, and $\phi$ is given by dot product with a vector in $F^S$, so $\phi$ is given by dot product with a vector in $F^X_{\fin}$. We have shown that $(F^X_{\count})^{\ast} = F^X_{\fin}$. $\square$ -But then $(F^X_{\count})^{\ast \ast} = (F^X_{\fin})^{\ast} = F^X$, and $F^X_{\count}$ is a proper subspace of $F^X$. This completes the proof. $\square$ -As Prof. West observes in his write up, one can identify a specific vector space $V$ for which this proof shows $V$ does not surject onto $V^{\ast \ast}$: The vector space $F^{\omega}_{\fin} \oplus F^{\omega_1}_{\count}$.<|endoftext|> -TITLE: Is the nth-power-sum graph connected? -QUESTION [6 upvotes]: This post was inspired by the Square-Sum Problem presented in Numberphile by Matt Parker. - He asked about Hamiltonianness for $n=2$, and we ask about connectedness for all $n \in \mathbb{N}^*$. -Given $n \in \mathbb{N}^*$, let $\mathcal{G}_n$ be the graph $(\mathbb{N}^*,\{ \{a,b\} \ | \ a+b \in S_n\})$, with $S_n = \{r^n | r\in \mathbb{N} \}$ . -Question: Is $\mathcal{G}_n$ connected? - -Checking: It is true for $n \le 5$. -Proof: For any $a \in \mathbb{N}^*$, there is $r\in \mathbb{N}$ such that $r^n \le a < (r+1)^n$. Then, $\{a,(r+1)^n-a\}$ is an edge of $\mathcal{G}_n$. Now, $(r+1)^n-aa_n$, there is $b0 and i<>t: - G.add_edge(i,t) - return G.is_connected() - -The result follows by the (minimal) computation below. $\square$ -sage: PowerSumGraph(13,6,2) -True -sage: PowerSumGraph(108,7,3) -True -sage: PowerSumGraph(2008,9,4) -True -sage: PowerSumGraph(49355,11,5) -True - - -The case $n=6$ works also, by the following (non-minimal) computation and $a_6 = 296476$. -sage: PowerSumGraph(1500000,13,6) -True - -The case $n=7$ is beyond my laptop capacity. For $n \ge 8$, the program should be modified because it would deal with integers beyond $2^{31}-1$, the maximum value for variables declared as int. - -REPLY [6 votes]: Yes, the graph $\mathcal{G}_n$ is connected, and its diameter is at most $n2^n$. To see this, write $s$ for $n2^{n-1}$, and fix any two vertices $a,b\in\mathbb{N}^*$. By Wright's solution of Waring's problem with proportionality conditions, for a large parameter $c\in\mathbb{N}$, we can find -$x_1,\dots x_s\in\mathbb{N}^*$ and $y_1,\dots y_s\in\mathbb{N}^*$ such that -$$ x_1^n+\dots+x_s^n=c+a\qquad\text{and}\qquad y_1^n+\dots+y_s^n=c+b,$$ -moreover the summands are asymptotically (as $c$ tends to infinity) -$$ x_1^n,y_s^n\sim\frac{c}{2s-1}\qquad\text{and}\qquad x_2^n,\dots,x_s^n,y_1^n,\dots,y_{s-1}^n\sim\frac{2c}{2s-1}.$$ -As a consequence, the following walk of length $2s$ connects $a$ and $b$: -$$ a\ \rightarrow\ x_1^n-a\ \rightarrow\ -x_1^n+y_1^n+a\ \rightarrow\ x_1^n-y_1^n+x_2^n-a\ \rightarrow\ \cdots\ \rightarrow\ b.$$ -Note that the edge sums in this walk are $x_1^n,y_1^n,\dots,x_s^n,y_s^n$, while each intermediate vertex is asymptotically $c/(2s-1)$, hence a positive integer for $c$ sufficiently large. The proof is complete.<|endoftext|> -TITLE: Reference request: Models of cuspidal representations of GL(n,k) where k is a finite field -QUESTION [8 upvotes]: Let $k=\mathbb{F}_q$ where $q$ is a prime power of odd cardinality. -Where could I find explicit models of all irreducible cuspidal (complex) representations of $GL_n(k)$ for $n\ge 3$? -I understand that the characters of such representations was constructed by J.A. Green “The characters of the finite general linear groups,” Trans. Amer. Math. Soc.,80, No. 2, 402–447 (1955). -If $n=2$, the cuspidal representations of $GL_2(k)$ can be constructed using Weil representations, which can be found in many online notes. But I could not find any explicit construction of cuspidal representations of $GL_n(k)$ when $n\ge 3$. - -REPLY [4 votes]: Gel′fand S.I., Representations of the general linear group over a finite field, Lie groups and their representations (Proc. Summer School on Group Representations of the Bolya: J ́anos Math. Soc., Budapest, 1971), 119–132, Halsted, New York, 1975. makes the action on the Kirilov model explicit.<|endoftext|> -TITLE: What is quantum algebra? -QUESTION [45 upvotes]: This might be a very naive question. But what is quantum algebra, really? -Wikipedia defines quantum algebra as "one of the top-level mathematics categories used by the arXiv". Surely this cannot be a satisfying definition. The arXiv admins didn't create a field of mathematics by choosing a name out of nowhere. -Wikipedia (and, in fact, the MathOverflow tag wiki) also lists some subjects: quantum groups, skein theories, operadic and diagrammatic algebra, quantum field theory. But again, I don't find this very satisfying, as I feel this doesn't tell me what the overarching idea of quantum algebra is. -(For example, inspired by the table of of contents of the Wikipedia article I could define algebraic topology to be "homotopy, homology, manifolds, knots and complexes". But first, I have certainly missed many subfields of algebraic topology, and second, this is missing the overarching idea, contained in the introduction of the Wikipedia article: algebraic topology is the use of "tools from abstract algebra to study topological spaces". It immediately makes the link behind all the themes I listed clearer, and if I encounter a new theme I can tell if it is AT or not using this criterion.) -This MO question is looking for the intuition behind quantum algebra and relations to quantum mechanics. The main thing I gathered from the answers (that I more-or-less knew already) is that "quantum = classical + ħ", or less informally that we are looking at noncommutative deformations of commutative, classical objects. But this doesn't seem to account for all of quantum algebra. For example, a TQFT is a functor from a category of cobordisms to some algebraic category. Where's ħ? Operadic algebra is also listed as one of the components of quantum algebra, but one can study operads a lot without talking about noncommutative deformations. In fact, I've seen and read many papers about operads listed in math.QA that don't seem to have anything to do with this picture. -In brief: What could be a one-sentence definition of quantum algebra? (In the spirit of the definition of algebraic topology above.) - -REPLY [11 votes]: I think that a modern realistic perception of the term "quantum algebra" has to be understood in its historical context, that is, the algebraic/geometric methods, originating from the study of the quantization problem in its various forms (first and second quantization , QFTs etc): -As far as i know, the term quantum algebra, has been introduced in Dirac's seminal paper "The fundamental equations of quantum mechanics", Proc. Roy. Soc. A, v.109, p.642-653, 1925 (a reprint can be found in Sources of Quantum Mechanics, ed. B.L. van der Waerden, p.307). It was short after Heisenberg had proposed his -revolutionary for the time- idea that quantum observables should correspond to hermitian matrices of -generally- infinite order. However, he considered the non-commutativity of the matrices as an obstacle in the further development of the idea. Heisenberg communicated his ideas to Fowler at Cambridge. Fowler, was by that time, the thesis advisor of Dirac and this is how the latter got involved. Dirac shortly proposed that the non-commutativity of quantum mechanical observables should be treated as a fundamental characteristic of the new theory to be developed. He also proposed that quantum observables $A$ and $B$ should belong in a non-commutative algebra, satisfying the relation -$$ -[A,B]=i\hbar \{A,B\} -$$ -as a "measure of departure" from commutativity. ($[.,.]$ stands for the commutator and $\{.,.\}$ for the classical Poisson bracket). A detailed account of the history of the development of the notion of quantum algebra together with references, historical and technical details, can be found at Varadarajan's Reflections on Quanta, Symmetries and Supersymmetries, ch.2. -During the next decades, the term quantum algebra started expanding and embracing new ideas and methods emerging from the studies of different aspects of the various quantization problems. Dirac's commutator was replaced by Moyal bracket (coinciding with Dirac's comm. modulo $\hbar^2$ terms) and this is how the deformation theory (already developed as a separate discipline at the level of assoc. and Lie algebras) entered the picture. Now quantum mechanical algebras of observables were viewed as deformations of the corresponding classical objects. Moshe Flato and his coworkers have been among the pioneers at that direction. -The rise of quantum groups and $q$-mathematics, expanded the term even more. Now whole new families of examples and methods arose, introducing new mathematical ideas and tools into the subject, such as hopf algebras, $q$-analytical tools, representation-theoretic methods, $q$-deformations of Weyl algebras etc. -The continuous development of Quantum Field theories together with the various technical and conceptual problems introduced by them, led to further expansions of the discipline of quantum algebras. Now, algebraic geometric, homological, homotopical and Category theoretical methods and notions got involved. The development of non-commutative geometry, also opened new directions of study. I am far from being an expert into such topics to provide further details but i have the feeling that almost everything inside "quantum algebras" has been in some way connected or at least originating (even in some distant sense) from the study of the quantization problems. -So, concluding, i would say that, although the requirement for a one-sentence definition of the topic of quantum algebra might seem superficial, a rough approximation (modulo my understanding of course) might be: - -The study of the algebraic/geometric theories, methods, techniques, notions and questions originating from the study of the various aspects of the quantization problem (broadly interpreted). - -P.S.: Inevitably, non-commutativity is a central topic in the frame of quantum algebra. In this sense, the above description, may be viewed to encompass even modern abstract tools and theories on the fundamentals and the properties of algebraic operations and structures. It is just that i feel a little sceptical, as to whether a modern "definition" of the field of quantum algebra should be build around the notion of non-commutativity itself.<|endoftext|> -TITLE: In a non-compact metric space, topological transitivity need not imply onto -QUESTION [8 upvotes]: I had asked this question on Mathematics Stack Exchange yesterday but it got no response so I'm asking here. - -Let $X$ be a compact metric space and $f:X \to X$ be continuous. If $f$ is topologically transitive. Then $f$ is onto. - -I'm trying to show that the compactness hypothesis cannot be removed. -I couldn't find any example of a non-compact metric space and a continuous function which is topologically transitive but not onto. - -Any hints will be appreciated. -Note: If $(X,f)$ is a dynamical system. Then $f$ is said to be topologically transitive if for every pair of non-empty open sets $U$ and $V$ in $X$ there exists $n \geq 1$ such that $f^n(U) \cap V\neq \emptyset.$ - -REPLY [11 votes]: By Birkhoff's theorem, a bounded linear operator on a Banach space is topologically transitive if and only if it is hypercyclic. Charles Read has developed a whole machinery for constructing non-surjective, hypercyclic operators on spaces of the form $\ell_1(X)$: - -C.J. Read, The invariant subspace problem for a class of Banach spaces, II. Hypercyclic operators. Israel J. Math., 63 (1) (1988), 1-40. - -Of course, possibly these are not the easiest counter-examples for your purposes, although I find them quite instructive.<|endoftext|> -TITLE: Relation Degree of Dualizing Sheaf and Euler Characteristic -QUESTION [7 upvotes]: Let $C$ be a curve over $k$ and $w_C$ it's dualizing sheaf. -If $dim_k H^0(C, \mathcal{O}_C) =1$ and $g:= H^0(C, \mathcal{O}_C)$ the arithmetic genus one easy computes $$deg(w_C) = 2g-2$$ where $deg$ is the map $deg:Pic(C) \to \mathbb{Z}, \mathcal{L} \mapsto \chi(\mathcal{L}) - \chi(\mathcal{O}_C)$. -Fothermore, using methods from simplicial cohomology one get the Euler characteristic $e(C) = 2-2g$ where $g$ is the topological genus of $C$. -I want to know a (maybe) geometrical but also a preferably deeper reason where the relation $deg(w_f) = -e(C)$ comes from, so why this invariants a so closed connected. - -REPLY [3 votes]: Another explanation for the equality $\mathrm{deg}(\omega_C) = -e(C)$, complementing that contained in Javier-Alvarez's MSE exposition linked by Francesco Polizzi, is Hodge theory. In short, Hodge theory gives an instance and a precise statement of how topological invariants are related to algebraic/holomorphic invariants. Specifically, Hodge theory tells you how to represent singular cochains as integrals with respect to holomorphic forms. More practically, Hodge theory expresses topological (singular) cohomology in terms of sheaf cohomology of $\mathcal{O}_C$ and $\Omega_C^1$. -But before I begin, let me clarify the setting in which I will be working in: - -My field will be $k = \mathbf{C}$, so that I can talk about topology. -The symbol $C$ refers to the projective scheme over $\mathbf{C}$. I will write $C(\mathbf{C})$ to denote the Riemann surface associated with $C$. -By "curve", I will mean "smooth projective curve over $\mathbf{C}$". Smoothness is required in general, as the relation $\mathrm{deg}(\omega_C) = -e(C)$ already fails for most nodal curves. - -For a smooth projective curve $C$, Hodge theory (together with Serre's GAGA) says that there are canonical isomorphisms -\begin{align*} - H^0(C(\mathbf{C}),\mathbf{C}) & \cong H^0(C,\mathcal{O}_C), \\ - H^1(C(\mathbf{C}),\mathbf{C}) & \cong H^1(C,\mathcal{O}_C) \oplus H^0(C,\Omega_C^1), \\ - H^2(C(\mathbf{C}),\mathbf{C}) & \cong H^1(C,\Omega_C^1), -\end{align*} -where - -the cohomology groups on the left are the topological (singular) cohomology groups of the Riemann surface $C(\mathbf{C})$ with coefficients in $\mathbf{C}$; -those on the right are sheaf cohomology groups computed on the scheme $C$; and -$\Omega_C^1$ is the sheaf of Kähler differentials of $C$. - -Thus we obtain the relation -\begin{align*} - e(C) & = \sum\nolimits_{i = 0}^2 (-1)^i \dim_{\mathbf{C}}(H^i(C(\mathbf{C},\mathbf{C})) \\ - & = \sum\nolimits_{i = 0}^1 (-1)^i \dim_{\mathbf{C}}(H^i(C,\mathcal{O}_C)) - - \sum\nolimits_{i = 0}^1 (-1)^i \dim_{\mathbf{C}}(H^i(C,\Omega_C^1)) \\ - & = \chi(\mathcal{O}_C) - \chi(\Omega_C^1) - = -\mathrm{deg}(\Omega_C^1). -\end{align*} -Since $C$ is smooth, the dualizing sheaf can be explicitly determined: by, say, Hartshorne, Algebraic Geometry, Corollary III.7.12, -$\omega_C \cong \Omega_C^1$. -Putting this into the relation above gives -$$ -\mathrm{deg}(\omega_C) = -e(C). -$$<|endoftext|> -TITLE: On the topological complexity of the set of winning strategies for Gale-Stewart Games -QUESTION [6 upvotes]: Given a set $A \subseteq \omega^\omega$, let $G_A$ denote the Gale-Stewart game with payoff set $A$ (so player $I$ wants the real built over the course of play to be in $A$ and player $II$ wants it not to be). Notice that the set of (not necessarily winning) strategies for player $I$ is simply the set of functions mapping finite sequences of natural numbers of even length to $\omega$ and the set of (not necessarily winning) strategies for $II$ is the same with "even" replaced by "odd". Thus, via some coding, in both cases if we wanted to, we could think of these sets as homeomorphic copies of Baire space itself. Now suppose that $A$ is determined for one of the players, say $I$. Then in the space of strategies for $I$, let $W(A)$ be the set of winning strategies. My question is about the relationship between the topological complexity of $A$ and $W(A)$. -Let me give a few naive examples to motivate what I mean. First, let $A = \omega^\omega$. In this case, not only does $I$ have a winning strategy for $G_A$ but of course any strategy that $I$ plays is winning. Therefore in this case $A \cong W(A) \cong \omega^\omega$. A (very) slightly less trivial example is as follows. Let $n \in \omega$ and let $U_n$ be the basic open set of all sequences whose first element is $n$. Then of course $I$ wins $G_{U_n}$ and the set $W(A)$ is the basic open set of all strategies whose first move is $n$. Therefore again $A$ and $W(A)$ have the same topological complexity. My main question is whether these naive examples are simply naive, or whether there is a general theorem to be mined from it. -Question 1: Is there a general theorem dictating the relationship between the topological complexity of $A$ and $W(A)$? What about if we restrict the possible $A$'s to "nice" sets (e.g. the Borels)? -Of course similar a similar question can be asked for measure: -Question 2: Is there a general theorem dictating the relationship between the measure theoretic properties of $A$ and $W(A)$? For instance, if $A$ is measurable, then is $W(A)$? What about the converse? Can one of $A$ and $W(A)$ be null and the other have positive measure? -I'm thinking of these questions in the context of ZFC but given the relationship between determinacy and inner model theory I would also be interested to hear if anything more interesting happens if we assume large cardinals or axioms related to games such as $AD^{L(\mathbb R)}$. - -REPLY [3 votes]: Abstractly, $W(A)$ is of the form $\forall^{\mathbb R} A \vee \forall^{\mathbb R} \lnot A$; in fact, if $I$ has a winning strategy for $A$, then $W(A)$ is $\forall^{\mathbb R} A$, and if $II$ has a winning strategy for $A$, then $W(A)$ is $\forall^{\mathbb R} \lnot A$. As Joel points out, though, $W(A)$ may be of much simpler complexity than $A$, as $A$ may contain some irrelevant "noise," e.g. $A$ may contain a copy of a Vitali set even though $A$ is determined, so like in most other cases of descriptive set theory it should be a question concerning pointclasses rather than individual sets. For example we now have that if $A$ is Borel, then $W(A)$ is coanalytic, if $A$ is $\Pi^1_n \cup \Sigma^1_n$, then $W(A)$ is $\Pi^1_{n+1}$ or even $\Pi^1_n$, etc. Does this answer your Question 1? -Concerning your Question 2, all coanalytic sets are Lebesgue measurable, so we get that $W(A)$ is always Lebesgue measurable if $A$ is Borel; also, if all $\Pi^1_n$ sets are determined, then all $\Pi^1_{n+1}$ sets are Lebesgue measurable, so that from this hypothesis we get that $W(A)$ is always Lebesgue measurable if $A$ is $\Pi^1_n$. But as above, $W(A)$ can be pretty simple even though $A$ itself is not contained in a model of determinacy, so there is no converse here. Finally, if $A$ is null, then $W(A)$ is null by Fubini; again, no converse here.<|endoftext|> -TITLE: Ways to add Aronszajn trees which are neither Souslin nor special -QUESTION [11 upvotes]: By an Aronszajn tree, I mean a tree of height $\omega_1$ with countable levels and no branch. Such a tree is Souslin if it has no uncountable antichains and special if it can be written as the countable union of antichains. It is well known that there is both a ccc forcing and a $\sigma$-closed forcing for adding Souslin trees and, given any Aronszajn tree $T$ (Souslin or otherwise) there is a ccc forcing for specializing $T$. Therefore we can add Souslin trees, and then specialize them, giving us new special Aronszajn trees. In this post I have a variety of questions about forcings which add Aronszajn trees that are neither special nor Souslin. The most basic of these is as follows. -Question 1: What are some of the most common ways of forcing to add an Aronszajn tree which is neither Souslin nor special? In this post Mohammad Golshani cites a theorem that immediately implies that such trees exist under certain $\diamondsuit$ hypotheses. Therefore forcings to add diamonds suffice but I'm curious whether there is a more explicit forcing where the generic is actually the tree. -My next question is about killing Souslin-ness (or whatever it's called) without specializing. -Question 2: Suppose $T$ is Souslin. Does there exist a forcing notion $\mathbb P$ such that in $V^\mathbb P$, $T$ is still Aronszajn but no longer Souslin and also not special? -I'm also curious about these questions in the case of fat Aronszajn trees: Aronszajn trees with the requirement of countable levels removed. The same ccc specializing forcing works to make such trees special so under MA all such trees are special. I'm curious whether non-special fat Aronszajn trees can be built by hand using forcing. -Question 3: What forcings will add a fat, non-special Aronszajn tree? - -REPLY [3 votes]: Regarding your comment before Question 3, you mean all such trees with size $<\mathfrak{m}$ are special. In particular, it doesn't say anything about trees of size $\geq 2^{\omega}$. Indeed, $T(\mathbb{R})$ where elements are bounded well ordered subsets of $\mathbb{R}$ ordered by end-extension is not special. It definitely has no uncountable branches. -For question 1, you can first add a Suslin tree (there are various ways of doing this), of which the most straightfoward one is to use countable well-pruned trees with end-extension (due to Jech). For the second step (which also gives yet another answer to the second question), you can pick $S\subset \omega_1$ that is stationary and co-stationary, and use a Shelah style forcing $Q(T,S)$ to make $T$ $S$-st-special (this means there exists a regressive function $f$ whose domain is $T\restriction S$ such that $t -TITLE: Around algebraic equivalence of cycles -QUESTION [5 upvotes]: Let $k$ be a finitely generated field, $X$ a smooth projective $k$-variety, $\ell$ a prime number, $\ell\in k^{\times}$, $r\ge 0$ an integer. -The Tate conjecture asserts surjectivity of the cycle class map: -$$c^r_{\ell}(X): Z_r(X)\otimes_{\mathbf{Z}}\mathbf{Q}_{\ell}\to H^{2r}_{\rm et}(X_{k^{\rm sep}},\mathbf{Q}_{\ell}(r))^{\text{Gal}(k^{\rm sep}/k)}$$ -$c^r_{\ell}(X)$ factors through the group of $r$-cycles modulo algebraic equivalence, $\text{NS}^r(X)\otimes_{\mathbf{Z}}\mathbf{Q}_{\ell}$. We denote by $\text{ns}_{\ell}^r(X)$ the resulting cycle map. - -It would seem the Tate conjecture would then follow from surjectivity of $\text{ns}^r_{\ell}(X)$. Am I missing something up there? Chow groups of cycles modulo rational equivalence are much larger than Néron Severi groups, so this is saying Tate cycles are expected to all be images of cycles modulo algebraic equivalence. Is this correct? -It is known (Thm. of the Base) that $\text{NS}^1(X)$ is finitely generated. Is this known for arbitrary $r\ge 0$? -Is anything known about $\ker(\text{CH}^r(X)\twoheadrightarrow\text{NS}^r(X))$? - -REPLY [5 votes]: A summary of some of these results is given in §19.3 of Fulton's Intersection theory [Ful98]. To address for example the questions you ask: - -If $A \stackrel f\twoheadrightarrow B \stackrel g \to C$ are maps, then the images of $g$ and $gf$ agree. It doesn't matter that $B$ is 'smaller'. -It is not true that $B^2(X) = \operatorname{CH}^2(X)/\!\sim_{\text{rat}}$ is finitely generated; in fact even $B^2(X) \otimes \mathbb Q$ need not be finite dimensional [Ful98, 19.3.3]; this is due to Clemens [Cle83]. The example is three-dimensional. It was not known at the time of writing of [Ful98] whether the torsion in $B^r(X)$ is finite; I don't know if this is known yet. -The kernel of $\operatorname{CH}^r(X) \to B^r(X)$ can sometimes be given the structure of an abelian variety, but sometimes it's bigger than that [Ful98, 19.3.4]. Even for $r = \dim(X)$ (i.e. $0$-cycles), this group can be very big [Ful98, 19.3.5]. - -The notation $B$ for what you call $\operatorname{NS}$ seems to be standard, but possibly not the only standard. - -References. -[Cle83] Clemens, Herbert, Homological equivalence, modulo algebraic equivalence, is not finitely generated. Publ. Math. Inst. Hautes Étud. Sci. 58, p. 231-250 (1983). Available online through Numdam. ZBL0529.14002. -[Ful98] Fulton, William, Intersection theory. Second edition. Ergebnisse der Mathematik und ihrer Grenzgebiete (3) 2. Springer-Verlag, Berlin, 1998. ZBL0885.14002.<|endoftext|> -TITLE: Finding a proof within a paper: reduced $K$-theory of Higson compactification of $[0,\infty)$ is uncountable -QUESTION [5 upvotes]: Emerson and Meyer's Paper "Dualizing the Coarse Assembly Map" (2006) states the following Proposition (5.1): -Let $X = [0,\infty)$ be the ray with its Euclidean metric coarse structure. -Then the reduced K-theory of the Higson compactification $\eta X$ of $X$ is uncountable. -The authors state that this is proved in the paper of A. N. Dranishnikov, J. Keesling and V. V. Uspenskij entitled "On the Higson corona of -uniformly contractible spaces" (1998). -However, I cannot seem to locate this proof within the reference, which is only 13 pages long. -It is probably the case that I do not understand enough to be able to find the statement from which this Proposition follows. Would someone more knowledgeable please point me to such a statement (if it exists) within the paper of Dranishnikov et. al.? -Many thanks! - -REPLY [2 votes]: It seems that Keesling's 1994 paper "The One-Dimensional Cech Cohomology of the Higson Compactification and Its Corona" contains the required result (Corollary 1). It may have been mis-cited.<|endoftext|> -TITLE: Bound for largest eigenvalue of symmetric matrices of uniform random variables over $[0,1]$ and fixed $1$s along diagonal and scattered $1$s -QUESTION [7 upvotes]: Given a $n\times n$ symmetric random matrix whose diagonal elements are all fixed as $1$. In addition, there are $k$ $1$s will be randomly scattered in upper triangular (of course, the corresponding places in the lower-triangle will be filled with $1$, and $2k < n^2-n$). All other elements are independent uniform random variables over $[0,1]$. - -Is there known bound (lower and upper) for the largest eigenvalue of such random matrices? - -If there is not, any suggestion of possible method (I can think of using Gershgorin circle) or reference to related materials is very much appreciated. -Gershgorin circle could help with the upper bound. For example, if we assume all those $1$s are in the same row, then we should be able to find the probabilistic bound for this case with Irwin–Hall distribution; but I currently have trouble dealing with the "randomly scattered" $1$s. -I am not familiar with the random matrix theory. I am not sure if there is anything from it can help this. - -REPLY [3 votes]: The diagonal elements just shift the spectrum (and the top eigenvalue) by $1$. So we may assume they are $0$. -You are essentially dealing with a symmetric matrix whose entries above the diagonal are iid, sum of a Bernoulli $\{0,1\}$ of parameter (=mean) $q=2k/n^2$ and of a uniform random variable on $[0,1]$. The mean is therefore $p=q+(1-q)/2$ and the variance is $\sigma^2=q+(1-q)/3-p^2$. Since, for any value of q, the mean is bounded below and so is the variance, you are dealing with the perturbation by $\sqrt{n}$ times a Wigner matrix of a matrix of rank $1$ and norm $pn$. Thus, your top eigenvalue will concentrate around $pn$ ($\pm O(\sqrt{n}))$ by estimates on the top eigenvalue of a Wigner matrix and -Weyl's inequalities. -I earlier referred to https://arxiv.org/pdf/math/0605624v1.pdf, but this is a very different scaling, so I scraped that answer.<|endoftext|> -TITLE: Is Visualization of Data a Subject of Mathematical Research? -QUESTION [5 upvotes]: Please excuse my naive question, but what kind of rôle does the visualization of (especially high-dimensional) data play in mathematical research? -I know, that it plays an important rôle in the analysis and interpretation of data and utilizes advanced mathematics, but, apart from being a servant to mathematical research, are there also examples, where mathematical research is solely dedicated to data-visualisation or, where data-visualisation brought up questions, that initiated important mathematical research? -A prominent example of mathematical research that has been initiated by data visualisation is Guthries observation, that apparently four colors suffice to color every map; hat observation had very fruitful consequences for mathematics. - - -Questions: - -what are further examples of problems that originated in data-visualization and became the subject of mathematical research? - -are there examples of mathematical research, that is dedicated to improving the visualization of data, i.e. to make features of interest, that are "burried" in the data, more "prominent"? - -REPLY [3 votes]: This is a rather open-ended question, so let me respond by offering some observations in the same spirit, focusing primarily on posets, graphs and simplicial complexes. -In a very general way, graphs and digraphs are used throughout mathematics to represent relations and other structures with metric, topological and enumerative-combinatorial properties, such as the diameter, connectedness and the number of subgraphs with prescribed properties reflecting some of the deeper or less obvious features of the represented structure. (These relations may originate from a mathematical model or be innate to a mathematical theory.) Advanced examples along these lines are provided by the order complexes of posets and Stanley-Reisner complexes of square-free ideals, where the complex both visualizes the underlying structure and brings forth its important features. In a different direction, diagrams in category theory and quivers in algebra, including advanced constructions such as the Auslander-Reiten quiver, are types of visualizations. -Here is a specific illustration already firmly within mathematical setting. The notion of Hasse diagram of a partially ordered set provides an important visualization of this structure that has been used to study its properties. For example, it is easy to see that the diamond lattice $M_3$ and the pentagon lattice $N_5$ are not distributive. In fact, a lattice is distributive if and only if it does not contain a sublattice isomorphic to $M_3$ or $N_5$. On a more rudimentary level, the fundamental concepts such as maximal and minimal elements, chains and maximal chains of a poset can be thought of as visualizations. Some research directions in group and module theory deal with the structure of groups/modules based on their properties of their subgroup/submodule lattice and one can argue that, for example, uniserial modules (where all submodules form a single chain) first became objects of interest because of this visualization aspect. Later, the success of this theory motivated investigating more complicated cases. -Finally, going beyond combinatorial and algebgraic structures, configuration spaces and moduli spaces of all kinds fit into a visualization paradigm: they "visualize" the totality of objects or "states of a system" and can be used to study generic or typical properties, statistics and often dynamics (when the configuration or moduli space arises from geometry).<|endoftext|> -TITLE: Set of translations of a real function having a dense linear span -QUESTION [5 upvotes]: Let $W$ be the space of continuous functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $\lim_{x\rightarrow \pm \infty} f(x)=0$, and consider the sup-norm topology on $W$. -Problem. does there exist $f\in W$ such that the set of translations of $f$ (i.e., the set of functions $f_i(x)=f(x+i)$, $i\in \mathbb{Z}$) have a dense linear span in $W$? -Having a dense linear span means that for every $g\in W$ and every $\epsilon>0$, there exists a finite subset $\Lambda \subseteq \mathbb{Z}$ and real numbers $c_i \in \mathbb{R}, i\in \Lambda$ such that: -$$\max_{x\in \mathbb{R}} \left |g(x)-\sum_{i \in \Lambda} c_if(x+i) \right |<\epsilon.$$ - -REPLY [7 votes]: Yes. This was proved by Atzmon and Olevski.<|endoftext|> -TITLE: Geometrically quantizing real Grassmannians -QUESTION [12 upvotes]: It seems that the Grassmannian of oriented 2-dimensional planes in $\mathbb{R}^n$ -$$ \mathrm{Gr}(n,2) = \frac{\mathrm{SO}(n)}{\mathrm{SO}(n-2) \times \mathrm{SO}(2)} $$ -has a symplectic structure that's invariant under the action of $\mathrm{SO}(n)$. My argument for this is as follows. Start with the cotangent bundle of $T^*\mathrm{S}^{n-1}$ and put the usual kinetic Hamiltonian -$$ H(x,p) = \frac{1}{2} \|p\|^2 , \qquad x \in \mathrm{S}^{n-1}, p \in T^*_x \mathrm{S}^{n-1}$$ -on this, getting $H \colon T^*\mathrm{S}^{n-1} \to \mathbb{R}$. This generates a flow on $T^*\mathrm{S}^{n-1}$ which corresponds to geodesic motion on the sphere. Now do symplectic reduction: form the submanifold $\{H = E\}$ for some constant $E > 0$, and mod out by the flow generated by $H$. This gives the space of oriented great circles in $\mathrm{S}^{n-1}$, and thanks to the theory of symplectic reduction this should have a symplectic structure. -But any oriented great circle in $\mathrm{S}^{n-1}$ arises by intersecting this sphere with unique oriented 2d subspace of $\mathbb{R}^n$. So, the space of oriented great circles in $\mathrm{S}^{n-1}$ should be diffeomorphic to $\mathrm{Gr}(n,2)$. -So, $\mathrm{Gr}(n,2)$ should have a $\mathrm{O}(n)$-invariant symplectic struture depending on $E > 0$. I guess changing the energy $E$ just rescales this symplectic structure. -Question 1. Is this correct so far? If so, it must be known. Do you know references? -Question 2. What's a nice simple formula for this symplectic structure on the real Grassmannian? -Question 3. Is this symplectic structure the imaginary part of an $\mathrm{SO}(n)$-invariant Kähler structure? -Question 4. If so, has someone geometrically quantized $\mathrm{Gr}(n,2)$ for values of $E$ making the symplectic structure integral? What Hilbert spaces do we get? -If it works, we should get some nice finite-dimensional unitary representations of $\mathrm{SO}(n)$. - -REPLY [2 votes]: Another way to view this: think of $T^*S^{n-1}$ as $\lbrace (q,p)\in\mathbb{R}^{2n}\,|\,\Vert q\Vert = 1, q.p=0\rbrace$, with symplectic structure $\omega((v_q,v_p),(w_q,w_p)) := v_q\cdot w_p - v_p\cdot w_q$. The $\operatorname{SO}(n)$-action is Hamiltonian, with $\operatorname{SO}(n)$-equivariant momentum map $J:T^*S^{n-1}\rightarrow \mathfrak{so}(n)^*$ given by -$$ -\langle J(q,p),\xi\rangle = p(\xi\cdot q)\qquad\textrm{for}\qquad\xi\in\mathfrak{so}(n). -$$ -Since the $\operatorname{SO}(n)$-action preserves the Hamiltonian, it drops to the reduced space $M_{\textrm{red}} = H^{-1}(E)/{\lbrace\textrm{Hamiltonian flow}\rbrace}$. Since the unreduced action acts transitively on $H^{-1}(E)$, the reduced action acts transitively on $M_{\textrm{red}}$. The $\operatorname{SO}(n)$-equivariance of the corresponding reduced momentum map $J_{\textrm{red}}:M_{\textrm{red}}\rightarrow \mathfrak{so}(n)^*$ -implies that $J_{\textrm{red}}$ maps $M_{\textrm{red}}$ to a coadjoint orbit $\mathcal{O}\subset\mathfrak{so}(n)^*$. Kostant's coadjoint orbit covering theorem then implies that $J_{\textrm{red}}$ is a symplectic covering map of $M_{\textrm{red}}$ onto $\mathcal{O}$ -, equipped with the (positive) Kostant-Kirillov-Souriau form -$$ -\omega^+_\mu(-\operatorname{ad}_\zeta^*\mu,-\operatorname{ad}_\chi^*\mu) := \langle\mu,[\zeta,\chi]\rangle. -$$ -In fact, in this case $J_{\textrm{red}}$ is a symplectomorphism. -To see things more concretely: identifying $\mathfrak{so}(n)$ with its dual via the trace form, we get the (still equivariant) Lie-algebra-valued momentum map $j:T^*S^{n-1}\rightarrow \mathfrak{so}(n)$ given by -$$ -j(q,p) = \frac{1}{2}(q p^\top - p q^\top)\in\mathfrak{so}(n). -$$ -Then for example taking the generic point $(q,p) = (e_1,\sqrt{2E} e_2)\in H^{-1}(E)$, this gives -$$ -j(q,p) = \begin{bmatrix} 0 & \sqrt{\frac{E}{2}} & \ldots & 0\\ --\sqrt{\frac{E}{2}} & 0 &\ldots & 0 \\ -\vdots \\ -0 & 0 &\ldots & 0\end{bmatrix} -$$ -Clearly the stabiliser of this element under the adjoint action of $\operatorname{SO}(n)$ is $\operatorname{SO}(2)\times\operatorname{SO}(n-2)$. Viewed in $T^*S^{n-1}$, the second factor acts trivially on $(q,p)$, while the first factor produces the great circles corresponding to the Hamiltonian flow. Hence the fibres of the quotient map $\pi:H^{-1}(E)\rightarrow M_{\textrm{red}}$ and the momentum map $j:H^{-1}(E)\rightarrow \mathcal{O}$ agree, which implies that the covering $j_{\textrm{red}}:M_{\textrm{red}}\rightarrow \mathcal{O}$ is actually a diffeomorphism. -It's straightforward now to put a complex structure on the coadjoint/adjoint orbit (the same construction works for the orbits of any compact Lie group). I will use adjoint orbits here. At any point $\xi\in \mathcal{O}\subset \mathfrak{so}(n)$, the operator $\operatorname{ad}_\xi:\operatorname{so}(n)\rightarrow\operatorname{so}(n)$ is skew-adjoint with respect to the trace form, and so has pure imaginary eigenvalues. Let -$$ -\mathfrak{n}_\xi^+ := \lbrace \zeta\in\mathfrak{so}(n)\,|\, \operatorname{ad}_\xi\zeta = i\lambda\, \zeta \quad\textrm{for some }\lambda>0\rbrace, -$$ -and define the polarization $F$ on $\mathcal{O}$ by -$$ -F_\xi := \lbrace \operatorname{ad}_\zeta \xi\,|\,\zeta\in\mathfrak{n}_\xi^+\rbrace. -$$ -At any point $\xi\in\mathcal{O}$, we have $\ker(\operatorname{ad}_\xi) = \mathfrak{so}(n)_\xi$ (the adjoint stabiliser algebra), and we can choose a Cartan subalgebra $\mathfrak{h}\subset\mathfrak{so}(n)_\xi$, from which we can introduce an ordering on weights. As Ben McKay mentioned, the usual Borel-Weil construction implies that the holomorphic sections with respect to the polarization $F$ give an irreducible representation with highest weight related to $\xi$. I would have to check my signs and conventions carefully, but I think in this case you get the dual to the irrep with highest weight $-\frac{i}{\hbar}\xi^\flat$, where $\cdot^\flat:\mathfrak{so}(n)\rightarrow \mathfrak{so}(n)^*$ is the duality with respect to the trace form (don't quote me on that, though :)).<|endoftext|> -TITLE: Are there examples of finite-dimensional weak Hopf C*-algebras with non-involutive antipode? -QUESTION [5 upvotes]: For finite-dimensional (non-weak) Hopf C*-algebras it is known that the antipode is always involutive, as claimed e.g. in https://arxiv.org/pdf/1007.5283.pdf. I couldn't find the same statement for weak Hopf C*-algebras, however. Are there counterexamples? - -REPLY [2 votes]: This is an answer to the question in the OPs answer. -I just took a brief look into - -Rehren - Weak C Hopf symmetry* in Quantum Groups Symposium at - "Group21", eds. H.-D. Doebner et al., Goslar 1996 Proceedings, Heron - Press, Sofia (1997), pp. 62-69. https://arxiv.org/abs/q-alg/9611007 - -which does the type III case. -The point is that $S^{-1}(q)=S^\ast(q^\ast)$, which is what the diagram on p84 in the mentioned article says.<|endoftext|> -TITLE: What about of periodic points of $\sum_{n=1}^\infty\frac{\mu(n)}{n}x^n$, $01$, but it's not needed). Using summation by parts, we get -$$\sum_{n=1}^\infty\frac{\mu(n)}{n}x^n=\sum_{n=1}^\infty(m(n)-m(n-1))x^n\\=\sum_{n=1}^\infty m(n)(x^n-x^{n+1})<\sum_{n=1}^\infty(x^n-x^{n+1})=x$$<|endoftext|> -TITLE: Geometric object corresponding to subalgebra generated by an ideal? -QUESTION [6 upvotes]: This question is closely related to What is the geometric object corresponding to a subalgebra in a polynomial ring. -There, it is asked, given a subalgebra of an algebra $S \subset R$ over a field $k$ what is the corresponding geometric object, in the sense of algebraic geometry. -The answer is that there are varieties $X,Y$ associated to $R,S$, respectively, along with a dense morphism $X \to Y$. -Here the term "variety" is a bit looser than often used. -My question is the same, but for a specific type of subalgebra. Given a $k$-algebra $R$ and an ideal $I \subset R$, let $S = k + I \subset R$ be the $k$-subalgebra generated by the ideal $I$. What is the geometric interpretation of the variety $\mathrm{Spec}(S)$ and the morphism $\mathrm{Spec}(R) \to \mathrm{Spec}(S)$? -I am happy to assume that $R$ is a finite type $k$-algebra, and even that it is reduced. I am most concerned about the case when $R = k[x_1, ..., x_n]$ is just a polynomial ring. - -REPLY [7 votes]: The image of the composition $S \to R \to R/I$ is equal to the constants $k \subset R/I$. This indicates that the morphism of the zero locus $V(I) = Z \to \mathrm{Spec}(R) \to \mathrm{Spec}(S)$ factors through the structure morphism $Z \to \mathrm{Spec}(k)$, and so the image must be a point. Therefore the morphism corresponding to the inclusion of subalgebras must be one in which the distinguished closed subvariety $Z$ is collapsed to a point. -Because $S \to R$ is injective, the morphism $\mathrm{Spec}(R) \to \mathrm{Spec}(S)$ will be dense. That the morphism is dense is actually equivalent to the kernel being contained in the nilradical. So I believe one should be able to show without much difficulty something like $\mathrm{Spec}(R) \to \mathrm{Spec}(S)$ is universal with respect to being a dense morphism to a reduced scheme, sending $Z$ to a point.<|endoftext|> -TITLE: Status of Thomason's idea for a symmetric monoidal model of stable homotopy - from his last paper -QUESTION [14 upvotes]: In 1995, Robert Thomason published “Symmetric monoidal categories model all connective spectra” in TAC. On page 2, he argues that symmetric monoidal categories are more convenient than “May’s coordinate-free spectra” for various reasons, one of which is that a symmetric monoidal structure is easier to obtain (this was before EKMM or the Hovey-Shipley-Smith paper on symmetric spectra). Thomason writes: - -As convincing evidence for this claim, I refer to my talk at the Colloque en l’honneur de Michel Zisman at l’Universite Paris VII in June 1993. There I used this alternate model of stable homotopy to give the first known construction of a smash product which is associative and commutative up to coherent natural isomorphism in the model category. This will be the subject of an article to appear. - -Unfortunately, he died later that year. Hence my question: - -Did anyone ever work out the construction of the smash product he had in mind? - -REPLY [3 votes]: As for connective spectra, a good smash product was first worked out by Manos Lydakis in the context of $\Gamma$-spaces. Thomason came to Bielefeld in the mid 1990s and spoke with Waldhausen's group about this matter. Lydakis carried it through.<|endoftext|> -TITLE: Fields of Definition of Elliptic Curves -QUESTION [6 upvotes]: I am currently studying the theory of complex multiplication and I find myself confused by the language in a lot of the literature. -In Silverman's Advanced Topics in the Arithmetic of Elliptic Curves, Silverman uses the term model without ever defining it (as far as I can see). What does he mean? F-isomorphism class? -In particular the proof of theorem II.2.3, he states -"We take a model for $E$ defined over $H=K(j(E))$". Why can one swap out $E$ with a model defined over $K(j(E))$. -Another example of this, is in Diamond, Darmon and Taylor's paper on Fermat's last theorem. In remark 1.3, it states that any elliptic curver $E/\mathbb{C}$ with CM is defined over an abelian extension of $K= \mathrm{End}_{\mathbb{C}}(E)\otimes E$. -I assume they are citing the fact that $K(j(E))$ is an abelian extension of $K$, but why is $E$ defined over $K(j(E))$? Is this even the supposed abelian extension $E$ is defined over? If not, which one is it? - -REPLY [11 votes]: What's usually meant when phrased this way is that within the $\overline K$-isomorphism class of $E$, there is an elliptic curve defined over $K(j(E))$. An indeed, for any elliptic curve $E$ defined over $\overline{\mathbb Q}$, there is an elliptic curve $E'$ defined over $\mathbb Q(j(E))$ that is $\overline{\mathbb Q}$-isomorphic to $E$. So that's the sort of model that one often takes. All this has nothing to do with CM, and is very elementary. -If $E$ has CM with $K=\text{End}_{\mathbb C}(E)\otimes\mathbb Q$, then part of the basic theory of CM is that $K(j(E))$ is an abelian extension of $K$, and indeed if $\text{End}_{\mathbb C}(E)$ is the full ring of integers of $K$, then $K(j(E))$ is the maximal abelian everywhere unramified extension of $K$ (the Hilbert class field of $K$). This is far less elementary, but sufficiently well-known that DDT probably didn't feel it necessary to give a reference. The fact that we can find a model for $E$ over $K(j(E))$ follows from the previous paragraph. We can actually find such a model over $\mathbb Q(j(E))$, but probably they want the endomorphisms to also be defined over the field, and thus need to take the compositum with $K$.<|endoftext|> -TITLE: A Markov consensus -QUESTION [6 upvotes]: Consider the following process. You start with $n$ nodes in different colors $c=c1,c2,...$ (representing an opinion). Say, $n=5, c=1,2,3,4,5$. Now each node checks which colors have weak majority (here, all :-) and takes a random one. Say, $c=1,3,3,5,1$. Only $1,3$ have majority. Rinse and repeat: Say, $c=3,3,1,1,3$. Next step, $c=3,3,3,3,3$, we have agreement, stop. Of course now it is interesting how many steps $E$ the process needs in the average (it's VERY fast). -Here are the values of $E$ for $1\leq n \leq 9$ via a simple (ehem) Markov chain calculation: $0,2 -,15/7 -,334/145 -,5575/2404 -,2562531/996556 -,374308417/151539984 -,20507023600358/7758785772837 -,37796077434852247/15034331949432832.$ -Clearly the numbers are evil, even the denominators. I would already be content with a much simpler question: What happens if $n\rightarrow\infty$? -a) $E\rightarrow\infty$ (very slowly), since it can hang up in ties arbitrarily often, -b) $E\rightarrow 2$ (very slowly), since it's extremely probable one color has majority due to random fluctuation, -c) $E\rightarrow C$ (very slowly), where $C$ is Reddmann's constant (hey, I want to have something named after me in math, I'm getting old :-) which is about $2.5-2.6$, judging by numeric simulations. (I went until $n=100000$ which is small compared to $\infty$ but my computer is a snail.) - -REPLY [5 votes]: (This is an extended comment, not an answer). - -Let $X_j$ be the number of votes for the $j$-th candidate in the first turn. -For large $n$, $X_j$ are roughly independent Poisson random variables with mean $1$. More precisely, one can obtain $X_j$ by choosing a random sample $Y_j$ of independent Poisson random variables and then add extra $n - \sum_{j = 1}^n Y_j$ random votes if this quantity is positive, or, otherwise, take away randomly chosen $\sum_{j = 1}^n Y_j - n$ votes. This is minor, as $|n - \sum_{j = 1}^n Y_j|$ is typically of order $O(\sqrt{n})$. (Correct me if I am wrong here). -On average, $n/(e k!)$ of the random variables $Y_j$ are equal to $k$. -If $k$ is extremely large and $e k! \ll n \ll e (k+1)!$, with high probability there will be no $Y_j$ greater than $k$ and there will be roughly $n / (e k!) \gg 1$ of $Y_j$ equal to $k$. This suggests that for such $n$ typically one turn will not be enough: $E \approx 3$. -On the other hand, if $k$ is very large and $n \approx e k!$, then with high probability there will be no $Y_j$ greater than $k$ and there will be roughly one $Y_j$ equal to $k$. Therefore I would expect that for such $n$ the winner will be chosen in one turn: $E \approx 2$. (Edit: exactly one of $Y_j$ is equal to $k$ with probability roughly $1/e$, so one should rather expect $E \approx 3 - 1/e$). -Therefore, my conjecture is: the limit does not really exist! - -Observe that $e \times 7! < 100000 < e \times 8!$ with $100000/(e \times 7!) \approx 7.3$ and $100000/(e \times 8!) \approx 0.91$. This shows that for $n = 100000$ is somehow in an intermediate regime. (However, $k = 7$ is not very large). -Can you repeat your simulations for $n = 109601 \approx e \times 8!$ and $n = 38750 \approx e \times 7! \times \sqrt{8}$? The former one should lead to $E \approx 2$ (edit: rather $E \approx 3 - 1/e$), while for the latter I expect $E \approx 3$. - -Edit #2: I did some numerical experiments to support the above claim about non-convergence of $E$. Here is what I got: - -This is the plot of the probability of a draw in the first round (which is roughly equal to $E - 2$) against $n$ in the logarithmic scale. Each data point was found by averaging $10^4$ repetitions. Bars correspond to $95\%$ confidence intervals (obtained by approximation by normal distribution). Red vertical lines mark $n = e k!$ for $k = 1, 2, \ldots$ -Looks like there is little hope that the Reddmann's constant exists. Good news, however, is that you can have even two of them: the upper and the lower Reddmann's constant (for the $\limsup$ and $\liminf$).<|endoftext|> -TITLE: Commutative group algebraic spaces -QUESTION [6 upvotes]: Is the category of commutative group algebraic spaces (commutative group objects in algebraic spaces) locally of finite type over a field, an abelian category? -I would benefit from a reference - -REPLY [9 votes]: The intervention of algebraic spaces is a red herring because algebraic space groups $G$ locally of finite type (and quasi-separated!) over a field $k$ are necessarily schemes. (I am assuming you are only interested in the quasi-separated case, though no motivating context for the question was given.) This is Lemma 4.2 in Artin's paper "Algebraization of Formal Moduli I" (where all algebraic spaces are taken to be quasi-separated, pervading the foundations such as for a reasonable theory of irreducible components); the proof is given near the end of section 4 uses results about biratonal group laws. So let's give a variant on that proof which avoids birational group laws (as that tool feels too heavy for the situation at hand, for which the real technical point is that only over the typically infinite-degree algebraic closure do we have enough rational points to make translation arguments). -Remark: The quasi-separatedness hypothesis is essential, even if we only consider algebraic space groups of finite type. For instance, if $k$ has characteristic 0 then the quotient $\mathbf{A}^1_k/\mathbf{Z}$ of the affine line modulo the constant $k$-group $\mathbf{Z}$ is an algebraic space group of finite type over $k$ that is not quasi-separated and is certainly not a scheme. -Coming back to the proof of Artin's Lemma 4.2, first note that $G^0$ is of finite type (same proof as for schemes, as Artin sketches). Moreover, it remains connected after any finite extension on $k$ since it is connected with a rational point (same proof as for schemes, using that $(G^0)_{k'} \to G^0$ is finite flat surjection with fiber ${\rm{Spec}}(k')$ over the identity $k$-point of $G^0$). In fact, as will be useful later, even $(G^0)_{\overline{k}}$ is connected (since any idempotent on this algebraic space of finite type descends to one on some $(G^0)_{k'}$ for some finite extension $k'/k$ inside $\overline{k}$, yet any such $(G^0)_{k'}$ is connected, so the descended idempotent is 0 or 1). -Similarly, for any connected component $E$ of $G$, $E_{k'}$ is closed and open in $G_{k'}$ with $E_{k'} \to E$ a finite flat surjection, so by picking $k'/k$ big enough to split the residue field over $k$ at some closed point of $E$ we get that all connected components of $E_{k'}$ have a $k'$-point and thus are translated copies of $(G^0)_{k'}$. -We claim each $E$ is a quasi-projective $k$-scheme (so in particular $G$ is a scheme). Grant this for $G^0$ for a moment. Fix $E$ and pick finite $k'/k$ so $E_{k'}$ has a $k'$-point in all connected components. Thus, as noted above, $E_{k'}$ is a disjoint union of finitely many copies of $(G^0)_{k'}$, so it is a quasi-projective scheme. But $E_{k'}$ as a $k$-scheme is equipped with evident descent data relative to the finite flat cover ${\rm{Spec}}(k') \to {\rm{Spec}}(k)$, and this is effective by quasi-projectivity with the descent also a quasi-projective over $k$ (see SGA1, VIII, 7.6). A further argument with descent for morphisms of algebraic spaces (left as an exercise) shows that this descent is identified with $E$, so $E$ is a scheme (granting the same for $G^0$). -Now it remains to show that $G^0$ is a quasi-projective scheme. -This is the step for which Artin appeals to the atom bomb of birational group laws, and we'll now make a bit more use of descent to provide an alternative procedure without nuclear weapons. It suffices to show $(G^0)_{k'}$ is a quasi-projective scheme for some finite extension $k'/k$, as then we can use a descent argument as above. By "chasing coefficients", it suffices to show $(G^0)_{\overline{k}}$ is a quasi-projective scheme (as we write $\overline{k}$ as the direct limit of finite subextensions over $k$ to descend an immersion $(G^0)_{\overline{k}} \hookrightarrow \mathbf{P}^n_{\overline{k}}$ to a map $(G^0)_{k'} \to \mathbf{P}^n_{k'}$ that is moreover an immersion for some finite extension $k'/k$; here we are using that $G^0$ is "finitely presented" over $k$, not just finite type, precisely because $G^0$ inherits quasi-separatedness from $G$). But $(G^0)_{\overline{k}}$ is connected (as we saw above!), so this finally allows us to replace $k$ with $\overline{k}$ so that $k$ is algebraically closed. We want to show over such $k$ that any locally finite type (and quasi-separated!) algebraic space group $H$ over $k$ is a scheme. -Every non-empty closed subspace of $H$ has a $k$-point since $k$ is algebraically closed, so it suffices to make an open subspace $U \subset H$ that is a scheme and contains all $k$-points (as then $(H-U)(k)$ is empty, forcing $H=U$ to be a scheme). It suffices to find a non-empty open subspace $V$ in $H$ that is a scheme, since that must have a $k$-point (as $k = \overline{k}$) and so can be translated to get an open subscheme around any $k$-point, the union of which (over all $k$-points of $H$) is a $U$ as desired. -The desired $V$ exists because (i) any quasi-separated algebraic space contains a dense open subspace that is a scheme (see II, 6.13 in D. Knutson's book Algebraic Spaces) and (ii) we are assuming that $H$ is quasi-separated.<|endoftext|> -TITLE: Automorphic quotient for quaternion algebras -QUESTION [5 upvotes]: Are automorphic quotient for quaternion algebras always compact (safe the totally split case)? -Is there any good reference for proof of this fact, or easy arguments to say do? - -REPLY [8 votes]: Weil's "Adeles and algebraic groups" proves this for general division algebras. It is labelled "Fujisaki's Lemma", and is a natural division-algebra generalization of the analogue for number fields themselves, namely, that $\mathbb J^1/k^\times$ is compact, where $\mathbb J^1$ is the collection of ideles of idele-norm $1$. For number fields, this compactness implies finiteness of (generalized) class number, and implies the (generalized) units theorem.<|endoftext|> -TITLE: Polymath-type projects for textbooks? -QUESTION [14 upvotes]: Why aren't there any "crowdsourced" projects for mathematical textbooks? - -Every year, many mathematicians put a lot of effort into crafting their own lecture notes or writing textbooks (or also research monographs). Why aren't there any open-source "crowdsourced" cooperative efforts towards book writing? The pros of this kind of collaboration are rather obvious. - -REPLY [12 votes]: The Homotopy Type Theory book is massively collaborative and regularly updated (through version control). It's also the definitive textbook in its field, and I think Voevodsky himself contributed to it. In particular, it is mentioned that: - -We have released the book under a permissive Creative Commons licence - which allows everyone to participate and improve it. - -I've found it to be particularly well-written and comprehensible.<|endoftext|> -TITLE: Main statement as theorem or corollary -QUESTION [27 upvotes]: In a text there will often be a few important results that are usually called Theorems. Intermediate statements are called Lemmas, and statements that follow immediately from previous results are called Corollaries. -There is a problem here, however. Especially in the context of theorem provers: A Theorem $A$ is often proved using induction. In order to perform this induction, the statement of the theorem needs to be strengthened to $B$ (otherwise the induction will not 'go through'). The original statement $A$ of the theorem will then follow trivially from $B$. We now have three possible naming conventions for this: - -Call $B$ a Lemma and $A$ a Theorem. This is attractive, because $A$ is the main statement of interest. The problem is, however, that the proof of $A$ is mainly done in $B$, which is now called an 'unimportant' Lemma. -Call $B$ a Theorem and $A$ a Corollary. This is attractive because $A$ follows immediately from $B$, hence a Corollary. Also, it makes clear that the proof of $B$ contains a lot of important steps. However, when looking over the text quickly, one might miss the main statement $A$, because it is phrased as a simple consequence of $B$. -A third option is to not have $B$ at all but only state Theorem $A$, and then perform the strengthening of the statement 'inline' in the proof via a cut. This is also inconvenient because it hides the technique used to prove $B$, which may be very important. - -I would be interested in how this is commonly solved. - -REPLY [2 votes]: You say "Especially in the context of theorem provers: ..." then go on to describe conflating content and presentation. The content is a dependency tree of theorems. The presentation uses words like "lemma", "theorem", and "corollary" to help organize the information for a reader. -If we are "in the context of theorem provers", we are discussing content, so the resolution is straightforward: everything is a theorem. Perhaps one wants to allow some presentation markup in one's theorem prover, but mixing content and presentation is an anti-pattern. -(In the context of presentation, the situation you are describing is the familiar pattern of having a denouement corollary after some complicated (sequence of) theorem(s) in the preceding work. Then the labelling indicates the amount of work the reader should expect to expend to understand the object. Until theorem provers start proving theorems about the difficulty in understanding a statement and proof assuming a reader model, this is not in the context of theorem provers.)<|endoftext|> -TITLE: What motivations for automorphic forms? -QUESTION [38 upvotes]: Automorphic forms are ubiquitous in modern number theory and stands as a mysterious Graal lying at the intersection of many fields, if not building valuable bridges between them. However, since this aim has been erected as one of the topmost paradigm of research in number theory, reasons for formulating motivations for studying automorphic forms may have faded. I am seeking here for such motivations and facts leading to making so many efforts toward the theory of automorphic forms. - -Why are we interested in automorphic forms/representations? - -There are many different levels of reading of this question, and it might be of use to both have in mind the possibility of some of them -- for we often stay at one particular of them, depending on our interests and culture --and the different treatment they require. -Type of audience - -(A) General Public (at most high school background, with no particular interest in science) -(B) Scientists (some undergraduate years or curiosity and interest toward science) -(C) Mathematicians (think of the question as an analog to Sarnak's book "Some applications of modular forms", but for automorphic forms/representations) -(D) Number Theorists (inner motivations, e.g. Langlands program, allowed) - -Type of argument - -(0) Meta (e.g. unification of different notions; "representations are a way to makes a group a group of motions, so endow it with a more geometrical structure, change the insight, or note that we know better many objects by their actions than by their elements", etc.) -(1) Outside motivations (e.g. PDE model many dynamics coming from physics or biology; high-dimensional spaces as efficient setting for robotics; formal logic leading to specification and verification, etc.) -(2) Math Facts -(e.g. historical reasons of why automorphic forms are powerful; "such fact reduces the study of a certain kind of objects to a simpler one", how it intervenes in a classification as a relevant part of the field, etc.) -(3) Math Hopes -(e.g. what a given result on automorphic forms would bring) - -This is a proposition and every comment or suggestion towards those classification are welcome. -Some comments on the classification - -the "Audience" classification can be superfluous because it can be an issue concerning language and culture (replacing "PDE" and "representation" by "evolution law" or "way to move" can allow to make an argument work for both C and A as well) more than contents, but not always -there is no necessary relation between the "Audience" tag and the "Argument" one. Even if some of those categories seems to be mutually exclusive (as A2), this is also part of the challenge to be able to present the whole diversity of motivations for automorphic forms to every audience. - -The purpose of this thread if both to gather ideas to answer the natural questions of mathematicians as well as friends about the reason of spending so much time working on those topics, and also to reinvent our field in facing different aspects and motivations for the automorphic world as well as having to put words on motivations that are sometimes far away from our mind (and I do not claim they are necessary). Those could also lead to natural introductions and motivations for students. -Ideas, precise references, detailed answers or specific examples are all welcome. -Note: Even if many examples motivating modular forms or Maass forms arise in the literature, I fail for months to find something for automorphic forms without specific distinction. - -REPLY [13 votes]: The specific issue of what automorphic forms on bigger groups than $GL(2)$ over $\mathbb Q$ (for example) may tell us about automorphic forms (and L-functions) for $GL(2,\mathbb Q)$ or $GL(1,k)$ for number fields $k$ does have at least a few good answers. First, about 1960 and a little before, Klingen's proof that zeta functions of totally real number fields $k$ (and L-functions of totally even characters on such fields) have good special values at positive even integers used the idea of pulling back holomorphic Hilbert modular Eisenstein series to elliptic modular forms. (I heard G. Shimura lecture on this c. 1975, and it was quite striking.) -Another example: already in the 1960s, J.-P. Serre and others saw that holomorphic-ness of symmetric-power L-functions for $GL(2)$ holomorphic modular forms would prove Ramanujan-conjecture-type results. How to prove that holomorphy? By finding an integral representation of such L-functions, and using that. This has met with varying degrees of limited success, e.g., in papers of H. Kim and F. Shahidi. -The previous example was grounded in the general pattern of Langlands-Shahidi treatment of L-functions in terms of constant terms of cuspidal-data Eisenstein series on (necessarily larger) reductive groups. The specifica cases where various Levi-Malcev components of parabolics were products of $GL2$'s or $SL2$'s produced several "higher" L-functions for $GL2$. -As variation on that, already in the Budapest conference in 1971, Piatetski-Shapiro observed that (what we often nowadays call) Gelfand pairs could produce Euler produces via integral representations (usually involving Eisenstein series). Various success-examples of this idea included work of PiatetskiShapiro-Rallis, Shimura, myself, M. Harris, S. Kudla, various collaborations among these people, and several others, beginning in the late 1970s. E.g., I was fortunate enough to stumble upon an integral representation for triple tensor product L-functions for $GL2$ in terms of an integral representation against Siegel Eisenstein series on $Sp(3)$ (or $Sp(6)$, if one prefers). M. Harris and S. Kudla found another such integral representation that covered special value results in the "other range" (in terms of P. Deligne's conjectures). -In yet other terms, Jacquet-Lapid-Rogawski (and several others) have demonstrated that a variety of L-functions appear as periods of Eisenstein series on "larger" reductive groups. (One novelty is using relative trace formulas to exhibit Euler products when the simpler "Gelfand pair" idea is not quite sufficient.)<|endoftext|> -TITLE: Geometric Lang conjecture - reference -QUESTION [11 upvotes]: In their paper "Uniformity of rational points", Caporaso, Harris, and Mazur asserted the following: -"The Geometric Lang conjecture has been proved for all surfaces with $c_1^2 > c_2$ ([B]), and has recently been announced for all surfaces ([LM])." -The referece [LM] in that paper (link here: http://www.ams.org/journals/jams/1997-10-01/S0894-0347-97-00195-1/S0894-0347-97-00195-1.pdf) is a preprint by S. Lu and M. Miyaoka, with no title. Their paper was published in 1997, so surely this preprint should be published by now, but I am unable to find it. Does anyone know the actual title of the paper, and perhaps what journal it is published in? - -REPLY [9 votes]: abx's comment was made while I was writing this, but I am posting it as an answer anyway. -There has not been a proof of this conjecture of Lang, which remains a wide open problem. Lu and Miyaoka's paper to which Caporaso, Harris and Mazur refer has to be this one (Bounding curves in algebraic surfaces by genus and Chern numbers, IMRN, 1995) and its sequel cited as [6] in higher dimensions. However, the main result of that paper, cf. Corollary 1 and the announced Theorem 0, has actually an (unnatural) smoothness condition on the rational or elliptic curves involved. -For a comparison you may recall that in the arithmetic case, for arithmetic surfaces (algebraic curves), Vojta has proved a weaker form of his conjecture that has an arithmetic discriminant term $d_a(P)$ in place of a field discriminant term and which, in contrast to the Vojta conjecture, is far too weak to be adequate for deriving any consequence about the $abc$ conjecture. -A related problem and another partial result on the geometric Lang conjecture can be found in this question of mine, and in its answer by Francesco Polizzi.<|endoftext|> -TITLE: A possibly surprising appearance of $\sqrt{2}.$ -QUESTION [20 upvotes]: Define $A=(a_n)$ and $B=(b_n)$ as follows: $a_0=1$, $a_1=2$, $b_0=3$, $b_1=4$, and $$a_n=a_1b_{n-1}-a_0b_{n-2} + 2n$$ for $n \geq 2$, where $A$ and $B$ are increasing and every positive integer occurs exactly once in $A$ or $B.$ Can someone prove that $$2n < a(n) - \sqrt{2} n < 3+2n$$ for $n \geq 2$ ? -Evidence: -$$A = (1, 2, 9, 12, 15, 18, 21, 26, 28, 33, 35, 40, 42, 47, 49, 54, 56,\dots)$$ -$$B = (3, 4, 5, 6, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 24, 25, \dots)$$ -I've checked that $0.207 < a_n - (2+\sqrt{2})n < 2.914$ for $2 <= n <= 18000.$ -This question is similar in form to Limit associated with complementary sequences. There P. Majer proves that $a_n - 4n$ is not bounded, whereas here, the claim is that $a_n - (2+\sqrt{2})n$ is bounded. - -REPLY [7 votes]: An illustration for the answer by Gjergji Zaimi: the mysterious sequence $\epsilon_1,\epsilon_2,...,\epsilon_{5000}$ - -And here, in response to the comment by André Henriques, is the plot of lengths of gaps between consecutive points in the set $\{\epsilon_k\mid10000 -TITLE: Substitutional modality -QUESTION [9 upvotes]: An informal definition of a logical truth is a sentence that's true in virtue of its form alone: $\phi$ is logically true iff all substitutions of $\phi$ that leave its logical vocabulary alone are true. -We might try to formulate a version of this idea in modal logic. Let $\mathcal{L}$ be a modal language, and let $v: \mathcal{L} \to \{0,1\}$ be a Boolean valuation (i.e. $v(A\wedge B) = min(v(A),v(B))$ and $v(\neg A)=1-v(A)$). Say that $v$ is proper if it additionally satisfies the constraint: - -$v(\Box A)=1$ if and only if $v(iA)=1$ for every substitution $i$ - -here we represent a substitution, $i$, as a function from letters to arbitrary sentences and write $iA$ for the result of applying that substitution to $A$. (If $S$ is a restricted set of substitutions, say that $v$ is an $S$-valuation if the corresponding biconditional with $i$ restricted to $S$ holds instead.) -This conception of modality validates some interesting principles: for example if $v(\Box A)=1$ then $v(iA)=1$ for every substitution $i$. In particular, for any given $j$, $v(i(jA)) =1$ for every $i$, since $i\circ j$ is also a substitution. So $v(\Box jA)=1$ for every $j$, and so $v(\Box\Box A)=1$. It follows that the S4 principle is true in every proper valuation. Indeed, one can show that every theorem of S4M is true in every proper valuation. (M is the McKinsey axiom, $\Box \Diamond A \to \Diamond \Box A$, and can be seen to be validated by considering substitutions that map letters to $\top$ and $\bot$.) -Note, however, that it's not obvious that there are any proper valuations. The bulleted claim is a constraint, not a definition, as it involves circularity. For example $v(\Box p) = 1$ iff $v(ip)$ is true for every $i$, and the circularity arises in cases where $ip =\Box p$. But the circularity is not vicious in this case, e.g. $v(\Box p)=0$ since $v(ip)=0$ when $ip= \bot$. I conjecture that the constraint is never vicious, and can always be satisfied. So I was wondering: - -Are there any proper valuations? - -I have some thoughts about constructing a valuation, but they haven't delivered anything so far. One useful fact to note is that if there is a proper valuation $v$, we can construct a Kripke model by letting $W$ be the set of substitutions, letting $i R (j\circ i)$ for all $i,j$, and letting $i \Vdash p$ iff $v(ip)=1$. Conversely, if there's a Kripke model on this frame satisfying $i \Vdash p$ iff $id\Vdash ip$ then we can construct a proper valuation by letting $v(\phi)=1$ iff $id\Vdash \phi$, where $id$ is the identity substitution. So this gives us another way of thinking about the problem. (There is also a topological reformulation of the problem, but I think that's enough for now.) -(Background: McKinsey talks about related notions of necessity here and investigates their logic. He constructs what I've called an $S$-valuation for a very restricted $S$. However, he doesn't seem to raise or recognize the issue with the unrestricted notion.) - -REPLY [8 votes]: $\def\ml{\mathrm{ML}}\let\LOR\bigvee\let\ET\bigwedge$The question you asked is a variant of Problem 42 in Friedman [1]. It also has an intuitionistic analogue, Problem 41, which asks if there exists a set $V$ of propositional formulas such that - -$A\land B\in V$ iff $A\in V$ and $B\in V$, - -$A\lor B\in V$ iff $A\in V$ or $B\in V$, - -$\bot\notin V$, and - -$A\to B\in V$ iff for every substitution $\sigma$, if $\sigma(A)\in V$, then $\sigma(B)\in V$. - - -Under some mild extra assumptions (if $V$ is closed under substitution, or just contains the schema $A\to(B\to A)$), it is easy to see that $V$ satisfies Friedman’s conditions if and only if it is a structurally complete intermediate logic with the disjunction property. -Both problems were solved affirmatively by Prucnal [2]. Concerning Problem 41, Prucnal proved structural completeness of Medvedev’s logic of finite problems $\ml$, which means that $V=\ml$ satisfies Friedman’s conditions. In fact, $\ml$ is the largest structurally complete intermediate logic with the disjunction property. We do not know any other such logic, and it is quite possible that $\ml$ is the only solution. -The largest modal companion of Medvedev’s logic, $\sigma\ml$, gives a solution to Problem 42 (hence to the question here). -Let me now explain the connection in more detail. First, recall that a rule -$A/B$ is admissible in a logic $L$ if for all substitutions $\sigma$, $\vdash_L\sigma(A)$ implies $\vdash_L\sigma(B)$; the rule is derivable in $L$ if $A\vdash_LB$. (In the case of normal modal logics, we take the global consequence relation as $\vdash_L$ here, so that $A\vdash_L\Box A$. Note that for $L\supseteq\mathrm{S4}$, we have the deduction theorem: $A\vdash_LB$ iff $\vdash_L\Box A\to B$ iff $\vdash_L\Box A\to\Box B$.) -A logic is structurally complete if all admissible rules are derivable. -A modal logic $L$ has the disjunction property if -$${}\vdash_L\Box A_1\lor\dots\lor\Box A_k\implies{}\vdash_LA_1\text{ or }\dots\text{ or }{}\vdash_L A_k.$$ -(As a special case, for $k=0$ this condition amounts to the consistency of $L$.) - -Proposition: For any Boolean valuation $v$, the following are equivalent: - -$v$ is proper - -$L=\{A:v(\Box A)=1\}$ is a structurally complete normal modal logic extending S4 satisfying the disjunction property. - - - -Proof: -$1\to2$: That $L$ is a normal extension of S4 is easy to show, and has been already observed in the question. For the DP, if $v(\Box(\LOR_i\Box A_i))=1$, then also $v(\LOR_i\Box A_i)=1$, hence $v(\Box A_i)=1$ for some $i$. -Assume that the rule $A/B$ is $L$-admissible; we want to show $\Box A\to\Box B\in L$, i.e., $v(\Box\sigma(A)\to\Box\sigma(B))=1$ for every substitution $\sigma$. But this follows by definition: if $v(\Box\sigma(A))=1$, then $\sigma(A)\in L$, thus $\sigma(B)\in L$ by admissibility, thus $v(\Box\sigma(B))=1$. -$2\to1$: We need to show -$$\vdash_LA\iff\forall\sigma\:v(\sigma(A))=1.$$ -Left-to-right: since $L$ is closed under substitution, it suffices to show $v(A)=1$. By writing $A$ in CNF and considering each conjunct separately, we may assume $A$ has the form -$$\ET_ip_i^{e_i}\land\ET_j\Box A_j\to\LOR_k\Box B_k,$$ -where $p_i$ are propositional variables, $e_i\in\{0,1\}$, and we put $p^1=p$, $p^0=\neg p$. Assume $v(\ET_j\Box A_j)=1$. Then $L$ derives each $A_j$, hence also $\Box A_j$; since it also derives $A$, it must derive -$$\ET_ip_i^{e_i}\to\LOR_k\Box B_k.$$ -For any $\{0,1\}$ assignment $a$, let $\sigma_a$ be the substitution defined by -$\sigma_a(p_i)=p_i^{a_i}$. It follows that $L$ derives -$$\sigma_a\Bigl(\ET_ip_i^{e_i}\to\LOR_k\Box B_k\Bigr)$$ -for each $a$, hence by combining them together, it derives -$$\LOR_k\LOR_a\Box\sigma_a(B_k).$$ -By the disjunction property, $L$ derives $\sigma_a(B_k)$ for some $k$ and $a$; but then $\vdash_LB_k$, as $\sigma_a$ is an involution. Thus, $v(\Box B_k)=1$ as needed. -Right-to-left: First, let us assume that $A$ is modalized, i.e., all occurrences of variables in $A$ are in the scope of some $\Box$. By considering the CNF, we may further assume it is of the form -$$\ET_j\Box A_j\to\LOR_k\Box B_k.$$ -Now, if $v(\sigma(A))=1$ for every $\sigma$, then the rule $\ET_jA_j/\LOR_k\Box B_k$ is admissible: for any $\sigma$, if $\ET_j\sigma(A_j)\in L$, then $v(\sigma(\ET_j\Box A_j))=1$, thus $v(\sigma(\LOR_k\Box B_k))=1$, thus $v(\Box\sigma(B_k))=1$ for some $k$, thus $\sigma(B_k)\in L$, thus $\LOR_k\Box\sigma(B_k)\in L$. By structural completeness, $\vdash_L\ET_j\Box A_j\to\LOR_k\Box B_k$, i.e., $\vdash_L A$. -I don’t have an elementary argument for the case when $A$ is not necessarily modalized, but it can be handled as follows. Let $L'=\{A:\forall\sigma\:v(\sigma (A))=1\}$. Then $L'$ is a quasi-normal extension of $L$. Using the machinery of Zakharyaschev’s canonical formulas (see e.g. [3]), one can show that $L'$ can be axiomatized (as a quasinormal logic) over S4 by modalized formulas. Since each of them is derivable in $L$ by the previous part of the proof, $L=L'$. QED -Now, it remains to show that structurally complete extensions of S4 with the disjunction property exist. As I already mentioned, let $\ml$ be Medvedev’s logic: it is defined semantically as the logic of the finite intuitionistic frames $M_n=\langle\mathcal P([n])\smallsetminus\{[n]\},{\subseteq}\rangle$ for $n\in\mathbb N$ (i.e., $M_n$ is the $n$-dimensional Boolean cube without its top element). $\ml$ is easy seen to have the disjunction property, and as proved by Prucnal, it is structurally complete. -Let $\sigma\ml$ be the largest modal companion of $\ml$. Note that $\sigma\ml$ is the logic of $\{M_n:n\in\mathbb N\}$ considered as modal frames. Since largest modal companions preserve the disjunction property (easy) and structural completeness (see Rybakov [4,Thm. 5.4.7]), $\sigma\ml$ has all the required properties: - -Proposition: A structurally complete normal modal logic extending S4 with the disjunction property exists. Thus, proper valuations exist. - -Note by the way that it is a long-standing open problem if $\ml$ (and $\sigma\ml$, for that matter) is decidable, or equivalently, recursively axiomatizable. (The semantic definition only guarantees it is co-r.e.) -References: -[1] Harvey Friedman, One hundred and two problems in mathematical logic, Journal of Symbolic Logic 40 (1975), no. 2, pp. 113–129, doi: 10.2307/2271891. -[2] Tadeusz Prucnal, On two problems of Harvey Friedman, Studia Logica 38 (1979), no. 3, pp. 247–262, doi: 10.1007/BF00405383. -[3] Alexander Chagrov and Michael Zakharyaschev, Modal logic, Oxford Logic Guides vol. 35, Oxford University Press, 1997. -[4] Vladimir Rybakov, Admissibility of logical inference rules, Studies in Logic and the Foundations of Mathematics vol. 136, Elsevier, 1997.<|endoftext|> -TITLE: Cyclotomic polynomials. -QUESTION [5 upvotes]: Let $\Phi_m(x)$ and $\Phi_n(x)$ be two different cyclotomic polynomials. Then $\Phi_m(x)$ and $\Phi_n(x)$ are coprime, so there are two polynomials $s(x), t(x)$ with, say, rational coefficients such that $s(x)\Phi_m(x)+t(x)\Phi_n(x)=1$. -Question. Can one find these $s$ and $t$ with integer coefficients? -I think the answer is "yes". - -REPLY [16 votes]: If this were true, then you would prove that $\Phi_m(x)$ and $\Phi_n(x)$ are coprime after reduction modulo $p$, which is far from true. For instance, $\Phi_4(x)=x^2+1$ and $\Phi_2(x)=x+1$ are not coprime modulo $2$. (Even $\Phi_2(x)$ and $\Phi_1(x)$ are not coprime modulo $2$, of course.)<|endoftext|> -TITLE: Cyclotomic polynomials 2 -QUESTION [16 upvotes]: My naive question may actually lead to something interesting. -Let $\Phi_m(x)$, $\Phi_n(x)$ be cyclotomic polynomials, $m -TITLE: KK-theoretical proof of Atiyah-Singer index theorem -QUESTION [8 upvotes]: Does anyone know of any detailed proof of the Atiyah-Singer Index Theorem using KK-theory/ Kasparov products? References to any papers and textbooks are greatly appreciated. Thanks! - -REPLY [3 votes]: I find the proof in the recent paper of Kasparov - -G. Kasparov. Elliptic and transversally elliptic index theory from the - viewpoint of KK-theory. J. Noncommut. Geom., 10(4):1303–1378, 2016 - -difficult but understandable. There are certainly more details than in Kasparov's Inventiones paper.<|endoftext|> -TITLE: Sha finiteness vs $\ell$-primary torsion -QUESTION [7 upvotes]: Where do I find a proof of the fact that over global function fields of characteristic $p>0$, finiteness of the Tate-Shafarevich group of an abelian variety is equivalent to finiteness of its $\ell$-primary torsion part for some (arbitrary) prime $\ell$ ($\ell = p$ allowed)? -What is the main idea? -One knows the Tate-Shafarevich group is torsion, and can write, for $K$ a function field and $A/K$ an abelian variety: -$$\text{Sha}(A/K) = \prod_{\ell}\text{Sha}(A/K)[\ell^{\infty}]$$ -so I guess one first shows that $\text{Sha}(A/K)[\ell^{\infty}]$ is trivial for almost all $\ell$, but then? - -REPLY [7 votes]: One shows that the finiteness of one $\ell$-primary component of Sha is equivalent to the Birch-Swinnerton-Dyer conjecture. $\mathrm{rk} A(K) = \mathrm{ord}_{s=1}L(A/K,s)$ is independent of $\ell$ and a non-zero rational number (the special $L$-value) has only finitely many prime divisors. -The references are: - -Peter Schneider, Zur Vermutung von Birch und Swinnerton-Dyer über globalen Funktionenkörpern, Math. Ann. 260 (1982), no. 4, 495–510 ($\ell \neq p$) -Werner Bauer, On the conjecture of Birch and Swinnerton-Dyer for abelian varieties over function fields in characteristic $p>0$, Invent. Math. 108 (1992), no. 2, 263–287 ($\ell = p$, good reduction) -Kazuya Kato and Fabien Trihan, On the conjectures of Birch and Swinnerton-Dyer in characteristic $p>0$, Invent. Math. 153 (2003), no. 3, 537–592 ($\ell = p$, general case) - -The finiteness of Sha is e.g. known for constant Abelian varieties: - -James Milne, The Tate-Šafarevič group of a constant abelian variety, Invent. Math. 6 (1968), 91–105: http://www.jmilne.org/math/articles/1968b.pdf (his PhD thesis supervised by Tate) - -There are similar results for higher dimensional bases: https://www.timokeller.name/BSDI.pdf (only for good reduction and $\ell \neq p$, contains also e.g. the finiteness of the prime-to-$p$ part of Sha of supersingular Abelian schemes over special base schemes) -The rough idea is as follows: The Kummer sequence $0 \to \mathscr{A}[\ell^n] \to \mathscr{A} \to \mathscr{A} \to 0$ (for the étale topology if $\ell \neq p$ and otherwise for the syntomic topology) gives a short exact sequence $$0 \to A(K) \otimes \mathbf{Z}_\ell \to H^1(X,T_\ell\mathscr{A}) \to T_\ell Ш(\mathscr{A}/X) \to 0.$$ Since $Ш(\mathscr{A}/X)[\ell^\infty]$ is of cofinite type, the $\ell$-adic Tate module is trivial iff the $\ell$-primary component is finite. The Hochschild-Serre spectral sequence degenerates by $\mathrm{cd}(\Gamma) = 1$ giving $$0 \to H^0(\bar{X},T_\ell\mathscr{A})_\Gamma \to H^1(X,T_\ell\mathscr{A}) \to H^1(\bar{X},T_\ell\mathscr{A})^\Gamma \to 0.$$ Now relate $H^1(\bar{X},T_\ell\mathscr{A})^\Gamma$ to the special $L$-value using Lemma 3.2 from [Bayer-Neukirch, On values of zeta functions and $l$-adic Euler characteristics, Invent. Math. 50 (1978/79), no. 1, 35–64]. ($X$ a model of the function field $K$, $\mathscr{A}/X$ the Néron model of $A/K$ [if it exists], $k = \mathbf{F}_q$ the finite ground field, $\bar{X} = X \times_k \bar{k}$ and $\Gamma = G_k$) -Another nice reference: Douglas Ulmer, Elliptic curves over function fields. Arithmetic of L-functions, 211–280, IAS/Park City Math. Ser., 18, Amer. Math. Soc., Providence, RI, 2011.<|endoftext|> -TITLE: Question on the 50th (known) Mersenne prime number -QUESTION [13 upvotes]: In a footnote to the list of known Mersenne prime numbers which can be found here, we read that the "ranking" therein is a provisional one since not all possible exponents between $37 \, 156 \, 667$ and $77 \, 232 \, 917$ have been eliminated/tested. -If we may infer from this that the individual(s) who recently found the largest known prime number had not previously discarded (quite understandably!) the possible compositeness of the $\pi(77 \, 232 \, 917)-\pi(74 \, 207 \, 281) - 1 = 166 \, 801$ members of the set $$\{M_{p} \colon p \in (74 \, 207 \, 281, 77 \, 232 \, 917), p \mbox{ is a prime number}\},$$ do we know what it was that prompted them to try their hand at establishing the primality of the very specific Mersenne number $M_{77 \, 232 \, 917}$ ? -Clearly enough, an analogous question can be formulated regarding the discovery of the 46th, 47th, 48th, and 49th known Mersenne prime numbers. If you know the answer to any of those allied questions, do not hesitate to enter it below (it might shed light on the case of the 50th known Mersenne prime). -Thanks in advance for your knowledgeable replies. - -REPLY [9 votes]: You can check the current status (updated hourly). Currently among the different stats are: - -All exponents below 46 251 389 have been tested and verified. -All exponents below 82 038 613 have been tested at least once.<|endoftext|> -TITLE: numerical radius of hyponormal operator -QUESTION [7 upvotes]: Let $F$ be a complex Hilbert space. We recall that an operator $S\in\mathcal{B}(F)$ is said to be hyponormal if $S^*S\geq SS^*$ (i.e. $\langle (S^*S-SS^*)z,z \rangle\geq 0$ for all $z\in F$). - -Assume that $S$ is hyponormal operator. Is $\omega(S)=\|S\|?$, with $\omega(S)$ denotes the numerical radius of $S$ and defined as: - $$\omega(S)=\displaystyle\sup_{\|x\|=1}|\langle Sx, x\rangle |.$$ - -Thank you. - -REPLY [8 votes]: Yes, the proof can be found in Stampfli, Joseph G., Hyponormal operators, Pacific J. Math. (12), no. 4 (1962), 1453--1458. -The proof is actually very simple. First of all, hyponormality may be stated as $\|Sx\|\geqslant \|S^{\ast}x\|$ for any $x\in F$. It follows that $\|Sx\|^2 = \langle Sx, Sx \rangle = \langle x, S^{\ast}Sx\rangle \leqslant \|x\|\cdot \|S^{\ast}Sx\| \leqslant \|x\|\cdot \|S^2x\|$, where we applied the definition to the vector $y:=Sx$. It follows that $\|S\|^2 \leqslant \|S^2\|$, so $\|S^2\|=\|S\|^2$. Then you proceed by induction to prove that $\|S^n\|=\|S\|^n$ for any $n$.<|endoftext|> -TITLE: Restrictions of the Stationary Tower forcing providing various critical points -QUESTION [7 upvotes]: Following Larson's "The Stationary Tower", let $\mathbb{P}_{<\delta}$ be the full stationary tower on $\delta$, and for a stationary $S\subset \mathcal{P}_\delta(V_\delta)$, $\mathbb{P}_{<\delta}^S$ is the restriction to $S$. -It is stated that "under fairly general assumptions on $S$, much of the basic theory of $\mathbb{P}_{<\delta}$ carries over to $\mathbb{P}_{<\delta}^S$", but the book then deals only with the case $S=\mathcal{P}_{\omega_1}(V_\delta)$. My questions are: - -What are the general assumptions? Is there a reference for the general case? -More specifically, if we let $S=\mathcal{P}_{\kappa}(V_\delta)$ for $\kappa<\delta$, will the same theorems still apply? I suppose we need $\kappa$ regular, does it need to be successor cardinal? -When $S=\mathcal{P}_{\omega_1}(V_\delta)$ and $\delta$ is Woodin, we get in the extension an embedding $j$ such that $crit(j)=\omega_1$ and $j(\omega_1)=\delta$. If $S=\mathcal{P}_{\kappa}(V_\delta)$, will we get $crit(j)=\kappa$ and $j(\kappa)=\delta$? If not, how (and for which $\kappa$) can we get such an embedding? - -REPLY [5 votes]: Here is a partial answer. Ideally, I'll think about it a little more and edit it. -Remark 2.7.17 of my book gives some pointers in the direction of your questions, and the note "Six lectures on the -stationary tower" on my webpage gives more information about the case $S = \mathcal{P}_{\kappa}(V_{\delta})$, for $\kappa$ a succesor cardinal. In both cases I leave a lot to the reader. I think everything in the six lectures applies to regular cardinals; I just didn't write it that way. I don't think there's a reference for the general case, but the Foreman-Magidor paper on -definable counterexamples to the continuum hypothesis covers some cases that I didn't, like the $\omega$-closed tower. -As for your third question, if $\kappa$ is a regular cardinal, then yes, $\kappa$ is the critical point of the embedding, and $j(\kappa) = \delta$. I think all the standard arguments carry over in this case. Even if $\kappa$ is singular, as I recall, you get an ultrapower which is closed under $\omega$-sequences, and possibly longer sequences, depending on the generic filter, but with critical point less than $\kappa$. I think this is done in one of the papers that I point to in Remark 2.7.17. -As for the question about arbitrary $S$, I suppose that one would like to characterize the ones that give you wellfounded ultrapowers. -This might be hard, but maybe one can prove something generalizing the situation for the $\mathcal{P}_{\kappa}(V_{\delta})$'s. -To start with, one would like to know what you need to prove Lemma 2.5.6 and Theorem 2.5.9 of my book, but even then this might not give you a characterization.<|endoftext|> -TITLE: Totally bounded spaces and axiom of choice -QUESTION [7 upvotes]: Wikipedia article on totally bounded spaces states "... the completion of a totally bounded space might not be compact in the absence of choice." Where is the axiom of choice used, and do you need it for metric spaces or only for general uniform spaces? - -REPLY [12 votes]: The issue here is that a metric space might not have non-trivial (read: not eventually constant) Cauchy sequences. For example, if the underlying space is a Dedekind finite set. -Indeed it is consistent that there is a dense subset of $[0,1]$ which is Dedekind finite. As a space with the inherited metric it is complete already and totally bounded, but it is not compact as it is not closed on $[0,1]$.<|endoftext|> -TITLE: Reference request: Partition function as a topological invariant of a QFT -QUESTION [8 upvotes]: I have read (mainly in the articles of Atiyah) that the partition function is the simplest topological invariant of a quantum field theory. -I have an arithmetic geometry background and know statistical physics quite well, but am a beginner in QFT. A reference for details on the above statement in mathematical language à la "Quantum Fields and Strings: A Course for Mathematicians" would be highly appreciated. -A sketch of the ideas involved would be even nicer. -If this question is too simple or vague, please let me know - I will transfer it to SE. - -REPLY [2 votes]: This is one case where you might just want to go back to the original article that proved that a partition function is metric-independent and hence a topological invariant: Edward Witten, Quantum field theory and the Jones polynomial (1989) --- see page 361. -For a sketch of the idea, this introduction by Marcos Marino seems useful (page 25+26).<|endoftext|> -TITLE: Minimal cardinality of a field where a polynomial has a root -QUESTION [13 upvotes]: Let $P(n)$ be the set of all monic polynomials of degree $n$ with integer coefficients, such that all coefficients have absolute value at most $2^n$. -Given a positive integer $n$ let us define $A(n)$ as the minimal number $m$ such that for any $f \in P(n)$ there exists a field $k$ with at most $m$ elements such that for some $x\in k$ we have $\bar f(x)=0$, where $\bar f$ is the reduction of $f$ modulo the characteristic of $k$. -Q: What is the asymptotic behaviour of $A(n)$? In particular is it true that $\liminf_{n} \frac{A(n)}{n} = 0$ ? - -REPLY [5 votes]: No to the final question, stealing an idea from Andreas Thom. For all primes $p$ and degrees $d$ with $p^d>m$, there is a degree $d$ monic polynomial over $\mathbb F_p$ that has no roots in any field of characteristic $p$ of order $\leq m$. Indeed, an irreducible polynomial of degree $d$ does the trick. -Applying this, for any degree $d$ with $2^d > m$, there is a degree $d$ monic polynomial over $\mathbb Z/ P_m \mathbb Z$ where $P_m= \prod_{p\textrm{ prime}, p \leq m} p $ which does not have any roots in any field of cardinality at most $m$. -We can now lift this to an integral polynomial, with coefficients $ m$ has $A(n)>m$. Obviously the first condition is the interesting one, so we have $$n \approx \log_2 P_m = \sum_{p \textrm{ prime}, p \leq m} \frac{\log p}{\log 2} \approx \frac{m}{\log 2}$$ by the prime number theorem. -So $A(n) $ is greater than, approximately, $n \log 2$<|endoftext|> -TITLE: "Towers" on singular cardinals with countable cofinality -QUESTION [9 upvotes]: Let $\lambda$ be a singular cardinal of countable cofinality. -Is there necessarily a sequence $\{A_\alpha\mid\alpha<\lambda^+\}$ of countable subsets of $\lambda$, such that $\alpha<\beta$ if and only if $A_\alpha\setminus A_\beta$ is finite? -In other words, we know that for $\omega$, there is an uncountable sequence of subsets which is increasing modulo finite changes. Is the same true for other cardinals of countable cofinality? - -REPLY [12 votes]: For $\lambda > 2^{\aleph_0}$, there is no such sequence. -Suppose $\lambda > 2^{\aleph_0}$. Because $2^{\aleph_0}$ cannot have countable cofinality, there is some $\kappa < \lambda$ with $2^{\aleph_0} < \kappa$. Consider the sequence $\{A_\alpha \cap A_\kappa \mid \alpha < \kappa\}$. For each particular $\alpha$, the sets $A_\alpha$ and $A_\alpha \cap A_\kappa$ differ by only finitely many elements. Each set of this form is a countable subset of the countable set $A_\kappa$, so there are fewer than $\kappa$ possibilities for the sets $A_\alpha \cap A_\kappa$. Thus we may find $\alpha < \beta < \kappa$ such that $A_\alpha \cap A_\kappa = A_\beta \cap A_\kappa$. But then $A_\alpha$ and $A_\beta$ differ by only finitely many elements, and in particular we have $A_\beta \setminus A_\alpha$ finite while $\alpha < \beta$, contrary to your condition.<|endoftext|> -TITLE: "skyscraper group scheme" -QUESTION [7 upvotes]: Is there a skyscraper group scheme? -Let $S$ be a DVR. Is there a group scheme $\mathcal{G}$ over $S$ which is generically {1} trivial i.e, identity group, but at the closed point some nontrivial group $G$? -For example: Let $C$ be a curve over $S$ whose generic fibre is smooth of genus $g$ with no automorphism but the closed fibre is a semistable curve(which has nontrivial automorphisms). Now consider the group of $S$-automorphisms of this curve $C$ over $S$? My question: Does this curve C over S has any non trivial S-automorphism? - -REPLY [5 votes]: Yes: the following examplpe is quite different from the example by Ariyan, although the generic and closed fibers are the same as his. -Start with the constant group scheme $A:=(\mathbb{Z}/2\mathbb{Z})_S$, and consider the closed subscheme which is the union of the zero section and the closed point of the other section. This is immediately seen to be a subgroup scheme of $A$, with generic fiber $0$ and closed fiber $\mathbb{Z}/2\mathbb{Z}$. It is finite over $S$ (in particular separated) but of course not flat. -You can construct infinitely many variants by replacing the closed point by any infinitesimal neighborhood of it. -If you take a stable curve $C$ over $S$, it it known that its automorphis functor $\mathscr{G}:={\underline{\mathrm{Aut}}}(C/S)$ is a finite unramified $S$-group scheme, hence it "looks like" the above example, rather than Ariyan's nonseparated one. In particular ($K$ being the fraction field), the restriction homomorphism $\mathscr{G}(S)\to \mathscr{G}(K)$ is bijective, so if $C_K$ has no $K$-automorphisms then if $C$ has no $S$-automorphisms. - -REPLY [3 votes]: Yes. -Let $\mathbb{A}^{1,2}$ be the affine line with a double origin. Consider the natural map $\mathbb{A}^{1,2}\to \mathbb{A}^1$. (To define this map, let $0_1$ and $0_2$ be the origins in $\mathbb{A}^{1,2}$. The above map sends any $x\neq 0_1, 0_2$ to $x$. It sends $0_1$ and $0_2$ to the origin in $\mathbb{A}^1$.) -The above morphism realizes $\mathbb{A}^{1,2}$ as a quasi-finite flat group scheme over $\mathbb{A}^1$. (It is not a separated group scheme over $\mathbb{A}^1$, of course.) -Note that the generic fibre of this group scheme is the trivial group. The fibre over the origin is the group $\mathbb{Z}/2\mathbb{Z}$. -To get to the situation you desire, let $\mathrm{Spec} \mathbb{C}[[t]]\to \mathbb{A}^{1}$ be a dominant map whose image contains the origin. Now base-change the group scheme $\mathbb{A}^{1,2}\to \mathbb{A}^1$ along this morphism to get a group scheme -$$ G\to \mathrm{Spec} \mathbb{C}[[t]]$$ with trivial generic fibre and a non-trivial special fibre. (Here $G=\mathbb{A}^{1,2}\times_{\mathbb{A}^1} \mathrm{Spec} \mathbb{C}[[t]]$.)<|endoftext|> -TITLE: varieties with no new cohomology in hyperplane sections -QUESTION [7 upvotes]: Let $X$ be a smooth closed subvariety in a complex projective space and $Z\subset X$ be a generic hyperplane section. Lefschetz hyperplane section Theorem says that -the restriction map $H^i(X)\to H^i(Z)$ is an isomorphism for $i2$ and the dual variety $X^*\subset(\mathbb P^N)^*$ is not a hypersurface (``new cohomology'' is spanned by vanishing cycles, and if $X^*$ is not a hypersurface, then there are no vanishing cycles since Lefschetz pencils have no degenerate fibers). -If $\dim X>1$ is odd, the converse is true. Indeed, in this case vanishing sycles have self-intersection $\pm 2$, hence they are never zero if they exist. -In small dimensions, a classification of varieties with small duals is known; see two papers by L.Ein titled ``On varieties with small dual varieties''.<|endoftext|> -TITLE: Instructions for using Coxeter 3.0 software -QUESTION [5 upvotes]: I am trying to use Coxeter 3.0 (http://www.liegroups.org/coxeter/coxeter3/english/coxeter3_e.html) to perform some computations for affine Weyl groups. I managed to install the program and get it running, but I cannot find any instructions. -I have figured out how to get the program to do some computations for finite Weyl groups (types A, B, etc), but have not figured out the commands for affine types. Does anyone have a link to instructions which explain how to do this, or can anyone tell me how to enter affine types into Coxeter 3.0? -For example, I can enter "type" -> "A" -> "2" to work with type A rank 2 in the finite case. But how do I enter affine type A (untwisted)? -I hope this is an appropriate place to post this request. - -REPLY [11 votes]: If you type "help" immediately on entering the program, you'll get a fairly long and useful introductory message. At whatever level you are, you are supposed to be able to type "help," and then the name of any command accessible at that level, to get a message about it. (Well, not all these help files exist.) The help files exist as text files in the directory "messages" (in the untarred file from the web site you mention). Instructions for entering Coxeter groups are in the file "wrongtype.mess." (I would have been tempted to call it "type.mess," but that's probably wrong for a reason explained in "unconventions.mess.") -In particular, the affine groups are entered in exactly the same way as their finite counterparts, using lower case letters: so -"type" -> "a" -> "3" -will get you the affine rank three group whose diagram is a triangle. -Also in the home directory from the tar file is "INTRO.tex," which produces a fifteen page mathematical introduction to what the software is doing and how. (Not so useful for finding the ignition switch.) -JianYi Shi at East China Normal University (and his students) have done a lot of work with affine Coxeter groups, including use of coxeter for computations of KL polynomials.<|endoftext|> -TITLE: It is true that $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$? -QUESTION [7 upvotes]: Let $H$ be a complex Hilbert space and $\mathcal{L}(H)$ be the algebra of all bounded linear operators on $E$. - -If $A,B\in \mathcal{L}(H)$, It is true that $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$? - -I try as follows: -Let $z\in \overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}$, then $z=\displaystyle\sum_{i=1}^dx_i\otimes y_i$, where $x_i\in\overline{\text{Im}(A)},\;y_i\in\overline{\text{Im}(B)}$ and $d\in\mathbb{N}$. So, there exists $(\alpha_i(n))_{n\in\mathbb{N}}\subset \text{Im}(A)$ and $(\beta_i(n))_{n\in\mathbb{N}}\subset \text{Im}(B)$ such that $x_i=\lim_{n\rightarrow\infty}\alpha_i(n)$ and $y_i=\displaystyle\lim_{n\rightarrow\infty}\beta_i(n)$. Moreover, $\alpha_i(n)=A\tilde{\alpha}_i(n),\;\beta_i(n)=B\tilde{\beta}_i(n)$ and -$$\alpha_i(n)\otimes\beta_i(n)=(A\otimes B)(\tilde{\alpha}_i(n)\otimes \tilde{\beta}_i(n))\subset \text{Im}(A\otimes B),$$ -where $(\tilde{\alpha}_i(n))_{n\in\mathbb{N}}\subset \mathcal{H}$ and $(\tilde{\beta}_i(n))_{n\in\mathbb{N}}\subset \mathcal{H}$. Let -$$z_n=\displaystyle\sum_{i=1}^d\alpha_i(n)\otimes \beta_i(n)\in \text{Im}(A\otimes B).$$ -Since $\|x\otimes y\|=\|x\|\|y\|$, then the map $(x,y)\longmapsto x\otimes y$ is continuous. So $z_n\longrightarrow z$ as $n\longrightarrow \infty$, this implies that $z\in \overline{\text{Im}(A\otimes B)}$. As a consequence, $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$. - -REPLY [7 votes]: The proof is okay. You did not specify the norm on the tensor product $H⊗H$ -in which you take the closure. I do not know which one you want. Most naturally this would be the Hilbert-space or $\ell^2$ tensor product (which also describes the space of Hilbert-Schmidt operators). But the proof works for all reasonable norms. Another one describes the space of compact operators. Some of these are called Schatten-norms. See the beginnig of here, or also section 7 of here (and references therein) for more information. -Added: -Details of the proof: -The Hilbert-Schmidt norm on $\mathcal H\otimes\mathcal H$ can be defined as: -$$ -\|z\|_2 = \inf \Big\{ \|(\|x_i\|)\|_{\ell^2}.\sup_{\|y'\|\le1}\|(\langle y_i,y'\rangle)\|_{\ell^2} : z = \sum x_i\otimes y_i \Big\} -$$ -Here the infimum is taken over all (finite) representations of $z$ in $\mathcal H\otimes\mathcal H$. This can be seen from section 7 of the second reference I gave above (quite complicated there); best are the references Chevet or Saphar therein which I do not have available now. -Given $z$ as in your proof above, choose a finite representation $z = \sum_{i=1}^{N(n)} x_i(n)\otimes y_i(n)$ such that -$$ -\|(\|x_i(n)\|)\|_{\ell^2}.\sup_{\|y'\|\le1}\|(\langle y_i(n),y'\rangle)\|_{\ell^2} -\le \|z\|_2 +\frac{\epsilon}n\,. -$$ -Now choose your $\alpha_i(n,m)$ and $\beta_i(n,m)$ in such a way that -$$ -\|(\|x_i(n)-\alpha_i(n,m)\|)\|_{\ell^2}<\frac{\epsilon}{nm}\,,\qquad -\|(\|y_i(n)-\beta_i(n,m)\|)\|_{\ell^2}<\frac{\epsilon}{nm}\,. -$$ -Then you can prove convergence in the $\ell^2$-tensor product.<|endoftext|> -TITLE: Formality of some inner RHom -QUESTION [13 upvotes]: Let $Y$ be a - scheme and $X$ be a closed subscheme. Consider $\underline{RHom}_{\mathcal O_Y}(\mathcal O_X,\mathcal O_X)$ (inner RHom). We can view this is as a sheaf of dg-algebras on $X$; when both $X$ and $Y$ are (for simplicity) of finite type over a field and smooth then its cohomology is the exterior algebra of the normal bundle to $X$ inside $Y$. -$\mathbf{Question:}$ What kind of results are known about the formality of this algebra in some nice cases? I am especially interested in the -case when $Y$ is symplectic and $X$ is Lagrangian in $Y$. - -REPLY [3 votes]: I don't think that a full answer to this question is known. Here are the results I'm aware of: - -if one forgets about the algebra structure there's a necessary and sufficient condition for $\mathbb{R}\underline{Hom}_{\mathcal O_Y}(\iota_*\mathcal O_X,\iota_*\mathcal O_X)$ to be formal as a sheaf of $\mathcal O_X$-modules. To view it as a sheaf of $\mathcal O_X$-modules, we rather consider $\mathbb{R}\underline{Hom}_{\mathcal O_X}(\iota^*\iota_*\mathcal O_X,\iota_*\mathcal O_X)$, where $\iota$ is the embedding of $X$ into $Y$. This is a result of Arikin and Calararu (see https://arxiv.org/pdf/1007.1671.pdf), and the sufficient condition is that the normal bundle extends to a bundle on the first infinitesimal neighborhood $X^{(1)}$ of $X$ into $Y$. It seems believed (but not proven) that the formality always holds as sheaves of $k$-modules, and even as sheaves of $\mathcal O_Y$-modules. -the only result I know about the formality as a sheaf of $k$-algebras is for the diagonal embedding $\Delta:X\to X\times X$. This is proven in a paper of mine together with Van den Bergh: https://arxiv.org/pdf/0708.2725.pdf (note that this can't hold over $\mathcal O_X$ as the homotopy for the commutativity of the product in the Hochschild cochain complex is not $\mathcal O_X$-linear). It is conjectured (but not known) that the result is more generally true for the case of the inclusion of a fixed point locus $X=Y^\sigma$ of an involution $\sigma:Y\to Y$. -I don't know about any formality result in the case of $X$ lagrangian in a symplectic $Y$. In this case one knows, after Behrend and Fantechi, that the Ext algebra comes equipped with a BV structure. This additionnal structure might be useful for approaching the question.<|endoftext|> -TITLE: uniquely determining a distribution using moments -QUESTION [5 upvotes]: Let $A$ be a parametric family of probability distributions that include all distributions in the form of $\phi(X)$ where $X\sim\mathcal{N}(0,\mathbf{I})$ is jointly Gaussian and $\phi:\mathbb{R}^d\to \mathbb{R}^d$ belongs to a parametric set of functions $G$. -To uniquely determine a distribution in $A$, how many moments do we need to know? For example, if $G$ is linear, A includes all Gaussian distributions. Thus, we need first and second moments to uniquely determine a distribution in the set $A$. I wonder if such a result can be extended for other $G$ such as polynomials with degree $k$. - -REPLY [4 votes]: Looks complicated in general, but here's a derivation that for $d=1$ and $k=2$, in the slightly further restricted case where $\phi(X)=aX^2+b$, three moments suffice. -Let $m_i=E(\phi(X)^k)$. -Recall $E(X^4)=3$ and $E(X^6)=15$. -We calculate -$$m_1=a+b$$ -$$m_2=3a^2+2ab+b^2$$ -with solution: -$$m_2=3a^2+2a(m_1-a)+(m_1-a)^2$$ -$$m_2=a^2+2am_1+m_1^2-2am_1 +a^2$$ -$$a^2=\frac12(m_2-m_1^2) =: u$$ -So far we don't know $a$ exactly. But -$$m_3=15a^3+9a^2b+3ab^2+b^3$$ -so we can use our value for $a^2$ to reduce -$$m_3=15a^3+9a^2(m_1-a)+3a(m_1-a)^2+(m_1-a)^3$$ -to a linear equation in $a$: -$$m_3=15ua+9um_1-9ua+3am_1^2-6um_1+3ua+m_1^3-3am_1^2+3um_1-ua$$ -$$m_3=8ua+6um_1+m_1^3$$ -and we get the somewhat appealing solution - -$$\begin{eqnarray*}a&=&\frac14\frac{m_3-m_1^3}{m_2-m_1^2}-\frac34m_1\\ b&=&m_1-a\end{eqnarray*}$$<|endoftext|> -TITLE: Equivalent descriptions of Coherent Groups -QUESTION [8 upvotes]: Attending a series of lectures, I have recently been exposed to the notion of Coherent groups, defined as following: -Def: A group $G$ is called Coherent if every finitely generated subgroup $H$ of $G$ is finitely presented. -Examples of such groups are free, surface groups, the fundamental group of 3-manifolds. In contrast $F_2 \times F_2$ (and any group which contains it) is not coherent. -I am just wondering if this notion has any equivalent description, possibly in terms of the module category of the group algebra $kG$, or any other area. -References for further reading are highly appreciated. - -REPLY [8 votes]: I do not think that there is a characterization in terms of modules over the group ring. All non-trivial proofs of coherence are geometric in nature. Probably the most non-trivial (so far) is the paper by Feighn and Handel "Mapping tori of free group automorphisms are coherent" Ann. of Math. (2) 149 (1999), no. 3, 1061–1077 where it is proved that ascending HNN extensions of free groups are coherent. This implies, in particular, that almost all 1-related groups with at least 3 generators are coherent (see MR2746769). The major outstanding problem in the area is coherence of $SL_3(\mathbb{Z})$.<|endoftext|> -TITLE: Rosenlicht's theorem and rationality questions -QUESTION [7 upvotes]: Let $G$ be a connected algebraic group over an algebraically closed field $\overline{k}$ acting on an irreducible variety $X$. A geometric quotient is a morphism of varieties $\pi: X \rightarrow X/\sim$ which on closed points (that is, as a morphism of classical varieties) satisfy the following: -(i): $\pi$ is a surjective open map, and the fibres are exactly the $G$-orbits of $X$. -(ii): for any open set $U$ of $X/\sim$, the ring homomorphism $\pi^{\ast}: \overline{k}[U] \rightarrow \overline{k}[\pi^{-1}U]$ is an isomorphism onto the $G$-fixed points of codomain. -Rosenlicht's theorem says that there exists a $G$-stable open subset of $X$ for which the geometric quotient exists. -Is there any generalization of Rosenlicht's theorem for when $G$ and $X$ are defined over an arbitrary field? The case I'm interested in is when $G$ and $X$ are geometrically connected subgroups of upper triangular unipotent matrices over a $p$-adic field $F$ (so all orbits are closed), and $X$ is normalized by $G$. - -REPLY [6 votes]: Jim Humphreys was right: No generalization is necessary. In Theorem 2 of -Rosenlicht, Maxwell: Some basic theorems on algebraic groups, Amer. J. Math. 78 (1956) 401--443, -the result is already stated over arbitrary base fields.<|endoftext|> -TITLE: Is the determinant the only multiplicative matrix function? -QUESTION [15 upvotes]: Is there a matrix invariant or property that is multiplicative, i.e., -$$f(AB) = f(A) f(B)$$ -other than the determinant? In addition, some matrix norms are submultiplicative, but is there a supermultiplicative property? - -REPLY [8 votes]: With the exception of $GL_2(\mathbb{F}_2)$, the commutator subgroup of $GL_n(k)$ is $SL_n(k)$ (for $k$ a field). So any multiplicative map from $GL_n(k)$ to an abelian group factors through determinant. This is with no hypotheses on continuity. -EDIT Johannes asks about noninvertible matrices. Let $M$ be an abelian monoid and $\mathrm{Mat}_{n \times n}(k) \to M$ a multiplicative map. Then $GL_n$ must map to the group of units of $M$, so $SL_n(k)$ must map to $1$ by the above (except for $n=2$, $k = \mathbb{F}_2$). Now, if $X$ and $Y \in \mathrm{Mat}_{n \times n}(k)$ are noninvertible matrices of the same rank then there are matrices $U$ and $V$ in $SL_n(k)$ with $UXV=Y$. So our function must be constant on functions of the same rank. Let $e_r$ be its value on matrices of rank $r$. Now, rank $n-1$ idempotents exist. So $e_{n-1}^2=e_{n-1}$. But any noninvertible matrix is a product of rank $n-1$ matrices, so $e_{n-k}$ is a power of $e_{n-1}$ and we deduce $e_{n-1} = e_{n-2} = \cdots = e_1 = e_0$. In short, a multiplicative map to an abelian monoid must take the same value on all noninvertible matrices.<|endoftext|> -TITLE: Is there a "killing" lemma for G-crossed braided fusion categories? -QUESTION [12 upvotes]: Edit: I found a serious flaw in the question and my answer, and I had to change a lot. The basic question is still there, but the details are a lot different. -Premodular categories -In braided spherical fusion (=premodular) categories $\mathcal{C}$ with braiding $c_{-,-}$, we can define the double braiding: -$$\gamma_{A,X}\colon A \otimes X \to A \otimes X$$ -$$\gamma_{A,X} := c_{X,A} \circ c_{A,X}$$ -Now we can define encirclings, where we braid an object around another and close the loop on the right hand side: -$$\Delta_{A,X}\colon A \to A$$ -$$\Delta_{A,X} = \operatorname{tr}_X(\gamma_{A,X})$$ -(The trace is defined via the pivotal structure.) -The "killing" lemma is quite well known. -Assume that $A$ is simple. Then: -$$\Omega := \bigoplus_{X \text{ simple}} d(X) X$$ -$$\Delta_{A,\Omega} = \left\{ \begin{matrix} d(\Omega) \cdot 1_A && \qquad A \text{ is in the symmetric centre} \\ 0 && \qquad \text{otherwise} \end{matrix}\right.$$ -$A$ is in the symmetric centre if braids trivially with every object. One says that $A$ is killed if it's not in the symmetric centre, hence the name of the lemma. -$G$-crossed braided fusion categories -Now consider $G$-crossed braided fusion categories. If you think of fusion categories as some sort of categorified group (algebra), then a $G$-crossed braided fusion category is a categorified crossed module. It is, amongst other things, a $G$-graded fusion category: -$$\mathcal{C} = \bigoplus_{g \in G} \mathcal{C}_g$$ -It also has a monoidal $G$-action: -$${}^g(-)\colon \mathcal{C}_h \to \mathcal{C}_{ghg^{-1}} \quad \forall g, h \in G$$ -And it has a crossed braiding: -$$ c_{A,X}\colon A \otimes X \to {}^gX \otimes A\quad \forall A \in \mathcal{C}_g$$ -This means that the double braiding is now not an endomorphism anymore: -$$ \gamma_{A,X}\colon A \otimes X \to {}^{ghg^{-1}}A \otimes {}^g X \quad\forall A \in \mathcal{C}_g, X \in \mathcal{C}_h$$ -You can only take the trace on the right hand side, and thus define encircling, if $X \cong {}^gX$. -The only canonical way I can think of is to demand that the grade of $A$ is trivial, i.e. $g = e$, since there is a coherence isomorphism $\epsilon_X\colon {}^eX \to X$. -Then we can define: -$$\Omega_h := \bigoplus_{X \text{ simple } \in \mathcal{C}_h} d(X) X$$ -$$\Delta_{A, \Omega_h} = \operatorname{tr}_{\Omega_h}\left((1_A \otimes \epsilon_{\Omega_h}) \circ \gamma_{A,\Omega_h} \right) \qquad \forall A \in \mathcal{C}_e$$ -But now source and target aren't equal: -$$\Delta_{A,\Omega_h}\colon A \to {}^hA$$ -In some cases we have $A \cong {}^hA$, and then one can ask whether the endomorphism is 0 or not, which leads me to my question: -Is there a similar lemma to the standard killing lemma in the $G$-crossed case? - -REPLY [2 votes]: Edit: -I used to believe that there is a possible generalisation stemming from work of Altschüler and Bruguières. (See Appendix C in Drinfeld, Gelaki, Nikshych, Ostrik - On braided fusion categories I). But that was based on a flawed assumption. (I didn't realise that $A$ must be in the trivial degree for it to work.) With the correct assumptions, the following can be said: -Sliding lemma -Like in braided spherical fusion categories, the sliding lemma is closely related to the killing lemma. -Lemma -Let $Y \in \mathcal{C}_g, A \in \mathcal{C}_e$. Up to coherences in the $G$-crossed fusion category, we have: -$$\Delta_{A,\Omega_h} \otimes 1_Y = c_{Y,A} \circ (\epsilon_Y \otimes \Delta_{A,\Omega_{hg^{-1}}}) \circ c_{A,Y}$$ -In words, we can slide the $Y$-strand over the $\Omega_h$-encircling so that it links with the $A$-strand. If you draw a picture, it will become clear. -Killing lemma -Recall that the trivial degree $\mathcal{C}_e$ is a braided spherical fusion category, so the classical killing lemma applies to $\Delta_{A,\Omega_e}$. -Lemma -Let $A \in \mathcal{C}_e$ be a simple object, and $\mathcal{C}_h \not\simeq 0$. Then: -$$\Delta_{A,\Omega_h}\colon A \to {}^hA$$ is nonzero, and thus an isomorphism, iff $A$ is in the symmetric centre of $\mathcal{C}_e$. -Proof -The statement is equivalent to the statement with $- \otimes 1_{\Omega_{h^{-1}}}$ applied to it (recall that $\mathcal{C}_h \not\simeq 0 \implies \Omega_h \neq 0 \implies \Omega_{h^{-1}} \cong \Omega_h^* \neq 0)$. By the sliding lemma, the left hand side is then equal to $c_{\Omega_{h^{-1}}, A} \circ (\epsilon_{\Omega_{h^{-1}}} \otimes \Delta_{A,\Omega_e}) \circ c_{A, \Omega_{h^{-1}}}$. Apply the killing lemma for $\Delta_{A,\Omega_e}$, yielding: -$$\Delta_{A,\Omega_h} \otimes 1_{\Omega_{h^{-1}}} = \left\{ \begin{matrix} d(\Omega_e) \cdot \gamma_{A,\Omega_{h^{-1}}} && \qquad A \text{ is in the symmetric centre of } \mathcal{C}_e \\ 0 && \qquad \text{otherwise} \end{matrix}\right.$$ -Observation -This seems to prove that if all degrees of $\mathcal{C}$ are nonzero (that is, the grading is "faithful"), all objects in the symmetric centre of $\mathcal{C}_e$ are automatically equivariant, i.e. possess a family of isomorphisms $A \cong {}^gA, \forall g \in G$. Surely, that result must have been known before? -Edit This has been commented on briefly by Etingof, Nikshych, Ostrik and Meir in Fusion categories and homotopy theory, (39).<|endoftext|> -TITLE: Trivial product of two matrices? -QUESTION [6 upvotes]: Is it possible to find two matrices $A$ and $B$, so that there does not exists a product of matrices $A$,$A^{-1}$,$B$,$B^{-1}$ that is equal to $Id$, under the condition that the product is irreducible, that is, it is is not trivial as a word (e.g. $AA^ {-1}B^{-1}B$). -If there is no simple answer I would be happy with some references to where I should start. - -REPLY [11 votes]: As far as I understand, the OP asks about existence of free subgroups with 2 generators in matrix groups. The simplest example (if one does not require the matrices to be orthogonal which the OP did not) is probably the free group generated by the matrices -$$\left[ \begin{array}{cc} 1 & 2 \\\ 0 & 1 \end{array} \right]$$ -and -$$\left[ \begin{array}{cc} 1 & 0 \\\ 2 & 1 \end{array} \right]\;,$$ -see The free group $F_2$ has index 12 in SL(2,$\mathbb{Z}$) - -REPLY [7 votes]: A brief comment, really, on R W's answer. Sanov's theorem states that the matrices -$$\begin{pmatrix} 1 & a \\ 0 & 1\end{pmatrix}$$ and $$\begin{pmatrix}1 & 0 \\ a & 0\end{pmatrix}$$ generate a free group as long as $a\geq 2,$ though $a>2, a \in \mathbb{N}$ gives a free group of infinite index in $SL(2, \mathbb{Z}).$<|endoftext|> -TITLE: Does $ M(x)=O(\sqrt{x}) $ if and only if the De Bruijn-Newman constant is negative? -QUESTION [5 upvotes]: The Riemann hypothesis is equivalent to the assertion that the De Bruijn-Newman constant $ \Lambda $ , as defined in https://www.sciencedirect.com/science/article/pii/S0001870809001133/pdf?md5=d2b0cbb38f79b80de06d8b9c99836fab&pid=1-s2.0-S0001870809001133-main.pdf&_valck=1, fulfills $ \Lambda\leq 0 $. On the other hand, Newman conjectured that $ \Lambda\geq 0 $, so that RH, if true, would be only barely so. So far it has been proven that $ -1.1\times 10^{-12}\lt\Lambda\lt 1/2 $. -Another assertion equivalent to RH is $ M(x)\ll_{\varepsilon}x^{1/2+\varepsilon} $ where $ M $ is the Mertens function defined as the summatory function of the Möbius function, and it has been proven that the stronger statement $ M(x)=O(\sqrt{x}) $ is incompatible with the great linear independence hypothesis, stating that, under RH, the imaginary parts of the standard L-functions are linearly independent over $ \mathbb{Q} $, the latter being widely believed to be true. -That way, the possibility that $M(x)=O(\sqrt{x}) $ may be too good to be true sounds similar to Newman's conjecture. -So my question is : is $ M(x)=O(\sqrt{x}) $ actually equivalent to $ \Lambda\lt 0 $? - -REPLY [14 votes]: The de Bruijn-Newman constant is nonnegative, as proved in this brand new preprint by Rodgers and Tao. It is also conjectured, but not proven yet, that $M(x)=O(\sqrt{x})$ is false, in which case it is actually equivalent to $\Lambda<0$. Time will tell.<|endoftext|> -TITLE: Degeneration of curves inside a family of surfaces -QUESTION [9 upvotes]: We are interested in understanding how a higher genus curve in a smooth surface of general type can degenerate into the non-normal locus of the limit of a degeneration of surfaces. More precisely: -Let $\mathscr{X} \to B$ be a family of surfaces over a smooth curve $B$ so that the general fiber $X_\eta$ is a smooth surface, the special fiber $X_0$ has semi log canonical singularities, and the canonical sheaf $\omega_{X_b}$ of every fiber is ample. This corresponds to a map from $B$ to the KSBA moduli space of stable surfaces. -Assume that the double locus of $X_0$ (i.e. the non-normal locus of $X_0$) has an irreducible component which is a rational curve $D$. -Question 1. Let $C_\eta \subset X_\eta$ be a smooth curve whose genus is $\geq 2$, and let $\mathscr{C} \to B$ be the family of curves obtains as the closure of $C_\eta$ inside $\mathscr{X}$. Denote by $C_0$ the special fiber $\mathscr{C} \cap X_0$. Can $C_0$ contain $D$ as an irreducible component? -We are additionally interested in how this can generalize to the case of pairs. In this setting, now suppose there is a divisor $\Delta \subset \mathscr{X}$ so that the pair $(X_b, \Delta_b)$ has slc singularities and the log canonical sheaf $\omega_{X_b} + \Delta_b$ is ample for every fiber, where $\Delta_b = X_b \cap \Delta$. Equivalently now there is a map from $B$ into the moduli space of stable pairs of dimension 2. -Question 2. If Question 1 is true, is it possible that $\Delta$ intersects $D$ (the component of $C_0$ inside the double locus of $X_0$) transversally at a smooth point of $C_0$ and that $\Delta \cap C_\eta \neq \emptyset$? - -REPLY [5 votes]: Let $S$ be the "usual" pinch point surface defined by $x^2t=y^2z$ in $\mathbb P^3$ and $T\subseteq \mathbb P^3$ an arbitrary general surface of degree $d-3\geq 2$. Let $X_0=S\cup T$. Note that then $\deg X_0=d$. Let $\ell\subseteq S$ be the double line and $H\subseteq \mathbb P^3$ a general surface of sufficiently high degree such that $\ell\subseteq H$ and let $C_0=H\cap X_0$. -Since $T$ is general, it is smooth and intersects $S$ transversally, so I think $X_0$ is even semi-smooth, but for sure slc. Now choose a smoothing of $X_0$ as a degree $d$ surface in $\mathbb P^3$, i.e., let $f:\mathscr X\to B$ be a family of degree $d$ surfaces in $\mathbb P^3$ parametrized by a smooth curve such that $X_0$ is a fiber of $f$ and the general fiber $X_\eta$ is smooth. Since $d\geq 5$, $\omega_{X_b}$ is an ample line bundle for every $b\in B$. -Now let $\mathscr C$ be the collection of curves $C_b=H\cap X_b$ for $b\in B$. The general member will be smooth and $C_0$ contains $\ell$ as an irreducible component. So this seems to answer your first question. -For the second, take the same example, but add a $\Delta$, i.e., let $\Gamma\subseteq \mathbb P^3$ be a high degree general surface and let $\Delta_b:=\Gamma\cap X_b$ and $\Delta\subseteq \mathscr X$ the preimage of $\Gamma$ in $\mathscr X$. Then $\Gamma$ and hence $\Delta$ intersects $\ell$ transversally at a point which is smooth on $C_0$ and since $\Gamma$ is ample, $\Delta\cap C_\eta\neq\emptyset$.<|endoftext|> -TITLE: Completely bounded maps approximately factoring through finite matrices -QUESTION [7 upvotes]: Let $A$, $B$ be two $C^\ast$-algebras and $\mathcal{F}(A,B)$ be the operator ideal of all completely bounded operators $T:A \to B$ for which there are uniformly bounded nets of completely bounded maps $(\psi_k)_k$ and $(\phi_k)_k$ -$$ - A \xrightarrow{\phi_k} M_{d_k}(\mathbb{C}) \xrightarrow{\psi_k} B -$$ -such that $\psi_k \circ \phi_k$ tends to $T$ in the pointwise norm topology. We can define a norm for $T$ as -$$ - \|T\|_{\mathcal{F}(A,B)} - = \inf \big\{ \sup_{k} \| \phi_k \|_{cb} \, \| \psi \|_{cb} : - \psi_k \text{ and } \phi_k \text{ as above } \big\} -$$ - -Question 1: Has this set of maps been studyied? Is it known to be equivalent to some other "natural" operator ideal? -Question 2: Is the property $id \in \mathcal{F}(A,A)$ equivalent to any known approximation property? - -Background. We have that -$$ -\text{Nuclearity} \subset \{id \in \mathcal{F}(A,A)\} \subset \text{CBAP}. -$$ -By [Smith] if we assume $id \in \mathcal{F}(A,A)$ and that both $\phi_k$ and $\psi_k$ are complete contractions then $A$ is nuclear. Because of this observation it can not be true that -$$ -\| id \|_{\mathcal{F}(A)} = \Lambda_{cb}(A), -$$ -where $\Lambda_{cb}(A)$ is the CBAP constant of $A$. Indeed, take a non-nuclear algebra $A$ with $\Lambda_{cb}(A) = 1$ and assume there are $\phi_k$, $\psi_k$ as above with $\| \psi_k \|_{cb} \| \phi_k \|_{cb} = 1$. If those maps are not contractions then $\| \psi_k \|_{cb} = \lambda$ and $\| \phi_k \|_{cb} =\lambda^{-1} $ and renormalization will give two contractions, which will force $A$ to be nuclear. Therefore, it does not seem likely that the approximation property of Question 2 above is equivalent to CBAP. -In a similar spirit, if we strengthen the condition so that $\psi_k$ is decomposable (i.e.: sum of cp maps) then $T$ is in the class of operators satisfying that -$$ - \| (id \otimes T): C \otimes_\min A \to C \otimes_\max B \| < \infty. -$$ -See [Pisier, Corollary 12.6]. -[Pisier] Pisier, Gilles, Introduction to operator space theory, London Mathematical Society Lecture Note Series 294. Cambridge: Cambridge University Press (ISBN 0-521-81165-1/pbk). vii, 478 p. (2003). ZBL1093.46001. -[Smith] -Smith, R.R., Completely contractive factorizations of $C^*$-algebras, J. Funct. Anal. 64, 330-337 (1985). ZBL0585.46050. - -REPLY [6 votes]: I don't have access to Pisier's memoir, but I think I can provide a rough sketch of the proof. -First step consists of showing that the assumptions imply that there exists a completely bounded projection $P:B(H) \to A^{\ast\ast}$. Then, and this is the hard part, one has to prove that it implies that $A^{\ast\ast}$ is injective, hence $A$ is nuclear. In fact, the assumptions imply that $A$ is exact and ijectivity of $A^{\ast\ast}$ immediately implies that $A$ has the WEP (weak expectation property), and it is not hard to show that combination of WEP and exactness (understood as existence of approximate factorisation of the embedding $A\subset B(H)$ through matix algebras) gives nuclearity. -In order to do that, we will work with maps $\phi_k^{\ast\ast}: A^{\ast\ast}\to M_{d_k}$ and $\psi_k: M_{d_k} \to A^{\ast\ast}$, whose composition converges to identity in point-ultraweak topology; it is not entirely clear to me that this is actually true. However, there is a way to produce a sequence that will tend to identity in point-ultraweak topology, using local reflexivity. Indeed, the assumptions say that $A$ has a strong form of completely bounded approximation property, therefore it is exact, hence locally reflexive. In particular, it means that whenever $X \subset A^{\ast\ast}$ is a finite dimensional subspace and $(\omega_1, \dots, \omega_n)\subset A^{\ast}$, then for any $\varepsilon > 0$ there is a completely contractive map $P: X \to A$ such that $|\omega_i(x-Px)| \leqslant \varepsilon \|x\|$. Suppose that we have $A\subset A^{\ast\ast} \subset B(H)$. If we compose such a $P$ with $T_k:= \psi_k \circ \phi_k$, then by by Arveson-Wittstock theorem these maps can be extended to maps from $B(H)$ to $A$, since $phi_k\circ P$ has values in $M_{d_k}$. These extensions may be viewied as completely bounded maps from $B(H)$ to $A^{\ast\ast}$, and we have $CB(B(H), (A^{\ast})^{\ast})\simeq B(H)\widehat{\otimes} A^{\ast}$, where $\widehat{\otimes}$ is the projective tensor product (of operator spaces). Therefore any bounded net would have a weak* cluster point. If we are careful enough, we may construct a net whose cluster point must be the identity on $A^{\ast\ast}$. -I will first describe the averaging approach of Christensen and Sinclair (see On von Neumann algebras which are complemented subspaces of $B(H)$, J. Funct. Anal. 122 (1994), no. 1, 91--102). The key is the following: suppose that $P:B(H) \to M$ is a completely bounded projection and $(u_i)$ is a sequence of elements in $M$ such that $\sum u_i^{\ast} u_i = \mathrm{Id}$ strongly. Then the ultraweakly closed convex hull of the maps of the form $x\mapsto \sum P(xu_{i}^{\ast}) u_i$ (indexed by such families of $u_i$'s) contains a completely bounded projection $Q$ which is a right $M$-module map. Doing it from the other side, we obtain an $M$-bimodule projection; this is enough by the results presented by Bunce and Paschke in Quasi expectations and amenable von Neumann algebras, Proc. Amer. Math. Soc. 71 (1978), 232--236. This is actually not very hard. If we assume that $M$ is a finite von Neumann algebra with trace $\tau$ then let $\varphi$ be a functional on $B(H)$ defined by $\varphi(x) :=\tau(Q(x))$, where $Q$ is a bimodular projection. Then $\varphi$ is unitarily invariant. We may assume that $\varphi$ is self-adjoint and then perform the Hahn-Jordan decompostion into positive and neqative parts, and uniqueness of this decomposition implies that both parts are unitarily invariant. If we take the positive part then we obtain a hypertrace on $B(H)$, which is sufficient for proving that $M$ is injective. The general case is handled by standard techniques, the semifinite case by approximation, and the type III case using the crossed product decompostion. -Pisier's (and Haagerup's?) approach is completely different, and he actually requires slightly less than complete boundedness, he requires that the projections works nicely with row and column spaces (see Projections from a von Neumann algebra onto a subalgebra, Bull. Soc. Math. France 123 (1995), 139--153). The main result is: let $M$ be a von Neumann algebra, and let $X_0$ ($X_1$) be $M^n$ equipped with the norm $\|(x_i)\|_{0}:=\|\sum x_i^{\ast}x_i\|_M^{\frac{1}{2}}$ ($\|(x_i)\|_{1}:=\|\sum x_i x_i^{\ast}\|_M^{\frac{1}{2}}$). The interpolation result is that $\|(x_i)\|_{\frac{1}{2}} = \|\sum \overline{x_i}\otimes x_i\|^{\frac{1}{2}}_{\overline{M}\otimes_{\mathrm{max}} M}$. In particular, even though general completely bounded maps do not interact nicely with the maximal tensor product, they do so if we restrict to ''positive definite'' tensors. -Suppose now that $N\subset M$ is a pair of von Neumann algebras such that there exists a completely bounded projection $P:M \to N$. By the interpolation result we have $\|\sum \overline{x_i}\otimes x_i\|_{\overline{M}\otimes_{\mathrm{max}} M} \leqslant \|P\|_{cb}^2 \|\sum \overline{x_i} \otimes x_i\|_{\overline{N} \otimes_{\mathrm{max}} N}$, for any tuple $(x_i)\subset N$. Note that if we denote $t=\sum \overline{x_i}\otimes x_i$, then $(t^{\ast}t)^m$ is a positive definite tensor as well for any $m$, and by $C^{\ast}$-identity we get the above inequality with constant $\|P\|_{cb}^{\frac{1}{m}}$ for any $m$, i.e. with constant $1$; it means that the inclusion $N \subset M$ is $\mathrm{max}^{+}$-injective. Then one can use the self-polar forms to show that in such case there exists a norm one projection from $M$ onto $N$. It would take some time to discuss it, but if anyone is interested I can elaborate on that. -The second approach seems way harder, but it gives more, e.g. Haagerup's characterisation of WEP algebras: a $C^{\ast}$-algebra $A$ is WEP iff the max and min norms coincide on positive definite tensors in $\overline{A}\otimes A$; this shows in particular that WEP depends only on the operator space structure of $A$. -Maybe a comment is in order; it seems to me that all of these results when something a priori completely bounded can be replaced by something completely contractive (or even completely positive), really come from interpreting various properties in terms of tensor products and then using the rigidity of the norm structure of a $C^{\ast}$-algebra, i.e. whenever we have equivalent norms they have to be equal. -I hope that this answer is helpful to someone; writing it certainly clarified matters a lot for me.<|endoftext|> -TITLE: Mathematical conjectures on which applications depend -QUESTION [78 upvotes]: What are some examples of mathematical conjectures that applied mathematicians assume to be true in applications, despite it being unknown whether or not they are true? - -REPLY [22 votes]: Although we have a proof now, I suspect grocers had intuitively stacked oranges in the most efficient way prior to the proof of Kepler's conjecture. I don't know for sure if any grocer was an applied mathematician. -https://www.newscientist.com/article/dn26041-proof-confirmed-of-400-year-old-fruit-stacking-problem/<|endoftext|> -TITLE: How would you organize a cycle of seminars aimed at learning together some basics of Derived Algebraic Geometry? -QUESTION [15 upvotes]: This question is similar to this one because it's asking about a possible roadmap towards learning some derived algebraic geometry (DAG). But it's also different, because the goal is not to form a research level maturity in DAG but just to go through an introduction to the main basic ideas of the subject. -Through this question I would like to gather some suggestions about running a cycle of a few self-learning seminars on DAG, with further info as follows: -1) It should be a cycle of around ten seminars (but this is kind of flexible), one per week, each of the duration of 50 minutes (with people belonging to the same institution). -2) Each seminar should be generally given by a different person. It should be reasonably independent from the technical details of the previous one, but it can assume the results and ideas outlined in all the previous seminars. -3) The background of the audience: "Hartshorne style" algebraic geometry, homological algebra, derived categories, derived functors, spectral sequences, some basics of algebraic stacks (not the theory of the cotangent complex), moduli spaces and some enumerative geometry, perhaps some elementary deformation theory. No background should be assumed in: homotopical algebra and simplicial methods, homotopy theory, model categories, higher category theory, topos theory, dg-categories, spectra (in the sense of algebraic topology). This doesn't mean that the subjects outside the above common background should not be present in the seminars: just that they can't be considered as understood (some seminars could be entirely devoted to introducing the basics of these prerequisites). -4) The seminars should emphasize the basic intuition, motivation, methods, and applications, rather than aiming at the greatest possible generality. -5) The "flavor" of DAG should be the most suitable for applications to "geometric" algebraic geometry (mainly over $\mathbb{C}$, as opposed to "arithmetic" algebraic geometry), not algebraic topology or homotopy theory. -6) It would be nice if at least one of the last seminars presented some actual application(s) of the theory: some instance in which DAG has been useful or even "unavoidable" to obtain some result(s). Applications to moduli spaces and enumerative geometry would be especially appreciated (I'm thinking maybe of the context of Joyce et al.'s construction of generalized DT invariants, and "shifted symplectic structures", but I'm not qualified to say more about that). - -REPLY [2 votes]: Here is a concrete example of a one week reading seminar that was organized on derived algebraic geometry: https://video.ethz.ch/events/2013/dag.html<|endoftext|> -TITLE: Does this Osgood-like condition imply continuity? -QUESTION [10 upvotes]: Let us consider a bounded, Borel function $F\colon \mathbb R^d \to \mathbb R^d$. Assume it satisfies the following -Osgood-like condition: -$$\tag{O} -\boxed{\vert \langle F(x) - F(y), x-y \rangle\vert \le \Vert x-y \Vert \rho( \Vert x-y \Vert)} \qquad \forall x,y \in \mathbb R^d, -$$ -where $\rho \colon [0,1) \to \mathbb [0,+\infty) $ is an Osgood modulus of continuity, i.e. a continuous, non-decreasing function with $\rho(0)=0$ and -$$ -\int_0^1 \frac{1}{\rho(s)} \, ds = +\infty. -$$ - -Q. Is it true that $F$ is continuous? Or, more precisely, is it true that $F$ is equivalent, up to null sets, to a continuous function? - -Beside the link above, something (mildly) related can also be found here. -Of course the case $d=1$ is trivial and the question makes sense essentially for $d \ge 2$. Thanks. -Update (the "linear" case). Quoting from here (pag. 2) - -[...] in the case when the modulus $\rho$ is linear, (O) implies that the symmetric part of the distributional derivative is bounded, hence Korn’s inequality gives that $F$ is equivalent, up to Lebesgue negligible sets, to a continuous function. - -I honestly do not see how to show rigorously this claim in the case when $\rho$ is linear. How to use Korn's inequality (I suspect a variant of this $L^2$ inequality is needed, with estimates in $L^\infty$ but do not know)? - -REPLY [11 votes]: You do not need such heavy high-tech as Korn's inequality or even Lebesgue measure theory for an elementary geometry homework. Let's say $F(0)=0$. -The first claim is that $F$ is bounded in some neighborhood of the origin. Indeed, just choose finitely many points $x_j$ such that $x_i-x_j$ span the space. Then for any $x$ in some fixed ball we have $\langle F(x)-F(x_j), x-x_j\rangle$ bounded by some constant whence $\langle F(x), x-x_j\rangle$ are bounded by a constant, whence $\langle F(x), x_i-x_j\rangle$ are bounded too and that is the end of this story. -Now suppose $x_j\to 0$ with $|F(x_j)|\ge\delta>0$. Then WLOG $F(x_j)\to u\ne 0$. Take any $x\ne 0$ close to $0$. We have -$$ -|\langle F(x)-0,x-0\rangle|\le |x|\rho(|x|) -$$ -and -$$ -|\langle F(x)-F(x_j),x-x_j\rangle|\le |x-x_j|\rho(|x-x_j|) -$$ -which, after passing to the limit, becomes -$$ -|\langle F(x)-u,x\rangle|\le |x|\rho(|x|) -$$ -Subtracting, we get -$$ -|\langle u,x\rangle|\le 2|x|\rho(|x|) -$$ -but this is absurd if $x$ is collinear with $u$ and $2\rho(|x|)<|u|$.<|endoftext|> -TITLE: Randomly covering a sphere -QUESTION [7 upvotes]: Let $S$ be the $n$-dimensional unit sphere in the Euclidean space. Further, -let $X_1,\ldots,X_k$ and $Y_1,\ldots,Y_m$ be iid $S$-valued random variables with common (unknown) distribution $\mu$. With $k$ fixed, what is the tightest value of -$$ -\mathbb{P} \left(\bigcap_{i=1}^k \bigcup_{j=1}^m \{ d(X_i, Y_j) \leq \varepsilon \} \right) > 1 - \delta, -$$ -with regard to $\delta$ for $\varepsilon > 0$? - -REPLY [8 votes]: There's a simple way to get upper and lower bounds that are given by the solution to essentially the same combinatorial problem. In particular, we throw $k$ red balls and $m$ blue balls into some bins with iid probability distributions over the bins, and we ask for the maximal probability that a non-empty bin has all red balls. Our upper bound on $\delta$ will be this probability when the number of bins is given by the size $N$ of some minimal $(\varepsilon/2)$-net of the unit sphere, and our lower bound on $\delta$ will be given by this probability when the number of bins is given by the size $M$ of some maximal $\varepsilon$-packing of the unit sphere. -To see this, recall that an upper bound on $\delta$ corresponds to bounding the probability that some $X_i$ is not close to any $Y_j$. Let $a_1,\ldots, a_N$ be an $(\varepsilon/2)$-net of the sphere. I.e., for every point $x$ on the sphere, there exists an $i$ such that $\|a_i - x\| \leq \varepsilon/2$. Clearly to upper bound $\delta$, it suffices to upper bound the probability that there exists an $X_j$ closest to some $a_i$ but no $Y_\ell$ close to $a_i$. We can then bound this by the probability that there exists an $X_j$ closest to some $a_i$ but not $Y_\ell$ closest to $a_i$. Notice that this exactly corresponds to the balls and bins game from above, with the number of bins equal to the size $N$ of the net. -On the other hand, let $b_1,\ldots, b_M$ be vectors on the unit sphere such that $\|b_i - b_j\| >\varepsilon$ for all $i \neq j$. We take the $X_i, Y_j$ to be uniformly random elements from $b_1,\ldots, b_M$ and notice that $Y_j$ is $\varepsilon$-close to $X_i$ if and only if $X_i = Y_j$. Notice that this exactly corresponds to the balls and bins game from above with the number of bins equal to the size $M$ of the packing. - -To finish the proof, we need to find an $(\varepsilon/2)$-net with small cardinality $N$, an $\varepsilon$-packing with large cardinality $M$, and then solve this balls and bins problem for the presumed answer. We'll handle the balls and bins problem last since it's least interesting. Constructing small nets and large packings is a deep problem, so if we wanted to analyze this to good accuracy, we'd be in trouble. (What I'm calling the size of a packing here is typically called a ``kissing number'' in the literature.) -But, if you just want to understand the asymptotics, then getting values of $N$ and $M$ that are good enough is pretty straightforward. In particular, a simple volume argument shows that we can take $N \leq V(1+\varepsilon/4) -/V(\varepsilon/4) = (1+4/\varepsilon)^n$, where $V(r)$ is the volume of a ball of radius $r$. (See, e.g., http://www-personal.umich.edu/~romanv/papers/non-asymptotic-rmt-plain.pdf Lemma 5.2.) -And, a similar volume argument shows that we can take $M \geq S(2)/S(\varepsilon) \geq \varepsilon^{-n}$, where $S(x)$ is the surface area of the spherical cap on the sphere of radius $1$ with maximal distance $x$. - -Finally, let's study the combinatorial problem. I.e., let's approximate the maximal probability that a non-empty bin has all red balls when we throw $k$ red balls and $m$ blue balls into $Q$ bins with iid probability distributions over the bins. We first observe that the maximum is achieved when the distribution is uniform over the bin. To see this, it suffices to compute this directly in the case $Q = 2$ and then to note that this implies the general case. (In the general case, we can isolate any pair of bins and argue that they must have the same probability by reducing to the $Q = 2$ case.) -For the $j$th bin, we have -$$\Pr[\text{$j$ is non-empty and red}] = \Pr[\text{no blue balls in $j$}] \cdot \Pr[\text{at least one red ball in $j$}] = (1-1/Q)^m \cdot (1-(1-1/Q)^k) \approx e^{-m/Q} \cdot \min\{ 1, k/Q\} -\; ,$$ -where $\approx$ here means that the approximation is good up to a constant factor in, say, $m$, $k$, and the probability. -Similarly, -$$ -\Pr[\text{exists a non-empty red bin}] \approx \min\Big\{1,\ \sum_j \Pr[\text{$j$ is non-empty and red}] \Big\} \\ -\approx \min \Big\{1,\ Q e^{-m/Q} \cdot \min\{ 1, k/Q\} \Big\} -\; . -$$ - -Putting everything together gives -$$ -C_1 \min \Big\{1,\ \varepsilon^{-n} e^{-C_2 m/\varepsilon^{-n}} \cdot \min\{ 1, k\varepsilon^n\} \Big\} \\ \leq \delta \leq \\ -C_3 \min \Big\{1,\ (1+4/\varepsilon)^n e^{-C_4 m/(1+4/\varepsilon)^n} \cdot \min\{ 1, k(1+4/\varepsilon)^{-n} \} \Big\} -\;. -$$ -Assuming we're in the more interesting case when the minima are not 1, then -$$ -C_1 k e^{-C_2 m/\varepsilon^{-n}} \leq \delta \leq -C_3 k e^{-C_4m/(1+4/\varepsilon)^n} -\;. -$$ -(Edit: On rereading this answer, I noticed that my approximations in the combinatorial case failed in the extreme case when $m$ is very large, so I added additional constants.)<|endoftext|> -TITLE: Elliptic operator on compact Hermitian manifold -QUESTION [7 upvotes]: Let $X^n$ be a compact complex manifold, and $\omega$ be a Hermitian metric on $X$. -Define an operator $P:=i\Lambda_\omega \bar{\partial} \partial$ on the space of the smooth function $C^\infty(X, \mathbb{C})$, where $\Lambda_\omega$ is the dual operator of $L_{\omega} = \omega \wedge \cdot$. -It is clear that $P$ is an elliptic operator and not self-adjoint in general. -In fact, $P^\ast$ (w.r.t $L^2$-inner product) can be expressed by $P^\ast = \frac{i}{(n-1)!} \ast_{\omega} \bar{\partial} \partial L^{n-1}_{\omega}.$ -In particular, the second principle symbol of $P$ is equal to the second principle symbol of $\Delta_{\bar{\partial}}$. -Combining this result and the theory of Fredholm operators provides that the $\mathrm{ind}(P)=0$. -Applying maximum principle and calculating locally, $\mathrm{ker}(P)=\mathbb{C}$, and function $f \in \mathrm{im}(P|_{C^\infty(X,\mathbb{R})})$ are not non-positive or not non-negative (ie. not constant sign) other than the zero function. -For the adjoint operator $P^\ast$, the dimension of the kernel can be obtained by $$\mathrm{dim}\, \mathrm{ker} (P^\ast) = \mathrm{dim}\, \mathrm{coker} (P) = \mathrm{dim}\, \mathrm{ker} (P) = 1.$$ -I would like to show that every real smooth function $f \in \mathrm{ker} (P^\ast|_{C^\infty(X, \mathrm{R})})$ are always non-positive or non-negative. -It is obvious to see that $\mathrm{ker} (P^\ast)$ is orthogonal complement to $\mathrm{im} (P)$ in the $L^2$-inner product. -In the appendix of the Kobayashi-Hitchin correspondence, Lübke and Teleman said that the orthogonal decomposition of $C^\infty(X, \mathbb{R}) = \mathrm{ker} (P^\ast|_{C^\infty(X, \mathrm{R})}) \oplus \mathrm{im} (P^\ast|_{C^\infty(X, \mathrm{R})})$ would provide that the result I want. -How can I gain that every real smooth function $f \in \mathrm{ker} (P^\ast|_{C^\infty(X, \mathrm{R})})$ are always non-positive or non-negative by the previous statement? - -REPLY [2 votes]: Here is the proof in Gauduchon's paper. Denote by $((.,.))$ the scalar product on various $L^2$ spaces. -Remark: If $f\in Ker(P^*)$, $f$ not identically zero, then $((f,1))\neq 0$. Indeed, if $((f,1))=0$, then $1\perp Ker(P^*)$, that is $1\in Im(P)$, therefore there exists $g\in {\mathcal C}^{\infty}(X, {\mathbb R})$ such that $P(g)=1$, and this is impossible since $g$ must have an absolute maximum on $X$, and at that point $P(g)\leq 0$. -If $f\in Ker(P^*)\setminus \{0\}$, we want to show that either $f\leq 0$ or else $f\geq 0$ so suppose that $f$ takes both negative and positive values on $X$. Then it is easy to find $h\in {\mathcal C}^{\infty}(X, {\mathbb R})$, $h>0$ such that $((hf,1))=0$. -If we consider the metric $h\omega$ which is conformal to $\omega$, and do all of the above with respect to $h\omega$, then it is easy to see that $Ker(P^*_h)$ is generated by $h^{1-n}f$ and that $((h^{1-n}f,1))_h=((hf,1))$ (here the index $h$ means with respect to the metric $h\omega$). Since $((hf,1))=0$, it follows that $((h^{1-n}f,1))_h=0$, and this contradicts the Remark above for the metric $h\omega$.<|endoftext|> -TITLE: Derived topological stacks? -QUESTION [9 upvotes]: I apologize for the vagueness of the following. -Informally, in the site of commutative rings, one roughly get the notion of a derived stack by swapping out the commmutative rings with its subcategory of simplicial objects. This is part of the story told by Toen and Vezzosi in their many expository accounts on derived geometry. -Now, there appears to be a fairly well-studied story for stacks defined in the site of topological spaces (topological stacks), notably in a series of papers by Noohi. Then it's natural to try to obtain a notion of a derived topological stack by replacing the source category by simplicial topological spaces. Has this been done by anyone? Is the story there trivial or uninteresting for some reason? - -REPLY [14 votes]: At Andre's suggestion, I'll turn my comments into an answer. If we correct the OP by asking about cosimplicial topological spaces, then we have to decide on a notion of equivalence for them, and the obvious one (tying in with DAG and derived differential geometry) is to ask for weak equivalence on the associated simplicial ring of $\mathbb{R}$-valued functions. But then it turns out that the derived structure is meaningless, as everything is equivalent to something with constant cosimplicial structure: -The reason for this is that the ring $A$ of $\mathbb{R}$-valued functions on a topological space has the structure of a $C^0$-ring, meaning that for every continuous function $f : \mathbb{R}^n \to \mathbb{R}$, we have a systematic way of evaluating $f(a_1, \ldots,a_n) \in A$ for all $a_1, \ldots, a_n \in A$. But any simplicial $C^0$-ring is discrete in the sense that $\pi_0A\simeq A$. -This can be proved by looking at the simplicial $C^0$-rings $P^n$ representing $\pi_n$ in the homotopy category. We find that the functor $\pi_n$ is identically $0$ because $\pi_n(P^n)=0$. For the standard cofibrant representative (given by $P^n_i=\mathbb{R}$ for $i -TITLE: Would an oracle for Rayo's function let you compute a model of $(V, \in)$? -QUESTION [18 upvotes]: Working in Kelly-morse set theory, let $R$ be an oracle that can compute Rayo's function. Can $R$ compute a countable model $M = (\mathbb N,\in_M)$ that is elementary equivalent to $(V, \in)$? - -REPLY [10 votes]: This is an excellent question! -Here are some steps in the positive direction. I claim that the Rayo function can compute -the theory of true arithmetic. Indeed, I claim more, that we can push this into the hyperarithmetic hierarchy. -To see this, let's consider just true arithmetic first. Let $R$ be the Rayo function; so $R(n)$ is -the smallest number not first-order definable in $V$ in the -language of set theory by an expression of size at most $n$. [This -definition is made relative to a fixed truth predicate, and it is -not sensible to speak of the Rayo function in contexts where there -isn't such a truth predicate. For example, we cannot refer to the -Rayo function in ZFC, but it is fine in GBC+ETR or KM.] -Now, I claim that we can compute recursively whether a given statement $\sigma$ -in the language of arithmetic is true or not in the standard model -$\langle\mathbb{N},+,\cdot,0,1,<\rangle$. The same idea appears in my solution to the question, The set of -largest numbers definable by formulas in different -lengths. -The algorithm is this: we -can compute atomic assertions directly, and we can reduce via -Boolean combinations. The only difficult case is to -check quantifiers $\exists m\ \varphi(m)$. But for this, I claim that it -is sufficient to check whether $\varphi(k)$ holds for any $k$ up to -$R(n)$, where $n$ is large enough to express the definition, "m is -the least natural number for which $\varphi(m)$ holds in -$\mathbb{N}$.'' The point is that if $\exists m\ \varphi(m)$ is -true, then the least such $m$ will be definable and therefore will -be smaller than $R(n)$ for that value of $n$. So we can use the Rayo function to reduce the infinite process of an existential quantifier into finitely many cases, since if none of those numbers works then we can be sure that there is no witness. -So the Rayo function $R$ computes $0^{(\omega)}$. -But actually, we can now push this further into the hyperarithmetic hierarchy. For example, we can compute $0^{(\omega+\omega)}$, which is the theory of the structure $\langle\mathbb{N},+,\cdot,0,1,<,0^{(\omega)}\rangle$. We just described how to compute atomic assertions in this structure, and now we can do the same trick again to get the theory of this structure, by using $R$ to bound the existential witnesses. -It seems to me that we can push this method much further, well into the hyperarithmetic hierarchy. But I'm not sure exactly how far. You want to push it all the way to $\text{Th}(V,\in)$, -which is quite a bit farther indeed.<|endoftext|> -TITLE: A generalization of Gradient vector fields and Curl of vector fields -QUESTION [6 upvotes]: Let $M$ be a smooth Riemannian manifold. The Riemannian metric enables us to equip the tangent bundle $TM$ with a symplectic structure $\omega$, which is the pullback of the standard symplectic $2$ form of the cotangent bundle. Let $X$ be a vector field on $M$. Then $X:M \to TM$ is a smooth map. - -What is a dynamical interpretation for vanishing $X^{*} (\omega)$, the pullback of $\omega$ via $X$. Is there a name for this property? For a given vector field, what kind of dynamical obstructions exist to have a Riemannian metric with vanishing the above $2$ form? - -In $2$ and $3$ dimensional Euclidean space the above $2$ form is closely related to "gradient vector fields" and Curl vector field, respectively. - -REPLY [10 votes]: This is equivalent to the fact that the image of the 1-form $X^\flat$ in $T^*M$ is a Lagrangian submanifold; equivalently, the 1-form $X^\flat$ is closed. So, locally, $X^\flat = df$ for some function $f$, or, $X=\operatorname{grad}^g(f)$.<|endoftext|> -TITLE: Diffeomorphisms on a real manifold whose derivative are holomorphic maps on the tangent bundle -QUESTION [7 upvotes]: Edit: According to the answers to the linked MSE question and the comment of Holonomia, I understand that the answer to the second question is that " Every tangent bundles is a complex manifold". - -Let $M$ be a compact real manifold. Assume that the tangent bundle $TM$ carries a holomorphic structure, namely it is equipped with a holomorphic atlas. We fix a holomorphic structure for $TM$. -We say that a map $f:M \to M$ has a holomorphic derivative if $Df:TM \to TM$ is a holomorphic map. - -The first Question: Is it true to say that the space of all diffeomorphism of $M$ with holomorphic derivative admits a structure of a finite dimensional Lie group? -The Second Question: What is an example of a manifold whose tangent bundle, as a manifold, does not admit a holomorphic atlas? - -I asked the latter question on MSE but I did not get any answer. -https://math.stackexchange.com/questions/2611104/some-questions-on-the-tangent-bundle-of-manifolds - -Further questions: -Added: How can one decide that a given map has holomorphic derivative? In this investigation, and motivated by CR equations, what kind of differential operators would appear? -As a particular example, we consider the Hopf map $p: S^3 \to S^2$. Is it a map with holomorphic derivative? Of course this question is meaningless if we do not fix a holomorphic structure for $TS^2$ and $TS^3$. So it is natural to ask: What is a precise holomorphic structure for these space? Can the holomorphic structure of the tangent bundle of a Riemann surface or a parallelizable manifold be determined explicitly? - -REPLY [4 votes]: I guess that you mean: $TM$ carries an almost complex structure $J:TM\to TM$ with $J^2=-1$. It is integrable to complex structure iff the Froelicher-Nijenhuis bracket $[J,J]$ vanishes. -First question: The Lie algebra of your group consists of all vector fields $X$ with $\mathcal L_XJ = [X,J]=0$. Since $J$ is invertible, $[X,J]=0$ is an elliptic equation of 1. order for $X$, so the solution space is finite dimensional on a compact manifold. -Second question: If $J$ exists, the manifold must be even dimensional and orientable. -Added: -So let us assume that $TM$ carries a complex structure. $TM$ is not compact, so the group of biholomorphic automorphismsm need not be finite dimensional. -First question: $\phi\mapsto T\phi=D\phi$ is an injective group homomorphism into the group of all those biholomophic diffeomorphisms of $TM$ which are also vector bundle homomorphisms; i.e., linear between fibers. Its Lie algebra consists of all vector fields which are holomorphic and linear when restricted to any fiber. To be holomorphic is an elliptic equation, and the linearity along fibers should imply that the corresponding solution space is finite dimensional.<|endoftext|> -TITLE: Definition of packing property -QUESTION [9 upvotes]: Definition 1: -A clutter $C$ is said to have the packing property if $C$ and all of its minors satisfy the König property. -where, -vertex cover of $C$ is a set of vertices that have non-empty intersection with all of the edges. The minimum cardinality of a vertex cover of $C$ will be denoted by $\alpha_0(C)$. A matching (or independent set) of $C$ is a set of pairwise disjoint edges. The maximum cardinality of a matching in $C$ will be denoted by $\beta_1(C)$. -A clutter $C$ is said to satisfy the König property if $\alpha_0(C) =\beta_1(C)$. -For a vertex $x \in V (C)$, the deletion $C \setminus x$ is formed by removing $x$ from the vertex set and deleting any edge in $C$ that contains $x$. The contraction $C/x$ is obtained by $V(C/x) = V (C) \setminus x$ - and $E \in E(C/x)$ if $x \notin E$ and either $E \in E(C)$ or $E \cup \{x\} \in E(C)$. -Any clutter formed by a sequence of deletions and contractions is called a minor of $C$. -Definition 2: -$$\min \{ \langle w,v \rangle \mid v \geq 0; vA^T \geq 1\} - =\max \{\langle y,1 \rangle\mid y \geq 0; Ay \leq w\}, $$ -where $A$ is an incidence matrix of $C$. -The packing property can also be restated in terms of the dual linear programming -system. An ideal has the packing property if and only if the dual linear programming -system as above has integral optimal solutions for all $(0, 1,\infty)$-vectors $w$, -that is, when entries of $w$ are all $0, 1$ or $\infty$. - -Question : How to prove Definition 1 <=> Definition 2? - -REPLY [8 votes]: That Def 1 and Def 2 are equivalent is a well-known Conjecture, still open as far as I know. Curiously, you can translate the whole conjecture to the language of commutative algebra, see for example page 26 of this survey. It is relatively easy to prove 2 implies 1, it is the other direction that's problematic.<|endoftext|> -TITLE: Is there a conformal diffeomorphism between R3 minus a line and R2 x S1? -QUESTION [12 upvotes]: There exists a conformal diffeomorphism between $\mathbb{R}^3$ and $S_3$ (less a point): -$$ -g = dr^2+r^2\left(d\theta^2 + \sin^2\theta\ d\phi^2\right) -$$ -$$ -r = R \tan \frac{\alpha}{2} -$$ -$$ -g = \frac{R^2}{4\cos^4\frac{\alpha}{2}}\left[d\alpha^2+\sin^2\alpha\left(d\theta^2 + \sin^2\theta\ d\phi^2\right)\right] -$$ -There also exists a conformal diffeomorphism between $\mathbb{R}^3$ (less a point) and the cylinder $\mathbb{R}\times S_2$: -$$ -r = R\ e^{x/R} -$$ -$$ -g=e^{2x/R}\left[dx^2+R^2\left(d\theta^2 + \sin^2\theta\ d\phi^2\right)\right] -$$ -There does not appear to be a conformal diffeomorphism between $\mathbb{R}^3$ (less a line) and the cylinder $\mathbb{R^2}\times S_1$. At least, my attempts to find one starting from $\mathbb{R^3}$ in cylindrical coordinates: -$$ -g = dz^2+ d\rho^2 +\rho^2 d\phi^2 -$$ -and remapping $(z, \rho)$ have failed. So my questions are: -1) Is it true that there is no such diffeomorphism? -2) If so, how do you show that? I think that local invariants like the Cotton tensor have nothing to say, because both $\mathbb{R}^3$ and $\mathbb{R^2}\times S_1$ are flat, and instead there is some kind of global obstruction. -Edited: -My original question was not sufficiently precise, it asked about existence of a conformal map, but I really meant existence of a conformal diffeomorphism. As Ben pointed out, there is the conformal map -$$ -\mathbb{R}^3 \to \mathbb{R}^2 \times S_1 -$$ -$$ -(x, y, z) \mapsto (x, y, \phi= \mathrm{mod}(z, 2\pi)) -$$ -but that is not injective and so not a diffeomorphism. - -REPLY [5 votes]: The map $\phi(x,y,\theta)=(x,y,e^{i \theta})$ is a locally isometric covering map, as it is given by taking the usual covering map $\Phi(\theta)=e^{i \theta}$, $\Phi \colon \mathbb{R} \to S^1$, $2 \pi$ periodic, and throwing in $x,y$. In particular, this map is a conformal map, but not a conformal diffeomorphism. Its lift is $\tilde\phi(x,y,\theta)=(x,y,\theta+2\pi)$, a conformal diffeomorphism. The metric is $dx^2+dy^2+dz^2$ where $z=2\pi \theta$. -Update: the question is now whether $\mathbb{R}^3$ minus a line is conformally diffeomorphic to $\mathbb{R}^2 \times S^1$. It is not: $\mathbb{R}^3$ minus a line has developing map taking it to $S^3$ minus a circle. (The map is your usual conformal map to $S^3$ as given in the question above.) On the other hand, $\mathbb{R}^2 \times S^1$ admits a conformal covering map (as given in my previous remarks) by $\mathbb{R}^3$, so its developing map has image $S^3$ minus a point. The same argument proves that $\mathbb{R}^3$ minus any closed set is not conformal to $\mathbb{R}^2 \times S^1$. See Richard Sharpe's book Differential Geometry to read about developing maps of conformal geometries.<|endoftext|> -TITLE: Does eigenvalue exist in a Hilbert space? -QUESTION [5 upvotes]: In a lecture on Quantum mechanics, the professor concluded that if $a$ is a linear operator with $[a, a^\dagger] = 1$, where $a^\dagger$ is the adjoint of $a$ and $[a, a^\dagger] = aa^\dagger - a^\dagger a$, and if $a^\dagger a$ has an eigenvalue, then vectors of form -\[ -| n \rangle = (a^\dagger)^n | 0 \rangle \quad ( n \in \mathbb Z_{ \ge 0 }) -\] -are all eigenvectors of $a^\dagger a$. ($| \lambda \rangle$ is the eigenvector of $a^\dagger a$ which belongs to the eigenvalue $\lambda$). -But he gives no reason why its eigenvalue exists. So, I asked him if an eigenvalue exists, and he said 'I cannot answer. Roughly speaking, it is some limit of the case of a finite dimensional complex vector space.' -My question is: -Let $H$ be an infinite dimensional Hilbert space, say $\ell^2$ space, whose inner product is denoted by $\langle \cdot, \cdot \rangle$. It is well-known that for a bounded linear operator $T \in \mathcal L(H)$, there corresponds a (bounded) linear operator $T^\dagger \in \mathcal L(H)$ such that $\langle x, Ty \rangle = \langle T^\dagger x, y \rangle$ for all $x, y \in H$. -Then, is it true that for any bounded linear operator $T \in \mathcal L(H)$ with $[T, T^\dagger] = 1$, there exists a scalar $\lambda \in \mathbb C$ and a nonzero vector $x \in H$ such that $T^\dagger T x = \lambda x$? -How could I deal with it? Is there a 'characteristic polynomial' of $T^\dagger T$? -Please give me the answer, or some references. - -REPLY [9 votes]: As noted in comments, there are no bounded operators whose commutator with their adjoint is a non-zero scalar. The prototype of unbounded operators with the desired property is the 1-dimensional Dirac operator $T=ix\pm i{d\over dx}$ on $L^2(\mathbb R)$. [Correction, thanks to @ChristianRemling for noticing my blunder!) The composition $T\circ T^*$ (not commutator, jeez) of it with its adjoint is (up to sign) $-\Delta+x^2+1$, essentially the quantum harmonic oscillator. The commutator is a constant (specifically, $\pm2$, the way I've written it. Whew...) Because the "confining potential" is present, this has compact resolvent, hence, purely discrete spectrum, and all the eigenvalues are indeed obtained as indicated. -In fact, the Stone-vonNeumann theorem essentially proves the uniqueness of such a "model" for the commutation relations, by proving that there is a unique irreducible repn of a Heisenberg group having a given central character. That is, up to some sort of isomorphism, the configuration of Hilbert space, $T$, and $T^*$ is uniquely determined. Thus, approximately, every such $T,T^*$ behave in the same way, so should have pure point spectrum, etc.<|endoftext|> -TITLE: What was the error in the proof of Roos' theorem? -QUESTION [32 upvotes]: Background: In 1961, Roos (who, sadly, apparently passed away just last month) purported to prove [1] that in an abelian category with exact countable products (AB4${}^\ast_\omega$), limits of inverse systems of epimorphisms are exact [2] [3]. The result stood for 41 years before Neeman published a counterexample in 2002. Influenced by Gabber, Roos published a corrected statement a few years later: in an abelian category with arbitrary coproducts (AB3) and exact countable products (AB4$^\ast_\omega$) which contains a generator, limits of inverse systems of epimorphisms are exact [4]. -Question: -In neither Neeman's paper nor in Roos' 2006 paper can I find a discussion of where in his 1961 paper Roos actually makes a mistake. Can anybody shed any light on this? - -Notes: -[1] Roos, Jan-Erik (1961). "Sur les foncteurs dérivés de lim. Applications". C. R. Acad. Sci. Paris. 252: 3702–3704. MR 0132091. -[2] By an "inverse system" $X^\bullet$, I mean a diagram $\dots \to X^n \to X^{n-1} \to \dots$ indexed by $\mathbb N$ (or $\mathbb Z$ -- it doesn't make a difference). By an "inverse system of epimorphisms" $X^\bullet$, I mean an inverse system where each map $X^n \twoheadrightarrow X^{n-1}$ is an epimorphism. In an abelian category $\mathcal C$ with countable products (AB3$^\ast_\omega$), there is a limit functor from inverse systems in $\mathcal C$, to $\mathcal C$, which is left exact. If the countable products in $\mathcal C$ are exact (AB4$^\ast_\omega$), then there is a natural two-term complex $\prod_n X^n \to \prod_n X^n$ which may be used to derive the limit functor (the resulting $\delta$-functor is effaceable via the map $X^\bullet \to \prod_{n \leq \bullet} X^n$ where the latter complex uses the projections for structure maps); because this complex has two terms, the derived functors $\varprojlim^d X^\bullet$ vanish for $d > 1$. When I say that $X^\bullet$ is exact, I mean that $\varprojlim^1 X^\bullet = 0$. I think I understand how to argue that if $X^\bullet$ is an epimorphic system in an AB4$^\ast_\omega$ category and $\varprojlim^1 X^\bullet = 0$, then the map $\varprojlim X^\bullet \twoheadrightarrow X$ is an epimorphism for each $n$ (actually it's the reverse implication that holds); when Roos' theorem holds, this is a formulation of the result which doesn't mention $\varprojlim^1$. -[3] There is a stronger statement about Mittag-Leffler sequences; let me stick with the statement about inverse systems of epimorphisms for simplicity. -[4] In the comments here Leonid Positselski points out that Roos' proof actually establishes that this holds more generally in any AB4$^\ast_\omega$ category with enough projective effacements (I hadn't heard of this concept until recently, but it's right there in Tohoku -- actually, enough local projective effacements suffices). In this form, the statement is actually quite straightforward to prove. - -REPLY [21 votes]: Taking a brief look at the Roos note we see that detailed proofs of the statements aren't provided. There is no argument in the note one could say is wrong. The paper with the corrected statement very carefully points out which parts of the original note do hold and which ones have to be modified and how and gives detailed arguments or precise references for the proofs of various results. See discussion at the start of Section 1 and at the start of Section 2. Thus, if you wanted to use this kind of result in another paper I would reference the newer paper. -In your question in your second footnote you have a sketch of an argument of something. But it suffers the same fate as the note of Roos of not being sufficiently written out to be sure if you have understood all possible pitfalls. I suggest you just write out completely what you were trying to say: the mathematics will guide you to the correct proofs and statements. The result should be longer than a Mathoverflow post; if not then you probably have not written it out completely. Good luck!<|endoftext|> -TITLE: Proofs shown to be wrong after formalization with proof assistant -QUESTION [150 upvotes]: Are there examples of originally widely accepted proofs that were later discovered to be wrong by attempting to formalize them using a proof assistant (e.g. Coq, Agda, Lean, Isabelle, HOL, Metamath, Mizar)? - -REPLY [25 votes]: I learned of the following stellar example from Lawrence Paulson. -Anthony Bordg, Yijun He, and Hanna Lachnitt have been involved in an ongoing effort to formalize quantum algorithms and results in quantum information theory in Isabelle/HOL. You can read about their efforts here. -In the course of their project, they naturally found themselves examining one of the seminal papers in the subject, Quantum games and quantum strategies, by J. Eisert, M. Wilkens, and M. Lewenstein. As of this writing, Google Scholar claims that this paper has nearly a thousand citations. Bordg, He, and Lachnitt found that there was a fundamental and unfixable error in one of the main results of the paper. They explain the details in an arXiv preprint.<|endoftext|> -TITLE: Proofs that the conformal group in dimension $\ge 3$ is a Lie group -QUESTION [8 upvotes]: Let $M$ be a smooth manifold of dimension $\ge 3$, equipped with a conformal structure (or a Riemannian metric). Then, the group of conformal diffeomorphisms is a finite dimensional Lie group. -A proof of this theorem can be found in "Transformation groups in differential geometry" by Kobayashi (Theorem 6.1, pg 143). - -Are there other proofs of this claim (perhaps simpler or more readable)? - -Kobayashi's theorem also bounds the dimension of the conformal group by - $\frac{1}{2}(n+1)(n+2)$, where $\dim M=n$. If I am only interested to know the group is a finite dim Lie group, and don't care about bounding its dimension, is there a simpler proof? -(I am also ready to assume $M$ is compact). - -REPLY [23 votes]: Answers to the OP's question depend on where the OP is willing to start. To prove that an abstractly defined group is (i.e., has the structure of) a Lie group, one will have to use something nontrivial, as this is not a trivial task, in general. -For example, É. Cartan's statement that the set of (smooth?, $C^k$?, analytic?) diffeomorphims of a manifold $M$ that preserve a (smooth? $C^k$?, analytic?) parallelization is (i.e., has a natural structure as) a Lie group, was taken to be obvious by him, but was not actually proved to modern standards of rigor until much later. It was the basis of all of his results about the group of symmetries of various geometric structures (including conformal geometry in dimensions at least $3$) being Lie groups. In particular, it implies that the group of self-isometries of a pseudo-Riemannian manifold is a Lie group. -If the OP is willing to assume that the group of self-isometries of a Riemannian manifold is a Lie group, then one can give an argument that the group of conformal symmetries of a manifold of dimension at least $3$ is a Lie group as well using only this, and, moreover, one can immediately see, based on Riemannian geometry, where the assumption that the dimension is at least 3 enters the picture. The idea is as follows: -If $g$ is a metric on an $n$-manifold $M$ and $u$ is a smooth function on $M$, then consider the expression $\mathrm{Ric}(e^{2u}g)$, that computes the Ricci curvature of the conformal metric $g' = e^{2u}g$: -$$ -\mathrm{Ric}(e^{2u}g) = \mathrm{Ric}(g)-(n{-}2)\left(\mathrm{d}u^2+|\nabla u|^2g\right) - (n{-}2)\,\mathrm{Hess}(u) + (\Delta u)\,g, -$$ -where $\mathrm{Hess}(u) = \nabla(\mathrm{d}u)$ is a quadratic form that expresses the second covariant derivatives of $u$ and whose trace is $-\Delta u$. -Thus, when $n\ge 3$, if one specifies the $1$-jet of $u$ at $p$, there will be a unique extension to a $2$-jet of $u$ at $p$ such that $\mathrm{Ric}(e^{2u}g)(p)=0$. (By contrast, when $n=2$, $\mathrm{Ric}(e^{2u}g) = (2K(g)+\Delta u)\,g$, where $K(g)$ is the Gauss curvature of $g$, so specifying $\mathrm{Ric}(e^{2u}g)(p)=0$ only detemines one coefficient of the $2$-jet of $u$ at $p$ instead of all $3$ coefficients.) -This observation can be used to show that, when $n\ge 3$, if we let $C\to M$ be the $\mathbb{R}^+$-bundle of all multiples of $g$, which depends only on the conformal class of $g$, then, on $\pi:J^1C\to M$, the bundle of $1$-jets of sections of $C$, there is a canonical $n$-plane bundle $D\subset T(J^1C)$ that is transverse to the fibers of $\pi:J^1C\to M$ and is such that the $1$-jet graph of a conformal metric $e^{2u}g$ is tangent to the $n$-plane bundle $D$ over a point $p\in M$ if and only if $\mathrm{Ric}(e^{2u}g)(p)=0$. Any conformal transformation of $\bigl(M,[g]\bigr)$ canonically induces a transformation of $J^1C$ that preserves this $n$-plane field $D$. This defines a canonical splitting $T(J^1C) = D\oplus \mathrm{ker}(\pi')$ that can then be used to construct a canonical metric $h$ on $J^1C$ that is preserved by any such transformation. Thus, the group of $[g]$-conformal transformations on $M$ is embedded in the group of self-isometries of $(J^1C,h)$ and hence is embedded as a (closed, it turns out) subgroup of a Lie group. Thus, in this way, it inherits the structure of a Lie group. -By the way, for entirely different reasons, the group of conformal transformations in dimension $2$ is also a Lie group, but it cannot be proved by the method above. Instead, one uses the Uniformization Theorem, which shows that any simply-connected Riemannian $2$-manifold is conformally equivalent to either the disk, the plane, or the sphere (with their standard conformal structures) and then appeals to elementary complex analysis to show that their conformal symmetries are Lie groups. -Finally, I cannot resist pointing out that the group of conformal transformations in dimension $1$ is, nowadays, not regarded as a Lie group, although Lie (and Cartan) regarded it as a Lie group, just infinite dimensional.<|endoftext|> -TITLE: Reasons behind assuming the existence of Siegel zeros can be used to prove something stronger than assuming GRH? -QUESTION [21 upvotes]: There are few results that I am aware of where one can prove something stronger by assuming the existence of Siegel zeros than by assuming the GRH. For example Heath-Brown proved the existence of Siegel zeros imply the twin prime conjecture while it is unknown under GRH. Also there is: Let $P(a,q)$ be the least prime $\equiv a \pmod q$. Then assuming GRH we have $P(a,q) \ll q^L$ where $L < 2 + \varepsilon$, but assuming the existence of the Sigel zeros we have $L < 2$. (This I just took from Good uses of Siegel zeros?) -I was interested in this phenomenon. I would greatly appreciate if anyone could explain why or give some ideas on why this is the case? -Also this is a question with subjective answer, but when there is a result of this type (where by assuming Siegel zeros one gets even stronger results than assuming GRH), do we generally expect it to be the truth? Thank you. -Any comments are appreciated. - -REPLY [46 votes]: Roughly speaking, GRH asserts that the Möbius function $\mu$ is "orthogonal" to all Dirichlet characters $\chi$, in the sense that correlations such as $\sum_{n \leq x} \mu(n) \overline{\chi(n)}$ are very small. This is the expected behaviour of the Möbius function, and through various standard analytic number theory manipulations one can also use GRH to control correlations between the von Mangoldt function $\Lambda(n)$ (which basically encodes primes) and various other functions, e.g. linear phases $e(\alpha n)$. On the other hand, GRH struggles to control self-correlations such as $\sum_{n \leq x} \mu(n) \mu(n+2)$ or $\sum_{n \leq x} \Lambda(n) \Lambda(n+2)$ (the latter being used to count twin primes). For instance, in the function field case GRH is a known theorem, but we are still unable to obtain an asymptotic (or even a lower bound of the right order of magnitude) for twin primes, though it is possible to establish the infinitude of twin primes in this case by more algebraic means. -In contrast, the existence of a Siegel zero means that there is a quadratic Dirichlet character $\chi$ with which $\mu$ has very high correlation (this can be quantified precisely using the "pretentious" approach to analytic number theory developed by Granville and Soundararajan). This would be very unusual behavior for $\mu$, and as such is actually a rather powerful piece of information. As a crude first approximation, a Siegel zero allows one to replace $\mu$ with $\chi$ with acceptable error (in practice one has to be more careful at small primes, though, for instance by imposing a suitable preliminary sieve). Thus for instance one could hope to approximate $\sum_{n \leq x} \mu(n) \mu(n+2)$ by something resembling $\sum_{n \leq x} \chi(n) \chi(n+2)$, which is relatively easy to bound nontrivially. Very roughly speaking, Heath-Brown's arguments proceed by similarly replacing the von Mangoldt function $\Lambda(n) = \sum_{d|n} \mu(d) \log \frac{n}{d}$ with the variant $f(n) := \sum_{d|n} \chi(d) \log \frac{n}{d}$, which is roughly of the same order of complexity as the divisor function; in particular sums such as $\sum_{n \leq x} f(n) f(n+2)$ are tractable enough by known methods (e.g. Kloosterman sum bounds) that they are a useful approximant to the otherwise intractable $\sum_{n \leq x} \Lambda(n) \Lambda(n+2)$.<|endoftext|> -TITLE: Does $End(V)$ remember $V$, where $V$ is a locally convex space? -QUESTION [9 upvotes]: Let $V$ be a locally convex topological vector space over $\mathbb C$, and let $A=\mathrm{End}(V)$ be its algebra of continuous linear endomorphisms (viewed just as a $\mathbb{C}$-algebra, not as a topological algebra). -Does the algebra $A$ remember $V$? -Namely, is it true that every representation of $A$ on a locally convex topological vector space $W$ for which the associated map $A\to \mathrm{End}(W)$ is an isomorphism is isomorphic to $V$ as an $A$-representation, and the isomorphism $V\cong W$ is a homeomorphism? - -REPLY [11 votes]: If $V$ is a Mackey space (for example, a Banach space), $\text{End} (V)$ coincides with $\text{End} (V_\sigma)$, where $V_\sigma$ denotes $V$ with the weak topology.<|endoftext|> -TITLE: Entropy of composition -QUESTION [12 upvotes]: I asked this at math.stackexchange.com, but got no answers. -Let $(X,B,\mu)$ be a probability space. Let $T,S:X→X$ be two measurable measure preserving maps that commute (i.e $TS=ST$). Let $A$ be a (countable measurable) partition of $X$. Show that $h(ST,A)≤h(S,A)+h(T,A)$. If $S=T$, it's rather easy. I couldnt get any further. Any help/reference will be appreciated. Thanks. - -REPLY [20 votes]: There is a good reason you were having difficulties in proving this. -This was an old question of Rohlin (MR0126526) which was first disproved in the topological setting by Goodwyn (MR0314023) and later independently by Thouvenot and Ornstein. An explicit example appears in the paper by Ornstein and Weiss (MR0910005; page 133); this paper should be relevant to you in many different ways if you are interested in such things. -That being given, there are many (smooth) contexts in which this property does hold (MR2690742); most notably consider commuting automorphisms of the torus (this is a simple exercise). -Let me know if you have trouble finding the papers.<|endoftext|> -TITLE: Simply generated sequences with mysterious differences -QUESTION [5 upvotes]: Suppose that $a_0 < a_1,$ $b_0 < b_1,$ and $$a_n=a_1b_{n-1}+a_0b_{n-2}+qn+r$$ for $n \geq 2$, where $a_0,a_1,b_0,b_1,q,r$ are integers such that $(a_n)$ and $(b_n)$ are increasing and ${(|a_n|)}$ and ${(|b_n|)}$ partition the positive integers. What can be proved about the cardinality of $$D=\{(a_n-a_{n-1},b_n-b_{n-1})\},$$ for $n \geq 0?$ -Experimental results: - -If $(a_0,a_1,b_0,b_1,q,r)=(-1,2,3,4,2,0)$, then $|D|=9$; see "Experimental fact" at A possibly surprising appearance of $\sqrt{2}.$ -If $(a_0,a_1,b_0,b_1,q,r)=(1,2,3,4,1,0)$, then $D=\{(1,1),(4,1),(4,2),(5,1),(6,1),(11,1)\}.$ -If $(a_0,a_1,b_0,b_1,q,r)=(3,4,1,2,1,-7)$, then $D=\{(1,1),(2,3),(8,1),(8,2),(11,1),(12,1),(16,2),(18,1)\}.$ - -Reasons for studying the set $D$ include these related questions: - -Is $(a_n-a_{n-1})$ ever linearly recurrent? -Let $d$ be a number that occurs infinitely many times in $(b_n-b_{n-1})$, and let $(p_n)$ be the sequence of numbers $k$ such that $b_k-b_{k-1}=d.$ Must $(p_n/n)$ converge? As an example, for $(a_0,a_1,b_0,b_1,q,r)=(-1,2,3,4,2,0)$, we have -$$(p_n) = (1,11,13,16,19,22,25,28,31,34,37,43,45,51,53,56,62,\dots),$$ -and it appears that $\lim_{n \to \infty} p_n/n = 1+\sqrt{2}.$ - -REPLY [4 votes]: This is less more than a long comment. It seems that a somehow simpler case is the when $a_{n+1}-a_n>1$ (at least eventually) which means $b_{n+1}-b_n\in\{1,2\}$ (eventually), and $a_{n+1}-a_n$ takes values in a finite set $\mathcal{A}$ of say $r$ integers greater or equal to $2$. Then the sequence $a_{n+1}-a_n$ is generated by a substitution map (like in the one described here or here, so that the entire sequence $a_{n+1}-a_n$, seen as an infinite string, has the form $p\cdot\tau(p)\cdot\tau(p)^2\cdot\tau(p)^3\dots$, where the dots stand for concatenation of words, and $p$ is a prefix. For such a substitution map $\tau$ one can write an associated $r\times r$ transition matrix $A:=(a_{ij})$ for the number of occurrences of each symbol in a transformed word: let $a_{ij}$ denote the number of occurrences of the letter $s_i$ in the word $\tau(s_j)$. Then, if in the vector $X$, $X_i$ is the number of occurrences of $s_i$ in a given word $w$, the coordinates of $AX$ give the number of occurrences of each $s_i$ in the word $\tau(w)$, so that, for $w=p$, the vector $A^k X$ gives the distribution of letters in $\tau(p^k)$; the length of $\tau(p^k)$ is the sum of coordinates of $A^{k-1} X$. This way the various asymptotics of $a_n$ can be easily related to the spectrum of $A$; in fact one use the standard techniques of finite Markov chains introducing a suitable Markov chain deduced from the map $\tau$. For instance, both quoted cases can be treated this way. -The case when $a_{n+1}-a_n$ can take the value $1$ frequently, so that $b_{n+1}-b_n$ assumes more values, seems less clear to me, but it may possibly be studied in the same lines.<|endoftext|> -TITLE: How can we show that $E\:\hat\otimes_\pi\:E$ is isomorphic to a subspace of $\left(E'\:\hat\otimes_\pi\:E'\right)'$? -QUESTION [5 upvotes]: Let - -$E$ be a $\mathbb R$-Banach space -$E\:\hat\otimes_\pi\:E$ denote the projective tensor product - - -How can we show that $E\:\hat\otimes_\pi\:E$ is isomorphic to a subspace of $\left(E'\:\hat\otimes_\pi\:E'\right)'$? - -Clearly, if $\mathfrak B(E'\times E')$ denotes the space of bounded bilinear forms on $E'\times E'$, then $\mathfrak B(E'\times E')$ is isometrically isomorphic to $\left(E'\:\hat\otimes_\pi\:E'\right)'$. -So, we could conclude, if we would be able to show that $E\:\hat\otimes_\pi\:E$ can be embedded into $\mathfrak B(E'\times E')$. -Actually, I know that $E\otimes E$ (the algebraic tensor product) can be embedded into $\mathcal B(E^\ast\times E^\ast)$ (the set of bilinear forms on the cartesian product of the algebraic dual space $E^\ast$ with itself). -A canonical choice for this embedding would be $$\sum_{i=1}^nx_i\otimes y_i\mapsto\left((\varphi,\psi)\mapsto\sum_{i=1}^n\varphi(x_i)\psi(y_i)\right)\tag1\;.$$ If $\iota$ denotes this embedding, then it's easy to see that $\iota$ is a bounded linear operator from $E\otimes_\pi E$ (the algebraic tensor product $E\otimes E$ equipped with the projective norm) to $\mathfrak B(E'\times E')$ and hence admits a unique extension to a bounded linear operator $\overline\iota$ from $E\:\hat\otimes_\pi\:E$ to $\mathfrak B(E'\times E')$. - -If this approach is sensible at all, the only thing I need to conclude is the injectivity of $\overline\iota$. How can we show that? - -REPLY [2 votes]: Let $P$ be one of the Banach spaces constructed by Theorem 3.2 in Pisier, "Counterexamples to a conjecture of Grothendieck". That is, $P \widehat\otimes P = P \check\otimes P$. Now, $P\check\otimes P$ embeds isometrically into $B(P',P) \subseteq B(P',P'') = (P' \widehat\otimes P')'$, and this embedding is exactly that described in the OP. So for $P$, we obtain a positive answer. -As Jochen Wengenroth says, if $E$ has the approximation property then $E\widehat\otimes E \rightarrow B(E',E'')$ is injective. It seems to me that for "nice" spaces this is unlikely to be bounded below.<|endoftext|> -TITLE: Another $2 \times 2$ matrix question -QUESTION [11 upvotes]: This question is similar to this previous one but I think it is harder. -Let $X$, $Y$, $Z$, and $W$ be $2\times 2$ Hermitian matrices. Can we always find $\theta,\phi \in [0,\pi/2]$ and $2\times 2$ unitaries $U$ and $V$ such that -$$SUXU^*S + CVYV^*C$$ and $$SUZU^*S + CVWV^*C,$$ -are both scalar multiples of the identity matrix, -where $S = {\rm diag}(\sin \theta, \sin \phi)$ and $C = {\rm diag}(\cos \theta, \cos\phi)$? -I'm guessing this is true. Maybe there is a homotopy reason as in the previous question. I think an answer in either direction would likely lead to a solution to this other previous question. -Edit: another way to ask this is, given $X$, $Y$, $Z$, and $W$, can we find $2\times 2$ matrices $A$ and $B$ such that $$AA^* + BB^*$$ $$AXA^* + BYB^*$$ $$AZA^* + BWB^*$$ are all scalar multiples of the identity (and the first one is not zero). -Edit 2: another reformulation. Let $\alpha$ and $\beta$ be the column vectors of $A^*$ and $\gamma$ and $\delta$ the column vectors of $B^*$ in the first edit. Then by looking at matrix entries of the expressions given there we see that the conditions become $$\langle X\alpha, \alpha\rangle + \langle Y\gamma, \gamma\rangle = \langle X\beta, \beta\rangle + \langle Y\delta, \delta\rangle$$ -and $$\langle X\alpha,\beta\rangle + \langle Y\gamma,\delta\rangle = 0,$$ -plus similar conditions with $Z, W$ or $I_2,I_2$ in place of $X,Y$. The problem is to find $\alpha,\beta,\gamma,\delta \in \mathbb{C}^2$, not all zero, satisfying these equations. - -REPLY [5 votes]: First let $P$ be the space of nonzero quadruples $(X,Y,Z,W)$ of Hermitian matrices, and let $Q$ be the set of sextuples $(X,Y,Z,W,A,B)$ where $(X,Y,Z,W)\in P$ and $(A,B)\neq(0,0)$ and the three matrices mentioned are multiples of the identity. There is a projection map $\pi\colon Q\to P$, and we want to know whether this is surjective. -Let $P_1$ be the subspace of $P$ where $\|X\|^2+\|Y\|^2+\|Z\|^2+\|W\|^2=1$, and let $Q_1$ be the subspace of $Q$ where $(X,Y,Z,W)\in P_1$ and $AA^*+BB^*=I$. Because our equations are appropriately homogeneous, it is equivalent to ask whether the projection $\pi\colon Q_1\to P_1$ is surjective. Note that $P_1$ and $Q_1$ are compact, so $\pi(Q_1)$ is closed in $P_1$. Thus, if there are any points not in $\pi(Q_1)$, then they form a nonempty open set, and we can hope to find a counterexample by random search. -For fixed $(X,Y,Z,W)\in P$ it works out that we are trying to solve 12 equations in 16 variables, so it is reasonable to hope for a solution. I generated 1000 random examples and in each case set four of the 16 variables arbitrarily and solved numerically for the remaining 12, with success in every case. This convinces me that the answer is probably positive. -Maple code is as follows: -with(LinearAlgebra): -r := () -> rand(-100..100)()/10.; -rand_hermitian := proc() - local a,b,c,d; - a := r(); b := r(); c := r(); d := r(); - <,>; -end: - -dagger := (M) -> map(conjugate,Transpose(M)); - -assume(a1::real,a2::real,a3::real,a4::real, - a5::real,a6::real,a7::real,a8::real, - b1::real,b2::real,b3::real,b4::real, - b5::real,b6::real,b7::real,b8::real); - -A := <,>; -B := <,>; - -for i from 1 to 1000 do - if modp(i,10) = 0 then print(i); fi; - W := rand_hermitian(); - X := rand_hermitian(); - Y := rand_hermitian(); - Z := rand_hermitian(); - MM := {A.dagger(A) + B.dagger(B), - A.X.dagger(A) + B.Y.dagger(B), - A.Z.dagger(A) + B.W.dagger(B)}: - MM := map(M -> op(expand([M[1,2],M[2,1],M[1,1]-M[2,2]])),MM): - MM := map(z -> op([Re(z),Im(z)]),MM) minus {0}: - sol := fsolve({a1-1,a8,b2,b8,op(MM)}); - if not(type(subs(sol,[a1,a2,a3,a4,a5,a6,a7,a8,b1,b2,b3,b4,b5,b6,b7,b8]) - [numeric$16])) then - print("problem"); - break; - fi; -od:<|endoftext|> -TITLE: What is the status of the assertion "There are arbitrarily large cardinals with the tree property"? -QUESTION [7 upvotes]: Of course, if you want your cardinals with the tree property to be strongly inaccessible, then you're asking about weakly compact cardinals. But what if you don't want them to be strongly inaccessible? -I see it's consistent that no successor cardinal has the tree property. I suspect this means it's consistent that no regular cardinal has the tree property? -As a non-set-theorist, I'm not sure whether this question is trivial, or hopeless! - -REPLY [2 votes]: Paul B. Larson mentions a partial result in A Brief History of Determinacy, page 48. Sadly he doesn't give a precise source. - -Foreman, Magidor and Schindler showed that - if there exist infinitely many cardinals $δ$ above the continuum such that the - tree property holds at $δ$ and at $δ^+$, then [the axiom of projective determinacy] holds. The hypothesis of this statement had been shown consistent relative to the existence of infinitely many supercompact cardinals by James Cummings and Foreman. It is not known whether the conclusion can be strengthened to $\text{AD}^{L(\mathbb{R})}$. - -So the tree property for all successors >$\omega_1$ has to be a very strong statement, at least as strong as the existence of infinitely many Woodin cardinals. This is the best partial/related result I know of. -EDIT: See this paper, which has the following abstract: - -Starting from a strong cardinal and a measurable cardinal above it, we construct a model of ZFC, in which, for every singular cardinal $δ$, $δ$ is strong limit, $2^δ = δ^{+++}$, and the tree property holds at $δ^{++}$. It answers a question of Friedman, Honzik and Stejskalova. We also produce, relative to the existence of a strong cardinal and two measurable cardinals above it, a model of ZFC in which the tree property holds at all regular even cardinals. The result answers questions of Friedman, Halilovic and Honzik . - -Which answers the question from the title, i.e. it is possible to have arbitrarily large cardinals with the tree property (relatively to very large cardinals).<|endoftext|> -TITLE: CLT for Bernoulli RV with negative correlation -QUESTION [5 upvotes]: Suppose $X_1,X_2,...$ are Bernoulli random variables with $P(X_i=1)=p_i$ and $X_i$ have negative correlation. Is there a CLT in this case, i.e. does $\frac{Z_n-(\Sigma^n_{i=1}p_i)}{\sqrt{n}}$ converge to a Gaussian in distribution? -And if we add the assumption that $\underset{n \longrightarrow \infty}{lim} \frac{\Sigma^n_{i=1}p_i}{n}=p\in(0,1)$, is it true then? - -REPLY [2 votes]: No, the CLT need not hold under these assumptions. Consider the following example: take $p=1/2$ for definiteness, and divide the (discrete) time into intervals $I_1=[1,2]$, $I_n=(2^{n-1}, 2^n]$, $n\geq 2$. On each interval, let $(X_k, k\in I_n)$ be the i.i.d. Bernoulli($1/2$) conditioned on $\sum_{k\in I_n}X_k=|I_n|/2$ (think about placing $|I_n|/2$ balls into $|I_n|$ urns at random); clearly, $(X_i,X_j)$ are negatively correlated when $i,j\in I_n$ for some $n$. Then the CLT doesn't hold since the sum becomes deterministic from time to time. -The correlations are not strictly negative though (because the $X$'s are independent if belong to different intervals), but you can probably make them so by some "small" perturbation.<|endoftext|> -TITLE: Convexity of distance-to-boundary function -QUESTION [6 upvotes]: Let $\Omega\subset\mathbb{R}^{n}$ be an open, -bounded convex domain. Denote $d_{\Omega}:\Omega\rightarrow\mathbb{R}$ -the distance-to-boundary function, that is, -$$ -d_{\Omega}\left(x\right):=\inf\left\{ \left|x-y\right|:y\in\partial\Omega\right\} . -$$ -It is well-known that if $\partial\Omega$ is smooth then $d_{\Omega}$ -is convex in a neighborhood of $\partial\Omega$ and $\left|\nabla d_{\Omega}\right|=1$ -in this neighborhood. -I would like to reduce the smoothness assumption of the boundary. -My question (suggestion) is: if $\partial\Omega$ is only Lipschitz, is it true -that $d_{\Omega}$ is convex in a neighborhood of $\partial\Omega$ -and $\left|\nabla d_{\Omega}\right|=1$ almost everywhere in this -neighborhood? -Any reference is welcome. - -REPLY [2 votes]: Here is another answer to the convexity (or more precisely concavity) part of the question, which I think is even more simple than the one Anton gave: -Let $p\in\Omega$, and $q$ be one of its closest points on $\partial\Omega$. Since $\Omega$ is convex, there exists a support hyperplane $H$ of $\Omega$ which passes through $q$. Furthermore $H$ is orthogonal to $pq$, since it supports the sphere $S_p$ centered at $p$ and passing through $q$. So $d_{\partial\Omega}(p)=d_H(p)$. Clearly $d_H(p)$ is not bigger than $d_{H'}(p)$ for any other support plane $H'$ of $\Omega$, because all these planes lie outside $S_p$. So -$d_{\partial\Omega}$ is the infimum of the distance functions to support hyperplanes of $\Omega$. These functions are linear and therefore concave. So $d_{\partial\Omega}$ is concave.<|endoftext|> -TITLE: Is there an efficient algorithm to check whether two matrices are the same up to row and column permutations? -QUESTION [7 upvotes]: Define $\mathcal M_n$ as the set of all $n\times n$ matrices with each entry either 1 or $x$. Two such matrices are equivalent iff they can be obtained from each other by swapping pairs of rows and columns (and possibly reflection). -For each line or column of a matrix $M\in\mathcal M_n$, we define its weight as the number of $x$'s occurring in it. The signature of $M$ is the set of the two vectors of row weights and column weights, wlog both in non-increasing order. (And wlog we'll reorder the rows and columns accordingly.) -We can further define two "incidence matrices", $R=R(M)$ and $C=C(M)$, where $r_{ij}$ is the number of $x$'s which are at the same position in both row $i$ and row $j$, likewise $c_{ij}$ for columns. For an equivalence class of a matrix in $\mathcal M_n$, those are well-defined up to simultaneous permutations of rows and columns (i.e. maintaining the entries on the main diagonal, which are the two sets of the signature). -By definition, equivalent matrices have the same incidence structure. I think the converse doesn't hold, at least not for $n$ big enough. For instance I am thinking of adjacency matrices of strongly regular graphs, which are not necessarily unique for a given parameter set. Now the strong regularity is not exactly captured by the incidence matrices, and those adjacency matrices are symmetric, so this is only a heuristic argument. For directed graphs with loops allowed, the equivalence classes w.r.t. swapping two lines or columns in the adjacency matrix seem rather hard to characterize. So probably graph theory won't be of much help for a rigorous argument. - -What would be a counterexample of the smallest size, i.e. two matrices which have the same incidence structure but are not equivalent? - -This question has occurred when I found a second extremal matrix for n=7 here. Both have the same incidence structure (any two lines have two $x$'s in common, and so have any two columns) and, up to sign, the same determinant, but very different (visible) symmetries. To reproduce them here, -$M_1=\begin{pmatrix} -x&1&1&\color{blue}x&1&x&x\\ -1&x&1&\color{blue}x&x&1&x\\ -1&1&x&\color{blue}x&x&x&1\\ -\color{blue}x&\color{blue}x&\color{blue}x&\color{blue}x&\color{blue}1&\color{blue}1&\color{blue}1\\ -1&x&x&\color{blue}1&1&x&x\\ -x&1&x&\color{blue}1&x&1&x\\ -x&x&1&\color{blue}1&x&x&1\\ -\end{pmatrix},\qquad M_2=\begin{pmatrix} -x&x&x&1&x&1&1\\ -1&x&x&x&1&x&1\\ -1&1&x&x&x&1&x\\ -x&1&1&x&x&x&1\\ -1&x&1&1&x&x&x\\ -x&1&x&1&1&x&x\\ -x&x&1&x&1&1&x -\end{pmatrix}.$ -The blue row and column of $M_1$ are not distinguishable from the others, as it is easy to obtain an identical matrix with the blue row and column elsewhere by certain swappings of lines and columns. - -But is there a somewhat efficient algorithm allowing to check whether a pair like $M_1$ and $M_2$ are equivalent? - -REPLY [6 votes]: Regarding the first question: there is a counterexample of size $91$. -There are four non-isomorphic finite projective planes of order $9$: https://doi.org/10.1016/0012-365X(91)90280-F, http://oeis.org/A001231. -Take the incidence matrices of non-isomorphic finite projective planes of order $9$, and replace each $1$ by $x$ and each $0$ by $1$. These matrices have $91$ rows and $91$ columns, and their incidence structure has all $r_{ij}$ and $c_{ij}$ equal to $1$.<|endoftext|> -TITLE: Can $\Delta_1$ (or $\Delta_0$)-elementary embeddings from $V$ to $V$ exist? -QUESTION [11 upvotes]: Suppose V is a model of Godel-Berney's set theory with the axiom of choice. A well-known result of Kunen says that there can be no elementary embedding $V$ to itself. This result further implies that there can be embedding from $V$ to $V$ which is $\Sigma_1$-elementary. -Is it consistent (with say Godel-Berney's set theory with the axiom of choice) that there is a $\Delta_1$-elementary embedding from $V$ to $V$? How about a $\Delta_0$-elementary embedding? - -REPLY [13 votes]: The answer is yes, the existence of such embeddings is equiconsistent over ZFC with a measurable cardinal. -If $\kappa$ is measurable, then there is a fully elementary embedding $j:V\to M$ into a transitive class $M$ with critical point $\kappa$. By composing this map with the inclusion $M\subset V$, we may view $j:V\to V$. The observation to make is that this remains $\Delta_0$-elementary, since every transitive class is $\Delta_0$-elementary in the universe $M\prec_{\Sigma_0}V$, and so we may view $j:V\to V$ as a composition of the original elementary embedding with this $\Delta_0$-elementary inclusion. -In fact, the embedding $j:V\to V$ is $\Delta_1$-elementary, since $M$ is actually $\Delta_1$-elementary in $V$, as $\Sigma_1$ assertions go up and $\Pi_1$ assertions go down. Since $M$ and $V$ are elementarily equivalent, they agree on the equivalence or non-equivalence of any two formulas. -(Note that there is a subtle issue about what exactly one means by $\Delta_1$-elementary, since there is no such thing as a $\Delta_1$-formula, unlike $\Delta_0$. For example, in the argument, it was important that $M$ and $V$ agreed on the equivalence of the $\Sigma_1$ nad $\Pi_1$ formulas in question. It can happen, however, that two models of ZFC do not necessarily agree like this, and there is a difference between $\Delta_1$ and provably-$\Delta_1$.) -Conversely, let us show that measurable cardinals are required. If $j:V\to V$ is a nontrivial $\Delta_0$-elementary embedding, then I claim, first, that there is a critical point, that is, a first ordinal that is moved. Notice that $j$ must take ordinals to ordinals. Assume toward contradiction that $j$ fixes all ordinals. Let $a$ be an $\in$-minimal set with $j(a)\neq a$. So $a\subset j(a)$ since $j(x)=x$ for all $x\in a$. By the axiom of choice, we may well-order $a$ with a relation $\lhd$ in some order-type $\gamma=|a|$. It follows that $j(\lhd)$ well-orders $j(a)$ in order-type $j(\gamma)=\gamma$. Furthermore, the $\beta^{th}$ element of $a$ gets mapped to the $j(\beta)=\beta^{th}$ element of $j(a)$ with respect to $j(\lhd)$. So $j$ is onto $j(a)$, and therefore $j(a)=j"a=a$, a contradiction. So we have shown that there must be a critical point $\kappa$, the least ordinal with $\kappa -TITLE: Is the derived category of $A$-dg-modules as a dg-category coincide with the ordinary definition of derived category? -QUESTION [11 upvotes]: Let $A$ be a unital dg-algebra over a base field $k$. We consider the category of (unbounded) right $A$-dg-modules with morphisms closed degree $0$ maps. We denote this category by dg-mod-$A$. We could consider homotopic morphisms and get the homotopy category [dg-mod-$A$]. Moreover we could invert quasi-isomorphisms in [dg-mod-$A$] and obtain its derived category. Let us denote this category by $D(A)$. -On the other hand, we could consider the dg-category DG-MOD-$A$: its objects are right $A$-dg-modules and its morphisms are chain complexes of maps between dg-modules. -We notice that there is a general construction of the derived category of a dg-category $\mathcal{C}$: We first consider the dg-category of right modules over $\mathcal{C}$, which is the dg-category of contravariant dg-functors from $\mathcal{C}$ to Ch$(k)$. Let us denote this dg-category by DGM-$\mathcal{C}$. We also consider the full dg-subcategory Acycl-$\mathcal{C}$, which consists of dg-functors $\mathcal{F}:\mathcal{C}\to \text{Ch}(k)$ such that $\mathcal{F}(c)$ is acyclic for any object $c\in \mathcal{C}$. Then the derived category of $\mathcal{C}$ is given by the Verdier quotient -$$ -[\text{DGM}-\mathcal{C}]/[\text{Acycl}-\mathcal{C}] -$$ -where $[\cdot]$ is the homotopy category of dg-categories. We denote the derived category of $\mathcal{C}$ by $D(\mathcal{C})$. We could show that the Yoneda functor induces a fully faithful functor $[\mathcal{C}]\to D(\mathcal{C})$. -Now go back to the dg-algebra $A$ and the dg-category DG-MOD-$A$. - - -My question is: is the derived category $D(A)$ equivalent to the derived category $D(\text{DG-MOD}-A)$? Why? - -REPLY [3 votes]: I think it is not good point of view to think that $A \mapsto H^0(Mod-A)[Qis^{-1}]$ is canonical construction for dg-algebras and $\mathcal{C} \mapsto H^0(Mod-\mathcal{C})/H^0(Acyc-\mathcal{C})$ is canonical construction for dg-categories. In fact, both constructions can be done for algebras and for categories in the same way. Third very useful model for $D(\mathcal{A})$ is $H^0(SF-\mathcal{A})$ - homotopy category of all semi-free dg-modules-$\mathcal{A}$ which just strictly full triangulated subcategory of $H^0(Mod-\mathcal{A})$. So I see there are 2 questions: - -Let $A$ is dg-algebra, why $D(A)$ and $D(Mod-A)$ canonically triangulated equivalent? -Let $\mathcal{C}$ is dg-category, why $H^0(Mod-\mathcal{C})[Qis^{-1}]$ and $H^0(Mod-\mathcal{C})/H^0(Acyc-\mathcal{C})$ canonically triangulated equivalent? - -Both questions are quite classical. -Let $F : A \to Mod-A$ canonical embedding of $A$ (which we consider as category with one object $*$ and $End(*) = A$) to $Mod-A$ as rank 1 free module-$A$. -First question is composition of following facts - -Let $SF-\mathcal{A} \subset Mod-\mathcal{A}$ full dg-category which consists all semi-free dg-modules. Then induced localization functor $H^0(SF-\mathcal{A}) \to D(A)$ is quasi-equivalence. (Every dg-module-$\mathcal{A}$ have semi-free resolution). Keller Deriving DG-categories 3.1 -Let $G : \mathcal{B} \to \mathcal{C}$ be full embedding of dg-categories. And let $G^* : SF-\mathcal{B} \to SF-\mathcal{C}$ be the extension dg-functor. Then the induced functor $H^0(G^*) : D(\mathcal{B}) \to D(\mathcal{C})$ is fully faithful. If, in addition, the category $H^0(\mathcal{C})$ is classically generated by $Ob \mathcal{B}$ then $H^0(G^*)$ is an equivalence. Proposition 1.15 in Luntz Orlov, "UNIQUENESS OF ENHANCEMENT FOR TRIANGULATED CATEGORIES" -$F$ is full embedding of classical generator of $SF-\mathcal{A}$. It is quite obvious, you can construct any semi-free modules from free using cone operation. Argument of such type must be also written in Keller Deriving DG-categories - -Your second question is more or less also implies from theorem 3 in Keller Deriving DG-categories.<|endoftext|> -TITLE: Convergence of sequence of inverse functions -QUESTION [5 upvotes]: I have a question: Consider two sequences of continuous, bijective functions $f_n$ and $g_n$ mapping $\mathbb{R}\to\mathbb{R}$. I know that for every compact $K\in\mathbb{R}$ that -$$\lim_{n\to\infty}\int_K |f_n(x)-g_n(x)| \mathrm d x =0.$$ -Furthermore the limit $f$ is surjective and continuous and -$$\lim_{n\to\infty}\int_K |f_n(x)-f(x)| \mathrm d x =0,$$ -Now, does this also hold for the inverse functions $f^{-1}_n,g^{-1}_n$, i.e. -$$\lim_{n\to\infty}\int_K |f_n^{-1}(x)-g_n^{-1}(x)| \mathrm d x =0?$$ -If not, is there at least a weaker form of convergence? Thanks for the answer :) - -REPLY [6 votes]: The statement is in fact true if we assume that $f_n$ and $g_n$ are bijective and continuous functions form $\mathbb{R}$ to $\mathbb{R}$, hence monotone, say increasing, and converge point-wise to a surjective function $f$ (hence monotone and continuous). -Indeed let $K\subset [a,b]\subset\mathbb{R}$. Since $f$ is surjective, there are $A\in\mathbb{R}$ and $B\in\mathbb{R}$ such that $f(A)b$ and by the point-wise convergence, there holds $f_n(A)b$, $g_n(B)>b$ eventually as $n\to\infty$. Now -$\int_a^b|f_n^{-1}(s)-g_n^{-1}(s)|ds$ is the two-dimensional measure of the set -$$ \big\{(s,t)\in[a,b]\times\mathbb{R} : \ f_n^{-1}(s)\wedge g_n^{-1}(s)\le t\le f_n^{-1}(s)\vee g_n^{-1}(s)\big\}$$ -which is eventually included in the set -$$ \big\{(s,t)\in\mathbb{R} \times[A,B] : \ f_n(t)\wedge g_n(t)\le s\le f_n(t)\vee g_n(t)\big\},$$ -whose Lebesgue measure is $\int_A^B|f_n(t)-g_n(t)|dt$. Since by assumption the latter converges to $0$, so does $\int_K|f_n^{-1}(s)-g_n^{-1}(s)|ds$. -$$*$$ -Also note that the function $f(x)=\lim f_n(x)$ is increasing, and being surjective, it is also continuous. Also, in these hypotheses $f_n$ and $g_n$ converge uniformly on compact sets, in fact, it would sufficient to assume $f_n\to f$ and $g_n\to f$ in $L^1_{\it loc}$.<|endoftext|> -TITLE: Proof assistant, Cura te ipsum -QUESTION [19 upvotes]: By a bona fide bug in a proof assistant I mean a software flaw which is serious enough to create a possibility of "proving" something which is actually false. This is not a purely academic problem https://cstheory.stackexchange.com/questions/37299/has-a-proof-checker-bug-ever-invalidated-a-major-proof. We mathematicians do not have a real reason to care about it for now, but I am convinced that this problem has a good potential to grow to scary proportions once a large scale program of formalization of mathematics is attempted. -Modern mathematics is more hierarchical then common software, and in a very nontrivial way. Because of this, a cleanup after fixing a bug may become rather painful. -I am curious about methods to avoid this which are based on mathematics (as opposed to some sort of management). One method I know about is verifying the code of a proof assistant using this same proof assistant. Strictly speaking, this is theoretically impossible due to the second Goedel theorem, but the way around it is to make the system stronger (for example, by adding a new axiom). There is a paper by J. Harrison about how it may be done for HOL light, J. Harrison, "Towards self-verification of HOL Light", Automated Reasoning, 2006 - Springer. More recent works in this direction are Myreen, Owens, Kumar, "Steps Towards Verified Implementations of HOL Light" , and Anand, Rahli, "Towards a Formally Verified Proof Assistant", ITP 2014: Interactive Theorem Proving. -Question 1. How far did it go? (All the above papers have the word ``towards'' in the title.) -Question 2. What can be said about formal verification of a proof assistant which may be of interest to mathematicians? (For example, are there nontrivial alternatives to, or variations of, self-verification?) -Remark. The question whether this or similar strategy actually makes a proof assistant perfectly bug-free (in the above sense) would be more appropriate on Computer Science SE, but I would not mind if anyone touches this topic. - -REPLY [10 votes]: See this dissertation by Ramana Kumar (Cambridge 2015). -http://www.sigplan.org/Awards/Dissertation/2017_kumar.pdf - -I present a proof of consistency of higher-order logic (HOL), in - particular for the entire inference system implemented by the kernel - of the HOL Light theorem prover [24]. The main lemma is a proof of - soundness against a new specification of the semantics of HOL. This - formalisation extends work by Harrison [23] towards self-verification - of HOL Light. Using the proof-grounded compilation technique, I show - how to produce a concrete implementation of a proof checker for HOL - based on the verified inference system. The result is a theorem prover - with very strong guarantees of correctness, and, as I will sketch, the - rare potential to verify its own concrete implementation in machine - code. - -On a different note regarding HOL-varieties, there is also HOL Zero. -http://proof-technologies.com/holzero/<|endoftext|> -TITLE: Why are they called "screen" distributions? -QUESTION [6 upvotes]: If $V$ is a vector space and $g$ is a symmetric degenerate bilinear form on $V$, every complementary subspace to the radical ${\rm rad}(V)$ is called a "screen subspace" of $V$: we have an orthogonal direct sum $V = {\rm rad}(V) \oplus SV$. -Likewise, if $M$ is a smooth manifold and $g$ is a symmetric and degenerate $(0,2)$-tensor field on $M$, any complementary distribution to ${\rm rad}(TM)$ is called a "screen distribution" of $M$: we have the orthogonal Whitney sum $TM = {\rm rad}(TM) \oplus S(TM)$. -These definitions are given by Bejancu and Duggal in several of their books and works. But I can't see any motivation for the name "screen". - -REPLY [3 votes]: Naturally, one should consider the quotient space $V/{\rm rad}(V)$ which consists of ${\rm rad}(V)$-rays (affine spaces parallel to ${\rm rad}(V)$). A screen space $SV$ intersects a ray in exactly one point, like a screen intersects the rays from a projector. Similarly on a manifold. - -REPLY [3 votes]: The word "screen" refers to lightlike dimensional reductions, a worldline in $(d+1)+1$ dimensional space-time is projected onto the screen $x^{d+1}=0$ and the projected $d+1$ dimensional curve is parameterized by Galilean time $t$. The projection is called the "shadow" on the "screen". For an introduction, see Classical aspects of lightlike dimensional reduction. - -A particular feature of the dimensional reduction is that relativistic dynamics projects to non-relativistic physics on the screen. The history goes back to Eisenhart's Dynamical Trajectories and Geodesics (1929), who discovered the equivalence of relativistic Poincaré invariance and nonrelativistic Galilei invariance in a lightlike reduced spacetime with one dimension less.<|endoftext|> -TITLE: Can the Turing degrees be linearly ordered? -QUESTION [9 upvotes]: Assuming the axiom of choice, every set can be linearly (indeed, well-) ordered. However, without choice this can fail, as witnessed most drastically by the consistency of amorphous sets. More reasonable failures of choice - e.g. via determinacy - tend to yield more reasonable behavior, but can still exclude certain structures from occurring. -I'm interested in what the situation with the Turing degrees can be, if we assume AD. That is: - -In ZF+DC+AD (plus whatever else is needed to get a good answer), can the set $\mathcal{D}$ of Turing degrees be linearly ordered? - -(Note that of course there is no demand that this ordering behave nicely with respect to $\le_T$, in any sense.) -I strongly suspect the answer is "no," but I don't immediately see how to prove it. -In fact, as far as I can tell very few sets of Turing degrees admit "definable" linear orderings - specifically, every example I can find is the image of some injective partial function $f$: $\subseteq2^\omega\rightarrow\mathcal{D}$. For example, the standard construction of a continuum-sized antichain of Turing degrees consists of building a continuous function $g:2^\omega\rightarrow 2^\omega$ such that $x\not=y\implies deg(g(x))\not=deg(g(y))$; we then push the lexicographic order on $2^\omega$ through the map $f=deg\circ g$. This raises the following question, especially assuming a negative answer to (1): - -Is there a reasonable extension of ZF which proves that every orderable set of Turing degrees is the image of some injective function from a subset of $2^\omega$? - - -As usual, we get closely related questions if we work in ZFC instead and restrict attention to (say) Borel sets and orders. I would also be interested in answers to these questions. - -REPLY [13 votes]: You can't linearly order the Vitali ($\mathcal{P}(\omega)/\mathrm{Fin}$) degrees if every set of reals has the property of Baire, since you can't even choose between complementary degrees. The set of $x$ which are chosen can't be meager or somewhere comeager, since below any initial segment you can find a pair of complements which are both in any given comeager set. -There's a continuous, Vitali-invariant map $c$ that sends mod-finite different subsets of $\omega$ to mutually Cohen-generic reals over $L$ (assuming $\mathcal{P}(\omega)^{L}$ is countable; this is overkill in any case), giving you an embedding of the Vitali degrees into the Turing degrees. Such a map can be induced by a pair of functions $f, g \colon \omega \to \mathrm{Fin}$ so that for all $x \subseteq \omega$, $c(x) = \bigcup\{ f(n) : n \in x\} \cup \bigcup \{ g(n) : n \not\in x\}$. This should answer the first question.<|endoftext|> -TITLE: Almost convex combinations in $\mathbb R^n$ -QUESTION [8 upvotes]: Working on some problems in the $C_p$-theory I discovered the following simple but amazing -Fact. For any subset $A\subset \mathbb R^n$, non-zero vector $a\in \bar A\subset\mathbb R^n$ and $\varepsilon>0$ there are points $a_1,\dots,a_n\in A$ and real numbers $t_1,\dots,t_n$ such that $a=\sum_{i=1}^nt_ia_i$ and $$1-\varepsilon<\sum_{i=1}^n t_i\le\sum_{i=1}^n|t_i|<1+\varepsilon.$$ -It seems that this fact (which can be easily proved by induction on $n$) is too elementary to be unknown. I would appreciate any reference (to a paper or textbook). Thank you. - -REPLY [3 votes]: I do not know the reference, but it looks that even more may be achieved: $t_1,\dots,t_{n-1}$ may be chosen close to 0 and $t_n$ close to 1. Indeed, if the span of $a\cup A$ is spanned by linearly independent vectors $a, a_1,\dots,a_k$, $k\leqslant n-1$, then choose $a_{n} $ close to $a$, we get $a_{n}=c a+c_1a_1+\dots+c_ka_k $, where $c$ is close to 1 and all $c_i$ close to 0, dividing by $c$ this gives a required representation for $a$.<|endoftext|> -TITLE: Two integral representations for $\zeta(3)$ from Zurab's integral and standard formulas for the gamma function -QUESTION [5 upvotes]: This morning I wrote with the help of a CAS, and integral representation for the Apéry's constant $\zeta(3)$ and some standard formulas two formulas involving this constant. I would like to know if these were in the litetature (my purpose is to know if if it possible to justify these and/or exploit to get more nice statements than mine, or some formula more interesting than mine using the way that I've used). -The integral representation wich I've combined is a formula that appears on [1], or see the first formula from the post in this site MathOverflow. I've combined with standard formulas that you can see in Google Books (currently this page is free view on Google Books) or your library as page 91 of [2] Section 6.3 The Gamma Function. - -Claim. It seems that the following formulas - $$\zeta(3)=\frac{1}{7\pi}\int_0^{\pi}x(\pi-x)\left(\psi\left(\frac{x+\pi}{2\pi}\right)-\psi\left(\frac{x}{2\pi}\right)\right)dx,\tag{1}$$ - where $\psi(s)$ denotes the digamma function, and - $$\zeta(3)=\frac{\pi^2}{7}\int_0^{\pi/4}\frac{\cos2x+\log\tan x}{\log^3\tan x}\cdot\frac{dx}{(\sin x)(\cos x)}\tag{2}$$ - are right. - -I found the formula $(1)$ in an attempt to combine Zurab's integral (the cited formula in MO's post) and Exercise 6.3.1 from [2] with the help of Wolfram Alpha online calculator; while than $(2)$ is a combination of Exercise 6.3.4 (from [2]) and Zurab's integral, combined with Exercise 6.3.2 (from [2]) and the help of mentioned online calculator. - -Questions. -1) Are known these formulas from the literature? Or are these a particular case of theorems from the literature? Then, please answer this Questions as a reference request and I try to search the articles and study those formulas. -2) In other case and with the purpose to improve my results... -... my attempt was to write explicitly the difference of the terms involving the digamma function using a series expansion, with the purpose to swap the integral sign and the infinite series that I've evoked, but my calculations are failed by some undeterminate term. I would like to know if it is feasible to expand such terms involving the digamma function and interchange the integral sign and the series, or well a different approach is better to rewrite $(1)$ as a formula for the Apéry's constant with a good mathematical meaning. -... my attempt was using the help of the mentioned online calculator (standard time of computation, my code and scarce knowledges of programming with this CAS) get the closed-form for $(2)$. But currently for me $(2)$ is a conjecture. Can you justify than $(2)$ is right? - Many thanks. - -Of course, if these formulas were not in the literature feel free to provide me general feedback about these formulas adding a comment. -References: -[1] Zurab Silagadze, Sums of Generalized Harmonic Series. For Kids from Five to Fifteen, RESONANCE, September 2015. -[2] Ram Murty, Problems in Analytic Number Theory, Second Edition, Sringer (2008). - -REPLY [7 votes]: For the second integral, with the substitutions $t=\tan x$, then $t=\exp u$ and taking advantage of the symmetry, it comes -\begin{align} -I_2&=\frac{\pi^2}{7}\int_0^{\pi/4}\frac{\cos2x+\log\tan x}{\log^3\tan x}\cdot\frac{dx}{(\sin x)(\cos x)}\\ -&=\frac{\pi^2}{7}\int_0^{1}\frac{\tfrac{1-t^2}{1+t^2}+\ln t}{\ln^3 t}\frac{dt}{t}\\ -&=\frac{\pi^2}{7}\int_{-\infty}^0 \frac{u-\tanh u}{u^3}\,du\\ -&=\frac{\pi^2}{14}\int_{-\infty}^\infty \frac{u-\tanh u}{u^3}\,du -\end{align} -This last integral can be calculated using the residue theorem, closing the contour by the upper half-circle. Poles are situated at $u=(2n+1)i\pi/2$ with $n\geq 0$: -\begin{align} -I_2&=2i\pi\frac{\pi^2}{14}\sum_{n\geq 0}\frac{-1}{\left( (2n+1)i\pi/2 \right)^3}\\ -&=\frac{8}{7}\sum_{n\geq 0}\frac{1}{ (2n+1)^3}\\ -&=\zeta(3) -\end{align} -For the first, integral, changing $x\to \pi -x$, one obtains: -\begin{align} -I_1&=\frac{1}{7\pi}\int_0^{\pi}x(\pi-x)\left(\psi\left(\frac{x+\pi}{2\pi}\right)-\psi\left(\frac{x}{2\pi}\right)\right)dx\\ -&=\frac{1}{7\pi}\int_0^{\pi}x(\pi-x)\left(\psi\left(1-\frac{x}{2\pi}\right)-\psi\left(\frac{1}{2}-\frac{x}{2\pi}\right)\right)dx -\end{align} -Summing both expressions, -\begin{equation} -2I_1=\frac{1}{7\pi}\int_0^{\pi}x(\pi-x)\left[\psi\left(\frac{1}{2}+\frac{x}{2\pi}\right)-\psi\left(\frac{x}{2\pi}\right)+\psi\left(1-\frac{x}{2\pi}\right)-\psi\left(\frac{1}{2}-\frac{x}{2\pi}\right)\right]dx -\end{equation} -Now, with the reflection formula for the digamma function -\begin{equation} -\psi\left(z\right)-\psi\left(1-z\right)=-\pi/\tan\left(\pi z\right) -\end{equation} -it can be written as -\begin{align} -I_1&=\frac{1}{14}\int_0^{\pi}x(\pi-x)\left[\tan\left( \frac{x}{2\pi} \right)+\frac{1}{\tan\left( \frac{x}{2\pi} \right)}\right]\,dx\\ -&=\frac{1}{7}\int_0^{\pi}\frac{x(\pi-x)}{\sin\left( \frac{x}{\pi} \right)}dx\\ -&=\frac{\pi^3}{7}\int_0^{1}\frac{t(1-t)}{\sin\left( t \right)}dt\\ -&=\zeta(3) -\end{align} -where we have used the representation given here.<|endoftext|> -TITLE: Quiver variety analogue of Grothendieck-Springer resolution -QUESTION [7 upvotes]: A standard example of Nakajima quiver varieties are type A Springer resolutions $\widetilde{\mathcal{N}} \to \mathcal{N}$. In the theory of Springer resolutions it is often beneficial to consider the full Grothendieck-Springer resolution, i.e. to work with the commutative diagram -$$\require{AMScd}\begin{CD} -\widetilde{\mathcal{N}} @>>> \widetilde{\mathfrak{g}} \\ -@VVV @VVV\\ -\mathcal{N} @>>> \mathfrak{g} -\end{CD}$$ -Is there a similar diagram where the left side is replaced by $\mathfrak{M} \to \mathfrak{M_0}$ for an arbitrary quiver variety. -To be clear, I do not expect that the analogue of the Grothendieck­-Springer resolution is necessarily a quiver variety itself, I merely ask whether to each (nice?) quiver variety one can associate a space that behaves similar to $\widetilde{\mathfrak{g}}$. I'm also intentionally vague about the word "similar", but for example a main feature of the above diagram is that it "lives over" $0 \hookrightarrow \mathfrak{h}//W \leftarrow \mathfrak{h}$. -Alternatively, why should I not expect such a thing to exist? - -REPLY [2 votes]: There are of course two moment maps to vary - the complex one and the real one. In most treatments of quiver varieties one fixes the complex level set to be zero and the real level set to a nonzero multiple of the identity, with the zero multiple giving the "quiver affine variety" $\mathfrak M_0$ (best not referred to as the "affine quiver variety" for fear of mis-association). -Very specifically, consider the $A_d$ quiver with only one framing vertex, attached to the first vertex, bearing dimension $n$. As you know, the various choices of dimension vector $(n_i)$ on the gauged vertices give $\mathfrak M=$ the various $d$-step (with possible repeats) flag varieties in $\mathbb C^n$. In this construction, one imposes the "preprojective" condition at each gauged vertex $v$, that the sum of all $2$-step paths $v\to w\to v$ is zero. It is fun to use this to derive that the invariant $X:$ frame $\to v_1\to$ frame satisfies $X^{d+1}=0$. Generically $X$ determines the point -in the quiver variety, but not always. -The difference now is to only ask that these sums be multiples $\varepsilon_i$ of the identity, instead of actually zero. Then $X$ satisfies instead $X(X-\varepsilon_1)(X-\varepsilon_1-\varepsilon_2)\cdots = 0$ if I recall correctly, and for generic $(\varepsilon_i)$ the invariant $X$ fully determines the point in the quiver variety, i.e. the quiver variety is an affine variety $GL(n)\big /\prod_i GL(n_i)$. Which is to say, varying the multiples $(\varepsilon_i)$ exactly recovers the Grothendieck-Springer family for $\mathfrak{gl}_n$, in the case that the dimensions on the gauge vertices are $n,n-1,n-2,\ldots,1$.<|endoftext|> -TITLE: Does a filtered A_N algebra give rise to a multiplicative spectral sequence? -QUESTION [7 upvotes]: The question is pretty much in the title. It is a classical fact that a filtered dga gives rise to a multiplicative spectral sequence. It is claimed in Remark 4.1 of https://arxiv.org/pdf/1410.6728.pdf that this generalizes to filtered $A_\infty$- algebras(see the page 27 of the same for this definition). One can define a notion of filtered $A_N$-algebras for $N \in \mathbb{N}^{\geq 0}$, $N \geq 3$ where $m_i$ are only defined for $i \leq N$ and the $A_{\infty}$ equations hold for $i \leq N$ (so the multiplication is still associative on the level of homology). -Question: Is the natural spectral sequence associated to a filtered $A_N$ algebra multiplicative? -I would guess the answer is yes, but could not find a reference for this fact and there seems to be some subtleties in the literature according to this earlier post: -Multiplicative structure on spectral sequence -Motivation: It seems like this might be an efficient way to prove that some spectral sequences that arise "in nature" are multiplicative, say by equipping the relevant (co) chain complex with an $A_3$ structure. - -REPLY [2 votes]: For the $d_r$-differentials to be derivations, i.e., to satisfy the Leibniz rule $d_r(x \cdot y) = d_r(x) \cdot y \pm x \cdot d_r(y)$ with $x \cdot y = m_2(x \otimes y)$, it is enough to have a filtered differential graded $A_2$-algebra. This follows from Sections 7 and 8 of Massey's 1954 paper "Products in Exact Couples", since the pairing $m_2 \colon A \otimes A \to A$ is a filtration-preserving chain map. -Assuming the Leibniz rule, if the pairing of $E_1$-terms (or $E_2$-terms) is associative and unital, then so are the induced pairings of $E_r$-terms for all greater $r$. -Aside: The "subtleties in the literature" concern exact couples that do not come from filtered chain complexes, e.g., those that arise from the homotopy groups of sequences of spectra. If a pairing of sequences is only defined in the stable homotopy category, then it is not generally clear that there will be a corresponding product in the spectral sequence.<|endoftext|> -TITLE: A primitive to $\mathrm{Vol}(x)\pm \mathrm{Vol}(y)$ on $(\mathbb{S}^n \times \mathbb{S}^n)\backslash \Delta$ -QUESTION [11 upvotes]: EDIT: I corrected some typos in my solution and wrote some details more precise (I was a bit sloppy first as I originally did not plan to discuss the full solution, but rather some ideas whether a solution can be constructed somewhat more naturally. E.g. using the Lie algebra and Lie group operations for $\mathbb{S}^3$.) -I am trying to solve the equation -$$ \mathrm{d}G(x,y) = \mathrm{Vol}(x)+(-1)^{n}\mathrm{Vol}(y):= H $$ -for $G\in \Omega^{n-1}\bigl((\mathbb{S}^n\times \mathbb{S}^n)\backslash \Delta=:M\bigr)$. Here $\mathrm{Vol}$ is the standard volume form on $\mathbb{S}^n$ and $\Delta$ the diagonal. The solution for $n=1$ is -$$ G(x,y)= - \alpha(x,y)\quad \text{for all }x,y\in \mathbb{S}^1 $$ -where $\alpha(x,y)$ is the counterclockwise angle from $x$ to $y$. I can find a solution for general $n\in \mathbb{N}$ as -$$ G = \int_{[0,1]} \psi^*H $$ -for a contraction $\psi: [0,1]\times N \rightarrow N$ of an open thickening $N$ of $M$ onto the antidiagonal $\overline{\Delta}:=\{(x,-x)\}$ (on $\overline{\Delta}$ holds namely $H=0$). Here I use a closed extension of $\mathrm{Vol}$ to $\mathbb{R}^{n+1}_{\neq 0}$. However, the solutions I obtain (in coordinates $x^i$ on $\mathbb{R}^{n+1}$) look too complicated and it is computationally hard to work with it further. -Does anybody have any idea how to solve this equation "nicely" e.g. by some natural construction of $G(x,y)$ similar to the case $n=1$? -I am mainly interested in the case $n=3$. Can the Lie group structure be used to construct a solution? -Thanks for any ideas! -Big picture: $H$ is a smooth integral kernel of the orthogonal projection to harmonic forms and $G$ is a (singular) integral kernel of the Green operator (:=the inverse of $\mathrm{d}$) -UPDATE1: As Robert Bryant suggests, the sign is indeed $(-1)^n$. -UPDATE2: Reading the nice answer of Robert Bryant, I just typed my complicated solution for a comparison. Perhaps it is the same except Robert using a clever notation: For every $n$ a solution is -$$ G(x,y)=\sum_{k=0}^{n-1} g_k(x\cdot y) \omega_k(x,y) $$ -where -$$ g_k(u) = \int_{0}^1 \frac{t^k(t-1)^{n-1-k}}{(2t(t-1)(1+u)+1)^{\frac{n+1}{2}}}\mathrm{dt} $$ -can be solved inductively and -$$ \omega_k(x,y) = \frac{1}{k!(n-1-k)!} \sum_{\sigma\in \mathbb{S}_{n+1}} (-1)^\sigma x_{\sigma_1}y_{\sigma_2}\mathrm{d}x_{\sigma_3}\ldots\mathrm{d}x_{\sigma_{2+k}}\mathrm{d}y_{\sigma_{3+k}}\ldots\mathrm{d}y_{\sigma_{n+1}}$$ -P.S. I am still tempted to say something like $G(x,y)((v_1,w_1),\ldots,(v_{n-1},w_{n-1}))$ is the angle between the hyperplane $\langle w_1,\ldots,w_{n-1}\rangle$ at $y$ and the parallel transport of the hyperplane $\langle v_1,\ldots,v_{n-1}\rangle$ from $x$ to $y$ along the shortest arc on the unique great circle through $x$ and $y$. This might be a complete nonsense as well... If my computations are correct then for $n=2$ holds -$$ G(e_1,\cos(\alpha)e_1 + \sin(\alpha)e_2) = \frac{\sin(\alpha)}{\cos(\alpha)-1}(\mathrm{d}y_3 - \mathrm{d} x_3 ). $$ - -REPLY [11 votes]: NB: I have edited my answer to remove the comment at the beginning about correcting the original sign of the OP in the formula for $H$, since the OP has now corrected the erroneous sign in the question. I have also added one remark at the end about the case when $n$ is even. -Now, one has -$$ -H = x^*\Omega + (-1)^n\,y^*\Omega -\tag1 -$$ -where $x:\mathbb{S}^n\times \mathbb{S}^n \to \mathbb{S}^n\subset \mathbb{R}^{n+1}$ and $y:\mathbb{S}^n\times \mathbb{S}^n \to \mathbb{S}^n \subset \mathbb{R}^{n+1}$ are the projections on the first and second factors and $\Omega$ is the $\mathrm{SO}(n{+}1)$-invariant volume form on $\mathbb{S}^n$ corresponding to a fixed orientation and inner product on $\mathbb{R}^{n+1}$. -Thus, to remove any sign ambiguities, fix an orientation and inner product on $\mathbb{R}^{n+1}$, and let $(u_0,\ldots,u_n)$ be an oriented orthonormal basis of $\mathbb{R}^{n+1}$, so that $U = u_0\wedge\cdots\wedge u_n$ is a basis of $\Lambda^{n+1}(\mathbb{R}^{n+1})$. Thus, for example, one has -$$ -\tfrac1{n!}\,x\wedge (\mathrm{d}x)^n = x^*\Omega\,U -\quad\text{and}\quad \tfrac1{n!}\,y\wedge (\mathrm{d}y)^n = y^*\Omega\,U -\tag2 -$$ -To save writing, for a form $\phi$ with values in $\Lambda^{n+1}(\mathbb{R}^{n+1})$, I will henceforth write $[\phi]$ for the scalar-valued differential form such that $\phi = [\phi]\, U$. Thus, $x^*\Omega = [\tfrac1{n!}\,x\wedge (\mathrm{d}x)^n]$, etc. -Using this notation, for example, one has, when $n=1$, that $\mathrm{d}G = H$ -where -$$ -G = [\,f(x{\cdot}y)\,x\wedge y\,],\tag3 -$$ -where $f:[-1,1)\to\mathbb{R}$ is the (unique) smooth function that satisfies -$$ -f(\cos t) = \frac{(\pi-t)}{\sin t}\quad\text{for}\quad 0 -TITLE: Paths in path component spaces -QUESTION [14 upvotes]: If $X$ is a topological space, one can naturally view the set $\pi_0(X)$ of path-components of $X$ as a quotient space of $X$ by collapsing each path-component to a point by a quotient map $q:X\to \pi_0(X)$. Of course, $q$ is a homeomorphism if and only if $X$ is totally path-disconnected. I'd like to know of criteria that ensure $\pi_0(X)$ is totally path-disconnected. -Interestingly, there are lots of spaces $X$ for which $\pi_0(X)$ is not totally path-disconnected. For instance, if $X=[0,1]\times [0,1]$ has the lexicographical ordering and is given the resulting order topology, then $q:X\to \pi_0(X)\cong [0,1]$ is just the projection onto the first coordinate. In fact, a little known gem due to Douglas Harris is that for every space $Y$, one can construct a paracompact Hausdorff space $S(Y)$ such that $\pi_0(S(Y))\cong Y$. -D. Harris, Every space is a path-component space, Pacific J. Math. 91 no. 1 (1980) 95-104. -However, the ordered square and spaces $S(Y)$ are not metrizable... -One must also be wary of separation axioms since if $X$ is the closed topologists sine curve, $\pi_0(X)$ is the two-point Sierpinski space, which is not $T_1$ and therefore not totally path-disconnected. -Question: If $X$ is a metric space and $\pi_0(X)$ is $T_1$, must $\pi_0(X)$ be totally path-disconnected? -Note: I'm interested in other variations of the question too where $X$ is separable or perhaps Polish and/or $\pi_0(X)$ is Hausdorff. -Update: I'm happy to say that this post motivated collaboration between Taras Banakh and myself. This MO-inspired work resulted in the following publication: -T.Banakh, J.Brazas, Realizing spaces as path-component spaces, Fundamenta Mathematicae 248 (2020) 79-89. DOI: 10.4064/fm529-12-2018 - -REPLY [8 votes]: There exists the following metric counterpart of the Harris result: -Theorem (Banakh, Vovk, Wojcik): Each metric space $X$ can be (canonically) identified with the space $\pi_0(\circledast(X))$ of path-components of some complete metric space $\circledast(X)$. -The space $\circledast(X)$ is a closed subspace of the complete oriented graph with vertices in the set $X$. -The above theorem follows from Theorem 7.3 and Proposition 7.4 of this paper published in Fund. Math. 212 (2011), 145-173. - -More generally, the Theorem holds for weakly first-countable spaces containing no countable connected subspaces. To formulate the general and more precise version of the Theorem, I need to recall some definition and results from this paper of Banakh, Vovk and Wojcik, which will be cited as [BVW]. -So, I thank Jeremy Brazas for opportunity to advertise this my paper [BVW] :) -Unfortunately, writing the paper [BVW] we did not know about the paper of Harris. After looking at the Harris' paper, I discovered that his construction of the space $S(X)$ (with $\pi_0(S(X))\cong X$) is similar to our construction of $\circledast(X)$ with $\pi_0(\circledast(X))\cong X$. But our space $\circledast(X)$ always is complete metric. As a result, our construction works only for premetric (= weakly first-countable) spaces. - -Let us recall that a topological space $X$ is weakly first-countable if to each point $x\in X$ one can assign a decreasing sequence $(B_n(x))_{n\in\mathbb N}$ of subsets containing $x$ such that a subset $U\subset X$ is open if and only if for each point $x\in U$ there exists $n\in\mathbb N$ such that $B_n(x)\subset U$. -By Proposition 3.7 in [BVW], a topological space $X$ is weakly first-countable if and only if the topology of $X$ is generated by a premetric $p$. -A premetric on a set $X$ is any function $p:X\times X\to[0,\infty)$ such that $p(x,x)=0$ for all $x\in X$. -The topology generated by a premetric $p$ on $X$ consists of all sets $U\subset X$ such that for every $x\in X$ there exists $\varepsilon>0$ such that the $\varepsilon$-ball $B(x;\varepsilon):=\{y\in X:p(x,y)<\varepsilon\}$ is contained in $U$. -A premetric space is a pair $(X,p)$ consisting of a set $X$ and a premetric $p$. -To each premetric space $X$ we shall assign some special complete metric space $\circledast(X)$, called the cobweb of the premetric space $X$. -The cobweb space is a closed subset of the complete oriented graph $\Gamma X$ over the set $X$. The complete oriented graph $\Gamma X$ is the set $X\cup\{(x,y,t)\in X\times X\times(0,1):x\ne y\}$ endowed with the path metric $d$ (in which every oriented edge $[x,y]=\{x,y\}\cup\{(x,y,t):0 -TITLE: Zeros of MacLaurin polynomials for the exponential function -QUESTION [19 upvotes]: Asked but never answered at MSE. -Let $\exp_n(z)$ denote the nth degree Taylor polynomial of $e^z$ : -$\exp_n(z) = 1 + z + z^2/2! + ... + z^n/n! \;$ . -The zeros of $\exp_n(z)$ were studied by Szego in the 1920's and -later by others. One of the consequences of Szego's results is -that the roots (after division by n) can come arbitrarily -close to the imaginary axis. -Question: Is it possible for exp$_n$(z) to have a root that lies -precisely on the imaginary axis? - -REPLY [16 votes]: Indeed the polynomial $\exp_n(z)$ has no purely imaginary zeros. Write -$$ -\exp_n(ix) = C_n(x) + i S_n(x) -$$ -in the obvious notation, with $C_n$ for truncations of cosine, and $S_n$ for truncations of sine. If there is a zero $ix$ for $\exp_n(z)$, then this $x$ must be a common root of $C_n$ and $S_n$, which are both polynomials in ${\Bbb Q}[x]$. Therefore $C_n$ and $S_n$ must have some non-trivial gcd in ${\Bbb Q}[x]$. Now we use the beautiful fact (going back to Schur) that $C_n(x)$ is irreducible, which completes our proof. I learnt of this result of Schur from these lovely notes of Keith Conrad; simply apply Theorem 1 there, or see Corollary 1 which mentions this explicitly.<|endoftext|> -TITLE: Defining Massey products as transgressions -QUESTION [15 upvotes]: Let $A$ be a dg algebra, and $x, z \in A$ cocycles. Let's consider the maps -$$ A \to A \oplus A \to A$$ -given by $y \mapsto (xy,yz)$ and $(u,v) \mapsto uz-xv$, respectively. We think of this as defining a double complex with three nonzero columns. (Here we need either to assume that $x,z$ are of degree $0$, or shift the grading in the second and third column.) The spectral sequence of a double complex will have an $E_1$-page given by $H(A) \to H(A) \oplus H(A) \to H(A)$. The $E_2$-page will have as its first column the kernel -$$ \{ y \in H(A) : xy = yz = 0\},$$ -and the $E_2$-differential gives a map to the third column, i.e. the cokernel -$$ H(A)/\left( x H(A) + H(A)z\right).$$ -This map is exactly the triple Massey product $\langle x,y,z\rangle$. -Question Is there a generalization of this construction to higher Massey products? I.e. can one write down a double complex with $n$ columns, such that the $E_{n-1}$-differential could have been used as an alternate definition of the $n$-fold Massey product? -To clarify what I'm after, there are of course many spectral sequences in which the differentials are known to be given by Massey products. What is special with the above construction is that it also produces the right indeterminacy: the triple Massey product $\langle x,y,z\rangle$ is in general well defined precisely as an element of $ H(A)/\left( x H(A) + H(A)z\right),$ which is also what you get from the spectral sequence. One tricky thing about trying to cook up a similar thing for higher products is that there is no similarly simple description of the indeterminacy for the higher products. - -REPLY [4 votes]: This question is rather old (and maybe you have found an answer to this), but https://arxiv.org/abs/1507.04691 gives an answer to what you were looking for, I think (page 16, at the middle): if $(A,d)$ is dg the differentials of the Eilenberg--Moore s.s. (look at the bar construction as a double complex) "are" the Massey products and the usual higher ones defined partially on $H^1(A)^{\otimes n}$ are given by a differential on $E_{n-1}$, concretely $E_{n-1}^{n,n} \longrightarrow E_{n-1}^{1,2}$.<|endoftext|> -TITLE: Almost last births in branching random walk -QUESTION [5 upvotes]: Consider a continuous time doubling branching random walk on the nonnegative integers. It starts with a particle at $0$. The initial particle dies after producing two offspring at $1$, each after an independent exponential$(1)$-distributed amount of time. In the same fashion, a particle born at $k$ will die upon producing two offspring at $k+1$ after, each after an independent exponential$(1)$-distributed amount of time. Suppose the last of the $2^n$ particles born at $n$ occurs at time $T$. This is known as the "Last-Birth Problem" and is fairly well studied. We are curious how many particles reach $n$ just before $T$. -To make this concrete let $X_n$ be the number of particles that reach $n$ at times in $[T-1,T)$. Is there a function $f(n)\to \infty$ such that $P(X_n > f(n)) \to 1$? -Perhaps this is false and only $O(1)$ particles arrive in the final second. However, there may be a huge wave of particles just in front of the last one. - -REPLY [3 votes]: Consider the binary tree of ancestry (thus, the vertices at depth $n$ are the particles that reach location $n$). Write on each edge $e$ between level $n-1$ to level $n$ the time it takes to generate the particle at level $n$, and call this variable $X_e$. If I understand correctly your model, these times are iid exponential (1). For any vertex $v$, let $S_v$ be the partial sum of $X_e$ for $e$ on the geodesic connecting the root to $v$. If I understand correctly, -$T=\max_{v: |v|=n} S_v$. The question you are asking is then a question on the extremal process of this branching random walk (how many particles in an interval of size $1$ to the left of $T$). By a result of Madaule, https://arxiv.org/pdf/1107.2543.pdf, the extremal process converges to -a decorated Poisson point process (with intensity $e^{-cx}$, some explicit $c$) viewed from its tip, and therefore $X_n$ -is a tight sequence. So $f(n)=O(1)$.<|endoftext|> -TITLE: Smallest tile to tessellate the hyperbolic plane -QUESTION [20 upvotes]: Is it known what the smallest tile (in terms of area) that can tessellate the hyperbolic plane is? In particular, it should tessellate the plane by itself. -I think it will be a Triangle group, but I'm not sure. -(In spherical geometry, the answer is that there is no smallest tile, because you can make bipyramids with arbitrarily small faces. I don't think this will be the case with the hyperbolic plane though.) - -REPLY [32 votes]: Binary Tiling -In fact, one can tile the hyperbolic plane with arbitrarily small tiles. There is a tiling of the hyperbolic plane (apparently due to Boroczky) by pentagons. - -The horizontal edges are horocycles in the upper half-space model of the hyperbolic plane, and the vertical lines geodesics. The edge at the top of each tile is half the length of the one at the bottom. One can make these arbitrarily thin, and hence have arbitrarily small area. - -REPLY [19 votes]: The tilings mentioned by Ian Agol are related to an action of a Baumslag-Solitar group $\{ a,b \bigg| b^{-1}a^2b=a \}$ on the hyperbolic plane. They have arbitrarily small area, but diameter uniformly bounded away from $0$. It is possible to tesselate the hyperbolic plane with a single tile with arbitrarily small diameter, too. Let there be $n$ arcs on top and $n+1$ arcs on the bottom. As $n \to \infty$ the distance between the top and bottom goes to $0$. The region is naturally related to a (non-faithful) action of a Baumslag-Solitar group $\{a,b \bigg|b^{-1}a^{n+1}b=a^n\}$ on the hyperbolic plane.<|endoftext|> -TITLE: Edge-transitive Cayley graphs of $S_n$ -QUESTION [10 upvotes]: I came across the following question which I haven't seen before: -Question. Fix $k\ge 3$. For infinitely many $n$, does there exists a generating set $\langle R_n \rangle = S_n$, $|R_n|=k$, such that the corresponding (undirected) Cayley graph $\Gamma(S_n,R_n)$ is edge-transitive? -Perhaps, there is a simple explicit combinatorial construction which would work for all large enough $n$ and some fixed $k$. Are there any references I am missing? -NOTE: Now that I see a nice answer by Brendan McKay, let me mention the reason for the question. I just learned the construction of edge-transitive expanders on $S_n$ for some sparse sequence of $n$, but my own simple construction to answer the question above worked only for $n=p+1$, where $p$ is a prime. - -REPLY [10 votes]: Here's a partial answer. -Take the generators to be a set of equal length cycles that are disjoint except that they have one point in common. For example $\langle (1,2,3,4), (1,5,6,7), (1,8,9,10)\rangle$. I believe that will generate $S_n$ except where the cycles have odd length (in which case they will generate $A_n$). I didn't prove that but I tried a few cases and expect it is easy. Moreover, it is obvious that the Cayley graph will have automorphisms swapping the generators around.<|endoftext|> -TITLE: What are the primes that are ramified? -QUESTION [10 upvotes]: Let $K$ be an imaginary quadratic field and $E$ be an elliptic curve with CM by $\mathcal{O}_K$. We know that the maximal unramified extension (Hilbert class field) $H/K$ is $K(j(E))$. Can we explicitly write down which are the primes that ramify in the ray class field ie $K(j(E), h(E[\mathfrak{p}]))$ for some $\mathfrak{p}$ where $h$ is the Weber function? -The only sort of information I have for now is the following: a prime $\mathfrak{q}$ splits completely in $K(j(E), h(E[\mathfrak{p}]))$ iff $\mathfrak{q}=(\alpha)$ with $\alpha\in \mathcal{O}_K$ and $\alpha\equiv 1 \mod \mathfrak{p}$. - -REPLY [11 votes]: All the information on the higher ramification groups can be derived from Theorem II.5.6, which TKe notes. -The Galois group of the ray class field is the group of fractional ideals relatively prime to $\mathfrak p$ modulo the principal ideals with generators $1$ mod $\mathfrak p$. This naturally maps to the class group, and the kernel consists of principal ideals, and is hence isomorphic to $(\mathcal O_K/\mathfrak p)^\times / \mathcal O_K^\times$. The ramification group at $\mathfrak p$ in the abelianization of the Galois group is $\mathcal O_{K_p}^\times$, and the maps is $\mathcal O_{K_p}^\times \to (\mathcal O_K/\mathfrak p)^\times / \mathcal O_K^\times$ is the obvious projection factoring through $(\mathcal O_K/\mathfrak p)^\times$. -Hence the ramification group of the extension is isomorphic to $(\mathcal O_K/\mathfrak p)^\times / \mathcal O_K^\times $ and is entirely tame - there are no higher ramification groups. -$\mathfrak p$ is ramified if and only if the natural map $ \mathcal O_K^\times \to (\mathcal O_K/\mathfrak p)^\times$ is not surjective. -This map can only be surjective when $\left|(\mathcal O_K/\mathfrak p)^\times \right| \leq \left| \mathcal O_K^\times\right|$, which happens: -if $K\neq \mathbb Q(i), \mathbb Q(\mu_3)$, only when $\mathfrak p$ is a split or ramified prime lying over $2$ or $3$. -If $K = \mathbb Q(i)$, only when $\mathfrak p$ is the prime lying over $2$ or one of the two primes lying over $5$. -If $K = \mathbb Q(\mu_3)$, only when $\mathfrak p = 2$, the prime lying over $3$, or one of the two primes lying over $7$. -In all these cases, it is easy to see that the map is indeed surjective and so $\mathfrak p$ does not ramify.<|endoftext|> -TITLE: Near-linear mappings from $\mathbb F_p$ to $\mathbb R$ -QUESTION [13 upvotes]: $\newcommand{\F}{{\mathbb F}}$ -$\newcommand{\R}{{\mathbb R}}$ -$\renewcommand{\phi}{\varphi}$ -Let $p\ge 5$ be a prime. -If the functions $\phi_1,\phi_2,\phi_3\colon\F_p\to\R$ satisfy $\phi_1(x)+\phi_2(y)=\phi_3(x+y)$ for all pairs $(x,y)\in\F_p^2$, then they are constant functions. The easiest way to see this is by comparing the images $A_i:=\mathrm{Im}(\phi_i)$: - $$ A_1+A_2 \subseteq A_3,\ A_3-A_1\subseteq A_2,\ A_3-A_2\subseteq A_1, $$ -whence $|A_1|=|A_2|=|A_3|=1$. - -Given that $\phi_1,\phi_2,\phi_3\colon\F_p\to\R$ are non-constant, what is the largest possible number of pairs $(x,y)\in\F_p^2$ satisfying - $$ \phi_1(x)+\phi_2(y)=\phi_3(x+y)? \tag{$\ast$} $$ - Equivalently, what is the smallest possible number of pairs $(x,y)\in\F_p^2$, for all choices of the non-constant functions $\phi_1,\phi_2,\phi_3$, such that ($\ast$) fails to hold? - -If $u,v,w\in\F_p$ are pairwise distinct and $w\ne u+v$ then, letting - $$ \phi_1=1_{\{u\}},\ \phi_2=1_{\{v\}},\ \phi_3=1_{\{w\}} $$ -(the indicator functions of the corresponding singletons), we have $3p-5$ pairs $(x,y)$ violating ($\ast$); is this the worst case? - -Is it true that for any non-constant functions $\phi_1,\phi_2,\phi_3\colon\F_p\to\R$, there are at least $3p-5$ pairs $(x,y)\in\F_p^2$ with - $$ \phi_1(x)+\phi_2(y) \ne\phi_3(x+y)? \tag{$\circ$}$$ - -What I can show, in two different ways, is that there are at least $p-1$ pairs $(x,y)$ satisfying ($\circ$). - -REPLY [6 votes]: Yes, for $p$ sufficiently large. -Assume that $\phi_1,\phi_2,\phi_3$ satisfy this condition for at most $3p-5$ values of $x,y$. -Hence for each $a$, all but at most $6p-10$ pairs $(x,y)$ satisfy -$$\phi_1(x+a) + \phi_2(y) - \phi_3(x+a+y) = 0 = \phi_1(x) + \phi_2(y+a)- \phi_3(x+a+y)$$ -in other words -$$\phi_1(x+a) - \phi_1(x) = \phi_2(y+a) - \phi_2(y)$$ -Now when we have the equation $$F(x)=G(y)$$ for all but at most $6p-10$ pairs $(x,y)$, then for some value of $y$ the equation must be satisfied for all but at most $6$ values of $x$. -So for all $a \in \mathbb F_p$, there is some $c_a$ with $\phi_1(x+a) - \phi_1(x) = c_a$ for all but at most $6$ values of $x$. -Now given $a,b$ there are at least $p-18$ values of $x$ where $\phi_1(x+a)-\phi_1(x) = c_a$, $\phi_1(x+a+b)-\phi_1(x+a)=c_b$, $\phi_2(x+a+b)-\phi_1(x) = c_{a+b}$, and hence as soon as $p>18$, $c_{a+b}=c_a+c_b$, $c$ is a homomorphism $\mathbb F_p \to \mathbb R$ and hence vanishes. -Hence for all $a$ in $\mathbb F_p$, $\phi_1(x+a)= \phi_1(x)$ for all but $6$ values of $x$. Hence $\phi_1(x)=\phi_1(y)$ for all but $6p$ values of $x,y$. Hence for some $y$, $\phi_1(x)=\phi_1(y)$ for all but $6$ values of $x$. -Applying the same logic to $\phi_1$ and $\phi_2$, there are constants $A,B,C$ such that $\phi_1(x)=A, \phi_2(x)=B,\phi_3(x)=C$, each with at most $6$ exceptions. So as soon as $p^3- 18 p^2 > 3p-5$, $A+B=C$. -Now let $\alpha,\beta,\gamma$ be the number of exceptions to these equations. There are $(\alpha+\beta+\gamma)p - (\alpha\beta+ \alpha\gamma+\beta\gamma)$ pairs $x,y$, where exactly one of $x,y,x+y$ is an exception, and for none of these pairs is $\phi_1(x)+\phi_2(y)=\phi_3(x+y)$. For $p$ sufficiently large, this is minimized by minimizing $\alpha+\beta+\gamma$, which is minimized when $\alpha=\beta=\gamma=1$, since all are at least one. -But in this case, the best we can do is your example.<|endoftext|> -TITLE: Maximal subgroups of simple groups with normal $2$-subgroups -QUESTION [10 upvotes]: Let $G$ be a finite non-abelian simple group. Question: - -Does there always exist a maximal subgroup $M$ of $G$ such that $M$ has a non-trivial normal elementary abelian $2$-subgroup? - -This seems to be the case for alternating groups (take $M$ a stabilizer of a $4$-set) and sporadic simple groups (look at tables in ATLAS). For groups of Lie type in characteristic $2$ it seems to me that one could choose $M$ to be a maximal parabolic subgroup. -Also if the answer to the question is no, what about the weaker question of $M$ containing a non-trivial normal abelian subgroup? And if the answer to either question is yes, is there a proof without classification? - -REPLY [10 votes]: Unfortunately I made a mistake in my original answer. The answer to the question is no, but the only finite simple groups that do not have a maximal subgroup with nontrivial normal 2-subgroup are the groups $L_p(3)$ with $p$ a prime and $p \ge 5$. -Let's go through the finite nonabelian simple groups. -As you say, for $G=A_n$ we can take $M$ to be the stabilizer of a set of size $4$. -You have checked the sporadic groups in the ATLAS. -If $G$ is of Lie type in characteristic $2$, then we can take $M$ to be a maximal parabolic subgroup. -So we can assume from now on that $G$ is of Lie type in odd characteristic. -The geometric-type maximal subgroups of the classical groups are listed in the book by Kleidman and Liebeck. Some of these are not maximal in small dimensions. The complete classification in dimensions up to $12$ was completed in a book by myself, Bray and Roney-Dougal. So we have enough information to check the result in groups of odd characteristic, and in fact imprimitive groups can be used for $M$ in most cases. -In some of the lists of $M$ below, there are a few small $n$ and $q$ for which $M$ is not maximal, but in each of these cases you can check that there is some other maximal subgroup that works - more details on request. -For $G=L_2(q)$ we have a dihedral subgroup of order $(q-1)$ or $(q+1)$. -Form now on, the subgroup $M$ is described as a subgroup of the relevant quasisimple matrix group, so we have to take it modulo scalars to get the corresponding maximal of the simple group. -In general, for $G=L_n(q)$ with $n \ge 3$ and $q \ge 5$, we can take $M$ to be the imprimitive group with structure $(q-1)^{n-1}:S_n$. -For $L_n(3)$, the subgroup with this structure is not maximal (it preserves an orthogonal form). If $n=rs$ is composite with $r \ge 2$ and $s>2$, then there is an imprimitive subgroup ${\rm GL}_r(3) \wr S_s$ of ${\rm GL}_n(3)$, and its intersection with ${\rm SL}_n(3)$ projects onto a maximal subgroup of $L_n(3)$ with a nontrivial normal $2$-subgroup. -Also $L_3(3)$ has $S_4$ as maximal subgroup and $L_4(3)$ has $S_4 \times S_4$, but for primes $p \ge 5$, $L_p(3)$ has no maximal with the required property. This can be seen from the complete list of its maximal subgroups, and I checked it directly by computer for $p=5,7,11$. -$G=U_n(q)$, we can take $M$ to be the imprimitive subgroupn with structure $(q+1)^{n-1}:S_n$, and this really is maximal for all $q \ge 3$. -$G=S_{2n}(q)$, take $M={\rm Sp}_2(q)^n:S_n$. -The orthogonal groups all have reducible maximal subgroups that are stabilizers of non-degenerate subspaces of dimension $2$ with nontrivial normal $2$-subgroups. -I know very little about the maximal subgroups of the exceptional groups of Lie type, but I looked at Table 5.2 of -M.W. Liebeck, J. Saxl and G.M. Seitz, “Subgroups of maximal rank in finite -exceptional groups of Lie type”, -Proc. London Math. Soc. -65 -(1992), 297-325. -In all cases there appears to be a maximal subgroup that looks like an imprimitive maximal of a classical groups, and has normal subgroups with structure $(q-1)^n$ or $(q+1)^n$, where $n$ is the Lie rank of the group.<|endoftext|> -TITLE: A residually finite modification of the wreath product -QUESTION [9 upvotes]: I have been looking for ways to construct examples of finitely generated residually finite groups that are poly-(locally virtually abelian) but not virtually solvable. -If $K$ is a finite non-solvable group, then the wreath product $K \wr \mathbb{Z}$ satisfies all of these properties, except that of residual finiteness. -While working on an unrelated problem, my co-author came up with a construction that looks like a residually finite version of the wreath product, which we then generalized. -Since neither of us are group theorists, we'd like to know whether it is a well-known construction, and possibly find a reference that we could cite in our paper. -Our literature searches haven't turned up anything, but it may be that we are just using the wrong keywords or sources. -The construction -Consider two residually finite countable groups $K$ and $H$ with $|K| \geq 2$. -Define the set -$$ -X = \bigcup_{\substack{A \trianglelefteq_{\text{f}} K \\ B \trianglelefteq_{\text{f}} H}} (K/A)^{H/B} -$$ -where $A$ ranges over the finite-index normal subgroups of $K$, and similarly for $B$ and $H$. -An element $f \in X$ is a function from $H/B$ to $K/A$. -Define a left action $\phi_K$ of $K$ on $X$ by -$$ -\phi_K(k, f)(g B) = - \begin{cases} - k \cdot f(g B), & \text{if $g \in B$,} \\ - f(g B), & \text{otherwise,} - \end{cases} -$$ -and a left action $\phi_H$ of $H$ on $X$ by -$$ -\phi_H(h, f)(g B) = f(h^{-1} g B). -$$ -Each set $(K/A)^{H/B}$ is invariant under both actions; the idea is that $K$ acts by multiplication on the element at the "origin", while $H$ acts by translating the coordinates. -We define $G = \langle \phi_K(K) \cup \phi_H(H) \rangle \leq \text{Sym}(X)$. -The resulting group $G$ is residually finite (since it splits $X$ into finite orbits) and finitely generated if both $K$ and $H$ are, and the maps $\phi_K : K \to G$ and $\phi_H : H \to G$ are injective homomorphisms. -Furthermore, $\phi_H$ has a retraction $\psi : G \to H$, and $\ker \psi$ is isomorphic to a subgroup of a direct product of copies of $K$. -If we choose $K = A_5$ and $H = \mathbb{Z}$, then $G$ is locally finite-by-$\mathbb{Z}$ and hence poly-(locally virtually abelian), and one can show that it is not virtually solvable (see the link above for a proof). - -REPLY [9 votes]: Edit: I expanded this post after reading more carefully your construction and reminding a relevant reference. So it splits into 3 parts. - -B.H. Neumann's examples -Comments on your construction -(my original post) A construction with A. Mann - -1 B.H. Neumann's construction -Reference: B.H. Neumann. Some remarks on infinite groups. J. London Math. Soc. 1(2) (1937) 120--127. -He considered a strictly increasing sequence $(m_n)$, $m_n\ge 5$ tending to infinity, of odd numbers, and $X_n$ a copy of $\{1,\dots,m_n\}$, and $X=\bigsqcup_n X_n$. Let $c$ act as the 3-cycle $(123)$ on each $X_n$, and $t$ act as the full cycle $(1,\dots,m_n)$ on all $X_n$. Consider $\Gamma=\langle t,c\rangle$. -This is residually finite (since orbits are all finite). -Neumann checks that $\Gamma$ has a homomorphism onto $\mathbf{Z}$, mapping $t$ to 1 and $c$ to 0. Moreover, its kernel contains a smaller normal subgroup, the finitely supported even permutations preserving all $X_n$ (that is, $\bigoplus_n\mathrm{Alt}_{m_n}$). The quotient is the semidirect product $\mathrm{Alt}(\mathbf{Z})\rtimes\mathbf{Z}$, where $\mathbf{Z}$ acts by shifing. -This construction is an illustration of the following fact: - -Every countable LEF group is quotient of a residually finite countable group by a locally finite FC-central normal group, which can be chosen of the form $\bigoplus_n F_n$ with each $F_n$ finite and normal in the whole group. - -Note that it also, by the way, provides examples of residually finite 2-generator groups that are (locally finite)-by-$\mathbf{Z}$ (hence poly-(virtually abelian)) and not virtually solvable (as your example and mine in (3) below), but unlike those, satisfy no group identity. -2 Comments on your construction -First let me rewrite it when $K$ is a simple group and $H=\mathbf{Z}$. Since the singletons $K/K$ play no role, let me remove them. So -$$X=\bigsqcup_{n\ge 1}K^n,$$ -where $\mathbf{Z}$ acts on $K^n$ by shifting, and $K$ acts on $K^n$ by left multiplication at 0 (that is, $k\cdot (k_0,k_1,\dots,k_{n-1})=(kk_0,k_1,\dots,k_{n-1})$. These two operations on $K^n$ simply provide a simply transitive action of the wreath product $K\rtimes\mathbf{Z}/n\mathbf{Z}$. -When putting things together, one gets a group which, modulo its subgroup of finitely supported permutations, is the wreath product $K\wr\mathbf{Z}$, another typical example of a finitely generated LEF group that is not residually finite. -Actually, this is exactly the kind of construction one gets when one tries to make explicit the proof of the LEF fact mentioned above. A variant would consist in choosing, instead of the simply transitive action of $K\wr\mathbf{Z}/n\mathbf{Z}$, rather its "natural" action, which is on the disjoint union of $n$ copies of $K$, or even, for $K=\mathrm{Alt}_5$ on $n$ copies of $\{1,2,3,4,5\}$. -Another consequence is that these groups are not isomorphic to mine described in (3). Indeed, by construction, your group has a nontrivial FC-center, since it contains nontrivial finitely supported permutations (for an explicit example, let $t$ be the generator of $\mathbf{Z}$, and choose in $K$ two non-commuting elements $b,c$; then $[tbt^{-1},c]$ is supported by the first copy). While my example in (3) has a trivial FC-center. - -3 In a note with Avinoam Mann (Some residually finite groups satisfying laws. Geometric group theory, 45–50, -Trends Math., Birkhäuser, Basel, 2007.), we constructed a few residually finite groups including the following one, which answers your initial question: -For $A$ an abelian group, consider the free group $W_A$ on the generators $(x_n)_{n\in A}$ in the variety generated by the alternating group $\mathrm{Alt}_5$ (or any non-solvable finite group). (If you're not familiar with varieties of groups, this means the quotient of the free group $F$ on $(x_n)_{n\in A}$ by the intersection of all kernels of homomorphisms $F\to\mathrm{Alt}_5$. For instance, $W_A$ has exponent 30. It embeds into some power of $\mathrm{Alt}_5$ and hence is locally finite. By shifting the generators, we obtain an automorphism of $W_{\mathbf{Z}}$ and hence a semidirect product $G=W_{\mathbf{Z}}\rtimes\mathbf{Z}$, generated by 2 elements ($x_0$ and the generator of $\mathbf{Z}$). -Since $G$ is (locally finite)-by-abelian, it is obviously poly-(locally virtually abelian) (in two steps). It admits the wreath product $\mathrm{Alt}_5\wr\mathbf{Z}$ as a quotient and hence is not virtually solvable. -That $G$ is residually finite is proved as in the proof of Proposition 6 of the linked paper (cf Remark 7). Namely, consider an element $wt^n\neq 1$ and let us construct a finite index subgroup in which it does not belong. This is trivial if $n\neq 0$, so consider $w\neq 1$ in $W_{\mathbf{Z}}$. It involves finitely many letters, say in $\{x_i:-m -TITLE: Saturated classes, generation by a set and pullbacks of categories -QUESTION [7 upvotes]: Assume that we have a pullback square -$$ -\begin{array}{ccc} -A & \rightarrow & B \\ -\downarrow & & \downarrow \\ -C & \rightarrow & D \\ -\end{array} -$$ -with all functors accessible, and all categories presentable (if needed, assume that $A \to C$ and $B \to D$ are left adjoints). -Suppose further that we have three saturated (that is, closed under retracts, pushouts and transfinite compositions) classes of arrows $S(B),S(C),S(D)$ of, -respectively, $B,C,D$. Assume that the functors $B \to D, C \to D$ preserve these classes. (It seems likely to me that the set $S(D)$ should not play any importance in what follows) -Form the set $S(A)$ by requiring that it consists of all arrows of $A$ that are sent to $S(B)$ and $S(C)$ by the respective functors. -Question: if $S(B),S(C),S(D)$ are generated by (small) sets, does it imply that $S(A)$ is generated by a set as well? -EDIT: it may be not obvious that the formed class S(A) is weakly saturated. Assume it is. -EDIT2: I do not know if the original question admits an affirmative answer. What is true is that, following a remark of Tim to this post, one can consider the situation with the right classes. Let me leave a statement -that may be of use for someone who googles this post up. -Lemma. Let $A$ be a presentable category together with a weak factorisation system $(L,R)$ having the property that - -$(L,R)$ is functorial (a closer inspection may be used to drop this), -The functor $A^{[1]} \to A^{[2]}$ associated to the weak factorisation system is accessible, -The right class $R$ viewed as a subcategory of $A^{[1]}$ is accessible and accessibly embedded. - -Then there exists a set $L_0 \subset L$ generating $(L,R)$. - -REPLY [2 votes]: Yes, see Theorem 3.2 in Makkai and Rosický's Cellular categories.<|endoftext|> -TITLE: Growth of stable homotopy groups of spheres -QUESTION [15 upvotes]: Let ${}_2\pi_n^S$ denote the $2$-power torsion subgroup of $n$th stable homotopy group of the sphere spectrum. Its order is a power of $2$: $$|{}_2\pi_n^S|=2^{k_n}.$$ -Question: What is known about growth of $k_n$? Is it polynomial? What is the best estimation? -Remark: Of course, any estimation on ${\rm Ext}^{s,t}_{\mathcal A_2}(\mathbb F_2,\mathbb F_2)$ implies an estimation on $k_n.$ The Lambda algebra gives an exponential estimation, it is not interesting. I'm not very familiar with the May spectral sequence. Does it give a better estimation? - -REPLY [5 votes]: A subexponential bound is available (using only things known 40 years ago). -Thanks to John Palmieri over here for pointing out that the $E_1$ term of the May spectral sequence is a commutative polynomial algebra and so ought to have graded dimension which counts some sort of partition, and for subsequently pointing out that on account of $h_{1,0}$, this observation must be supplemented with some information about vanishing lines in the Adams spectral sequence. -Indeed, the May $E_1$ term $V^{\ast\ast\ast}$ is a polynomial algebra in $h_{ij}$ for $i\geq 1, j \geq 0$, with tridegree $|h_{ij}| =(s,t,u) = (1,2^j(2^i-1),i)$, i.e. bidegree $(s,t) = (1,2^j(2^i-1))$ in the Adams $E_2$, i.e. degree $t-s = 2^j(2^i-1)-1$ in the stable stems. Note that $h_{1,0}$ has bidegree $(s,t) = (1,1)$; all other $h_{i,j}$'s have $t-s > 0$. -Let $W^{\ast\ast\ast} \subseteq V^{\ast\ast\ast}$ be the subalgebra generated by the $h_{i,j}$'s other than $h_{1,0}$. Keeping just the last grading $k = t-s$, we see that $\dim W^k$ counts the number of ways of partitioning $k$ using positive integers of the form $2^j(2^i-1)-1$, i.e. positive integers whose binary expression contains exactly one zero (since the numbers $i$ and $j$ are uniquely determined by the quantity $2^j(2^i-1)-1$). This is less than the total number of partitions, and hence subexponential. The estimate via total partitions tells us that $\dim W^k \leq \exp(c\sqrt{k})$ for some $c>0$, but since the allowed parts for partitioning are exponentially sparse like in the Steenrod algebra, I'd guess that an upper bound of the form $\dim W^k \leq \exp(c(\log k)^2)$ actually follows from this if one works through the combinatorics, just as it does with the Steenrod algebra. -Adding back in the generator $h_{1,0}$, we see that $V^k$ is infinite-dimensional for all $k$. But because the $s$-grading of $h_{1,0}$ is still positive, we can use the fact that the $E_2$ page of the Adams spectral sequence (which I'm just calling $Ext^{\ast,\ast}$) has a vanishing line, in the sense that $Ext^{s,t} = 0$ for $0 < t-s < 2s + d$ for some constant $d$. I believe this implies that every element of $Ext^{s,t}$ with $t-s = k$ can be written as a sum of monomials $h_{1,0}^a\prod h_{i,j}^{b_{i,j}}$ where $a < k/2 - d/2$, so that we essentially have $\dim (\oplus_{t-s = k} Ext^{s,t}) \leq \sum_{a=0}^{k/2} \exp(c\sqrt{k-a})$. Because subexponential functions are closed under integration, it follows that the Adams $E_2$ page already has subexponential growth, and hence that $\log_2 |\pi_k\mathbb S_{(2)}|$ likewise has subexponential growth. -There is an odd primary analog of this too. See Ravenel's green book for a version of the May spectral sequence at odd primes where the $E_1$ term is a commutative polynomial algebra. -The dimensions of the graded parts of $W^k$ start off, if I coded things correctly, as: - - - - -degree $k$ -bound on $\dim W^k$ - - - - -0 -1 - - -1 -1 - - -2 -2 - - -3 -3 - - -4 -5 - - -5 -7 - - -6 -11 - - -7 -15 - - -8 -21 - - -9 -28 - - -10 -38 - - -11 -49 - - -12 -65 - - -13 -83 - - -14 -107 - - -15 -136 - - -16 -172 - - -17 -215 - - -18 -269 - - -19 -332 - - - - -This sequence must be well-known, as I think it is the basis for Bruner's Ext software. So I'm surprised that I can't find it in OEIS. Perhaps I have made a mistake. - -Here is an earlier version of this answer: -Thanks to Nicholas Kuhn over here for pointing out that the dimension of the Lambda algebra can be used to bound the size of the $E_2$ term of the Adams spectral sequence. -The dimensions of the graded pieces of the Lambda algebra appear in the OEIS: https://oeis.org/A049285 and a bit of searching reveals that Tangora computed the asymptotics in Level number sequences of trees and the Lambda algebra, where I think he also considers the odd-primary case. Unfortunately, the results are stated in terms of generating functions, and a quick look has not allowed me to find a place where he actually states the asymptotic upper bound he gets on the dimension of the Lambda algebra. But his work is based on Flajolet and Prodinger's Level number sequences for trees, and if I'm deciphering things correctly, it looks like the bound is still exponential.<|endoftext|> -TITLE: Injection from $\aleph_2$ into the power set of $\mathbb{R}$ -QUESTION [10 upvotes]: Assume $\mathsf{ZF} + \mathsf{DC}$. Must there exist an injection from $\aleph_2$ to $\mathcal{P}(\mathbb{R})$? If not, what is the consistency strength of the nonexistence of such an injection? -I hope I'm not missing something obvious. Here are my thoughts so far: - -There is always a surjection from $\mathcal{P}(\mathbb{R})$ to $\aleph_2$. -$\mathsf{ZFC}$, or more generally the existence of a wellordering of $\mathcal{P}(\mathbb{R})$, implies there is such an injection. -$\mathsf{ZF} + \mathsf{AD}$ implies the existence of a surjection from $\mathbb{R}$ to $\aleph_2$, which implies there is such an injection. -If there is an inner model of $\mathsf{ZFC}$ satisfying $(\aleph_1^V)^+ = \aleph_2^V$, for example in $V(\mathbb{R}^{V[G]})$ where $\kappa$ is inaccessible and $G \subset \operatorname{Col}(\omega,\mathord{<}\kappa)$ is a $V$-generic filter, then there is such an injection. - -REPLY [9 votes]: The consistency strength is just that of ZFC. -Start with GCH, and do a symmetric collapse of $\aleph_{\omega_1}$ to be $\aleph_2$. Since everything is $\sigma$-closed, you get DC for free, and the reals can even be well ordered. -But any injection from $\aleph_2$ into $\cal P(\Bbb R)$ would have had to be adjoined by some bounded part of the iteration (by homogeneity), so none exist, as with the usual arguments for $L(\Bbb R)$.<|endoftext|> -TITLE: Hodge Numbers and Leray Spectral Sequence -QUESTION [7 upvotes]: Mark Gross' notes survey of SYZ fibrations and toric degenerations begin by explaining why dual torus fibrations interchange Hodge numbers. But he defined the Hodge numbers in an unusual way -$$h^{p,q}(X) = \text{dim}_\mathbb{R}H^p(B,R^qf_\ast \mathbb{R}),$$ -where $f: X \to B$ is a torus fibration, $B$ is a real $3$-manifold, and $X$ is a complex Calabi-Yau threefold. -I thought the Hodge numbers were $h^{p,q}(X) = \text{dim}_\mathbb{R}H^p(X,\Omega^q_X).$ Why ought these numbers agree? -Reference. https://arxiv.org/pdf/0802.3407.pdf - -REPLY [12 votes]: I don't think I defined the Hodge numbers in this way. Rather, the argument in Section 1 shows that the Hodge numbers agree with the dimensions of -the terms in the $E_2$ page of the Leray spectral sequence. The argument -given there only works in three dimensions (and some strong assumptions -on the fibration) because in the Calabi-Yau 3-fold cases, the Hodge -numbers are in fact topological invariants, i.e., $h^{1,1}=b_2$ and -$h^{1,2}=b_3/2 -1$. -In other work, again with some assumptions, it is shown that the groups -$H^p(B, R^qf_*{\bf Z})$ agree with the graded pieces of the weight filtration for the limiting mixed Hodge structure associated with a toric degeneration.<|endoftext|> -TITLE: Second order recurrence relation for third order polynomial root -QUESTION [6 upvotes]: Consider this recurrence relation: -$$ -\begin{eqnarray*} -f_0&=&1\\ - f_n&=& -\sum_{m=0}^{n-1} \frac{\left(\frac{m+3}{2}\right)_{m-1}}{\left(\frac{m+2}{2}\right)_m} f_{n-m-1} f_m\ \ \ \text{for $1\leq n$.} -\end{eqnarray*} -$$ -where the Pochhammer symbol denotes the rising factorial. The generating function $f(z)=\sum_{n=0}^\infty f_nz^n$ seems to be a root of -$$ -0=12 f^3 z^2- (f-1)^2 (f+2) -$$ -I have checked this to be true for the first 600 terms. However, I have been unable to come up with a proof. Do you have any ideas on how I might show this to be true? -Cheers, -Petter - -REPLY [9 votes]: This is sequence A244038 in OEIS after scaling by $3^n$, so $f_n=(4/3)^n\binom{3n/2}n$. The fact that it satisfies a cubic equation -is certainly a well-known result in hypergeometric functions. -EDIT: remove the "well-known": set $F(x)=\sum_{n\ge0}\binom{3n/2}{n}x^n$. -Then $$F(x)=_2F_1(1/3,2/3;1/2;27x^2/4)+(3x/2)_2F_1(5/6,7/6;3/2,27x^2/4)$$ -(which can probably be slightly simplified using contiguity relations), -but can a hypergeometric expert explain why $(27x^2/4-1)F^3+3F-2=0$ ?<|endoftext|> -TITLE: Relative volume increase of $\delta$-fattening of a compact set -QUESTION [6 upvotes]: For a non-empty, compact set $A \subseteq \mathbb{R}^n$, the $\delta$-fattening of $A$, $A_\delta$, is defined to be the set -$$ -A_\delta = \cup_{a \in A} B_{\delta}(a), -$$ -where $B_\delta(a)$ denotes the closed ball centered at $a$ with radius $\delta$. -Is it possible to establish an upper bound on $\mu(A_\delta)$ in terms of $\mu(A)$, where $\mu$ is the Lebesgue measure over $\mathbb R^n$? -In a previous post, it was claimed that: -Claim -Let $A$ be a nonempty compact subset of $\mathbb R^n$ with $\mu(A)>0$. Then for all $\delta>0$ -$$\mu(A_\delta)\le \left(1+\delta\,\frac{\lambda(\partial A)}{n\,\mu(A)}\right)^n\mu(A)\tag1,$$ where $A_\delta$ is the $\delta$-fattening of the set $A$, and $\lambda(\partial A)$ is the Minkowski content -$$\lambda(\partial A)=\liminf_{\delta\to 0}\delta^{-1}(\mu(A_\delta)-\mu(A))\tag2.$$ -However, the proof utilized the fact that $f(\delta) = \left(\mu(A_\delta)/\mu(A)\right)^{1/n}$ is concave, which per this post is not true. Can the claim above or a similar inequality be established? - -REPLY [4 votes]: The claim is false in general. The example given by George Lowther disproves it. Indeed, in that example, $n=2$ and $A$ is the union of the unit disk and a one-point set at distance $R>1$ from the origin, so that $\mu(A_\delta)=f(\delta):=\pi[(1+\delta)^2+\delta^2]$ for $\delta\in[0,(R-1)/2]$ and $\lambda(\partial A)=f'(0)=2\pi$. -So, -the ratio of the RHS of your proposed inequality (1) to its LHS is -$$\frac{(1+\delta)^2}{(1+\delta)^2+\delta^2}<1$$ -for $\delta\in(0,(R-1)/2]$.<|endoftext|> -TITLE: When does prime elements remain prime in certain integral extension -QUESTION [8 upvotes]: Let $R$ be an integral domain and $\bar R$ denote its integral closure in the fraction field (i.e. normalization). If $p\in R$ is a prime element in $R$, then does $p$ remain prime in $\bar R$ also ? -If this is not true in general, then what if we also assume $R$ is Noetherian ? -By a prime element in an integral domain $R$, I mean a non-zero non unit $p\in R$ such that $p |ab $ for some $a,b \in R$ implies $p|a$ or $p|b$ i.e. if $pR$ is a prime ideal in $R$ . I can see that $p$ still remains a non-unit in $\bar R$ , but I'm unable to say anything about the ideal $p\bar R$. -UPDATE : The claim is true for any Noetherian domain. This is Lemma 4.7 in ; On finite generation of $R$-subalgebras of $R[X]$ , Amartya K. Dutta; Nobuharu Onoda; Journal of Algebra 320 (2008) 57- 80. On finite generation of R-subalgebras of R[X] - ScienceDirect -https://www.sciencedirect.com › pii - -REPLY [7 votes]: This is true when $R$ is reasonable. The properties that I use are: - -$R$ is Noetherian; -$\tilde R$ is Noetherian; -$\tilde R$ is catenary and equidimensional (i.e. every maximal chain $0 = \mathfrak q_0 \subsetneq \ldots \subsetneq \mathfrak q_d$ of prime ideals in $\tilde R$ has the same length). - -For example, these are all satisfied if $R$ is of finite type over a field or over $\mathbb Z$. It might be possible to weaken some of these hypotheses. -Lemma. Let $f\colon R \to S$ be an integral ring map, let $\mathfrak q \subseteq S$ be a prime, and let $\mathfrak p = f^{-1}(\mathfrak q)$. Then -$$\dim R/\mathfrak p = \dim S/\mathfrak q.$$ -Proof. This is poset-theoretic, using only the going up theorem for integral maps [AM, Thm. 5.11]. Indeed, the going up theorem implies that a chain $\mathfrak p = \mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \ldots$ of primes of $R$ containing $\mathfrak p$ can be lifted to some chain $\mathfrak q = \mathfrak q_0 \subsetneq \mathfrak q_1 \subsetneq \ldots$ of $S$, whence $\dim S/\mathfrak q \geq \dim R/\mathfrak p$. -Conversely, if $\mathfrak q_1 \subseteq \mathfrak q_2$ are primes of $S$ with $f^{-1}(\mathfrak q_1) = f^{-1}(\mathfrak q_2) = \mathfrak p$, then we must have $\mathfrak q_1 = \mathfrak q_2$. Indeed, they correspond to primes in the integral ring map $\kappa(\mathfrak p) \to S \otimes_R \kappa(\mathfrak p)$, and there are no inclusions between prime ideals of $S \otimes_R \kappa(\mathfrak p)$ [Tag 00GS(3)]. Hence, the inverse image of a chain $\mathfrak q = \mathfrak q_0 \subsetneq \mathfrak q_1 \subsetneq \ldots$ of primes of $S$ containing $\mathfrak q$ is a strict chain $\mathfrak p = \mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \ldots$ of primes of $R$ containing $\mathfrak p$, whence $\dim R/\mathfrak p \geq \dim S/\mathfrak q$. $\square$ -Remark. In the proof below, we want to relate the heights of $\mathfrak q$ and $\mathfrak p$ as in the lemma. We can do this under assumption (3), for this forces $\operatorname{ht}(\mathfrak p) = \dim R - \dim R/\mathfrak p$ (and similarly for $\mathfrak q$). - - -Proposition. Let $R$ be a domain satisfying properties (1)-(3) above. If $p \in R$ is a prime element, then $p$ is a prime element in $\tilde R$. - - -Proof. By assumption, $\mathfrak p = (p)$ is a prime ideal. By Krull's Hauptidealsatz [AM, Cor. 11.17], this implies that $\mathfrak p$ has height $1$, i.e. $R_\mathfrak p$ is a $1$-dimensional domain. Since its maximal ideal $\mathfrak pR_\mathfrak p$ is principal, we conclude that $R_\mathfrak p$ is a DVR [AM, Prop. 9.2] with uniformiser $p$; in particular $R_\mathfrak p$ is normal. -On the other hand, normalisation commutes with localisation [AM, Prop. 5.12]. Thus, -$$(\tilde R)_\mathfrak p = (R_\mathfrak p)^\sim = R_\mathfrak p,$$ -since $R_\mathfrak p$ is normal. That is, the natural map $R \to \tilde R$ becomes an isomorphism when tensoring with $R_\mathfrak p$, hence also when tensoring with $\kappa(\mathfrak p) = R_\mathfrak p/\mathfrak pR_\mathfrak p$. The primes of $\tilde R \otimes_R \kappa(\mathfrak p)$ are the primes of $\tilde R$ lying over $\mathfrak p$ [AM, Exc. 3.21(iv)], so we conclude that there is a unique such prime $\mathfrak q$. Note that $\mathfrak q$ is minimal over $\mathfrak p\tilde R$, hence has height $1$ by Krull's Hauptidealsatz. -If $\mathfrak r \subseteq \tilde R$ is another height $1$ prime, then $p \not\in \mathfrak r$. Indeed, if $p \in \mathfrak r$, then $\mathfrak p' = \mathfrak r \cap R$ contains $\mathfrak p$. Applying the lemma and the remark above, we conclude that $\operatorname{ht}(\mathfrak p') = \operatorname{ht}(\mathfrak r) = 1$. Hence $\mathfrak p' = \mathfrak p$ since $\mathfrak p \subseteq \mathfrak p'$ and both have height $1$. -Hence, for a height $1$ prime $\mathfrak r \subseteq \tilde R$, we have -$$v_{\mathfrak r}(p) = \left\{\begin{array}{cc} 1, & \mathfrak r = \mathfrak q,\\ 0, & \mathfrak r \neq \mathfrak q, \end{array}\right.$$ -since $p$ is a uniformiser of the DVR $\tilde R_\mathfrak q \cong R_\mathfrak p$. If $q \in \mathfrak q$, then $v_\mathfrak r(q) \geq v_\mathfrak r(p)$ for all height $1$ primes $\mathfrak r \subseteq \tilde R$. Hence, $\frac{q}{p} \in \tilde R$ [Eis, Cor. 11.4], which shows that $\mathfrak q \subseteq (p)$. The reverse inclusion follows since $\mathfrak q \cap R = \mathfrak p$, hence $(p) = \mathfrak q$ is prime. $\square$ -Remark. In geometric language, we proved: - -There is a unique irreducible divisor $V(\mathfrak q) \subseteq \operatorname{Spec} \tilde R$ dominating the irreducible divisor $V(\mathfrak p) \subseteq \operatorname{Spec} R$; -The locus $V(p) \subseteq \operatorname{Spec} \tilde R$ does not split off a new component of higher codimension; -The uniformiser $p$ for the divisor $V(\mathfrak p) \subseteq \operatorname{Spec} R$ remains a uniformiser for $V(\mathfrak q) \subseteq \operatorname{Spec} \tilde R$ (there is no ramification). - - -References. -[AM] Atiyah, M.F.; Macdonald, I.G., Introduction to commutative algebra. Addison-Wesley Publishing Company (1969). ZBL0175.03601. -[Eis] Eisenbud, D., Commutative algebra with a view toward algebraic geometry. Graduate Texts in Mathematics 150, Springer-Verlag (1995). ZBL0819.13001.<|endoftext|> -TITLE: Equi-Hölder embeddings of compact metric spaces of finite packing dimension into $\ell_2$ -QUESTION [8 upvotes]: Problem. Does a compact metric space of finite packing dimension admit an equi-Hölder embedding into a Hilbert space? -A map $f:X\to Y$ between metric spaces $(X,d_X)$, $(Y,d_Y)$ is called equi-Hölder embedding if there are positive real constants $c,C,\alpha$ such that -$$c\cdot d_X(x,y)^\alpha\le d_Y(f(x),f(y))\le C\cdot d_X(x,y)^\alpha$$ -for all $x,y\in X$. -The packing dimension of a compact metric space $(X,d)$ in the (finite or infinite) number -$$Dim(X)=\limsup_{\varepsilon\to 0}\frac{\ln N_\varepsilon(X)}{\ln(1/\varepsilon)},$$ -where $N_\varepsilon(X)$ is the cardinality of the smallest cover of $X$ by subsets of diameter $\le\varepsilon$. -Remark. By the Assouad Embedding Theorem, a metric space $X$ admits an equi-Hölder embedding in a finite-dimensional Hilbert space if and only if $X$ is doubling. It can be shown that doubling metric spaces have finite packing dimension. - -REPLY [3 votes]: Not an answer, but related: -In [1] (see also [2]) the authors prove the following: -If $X$ has packing dimension (in your sense; they call it "fractal dimension") less than $m/2$, then $X$ has an embedding $f$ into $\mathbb{R}^m$ such that -$$ L d(x,y)^\gamma \leq |f(x)-f(y)| \leq d(x,y) $$ -for all $x,y\in X$, where $L>0$ and $\gamma\geq 1$. -This is not the same as what you ask, but it is the only result of this type that I know that assumes finite packing dimension rather than finite Assouad dimension. -[1] C. Foias, E. Olson, E. Finite fractal dimension and Hölder-Lipschitz parametrization. Indiana Univ. Math. J. 45 (1996), no. 3, 603–616. -[2] J. Robinson, -Dimensions, embeddings, and attractors. -Cambridge Tracts in Mathematics, 186. Cambridge University Press, Cambridge, 2011.<|endoftext|> -TITLE: On HTT's Lemma 3.3.4.1 -QUESTION [10 upvotes]: While studying the book Higher Topos Theory I have encountered some difficulty with Lemma 3.3.4.1, which says that the pullback along a cartesian fibration of a map q such that $q^{op}$ is cofinal is a cartesian equivalence (i.e. a weak equivalence in the model structure on marked simplicial sets) once we declare the marked 1-simplices in both the domain and codomain to be the cartesian edges with respect to the appropriate cartesian fibrations. -But this seems to imply that such a q is a weak categorical equivalence, i.e. a weak equivalence in the Joyal model structure, at least when it's between fibrant objects. -This essentially because we can consider the pullback along the identity map on the codomain of q, which surely is a cartesian fibration, and then use the fact that the cartesian model structure on marked simplicial sets is Quillen equivalent to the Joyal one. -Now, this fact is clearly false, since we can pick one of the two maps $\Delta^0 \to \Delta^1$ which is not a categorical equivalence. -Am I missing something or is there a problem in the proof? I think, if ever, the problem might lie in the commutative square that appears in the proof, and in the statement that the upper horizontal arrow is a Joyal equivalence. - -REPLY [12 votes]: The lemma is true as stated. The Cartesian equivalence in question does not imply an equivalence in the Joyal model structure after forgetting the markings; rather, it implies an equivalence after formally inverting the markings in question. In more detail: - -The forgetful functor $\mathrm{Set}^+_{\Delta} \to \mathrm{Set}_{\Delta}$ is a (right) Quillen equivalence with the Cartesian model structure on the left and the Joyal model structure on the right. But one must take care that the derived functor is computed on fibrant objects. Fibrant objects on the source are $\infty$-categories $\mathcal{C}$ marked by their equivalences. So, given a pair $(\mathcal{C}, \mathcal{E})$, the corresponding $\infty$-category under this equivalence is really a model for the localization $\mathcal{C}[\mathcal{E}^{-1}]$. -For the case in question, $X$ is marked by the $p$-cartesian morphisms, and $X'$ by the $p'$-cartesian morphisms. So the fact that $X' \to X$ is a Cartesian equivalence means that the functor becomes an equivalence after inverting those morphisms. -The example with the identity map is actually a great example: if $X = S$ then every morphism is $p$-cartesian. So the lemma now says: if $S' \to S$ is initial, then it induces an equivalence after inverting all the morphisms. Another way of saying that is that it induces an equivalence on 'classifying spaces', or that it induces an equivalence in the Quillen model structure- and that's true! Initial and final maps are homotopy equivalences. - -The intuition is supposed to be that, if $p: X \to S$ is Cartesian, $X$ is like an oplax colimit of the diagram $S^{op} \to \mathsf{Cat}_{\infty}$ (this has been made precise by Gepner-Haugseng-Nikolaus). The difference between an oplax colimit and a colimit is that some comparison morphisms are just morphisms instead of equivalences, so you fix that by inverting them. So (as is shown in different language in this section of HTT) the localization $X[\mathcal{E}^{-1}]$ is meant to model the colimit of an $S^{op}$ shaped diagram. But colimits may be computed after restricting along a final map $S^{'op} \to S^{op}$, whence the lemma. (Obviously this reasoning is circular, since the point of the section is to justify the claim that $X[\mathcal{E}^{-1}]$ models the colimit, but hopefully the lemma is more believable now.)<|endoftext|> -TITLE: The Tits classes of simply connected simple real groups -QUESTION [6 upvotes]: Let $G$ be a simply connected semisimple group over a perfect field $k$ (at the moment I am interested in the case $k=\mathbb R$). -Then $G$ is an inner form of a quasi-split $k$-group $G_{\rm qs}$: -there exists a quasi-split form $G_{\rm qs}$ of $G$ and a 1-cocycle $c\in Z^1(k,\overline{G}_{\rm qs})$ such that -$G=\,_c G_{\rm qs}$, the inner twist of $G_{\rm qs}$ by the cocycle $c$, where $\overline{G}_{\rm qs}=G_{\rm qs}/Z_{\rm qs}$ and $Z_{\rm qs}=Z(G_{\rm qs})$. -Let $\xi=[c]\in H^1(k,\overline{G}_{\rm qs})$, the cohomology class of the cocycle $c$. -Let -$$\Delta\colon H^1(k,\overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ -denote the connecting map from the cohomology exact sequence -$$H^1(k,Z_{\rm qs})\to H^1(k,G_{\rm qs})\to H^1(k, \overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ -induced by the short exact sequence -$$ 1\to Z_{\rm qs}\to G_{\rm qs}\to \overline{G}_{\rm qs}\to 1.$$ -By definition, the Tits class of $G$ is -$$t_G=\Delta(\xi)\in H^2(k,Z_{\rm qs}),$$ -see The Book of Involutions, (31.7). -Note that $Z_{\rm qs}=Z:=Z(G)$. - -Question 1. Is there anywhere in the literature or in the Internet a table of $t_G$ for all simply connected absolutely simple ${\mathbb R}$-groups $G$? - -For all connected Dynkin diagram except for ${\mathsf D}_{2m}$ (and also for ${\mathsf D}_{2m}$ when $G_{\rm qs}$ is not split) -we have either $H^2({\mathbb R},Z_{\rm qs})=0$ or $H^2({\mathbb R},Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}$. -From the cohomology exact sequence we see that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, $\xi$ comes from $H^1({\mathbb R}, G)$. -I have a table of strong inner forms of $G_{\rm qs}$, and so I know when $t_G=0$; for non-strong inner forms we have $t_G\neq 0$, hence $t_G=1+2{\mathbb Z}\in {\mathbb Z}/2{\mathbb Z}$. -However, for ${\mathsf D}_{2m}$, in the case when $G_{\rm qs}$ is split, namely $G_{\rm qs}\simeq {\bf Spin}(2m,2m)$, we have $H^2({\mathbb R}, Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}\times{\mathbb Z}/2{\mathbb Z}$. -I know that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, if $G\simeq{\bf Spin}(2m+4q, 2m-4q)$, -but I need an explicit formula for the cases ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and the quaternionic form ${\bf Spin}^*(4m)$. -Thus Question 1 reduces to the following question: - -Question 2. What are the Tits classes for the simple ${\mathbb R}$-groups ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and ${\bf Spin}^*(4m)$ ? - -In order to formulate an answer to Question 2, one needs an explicit description of the center of the split ${\mathbb R}$-group ${\bf Spin}(2m,2m)$. -Such a description is given, for example, in Brian Conrad's cheat sheet. - -REPLY [2 votes]: Question 1: I haven't seen an explicit table in the literature of the Tits classes for simple $R$-groups. -That said, such a table can be constructed from tables in the literature. Specifically, the Tits class is determined by the Tits algebras corresponding to the minuscule dominant weights by Proposition 7 in my paper Outer automorphisms of algebraic groups and determining groups by their maximal tori -(Michigan Mathematical Journal 61 #2 (2012), 227-237). -These Tits algebras are described for each simple type in general terms for any $k$ in section 27 of The Book of Involutions, and their precise values for $k = \mathbb{R}$ are available in several references, for example Tits's Tabellen zu den einfachen Lie Gruppen und ihren Darstellungen. -Question 2: Victor answered this question in the language of Tits algebras, not the Tits class, so I just translate his answer. Identify the projections on the two components of $H^2(\mathbb{R}, Z_{qs}) \cong \mathbb{Z}/2 \times \mathbb{Z}/2$ with the highest weights of the half-spin representations as in section 4.3 of Tits's paper Représentations linéaires irréductibles d'un groupe réductif sur un corps quelconque; this agrees with the identification of $Z_{qs}$ with $\mu_2 \times \mu_2$ that you refer to. For $SO(2m + 4q + 2, 2m - 4q - 2)$, both half-spin representations are quaternionic, so the Tits class is $(1,1)$. For $SO^*(4m)$, one half-spin representation is quaternionic and the other is real, so the Tits class is $(0,1)$ or $(1,0)$.<|endoftext|> -TITLE: Topological embeddings of real projective space in euclidean space -QUESTION [20 upvotes]: I was wondering whether the real projective space $\Bbb{R}P^n$ embeds topologically into $\Bbb{R}^{n+1}$ for odd $n$. -It certainly doesn't for even $n$ because of Alexander duality. Also it doesn't embed smoothly for any $n$. I will prove the two above statements in my algebraic topological class, but I couldn't find anything in the literature for topological embeddings in the odd case. - -REPLY [14 votes]: The first page of W.~Massey's paper On the imbeddability of the real projective spaces in Euclidean space states that $\mathbb RP^n$ with $n>1$ cannot be imbedded topologically in $\mathbb R^{n+1}$ because its mod $2$ cohomology algebra does not satisfy a certain condition given by R. Thom. -The reference points to Theorem V.15 (p.180) of Thom's paper -Espaces fibrés en sphères et carrés de Steenrod, -Annales scientifiques de l'École Normale Supérieure, Série 3 : Volume 69 (1952) , p. 109-182. -EDIT: Actually, a proof can be found in Is it true that all real projective space $RP^n$ can not be smoothly embedded in $R^{n+1}$ for n >1.<|endoftext|> -TITLE: Online Interactive mathematics games for mathematicians or mathematicians-to-be! -QUESTION [6 upvotes]: I am aware that this is not a research level mathematics question. Also, it would not have a single answer (if any) and yet, I hoping that it will be considered as a community wiki question worthy of MO attention. -About four months ago, I have created a website in honor of Maryam Mirzakhani. Timothy Gowers has kindly introduced it as follows. - -[It is] a website in memory of Maryam Mirzahkani. I don’t mean that it - is a memorial website: rather, it has taken an important aspect of her - life — her interaction with other like-minded mathematicians at a - young age — and aims to facilitate such interactions for others, by - the setting up of maths4maryams groups. - -It is just a start of an idea and with my nonexisting website making ability it is growing very slowly. I am hoping that the members of each group and the groups find their own way of mathematically interacting with each other without any intervention by the admins of the website (that is currently one!). However, I thought it would be nice if I could add one or two online interactive games on the main page of the website to break the ice. -Considering that the users of the website Mathematics 4 Maryams are mathematically inclined, I am wondering if there is any online interactive (mathematics) game you might want to suggest. - -REPLY [4 votes]: I'm not sure if this is the sort of thing that you have in mind, but there's a text-based interactive adventure game called A Beauty Cold and Austere. It was written by Mike Spivey and is highly mathematical.<|endoftext|> -TITLE: Smallest tile to *isohedrally* tessellate the hyperbolic plane -QUESTION [5 upvotes]: Is there a smallest tile (in terms of diameter) that isohedrally tessellates the hyperbolic plane? -In this question, we ask the same question without the isohedral requirement, and the answer was no. We also asked about area instead of diameter (which turned out to not matter). I'm wondering if the isohedral requirement will lead to there being a smallest tile. - -REPLY [7 votes]: To clarify, I'm assuming that by isohedral, you mean a tiling of the hyperbolic plane in which the symmetry group acts transitively on the tiles. In this case, I can prove that the $(2,3,7)$-triangle group tiling by triangles with angles $\pi/2,\pi/3,\pi/7$ achieves the minimal diameter (as suggested in the comments). - -First, to answer the substance of the question, one may prove that a minimal diameter isohedral tile exists. The bound on the diameter bounds the area of the quotient orbifold. Moreover, the orbifold must be of diameter bounded by that of the fundamental domain. One may worry that a sequence of diameters of isohedral tiles approaching the infimum might not converge to the diameter of an isohedral tile. However, one may argue by the compactness of the Hausdorff metric for hyperbolic 2-orbifolds of bounded diameter and their bounded diameter fundamental domains that the sequence converges, and hence a minimal diameter will be realized. -More explicitly, let's show that the $(\pi/2,\pi/3,\pi/7)$-triangle achieves the infimal diameter of isohedral tiles. -The diameter of this right hyperbolic triangle is the hypotenuse $l$ (projection to either of the legs is length-decreasing) satisfying $\cosh(l)=\cot(\pi/3)\cot(\pi/7)$ by the hyperbolic law of cosines. Then we may bound the area of a fundamental domain of diameter at most $l$, since it lies in a disk of radius $\leq l$. In fact, we may do a bit better: let $F$ be a compact fundamental domain, then there is a pair of points $x,y \in F$ such that $d(x,y)=diam(F)=d$. Then $F\subset D(x,d)\cap D(y,d)$, the intersection of two disks of radius $d$ whose centers are distance $d$ apart. - -A computation in hyperbolic geometry gives the area of this lune as $4\delta (\cosh d +1/2) -2\pi$, where $\cos \delta = 1-\frac{1}{1+\cosh d}$. Clearly this area is monotonic in terms of $d$. Hence the area must be $\leq 0.472521$, which we get by setting $\cosh d= \cot(\pi/3)\cot(\pi/7)$, the diameter of the $2,3,7$-triangle. -This area is less than the minimal limit area for 2-orbifolds $\pi/6\approx 0.523599$. -There are a few possible orbifold types with area $\leq 0.523599$, which one may consider by computing the possible euler characteristics and use the Gauss-Bonnet theorem. -In orbifold notation, the possibilities are -$pqr, *pqr$ with $1/p+1/q+1/r <1$, and $p*q$ (with certain restrictions). However, the $pqr$ orbifolds are turnovers which double cover the $*pqr$ triangle orbifolds. Hence their diameter will be at least as big as that of the $*pqr$ orbifold. Similarly, the $p*q$ orbifolds are index two covers of the $*2 p (2q)$ triangle orbifolds. Thus, we need only consider the case of triangle orbifolds. -For triangle orbifolds, the minimal diameter is that of the triangle fundamental domain, since the diameter of any fundamental domain will be greater than or equal to the diameter of the orbifold, which is the same as the diameter of the triangle. -One may then argue that the hyperbolic triangle with angles $(\pi/p,\pi/q,\pi/r)$ will have minimal diameter for $(p,q,r)=(2,3,7)$. -The diameter is the maximum of the edge lengths. One may easily see -that a triangle with angles $(\alpha,\beta,\gamma)$ fits inside any triangle with angles $(\alpha',\beta,\gamma)$, where then angles are acute, and $\alpha' \leq \alpha$. Along the edge opposite to $\alpha$, slide the adjacent edges apart to decrease the angle to $\alpha'$, and making the triangle larger. - -Hence one wants to maximize the angles to minimize the diameter. -For triangle tiles, we therefore want to minimize $p,q,r$ with $1/p+1/q+1/r <1$. Decreasing any one of these by one will either make one of them $1$, or will give the angles of a spherical or euclidean triangle. Therefore the minimal such triples are $(2,3,7), (2,4,5), (3,4,4), (3,3,4)$, obtained from the euclidean triangles and spherical triangles by increasing one coordinate, and then eliminating non-minimal triples. The triples $(3,4,4)$ and $(3,3,4)$ give triangle orbifolds covering orbifolds $(2,4,6)$ and $(2,3,8)$ respectively, so have larger diameter, and hence are not minimal. Then a computation shows that the right triangle with angles $(\pi/2,\pi/4,\pi/5)$ has hypotenuse (and hence diameter) $\cosh^{-1}(\cot(\pi/4)\cot(\pi/5)) > -\cosh^{-1}(\cot(\pi/3)\cot(\pi/7))$, so is not minimal. Thus, the $(2,3,7)$ triangle has minimal diameter.<|endoftext|> -TITLE: How do analysts think about functions with poles at all roots of unity? -QUESTION [11 upvotes]: In branches of algebra impinging on the enumeration of partitions, one often encounters formulas like -$$\prod_i \left( \frac{1}{1-q^i} \right)^{n_i}$$ -for some integers $n_i$. E.g., with $n_i = 1$, this counts partitions, and with $n_i = i$, plane partitions. -Such formulas are usually understood as formal products, but if you tried to take them seriously as analytic functions, you would be thinking about functions with poles at all roots of unity. - -Is there a branch of analysis which studies such functions? Does that theory have consequences in combinatorics? - -(I don’t know even enough analysis to know what tag to put.) - -REPLY [14 votes]: One general warning about interpretation as analytic functions: -Just because an infinite product formula looks like it tells you where the zeros and poles are, it might not work the way it does for finite products. -For example, we can consider a similar-looking infinite product -$$\prod_{i\geq 0}\frac1{1+q^{2^i}} = \frac{1}{1+q+q^2+q^3+\cdots} = 1-q,$$ -which looks like it should have poles at all $2^i$-th roots of -1, but in fact it doesn't have any poles in its analytic continuation. -(My apologies if this was already understood -- is there a less-obvious reason for the claim about having poles at all roots of unity?) - -One analytic result about such combinatorially-inspired functions is the Polya-Carlson theorem, which gives a surprisingly strong dichotomy: -if a power series $f(q) = \sum_{n\geq 0}a_nq^n$ has integer coefficients and radius of convergence 1 (so it defines an analytic function on the open unit disk), then either -a) the function cannot be analytically continued past the unit circle, or -b) the function is rational, of the form $f(q) =\frac{g(q)}{(1-q^m)^n}$ for -polynomial $g(q)$ and -$m,n\in \mathbb N$.<|endoftext|> -TITLE: When is a bilinear form equivalent to a trace form? -QUESTION [10 upvotes]: Associated to a finite, separable field extension $L/K$, there is a natural nondegenerate bilinear form, the trace form, defined by $$\langle x,y \rangle := \mathrm{Tr}_{L/K}(xy)$$ -Now, given a finite dimensional $K$-vector space $V$ with a nondegenerate bilinear form $\langle,\rangle$ what are some interesting/useful necessary or sufficient conditions for $\langle, \rangle$ to be equal to the bilinear form associated to a finite separable field extension $L/K$? Or more generally to some algebra $A/K$? - -REPLY [7 votes]: In general, this is a difficult question. -The answer is completely known if $K$ is a number field. I'm going to rephrase the result in terms of quadratic forms, but it is the same, really. -Before that, i'm gonna give some necessary conditions, valid over an arbitrary field. -Recall first that, if $q$ is a non degenerate quadratic form over an arbitrary field $K$ of characteristic different from $2$, it is isomorphic to a diagonal form $\langle a_1,\ldots,a_n\rangle$, $a_i\in K^\times$. -If $K$ is an ordered field, the signature of $q$ with respect to this ordering is the number of positive $a_i's$ minus the number of negative ones. This is an integer which does not depend on the choice of the diagonalization (and not a pair of natural integers). -We say that a quadratic form is positive if all its signatures (w.r.t. the orderings of $K$) are non-negative. -Of course, if $K$ has no ordering, any quadratic form is positive. -One may show that the trace form of a separable extension $E/K$ (or more generally of an etale $K$-algebra), that i will denote by $q_E$, satisfies the following conditions (here $K$ is an arbitrary field of characteristic different from $2$): - -the quadratic form $q_E$ positive. -If $n=\dim_K(E)$, then $q_E\simeq r_n\perp q'$, where -$r_n$ is defined as follows: -Write $n=\displaystyle\sum_{i=1}^h 2^{m_i},0\leq m_1<\cdots -TITLE: On Street's "australian conspectus" -QUESTION [8 upvotes]: Skimming the australian conspectus of higher category theory I noticed I have a few questions, both mathematical and historical. - -At about the middle of page 6, Kock-Zoberlein monads are defined as "strict monoidal 2-functors $\text{Ord}_\text{fin}\to [{\cal K,K}]$", where $\text{Ord}_\text{fin}$ is the 2-category of finite ordinals, monotone maps and pointwise order between maps. This definition surprises me, as I thought KZ-monads were the 2-dimensional analogue of idempotent monad. Where is the idempotency here? -p. 7: "Gray then pointed out that, for $\mathcal V = [\Delta°, Set]$ (the category of simplicial sets), homotopy limits of $\cal V$-functors could be obtained as limits weighted by the composite $A\xrightarrow{L_A}{\bf Cat}\xrightarrow{N} [\Delta°, Set]$." Really? Wasn't this first outlined in Bousfield and Kan's book on homotopy limits and completions? -p. 8: "In sheaf theory there are various ways of approaching the associated sheaf. Grothendieck used a so-called “L” construction. Applying L to a presheaf gave a separated presheaf (some “unit” map became a monomorphism) then applying it again gave the associated sheaf (the map became an isomorphism). I found that essentially the same L works for stacks. This time one application of L makes the unit map faithful , two applications make it fully faithful , and the associated stack is obtained after three applications when the map becomes an equivalence ." I've always found the associated sheaf construction quite striking: it's a left exact localization which is "quadratic" in the sense that it is $L^2$ for some endofunctor $L$. Here Street is telling that $L^3$ works for stacks as it gradually builds faithfulness, fully faithfulness, and essential surjectivity of a map. What's going on? Why is it so? - -REPLY [5 votes]: For 2., there is a sense in which you are right but that is `in hindsight'. The nature of homotopy limits from a categorical point of view was not clear (at least to me). There were several approaches knocking around and sorting them out categorically was not clear. I remember reading B & K and trying to link it with Vogt's ideas. I learnt a lot from John Gray's article and from there Bourn and Cordier started clarifying things, but the ideas and terminology of weighted limits were slow to be fully understood outside a small group of enriched category theorists. Kelly's book appeared in 1982 but the ideas, although available earlier than that took time to be absorbed, and then Gray's insights on B & K's holims became clearer (to me). I am unable to comment on B & K's insights, but I always felt they came from work on the derived functors of the limit functor and various other situations outside a purely categorical context. They were doing homotopy theory and using categorical constructions, not doing category theory. Were they aware of indexed / limits etc.? I do not know. -I chatted to John Gray about this but he had realised the observation that Ross mentioned whilst trying to understand what B & K were doing from a categorical viewpoint and had compared their approach with some points I had made in an article in the Cahiers. (I also noted that Illusie in his thesis had developed the total derived functor of the limit functor independently, but that is not an answer to your question, just a comment in passing!)<|endoftext|> -TITLE: Is $V=\textsf{HOD}\not\Rightarrow\textsf{GCH}$ consistent? -QUESTION [5 upvotes]: Whenever $M$ is some fine-structural $L$-like model we can prove the implication $V=M\Rightarrow\textsf{GCH}$. For $L$ this is due to Gödel, and for the modern extender models it follows simply by construction. The most recent direction in the inner model programme searches for fine-structural models of the form $\textsf{HOD}^N$ with $N$ some "nice" model (most recently this includes $N=M_n(x,g)$ for a Turing cone of reals $x$ and $g$ generic for making its least $N$-inaccessible countable). My question is then whether $V=\textsf{HOD}\Rightarrow\textsf{GCH}$ has been shown to be unprovable in general (modulo large cardinals of course)? Is there some pathological $\textsf{HOD}$ which "obviously" doesn't satisfy $\textsf{GCH}$? Also, I'm not even sure if $\textsf{HOD}^{\textsf{HOD}^N}=\textsf{HOD}^N$ holds for these models, as this (apparently) doesn't hold in general. - -REPLY [12 votes]: Yes, one can produce a model of $ZFC+V=HOD$ in which the $CH$ fails. This should follow from Consistency results about ordinal definability. -In fact, I think the arguments of my paper HOD, V and the GCH can be used to produce a model of $ZFC+V=HOD$ in which the $GCH$ fails everywhere. -Also by a result of Apter (see Large cardinals need not be large in HOD for a proof), if there are class many Laver indestructible supercompact cardinals, then $V=HOD$ holds. Note that in such a model $GCH$ must fail at class many regular cardinals.<|endoftext|> -TITLE: Sophus Lie's contribution to solution of problems of variational type as in Euler and Lagrange -QUESTION [7 upvotes]: The original impetus for Sophus Lie's work was apparently to streamline the solution of certain problems of variational type such as those treated in the work of Euler and Lagrange. This presumably involves solution of differential equations involving groups of symmetries. Is there a reference that explains this historical background? - -REPLY [7 votes]: In - -Vladimir Itskov: Orbit Reduction of Contact Ideals. Contemporary Mathematics. Vol. 285, 2001 - -one reads on page 172: - -As first observed by Sophus Lie [8], the Euler-Lagrange equations of every invariant variational problem can be written in terms of the differential invariants of the group action. In other words, the Euler-Lagrange equations of a group-invariant variational problem can be pushed forward to the orbit space. Surprisingly, up to date there was no general understanding of the meaning of the pushed forward equations on the orbit space, nor there was a general algorithm of producing the group-invariant Euler-Lagrange equations. - -In the above, the "nor there was a" seems not quite correct English and should be 'nor was there a'; the reference '[8]' is: - -Sophus Lie Über Integralinvarianten und ihre Verwertung für die Theorie der Differentialgleichungen, Leipziger Berichte 49, 369-410, 1897. - -and a slightly commented/edited/expanded/corrected version of the above seems to be available here; (I don't know what the referent of the initials 'G.S.' in that document is.)<|endoftext|> -TITLE: A diagram for understanding action/coaction compatibility in a Yetter-Drinfeld module -QUESTION [7 upvotes]: For a Hopf algebra $H$ with antipode $S$, let $M$ be a left $H$-module with the action $h \otimes m \mapsto \rho(h,m)$, and also a left $H$-comodule with coaction $\delta \colon m \mapsto m^{(-1)} \otimes m^{(0)}$. For $M$ to be a Yetter-Drinfeld module, it must satisfy the compatibility condition -$$ - \delta(\rho(h,m)) = h_{(1)}m^{(-1)} S(h_{(3)}) \otimes \rho(h_{(2)},m^{(0)})\,. -$$ -Is there a natural commutative diagram to draw that illustrates this compatibility condition? I've included my attempt below in a CW answer. Also, is there any more reason behind this condition besides "it's the condition we need to be true to get the nice braiding to work out in the Yetter-Drinfeld category?". - -REPLY [3 votes]: The equivalent way of writing the Yetter-Drinfeld compatibility condition between left action and left coaction described in this post can be easily translated into a commutative diagram. It looks like this -$$ -(\nabla\otimes\rho)(1\otimes\tau\otimes 1)(\Delta\otimes\delta)=(\nabla^\mathrm{op}\otimes 1)(1\otimes\delta)(1\otimes\rho)(\Delta^\mathrm{op}\otimes 1)\,. -$$ -Similary equations work for ${}_H\mathcal{YD}^H$ instead of ${}_H^H\mathcal{YD}$, see for example Kassel, Definition IX.5.1, page 220. These equations tell us how $\rho$ and $\delta$ "commute". We need precisely this condition to compute the Drinfeld center of ${}_H\mathrm{Mod}$.<|endoftext|> -TITLE: Who was Heinrich Hake? -QUESTION [21 upvotes]: Hake's Theorem, due to Heinrich Hake of Düsseldorf in 1921, says that an improper Henstock–Kurzweil integral (aka generalized Riemann integral, gauge integral, Perron integral, or Denjoy integral) on a bounded interval is already proper. That is, if $f$ is defined on a half-open interval $[a,c)$, $f$ is HK-integrable on $[a,b]$ for each $b$ satisfying $a \leq b < c$, and the limit of $\int_a^b f$ as $b \to c^-$ exists, then we may define $f(c)$ however we like and find that $f$ is integrable on $[a,c]$ and that $\int_a^c f$ equals the aforementioned limit. (The converse is also true; if $\int_a^c f$ exists, then it may be calculated as a limit.) -I can't find anything about this Hake. Most references just say ‘Hake's theorem’, a few say ‘H. Hake’, and Bartle's book on the HK integral says ‘Heinrich Hake in 1921’. Bartle also gives a reference, an article in Mathematische Annalen, whose byline says ‘Heinrich Hake in Düsseldorf’. And that's it. -Various online search attempts give me references to this theorem, the contemporary Düsseldorf telephone directory, and the 18th-century law professor Ludolf Heinrich Hake, but nothing more about the 20th-century mathematician who proved the theorem. Does anybody know anything about him? - -REPLY [27 votes]: The Deutsche Mathematiker-Vereinigung has a biographical entry: Heinrich Hake was born August 4, 1891 in Düsseldorf. He attended high school in Attendorn (Westfalen), studied in Göttingen and Bonn, where he completed his studies in 1914. After military service he conducted Ph.D. research with prof. Hans Hahn in Bonn, where he defended his dissertation on 2 March 1921, entitled: Über de la Vallée Poussins Ober- und Unterfunktionen einfacher Integrale und die Integraldefinition von Perron. It was published in Math. Ann. 83 (1921) 119-148. -No date of death is listed, but it must have been after 1945. -I note that Bonn University marked the 125 birthday of Heinrich Hake in a memorial calendar.<|endoftext|> -TITLE: Elementary (English) reference for the cotangent complex? -QUESTION [8 upvotes]: I'm trying to understand cotangent complexes and their role in deformation theory, and later the statement that they're somehow natural in a derived scheme/stack. - -I understand that the standard reference is Illusie's two books, unfortunately my French abilities are lacking -I'm aware of the stacks project treatment. It looks quite terse. Is there a more elementary treatment? -I think the nlab article is amazing, shame it's rather brief. - -People who understand the cotangent complex: how did you first learn it? - -REPLY [4 votes]: The homotopy-theoretic way to look at the cotangent complex is as the left derived functor of the Kahler differentials functor. Now: - -This statement makes sense just in the context of homological algebra. i.e. in some category of chain complexes of modules. This route to constructing the cotangent complex is the subject of section 8.8 in Weibel's homological algebra book. -More generally we have notions of derived functors in any model category, a good reference is Dwyer and Spalinski. Constructing the cotangent complex at this level of generality is the subject of a very nice note by Zhou.<|endoftext|> -TITLE: Iterated forcing with distributive forcing notions -QUESTION [9 upvotes]: Assume $(\kappa_n| n<\omega)$ is an increasing sequence of inaccessible cardinals with $\kappa_\omega=\sup_{n<\omega}\kappa_n.$. Let -$((\mathbb{P}_n| n \leq \omega), (\dot{\mathbb{Q}}_n | n<\omega))$ -be a full support iteration of forcing notions, where for each $n< \omega$, we have $\Vdash_{\mathbb{P}_n}$``$\dot{\mathbb{Q}}_n$ has size < $\kappa_{n+1}$ and does not change $\dot{V}_{\kappa_n}$''. - -Question (a) Does forcing with $\mathbb{P}_\omega$ add a new real$?$ -(b) Does forcing with $\mathbb{P}_\omega$ collapse $\kappa_\omega?$ - -Remark. I assume each $\kappa_n$ is inaccessible but not Mahlo. - -REPLY [2 votes]: One can get at least a counter example for (2) from countably many measurable cardinals. -Let $\mu_n$ be a measurable cardinal between $\kappa_n$ and $\kappa_{n+1}$. Let $\kappa_{\omega} = \sup \kappa_n = \sup \mu_n$. Let me assume that there is no inner model with a Woodin cardinal. -Let $\mathbb{Q}_n$ be the Prikry forcing for singularizing $\mu_n$ (as defined in the generic extension by $\mathbb{P}_n$). Let me claim that the full support iteration $\mathbb{P}_\omega$ collapses $\kappa_{\omega}^+$ and does not add reals. -Let $\langle P_n \mid n < \omega\rangle$ be the sequence of the generic Prikry sequences. -Claim: For $r \in {}^\omega \omega$ let $g_r(n) = P_n(r(n))$. Then $\left(\prod \mu_n\right)^V$ and $\{g_r \mid r\in {}^\omega \omega\}$ are interleaved. -Proof: First, let us note that $\mathbb{P}_\omega$ does not add reals. Indeed, let $\dot{r}$ be a name for a new real and let $p \in \mathbb{P}_\omega$. Let us define by induction a sequence of conditions $p_n$ such that for all $n$: $p_n \restriction n \leq^* p_{n-1} \restriction n$, $p_n \leq p_{n-1}$ and $p_n$ decides the value of $n \in \dot{r}$. This is done using the Prikry property of $\mathbb{P}_n$. -Next, let $f\in \prod \mu_n \cap V$. Using the Prikry Property of the finite iterations we can show that for every condition $p$ there is a direct extension $p'$ such that for all $n$, $p'$ decides the length of the stem of $p'(n)$. Let us extend $p'$ in all coordinates in order to obtain a function which dominates $f$ everywhere. -On the other hand, if $r\in {}^\omega \omega$, let $p$ be a sufficiently strong condition so that $p$ decides the length of its stems and $\mathrm{len}\ \mathrm{stem}(p(n)) > r(n)$. Now the the name for the ordinal $g_r(n)$ is a $\mathbb{P}_n$-name. Since this is a small forcing (of size $<\mu_n$), there is some $\beta_n < \mu_n$ such that $\Vdash \dot{g}_r(n) < \beta_n$. Let $f(n) = \beta_n$, then clearly $p$ forces that $f$ dominates $g_r$. QED -We conclude that the cofinality of $\kappa_{\omega}^{+}$ in the generic extension is at most the continuum. By the weak covering lemma, $\kappa_\omega$ cannot be a cardinal in the generic extension.<|endoftext|> -TITLE: regulator of an elliptic curve rational/irrational/transcendental? -QUESTION [5 upvotes]: Let $K$ be a number field and $E/K$ an elliptic curve (or abelian variety) with $\mathrm{rk}\,E(K) > 0$. Can/will the elliptic (abelian) regulator $\mathrm{Reg}(E/K)$ be rational/irrational/transcendental? - -REPLY [6 votes]: There is currently no example of a number field $K$, an elliptic curve $E/K$, and a non-torsion point $P\in E(K)$, for which it is known that either $\hat h_E(P)$ or $\hat H_E(P)$ is not rational. However, there is an old result of Daniel Bertrand in which he shows in certain cases that the $p$-adic canonical height is transcendental over $\mathbb Q$, which is in fact how he proves that it's non-zero!<|endoftext|> -TITLE: New binomial coefficient identity? -QUESTION [21 upvotes]: Is the following identity known? -$$\sum\limits_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n+k}{n-k}\binom{2k}{k}= -\frac{1}{2n+1}$$ -I have not found it in the following book: - -Henry Wadsworth Gould, Combinatorial identities: a standardized set of tables listing 500 binomial coefficient summations, Morgantown, West Virginia, 1972. - -REPLY [7 votes]: Use $\binom{n+k}{k}\binom{n}k$ in the sum. Define the functions -$$F(n,k)=(-1)^k\frac{2n+1}{2k+1}\binom{n+k}k\binom{n}{k}, \qquad -G(n,k)=2\cdot(-1)^{k-1}\binom{n+k}{k-1}\binom{n}{k-1}.$$ -Then $F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$. Sum over all integers $k$ to obtain $$f(n+1)-f(n)=0$$ where $f(n)=\sum_kF(n,k)$ is your sum. Since $f(0)=1$, the identity follows. -This method is called the Wilf-Zeilberger technique of summation routine.<|endoftext|> -TITLE: Which Banach spaces are absolute Lipschitz extensors for compacta? -QUESTION [5 upvotes]: A metric space $X$ is defined to be an absolute Lipschitz extensor for compacta if each Lipschitz map $f:K\to K$ defined on a compact subset $K\subset X$ extends to a Lipschitz map $\bar f: X\to X$. -Question. Is each Banach space an absolute Lipschitz extensors for compacta? -I admit that the answer to this question is negative. In this case the Question can be refined to -Problem. Which Banach spaces are absolute Lipschitz extensors for compacta? -Remark. The class of absolute Lipschitz extensors for compacta includes all absolute Lipschitz retracts, so it includes all Hilbert spaces and all Banach spaces $C_u(M)$ of bounded continuous functions on a metric space $M$. What about uniformly rotund Banach spaces (are they absolute Lipschitz extensors for compacta)? Are the Banach spaces $L_p(\mu)$ absolute Lipschitz extensors for compacta? - -REPLY [6 votes]: Here is one way you can prove that a space $X$ is not an absolute extensor for compacta: Find sequences $(E_n)$ and $(F_n)$ of finite dimensional subspaces of $X$ and a constant $C$ so that for every $n$ there is a linear isomorphism $T_n$ from $E_n$ onto $F_n$ s.t. $\|T_n\|=1$, $\|T_n^{-1}\|\le C$, each $F_n$ is the range of a projection of norm at most $C$, but the $E_n$ are not uniformly complemented. If the restriction of $T_n $ to the unit ball $B_{E_n}$ of $E_n$ has a $D$ Lipschitz extension from $B_X$ into $X$, then it has a $CD$-Lipschitz extension into $F_n$, and the positively homogenous extension of this mapping is a $3CD$-Lipschitz extension of $T_n$ to a mapping from $X$ into $F_n$. By Lindenstrauss' 1964 paper on non linear projections (see the book of Benyamini-Lindenstrauss), there is then a linear extension $S_n:X \to F_n$ with $\|S_n\|\le 3DC$ so that $T_n^{-1}S_n$ is a projection from $X$ onto $E_n$ having norm at most $3DC^2$. So if such an X is an absolute extensor for compacta, there is no uniform Lipschitz bound on extensions of non expansive linear mappings from finite dimensional subspaces. -It remains to show that if a space $X$ is an absolute extensor for compacta, then there is a uniform bound on extensions of non expansive linear mappings defined on finite dimensional subspaces. Suppose there is no such uniformity for $X$. Notice that then if $Y$ is a finite codimensional subspace of $X$, then there also is no uniformity for extensions of non expansive linear mappings defined on unit balls of finite dimensional subspaces of $Y$ into $X$. Now use the Mazur technique for constructing basic sequences to build a finite dimensional Schauder decomposition $(E_n)$ for some subspace of $X$ and non expansive linear mappings $f_n: E_n \to X$ s.t. any extension of $f_n$ to $X$ has Lipschitz constant at least $n $. Let $K = \cup_n n^{-1} B_{E_n}$ and define $F:K\to X$ by $f(x) = f_n(x)$ if $x\in n^{-1} E_n$. Then $K$ is compact and $f$ is Lipschitz and $f$ has no Lipschitz extension to $X$. -There are many spaces that contain sequences $(E_n)$ and $(F_n)$ of finite dimensional subspaces for which there is a constant $C$ so that for every $n$ there is a linear isomorphism $T_n$ from $E_n$ onto $F_n$ s.t. $\|T_n\|=1$, $\|T_n^{-1}\|\le C$, each $F_n$ is the range of a projection of norm at most $C$, but the $E_n$ are not uniformly complemented. -If $X$ is $ \ell_p$ or $L_p$, $1\le p \not=2 <\infty$, then $E_n$ can be taken to be uniformly isomorphic to $\ell_p^n$. If $X$ is super reflexive (or even just has non trivial type) but does not have type $p$ for some $p<2$, then $E_n$ can be taken to be uniformly isomorphic to $\ell_2^n$. These results are fairly deep, BTW, but well known to researchers in Banach space theory.<|endoftext|> -TITLE: On a pattern for upside-down Ramanujan pi formulas -QUESTION [20 upvotes]: Define, -$$\lambda_n =\frac{(\tfrac12)_n}{(1)_n} =\frac{(\tfrac12)_n}{n!} =\frac{\tbinom{2n}{n}}{2^{2n}} =\binom{n-\tfrac12}{n}$$ -with Pochhammer symbol $(x)_n$ and binomial $\tbinom{n}{k}$. I noticed that the following 14 formulas have a nice "affinity". - -Level 3: - -$$\sum_{n=0}^\infty \lambda_n^3\, \frac{6n+1}{2^{2n}} =\frac{2^2}{\pi}\tag1$$ -$$\sum_{\color{red}{n=1}}^\infty \frac1{n^3\lambda_n^3}\, \frac{6\big(n-\tfrac12\big)+1}{2^{2n}} =\pi^2\tag2$$ - -$$\sum_{n=0}^\infty \lambda_n^3\, \frac{42n+5}{2^{6n}} =\frac{2^4}{\pi}\tag3$$ -$$\sum_{\color{red}{n=1}}^\infty \frac1{n^3\lambda_n^3}\, \frac{42\big(n-\tfrac12\big)+5}{2^{6n}} =\frac{\pi^2}3\tag4$$ - -$$\sum_{n=0}^\infty \lambda_n^3\, \frac{4n+1}{(-1)^{n}} =\frac{2}{\pi}\tag i$$ -$$\sum_{\color{red}{n=1}}^\infty \frac1{n^3\lambda_n^3}\, \frac{4\big(n-\tfrac12\big)+1}{(-1)^n} =-16G\tag{ii}$$ - -$$\sum_{n=0}^\infty \lambda_n^3\, \frac{6n+1}{(-2^3)^{n}} =\frac{2\sqrt2}{\pi}\tag{iii}$$ -$$\sum_{\color{red}{n=1}}^\infty \frac1{n^3\lambda_n^3}\, \frac{6\big(n-\tfrac12\big)+1}{(-2^3)^n} =-4G\tag{iv}$$ -with Catalan's constant $G$. - -Level 5: - -$$\sum_{n=0}^\infty \lambda_n^5\, \frac{20n^2+8n+1}{(-2^2)^n}=\frac{2^3}{\pi^2}\tag5$$ -$$\sum_{\color{red}{n=1}}^\infty \frac1{n^5\lambda_n^5}\, \frac{20\big(n-\tfrac12\big)^2+8\big(n-\tfrac12\big)+1}{(-2^2)^n} =-56\zeta(3)\tag6$$ - -$$\sum_{n=0}^\infty \lambda_n^5\, \frac{205n^2+45n+\tfrac{13}4}{(-2^{10})^n}=\frac{2^5}{\pi^2}\tag7$$ -$$\sum_{\color{red}{n=1}}^\infty \frac1{n^5\lambda_n^5}\, \frac{205\big(n-\tfrac12\big)^2+45\big(n-\tfrac12\big)+\tfrac{13}4}{(-2^{10})^n} =-2\zeta(3)\tag8$$ -with Apery's constant $\zeta(3)$. - -Level 7: - -$$\sum_{n=0}^\infty \lambda_n^7\, \frac{84n^3+38n^2+7n+\tfrac12}{2^{6n}} =\frac{2^4}{\pi^3}\tag9$$ -$$\sum_{\color{red}{n=1}}^\infty \frac1{n^7\lambda_n^7}\, \frac{84\big(n-\tfrac12\big)^3+38\big(n-\tfrac12\big)^2+7\big(n-\tfrac12\big)+\tfrac12}{2^{6n}} =\frac{\pi^4}2\tag{10}$$ - -Most of these are scattered throughout the literature in various guises. See, for example, Guillera and Rogers' paper "Ramanujan Series Upside Down" which focuses on level 3. The level 3 formulas for 1/pi were found by Ramanujan and can be explained by modular forms, while $(9)$ is by Gourevitch and $(10)$, in a different guise, is by MO user zy_. In this post, he remarked that Guillera, in private correspondence, considered it as new. (Note that its partner was found by Gourevitch and included way back in a 2003 paper by Guillera.) - -Q: What is the unifying theory for these ten formulas, and can we find paired examples for higher levels, like for $\zeta(5)$? (There is a Ramanujan-type formula for $\zeta(5)$ found by zy_ in the post cited, but it does not use $\lambda_n$ and doesn't seem to have a "partner".) - -REPLY [10 votes]: All these pairs of formulas are examples of the transformation $n \to n + \frac12$ removing a factor which does not depend on n. The upside-down transformation is essentially $n \to -n$ (therefore it changes $z$ to $z^{-1}$) reinterpreting $(a)_{-n}$ as $\frac{(-1)^n}{(1-a)_n}$ if $a \neq 1$, and $(1)_{-n}$ as $\frac{n(-1)^n}{(1)_n}$ which preserves formally the recurrence $\Gamma(x+1)=x \, \Gamma(x)$ (see Chapter 7 of the book A=B by Petkovsek, Wilf, Zeilberger) and another application to the WZ-method in the Section 4 of this paper). -The very nice formula for $\zeta(5)$ discovered by zy_ allows us to discover a new "divergent" (convergent by analytic continuation) Ramanujan-like series for $1/\pi^4$ by using the upside-down-transformation. -The transformation $n \to n+\frac12$ applied to $\lambda(n)$ essentially inverts $\lambda(n)$ giving $\frac{1}{\lambda(n)}$ but does not invert $z^n$. Hence it is not an upside-down transformation. This explains why the pattern observed in the post cannot be generalized in the way pointed out by the author. -In the Appendix of this paper there are examples of the "upside-down" technique. Another example is in the Addendum of this paper. In this unpublished file there are many examples of the transformation $n \to n + \frac12$.<|endoftext|> -TITLE: What is the smallest sphere whose surface includes 100 integer points? -QUESTION [9 upvotes]: Let $S(r)$ be the surface of the origin-centered sphere in $\mathbb{R}^3$. -A point is an integer point if all its coordinates are integers. - -What is the smallest radius $r_n$ such that $S(r_n)$ includes $\ge n$ integer points? - -What is the growth rate of $r_n$ with respect to $n$? -Is there an algorithm that could compute $r_n$ for a specific given $n$? -It is known that rational points (all coordinates rational) are dense on $S(1)$: -see, e.g., the MO question -Rational points on a sphere in $\mathbb{R}^d$. -One possible approach is via rational points of bounded height. -The height of a rational $a/b$ in lowest terms is max$(|a|,|b|)$, and -the height of a rational point is the max of the heights of its coordinates. - -          - - -          - -Rationals of height $\le 2048$ on sphere. -Image due to Stefan Kohl in this answer. - - -Choose an appropriate $h_\max$, perhaps using an -estimate -of the number of rational points of height at most $h_\max$ on $S(1)$. -Then scale all coordinates by the LCM of the points' denominators. -For example, for $h_\max=10$, scaling by $2^3 \cdot 3^2 \cdot 5 \cdot 7 = 2520$ -would suffice to clear all denominators, -so that $S(2520)$ includes all those points at integer coordinates. -But this would not necessarily result in the minimum $r_n$ for a given $n$. -It would likely be better to use rational points that result in a small LCM. -Exact calculations on a sphere -(for example, computing Voronoi diagrams on a sphere) often need integer points -of bounded size. - -Update. The exact answer to the question in the title, -due to Dap and Gerhard Paseman, is -that a sphere of radius $\sqrt{74} \approx 8.6$ includes $120$ integer points -on its surface, -and smaller spheres include fewer than $100$ points. - -REPLY [5 votes]: The sphere centred on $(1,1,1)/2$ and radius $\sqrt{131}/2\approx 5.723$ contains $24+48+48=120$ points with integer coordinates, thanks to $$2\{\pm x,\pm y,\pm z\}-(1,1,1)=\{9,5,5\},\{9,7,1\},\{11,3,1\}.$$ -Successive best $a,n$ where the sphere centred on $(1,1,1)/2$ and radius $\sqrt{a}/2$ contains $n$ points with integer coordinates: -(3,8),(11,24),(27,32),(35,48),(59,72),(131,120),(251,168),(299,192),(419,216),(611,240),(659,264),(731,288),(899, -336),(971,360),(1091,408),(1691,432),(1739,480),(1811,552),(2219,576),(2651,624),(2939,696),(3251,744),(4091,792),( -4259,840),(4619,864),(5099,936),(5771,1056),(6971,1080),(7619,1104),(8291,1128),(8531,1200),(9539,1320),(11051,1488),( -12011,1560),(13859,1608),(14339,1680),(15539,1728),(18851,1776),(19211,1800),(19379,1944),(20459,1968),(22571,2088),( -25091,2112),(25451,2160),(26171,2184),(26771,2352),(28019,2376),(31379,2472),(31979,2496),(33491,2592),(36539,2736),( -38099,2784),(38939,2832),(39731,3024),(42059,3120),(49139,3144),(51939,3168),(53819,3192),(55571,3360),(58211,3432),( -59219,3672),(65771,3696),(66491,3960),(74051,4200),(87779,4416) -GAP code: -a:=[];R:=89999;for i in [1..R] do a[i]:=0;od; -for x in [1,3..299] do - for y in [1,3..x] do - for z in [1,3..y] do -r:=x*x+y*y+z*z; if r<=R then -if x>y and y>z and z>0 then n:=48; - else if (x>z and z>0) or (x>y and z=0) then n:=24; - else if x=z then n:=8; - else if y=0 then n:=6;fi;fi;fi;fi; -a[r]:=a[r]+n;fi;od;od;od; -b:=0;for i in [1..R] do if a[i]>b then b:=a[i];Print("(",i,",",b,"),");fi;od;<|endoftext|> -TITLE: Spectrifications of spacewise homotopy equivalences -QUESTION [5 upvotes]: One can show that spectrifications of maps between $\Sigma$-cofibrant prespectra that are spacewise homotopy equivalences are homotopy equivalences of spectra. I'm interested in the following: -Does this result hold if only the source prespectrum is assumed to be $\Sigma$-cofibrant? -The reason for my interest is the following. The adjoint from prespectra to inclusion prespectra is difficult to deal with. I see that sometimes people replace a prespectrum by a thickened one, via a cylinder construction, as in Chapter X, Section 5 of Elmendorf-Kriz-Mandell-May's "Rings, modules and algebras in stable homotopy theory", to get a $\Sigma$-cofibrant prespectrum so that we have a nicer description of the spectrification. This is done, for example, when defining the topological Hochschild spectrum in Hesselholt-Madsen's "On the $K$-theory of finite algebras over Witt vectors of perfect fields". Say the original prespectrum is $t$, the cylinder construction yielding a $\Sigma$-cofibrant prespectrum is denoted by $K$ and spectrification is denoted by $L$. We have a map of prespectra $Kt \to t$ which is a spacewise homotopy equivalence. I would hope that $LKt$ and $Lt$ are related in a nice way so as to say that the intermediate thickening didn't essentially change anything. The "nice way" I'm hoping for is that one can deduce that the induced map $LKt \to Lt$ is a homotopy equivalence (or perhaps weak homotopy equivalence). - -REPLY [9 votes]: This is false. Unfortunately, lack of cofibrancy in this kind of situation can mean an almost entire loss of control over what happens to the spectrification. Let's give an example. -Here is a prespectrum $S$: it associates to an inner product space $V$ the one-point compactification $S^V = V \cup \{\infty\}$. -For every $V$, there is a subspace $D(V)$ which is the complement of the open ball of radius $1/\dim(V)$, and a quotient space $T(V) = S^V / D(V)$. This quotient is the one-point compatictification of the ball of radius $1/\dim(V)$, so $T(V)$ is a sphere where we have collapsed all the long vectors. We have natural structure maps $S^W \wedge T(V) \to T(W \oplus V)$ that are compatible across $V$ and $W$ (the right-hand term is a larger quotient of $S^{W \oplus V}$ than the left-hand one). These assemble together into a prespectrum $T$ and there's a natural map $S \to T$ which is levelwise a homotopy equivalence. Moreover, $S$ is $\Sigma$-cofibrant (it's the suspension prespectrum of $S^0$). -$S$ will spectrify to the sphere spectrum. However, $T$ spectrifies to something entirely different. Any map from $T$ to a spectrum $Y$ (whose structure maps are homeomorphisms) will factor through -$$T(V) \to {\rm colim}_{W} \Omega^W T(W \oplus V) \to Y(V)$$ -and any nonzero vector $v$ in $T(V)$ gets sent to a loop through vectors of length strictly greater than or equal to $||v||$, and these become equivalent to the point $\infty$ in the colimit because we are progressively identifying vectors of shorter and shorter length with $\infty$. -Long story short, the spectrification of $T$ factors through a prespectrum $W$ which sends each $V$ to the Sierpinski two-point space, and the map $S \to W$ induces zero on homotopy groups.<|endoftext|> -TITLE: Acyclic Finite Groups -QUESTION [10 upvotes]: A group is called acyclic if its classifying space has the same homology of a point. Examples of acyclic groups include Higman's group with four generators and relations, also known for being an example of an infinite, finitely generated group with no finite quotients, and "SQ-Universal": -$$\langle x_0, x_1,x_2,x_3\mid x_{i+1}x_i x_{i+1}= x_i^2 \hskip .1 in \mathrm{for}\hskip .1 in i=0,\ldots, 3\rangle$$ -and the group of bijections of an infinite countable set. -Is there an example of a finite acyiclic group? Or a reason why such a group must be infinite? - -REPLY [21 votes]: An acyclic finite group is trivial. In fact something even stronger is true. See Culler, Marc Homology equivalent finite groups are isomorphic. Proc. Amer. Math. Soc. 72 (1978), no. 1, 218–220.<|endoftext|> -TITLE: Curves in del Pezzo surfaces satisfying certain intersection inequality -QUESTION [5 upvotes]: Let $X$ be a del Pezzo surface (over $\mathbb{C}$), which is obtained by a blow up $\pi: X \rightarrow \mathbb{P}^{2}$ in a collection of points. Let $H$ be the hyperplane class of $\mathbb{P}^{2}$. -Question: Consider smooth, irreducible curves $\Sigma$ satisfying the inequality $-K_{X}\cdot \Sigma > \frac{1}{2}\Sigma \cdot \Sigma + \pi^{*}H \cdot \Sigma $. -Is the genus of such curves bounded above? - -REPLY [6 votes]: Here is one proof that the genus is bounded (although presumably a more elegant solution exists). -Write $\Sigma=d\pi^*H - \sum_{i=1}^nm_iE_i$ in the usual basis for $\operatorname{NS}(X)$ (i.e. $\Sigma$ is the strict transform of a curve of degree $d$ in $\mathbb{P}^2$, with multiplicity $m_i$ at each of the points which is blown up). -We can rearrange the inequality into the form $\tfrac12(\Sigma^2 + K_X\Sigma) < \tfrac12(d-\sum m_i)$. Then, from the usual genus formula for $\Sigma\subset X$, we have -\begin{equation*} -g_\Sigma = 1 + \tfrac12(\Sigma^2 + K_X\Sigma) < 1 + \tfrac12\left(d-\sum_{i=1}^n m_i\right) -\end{equation*} -so we may assume that $d > 2 + \sum m_i$, else $g_\Sigma \leq 2$. -Now we can prove the result by induction on $n$ (the number of points blown up on $\mathbb{P}^2$). If $n=0$ it is easy to see that the inequality becomes $3d > \tfrac12d^2 + d$, hence $d<4$ and $g\leq1$. -By the induction hypothesis, if $n>0$ then we may assume each $m_i\geq 1$ (else blow down the point with $m_i=0$ and reduce to the $n-1$ case). The inequality is -\begin{equation*} -3d - \sum_{i=1}^nm_i > \frac12 \left(d^2 - \sum_{i=1}^n m_i^2 \right) + d -\end{equation*} -which can be rearranged as -\begin{equation*} -(d - 2)^2 < \sum_{i=1}^n (m_i-1)^2 + 4 - n -\end{equation*} -But, since $m_i\geq 1 \: \forall i$, we have $\sum (m_i-1)^2 \leq \left(\sum(m_i-1)\right)^2$ and since $\sum m_i < d - 2$ we have -\begin{equation*} -(d - 2)^2 < (d-2-n)^2 + 4 - n -\end{equation*} -which gives the bound $d < \frac{n^2+3n+4}{2n}$, and hence a bound for $g$.<|endoftext|> -TITLE: Rational Canonical Form over $\mathbb{Z}/p^k\mathbb{Z}$ -QUESTION [8 upvotes]: We have a structure theorem for f.g. modules over $R$, whenever $R$ is a PID. In the case of $R=\mathbb{Z}/p\mathbb{Z}[x]$ ($p$ a prime), the structure theorem can be used to obtain the rational canonical form for matrices over the finite field $\mathbb{Z}/p\mathbb{Z}$. - -I am interested in some kind of canonical form for matrices over $\mathbb{Z}/p^k\mathbb{Z}$. Is there such a canonical form in the literature? -This question naturally leads to a more concrete one: What is known about the structure of f.g. modules over $\mathbb{Z}/p^k\mathbb{Z}[x]$? - -If I have a matrix $A \in \mathrm{Mat}_n(\mathbb{Z}/p^k\mathbb{Z})$, the relevant $\mathbb{Z}/p^k\mathbb{Z}[x]$-module is the ``vector space'' $(\mathbb{Z}/p^k\mathbb{Z})^n$, on which $x$ acts as multiplication by $A$. - -REPLY [10 votes]: The problem is open, and not because nobody tried. For instance, it is known that the number of similarity classes in $M_n(\mathbf Z/p^2 \mathbf Z)$ is equal to the number of simultaneous conjugacy classes of pairs of commuting matrices in $M_n(\mathbf Z/p\mathbf Z)$ (S. J AMBOR and W. PLESKEN , Normal forms for matrices over uniserial rings of length two, J. Algebra 358 (2012), 250–256). For small values of $n$, exact formulae are given in my paper in IUMJ Similarity of matrices over local rings of length two with Pooja Singla and Steven Spallone, which is also available on the arXiv. -From the classification/structure theorem point of view, these are wild problems, so you should not expect a good answer. You will find a discussion of the history and other results on this problem in the intro to my paper mentioned above.<|endoftext|> -TITLE: Computing Tamagawa number of torus in Quaternion algebra -QUESTION [6 upvotes]: Consider a rational Quaternion algebra $M$ over $\mathbb{Q}$ that does not split at $\infty$. For example take the rational Hamilton quaternions $M=\mathbb{Q}(-1,-1)$. -For the adele ring $\mathbb{A}$ we define $M(\mathbb{A}):= M\otimes_{\mathbb{Z}} \mathbb{A}$ and $G=G(\mathbb{A}):= M(\mathbb{A})^\times / \mathbb{A}^\times$. Analogous we can define $G(\mathbb{Q})$. This is then a cocompact lattice in $G$. For a $\gamma \in G(\mathbb{Q})$ we can consider the centralizer $G_\gamma$ of $\gamma $ in $G$. If $\gamma$ is a generic semisimple element and I am not mistaken, $G_\gamma$ is an algebraic torus defined over $\mathbb{Q}$. -Is there a way to compute the Tamagawa number of these tori? Thanks a lot in advance. - -REPLY [6 votes]: Here are some more details. As John Voight said, the quaternion algebra is kind of irrelevant here. If $\gamma$ is a regular semisimple element, then its centralizer is a torus ${\mathbf T}$ over ${\mathbb Q}$ satisfying -$${\mathbf T}({\mathbb Q}) = K^\times / {\mathbb Q}^\times,$$ -where $K = {\mathbb Q}(\gamma)$ is the centralizer of $\gamma$ in the quaternion algebra. In terms of algebraic groups, ${\mathbf T}$ fits into a short exact sequence of tori, -$$1 \rightarrow {\mathbf G}_m \rightarrow {\mathbf R}_{K/{\mathbb Q}} {\mathbf G}_m \rightarrow {\mathbf T} \rightarrow 1.$$ -There aren't too many rank-one tori. Since ${\mathbf T}$ is nonsplit, the character lattice of ${\mathbf T}$ is ${\mathbb Z}$, with the unique nontrivial action of $Gal(K/{\mathbb Q})$. Thus ${\mathbf T}$ is also isomorphic to the norm-one torus ${\mathbf R}_{K/{\mathbb Q}}^1 {\mathbf G}_m$. -It happens that norm-one tori arising from cyclic extensions have trivial Sha. This is in Platonov and Rapinchuk, I think, and probably goes back to Tate or something. The basic idea is that Sha captures the failure of the Hasse principle. In the context above, -$$Sha({\mathbf T}) \cong \frac{\left( {\mathbb Q}^\times \cap N_{K/{\mathbb Q}} {\mathbb A}_K^\times \right) }{ N_{K / {\mathbb Q}} K^\times}.$$ -This is trivial, by the Hasse Norm Theorem (since $K / {\mathbb Q}$ is cyclic). -So we have -$$Tam({\mathbf T}) = \frac{\# Pic({\mathbf T})}{\# Sha({\mathbf T})} = \# Pic({\mathbf T}).$$ -Explicitly, the Picard group $Pic({\mathbf T})$ in this context is the group $H^1(Gal(K/{\mathbb Q}), {\mathbb Z})$, where the order-two Galois group acts by the nontrivial automorphism. This is a group of order two. Hence -$$Tam({\mathbf T}) = 2.$$<|endoftext|> -TITLE: Dominance relation among Cartan matrices implies containment of root systems: Is this known? -QUESTION [10 upvotes]: Suppose $A$ and $A'$ are symmetrizable (generalized) Cartan matrices, in the sense of Kac's book Infinite-dimensional Lie algebras. Say $A$ dominates $A'$ if every entry of $A$ has weakly greater absolute value than the corresponding entry of $A'$. Write $\Phi(A)$ for the Kac-Moody root system associated to $A$. I can prove the following: -Proposition. If $A$ dominates $A'$ and $\Phi(A)$ and $\Phi(A')$ are constructed with the same ordered set of simple roots, then $\Phi(A')\subseteq\Phi(A)$. -A few comments: - -We need the full root system here, including the imaginary roots (if there are any). -Yes, this looks like it is false. For example, there is no containment relation among the root systems of types $A_2$ and $B_2$, right? Right, if we embed them into the same vector space so that the two associated Euclidean metrics coincide. But wrong if we do what the proposition says. A useful rephrasing of the proposition: Choosing simple roots any way we please, the set of simple root coordinates of roots in $\Phi(A')$ is contained in the set of simple root coordinates of roots in $\Phi(A)$. - -So, the question (as already asked in the title): Is this known? -On the one hand, the proof is fairly simple, using the Serre relations, so maybe it's known. On the other hand, it may not be very Lie-theoretically natural (for example, I don't think there is a corresponding subalgebra relationship), so maybe it's not known. It is very combinatorially natural and is connected to some interesting facts about the weak order on finite Coxeter groups and to some interesting phenomena surrounding dominance relations on exchange matrices (in the sense of cluster algebras). - -REPLY [2 votes]: I suggest to have a look at Lemma 3.5 in https://arxiv.org/abs/1509.01976<|endoftext|> -TITLE: Criterion for alternation of the linking form -QUESTION [5 upvotes]: I was recently informed by a source of the following fact: -Theorem 1: The linking form on an orientable smooth 5-manifold $M$ is alternating if and only if $M$ is spin$^{\mathbb{C}}$. -Question 1: Does anybody know a reference/attribution for this fact? I've poked around online and in the library, and asked my usual go-to experts, but found nothing. (This page gives a reference to a criterion of Wall in the simply-connected case, but not the general one.) -I'm not in a position to be able to ask my source for a reference, but I'm pretty confident that Theorem 1 is correct because I think I have a proof. In fact the proof gives a criterion for the linking form on any orientable, odd-dimensional topological manifold to be alternating, which specializes to the above fact for smooth 5-manifolds. -Question 2: Is such a criterion already known? Written down? Is it remotely interesting? - -REPLY [4 votes]: The final sentence of Theorem 1 of Browder's Remark on the Poincaré Duality Theorem states that a five-dimensional space $X$ with Poincaré duality has $H_2(X) = F + T + T + \mathbb{Z}_2$, where $F$ is free and $T$ is torsion, if and only if $w_3 \neq 0$. If $b$ denotes the linking form and $x$ is the generator of $\mathbb{Z}_2$, then $b(x, x) = \frac{1}{2}$ so $b$ is not alternating. Therefore, if $b$ is alternating, then $w_3 = 0$ which is equivalent to $w_2$ having an integral lift (which is equivalent to spin${}^c$ in the smooth case). -I don't know of a reference for the converse. However, if $H_2(X)$ has no two-torsion, then it is spin${}^c$ and the fact that the linking form is alternating follows from the classification of non-singular anti-symmetric linking forms; see here.<|endoftext|> -TITLE: is there any non prime (pseudoprime) that holds true for this test, or any prime that fall out of this test? -QUESTION [5 upvotes]: The sequence is defined by the following formula $d_{n + 3} = d_{n + 2} + 2d_{n + 1} - d_n$ where $d_1 = 0, d_2 = 1, d_3 = 2$, $\{0, 1, 2, 4, 7, 13, 23,42, ...\}$ if this sequence is calculated over finite field $p$ $$(d_{n + 3} = d_{n + 2} + 2d_{n + 1} - d_n)\,mod\,\text{p}$$ It follows that if $p$ is prime, then the last three terms $d_{p - 1},\; d_p$, and $d_{p + 1}$ fall within one of the following ending patterns -$\;\{4, 2, 1\}$ , $\{p - 1, 0, 1\}$, or $\{p - 6, p - 3, p - 2\}$. this is true for all prime up to tested range, $60000$ - -my questions - - -why only primes follow these patterns? -is there any non prime(s) following this patterns? -is there any published work like this and in the same pattern. To see if i can study more about it - -REPLY [4 votes]: Prime $p=7$ fails the test, but this is the only exception below $10^7$. This is likely explained by the characteristic polynomial $x^3 - x^2 - 2x + 1$ having discriminant $49=7^2$. -Also, there are nine pseudoprimes below $10^7$: -$$530881,~ 597871,~ 1152271,~ 3057601,~ 3581761,~ 3698241,~ 5444489,~ 5968873,~ 6868261.$$<|endoftext|> -TITLE: Constructing ($\infty, 1)$-category from Morse theory on a manifold -QUESTION [15 upvotes]: It is well known that a topological space is more or less the same as an $\infty$-groupoid. I'm wondering if there is an analogous construction which starts with a manifold endowed with Morse theory (either a function or a gradient-like vector field) and leads to an $(\infty, 1)$-category. The idea should be that morphisms are paths going "downwards". -One can easily come up with at least two definitions, which are obviously non-equivalent: the first one is taking critical points as objects and the space of gradient trajectories as morphisms, and the second one is taking the set of all points as objects and nondecreasing continuous paths as morphisms. -So this boils down to the following: is there any precise sense in which these categories are equivalent? And a minimal question is: -Is there any precise sense in which the two following categories are equivalent - the category with objects $0$ and $1$ and one non-identity morphism from $0$ to $1$, and the category with set of objects $[0, 1]$ and unique morphism between $a$ and $b$ iff $a \leq b$? - -REPLY [6 votes]: This problem has a long history going back at least to the paper of Cohen, Jones and Segal (available here: http://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0883-04.pdf). -In terms of your first definition, -the answer is that if $f:M \to \Bbb R$ is a Morse-Smale function on a closed Riemannian manifold, then there is indeed a topological category whose objects are the critical points and where morphisms are given by sequences of contiguous flow lines connecting the critical points. -There are a number of formidable technical difficulties in getting this to work since the composition law needs to be continuous and associative. Many of the papers in this area contain gaps. -However, I would like to unabashedly credit Lizhen Qin (my student) with settling this question. See the papers -http://front.math.ucdavis.edu/1107.5527 -http://front.math.ucdavis.edu/1102.2838 -http://front.math.ucdavis.edu/1012.3643<|endoftext|> -TITLE: Gaps in Squarefree numbers -QUESTION [8 upvotes]: The density of squarefree numbers is well known. -I am wondering about the squarefree numbers in $[2,n]$ namely the second and higher moments of the gaps. Is anything known about these? - -REPLY [17 votes]: A number of authors e.g. Hooley, Filaseta, Trifonov have considered this problem of moments of gaps between square-free numbers. For example, Filaseta and Trifonov (paper in Proc. London Math. Soc. (1996); see Theorem 4) showed that for all $0\le \gamma < 43/13=3.30769\ldots$ one has -$$ -\sum_{s_{n+1}\le x} (s_{n+1}-s_n)^{\gamma} \sim B(\gamma) x, -$$ -for some constant $B(\gamma)$. Here $s_n$ denotes the $n$-th square-free number.<|endoftext|> -TITLE: Balanced circle packing -QUESTION [9 upvotes]: In the Euclidean plane, given is a collection of $k$ circular disks $D_1,…,D_k$ of radii $r_1,…, r_k$, supplied with weights $w_1,…,w_k$, assuming that each circle’s center of gravity coincides with the circle’s center (in general, the weights are independent from the radii). If the circles are packed in a circular container $D$ so that the center of gravity of their configuration coincides with the center of the container, then we say that the packing is balanced, and it is called the tightest balanced packing if the radius of the container is the smallest possible. The natural problem is, for a given finite family of weighted circular disks, to determine its tightest balanced packing. Generalizations to other shapes and higher dimensions are natural. -Question 1. Does anyone know of any references on this subject? -Some trivial inequalities. Let $r_b$ denote the minimum "balanced" radius, while $r$ - the minimum packing radius, no weights. Obviously, $r\le r_b$. But obviously as well, $r_b\le 2r$. A bit less obviously, $r_b<2r$ while $r_b$ can be arbitrarily close to $2r$. -Question 2. The natural case in which the weight of a disk coincides with its area, resp. volume, is of special interest: how much in this case can the radius of the container for a tightest balanced packing differ from the radius in the tightest packing without the balance requirement? To be more precise: what is the least upper bound on the ratio ${r_b}\over{r}$? - -REPLY [3 votes]: Here are examples of balanced circle packings whose touching graphs are disconnected: - -Circles of same color are congruent and of equal weight. The drawing on the right shows how recursion can produce an arbitrarily large number of connected components. The very same number of circles and with the same arrangement pattern works in every dimension $d\ge2$; just replace each circle, including the container, with a concentric $d$-dimensional ball of the same radius. -Another example: - -$D_1$ and $D_2$ touch the boundary of $D$ and each other, but $D_i$ (for $i>2$) touches neither the boundary of $D$ nor any of the other disks.<|endoftext|> -TITLE: Is there any reason to use paracontrolled calculus over regularity structures? -QUESTION [8 upvotes]: Paracontrolled calculus was developed by Gubinelli, Imkeller and Perkowski as a way of treating singular stochastic PDEs such as KPZ, $\Phi_3^4$ or PAM, around the same time regularity structures were introduced by Hairer. -Regularity structures are now the standard treatment of singular stochastic PDEs although I have been reading that there are perhaps potentional benefits to paracontrolled calculus. This paper for example says: - -We develop in this work a general version of paracontrolled calculus that allows to treat analytically within this paradigm some singular partial differential equations with the same efficiency as regularity structures, with the benefit that there is no need to introduce the algebraic apparatus inherent to the latter theory. - -I'm not sure that this is such a large benefit. Is this the only reason to work with paracontrolled calculus, or are there other benefits? Is there anything you can do in paracontrolled calculus that you can't in regularity structures? - -REPLY [16 votes]: I don't think that the reason given in the paper by Bailleul and Bernicot is a good one. Basically, they treat an example which is simple enough so that it is still manageable to describe the various bits and pieces needed to control their solutions "by hand" instead of combining them into a single object in a more coherent way. -This being said, there are certainly situations in which paracontrolled calculus is more efficient than regularity structures thanks to the fact that it provides a more "global" decomposition of the solution. (Think of Fourier decomposition vs Taylor expansion.) For example, it is not clear how one would implement the strategy of proof for the result of this article by Weber and Mourrat in the framework of regularity structures. It also has the advantage of relying solely on concepts that are quite well-known and have been used in other context for a long time, which makes it possible to reuse more existing results. (For example it makes use of classical function spaces, so that embedding theorems and functional inequalities can just be taken "off the shelf". In the theory of regularity structures there are analogues of most function spaces that share many properties with their classical counterparts, but this always has to be proven first...)<|endoftext|> -TITLE: Characterizing positivity of formal group laws -QUESTION [40 upvotes]: The formal group law associated with a generating function $f(x) = x + \sum_{n=2}^\infty a_n \frac{x^n}{n!}$ is $$f(f^{-1}(x) + f^{-1}(y)).$$ In my thesis, I found a large number of examples of formal group laws that have combinatorial interpretations and thus have nonnegative coefficients. In Sec 9.1 I conjectured the following characterization for positivity of a formal group law: -Conjecture. $f(f^{-1}(x) + f^{-1}(y))$ has nonnegative coefficients if and only if $$\phi(x) = \frac{1}{\frac{d}{dx} f^{-1}(x)}$$ has nonnegative coefficients. -At least one direction is easy: The positivity of the FGL implies positivity of $\phi(x)$. -I have not been able to prove the converse, but there is some evidence. Start with $$\phi(x) = 1 + t_1x + t_2\frac{x^2}{2!} + t_3\frac{x^3}{3!} + \cdots$$ for indeterminates $t_i$ and define $f(x)$ by $f(0) = 0$, $1/(f^{-1})'(x) = \phi(x)$, or equivalently, $f'(x) = f(\phi(x))$. Then we can compute the coefficients of $f(f^{-1}(x) + f^{-1}(y))$ and they seem to all be polynomials with nonnegative coefficients in the variables $t_i$. -Often it is more illuminating to consider the slightly more general symmetric function $$F = f(f^{-1}(x_1) + f^{-1}(x_2) + \cdots).$$ The expansion of $F$ in the monomial basis of the ring of symmetric functions is -\begin{align*} -F = m_1 &+ (2t_1)\frac{m_{11}}{2!} + (3t_2)\frac{m_{21}}{3!} + (6t_1^2 + 6t_2)\frac{m_{111}}{3!}\\ - &+ (4t_3)\frac{m_{31}}{4!} + (12t_1t_2 + 6t_3)\frac{m_{22}}{4!} + (36t_1t_2 + 12t_3)\frac{m_{211}}{4!} \\ -&+ (24t_1^3 + 96t_1t_2 + 24t_3)\frac{m_{1111}}{4!} + (5t_4)\frac{m_{41}}{5!} + (30t_2^2 + 30t_1t_3 + 10t_4)\frac{m_{32}}{5!}\\ &+ (60t_2^2 + 80t_1t_3 + 20t_4)\frac{m_{311}}{5!} + (120t_1^2t_2 + 120t_2^2 + 150t_1t_3 + 30t_4)\frac{m_{221}}{5!}\\ -&+ (420t_1^2t_2 + 240t_2^2 + 360t_1t_3 + 60t_4)\frac{m_{2111}}{5!} \\ -&+ (120t_1^4 + 1320t_1^2t_2 + 480t_2^2 + 840t_1t_3 + 120t_4)\frac{m_{11111}}{5!}\\ -&+ \cdots -\end{align*} -Note that $f(x)$ here has a combinatorial interpretation due to Bergeron-Flajolet-Salvy: $f(x)$ is the exponential generating function for increasing trees weighted by their degree sequence in the variables $t_i$. So there is reason to think that there is a combinatorial interpretation of $F$ in terms of increasing trees. -An interesting special case if $\phi(x) = 1 + x^2$, so that $f(x) = \tan(x)$. Then the associated formal group law is a sum of Schur functions of staircase-ribbon shape: $$f(f^{-1}(x_1) + f^{-1}(x_2) + \cdots ) = \sum_{n=1}^\infty s_{\delta_{n} / \delta_{n-2}}$$ where $\delta_n$ is the partition $(n,n-1, n-2, \ldots, 1)$. (See Ardila-Serrano, Prop 3.4.) This can also be interpreted in terms of binary increasing trees. -In many examples given in my thesis I found that there was a combinatorial interpretation of the FGL in terms of chromatic symmetric functions, but I was not able to apply those methods to this more general case. -Edit: Tom Copeland suggested I share some of the Sage code I used to generate these coefficients. Here is a Jupyter notebook in CoCalc that shows the computations. - -REPLY [16 votes]: Given $\phi(x)\in\mathbb{R}[[x]]$, with $\phi(0)=1$, we have defined $g(x):=\int^x_0{dt\over \phi(t)}$, $f:=g^{-1}$ and $$F(x,y)=f\big(g(x)+g(y)\big)=\sum_{n=0}^\infty \psi_n(x) {y^n\over n!}\in\mathbb{R}[[x,y]].$$ Let's write a recursion for the coefficient sequence $\psi_n=\partial_y^nF(x,0)\in\mathbb{R}[[x]]$, solving by series the differential equation satisfied by $F$, -$$\cases{\phi(x)\, F_x(x,y)=\phi(F(x,y))\\ F(x,0)=x\ .}$$ -One finds $\psi_0=x$, $\psi_1=\phi,\dots$ . Let's take $\partial_y^n$ at $ {y=0}$ on both sides. Faà di Bruno: -$$\partial_y^n\big( \phi\circ F\big)\big|_{y=0}=\Big(\sum_{\alpha\in\operatorname{par}[n]} \phi_y^{(|\alpha|)} (F)\, \prod_{s\in\alpha} \partial_y^{|s|}F \Big) \ \Big|_{y=0} =\sum_{\alpha\in\operatorname{par}[n]} \phi^{(|\alpha|)}(x) \prod_{s\in\alpha} \psi_{|s|}(x) ,$$ -(Legenda: The sum is indexed on the set of all partitions of $[n]:=\{1,2,\dots,n\}$, and $|\cdot|$ denotes cardinality. The latter equality comes from $F(x,0)=x$ and $\partial_y^{j}F(x,y)\big|_{y=0}=\psi_j$). -Now we isolate the term $\phi'\psi_n$, that corresponds to the partition $\alpha$ into a single class, from the terms of the sum indexed on the set of non-trivial partitions, with $|\alpha|>1$, denoted $\operatorname{par}^*[n]$. Note that each of these terms contains more than one factor $\psi_j$. -$$\phi\psi'_n -\phi' \psi_n =\sum_{\alpha\in\operatorname{par}^*[n]} \phi^{(|\alpha|)} \prod_{s\in\alpha} \psi_{|s|} .$$ -Multiplying by the integrating factor $\phi^{-2}$ , and since $\psi_n(0)=0$, for $n>1$ -$$ \psi_n(x) =\phi(x)\int_0^x\big(\!\sum_{\alpha\in\operatorname{par}^*[n]} \phi^{(|\alpha|)} \prod_{s\in\alpha} \psi_{|s|}\,\big)\phi^{-2}\ dt .$$ -It is now clear by complete induction that for any $n\ge1$, $\psi_n$ is equal to $\phi$ times a series with positive coefficients, proving your conjecture.<|endoftext|> -TITLE: What percentage of published mathematics papers are correct? -QUESTION [26 upvotes]: As described in Larcombe and Ridd, estimates for the percentage of published papers found to be reliable or reproducible in sciences such as biomedical science, have been as low as 50%, 25% and even 11%. -My question is: In mathematics, what percentage of published papers are correct? (i.e. all new theorems presented in the paper are correct, and can thus be relied on). Have any studies been done? Perhaps estimates could be made, based on the number of published corrections? -Note that while the problem of finding the exact answer to this question, for some given set of journals, might be beyond our present capability, the question itself, is objective - not a matter of opinion. - -REPLY [42 votes]: This graph from Errors and Corrections in Mathematics Literature indicates about 1.4% of published mathematics papers were followed by a correction. - - -Corrections as percent of journal documents for subject areas of - Scopus. Three areas are shown both inclusive and exclusive of - interdisciplinary work. Comments are undercounted for journals that - choose to identify them as “letters”. - -There is also a breakdown per mathematical topic: - - -Error correction rates as percent of articles in the primary Mathematics Subject - Classifications from Zentralblatt. The eleven subjects in bold - contribute 51 percent of publications.<|endoftext|> -TITLE: The L-function of Q(-1/2) and the "number of prime $p\equiv 3$ divisors" function -QUESTION [6 upvotes]: In the framework of classical motives, there is no such thing as a motive $\mathbb Q(-\tfrac 12)$, i.e. a tensor root of $\mathbb Q(-1)$. There is one, however, in a more general setting of "exponential motives", over any number field containing $\sqrt{-1}$. There is even a motive $M$ over $\mathbb Q$ whose tensor square is $\mathbb Q(-1) \otimes H$, where $H$ is the Artin motive of the quadratic character associated with $\mathbb Q(i)|\mathbb Q$. These exponential motive have attached $L$-functions to them, and my question is about the $L$-function of $M$. Here it is, first as an Euler product: -$$L(s) = \prod_p\frac{1}{1-g_pp^{-s}}$$ -with -$$g_p = \begin{cases} -0 & \mbox{if $p = 2$}\\ -\sqrt p & \mbox{if $p\equiv 1 \bmod 4$}\\ -i\sqrt p & \mbox{if $p\equiv 3 \bmod 4$.} -\end{cases}$$ -These $g_p$ arise as Gauss sums. The Euler product converges absolutely for $\Re(s)>\tfrac 32$. Here is the same $L$-function written as a Dirichlet series: -$$L(s) = \sum_{n\geq 1 \:\mathrm{odd}}\frac{i^{r(n)}}{n^{s-1/2}}$$ -where $r(n)$ stands for -$$r(n) = \sum_{p \equiv 3 \:(4)}v_p(n),$$ -the number of prime divisors $p\equiv 3\:(4)$ of $n$. The first question to ask about $L(s)$ is whether it has a singularity at $s=\tfrac32$ (it probably does), and if so, what kind. So I tried to figure out how much "cancellation" there is in the not-absolutely-converging series -$$\sum_{n\geq 1 \:\mathrm{odd}}\frac{i^{r(n)}}{n}$$ -We need to understand the class of $r(n)$ modulo $4$. So, for $r = 0,1,2,3$ and $x>0$, define the following sets of odd integers: -$$A(r,x) = \{n <\:\mathrm{odd}\:|\:\: r(n)\equiv r \bmod 4\}$$ -If for large $x$ these four sets have about the same cardinality, the sum might converge. I have no doubt that for $r=0,1,2,3$, the limit -$$A(r) := \lim_{x\to \infty}\frac {2 \#A(r,x)}x$$ -exists, and because of the parity of $r(n)$ depending only on $n \bmod 4$ we have $A(0)+A(2)=\tfrac 12$ and $A(1)+A(3)=\tfrac 12$. Now (lack any better ideas) I have done a count by computer, and to my amazement it seems that $A(0)$ is not $\tfrac 14$. Specifically, I counted the number of odd integers $n$ between $N$ and $N+10^8$ for which $r(n)$ is congruent to $0, 1, 2, 3$. The numerical calculation suggests the following approximate values -$$\begin{array}{l|llll} - & 4A(0) & 4A(1) & 4A(2) & 4A(3) \\ \hline -N=10^9 & 0.8201 & 1.3632 & 1.1799 & 0.6368\\ -N=10^{10} & 0.8114 & 1.3375 & 1.1886 & 0.6625\\ -N=10^{11} & 0.8048 & 1.3143 & 1.1952 & 0.6857\\ -N=10^{12} & 0.8000 & 1.2937 & 1.2000 & 0.7063 -\end{array}$$ -Equidistribution would mean that all entries in the table are close to $1$. My question can now be formulated most eloquently as: - -What the heck is going on here? - -Why are there fewer odd integres with $r(n)\equiv 0\bmod 4$ than there are with $r(n)\equiv 2\bmod 4$? And why does the situation get worse for large integers, when intuitively the contrary should happen? - -REPLY [5 votes]: For your first question, note that one can write your $L$-series as -$$ -L(s) = \zeta\big(s-\tfrac12\big)^{(1+i)/2} L\big(s-\tfrac12,\chi_{-4}\big)^{(1-i)/2} M(s), -$$ -where $\chi_{-4}$ is the nontrivial character modulo $4$, and $M(s)$ is an Euler product that converges absolutely for $\Re s>1$. Therefore $L(s)$ does have a singularity at $s=\frac32$, one that looks like $c/(s-1)^{(1+i)/2}$ (that is, a logarithm-style singularity). -As for the eloquent question: remember that the average value, for $n\le x$, of $r(n)$ is about $\frac12\log\log x$, and the variance of $r(n)$ is also about $\frac12\log\log x$ (by the analogue of Hardy-Ramanujan's theorem about $\omega(n)$). For numbers $x$ between $10^9$ and $10^{12}$, the quantity $\frac12\log\log x$ is between $1.51$ and $1.66$. Therefore it it simply much more likely for numbers in this range to have $1$ or $2$ prime factors that are $3\pmod4$ than $0$ or $\ge3$ such prime factors. Of course the convergence to this average value (and indeed to a normal distribution, by a standard Erdös–Kac argument) is very slow; but I suspect you would see the opposite behavior in your histograms for $x$ near $e^{e^{7}} \approx 10^{476}$. -Ultimately, one can prove an analogue of the Selberg-Sathe theorem, which concerns $\#\{n\le x\colon \omega(n) = k\}$ uniformly for $k<1.5\log\log x$ say, for $\#\{n\le x\colon r(n) = k\}$, using the Selberg-Delange method (as in Tenenbaum's book for example). From this it follows, simply by sorting (according to their residue class modulo $4$) the integers $k$ near the average value of $\omega(n)$ or $r(n)$, that each of your $A(r)$ does converge to $\frac14$. It's just that the convergence is extremely slow, and does oscillate macroscopically along the way.<|endoftext|> -TITLE: Is $[0,1]$ a disjoint union of $\aleph_1$ compact subsets with empty interior? -QUESTION [11 upvotes]: Is $[0,1]$ a disjoint union of $\aleph_1$ compact subsets with empty interior? - -The answer is obviously yes assuming the continuum hypothesis. Also, by Baire's lemma, the answer is negative if one replaces $\aleph_1$ with $\aleph_0$. -Does anybody know if this question has a definite answer or whether this is independent of ZFC? - -REPLY [19 votes]: It is independent of ZFC. As you mention, CH implies the answer is yes. A different axiom, MA+$\neg$CH, implies the answer is no. -A little more precisely, there is a cardinal number denoted $\mathrm{cov}(\mathcal M)$, one of the so-called ``small cardinals'', that is defined to be the smallest number of meager sets needed to cover $[0,1]$. MA implies this cardinal is equal to the continuum, and if that is the case (and CH fails) then the answer to your question is no, because a partition of $[0,1]$ as you describe is also a covering of $[0,1]$ with meager sets. -But this is not the end of the story. There is a difference between covering and partitioning, and in general it seems harder to partition $[0,1]$ into copies of the Cantor space than to cover it with copies of the Cantor space. I have a paper with Arnie Miller (link) where we look at this question (and some similar ones), and show that it is consistent with the continuum being very large that for every $\kappa < \mathfrak{c}$ there is a partition of $[0,1]$ into exactly $\kappa$ copies of the Cantor space. -EDIT: I just realized that Taras Banakh has asked a very similar question here. The statement of his question actually contains an answer to your question, and you can learn a lot more about this question by looking at his question, the comments under it, and the answer I posted there.<|endoftext|> -TITLE: Wasserstein distance and the Kantorovich-Rubinstein duality -QUESTION [6 upvotes]: The only few references I could find on this topic are either amateur blog posts (http://n.ethz.ch/~gbasso/download/A%20Hitchhikers%20guide%20to%20Wasserstein/A%20Hitchhikers%20guide%20to%20Wasserstein.pdf and https://vincentherrmann.github.io/blog/wasserstein/) or the 1000 page tome by Cedric Villani, (http://cedricvillani.org/wp-content/uploads/2012/08/preprint-1.pdf) -Is there any expository reference/lecture notes about this which is somewhere between the two kinds of references above? Any paper by maybe Cedric Villani himself which covers all these grounds? - -REPLY [2 votes]: Kantorovich himself wrote a very detailed exposition which constitutes Section 8.4 of Functional Analysis by Akilov and Kantorovich<|endoftext|> -TITLE: Atiyah Bott-Shapiro orientation Vs Anderson-Brown-Peterson Splitting -QUESTION [9 upvotes]: Are the Atiyah-Bott-Shapiro Orientation and the Anderson-Brown-Peterson Splitting compatible in any sense? -The first guess is that the ABS-Orientation is related to the projections on the $BO\langle 4(n(J))\rangle$, respectively $BO\langle 4(n(J))-2\rangle$ factors. - -REPLY [8 votes]: According to Michael Hopkins, Mark Hovey, Spin cobordism determines real K-theory, Mathematische Zeitschrift 210.1 (1992): 181-196, 4th page of the pdf file, the Atiyah-Bott-Shapiro Orientation is just one of the Anderson-Brown-Peterson Splitting map $\pi ^0$. Now, according to the paper by Anderson-Brown-Peterson, the discussion between Theorem 1.3 and Corollary 1.4, this map is just the projection to the "bottom summand" $BO\langle 0 \rangle $.<|endoftext|> -TITLE: On the isomorphism problem of enveloping algebras -QUESTION [12 upvotes]: Let $\mathfrak{g}$ and $\mathfrak{g}'$ be Lie algebras. It is known that if $U(\mathfrak{g})\cong U(\mathfrak{g}')$ as associative algebras, then it is not necessarily true that $\mathfrak{g}\cong \mathfrak{g}'$ as Lie algebras. -I am looking for examples such that $U(\mathfrak{g})\cong U(\mathfrak{g}')$ as algebras but $\mathfrak{g}\not\cong \mathfrak{g}'$ as Lie algebras (over an algebraically closed field). Moreover, are there examples such that the categories $U(\mathfrak{g})-\text{Mod}$ and $U(\mathfrak{g}')-\text{Mod}$ are not monoidally equivalent? -I'm not very familiar with the isomorphism problem for enveloping algebras, a quick google search only gave me counterexamples in positive characteristic. I'd be very happy with examples in characteristic zero (infinite dimensions are allowed). I'm more into the monoidal stuff and might figure out myself whether the representation categories are monoidally equivalent. -Edit: I'm asking this because I naturally encountered a quantized version of this problem. Obviously the categories $U(\mathfrak{g})-\text{Mod}$ and $U(\mathfrak{g}')-\text{Mod}$ are Morita equivalent but there is more information here. First of all $U(\mathfrak{g})\cong U(\mathfrak{g}')$ as algebras which clearly is stronger but they are also enveloping algebras of Lie algebras, further restricting possibilities. In the quantized version I'm looking at, I suspect the representation rings of both categories to be the same making the difference in the monoidal structure very subtle. So I'm wondering whether anything on this subject is known in the non-quantized world. - -REPLY [8 votes]: In the recent paper Lie, associative, and commutative quasi-isomorphism, R. Campos, D. Petersen, F. Wierstra, and I settled the question above for nilpotent Lie algebras: if two nilpotent Lie algebras have universal enveloping algebras that are isomorphic as unital associative algebras, then the two Lie algebras also are isomorphic. -In fact, we proved a more general result in the differential graded context: - -Theorem B: Let $\mathfrak{g}, \mathfrak{h}$ be two dg Lie algebras. If $U\mathfrak{g}$ and $U\mathfrak{h}$ are quasi-isomorphic as unital associative dg algebras, then the homotopy completions $\mathfrak{g}^{\wedge h}$ and $\mathfrak{h}^{\wedge h}$ are quasi-isomorphic as dg Lie algebras. - -This has the statement above as a corollary, since one can show that a Lie algebra that is either strictly positively graded or non-negatively gradedand nilpotent is always quasi-isomorphic to its homotopy completion (in the language of the paper, it is homotopy complete). There are other interesting implications of this result in rational homotopy theory. -In my view (my coauthors might disagree) the spirit of the proof is mostly deformation theoretical, but operad theory play a big supporting role. For those who are interested in the structure of the proof without the technical details, we give a sketch of the arguments in paragraphs 0.27-0.31. - -In a previous version of this answer and of the paper, we claimed that the more general statement that if two dg Lie algebras have universal enveloping algebras that are quasi-isomorphic as associative dg algebras, then the two dg Lie algebras are themselves quasi-isomorphic. Unfortunately, the proof had a gap that we were not able to fix. This more general statement remains open.<|endoftext|> -TITLE: Chain-rule and change of variables in BV/Sobolev -QUESTION [5 upvotes]: A lot of results are available for the following chain-rule problem: -(CRP1) Let $f\colon \mathbb R \to \mathbb R$ be a $C^1$/Lipschitz function and let $g \colon \mathbb R^d \to \mathbb R$ be a weakly differentiable function (e.g. $W_{\rm loc}^{1,p}$ or $BV_{\rm loc}$). Then the function $f \circ g$ is weakly differentiable as well and explicit chain rule formulas hold, like for instance in the Sobolev setting -$$ -(f \circ g)'(x) = f'(g(x)) g'(x) -$$ -a.e. with respect to Lebesgue measure (with some standards caveat when $f$ is Lipschitz). -I am wondering for the other way round, i.e. -(CRP2) Let $f\colon \mathbb R \to \mathbb R^d$ be a $C^1$/Lipschitz function and let $g \colon \mathbb R^d \to \mathbb R$ be a weakly differentiable function (e.g. $W_{\rm loc}^{1,p}$ or $BV_{\rm loc}$). What can we say about the function $g \circ f \colon \mathbb R \to \mathbb R$? For instance in the Sobolev setting it seems to me that the formula -$$ -(g \circ f)'(x) = \nabla g(f(x)) \cdot f'(x) -$$ -(a.e. with respect to Lebesgue measure) makes sense, doesn't it? Are there any references about this topic? -Thanks. - -REPLY [2 votes]: Here is a related example from: -P. Hajlasz, Sobolev mappings: Lipschitz density is not a bi-Lipschitz invariant of the target. Geom. Funct. Anal. 17 (2007), 435-467. -Theorem. There is a Lipschitz function -$\varphi\in {\rm Lip}\, (\mathbb{R}^2)$ -with compact support such that the bounded operator -$\Phi:W^{1,p}([0,1],\mathbb{R}^2)\to W^{1,p}([0,1])$ defined as -composition $\Phi(u)=\varphi\circ u$ is not continuous for any $1\leq -p<\infty$. -The operator is bounded in the sense that the Sobolev norm of $\varphi\circ u$ is bounded by constant times that of $u$. However, the operator is not continuous as a mapping between Banach spaces $W^{1,p}([0,1],\mathbb{R}^2)$ and $W^{1,p}([0,1])$. -However as was proved in -M. Marcus,V. J. Mizel, Every superposition operator mapping one Sobolev space -into another is continuous. J. Funct. Anal. 33 (1979), 217-229, -the composition operator in the case in which $\varphi$ is a Lipschitz function on $\mathbb{R}$ is continuous. -Another reference for understanding the chain rule for Sobolev and BV functions is: -Ambrosio, L.; Dal Maso, G. A general chain rule for distributional derivatives. Proc. Amer. Math. Soc. 108 (1990), no. 3, 691-702.<|endoftext|> -TITLE: zeros of holomorphic function in n variables -QUESTION [5 upvotes]: Conjecture: Let $f:{\mathbb C}^n\rightarrow{\mathbb C}$ be an entire function in $n$ complex variables. Assume -that for every $x\in{\mathbb R}^n$ there exists a $y_x\in{\mathbb R}^n$ such that -$f(x+iy_x)\equiv f(x_1+iy_{x,1},\ldots,x_n+iy_{x,n})=0$. Then $f\equiv 0$. (We don't assume continuity -of $x\mapsto y_x$.) -Remarks: - -The conjecture is true for $n=1$: If $f$ satisfies the hypothesis, it has -uncountably many different zeros, namely $\{ x+iy_x\ | \ x\in{\mathbb R}\}$. Since ${\mathbb C}$ is -$\sigma$-compact, there exists a compact $K\subset{\mathbb C}$ containing uncountably many zeros of -$f$. Thus the set of zeros of $f$ has an accumulation point in $K$, and by a well-known result $f$ -vanishes identically. -The argument for $n=1$ does not adapt trivially to higher $n$ since holomorphic functions of -severable variables have no isolated zeros. -In the conjecture it is important that $x$ consists of the real and $y$ of the imaginary parts of -the variables. For example, the following is not true: Assuming that -$f:{\mathbb C}^2\rightarrow{\mathbb C}$ is entire and for every $z_1=x_1+iy_1$ there is a -$z_2=x_2+iy_2$ such that $f(z_1,z_2)=0$, we have $f\equiv 0$. A trivial counterexample is -$f(z_1,z_2)=z_1-z_2$. -Most books on holomorphic functions in several variables devote some attention to the set of -zeros of such a function. Usually this leads to a proof of Weierstrass' preparation theorem, where -the discussion ends. This is of no immediate help to me, since I would actually be happy with a -proof of the conjecture for polynomials in $n$ complex variables! In this case, -Weierstrass' theorem does not give any new information. - -REPLY [8 votes]: The conjecture is obviously false even for $n=2$. Check $f(z,w)=(w-z^2)(z-(w+1)^2)$. Write $z=x+iy$ and $w=u+iv$. Given $x$ and $u$, I can make first term zero unless $u>x^2$. I can make the second term zero unless $x>(u+1)^2$. Since both inequalities can not be true, we are done.<|endoftext|> -TITLE: Is the sum of a Darboux function and a polynomial necessarily a Darboux function? -QUESTION [6 upvotes]: A function $f: \mathbb{R} \to \mathbb{R}$ is called a Darboux function if and only if it maps every connected subset of $\mathbb{R}$ to a connected set. -As an example : -We know that (a.k.a., the Intermediate Value Property of continuous functions) all continuous functions are Darboux functions, further all the functions that are derivatives of some other function are also Darboux functions. -Question : -Given any Darboux function $f: \mathbb{R} \to \mathbb{R}$ and a polynomial $p(x) \in \mathbb{R} [ x ]$ then is the function $f+p$ always a Darboux function ? -Note that : (from the work of other mathematicians we know that : ) -Sum of two Darboux functions need not be a Darboux function. -In fact, a theorem of Sierpinski says that every function $f: \mathbb{R} \to \mathbb{R}$ can be written as a sum of two Darboux functions. -Further, -Here ( http://matwbn.icm.edu.pl/ksiazki/fm/fm146/fm14622.pdf ) in a mathematics paper by Juris Steprans it says that the sum of Darboux functions and continuous (actually differentiable functions are considered here) need not be Darboux functions. -The source of thought for the original problem is the two methods to solve the following problem (it appeared as a problem in the Freshman exam in mathematics at Cornell university in some past year). -Prove that there is no continuous function $f: \mathbb{R} \to \mathbb{R}$ such that for all rational numbers $x$ we have $f(x)$ is irrational and for all irrational numbers $x$ we have $f(x)$ is rational. -First way : -Assume the contrary, (that is assume there exists such a continuous function $f: \mathbb{R} \to \mathbb{R}$ that takes every rational number to irrationals and every irrational number to rationals) -Consider the functions, $h_1 : \mathbb{R} \to \mathbb{R}$ and $h_2 : \mathbb{R} \to \mathbb{R}$ defined by $h_1(x)= f(x) - x$ and $h_2(x) = f(x) + x$ for all real numbers $x$. -Then by assumption $h_1,h_2$ are continuous and have only irrational numbers in their range. Thus by Intermediate Value Property (a.k.a., Darboux property of continuous functions) they must be constant and thus their sum is a constant function. But observe $h_1+h_2 = 2f$ , thus forcing $f$ to be constant and hence a contradiction (since, $f$ has both rationals and irrationals in its range). -This solves the problem. -Another solution : Assume the contrary, -Observe that their must exist irrationals $a,b$ with $a -TITLE: Why is the first integral Pontryagin class a homeomorphism invariant? -QUESTION [14 upvotes]: At the end of his 1956 paper On Manifolds Homeomorphic to the 7-Sphere, Milnor shows that either - -There exists a closed topological 8-manifold with no smooth structure; or -The first Pontryagin class $p_1$ of an open smooth 8-manifold is not a topological invariant. - -We now know that 1 is true. -In section 4.4 of The Novikov Conjecture - Geometry and Algebra, Kreck and Lück outline a proof that $p_i$ is not a homeomorphism invariant for $i>1$ and claim that $p_1$ is a homeomorphism invariant but do not supply a proof. They reference a manuscript (On the topological invariance of Pontrjagin classes) of Kreck and say that it is in preparation, but said manuscript does not seem to have been published since. -Does anyone have a reference or proof for $p_1$ being a topological invariant? - -REPLY [6 votes]: The topological invariance of the first Pontryagin class is proved in the paper - -B.L. Sharma. Topologically invariant integral characteristic classes. Topology Appl. 21 (1985), no. 2, 135–146. (link to Elsevier website) - -In low degrees, Sharma computed the smallest multiples of the Pontryagin classes which are topological invariants, and Theorem 1.6 of his paper states that that multiple is 1 for the first Pontryagin class. -(Note: I found that via a discussion Section 22 of Rudyak's "Piecewise linear structures on manifolds", (arxiv link))<|endoftext|> -TITLE: Does Helly's theorem hold in the hyperbolic plane? -QUESTION [10 upvotes]: The Helly theorem in the Euclidean plane asserts that if $S_1, \dots, S_n$ are $n \ge 3$ convex subsets such that $S_i \cap S_j \cap S_k \ne \emptyset$ for all distinct triples $i,j,k$, then the total intersection $\bigcap_{i = 1}^n S_i$ is also nonempty. -I'm wondering if the same theorem is true in the hyperbolic plane (for concreteness, let's assume the Poincaré disk model). My understanding is that if the analogue of Radon's theorem is true in this setting, then Helly follows axiomatically. -Radon's theorem in the Euclidean plane asserts that given any four points $x_1, \dots, x_4$, there is a partition into two nonempty subsets such that the convex hulls intersect. The proof I know uses the affine structure on the Euclidean plane and so doesn't seem to port directly into hyperbolic space. On the other hand, I can verify that the Radon property holds for all the collections of four points I've looked at... - -REPLY [17 votes]: I don't understand your reference to the model (since the geometry of the hyperbolic plane does not depend on any model), but, in fact, the Beltrami-Klein model demonstrates that any qualitative statement about convex sets in the Euclidean plane holds in the Hyperbolic plane and vice versa, since the model maps convex sets to convex sets. -EDIT This has (almost) absolutely nothing to do with the above, but the topological version of Helly's theorem goes back to at least Debrunner (and it is a Monthly paper, so is human-readable), no need to allude to Farb's paper. - -REPLY [9 votes]: The original proof of Helly's theorem was topological and only uses basic homological properties of convex sets. It generalizes to all sorts of contexts, including the one you are interested in. Here is a general statement of what it can do. A homology cell is a topological space whose reduced singular homology is the same as that of a point (this implies in particular that it is nonempty). -Theorem: Let $X$ be a normal topological space such that for some $n \geq 1$, every open set $Y \subset X$ satisfies $H_q(Y)=0$ for $q \geq n$. Let $X_1,\ldots,X_k$ be a collection of closed homology cells in $X$. Assume that the intersection of any $r$ of the $X_i$ is nonempty for all $r \leq n+1$ and is a homology cell for $r \leq n$. Then the intersection of all the $X_i$ is a homology cell (and in particular is nonempty). -A discussion of this with references is in Section 3 of -B. Farb, -Group actions and Helly's theorem, -Adv. Math. 222 (2009), no. 5, 1574–1588.<|endoftext|> -TITLE: Distribution of square roots mod 1 -QUESTION [53 upvotes]: I was wondering about the distribution of $\sqrt{p}$ mod $1$ this morning, as one does while brushing one's teeth. I remembered the paper of Elkies and McMullen (Duke Math. J. 123 (2004), no. 1, 95–139.) about $\sqrt{n}$ mod $1$, but hadn't really thought about it before. -Question 1: is $\sqrt{p}$ equidistributed mod $1$, as $p$ varies over all prime numbers? Is this known? Within range of current techniques? -Question 2: What about subtler statistics of $\sqrt{p}$ (and $\sqrt{n}$) mod $1$? -I made three plots, giving histograms of $\sqrt{n}$ mod $1$ (for natural numbers up to 100,000) and $\sqrt{p}$ mod $1$ (for primes up to 1 million) and (for comparison) a histogram of 100,000 samples drawn uniformly at random from {0,1,...,999}. Here they are for your enjoyment. - - - -There's some wild stuff going on, I think! -Question 2(a): What's up with these sharp peak/valleys at rational numbers, in the distribution of $\sqrt{n}$ mod $1$? They are especially prominent at fractions of the form $a / 2^{e}$. How tall are these peaks near rational numbers? They persist when sampling from the primes, i.e., in the distribution of $\sqrt{p}$ mod $1$ too. -Question 2(b): Outside of those funky spots in 2(a), the distribution of $\sqrt{n}$ mod $1$ is far flatter than one would expect, e.g., from samples drawn uniformly at random as displayed in the bottom histogram. This must have been noticed and quantified before... what's the relevant quantitative result here? -Question 2(c): The distribution of $\sqrt{p}$ mod 1 displays the same funky spots near rational numbers, but otherwise seems much closer (in noise-volume) to the random samples at the bottom. Maybe for a larger sample, the funkiness goes away... I don't know. Explanations or conjectures are welcome. -Question 3: These seem like natural images to look at. If you know a reference where others have drawn such pictures or studied similar phenomena, I'd love to take a look! --------------Update after answers below----------------- -It looks like the answer to Question 1 is YES. Lucia's answer below explains this, and also some of the flatness evident in the $\sqrt{n}$ distribution mod 1. -Igor and Aaron discuss the "spikes" around rational numbers. This seems related to binning: if our bins have width 1/1000, we see spikes at multiples of 1/2, 1/4, 1/5, 1/10, etc., related to divisors of 1000. Here's a new picture, which might help us understand the behavior of the distribution of $\sqrt{n}$ mod 1 near rational numbers. I've intentionally drawn the bins so that their endpoints lie on rational numbers with denominator up to 60. (I call this Farey-binning). This seems to bring the "spikes" around rational numbers down to the same size (independent of denominator). -I think I'll accept Lucia's answer soon, because it answers the most direct Question 1. But more insights are welcome. - -REPLY [4 votes]: It is known that the distribution of the pair correlations of $\sqrt{n}$ mod 1 is asympotitically Poissonian (that is, "random"). See [1]. Note that having Poissonian pair correlations implies being equidistributed [2], so this is a stronger result than the equidistribution result. -I don't know if a similar result is known for $\sqrt{p}$. Also, I don't know about results for higher correlations or neighbor spacings. -You might also want to check the related paper [3]. -[1] El-Baz, Daniel; Marklof, Jens; Vinogradov, Ilya. The two-point correlation function of the fractional parts of \sqrt{n} is Poisson. Proc. Amer. Math. Soc. 143 (2015), no. 7, 2815–2828. -[2] https://arxiv.org/abs/1612.05495 -[3] https://www.maths.bris.ac.uk/~majm/bib/nato.pdf<|endoftext|> -TITLE: Smooth projective models of Severi-Brauer varieties over a DVR are also Severi-Brauer varieties -QUESTION [12 upvotes]: Let $R$ be a DVR with uniformizer $\pi$, fraction field $K$ and residue field $k$. Let $X/K$ be a Severi-Brauer variety and $\mathscr X/R$ a smooth, projective model of it. Is it true that $\mathscr X_k/k$ is also a Severi-Brauer variety? What if we only assume that $\mathscr X/R$ is smooth and proper? -One way to try and prove this would be to show that if $X$ is trivial as a Severi-Brauer variety, then $\mathscr X \cong \mathbb P^n_R$ over $R$. Is this true? -(Note: I asked this a few hours ago on mathstackexchange but on the suggestion of someone, I am posting it here. I have deleted the stackexchange post.) - -REPLY [16 votes]: I wrote up some notes on this in 2004. There have been some developments since then that I will indicate below. -Denote the smooth, proper morphism as follows, $$\pi:\mathcal{X}\to \text{Spec}\ R.$$ Since $\pi$ is flat and proper, also the fiber product morphism, $$\mathbb{P}^n\times_{\text{Spec}\ R}\mathcal{X}\to \text{Spec}\ R,$$ is also flat and proper. For a flat, proper morphism, Michael Artin proved representability of the relative Hilbert functor by an algebraic space that is locally finitely presented and separated over the base. Thus, there exists a universal pair $$(\rho:I\to \text{Spec}\ R,\phi:I\times_{\text{Spec}\ R}\mathbb{P}^n \xrightarrow{\cong} I\times_{\text{Spec}\ R}\mathcal{X})$$ of a separated, locally finitely presented scheme $I$ over $\text{Spec}\ R$ and an isomorphism $\phi$ of $I$-schemes, and this is compatible with arbitrary base change of $\text{Spec}\ R$. -Since $\text{Ext}^2_{\mathcal{O}_{\mathbb{P}^n}}(\Omega_{\mathbb{P}^n_F/F},\mathcal{O}_{\mathbb{P}^n_F})$ equals $0$, the morphism $\rho$ is smooth. Thus, the image of $\rho$ is an open subset, and the restriction of $\rho$ over this open subset is a (smooth) torsor for the (smooth) automorphism group scheme $\text{Aut}(\mathbb{P}^n)=\textbf{PGL}_{n+1}$. In particular, if we restrict over the open image of $\rho$, then $\pi$ is étale locally isomorphic to $\mathbb{P}^n_R$. -Question 1. Is $\rho$ surjective on points? -This can be checked after surjective base change of $\text{Spec}\ R$. Thus, without loss of generality, assume that the generic fiber $X_K$ is $K$-isomorphic to $\mathbb{P}^n_K$. Let $H\subset X_K$ be an effective Cartier divisor whose associated invertible sheaf generates the Picard group. Since $\mathcal{X}$ is smooth over $\text{Spec}\ R$, also $\mathcal{X}$ is regular. Thus the closure $\mathcal{H}$ of $H$ in $\mathcal{X}$ is a Cartier divisor that is flat over $\text{Spec}\ R$. Denote by $\mathcal{L}$ the associated invertible sheaf (the dual of the ideal sheaf of $\mathcal{H}$). The restriction of $\mathcal{L}$ to $X_K$ is an ample generator of $\text{Pic}(X_K)$. -Lemma 2. Let $\pi:\mathcal{X}\to \text{Spec}\ R$ be a proper, flat morphism of schemes, and let $\mathcal{L}$ be an invertible sheaf on $\mathcal{X}$ whose restriction to $X_K$ is a generator of the Picard group. If the closed fiber $X_k$ is integral, then every invertible sheaf on $\mathcal{X}$ is isomorphic to $\mathcal{L}^{\otimes m}$ for some integer $m$. -Proof. Since $\pi$ is proper and flat with integral closed fiber, also $\mathcal{X}$ is integral. Thus, every invertible sheaf on $\mathcal{X}$ is $R$-flat. So the pushforward by $\pi$ is a finitely generated, torsion-free $R$-module, i.e., it is a finite, free $R$-module. -For every invertible sheaf $\mathcal{M}$ on $\mathcal{X}$, since $\mathcal{L}|_{X_K}$ generates $\text{Pic}(X_K)$, there exists an integer $m$ such that $\mathcal{M}|_{X_K}$ is isomorphic to $\mathcal{L}^{\otimes m}|_{X_K}$. The pushforward $\pi_*\textit{Hom}_{\mathcal{O}_X}(\mathcal{M},\mathcal{L}^{\otimes m})$ is a finite, free $R$-module whose $K$-fiber equals $H^0(X_K,\mathcal{O}_{X_K})$. Thus, there exists an element of this $R$-module whose $K$-germ is an invertible element of $H^0(X_K,\mathcal{O}_{X_K})$. This element corresponds to a homomorphism of invertible sheaves, $$\alpha:\mathcal{M} \to \mathcal{L}^{\otimes m},$$ that is an isomorphism on $X_K$. Thus, the support of the cokernel of $\alpha$ is an effective Cartier divisor on $\mathcal{X}$ that is disjoint from $X_K$. -Since the closed fiber $X_k$ is integral, the support of the cokernel equals $r\underline{X}_k$ for some integer $r\geq 0$. Thus $\mathcal{M}$ is isomorphic to $\mathcal{L}^{\otimes m}(-r\underline{X}_k)$ for some integer $r$. Since $\mathcal{O}(-\underline{X}_k)$ is the pullback by $\pi$ of the maximal ideal of $R$, and since this is a free $R$-module, also $\mathcal{O}(-\underline{X}_k)$ is a free $\mathcal{O}_{\mathcal{X}}$-module. Therefore, $\mathcal{M}$ is isomorphic to $\mathcal{L}^{\otimes m}$. QED -Proposition 3. With the same hypotheses as in Lemma 1, assume also that $\mathcal{X}$ is regular. Then either the invertible sheaf $\mathcal{L}$ is $\pi$-ample or the invertible sheaf $\mathcal{L}^\vee$ is $\pi$-ample. -Proof. This is automatic if $\pi$ is finite. Hence assume that $\pi$ has positive fiber dimension. By Chow's Lemma, a blowing up of $X_K$ is projective. This blowing up is covered by complete intersection curves whose pushforwards to $X_K$ are numerically equivalent. Up to replacing $\mathcal{L}$ by $\mathcal{L}^\vee$, assume that the degree of $\mathcal{L}$ on these curves is nonnegative. The claim is that for every integer $m$ that is sufficient positive and divisible, the invertible sheaf $\mathcal{L}^{\otimes m}$ is globally generated and the complete linear system defines a finite morphism to projective space. -For every open affine $U \subset \mathcal{X}$ whose closed fiber $U_k$ is nonempty, for every pair $p,q\in U_k$ of distinct closed points, there exists $f\in H^0(U,\mathcal{O}_{\mathcal{X}})$ that is zero on $p$ and that is nonzero on $q$. The zero scheme $\text{Zero}(f)$ is a Cartier divisor in $U$. The closure in $\mathcal{X}$ of $\text{Zero}(f)$ is a Cartier divisor $D$ that contains $p$, yet does not contain $q$. By the previous lemma, $\mathcal{O}(\underline{D})$ is isomorphic to $\mathcal{L}^{\otimes m}$ for some integer $m$. Since $D$ is effective, its restriction to $X_K$ is effective. Thus, the intersection number with the covering family of curves is positive. Therefore, the integer $m$ is also positive. Via this isomorphism, there exists a global section $s$ of $\mathcal{L}^{\otimes m}$ that vanishes at $p$ yet does not vanish at $q$. Note that for every positive integer $d$, also $s^d$ is a global section of $\mathcal{L}^{\otimes md}$ that vanishes at $p$ yet does not vanish at $q$. -Thus, for every integer $d\geq 1$, for the base locus $B_d$ of the complete linear system of $\mathcal{L}^{\otimes d}$, for every $q\in B_d$, there exists an integer $m\geq 1$ such that $B_{dm}$ is contained in $B_d$ yet does not contain $q$. By Noetherian induction, for all integers $m$ that are sufficiently positive and divisible, $B_m$ is empty. Moreover, for every closed point $p\in X_k$, for every open affine $U$ neighborhood of $p$, for every closed point $q\in U_k$, there exists an integer $m\geq 1$ such that the associated $R$-morphism $$\phi_m:\mathcal{X}\to \mathbb{P}H^0(\mathcal{X},\mathcal{L}^{\otimes m}),$$ does not contain $(p,q)$ in the fiber product $$\mathcal{X}\times_{\mathbb{P}H^0(\mathcal{L}^m)}\mathcal{X}\subset \mathcal{X}\times_{\text{Spec}\ R}\mathcal{X}.$$ Thus, by another Noetherian induction argument, for all integers $m$ that are sufficiently positive and divisible, every irreducible component of the fiber product projects finitely to $\mathcal{X}$ under projection on the first factor. Thus, $\phi_m$ is a finite morphism, and $\mathcal{L}^{\otimes m}$ is ample. QED -Nota bene. There is a vast generalization of Lemma 2 for specializations of rationally connected varieties in the following article of Gounelas and Javanpeykar. -Frank Gounelas, Ariyan Javanpeykar -Invariants of Fano varieties in families -https://arxiv.org/pdf/1703.05735.pdf -Corollary 4. With the same hypotheses as in Proposition 3, for an $R$-flat closed subscheme $\mathcal{Y}\subset \mathcal{X}$ of pure relative dimension $e$, if the $\mathcal{L}$-Hilbert polynomial of $\mathcal{Y}$ equals $P_{\mathcal{Y},\mathcal{L}}(t) = 1\cdot (t^e/e!) + O(t^{e-1})$, then the closed fiber $Y_k$ of the normalization of $\mathcal{Y}$ is geometrically integral. -Proof. Since $\mathcal{Y}$ is $R$-flat, the Hilbert polynomial of the closed fiber equals the Hilbert polynomial of the generic fiber. Since the normalization of $\mathcal{Y}$ is $S_2$, the closed fiber $Y_k$ is $S_1$. Thus, to prove that $Y_k$ is geometrically integral, it suffices to prove that it is geometrically irreducible and reduced at the generic point. The leading coefficient of the Hilbert polynomial is additive for irreducible components and multiplicative for multiplicity at a generic point. Since the coefficient equals $1$, and since $\mathcal{L}$ is ample (so that the leading coefficient of every irreducible component with its reduced structure is a positive integer), it follows that $Y_k$ is geometrically irreducible and has multiplicity $1$ at every generic point. QED -Corollary 5. With the same hypotheses as in Proposition 3, assume further that $(X_K,\mathcal{L}|_{X_K})$ is $K$-isomorphic to $(\mathbb{P}^n_K,\mathcal{O}(1))$ for $n\geq 1$. Let $\mathcal{Y}$ be an $R$-flat Cartier divisor in $\mathcal{X}$ in the linear system of $\mathcal{L}$. Then $\mathcal{Y}$ is smooth over $\text{Spec}\ R$. -Proof. Denote by $\mathcal{Y}_{\text{sm}}$ the maximal open subscheme of $\mathcal{Y}$ that is smooth over $\text{Spec}\ R$. Let $\mathcal{Y}_{\text{sing}}$ denote the closed complement of this open in $\mathcal{Y}$. The claim, to be proved by contradiction, is that $\mathcal{Y}_{\text{sing}}$ is empty. These are both compatible with arbitrary base change of $\text{Spec}\ R$. Thus, if $\mathcal{Y}_{\text{sing}}$ is nonempty, then after faithfully flat base change of $\text{Spec}\ R$ assume that there exists a $k$-point $p$ of $\mathcal{Y}_{\text{sing}}$ and a $k$-point $q$ of $X_k\setminus \mathcal{Y}_{\text{sing}}$. After further base change, assume that these are the specializations of $K$-points $p_K\in Y_K$ and $q_K \in X_K\setminus Y_K$. -Since $(X_K,\mathcal{L}|_{X_K})$ is isomorphic to $(\mathbb{P}^n_K,\mathcal{O}(1))$ there exists a reduced, closed $K$-curve $C_K$ in $X_K$ containing the $K$-points $p_K$ and $q_K$ having $\mathcal{L}$-degree equal to $1$. This curve is the image of a closed immersion of $K$-schemes, $$u_K:(\mathbb{P}^1_K,0,\infty)\mapsto (X_K,p_\eta,q_\eta),$$ with $u_K^*\mathcal{L}$ isomorphic to $\mathcal{O}(1)$. -Form the closure of $C_K$ in $\mathcal{X}$, and then form the normalization of this closed subscheme of $\mathcal{X}$. Denote this normalization as follows, $$(\rho:\mathcal{C}\to \text{Spec}\ R,u:\mathcal{C}\to \mathcal{X}).$$ By the previous corollary, the closed fiber $C_k$ is geometrically integral. As a geometrically integral specialization of a smooth curve of genus $0$, also $C_k$ is smooth of genus $0$ (this can fail in higher genus, e.g., a smooth plane cubic can specialize to an irreducible, nodal plane cubic). The inverse image of $\mathcal{Y}$ is an effective Cartier divisor in $\mathcal{C}$ that does not contain $q$, and thus it does not contain the entire closed fiber $C_k$. Thus, this Cartier divisor in $\mathcal{C}$ is $R$-flat. Since the degree of this Cartier divisor on $C_K$ equals $1$, also the degree on $C_k$ equals $1$. However, the intersection multiplicity at $p$ is at least as large as the multiplicity of the Cartier divisor $\mathcal{Y}$ at $p$. Thus, this Cartier divisor in the closed fiber $X_k$ has multiplicity $1$ at $p$, i.e., it is smooth at $p$. This contradicts that $p$ is contained in $\mathcal{Y}_{\text{sing}}$. This contradiction proves that $\mathcal{Y}$ is smooth over $\text{Spec}\ R$. QED -Let $\mathcal{Y}$ be the closure in $\mathcal{X}$ of a Cartier divisor in $X_K$ that is in the linear system of $\mathcal{L}|_{X_K}$. -Proposition 6. With hypotheses as in Corollary 5, the restriction homomorphism, $$r:H^0(\mathcal{X},\mathcal{L})\to H^0(\mathcal{Y},\mathcal{L}|_{\mathcal{Y}}),$$ is surjective, the complete linear system of $\mathcal{L}$ is globally generated, and the associated morphism of the complete linear system is an isomorphism to $\mathbb{P}^n_R$. -Proof. This is proved by induction on $n$. The base case is when $n$ equals $1$. Every smooth, proper curve of genus $0$ is geometrically isomorphic to $\mathbb{P}^1$. Thus, by way of induction, assume that $n\geq 2$ and assume that the result is true for smaller values of $n$. In particular, by Corollary 5, the pair $(\mathcal{Y},\mathcal{L}|_{\mathcal{Y}})$ satisfies the hypotheses for $n-1$. By the induction hypothesis, this pair is isomorphic to $(\mathbb{P}^{n-1}_R,\mathcal{O}(1))$. After a further finite, flat base change, assume that there exists a section $\sigma$ of $\pi$ whose image is disjoint from $\mathcal{Y}$, i.e., the $k$-fiber is not contained in $\mathcal{Y}_k$. -Let $H_k\subset \mathcal{Y}_k$ be a Cartier divisor in the linear system of $\mathcal{L}|_{\mathcal{Y}_k}$. Since $(\mathcal{Y},\mathcal{L}|_{\mathcal{Y}})$ is $R$-isomorphic to $(\mathbb{P}^{n-1}_R,\mathcal{O}(1))$, there exists a lift of this Cartier divisor to an $R$-flat Cartier divisor $\mathcal{H}$ in $\mathcal{Y}$. In the generic fiber $X_K$, the subset $\mathcal{H}_K$ is a codimension $2$ linear subvariety, and $\sigma_K$ is a $K$-point that is disjoint. Thus, there exists a unique Cartier divisor $D_K$ in the linear system of $\mathcal{L}|_{X_K}$ that contains $\mathcal{H}_K$ and $\sigma$. By the previous corollary, the closure $\mathcal{D}$ in $\mathcal{X}$ of $D_K$ is a smooth Cartier divisor in the linear system of $\mathcal{L}$ that contains $\mathcal{H}$ and that contains $\sigma$. -If the intersection of $\mathcal{D}$ with $\mathcal{Y}_k$ is strictly larger than $\mathcal{H}_k$, then it completely contains the Cartier divisor $\mathcal{Y}_k$. Since $\mathcal{D}_k$ is irreducible, this implies that $\mathcal{D}_k$ equals $\mathcal{Y}_k$. This contradicts that $\mathcal{D}$ contains $\sigma$, since $\sigma$ is disjoint from $\mathcal{Y}$. Therefore, the restriction of $\mathcal{D}$ to $\mathcal{Y}$ equals $\mathcal{H}$. -As the divisors $\mathcal{H}$ vary over a bases for the complete linear system of $\mathcal{L}|_{\mathcal{Y}}$, the divisors $\mathcal{D}$ give a subsystem of the complete linear system of $\mathcal{L}$ that restricts isomorphically to the complete linear system of $\mathcal{L}|_{\mathcal{Y}}$. In particular, the base locus of this linear subsystem of $\mathcal{L}$ is disjoint from $\mathcal{Y}$, since the base locus of $\mathcal{L}|_{\mathcal{Y}}$ is empty. Consider the linear subsystem of $\mathcal{L}$ generated by this linear system together with the Cartier divisor $\mathcal{Y}$. This has empty base locus, and thus defines an $R$-morphism, $$\phi:\mathcal{X}\to \mathbb{P}^n_R.$$ Since $\mathcal{L}$ is ample, this morphism is finite. Moreover, since $\phi$ is an isomorphism on $K$-fibers, $\phi$ is birational. Since $\mathbb{P}^n_R$ is normal (and even regular), it follows from Zariski's Main Theorem that $\phi$ is an isomorphism. QED -Remark. If the residue field has characteristic $0$, then this result, often called "deformation-in-the-large", also holds if the geometric generic fiber is a smooth quadric hypersurface in $\mathbb{P}^n$ (the Ph.D. thesis of Jun-Muk Hwang), or if the geometric generic fiber is a projective homogeneous variety of cominuscule type (joint result of Jun-Muk Hwang and Ngaiming Mok). -MR1608587 (99b:32027) -Hwang, Jun-Muk(KR-SNU); Mok, Ngaiming(PRC-HK) -Rigidity of irreducible Hermitian symmetric spaces of the compact type under Kähler deformation. -Invent. Math. 131 (1998), no. 2, 393–418. -https://arxiv.org/abs/math/9604227 -This has been extended to positive characterstic and mixed characteristic by Jan Gutt (using totally different techniques that do not reduce to Kobayashi-Ochiai as in Hwang-Mok). Precisely, for a fixed cominuscule type $(G,P)$, for a fixed integer $m$, there exists an explicit integer $p_0(G,P,m)$ depending on certain Schubert calculus computations such that for every prime $p\geq p_0(G,P,m)$, if the residue characterstic of $k$ equals $p$ (or $0$) and if $\mathcal{L}^{\otimes m}|_{\mathcal{X}_k}$ is very ample, then also $\mathcal{X}_k$ is a projective homogeneous variety of the same cominuscule type $(G,P)$. -Jan Gutt -On the extension theorem of Hwang and Mok -Journal für die reine und angewandte Mathematik (Crelles Journal), -ISSN (Online) 1435-5345, ISSN (Print) 0075-4102,<|endoftext|> -TITLE: What do "pivotal" and "spherical" mean for (unitary) fusion categories on the level of the $F$-symbols? -QUESTION [7 upvotes]: For me, a fusion category (over $\mathbb{C}$) is just a tensor $F$ (the associator, with $6$ simple-object labels and $4$ fusion space indices) and a tensor $d$ (the quantum dimensions, with one simple-object label), such that the $F$-tensor fulfils the pentagon equation. (This seems to be the most useful language for physics.) -If furthermore the fusion category is unitary, this means that (one can find a basis for the fusion spaces such that) the $F$-tensor is unitary when interpreted as a linear map between two groups of its indices. -What extra structure/conditions do I have to add in this tensor language to make the (unitary) fusion category pivotal/spherical? -For example, it seems that a unitary fusion category is spherical if the $F$-tensor is symmetric under the full tetrahedral symmetry group (after normalizing with the square roots of the quantum dimensions of four of its simple-object labels). This seems not to be the most general case, however. - -REPLY [7 votes]: This is really an answer to the question in the comment of Noah's answer. -Let $F$ be a solution to the pentagon equations for some fusion category $\mathcal C$. Pivotal structures on $\mathcal C$ should be in $1-1$ correspondence with solutions to the polynomial equations -\begin{equation} -\epsilon_c^{-1} \epsilon_b \epsilon_a = \sum_{s=1}^{N_{b c^*}^{a^*}}\sum_{t=1}^{N_{c^* a}^{b^*}} -F_{abc^*}^{\mathbf 1}\lbrack\begin{smallmatrix}i & c & 1 \\ -1 & a^* & s\end{smallmatrix}\rbrack -F_{bc^*a}^{\mathbf 1}\lbrack\begin{smallmatrix}s & a^* & 1 \\ -1 & b^* & t\end{smallmatrix}\rbrack -F_{c^* ab}^{\mathbf 1}\lbrack\begin{smallmatrix}t & b^* & 1 \\ -1 & c & i\end{smallmatrix}\rbrack -\end{equation} -for all $i \in \lbrace 1,\ldots, N_{ab}^c\rbrace$, where $N_{ab}^c$ is the dimension of $hom(a \otimes b,c)$. The $\epsilon_a$ are referred to as pivotal coefficients. There are various ways in which the pivotal coefficients and quantum dimensions for a given solution can be played around with, given our initial solution $F$ to the pentagon equations. For ease, we'll go ahead and assume that our $F$ matrices use basis choices which agree with those in 1305.2229 (which are pretty standard choices), in which case the left and right quantum dimensions of an object $a$ are related to $(F,\epsilon)$ via -\begin{align} -q_l(a) &= \epsilon_a\left ( -F_{a^*a a^*}^{a^*}\lbrack\begin{smallmatrix}1 & 1 & 1 \\ -1 & 1 & 1\end{smallmatrix}\rbrack\right)^{-1}, & -q_r(a) &= \left (\epsilon_a -F_{a a^* a}^{a}\lbrack\begin{smallmatrix}1 & 1 & 1 \\ -1 & 1 & 1\end{smallmatrix}\rbrack\right)^{-1}. -\end{align}<|endoftext|> -TITLE: Is the axiom $\Diamond\Box\varphi\to\Box\Diamond\varphi$ in c.c.c. forcing potentialism equivalent to the productivity of c.c.c. forcing? -QUESTION [22 upvotes]: This question arose in connection with a lecture series on -Potentialism -that I have just completed here in Hejnice in the Czech Republic at -the Winter School 2018 (see -Slides). Several of us discussed the question late into the evening after my talks, and still we have no resolution. -Consider the context of c.c.c. forcing over the set-theoretic universe $V$. Let us interpret the modalities by c.c.c. forceability, so $\Diamond\varphi$ means that $\varphi$ is c.c.c. -forceable, and $\Box\varphi$ means that $\varphi$ holds in all c.c.c. -extensions. -It is easy to see that under Martin's axiom $\text{MA}_{\omega_1}$, -every instance of the $(.2)_{ccc}$ axiom scheme -$$\Diamond\Box\varphi(a)\to\Box\Diamond\varphi(a)$$ holds, where -$a$ is a parameter of hereditary size at most $\omega_1$. This -scheme asserts that any statement that is c.c.c. possibly necessary is c.c.c. necessarily possible. Since it follows from MA, but does not hold in $L$, for example, one can view the axiom as a weak version of MA. -For example, the $(.2)_{ccc}$ axiom scheme implies that there are no Suslin trees. To see this, suppose $T$ is a Suslin tree and consider the statement "$T$ has a branch." This is c.c.c. possibly necessary, since we could force with $T$, but it is not c.c.c. - necessarily possible, since we could force instead to specialize $T$, - which would prevent it from getting a branch in any c.c.c. - extension. -The $(.2)_{ccc}$ scheme holds provided merely that the product -$\mathbb{P}\times\mathbb{Q}$ of two c.c.c. forcing notions is still -c.c.c., which is a consequence of $\text{MA}_{\omega_1}$. To see this, suppose that $\Diamond\Box\varphi(a)$ holds. So there -is a c.c.c. forcing notion $\mathbb{P}$ forcing $\Box\varphi(a)$. -Now, for any c.c.c. forcing notion $\mathbb{Q}$, the product -$\mathbb{P}\times\mathbb{Q}$ is c.c.c., and so $\mathbb{P}$ remains -c.c.c. after forcing with $\mathbb{Q}$. The product extension will -satisfy $\varphi(a)$, since it can be viewed as an extension of -$V^{\mathbb{P}}$, and so $\Box\Diamond\varphi(a)$ holds in $V$, as -desired. - -Question. Is the (.2) axiom scheme for c.c.c. forcing - $$\Diamond\Box\varphi(a)\to\Box\Diamond\varphi(a)$$ equivalent to the assertion that the product of c.c.c. - forcing is necessarily c.c.c.? - -We allow any $\varphi$ in the language of set theory and any -parameter $a$ of hereditary size at most $\omega_1$. -The question is extremely natural, since the productivity of c.c.c. -forcing is closely connected with directedness of the (mutually -generic) c.c.c. extensions, and the finite directed pre-orders are -complete for S4.2, whose central axiom is precisely the scheme in -question. -Let me make several observations. - -If one assumes the c.c.c. maximality principle, which is the -S5 scheme $$\Diamond\Box\varphi(a)\to\varphi(a),$$ -then indeed the -product of c.c.c. forcing is c.c.c., and the reason is that if a product $\mathbb{P}\times\mathbb{Q}$ of -c.c.c. forcing was not c.c.c., then there is an instance of size $\omega_1$ and the assertion "$\mathbb{P}$ is -not c.c.c." would be possibly necessary by the c.c.c. forcing -$\mathbb{Q}$, and so it would have to be already true by the -maximality principle, contradicting the assumption that -$\mathbb{P}$ was c.c.c. -The (.2) axiom is exactly equivalent to the assertion that there are no railway switches, to use the terminology of my talk. So the question is asking whether you can construct a railway switch from any violation of c.c.c. productivity. - -Thus, the question appears to be fundamentally connected with the -question: if a c.c.c. forcing notion $\mathbb{P}$ is not c.c.c. -necessarily c.c.c., then must there be fundamentally two different -and incompatible ways to c.c.c. destroy the c.c.c.ness of -$\mathbb{P}$? If so, then this would amount to an affirmative -answer to the question, because the statement that $\mathbb{P}$ was -not c.c.c. because of the first way would be c.c.c. possibly -necessary, but not c.c.c. necessarily possible, because we could -destroy it the other way. -Lastly, here is a slightly weaker version of the question. - - -Weaker Question. Does the (.3) axiom scheme for c.c.c. forcing, with parameters of hereditary size at most $\omega_1$ $$\Diamond\varphi(a)\wedge\Diamond\psi(a)\to\Diamond[(\varphi(a)\wedge\Diamond\psi(a))\vee(\psi(a)\wedge\Diamond\varphi(a))]$$ imply the productivity of c.c.c. forcing? - - -The point is that S4.3 implies S4.2, and so this question may be easier to answer. The axiom (.3) should be thought of as expressing the linearity of truth, since it says that if two statements are c.c.c. forceable, then one of them can be viewed as happening before the other. This is weaker than the maximality principle, and stronger than (.2), but still I don't know how to prove it implies the product of c.c.c. forcing is c.c.c. - -REPLY [6 votes]: Failure of $\sf CH$ -It seems I was a bit off on my initial go at showing $(.2)_{ccc}$ entails the failure of $\mathsf{CH}$. Here is the correct argument. -Proposition 1: Assuming $(.2)_{ccc}$. - -If $K\subset [\omega_1]^2$ is uncountable and c.c.c. then the graph $L = [\omega_1]^2 \backslash K$ is not powerfully c.c.c. - -(in particular $\mathsf{CH}$ fails, see Galvin1977; also, if I understand correctly, this statement is an odd mix of some of the c.c.c. partition relations listed here LarTodo2001, p. 87.) -Proof: Consider the statement $\varphi(a) := (\exists H)(\vert H \vert > \omega$, and $[H]^2 \subset a)$, then for each uncountable c.c.c. graph $K$, we have $\diamondsuit \square \varphi(K)$ (as witnessed by the partial order $\mathbb{P}_{K}$ of finite $K$-cliques, which is assumed to be c.c.c.) Applying $(.2)_{ccc}$ we must conclude $\square \diamondsuit \varphi(K)$. -To establish the result, note that if the finite support product of countably many copies of $\mathbb{P}_L$ is c.c.c., then in the corresponding extension, $\omega_1$ is covered by countably many $L$-homogeneous sets. It follows that in no further $\omega_1$-preserving forcing extension, can the statement $\varphi(K)$ hold. $\square$ -Remark: the use of "not powerfully c.c.c." in the above was a bit of a cop-out since I spent far to long treading water on simply "not c.c.c."; however it does seem to be important. - -Partial Result on Products. -It turns out, under $(.2)_{ccc}$, there is a connection between being powerfully c.c.c. and productively c.c.c.; in particular -Proposition 2: Assuming $(.2)_{ccc}$: - -If $\mathbb{P}$ is powerfully c.c.c. (that is, each finite power of $\mathbb{P}$ is c.c.c.), then $\mathbb{P}$ is productively c.c.c. (i.e. the product of $\mathbb{P}$ with any other c.c.c. forcing, is again c.c.c.) - -Proof: Given an order relation $a \subset \omega^{V}_1 \times \omega^{V}_1$ defined on $\omega^{V}_1$, let $\varphi(a)$ be the statement -$\varphi(a):= \left(\exists \mathcal{C} \subset \mathcal{P}(\omega^V_1)\right)(| \mathcal{C} | \le \omega$, each $C \in \mathcal{C}$ is centered with respect to $\le_a$, and $\omega^V_1 \subset \cup\mathcal{C})$ -(it should be clear that $\varphi(a)$ holds, iff, the partial order $(\omega^V_1, \le_a)$ is $\sigma$-centered.) -Now, for each powerfully c.c.c. order relation $R \subset \omega^V_1 \times \omega^V_1$, the statement $\diamondsuit \square \varphi(R)$ holds (since by assumption, the finite support product of countably many copies of the partial order $(\omega^V_1, \le_R)$ is c.c.c. and in the extension generated by this product, the $\Sigma_1$ statement $\varphi(R)$ holds.) As such, from $(.2)_{ccc}$, we can conclude $\square \diamondsuit \varphi(R)$. -To establish the result, fix some c.c.c. partial order $\mathbb{P}$. Applying our interpretation of the modal operators $\square$ and $\diamondsuit$, we are guaranteed the existence of a $\mathbb{P}$-name $\dot{\mathbb{Q}}$, such that, the iteration $\mathbb{R}=\mathbb{P}\ast \dot{\mathbb{Q}}$ is c.c.c. and forces $\varphi(R)$. -Therefore, fixing an $\mathbb{R}$-name $\dot{\mathcal{C}}$ witnessing $\varphi(R)$ and arbitrary $\mathbb{P}$-name $\dot{A}$ forced by $1_\mathbb{P}$ to be an $R$ anti-chain, the following hold: - -$\Vdash_\mathbb{R} (\forall C \in\dot{\mathcal{C}})( |C \cap \dot{A}| < \omega)$ -$\Vdash_\mathbb{R} |\mathcal{\dot{C}}| < \dot{\omega}_1=\omega^V_1$, and -$\Vdash_\mathbb{R} \dot{A} \subset \bigcup \mathcal{\dot{C}}$ - -and combine to imply $\Vdash_\mathbb{P} | \dot{A} | < \dot{\omega}_1 = \omega^V_1$. Hence $\Vdash_{\mathbb{P}} (\omega^V_1, \le_\check{R}) \text{ is c.c.c. }$ and the result follows. $\square$ -Remark: It's worth pointing out that under $\sf CH$, powerfully c.c.c. does not imply productively c.c.c. ( Galvin1977 ) - -The following dichotomy is an interesting corollary to Proposition 2. -Product Dichotomy: Assuming $(.2)_{ccc}$: -For every partial order $\mathbb{P}$: Exactly one of the following holds, - -The product of $\mathbb{P}$ with any c.c.c. partial order is c.c.c. - -Or, - -Some finite power of $\mathbb{P}$ is not c.c.c. - -Proof: Without loss of generality assume $\mathbb{P}$ is c.c.c. and apply Proposition 2. - -In light of the previous corollary, the following seems like a natural question. -Question: Does $(.2)_{ccc}$ imply every c.c.c partial order is powerfully c.c.c.? -As pointed out by @JoelDavidHamkins, it was already enough to just consider the square of c.c.c. partial orders.<|endoftext|> -TITLE: A question about ordinal analysis -QUESTION [5 upvotes]: I have several questions related to ordinal analysis. -According to [1], here are the proof-theoretic ordinal of some well-known theories (using $|T|$ do denotate the proof-theoretic ordinal of $T$): - -$ |\text{ATR}_0|=\Gamma_0,|\text{ATR}|=\Gamma_{\varepsilon_0} $ -$ |\Pi^1_0-\text{CA}_0|=\varepsilon_0,|\Pi^1_0-\text{CA}|=\varepsilon_{\varepsilon_0}<\varphi(\varepsilon_0,0)=|\Delta^1_1-\text{CA}|$ -$|\Pi^1_1-\text{CA}_0|=|\Delta^1_2-\text{CA}_0|=\psi(\Omega_\omega)$ -$|\Pi^1_1-\text{CA}|=\psi(\Omega_\omega\varepsilon_0)<\psi(\Omega_{\varepsilon_0})=|\Delta^1_2-\text{CA}|$ - -(Notice when there is a 0 subscript and when there isn't. Of course $\Pi^1_0-\text{CA}_0$ is $\text{ACA}_0$.) -I have never seen $|\Delta^1_1-\text{CA}_0|$ mentioned anywhere, but seeing the $\Delta^1_2$ case I suppose it has proof-theoretic ordinal $\varepsilon_0$. -More generally the following equality seem a reasonable conjecture: - $ |\Pi^1_n-\text{CA}_0| = |\Delta^1_{n+1}-\text{CA}_0| < |\Pi^1_n-\text{CA}|<|\Delta^1_{n+1}-\text{CA}| $ for all $n$. Does anybody has a proof or reference for this? -Also every time the 0 subscript is dropped, the ordinal $\varepsilon_0$ appears in the proof-theoretic ordinal of the resulting theory. Is there a simple reason or intuition why this happens? - -REPLY [6 votes]: First of all, note that we don't (yet) have ordinal analyses of subsystems of second order arithmetic beyond $\Pi^1_2$-CA$_0$. -Still, we can say something about the pattern you indicate using known results about these systems. A good reference is the book by Stephen G. Simpson: Subsystems of Second Order Arithmetic. -There we find for instance: - -$\Delta^1_{k+3}$-CA$_0$ is a conservative extension of $\Pi^1_{k+2}$-CA$_0$ for $\Pi^1_4$-sentences. (Cor. IX.4.12) -$\Delta^1_{k+2}$-CA proves the existence of a countable coded $\beta$-model of $\Pi^1_{k+1}$-CA$_0$. (Cor. VII.7.9) -$\Pi^1_{k+1}$-CA$_0$ proves the existence of a countable coded $\beta$-model of $\Delta^1_{k+1}$-CA$_0$. (Cor. VII.7.9) - -Finally, let me speculate a bit on your last question. I don't think there's a formal theorem to this effect. (At least not with current technology.) The subsystem $T$ without the subscript $0$ differs from $T_0$ by adding full induction. When we have all the tools for an ordinal analysis of $T_0$, we can often get one for $T$ by (in some sense) iterating our procedure for $T_0$ along $\varepsilon_0$ to do the cut-elimination with full induction. But it's not so straight-forward, as the exact steps differ widely in each case, for instance if collapsing is involved. Have a look at Pohlers' chapter on Subsystems of set theory and second-order number theory in the Handbook of Proof Theory. There you'll see some instances of this pattern, at least as far as the upper bounds are concerned. -Perhaps a more general statement could have been made using Girard's $\Pi^1_n$-logic, but that's the realm of wild speculation on my part, and I don't think this was ever developed enough, particularly in connection with subsystems of second order arithmetic.<|endoftext|> -TITLE: Uniqueness of the wonderful compactification of a semi-simple group -QUESTION [5 upvotes]: Let $G$ be a semi-simple group over an algebraically closed field of characteristic zero. In which cases there is a unique wonderful compactification of $G$ (modulo isomorphism)? -For instance, is the wonderful compactification of $SL(n,\mathbb{C})$ unique? - -REPLY [7 votes]: I assume you mean the variety $G$ considered as a $G \times G$ variety via the action $(g,h) \cdot x = gxh^{-1}$, which is the standard interpretation in the literature. The variety $G$ is spherical as a $G \times G$ variety, meaning that it contains a dense $B \times B$-orbit (where $B$ is a Borel subgroup of $G$). This statement is the content of the well-known Bruhat decomposition of $G$. -There is a more general statement: If $X$ is a spherical $G$-variety, where $G$ is reductive algebraic group over an algebraically closed field of any characteristic, then if $X$ admits a wonderful compactification, it is unique up to $G$-isomorphism. -The variety $G$ does have a wonderful compactification, which is therefore unique up to $G \times G$-isomorphism. The existence of the wonderful compactification of $G$ goes back to the original work on wonderful compactifications, Complete symmetric varieties by De Concini and Procesi (MSN). I don't know where the first proof of uniqueness appears, but a good introduction to the subject that includes a full proof is Pezzini's survey article, Lectures on spherical and wonderful varieties. -Here's a sketch of the main ideas involved. Any wonderful compactification of $G$, which I will denote $Y$, contains a big cell $Y_0$ with the following properties: - -$Y_0$ is dense in $Y$ -$Y_0$ is $B \times B$-invariant and isomorphic to affine space -$G \cdot Y_0 = Y$ - -So suppose $Y$ and $Y'$ were wonderful compactifications of $G$ with big cells $Y_0$ and $Y'_0$. Let $g, g' \in G$ and consider the unique birational $G \times G$-equivariant map $f: Y \dashrightarrow Y'$, i.e. $f(g_1 g g_2^{-1}) = g_1 g' g_2^{-1}$. This map induces an isomorphism $Y_0 \cong Y'_0$ because they have isomorphic coordinate rings (an argument is needed why the map can be extended to the big cell). Then $f$ extends to an isomorphism $Y \cong Y'$ since $(G \times G) \cdot Y_0 = Y$.<|endoftext|> -TITLE: What kind of arithmetic information does the ring of integers in an infinite extension carry? -QUESTION [11 upvotes]: The fact that the ring of integers in a finite extension of $\Bbb Q$ is a Dedekind domain and purely algebraic properties of Dedekind domains are absolutely essential for algebraic number theory. So it seems natural to ask about infinite extensions next. -While the (not necessarily absolute) Galois groups of infinite algebraic extensions play a role e.g. in Iwasawa theory, it seems to me that the rings of integers of infinite algebraic extensions receive little attention. (Though this may just be a lack of knowledge on my part.) -There is of course the classical result due to Dedekind that the ring of all algebraic integers is a Bezout domain. There are also notions in commutative algebra to describe what kind of ring the integral closure in an infinite algebraic extension will be. (The fact that one may call Prüfer domains "arithmetical domains" makes this question seem more pertinent.) -But while knowing for example the class group of the ring of integers in a finite extension can have striking consequences, it is not obvious to me what kind of arithmetic information we get for infinite algebraic extensions when we know for example class group of the ring of integers or that it is a Bezout domain. And I've never seen general results about Prüfer or Bezout domains applied in algebraic number theory. -So my question is the question in the title: What kind of arithmetic information can we get out of algebraic structure of the ring of integers in an algebraic extension? Do you know examples where results about Prüfer domains/Bezout domains/Krull domains etc. are applied to number theory? - -REPLY [3 votes]: Picking up on the phrase "arithmetic information" in your question, let me give a brief answer coming from logic, although I recognize that this is likely not the answer for which you are looking. -Logicians often measure the arithmetic information content of their mathematical structures by investigating which Turing degrees are encoded within those structures. To my way of thinking, anyone -taking the phrase "arithmetic information" seriously will end up talking about the Turing degrees. -For any nonstandard model $\mathcal{M}$ of Peano arithmetic PA, we -have the corresponding nonstandard ordered field of rational -numbers $\newcommand\Q{\mathbb{Q}}\Q^*$, which has the nonstandard -integers $\newcommand\Z{\mathbb{Z}}\Z^*$ sitting inside as a discretely ordered -ring, of which it is the quotient field. (But I'm not actually sure -whether $\Q^*$ can be algebraic over $\Q$.) -Logicians and researchers in the field known as models of PA -measure the arithmetic information content of such a structure by -means of the standard -system -of the model. -The standard system is the set of subsets -$A\subseteq\newcommand\N{\mathbb{N}}\N$ that are coded in the -model, in the sense that there is an element $n\in \N^*$, usually -nonstandard, such that $i\in A$ if and only if the $i^{th}$ prime -divides $n$ in $\N^*$. There are a huge variety of coding methods, -and all of them are equivalent for the purpose of defining the -standard system. -For example, a set $A$ is in the standard system of the model if -and only if there is a diophantine equation $p(x,\vec x)=0$ with -coefficients in $\Z^*$ such that $i\in A\iff p(i,\vec x)=0$ has a -solution in $\Z^*$, for $i\in\N$. -In countable models, a collection $S$ of sets $A\subseteq\N$ is the -standard system of a model of PA if and only if $S$ is a Boolean -algebra, closed under Turing reduction, and whenever $T\subset 2^{<\N}$ is an -infinite binary tree coded in $S$, then there is a branch -through $T$ in $S$. -The standard system of a nonstandard model of arithmetic measures -the arithmetic information content of the model because it -identifies upper bounds on which kind of arithmetic sets could be -defined in the model. -For example, some nonstandard models of arithmetic have their -standard systems consisting entirely of low -sets, which -have comparatively little information content; meanwhile, others -have a standard system closed under the Turing jump, and so their -information content exceeds the halting problem and much more. -A simple compactness argument shows that some nonstandard models of -PA have a standard system containing all sets of natural numbers. -It remains an open question, however, whether there is a Borel -definable such nonstandard model of PA, whose standard system -includes all sets $A\subset\N$.<|endoftext|> -TITLE: Origin of definitions of ramified Hecke operators -QUESTION [8 upvotes]: Consider a classical space $M_k(N)$ of elliptic modular forms of weight $k$ for $\Gamma_0(N)$. The definition of an unramified Hecke operator $T_{p^m}$ in terms of double cosets is the disjoint union of double cosets -$$\Gamma_0(N) \begin{pmatrix} a&b \\ c&d \end{pmatrix} \Gamma_0(N),$$ -where $a, b, c, d \in \mathbb Z$ with $ad-bc = p^m$ and $c \equiv 0$ mod $N$. The usual definition for a ramified Hecke operator $T_{p^m}$ ($p | N$), the union of double cosets as above, now with the additional restriction that $p \nmid a$. -On the other hand, Eichler in his work on the basis problem worked with ramified -Hecke operators without this restriction $p \nmid a$. -To me, the usual definition seems rather ad hoc, and the one Eichler uses seems more natural from the point of view of orders and ideals. (Note these definitions give different operators.) Of course, the usual definition is nice because it acts nicely on Fourier coefficients. -Question: Where were these ramified Hecke operators first defined? -For bonus credit: how did these definitions actually come about? - -REPLY [7 votes]: These operators certainly appeared in the 1970 paper by Atkin and Lehner: - -Atkin, A. O. L.; Lehner, J. Hecke operators on $\Gamma_0(m)$. Math. Ann. 185 - (1970), 134–160. - -I don't know for sure that this is the first time these operators appeared in print, but the Mathematical Reviews entry for this paper, by Rankin, seems to suggest so; the concluding paragraph of the review is - -These results show that it is possible to obtain a satisfactory theory - of Hecke operators on $\Gamma_0(m)$ not only for primes $p \nmid m$, but also for - primes that do divide the level m.<|endoftext|> -TITLE: Generic supercuspidal representations of $\operatorname{GL}_n$ can be defined by integrals over $U$ -QUESTION [5 upvotes]: Let $(V,\pi)$ be an irreducible, admissible, supercuspidal representation of $G = \operatorname{GL}_n(F)$ for $F$ a $p$-adic field. Let $B = TU$ be the usual Borel subgroup, maximal torus, and unipotent radical of $G$. Let $f_{v^{\ast},v}(g) = \langle v^{\ast}, \pi(g)v \rangle$ be a matrix coefficient for $v \in V$ and $v^{\ast} \in V^{\ast \infty}$. Then $f = f_{v^{\ast},v}$ is locally constant and compactly supported modulo the center $Z$ of $G$. If $\chi$ is a generic character of $U$, the integral -$$\int\limits_U f(ug)\chi(u^{-1}) du $$ -can be shown to converge absolutely for every $g \in G$. Also, a change of variables shows that for fixed $v^{\ast}$, -$$v \mapsto \int\limits_U f_{v^{\ast},v}(u)\chi(u^{-1})du \tag{1} $$ -defines a linear functional $\lambda: V \rightarrow \mathbb{C}$ which satisfies $\lambda(\pi(u_1)v) = \chi(u_1)\lambda(v)$ for all $u_1 \in U$. However, this linear functional might be the zero functional. -In general, $\pi$ is called generic if there exists a nonzero linear functional satisfying the property of the previous paragraph for $\chi$. If $\pi$ is generic for $\chi$, it is also generic for every other generic character. - -1 . Is every irreducible, admissible supercuspidal representation of $G$ generic? -2 . If $\pi$ is generic, does there exist a smooth linear functional $v^{\ast}$ such that the map (1) is not the zero map? In other words, if there is a nonzero Whittaker functional, can it always be defined by an integral? - -REPLY [9 votes]: The answer to both questions is yes. - -All irreducible supercuspidal representations of ${\rm GL}(N,F)$ are generic. See e.g. I. M. Gelfand and D. A. Kajdan, -Representations of the group ${\rm GL}(n,K)$ where $K$ is a local field, -Lie groups and their representations. -All ingredients to prove 2. are in - -Paskunas, Vytautas; Stevens, Shaun On the realization of maximal simple types and epsilon factors of pairs. Amer. J. Math. 130 (2008), no. 5, 1211–1261. -First it is known by Bushnell and Kutzko that any irreducible supercuspidal representation of $G$ is of the form $\pi ={\rm ind}_J^G \Lambda$ (compactly induced representation), where $(J,\Lambda )$ is a maximal simple type in the sense of B. and K. Paskunas and Stevens prove that one can arrange the data so that $Hom_{U\cap J} (\Lambda , \chi ) \not= 0$ (1). It is a classical fact that if $c$ is a coefficient of $\Lambda$, then viewed as a fonction on $G$ (extend by $0$ off $J$), $c$ is a coefficient of $\pi$ (of the form $f_{v* ,v}$) (cf. e.g. Carayol Ann. Scient. ENS). Now writing condition (1) as the non-vanishing of an integral, you get exactly what you want. -Tell me if you want some more detail.<|endoftext|> -TITLE: A map from a symmetric product of a curve to its Jacobian -QUESTION [5 upvotes]: Let $C$ be a smooth projective curve over an algebraically closed field $k$, of genus $g$. -It is well known that, after fixing a point $p_0$, the map $C^{(n)}\to J$ sending $\{a_1,\dots,a_n\}$ to $[a_1+\dots+a_n-np_0]$, from the n-th symmetric product of the curve to its Jacobian, is an algebraic projective bundle (for $n>2g-2$). -Consider the map $C^{(n)}\times C^{(n)}\to J$, sending $(\{a_1,\dots,a_n\},\{b_1,\dots,b_n\})$ to $[a_1+\dots+a_n-b_1-\dots-b_n]$. -Is this map also a fiber bundle? - -REPLY [5 votes]: Let $J_n = Pic(C)_n$ --- the moduli space of line bundles of degree $n$ on $C$. Then there is a map -$$ -C^{(n)} \to J^n,\qquad \{a_1,\dots,a_n\} \mapsto O(a_1+\dots+a_n). -$$ -This is a slightly more canonical version of the map you considered, in particular it is a projective bundle for $n > 2g - 2$. -The map you are interested in can be written as the composition -$$ -C^{(n)} \times C^{(n)} \to J_n \times J_n \cong J_n \times J_{-n} \to J_0. -$$ -Here the first map is the product of to projective bundles (its fiber is a product of two projective spaces), the second is an isomorphism (given by dualization of a line bundle in the second factor), and the third is a (trivial) abelian fibration. -So, altogether, the fibers of your map are fiber products of two projective bundles over an abelian variety.<|endoftext|> -TITLE: Tensor products of $\infty$-algebras over operads -QUESTION [10 upvotes]: Let $A$ and $B$ be $A_\infty$-algebras. It's true, but it's a quite nontrivial fact, that the tensor product $A \otimes B$ can be given the structure of $A_\infty$-algebra, too. What is much easier to prove is that the tensor product of an $A_\infty$-algebra with a dg-algebra is again an $A_\infty$-algebra. -One can ask similarly whether the tensor product of a $C_\infty$-algebra and an $L_\infty$-algebra is an $L_\infty$-algebra, generalizing the simple fact that the tensor product of a commutative algebra and a Lie algebra is again a Lie algebra. But in fact I can't even prove what should be a much simpler statement: that the tensor product of a $C_\infty$-algebra with a Lie algebra is an $L_\infty$-algebra. Is this true? -Remark. What is quite clear is that the tensor product of a commutative algebra and an $L_\infty$-algebra is an $L_\infty$-algebra. - -REPLY [3 votes]: If you know the statement for strict algebras, then you "know" the statement for $\infty$-algebras by general nonsense. For example, the fact that a Lie algebra tensored with a commutative algebra again has a canonical Lie structure gives you an operad map $L \to L \otimes C$ (where I use the Boardman--Vogt tensor product, and in this case the map is an iso, but I don't need it to be). I also have surjective quasiisomorphisms $L_\infty \to L$ and $C_\infty \to C$. The tensor product of surjective quasiisomorphisms is again a surjective quasiisomorphism, so I get a surjective quasiisomoprhism $L_\infty \otimes C_\infty \to L \otimes C$. On the other hand, the cofibrancy of $L_\infty$ (in the model in which surjective quasiisomorphisms are acyclic fibrations) implies that I can lift the map $L_\infty \to L \otimes C$ against the map $L_\infty \otimes C_\infty \to L \otimes C$. After doing this lifting, I achieve the map $L_\infty \to L_\infty \otimes C_\infty$. Pulling back along this map gives me an $L_\infty$ structure on the tensor product of $L_\infty$ and $C_\infty$ algberas. -So the question you ask boils down to understanding the map $L_\infty \to L_\infty \otimes C_\infty$. The general nonsense says that it is not strictly canonical, but only canonical up to a contractible space of homotopies. However, it is not too hard to write down an explicit map in terms of "sums over diagrams". Perhaps you are asking for an explicit such sum?<|endoftext|> -TITLE: Rotation number of composition -QUESTION [5 upvotes]: Let $f,g:S^1 \to S^1$ be orientation-preserving homeomorphisms. Consider the lift $F,G:\mathbb R \to \mathbb R$. Let $\rho(G)$ and $\rho(F)$ be a rotation numbers. What we can say about rotation number of their composition? i.e about $\rho(F\circ G)$. I know that if $F$ and $G$ are commute then $\rho(F\circ G)=\rho(F)+\rho(G)$. But what if not? For example if $\rho(G) = \rho(F)=0$ then what can we say about $\rho(F\circ G)$? - -REPLY [3 votes]: @user3421341,Theorem 3.9 in the paper of Calegari and Walker gives a complete answer to your question, for arbitrary values of $\rho(F)$ and $\rho(G)$.<|endoftext|> -TITLE: Equivalence between existence of limits and products+equalizers, for derivators -QUESTION [5 upvotes]: $\require{AMScd}\def\D{\mathbb{D}}\def\prepull{\vcenter{\lrcorner}}$ -It is well known that for a category $\cal C$ the existence of finite limits is equivalent to the existence of finite products and equalizers, or to the existence of a terminal object and pullbacks. -What about (pre)derivators? Is it true that, for example, the existence of finite homotopy limits (right Kan extensions), the existence of homotopy pushouts, and homotopy equalizers, can be interchanged? -More precisely, can one infer from the existence of all homotopy limits of shape $\prepull$, and from the presence of terminal objects, the existence of all homotopy limits of shape $\rightrightarrows$, and from this, in turn, the existence of all finite limits (homotopy limits indexed over a homotopy finite diagram)? -A rather naive attempt for such a a statement is: given $X\in\D(\prepull)$ there is a diagram $\bar X\in \D(\rightrightarrows)$ such that $t_{1,*}X\cong t_{2,*}\bar X$, if -$$ -\begin{CD} -@. \D(\prepull)\\ -@. @VVt_{1,*}V\\ -\D(\rightrightarrows) @>>t_{2,*}> \D(e) -\end{CD} -$$ denote the respective homotopy limit functors. Conversely, given $\bar X$ there is $X$ satisfying the same condition. -I suspect this naive approach fails, though, especially because the proof of the fact that the existence of homotopy pushouts and finite homotopy colimits are equivalent appears here as 7.1, and relies on high-tech arguments. Moreover, this correspondence, if true and pushed to the highest coherence level, seems to set up an equivalence of derivators $\D^\prepull\cong \D^\rightrightarrows$, which is hard to believe. - -REPLY [4 votes]: $\require{AMScd}\def\D{\mathbb{D}}\def\prepull{\vcenter{\lrcorner}}$ -The result you mention from the paper by Ponto-Shulman does not say exactly what you stated. The setting is that of derivators (so you already have all the homotopy Kan extensions indexed by small cats in some fixed category of diagrams). In that setting they prove the following: -Theorem 7.1 (Ponto-Shulman). -Let $A$ be a homotopy finite category. - -If $\mathbb D$ is a derivator and $\mathbb E \subseteq \mathbb D(e)$ contains the initial object and is closed under [homotopy] pushouts, then it is closed under [homotopy] $A$-colimits; -If $F \colon \mathbb D \to\mathbb E$ is a morphism of derivators that preserves the initial object and [homotopy] -pushouts, then it preserves [homotopy] $A$-colimits. - -The idea of their proof is that you can actually construct homotopy $A$-colimits via homotopy pushouts. You are actually asking about the dual setting, which of course holds similarly. In what follows I'll show that in the dual of the above theorem one can add to the characterization of classes closed under taking homotopy limits via homotopy pullbacks, also a characterization that uses homotopy equalizers. The reason for that is simple: homotopy pullbacks can be computed via homotopy equalizers: -Consider the following categories: - -$Ob(\prepull)=\{0,1,2\}$, $\hom_\prepull(0,1)=\{a\}$, $\hom_\prepull(2,1)=\{b\}$; -$Ob(\rightrightarrows)=\{0,1\}$, $\hom_{\rightrightarrows}(0,1)=\{a,b\}$; -$Ob(\square)=\{-\infty,0,1,2\}$, $\hom_\square(0,1)=\{a\}$, $\hom_\square(2,1)=\{b\}$, $\hom_\square(-\infty,0)=\{c\}$, $\hom_\square(-\infty,2)=\{d\}$; -$Ob(\to\rightrightarrows)=\{-\infty,0,1\}$, $\hom_{\to\rightrightarrows}(0,1)=\{a,b\}$, $\hom_{\to\rightrightarrows}(-\infty,0)=\{c\}$. - -Consider also the following functors: - -$i_{\prepull}\colon\prepull\to \square$ is the obvious inclusion; -$i_{\rightrightarrows}\colon\{\rightrightarrows\}\to \{\to\rightrightarrows\}$ is the obvious inclusion; -$q_{\prepull}\colon \prepull\to \{\rightrightarrows\}$ is such that $q_\prepull(0)=0=q_\prepull(2)$, $q_\prepull(1)=1$; -$q_{\square}\colon \square\to \{\to\rightrightarrows\}$ is such that $q_\square(-\infty)=-\infty$, $q_\square(0)=0=q_\square(2)$, $q_\square(1)=1$. (Of course, $q_\square(c)=c=q_\square(d)$, $q_\square(a)=a$ and $q_\square(b)=b$). - -Given a derivator $\mathbb D$ and $X\in \mathbb D(\prepull)$, I claim that -$$ -(-\infty)^*(i_{\rightrightarrows})_*(q_\prepull)_*X\cong (-\infty)^*(i_\prepull)_*X. -$$ -In other words, the object $\bar X$ you are asking for in your question is $(q_\prepull)_*X$. To prove the above claim you have to notice that $q_{\square} i_{\prepull}=i_{\rightrightarrows}q_{\prepull}$, hence $(q_{\square})_* (i_{\prepull})_*\cong(q_{\square} i_{\prepull})_*=(i_{\rightrightarrows}q_{\prepull})_*\cong (i_{\rightrightarrows})_*(q_{\prepull})_*$. Thus, we are reduced to verify that $(-\infty)^*(q_{\square})_*\cong (-\infty)^*$. Let $Y\in \mathbb D(\square)$, to compute $(-\infty)^*(q_{\square})_*Y$ one can use the axiom (Der.4) (i.e., Kan extensions are pointwise in a derivator), so that -$$ -(-\infty)^*(q_{\square})_*Y\cong \mathrm{Hocolim}_{-\infty/q_{\square}} \mathrm{pr}^*Y -$$ -where $\mathrm{pr}\colon -\infty/q_{\square}\to \square$ is the obvious projection. So let us describe the category $-\infty/q_{\square}$: - -$Ob(-\infty/q_{\square})=\{(-\infty,id),(-\infty,c),(-\infty,ac),(-\infty,bc)\}$ -the object $(-\infty,id)$ is clearly initial. - -Hence, -$$ -\mathrm{Hocolim}_{-\infty/q_{\square}} \mathrm{pr}^*Y\cong (-\infty,id)^*\mathrm{pr}^*Y\cong (-\infty)^*Y, -$$ -where the first isomorphism holds by the well-known fact that taking the homotopy limit indexed by a category with an initial object is the same as evaluating at the initial object. This concludes the proof. -Let me add a final remark. In the comments to the question, Marc Hoyois suggests you to look at Proposition 4.4.3.2 in Lurie's Higher Topos Theory. If you look inside the proof of that result (dualizing to your setting), you will see that it actually gives a proof that homotopy pullbacks can be constructed using equalizers in the setting of $(\infty,1)$-categories. In fact, if you follow the steps in that proof (adapted to the setting of derivators), you start with an object $X\in \mathbb D(\prepull)$ and construct a "incoherent" diagram in $\mathbb D(e)^{\rightrightarrows}$ of the form: -$$ -X_0\times X_1\rightrightarrows X_2 -$$ -This is exactly the underlying diagram of the object $(q_{\prepull})_*X$ constructed above. - -Edit: -Let me be very precise on the exact statement of the theorem one gets with the above argument: -Theorem. -Let $A$ be a homotopy finite category. If $F \colon \mathbb D \to\mathbb E$ is a morphism of derivators that preserves finite products and homotopy equalizers, then it preserves homotopy $A$-colimits. -Proof. By Ponto-Shulman it is enough to verify that $F$ preserves pullbacks, that is, given $X\in \D(\prepull)$, there is a natural isomorphism -$$ -F((-\infty)^*(i_{\prepull})_*X)\cong (-\infty)^*(i_{\prepull})_*F(X). -$$ -Now, we know that $(-\infty)^*(i_{\prepull})_*X\cong (-\infty)^*(i_{\rightrightarrows})_*(q_\prepull)_*X$ and, also using that $F$ commutes with $(-\infty)^*i_{\rightrightarrows}$, we get -\begin{align*} -F((-\infty)^*(i_{\prepull})_*X)&\cong F((-\infty)^*(i_{\rightrightarrows})_*(q_\prepull)_*X)\\ -&\cong (-\infty)^*(i_{\rightrightarrows})_*F((q_\prepull)_*X) -\end{align*} -To conclude it is enough to show that the canonical morphism -$$ -F(q_\prepull)_*X\to (q_\prepull)_*F(X) -$$ -is an isomorphism. Since, in a derivator, a morphism is an iso if and only if its components are isos (this is usually the axiom (Der.2)), it is enough to look at the underlying diagrams. With some standard computation with the axiom (Der.4), one gets that the underlying diagram of $F(q_\prepull)_*X$ is $(F(X_0\times X_1)\rightrightarrows F(X_2))$, while the underlying diagram of $(q_\prepull)_*F(X)$ is $(F(X_0)\times F(X_1)\rightrightarrows F(X_2))$. Since $F$ also commutes with finite products, these diagrams are isomorphic, concluding our proof.\\\<|endoftext|> -TITLE: Reference Request - Recovering a function from its definite integrals (inverse problem) -QUESTION [6 upvotes]: I'm having a difficult time finding any theory on an inverse problem I've come up against. Let's say I have an unknown function $f:[0,1] \rightarrow \mathbb{R}$, and I know $\int_{a}^{b} f$ for some collection $A$ of pairs $(a,b)\in[0,1]^2$. I'm looking for pointers to any material that discusses condtions on $f$ and $A$ that are sufficient recover (all of? some of?) the values of $f$. Google searches just keep turning up elementary-calculus-help-type pages. I'm a beginning graduate student, if it matters. Thanks. -Edit: I'm actually looking for something broader than I asked for. I know that recovering the values of $f$ is a lot to ask and is very unlikely unless $A=[0,1]^2$; I'm also looking for approximations to $f$, anything that can be said about its properties/behaviour, etc. when $A\subsetneqq [0,1]^2$. - -REPLY [4 votes]: Here is how you make an inverse problem of this problem: Choose a space $X$ for the function $f$ you are looking for (e.g. $L^2(0,1)$ to work in Hilbert spaces, but other spaces may be more suitable, depending on your needs). -I assume that you only have finitely many definite integrals (since I assume that this is a practical problem where the definite integrals come from measurements). -Now let us denote your tuples as $(a_1,b_1),\dots (a_N,b_N)$. You forward operator is -$$\newcommand{\RR}{\mathbb{R}} -K:X\to\RR^N -$$ -mapping $f$ to the $N$-vector with components $\int_{a_i}^{b_i}f(x)\,dx$. So you are given a vector $g\in\RR^N$ and want some solution to -$$ -Kf = g. -$$ -Now you are in business with the standard theory for linear inverse problems. -You have some of the usual problems coming with an inverse problem: Non-uniqueness (the operator is not injective) and probably instability in some sense (depending on you data and values $(a_i,b_i)$). (As far as I see, non-solvability should not be an issue as $K$ should be surjective for meaningful tupels $(a_i,b_i)$). -To deal with non-uniqueness: You may view this as an advantage as you can choose among all solutions of $Kf=g$. To pick one, you can choose regularization functional $R:X\to [0,\infty]$ and define a minimum-$R$-solution as solution of -$$ -\min\{R(f)\mid f\in X,\ Kf=g\}. -$$ -From a computational point of view, convex functional $R$ are beneficial and you can choose $R$ to impose some structure on your solution, e.g. $R(f) = \int_0^1 |f'(x)|^2\, dx$ imposes some smoothness (effectively this means that you constrain your solutions to the Sobolev space $H^1$). The most straight-forward choice would be $R(f) = \int_0^1 |f(x)|^2\, dx$ which should produce a linear equality as optimality condition (and you are effectively computing the Moore-Penrose pseudo-inverse). I could say more about regularizing functionals if needed. -If your data vector $g$ is also uncertain, i.e. it may be given by measurement data with an error, you may want to relax your problem and look for solutions of -$$ -\min\{R(f)\mid d(Kf,g)\leq\delta\} -$$ -for some discrepancy functional $d$ and some value $\delta>0$. Both should be related to the error in your data. Note that this is in some way equivalent to (generalized) Tikhonov regularization which would be solving -$$ -\min_f d(Kf,g) + \lambda R(f) -$$ -for some regularization parameter $\lambda>0$. The most simple case of this would be standard Tikhonov regularization in Hilbert spaces: -$$ -\min_f \|Kf-g\|_{2}^2 + \lambda\|f\|_{L^2(0,1)}^2 -$$ -leading to the linear optimality condition -$$ -K^*(Kf-g) + \lambda f = 0. -$$ -The adjoint operator $K^*:\RR^N\to L^2(0,1)$ is given by -$$ -K^*g = \sum_{i=1}^N g_i\chi_{[a_i,b_i]} -$$ -(where $\chi_{[a_i,b_i]}$ is the characteristic function of $[a_i,b_i]$). -So the optimality condition is actually -$$ -\sum_i \left[\langle f,\chi_{[a_i,b_i]}\rangle - g_i\right]\chi_{[a_i,b_i]} + \lambda f = 0. -$$ -This shows that the regularized solution is also a linear combination of the characteristic functions $\chi_{[a_i,b_i]}$ and thus, we still get a finite dimensional linear problem for the coefficients. -If you want some smoothness, try $R(f) = \int_0^1 |f'(x)|^2\, dx$. This would give an optimality conditions like -$$ -\sum_i \left[\langle f,\chi_{[a_i,b_i]}\rangle - g_i\right]\chi_{[a_i,b_i]} - \lambda f'' = 0 -$$ -and thus the solution is piecewise quadratic.<|endoftext|> -TITLE: A natural residue formula -QUESTION [9 upvotes]: A residue formula -I have strong evindence to believe that the following identity holds: -$$ -\frac{n!}{2\pi i}\oint_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}} = d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right),\quad a,d,n+1 \in \mathbb{Z}_{\ge 1}. -$$ - -Have you seen this formula before? Any ideas on how to prove it? - -Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement. -Verification for a finite sample and proof for small $n$ -I verified this for thousands of particular values of $(a,d,n)$ and proved it for all $(a,d)$ with small $n$. -To get a proof for small $n$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $n$ and I don't think a proof for all $n$ will emerge from this line of attack unless you are a brilliant combinatorialist. -In order to check this equality for particular values of $(a,d,n)$, it makes a huge difference in computation time to combine Cauchy's integral formula with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,n)$. - -REPLY [20 votes]: Let us consider the function $w=z^d$ around $z=1$. It is holomorphic and locally invertible with a holomorphic inverse. So we can think of $z$ as $w^{1/d}$ where the latter symbol means a branch of the $d$-th root function taking $1$ at $1$. With this notation, the integral can be written as -$$\frac{n!}{2\pi i}\oint_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}}=\frac{n!}{2\pi i}\oint\frac{1}{d}\frac{w^{(a-d)/d} \mathrm{d}w}{(w-1)^{n+1}}.$$ -Here, we understand $w^{(a-d)/d}$ as the $(a-d)$-th power of $z=w^{1/d}$, and the integration is on the closed curve $\{w=(1+\epsilon\cdot e^{i t})^d:\ 0\leq t\leq2\pi\}$ with winding number $1$ around $w=1$. Hence, by Cauchy's integral formula, the right hand side equals $1/d$ times the $n$-th derivative of $w^{(a-d)/d}$ at $w=1$: -$$\frac{n!}{2\pi i}\oint_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}}=\frac{1}{d}\prod_{j=1}^n\left(\frac{a-d}{d}+1-j\right)=d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right).$$ -P.S. I have not seen this formula before.<|endoftext|> -TITLE: Chromatic number of a topological space -QUESTION [35 upvotes]: Here is a question I asked myself years ago. Since it is not really in my field, I hope to find some (partial) answers here... Since it was unclear, I precise that I am looking for an answer in ZFC, so using the axiom of choice if needed. -The chromatic number of a graph $G$ is the minimal cardinal of a partition of $G$ into independant sets, i.e. sets all of whose connected components are trivial. This definition of independant sets suggests that a good analogue of them in the setting of topological spaces would be totally disconected sets. This motivate the following definition: - -Definition. Define the chromatic number of a topological space $X$, denoted by $\chi(X)$, as the minimal cardinal of a partition of $X$ into totally disconnected sets. - -My main question is the following: - -Question. What is the chromatic number of $\mathbb{R}^n$? - -Some remarks. If $X$ can be embedded in $Y$, then $\chi(X) \leqslant \chi(Y)$. We also have that $\chi(X \times Y) \leqslant \chi(X)\chi(Y)$, because if $(A_i)_{i \in I}$ and $(B_j)_{j \in J}$ are respective partitions of $X$ and $Y$ into totally disconnected sets, then $(A_i \times B_j)_{(i, j) \in I \times J}$ is so for $X \times Y$. We also have that $\chi(\mathbb{R}) = 2$, witnessed by the partition $\{\mathbb{Q}, \mathbb{I}\}$, where $\mathbb{I}$ denote the set of irrational numbers. So $\chi(\mathbb{R}^n) \leqslant 2^n$. -But we have better. Actually, $\chi(\mathbb{R}^2) \leqslant 3$, witnessed by the partition $\{\mathbb{Q}^2, (\mathbb{Q} \times \mathbb{I} \cup \mathbb{I}\times \mathbb{Q}), \mathbb{I}^2\}$ (where $\mathbb{Q} \times \mathbb{I} \cup \mathbb{I}\times \mathbb{Q}$ is totally disconnected because it embeds into $\mathbb{I}^2$ via $(x, y) \mapsto (x + y, x - y)$). So $\chi(\mathbb{R}^{2n}) \leqslant 3^n$ and $\chi(\mathbb{R}^{2n + 1}) \leqslant 2 \times 3^n$. -My conjecture is that $\chi(\mathbb{R}^n) = n + 1$. For the upper bound, I suspect that some partition of the form $\{A_0, \ldots A_n\}$ where $A_i$ is the set of elements of $\mathbb{R}^n$ having exactly $i$ rational coordinates could work, but I don't manage to prove anything. For the lower bound, I only have an intuition given by the following "image": if you consider $A \subseteq \mathbb{R}^n$ totally disconnected, then it sounds reasonable to think that there exists a set $B\subseteq \mathbb{R}^n$, homeomorphic to $\mathbb{R}^{n - 1}$, passing "between" the points of $A$; however, this is just an inutuition and I don't know if it is true. I suspect that some algebraic topology would be needed to prove that, but I have almost no experience in this field so I couldn't investigate more; I am not even able to show that $\chi(\mathbb{R}^n) > 2$ for some $n$. -An other possibility would be that $\chi(\mathbb{R}^n) = 2$ for every $n$, because of some "weird" colouring. What makes me suspect that is that if we replace "connected" by "arcwise connected" in the definition of the chromatic number, then it becomes easy to build a partition of $\mathbb{R}^n$ into two parts each of which having no non-trivial arcwise connected subset, by a diagonal argument using the axiom of choice. However, this proof uses the fact that there are exactly $\mathfrak{c}$ arcs $\gamma : [0, 1] \longrightarrow \mathbb{R}^n$. To do the same proof for connected sets, we would need the existence of a family of $\mathfrak{c}$ non-trivial connected subsets of $\mathbb{R}^n$ such that every non-trivial connected subset of $\mathbb{R}^n$ has a subset in this family, and I don't think that such a family exists (but I don't know). -In case where we manage to prove that $\chi(\mathbb{R}^n) = 2$ thanks to some construction using the axiom of choice, it would be interesting to investigate what happens (still supposing choice) if we impose some restriction on the complexity of the sets in the partition, for example to be Borel. - -REPLY [31 votes]: The chromatic number $\chi(X)$ of a topological space $X$ is related to the separation dimension $t(X)$ introduced and studied by Steinke. -The separation dimension $t(X)$ is defined inductively: -$\bullet$ $t(\emptyset)=-1$ -$\bullet$ $t(X)=0$ for any space $X$ of cardinality $|X|=1$; -$\bullet$ if $|X|\ge 2$, then $t(X)\le n$ for some $n\in\mathbb N$ if for each subspace $M\subset X$ with $|M|\ge 2$ there exists a set $A\subset M$ such that $t(A)\omega$ was proved by Krasinkiewicz (see also this paper).<|endoftext|> -TITLE: Annihilating random walkers -QUESTION [15 upvotes]: Suppose there are several walkers moving randomly on $\mathbb{Z}^2$, -each taking a $(\pm 1,\pm 1)$ step at each time unit. -Whenever two walkers move to the same point, they -annihilate one another. This deletion always occurs in pairs, -so if $3$ walkers move to the same point, $2$ are deleted and $1$ remains, -whereas if $4$ move to one point, all $4$ are deleted. -At each time step, a new walker is created at the origin if it is not -occupied. If it is occupied, the new walker annihilates that origin occupier, -and no new walker is created at that time step. - -          - - -          - -After $n=50$ steps, $10$ walkers remain. -Green: newly created. Red: pair annihilation. - - - -Q1. - If the process is run from time $0$ to time $n$, - and from then on no more new walkers are created at the origin - but the simulation otherwise continues, - how many walkers are expected to exist as $t \to \infty$? - -I believe the answer is $0$, -essentially because of -Polya's recurrence theorem -applied to pairs of walkers. - -Q2. - If the process is run forever, continuing to create new - walkers at the origin if unoccupied, what is the - expected number of remaining walkers as $t \to \infty$? - How does this number grow (or shrink) with respect to $t$? - -Here I have little intuition, and would appreciate an analysis. - -REPLY [8 votes]: With a "physicist approach", I would write down the following equation for $f(x,t)$ that should represent the "density" of walker around $x$ at time $t$: $$\partial_t f =\Delta f -\alpha f^2 +\delta_0 $$ -with $\partial_t f =\Delta f+\delta_0$ is the diffusion equation with source at $0$ and $\alpha f^2$ the collisions term which is the density of two walker on the same site with an "independent hypothesis". (3 walker collisions term are neglected) -All this have to be rigourously derivate and it would probably need many pages to do so. However I would be very confident that this equation is the good one. -One can then try to solve the equation which is by symetrie, rotational invariant $$\partial_tf=\frac{1}{r}\partial_rf+\partial_{rr}f-\alpha f^2+\delta_0 $$ -On $]0,\infty]$ the stationnary solution is $$\frac{4}{\alpha r^2} $$ -One can then expect that $f\rightarrow \frac{4}{\alpha r}$ for $t\rightarrow \infty$. Because $$\int_1^\infty r(\frac{4}{\alpha r^2})=\infty $$ -The number of walker should grow to $\infty$. -I would also suspect that the limiting solution should be invariante with $r\rightarrow \lambda r$ and $\rightarrow \lambda^2 t$. In that case the number of walker should behave like $$N\sim a*\log(\sqrt{t}) $$ -(Remark that one should also be able to proved this last statement by grand deviation arguments as the probability that a walker goes to $\infty$.)<|endoftext|> -TITLE: Finiteness of Galois cohomology -QUESTION [9 upvotes]: Let $k$ be a field, $X$ a smooth projective variety over $k$, $\overline{X} := X\times_k {k}^{\rm sep}$ for a separable closure ${k}^{\rm sep}$ of $k$, $\ell$ a prime with $\ell\in k^{\times}$. - -Are the Galois cohomology groups $$H^i(\text{Gal}({k}^{\rm sep}/k),H^j_{\rm ét}(\overline{X},\mathbf{Z}_{\ell}))$$ - finite for $i\ge 1$, $j\ge 0$? Are they torsion? - -I would appreciate to get some references. -I have seen stated that $H^1(\text{Gal}({k}^{\rm sep}/k),H^j_{\rm ét}(\overline{X},\mathbf{Z}_{\ell}))$ is finite for $j\ge 0$ and $k$ a finite field, but I had the impression this should be a general fact about cohomology of profinite groups with coefficients in finitely generated $\mathbf{Z}_{\ell}$-modules with appropriate action. - -REPLY [7 votes]: I'm not sure what's going on with this question, but let me drop a few lines to summarize what the official position should be. -First off, I can't think about any situation these groups may occur beyond that of a Hochschild-Serre spectral sequence, so I hope group cohomology here is meant to be continuous group cohomology, and $\mathbf{Z}_{\ell}$-cohomology must then mean continuous étale cohomology. -The given answer (the accepted one) is incorrect. -In the functorial short exact sequences -$$0\to {\lim}^1 H^{i-1}(X_{\rm ét},\mathbf{Z}/(\ell^n))\to H^i\to \lim H^i(X_{\rm ét},\mathbf{Z}/(\ell^n))\to 0$$ -where $H^i$ is either $H^i(X_{\rm proét},\underline{\mathbf{Z}}_{\ell})$ or Jannsen's continuous étale cohomology $H^i_{\rm cont}(X,\{\mathbf{Z}/(\ell^n)\})$ (since they agree for any $X$) the ${\lim}^1$ term vanishes as soon as $H^{i-1}(X_{\rm ét},\mathbf{Z}/(\ell^n))$ is finite, by Mittag-Leffler. -In particular, if $X$ is defined over a separably closed field and proper (as in the OP's assumptions), $H^i$ agrees with usual $\ell$-adic cohomology. -Will's example is the typical way to show that geometric $\ell$-adic cohomology is usually not a discrete Galois representation, so any argument trying to infer $H^i(\text{Gal}(k^{\rm sep}/k), H^j)$ is torsion for $i>0$ and $j\ge 0$ from discreteness of $H^j$ is wrong. -Also, it is wrong to say that the same argument as for geometric étale cohomology of abelian sheaves, showing the Galois action is discrete, works for proétale cohomology. One can define the action abstractly and the concrete way as in the accepted answer, but these two actions don't generally agree anymore. It boils down to the fact that evaluation at usual geometric points does not give a conservative family of fiber functors on the proétale topos. -Both questions asked by the OP have negative answer, regardless of what $H^i$ is meant to be, among the possibilities discussed here. -Maybe the OP was meaning to ask something different?<|endoftext|> -TITLE: Action that is Bourbaki proper but not Palais proper -QUESTION [14 upvotes]: I'm working with different definitions of proper action (Cartan, Bourbaki and Palais) and the relation between them. All the spaces I'm working with are $T_{3.5}$, the definitions are: -If $U$ and $V$ are subsets of a $G$-space $X$ then we say that $U$ is thin relative to $V$ if $\{g \in G \; : \; gU \cap V \neq \emptyset \}$ has compact closure in $G$. If $U$ is thin relative to itself then we say that $U$ is thin. A subset $S$ of a $G$-space $X$ is small if every point of $X$ has a neighborhood wich is thin relative to $S$. -Bourbaki proper. Let $X$ be a $G$-space. We say that the action is Bourbaki proper (B-proper) if the function $\delta: G \times X \to X \times X$, $\delta(g,x) = (x,gx)$, is perfect (i. e. closed and the fibers of points $\delta^{-1}((x,x))$ are compact). If $G$ is locally compact, the action on $X$ is Bourbaki proper if and only if for every $x, y \in X$ there are neighborhoods $V_x$ and $V_y$ in $X$ of $x, y$, that are thin relative. -Palais proper. Let $X$ be a $G$-space with $G$ locally compact. We say that the action is Palais proper (P-proper) if each point of $X$ has a small neighborhood. -I need and example of an action that is B-proper but not P-proper . Some facts that could help are that P-proper implies B-proper and if the action is P-proper then the orbit space $X/G$ is $T_{3.5}$, like $X$, but if the action is B-proper then $X/G$ is only $T_2$. -Also if $X$ is locally compact then the two definitions are equivalent. Then, to build the example we need a $G$-space with $X$ not locally compact but that satisfy the axiom $T_{3.5}$ (or $T_{3}$) whose orbit space is $T_2$ but not $T_{3.5}$ $(T_{3})$. Some examples of space that are $T_{3.5}$ but not locally compact (where I have been looking for the example) are the Moore plane (Niemytski plane), Sorgenfrey line and infinite dimensional vector spaces (I have been trying with Hilbert cube). - -REPLY [2 votes]: Here is another example: Consider a compact finite dimensional smooth manifold $M$, and let $\text{Met}(M)$ be the space of all smooth Riemannian metrics on $M$, -an open subset in the Frechet space $\Gamma(S^2T^*M)$. Then the regular Frechet Lie group $\text{Diff}(M)$ of all diffeomorphisms of $M$ acts smoothly on $\text{Met}(M)$ by pullback. The quotient space $\text{Met}(M)/\text{Diff}(M)$ is Hausdorff in the quotient topology. The action is free on generic Riemannian metrics, and is the isometry group of a metric $g$ in general, which is a compact Lie group in general, since $M$ is compact. Thus this action is B-proper, but not P-proper, since the group $\text{Diff}(M)$ is not locally compact.<|endoftext|> -TITLE: closed geodesics are dense in both hyperbolic surface and unit tangent bundle of hyperbolic surface -QUESTION [5 upvotes]: Could you please recommend me some references for proofs of this fact: "closed geodesics are dense in both hyperbolic surface and unit tangent bundle of hyperbolic surface". Thanks in advance! - -REPLY [7 votes]: See the paper MR1763264 by Marcos Salvai. Here is a link to the paper.<|endoftext|> -TITLE: Generalization of results from specific algebraic theories to Universal Algebra -QUESTION [7 upvotes]: I'm relatively new to universal algebra, but it seems that lots of theorems from specific algebraic theories (groups, rings) can be stated in the context of universal algebra, perhaps I'm wrong. -Examples being Jordan-Holder, Lasker-Noether, Hilbert Nullstellensatz (maybe not?) and many more, where one can talk about sobobjects, and congruences instead of "subgroups" and "ideals". -Sometimes (obviously) those theorems fail in the universal context, but some of the results stand, and even more so: We might be able to measure the failure of those theorems in the universal context. -I hardly see those kind of deep results in introductions to universal algebra even though they can give a strong "justification" for the study of the field. -Therefore I have a few questions: - -Is there a specific reason not to generalize those results? -Is there a reason such results are not thought in Universal Algebra introductions (at least all the ones I found)? -Is these sort of generalizations have open research problems that cause difficulties (relates to the first question)? - -REPLY [3 votes]: Many of the results you mention form part of the categorical approach to (parts of) Universal Algebra. There is a lot of active research in this area, especially in the study of semi-abelian categories, Malt'sev categories, etc. Although this is not main stream Universal algebra and some universal algebraists may not feel it deserves mention as part of that subject, it may help to look at the book by Bourn and Borceux mentioned in the n-Lab article on semi-abelian categories.<|endoftext|> -TITLE: Rim hook decomposition and volume of moduli spaces -QUESTION [24 upvotes]: I did some computer experiments, counting the number of rim-hook decompositions (aka border-strip decompositions) of rectangles of shape $2n \times n$, where each strip has size $n$. -Here are 12 of the 1379 such decompositions for $n=4$: - -Let $a_n$ be the number of such decompositions. -I conjecture that $a_n$ is given by A115047 in OEIS, -that is, $a(0)=1$ and -$$ -a(n) = \sum_{i=1}^n - \binom{n - 1}{i - 1} \binom{n + 3}{i + 1} a(i - 1) a(n - i) \frac{ i(n - i + 1)}{2 (n + 2)}, -$$ -or $1, 5, 61, 1379, 49946, 2648967,...$. -Edit(2018-02-22): My student have checked that the sequences agree up to $n=12$, which is quite compelling. -The interesting part is that there is no such reference in OEIS. This entry regards Weil-Petersson volumes of muduli spaces of an $n$-punctured Riemann sphere, which is quite far from my field. -Q: Can we prove the recursion above? -I think this (conjectured) connection is interesting, giving a combinatorial interpretation of moduli space volumes. -Perhaps this can be extended to other genus? -EDIT: Using the strategy in the formula given in the answer below, -this conjecture is now proved in this preprint. -Edit II Paper is now published in the Journal of Integer sequences. -The answer to the original question is therefore, YES. - -REPLY [7 votes]: Abacus is quite useful to count objects related to hooks. For partition $((2n)^n)$, the $n$-abacus is $\{2n, \dots, 3n-1\}$ on $n$ runners. Thus, the number of labelled rim hook decompositions (labelled by the order of removal) equals to the number of permutations of $x_1, \dots, x_n, y_1, \dots, y_n$ such that $x_i$ appears before $y_i$ for every $i$. -In order to count unlabelled rim hook decompositions, we need to find out when the order of two consecutive rim hook removal can be swapped. Then, we can add some constraint that allows one type of removal, but forbids the other type of removal. All labelled rim hook decompositions with the new constraint are in bijection with unlabelled rim hook decompositions. -If we translate the new constraint we get to permutations of $x_i$'s and $y_i$'s, that means: the permutation does not have consecutive $x_i$ and $y_j$ such that $i \gt j$. -It is then straightforward to use inclusion-exclusion to count the number of permutations satisfying both requirements. I get a complicated formula for $a_n$, and the formula matches A115047 for $n$ up to $60$. I believe that it should not be too hard to prove that the formula for $a_n$ satisfies the recursion. - -Edit (Mar. 3, 2018) -The $n$-abacus of a partition $\lambda = (\lambda_1, \dots, \lambda_m)$ consists of $n$ runners and $m$ beeds located at $\{\lambda_i + m - i\}$, and the map from beeds to runners is given by $\mathbb{N} \to \mathbb{N} / n \mathbb{N}$. Removing an $n$-hook corresponds to the move of a beed along its runner to a smaller adjacent unoccupied location. A labelled rim hook decomposition is just a sequence of such moves so that no further moves can be made. -For the partition $((2n)^n)$, let $x_i$ be the move of the beed at $2 n + i - 1$ to $n + i - 1$, and let $y_i$ be the move of the beed at $n + i - 1$ to $i - 1$. Two hook removal at $b$ and $b'$ can be swapped if and only if $|b - b'| \gt n$. For this particular partition, that means $x_i y_j$ or $y_j x_i$ such that $i \gt j$. Therefore, we can reformulate the problem of finding $a_n$ to a problem of counting permutations of $x_i$'s and $y_i$'s. For general partitions, the same method works, and we just need to consider possibly larger "alphabet" and longer "forbidden words". -Below is one method of counting such permutations. There must be a much better way to count them. -Suppose that we know some consecutive occurrences of $x_i y_j$'s. Consider the bipartite graph with vertices $x_i$'s and $y_i$'s and edges $x_i y_i$ for all $i$ and $x_i y_j$ for all known consecutive $x_i y_j$ with $i \gt j$. This bipartite graph is a disjoint union of paths, otherwise it would not be legitimate. Let $p$ be the partition such that the parts of $2p$ are the sizes of the connected components of the bipartite graph mentioned above. Therefore, by inclusion-exclusion, we have the formula -$$ a_n = \sum_{p \vdash n} (-1)^{|p - 1|} \frac{1}{|m|!} {|m| \choose m} {|p| \choose p} {|p + 1| \choose p + 1},$$ -where $p$ runs over partitions of $n$, $m := (m_1, \dots, m_n)$ with $m_i$ being the multiplicity of $i$ in $p$, $p + c$ is the addition of $c$ to each part of $p$, $|p|$ is the sum of parts in $p$ and ${|p| \choose p}$ is the multinomial coefficient.<|endoftext|> -TITLE: Gaussian integrals over the space of symmetric matrices -QUESTION [9 upvotes]: Let $S\in\mathcal S_N$ be a $N\times N$ symmetric matrix over the reals, and introduce the (normalised) gaussian measure -$$ -\mathrm d\mu(S):=2^{-\frac 12N}\pi^{-\frac14N(N+1)}\exp\left[-\frac12\operatorname{tr}(S^2)\right]\mathrm dS -$$ -where -$$ -\mathrm dS:=\prod_{i\le j\le N}\mathrm dS_{ij} -$$ -I am interested in the integral -$$ -p_n(N):=\int_{\mathcal S_N}\operatorname{tr}(S^{2n})\mathrm d\mu(S) -$$ -which is in fact a polynomial in $N$ of degree $n+1$. It can be shown that the coefficients of these polynomials are given by a certain sum over all possible gluings of the $2n$-star (cf. Ref.1, exercise 3.2.12). -These polynomials can be computed explicitly (for example, with the aid of the Weyl integration formula, cf. this MO post, although I found it much more efficient to compute the integrals explicitly by means of Wick's theorem). The first few read -\begin{equation} -\begin{aligned} -p_0(n) &= n\\ -p_1(n) &= \frac12 n (1 + n)\\ -p_2(n) &= \frac14 n (5 + 5 n + 2 n^2)\\ -p_3(n) &= \frac18 n (41 + 52 n + 22 n^2 + 5 n^3)\\ -p_4(n) &= \frac{1}{16} n (509 + 690 n + 374 n^2 + 93 n^3 + 14 n^4)\\ -p_5(n) &= - \frac{1}{32} n (8229 + 12143 n + 7150 n^2 + 2290 n^3 + 386 n^4 + 42 n^5)\\ -p_6(n) &= - \frac{1}{64} n (166377 + 258479 n + 167148 n^2 + 58760 n^3 + 12798 n^4 + - 1586 n^5 + 132 n^6) -\end{aligned} -\end{equation} -My question: is there any known formula to compute these polynomials efficiently (say, in the form of an explicit recurrence relation, or a closed-form generating function, etc.)? -Note that $p_n(1)=(2n-1)!!$, which is not too difficult to prove. -If we integrate over hermitian matrices instead of symmetric matrices, the answer to the question above is affirmative (the recurrence relation and the generating functional can be found in Ref.1 §1). In the symmetric case, the best I could find was Ref.2, which is not nearly as explicit as I would like (if the recurrence relation or the explicit generating function for $p_n(N)$ is somewhere there, I couldn't find it myself). -As per Wick's theorem, I am lead to believe it should be useful to express the polynomials $p_n(N)$ using as a basis the bionomial coefficients instead of the powers $\{1,N,N^2,\dots\}$: -$$ -p_n(N):=\sum_{j=1}^{n+1}\frac{1}{2^n}\alpha_{n,j}\binom{N}{j} -$$ -where the coefficients can be arranged in a Pascal-like triangle: -\begin{equation} -\begin{aligned} -&\hspace{127pt}1\\ -&\hspace{100pt}1\hspace{50pt} 1\\ -&\hspace{80pt}6\hspace{40pt} 11\hspace{40pt} 6\\ -&\hspace{60pt}60\quad\qquad 153\quad\qquad 156\quad\qquad 60\\ -&\qquad\qquad840\qquad 2673\qquad 3846\qquad 2796\qquad 840\\ -&\qquad15120\quad 56715\quad 102960\quad 106560\quad 60960\quad 15120\\ -&332640\ 1420695\ 3066390\ 4032360\ 3304080\ 1568880\ 332640\\ -\end{aligned} -\end{equation} -I couldn't find a recurrence relation for the $\alpha_{n,j}$ coefficients either, so I'm not sure this representation is actually more useful than the previous one, but I wanted to include it as well just in case. Note that the triangle is almost symmetric, so that may mean something. -Note: OEIS was unable to recognise any of these sequences/polynomials. Could it be useful to submit them? - -References. - -Sergei K. Lando, Alexander K. Zvonkin - Graphs on Surfaces and Their Applications (Springer, 2004). -I.P. Goulden, D.M. Jackson, Maps in locally orientable surfaces, and integrals over real symmetric matrices, Canadian J. Math. 49, 1997, 865-882. - -REPLY [11 votes]: A recursion formula for the moments of the Gaussian orthogonal ensemble, M. Ledoux (2009). -The desired recursion formula for the moment $b_p^N\equiv E\,[\,{\rm tr}\,(S_N^{2p})]$ is - -I notice a difference in normalization, you'll want to divide $b_p^N$ by $2^p$.<|endoftext|> -TITLE: An orientable non-spin${}^c$ manifold with a spin${}^c$ covering space -QUESTION [17 upvotes]: Is there a closed, smooth, orientable manifold which is not spin${}^c$ but has a finite cover which is spin${}^c$? - -Such examples exist when spin${}^c$ is replaced by spin: an Enriques surface is not spin but it is double covered by a K3 surface which is spin. -Every orientable manifold of dimension at most four is spin${}^c$, so any such examples must have dimension at least five. The only non-spin${}^c$ manifold I know of is the Wu manifold $SU(3)/SO(3)$. Of course taking products or connected sums with other manifolds provides more examples. - -REPLY [7 votes]: The Dold manifold $P(1,2)$ is a simple example. It is defined by quotienting the ${\rm Spin}^c$ 5-manifold $S^1\times \mathbb{CP}^2$ via the (fixed points free and orientation preserving) involution $(x,z) \mapsto (-x,\bar z)$. In general this construction works for every $P(m,n)=S^m \times \mathbb{CP}^n$ with $m$ odd and $n$ even. -See this paper of Plymen from 1986, Section 4.7<|endoftext|> -TITLE: A Minkowski-like inequality -QUESTION [5 upvotes]: Assume that $X$ and $Y$ are two arbitrary non-negative random variables. Is the following inequality true for $1\leq\alpha\leq 2$? -\begin{align} -\left(\mathbb{E}\left[X^\alpha\right]-\mathbb{E}\left[X\right]^\alpha\right)^{\frac{1}{\alpha}}+\left(\mathbb{E}\left[Y^\alpha\right]-\mathbb{E}\left[Y\right]^\alpha\right)^{\frac{1}{\alpha}}\geq \left(\mathbb{E}\left[(X+Y)^\alpha\right]-\left(\mathbb{E}\left[X+Y\right]\right)^\alpha\right)^{\frac{1}{\alpha}} -\end{align} - -REPLY [4 votes]: The answer to this question is yes. -Indeed, let $a:=\alpha$, and suppose $1\le a\le2$. Without loss of generality (wlog), $12$. Indeed, then, for instance, $F(\mu)=\frac12-(\frac12)^{a-1}>0$ if $Y=1$ and $P(X=0)=P(X=1)=1/2$, which implies $h'(0)>0$, which implies $h(t)>0$ for small enough $t>0$.<|endoftext|> -TITLE: An axiom for collecting proper classes -QUESTION [5 upvotes]: I'm currently working on some universal algebra using proper classes (in MK class theory), and I repeatedly run into situations where I want to collect together some proper classes as the members of a new algebraic structure. -For example, the construction given here yields a bunch of equivalence classes $\equiv/(m_0,m_1)$ for $(m_0,m_1)\in\mathbb{M}^2$, and if $\mathbb{M}$ is a proper class then each of these equivalence classes is a proper class. Despite this we would like to treat $\mathbb{M}^2/\equiv$ as a group, so to get around the problem we can technically collect together one representative from each equivalence class and be good to go with the class of representatives serving as the new group. -This feels clunky to me from a logical standpoint every time I have to do it, and it makes me wonder if there may be situations where we want to collect up some proper classes which do not admit representatives so simply. Consequently, I would like to add an axiom to MK which allows me to dictate one of two things: - -I can collect together proper classes into 'hyper classes' under certain circumstances. This seems like the direct and obvious route at first; it is suggested by Andreas Blass under similar circumstances, however this would require an extension of the formal language of MK to include a relation symbol between proper classes and hyper classes that is not $\in$, and this bothers me. (this may bother me less as I learn more logic/model theory) -Classes which are definable as equivalence classes using the first order language of sets and a relation $\mathcal{R}$ on a class $\mathbb{B}$ (possibly a proper class) are sets. - -The second one is more along the lines of what I want, but this form of it is inconsistent. For example, if we let $\mathfrak{G}(O_n)$ be the Grothendieck ring of the ordinals as defined here for $\omega_1$ (just replace $\omega_1$ with $O_n$ in the construction), we technically need to take a quotient of $\mathfrak{G}(O_n)$ by the equivalence relation $$\equiv=\{(\alpha-\beta,\gamma-\zeta):\alpha+\zeta=\gamma+\beta\}$$ to remove extra 'difference representations' of the same element in $\mathfrak{G}(O_n)$. We now have that $\equiv/(\alpha-\beta)$ is a set for any choice of $\alpha,\beta$ under this new axiom, but $$\equiv/(0-0)=\{\alpha-\alpha:\alpha\in O_n\}$$ is trivially in bijection with $O_n$, a proper class. - -In summary, my question is: - -What is a concise axiom that (when added to MK class theory) would allow us to collect together proper classes satisfying certain algebraic formulae as though they were sets? - -If an extension of the language of sets with an additional relation for hyper classes seems the most concise option, I would be open to an argument in favor of that as well. -It also seems from the discussion here that I may want to simply consider collecting these objects together in the meta-theory, but I do not know what pitfalls await me if I try this apparently bold route. - -REPLY [6 votes]: Unless you are committed to beginning with MK class theory, which is not conservative over ZFC, I suspect you can get all you want working in Ackermann's set theory as developed by W. Reinhardt in Ackermann's set theory equals ZF, Ann of Math Log 2, pp. 189-249. -There is a nice overview of the theory by Azriel Levy in The Role of Classes in Set Theory, which appears both as chapter of Foundations of Set Theory (Second Revised Edition), A. Fraenkel, Y. Bar-Hillel and A. Levy, North-Holland Publishing Co. (1973) and as a chapter of Sets and Classes (G.H. Müller ed), North-Holland Publishing Co. (1976). -In Reinhardt's version of Ackermann's theory, which is conservative over ZFC (as well as over NBG with Global Choice), given a class $A$ having the power of $On$ one can form $P(A), PP(A), PPP(A), ...$, where $P(A)$ is the power class of $A$. -For some reason Ackermann's theory has not received much attention. Perhaps Joel or someone else knowledgable about such matters can explain why this has been the case. -Edit. I wrote this before I saw Joel's comment.<|endoftext|> -TITLE: Basis problem of modular forms arising from quaternion algebra -QUESTION [8 upvotes]: This question is a follow-up of an old question posted on MathOverflow. -Motivation: The exact equations of modular curves $X_0^{+}(p)=X_0(p)/w_p$($p>13$ is a prime number,$w_p$ is the Fricke involution) are intriguing objects. One can construct exact equations from modular forms of weight 2 on $\Gamma_0(p)$ from a quaternion algebra $A(p)$ over $\mathbb{Q}$ ramified at $p$ and $\infty$. Let $O$ be a maximal order of $A(p)$ and $I$ be a left $O$-ideal. The theta function determined by $I$ $$\theta_I(\tau)=\sum_{x\in I}e^{2\pi i\tau\frac{N(x)}{N(I)}}$$ is always a modular form of weight 2 on $\Gamma_0(p)$ with Fricke eigenvalue $-1$(see A. Pizer's paper). -Experiment: According to the Proposition 2.17 of Pizer's paper, the number of different theta functions determined by left ideals of $A(p)$ is bounded by the type number $T(p)$, which can be computed explicitly in another paper of Pizer (p. 94). A numerical computation with MAGMA seems to suggest that the number of distinct theta functions associated to the left $O$-ideals of $A(p)$ always reaches the upper bound, and all these theta functions are linear independent. -Question: Is there any example that the linear independence of theta functions fails for a certain prime number $p>13$? If not, is there a reference for the proof of this fact? -Update: The conjecture is likely to be false for $p=227$. - -REPLY [8 votes]: Hecke conjectured that $\theta_I$ form a basis for the space $S_2(p)$, but this was found to fail for $p=37$ by Eichler. In fact, Gross realized that whenever you get vanishing central $L$-values you get a linear relation among theta series. This happens in $S_2(37)$ since there is an elliptic curve with root number $-1$ of conductor 37. -However, Eichler showed you can find a basis for the space of modular forms among the theta series attached to lattices $J^{-1}I$ as $I, J$ vary over left $O$-ideals (so you range over left $O'$-ideals where $O'$ ranges over all maximal orders). The precise linear relations are mysterious however. See for instance the introduction to Hijikata, Pizer and Shemanske's Memoirs article on the basis problem.<|endoftext|> -TITLE: Cofiber of the inclusion of an $E_0$-algebra $M$ into the free $E_k$-algebra generated by it -QUESTION [8 upvotes]: Let $\mathcal{C}$ be the $E_k$-monoidal $\infty$-category of left modules over a fixed connective $E_{k+1}$-ring spectrum $A$. Suppose that $M$ is an object of $\mathcal{C}$ which is an $E_0$-algebra, in other words $M$ admits a unit map $A\to M$. There is an adjunction $U\colon Alg_{E_k}(\mathcal{C})\leftrightarrows Alg_{E_0}(\mathcal{C})\colon F$ by which we can construct the free $E_k$-algebra on $M$. Note that the $E_k$ algebra generated by the $E_0$-algebra $M$, i.e. with fixed unit, will not generally be equivalent to the $E_k$-algebra generated by $M$ thought of as object of $\mathcal{C}$ (an algebra for the trivial operad). -The unit of the adjunction gives a map of $E_0$-algebras $u\colon M\to F(M)$. I do not have a good description of $F(M)$ in general, but I am interested in showing that if the cofiber of the unit map $A\to M$ is $d$-connected (so $\pi_\ast(M/A)=0$ for $\ast0$, and produce $N/M$ as $colim(M^{\otimes_An}/M)$ for $n>0$. And here I'm being really shady by just writing that as some kind of quotient, because there are a lot of maps from $M$ to $M^{\otimes_An}$ and we're equalizing all of them. Anyway, while $N/M$ is probably quite complicated, it seems like we can say that, at the very least, it is the colimit of a diagram of $2d$-connected spaces, since $M$ is $d$-connected and every object of that diagram is of the form $M^{\otimes_A n}$ for $n>1$, with possibly a bunch of cells attached when we mod out by all the maps from $M$. Thus at least up to degree $2d-1$, $M$ and $F(M)$ are equivalent. - -REPLY [4 votes]: Let me try to turn my comment into an answer. Again, the intuition is to filter $\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$ by arity, and the result (at least in slightly less generality) is certainly classical, but I'll give a proof in the language you use above. I will also make the following assumptions (which hopefully you are making too): - -The $\mathbb{E}_{k+1}$-ring $A$ is connective. -Let $\overline{M}$ denote the cofiber of the unit $A \to M$. Then $\overline{M}$ is $d$-connective (i.e. $(d-1)$-connected). - - -Proof: I claim that there is a filtration $\{F_k\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M\}$ of the induced algebra such that - -The associated graded is $\bigoplus_{n\ge 0} \mathrm{Sym}^n_{\mathbb{E}_k}(\overline{M})$ -$F_1=M$ and the map $F_1=M \to \mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$ is the standard one. - -I'll build that in a minute but first let's check that it proves the result. It suffices to check that $\mathrm{Sym}^n_{\mathbb{E}_k}(\overline{M})$ is at least $2d$-connective as soon as $n\ge 2$. But, by definition, this object is a colimit over a diagram whose objects all look like $\overline{M}^{\otimes_An}$. Since the $t$-structure is compatible with the tensor product (since $A$ is connective), these are $nd$-connective, and the subcategory of $nd$-connective objects in closed under colimits (since connectivity is checked on underlying spectra and it's true there). So the result is proved assuming we have the filtration. Now let's build it. -First recall the construction of $\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$. We're meant to consider all the ways to build tensor powers like $M^{\otimes_An}$ (one for each rectilinear embedding $\coprod_n I^k \hookrightarrow I$) and then glue things together via the various maps $M^{\otimes_An} \to M^{\otimes_Am}$ induced by the unit when $n\le m$. Explicitly, let $D$ denote the $\infty$-category whose objects are rectilinear embeddings $\coprod_n I^k \hookrightarrow I$ ($n$ arbitrary) and whose morphisms are embeddings $\coprod_n I^k \hookrightarrow \coprod_m I^k$ compatible with the structure map up to isotopy and which are injective on path components. (A rigorous definition is the fiber product in HA.3.1.3.1 in the case $\mathcal{C} = \mathsf{LMod}_A$, $\mathcal{A}^{\otimes} = \mathbb{E}_0^{\otimes}$, and $\mathcal{B}^{\otimes} = \mathbb{E}_k^{\otimes}$). Then $M$ extends to a diagram $M^{(-)}: D \to \mathsf{LMod}_A$ and the module underlying $\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$ is the colimit of this diagram. -Now, $D$ admits an evident filtration by $n$, $D_{\le n} \subset D_{\le n+1}$. Restricting and taking colimits defines a filtration $\{F_k\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M\}$ and I want to show it has the two properties above. The second should be straightforward from the definition, so I'll just do the first. -Notice that there is a natural map $F_n/F_{n-1} \to \mathrm{Sym}^n_{\mathbb{E}_k}(\overline{M})$ coming from the natural maps $M^{\otimes_An} \to \overline{M}^{\otimes_An}$. We want this to be an equivalence. We'll prove this by induction on $n$. When $n=1$ the result is true by inspection: we're looking at the map $F_0=A \to F_1=M$ and taking the cofiber to get $\overline{M}$. I think one can prove the general case by induction on $n$, but I would have to come back and write the details if you want them.<|endoftext|> -TITLE: Divergence of Green function of random walks at spectral radius -QUESTION [5 upvotes]: Let $P=(p(x,y))_{x, y\in N}$ be the transition matrix over countable states $N$. -Consider the generating Green function $G(x, y|t)=\sum_{0}^{\infty} p^n(x, y) t^n$, where $p^n(x,y)$ is the $(x,y)$-entry of the matrix $P^n$. -The spectral radius is given by $\rho(P)=\limsup_{n\to\infty} p^n(x,y)^{1/n}$. It is clear that $G(x,y|t)$ has convergence radius $1/\rho(P)$. -I'm now interested in when $G(x,y|t)$ diverges at $t=1/\rho(P)$. In particular, I'm considering that $P$ induces a random walk on a non-amenable group $N$. Does there any results related to this divergence in some interesting class of groups? -I'm newbie to random walk on graphs. Any comments would be extremely welcome! - -REPLY [5 votes]: Concerning random walk on non-amenable groups, the Green function converges at the spectral radius: this is a result of Guivarc'h, quoted -in Wolgang Woess's book (Random walks on infinite graphs and groups), chapter IIB (in particular Theorem 7.8). -For a bit more explicit statement, one can laso look at the introduction of this paper of Gouezel and Lalley: -https://arxiv.org/pdf/1107.5591.pdf<|endoftext|> -TITLE: Torsion in Deligne cohomology -QUESTION [9 upvotes]: Let $X$ be a smooth projective complex analytic space, $i,p\ge 0$ integers, $\mathbf{Z}(p)_{\mathcal{D}}$ the Deligne complex of $X$, $H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ its hypercohomology. -What properties does the subgroup of torsion elements of $H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ have? - -For instance, is it finite? Does it contain divisible elements? - -Since $H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ is an extension of a finitely generated $\mathbf{Z}$-module by a quotient of a graded in the Hodge filtration on de Rham cohomology of $X$, one expects the answer depends on $i,p$. -More precisely, the question might as well be asked about the quotient group -$$J^{i,p}(X/\mathbf{C}) := \frac{H^i_{\rm dR}(X/\mathbf{C})}{F^pH^i_{\rm dR}(X/\mathbf{C}) + H^i(X,\mathbf{Z}(p))}$$ - -What if one restricts attention to the set $J^{i,p}(X/k)$ of those classes in $J^{i,p}(X/\mathbf{C})$ that come from algebraic cycles on the algebraization of $X$, and are defined on a fixed subfield $k\subset\mathbf{C}$? - -More precisely, if $\mathcal{X}$ is a smooth projective algebraic $k$-variety such that $(\mathcal{X}\otimes_k\mathbf{C})^{\rm an}\simeq X$, and $c : H^i_{\mathcal{M}}(\mathcal{X},\mathbf{Z}(p))\to H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ is the cycle map, define $J^{i,p}(X/k)$ to be -$$J^{i,p}(X/k) := J^{i,p}(X/\mathbf{C})\times_{H^i_{\mathcal{D}}(X,\mathbf{Z}(p))}H^i_{\mathcal{M}}(\mathcal{X},\mathbf{Z}(p))$$ -What can be said about the torsion subgroup of $J^{i,p}(X/k)$? -If $k$ is algebraically closed, do we have -$$J^{i,p}(X/k)_{\rm tor} = J^{i,p}(X/\mathbf{C})_{\rm tor}\ ?$$ -Example: the case $i=2,p=1.$ -As suggested in the comment, one can think about the case $i = 2$, $j=1$ first. Here $\mathbf{Z}(1)_{\mathcal{D}} \simeq\mathbf{G}_{\rm m}[-1]$ and $H^2_{\mathcal{D}}(X,\mathbf{Z}(1)) = H^1(X,\mathbf{G}_{\rm m}) = \text{Pic}(X)$. In this case, we have an exact sequence: -$$H^1(X,\mathbf{G}_{\rm a}) \to \text{Pic}(X)\xrightarrow{c_1} H^2(X,\mathbf{Z}(1))$$ -that identifies the extension -$$0\to J^{i,p}(X/\mathbf{C})\to H^i_{\mathcal{D}}(X,\mathbf{Z}(p))\to\text{Hdg}^{i,p}(X/\mathbf{C})\to 0$$ -with -$$0\to \text{Pic}^0(X)\to\text{Pic}(X)\to\text{NS}(X)\to 0$$ -whence $J^{2,1}(X/\mathbf{C}) = \text{Pic}^0(X)$. By GAGA we have $\text{Pic}^0(X)\simeq\text{Pic}^0(\mathcal{X}_{\mathbf{C}})$ and since $\mathbf{C}$ is separably closed $$\text{Pic}^0(\mathcal{X}_{\mathbf{C}}) = \underline{\text{Pic}}^0_{\mathcal{X}_{\mathbf{C}}/\mathbf{C}}(\mathbf{C}) = (\underline{\text{Pic}}^0_{\mathcal{X}/k}\times_k\mathbf{C})(\mathbf{C}).$$ -If $k$ is separably closed too, then indeed the torsion subgroup of $\text{Pic}^0(X)$ agrees with that of $\underline{\text{Pic}}^0_{\mathcal{X}/k}(k)$, since the kernel of multiplication by $n$ on $\underline{\text{Pic}}_{\mathcal{X}/k}^0$ is an étale group scheme for every $n$ (since $k$ of characteristic zero). -Since $\underline{\text{Pic}}^0_{\mathcal{X}/k}(k) = J^{2,1}(X/k)$, we indeed get -$$J^{2,1}(X/\mathbf{C})_{\rm tor} = J^{2,1}(X/k)_{\rm tor}.$$ - -REPLY [4 votes]: The following is not an answer, since the last comment to your question already answers most of it. -It is, rather, a "better question". -What you really want to ask is: let $f: X\to\text{Spec}(k)$ be a proper, geometrically connected, geometrically reduced scheme over a perfect field. -For integers $p,q$, $p\ge 0$, does there exist a $k$-group scheme $P^{p,q}_{X/k}$?, such that: - -$P^{p,q}_{X/k}$ is locally of finite type over $k$ and quasi-separated. -The component of the identity $(P^{p,q}_{X/k})^0$ is connected, finite type and separated. -Formation of $P^{p,q}_{X/k}$ commutes with separable field extensions on $k$. -If $X$ is smooth over $k$, then $(P^{p,q}_{X/k})^0$ is smooth. -If $X$ is smooth and projective over $k$, then $\text{NS}^{p,q}(X) := P^{p,q}_{X/k}/(P^{p,q}_{X/k})^0$ is a constant étale $k$-group scheme whose value group is finitely generated. -If $X$ is smooth and projective over $k$, and $k$ is a subfield of $\mathbf{C}$, then -$$(P^{p,q}_{X/k}\times_k\mathbf{C})(\mathbf{C}) = H^p_{\mathcal{D}}(X_{\mathbf{C}}^{\rm an},\mathbf{Z}(q)),\ \ ((P^{p,q}_{X/k})^0\times_k\mathbf{C})(\mathbf{C}) = J^{p,q}(X^{\rm an}_{\mathbf{C}}/\mathbf{C}),\ \ (\text{NS}^{p,q}(X)\times_k\mathbf{C})(\mathbf{C}) = \text{Hdg}^{p,q}(X_{\mathbf{C}}^{\rm an}/\mathbf{C}).$$ - -In other words, you're actually looking for "higher Picard functors", representable under the assumptions stated at the beginning of this post, such that their connected component of the identity is an extension of a connected smooth affine $k$-group scheme by an abelian $k$-variety, and such that an analog of the theorem of the base holds true. -For $q=1$, this problem is related to the question whether $R^{p-1}f_{\rm fppf, *}\mathbf{G}_m$ is representable, which, for $p=2$, we now to be the case by work of Artin. Even the case $q=1$, for arbitrary $p\ge 2$, is extremely hard. A seemingly easier question is whether $R^{p-1}f_{\rm fppf, *}\mu_n$ is representable, and even this question (posed by Artin when $n$ is prime) is very hard too, and answered positively only in very special cases (mostly by work of Lieblich). -If the above dream were true, then your question becomes indeed very interesting, if asked about $(P^{p,q}_{X/k})^0(k^{\rm sep})^{\text{Gal}(k^{\rm sep}/k)}$ instead of $J^{p,q}(X_{\mathbf{C}}/\mathbf{C})$, and when $k$ is a finitely generated field. -The functors $P^{p,q}_{X/k}$, which I'm not going to define here, occurred at some point in my work, where I was able to show their representability in some special cases I was interested in (by direct construction, ie. not via Artin's axioms). The general question, open for now, seems to be very hard (if true at all).<|endoftext|> -TITLE: Is a Mahlo cardinal also a stationary limit of m-inaccessible cardinals? -QUESTION [5 upvotes]: I have been trying for some time to get a grip on how large Mahlo cardinals are, but am finding the definition rather unsuggestive. -Let $\kappa$ be the smallest Mahlo cardinal. By definition, the set of inaccessible cardinals smaller than $\kappa$ is stationary in $\kappa$. Using this, I can prove (I think) that the smallest 1-inaccessible cardinal is strictly smaller than $\kappa$. -But can I do better than this ? -Is the set of 1-inaccessible cardinals smaller than $\kappa$ also stationary ? -If so, can I somehow iterate the argument to prove that for every ordinal m < $\kappa$, the set of m-inaccessible cardinals < $\kappa$ is stationary in $\kappa$ ? -The last property would imply that $\kappa$ is the $\kappa$-th m-inaccessible cardinal for every m < $\kappa$ and therefore, truly gigantic, so I am very interested to know if this is true. -Apologies in advance if this is too basic for MathOverflow. I haven't seen an explicit proof or disproof of this property anywhere. - -REPLY [8 votes]: Yes. Erin Carmody gives a good account of this in her dissertation. - -Erin Carmody, Force to change large cardinal strength, arXiv:1506.03432, 2015. - -If you see the material leading up to her theorem 11, she first develops the degrees of of inaccessibility beyond $\alpha$-inaccessible and into things such as hyperinaccessible, richly inaccessible, utterly inaccessible and so on, ultimately defining a formal language of terms in $\Omega$, such as $(\Omega^3\cdot 5+\Omega)$-inaccessible. And then she proves that if $\kappa$ is Mahlo, then it is $t$-inaccessible for any term $t$ in her language and indeed a stationary limit of such $t$-inaccessible cardinals. -This hierarchy goes way beyond merely $\kappa$-inaccessible, and so it seems to be the answer to your question. -One basic observation that will get you a long way is the following. -Theorem. If $\kappa$ is Mahlo, then for any $A\subset\kappa$ there is a stationary class $S$ of inaccessible cardinals $\gamma$ for which $\langle V_\gamma,\in,A\cap\gamma\rangle\prec\langle V_\kappa,\in,A\rangle$. -This is just because the class of ordinals like that is club, and so by Mahloness it meets the regular cardinals. Now, as $A$ gets thinner, these $\gamma$ show that there are increasing degrees of hyperinaccessibility realized below the Mahlo cardinal $\kappa$. Any expressible property of $\kappa$ in $V_\kappa$ will hold on a stationary set of $\gamma$ below $\kappa$. -The basic idea is due to Mahlo, I believe. In Erin's presentation, she shows that the $t$-inaccessible cardinals form a set in a certain normal filter, which therefore form a stationary set.<|endoftext|> -TITLE: Counting smooth numbers in short intervals -QUESTION [6 upvotes]: I am reading a few papers about counting smooth numbers in the interval $[x, x+\sqrt{x}]$, including the work of Harman, and Matomaki. -Both authors mentioned that the Dirichlet polynomial techniques break down when the length of the interval is about $\sqrt{x}$. Why is that the case? In particular, does this defect come from the method itself or the same barrier applies to all natural methods (e.g. sieve methods) to attack this question? - -REPLY [2 votes]: Suppose that $F(s) = \sum_{n \sim x} a_n n^{-s}$ where $a_n = 1$ if $n$ is $x^{\varepsilon}$ smooth and $a_n = 0$ otherwise. We want to understand the distribution of the $a_n$'s in a short interval of length $h$. Then by the usual Perron formula, -$$ -\sum_{x < n < x + h} a_n = \frac{1}{2\pi i} \int_{-T}^{T} F(1 + it) \cdot \frac{(x + h)^{1 + it} - x^{1 + it}}{1 + it} dt + O(x / T). -$$ -After a few simplifications we see that we roughly have to choose $T = x / h$ and that the main integral simplifies to something like -$$ -h \int_{-x/h}^{x/h} F(1 + it) x^{it} dt -$$ -At this point most methods take absolute values. However once we take absolute values the best bounds for $F$ that we can hope for is $|F(1 + it)| \ll 1/\sqrt{x}$. This corresponds to having perfect square-root cancellation at every point. In such a (most optimistic scenario), the bound that we get for the above is at best -$$ -h \times \frac{\sqrt{x}}{h} = \sqrt{x} -$$ -This means that as soon as we take absolute values in the Perron integral we forfeit the possibility to go into intervals shorter than $\sqrt{x}$. To go into shorter intervals one needs to somehow exploit cancellations in the Perron integral. At this stage one is rather advised to use other Fourier techniques (for instance exponential sums).<|endoftext|> -TITLE: What are the typical forcings to shoot a club through a stationary subset of $[\lambda]^\omega$ -QUESTION [9 upvotes]: Let $\lambda\geq \omega_2$ be a regular cardinal and $S\subset[\lambda]^\omega$ be a stationary set. I'm looking for a property of $S$, say "shootable", such that there exists a forcing extension preserving $\lambda, \omega_1$ as cardinals that shoots a club into $S$. I've encountered ad hoc examples, but I'd really hope if there are more explicit descriptions of a stationary set being "shootable" and given that a canonically defined reasonable forcing to shoot a club through it. I'm flexible with any cardinal arithmetic assumptions. - -REPLY [7 votes]: There is a notion of fat stationary subset of $P_\kappa(\lambda)$ and a natural forcing for shooting a club through it, that you can find in the thesis Fat subsets of $P_{\kappa}(\lambda)$ by Ivan Zaigralin. In particular look at chapter 2 of the thesis.<|endoftext|> -TITLE: The number of involutions in a permutation group -QUESTION [14 upvotes]: If $G$ is a group let $I(G)$ be the number of involutions (elements of order 2) in $G$. My question is then easily stated: does there exists a constant $C > 1$ such that for every $n \ge 1$ and every subgroup $G \subset S_n$ of the symmetric group $S_n$ we have -$$ -C^{-n} |G|^{\frac 1 2} \le I(G)+ 1 \le C^n |G|^{\frac 1 2}. -$$ -Some remarks : -*This is true for primitive groups, since those groups are either very small (of size at most $D^n$ for some absolute $D$) or they have to be the full symmetric or alternating groups (for which the number of involutions is precisely known). -*Using iterated wreath products decompositions and the previous remark it is possible to show something like -$$ -C^{-n\log\log(n)} |G|^{\frac 1 2} \le I(G)+ 1 \le C^{n\log\log(n)} |G|^{\frac 1 2}. -$$ -(This is actually good enough for the application I have in mind but I was wondering whether a sharper result to which I could refer existed). -*The lower bound would be sharp, since for example a 3-Sylow of $S_n$ is of size roughly $3^{n/2}$ as $n$ goes to infinity and contains no involutions. -*This question: Number of involutions in a finite group seems like it could be relevant but estimating the number of conjugacy classes in this setting seems to have to be rather involved. -Edited to add : -*I am interested only in the exponential aspect of the bound, but one might also ask for optimal $c < 1 < C$ such that -$$ -c^n |G|^{\frac 1 2} \le I(G)+ 1 \le C^n |G|^{\frac 1 2}. -$$ -(see Yves' comments below). -*With this notation the third point above (which I edited for clarity) gives an upper bound $< 1$ for $c$; looking at subgroups of exponent 2 in $S_n$ (for example $(\mathbb Z/2\mathbb Z)^{n/2}$) also gives a lower bound $> 1$ for $C$. An upper bound for $C$ is given by Geoff Robinson's answer, now we only lack a lower bound $>0$ for $c$. - -REPLY [6 votes]: I can answer in one direction. L.G. Kovacs and I proved ( around 1993) that any finite subgroup $G$ of $S_{n}$ has at most $5^{n-1}$ conjugacy classes. Later authors showed that it is possible to replace $5$ by a smaller constant, but the existence of such a constant seems to be enough for this question. It follows that any subgroup $G$ of $S_{n}$ has at most $\sqrt{5}^{n-1}|G|^{\frac{1}{2}} -1$ involutions, using properties of the Frobenius-Schur indicator, as in the question you refer to. -Later edit: It might be worth recasting the problem ( this is not so relevant for the direction already proved above): if the group $G$ above has no involutions, ( equivalently, has odd order), then $G$ is certainly solvable, so we have $|G| \leq 24^{\frac{n-1}{3}}$ by a result of J.D. Dixon ( again, the constant can be improved here, but its existence is enough for present purposes). This shows that $c = 24^{\frac{1-n}{6}}$ works above when $|G|$ is odd, and clearly $C = 1$ works here too ( for the other direction). -So we only now need to consider the case when $G$ has even order. It is well known that every $2$-subgroup of $S_{n}$ has order at most $2^{n-1}$, so that a Sylow $2$-subgroup of $G$ has order at most $2^{n-1}$ and, in particular, $G$ certainly has at most $2^{n-1}$ conjugacy classes of involutions. -If there is a positive constant $c$ as asked for in the later version of the question, then $G$ has an involution $t$ such that $[G:C_{G}(t)] > \left(\frac{c}{2} \right)^{n}|G|^{\frac{1}{2}}$ and hence $|C_{G}(t)| < \left(\frac{2}{c} \right)^{n}|G|^{\frac{1}{2}}.$ -It follows that the question is equivalent to asking whether it is true that there are positive (finite) constants $d,D$ such that whenever $G$ is an even order -subgroup of $S_{n},$ there are involutions $t,u \in G$ such that -$|C_{G}(t)| \leq d^{n}|G|^{\frac{1}{2}}$ and $|C_{G}(u)| \geq D^{n}|G|^{\frac{1}{2}}.$<|endoftext|> -TITLE: Reduction mod $n$ of symplectic group -QUESTION [9 upvotes]: Let $g,n$ be positive integers, is there a reference that $\mathrm{Sp}(2g,\mathbb{Z})\to\mathrm{Sp}(2g,\mathbb{Z}/n\mathbb{Z})$ is surjection? -The only reference I could find is lemma 5.16 in Deligne–Mumford, but to be honest, I don't quite understand the argument: Why is the reduction map surjective on unipotent elements? -Deligne and Mumford - The irreducibility of the space of curves of given genus - -REPLY [13 votes]: According to Demazure, when he was a graduate student he asked Serre about the surjectivity of $G(\mathbf{Z}) \to G(\mathbf{Z}/(n))$ for all $n>0$ with $G = {\rm{Sp}}_{2g}$ or maybe ${\rm{SL}}_N$ (I can't remember which he said it was). Serre told him that he should ask Grothendieck for the "right" answer. Grothendieck told him that this was the "wrong question", and to approach such matters from the "right" point of view was the original goal of the SGA3 seminar on reductive group schemes over rings, etc. (killing a fly with a sledgehammer?). -That being said, a slick perspective which transforms the problem into something more robust over fields goes as follows. Let $\mathbf{A}_f$ be the ring of finite adeles for $\mathbf{Q}$ (this is a $\mathbf{Q}$-algebra, namely $\mathbf{Q} \otimes_{\mathbf{Z}} \widehat{\mathbf{Z}}$ in which $\widehat{\mathbf{Z}}$ is a compact open subring), so $K = G(\widehat{\mathbf{Z}})$ is a compact open subgroup of $G(\mathbf{A}_f)$ which meets $G(\mathbf{Q})$ in $G(\mathbf{Z})$. Since $G$ is $\mathbf{Z}$-smooth, the reduction map $K = G(\widehat{\mathbf{Z}}) \to G(\mathbf{Z}/(n))$ is surjective for Hensel's Lemma reasons and has open kernel. Thus, the original surjectivity question has an affirmative answer if $G(\mathbf{Q})$ is dense in $G(\mathbf{A}_f)$ (as this implies $G(\mathbf{Z})$ is dense in $G(\widehat{\mathbf{Z}})=K$, so $G(\mathbf{Z})$ maps onto $K/K'$ for any open normal subgroup $K'$ of $K$). -In other words, we can now forget about $\mathbf{Z}$-structures and instead attack a question entirely in terms of the $\mathbf{Q}$-group $H = G_{\mathbf{Q}}$: when is $H(\mathbf{Q})$ dense in $H(\mathbf{A}_f)$? We'd like an affirmative answer for at least $H$ equal to ${\rm{SL}}_N$ and ${\rm{Sp}}_{2g}$. In the case of $H=\mathbf{G}_{\rm{a}}$, such density holds and is just a special case of the classical "strong approximation" property for adele rings of global fields. This makes contact with Paul Broussous' comment about unipotent groups when $H$ is a split simply connected $\mathbf{Q}$-group due to the way in which such groups have an "open cell" expressed in terms of certain ${\rm{SL}}_2$-subgroups (especially the direct product structure for a split maximal torus in terms of a basis of simple coroots for the cocharacter lattice, one way of characterizing the simply connected case). -For an actual argument along these lines (i.e., bootstrapping from strong approximation for adele rings using the structure theory of split simply connected groups over fields), see http://math.stanford.edu/~conrad/248BPage/handouts/strongapprox.pdf (and note how it makes essential use of working over a field rather than over a ring of integers). This has a vast generalization to arbitrary simply connected groups over global fields relative to a suitable finite non-empty set of places (such as the place $\{\infty\}$ for $\mathbf{Q}$ when the group is split) via the "strong approximation theorem" for such groups, but that lies far deeper than the split case discussed above.<|endoftext|> -TITLE: Is height preserved in a normalization? -QUESTION [12 upvotes]: Let $R$ be a domain and $\tilde R$ its integral closure in its fraction field: $R\subset \tilde R\subset Frac(R)$. -Is it true that a prime ideal $ \tilde {\mathfrak p} \subset \tilde R$ and its trace $\mathfrak p= \tilde {\mathfrak p}\cap R\subset R$ are related by the equality of heights $$ ht(\tilde {\mathfrak p})=ht(\mathfrak p)\quad (?)$$ -This is true for example if $R$ is finitely generated over a field, since then we have the relation $$\operatorname {dim }(R)=\operatorname {dim }(R/\mathfrak p)+ht(\mathfrak p)$$ and the similar relation $$\operatorname {dim} (\tilde {R})=\operatorname {dim }(\tilde R/\tilde {\mathfrak p})+ht(\tilde {\mathfrak p})$$ -Since dimension is conserved in integral ring extensions we have $$\operatorname {dim }(\tilde R)=\operatorname {dim }(R)\quad, \quad \operatorname {dim }(\tilde R/\tilde {\mathfrak p})=\operatorname {dim }(R/\mathfrak p)$$ from which the questioned equality $ ht(\tilde {\mathfrak p})=ht(\mathfrak p)$ follows. -But is the equality $(?)$ true in general, i.e. without the hypothesis of finite generation over a field? -[The motivation for my question comes in part from this answer and the comments it provoked] - -REPLY [13 votes]: The answer for a general commutative Noetherian domain is "no". The following is adapted from a paper I wrote with Jay Shapiro a few years back. -Consider Nagata's book Local Rings, Appendix, Example 2 (see also Stacks Project, tag 02JE). I'm changing notation and specializing a bit (by setting his parameter $m$ to $0$), but it's basically like this. He first constructs $S$, a normal Noetherian domain with exactly two maximal ideals ${\frak m}_1, {\frak m}_2$, such that height$({\frak m}_i)=i$ for $i=1,2$, and a field $k$ along with an injective ring map $k \hookrightarrow S$ such that the composite maps $k \rightarrow S/{\frak m}_1$, $k \rightarrow S/{\frak m}_2$ are isomorphisms. (I think this is already impossible for excellent $k$-algebras.) Then let $J := {\frak m}_1 \cap {\frak m}_2$ and $R := k+J$. Then $S$ is module-finite over $R$ because of how pullbacks of rings work, and it is elementary that their fraction fields coincide, so $S=\tilde R$, and by the Eakin-Nagata theorem $R$ is Noetherian. -But then ${\frak m}_1 \cap R = J$, which has height 2 because it is the unique maximal ideal of the two-dimensional local ring $R$, even though ${\frak m}_1$ has height 1. -On the other hand, if $R$ is universally catenary, then the answer is "yes": height is preserved when contracting primes from $\tilde R$ to $R$. This follows from two results of Ratliff. First, use [Notes on three integral dependence theorems, J. Algebra, 1980, Corollary 2.5], which says that all we need to check is that for any $f\in \widetilde{R[X]}$, height is preserved when contracting primes from $R[X,f]$ to $R[X]$. But since $R[X]$ is universally catenary, Ratliff's theorem on (given as Theorem 15.6 in Matsumura's Commutative Ring Theory) says that the Dimension Formula (the one that involves transcendence degrees) holds between $R[X]$ and $R[X,f]$ (since $R[X,f]$ is finitely generated as an $R$-algebra). Since both of the transcendence degrees involved are then 0, it follows that contraction of primes between these two rings preserves height.<|endoftext|> -TITLE: How to specify a compact topological 4-manifold with a finite amount of data -QUESTION [36 upvotes]: Finite data for 4-manifolds: I have a somewhat hazy memory of seeing a Ph.D. thesis around 2005 which showed that any compact topological 4-manifold $M$ can be specified by a finite amount of data. The very rough idea is to exploit the fact that $M \setminus pt$ ($M$ minus a point) can be triangulated and explicitly give such a triangulation for a large initial segment $X$ of $M \setminus pt$, and an even larger initial segment $Y$ of $M \setminus pt$, together with a certificate that $Y \setminus X$ contains a topologically flat embedded S^3 cutting off the end of $M \setminus pt$. This data allows you to build the closed topological manifold $M$ and argue that the result is unique. The important detail is how to give the required certificate. Over the years several people have asked me about this “finite data for 4-manifold” question. I would like to locate the reference, or failing that find time to write up a proof. - -REPLY [4 votes]: In my understanding, I guess that the following strategy could be be attempted. Let $M$ be a closed connected orientable 4-manifold (while nonorientable 4-manifolds are doubly covered by orientable ones). Surgerying $M$ along finitely many embedded tame circles produces a simply connected 4-manifold $M'$. Then, $M$ can be recovered from a certain simply connected manifold by surgerying finitely many pairwise disjoint locally flat 2-spheres with trivial normal bundle. Now, simply connected 4-manifolds are classified by Mike Freedman's celebrated theorem in terms of the intersection form and the Kirby-Siebenmann invariant (which amounts to finite data). It remains to understand how much data we need to determine the embedded spheres to be surgered in a simply connected 4-manifold. Of course, we need more than homotopy (= homology since $\pi_1 = 0$) classes.<|endoftext|> -TITLE: A definition of topology using monads (a.k.a. halos) -QUESTION [8 upvotes]: In nonstandard analysis, there is a way of studying topological spaces known as "monads" (more commonly known as halos, as it turns out). The monad of a point $x$ (written $\mu(x))$ is the set of all points that are "near" it. -In particular, in a topological space $(X,T)$, the monad of $x \in X$ is defined as $$\mu (x)=\bigcap\{{}^*O:O \in T, x \in O \}$$ (see "On Nonstandard Topology"). -or the intersection of the open sets that contain $x$. For example, the monad of $0$ (given the normal topology on $\mathbb R$) is the set of infinitesimals (since any open set that contains $0$ also contains every infinitesimal). -This provides an interesting way of defining various concepts in topology: - -A set $S$ is open iff $\mu(s) \subseteq {}^*S$ for all $s \in S$ (an ideal point near a point in an open set is in the open set). -A set $S$ is closed iff $\mu(s) \cap {}^*S \neq \emptyset$ implies $s \in S$ for all $s \in X$ (any point near an ideal point in a closed set is in that closed set). -$(X,T)$ is compact iff for all $z \in {}^*X$, there exists $x \in X$ such that $z \in \mu (x)$ (every ideal point is near a point in a compact set). -$(X,T)$ is hausdorff iff $\mu(x) \cap \mu (y)=\emptyset$ for all $x,y \in X, x \neq y$ (no ideal point is near two different points in a hausdorff space). -The function $f$ from $(X,T)$ to $(Y,U)$ is continuous iff $f(\mu (x)) \subseteq \bar \mu (f (x))$ for all $x \in X$, assuming that $\mu$ is $(X,T)$'s monad function, and $\bar \mu$ is $(Y,T)$'s monad function. -The function $f$ is a local homeomorphism iff $f(\mu (x)) = \bar \mu (f (x))$ for all $x \in X$ (and a homeomorphism if $f$ is bijective). - -This has me wondering, has any one defined the concept of topology in terms of monads? The monads of a space obviously uniquely determine it, since they can be used to recover it. So, we technically could say that a topology on $X$ is a $\mu$ such that the open sets corresponding to $\mu$ satisfy the topology axioms. If we did that though, we might as well just use the regular definition! That said, there is a probably a natural definition in terms of $\mu$ making no mention of open sets (and then open sets are later a defined concept, instead of given). - -REPLY [3 votes]: Two researchers in the 1980s have independently discovered the necessary axioms for defining a topology out of halo axioms. The relevant papers are: -Vakil, N. Monadic binary relations and the monad systems at near-standard points. The journal of Symbolic Logic, 52(3):689-697, Sep 1987. (link) -and -Tewfik, S. General topology. In Diener, F. and Diener, M., editors. Nonstandard Analysis in Practice, pp. 109-144. Springer-Verlag Berlin Heidelberg, Berlin, 1995. (link). -I accessed these resources via my university so you may not be able to see the material directly in the links above, unfortunately. -In these papers it is proved that in order for a standard topology to be uniquely determined, it is necessary and sufficient that the infinitesimal closeness relation of a space be monadic, locally reflexive, and locally transitive. I will elaborate on the meaning of these axioms a bit more. -Monadicity -This axiom requires that there must be a filter in $X \times X$ whose intersection monad is the infinitesimal closeness relation in *$X$. In other words, if every point $x$ in *$X$ has its halo $\mu(x) \subset$ *$X$, then the relation -$$W = \left \{ (x, y) \in \text{*} \left( X \times X \right) \mid y \in \mu(x) \right \} $$ -should be able to be written as -$$W=\bigcap_{F \in \mathcal{F}}{\text{*}F}$$ -for some filter $\mathcal{F}$ in $X \times X$. (In order to guarantee that all filters have a nonempty intersection monad, a model with sufficient saturation properties must be assumed.) -Tewfik chooses a slightly different approach and requires that the relation be expressed as a halic formula. Since I don't know much about IST I don't really understand his definition, but considering what he says in the book I believe it to be more or less equivalent to the one presented above. -Note that in order for the above axiom to make sense, the halo system must be defined for all points in *$X$, not only the standard ones in $X$. -Intuitively, this axiom says that if two points are infinitesimally close to each other then we should be able to approximate them with a collection of ever-decreasing sets, kind of like when we approximated with the collection $\left \{ x \in \text{*} \mathbb{R} \mid \lvert x-a \rvert < \frac{1}{n}\right \} \left(n \in \mathbb{N} \right)$ points infinitesimally close to $a \in \mathbb{R}$. -Local reflexivity -This axiom states that $x \in \mu(x)$ for all $x \in X$. Tewfik noted that this automatically guarantees reflexivity ($x \in \mu(x)$ for all $x \in \text{*}X$), so it is sufficient only to state the local proposition as an axiom. The intuition is obvious; we should be able to say of any point that it is infinitesimally close to itself. -Local transitivity -This axiom requires that for all $x \in X$, $y \in \mu(x)$ and $z \in \mu(y)$ should imply $z \in \mu(x)$. From this axiom we can deduce the fourth neighborhood axiom of topology (namely, that any neighborhood contains some smaller open neighborhood). This allows us to prove that if a halo system $\mu$ defines a topology $T$, then we can work backwards from $T$ to retrieve precisely the halo system with which we started: $$\mu(x)=\bigcap{\left \{\text{*}O: O \in T, x \in \text{*}O \right \}} \, (x \in X) \text{.}$$ -Therefore we can be assured that every halo system satisfying the above axioms uniquely define a topology in the classical sense, so this nonstandard characterization of a topology really coincides with the classical one. -The intuition behind 'transitivity' is fairly acceptable, I think. From the notion of 'closeness' we get the impression that if some point $A$ is close to another point $B$ that is again close to $C$, then probably $A$ should be close to $C$ as well. For the 'local' part I haven't been able to find a persuasive justification, however. -This axiomatization is elegant, intuitive, and can be readily generalized to other settings (for example, it is well known that we can characterize uniform spaces by requiring its infinitesimal closeness relation to be a monadic equivalence relation). I think it deserves to be known better. I wonder why the relevant papers seem to have been largely forgotten or dismissed. -Edit: -Originally, there was a paragraph in the section 'monadicity' that said, - -Alternatively, one could define halos $\mu(x)$ for only $x \in X$, -state as an axiom that for each $x \in X$ there must be a -(neighborhood) filter $\mathcal{N}(x)$ whose intersection monad is -precisely $$\mu(x)=\bigcap_{N \in \mathcal{N}(x)}{\text{*}N}\text{,}$$ -and then extrapolate this to define halos of nonstandard points $x \in \text{*}X$ as $$\mu(x)=\bigcap_{\text{*}N \in \text{*}\mathcal{N}(x)}{\text{*}N}\text{.}$$ (This is not Vakil and -Tewfik's idea. I worked this formulation out just because I'm not used -to dealing with binary relations, and also because I found it quite -confusing that the definition of halos should include all nonstandard -points.) - -This formulation in general is not equivalent to Vakil and Tewfik's approach. According to their approach, multiple halo systems can define the same topology if they agree with one another on every standard point. My way of 'extrapolating' therefore amounts to arbitrarily choosing a single halo system for each topology. In fact, it is equivalent to requiring that -$$\mu(x)=\bigcap{\left \{\text{*}O: O \in T, x \in \text{*}O \right \}} \text{,}$$ -even for nonstandard points $x \in \text{*} X$. A fairly natural choice, but still arbitrary. If one wishes to replicate Vakil and Tewfik's condition, then one can impose the existence of a 'set of indices' $I$, and require that for each $x \in X$ its neighborhood filter admits a basis $N(i, x) \, (i \in I)$. Now if we extrapolate this to -$$\mu(x)=\bigcap_{i \in I}{\text{*}N(i, x)} \, (x \in \text{*}X) \text{,}$$ -one can easily show that -$$W = \left \{ (x, y) \in \text{*} \left( X \times X \right) \mid y \in \mu(x) \right \} $$ -is the intersection monad of the filter generated by the basis $\left\{(x, y) \mid y \in N(i, x) \right\} \, (i \in I)$. The result of the extrapolation will depend on the choice of basis, however.<|endoftext|> -TITLE: Exercise 1.1.(c) in Hartshorne's Deformation Theory -QUESTION [5 upvotes]: Exercise 1.1.(c) in Hartshorne's Deformation Theory: - -Over an algebraically closed field $k$, we define a curve in $\mathbb P^2_k$ to be the closed subscheme, defined by a homogeneous polynomial $f(x,y,z)$ of degree $d$ in the coordinate ring $S=k[x,y,z]$. -(c) For any finitely generated $k$-algebra $A$, we define a family of curves of degree $d$ in $\mathbb P^2$ over $A$ to be a closed subscheme $X\subseteq\mathbb P^2_A$, flat over $A$, whose fibers above closed points of $\mathrm{Spec}\,A$ are curves in $\mathbb P^2$. Show that the ideal $I_X\subseteq A[x,y,z]$ is generated by a single homogeneous polynomial $f$ of degree $d$ in $A[x,y,z]$. - -My attempts: the condition on fibers above closed points is equivalent to the condition that for any $\mathfrak m\in\mathrm{Specm}\,A$ the ideal $(I_X,\mathfrak m)/\mathfrak m$ is principal. Equivalently, for any $\mathfrak m$ there is $f_{\mathfrak m}\in I_X$ such that $I_X\subset (f_{\mathfrak m},\mathfrak m)$. Also, it is clearly sufficient to prove that $I_X$ contains a homogeneous element of degree $d$. - -REPLY [17 votes]: I agree. Thanks for discovering the error. And by the way there is another error on the same page, line -1, there is a -2 that should be a -4. -Robin Hartshorne<|endoftext|> -TITLE: Injective additive grids -QUESTION [7 upvotes]: We call a map $f:{\mathbb Z}\times {\mathbb Z} \to {\mathbb Z}$ an additive grid if for all $x,y \in {\mathbb Z}$ we have that $f(x,y)$ is the sum of the neighboring values, that is, $$f(x,y) = f(x-1,y)+f(x+1,y) + f(x,y-1) + f(x,y+1).$$ -To me it feels like there cannot be injective additive grids - but I haven't been able to come up with a proof. So: are there injective additive grids? - -REPLY [2 votes]: Yes, such a grid can be constructed easily by induction. -As can be seen from the answer given to your previous question, the adjacent rows can be chosen arbitrarily. -Denote the numbers in these rows by $\ldots,x_{-1},x_0,x_1,\ldots$ and $\ldots,y_{-1},y_0,y_1,\ldots$. -How does the rest of the grid change if we change $x_0$? -The elements that are in the infinite triangle whose sides are the halflines that go North-West and North-East from $x_0$. -The coefficients can be also computed, along the sides they are always $\pm x_0$, while inside the triangle some other numbers (possibly $0$). -Let us pick the numbers in the order $x_0,x_1,x_{-1},x_2,\ldots$ such that they grow very fast. -I claim that this guarantees that all elements of the grid are different. -(Let's ignore the $y_i$'s - it's easy to make the argument work for them too.) -Suppose that there are two numbers that are equal. -For both, take the most significant $x_i$ that influenced them. -This will be one that is diagonally South-West or South-East. -If these $x_i$ differ, the numbers also differ. -If they are the same, then the coefficient of how $x_i$ influences are numbers might be still different. -If it's also the same, take the second most significant $x_j$ etc. -Sooner or later we get a different $x_j$ or coefficient, and we are done. -Note that in the above argument we have used that for any $k$ there are finitely many numbers whose most significant $x_i$ has $|i|\le k$.<|endoftext|> -TITLE: How does Saito's treatment of the conductor and discriminant reconcile with an elliptic curve? -QUESTION [18 upvotes]: Saito (1988) gives a proof that -$$\textrm{Art}(M/R) = \nu(\Delta)$$ -Here, $M$ is the minimal regular projective model of a projective smooth and geometrically connected curve $C$ of positive genus over the field of fractions of $R$, a ring with perfect residue field. The $\Delta$ here is the "discriminant" of $M$ which measures defects in a functorial isomorphism described in this answer. And $\textrm{Art}(M/R)$ is the Artin Conductor. -I know at some point, somewhere in the literature, someone took this result and applied it to elliptic curves and realized that when $C$ is an elliptic curve, $\Delta$ above represents the discriminant of the minimal Weierstrass equation of the elliptic curve and that $\textrm{Art}(M/R)$ is the standard global conductor of the elliptic curve. -So my question is: how exactly that is done? How does one take the definition of the discriminant and conductor in Saito's paper and in the linked answer and prove that when looking at an elliptic curve, they are the regular discriminants and conductors? -Or am I really confused? -EDIT: Watson Ladd has provided an amazing paper by Ogg that attempts a case-by-case proof of the result specialized to an elliptic curve. And upon researching, I've found that many people claim Saito (1988)'s general result proved Ogg's result "fully". So I guess my question can also be restated as how you get from Saito's result to Ogg's formula. - -REPLY [15 votes]: This is a great question, but I don't think there is an easy answer. -Saito himself proves on p.156 (Cor. 2) that his results imply Ogg's formula, including the missing -case of 2-adic fields. However, the proofs are quite condensed and the underlying technology very advanced. -Relative dimension 0. -Saito's approach is motivated by the classical relation between the different, discriminant and -conductor for number fields or local fields. Say $K/{\mathbb Q}_p$ is finite, and -$$ - f: X=\textrm{Spec }{\mathcal O}_K \longrightarrow - \textrm{Spec }{\mathbb Z}_p=S. -$$ -Classically, there is the different (ideal upstairs) -$$ - \delta = \{\,x\in K\>|\>\textrm{Tr}(x{\mathcal O}_K)\subset{\mathbb Z}_p\,\}^{-1} - \>\>\subset {\mathcal O}_K, -$$ -the discriminant (ideal downstairs) -$$ - \Delta = (\det \textrm{Tr} (x_i x_j)_{ij})\subset {\mathbb Z}_p, - \quad\qquad x_1,...,x_n\textrm{ any }{\mathbb Z}_p\textrm{-basis of }{\mathcal O}_K -$$ -and the Artin conductor ${\mathfrak f}({\mathcal O}_K/{\mathbb Z}_p)$ (also an ideal downstairs). -The conductor-discriminant formula in this case says -$$ - \textrm{order }(\Delta) = \textrm{order }\textrm{Norm}_{K/{\mathbb Q}_p}(\delta) = \textrm{order }{\mathfrak f}({\mathcal O}_K/{\mathbb Z}_p), -$$ -where order is just the valuation: order$(p^n{\mathbb Z}_p)=n$. -Saito interprets these sheaf-theoretically as follows: $\Delta$ is a homomorphism of invertible -${\mathcal O}_S$-modules -$$ - \Delta: (\textrm{det }f_*{\mathcal O}_X)^{\otimes 2} \rightarrow {\mathcal O}_S, \qquad - (x_1\wedge...\wedge x_n)\otimes (y_1\wedge...\wedge y_n) \mapsto \det(\mathrm{Tr}_{X/S}(x_iy_j)), -$$ -which is an isomorphism on the generic fibre; the classical discriminant is its order = length of the cokernel -on the special fibre. -Then, $-\textrm{order }{\mathfrak f}({\mathcal O}_K/{\mathbb Z}_p)=\textrm{Art}(X/S)$; -this can be defined for any relative scheme over $S$ in terms of l-adic etale cohomology, -$$ - \textrm{Art}(X/S) = \chi_{et}(\textrm{generic fibre}) - \chi_{et}(\textrm{special fibre}) - - (\textrm{Swan conductor}). -$$ -Finally, the different happens to be the localised first Chern class -$$ - \delta = c_1(\Omega_{X/S}) = c_1({\mathcal O}_X\to \omega_{X/S}). -$$ -Relative dimension 1. -Now we move to one dimension higher, so that $S$ is the same, but $X$ is now a regular model of, say, -an elliptic curve $E/{\mathbb Q}_p$. -The conductor $\textrm{Art}(X/S)$ is still defined, and it is essentially the conductor of $E$, -except that $\chi_{et}(\textrm{special fibre})$ has an $H^2$-contribution from the irreducible components -of the special fibre. To be precise, as explained in Liu Prop. 1 -(or using Bloch Lemma 1.2(i)), -$$ - -\textrm{Art}(X/S) = n + f - 1, -$$ -where $f$ is the the classical conductor exponent of an elliptic curve, and $n$ is the number of components -of the special fibre of the regular model $X$. So, Ogg's formula in this language reads -$$ - -\textrm{Art}(X/S) = \textrm{ord }\Delta_{min}, -$$ -where $\Delta_{min}$ is the discriminant of the minimal Weierstrass model. -So Ogg's formula is like "conductor=discriminant" formula in the number field setting, and Saito proves it -through "conductor=different=discriminant". To be precise, there are three equalities -$$ - -\textrm{Art}(X/S) = -\deg c_1(\Omega^1_{X/S}) = \textrm{ord }\Delta_{Del} = \textrm{ord }\Delta_{min} -$$ -The first one, "conductor=different" was done by Bloch here (I have no access to this) -and here in 1987. Then $\Delta_{Del}$ is the Deligne discriminant, defined by -Deligne in a letter to Quillen. Analogously to the -sheaf-theoretic interpretation of the discriminant in the relative dimension 0 case, -Deligne constructs a canonical map -$$ - \Delta: \det(Rf_*(\omega_{X/S}^{\otimes 2})) \rightarrow \det(Rf_*(\omega_{X/S}))^{\otimes 13}). -$$ -It is again an isomorphism on the generic fibre, and Saito calls the (Deligne) discriminant the order -of this map on the special fibre. The second equality, "different=discriminant", is the main result of -Saito's paper, and it is very technical. And finally, Saito on pp.155-156 proves that for elliptic curves, -$\textrm{ord }\Delta_{Del} = \textrm{ord }\Delta_{min}$ (third equality), using properties of minimal -Weierstrass models (at most one singular point), Neron models and existence of a section for models of -elliptic curves. -Personal note. -It would be amazing if someone deciphered Saito's proof and wrote it down in elementary terms. -I don't think such a treatment exists, though there are very nice papers by Liu and by Eriksson -on conductors and discriminants. -They treat genus 2 curves and plane curves, respectively, and they are much more accessible.<|endoftext|> -TITLE: What is $\int_{0}^{z} e^{-a^{2} x^{2}} {\rm erf}(bx)\, dx$? -QUESTION [7 upvotes]: The integral $$\int_{0}^{z} e^{-a^{2} x^{2}} {\rm erf}(bx)\, dx$$ is related to the convolution of two half-normal distributions. This can be inferred from this question on MSE. The following expression for the definite version of this integral is known. See, for instance, "A Table of Integrals of the Error Functions" by Edward W. Ng and Murray Geller, equation 4.3.2 (link here). It is as follows: $$ \int_{0}^{\infty} e^{-a^{2} x^{2} } {\rm erf} (bx) \,dx = \frac{\sqrt{\pi}}{2a} - \frac{1}{a \sqrt{\pi}} \tan^{-1} \big{(} \frac{a}{b} \big{)} .$$ -However, I don't know about a similar expression for the indefinite integral. Do you know about a paper or other source that adresses the indefinite integral? Or do you know how to compute it yourself? -The reason I ask about this integral is because I'd like to compute the convolution of two half-normal distributions. Please let me know if you have a reference in which the convolution of two half-normal distributions with unequal variances is calculated. - -REPLY [10 votes]: This indefinite integral is a special function, called Owen's T: -$$\int_0^z e^{-a^2 x^2}{\rm erf}\,(bx)\,dx=\frac{\arctan(b/a)}{a\sqrt\pi}-\frac{2\sqrt\pi}{a} T\left(\sqrt{2} az,b/a\right)$$ - -Here is the requested derivation: -$$\int_0^z e^{-a^2x^2}{\rm erf}\,(bx)\,dx=\frac{2}{\sqrt\pi}\int_0^z e^{-a^2x^2}\left(\int_0^{bx}e^{-t^2}dt\right)dx=$$ -$$\frac{2}{\sqrt\pi}\int_0^z e^{-a^2x^2}\left(\int_0^{b}e^{-(yx)^2}x\,dy\right)dx=$$ -$$\frac{2}{\sqrt\pi}\int_0^b\left(\int_0^z e^{-(a^2+y^2)x^2}xdx\right)dy=\\ -\frac{1}{\sqrt\pi}\int_0^b\frac{e^{-(a^2+y^2)z^2}-1}{a^2+y^2}\,dy=$$ -$$\frac{1}{a\sqrt\pi}\left[ \arctan(b/a)-2\pi T(\sqrt{2}az,b/a)\right]$$ -with the definition $T(z,b)=\frac{1}{2\pi}\int_0^b(1+y^2)^{-1}\exp[-\tfrac{1}{2}z^2(1+y^2)]\,dy$ of Owen's T-function.<|endoftext|> -TITLE: Example of an abelian category with enough projectives and injectives which are not dual -QUESTION [9 upvotes]: For trying to understand how general a certain theorem is, I'm looking for an example of an essentially small abelian category which has enough projectives and enough injectives, but whose category of projectives is not equivalent to the category of injectives. -By a result of Auslander, each such category can be written as the category $\operatorname{mod} \operatorname{proj} \mathcal{A}$ of coherent/finitely presented functors (likewise as $\operatorname{mod}\operatorname{inj}\mathcal{A}$). Typical examples up to this point include the category of finite dimensional modules over a finite dimensional algebra, or generalisations of this, e.g. what is sometimes called a dualising $k$-variety, where the category is a $k$-category for a ground field and the $k$-duality provides an equivalence. - -REPLY [22 votes]: The category of countable abelian groups is an essentially small abelian category, and has enough projectives and injectives (the countable free abelian groups and the countable divisible groups respectively). However, there is an injective with endomorphism ring $\mathbb{Q}$, but no such projective, so the categories of projectives and injectives can't be equivalent or dual.<|endoftext|> -TITLE: Is this proof of Perron's theorem correct, and if so is it original? -QUESTION [45 upvotes]: A few years ago, I came up with this proof of Perron's theorem for a class presentation: -https://pi.math.cornell.edu/~web6720/Perron-Frobenius_Hannah%20Cairns.pdf -I've written an outline of it below so that you don't have to read a link. -It's close in spirit to Wielandt's proof using $$\rho := \sup_{\substack{x \ge 0\\|x| = 1}} \min_j {|{\sum x_i A_{ij}}| \over x_j},$$ but I think it's simpler. (In particular, you don't have to divide by anything or take the sup min of anything.) The exception is the part where you prove that the spectral radius has only one eigenvector, which is exactly the same as Wieland's proof. -I believe it's correct, but it hasn't passed through any kind of verification process aside from being presented in class. And I've had a couple people write me about it, and, for God knows what reason, it comes up on Google in the first couple pages if you search for "Perron-Frobenius." -So I'd appreciate it if you would look at this and see if you see anything wrong with it. And if you don't, I'd like to know if it's original, because if so then I get to feel proud of myself. -Here is the proof: - -Let $A > 0$ be a positive $n \times n$ matrix with eigenvalues $\lambda_1, \ldots, \lambda_n$, counted with multiplicity. Let $\rho = \max |\lambda_i|$ be the spectral radius. We want to prove that $\rho$ is a simple eigenvalue of $A$ with a positive eigenvector, and that every other eigenvalue is strictly smaller in absolute value. - -Let $\lambda$ be an eigenvalue with $|\lambda| = \rho$, and finally let $\psi$ an eigenvector for $\lambda$. Consider $$\Psi := |\psi| = (|\psi_1|, \ldots, |\psi_n|).$$ -Then $A \Psi = A |\psi| \ge |A \psi| = |\lambda \psi| = \rho |\psi| = \rho \Psi$, where "$x \ge y$" means that each coordinate of $x$ is greater than or equal to each coordinate of $y$. - -Suppose $A \Psi \ne \rho \Psi$. Then by positivity we have $A^2 \Psi > \rho A \Psi$, which means that by continuity there is some $\varepsilon > 0$ with $A^2 \Psi \ge (\rho + \varepsilon) A \Psi$. Therefore \begin{align*}A^{n+1} \Psi &\ge (\rho + \varepsilon) A^n \Psi \\&\cdots\\&\ge (\rho + \varepsilon)^n A \Psi \ge 0\end{align*} -and taking norms we get $\Vert A^{n+1} \Psi \Vert_1 \ge (\rho + \varepsilon)^n \Vert A \Psi \Vert_1$, so the operator 1-norm of $A^n$ is at least $(\rho + \varepsilon)^n$, which is a contradiction with Gelfand's formula $\lim \Vert A^n \Vert^{1/n} = \rho$. - -Therefore $A \Psi = \rho \Psi$ and $\rho$ is an eigenvalue with positive eigenvector $\rho \Psi = A \Psi > 0$. - -Suppose there is an eigenvalue $\lambda$ with $|\lambda| = \rho$. Let $\psi$ be an eigenvector for $\lambda$. We have seen above that $A \Psi = \rho \Psi = |A \psi|$ or $\sum_j A_{ij} |\psi_j| = |\sum_{ij} A_{ij} \psi_j|$. Fix an index $i$. Then $A_{ij} > 0$ for each row $j$, so $\sum_{ij} A_{ij} \psi_j$ is a weighted sum of $\psi_j$ where all the weights are positive, and its absolute value is the weighted sum of $|\psi_j|$ with the same weights. Those two things can only be equal if all the summands $\psi_j$ all have the same complex argument, so $\psi = e^{i\theta} \psi'$ where $\psi' \ge 0$, and $\lambda \psi' = A \psi' > 0$, so $\lambda > 0$. Therefore $\lambda = \rho$. - -Now we know that every eigenvalue with $|\lambda| = \rho$ is $\rho$, and it has one positive eigenvector (and possibly more), but we don't know how many times $\rho$ appears in the list of eigenvalues. That is, we don't know whether it's simple or not. - -We can prove that $\rho$ has only one eigenvector by the same argument in Wielandt's proof. We know $\Psi$ is a positive eigenvector. Suppose that there is another linearly independent eigenvector $\psi$. We can pick $\psi$ to be real (because $\mathop{\rm Re} \psi$ and $\mathop{\rm Im} \psi$ are eigenvectors or zero and at least one is linearly independent of $\Psi$). Choose $c$ so $\Psi + c \psi$ is nonnegative and has one zero entry. Then $\rho (\Psi + c \psi) = A(\Psi + c \psi) > 0$ by positivity, but it has one zero entry, which is a contradiction. So there's no other linearly independent eigenvector. - -Now that we know there's only one eigenvector, we can prove that $\rho$ is a simple eigenvalue. By the previous reasoning, there is a positive left eigenvector $\Pi$ of $\rho$, so $\Pi A = \rho A$. Then $\Pi > 0$ and $\Psi > 0$, so $\Pi \Psi \ne 0$. Then $\Pi^0 := \{x: \Pi x = 0\}$ is an $(n-1)$-dimensional subspace of $\mathbb R^n$ and $\Psi \notin \Pi^0$, so we can decompose $\mathbb R^n$ into the direct sum $$\mathbb R^n = \mathop{\text{span}}\{\Psi\} \oplus \Pi^0.$$ - -Both of these spaces are invariant under $A$, because $A \Psi = \rho \Psi$ and $\Pi A x = \rho \Pi x = 0$. Let $x_2, \ldots, x_n$ be a basis of $\Pi^0$. Let $$X = \begin{bmatrix}\Psi&x_2&x_3&\cdots&x_n\end{bmatrix}.$$ -Then the invariance means that -$$X^{-1}AX = \begin{bmatrix}\rho&0\\0&Y\end{bmatrix}$$ where the top right $0$ says $\Pi^0$ is invariant under $A$ and the lower left $0$ says $\mathop{\text{span}}\{\Psi\}$ is invariant under $A$. Here $Y$ is some unknown $(n-1) \times (n-1)$ matrix. - -$A$ is similar to the above block matrix, so the eigenvalues of $A$ are $\rho$ followed by the eigenvalues of $Y$. If $\rho$ is not a simple eigenvalue, then it must be an eigenvalue of $Y$. - -Suppose $\rho$ is an eigenvalue of $Y$. Let $\psi'$ be an eigenvector with $Y \psi' = \rho \psi'$. Then $A X {0 \choose \psi'} = \rho X {0 \choose \psi'}$ and $X{0 \choose \psi'}$ is linearly independent of $\Psi = X {1 \choose \mathbf{0}}$. We've already proved that $A$ has only one eigenvector for $\rho$, so that is impossible. Therefore, $\rho$ is not an eigenvalue of $Y$, so $\rho$ is a simple eigenvalue of $A$. That's the last thing we had to prove. - -Extending to $A \ge 0$ with $A^n > 0$ works as usual. - - -Thanks! - -REPLY [3 votes]: This is a nice argument that you could use in some teaching notes. As for whether it's publishable, there are two questions. -Are you proving a completely unknown statement? -Are you using any new arguments, techniques, or ideas in the proof, or is your proof in some sense ''morally'' the same as existing proofs in the literature? -If the answer is no to both of those, you probably won't be able to publish the article in a journal, although you can of course employ it in lecture notes and so on. -You may be able to publish the proof as a part of an article in a more pedagogical journal like American Mathematical Monthly, but these journals typically receive many submissions and are very competitive to get into, so you will have to make sure the article is polished and that the novelty of the article is carefully explained. -Edit: I see that the OP has indeed now had an article based on the above proof published in the American Mathematical Monthly, which is great news!<|endoftext|> -TITLE: Determinant of identity matrix plus Hilbert matrix -QUESTION [12 upvotes]: I am looking for the determinant -$$ \det(I_n + H_n) $$ -where $I_n$ is the $n \times n$ identity matrix and $H_n$ is the $n \times n$ Hilbert matrix, whose entries are given by -$$ [H_n]_{ij} = \frac{1}{i+j-1}, \qquad\qquad 1 \le i,j \le n $$ -Is anything known about this determinant for finite $n$ or about its asymptotic behaviour for $n \rightarrow \infty$? -More generally, are there results about the determinant of "identity plus Hankel" matrices or their asymptotic behaviour? - -REPLY [5 votes]: If you look at $H=\frac 1{(i+j-1)\pi}$ instead, then $\det(1 + H)\sim n^{3/8}$, as $n\to\infty$. This is basically proven in arXiv:1808.08009, arXiv:1905.03154<|endoftext|> -TITLE: preservation of forcing rigidity in iterations -QUESTION [5 upvotes]: Say that a partial order $P$ is forcing-rigid in a model $V$ if whenever $G \subseteq P$ is generic over $V$, then in $V[G]$, $G$ is the only filter which is $P$-generic over $V$. This implies there are no nontrivial automorphisms of $P$. -If $P$ is forcing-rigid and $P$ forces "$\dot Q$ is forcing-rigid," then is $P * \dot Q$ forcing-rigid? - -REPLY [6 votes]: No, rigidity is not preserved in iterations. In particular, -Proposition: If $T$ is a rigid Souslin-tree with the property that -$$1 \Vdash_T T_{s} =\{ t \in T : t\le_T s \}\text{ is rigid and Souslin for every }s \not\in \dot{G}$$ -(rigid and Souslin off-the-generic-branch in the terminology of FuchsHam2008.) Then, - -$1\Vdash_T $" $\check{T}$ is rigid and totally-proper." - -Proof: Assume $T$ satisfies the required property. -To see that $T$ remains proper, fix some countable elementary sub-model $M\prec H(\lambda)$ with $T \in M$ and let $t\in T\cap M$ and $\dot{s} \in M^{T}$ be such that $t \Vdash \dot{s} \in \check{T}$. Noting that $T$ is c.c.c., we can find some $\dot{s}_0 \in M^T$ such that $t \Vdash_T \dot{s}_0 \le_T \dot{s}$ and $\dot{s}_0 \not\in \dot{G}$. Now let $\dot{D} \in M^{T}$ be a $T$-name for a dense-open subset of $T$ and $(u,\dot{v})\le (t,\dot{s}_0)$ be any extension with $Lev_{T}(u) \ge \delta = M\cap \omega_1$ and $u \Vdash Lev_{T}(\dot{v}) \ge \check{\delta}$. -Then, we must have $ u \Vdash (\exists r \in \dot{D} \cap \check{M})(\dot{v} \le r)$ (since otherwise, a standard reflection argument yields $u \Vdash (\exists a \in T \cap M)(a \not\in \dot{G}$ and $T_a$ is not Souslin$)$). It follows that $u \Vdash_T \dot{v}$ is totally $(M[\dot{G}], \check{T})$-generic; and so $1 \Vdash_{T} \check{T}$ is totally-proper. -To see that $T$ remains rigid, note that if $t \Vdash_{T} \check{T}$ is not rigid, then for some $s \in T$ with $s \perp t$, we must have $t \Vdash_{T} \check{T}_s$ is not rigid, or $\check{T}_s$ is not c.c.c. $\square$. -Remark: To see that this provides a counter-example, note that the two-step iteration of $T$ with itself is isomorphic to the square $T^2$ which admits the non-trivial automorphism $(s,t)\rightarrow (t,s)$. (being totally-proper in the extension didn't really matter, I just thought it was worth pointing out.)<|endoftext|> -TITLE: Galois representation associated to CM-newforms -QUESTION [5 upvotes]: Let $f(z)=\sum_{n\ge 1}a(n)e(nz)$, be a newform of CM-type, and let $\psi_f$ be the associated Hecke character, so that, -$$ -f(z)=\sum_{\mathfrak{a}}\psi_f(\mathfrak{a})e(N(\mathfrak{a})z), -$$ -and let $\rho_{\lambda,f}$ be the associated Galois representation. -Let $\frak{p}$ be a prime ideal of the field by which $f$ has CM. My question is: -How to prove that the characteristic polynomial of $\rho_{\lambda,f}(\mathrm{Frob}_{\mathfrak{p}})$ satisfying -$$ -(x-\psi_f(\mathfrak{p}))(x-\psi(\mathfrak{p'}))? -$$ -where $\mathfrak{p'}$ the conjugate of $\mathfrak{p}$. - -REPLY [8 votes]: Let me abbreviate $\rho_{\lambda,f}$ as $\rho$, and $\psi_f$ as $\psi$. -By definition, $L(s,\rho)=L(s,f)=L(s,\psi)$. The equality of the Euler factors of $L(s,\rho)$ and $L(s,\psi)$ at the split prime $p=\mathfrak{p}\mathfrak{p}'$ means that -$$\det(1-\rho(\mathrm{Frob}_{\mathfrak{p}})p^{-s})=(1-\psi(\mathfrak{p})p^{-s})(1-\psi(\mathfrak{p}')p^{-s}).$$ -Multiplying both sides by $p^{2s}$ and renaming $p^s$ to $x$, we get -$$\det(x-\rho(\mathrm{Frob}_{\mathfrak{p}}))=(x-\psi(\mathfrak{p}))(x-\psi(\mathfrak{p}')).$$<|endoftext|> -TITLE: Precise formulation of conjectures on orders of vanishing? -QUESTION [12 upvotes]: Let $X$ be a smooth and proper scheme over $\text{Spec}(\mathbf{Z})$. -C. Soulé has conjectures about special values of the completed zeta function of $X$, $\zeta(X,s)$, which were first reformulated by Bloch in terms of his higher Chow groups as follows: -$$-\text{ord}_{s=\text{dim}(X)-n}\zeta(X,s) = \sum_{i}(-1)^i\text{rk}\text{CH}^n(X,i).$$ -However, in the literature (see this paper by Flach and Morin, for example) one can find formulas for the order of vanishing of $\zeta(X,s)$ at integers (and for its special values) that do not involve Bloch's higher Chow groups, but rather an Arakelov version of higher Chow groups. -It seems these latter ones are typically much larger than usual higher Chow groups, so I am wondering what conjectural statement is actually expected to be true. -After some time searching the literature, I could not pinpoint which between Bloch's original statement and the Flach-Morin statement, say, was meant to be the correct conjecture, and the lack of a succinct source clearly stating such expectations is very confusing. -Specifically, can anyone please clarify what, in the expected formula: -$$\text{ord}_{s=n}\zeta(X,s) = \sum_i(-1)^i\text{rk}_{\mathbf{Z}}{C}^n_i$$ -(up to a sign) $C^n_i$ are truly meant to be in terms of higher Chow groups and their variants? (Bloch's higher Chow groups? Arakelov Chow groups? Real Deligne cohomology groups?) -More precise questions: - -Let $X$ be a smooth projective variety defined over $\mathbf{Q}$. What is the relation between the ranks of all the variants of $\text{CH}^n(X,i)$ used in potential formulas for $\text{ord}_{s=n}\zeta(X,s)$? - -We denote by $\widehat{\text{CH}}^n(X,i)$ the Arakelov Chow groups of Bloch and Soulé, and by $H^{2n-i}_{\mathcal{D}}(X,\mathbf{R}(n))$ the real Deligne cohomology of $X$. - -Are the cycle maps $\text{CH}^n(X,i)\to\widehat{\text{CH}}^n(X,i)$ and $\text{CH}^n(X,i)\to H^{2n-i}_{\mathcal{D}}(X,\mathbf{R}(n))$ expected to be isomorphisms rationally? (surely not in the first case) -What features is the integral cycle map - -$$\text{CH}^n(X,i)\to H^{2n-i}_{\mathcal{D}}(X,\mathbf{Z}(n))$$ -expected to have? (ie. finite kernel, discrete image?) and for which $i,n$? - -REPLY [2 votes]: You can find the relation between Soulé's pole order conjecture and the formulation in terms of arithmetic Chow groups in a paper of mine "Special L-values of geometric motives" (Asian Journal of Mathematics (2017), Vol. 21 (2) pp. 225–264). Briefly, I raise there the conjecture that motivic homology (of which Chow groups are instances) is related to what we call Arakelov motivic cohomology (of which the groups $\widehat{CH}(X)$ are instances) by means of a conjecturally perfect pairing of real vector spaces. I also show in this paper that the perfectness of this pairing is equivalent to the conjunction of Soulé's, Beilinson's and Tate's conjecture, in the presence of the "usual" conjectures on mixed motives. Thus, if you believe this perfectness conjecture (which is, in my opinion, a conceptually appealing statement), you can phrase the pole order conjecture equivalently in terms of usual Chow groups or in terms of arithmetic Chow groups. -In addition to David Loeffler's comment, let me also clarify that the literature offers two different types of arithmetic Chow groups. Gillet-Soulé introduce one which sits in an exact sequence roughly of the form -$$\dots \to \text{first higher }CH \to A \to \widetilde{CH} \to CH \to 0,$$ -where $A$ is a real vector space of the type "all differential forms (of some degree) modulo boundaries". In particular, these spaces are usually infinite-dimensional and their group $\widetilde{CH}$ is indeed much bigger than $CH$. These groups are not the ones featuring in the formulation of the conjecture in Flach-Morin and my work you mention. Instead it is a subgroup $\widehat{CH} \subset \widetilde{CH}$ (which is also considered by Gillet-Soulé), defined in a way so that there are exact sequences -$$\dots \to H^*_D(X, *) \to \widehat{CH} \to CH \to H^*_D(X, *) \to \dots$$ -i.e., these groups measure precisely the extent to which the Beilinson regulator is not an isomorphism.<|endoftext|> -TITLE: A model of ZF without a well-ordering of the reals in which any two sets of reals are comparable -QUESTION [11 upvotes]: Let us say that two sets $A$ and $B$ are comparable if there is an injection from $A$ to $B$ or there is an injection from $B$ to $A$. Obviously, in a model of ${\rm ZFC}$ any two sets are comparable by comparing their cardinalities. But this is not necessarily the case in a model of ${\rm ZF}$. For example in Cohen's classical symmetric submodel showing the independence of the Axiom of Choice from ${\rm ZF}$, there is a set of reals which is incomparable with $\omega$. -Is there a known construction of a model of ${\rm ZF}$ in which the reals are not well-ordered, but any two sets of reals are comparable? - -REPLY [8 votes]: Yes. The perfect set property will ensure every set of reals is countable or has size continuum. - -Solovay, R.M., A model of set-theory in which every set of reals is Lebesgue measurable, Ann. Math. (2) 92, 1-56 (1970). ZBL0207.00905. - -In the paper Solovay proved that from the consistency of an inaccessible cardinal it is consistent that every set of reals is countable or has the perfect set property. Of course, there you need the inaccessible cardinal if you want Dependent Choice. If you're willing to forego that, you can forego the inaccessible cardinal, as shown by Truss. - -Truss, John, Models of set theory containing many perfect sets, Ann. Math. Logic 7, 197-219 (1974). ZBL0302.02024. - -There he shows that you can always push this up in the sense that you can have that every set of reals is well-orderable or has a perfect subset and you can pretty much have no limit on what "well-orderable" means in that case. (Although I believe Solovay also does that at the end of his paper, but I do not recall at the moment.)<|endoftext|> -TITLE: Schur's Lemma for Quantized Universal Enveloping Algebra -QUESTION [5 upvotes]: Let $U_q(\mathfrak{g})$ (defined over $\mathbb{C}(q)$) be the quantized universal enveloping algebra of a simple Lie algebra $\mathfrak{g}$. Let $M$ a finite-dimensional simple left $U_q(\mathfrak{g})$-module. Is it true that $\dim_{\mathbb{C}(q)}\mathrm{End}_{U_q(\mathfrak{g})}(M)=1$? How to sketch a proof? The problem is that the field $\mathbb{C}(q)$ is not algebraically closed. - -REPLY [4 votes]: This actually requires a fair amount of the structure theory of the representations of quantum groups. If you have a simple, then after a finite field extension, you can diagonalize the operators $K_i$. This shows that there must be a highest weight vector $v$ of a given weight. Thus your simple (after a field extension) is the unique simple quotient of a Verma module. Therefore, your original simple is defined over $\mathbb{C}(q)$ if and only if the weight of $K_i$ lies in $\mathbb{C}(q)$. This happens because if you're finite dimensional, then $E^{(n)}F^{(n)}$ acts by the quantum binomial coefficient $\binom{K_i}{n}$. This splits completely as a polynomial in $K_i$, so the simple is defined over $\mathbb{C}(q)$.<|endoftext|> -TITLE: How to think about $\mathbf{Z}(n)_{\mathcal{M}}$ -QUESTION [7 upvotes]: One definition of motivic cohomology for smooth schemes $X$ over a field, is via Friedlander-Suslin complexes. -A refresher (you may skip to the question at the bottom) -One defines -(1) $z_n(X,d) :=$ free abelian group generated by all reduced, irreducible closed $k$-subschemes $W\subset X\times(\mathbf{P}^1_k)^d\times\Delta_k^n$ that, with the projection onto $X$ finite surjective. This is a simplicial abelian group, functorial on smooth $k$-schemes with respect to arbitrary morphisms. -(2) $z_n^{\infty}(X,d) :=$ the sum of $(i_{\infty, j})_*z_n(X,d-1)$ for $j=1,\ldots, d$, where $i_{\infty,j} : (\mathbf{P}^1_k)^{d-1}\to(\mathbf{P}^1_k)^d$ inserts $\infty$ at the $j$-th spot. -(3) $\mathbf{Z}(d)_{X,\mathcal{M}} := z_{\bullet}(X,d)/z_{\bullet}^{\infty}(X,d)$. -It turns out the Zariski, resp. étale hypercohomology of the complex $\mathbf{Z}(d)_{X,\mathcal{M}}$ agrees with $H^*_M(X,\mathbf{Z}(d))$, resp. $H^*_L(X,\mathbf{Z}(d))$. -The construction $\mathbf{Z}(d)_{X,\mathcal{M}}$ sort of consists of singular cohomology "twisted" by higher spheres $S^{2d}$, where one uses the "smash product" $(\mathbf{P}^1_k)^{\wedge d}$ as a replacement for $S^{2d}$. If this smash product existed, the Suslin complex of $X\times (\mathbf{P}^1_k)^d$ would be $\mathbf{Z}(d)_{X,\mathcal{M}}$. - -QUESTION - What is the precise topological analog of the Friedlander-Suslin complex $\mathbf{Z}(d)_{X,\mathcal{M}}$? That is, if $X$ is a topological manifold, then $\mathbf{Z}(0)_{X,\mathcal{M}}$ is the usual singular cochain complex. What should $\mathbf{Z}(d)_{X,\mathcal{M}}$ be? Some singular cohomology of a pair? (if so, what pair?) - -EDIT: followup question here - -REPLY [11 votes]: [All cohomology will be reduced cohomology for ease of notation]. -There is no analog for classical homotopy theory. This is related to the fact that the Picard group of the category of spectra is $\mathbb{Z}$ (so the only twists are shifts in degree). -But not all is lost. -Let us enter the more exotic, but still quite familiar, world of $C_2$-equivariant homotopy theory. Its objects are spaces equipped with a $C_2$-action, morphisms are $C_2$-equivariant morphisms and homotopies are $C_2$-equivariant homotopies. In particular, if we have a homotopy equivalence it induces an equivalence both on the underlying space and on the fixed points. The analog of $\mathbb{Z}(0)$ here is given by the cohomology for the constant Mackey functor $\underline{\mathbb{Z}}$ which can, if the action is nice [think $C_2$-manifolds], be defined as -$$H^*(X;\underline{\mathbb{Z}}):=H^*(X/C_2;\mathbb{Z})\,.$$ -(note for example that if $X$ has trivial $C_2$-action, this is just ordinary cohomology). In the equivariant world we are equipped with the classical shifts -$$H^*(X\wedge S^n;\underline{\mathbb{Z}})\cong H^{*-n}(X;\underline{\mathbb{Z}})\,,$$ -but no one prevents us to use more exotic shifts. Let $S^\rho$ the 1-point compactification of the regular representation of $C_2$ (and $S^{n\rho}=(S^\rho)^{\wedge n}$). Then we can define -$$H^*(X;\underline{\mathbb{Z}}(n)):=H^*(X\wedge S^{n\rho};\underline{\mathbb{Z}})\,.$$ -(This can in fact be defined also for negative $n$, but I don't want to enter into the details here). -The analogy is more taut than one could at first think. In fact, there is a "realization" functor from smooth schemes over $\mathbb{R}$ to $C_2$-equivariant spaces, sending $X$ to $X(\mathbb{C})$ with the conjugation action. This sends $\mathbb{P}^1$ to $S^\rho$, and you obtain a map -$$H^*_{mot}(X;\mathbb{Z}(n))\to H^*(X(\mathbb{C})\wedge S^{n\rho};\underline{\mathbb{Z}})$$ -In fact, $H\underline{\mathbb{Z}}$ is the initial $C_2$-cohomology theory receiving such a map. -References -For the definition and the basic properties of $H^*(-;\underline{\mathbb{Z}})$ I like section 3 (Mackey functors, homology and homotopy) of -Hill, M.A.; Hopkins, M.J.; Ravenel, D.C., On the nonexistence of elements of Kervaire invariant one, Ann. Math. (2) 184, No. 1, 1-262 (2016). ZBL1366.55007. -For the connection with real realization, real realization is defined (although it is not the most modern or elegant presentation) in section 3.3.3 of -Morel, Fabien; Voevodsky, Vladimir, $\mathbb{A}^1$-homotopy theory of schemes, Publ. Math., Inst. Hautes Étud. Sci. 90, 45-143 (1999). ZBL0983.14007. -Finally, the fact that the real realization sends the motivic cohomology spectrum to $H\underline{\mathbb{Z}}$, is theorem 4.17 of (thanks to Drew Heard for this one!) -Heller, J.; Ormsby, K., Galois equivariance and stable motivic homotopy theory, Trans. Am. Math. Soc. 368, No. 11, 8047-8077 (2016). ZBL1346.14049. -Moreover theorem 4.18 gives you a range in which the natural map $$H^p(X;\mathbb{Z}/n(q))\to H^p(X(\mathbb{C})_+\wedge S^{q\rho};\underline{\mathbb{Z}/n})$$ -is an isomorphism (beware that their indexing conventions for cohomology differ from the ones in this answer)<|endoftext|> -TITLE: Retracting a wedge of spheres off a homotopy fiber -QUESTION [9 upvotes]: There is a general principle that, for finite simply-connected CW complexes, things that are true rationally are usually true once you localize away from a finite list of primes. -I'm interested in the possibility that a wedge $S^a \vee S^b$ might be a retract of a space, and I'll be satisfied if there is such a retract only after localizing away from a finite list of primes. (This is the case, -for example, for any suspension $\Sigma X$ of a finite complex $X$.) -My question is this: - -Suppose $X$ and $Y$ are finite simply-connected complexes, and neither - of them has a wedge of two spheres as a retract, even after inverting - any finite set of primes. Is it possible that there is a map $f:X\to Y$ - whose homotopy fiber $F_f$ does have a wedge of two spheres as a - retract? - -REPLY [7 votes]: Take the homotopy fibre of the inclusion $f:\mathbb{C}P^2\vee\mathbb{C}P^2\hookrightarrow \mathbb{C}P^2\times\mathbb{C}P^2$. Both of these spaces are simple connected and neither has a wedge of spheres as a retract, even after inverting any finite collection of primes (unless I've overlooked something). The homotopy fibre $F$ of the map $f$ has the homotopy type -$F_f\simeq \Omega\mathbb{C}P^2\ast\Omega\mathbb{C}P^2\simeq(S^1\times\Omega S^5)\ast (S^1\times\Omega S^5)\simeq S^3\vee\Sigma^2\Omega S^5\vee \Sigma^2\Omega S^5\vee\dots$ -Hopefully there should be enough spheres for you here. Already we have an $S^3$ retracting off, and using the James splitting we get $\Sigma^2\Omega S^5\simeq \bigvee^\infty_{i=1} S^{4i+2}$. So (at least) $S^3\vee S^6$ indeed retracts off of $F_f$, and this is integral. In fact $F_f$ has the integral homotopy type of an infinite wedge of spheres.<|endoftext|> -TITLE: Connections between representations of $\operatorname{SL}_n$ and $\operatorname{GL}_n$ -QUESTION [8 upvotes]: Let $G = \operatorname{GL}_n(F)$ for a $p$-adic field $F$, and let $G_D = \operatorname{SL}_n(F)$. I am wondering if there is a connection between irreducible, admissible representations of $G$ and of $G_D$. -If $(\pi,V)$ is one for $G$, then I am not sure whether the restriction of $\pi$ to $G_D$ remains irreducible and admissible. -On the other hand, if we begin with $(\pi,V)$ as an irreducible admissible representation of $G_D$, can we associate this to an irreducible admissible representation of $G$ by some sort of procedure involving induced representations? I would not expect -$$\operatorname{c-Ind}_{G_D}^G \pi = \operatorname{Ind}_{G_D}^G \pi$$ -to hold in general, and $\operatorname{c-Ind}_{G_D}^G \pi$, though admissible, need not be irreducible. I would hope at best that there is some irreducible admissible subrepresentation $\sigma$ of $\operatorname{c-Ind}_{G_D}^G \pi$ for which $\sigma|_{G_D} \cong \pi$. - -REPLY [5 votes]: The answer to your questions (with proofs) may be found in -C.J. Bushnell, P.C. Kutzko, The admissible dual of SL(N). I -Annales scientifiques de l'École Normale Supérieure, Série 4 : Volume 26 (1993) no. 2 , p. 261--280 -First if $\pi$ is an irreducible smooth (complex) representation of $G$, then $\pi_{\mid G_{D}}$ is a finite direct sum of irreducible smooth representations of $G_D$. This is loc. cit. Prop. (1.7)(i), page 267. - Second, any irreducible smooth representation of $G_D$ is a component of $\pi_{\mid G_{D}}'$ for some irreducible smooth representation $\pi'$ of $G$. This is loc. cit. Prop (1.17)(i), page 270. -Moreover the restriction functor ${\rm Rep}(G)\longrightarrow {\rm Rep}(G_D)$ induces a surjective map on the sets of supercuspidal irreducible representations (resp. discrete series irreducible representations). This is loc. cit. Prop. 1.20, page 271.<|endoftext|> -TITLE: Polynomials for which $f''$ divides $f$ -QUESTION [14 upvotes]: Let $n \geq 2$ and let $a < b$ be real numbers. Then it is easy to see that there is a unique up to scale polynomial $f(x)$ of degree $n$ such that -$$f(x) = \frac{(x-a)(x-b)}{n(n-1)} f''(x).$$ - -Have these polynomials been studied? Do they have a standard name? -Is it true that all the roots of $f(x)$ are real numbers in $[a,b]$? - -Here are the polynomials $f$, with $a=-1$ and $b=1$, for $2 \leq n \leq 10$: --1 + x^2, --x + x^3, -1/5 - (6 x^2)/5 + x^4, -(3 x)/7 - (10 x^3)/7 + x^5, --(1/21) + (5 x^2)/7 - (5 x^4)/3 + x^6, --((5 x)/33) + (35 x^3)/33 - (21 x^5)/11 + x^7, -5/429 - (140 x^2)/429 + (210 x^4)/143 - (28 x^6)/13 + x^8, -(7 x)/143 - (84 x^3)/143 + (126 x^5)/65 - (12 x^7)/5 + x^9, --(7/2431) + (315 x^2)/2431 - (210 x^4)/221 + (42 x^6)/17 - (45 x^8)/17 + x^10 - -REPLY [15 votes]: Recording a CW-answer to take this off the unanswered list. The Gegenbauer polynomials are defined by the differential equation -$$(1-x^2) g'' - (2 \alpha+1) x g' +n(n+2 \alpha) g =0.$$ -Putting $\alpha = -1/2$, we get -$$g=\frac{(x+1)(x-1)}{n(n-1)} g''.$$ -If we want some other values for $a$ and $b$, we can put $f(x) = g(\ell(x))$ where $\ell$ is the affine linear function with $\ell(a) = -1$ and $\ell(b)=1$. -The Gegenbauer polynomials are orthogonal with respect to the weight $(1-x^2)^{\alpha - 1/2}$, so $(1-x^2)^{-1}$ in our case, so standard results on orthogonal polynomials tell us that the roots are in $[-1,1]$. If you are worried about the poles at $x = \pm 1$, then put $g_n(x) = (1-x^2) h_n(x)$, and the $h$'s are orthogonal with respect to $(1-x^2)$.<|endoftext|> -TITLE: Rational normal curves and tangent lines -QUESTION [6 upvotes]: Let $C,\Gamma\subset\mathbb{P}^n$ be degree $n$ rational normal curves in $\mathbb{P}^n$, such that for any $p\in C$ the tangent line $T_pC$ of $C$ at $p$ is tangent to $\Gamma$ as well. This means that there exists a point $q\in\Gamma$ such that $T_q\Gamma = T_pC$. -Clearly, if $n = 2$ this means that $C$ and $\Gamma$ are two conics having the same dual conic i.e. $C^{*} = \Gamma^{*}$, and therefore $C = \Gamma$. -Is it true that this property (any tangent line of $C$ is tangent to $\Gamma$ as well) forces $C = \Gamma$ ? - -REPLY [4 votes]: I want to suggest a simpler argument than Jason's, but I am not sure whether it works in positive characteristic or not. -Assume $n > 2$. -Consider the surface $S(C)$ swept by tangent lines to $C$. It is easy to see that the surface is not normal, its normalization is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$, its map to $\mathbb{P}^n$ is given by (an incomplete) linear system $\mathcal{O}(1,n-1)$, such that $C$ is the image of the diagonal in $\mathbb{P}^1 \times \mathbb{P}^1$ and is equal to the singular locus of the surface. -Now, assume that every tangent to $C$ is also tangent to $\Gamma$. Then $S(C) \subset S(\Gamma)$, and since both surfaces are irreducible, we have $S(C) = S(\Gamma)$. But then $C = Sing(S(C)) = Sing(S(\Gamma)) = \Gamma$.<|endoftext|> -TITLE: Dominating families in bigger cardinals -QUESTION [15 upvotes]: A dominating family on $\omega^\omega$ is a set $\mathcal D \subset \omega^\omega$ such that for every $f \in \omega^\omega$ there exists $g \in \mathcal D$ such that $f<^* g$ (that is, $f(n) -TITLE: Clarifying an interpretation of algebraic spaces -QUESTION [7 upvotes]: From several lecture notes and some posts, people claim that while schemes are constructed by gluing affine schemes over the Zariski topology, algebraic spaces are constructed by gluing affine schemes over the étale topology, which I do not really understand. Could someone explain this point carefully? Examples? - -REPLY [3 votes]: If I remember correctly, this goes roughly as follows. Consider the category $\mathcal C=\operatorname{Rings}^{op}$, first endowed with the Zariski topology. You can consider sheaves on this site that are locally covered by representable sheaves. Such sheaves form a category equivalent to the category of schemes. -As you can guess, if you now consider $\mathcal C$ endowed with the étale topology, you will get a category equivalent to the category of algebraic spaces. -ps : I am looking for a reference. The best I found by now : -Commutative rings to algebraic spaces in one jump? -Erratum : as nfdc23 points out, some condition on the diagonal is missing. The correct definition that I copy from Chris Schommer-Pries answer here -Quasi-separatedness for Algebraic Spaces -is the following -Definition: An algebraic space over $S$ is a functor $X : (Sch/S)^{op} \to S_{et}$ such -that - -$X$ is a sheaf on the big étale topology on S, -$\Delta : X \to X \times_S X$ is representable, and -there exists an $S$-scheme $U \to S$ and a surjective étale morphism $U \to X$. - -This is Definition 5.1.10 in Olsson's book Algebraic Spaces and Stacks -https://bookstore.ams.org/coll-62/ . -In remark 5.1.11 he remarks that Knutson's definition includes the fact that $\Delta$ is quasi-compact. -The same definition and more information can be found in the stacks project : -see https://stacks.math.columbia.edu/tag/025Y and https://stacks.math.columbia.edu/tag/076M .<|endoftext|> -TITLE: Some T2 spaces must have a small dense? -QUESTION [5 upvotes]: If a Hausdorff space $\ X\ $ admits a dense subset $ A \hookrightarrow X\ $ such that -$$|X|^{|A|}\ =\ |X|$$ -then indeed $$|X| \leq |\text{End}_{\text{Top}}(X)| \leq |X|^{|A|}\ = \ |X|.$$ -It is the case of $\mathbb{Q} \hookrightarrow \mathbb{R}$. Thus, if there is a small enough dense subspace, there are not so many endomorphisms. - -Is the converse true? - -Namely, suppose $|\text{End}_{\text{Top}}(X)| = \ |X|.$ Is it true that there exists a dense subset $A$ such that $|X|^{|A|}\ =\ |X|$? - -This question generated some attention on MSE, but it did not receive any answer, thus I am reposting it here. - -REPLY [9 votes]: Is the converse true? -No. -The paper Constructions and Applications of Rigid Spaces, I, Advances in Mathematics 29 (1978), 89-130, by Kannan and Rajagopalan, describes a countaby infinite Hausdorff space $X$ such that the only continuous maps $f\colon X\to X$ are the constant maps and the identity map. Such a space satisfies -$|\text{End}_{\text{Top}}(X)| = \omega = \ |X|.$ -If $A\subseteq X$ is a subset such that $|X|^{|A|}=|X|=\omega$, then $A$ is forced to be finite, and therefore $A$ cannot be dense.<|endoftext|> -TITLE: How to think about infinite generatedness of motivic cohomology -QUESTION [11 upvotes]: In this question I previously asked how to think about the motivic complex $\mathbf{Z}(1)_{\mathcal{M}}$, whose Zariski hypercohomology should morally be the "singular cohomology" $H^*((-)\wedge S^{2n},\mathbf{Z})$, or the "singular cohomology of a pair" $H^*((-)\times\mathbf{P}^n_k,(-)\times\mathbf{P}^{n-1}_k)$. -Denis Nardin's answer explains a suggestive analogy, and here is a followup. -It happens that several motivic cohomology groups are infinitely generated, even if $X$ is a smooth projective variety over a field (even for $k = \mathbf{Q}$ and $X = \text{Spec}(k)$). -How to think about this? Is there an analogy with infinite generatedness of the singular cohomology of a pair, for instance? -In other words, what should be the "moral reason" why this infinite generatedness occurs? Any analogy coming from algebraic topology would be greatly clarifying. - -REPLY [14 votes]: While waiting for someone more competent than me to answer, let me turn the question right back to you. Why should motivic cohomology be finitely generated? -The answer is, of course, that there's no reason for it. And it is not. Let us take a look at a special example -The Picard group -Let us fix the ground field to be $\mathbb{C}$. The Picard group of a variety is the group of isomorphism classes of line bundles with multiplication given by the tensor. It turns out to be a very "homotopical" invariant. We can consider various variants of it: topological Picard group, analytic Picard group, algebraic Picard group..., depending on what kind of line bundles we are interested in. -It is well known that the topological Picard group of a space $X$ is just $H^2(X;\mathbb{Z})$ (this comes from the identification of $\mathbb{CP}^\infty$ with $K(\mathbb{Z},2)$). -In the algebraic world, something similar happens. The Picard group is equivalent to $H^2_{mot}(X;\mathbb{Z}(1))$. But in this case we know the Picard group cannot be finitely generated: there is a map from the algebraic Picard group to the topological Picard group (called the first Chern class) but, for example, if $X$ is a curve its kernel is given by the points of an abelian variety, the Jacobian of $X$. This is a massively nonfinitely generated abelian group, in fact it is isomorphic to $(\mathbb{R}/\mathbb{Z})^{2g}$ where $g$ is the genus of $X$.<|endoftext|> -TITLE: Maximal subideal of an ideal -QUESTION [6 upvotes]: For a commutative ring $R$ with unity, I am looking for an equivalent condition for an ideal $T$ to have the property that $T$ contains a unique maximal proper subideal, equivalently, the sum of proper subideals of $T$ is not equal to $T$. - -REPLY [9 votes]: Claim: -Let $T$ be an ideal in a commutative ring. If $T$ contains -a unique maximal proper subideal, then $T$ is principal. -Proof: Indeed take any element outside the unique maximal proper subideal. The ideal generated by that cannot be a proper subideal and hence has to be $T$. -So, this suggests that this is a very special property. For instance, that maximal subideal would have to contain the square of any generator of $T$ and if those generators are not zero-divisors, then they are associate (i.e. they can only differ by a unit multiple). If any generator is a zero-divisor then every element in $T$ is a zero-divisor. I think you can continue this line of thought to derive more special properties when this holds. -There is also a partial converse: -Claim: -Let $T$ be a non-zero principal ideal in a commutative ring. Then every proper subideal of $T$ is contained in an ideal that is maximal among proper subideals in $T$. -(By the usual proof of existence of maximal ideals) -Proof: Let $T\subseteq R$ be a principal ideal generated by $t\in T$, $T'\subsetneq T$ a proper subideal, and consider the set of ideals $I\subseteq T$ such that $T'\subseteq I$, $t\not\in I$ which is the same as the set of proper subideals of $T$ containing $T'$. If there is a chain of such ideals $\ldots\subseteq I_\lambda\subseteq \dots$ with $\lambda\in \Lambda$, then $t\not\in J:=\cup_{\lambda\in\Lambda}I_\lambda$ is an upper bound of the chain and hence the statement follows by Zorn's lemma. -However, of course, this does not mean that there is a unique one... -On the positive side, here is a situation where what you are hoping for holds: -Claim: If $R$ is a DVR, then the ideals in $R$ form a totally ordered set. -Proof: This follows essentially from the definition of a DVR. -So, I suppose one can conclude that this holds sometime, but for most ideals in most rings it does not.<|endoftext|> -TITLE: Relation between the Hochschild cohomology of group algebras and groupoids -QUESTION [6 upvotes]: Is there a known relation between the Hochschild cohomology of group algebras and cohomology of groupoids? -Clarification: It is known that 1-dimensional Hochschild cohomology of the Group algebra C[G] is isomorphic to so-called external derivations of C[G]. -On the other hand, it is known that the space of derivations can be identified with so-called characters on a groupoid aG of adjoint actions of the group G. -Therefore 1-dimensional Hochschild cohomology can be identified with 1-dimensional cohomology of the Cayley complex of the groupoid aG in the case when the group G is a finite presented group. I would like to know if a similar identification is known for the Hochschild cohomologies of higher dimensions - -REPLY [3 votes]: For every $\mathbb{Z}G$-bimodule $M$ there exists a $G$-module $U(M)$ such that the Hochschild cohomology of $\mathbb{Z}G$ with coefficients in $M$ is naturally isomorphic to the group cohomology of $G$ with coefficients in $U(M)$. -To see this observe that there is an adjunction between $G$-modules and $\mathbb{Z}G$-bimodules defined as follows: the right adjoint -$$U: \mathbb{Z}G-{\rm BiMod} \to G-{\rm Mod}$$ -sends a $\mathbb{Z}G$-bimodule $M$ to the $G$-module $M$ with the action given by $(g,m) \mapsto gmg^{-1}$, and the left adjoint $L: G-{\rm Mod} \to \mathbb{Z}G-{\rm BiMod}$ sends a $G$-module $M$ to the $\mathbb{Z}G$-bimodule $$L(M) := M \otimes \mathbb{Z}G = \oplus_{g \in G}M\left$$ -where the right action is given by $(\sum_g m_g\left,h) \mapsto \sum_g m_g\left$ and the left action is given by $(h,\sum_g m_g\left) \mapsto \sum_gh(m_g)\left$. It then follows from general categorical considerations that for every $\mathbb{Z}G$-bimodule $M$ there is a natural isomorphism -$$ {\rm HH}^n(\mathbb{Z}G,M) = {\rm Ext}_{\mathbb{Z} G-{\rm BiMod}}(\mathbb{Z}G,M) = {\rm Ext}_{\mathbb{Z} G-{\rm BiMod}}(L(\mathbb{Z}),M) \cong $$ $$ {\rm Ext}_{G-{\rm Mod}}(\mathbb{Z},U(M)) = {\rm H}^n(G,U(M)). $$ -In particular, the Hochschild cohomology ${\rm HH}^n(\mathbb{Z}G) := {\rm HH}^n(\mathbb{Z}G,\mathbb{Z}G)$ is naturally isomorphic to the group cohomology of $G$ with coefficients in the $G$-module $\mathbb{Z}G$ equipped with the conjugation action. -When $G$ is finite the group cohomology $H^n(G, \mathbb{Z}G)$ is isomorphic to the cohomology $H^n(aG,\mathbb{Z})$ of the adjoint groupoid $aG$, and so the connection that you state holds in higher dimensions. I'm not sure what happens when $G$ is infinite. -Remark (of a somewhat unrelated nature): -The connection above comes from interpreting the group cohomology of $G$ as the Quillen cohomology of the ($\infty$-)-groupoid ${\bf B}G$ and the Hochschild cohomology of $\mathbb{Z}G$ as the Quillen cohomology the dg-category ${\bf B}(\mathbb{Z}G)$. The connection then arises from the natural adjunction between $\infty$-groupoids and dg-categories.<|endoftext|> -TITLE: Families of subsets whose characteristic vectors are spanning sets -QUESTION [5 upvotes]: Let $X$ be a finite set and $\mathbb CX$ be a vector space with basis $X$. If $Y\subseteq X$ is a subset, then by the characteristic vector of $Y$ I mean $\sum_{y\in Y}y$. - -My question is: - -Is there a standard name in the literature for a family of subsets of $X$ whose characteristic vectors span $\mathbb CX$? -Are there any decent combinatorial characterizations of families of subsets satisfying 1.? - -REPLY [3 votes]: Ad 2. Without loss of generality let $X$ be the finite ordinal $n$. Then $\mathbb{C}X\cong\mathbb{C}^n$. Now, - -because of your definition of characteristic vectors, -because the supports of the characteristic vectors correspond to the sets in the family, -because you asked the question in the context of vector spaces, -because every generating set of a vector space contains a basis1, - -it follows that your question is at least as difficult as: - -Are there any decent combinatorial characterizations of nonsingular zero-one matrices?2 - -While of course the term 'decent' is not mathematically defined, all hope of what most people would call a 'decent' characterization was lost by 1967, at the latest, with the publication of: - -János Komlós: On the determinant of (0,1)-matrices. Studia Scientiarum Mathematicarum Hungarica 2 (1967) 7-21 - -which seems to have been the first proof that - -Almost all finite square zero-one matrices are nonsingular (in characteristic zero). $\hspace{3em}$ (1) - -By interpreting the zero-one-vectors as subsets of the set $X$, you can read Komlós' article as an answer why most families of subsets of $X$ are generating-sets of $\mathbb{C}X$, if only the number of sets in the family is at least $n=\mathrm{dim}_{\mathbb{C}}\ \mathbb{C}X$, and hence such generating sets cannot be characterized. -In that sense, Komlós' 1967 article is some sort of negative answer to your question no 2. -If later work of Komlós, and in several later publications of Bourgain, Tao, Vu and Wood, theorem (1) was quantified by giving upper bounds for the probability that a randomly chosen such matrix is singular. Detailing these difficult results (wich roughly speaking show that singular zero-one matrices are exponentially rare w.r.t. dimension $n$) would seem excessive for an answer to your question; what matters here is that - -Any characterization of the kind you were asking for would have to account for the fact that if you select the "family of at least $\lvert X\rvert$ subsets of $X$" randomly, then asymptotically almost surely you'll obtain one such "spanning set". - -In an intuitive sense, this is dashes all hopes of a decent structural characterization in the traditional sense. -That said, it is impossible to predict what the future will bring, and of course there are ideas like rough structure and classification, which might be more adequate to such a question than the traditional structure theorems of algebra, which decompose a structure into simpler structures of a deterministic specified kind. -One could also perhaps hope to define some kind of 'distance' $\mathrm{d}$ of the "family" from the 'diagonal family' $\{ \{i\}\colon i\in X \}$, which arguably is the simplest generating set, and then try to prove quantitative results to the effect of 'there are roughly $h(d)$ generating families $\mathcal{F}$ of distance $\mathrm{d}(\mathcal{F})=d$', or even dream of structural results. -Ad 1. While non-existence is difficult to prove, I am reasonably confident to state that no attested technical term exists. Of course, trivially, every family of sets of the kind you require must be an edge cover of the ground set $X$, in the sense of hypergraph theory, but this obvious necessary condition is far from a characterization. There are some vaguely related terms, like 'sparsity patters' of nonsingular matrices, or 'sign patterns' of nonsingular matrices. -${}$___________________________________________________________ -1 This is of course not true for modules. -2 As usual, 'nonsingular zero-one matrix' without further qualifications means 'zero-one matrix which is nonsingular in characteristic zero', i.e., 'zero-one matrix whose determinant is not zero when computed in the ring $\mathbb{Z}$.<|endoftext|> -TITLE: Cardinalities of which there exists partitions of a set containing elements of the same size -QUESTION [8 upvotes]: Let $A$ be a nonempty set. Then we call a nonempty set $p(A)$ to be a partition set of $A$ if and only if all the following are true : - -$p(A)$ is a subset of the power set of $A$. -The elements of $p(A)$ are pairwise disjoint. -Every element of $A$ is present in some element of $p(A)$. - -Now, for a set $A$ and a set $D$ both nonempty we say that $D$ divides $A$ if and only if there is a partition set $p_D(A)$ of $A$ such that every element of $p_D(A)$ is bijective to $D$. -Question -Suppose given two infinite sets $A,B$. Is it true that : either $A$ divides $B$ or $B$ divides $A$ ? - -REPLY [16 votes]: This is equivalent to the axiom of choice. -If the axiom of choice holds, then given $A$ and $B$ which are infinite, then $|A\times B|=\max\{|A|,|B|\}$. So let's say $|A|$ is the maximal one, then this means there is a bijection between $A$ and $A\times B$, so we can partition $A$ to sets of size $|B|$ by considering $\{a\}\times B$ as a partition of $A\times B$. -In the other direction, assume that any two sets one divides the other. Take $A$ to be some infinite set, and let $B$ be the least ordinal such that there is no injection from that ordinal into $A$. This ordinal exists, as per Hartogs' theorem. Now since $B$ does not inject into $A$, this means that you cannot partition $A$ into sets of size $B$. But now we can partition $B$ into sets of size $A$, so pick one of these sets, and this gives us an injection from $A$ into $B$ which well-orders $A$. So every set can be well-ordered, and therefore the axiom of choice holds.<|endoftext|> -TITLE: Non-zero homotopy/homology in diffeomorphism groups -QUESTION [12 upvotes]: Let $M$ be a (possibly simply connected) compact manifold $M$. Are there always non-zero classes in the homotopy or homology of $\mathrm{Diff}(M)$ that directly arise from the topology of $M$ itself? -As an example of the type of answers I am looking for I construct non-zero classes in the homotopy and homology of the loop space $\Omega(M)$, which come from the topology of $M$. -Let $M$ be simply connected. Then there is a smallest positive dimension $d$ where $H^d(M)$ is nonzero. Hence $\pi_d(M)\cong H_d(M)$ is non-zero by Hurewicz' Theorem. The long exact sequence in homotopy of the pathspace fibration shows that $\pi_{d-1}(\Omega M)\cong \pi_d(M)$. Applying Hurewicz' Theorem again we see that $H_{d-1}(\Omega M)\cong \pi_{d-1}(\Omega M)\cong \pi_d(M)\cong H_d(M)$. Thus the homology and homotopy have non-trivial elements that come from the topology of $M$. - -REPLY [6 votes]: The diffeomorphism groups $\text{Diff}(M)$ are sensitive to stabilization, say replacing $M$ by $M \times [0,1]$, so the direct contribution of the homotopy type of $M$ to $\text{Diff}(M)$ can be obscure. If you instead look at the concordance = pseudoisotopy spaces $$P(M) = \text{Diff}(M \times [0,1] \ \text{rel}\ M \times \{0\}),$$ then the stabilization maps $P(M) \to P(M \times [0,1])$ get highly connected as the dimension of $M$ grows (by Kiyoshi Igusa's stability theorem), hence the low-dimensional homotopy and (co-)homology of $P(M)$ agrees with that of the stable pseudoisotopy space $$\mathscr{P}(M) = \text{colim}_n P(M \times [0,1]^n).$$ The homotopy type of $M$, being the space of points in $M$, and the homotopy type of the free loop space $\mathscr{L}M = Map(S^1, M)$, being the space of closed loops in $M$, both contribute to $\mathscr{P}(M)$, basically through maps $$\mathscr{P}(*) \times M \to \mathscr{P}(M)$$ and $$\mathscr{P}(S^1) \times \mathscr{L}M \to \mathscr{P}(M).$$ See the paper -https://projecteuclid.org/download/pdf_1/euclid.jdg/1214447541 -of Tom Farrell and Lowell Jones. There is a naturally defined involution on $P(M)$, and by the work of Allen Hatcher, Michael Weiss and Bruce Williams you can use it to largely recover $\text{Diff}(M)$ from $P(M)$. A more precise statement involves the block diffeomorphism group $\widetilde{\text{Diff}}(M)$, which is quite well understood by surgery theory. The survey "Automorphisms of manifolds" by Weiss and Williams might be a good source. By the stable parametrized $h$-cobordism theorem, written up by Friedhelm Waldhausen, Bjørn Jahren and myself, the spaces $\mathscr{P}(*)$ and $\mathscr{P}(S^1)$ are very close to Waldhausen's algebraic $K$-theory spaces $A(*)$ and $A(S^1)$, which agree with the algebraic $K$-theory spaces of the ring spectra $S$ and $S[\mathbb{Z}]$, respectively. I have some papers on $K(S)$, and Lars Hesselholt has more information about $K(S[\mathbb{Z}])$. I think this is one of the main reasons to be interested in the algebraic $K$-theory of ring spectra.<|endoftext|> -TITLE: Proportion of numbers with prime divisors from restricted set -QUESTION [7 upvotes]: Let $X$ be large, and let $\mathcal{P} \subset \{1, \dots, X\}$ be a set of primes. What is a good upper bound for -$$ -\sum_{\substack{1 \leq n \leq X,\\ p \nmid n \text{ for all }p \in \mathcal{P}}} 1. -$$ -From standard arguments in sieve theory (Brun's sieve, I think) one obtains that the sum in question is -$$ -\ll X \prod_{p \in \mathcal{P}} \left(1 - \frac{1}{p} \right). -$$ -However, if $\mathcal{P}$ is close to maximal (close to containing all primes in the range $\{1, \dots, X\}$), then this estimate seems to be far from being optimal. For example, when $\mathcal{P}$ contains all the primes in this range, then the sum equals 1, since not other number in $\{1, \dots, X\}$ is coprime to all primes in this range, while the upper bound above is around $X/\log X$. -That was a trivial example of course, but what happens when $\mathcal{P}$ contains "most" primes in the given range? Assume, for example, that $\mathcal{P}$ is so large that $\sum_{p \in \mathcal{P}} p^{-1} \geq (1-\varepsilon) \log \log X$, for fixed $\varepsilon$. Then what is a good upper bound for the sum above? (As noted, here $\varepsilon$ is fixed and "small", and $X \to \infty$.) - -REPLY [9 votes]: This question is addressed by Granville, Koukoulopoulos, and Matomaki (paper in Duke) and its sequel by Matomaki and Shao. These results show that for the sieve bound is of the right size, then ${\mathcal P}$ should omit a good number of large primes. In other words the situation where one does much better than the sieve bound is when all the large primes are in ${\mathcal P}$ -- think of the situation of smooth numbers. Conditions like $\sum_{p\in {\mathcal P}} 1/p \ge (1-\epsilon)\log \log X$ are too coarse in this problem.<|endoftext|> -TITLE: Question on condition for a sheaf to be locally free in Orlov 2004 -QUESTION [6 upvotes]: In "Triangulated Categories of Singularities and D-Branes in Landau-Ginzburg Models", Orlov twice mentions the following criterion for a sheaf $P_1$ to be locally free: -If for all closed points $t:x \hookrightarrow X$ we have $Ext^i(P_1, t_* \mathscr{O}_x)=0$ for all $i>0$, then $P_1$ is a locally free sheaf. -I cannot prove this nor find a reference. My only thought is to use adjunction to make this problem local; that is first work with $Ext^i(t^{-1}(P_1), \mathscr{O}_x)$, and then take a left resolution $P^{\cdot} \xrightarrow{\sim} P_1$. Next I'd try to use a spectral sequence such as $$ -E_1^{i, j} = -Ext_\mathcal{O}^j(P^{i}, \mathcal{O}_x) -\Rightarrow -Ext_\mathcal{O}^{i + j}(P^{\cdot}, \mathcal{O}_x)=Ext^{i+j}(P_1, \mathcal{O}_x). -$$ -but I'm not sure this spectral sequence is valid (I'm trying to use the 2nd spectral sequence on https://stacks.math.columbia.edu/tag/07A9 but derived in the first factor instead of the second; hopefully this introduces a change in sign). -Does anyone have a proof or a reference? - -REPLY [7 votes]: The question is local, so it is enough to show that if $A$ is a Noetherian local ring with maximal ideal $\mathfrak{m}$ and $M$ is a finitely generated module such that $Ext^i(M,A/\mathfrak{m}) = 0$ for $i > 0$ then $M$ is free. Let $n = \dim(M/\mathfrak{m}M)$ and let $f:A^{\oplus n} \to M$ be a homomorphism that induces an isomorphism modulo $\mathfrak{m}$. Then $f$ is surjective by Nakayama lemma and setting $N = Ker(f)$ we have an exact sequence -$$ -0 \to N \to A^{\oplus n} \to M \to 0. -$$ -Since -$$ -Hom(M,A/\mathfrak{m}) \cong Hom(M/\mathfrak{m}M,A/\mathfrak{m}) \cong (A/\mathfrak{m})^{\oplus n} \cong Hom(A^{\oplus n},A/\mathfrak{m}) -$$ -and $Ext^1(M,A/\mathfrak{m}) = 0$, it follows that $Hom(N,A/\mathfrak{m}) = 0$, hence $N/\mathfrak{m}N = 0$, hence by Nakayama lemma $N = 0$, hence $M$ is free.<|endoftext|> -TITLE: Why is the theorem of the base mostly cited only for smooth proper varieties -QUESTION [10 upvotes]: This is a very soft question, and I'm not sure what I expect as an answer. -In SGA6, Expose XIII, Theoreme 5.1 it is proven that, if $X$ is a proper scheme over a field $k$, then $NS(X)$ is finitely generated. Here $NS(X) := \mathrm{Pic}_{X/k}(k)/\mathrm{Pic}_{X/k}^0(k)$. -However, on wikipedia's page for the "theorem of the base" this theorem is only stated for smooth projective varieties. In an earlier question on MO the same happens; see Modern Proof of the Theorem of the Base. A quick google search gives a lot of papers citing this result from SGA6, but most authors impose smoothness when citing SGA6. Do note that Section 5.3 in these notes math.stanford.edu/~conrad/249CS15Page/handouts/abvarnotes.pdf states that $NS(X)$ is finitely generated for proper geometrically integral, not necessarily smooth, schemes. -What is the reason that the theorem of the base from SGA6 is mostly cited for smooth proper schemes? Does the Neron-Severi group exhibit pathological behaviour if $X$ is singular? - -REPLY [5 votes]: The comment by nfdc23 answers the question: -"There's no good reason, and in particular nothing pathological for the non-smooth case. Perhaps some paper working with smooth varieties stated the result in the relevant context and someone getting it from there copied the SGA6 reference without reading it and carried over the smoothness for safety, and then it spread due to people continuing to copy the citation from papers without reading the original reference. Near the end of 8.4 in the book Neron Models the result is stated in the same generality as in SGA6 (in particular, no parasitic smoothness hypotheses)."<|endoftext|> -TITLE: Does $\diamondsuit(\kappa)$ provably hold at Woodins or inaccessible Jónssons $\kappa$? -QUESTION [11 upvotes]: Usually the question whether the diamond principle $\diamondsuit(\kappa)$ holds for some large cardinal $\kappa$ only concerns large cardinal notions of very low consistency (among the weakly compacts). Partly since it does hold for all subtle cardinals, which are only barely stronger than the weakly compacts, and pretty much every large cardinal notion below a weakly compact has been shown to consistently not satisfy it (see Failure of diamond at large cardinals and Ben Neria ('17)). -That subtle cardinals satisfy diamond of course means that almost all large cardinals do satisfy it as well, but there are some strange ones lying around though, including Woodin cardinals and inaccessible Jónsson cardinals. Is anything known about diamond holding for any of these two? - -REPLY [2 votes]: Assuming that by "inaccessible Jónsson" you meant a regular limit cardinal of uncountable cofinality which is Jónsson; then using the arguments of [1] (Theorem 15 p.115), we have - -if $\mathbb{P}$ is c.c.c. and $\kappa$ is Jónsson then for any $V$-generic $G\subset \mathbb{P}$, $V[G]\vDash$ "$\kappa$ is Jónsson". - -In particular, if $\kappa$ is Jónsson, and $G\subset \mathbb{P}=\mathsf{Fn}(\kappa^{+}, 2)$ is $V$-generic, then - -$V[G] \vDash $ "$\kappa < 2^{\aleph_0}$ and $\kappa$ is Jónsson." - -hence $V[G] \vDash \neg \diamondsuit_\kappa$ and $\kappa$ is Jónsson. Moreover, If we started with $\kappa$ which was a regular limit cardinal then the same holds for $\kappa$ in $V[G]$. -[1] Devlin, Keith J., Some weak versions of large cardinal axioms, Ann. Math. Logic 5, 291-325 (1973). ZBL0279.02051.<|endoftext|> -TITLE: Why can’t you use cyclotomic polynomials to factor big numbers really quickly? -QUESTION [19 upvotes]: Two simple remarks: - -The polynomial $x^k-1$ can be factorised over the integers as a product of (irreducible) cyclotomic polynomials: $$x^k-1 = \prod_{d|k}\Phi_d(x).$$ -If we choose $k$ to be a number that has a lot of divisors, then $x^k-1$ will have a lot of factors. For example, if $k$ is a product of $b$ distinct primes then $x^k-1$ has $2^b$ factors. -Suppose we are given some largeish natural number $n$, and we want to factorise it. One way to find a factor of $n$ would be to compute the product $a$ of lots of different numbers $a_i$, modulo $n,$ and then compute $\gcd(n, a)$. If we are lucky and some of the $a_i$ are factors of $n$ but $a$ is not a multiple of $n$, then the gcd will be a non-trivial factor of $n$. - -Putting these together, one might imagine that a good way to find factors of $n$ would be to compute $\gcd(n, x^k-1\mathrm{\ mod\ } n)$ with $k$ a product of distinct primes and $x>1$. This is equivalent to testing $\Phi_d(x)$ for a common factor with $n$ for each of the exponentially-many divisors $d|k$. So it might naively be hoped that one could thereby factorise $n$ in time $O(\log n)$ or thereabouts. -Of course this does not actually work – one cannot thus factorise enormous numbers in the blink of an eye. I justify this claim on the non-mathematical grounds that a) if such a simple method worked then someone would have noticed by now; and the heuristic grounds that b) I tried it, and it didn’t. -What I would like to know is why it doesn’t work. Presumably the problem is that the $2^b$ values $\Phi_d(x)$ are distributed in a sufficiently non-uniform way to stymie the procedure. What’s going on here? -I can’t see much hint of this non-uniformity in examples that are small enough to compute all the $\Phi_d(x)$ explicitly in a reasonable amount of time. For example, if $n=61\times71=4331$ and $k=9699690$ is the product of the first eight primes, then the expression $\Phi_d(2)$ takes 244 different values as $d$ ranges over the $2^8=256$ divisors of $k$. (And as it happens, $\Phi_{35}(2)$ is a multiple of 71, and so computing $\gcd(4331, 2^{9699690}-1\mathrm{\ mod\ } 4331)$ reveals this factor.) - -Added: Thanks to David E Speyer for a clear concise answer. Just to make it explicit, the “non-uniformity” at play here is that each prime factor of $\Phi_d(x)$ is congruent to 1 modulo $d$. - -REPLY [6 votes]: Cyclotomic Factoring is a class of integer factoring algorithms. It is based on the Euler congruence $x^{\varphi(n)}-1 \equiv 0 \mod n$. There are a few well known special cases. - -The $p-1$-Pollard algorithm. -The $p+$-William algorithm. -Aurifeuillian Factoring, this is the oldest. -The time complexity is $O(n^{1/4})$ arithmetic operations. Many authors have tried to obtain $O(n^{1/5})$, but no major reduction in complexity have been achieved in the last few decades. -There is a large literature on this topic, a few are listed. -A. Granville, Aurifeuillian factors, Math Comp. Vol. 75, 2006. -E. Bach, Factoring with Cyclotomic Polynomials, Math. Comp. 1989.<|endoftext|> -TITLE: Regularized linear vs. RKHS-regression -QUESTION [8 upvotes]: I'm studying the difference between regularization in RKHS regression and linear regression, but I have a hard time grasping the crucial difference between the two. -Given input-output pairs $(x_i,y_i)$, I want to estimate a function $f(\cdot)$ as follows -\begin{equation}f(x)\approx u(x)=\sum_{i=1}^m \alpha_i K(x,x_i),\end{equation} -where $K(\cdot,\cdot)$ is a kernel function. The coefficients $\alpha_m$ can either be found by solving -\begin{equation} - {\displaystyle \min _{\alpha\in R^{n}}{\frac {1}{n}}\|Y-K\alpha\|_{R^{n}}^{2}+\lambda \alpha^{T}K\alpha},\end{equation} -where, with some abuse of notation, the $i,j$'th entry of the kernel matrix $K$ is ${\displaystyle K(x_{i},x_{j})} $. This gives -\begin{equation} -\alpha^*=(K+\lambda nI)^{-1}Y. -\end{equation} -Alternatively, we could treat the problem as a normal ridge regression/linear regression problem: -\begin{equation} - {\displaystyle \min _{\alpha\in R^{n}}{\frac {1}{n}}\|Y-K\alpha\|_{R^{n}}^{2}+\lambda \alpha^{T}\alpha},\end{equation} -with solution -\begin{equation} -{\alpha^*=(K^{T}K +\lambda nI)^{-1}K^{T}Y}. -\end{equation} -What would be the crucial difference between these two approaches and their solutions? - -REPLY [2 votes]: Both of the penalties can be thought of as arising from the linear regression setting in a Bayesian framework with predictor matrix $K$ and a Gaussian prior over the vector $\alpha$, centered at zero with prior variance $V$. -In the ridge regression case $V = n^{-1}\lambda^{-1}I$ and in the other case $V = n^{-1}\lambda^{-1}K^{-1}$ (as a kernel matrix $K$ is symmetric and PSD; I'm also assuming it is invertible). This follows just by equating terms; the posterior mean has the form $(K^tK + V^{-1})^{-1}K^tY$. Plugging in $V = n^{-1}\lambda^{-1}K^{-1}$ gives $$(K^tK + n\lambda K)^{-1}K^tY = (K^t + n\lambda I)^{-1}K^{-1}K^tY = (K + n\lambda I)^{-1}Y.$$ -Anyway, this is all just definitions, but the perspective might be intuition-boosting: the RKHS version stipulates explicitly that the prior over alpha has higher precision (more regularization) along directions of high variation as defined by the kernel function.<|endoftext|> -TITLE: A group of type F that is an extension of type F-by-type F -QUESTION [11 upvotes]: Let us first recall that a group of type $F$ is a group admitting a compact classifying space. -Let $K$ and $Q$ be groups of type $F$. Consider the family $\mathcal{G}(K, Q)$ consisting of groups $G$ of type $F$ for which there exists a short exact sequence -$$ -1 \to K \to G \to Q \to 1. -$$ -Is it true that all groups in $\mathcal{G}(K, Q)$ have the same cohomological dimension? -If not, please give me a counterexample. - -REPLY [6 votes]: Here is a partial answer. Theorem 5.5 of Homological Dimension of Discrete Groups by Bieri implies that the cohomological dimension of $G$ is the sum of the cohomological dimensions of $K$ and $Q$ provided $H^n(K,ZK)$ is free abelian where n is the cohomological dimension of $K$. He also only needs $K$ and $Q$ of type FP.<|endoftext|> -TITLE: A (reverse)-Minkowski type inequality for symmetric sums -QUESTION [10 upvotes]: Let $(u_1, u_2, u_3, u_4)$ and $(v_1, v_2, v_3, v_4)$ be vectors in $\mathbb R_+^4$. Is the following inequality true? -\begin{align*} -\left(\sum_{{[4] \choose 3}} \sqrt{u_i u_j u_k}\right)^{2/3} + \left(\sum_{{[4] \choose 3}} \sqrt{v_i v_j v_k}\right)^{2/3} \leq \left(\sum_{{[4] \choose 3}} \sqrt{(v_i+u_i) (v_j+u_j) (v_k+u_k)}\right)^{2/3} -\end{align*} -Here $\sum_{{[4] \choose 3}}$ refers to $\sum_{1\leq i < j < k \leq 4}$. -I ran a million Matlab simulations for random vectors and it did not yield any counterexample. -Note: Previously asked on MSE (https://math.stackexchange.com/questions/2647608/a-minkowski-like-inequality-for-symmetric-sums) - -REPLY [4 votes]: The said claim follows from the following general result on elementary symmetric polynomials, denoted $e_k$ below. -$\newcommand{\vx}{\mathbf{x}}\newcommand{\vy}{\mathbf{y}}$ - -Theorem A (S. 2018). $\,$ Let $p\in (0,1)$ and $x \in \mathbb{R}_+^n$. The map - \begin{equation*} -\phi_{k,n}(x) := x \mapsto \left[\frac{e_k(x_1^p,\ldots,x_n^p)}{e_{k-1}(x_1^p,\ldots,x_n^p)}\right]^{1/p}, -\end{equation*} - is concave. Recently, I typed up a proof to this inequality (and a few others). Please see this preprint for a proof. - -As a corollary, we obtain a concavity result that implies the OP's conjectured inequality as a special case (using positive homogeneity). - -Corollary. Let $p\in (0,1)$ and $\vx \in \mathbb{R}_+^n$. Then, $[e_k(\vx^p)]^{1/pk}$ is concave (we write $\vx^p \equiv (x_1^p,\ldots,x_n^p)$. - \begin{align*} - [e_k((\vx + \vy)^p)]^{1/pk} &= - \left[\frac{e_k((\vx+\vy)^p)}{e_{k-1}((\vx+\vy)^p)}\cdot \frac{e_{k-1}((\vx+\vy)^p)}{e_{k-2}((\vx+\vy)^p)}\cdots \frac{e_1((\vx+\vy)^p)}{e_0((\vx+\vy)^p)} \right]^{1/pk}\\ - &= \left[\phi_{k,n}(\vx+\vy)\phi_{k-1,n}(\vx+\vy)\cdots\phi_{1,n}(\vx+\vy)\right]^{1/k}\\ - &\ge \left[\left(\phi_{k,n}(\vx)+\phi_{k,n}(\vy)\right)\cdots\left(\phi_{1,n}(\vx)+\phi_{1,n}(\vy)\right)\right]^{1/k}\\ - &\ge \prod_{j=1}^k[\phi_{j,n}(\vx)]^{1/k} + \prod_{j=1}^k[\phi_{j,n}(\vy)]^{1/k}\\ - &= [e_k(\vx^p)]^{1/pk} + [e_k(\vy^p)]^{1/pk}, -\end{align*} - where the first inequality follows from Theorem A and the second is just Minkowski.<|endoftext|> -TITLE: Godbillon's proof of strong excision in de Rham cohomology with compact support -QUESTION [5 upvotes]: In Godbillon's book "Topologie Algébrique", chapter "de Rham cohomology with compact support", he states thm 3.1: - -Let $M$ be a smooth manifold, $N$ a closed (as subset) submanifold, the injections $i : M\setminus N \to M$ and $j : (M,\emptyset) \to (M,N)$, then the homomorphism $i_* : H_c(M\setminus N) \to H_c(M)$ (which is extension by $0$ of a form with compact support on $M\setminus N$) factors through - $H_c(M\setminus N) \xrightarrow{i'_*} H_c(M,N) \xrightarrow{j^*} H_c(M)$, i.e. $i_* = j^* \circ i'_*$. -Thm 3.1: $H_c(M\setminus N) \xrightarrow{i'_*} H_c(M,N)$ is an isomorphism. - -His proof uses a version (not the usual one) of the "Tubular Neighborough theorem" : $M$ and $N$ as before, there exists a closed submanifold $T$ of $M$ having the same dimension, such as 1) $N \subset \mathring{T}$, 2) there is a proper smooth retraction $r : T \to N$, 3) with the injection $k : N \to T$, $k \circ r$ is smoothly and properly homotopic to $Id_T$. -So here $k^* : H_c(T) \to H_c(N)$ is an isomorphism. -Then for the surjectivity of $i'_*$ he goes : let $\alpha \in Z_c(M,N)$. $\alpha$ being null on $N$ there is a form $\beta \in \Omega_c(T)$ such that $\alpha = d\,\beta$ (me : because $k^*$ is an isomorphism, but $\beta$ is not null on $N$ a priori). Let $\gamma \in \Omega_c(M)$ such that $\gamma = \beta$ on $T$ (me : because here, with $l : T \to M$ the injection, $l^* : H_c(M) \to H_c(T)$ is surjective). So $\alpha' = \alpha - d\;\gamma$ is a closed form with compact support, cohomolog to $\alpha$, null on $T$ (* so $Supp(\alpha') \subset M \setminus N$*) and $\alpha' \in Z_c(M\setminus N)$ completing the proof. -I am OK with its "notation shortcuts", but we're looking for a form in $Z_c(M\setminus N)$ whose image by $i'_*$ (which is extension by $0$) is cohomologous to $\alpha$ relative to $N$. He shows that $\alpha'$ is cohomologous to $\alpha$ in $\Omega_c(M)$, but not that it is cohomologous relative to $N$, i.e. in $\Omega_c(M,N)$, since there is no reason why $\beta$ or $\gamma$ should be null on $N$. -Is Godbillon's proof flawed, or am I missing something obvious ? - -REPLY [2 votes]: Answering my own question. I do think that Godbillon's argument is incomplete. In fact it is possible to find a form $\beta \in \Omega_c(T)$ such that $d\,\beta = \alpha \mathclose{}|\mathopen{}_T$ and which is null on $N$. So $\gamma$ is also null on $N$, and $\alpha'$ is cohomologous to $\alpha$ in $\Omega_c(M,N)$. -The missing part comes from the following theorem : if $M$ is a manifold, $P$ a closed submanifold, $i : P \hookrightarrow M$, and if there is a strong proper deformation retract $r : M \to P$, then if $\alpha \in \Omega_c(M,N)$, it has a primitive $\beta \in \Omega_c(M,N)$ i.e. which is also null on $N$. -The proof results from the usual proof of homotopy invariance of de Rham cohomology, where it is shown that for $f,g : M \to N$ ($N$ another manifold) being homotopic with $H : M \times I \to N$, we have for $\omega \in \Omega^k(N)$ : -$$(g^* - f^*) \omega = d\,L^k(\omega) + L^{k+1}(d\,\omega)$$ with -$$L^k : \Omega^k(N) \to \Omega^{k-1}(M) \quad \omega \mapsto \int_0^1 J_t^*(\iota_{\partial_t}H^* \omega) dt$$ -Next it is easily shown that if $H$ is a proper homotopy, then -$$ L^k(\Omega_c^k(N)) \subset \Omega_c^{k-1}(M)$$ -With the data of the question, setting $M \equiv T$, $P \equiv N$, $f \equiv i \circ r$ and $g \equiv Id_T$, $i: N \hookrightarrow T$, we get for $\alpha\mathclose{}|\mathopen{}_T \in Z^k_c(T,N)$ i.e. $i^* \alpha\mathclose{}|\mathopen{}_T = 0$ -$$\alpha\mathclose{}|\mathopen{}_T = d\,\beta, \quad \beta = L^k(\alpha\mathclose{}|\mathopen{}_T)$$ -and the formula for $L^k$ shows that $\beta$ is also null on $N$.<|endoftext|> -TITLE: Subgroup of hyperbolic group generated by non-torsion elements -QUESTION [14 upvotes]: Let $G$ be a hyperbolic group. I know that it is an open problem whether $G$ has a torsion-free subgroup of finite index. But if we let $N$ be the subgroup of $G$ generated by its non-torsion elements, then is $G/N$ necessarily finite? -If not, $G/N$ would be a finitely generated infinite torsion group with finitely many conjugacy classes. - -REPLY [10 votes]: Here is how I would prove it using ping-pong. The relevant form of ping-pong is the following proposition: -Proposition. Let $X$ be a set and $G_1, G_2$ be groups acting on $X$ such that there exist two subsets $X_1, X_2\subset X$ with disjoint complements and nonempty intersection, such that -$$ -\forall g_i\in G_i -\{1\}, g_i(X_i)\cap X_i=\emptyset. -$$ -Then the group $\langle G_1, G_2\rangle$ of bijections of $X$ generated by $G_1, G_2$ is naturally isomorphic to $G_1\star G_2$. -Now, back to our hyperbolic group $G$. I will assume that $G$ is a non elementary hyperbolic group whose finite radical $W(G)$ is trivial. (Otherwise, see Yves' answer.) Given $g\in G$, let $Fix(g)$ denote the fixed-point set of $g$ in $\partial G$ (the Gromov boundary of $G$). -The following lemma answers Yves' question: -Lemma 1. If $g\in G-\{1\}$ has finite order then $Fix(g)$ has empty interior. -Proof. Suppose not. Then, since $W(G)=1$, $Fix(g)\ne \partial G$. -Recall: -Theorem. The set of pairs $(p_+, p_-)$ of attractive/repulsive fixed points of infinite order elements of $G$ is dense in $\partial G\times \partial G$. -Hence, one can find such fixed pair for an element $h\in G$ such that $p_+\notin Fix(g), p_-\in Fix(g)$. Then the sequence of conjugates -$$ -g_n=h^n g h^{-n}, n\in {\mathbb N}, -$$ -contains infinitely many distinct elements (since the sequence of complements to $Fix(g_n)$ in $\partial G$ converges to the singleton $p_+$). At the same time, the sequence $(h^{-n})_{n\in {\mathbb N}}$ is uniformly close to the quasiconvex hull of $Fix(g)$. Since $g$ acts on this quasiconvex hull with bounded displacement, we obtain that the sequence of word norms $|g_n|$ is bounded. A contradiction. qed -Lemma 2. Let $g\in G$ be nontrivial, generating a finite cyclic group $C< G$. Let $h\in G$ be an infinite order element such that -$$ -g^k(Fix(h))\cap Fix(h)=\emptyset -$$ -for all $g^k\in C-\{1\}$. Then there exists $n$ such that the subgroup $\langle g, h^n\rangle$ of $G$ generated by $g$ and $h^n$ is naturally isomorphic to $C \star \langle h^n \rangle$. -Proof. Let $U$ be a small neighborhood of $Fix(h)$ in $\partial G$ such that -$$ -g^k(U)\cap U=\emptyset -$$ -for all $g^k\in C-\{1\}$ (such a neighborhood existents since $C$ is finite). Let $V= \partial G -U$. Then there exists $n\in {\mathbb N}$ such that for all nontrivial elements $f\in \langle h^n\rangle$ we have -$$ -f(V)\cap V=\emptyset. -$$ -It now follows from Tits' ping-pong that the subgroup $\langle g, h^n\rangle$ is naturally isomorphic to $C \star \langle h^n\rangle $. qed -Lemma 3. Given a nontrivial finite order element $g\in G$, there exists an infinite order element $h\in G$ such that $$ -g^k(Fix(h))\cap Fix(h)=\emptyset -$$ -for all $g^k\in \langle g \rangle -\{1\}$. -Proof. Let $U$ be a nonempty open subset of $\partial G -U$. Since $C=\langle g\rangle $ is finite, we can choose $U$ such that -for all nontrivial elements $f\in C$ we have -$$ -f(U)\cap U=\emptyset. -$$ -(For instance, one can use Lemma 1 here.) -On the other hand, by minimality of the action of $G$ on $\partial G$, -there exists an infinite order element $h\in G$ such that $Fix(h)\subset U$. (This also follows from the Theorem stated above.) qed -Now, Lemmas 2 and 3 imply: -Corollary. Let $g\in G$ be a nontrivial finite order element. -Then there exists an infinite order element $h\in G$ such that the subgroup $\langle g, h\rangle$ of $G$ is naturally isomorphic to $\langle g\rangle \star \langle h\rangle$. In particular, the product $gh$ also has infinite order.<|endoftext|> -TITLE: Supremum norm of certain quantity -QUESTION [5 upvotes]: Is there any easy way of finding supremum of the quantity $$\sum_{i,j=1}^n|z_i-z_j|,$$ where $|z_i|=1$ for $1\leq i\leq n$ ? We are considering complex variables of course. - -REPLY [7 votes]: The following argument appears on p.156 of - -L. Fejes Toth, Regular figures, A Pergamon Press Book, The Macmillan - Co., New York, 1964. - -Assume $z_{1},\ldots,z_{n}$ are ordered on the circle and let $S=\sum_{1\leq i -TITLE: Vanishing natural transformation exact triangle -QUESTION [10 upvotes]: This question is a follow-up to this question I asked some time ago. Let $X$ be a smooth projective variety of dimension $n$ over $\mathbb{C}$. Let $\omega \in H^{n}(X,K_X)$, $\omega \neq 0$. Let -$$A \longrightarrow B \longrightarrow C \longrightarrow A[1]$$ -be an exact triangle in $D^b(X)$. Assume that the morphism: -$$\mathrm{id}_A \otimes \omega : A \longrightarrow A \otimes K_X[n]$$ -vanishes in $\mathrm{Hom}(A,A\otimes K_X[n])$ and that the morphism: -$$\mathrm{id}_B \otimes \omega : B \longrightarrow B \otimes K_X [n]$$ -vanishes in $\mathrm{Hom}(B,B\otimes K_X[n])$. I'd like to know if the morphism: -$$\mathrm{id}_C \otimes \omega : C \longrightarrow C \otimes K_X[n]$$ -vanishes in $\mathrm{Hom}(C,C\otimes K_X[n])$? -I know that there is no five lemma for zero-morphisms and that if this statement is true it is not just a formal consequence of the definition of a triangulated category (see the answer to the question I am referring to in the intro). -On the other hand, seen the very special nature of the involved morphisms, I tought it might be true. Or perhaps this is plain wrong and a counter-example would be greatly appreciated. -Thanks in advance! - -REPLY [4 votes]: So I think I have an argument which shows that if $\mathrm{id}_{A} \otimes \omega : A \longrightarrow A \otimes K_X[n]$ is zero and $\mathrm{id}_{B} \otimes \omega : B \longrightarrow B \otimes K_X[n]$ is zero then $\mathrm{id}_{C} \otimes \omega : C \longrightarrow C \otimes K_X[n]$ is not necessarily zero. This argument uses crucially the result of Thomason that was brought to my attention by Jeremy Rickard (who I heartily thank for that). -Let $X$ be a K3 surface and let $\mathcal{T}$ be the cone of the natural map: -$$\mathcal{O}_X \longrightarrow \mathcal{O}_X[2]$$ -induced by $\omega$. The object $\mathcal{T}$ has rank $0$ in $D^b(X)$ and one easily computes that $\mathrm{End}(\mathcal{T}, \mathcal{T}) = \mathbb{C}$. Assume that the morphism $\mathrm{id}_{\mathcal{T}} \otimes \omega : \mathcal{T} \longrightarrow \mathcal{T}[2]$ is not zero. Then it is a generator of $\mathrm{Ext}^2(\mathcal{T},\mathcal{T}) = \mathrm{End}(\mathcal{T})^* = \mathbb{C}$. This is impossible. Indeed, we have $$\mathrm{Trace}(\mathrm{id}_{\mathcal{T}} \otimes \omega) = \mathrm{rank}(\mathcal{T}).\omega = 0$$ but Serre duality implies that: -$$\mathrm{Trace} : \mathrm{Ext}^2(\mathcal{T}, \mathcal{T}) \longrightarrow H^2(X,\mathcal{O}_X)$$ -is surjective. This shows that $\mathrm{id}_{\mathcal{T}} \otimes \omega = 0$ in $\mathrm{Ext}^{2}(\mathcal{T}, \mathcal{T})$. -Now, assume that for any exact triangle $A \longrightarrow B \longrightarrow C$ in $D^b(X)$ such that $\mathrm{id}_{A} \otimes \omega : A \longrightarrow A[2]$ is zero and $\mathrm{id}_{B} \otimes \omega : B \longrightarrow B[2]$ is zero, we have $\mathrm{id}_{C} \otimes \omega : C \longrightarrow C[2]$ is also zero. (I denote this hypothesis by $(*)$). -Then, the subcategory of $D^b(X)$ consisting of objects $S$ such $\mathrm{id}_{S} \otimes \omega$ is zero would be a thick $\otimes$-triangulated subcategory of $D^b(X)$ (see the comment of Jeremy Rickard to the question). -It has been checked that $\mathcal{T}$ is in this category and the cohomological support of $\mathcal{T}$ is $X$. Hence, by Thomason classification of thick $\otimes$-triangulated subcategories of $D^b(X)$, we deduce that for any $S \in D^b(X)$ such that the cohomological support of $S$ is $X$, we have the vanishing of $\mathrm{id}_{S} \otimes \omega : S \longrightarrow S \otimes[2]$. This is clearly absurd and hypothesis $(*)$ is incorrect.<|endoftext|> -TITLE: Checking positive semi-definiteness of integer matrix -QUESTION [7 upvotes]: Key Problem : Is there any theorem about eigenvalues or positive semi-definiteness of small size matrices with small integer elements? - - -I have to check positive semi-definiteness of many symmetric matrices with integer elements. First I used eigenvalues, but floating point round error happens : eig(in numpy) sometimes gives small negative eigenvalues when a matrix is actually positive semi-definite. -I know Sylvester's criterion for positive semi-definiteness can avoid this problem. But I really don't want to use it since it requires computation of determinant of all principal minors, and I have to deal with really many matrices($ > 10^{20}$). I have to do anything to reduce number of calculations. - -All the elements are integer and have small absolute values($\ <3 $), sizes of matrices are also small($\ < 10 \times 10 $). It seems pretty good condition, so I believe someone already researched this kind of matrices. Does anyone know useful theorems for this situation? -P.S: I'm pretty newbie in English Internet community and not native English speaker. So if you find something awkward, pardon me and let me know. - -REPLY [5 votes]: For small symmetric matrices, you could look at the characteristic polynomial. -The real symmetric matrix $A$ is positive semidefinite iff the coefficients of the characteristic polynomial are alternating in sign. For $n \times n$ matrices this gives you $n$ integer expressions to check. - -REPLY [2 votes]: I wrote a program that takes a square integer matrix $H$ and produces square rational $P$ such that $P^T H P = D$ is diagonal and rational. In case it matters, $\det P = \pm 1.$ The program outputs in Latex. By Sylvester's Law of Inertia, $H$ is positive definite if and only if $D$ is positive definite, and there is no approximation involved. It is really just repeated completing the square made up into a reverse direction algorithm by parties unknown. The algorithm is given in detail at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr -The way I like to write this, we introduce one new "elementary" matrix at a time and keep track of various things. In the most fortunate outcome, only one type of matrix is used and $P$ is upper triangular, but this does not always happen. -Oh, this works fine with semi-definite matrices. If the diagonal $D$ has some positive entries and some (diagonal) zero entries, then $H$ is positive semi-definite. No guesswork. -Here is the input -~jagy@phobeusjunior:~$ ./matrix_congruence 5 -input row number 1 here 0 1 2 3 4 -0 1 2 3 4 -input row number 2 here 1 5 6 7 8 -1 5 6 7 8 -input row number 3 here 2 6 9 10 11 -2 6 9 10 11 -input row number 4 here 3 7 10 12 13 -3 7 10 12 13 -input row number 5 here 4 8 11 13 14 -4 8 11 13 14 - -$$ P^T H P = D $$ -$$\left( -\begin{array}{rrrrr} -0 & 1 & 0 & 0 & 0 \\ -1 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -4 & - 2 & 1 & 0 & 0 \\ - \frac{ 8 }{ 5 } & \frac{ 1 }{ 5 } & - \frac{ 8 }{ 5 } & 1 & 0 \\ - \frac{ 8 }{ 11 } & \frac{ 1 }{ 11 } & \frac{ 3 }{ 11 } & - \frac{ 17 }{ 11 } & 1 \\ -\end{array} -\right) -\left( -\begin{array}{rrrrr} -0 & 1 & 2 & 3 & 4 \\ -1 & 5 & 6 & 7 & 8 \\ -2 & 6 & 9 & 10 & 11 \\ -3 & 7 & 10 & 12 & 13 \\ -4 & 8 & 11 & 13 & 14 \\ -\end{array} -\right) -\left( -\begin{array}{rrrrr} -0 & 1 & 4 & \frac{ 8 }{ 5 } & \frac{ 8 }{ 11 } \\ -1 & - \frac{ 1 }{ 5 } & - 2 & \frac{ 1 }{ 5 } & \frac{ 1 }{ 11 } \\ -0 & 0 & 1 & - \frac{ 8 }{ 5 } & \frac{ 3 }{ 11 } \\ -0 & 0 & 0 & 1 & - \frac{ 17 }{ 11 } \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) - = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -0 & 0 & 5 & 0 & 0 \\ -0 & 0 & 0 & \frac{ 11 }{ 5 } & 0 \\ -0 & 0 & 0 & 0 & \frac{ 6 }{ 11 } \\ -\end{array} -\right) -$$ -$$ D_0 = H $$ -$$ E_j^T D_{j-1} E_j = D_j $$ -$$ P_{j-1} E_j = P_j $$ -$$ E_j^{-1} Q_{j-1} = Q_j $$ -$$ P_j Q_j = I $$ -$$ P_j^T H P_j = D_j $$ -$$ Q_j^T D_j Q_j = H $$ -$$ H = \left( -\begin{array}{rrrrr} -0 & 1 & 2 & 3 & 4 \\ -1 & 5 & 6 & 7 & 8 \\ -2 & 6 & 9 & 10 & 11 \\ -3 & 7 & 10 & 12 & 13 \\ -4 & 8 & 11 & 13 & 14 \\ -\end{array} -\right) -$$ -============================================== -$$ E_{1} = \left( -\begin{array}{rrrrr} -0 & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{1} = \left( -\begin{array}{rrrrr} -0 & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{1} = \left( -\begin{array}{rrrrr} -0 & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{1} = \left( -\begin{array}{rrrrr} -5 & 1 & 6 & 7 & 8 \\ -1 & 0 & 2 & 3 & 4 \\ -6 & 2 & 9 & 10 & 11 \\ -7 & 3 & 10 & 12 & 13 \\ -8 & 4 & 11 & 13 & 14 \\ -\end{array} -\right) -$$ -============================================== -$$ E_{2} = \left( -\begin{array}{rrrrr} -1 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -0 & 1 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{2} = \left( -\begin{array}{rrrrr} -0 & 1 & 0 & 0 & 0 \\ -1 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{2} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{2} = \left( -\begin{array}{rrrrr} -5 & 0 & 6 & 7 & 8 \\ -0 & - \frac{ 1 }{ 5 } & \frac{ 4 }{ 5 } & \frac{ 8 }{ 5 } & \frac{ 12 }{ 5 } \\ -6 & \frac{ 4 }{ 5 } & 9 & 10 & 11 \\ -7 & \frac{ 8 }{ 5 } & 10 & 12 & 13 \\ -8 & \frac{ 12 }{ 5 } & 11 & 13 & 14 \\ -\end{array} -\right) -$$ -============================================== -$$ E_{3} = \left( -\begin{array}{rrrrr} -1 & 0 & - \frac{ 6 }{ 5 } & 0 & 0 \\ -0 & 1 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{3} = \left( -\begin{array}{rrrrr} -0 & 1 & 0 & 0 & 0 \\ -1 & - \frac{ 1 }{ 5 } & - \frac{ 6 }{ 5 } & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{3} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{3} = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 7 & 8 \\ -0 & - \frac{ 1 }{ 5 } & \frac{ 4 }{ 5 } & \frac{ 8 }{ 5 } & \frac{ 12 }{ 5 } \\ -0 & \frac{ 4 }{ 5 } & \frac{ 9 }{ 5 } & \frac{ 8 }{ 5 } & \frac{ 7 }{ 5 } \\ -7 & \frac{ 8 }{ 5 } & \frac{ 8 }{ 5 } & 12 & 13 \\ -8 & \frac{ 12 }{ 5 } & \frac{ 7 }{ 5 } & 13 & 14 \\ -\end{array} -\right) -$$ -============================================== -$$ E_{4} = \left( -\begin{array}{rrrrr} -1 & 0 & 0 & - \frac{ 7 }{ 5 } & 0 \\ -0 & 1 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{4} = \left( -\begin{array}{rrrrr} -0 & 1 & 0 & 0 & 0 \\ -1 & - \frac{ 1 }{ 5 } & - \frac{ 6 }{ 5 } & - \frac{ 7 }{ 5 } & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{4} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & \frac{ 7 }{ 5 } & 0 \\ -1 & 0 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{4} = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 8 \\ -0 & - \frac{ 1 }{ 5 } & \frac{ 4 }{ 5 } & \frac{ 8 }{ 5 } & \frac{ 12 }{ 5 } \\ -0 & \frac{ 4 }{ 5 } & \frac{ 9 }{ 5 } & \frac{ 8 }{ 5 } & \frac{ 7 }{ 5 } \\ -0 & \frac{ 8 }{ 5 } & \frac{ 8 }{ 5 } & \frac{ 11 }{ 5 } & \frac{ 9 }{ 5 } \\ -8 & \frac{ 12 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 9 }{ 5 } & 14 \\ -\end{array} -\right) -$$ -============================================== -$$ E_{5} = \left( -\begin{array}{rrrrr} -1 & 0 & 0 & 0 & - \frac{ 8 }{ 5 } \\ -0 & 1 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{5} = \left( -\begin{array}{rrrrr} -0 & 1 & 0 & 0 & 0 \\ -1 & - \frac{ 1 }{ 5 } & - \frac{ 6 }{ 5 } & - \frac{ 7 }{ 5 } & - \frac{ 8 }{ 5 } \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{5} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 8 }{ 5 } \\ -1 & 0 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{5} = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & \frac{ 4 }{ 5 } & \frac{ 8 }{ 5 } & \frac{ 12 }{ 5 } \\ -0 & \frac{ 4 }{ 5 } & \frac{ 9 }{ 5 } & \frac{ 8 }{ 5 } & \frac{ 7 }{ 5 } \\ -0 & \frac{ 8 }{ 5 } & \frac{ 8 }{ 5 } & \frac{ 11 }{ 5 } & \frac{ 9 }{ 5 } \\ -0 & \frac{ 12 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 9 }{ 5 } & \frac{ 6 }{ 5 } \\ -\end{array} -\right) -$$ -============================================== -$$ E_{6} = \left( -\begin{array}{rrrrr} -1 & 0 & 0 & 0 & 0 \\ -0 & 1 & 4 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{6} = \left( -\begin{array}{rrrrr} -0 & 1 & 4 & 0 & 0 \\ -1 & - \frac{ 1 }{ 5 } & - 2 & - \frac{ 7 }{ 5 } & - \frac{ 8 }{ 5 } \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{6} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 8 }{ 5 } \\ -1 & 0 & - 4 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{6} = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & 0 & \frac{ 8 }{ 5 } & \frac{ 12 }{ 5 } \\ -0 & 0 & 5 & 8 & 11 \\ -0 & \frac{ 8 }{ 5 } & 8 & \frac{ 11 }{ 5 } & \frac{ 9 }{ 5 } \\ -0 & \frac{ 12 }{ 5 } & 11 & \frac{ 9 }{ 5 } & \frac{ 6 }{ 5 } \\ -\end{array} -\right) -$$ -============================================== -$$ E_{7} = \left( -\begin{array}{rrrrr} -1 & 0 & 0 & 0 & 0 \\ -0 & 1 & 0 & 8 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{7} = \left( -\begin{array}{rrrrr} -0 & 1 & 4 & 8 & 0 \\ -1 & - \frac{ 1 }{ 5 } & - 2 & - 3 & - \frac{ 8 }{ 5 } \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{7} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 8 }{ 5 } \\ -1 & 0 & - 4 & - 8 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{7} = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & 0 & 0 & \frac{ 12 }{ 5 } \\ -0 & 0 & 5 & 8 & 11 \\ -0 & 0 & 8 & 15 & 21 \\ -0 & \frac{ 12 }{ 5 } & 11 & 21 & \frac{ 6 }{ 5 } \\ -\end{array} -\right) -$$ -============================================== -$$ E_{8} = \left( -\begin{array}{rrrrr} -1 & 0 & 0 & 0 & 0 \\ -0 & 1 & 0 & 0 & 12 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{8} = \left( -\begin{array}{rrrrr} -0 & 1 & 4 & 8 & 12 \\ -1 & - \frac{ 1 }{ 5 } & - 2 & - 3 & - 4 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{8} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 8 }{ 5 } \\ -1 & 0 & - 4 & - 8 & - 12 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{8} = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -0 & 0 & 5 & 8 & 11 \\ -0 & 0 & 8 & 15 & 21 \\ -0 & 0 & 11 & 21 & 30 \\ -\end{array} -\right) -$$ -============================================== -$$ E_{9} = \left( -\begin{array}{rrrrr} -1 & 0 & 0 & 0 & 0 \\ -0 & 1 & 0 & 0 & 0 \\ -0 & 0 & 1 & - \frac{ 8 }{ 5 } & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{9} = \left( -\begin{array}{rrrrr} -0 & 1 & 4 & \frac{ 8 }{ 5 } & 12 \\ -1 & - \frac{ 1 }{ 5 } & - 2 & \frac{ 1 }{ 5 } & - 4 \\ -0 & 0 & 1 & - \frac{ 8 }{ 5 } & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{9} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 8 }{ 5 } \\ -1 & 0 & - 4 & - 8 & - 12 \\ -0 & 0 & 1 & \frac{ 8 }{ 5 } & 0 \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{9} = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -0 & 0 & 5 & 0 & 11 \\ -0 & 0 & 0 & \frac{ 11 }{ 5 } & \frac{ 17 }{ 5 } \\ -0 & 0 & 11 & \frac{ 17 }{ 5 } & 30 \\ -\end{array} -\right) -$$ -============================================== -$$ E_{10} = \left( -\begin{array}{rrrrr} -1 & 0 & 0 & 0 & 0 \\ -0 & 1 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & - \frac{ 11 }{ 5 } \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{10} = \left( -\begin{array}{rrrrr} -0 & 1 & 4 & \frac{ 8 }{ 5 } & \frac{ 16 }{ 5 } \\ -1 & - \frac{ 1 }{ 5 } & - 2 & \frac{ 1 }{ 5 } & \frac{ 2 }{ 5 } \\ -0 & 0 & 1 & - \frac{ 8 }{ 5 } & - \frac{ 11 }{ 5 } \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{10} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 8 }{ 5 } \\ -1 & 0 & - 4 & - 8 & - 12 \\ -0 & 0 & 1 & \frac{ 8 }{ 5 } & \frac{ 11 }{ 5 } \\ -0 & 0 & 0 & 1 & 0 \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{10} = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -0 & 0 & 5 & 0 & 0 \\ -0 & 0 & 0 & \frac{ 11 }{ 5 } & \frac{ 17 }{ 5 } \\ -0 & 0 & 0 & \frac{ 17 }{ 5 } & \frac{ 29 }{ 5 } \\ -\end{array} -\right) -$$ -============================================== -$$ E_{11} = \left( -\begin{array}{rrrrr} -1 & 0 & 0 & 0 & 0 \\ -0 & 1 & 0 & 0 & 0 \\ -0 & 0 & 1 & 0 & 0 \\ -0 & 0 & 0 & 1 & - \frac{ 17 }{ 11 } \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -$$ -$$ P_{11} = \left( -\begin{array}{rrrrr} -0 & 1 & 4 & \frac{ 8 }{ 5 } & \frac{ 8 }{ 11 } \\ -1 & - \frac{ 1 }{ 5 } & - 2 & \frac{ 1 }{ 5 } & \frac{ 1 }{ 11 } \\ -0 & 0 & 1 & - \frac{ 8 }{ 5 } & \frac{ 3 }{ 11 } \\ -0 & 0 & 0 & 1 & - \frac{ 17 }{ 11 } \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; Q_{11} = \left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 8 }{ 5 } \\ -1 & 0 & - 4 & - 8 & - 12 \\ -0 & 0 & 1 & \frac{ 8 }{ 5 } & \frac{ 11 }{ 5 } \\ -0 & 0 & 0 & 1 & \frac{ 17 }{ 11 } \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) -, \; \; \; D_{11} = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -0 & 0 & 5 & 0 & 0 \\ -0 & 0 & 0 & \frac{ 11 }{ 5 } & 0 \\ -0 & 0 & 0 & 0 & \frac{ 6 }{ 11 } \\ -\end{array} -\right) -$$ -============================================== -$$ P^T H P = D $$ -$$\left( -\begin{array}{rrrrr} -0 & 1 & 0 & 0 & 0 \\ -1 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -4 & - 2 & 1 & 0 & 0 \\ - \frac{ 8 }{ 5 } & \frac{ 1 }{ 5 } & - \frac{ 8 }{ 5 } & 1 & 0 \\ - \frac{ 8 }{ 11 } & \frac{ 1 }{ 11 } & \frac{ 3 }{ 11 } & - \frac{ 17 }{ 11 } & 1 \\ -\end{array} -\right) -\left( -\begin{array}{rrrrr} -0 & 1 & 2 & 3 & 4 \\ -1 & 5 & 6 & 7 & 8 \\ -2 & 6 & 9 & 10 & 11 \\ -3 & 7 & 10 & 12 & 13 \\ -4 & 8 & 11 & 13 & 14 \\ -\end{array} -\right) -\left( -\begin{array}{rrrrr} -0 & 1 & 4 & \frac{ 8 }{ 5 } & \frac{ 8 }{ 11 } \\ -1 & - \frac{ 1 }{ 5 } & - 2 & \frac{ 1 }{ 5 } & \frac{ 1 }{ 11 } \\ -0 & 0 & 1 & - \frac{ 8 }{ 5 } & \frac{ 3 }{ 11 } \\ -0 & 0 & 0 & 1 & - \frac{ 17 }{ 11 } \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) - = \left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -0 & 0 & 5 & 0 & 0 \\ -0 & 0 & 0 & \frac{ 11 }{ 5 } & 0 \\ -0 & 0 & 0 & 0 & \frac{ 6 }{ 11 } \\ -\end{array} -\right) -$$ -$$ Q^T D Q = H $$ -$$\left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ - \frac{ 6 }{ 5 } & - 4 & 1 & 0 & 0 \\ - \frac{ 7 }{ 5 } & - 8 & \frac{ 8 }{ 5 } & 1 & 0 \\ - \frac{ 8 }{ 5 } & - 12 & \frac{ 11 }{ 5 } & \frac{ 17 }{ 11 } & 1 \\ -\end{array} -\right) -\left( -\begin{array}{rrrrr} -5 & 0 & 0 & 0 & 0 \\ -0 & - \frac{ 1 }{ 5 } & 0 & 0 & 0 \\ -0 & 0 & 5 & 0 & 0 \\ -0 & 0 & 0 & \frac{ 11 }{ 5 } & 0 \\ -0 & 0 & 0 & 0 & \frac{ 6 }{ 11 } \\ -\end{array} -\right) -\left( -\begin{array}{rrrrr} - \frac{ 1 }{ 5 } & 1 & \frac{ 6 }{ 5 } & \frac{ 7 }{ 5 } & \frac{ 8 }{ 5 } \\ -1 & 0 & - 4 & - 8 & - 12 \\ -0 & 0 & 1 & \frac{ 8 }{ 5 } & \frac{ 11 }{ 5 } \\ -0 & 0 & 0 & 1 & \frac{ 17 }{ 11 } \\ -0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right) - = \left( -\begin{array}{rrrrr} -0 & 1 & 2 & 3 & 4 \\ -1 & 5 & 6 & 7 & 8 \\ -2 & 6 & 9 & 10 & 11 \\ -3 & 7 & 10 & 12 & 13 \\ -4 & 8 & 11 & 13 & 14 \\ -\end{array} -\right) -$$<|endoftext|> -TITLE: Homotopy equivalence between two basepoints of the etale homotopy type of the one-torus -QUESTION [6 upvotes]: Let $T = \mathbb{G}_m$ be the torus, and let $\tilde{T}$ be its étale universal cover (a pro-object in schemes of finite type). Then both $T$ and $\tilde{T}$ have a well-defined étale homotopy type. Explicitly, the homotopy types are $ét(T) = B\hat{\mathbb{Z}}$ and $ét(\tilde{T}) = *,$ for $*$ the point. In particular, the natural covering map $$\pi:\tilde{T}\to T$$ gives a basepoint (in a suitable homotopy sense) $$ét(\pi)$$ of $ét(T)$. Now group structure on $T$ lets us define a new point $\pi^2 := \pi*\pi,$ which is the composition of $\pi$ with the squaring map $[2]:T\to T$. While both $\pi, \pi^2:\tilde{T}\to T$ realize $\tilde{T}$ as a universal cover of $T$, they are not equal (as can be seen e.g. by looking at the map on tangent spaces at $1$). On the other hand, since $B\hat{\mathbb{Z}}$ is (I think?) connected, the maps $$ét(\pi), ét(\pi^2):ét(\tilde{T})\to ét(T)$$ should be homotopy equivalent in a suitable homotopy category. -Question is there a way to see the homotopy equivalence between $ét(\pi)$ and $ét(\pi^2)$ explicitly? Here by "explicitly", I mean as an interval in the mapping space between natural topological models, or an interval in some other model category. Edit: I'd also like for the functor from varieties to etale types to take etale maps to fibrations, so that in particular the etale types of $\pi, \pi^2$ are not a priori equal. - -REPLY [2 votes]: There is a completely explicit way to do it in this case I think. The Etale homotopy type of $T$ is represented by the pro-object $\{B \mathbb{Z}/n\}_{n \in \mathbb{N}^\times}$, ignoring for a moment the Galois action and assumeing that the field is of characteristic $0$ for simplicity. The object -$\tilde{T}$ is very similar in nature. It is represented by the pro-object -$(T,n)_{n \in \mathbb{N}^\times}$ with structure maps -$(T,nm) \stackrel{[m]}{\to}(T,n)$. The map $\pi: \tilde{T} \to T$ is given by -$(T,n) \stackrel{[n]}{\to} T$. Here $[n]$ is rising to the $n$-th power. -The map $\pi * \pi$ is given by $(T,n) \stackrel{[2n]}{\to} T$. -We want to show that the two maps are homotopic. For this we need to pull back the pro-cover $(T \stackrel{[n]}{\to} T)_{n \in \mathbb{N}^\times}$ along the two maps $\pi$ and $\pi * \pi$. For both you just get the same pro-cover over every copy $(T,n)$ because the pull-back $[n]$ along $[m]$ is just $[n]$. -After passing to the corresponding simplicial sets, we end up with the following problem in topology: -Consider the pro-space indexed by $(\mathbb{N}^\times)^2$ given by -$X_{m,n} = B\mathbb{Z}/n$, with structure maps given by the reduction maps -$\mathbb{Z}/nk \to \mathbb{Z}/n$ in the second coordinate and by -the multiplication my $k$ maps $\mathbb{Z}/n \to \mathbb{Z}/n$ in the first coordinate. We need to show that the map $X_{n,m} \to X_{n,m}$ given by multiplication by $2$ is homotopic to the identity, in an explicit way. -Now, you can note that the maps in this story are all naturally based. So the connectivity of the target is somehow irrelevant, we only need to know explicitly how to contract the source. But there is a null-homotopy -$(X_{n,m}) \to (X_{n,m})$ given by the map -$(X_{n,m} \times \Delta^1) \to (X_{n,m})$ given by sending the edge of $\Delta^1$ to the coherent collection of compositions $X_{n,nm} \to X_{n,m} \to X_{n,m}$ where the first map is the structure map (i.e. multiplication by $n=0$) and the second map is the identity (the fact that this gives an edge in the hom space between the pro-objects follows from the identity $Pro-Hom(X_i,Y_j) = holim_j hocolim_i Him(X_i,Y_j)$). Call this simplex $\sigma$, we get that -$\partial_0 \sigma = Id_X$ while $\partial_1 \sigma$ factor through a point-system. This gives an explicit contraction of $X$ and by composing with $\pi$ and $[2]\pi$ we get a homotopy between them.<|endoftext|> -TITLE: Hecke algebra of GL(2,F) -QUESTION [8 upvotes]: I was studying about the Hecke algebra from Bernstein's notes on p-adic representation theory and various other sources. First a disclaimer: everything below is fairly new to me so please feel free to correct me in the probably various places I am wrong) -I am trying to make some basic computations, like for example compute the whole algebra probably, $H_K$ (the left-invariant on $K$ distributions) and the center (which in the rest of literature is what is usually denoted by $H_K$ I think, essentially the bi-$K$-invariant distributions). Now the center can be computed by the Satake isomorphism (and it is isomorphic to $\mathbb{C}[z_1^{\pm},z_2^{\pm}]^{S_2}$). For the rest I really could not come up with an explicit computation. -Now searching around, I found that most sources do not define the Hecke algebra as the locally constant compactly supported distributions, but by a purely algebraic definition with some generators over the Weyl group. I really cannot understand this definition. -Can you provide me with some source that explains this explicit presentation of the Hecke algebra, and why is it the same as Bernstein's one? - -REPLY [5 votes]: Let $G$ be a reductive $p$-adic group. First fixing a Haar measure on $G$, you can identify the algebra of distributions of $G$ with the ("big") Hecke algebra $H(G)$ of locally constant complex functions with compact support equipped with convolution $\star$. If $K$ is any compact open subgroup, the bi-$K$-invariant functions form a subalgebra $H(G,K)$, called the $K$-spherical Hecke algebra. More generally if $\rho$ is an irreducible smooth representation of $K$, you may form the subalgebra $H(G,\rho ) = e_\rho \star H(G)\star e_\rho$, where $e_\rho \in H(G)$ is the idempotent attached to $\rho$. Note that $H(G,K)$ corresponds to the particular case where $e_\rho$ is the trivial character. -All these algebras are called Hecke algebras. On the other hand there are standard Hecke algebras defined in an algebraic manner via generators and relations, like Iwahori-Hecke algebras. -You cannot expect to understand any sort of Hecke algebra in a naive approach. -Any progress in the description of these algebras rely on a very fine understanding of the structure of $G$ (in particular of the double coset set $K\backslash G/K$) as well as a deep understanding of the structure of the representations. In general one wants to find an explicit isomorphism between an Hecke algebra that one wants to understand with a standard Hecke algebra. -Historically the first results concerned the spherical Hecke algebra $H(G,K)$, where $K$ is a "special" maximal subgroup of $G$ (because of its importance in the theory of automorphic forms) via the Satake isomorphism and the Iwahori Hecke algebra (when $K$ is an Iwahori subgroup), because of its importance in the study of unramified principal series. -The general understanding of the Hecke algebras that arise from reductive $p$-adic groups is one of the aims of Type Theory as conceptualized by Bushnell and Kutzko. It turns out that a lot of those Hecke algebras are in fact isomorphic to (or at least Morita equivalent to) standard Hecke algebras! But this is not trivial: this is the result of decades of research starting in the $50$'s.<|endoftext|> -TITLE: What is the translation of this ancient Greek verb πυθαγοριζει -QUESTION [8 upvotes]: Here it is used in a sentence - -It is therefore a priori probable that Plato πυθαγοριζει in the passage where he says that between two planes one mean suffices, but to connect two solids, two means are necessary. - Sir Thomas Heath: A History of Greek Mathematics, Volume I: From Thales to Euclid, page 89 - -I realize this is not an equation question, but this forum might seems an appropriate place to ask a question about words used in discussions of number theory - -REPLY [14 votes]: The verb is πυθαγορίζω (infinitive πυθαγορίζειν), formed by adding the derivative suffix -ιζω to the name of Pythagoras (Πυθαγόρας). So it just means "to be [or, in the context you quote, presumably, to act as] a disciple of Pythagoras"; or, to use an English equivalent, "to Pythagorize". -Sources: - -The "Pythagorize" entry from Oxford Dictionaries (I'm not sure whether this coincides with the OED), which gives the etymology as "from ancient Greek Πυθαγορίζειν to be a disciple of Pythagoras from Πυθαγόρας, the name of Pythagoras + -ίζειν". -Perseus Greek word study tool [click on "LSJ"] translates it as "to be a disciple of Pythagoras", quoting uses: Antiph.226.8, Alex.220.1. -The entry for the suffix -ίζω in Wiktionary.<|endoftext|> -TITLE: weak closedness of the unit ball for a dual pair of Banach space -QUESTION [5 upvotes]: Suppose that $\langle X,Y\rangle$ is a dual pair of Banach spaces satisfying $|\langle x,y\rangle|\leq \Vert x\Vert\Vert y\Vert$ for all $x\in X$, $y\in Y$. -Is it true that the unit ball of $X$ is $\sigma(X,Y)$-closed? - -REPLY [10 votes]: Okay, even assuming that $Y$ separates points the answer is still no. Take $X = l^1$ and let $Y$ be the set of elements $(a_n)$ of $l^\infty$ which satisfy $\lim a_n = 2a_1$. It's easy to see that $Y$ separates the points of $X$, but $e_n \to 2e_1$ weakly where $(e_n)$ is the standard basis of $l^1$. So $2e_1$ is in the weak closure of the unit ball.<|endoftext|> -TITLE: Dependence of the Hölder exponent in De Giorgi-Nash-Moser -QUESTION [5 upvotes]: I am curious about the Hölder exponent obtained by the De Giorgi-Nash-Moser theory, as a function of the ellipticity. -More precisely: suppose $u$ satisfies weakly -$$ -D_i(a^{ij}D_ju)=f -$$ -on the $d$-dimensional ball of radius $R$, with $0$ Dirichlet boundary conditions, with the matrix $(a^{ij})_{i,j=1..d}$ bounded from below and above by $\lambda I$ and $\Lambda I$, respectively, $\lambda>0$. For the right-hand side, let's just take $f\in L_\infty$. Then we know that $u\in C^{\alpha}(B_R)$, with some $\alpha=\alpha(d, \Lambda/\lambda)$, my question is whether something about the dependence on $\Lambda/\lambda$ is known. The analogous question for interior regularity (which might be easier) could also be of interest. Thanks! - -REPLY [7 votes]: (By scaling we can take $\lambda=1$ for simplicity. I will also take $f=0$, and discuss just the interior estimates.) -The precise exponent is known in two dimensions (see Piccinini & Spagnolo 1972). It is $\Lambda^{-1/2}$. You can find the "worst" coefficient field in the obvious way: the diffusion matrix $a(x)$ has eigenvalue $\Lambda$ in the $x$ direction, and eigenvalue $\lambda=1$ in directions orthogonal to $x$. You can prove this is the worst as Piccinini and Spagnolo do, with a beautiful monotonicity formula type computation. -In higher dimensions, the precise exponent is not known. It is however conjectured based on the same radial example, extended in the obvious way in $d>2$. The exact exponent should be (according to my personal notes, hopefully I have not made a mistake) -\begin{equation*} -\alpha(d,\Lambda) = \frac12 \left( -(d-2) + \sqrt{(d-2)^2 + \frac{4(d-1)}{\Lambda} } \right) . -\end{equation*} -Okay, forget about the exact formula, the point is that this scales like $c(d) \Lambda^{-1}$ for $\Lambda \gg1$ in all dimensions $d>2$. -Ok, what is the best exponent that can be proved? The best exponent which has been proved is this: -\begin{equation*} -\alpha(d,\Lambda) = \exp \left( - C \Lambda^{1/2} \right). -\end{equation*} -This is the best exponent one could ever hope to have based on current methods! Indeed, all known methods of proof use a decay of oscillation iteration to get the $C^{\alpha}$ estimate (except in $d=2$, more on that below). The best decay of oscillation estimate is related to the best constant in the Harnack inequality, and this is known to be $\exp( C \Lambda^{1/2})$. There are very easy examples you can make to confirm you can do no better. The Harnack inequality with this constant was proved in an amazing paper by Giusti and Bombieri (Inventiones, also 1972). -So, to summarize: in $d>2$, we cannot prove the estimate with the conjectured optimal exponent because we do not possess a proof powerful enough to do so. All known proofs cannot get there. The bound we have is hopeless pathetic next to what we think is true, so the situation is bad, very bad! Anything better than the Giusti-Bombieri exponent would be, in my opinion, a major breakthrough. -But then what is going on in $d=2$? There is a proof of the conjectured bound, so do we have a better proof? Why yes-- indeed we do. In $d=2$, the De Giorgi-Nash $C^{\alpha}$ estimate has an easy proof. It follows from the hole-filling argument, which is just the Caccioppoli inequality (i.e., the usual $L^2$ energy estimate), plus the Sobolev inequality. This is basically because the Sobolev inequality says that in $d=2$ the space $H^1$ almost embeds into $L^\infty$. Since you can actually do a tiny bit better than $H^1$ by hole filling, you can get $C^\alpha$ without any ideas of De Giorgi. The simple hole filling computation already gives an exponent that scales like $\Lambda^{-1/2}$. What the Piccinini & Spagnolo paper does is optimize this computation, replacing an iteration in dyadic balls by a more careful monotonicity computation. -So $d=2$ should be expected to be much easier, and in hindsight it is maybe not so surprising that we can get the optimal exponent in this case.<|endoftext|> -TITLE: linearly dependent symmetric matrices -QUESTION [5 upvotes]: Let $k$ be an infinite field. Assume that $n,m$ are two positive integers such that $n>m$. Consider symmetric matrices $A_1,\dots,A_m$ of size $n\times n$. Suppose for each $i=1,\dots,m$, every column of the matrix $A_i$ has a non-zero entry which is not on the main diagonal. Is the following implication correct? -If, for every vector $v \in k^n$ the vectors $A_1v,\dots,A_mv$ are linearly dependent, then the matrices $A_1,\dots,A_m$ are linearly dependent. - -REPLY [5 votes]: The implication does not hold. -Counterexample: Let $(n,m)=(4,3)$. We consider the subspace -$$M=\left\lbrace\left(\begin{matrix} -a&b&b&b\\ -b&c&c&c\\ -b&c&c&c\\ -b&c&c&c\end{matrix}\right)\colon a,b,c\in k\right\rbrace\subseteq\operatorname{Mat_{4\times 4}(k)}.$$ -Note that every matrix in $M$ is symmetric by construction. We claim that there are linearly independent elements $A_1,A_2,A_3\in M$ such that every column of every matrix contains a non-diagonal element which is non-zero. This is equivalent to a choice of $3$ linearly independent vectors in $k^3$ of the form $(a,b,c)$ with $b\neq 0$, which is always possible. -Let $v\in k^4$. For every $A\in M$ the vector $Av\in k^4$ lies in the $2$-dimensional subspace $\{(x,y,y,y)\colon x,y\in k\}\subseteq k^4$. In particular, $A_1v,A_2v,A_3v$ are linearly dependent.<|endoftext|> -TITLE: Which $K$-groups $K(C^*_r(G))$ are computed? -QUESTION [11 upvotes]: We have the Pimsner-Voiculescu exact sequences and the Baum-Connes map -for possible computation of the $K$-theory of the reduced group $C^*$-algebra $C^*_r(G)$ for a topological, locally compact, second-countable Hausdorff group $G$. -Up to now I have not seen much computations of $K(C^*_r(G))$. -Has anyone references to such computations, in particular in computing the left hand side of the Baum-Connes map, under the Chern map. -That is, computations of the Czech cohomology groups -$$\lim_{X \subseteq \underline EG} H(X,G)$$ -(something like that). - -REPLY [8 votes]: Here are some known computations for infinite discrete groups. Basically, most of these proceed by computing the equivariant K-homology of the classifying space of proper actions and deduce the computation for K-theory of the group $C^\ast$-algebra via the assembly map. -For the Bianchi groups: - -A.D. Rahm. On the equivariant K-homology of ${\rm PSL}_2$ of the imaginary quadratic integers. Ann. Inst. Fourier 66 (2016), 1667-1689. (link to journal page) - -Computations for Heisenberg-type groups have been established in the thesis of Olivier Isely (link) -Right-angled Coxeter groups: - -R. Sanchez-Garcia: Equivariant K-homology for some Coxeter groups. J. London Math. Soc. 75 (2007), 773-790. (link to arXiv) - -For hyperbolic reflection groups: - -J-F. Lafont, I.J. Ortiz, A.D. Rahm, R.J. Sanchez-Garcia: Equivariant K-homology for hyperbolic reflection groups. arXiv:1707.05133 (link to arXiv) - -The last paper also contains discussion and many further literature references to further computations of K-theory of group $C^\ast$-algebras, most notably by Wolfgang Lück and collaborators. There is also a book in progress on the isomorphism conjectures which contains a chapter on computations, see Wolfgang Lück's homepage. - -REPLY [5 votes]: There are some recent papers of Valette and coauthors which give explicit calculations/descriptions of the K-theory for some of the Baumslag-Solitar groups, namely BS($1,n$) (Pooya and Valette, arXiv 1604.05607), and also for certain lamplighters over ${\bf Z}$ (Flores, Pooya and Valette, arXiv 1610.02798). In both cases, the authors determine the LHS and the RHS of the Baum-Connes "picture" separately, and then verify explicitly that the BC map is an isomorphism. -(Apologies if these examples are covered in the references already provided by Matthias Wendt.)<|endoftext|> -TITLE: History of the classification of mathematical subjects -QUESTION [11 upvotes]: I would like to know if there are sources on the history of the classification of mathematical subjects. -Gérard Lang - -REPLY [6 votes]: Here is one such historical overview: -Mathematics in library subject classification systems, by Craig Fraser (2016). (Springer link) - -Insofar as library science is concerned, modern classification of - mathematical subjects occurred within the larger framework of library - classification, a vast project receiving sustained attention in the - period from 1870 to 1920. The work of the library cataloguers was - carried out against the background of a broad nineteenth-century - interest in the classification of knowledge. We explore different - views during this period concerning the position of mathematics in the - overall scheme of knowledge, the scope of mathematics, and the - internal organization of the different parts of mathematics. We - examine how mathematical books were classified, from the most general - level down to the level of particular subject areas in analysis. The - focus is on the Library of Congress classification system in its - various iterations from 1905 to the present.<|endoftext|> -TITLE: Explicit local normal form symplectomorphism at torus fixed point of a coadjoint orbit -QUESTION [6 upvotes]: Let $K$ be a compact, connected (probably also simple) Lie group and with a maximal torus $T$. Regular coadjoint orbits $\mathcal{O}_{\lambda} \cong K/T$, parameterized by a regular element $\lambda \in \mathfrak{t}^*$, have several properties: - -They are symplectic, equipped with KKS forms. -They are Hamiltonian $T$-manifolds with respect to the coadjoint action of $T$, generated by the canonical projection map $\mathcal{O}_{\lambda} \to \mathfrak{t}^*$. -The fixed points of the $T$-action are precisely the elements of $\mathcal{O}_{\lambda} \cap \mathfrak{t}^*$ (where $\mathfrak{t}^*$ is identified with the subspace $(\mathfrak{k}^*)^T$). -Using the killing form (or whatever scalar multiple of it that you like), we can make an identification $T_{\lambda}\mathcal{O}_{\lambda} \cong \mathfrak{t}^{\perp}$. One has that $\mathfrak{t}^{\perp}$ is invariant under the adjoint action of $T$ and coincides under this identification with the isotropy action of $T$ at the fixed point $\lambda$. - -The local normal form theorem for Hamiltonian torus actions tells us that there exists a symplectomorphism $\varphi$ from a neighbourhood of 0 in the symplectic vector space $T_{\lambda}\mathcal{O}_{\lambda}$ to a neighbourhood of $\lambda$ in $\mathcal{O}_{\lambda}$ such that: - -$\varphi$ is $T$-equivariant. -$\varphi$ intertwines the moment maps, i.e. if $\mu = \phi + \lambda$ denotes the quadratic moment map for the isotropy action of $T$ on $T_{\lambda}\mathcal{O}_{\lambda}$ shifted by $\lambda$, and $\text{pr}\colon \mathcal{O}_{\lambda} \to \mathfrak{t}^*$, then -$$ \text{pr} \circ \varphi = \mu.$$ - -Since everything in this example is fairly explicit, it seems reasonable to ask for an explicit formula for the map $\varphi$. - -What is a general formula for $\varphi$? - -There is an ``obvious candidate'' that looks good at first: one can easily define the map -$$\psi\colon \mathfrak{t}^{\perp} \to \mathcal{O}_{\lambda}, \, \psi(X) = Ad^*_{e^X}\lambda.$$ -This map is $T$-equivariant, but it fails to be symplectic (which can be demonstrated easily for coadjoint orbits of $SU(2)$). -(As a side note, the killing form restricted to $\mathfrak{t}^{\perp}$ and then extended $\mathcal{O}_{\lambda}$ by $K$-invariance makes $\mathcal{O}_{\lambda}$ a reductive homogeneous space, and $\psi$ is in fact the exponential map for this metric.) -I believe that, essentially, the map $\psi$ fails to be symplectic because $e^Xe^Y \neq e^{X+Y}$ if $[X,Y]\neq 0$ (which, if you are not careful means that you can accidentally ``prove'' this map is symplectic and spend an afternoon being very confused), which suggests that maybe there is some sort of CBH-related correction to $\psi$ that makes it symplectic (I am really just taking a guess here). -Edit: It will be sufficient to find a $T$-equivariant map $\rho\colon \mathfrak{t}^{\perp} \to K$ such that -$$\varphi(Y) := Ad_{\rho(Y)}\lambda$$ -satisfies the condition -$$ d(\varphi)_Y(ad_X\lambda) = ad_{Ad_{\rho(Y)}X}\varphi(Y).$$ -Edit 2: Unpacking the proof of local normal forms, one sees that there exists a map $\phi\colon U \to \mathfrak{t}^{\perp}$ defined on a neighbourhood $U$ of the origin in $\mathfrak{t}^{\perp}$ that fixes the origin and is a diffeomorphism onto it's image, such that the map -$$\varphi(Y) := Ad_{e^{\phi(Y)}}\lambda$$ -is a symplectomorphism. $\phi$ is the time-1 flow of a moser vector field. It's not clear from this perspective whether an explicit formula for $\phi$ is a reasonable thing to hope for. -Edit 3: I found a partial answer that I've posted below, but I am interested if anyone can expand on it. - -REPLY [2 votes]: Here is a partial answer. -The map $\mathfrak{t}^{\perp} \to T_{\lambda}\mathcal{O}_{\lambda}$, $X \mapsto ad_X(\lambda)$, endows $\mathfrak{t}^{\perp}$ with a linear symplectic form -$$ \omega_0(X,Y) = \langle \lambda, [X,Y]\rangle.$$ -Based on edit 2 above, we know there exists a $T$-equivariant map $\phi\colon \mathfrak{t}^{\perp} \to \mathfrak{t}^{\perp}$ (at least, defined and invertible near 0) such that $\phi(0) = 0$ and the composition $$\varphi(Y) := Ad_{e^{\phi(Y)}}\lambda$$ - is a symplectomorphism. - -Example ($K = SU(2)$) Here we can compute using cylindrical coordinates on the coadjoint orbit and polar coordinates on - $\mathfrak{t}^{\perp}$ to show that $$\phi\left(\left(\begin{array}{cc} 0 & - \rho e^{i\theta} \\ -\rho e^{-i\theta} & 0 \end{array}\right)\right) = - \left(\begin{array}{cc} 0 & \arcsin(\rho) e^{i\theta} \\ - -\arcsin(\rho) e^{-i\theta} & 0 \end{array}\right).$$ - -Let $\pi\colon K \to \mathcal{O}_{\lambda}$ be the map $k \mapsto Ad_k\lambda$. We know that - -$d(\pi)_e Y = ad_Y\lambda$ -$d(\pi)_k = d(Ad_k)_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{k^{-1}})_k$ (where $\mathcal{L}_k\colon K \to K$ is left multiplication). -$d(\mathcal{L}_{k^{-1}})_k$ is simply the (left-invariant) Maurer-Cartan form $\theta$ evaluated at $k$ -Because it is a linear map, $d(Ad_k)_{\lambda} = Ad_k$. We also know that $Ad_kad_X \lambda = ad_{Ad_kX}Ad_k\lambda$. - -Thus: -\begin{equation} - \begin{split} - d(\varphi)_Y & = d(\pi)_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ - & = d(Ad_{e^{\phi(Y)}})_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{e^{-\phi(Y)}})_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ - & = Ad_{e^{\phi(Y)}}\left(d(\pi)_e \circ \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)\\ - & = Ad_{e^{\phi(Y)}}\left(ad_{\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y}\lambda\right)\\ - & = ad_{Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)}\left(Ad_{e^{\phi(Y)}}\lambda\right). - \end{split} -\end{equation} -The map $\varphi$ is a symplectomorphism if $\varphi^* \omega = \omega_0$. Explicitly, for $X,Z \in \mathfrak{t}^{\perp}$, -$$(\varphi^*\omega)_Y(X,Z) = \omega_{\varphi(Y)}\left(d(\varphi)_Y X,d(\varphi)_Y Z\right)$$ -which by the previous equation -$$ = \langle Ad_{e^{\phi(Y)}}\lambda, [Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X\right), Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z\right)]\rangle$$ -$$ = \langle \lambda, [\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X, \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z]\rangle.$$ -This simplifies by substituting the formula for the derivative of the exponential map: -$$\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)} = e^{-\phi(Y)}e^{\phi(Y)}\left(\frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}\right) = \frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}$$ -to give -$$(\varphi^*\omega_{\lambda})_Y(X,Z) = \bigg\langle \lambda, \frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}[ d\phi_Y X, d\phi_Y Z]\bigg\rangle.$$ - -Thus, our equation for $\phi$ is \begin{equation} \bigg\langle - \lambda, \frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}[ d\phi_Y X, d\phi_Y - Z]\bigg\rangle = \langle \lambda, [X,Z]\rangle. \end{equation} - -The series expansion of the LHS will have some terms that disappear because they pair with $\lambda$ to zero, but I don't have a general formula for the ones that remain.<|endoftext|> -TITLE: What is known about the relationship between Fermat's last theorem and Peano Arithmetic? -QUESTION [14 upvotes]: As far as I know, whether Fermat's Last Theorem is provable in Peano Arithmetic is an open problem. What is known about this problem? -In particular, what is known about the arithmetic systems $PA + \text{Fermat's Last Theorem}$ and $PA + \lnot \text{Fermat's Last Theorem}$? -Note in particular that if $PA \vdash \text{Fermat's Last Theorem}$, the first system will just be $PA$ and the second system with be inconsistent. If FLT is independent of PA, then both will be consistient systems distinct from $PA$. $PA \vdash \lnot \text{Fermat's Last Theorem}$ is not possible, since $PA$ and Fermat's last theorem are both true in $\mathbb N$. - -REPLY [24 votes]: The main reference for this topic is Angus Macintyre's appendix to Chapter 1 ("The Impact of Gödel's Incompleteness Theorems on Mathematics") of Kurt Gödel and the Foundations of Mathematics: Horizons of Truth (Cambridge University Press, 2011). -There are a a couple of reasons why one might wonder whether the proof of FLT can be formalized in PA. - -The modularity theorem, as well as various arguments in the proof of the modularity theorem, appears at first glance to quantify over higher-order structures (sets of integers, sets of sets of integers, etc.), and so one might wonder whether all the necessary concepts can even be stated in the first-order language of arithmetic. -Even if the modularity theorem and all other crucial intermediate statements along the way to FLT can be formulated arithmetically, there is no a priori guarantee that we don't need to appeal to axioms with greater logical strength than PA. - -The "mathematician in the street" is typically more worried about #2. After all, doesn't the proof of FLT use all kinds of fancy-shmancy high-powered mathematics, and isn't PA "elementary"? Maybe one needs powerful axioms to carry out all those "advanced" arguments. While this is a theoretical possibility, there isn't any hint in the proof of FLT that strong axioms are needed. Experts can "smell" the intrusion of a strong axiom from a long way off; for example, the primary reason that the Robertson–Seymour graph minor theorem requires more than PA is that it invokes a highly sophisticated argument by induction at a crucial point. The proof of FLT doesn't give off this sort of odor anywhere. Of course, a smell test is not a proof, but the smart money is that if we run into difficulties trying to prove FLT in PA, it won't be because we need some strong arithmetic axiom. For more on this topic, especially the oft-discussed axiom of universes, see Colin McLarty's 2010 paper in the Bulletin of Symbolic Logic as well as his more recent arXiv preprint. -Issue #1, of whether all the higher-order apparatus in which the proof of FLT is couched is intrinsically necessary, is what Macintyre mainly concerns himself with. His appendix is essentially a ten-page sketch of an argument that all the relevant concepts and intermediate results on the road towards proving FLT can be replaced by "finite approximations" using standard techniques. As Macintyre says, - -There is no possibility of giving a detailed account in a few pages. I hope nevertheless that the present account will convince all except professional skeptics that [the modularity theorem] is really $\Pi_0^1$. - -In short, to say that it is an "open problem" whether FLT is provable in PA, while technically correct, is somewhat misleading. Demonstrating in full detail that FLT is a theorem of PA is more akin to an engineering project, where we more or less know how to proceed and are pretty sure—though of course not mathematically certain—that the project can be carried out if we just work hard enough at it.<|endoftext|> -TITLE: On faces of convex sets of positive semidefinite matrices -QUESTION [5 upvotes]: A face $F$ of a convex set $C$ is a set such that, if $x \in F$ is a convex combination of other elements in $C$, then they must also be in $F$. I will denote by $F(C,x)$ the smallest face of $C$ which contains $x$. -The faces of the positive semidefinite cone $H_+ = \text{conv} \{ xx^* \}$ in the real vector space of Hermitian matrices are well characterized, and we know that $F(H_+, A) = \left\{ B : \text{ker}(A) \subset \text{ker}(B) \right\}$. -I am interested in the faces of subsets of this cone of the form -$$ C_p = \text{conv} \{ vv^* : \|v\|_p \leq 1 \} $$ -where conv denotes the convex hull and $\|\|_p$ is any $\ell_p$ norm. I am specifically looking at $p=1$ and $p=\infty$, but more general results could also be interesting. -In particular, is there a way to characterize the smallest face containing a given point, and find its dimension? Similarly as for $H_+$, every matrix $A \in C_p$ certainly belongs to the face given by -$$ \text{conv} \{ vv^* : \|v\|_p \leq 1,\, \text{ker}(A) \subset \text{ker}(vv^*) \} $$ -but now this face is not necessarily the smallest (take e.g. the extremal points of $C_p$, which we know to lie in a 0-dimensional face). Could it be that this face is still the smallest for any non-extremal point? -I am curious about this because an interesting corollary of the characterization of the faces of $H_+$ is that we have $\text{dim}\, F(H_+, A) = \text{rank}(A)^2$, which means that there are faces of dimension 1 (the extreme rays generated by rank-one matrices) and faces of dimension 4 (generated by rank-two matrices), but no faces of dimension 2 or 3. Can anything similar be said about the sets $C_p$? - -REPLY [3 votes]: The space $H_+$ of positive semidefinite Hermitian $n\times n$ matrices is invariant under the action by conjugation of the group $SU(n)$, which is isometric for the $\ell^2$ metric. Here the space of diagonal matrices with non-negative entries is a submanifold which meets each orbit orthogonally: this is called a polar action. The orbit space is the Weyl chamber of diagonal matrices with (say) decreasingly ordered non-negative entries. The faces of the cone are not the orbits, but the strata of the orbit type stratification. Each connected component of such an orbit type stratum is projected onto a face of the Weyl chamber, so its dimension is the dimension of the face of the Weyl chamber plus the dimension of a typical orbit (= $SU(n)/\text{Isotropy group}$) which is easy to compute. -See 7.1, section 30, and 34.16 of here for background and more information<|endoftext|> -TITLE: Regulator of abelian extensions of Q -QUESTION [6 upvotes]: Let $K = \mathbb Q(\mu_m)$ and $\zeta_K$ it's Dedekind zeta function. We know from the class number formula that, around $0$: -$$\zeta_K(s) \sim s^{r_1+r_2-1}h(K)R(K)/w(K) $$ -where $h,R,w$ stand for the size of the class group, the regulator and the size of the subgroup of roots of unity. -On the other hand, we have the decomposition: -$$\zeta_K(s) = \prod_\chi L(s,\chi)$$ -where the product is over suitable Dirichlet characters and moreover, $L(0,\chi) = -B_{1,\chi}$ - the generalized Bernoulli numbers. -On the other hand, $R(K)$ should not be algebraic and there should be a corresponding transcendental contribution from $L(0,\chi)$ and the explicit formula for $B_{1\chi}$ shows that this comes only from those $\chi$ such that $L(s,\chi) = 0. $ -Question 1: In the case that $L(s,\chi) = 0$, is it possible to say what the first non zero Taylor coefficient is? -Question 2: Can we decompose (even if it is only conjecturally) $h(K),R(K),w(K)$ into factors corresponding to each Dirichlet character appearing in the decomposition of $\zeta_K$ . What about the order of vanishing of $L(0,\chi)$? - -REPLY [9 votes]: All of your questions have comprehensive and beautiful answers: providing them is exactly what the subject of Iwasawa theory was developed to do. You should perhaps read either Washington's Cyclotomic Fields or Coates and Sujatha's Cyclotomic Fields and Zeta Values. -The order of vanishing of $L(s, \chi)$ at $s = 0$, and its leading term there, is described quite precisely by the results of Chapter 4 of Washington, particularly Corollary 4.4 and Theorem 4.9. (These are stated in terms of values at s = 1, but the functional equation given just after Theorem 4.5 relates $L(s, \chi) $and $L(1-s, \bar\chi)$.) From these results it follows that: - -if $\chi(-1) = -1$, or $\chi$ is trivial, then $L(\chi, 0)$ is a non-zero algebraic number. -if $\chi(-1) = +1$ and $\chi$ is non-trivial, then $L(\chi, 0) = 0$ and $L'(\chi, 0)$ is a non-zero $\overline{\mathbf{Q}}$-linear combination of logarithms of algebraic numbers (more specifically, of cyclotomic units). - -As for decomposing $h(K)$ etc into pieces corresponding to Dirichlet characters, you might want to look at Chapter 10 -- in particular Corollary 10.15 and the remark following it, which explain such a decomposition for the p-part of the class group when $K = \mathbf{Q}(\zeta_p)$.<|endoftext|> -TITLE: Properties of Zero Line-Sum Matrices -QUESTION [7 upvotes]: By a Zero Line-Sum (ZLS) matrix I mean matrices with the property, that each row sum and each column sum equals zero: -$$A\in\mathbb{R}^{m\times n}:\ \sum_{i=1}^{n}a_{ij}=\sum_{j=1}^{m}a_{ij}=0$$ -These can be thought of as being the difference of two "ordinary" doubly stochastic matrices. -ZLS matrices obviously don't have full rank but, as not all rank-deficient matrices have the property of being ZLS, I wonder if ZLS matrices are also special in different aspects and also w.r.t. the mappings they define. - -Question: - Have ZLS matrices already been investigated, and what are non-trivial special properties of them that have been identified? - I am looking for information on matrices, that equal the difference of two doubly stochastic matrices and, on the special properties of the transformations they define. - -Clarification in response to Jochen Glueck's correct remarks: -I use the term "projection" in a formally not correct way, namely meaning any mapping to a lower-dimensional space. -Remark: I have edited this question to replace the former "0-Stochastic projection matrix" with the "Zero Line-Sum" matrix following the suggestion of Gerry Myerson; "line" in this context is the common term for row and column. - -REPLY [3 votes]: Below find six suggestions. -Case of general ZLS matrices. -Examples of properties are: - -All cofactors of a square ZLS are equal. -Each square ZLS matrix of dimension $n$ has eigenvector $1^{n\times 1}$ with eigenvalue $0$. -If an $n\times n$ ZLS $A$ is moreover symmetric, and if $\mathrm{Sp}(\cdot)$ denotes the spectrum of a matrix, then each cofactor of $A$ equals $\frac{1}{n}\prod_{\lambda\in\mathrm{Sp(A)\setminus\{0\}}}\lambda$. -Rectangular ZLS matrices over a field $K$ form a $K$-vector space in the obvious way, and according to this thread its dimension is $(m-1)(n-1)$, and if $m=n$, then according to that thread its orthogonal complement w.r.t. the Frobenius inner product consists solely of sums of two rank-one matrices, more precisely, this orthogonal complement consists solely of matrices of the form $v\cdot 1^{1\times n} + 1^{n\times 1}\cdot w^{\text{t}}$. -Some relevant Lie theoretic consideration can be found in - - -Andreas Boukas, Philip Feinsilver, Anargyros Felouris, On the Lie structure of zero sum and related matrices. Random Oper. Stoch. Equ. 23, No. 4, 209-218 (2015) - - -Case of matrices which are differences of two doubly-stochastic matrices - -If $A$ is the sum of two symmetric stochastic matrices, you may find it profitable to look in the direction of Horn's conjecture. Starting points could be R Bhatia: Algebraic Geometry Solves an Old Matrix Theorem. Resonance. December 1999, or this MO thread.<|endoftext|> -TITLE: Are the following subsets of a Hilbert space always homeomorphic? -QUESTION [8 upvotes]: Let $F$ be a infinite-dimensional complex Hilbert space, with inner product $\langle\cdot\;| \;\cdot\rangle$, the norm $\|\cdot\|$, the 1-sphere $S(0,1)=\{x\in F;\;\|x\|=1\}$ and let $\mathcal{B}(F)$ be the algebra of all bounded linear operators on $F$. - -Let $M\in \mathcal{B}(F)$ be a bounded operator. Suppose - -that $M\in \mathcal{B}(F)^+$, i.e., $\langle Mx,x\rangle\geq0$ for all $x\in F$, and - -that $M$ is an injective operator on $F$. - - -Consider -$$S_M(0,1)=\{x\in F:\;\langle Mx, x\rangle=1\}.$$ -Is $S_M(0,1)$ always homeomorphic to the 1-sphere $S(0,1)$? - -REPLY [7 votes]: The topological equivalence of the set $S_M:=\{x\in F:\langle Mx,x\rangle=1\}$ and the unit sphere $S:=\{x\in F:\|x\|=1\}$ can be proved as follows. -The assumptions on $M$ and the spectral theorem (or just the equality $\langle Mx,x\rangle=\langle \sqrt{M}x,\sqrt{M}x\rangle$) imply that $\langle Mx,x\rangle>0$ for any non-zero vector $x\in F$. Then the map $$h:S\to S_M,\;h:x\mapsto \frac{x}{\sqrt{\langle Mx,x\rangle}},$$ is a homeomorphism with inverse $$h^{-1}:S_M\to S,\;\;h^{-1}:y\mapsto \frac{y}{\|y\|}.$$ -Acknowledgement. I would like to thank Nik Weaver for his very helpful comments, which allowed to simplify the initial answer (which infolved a powerful machinery of infinite-dimensional topology) to the present (almost) trivial form. - -Added after comments of @MathUsers: For a positive (but not necessarily injective) opeartor $M$ on a Hilbert space $F$ the set $S_M:=\{x\in F:\langle Mx,x\rangle=1\}$ is homeomorphic to the product of the sphere in a Hilbert space and a Hilbert space (of suitable dimensions). This can be shown as follows. -Using the Spectral Theorem, show that the Hilbert space $F$ admits an orthogonal projector $P:F\to Y$ onto its subspace $Y\subset F$ such that -$M=M\circ P=P\circ M$ and $\langle My,y\rangle>0$ for every $y\in Y\setminus\{0\}$. Write $F$ as the orthogonal sum $F=X\oplus Y$ where $X=P^{-1}(0)$ is the kernel of the projector $P$. Let $S_Y=\{y\in Y:\|y\|=1\}$ be the unit sphere in the Hilbert space $Y$. -Theorem. The space $S_M:=\{x\in F:\langle Mx,x\rangle=1\}$ is homeomorphic to $X\times S_Y$. -Proof. The map $$h:X\times S_Y\to S_M,\;\;h:(x,y)\mapsto x+\frac{y}{\sqrt{\langle My,y\rangle}}$$is a homeomorphism with the inverse $$h^{-1}:z\mapsto (z-Pz,\tfrac{Pz}{\|Pz\|}).$$ -Corollary. The space $S_M$ is homeomorphic to the unit sphere $S$ in $F$ if and only if the positive opeartor $M$ has infinite-dimensional range.<|endoftext|> -TITLE: Axioms for constructive Euclidean geometry -QUESTION [15 upvotes]: In the summer I will be teaching a course in (plane) Euclidean geometry to future high school teachers and I am looking for a suitable axiom system (unlike College (Euclidean) geometry textbook recommendations where books are discussed). -Since the students should do a lot of ruler-and-compass constructions, I would like the axiom system to make the possibility of these constructions as obvious as possible (much like Birkhoff's system makes constructions with marked ruler and protractor natural). In particular, I would like to see easily (postulate?) that two circles intersect unless the triangle inequality says they don't (I think the proof would usually involve introducing coordinates and using the intermediate value theorem more or less explicitly). On the other hand I don't think I actually need completeness. -In algebraic terms I would like to axiomatize $K^2$ where $\mathbb{Q} < K < \mathbb{R}$ and $K$ is quadratically closed. -So the questions are: - -Is there a good reference for such an axiom system? Is it safe to replace the completeness axiom by "two circles intersect unless they obviously don't" in an appropriate system? [Of course, to say "they obviously don't" I need the axiom system to be based on metric rather than in-betweenness, which is fine with me.] -Is there any need for completeness when doing basic Euclidean geometry? - -[One statement I can think of that does need completeness is: any finite set of points is contained in a unique minimal disc. But the proof uses completeness explicitly and thus is beyond high school teachers anyway.] -Edit: when I say "circle" I should of course say "circle whose radius is a distance of two points", since, for example, a circle with transcendental radius in $\bar{Q}^2$ is empty. - -REPLY [3 votes]: The set of geometric tools that are used in Euclidean geometry consists only of a pencil, ruler, and compass. Using this toolset, we cannot construct all points on the R^2 plane. It is well known that some geometry problems (such as the angle trisection problem) are unsolvable with this toolset. -David Hilbert, who proposed the first formal system of axioms for Euclidean geometry, used a different set of tools. Namely, he used some imaginary tools to transfer both segments and angles on the plane. It is worth noting that in the original Euclidean geometry, these transfers are performed only with the help of a ruler and a compass. Although Hilbert geometry is very similar in its results to Euclidean one, it nevertheless differs significantly from it. A striking feature of the Hilbert system of axioms is the complete absence of circles. For this reason, it is impossible not only to trisect an angle but also to intersect two circles. In other words, it turned out that in Hilbert geometry one can construct even fewer points than in Euclidean geometry. When Hilbert's contemporaries called his attention to this fact, he added to his system of axioms the so-called "completeness axiom", the essence of which is reduced to the following statement: "Let there exist all points of the plane R ^ 2 that I could not construct". In other words, Hilbert cannot construct the intersection point of two circles with his tools, but he can prove that it exists using the axiom of completeness. -The followers of Hilbert circumvent this difficulty by additional principles. For example, Hartshorne directly adds to Hilbert's system of axioms the axiom of the intersection of two circles (E). Now in such an extended geometry "Hilbert + E" it is possible to construct all the points of Euclidean plane. However, the following question arises. Since the addition of the circle-circle intersection axiom (E) expands significantly the constructive capabilities of Hilbert's geometry, then maybe his original tools for transferring segments and angles turn out to be superfluous in this extended geometry? This indeed is the case! However, to eliminate Hilbert's tools, some of his other axioms must also be modified. -There are also other known flaws in Hilbert's geometry. For example, Hilbert's treatment of angles cannot be considered completely satisfactory. The point is that Hilbert only defines interior angles (greater than null angle and less than straight angle). As a result of this truncated definition, it is not possible to universally define the addition of two arbitrary interior angles. The sum of two interior angles in Hilbert theory is determined only iff the resulting angle is also interior. Thus, the addition of angles in Hilbert theory does not form a group. Another drawback is that Hilbert does not give a clear definition of triangle orientation on the plane, although he uses this concept to define polygon areas. -In my recent work - -Evgeny V. Ivashkevich, On Constructive-Deductive Method For Plane Euclidean Geometry, arXiv:1903.05175 - -I abandoned Hilbert's tools and replaced them with the Euclidean ones (ruler and compass). As a result, the constructive part of my development remains faithful to Euclid. At the same time, I have modified some of Hilbert's axioms in such a way as to give a modern definition of order and addition of angles (both convex and reflex) and of triangles orientation on the plane.<|endoftext|> -TITLE: Gabriel-Ulmer duality for $\infty$-categories -QUESTION [11 upvotes]: Gabriel-Ulmer duality states that 2-categories $\mathrm{Lex}$ (of small finitely complete categories and functors preserving finite limits) and $\mathrm{LFP}$ (of locally finitely presentable categories and finitary right adjoints) are dual. This duality should be true in the setting of $\infty$-categories if we replace $\mathrm{Lex}$ with the $\infty$-category of finitely complete and idempotent complete $\infty$-categories and $\mathrm{LFP}$ with the $\infty$-category of locally finitely presentable $\infty$-categories. Is there a proof of this fact in the literature? - -REPLY [11 votes]: I'm not aware of anyone writing the proof down, but I think we can patch it together as an easy consequence of several facts in Lurie's Higher Topos Theory (henceforth HTT). -The statement, as I understand it, is that the functor -$$\mathrm{Fun}^{lex}(-,\mathrm{Space}):(\mathrm{Cat}_\infty^{lex,\natural})^{op}\to\mathrm{Pr}^R_\omega$$ -is an equivalence of $\infty$-categories (in fact we'd like it to be an equivalence of $(\infty,2)$-categories, but I'm not going to address that now), where - -$\mathrm{Cat}_\infty^{lex,\natural}$ is the $\infty$-category of small idempotent complete and finitely complete $\infty$-categories and right exact functors; -$\mathrm{Pr}^R_\omega$ is the $\infty$-category of presentable compactly generated $\infty$-categories and $\omega$-accessible right adjoint functors ($\omega$-accessible just means that the functor preserve cofiltered limits). - -I'm going to factor this functor as the composition of functors that HTT proves to be equivalences. We can write it as -$$ (\mathrm{Cat}_\infty^{lex,\natural})^{op}\xrightarrow{(-)^{op}}(\mathrm{Cat}_\infty^{rex,\natural})^{op}\xrightarrow{Ind(-)}(\mathrm{Pr}^L_\omega)^{op}\cong \mathrm{Pr}^R_\omega$$ -Note that by HTT.5.3.5.4, $\mathrm{Ind}(C)=\mathrm{Fun}^{lex}(C^{op},\mathrm{Space})$, so this composition is exactly the functor we wanted to study. -Now let us check that all functors in this diagram are equivalences: - -The first functor simply sends an idempotent complete, finitely complete ∞-category to its opposite, which is an idempotent complete and finitely cocomplete ∞-category. -The second functor is the (inverse of the) equivalence of HTT.5.5.7.8 for $\kappa=\omega$ -The third functor is just the equivalence of HTT.5.5.3.4 that sends every presentable ∞-category to itself and every functor to its right adjoint (in fact it is the restriction of this to the subcategory of compactly generated ∞-categories and $\omega$-accessible functors). - -Finally, I'd like to remark that the first and third functors are also evidently an equivalence of $(\infty,2)$-categories (once you precisely defined what you mean by $op$). I believe the second functor is too, once you have stared carefully at HTT.5.3.5.10, but that would require a little bit of attention.<|endoftext|> -TITLE: Countable support iteration of proper forcings and the tree property -QUESTION [9 upvotes]: I'm mainly concerned with countable support iterations of proper forcings that add reals of some large cardinal length. It is known that countable support iteration of Sacks forcing/Cohen forcing of weakly compact ($\kappa$) length forces $\kappa=\omega_2$ has the tree property. Is there a general theorem, like: for any $\langle P_i, \dot{Q}_j: i\leq \kappa, j<\kappa\rangle$ countable support iteration of proper forcings that add reals for some large cardinal $\kappa$ (weakly compact), then the tree property at $\omega_2$ holds in the forcing extension? Note $2^\omega=\kappa=\omega_2$ in the extension. -Edit: for larger cardinals as Sean Cox pointed out, the answer is positive. - -REPLY [8 votes]: If $\kappa$ is huge, then any countable support iteration $\mathbb{P}$ of proper forcing up to $\kappa$ (where each component is of size $<\kappa$) that forces $2^\omega = \kappa = \omega_2$, must also force the tree property at $\omega_2$. -To see this, suppose $j: V \to N$ is a huge embedding with critical point $\kappa$. The assumptions on $\mathbb{P}$ ensure that if $G$ is $(V,\mathbb{P})$-generic, the quotient $j(\mathbb{P})/G$ is proper from the point of view of $N[G]$, and by closure of $N$ in $V$, this quotient is proper from the point of view of $V[G]$ as well (because $V[G]$ sees that it really is a CS iteration of proper forcings). Then $V[G]$ sees that the poset $j(\mathbb{P})/G$ is a proper forcing that introduces a generic elementary embedding with critical point $\omega_2$. By the proof of Theorem 5 in my Chang's Conjecture and semiproperness of nonreasonable posets, this implies that $V[G]$ satisfies a strong form of Chang's Conjecture that I call $\text{SCC}^{\text{cof}}_{\text{gap}}$ (this is just a minor variation of an argument of Hiroshi Sakai). By a result of Torres-Perez and Wu (``Strong Chang’s Conjecture and the tree property at $\omega_2$"), together with failure of CH this implies the tree property at $\omega_2$. -(Edit: I see now that, due to a newer theorem of Torres-Perez and Wu in "Strong Chang's Conjecture, Semi-Stationary Reflection, -the Strong Tree Property and two-cardinal square principles", in $V[G]$ you actually get the Strong Tree Property at $(\omega_2,\lambda)$ for all $\lambda < j(\kappa)$, since the kind of parameterized Strong Chang's Conjecture they use in that paper holds in $V[G]$ below $j(\kappa)$. In particular, $V_{j(\kappa)}[G]$ models the full Strong Tree Property. Also, as Jing pointed out in the comments below, you don't really need hugeness for this; strong compactness of $\kappa$ is enough. NOTE: the abstract in the Torres-Perez and Wu paper fails to mention failure of CH, which is of course required for their proof in Section 3). -Even if $\mathbb{P}$ is an RCS iteration of semiproper posets (each of size $<\kappa$), you get the same result. The only difference is that the generic elementary embedding is obtained by a semiproper (rather than proper) forcing, but that's still enough (by Sakai's argument) to get what I call $\text{SCC}^{\text{cof}}$ and apply the Torres-Perez and Wu theorem. -Also note that in many cases, measurability of $\kappa$ is enough to run the argument. You just need that the individual posets used in the $j(\mathbb{P})/G$ iteration are not only proper in the generic extension of $N$, but also in the corresponding generic extension of $V$. There should be plenty of scenarios where $\kappa$-closure of $N$ in $V$ (rather than $j(\kappa)$-closure) is enough to obtain that (and similarly for the RCS iteration of semiproper).<|endoftext|> -TITLE: The geometry of $\mathbb{R}^n$ -QUESTION [13 upvotes]: Let $X,Y$ be finite-dimensional real normed spaces. Consider the set of linear operators $L(X,Y)$ between the two spaces. -Then we define the set of equivalence classes -$$G(X,Y):=\left\{[T]; T,S \in L(X,Y) \text{ are equiv., if there is an isometry P on Y with }PT=S \right\}.$$ -I ask: Does $$\sup_{x \in X} \left\lvert \left\lVert Tx \right\rVert-\left\lVert Sx \right\rVert \right\rvert =0$$ imply that $[T]=[S]$? - -REPLY [12 votes]: The answer is no, in general. -In order to construct a counterexample, let $X = Y = \mathbb{R}^n$ for any $n \ge 2$ and endow this space with the $p$-norm for your favourite $p \in [1,\infty] \setminus \{2\}$. The point about this choice of the norm is that the (linear) isometries on $\mathbb{R}^n$ with respect to this norm are exactly the signed permutation matrices, i.e. permutation matrices for which some of the one's have been replaced with $-1$ (see this post). -Now, fix an arbitrary linear functional $\varphi: \mathbb{R}^n \to \mathbb{R}$ of norm $1$. Let $e_1 \in \mathbb{R}^n$ be the first canonical unit vector and let $f \in \mathbb{R}^n$ be a vector of norm $1$ such that neither $f$ nor $-f$ is a canonical unit vector. We define $Tx = \langle \varphi, x\rangle e_1$ and $Sx = \langle \varphi, x\rangle f$ for each $x \in \mathbb{R}^n$. -Then we have $\|Tx\| = \|Sx\| = |\langle \varphi, x \rangle|$ for all $x \in \mathbb{R}^n$. However, if we had $PT = S$ for an isometry $P$, this would imply $\langle \varphi,x\rangle P e_1 = \langle \varphi, x \rangle f$ for all $x \in \mathbb{R^n}$. By plugging in any $x \in \mathbb{R}^n$ which is not in the kernel of $\varphi$, we conclude that $Pe_1 = f$. But his cannot be true since $P$ is a signed permutation matrix. -EDIT If think it's worthwhile to add the following general result which answers the question in a more complete way: -Theorem 1. The answer to the question is "yes" if and only if $Y$ is (isometrically) a Hilbert space. -For the proof, we need the following result: -Theorem 2. A finite dimensional Banach space $Y$ is isometrically a Hilbert space if and only if the group of all linear isometries on $Y$ acts transitively on the unit sphere of $Y$ (i.e. for all vectors $y_1,y_2 \in Y$ of norm $1$ we can find a linear isometry $P$ on $Y$ such that $Py_1 = y_2$). -It is interesting to note that it is open whether the same result remains true for separable Banach spaces; see for instance the article [Ferenczi and Rosendal: On isometry groups and maximal symmetry (2013)] for details. The finite dimensional result in Theorem 2 is attributed to Mazur in [op. cit., page 1773]. -Proof of Theorem 1. "$\Rightarrow$" Assume that $Y$ is not a Hilbert space. By Theorem 2 we can find two vectors $y_1,y_2 \in X$ such that $Py_1 \not= y_2$ for any isometry $P$ on $Y$. Now, one can mimic the construction from the first part of this answer by replacing the vectors $e_1$ and $f$ with $y_1$ and $y_2$, respectively. -"$\Leftarrow$" Let $Y$ be a Hilbert space. Since $\|Tx\| = \|Sx\|$ for all $x,y \in X$, the operators $T$ and $S$ have the same kernel $K$. Choose a vector subspace $V \subseteq X$ such that $X = K \oplus V$. Then the restrictions $T|_V: V \to Y$ and $S|_V: V \to Y$ are injective and the mapping $J := S|_V \; - (T|_V)^{-1}: \operatorname{rg}T \to \operatorname{rg}S$ is a bijective isometry from the range of $T$ to the range of $S$. -Since $\operatorname{rg}T$ and $\operatorname{rg}S$ have the same dimension, so do their orthogonal complements in $Y$. Thus, there exists an isometry $R: (\operatorname{rg}T)^\perp \to (\operatorname{rg}S)^\perp$. The operator $P := J \oplus R$ is an isometry on $Y$ which fulfils $PT = S$. - -REPLY [2 votes]: The answer is no in general. E.g., let $X=\mathbb R$ with the standard norm and $Y=\mathbb R^2$ with the norm given by the formula $\|y\|=|y_1|\vee|y_2-y_1|$. Let $S$ and $T$ be defined by the formulas $S1:=(1,0)$ and $T1:=(0,1)$. Then $\|Tx\|=\|Sx\|$ for all $x\in\mathbb R$. -If $P$ is an isometry of $Y$ such that $PS=T$, then $P(1,0)=(0,1)$ and $P(0,1)=(a,b)$, where $\|(a,b)\|=1$. Further, then would have $|y_1|\vee|y_2-y_1|[=\|y\|=\|Py\|]=|y_2a|\vee|y_1+y_2(b-a)|$ for all real $y_1,y_2$. -But this is impossible for any real $a,b$. Indeed, letting $y_1=y$ and $y_2=1$, we would have $|y|\vee|y-1|=|a|\vee|y+b-a|$ for all real $y$. Letting here $y\to\infty$, we would get $y=y+b-a$, so that $b=a$ and hence $|y|\vee|y-1|=|a|\vee|y|$ for all real $y$. Letting here $y\to-\infty$, we would get -$1-y=-y$, a contradiction.<|endoftext|> -TITLE: Ramsey type theorem -QUESTION [6 upvotes]: Let $\mathcal{P}(\{0,\dotsc,7\})$ denote the power set of $\{0,\dotsc,7\}$. -Is the following true? - -For any function $f: \mathcal{P}(\{0,\dotsc,7\})\rightarrow\{0,1\}$ there exists $0\leq k\leq 3$ and three mutually disjoint sets $A,B,C\subseteq \{0,\dotsc,7\}$ such that - $\min A = 2k, \min B = 2k+1,$ and $$f(C) = f(A\cup C) = f(B\cup C) = f(A\cup B\cup C)?$$ - -REPLY [8 votes]: Yes, your conjecture is true. -Suppose otherwise. Then there exists a counterexample $f : \mathcal{P}(8) \rightarrow \{0, 1\}$. For each set $X \in \mathcal{P}(8)$, let the proposition $P_X$ denote $f(X) = 1$. -There are $5440$ different choices of the tuple $(A, B, C) \in \mathcal{P}(8)^3$ satisfying your constraints. For each such tuple, we obtain two clauses which must be true of the counterexample $f$: -$$ (\neg P_C \lor \neg P_{A \cup C} \lor \neg P_{B \cup C} \lor \neg P_{A \cup B \cup C}) $$ -$$ (P_C \lor P_{A \cup C} \lor P_{B \cup C} \lor P_{A \cup B \cup C}) $$ -This gives a succinct list of $10880$ clauses which must be true of the $256$ primitive propositions $\{ P_X : X \in \mathcal{P}(8) \}$. -Inputting this list of clauses into the SAT solver Glucose gives the response 'UNSAT' (meaning 'unsatisfiable'), so no such counterexample exists. It also exports a verifiable certificate of unsatisfiability which can be checked in polynomial time. -The proof is somewhat unilluminating, because it doesn't give any indication as to why your conjecture is true, just that it is.<|endoftext|> -TITLE: Injectivity of a Fredholm operator -QUESTION [6 upvotes]: While doing my study on the boundary-crossing time of a stochastic process, I happened to deal with the following question which is somehow related to Fredholm theory. -Question : Suppose $K$ is continuous non-negative function on $[0,T]^2$ such that $K(t,s)= 0 $ if and only if $s \ge t$. Prove or disprove the following statement: -"If $f$ is integrable function on $[0,T]$, then $ \int_{[0,T]} K(t,s)f(s)ds= 0 \hspace{3mm} \forall t \iff f= 0 $ a.s" -Thought : This statement is reasonable because for each $t$, $K(t, \cdot)$ can mimic the role of the indicator function $\mathbb{1}_{[0,t]}$ . -I'm essentially looking for an answer helping me proceed with that question, for example, an approach, or a condition to be added in order to make the question more agreeable,etc. Nevertheless, no matter how your help is, I always deeply appreciate it. -Thank you. - -REPLY [7 votes]: Surprisingly (to me), the statement is false. My counterexample is a little messy, but the idea is fairly simple. -Take $T = 1$ and set $a_n = \frac{1}{n}$ and $b_n = 1 - \frac{1}{n}$ for $n \in \mathbb{N}$. Define $f$ by setting $f(t) = 1$ on the intervals $[a_{2n+2}, a_{2n+1})$ and $f(t) = -1$ on the intervals $[a_{2n+1},a_{2n})$. The point of using $\frac{1}{n}$ is that the lengths of adjacent intervals are approximately equal for large $n$ (i.e., their ratio goes to $1$). We seek a family of functions $k_t(s)$ which is continuous in both $s$ and $t$, which satisfies $k_t^{-1}(\{0\}) = [t, 1]$, and such that $\int k_t(s)f(s) = 0$ for all $t$. -First we construct $k_1$. We impose the condition that $k_1(a_{2n}) = b_{2n}$ for all $n$. Then we define $k_1$ on $[a_{2n+2}, a_{2n}]$ to be any continuous function which satisfies $\int_{a_{2n+2}}^{a_{2n}} k_1f = 0$ (i.e., $\int_{a_{2n+2}}^{a_{2n+1}} k_1 = \int_{a_{2n+1}}^{a_{2n}} k_1$) and satisfies $b_{2n} \leq k_1 \leq \alpha_nb_{2n}$ where $\alpha_n \to 1$ as $n \to \infty$. (Possible because successive intervals have approximately the same length.) Then $k_1$ is continuous, its integral against $f$ is zero, and $k_1^{-1}(\{0\}) = \{1\}$. -Next, for each $n \geq 1$ let $k_{a_{2n}}(s) = k_1(s) - b_{2n}$ for $s \in [0,a_{2n}]$ and let it be $0$ for $s > a_{2n}$. These functions also have the desired properties. -To complete the construction I have to show that you can continuously interpolate $k_{a_{2n}}$ and $k_{a_{2n + 2}}$ while preserving the desired properties. You can do this by letting $k_t(s) = k_1(s) - (1-t)$ for $s \in [0,a_{2n+4}]$ (this ensures that $\int_0^{a_{2n+4}} k_t f = 0$) and dealing with the $s \in [a_{2n+4},a_{2n}]$ case separately. On the latter interval I don't have a simple formula, but I think it's clear that this can be done. If I had to write down an explicit formula I'd make each $k_t$ piecewise linear so you're just varying a couple of values of $k_t$ within $[a_{2n+4},a_{2n}]$ in a way that keeps the integral against $f$ zero and makes $k_t$ to go zero at $s = t$. -Well, that's a sketch of a counterexample. -tl;dr: for a counterexample to work, $f$ must take both positive and negative values in any neighborhood of $0$. Once such an $f$ is chosen, find functions $k_{1/n}$ which are continuous, nonzero on $[0,1/n)$ and zero on $[1/n,1]$ and whose integral against $f$ is zero. Finally, continuously interpolate between $k_{1/n}$ and $k_{1/(n+1)}$ to get functions $k_t$ with the same properties. You have to be a little careful in the last step, but there's no serious obstruction.<|endoftext|> -TITLE: Is there a relationship between a quotient group of the fundamental group of X and the fundamental group of a quotient topology of X? -QUESTION [17 upvotes]: Let ($X$, $x_0$) be a topological space with a base point, and denote the fundamental group of $X$ as $\pi_1(X)$. Let $N$ be a normal subgroup of $\pi_1(X)$. -Does there necessarily exist an equivalence relation $\thicksim$ on $X$ such that $\pi_1(X/\thicksim)$ is isomorphic to $\pi_1(X)/N$? - -REPLY [2 votes]: Not exactly an answer to your question, but it might be interesting anyway. Suppose your space $X$ is nice enough (path connected, locally path connected, semi-locally simply connected). Then under the Galois correspondence, the normal subgroup $N \triangleleft \pi_1(X)$ corresponds to a path-connected cover $p : Y \to X$ such that $\pi_1(Y) = N$ and $p_* : \pi_1(Y) \to \pi_1(X)$ is the inclusion. -Consider the mapping cylinder $M_p = (Y \times [0,1]) \cup_p X$. The covering $p : Y \to X$ factors as $Y \xrightarrow{\tilde{p}} M_p \xrightarrow{r} X$, where $r$ is a homotopy equivalence (a deformation retract). Therefore you have $\pi_1(M_p) = \pi_1(X)$ and $\tilde{p}_*$ induces the same map as $p_*$, i.e. the inclusion. -Finally, you have the long exact sequence of relative homotopy groups: -$$\dots \to \pi_1(Y) \to \pi_1(M_p) \to \pi_1(M_p, Y) \to \pi_0(Y) \to \pi_0(X) \to \dots$$ -and since $X$ and $Y$ are connected, $\pi_0(Y) \to \pi_0(X)$ is an isomorphism. Thus -$$\pi_1(M_p, Y) = \pi_1(M_p) / \pi_1(Y) = \pi_1(X)/N.$$ -Finally if your spaces are super-nice (CW-complexes), then there is a surjection $\pi_1(M_p, Y) \to \pi_1(M_p / Y)$, where $M_p / Y$ is the mapping cone of $p$. So you can view $\pi_1(X)/Y$ as a quotient of the fundamental group of a quotient of a space which deformation retracts onto $X$. I don't know (think) that you can say something better than that.<|endoftext|> -TITLE: First Chern class of a specific line bundle -QUESTION [7 upvotes]: Let $E$ be a spin$^c$ bundle and $spin^c(E)$ the corresponding $spin^c(n)$-principial bundle. Let $g_{U,V}: U \cap V \to spin^c(n)$ denote transition functions for this principial bundle and consider the map $\nu:spin^c(n) \to \mathbb{T}$ defined by $\nu(w)=w^!w$ where $(v_1 \cdot ... \cdot v_r)^!=v_r \cdot ... \cdot v_1$. We can form the composition $\nu \circ g_{U,V}:U \cap V \to \mathbb{T}$. As this satisfies cocycle property we can form the line bundle $L_E$. -Let us assume that $E$ is $spin^c$ but is not spin. - -How to prove that the first Chern class of $L_E$ is odd (in the sense that $j^*(c_1(L_E)) \neq 0 $ where $j^*$ is mod 2 reduction of coefficients)? - -REPLY [6 votes]: One way to prove this is to first check that it is true when $E$ is the spin$^c$ bundle canonically associated to a complex vector bundle $F$ and then argue that complex and spin$^c$ bundles have the essentially same characteristic classes in degree up to 4. -To associate a spin$^c$-bundle to a complex vector bundle, consider the inclusion $\rho: U \hookrightarrow Spin^c$ obtained by lifting $i \times \det : U \hookrightarrow SO \times U(1)$ by the double cover $Spin^c \to SO \times U(1)$ (where $i : U \hookrightarrow SO$ is the obvious inclusion). Given a principal $U$-bundle $F$, we can thus consider $E = F \times_\rho Spin^c$, i.e. the quotient $(F \times Spin^c)/U$ where $h \in U$ acts by $(f,g) \mapsto (f\cdot h, \,\rho(h^{-1})g)$. Then we get a well-defined action of $Spin^c$ on $E$ by setting $[(f,g)]\cdot h \mapsto [(f, g\rho(h))]$ for $h \in Spin^c$, and this makes $E$ a principal $Spin^c$-bundle. -In a similar way, the fundamental line bundle $L_E$ of any spin$^c$ bundle $E$ is the $U(1)$-bundle associated to $E$ by the homomorphism $\psi : Spin^c \to U(1)$ defined by composing the double cover of $SO \times U(1)$ with projection to the right factor. Since $\psi \circ \rho : U \to U(1)$ is precisely $\det$, the fundamental line bundle of the spin$^c$ bundle associated to a complex vector bundle is simply $\det F$. Hence $c_1(L_E) = c_1(\det F) = c_1(F)$ in this case, and it is well-known that $w_2(E) = c_1(F)$ mod 2. -For the final part, note that the classifying map $f : BU \to BSpin^c$ for the universal complex bundle $EU$ as a spin$^c$ bundle can be represented as a fibre bundle with fibres $Spin^c/U$. For if $ESpin^c$ is a contractible space with a free $Spin^c$ action, then the subgroup $U$ also acts freely. Hence we can take $ESpin^c/U$ as a representative of $BU$, and then $f$ is simply the projection $ESpin^c/U \to ESpin^c/Spin^c$, whose fibres are $Spin^c/U$. -This fibre has a transitive action by $Spin$ with stabiliser $SU$, so it's homeomorphic to $Spin/SU$. In degree $0 < i \leq 5$, the homotopy groups are -$$ \pi_i(Spin/SU) = \pi_i(Spin(6)/SU(3)) = \pi_i(SU(4)/SU(3) = \pi_i(S^7) = 0, $$ implying that $f$ is a 5-equivalence (i.e. $f_* : \pi_i(BU) \to \pi_i(BSpin^c)$ is an isomorphism for $i < 5$, and surjective for $i = 5$). Therefore $f^* : H^i(BSpin^c) \to H^i(BU)$ is an isomorphism for $i \leq 4$. Hence any relation between spin$^c$ characteristic classes of degree up to $4$ that is valid for complex vector bundles is valid for any spin$^c$ bundle.<|endoftext|> -TITLE: Spinor bundle tensored with certain line bundle gives the dual spinor bundle -QUESTION [6 upvotes]: Let $E$ be a $spin^c$ bundle and $L_E$ be a (complex) line bundle defined using transition functions $\nu \circ g_{U,V}$ where $\nu:spin^c(n) \to \mathbb{T}$ is map such that $\ker \nu=spin(n)$ and $g_{U,V}$ are transition functions for $spin^c(n)$-principial bundle $spin(E)$ (see also here). Let $S$ be a spinor bundle i.e. vector bundle constructed from the system of transition functions $c \circ g_{U,V}$ where $c$ is irreducible representation of Clifford algebra. - -How to prove that the bundle $L_E$ satsifies the following $S \otimes L_E \cong S^*$ (where $V^*$ denotes the dual bundle). - -I tried to show this using tranistion functions: the best situation would be if transition functions from both sides exactly coincide. This is equivalent to -$$c(g_{U,V}(x)) \otimes \nu (g_{U,V}(x))=([c(g_{U,V}(x))]^t)^{-1}.$$ - -REPLY [4 votes]: Let $c : \mathbb{C}\mathrm{l}_n \to \operatorname{End}(S_n)$ be your irrep, so that the “dual” irrep $c^\ast : \mathbb{C}\mathrm{l}_n \to \operatorname{End}(S_n^\ast)$ is defined by -$$ - \forall x \in \mathbb{C}\mathrm{l}_n, \quad c^\ast(x) := c(x^!)^t. -$$ -Thus, if $\pi := c\vert_{\operatorname{Spin}^\mathbb{C}(n)} : \operatorname{Spin}^\mathbb{C}(n) \to U(S_n)$, then $\pi^\ast : \operatorname{Spin}^\mathbb{C}(n) \to U(S_n^\ast)$ is given by -$$ - \forall \lambda \in \mathbb{T}, \; \forall x \in \operatorname{Spin}(n), \quad \pi^\ast(\lambda x) = c((\lambda x)^{-1})^t = c^\ast(\lambda^{-1}x), -$$ -so that the induced representation $$\operatorname{Hom}(\pi,\pi^\ast) : \operatorname{Spin}^\mathbb{C}(n) \to U(\operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast)) = \mathbb{T}\operatorname{id}_{\operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast)}$$ is given by -$$ - \forall \lambda \in \mathbb{T}, \; \forall x \in \operatorname{Spin}(n), \; \forall T \in \operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast), \\ \operatorname{Hom}(\pi,\pi^\ast)(\lambda x)T := \pi^\ast(\lambda x) \circ T \circ \pi(\lambda x)^{-1} = c^\ast(\lambda^{-1} x) \circ T \circ c(\lambda^{-1}x^!) = c^\ast\left(\nu((\lambda x)^{-1})\right) \circ T. -$$ -Since -$$ - S \cong \operatorname{Spin}(E) \times_\pi S_n, \quad S^\ast \cong \operatorname{Spin}(E) \times_{\pi^\ast} S_n^\ast, -$$ -you can now use local trivialisations of $\operatorname{Spin}(E)$ to show that -$$ - \operatorname{Hom}_{\mathbb{C}\mathrm{l}(E)}(S,S^\ast) \cong \operatorname{Spin}(E) \times_{\nu^{-1}} \mathbb{C} \cong L_E^\ast, -$$ -and hence, by the natural isomorphism $S_n^\ast \cong S_n \otimes \operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast)$, that -$$ - S^\ast \cong S \otimes \operatorname{Hom}_{\mathbb{C}\mathrm{l}(E)}(S,S^\ast) \cong S \otimes L_E^\ast. -$$<|endoftext|> -TITLE: Regular $p$-norm of a matrix -QUESTION [9 upvotes]: Let $n \in \mathbb{N}$ and $p \in [1,\infty]$ be fixed and endow $\mathbb{C}^n$ with the $p$-norm $\|\cdot\|_p$. For every matrix $A \in \mathbb{C}^{n \times n}$ we denote the operator norm of $A$ as an operator on $\mathbb{C}^n$ by $\|A\|_p$, too. Moreover, let $|A|$ denote the matrix whose entries are the absolute values of the entries of $A$. -The number $\| \,|A|\,\|_p$ is sometimes called the regular norm of $A$ (in particular in Banach lattice theory, where a norm on the space of regular operators is constructed this way). -We clearly have $\|A\|_p \le \| \,|A|\,\|_p$, and equality holds for $p = 1$ and $p = \infty$. For general $p$, Mark Meckes explained in his answer to this question that the Riesz-Thorin theorem implies the estimate -\begin{align*} - \| \,|A|\,\|_p \le c_{n,p} \|A\|_p, \tag{E} -\end{align*} -where $c_{n,p} = n^{\frac{2}{p}(1 - \frac{1}{p})}$. -Question: Is the estimate (E) sharp? If not, what is the optimal constant for the estimate (depending on $n$ and $p$)? -Remarks: - -Clearly, the inequality (E) is sharp (and actually an equality) for $p = 1$ and $p = \infty$. -Let $U \in \mathbb{C}^{n\times n}$ be such that $\frac{1}{\sqrt{n}}U$ is a unitary matrix and such that each entry of $U$ has modulus $1$ (such a matrix exists for every $n$; for instance, we can choose $U$ to be the transformation matrix of the Fourier transform on $\mathbb{Z}/n\mathbb{Z}$). Then one can prove by employing Hölder's inequality several times that $\|\, |U| \, \|_p = n$ for each $p \in [1,\infty]$, and that $\|U\|_p = n^{1/p}$ for $p \in [1,2]$, while $\|U\|_p = n^{1-1/p}$ for $p \in [2,\infty]$. Thus, we obtain $\|\,|U|\,\|_p = \tilde c_{n,p} \|U\|_p$, where -\begin{align*} - \tilde c_{n,p} = - \begin{cases} - n^{1-1/p} \quad & \text{for } p \in [1,2], \\ - n^{1/p} \quad & \text{for } p \in [2,\infty]. - \end{cases} -\end{align*} -We have $\tilde c_{n,p} \le c_{n,p}$, and the existence of $U$ shows that the optimal constant for (E) is somewhere in between. -Note that $\tilde c_{n,2} = c_{n,2} = n^{1/2}$, so we know that the estimate (E) is sharp for $p=2$; this is also mentioned in Mark Meckes' answer quoted above. - -Note: A loosely related question concerned with the Schatten $p$-norm of a matrix can be found here. -Edits made 2018-06-03: - -I've replaced the real scalar field with $\mathbb{C}$ since this seems to be more appropriate for the question. -I've added the example matrix $U$ to show that the number $\tilde c_{n,p}$ is a lower bound for the optimal constant. - -REPLY [6 votes]: It is trivially $n^{1/p}$ for $p>2$ (and, therefore, $n^{1-\frac 1p}$ for $1 -TITLE: Matrix elements of exponential of tridiagonal matrices -QUESTION [7 upvotes]: Is there a way to compute one matrix element of the exponential of a tridiagonal matrix without having to compute the rest of the elements? - -Motivation: I'm trying to find the first passage time distribution from a master equation. I can impose an absorbing boundary at the threshold $n$, and the master equation with the new boundary condition is of the form $\frac{dp}{dt}=A p$ for a tridiagonal $n\times n$ matrix $A$. Then the first passage time distribution can be written as a particular matrix element of $\exp(A\,t)$. It takes forever for Mathematica to compute $\exp(A\,t)$ for large $n$, so I was wondering if there is a way to compute only the desired matrix element and not the whole matrix. - -REPLY [12 votes]: Yes! Most methods to compute exponentials of large sparse matrices are based on computing directly $\exp(A)b$ for a given vector $b$ rather than the full matrix $\exp(A)$. Just take $b$ as a vector of the canonical basis, and you're set. -The basic idea is that these algorithms are based on approximating the exponential with a rational function which is then expanded into a sum of partial fractions -$$\exp(A)b \approx q(A)^{-1}p(A)b = \sum_{i=1}^k \omega_k (A-\tau_k I)^{-1} b.$$ -Hence the problem is reduced to solving several shifted linear systems of the form $(A-\tau_k I)x_k=b$; this should work particularly well for a tridiagonal matrix, for which linear systems can be solved cheaply. -This technique is often combined with ideas from Krylov subspace methods, especially rational Krylov methods. -See for instance a chapter in the review https://doi.org/10.1017/S0962492910000036, https://doi.org/10.1137/100788860, or http://onlinelibrary.wiley.com/doi/10.1002/gamm.201310002/full, as well as the Matlab code in Stefan Güttel's webpage which you can use to compute it. -All these algorithms are approximate (but in the end, what is not approximate on a computer?), so you may encounter trouble if the element that you wish to compute is many orders of magnitudes smaller than $\|\exp(A)\|$. In that case, you may be interested in a bound on the decay rate of off-diagonal entries in https://arxiv.org/abs/1501.07376.<|endoftext|> -TITLE: Moy-Prasad and lattice stablizers -QUESTION [5 upvotes]: Consider $\mathrm{SO}(5)$, or maybe $\mathrm{SO}(n)$ over your favorite locally compact non-Archimedian field of characteristic $0$. There are two interesting families of compact open subgroups. The first is those coming from the Moy-Prasad filtration. The second are the stabilizers of lattices. My question is as follows: is it the case that a lattice stabilizer is a subgroup of an arbitrary step in the Moy-Prasad filtration? -I know the top of the Moy-Prasad filtration is a maximial parahoric, which is known (by Gan-Yu) to be a subgroup of the stabilizer of a lattice. But I don't know much more than that. - -REPLY [4 votes]: Let $F$ be a non-archimedean locallly compact field (of odd residue characteristic). Let $G$ be a classical group attached to a non-degenerate (symmetric, symplectic, or hermitian) form on an $F$-space $V$. Let $G'={\rm Aut}_F(V)$, a general linear group. Then Bruhat and Tits proved (Bull. SMF 1987) that the building $X$ of $G$ embeds in the building $X'$ of $G'$ in a $G$-equivariant way, the image being the fixed point set of a certain involution. A lattice $L$ in $V$ corresponds to a point $x\in X\subset X'$. Therefore its stabilizer $G_L$ is the stabilizer of $x\in X$ in $G$. It follows that $G_L$ is non necessarily a parahoric subgroup of $G$, but that its connected component $G_L^0$ is a parahoric subgroup. -More generally we have the notion of a self dual lattice sequence in $V$. There are decreasing sequences of lattices satisfying a certain symmetry condition relative to the form defining $G$. To such a sequence $\Lambda$ one attaches : --- a point $x=x_\Lambda$ in $X$, --- a filtration of the Lie algebra of G and a filtration of the parahoric subgroup attached to $x$. -It turns out that these filtrations are nothing other than the filtrations defined by Moy and Prasad. The proof is due to Bertrand Lemaire: -Lemaire, Bertrand Comparison of lattice filtrations and Moy-Prasad filtrations for classical groups. J. Lie Theory 19 (2009), no. 1, 29--54.<|endoftext|> -TITLE: Specialization map étale cohomology -QUESTION [7 upvotes]: Let $R$ be a henselian dvr, $s,\eta\in\text{Spec}(R)$ the closed and generic points, and $f : X\to \text{Spec}(R)$ a proper smooth scheme. -For a prime $\ell$ invertible on $R$, is there a specialization map -$$sp^i_{\eta,s} : R^if_*(\mu_{\ell^n})_{s}\to R^if_*(\mu_{\ell^n})_{\eta}$$ -(with $\eta$ and $s$, not the geometric points over them)? -Is it an isomorphism, injective, surjective? - -REPLY [8 votes]: If $\ell$ is invertible in $R$, then $R^i f_* (\mu_{\ell^n}) $ is a locally constant sheaf on $R$ by smooth and proper base change. Hence it is a represenation of the fundamental group of $R$, which is equal to the Galois group of the residue field, because $R$ is Henselian. -This fundamental group is easily seen to be a quotient of the Galois groups of both $s$ and $\eta$, and hence the "stalks" at $s$ and $\eta$, which are the Galois-invariants of the geometric stalks, are both equal to the $\pi_1$-invariants, hence naturally isomorphic.<|endoftext|> -TITLE: Does exist a Kahler-Einstein metric on the blow-up of $\mathbb{P}^3$ along a smooth plane cubic? -QUESTION [5 upvotes]: This might be well known for the experts but I am not able to find a reference. I was wondering if there exists a Kahler-Einstein metric on the Fano threefold given by blow-up of $\mathbb{P}^3$ along a smooth plane cubic or not. I believe that the answer shoud be "no" by Matsushima's criterion saying that if the automorphism group of the variety is not reductive then there is no such metric. -I was thinking that the following might be a (sketch of) proof: The automorphisms of $\mathbb{P}^3$ fixing a plane (say $x_3=0$) are of the form -$$\begin{pmatrix} -* & * & * & * \\ -* & * & * & * \\ -* & * & * & * \\ -0 & 0 & 0 & * -\end{pmatrix} $$ -This group has projective dimension 12 and since 9 points determines a cubic in $\mathbb{P}^2$ we have that the automorphism group of the blow-up is 3-dimensional. On the other hand, the 3-dimensional unipotent subgroup -$$\begin{pmatrix} -1 & 0 & 0 & * \\ -0 & 1 & 0 & * \\ -0 & 0 & 1 & * \\ -0 & 0 & 0 & 1 -\end{pmatrix} $$ -acts trivially on the plane $x_3=0$, so in particular fixes the 9 points determining the cubic and hence lifts to the automorphism group of the blow-up. Since they have the same dimension (but the latter might be non connected), the automorphism group if a finite extension of a unipotent group, hence not reductive. -Am I right? Sorry for being sketchy! Thanks a lot in advance for any comment, I just started to introduce myself to the subject. -PS: By the way, does the fact that $\operatorname{Aut}(X)$ is reductive implies that the connected component of the identity $\operatorname{Aut}^0(X)$ is reductive as well? - -REPLY [8 votes]: No, there are no Kähler—Einstein metrics on the blow-up of $\mathbb{P}^3$ along a plane cubic. -Let $X$ be the blow-up of $\mathbb{P}^3$ along a plane cubic. Then the alpha-invariant $\alpha(X)=1/4$ by Cheltsov—Shramov[1]. On the other hand, the only K-semistable smooth Fano 3-fold with $\alpha=1/4$ is $\mathbb{P}^3$ by my result[2]. So $X$ could not be K-semistable, hence not KE. -There is also another earlier proof by Kento Fujita[3], which shows that the blow-up of $\mathbb{P}^3$ along a plane cubic (which is referred to as No. 28 in Table 2 of Mori--Mukai) is not K-semistable. -Using the same methods you can show that the blow-up of $\mathbb{P}^3$ along a conic or a line is not K-semistable. -References: -[1] Bottom line, Page 952, Log canonical thresholds of smooth Fano threefolds, Russian Math. Surveys 63:5 859--958 -[2] K-semistable Fano manifolds with the smallest alpha invariant, Internat. J. Math. 28 (2017), no. 6, 1750044, 9pp -[3] On K-stability and the volume functions of Q-Fano varieties, Proc. LMS 113.5 (2016): 541--582<|endoftext|> -TITLE: Vanishing of H-cohomology -QUESTION [8 upvotes]: This looks elementary, but somehow I am stuck, please bear with me: -Let $H$ be a differential 3-form, nowhere vanishing, but not necessarily closed. What is a sufficient condition that the sequence of forming wedge products with $H$ is exact, i.e. that -$$ - \mathrm{ker}(H \wedge(-))\,/\,\mathrm{im}(H \wedge(-)) \;=\; 0 -$$ -? -A necessary condition is clearly that $H \neq \alpha_1 \wedge \beta_2$ for a 1-form $\alpha_1$. When is this sufficient? -If $H$ is closed, this is asking for vanishing "$H$-cohomology" in the terminology of arXiv:0501406. -I am really interested in the variant where $H$ is not a differential form, but an element of bi-degree $(3,\mathrm{even})$ in the Chevalley-Eilenberg algebra of a super Lie algebra, specifically the element in (28) of hep-th/0406020. -But that shouldn't make make much of a difference of principle. More generally I could ask this question for any free graded-commutative algebra in characteristic zero. - -REPLY [4 votes]: I have not thought about the variant problem, but in the ordinary differential form case, I do not believe it is possible to have vanishing $H$-cohomology on a (finite-type) smooth manifold $M$, regardless of any conditions on $H$ such as nowhere vanishing, closed, etc. -We proceed by contradiction, supposing that the $H$-cohomology does vanish on $M$. Note that as a differential form, one can write $H$ as a finite sum $H = \sum_{i=1}^{k}\beta_i \wedge \alpha_i$ (one establishes finiteness via a partition of unity argument on a finite covering by coordinate patches). Our key claim is the following: -Claim: For any sequece $1 \leq i_1 < \ldots < i_{\ell} \leq k$, there exist differential forms $\gamma_{i_1\cdots i_{\ell}} \in \Omega^{k-3-\ell}(M)$ (in particular they are zero if $\ell > k-3$) such that the differential forms $\alpha_{\widehat{i_1\cdots i_{\ell}}} := \alpha_1 \wedge \cdots \wedge \widehat{\alpha_{i_1}} \wedge \cdots \wedge \widehat{\alpha_{i_{\ell}}} \wedge \cdots \wedge \alpha_k$ satisfy $$\alpha_{\widehat{i_1\cdots i_{\ell}}} = (-1)^{i_1+\cdots+i_{\ell}+\ell} \left(H \wedge \gamma_{i_1\cdots i_{\ell}} + \sum_{j=1}^{\ell}(-1)^{j+1}\beta_{i_j}\wedge \gamma_{i_1\cdots\widehat{i_j}\cdots i_{\ell}}\right).$$ -Remark on Notation: If $\ell = 0$, then we just take $\alpha_{\widehat{\emptyset}} = \alpha_1 \wedge \cdots \wedge \alpha_k$. Similarly, we take $1 = \alpha_{\widehat{1,2,\ldots,k}}$. -Claim implies result: For the case of $1 = \alpha_{\widehat{1,2,\ldots,k}}$, we see that the right-hand side of the equation is just $0$ since the the corresponding $\gamma$ terms are in negative degree. This yields a contradiction. (One could also have taken $\ell = k-1$ instead of $k$ and assumed that the decomposition of $H$ had minimal $k$.) -Proof of claim: We proceed by induction on $\ell$, beginning with the case of $\ell = 0$. We see that $$H \wedge (\alpha_1 \wedge \cdots \wedge \alpha_k) = 0,$$ and so by vanishing of $H$-cohomology, indeed we find $\alpha_{\widehat{\emptyset}} = H \wedge \gamma_{\emptyset}$. -Now suppose that we have that the formula holds for all $\ell < \ell_0$. Then -\begin{eqnarray*} -H \wedge \alpha_{\widehat{i_1\cdots i_{\ell_0}}} &=& \sum_{i=1}^{k} \beta_i \wedge \alpha_i \wedge \alpha_{\widehat{i_1\cdots i_{\ell_0}}} \\ -&=& \sum_{j=1}^{\ell_0} (-1)^{i_j+j}\beta_{i_j} \wedge \alpha_{\widehat{i_1\cdots \widehat{i_j}\cdots i_{\ell_0}}} \\ -&=& \sum_{j=1}^{\ell_0} (-1)^{i_1 + \cdots + i_{\ell_0}+\ell_0+j+1}\beta_{i_j} \wedge \\ && \left(H \wedge \gamma_{i_1\cdots \widehat{i_j}\cdots i_{\ell_0}}+\sum_{k=1}^{j-1}(-1)^{k+1}\beta_{i_k}\wedge \gamma_{i_1\cdots \widehat{i_k} \cdots \widehat{i_j} \cdots i_{\ell_0}} + \sum_{k=j+1}^{\ell_0}(-1)^k\beta_{i_k}\wedge \gamma_{i_1\cdots \widehat{i_j} \cdots \widehat{i_k} \cdots i_{\ell_0}}\right) \\ -&=&(-1)^{i_1 + \cdots + i_{\ell_0}+\ell_0}\sum_{j=1}^{\ell_0} H \wedge (-1)^{j+1}\beta_{i_j} \wedge \gamma_{i_1\cdots \widehat{i_j}\cdots i_{\ell_0}} -\end{eqnarray*} -By vanishing of $H$-cohomology, the inductive step now follows.<|endoftext|> -TITLE: Localization and intersection -QUESTION [9 upvotes]: It is very well known that if $\mathfrak p_1, \ldots,\mathfrak p_n$ are prime ideal of an integral domain $A$, then we have the equality$$S^{-1}A=\bigcap_{i=1}^n A_{\mathfrak{p}_i},$$ where $S:=A\setminus\cup_i\mathfrak p_i$. -For a proof of the above, see for example this. -Now I'm asking myself if the equality above holds true for any family (infinite) of prime ideals. Any reference or counterexample would be appreciated. - -REPLY [6 votes]: With a view to understand to which extent OP's identity can fail, I am sharing these easy positive results. The emphasis is on the fact that OP's question tightly relates to (generalizations of) the prime avoidance lemma for ideals of denominators. -Let $A$ be an integral domain. Let $(\mathfrak{p}_i)_{i \in I }$ be an arbitrary family of prime ideals of $A$. Let $S = A \setminus \bigcup_{i \in I} \mathfrak{p}_i$. - -Claim 1. The inclusion $S^{-1}A \subset \bigcap_{i \in I} A_{\mathfrak{p}_i}$ always holds. If moreover $\{\mathfrak{p}_i\}_{i \in I}$ contains the set of all maximal ideals of $A$, then $A = S^{-1}A = \bigcap_{i \in I} A_{\mathfrak{p}_i}$. -Proof. Elementary, see e.g., [1, Theorem 4.7]. - -Let $R$ be an arbitrary commutative ring with identity. We say that $R$ satisfies the prime avoidance lemma for unions of cardinality $\kappa$ if for every ideal $I$ of $R$ and every index set $K$ of cardinality $\kappa$, the inclusion $I \subset \bigcup_{k \in K} P_k$ implies $I \subset P_l$ for some $l \in K$, where the ideals $P_k$ are arbitrary prime ideals of $R$. -Note that the prime avoidance lemma for countable unions holds for some interesting classes of rings, see for instance the answer of Neil Epstein to this MO post. For more on prime avoidance, I recommend [Exercises 3.16 to 3.20, 2]. The bleeding-edge results on infinite prime avoidance are as of today in [3]. - -Claim 2. If $A$ is GCD domain, or satisfies the prime avoidance lemma for unions of cardinality $\vert I \vert$, then $S^{-1}A = \bigcap_{i \in I} A_{\mathfrak{p}_i}$. -Proof. Let $x \in \bigcap_{i \in I} A_{\mathfrak{p}_i}$ and let $\mathfrak{d} = \{ d \in A \, \vert \, dx \in A \}$ be the ideal of denominators of $x$. Then we have $\mathfrak{d} \not\subset \mathfrak{p}_i$ for every $i \in I$. If $A$ is a GCD domain, then $\mathfrak{d}$ is a principal ideal generated by some $d_x \in S$, thus $x \in S^{-1}A$. If $A$ satisfies the prime avoidance lemma for unions of cardinality $\vert I \vert$, then we certainly have $\mathfrak{d} \cap S \neq \emptyset$, so that $x$ has a denominator in $S$. - - -[1] H. Matsumura, "Commutative Ring Theory", 1986. -[2] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 1995. -[3] J. Chen, "Infinite prime avoidance", 2017.<|endoftext|> -TITLE: Consistency strength of Berkeley cardinals -QUESTION [8 upvotes]: From Cantor's Attic: - -A cardinal κ is a Berkeley cardinal, if for any transitive set $M$ with $κ∈M$ and any ordinal $α<κ$ there is an elementary embedding $j : M → M$ with $\alpha<\text{crit}(j)<\kappa$. These cardinals are defined in the context of ZF set theory without the axiom of choice. - -I have seen claimed several times that the existence of a Berkeley is stronger than that of a Reinhardt cardinal (let's say in NBG without Choice, as formalizing existence of Reinhardts in ZF takes some extra consideration). I can't however find any proof of this claim. -Of course if Berkeley-ness were about elementary embeddings of arbitrary classes containing $\kappa$, then it would be obvious, but we are only talking about transitive set models here, so I don't really see what to do. Can anyone provide a proof that Berkeley are stronger than Reinhardts? What about Super Reinhardts? - -REPLY [7 votes]: The following is proved by Bagaria-Koellner-Woodin: -Theorem. If $δ_0$ is the least Berkeley cardinal, then there exists $γ < δ_0$ such that $(V_γ,V_{γ+1})\models ZF_2+$“There exists a Reinhardt card., witnessed by $j$, and an $ω$-huge card. above $κ_ω(j)$”. -See Large Cardinals beyond Choice. Let me sketch the proof, skipping some details: -First, one can show that for a tail of ordinals $\beta$, if $j: V_\beta \to V_\beta$ is an elementary embedding with $crit(j)< \delta_0$, then $j(\delta_0)=\delta_0$ and $\{\alpha < \delta_0: j(\alpha)=\alpha \}$ is cofinal in $\delta_0$. Fix such a $\beta$ and $j$. Let $\kappa=crit(j), \lambda=κ_ω(j)$ -Since $\delta_0$ is a Berkeley cardinal there are embeddings $j': V_\beta \to V_\beta$ with $\lambda < -crit(j') < \delta_0$. Also one can show $\kappa_\omega(j')< \delta_0$ holds for any such $j'$. -So there are $\lambda'< \delta_0$ and $j': V_{\lambda'} \to V_{\lambda'}$ with $κ_ω(j')=\lambda'$ and -$crit(j') > \lambda$. Let $\lambda'$ be the least ordinal with this property and let $\gamma$ - be the least -strongly inaccessible cardinal greater than $\lambda'$. Then $j(\lambda') = \lambda'$ and $j(\gamma) = \gamma.$ - Thus, $j\restriction V_\gamma$ -and $j'$ witness that $(V_γ,V_{γ+1})\models ZF_2+$“There exists a Reinhardt card., witnessed by $j$, and an $ω$-huge card. above $κ_ω(j)$”.<|endoftext|> -TITLE: Integral complete 4-partite graphs -QUESTION [8 upvotes]: For given block sizes $a0$ are enough because we can -clear denominators to make them integral, and then permute to get -$a -TITLE: Is every true statement independent of $PA$ equivalent to some consistency statement? -QUESTION [7 upvotes]: Most true statements independent of PA that I know of is equivalent to some consistency statement. For example - -Con(PA), Con(PA + Con(PA)), Con(PA + Con(PA) + Con (PA + Con(PA)), $\dots$ -Goodstein's theorem is equivalent to Con(PA) -Any conjunction or disjunction of the above. - -Is every true statement independent of PA equivalent to some consistency statement? -By "equivalent to some consistency statement", I mean that $PA \vdash S \iff Con(T)$, for some theory $T$. Also, $T$ should be either finite, or specified by a Turing machine that outputs its axioms (and such that PA proves that the Turing machine never stops outputting statements), so that the description of $T$ doesn't throw PA off. -EDIT: In particular, are there are $\Pi^0_1$ examples? - -REPLY [7 votes]: $1$-consistency of $PA$ is a true $\Pi_3$ sentence which is not provable in $PA$+{all true $\Pi_1$ sentences} (see this article). Simple (iterated) consistency statements (as you mentioned above) are all (true) $\Pi_1$ sentences, so it is not equivalent to any $\Pi_1$ sentence.<|endoftext|> -TITLE: What is the distance between two points on the Berger metric of the squashed three-sphere? -QUESTION [7 upvotes]: The Berger metric on a "squashed" three-sphere is given (in Euler coordinates) by -4 $ds^2 = \lambda^2 (d \tau + \cos \theta d \phi)^2 + d \theta^2 + \sin^2 \theta d \phi^2$. -See for example Eq. 1.11 of this paper. This formula describes the geodesic distance between two infinitesimally separated points. -My question: what is the geodesic distance between any two points on the Berger sphere? -What is the distance as a function of the coordinates $\tau_1,\phi_1,\theta_1$ and $\tau_2,\phi_2,\theta_2$ of the two points? - -REPLY [2 votes]: Geodesics in Berger sphere become circle arcs in the quotient -$$\tfrac12\cdot\mathbb{S}^2=\mathbb{C}\mathrm{P}^1=\mathbb{S}^3/\mathbb{S}^1.$$ -Once you see it, all calculations should be easy. (But I do not want to calculate.)<|endoftext|> -TITLE: Nice S¹-action implies existence of unconditional basis? -QUESTION [6 upvotes]: Let $V$ be a Banach space equipped with a continuous linear action of $S^1$ (meaning, the map $S^1\times V\to V$ is continuous). Assume that all the eigenspaces of the $S^1$-action are finite dimensional. - -Does $V$ then admit an unconditional basis consisting of eigenvectors for the $S^1$-action? - -If the answer is yes, then I'm also interested in the corresponding question when $V$ is Frechet, or when $V$ is a general complete locally convex topological vector space. - -REPLY [3 votes]: I think that the answer to my question is no: -The space $L^1(S^1)$ (equipped with its obvious $S^1$-action) satisfies all my assumptions, but it has no unconditional basis: https://en.wikipedia.org/wiki/Schauder_basis#Unconditionality -In particular, the functions $\theta\mapsto e^{in\theta}$ do not form an unconditional basis.<|endoftext|> -TITLE: Obstruction to a general S^1-action -QUESTION [17 upvotes]: Question: Suppose you have a simply connected, closed, orientable, smooth manifold $M$. What are some restrictions to the existence of a smooth (non-trivial) $S^{1}$-action on $M$? -Note: 1.There are some well known restrictions in the case that we require this action to have isolated fixed points (or preserve some additional structures), but I am not aware of any obstruction in this generality. - -I am aware of the classification of manifolds with $S^{1}$-actions in dimension $2$ and $3$. In higher dimensions (say $\geq 6$) it seems to become less clear. -I am particularly interested in dimension 6. I would like to get a picture of which manifolds in this dimension (satisying the above restrictions), have a smooth $S^{1}$-action. - -Edit: Out of curiosity, using Puppe's methods, is it possible to construct a simply connected, smooth, complex projective 3-fold such that the underlying smooth manifold has no circle actions? -Edit 2 It turns out that (the underlying smooth 6-manifold of) any smooth hypersurface in $\mathbb{C}\mathbb{P}^4$ of degree atleast 3 has no smooth (non-trivial) circle actions. It follows from the main result of https://arxiv.org/pdf/1108.5327.pdf. - -REPLY [10 votes]: V. Puppe, in Simply connected manifolds without $S^1$-symmetry. Algebraic topology and transformation groups, Proc. Conf., Göttingen/FRG 1987, Lect. Notes Math. 1361, 261-268 (1988) proved the following: - -There exist a simply connected, closed, oriented smooth 6-dimensional manifold $M$ such that no closed, orientable manifold with the same rational cohomology algebra as $M$ admits a non-trivial circle action.<|endoftext|> -TITLE: Trace-class operator satisfies $\sum |\lambda_n|<\infty$? -QUESTION [13 upvotes]: Here's an "exercise" which I thought should be easy, but which I find myself unable to do. -Let $V$ be a Banach space. -Recall that an operator $f:V\to V$ is trace-class if it is in the image of the natural map $V\otimes_\pi V'\to \mathcal L(V,V)$, where $\otimes_\pi$ denotes the projective tensor product, and $V'=\mathcal L(V,\mathbb C)$ is the continuous dual of $V$. - -Exercise:  [Edit: this is probably FALSE; thank you Mateusz Wasilewski] - Let $f:V\to V$ be a trace-class linear map which is "diagonalizable", in the sense that the linear span of its eigenspaces is dense in $V$. - Write $\lambda_1,\lambda_2,\ldots$ for the eigenvalues of $f$ (enumerated with multiplicities). Then $$\sum |\lambda_n|<\infty.$$ - -Given that my previous "exercise" turned out to be probably false, here's a new one: - -Exercise v2: - Let $f:V\to V$ be a trace-class linear map which is diagonalizable (in the same sense as above), and whose eigenvalues $\lambda_1,\lambda_2,\ldots$ are all positive real numbers. Then $$\sum \lambda_n<\infty.$$ - -Can someone please help me prove it? -And if this second exercise also too difficult, here's an even simpler one (which I also don't know how to prove): - -Exercise v3: - Let $f:V\to V$ be a trace-class linear map which is diagonalizable (in the same sense as above), with eigenvalues $\lambda_1,\lambda_2,\ldots$. Then $$\lim_{n\to\infty} \lambda_n=0.$$ - -PS: I'm also interested in the same question when $V$ is Frechet, or when $V$ is a general complete locally convex topological vector space. - -REPLY [5 votes]: As I mentioned in a comment, exercise 3 has a positive answer: Nuclear operators are absolutely $2$-summing, and $2$-summing operators have $2$-summable eigenvalues (see, e.g., Tomczak's book). -Exercise 2 has a negative answer, but I am not satisfied with the example and do not know what happens in various classical spaces. First, note that it is sufficient to find for each $n$ a space $X_n$ on which there is a rank $n$ projection $P_n$ that has nuclear norm of order $n^{1/2}$. One can on $(I-P_n)X_n$ construct a diagonalizable (in the sense of the OP) operator $U_n$ with positive eigenvalues and nuclear norm as small as desired, so $S_n :=P_n + U_n(I-P_n)$ will have nuclear norm of order $n^{1/2}$ but the eigenvalues sum to about $n$. Multiply the $S_n$ by suitable constants and take an $\ell_2$ direct sum to see that exercise 2 has a negative answer. -The problem is that on a space that has the approximation property, the nuclear norm dominates the trace for finite rank operators, so the desired $X_n$ must fail the approximation property. Fortunately (or unfortunately, depending on one's point of view) Pisier constructed a space on which the nuclear norm of every finite rank operator is of order the norm of the operator. Now use the fact that in every Banach space there is a projection of norm at most $n^{1/2}$ onto every $n$ dimensional subspace (also in Tomczak's book).<|endoftext|> -TITLE: Measurable maximal independent set in infinite graph of bounded degree -QUESTION [6 upvotes]: We have a graph $G$. The vertices of $G$ are a measurable subset of $\mathbb{R}^n$ for some $n$. The degree of each vertex is bounded by some absolute finite constant $K$. -Q1. Does $G$ have a maximal independent set? (This is a set of mutually non-adjacent vertices such that each vertex not in the set is adjacent to at least one vertex in the set.) -Q2. If the answer to Q1 is "yes", can we find a maximal independent set which is measurable? -As you can probably tell, my set theory is weak. - -REPLY [6 votes]: A relevant statement is Proposition 4.2. in - -[KST1999] A. S. Kechris, S. Solecki, S. Todorcevic, Borel Chromatic Numbers. Advances in Mathematics 141, 1-44 (1999) - -In short, the proposition shows that if you settle for Borel sets instead of arbitrary measurable sets, then Q1 and even Q2 seem(problem) to admit of an affirmative answer, and then even in greater generality (only local countability, not boundedness, of the vertex degrees required).(caution) -In detail, - -Proposition 4.2 ([KST1999, p. 12]) If $X$ is a standard Borel space, and if $R\subset X\times X$ is a symmetric irreflexive relation, and if -(range) for each Borel set $Y\subseteq X$ the set$\quad$ $R(Y) := \{x\in X\colon\ \exists y\qquad y\in Y\quad\wedge\quad (y,x)\in R\}$ is a Borel set, too, and -(chrom) if the Borel chromatic number of $(X,R)$ is countable, -then $X$ contains a maximal $R$-independent set, which moreover is Borel (and hence measurable). - -To prepare for a conditional answer to your question, one also should quote (many thanks to Francis Adams for pointing this out): - -Proposition 4.5 ([KST1999, p. 13]) If $X$ is a standard Borel space, and if $R\subset X\times X$ is an irreflexive relation (not necessarily symmetric), and if -(preimage) for each Borel set $Y\subseteq X$, the set $R^{-1}(Y):=\{x\in X\colon\ \exists y\qquad y\in Y\quad\wedge\quad (x,y)\in R\}$ is a Borel set, too, an -(degree) (*) each $x\in X$ has finite degree in the undirected graph obtained from $R$ by symmetrizing the relation, -then the Borel chromatic number of $(X,R)$ is countable. - -Now to your question proper. Suppose $R$ is any irreflexive symmetric relation on, not only a measurable, but a Borel vertex-set. (This assumption is why this answer is only a conditional answer to your question; arbitrary measurable sets are not covered, and would not allow to conclude that the maximal independent set is measurable, as exemplified by Han's example at 2018-02-19 03:22:58Z). Suppose that as in the OP, $R$ has bounded vertex-degree; then (degree) in Proposition 4.5 holds. -Since $R$ is symmetric, for any $Y\subseteq X$ we have $R(Y)=R^{-1}(Y)$. -Now we can conclude the following: - -if we could(problem) show that under the hypotheses in the OP it follows that (range)(* *) is true, -then first Proposition 4.5 would imply that (chrom) is true, whereupon Proposition 4.2 then would imply that $X$ contains a maximal $R$-independent set, which moreover is Borel (and hence measurable), answering the OP's question affirmatively. - -${}$_________________________ -(*) In stating the following condition, op. cit. strangely first passes to the undirected graph underlying $R$, but then goes on to speak of an "outdegree" (cf. op. cit. Proposition 4.5); I find this confusing and can't explain this. I take the "out-degree" to simply mean "degree", since after all one is speaking of an undirected graph here. -(* *) Because of $R(Y)=R^{-1}(Y)$, we have (range) $\leftrightarrow$ (premimage). -(caution) An earlier version of this answer erroneously stated that one may then also conclude that the maximal independent set was (topologically-) discrete; this particular additional statement was false. (And quite nonsensical, since in Proposition 4.2 no topology is specified.) I was misled by the unusual usage of 'discrete' in op. cit. (and by my predjudices: I was thinking of the Lebesgue $\sigma$-algebra on $\mathbb{R}^d$, which makes one think of topology), but Propostion 4.2 in op. cit. is more general and does not use 'discrete' in the topological sense: unusually, 'discrete' in op. cit. is nothing but a synonym for what in graph-theory is called 'independent' (i.e., set no two members of which are connected by an edge). My error is understandable: in the setting of Borel-spaces, taking 'discrete' to have its usual topological meaning is quite reasonable, so why would one rummage through the article to find the definition of 'discrete'? -However, my statement that the maximal independent set could in addition be assumed to be discrete was not only not the intended one, but it is also materially false: a counterexample is the graph with vertex-set equal to the unit-circle, and two points adjacent if and only if one can be carried to the other by rotating the circle by a rational angle. (This is precisely the complement of the graph defined by Han at 2018-02-19 03:22:58Z.) -In this graph, the set $\{ (\cos(\varphi),\sin(\varphi))\colon\quad \varphi\in\mathbb{Q}\cap[0,2\pi)\}$ is a maximal independent set. Of course---being countable---it is Borel (as it must by Proposition 4.2.) but, needless to say, it is not (topologially-)discrete. -(problem) The problem is that, some helpful comments notwithstanding, I still don't see whether the following is true - -(problem) For any standard Borel space $X$, any Borel set $Y\subseteq X$, and any irreflexive symmetric relation $R\subset X\times X$ with finite degree at each point, does it follow that he set $R(Y) = \{x\in X\colon\ \exists y\qquad y\in Y\quad\wedge\quad (y,x)\in R\}$ is a Borel set, too? - -I did not look long into this matter, but in view of some similar results depending on the topology, I would rather suspect that (problem) has negative answer. Some results I looked at involve images of Borel sets under finite-to-one functions, which seems somewhat related, but $R$ need not be a function, and moreover some of the relevant results make conditions on topologies, which one does not have available here at all.<|endoftext|> -TITLE: $K[[X_1,...]]$ is a UFD (Nishimura's Theorem) -QUESTION [11 upvotes]: Let us define the infinitely-many-variable formal power series ring -$$ -K[[X_1,\ldots]] \colon= \underset{m \geq 1}{\varprojlim}\,K[[X_1,\ldots,X_m]]. -$$ -$K[[X_1,\ldots]]$ is known to be a UFD by a theorem of Nishimura (c.f. On the unique factorisation theorem for formal power series, Journal Math. Kyoto. Univ. Vol 7. No 2. 1967, 151-160). -Now let us choose an irreducible element $f \in K[[X_1,\ldots]]$ and consider the image $f_m \in K[[X_1,\ldots,X_m]]$ of $f$ by the natural quotient ring homomorphism $K[[X_1,\ldots]] \twoheadrightarrow K[[X_1,\ldots,X_m]]$. -Q. Is $f_m$ also irreducible for $m \gg 0$? - -REPLY [4 votes]: Permit me to make the following bibliographic remark: the very same article of Nishimura which was cited by OP, already contains an affirmative answer to the OP's question: (1) on page 157 of Nishimura's 1967 article one reads - - - -Nishimura's proof, which seems self-contained and recommendable reading, uses too many preliminary results to be conveniently summarizable (by me). I tried to write an exposition, but that attempt foundered on my not understanding Nishiguro's argument in few places (which does imply anything for Nishimura's proof of course). -${}$___________________________ -(1) So the OP either did not read or did not trust Nishigura's article in its entirety (and thought it more polite to not mention any issues that there may or may not be with Ishimura's argument); of course there isn't anything wrong with that; it's perfectly fine to quote Nishigura's article for the purpose of giving a reference for the UFD-claim only. I simply think it should be pointed out for completeness that Nishimura's article seems to already contain an answer.<|endoftext|> -TITLE: What is this quotient of the triangle 2-3-7 group? -QUESTION [15 upvotes]: I have been working with Hurwitz groups, and I came across the group $G := \langle a, b \ | \ a^2, b^3, (ab)^7, ([a,b]^2ab)^6 \rangle$. I'm trying to figure out exactly what this group is. I know it contains two copies of ${\rm PSL}(2,13)$, and there seems to be a very large 2-group in there as well, of order $2^{28}$ at least. Is this group even finite, and if so, what is the order? Does it contain any other simple groups? - -REPLY [24 votes]: It is infinite. -As you pointed out yourself, the kernels of the maps onto ${\rm PSL}(2,13)$ have $2$-quotients of apparently unbounded order, which can be computed using the $p$-quotient algorithm. -I also found homomorphisms onto the Janko sporadic groups J1 and J2. Then I found a subgroup of index $525$ in J2 of which the inverse image in $G$ has infinite abelianization. Here are some Magma commands to do some of these computations. -> G := Group; -> SQ := SimpleQuotients(G,1000000: Limit:=10 ); -> ChiefFactors(Image(SQ[1][1])); - G - | A(1, 13) = L(2, 13) - 1 -> P := pQuotient(Kernel(SQ[1][1]), 2, 3 : Print:=1); - -Lower exponent-2 central series for $ - -Group: $ to lower exponent-2 central class 1 has order 2^14 - -Group: $ to lower exponent-2 central class 2 has order 2^42 - -Group: $ to lower exponent-2 central class 3 has order 2^70 - -> ChiefFactors(Image(SQ[2][1])); - G - | J1 - 1 -> h := SQ[3][1]; -> I := Image(h); -> ChiefFactors(I); - G - | J2 - 1 -> L := LowIndexSubgroups(I,1000); -> [Index(I,l): l in L]; -[ 840, 560, 280, 525, 315, 100, 1 ] -> p := CosetAction(I,L[4]); -> S := sub< G | h*p >; -> Index(G,S); -525 -> AbelianQuotientInvariants(S); -[ 6, 0 ]<|endoftext|> -TITLE: Haar measure on $\mathrm{SL}_3(\mathbb{Z}) \backslash \mathrm{SL}_3(\mathbb{R}) / \mathrm{SO}_3(\mathbb{R})$ -QUESTION [6 upvotes]: The bi-invariant Haar measure on the quotient $\mathrm{SL}_2(\mathbb{Z}) \backslash \mathrm{SL}_2(\mathbb{R}) / \mathrm{SO}_2(\mathbb{R})$ represents the moduli space of rank two real lattices modulo rotation and is easy to write down. An element of the quotient can be written (using the Iwasawa decomposition) as: -\[\begin{pmatrix} y^{1/2} & xy^{-1/2} \\ -0 & y^{-1/2} \end{pmatrix}\] -with $z=x+iy$ in a fundamental domain of the action of $\mathrm{SL}_2(\mathbb{Z})$ over the upper-half plane, such as $D = \{(x,y) : x^2+y^2 \geq 1,|x| \leq 1/2,y > 0\}$. -In these coordinates, the invariant measure is a scale of the hyperbolic measure, namely -\[\frac{3}{\pi} \frac{dx \, dy}{y^2}.\] -Is there a similar nice parametrisation for the rank 3 case (namely explicating the bi-invariant Haar measure of $\mathrm{SL}_3(\mathbb{Z}) \backslash \mathrm{SL}_3(\mathbb{R}) / \mathrm{SO}_3(\mathbb{R})$? - -REPLY [14 votes]: Sure, and you can even do this for $\mathrm{SL}_n$; I believe this goes back to Siegel. A good hands-on reference is Chapter 1 of Automorphic Forms and $L$-Functions for the group $\mathrm{GL}(n,\mathbb{R})$ by Dorian Goldfeld. -More precisely, we have the Iwasawa decomposition $z = xy$ for $z \in \mathrm{SL}_n(\mathbb{R}) / \mathrm{SO}_n(\mathbb{R})$, where -\[x = \begin{pmatrix} 1 & x_{1,2} & x_{1,3} & \cdots & x_{1,n} \\ - & 1 & x_{2,3} & \cdots & x_{2,n} \\ - & & \ddots & & \vdots \\ - & & & 1 & x_{n-1,n} \\ - & & & & 1 -\end{pmatrix}, \qquad -y = \begin{pmatrix} y_1 \cdots y_{n - 1} t & & & \\ - & \ddots & & \\ - & & y_1 t & \\ - & & & t -\end{pmatrix} -\] -where $x_{j,k} \in \mathbb{R}$, $y_{\ell} \in (0,\infty)$, and $t = \prod_{\ell = 1}^{n - 1} y_{\ell}^{\frac{\ell}{n} - 1}$ (so that $\det z = \det x \det y = 1$). Then the Haar measure is -\[dz = \prod_{1 \leq j < k \leq n} dx_{j,k} \prod_{\ell = 1}^{n - 1} y_{\ell}^{-\ell(n - \ell) - 1} \, dy_{\ell},\] -which gives $\mathrm{SL}_n(\mathbb{Z}) \backslash \mathrm{SL}_n(\mathbb{R}) / \mathrm{SO}_n(\mathbb{R})$ volume -\[n 2^{n - 1} \prod_{m = 2}^{n} \frac{\zeta(m)}{\mathrm{vol}(S^{m - 1})},\] -where $\zeta(s)$ denotes the Riemann zeta function and $S^{n - 1}$ denotes the $n - 1$-sphere, which has volume $2\pi^{n/2} / \Gamma(n/2)$. -In particular, the volume of $\mathrm{SL}_2(\mathbb{Z}) \backslash \mathrm{SL}_2(\mathbb{R}) / \mathrm{SO}_2(\mathbb{R})$ with respect to the Haar measure $dz = y^{-2} \, dx \, dy$ is -\[4 \frac{\zeta(2)}{\mathrm{vol}(S^1)} = \frac{\pi}{3},\] -as $\zeta(2) = \pi^2/6$, while the volume of $\mathrm{SL}_3(\mathbb{Z}) \backslash \mathrm{SL}_3(\mathbb{R}) / \mathrm{SO}_3(\mathbb{R})$ is -\[\frac{\zeta(3)}{4},\] -since $\Gamma(3/2) = \sqrt{\pi}/2$; note that there is no nice closed-form expression for $\zeta(3)$.<|endoftext|> -TITLE: Is every endomorphism of the sheaf of holomorphic functions on a disk a differential operator? -QUESTION [10 upvotes]: Let $D= \{z\in \mathbb{C}:|z| < 1\}$ be the unit disk. And consider the sheaf of holomorphic functions $\mathcal{O}_{D}$. - -Question (?) : Is there a sheaf endomorphisms $\phi : \mathcal{O}_D \to \mathcal{O}_D$ which is not a (possibly infinite order) differential operator. I.e. not of the form: -$$\phi=\Sigma_{n=0}^{\infty} b_n(z) \partial^n$$ -Where $\partial =\frac{d}{d z}$ and $b_n \in \mathcal{O}_D$ . - -EDIT: Suppose I require that $\phi$ be continuous w.r.t. to the natural frechet topology on $\mathcal{O}_D$ coming from uniform convergence on compact subsets, does the answer change? - -REPLY [4 votes]: Probably this paper answers your question negatively. -In particular it shows that if $X$ is an open subset of $\mathbb{C}^n$ and the sheaf $\mathcal{O}$ of holomorphic functions on $X$ is given the topology of compact convergence then every continous endomorphism of $\mathcal{O}$ is given by a convergent $\mathcal{O}$-linear sum of operators $\partial_1^{\alpha_{1}}\cdots \partial_n^{\alpha_n}$ with $\alpha_i\in \mathbb{N}$.<|endoftext|> -TITLE: Nonnegative matrices and singular values -QUESTION [5 upvotes]: I would like to prove (or prove it is not true with a counter example) the following result: -Let $A$, $B$ be two squares matrices of size $n\times n$ with positive entries. -If $A \leq B$, then $\sum_{i = 1}^n \sigma_i(A) \leq \sum_{i = 1}^n \sigma_i(B)$. -The sign $\leq$ between matrices is meant elementwise and the notation $\sigma_i(A)$ is used to denote the i-th singular value of $A$. The Perron-Frobenius theorem only provides information on the spectral radius, while the above statement involves all singular values. - -REPLY [8 votes]: The inequality is not true, in general. Here is a counterexample: -Let $B = - \begin{pmatrix} - 1 & 1 \\ 1 & 1 - \end{pmatrix} -$ -and let $A = - \begin{pmatrix} - 1 & 1 \\ 0 & 1 - \end{pmatrix} -$ (my favourite counterexample matrix). -The singular values of $B$ are $2$ and $0$. A short computation shows that the singular values of $A$ are $(\frac{3 + \sqrt{5}}{2})^{1/2}$ and $(\frac{3 - \sqrt{5}}{2})^{1/2}$; the sum of those two values is strictly larger than $2$ (approximately $2.236$, but admittedly I used a calculator to check this...)<|endoftext|> -TITLE: 2-adic valuation of $L(0,\chi)$ for a Dirichlet character -QUESTION [6 upvotes]: Let $\chi : (\mathbb Z/f\mathbb Z)^\times \to K = \mathbb Q(\mu_{\phi(f)})$ be a primitive Dirichlet character. Assume moreover that it is not quadratic, that is, $\chi^2$ is not the trivial character. Let $\pi_1,\dots,\pi_g$ be the primes lying over $2$ and $v_1,\dots,v_g$ be the corresponding valuations. Recall that: -$$L(0,\chi) = \frac1f\sum_{n=1}^fn\chi(n)$$ -Experimentally (upto conductor 200), I find that there always exists some $k$ such that $v_k(L(0,\chi)) > 0$. Does anyone know a proof? -Note that it is not true that $2 | L(0,\chi)$. For instance, for for the character of conductor $5$ mapping $2 \to i$, we have $L(0,\chi) = (i+3)/5$. There are lots of other examples. -Note also that we do require the condition that $\chi$ is non quadratic. For instance, if $f = p \equiv 3 \pmod4$ and $\chi$ is quadratic, then: -$$pL(0,\chi) \equiv (p-1)/2 \equiv 1 \pmod 2.$$ -I asked this question a few hours before on stackexchange but at the suggestion of someone, I am posting it here. I have deleted the question on stackexchange. - -REPLY [9 votes]: Your question is entirely answered and with a lot of additional detail by Corollary 11.4.2 of my book Springer Graduate Texts in Math GTM 240. This must be in the literature outside of my book, but I do not know a reference.<|endoftext|> -TITLE: maximal distance of nearby iid unifrom random variables -QUESTION [6 upvotes]: Question: Let $X_1, \ldots ,X_n$ be $n$ iid uniformly distributed random variables, i.e., $X_j \sim \mathcal{U}(0,1)$ for each $j=1,\ldots ,n$. What is the PDF of the maximal distance between to nearby elements? More formally, if we denote $X_{(k)}$ as the location of the $k$-th smallest element, what is the PDF of $\max\limits_{k=1,\ldots ,n-1} d_k$, where $d_k := X_{(k+1)} - X_{(k)}$. -The motivation for this question is approximation and interpolation on random grids. -What I know: Each $X_{(k)}$ is distributed by a Beta distribution, but so far I was not able to go on from there. - -REPLY [4 votes]: Let us just adapt the reasoning at stats-SE. Let $d_0:=X_{(1)}$ and $d_n:=1-X_{(n)}$. -Let $\Delta_j:=d_{j-1}$ for $j\in[n+1]:=\{1,\dots,n+1\}$, so that -\begin{equation} - \max_{1\le k\le n-1}d_k=\max_{2\le j\le n}\Delta_j. -\end{equation} -The first key observation at stats-SE is that the $\Delta_j$'s are exchangeable, since $(\Delta_1,\dots,\Delta_{n+1})$ equals $(E_1,\dots,E_{n+1})/S$ in distribution, where $E_1,\dots,E_{n+1}$ are iid standard exponential r.v.'s and $S:=E_1+\dots+E_{n+1}$ (see e.g. formula (2.2.1)). The second key observation at stats-SE is that -for any $J$ in the set $\binom{[n+1]}r$ of subsets of cardinality $r$ of the set $[n+1]$ and any $x\ge0$ -\begin{equation*} - P(\Delta_j>x\ \forall j\in J)=P(\Delta_j>x\ \forall j\in[r])=(1-rx)_+^n, -\end{equation*} -where $u_+^n:=\max(0,u)^n$. -So, by the inclusion-exclusion formula, we obtain the cdf of $\max_{1\le k\le n-1}d_k$: -\begin{align*} - P(\max_{1\le k\le n-1}d_k\le x)&=1-P(\max_{2\le j\le n}\Delta_j> x) \\ - &=1-P(\max_{1\le j\le n-1}\Delta_j> x) \\ - &=\sum_{r=0}^{n-1}(-1)^r \sum_{J\in\binom{[n-1]}r}P(\Delta_j>x\ \forall j\in J) \\ - &=\sum_{r=0}^{n-1}(-1)^r \binom{n-1}r (1-rx)_+^n -\end{align*} -for all $x\ge0$. -Addition in response to a comment by the OP: Let -\begin{equation*} - M_n:=\max_{1\le k\le n-1}d_k. -\end{equation*} -Then, using the above expression for the cdf of $M_n$, we have -\begin{multline*} - E M_n^2=\int_0^\infty 2x\,P(M_n>x)dx=\sum_{r=1}^{n-1}(-1)^{r-1} \binom{n-1}r \int_0^{1/r} 2x\,(1-rx)^n\,dx \\ - = \sum_{r=1}^{n-1}(-1)^{r-1} \binom{n-1}r\,\frac{2}{r^2(n+2)(n+1)} - =\frac{\psi(n)^2+2 \gamma \psi(n)- \psi'(n)+\pi ^2/6+\gamma ^2}{(n+2)(n+1)}, -\end{multline*} -where $\gamma$ is the Euler constant and $\psi:=\Gamma'/\Gamma$ is the digamma function; the latter expression for $E M_n^2$ was obtained by using a computer algebra package. We have $\psi(n)\sim\ln n$ and $\psi'(n)\to0$ as $n\to\infty$ wiki: Polygamma function, whence -\begin{equation*} - E M_n^2\sim\frac{\ln^2 n}{n^2}. -\end{equation*} - -Alternatively, without using a computer algebra package, we can obtain exact and asymptotic expressions for $E M_n^2$ as follows: using the formula $\frac1{r^2}=\int_0^\infty xe^{-rx}dx$ for $r>0$, then using the binomial formula, then making the substitution $u=1-e^{-x}$, and then integrating by parts, we have -\begin{multline*} - \frac{(n+2)(n+1)}{2}\,E M_n^2 - = \sum_{r=1}^{n-1}(-1)^{r-1} \binom{n-1}r\,\frac1{r^2} - = -\int_0^\infty x\,dx\sum_{r=1}^{n-1} \binom{n-1}r\,(-e^{-x})^r \\ - = \int_0^\infty x\,dx\,(1-(1-e^{-x})^{n-1}) - = -\int_0^1\ln(1-u)\,\frac{1-u^{n-1}}{1-u}\,du \\ - = -\sum_{j=0}^{n-2}\int_0^1\ln(1-u)\,u^j\,du - = \sum_{j=0}^{n-2}\frac1{j+1}\int_0^1\,\frac{1-u^{j+1}}{1-u}\,du - = \sum_{j=0}^{n-2}\frac1{j+1}\sum_{k=0}^j\int_0^1\,u^k\,du \\ - = \sum_{j=0}^{n-2}\frac{H_{j+1}}{j+1} - = \sum_{k=1}^{n-1}\frac{H_k}k, -\end{multline*} -so that -\begin{equation} - E M_n^2=\frac2{(n+2)(n+1)}\,\sum_{k=1}^{n-1}\frac{H_k}k, -\end{equation} -where $H_k$ is the $k$th harmonic number. Since $H_k\sim\ln k$ as $k\to\infty$, we have -\begin{equation} - E M_n^2\sim\frac2{n^2}\,\int_1^n\frac{\ln x}x\,dx=\frac{\ln^2 n}{n^2} -\end{equation} -as $n\to\infty$, the same result as before.<|endoftext|> -TITLE: Reasoning Using Countable Subsets of Real Numbers -QUESTION [6 upvotes]: The purpose of my question is trying to understand whether, in some cases, we can achieve greater certainty of reasoning (say when dealing with statements about natural numbers, integers or rational numbers) by using countable subsets of reals instead of using reals as a whole. I will try to explain my question in some detail, because otherwise it might seem completely unmotivated (and perhaps not making much sense). -For concreteness, consider a certain countable subset of real numbers $R$, which we might call the set of $r$-numbers (for the lack of a better word). As one specific case, call $A$ the collection of all "arithmetical functions" (functions computable using some $0^{(n)}$ oracle for some finite ordinal $n$). Some functions within $A$ are accepted as members of the set $R$ (the collection of $r$-numbers). Let's just briefly decide the format when a given function $f:\mathbb{N} \rightarrow \mathbb{N}$ is accepted as an $r$-number: -$\\f(0) \le 1$ (interpreted as sign) -$f(a) \le 9$ when $a \ge 2$ (interpreted as decimal expansion) -Now for $r$-numbers, we define the operations of equality, comparison, addition, subtraction, multiplication and division using appropriate (infinitary) modifications of highschool algorithm. We can also state the "psuedo-completeness" property: -"Every non-empty "arithmetical collection" of $r$-numbers, that is bounded above, has a least upper bound which is also an $r$-number." Here a non-empty "arithmetical collection" being defined by a suitable "arithmetical function" of the form $f:\mathbb{N^2} \rightarrow \mathbb{N}$ (each row of the function being interpreted as an $r$-number). - -Now it possibly happens often enough that when reasoning about a statement involving naturals,integers or rationals that mathematicians "switch" to some form of reasoning involving continuous objects. After after some reasoning steps, they possibly switch back to the desired result involving naturals, integers or rationals. -Suppose that during some argument (involving $\mathbb{N}$) one switches to real numbers and then back to discrete domain (before completing the argument). My question is, can one give examples where a switch to $\mathbb{R}$ can be "replaced" by a switch to a suitable "countable subset of $\mathbb{R}$" without having to change the argument entirely/substantially. My question is not limited to specific set $R$ I described (it was just meant as an example) .... any bigger countable subset of reals can be used. -P.S. In case the question is received well-enough, I am afraid I am not qualified enough to judge the answers. I can't think of any such examples (possibly due to lack of knowledge or perhaps that finding such examples is little harder). - -REPLY [8 votes]: Suppose that during some argument (involving ℕ) one switches to real numbers and then back to discrete domain (before completing the argument). My question is, can one give examples where a switch to ℝ can be "replaced" by a switch to a suitable "countable subset of ℝ" without having to change the argument entirely/substantially. - -A large portion of reverse mathematics consists of doing precisely this (see Simpson's "Subsystems of Second Order Arithmetic"): taking statements which appear to involve the reals (or other equivalent sets, like the power set of the natural numbers) and showing that the same results follow when using only certain axioms about the existence of reals. -Reverse math is usually presented axiomatically, but it's common to think in terms of $\omega$-models: to relate provability in the formal theory $ACA_0$ with those statements which hold when we only use reals which are definable by an arithmetic formula, and so on.<|endoftext|> -TITLE: Tate-Shafarevich group over number fields -QUESTION [5 upvotes]: Let $A$ be an abelian variety over a number field $K$, $\text{Sha}(A/K)$ its Tate-Shafarevich group, $\ell$ a prime. -Is it known that the $\ell$-primary torsion subgroup $\text{Sha}(A/K)\{\ell\}$ is trivial for almost all primes $\ell$? - -REPLY [16 votes]: No. -It is always difficult to "prove" that something is "not known", but this may do: I claim it is not even known when $K=\mathbb Q$, $A$ is an elliptic curve $E$. In fact in this case, the result you ask for is not even known when $E$ has CM. In fact, even in this very special case, it is not known that the $l$-primary torsion subgroup of the Sha is finite for almost all primes $l$, while you ask for trivial instead of finite. Actually, it is not even known that this $l$-primary torsion subgroup is finite for infinitely many primes $l$. See for example Theorem 1.1 of Coates-Liang-Sujatha here in https://link.springer.com/article/10.1007/s00032-010-0127-2<|endoftext|> -TITLE: Calculating topological index -QUESTION [6 upvotes]: Consider the space $X=BSL(8,\mathbb{C})/(\mathbb{Z}/2)$. The topological Brauer group of $X$ is given by $Br_{top}(X)=Tor(H^{3}(X;\mathbb{Z}))=\mathbb{Z}/2$. I'm studying concepts of topological period and index of a class in $Br_{top}(X)$ and I would like to calculate $ind_{top}(\alpha)$ where $\alpha$ is the nontrivial class in the topological Brauer group of the space given above. -My question is: What should I know about this space that can help me to calculate the index?. -It's known that $per_{top}(\alpha)|ind_{top}(\alpha)$ and that they have the same prime divisors. Since $per_{top}(\alpha)=2$, then $ind_{top}(\alpha)=2^{m}$ for some positive integer $m>0$. - -REPLY [7 votes]: The index in this case is $8$. You can see that it divides $8$ as your space $X$ supports a tautological degree $8$ topological Azumaya algebra given by the map $B(SL_8/\mu_2) \rightarrow BPGL_8$. If the index were lower, it would be very strange: it would imply that if $Y$ is any space with $H^2(Y,\mathbb{Z})=0$ and a period $2$ Brauer class with a degree $8$ representative, then in fact the index would be less than $8$. Intuition suggests this is not the case. -To prove the claim above, you can argue as in Section $6$ of this paper: https://arxiv.org/abs/1208.4430. Basically, the $5$-skeleton of $B(SL_8/\mu_2)$ is $5$-equivalent to the $5$-skeleton of $K(\mathbb{Z}/2,2)\times K(\mathbb{Z},4)$. The paper above shows that the index of the tautological non-zero period $2$ Brauer class on the $5$-skeleton of $K(\mathbb{Z}/2,2)$ is $8$. -Another approach is to use the incompressibility of classifying spaces as explained in Jackowski-McClure-Oliver's work on maps between classifying spaces.<|endoftext|> -TITLE: If $X$ and $Y$ are homotopy equivalent, then are $X \times \mathbb{R}^{\infty}$ and $Y \times \mathbb{R}^{\infty}$ homeomorphic? -QUESTION [38 upvotes]: Let $X$ and $Y$ be reasonable spaces. Since $\mathbb{R}^{\infty}$ is contractible, -$$ -X \times \mathbb{R}^{\infty} \cong Y \times \mathbb{R}^{\infty} \;\;\; \implies \;\;\; X \simeq Y. -$$ -Is the converse also true? -My vague intuition: the factors of $\mathbb{R}^{\infty}$ provide so much extra room that there will never be a geometric obstruction to producing a homeomorphism. Evidently, there is no homotopy-theoretic obstruction, so maybe the converse is true. -On the other hand, I really have no idea and could be missing something basic. For example, the plane with two punctures is homotopy equivalent to a wedge of two circles. However, I do not know about a homeomorphism -$$ -(\mathbb{C} - \{0, 1\}) \times \mathbb{R}^{\infty} \overset{?}{\cong} (S^1 \vee S^1) \times \mathbb{R}^{\infty}. -$$ - -Clarification about the meaning of $\mathbb{R}^{\infty}$ and the intent of the question -When I wrote the question, I had in mind the infinite union $\cup_n \mathbb{R}^n$ inside the product $\mathbb{R}^{\mathbb{N}}$. However, since I would like the answer to the question to be "yes," I am also interested in other versions of $\mathbb{R}^{\infty}$. -I asked the question because of an algebraic limiting construction in a paper I'm writing, and I felt that a topological version of the limit would be satisfying. The algebraic version is already working for spaces like $X \times \mathbb{C}^n$ for large-enough $n$, and converges algebraically to some limiting group, but this doesn't give too many hints about the topology I should use on $\mathbb{C}^{\infty}$, or if the limit can even be considered topologically. -My application involves singular homology, and the direct limit topology is well-suited to this application, but other choices may be as well. - -REPLY [44 votes]: This question is answered by two classical theorems of infinite-dimensional topology, which can be found in the books of Bessaga and Pelczynski, Chigogidze or Sakai. -Factor Theorem. For any Polish absolute neighborhood retract $X$ (= neighborhood retract of $\mathbb R^\omega$) the product $X\times\mathbb R^\omega$ is an $\ell_2$-manifold. -Classification Theorem. Two $\ell_2$-manifolds are homeomorphic if and only if they are homotopy equivalent. -So, the reasonable space in your question should read as a Polish absolute neighborhood retract. By the way, this class of spaces includes all (countable locally) finite simplicial complexes, mentioned in the answer of Igor Rivin. - -Remark on possible generalizations. Chapter IX of the book of Bessaga and Pelczynski contains generalizations of the above two theorems to manifolds modeled on normed spaces $E$, which are homeomorphic to $E^\omega$ or $E^{\omega}_0:=\{(x_n)_{n\in\omega}\in E^\omega:\exists n\in\omega\;\forall k\ge n\;(x_k=0)\}$. -Chapter 7 of Chigogidze's book contains generalizations of the Factor and Classification Theorems to manifolds modeled on uncountable products of lines. -Chapter 5 of Sakai's book contains generalizations of the Factor and Classification Theorems to manifolds modeled on the direct limits $\mathbb R^\infty$ and $Q^\infty$ of Euclidean spaces and Hilbert cubes, respectively. -So, typical generalizations of Factor and Classification Theorems look as follows: - -Theorem. Let $E$ be a reasonable model space (usually it is an infinite-dimensional locally convex space with some additional properties). -$\bullet$ For any neighborhood retract $X$ of $E$ the product $X\times E$ is an $E$-manifold. -$\bullet$ Two $E$-manifolds are homeomorphic if and only if they are homotopically equivalent. - - -Depending on the meaning of your model space $\mathbb R^\infty$ the meaning of a reasonable space also changes. If $\mathbb R^\infty$ is the direct limit of Euclidean spaces, then a reasonable space means an absolute neighborhood extensors which is a direct limits of finite dimensional compacta. If $\mathbb R^\infty$ is the union of the Euclidean spaces in the countable product of lines, then a reasonable space is a locally contractible space which is the countable union of finite-dimensional compact subsets.<|endoftext|> -TITLE: Mysterious symmetry - in search for a bijection -QUESTION [30 upvotes]: I have a mysterious symmetry that I have not managed to prove. -First some definitions (see picture below) -Fix a partition that fit in a staircase shape with $n$ rows. -There are $Catalan(n)$ such shapes. -We can represent this with a diagram $D$, as below, -where the gray squares is the partition. -The yellow squares are enumerated $1,\dotsc,n$ from top to bottom, -and thus any permutation in $S_n$ is seen as a labeling of the yellow squares. - -Let $a+b=n$. Given a permutation in $S_n$ seen as a labeling of the yellow squares, the first block is the squares with labels $1,\dotsc,a$ -and the second block is the remaining $b$ squares. -Thus, the blue labels is the first block, and the red labels is the second block in my example. -A permutation $\sigma \in S_n$ is called $(D,a,b)$-good if the following holds: - -The smallest label in each of the two blocks appear lowest in its block. -If $i$ and $i+1$ are in the same block, and $i$ below $i+1$, -then the square in the same row as $i+1$ and same column as $i$ must be white. - -In the diagram, the permutation $342615$ is shown, -and one can verify that it is $(D,4,2)$-good. -The white squares that has to be white due to the second condition -has been marked with bullets. -Let $Good(D,a,b)$ denote the set of $(D,a,b)$-good permutations. -Warmup exercise -Show that $|Good(D,a,b)|=|Good(D,b,a)|$. -Finally, we define the ascent, $asc_D$-statistic on permutations as follows: -For every white square $S$, we let $S_1$ be the index of the yellow square in the same row, and $S_2$ be the index of the yellow square in the same column. -Then $asc_D(\sigma)$ is the number of white squares $S$, -such that $\sigma(S_1)<\sigma(S_2)$. -In our diagram, $asc_D(342615) = 1+1+2+0+1 = 5$, -where the terms are contributions from each row. -My problem -Show (bijectively) that for every diagram $D$ and choice of $a+b=n$, -$$ -\sum_{\sigma \in Good(D,a,b)} q^{asc_D(\sigma)} -=\sum_{\sigma \in Good(D,b,a)} q^{asc_D(\sigma)}. -$$ -For the diagram $D$ here, we have that $|Good(D,4,2)|=|Good(D,2,4)|=20$, -and that both sums above become -$$1 + 3 q + 4 q^2 + 4 q^3 + 4 q^4 + 3 q^5 + q^6.$$ -Comments: For some diagrams $D$, it is straightforward -to produce a bijection, in particular the case when $D$ has no gray squares. -One would hope that a bijection would the number ascents 'within blocks', -that is, ascents where $S_1$ and $S_2$ belong to the same block. -However, this cannot be done for general $D$. -One can generalize the problem to permutations with more than two blocks, -but the 2-block case implies the general case. -I am quite confident this result follows (non-bijectively) from a result by C. Athanasiadis, but it requires several messy steps. -Motivation -This is related to the $p_\lambda$-expansion of certain LLT polynomials. - -REPLY [7 votes]: Oliver Pechenik and I have some partial progress to report. Maybe someone else can see how to supply the remaining missing ingredients. -First, let's establish some notation. -We say your ascent statistic enumerates the $D$-inversions of the permutation $\sigma$, denoted $\iota_D(\sigma)$. -This makes sense since $\iota_\varnothing(\sigma)$ counts the usual inversions of $\sigma$. -Let $G_D(a,b)$ be your set $Good(D,a,b)$ and define the $D$-inversion generating function $g_{D,a,b}(q) = \sum_{\sigma \in G_D(a,b)} q^{\iota_D(\sigma)}$. -As mentioned in my comment, the generating function $g_{D,a,b}$ appears to have coefficient symmetry. -Let $\delta_{n}$ be the staircase partition $(n-1,n-2,\dots,1)$. -Conjecture 1: -Let $a+b = n$ and $D \subset \delta_n$ and write -$$ -g_{D,a,b}(q) = h_0 + h_1 q + h_2 q^2 + \dots + h_m q^m, -$$ -where $h_m \neq 0$. -Then $h_i = h_{m-i}$ for all $i$. -We have extensively (but not exhaustively) computer tested this conjecture and are convinced that it is true. -Up to this unproved conjecture, we have a complete proof of your desired $a/b$ symmetry. -More specifically, we can show the following. -Theorem 2: -For $a+b = n$ and $D \subset \delta_{n-1}$, let $m$ be the degree of $g_{D,a,b}(q)$. -There is an explicit weight-reversing bijection $\Phi_{a,b} : G_D(a,b) \to G_D(b,a)$ with -$$ -\iota_D(\sigma) = m - \iota_D(\Phi(\sigma)) -$$ -for all $\sigma$. -Note that the statement you wanted follows directly from combining the theorem and the conjecture above. -Before sketching the proof of Theorem 2, let me outline an example. -We begin with $\sigma = 14237568 \in G_{(3,2)}(3,5)$, which has the following diagram: -              -              -              -              -   . -Here, entries with values in $[a]$ are underlined (blue above) and those with values in $[a+1,n]$ are overlined (red above). -The indices corresponding to underlined and overlined entries are $A=\{1,2,4\}$ and $B = \{3,5,6,7,8\}$, respectively. -The $\blacksquare$'s are cells that cannot correspond to $D$-inversions due to our choice of $A$ and $B$, while the $\square$'s are cells that must have $D$-inversions because of $A$ and $B$. -The $\bullet$'s are the same as in Per's diagram, while the entry marked $\star$ cannot correspond to a $D$-inversion because of the $\bullet$ above it. -Finally, the entries with $\blacktriangle$ or $\triangle$ could correspond to $D$-inversions (unfilled) or not (filled). -Our map proceeds as follows: - -Exchange $A$ and $B$, preserving relative order (this is $\Psi_{a,b}$ below), resulting in $61784235$. - -              -              -              -              -   . -This has the effect of exchanging $\blacksquare$'s and $\square$'s, while preserving each other entry. - -Within $A$ and $B$, we can transform the permutation so that the number of $\triangle$'s and $\blacktriangle$'s in each row is exchanged, resulting in $\tau = 61872453 \in G_D(b,a)$. - -              -              -              -              -   . -This involves two applications of the map $\Phi_{n,0}$ below; one to the entries in $A$ and one to the entries in $B$. -Note the position of triangles within each row changes, but the total number is preserved. -Here, $\iota_D(\sigma) = 4$ while $\iota_D(\tau) = 13$. -Note $4+13 = 17 = \binom{7}{2} - (|D|-1)$, which is the maximum possible number of $D$-inversions in $G_D(a,b)$ (or $G_D(b,a)$), as required by Theorem 2. -As a first step towards a proof, we sketch a bijective proof of Conjecture 1 in the special case $b=0$. -We say $(i,j)$ is a $D$-inversion if $(i,j) \notin D$ and $\sigma_j > \sigma_{n+1-i}$. -The $D$-code is $\textbf{c}^D(\sigma) = (c_1^D(\sigma),\dots,c_{n-1}^D(\sigma))$ where $c_k^D(\sigma)$ is the number of $D$-inversions of $\sigma$ of the form $(k,\ell)$. -Note $\textbf{c}_\varnothing$ is ordinary Lehmer code. -Also, $\iota_D(\sigma) = \sum c_i^D(\sigma)$. -For $\sigma \in G_D(n,0)$, observe that $\sigma_1 = 1$. -Note also that $G_D(n,0) = \varnothing$ if $D \not \subset \delta_{n-1}$. -Lemma 3: -For $D \subset \delta_{n-1}$, there is a unique $\tau\in G_D(n,0)$ that is maximal with respect to $\iota_D$. -Moreover, $c_i^D(\tau) = n-i - D_i$. -Proof: -We construct $\tau$ recursively. -Set $\tau_1 = 1$. -For each value $i \in [2,n]$, place $i$ as far to the right as possible in $\tau$, satisfying the constraint that the cell corresponding to $i$ and $i-1$ in $\delta_n$ is not in $D$. -It is easy to check that the resulting permutation has the desired $D$-code. -To see $\tau$ is unique, we first provide an upper bound on the code of $\sigma \in G_D(n,0)$. -Let $(i_1,j_1), \dots, (i_k,j_k)$ be the maximally northwest cells of $\delta_n$ that satisfy $\sigma_{i_1} = \sigma_{n+1-j_1} + 1$ (these are some of the cells labeled with $\bullet$ in Per's diagram). -We can order these so that $j_1 < \dots < j_k$, hence $i_1 > \dots > i_k$. -For $k \in [n-1]$, let $\ell$ be maximal so that $k \geq i_\ell$. -Then $(k,j_\ell) \in \delta_n \setminus D$ and cannot correspond to a $D$-inversion. -Therefore, $c^D_k(\sigma) \leq n-k-D_k$. -On reflection, it becomes apparent that $\tau$ is the unique permutation attaining this bound for each $k$. -Let $\textbf{c}^D_M$ be the code defined in Lemma 3. -Next, we show the $D$-code functions similarly to the ordinary Lehmer code. -Lemma 4: -For $D \subset \delta_{n-1}$ and $\textbf{b} = (b_1,\dots,b_k)$ with $b_i \leq n-i-D_i$, there is a unique permutation $\sigma \in G_D(n,0)$ with $\textbf{c}_D(\sigma) = \textbf{b}$. -Proof: -One can recurse down from the maximal $\tau$ constructed in Lemma 1, picking off one inversion at a time. The details are a bit technical, but we seem to understand the algorithm pictorially. -Corollary 5: -The function $g_{D,n,0}(q)$ is coefficient symmetric. -Proof: -By Lemma 4, we can define the bijection $\Phi_{n,0}: G_D(n,0) \to G_D(n,0)$ where $\textbf{c}^D(\Phi_{n,0}(\sigma)) = \textbf{c}^D_M - \textbf{c}^D(\sigma)$. -The bijection $\Phi_{n,0}$ is one of the key tools in our proof of Theorem 2. -The other tool is a solution to the warm-up exercise. -For sets $A \sqcup B = [n]$ with $|A| = a$ and $|B| = b$, let $G_D(A,B)$ be the set of permutations in $G_D(a,b)$ with $\sigma^{-1}(A)=[a]$ (so $\sigma^{-1}(B) = [a+1,b]$). -The corresponding $D$-inversion generating function is denoted $g_{D,A,B}(q)$. -Clearly, -$$ -G_D(a,b) = \bigsqcup_{A \sqcup B = [n]} G_D(A,B) -\ \ \ \mbox{and}\ \ \ -g_{D,a,b}(q) = \sum_{A \sqcup B = [n]} g_{D,A,B}(q), -$$ -where $|A| = a $ and $|B| = b$. -For fixed $A,B$ and $D$, let $\psi_{A,B}$ be the number of $D$-inversions corresponding to $(i,j) \in \delta_n\setminus D$ where $n+1-i \in A$ and $j \in B$. -Lemma 6: -For $A \sqcup B = [n]$, $g_{D,A,B}(q)$ is coefficient symmetric. -Proof: -The generating function $g_{D,A,B}(q)$ factors as -$$ -g_{D,A,B} = g_{D',a,0} \cdot g_{D'',b,0} \cdot q^{\psi_{A,B}}, -$$ -where $D'$ is given by intersecting $D$ with the rows and columns of $\delta_n$ containing an element of $[a]$ and $D''$ is similarly given by intersecting $D$ with those rows and columns containing an element of $[a+1,n]$. Since each factor is coefficient symmetric by Corollary 5, we are done. -We define the map $\Psi_{a,b}:S_n \to S_n$ where for $\tau = \Psi_{a,b}(\sigma)$ we have -$$ -\tau_i = \begin{cases} -\sigma_i+b, & \text{if } \sigma_i \leq a;\\ -\sigma_i-a, & \text{if } \sigma_i > a. -\end{cases} -$$ -We now prove the warmup exercise. -Lemma 7: -The map $\Psi_{a,b}$ is a bijection from $G_D(a,b)$ to $G_D(b,a)$. -Proof: -For $A\sqcup B = [n]$ with $|A|=a,|B|=b$, observe that $\Psi_{a,b}$ is a bijection from $G_D(A,B)$ to $G_D(B,A)$. -The lemma follows. -The effect of $\Psi_{a,b}$ on $\iota_D$ is the key to proving Theorem 2. -Lemma 8: -Fix $D \subset \delta_n$ and $A \sqcup B = [n]$ with $|A| = a$ and $|B| = b$. -For $\sigma \in G_D(A,B)$ we have $\iota_D(\Psi_{a,b}(\sigma)) = \psi_{B,A} - \psi_{A,B} + \iota_D(\sigma)$. -Proof: -After applying $\Psi_{a,b}$, we lose the $\psi_{A,B}$ $D$-inversions in $\sigma$ that are forced by $A$ and $B$ but gain the $\psi_{B,A}$ $D$-inversions forced by $B$ and $A$. -Lemma 9: -Let $D \subset \delta_n$ and $A \sqcup B = [n]$ with $|A|=a$ and $|B|=b$. -Define $D'$ and $D''$ as in the proof of Lemma 6 and fix $\ell = \binom{a-1}{2} -|D'| + \binom{b-1}{2}-|D''|$. -Then $\psi_{A,B} + \psi_{B,A} +\ell = m$ where $m$ is the max degree of $g_{D,a,b}$. -Proof: -Note the quantity $\ell$ is the max degree coming from inversions not forced by $A$ and $B$. -Some cells in $\delta_n - D$ will never correspond to inversions for reasons similar to those in Lemma 3. -The remaining cells in $\delta_n - D$ correspond either to an inversion forced by $A$ and $B$ or an inversion forced by $B$ and $A$. -Proof of Theorem 2: -Let $\sigma \in G_D(a,b)$ with $\sigma^{-1}([a]) = A$ and $\sigma^{-1}([a+1,n]) = B$ (so $\sigma \in G_D(A,B)$, as well). -Moreover, let $D'$ and $D''$ be as in Lemma 6. -We define the map $\Phi_{a,b}$ as follows: - -First, apply $\Phi^{D'}_{a,0}$ to the subword of $\sigma$ indexed by $A$ and $\Phi^{D''}_{b,0}$ to the subword of $\sigma$ indexed by $B$, obtaining $\tau \in G_D(A,B)$. -Then, apply $\Psi_{a,b}$ to $\tau$, obtaining $\rho$. - -We now show $\Phi_{a,b}$ has the desired properties. -First, observe that step 1 sends $\iota_D(\sigma)$ to its complement in $g_{D,A,B}(\sigma)$, as per Lemma 6. -Next, we combine Lemma 8 and Lemma 9 to show $\iota_D(\rho) = m - \iota_D(\sigma)$. -The theorem follows. -Some remarks: - -Symmetry cannot be proved for general $G_D(a,b)$ using code containment. -For example, $G_{(3,1)}(3,2)$ has four elements with $(3,1)$-codes $(0,1,2,0)$, $(0,1,0,1)$, $(0,0,1,0)$ and $(0,0,0,0)$. -There might be a simple proof of Conjecture 1 using the -$b=0$ case. We thought we had one, but couldn't remember it two days -later. -Any proof of symmetry may apply equally well to the case where there -are more than two blocks. In general, it appears the max degree of -$G_D(a_1,a_2,\dots,a_k)$ is $\binom{n}{2} - |D| - (n+1-k)$.<|endoftext|> -TITLE: Do pseudo-differential operators form a sheaf of algebras? -QUESTION [6 upvotes]: Let $M$ be a smooth manifold. -I have been trying to figure out from the literature I know whether (any flavor of) pseudo-differential operators form a sheaf of algebras (w.r.t. the usual topology on $M$). Sadly the best results I could find showed at best that pseudo-differential form a sheaf of left $C^{\infty}$-modules without treating the question of whether the multiplication is well defined nevermind associative. So my question is rather simple: - -Do pseudo-differential operators form a sheaf of (associative?) $\mathbb{C}$-algebras on $M$? - If not, what fails? - -I'm pretty sure that if one considers pseudo-differential operators modulo smothing operators than these do form a sheaf of algebras (However a precise reference here would be welcome). As for the entire space of pseudo-differential operators this seems rather non-trivial to me (I'm actually not entirely sure whether this should be true or not). I apologize if this is too elementary for this site. - -REPLY [3 votes]: Pseudodifferential operators are operators whose Schwartz kernel -is smooth outside of the diagonal and is conormal with respect to the diagonal -(differentiating finitely many times with respect to smooth vector -fields tangent to the diagonal produces a distribution in a certain Besov space). -From this definition one immediately sees that pseudodifferential operators do form a presheaf of abelian groups, albeit in a nonobvious way: a Schwartz kernel on M×M -can be restricted to U×U. -But there is no way this could be a sheaf: knowing restrictions to U×U -and V×V such that U and V cover M does not tell what happens on M×M -outside of U×U and V×V. -Sheafification produces germ equivalence classes of conormal distributions -near the diagonal. -This is a larger sheaf than equivalence class of pseudodifferential -operators modulo smoothing operators (which do form a sheaf for trivial reasons). -Indeed, all germs of smooth functions near the diagonal -produce different elements in the sheafification, -but become equal once one mods out smoothing operators. -The above talks about sheaves of abelian groups. -Pseudodifferential operators cannot always be composed -(and accordingly do not form presheaves of algebras); -one needs conditions like properness of support. -However, the sheafification and the equivalence classes modulo smoothing operators both do form sheaves of algebras.<|endoftext|> -TITLE: Is every algebraic space a 1-geometric stack? -QUESTION [6 upvotes]: In many references (Toen, Higher and derived stacks: a global overview, Toen, Vezzosi, Homotopical algebraic geometry II, and so on), the definition of $n$-geometric stack appears. -In the non-derived case, the definition starts by declaring affine schemes as $(-1)$-geometric stacks and inductively defines $n$-geometric stacks by some procedure. -Toen-Vezzosi, HAG II, Remark 2.1.1.5 says that algebraic spaces and schemes are automatically 1-geometric stacks. I can check that schemes are 1-geometric. (It does not depend on whether a scheme is separated or not.) But I can't check easily that algebraic spaces are 1-geometric stacks. - -REPLY [12 votes]: Always check the definitions being used in the reference - there are even significant differences between different arXiv versions of HAG2. Once you know you have an epimorphism from a union of affines, saying that $X$ is $n$-geometric in the HAG2 v7 sense basically amounts to saying that the higher diagonal -$$ -X \to \mathrm{map}(S^{n},X) -$$ -is affine. Thus $0$-geometric is equivalent to semi-separated, and any algebraic space $X$ is $1$-geometric because $\mathrm{map}(S^1,X)\cong X$. -EDIT: in response to keaton's comment, the condition isn't quite equivalent to $n$-geometricity, as there are some epimorphism conditions to check, but arises inductively because -$$ -X\times^h_{\mathrm{map}(S^{n},X)}X \cong \mathrm{map}(S^{n+1},X). -$$ -FURTHER EDIT with more details now I have time: -If you take affines $U,V$ etale over $X$, you want to show that $U\times_XV$ is $0$-geometric, and you already know that it is a scheme. Since the map $U\times_XV \to U \times V$ is a pullback of the diagonal $X \to X \times X$ of $X$, the relative diagonal $U\times_XV \to (U\times_XV)\times^h_{U \times V}(U\times_XV)$ is a pullback of $X \to \mathrm{map}(S^1,X)$, so is an isomorphism. The (absolute) diagonal of $U\times_XV$ is then a pullback of the diagonal of $U\times V$, hence affine.<|endoftext|> -TITLE: Points of abelian varieties over purely transcendental extensions -QUESTION [11 upvotes]: I heard about the result in the theory of abelian varieties which says the following: given an abelian variety $X$ defined over a field $k$ and a purely transcendental extension $k\subset L\subset L'$ we have that the group $X(L)$ of $L$-points is the same as the group $X(L')$. Could you give me (the reference for) the proof of this result? - -REPLY [16 votes]: This follows from the following well-known lemma. -Lemma. Let $A$ be an abelian variety over $k$. Then any map $f \colon \mathbb P^1 \to A$ is constant. -Proof 1. The map $f$ induces a map on the Albanese $f_* \colon \operatorname{Alb}_{\mathbb P^1} \to \operatorname{Alb}_A$ sitting in a commutative diagram -$$\begin{array}{ccc}\mathbb P^1 & \stackrel{f}\longrightarrow & A \\ \downarrow & & \downarrow \\ \operatorname{Alb}_{\mathbb P^1} & \stackrel{f_*}\longrightarrow &\ \operatorname{Alb}_A. \end{array}$$ -But $\operatorname{Alb}_{\mathbb P^1}$ is trivial, and the map $A \to \operatorname{Alb}_A$ is an isomorphism. $\square$ -Proof 2. After translating on $A$, we may assume $f(0) = 0$. The point $0\in \mathbb P^1$ is the identity element for two different group structures: the open subscheme $\mathbb P^1 - \infty$ is isomorphic to $\mathbb G_a$, and $\mathbb P^1 - \{\infty, 1\}$ is isomorphic to $\mathbb G_m$. Then $f|_{\mathbb G_a}$ and $f|_{\mathbb G_m}$ are both group homomorphisms, which is impossible unless $f$ is constant. $\square$ -Proof 3. There is a third proof using triviality of $\Omega_A$ and the fact that $\Omega_{\mathbb P^1}$ has no global sections. See e.g. Bhatt's notes, Cor. 3.6. $\square$ -Corollary. Let $A$ be an abelian variety over $k$, and let $C$ be a rational curve (not necessarily smooth or proper). Then any map $f \colon C \to A$ is constant. -Proof. By assumption, there is an open $U \subseteq C$ that is isomorphic to an open in $\mathbb P^1$. Moreover, $f|_U$ extends uniquely to a morphism $\mathbb P^1 \to A$ because $A$ is proper. Hence, $f|_U$ is constant. Because $U$ is dense, this forces $f$ to be constant. $\square$ -Corollary. Let $A$ be an abelian variety over $k$, and let $Y$ be a rational variety (not necessarily smooth or proper). Then any morphism $f \colon Y \to A$ is constant. -Proof. It suffices to prove that $f$ is constant on a big open, so we may replace $Y$ by an open $U \subseteq Y$ that is isomorphic to an open in $\mathbb P^n_k$. Then $Y$ is covered by (not necessarily proper) rational curves $C \subseteq Y$. By the corollary above, we conclude that $f$ is constant on all of those, hence $f$ is constant. $\square$ -The answer to the OP now follows: $L$-points of $X$ are the same as the $L$-points of $X_L = X \times_k L$, so we may assume that $L = k$. Then a morphism $\operatorname{Spec} L' \to A$ spreads out to a rational variety $Y$ with function field $L'$, hence is constant (i.e. comes from a $k$-point) by the arguments above.<|endoftext|> -TITLE: Non-split Aut(G) $\to$ Out(G)? -QUESTION [14 upvotes]: I'm looking for examples of outer automorphisms of a finite group $G$ which do not lift to automorphisms (i.e. non-split quotient map $\mathrm{Aut}(G)\to \mathrm{Out}(G)$, where $\mathrm{Out}(G) = \mathrm{Aut}(G)/\mathrm{Inn}(G)).$ -I'm aware of the well-known $A_6$ (also $S_6$?) example, as explained in this Wikipedia article. The article also mentions $PSL(2.q^2)$ for $q$ odd. -Are these the smallest examples? -Is there a complete (or at least longer) list of examples somewhere? - -REPLY [6 votes]: A variation on Geoff's comment: Griess has proved that it's usually nonsplit for extraspecial $2$-groups: - -Robert L. Griess, Jr., Automorphisms of extra special groups and - nonvanishing degree 2 cohomology, Pacific Journal of - Mathematics Vol. 48, No. 2 (1973) pp 403-422 doi:10.2140/pjm.1973.48.403, - (pdf)<|endoftext|> -TITLE: What are the dualizable objects with respect to pushout-product? -QUESTION [6 upvotes]: If $(\mathcal C, \otimes)$ is a monoidal category, the pushout product turns the arrow-category ${\mathcal C}^{[1]}$ into a monoidal category. -Adding more and more properties to $\otimes$ results in analogous properties being enjoyed by $\widehat{\otimes}$: for example, if $\otimes$ is symmetric, then so is $\widehat\otimes$, and if $(\mathcal C, \otimes)$ is left/right closed then so is $(\mathcal C^{[1]}, \widehat\otimes)$. Now, - -Assume that $(\mathcal C, \otimes)$ is compact closed. It is not difficult to show that if $f \in\mathcal C^{[1]}$ (say $f :A\to B$ to fix notation) is invertible, then $f^\dagger = (f^{-1})^* : A^*\to B^*$ serves as its dual. But is this necessary? In poor words, if all objects are $\otimes$-dualizable, then an arrow is $\widehat\otimes$-dualizable iff it is invertible? - -(the best guess I've been given is that $0\to X$ is dualizable even for nonzero objects -a compact closed category has a zero object, so the notation-; this is reasonable and I expect it, nevertheless I get lost in diagrams when I try to prove that there are suitable unit and counit). - -REPLY [6 votes]: Assuming that $C$ has finite colimits preserved in each variable by $\otimes$ (as is necessary for the pushout product to exist and be associative and unital), the functor $X\mapsto (\emptyset\to X)$ embeds $C$ strong-monoidally in $C^{[1]}$. In particular, it preserves dualizable objects, so if $C$ is compact closed then all objects of the form $\emptyset\to X$ are dualizable in $C^{[1]}$.<|endoftext|> -TITLE: How many Tic Tac Toe games are possible? -QUESTION [5 upvotes]: Consider the average game of Tic Tac Toe or Noughts and Crosses. The game is played on a 3 by 3 two dimentional board. The game is played by two people and each person is allowed to only add one type of piece to the board - $\bigcirc$ or $\large\times$, where the person with the crosses always starts first, the players take turns, and whoever puts three pieces in a row (either horizontally,vertically or diagonally) first, wins. -The fact that some games finish without filling the entire board is where this gets tricky. -So my questions are: -How many possible games of Tic Tac Toe, which finish at the ninth move, are there? (aka games that fill up the entire board.) -How many possible games of Tic Tac Toe, which finish before the ninth move, are there? (aka games that have at least 3 pieces of one type in a row 'stricked', where there are still empty spaces on the board.) -*Not that here games which are a rotations of other games count as different games. - -REPLY [12 votes]: There is a fairy detailed computation on this page of Henry Bottomley's. There seem to be 81792 games ending in a win +46080 games ending in a draw = 127872 games ending on the 9th move (out of 255168 games possible).<|endoftext|> -TITLE: Evaluating the sum of $kl^2$ over $p,q,k,l$ such that $pk +ql = n$ -QUESTION [11 upvotes]: I have come across the following sum: -$$\sum_{\substack{p, q, k, l \in \mathbb{N} \\ k > l \\ pk + ql = n}}kl^2$$ -and I am trying to simplify it, hoping to get a nice formula in terms of $n$ and some arithmetic functions of $n$. -For instance, the following sum can be evaluated as follows: -\begin{align*} - \sum_{\substack{p, q, k, l \in \mathbb{N} \\ pk + ql = n}}kl &= \sum_{\substack{\alpha, \beta \in \mathbb{N}\\ \alpha+ \beta = n}}\sum_{k | \alpha}k \sum_{l | \beta}l \\ -&= \sum_{\substack{\alpha, \beta \in \mathbb{N}\\ \alpha+ \beta = n}} \sigma_1(\alpha) \sigma_1(\beta)\\ -&= (\sigma_1 \Delta \sigma_1)(n) -\end{align*} -where $\sigma_1(n)$ is the sum of divisors of $n$ and $\Delta$ is the discrete convolution. -Ramanujan has a formula for the discrete convolution of $\sigma_1$ with itself given by -$$(\sigma_1 \Delta \sigma_1)(n) = \frac{5}{12}\sigma_3(n) + \frac{1}{12}\sigma_1(n) - \frac{1}{2} n \sigma_1(n)$$ -where $\sigma_3(n)$ is the sum of the cubes of the divisors of $n$. -Any thoughts on how one might proceed with the sum in the beginning of the question? The asymmetry between $k$ and $l$ is causing some problems so the same approach as for the second sum does not quite work. -I would be very much grateful for any suggestions. Thanks! - -REPLY [3 votes]: Sums of this type can be calculated using standard tools from analytic number theory (Kloosterman sums + Poisson summation). This sum with additional restrictions arose in Frobenius problem. For precise statement see Lemma 8 (page 828) in the article On the Frobenius problem for three arguments (2016) by I. Vorob'ev. -In my earlier work The solution of Arnold's problem on the weak asymptotics of Frobenius numbers with three arguments (2009) the same tools used for calculation of more simple sums -$$\sum_{\substack{p, q, k, l \in \mathbb{N} \\ pk + ql = n\\\text{+additional restrictions}}}k $$ -but this technique allows to replace the summand $k$ by more or less arbitrary smooth function of $p,q,k,l.$ -If you need sharper error terms then you should apply more advanced tools like cancellation of Kloosterman sums, Kuznetsov trace formula and averaging over spectrum of Laplace operator from An Additive Divisor Problem by J. M. Deshouillers and H. Iwaniec.<|endoftext|> -TITLE: Symmetric powers of Ramanujan tau-function -QUESTION [6 upvotes]: Let $\Delta(z)$ be the modular form associated with Ramanujan $\tau$-function. -For any $k=2,3,...$, $Sym^k\Delta$ is conjectured to be an automorphic form on $\mathrm{GL}(k+1)$ and $L(s, Sym^k\Delta)$ is conjectured to be holomorphic and satisfies a functional equation. -Question 1: For which $k$, has $Sym^k\Delta$ been proved to be automorphic form on $\mathrm{GL}(k+1)$? -Question 2: For which $k$, has $L(s,Sym^k\Delta)$ been proved to be homorphic and satisfies a functional equation? -I know certainly that both questions is YES for $k=2,3,4$ (Gelbart-Jacquet, Kim-Shahidi, Shahidi). But I heard there is progress beyond that because $\Delta$ is a modular form rather than arbitrary automorphic representation on $\mathrm{GL}(2)$. - -REPLY [10 votes]: According to this exciting new preprint of Newton and Thorne (https://arxiv.org/abs/1912.11261), the answer to both Questions 1 and 2 is "all $k\geq 1$".<|endoftext|> -TITLE: Applications of "model-theoretic" forcing -QUESTION [8 upvotes]: The notion of forcing was invented by Paul Cohen, who used -it to prove the independence of the Continuum Hypothesis. He constructed a model of set theory in which the CH fails, thus showing that CH is not -provable from ZF. Forcing was adapted from set theory to -model theory by Abraham Robinson. Robinson developed two types -of model-theoretic forcing, finite forcing and infinite forcing. -I would like to know the recent applications of model-theoretic forcing. -Any reference will be appreciated. - -REPLY [6 votes]: Model theoretic forcing is extended to wider languages with some applications, for example: -1- Model theoretic forcing in analysis which gives an exposition of forcing for metric structures. -2- Shelah extended it to the context of abstract elementary classes, see for example his book Classification theory for abstract elementary classes, Section IV.2. -You may also look at Vasey's thesis Superstability and Categoricity in Abstract Elementary Classes. -3- Andres Villaveces has several papers, where he defines some variants of forcing. See for example Sheaves of G-structures and generic G-models and Sheavesofmetricstructures and ... -4- You may be also interested in the work of Xavier Caicedo Logic of sheaves of structures. In this paper, Caicedo introduces a concept of sheaves of structures over a topological space, and a notion of model-theoretic forcing for such structures. This is a very general approach to model-theoretic forcing. For an English exposition, you may look at The logic of sheaves, sheaf forcing and the independence of the Continuum Hypothesis.<|endoftext|> -TITLE: Topological Classification of Four-Manifolds -QUESTION [7 upvotes]: It is known that the topological classification of a closed Riemann surface is determined by its genus. Similar statements are proven for other compact Riemann surfaces with boundaries/marked points. I was wondering about the similar classification for a general compact four-manifolds possibly with boundaries or even open four-manifolds. More concretely, I am wondering to which extent the work of Michael Freedman classifies the topological four-manifolds, and what information is required to uniquely specify the topological class of a compact/non-compact four-manifold. - -REPLY [6 votes]: As Paul Siegel pointed out, the fundamental group of a smooth closed orientable 4-manifold can be an arbitrary finitely presented group, and for this reason a general classification is not possible, unless, possibly, for some classes of fundamental groups, e.g. the trivial group, by Freedman's work. However, the known classifications, even with this restriction, are in the topological category, while there is no complete classification, even for a fixed topological type, in the smooth (or PL) category. For example, nothing is known about the smooth classification of smooth 4-manifolds homeomorphic to the 4-sphere (the smooth 4-dimensional Poincaré conjecture says that there is just one, namely the standard $S^4$). -Regarding your last question, in the smooth category there are handle decompositions, which allow to build compact smooth 4-manifolds as the union of finitely many $k$-handles, $0\leq k\leq 4$, where a 4-dimensional $k$-handle ($k$ is the index of the handle) is a copy of $B^k \times B^{4-k}$ attached along $S^{k-1} \times B^{4-k}$ to the boundary of a given 4-manifold. If the manifold is closed and connected, then there is such a handle decomposition with only one 0-handle and one 4-handle. A theorem of Poenaru and Laudenbach helps in making this presentation effective: any smooth closed connected oriented 4-manifold can be reconstructed, uniquely up to diffeomorphisms, from a handle decomposition where only the handles up to index 2 are given (in other words, you do not need to know 3- and 4-handles to determine the closed manifold). There is also a nonorientable version of this. -This information can be encoded in a Kirby diagram, which provides a finite presentation of any smooth closed 4-manifolds. In the TOP category more work is needed, see the MO question mentioned by Igor.<|endoftext|> -TITLE: Homotopy cosheaf? -QUESTION [6 upvotes]: Let $C$ be a site, $\mathbf{S}$ some ($\infty$-? homotopy?) category of spaces. - -Question. What do you call a (covariant!) functor $F:C\to \mathbf{S}$ enjoying the following property: for every hypercovering $a\colon V_\bullet \to U$ in $C$, the induced map $$ {\rm hocolim}\, F(V_\bullet)\longrightarrow F(U) $$ is a homotopy equivalence? - -Have such things been studied? References are welcome. - -REPLY [6 votes]: This is just a sheaf valued in the ∞-category $\mathrm{Space}^{op}$. It is usually called a cosheaf. A place where this kind of thing shows up is in factorization algebras, that can be described as particular cosheaves over the Ran space. -I don't think they behave significantly differently from sheaves in any other complete ∞-category, so I do not believe there are specific references for them.<|endoftext|> -TITLE: Equivalent of Lusin's Theorem in Borel setting -QUESTION [5 upvotes]: Let $X$ be a Polish space, $\mathcal B$ the sigma-algebra -of Borel sets. Let $E$ be an -aperiodic countable Borel equivalence relation on -$X \times X$ (this means that every class of equivalence -is countably infinite). A set $C\in \mathcal B$ is called a complete -section for $E$, if $\forall x \in X$ $\exists y \in C$ such that -$(x, y) \in E$. -$\textit{Question:}$ Let $f : X \to \mathbb R$ be a Borel real-valued function. -Is there a complete section $C$ such the restriction of $f$ on the set -$C$, $f|_C$, is a continuous function. - -REPLY [4 votes]: The answer is No. -A suitable counterexample can be constructed as follows. -On the real line $\mathbb R$ consider the equivalence relation $E=\{(x,y)\in\mathbb R\times \mathbb R:x-y\in\mathbb Q\}$. -Fix a countable base $\{U_n\}_{n\in\omega}$ of the topology on the real line. Take a countable set $X=\{x_n\}_{n\in\omega}\subset\mathbb R$ such that $x_n-x_m\notin \mathbb Q$ for any distinct $n,m\in\omega$. -In the real line $\mathbb R$ consider the $G_\delta$-set -$$G:=\mathbb R\setminus\bigcup_{n\in\omega}(\mathbb R\setminus U_n)\cap(x_n+\mathbb Q)=(\mathbb R\setminus(X+\mathbb Q))\cup\bigcup_{n\in\omega}U_n\cap(x_n+\mathbb Q)$$ and the Borel function $f:G\to\mathbb R$ -$$f(x)=\begin{cases} -n&\mbox{if $x\in U_n\cap (x_n+\mathbb Q)$ for some $n\in\omega$};\\ -0&\mbox{otherwise}. -\end{cases} -$$ -It is easy to see that the equivalence relation $E_G:=E\cap(G\times G)$ on $G$ has -no complete section $C$ with continuous restriction $f|C$; moreover, for any complete section $C$ of $E_G$, the restriction $f|C$ has no continuity points. - -There are (a bit more involved) counterexamples even for very good equivalence relations. -To construct a suitable example, take the convergent sequence $S:=\{0\}\cup\{2^{-n}:n\ge0\}\subset\mathbb R$ and on the compact zero-dimensional space $X:=S^\omega\times S$ consider the equivalence relation -$$E=\{((x,y),(x',y'))\in X\times X:x=x'\}.$$ Observe that the quotient map $q:X\to X/E=S^\omega$ is open and closed. -Now take any bijective map $p:S\to D$ to a countable space $D$. Its countable power $P:S^\omega\to D^\omega$, $P:(x_n)_{n\in\omega}\mapsto (p(x_n))_{n\in\omega}$, is known as the Pawlikowski function and is a standard example of a Borel function of the first Baire class, which is not $\sigma$-continuous. -More precisely, the Pawlikowski function has the property that a subset $C\subset S^\omega$ is nowhere dense in $S^\omega$ if the restriction $P|C$ is continuous. -We claim that the Borel function $f:=P\circ q:X\to D^\omega$ and the equivalence relation $E$ yield a counterexample to the question posed by Shrey. Indeed, assume that the equivalence relation $E$ has a complete section $C$ with continuous restriction $f|C$. For every $s\in S$ let $C_s:=\{x\in S^\omega:(x,s)\in C\}$ and observe that the continuity of the restriction $f|C$ implies the continuity of the map $P|C_s$ and hence nowhere density of $C_s$ in $S^\omega$. Since the space $S^\omega$ is compact and hence Baire, we conclude that $\bigcup_{s\in S}C_s\ne S^\omega$, which contradicts the choice of $C$ as a complete section of $E$.<|endoftext|> -TITLE: When do automorphisms on open subsets extend -QUESTION [7 upvotes]: Let $X$ be a normal affine variety of dimension at least two over $\mathbb{C}$ and let $U\subset X$ be a dense open. Assume that $\mathrm{codim}(X\setminus U) \geq 2$. -I think Hartog's lemma implies that every automorphism of $U$ extends to an automorphism of $X$. Is that true? - - -Do we have a surjective homomorphism $\mathrm{Aut}(X) \to \mathrm{Aut}(U)$? - -REPLY [3 votes]: The answer to your second question is no. Let $X$ be a three-dimensional algebraic torus, then the automorphism group of $X$ is the extension of $GL_3(\mathbb Z)$ by $X$ (acting by translations. Let $U$ be the complement of three points in general positionin a one-dimensional subtorus. Then the automorphism group of $U$ is the subgroup of $GL_3(\mathbb Z)$ fixing a vector. If there were a surjective homomorphism $Aut(X) \to Aut(U)$, the image of the torus would be a normal abelian subgroup, hence when sent to $GL_2(\mathbb Z)$ it would lie in $\pm 1$, but there is no surjective homomorphism $GL_3(\mathbb Z) \to PGL_2(\mathbb Z)$.<|endoftext|> -TITLE: Where can I find Rademacher's wrong disproof of the Riemann Hypothesis? -QUESTION [13 upvotes]: Mathematical folklore has it that the famous algebraist Hans Rademacher once came up with a wrong disproof of the Riemann Hypothesis, which was initially believed by another famous mathematician, Carl Siegel. I vaguely remember that they say Rademacher's error was that he mistakenly assumed that logarithms of complex numbers are uni-valued. -But where can I find this particular work of Rademacher? -I would want to go over it in detail and I imagine something could be learned from it. A Google search didn't yield anything meaningful. - -REPLY [8 votes]: This does not directly answer the question, but gives the details of the story, including the source of the error. -Time Magazine, Monday April 30, 1945: - The year is erroneously given as 1943 in the book linked to below - -A sure way for any mathematician to achieve immortal fame would be to - prove or disprove the Riemann hypothesis. This baffling theory, which - deals with prime numbers, is usually stated in Riemann's symbolism as - follows: "All the nontrivial zeros of the zeta function of s, a - complex variable, lie on the line where $\sigma$ is ½ -- ($\sigma$ - being the real part of s)." The theory was propounded in 1859 by - Georg Friedrich Bernhard Riemann (who revolutionized geometry and laid - the foundations for Einstein's theory of relativity). No layman has - ever been able to understand it and no mathematician has ever proved - it. -One day last month electrifying news arrived at the University of - Chicago office of Dr. Adrian A. Albert, editor of the Transactions of - the American Mathematical Society. A wire from the society's - secretary, University of Pennsylvania Professor John R. Kline, asked - Editor Albert to stop the presses: a paper disproving the Riemann - hypothesis was on the way. Its author: Professor Hans Adolf - Rademacher, a refugee German mathematician now at Penn. -On the heels of the telegram came a letter from Professor Rademacher - himself, reporting that his calculations had been checked and - confirmed by famed Mathematician Carl Siegel of Princeton's Institute - for Advanced Study. Editor Albert got ready to publish the historic - paper in the May issue. U.S. mathematicians, hearing the wildfire - rumor, held their breath. Alas for drama, last week the issue went to - press without the Rademacher article. At the last moment the professor - wired meekly that it was all a mistake; on rechecking. Mathematician - Siegel had discovered a flaw (undisclosed) in the Rademacher - reasoning. U.S. mathematicians felt much like the morning after a - phony armistice celebration. Sighed Editor Albert: ''The whole thing - certainly raised a lot of false hopes." - -The "undisclosed" flaw found by Siegel is identified on page 109 of The Riemann Hypothesis: The Greatest Unsolved Problem in Mathematics:<|endoftext|> -TITLE: Explicit Formula of Delsarte's Linear Programming Upper Bound for $A_q(n,3)$ -QUESTION [5 upvotes]: The problem of giving an explicit formula for $A_q(n,d)$ is sometimes referred to as "the main problem in coding theory." The value of $A_q(n,d)$ is given by the maximum number of codewords in a q-ary code of length $n$ and distance $d$. More specifically, let the hamming weight of an element of $\mathbb{F}_q^n$ be its $l_0$-pseudonorm, the number of non-zero components, and the hamming distance between two elements $f,g$ the weight of their difference $d(f,g)$. Then $A_q(n,d)$ is the largest set $S \subset F_q^n$ s.t. for two elements $f,g \in S$, $d(f,g)\geq d$. -There are a number of famous upper bounds on $A_q(n,d)$, including Hamming's sphere packing bound. The best are given by a linear programming approach (now improved to a semi-definite programming approach) given by Delsarte in the late 70s. I have recently been searching for an explicit formula for Delsarte's Linear Programming Upper Bound for $A_q(n,3)$ in the literature, which correspond to single error correcting codes, and have not had much luck for non-binary codes. For binary codes this appears to be well known, and shown as early as 1977 by Best and Brouwer. -Non-binary codes seem to be a completely different story. There is a paper called "Some upper bounds for codes derived from Delsartes inequalities for -Hamming schemes" by C. Roos and C. de Vroedt, which the authors claim deals with the q-ary case, but I have not been able to find a copy. There appears to have been a very large amount of work in this field so I would be shocked if no such formula exists (well, at least a formula for some special cases of n,q). -Is there a body of work in this area I am missing? Do such formulae exist? -Note: I have also posted this question on the TCS SE here: https://cstheory.stackexchange.com/questions/40238/explicit-formula-of-delsartes-linear-programming-upper-bound-for-a-qn-3 -The results on $A_q(n,d)$ are often published in top combinatorics journals (Journal of Combinatorial Theory, Series A, for instance), and so I think it is appropriate and hopefully of interest to MO users as well. - -REPLY [2 votes]: The paper "Some upper bounds for codes derived from Delsarte's inequalities for Hamming schemes" by C. Roos and C. de Vroedt does give a formula for Delsarte's bound. Or at least according to Mathematical Reviews it provides such a formula. I too was unable to locate the paper, but I did find the MR which states the following formula for the bound denoted by $D(n,3,q)$. -$$D(n,3,q) = q^n\frac{\lambda n - a(q-a)}{(\lambda n + a)(\lambda n + a - q)}$$ -Where $\lambda = q-1$, $n \equiv a \pmod q$, and $1 \leq a \leq q$. The formula is said to hold for $q > 2$ and $n$ sufficiently large.<|endoftext|> -TITLE: Equidistribution of the partition function -QUESTION [13 upvotes]: Experiments with the partition function $p(n)$ (for $n = 1, \dotsc, 100000$) seem to indicate that $p(n)$ is uniformly distributed modulo $q$ for every small prime $q.$ Is this known? Is it a special case of some general conjecture? -EDIT -As pointed out by him whose name cannot be pronounced in the comments, the above is not quite true: -For $q=5,$ the density of $0$ is twice as high as that of other residues (which are equally likely). -For $q=7,$ all nonzero residues are equally likely, but $0$ is more than twice as likely as the others. -For $q=11,$ the residue $0$ is exactly twice as likely as the others (all of which are equally likely). - -REPLY [15 votes]: I believe this is not known. Ahlgren and Boylan proved in 2003 that $p(n)$ hits every mod $q$ residue class infinitely often, assuming $q\neq 3$ is a prime. They also gave a lower density estimate for $q\geq 5$, but this is too weak to imply actual positive lower density (except for the zero residue class).<|endoftext|> -TITLE: Model theory of the restricted complex analytic functions -QUESTION [7 upvotes]: Let $\mathbb{C}_{an}$ be the expansion of the structure $(\mathbb{C}; +,-,×,0,1)$ by adding the restricted complex analytic functions. This is the complex analog of the familiar $\mathbb{R}_{an}$ in O-minimality. -What do we know about the model theory of $\mathbb{C}_{an}$? Model-completeness? Quantifier-elimination? etc. -PS. It seems Rückert's Nullstellensatz gives us model-completeness (This came to my mind just now) Is this true? - -REPLY [9 votes]: The structure $\mathbb C_{an}$ is bi-interpretable with $\mathbb R_{an}$. To see this, take for example the restriction of the complex exponential function to the unit square S with corners 0,1,i,1+i. Then from this restriction you can define S as the set of points where the restricted function takes a non-zero value. Then the set of complex numbers $x$ satisfying $\forall y[y \in S \to xy \in S]$ is the unit interval [0,1]. Then taking $\pm$ and $1/x$ you can define $\mathbb R$. -So while many people do study restricted holomorphic functions, they do so usually in the context of o-minimality.<|endoftext|> -TITLE: Derived completion of complexes -QUESTION [5 upvotes]: Suppose $K$ is a bounded above complex of free abelian groups, and take its derived $\ell$-adic completion $K^{\wedge,\ell} = R\lim (K/\ell^n)$ in the derived category, for $\ell$ a prime. -If $K\to L$ is a map in the derived category, such that $K^{\wedge,\ell}\to L^{\wedge,\ell}$ is an isomorphism for every $\ell$, is $K\to L$ an isomorphism? (in the derived category) - -REPLY [11 votes]: As I mentioned in a comment, $K \to L$ must also be an isomorphism after rationalizing. For example, if $0 \to R \to F \to \Bbb Q \to 0$ is a free resolution of $\Bbb Q$, then let -$$ -K = \dots \to 0 \to 0 \to R \to F \to 0 \to 0 \to \dots -$$ -and $L$ be the zero complex. The map $K \to L$ is an isomorphism in the derived category after completing at any prime ($K/\ell^n$ has trivial homology for any $n$), but is not an isomorphism in the derived category itself. -However, if you add the condition that you have an isomorphism after rationalizing, the statement is true. Here is one way that you can prove that. -For any free abelian group $F$, there is an arithmetic square -$$ -\require{AMScd} -\begin{CD} - F @>>> \prod_\ell F^{\wedge,\ell}\\ - @VVV @VVV\\ - F \otimes \Bbb Q @>>> \left(\prod_\ell F^{\wedge,\ell}\right) \otimes \Bbb Q -\end{CD} -$$ -which we can re-express as a natural sequence -$$ -0 \to F \longrightarrow (F \otimes \Bbb Q) \oplus \prod_\ell F^{\wedge,\ell} \longrightarrow -\left(\prod_\ell F^{\wedge,\ell}\right) \otimes \Bbb Q \to 0 -$$ -that is exact whenever $F$ is free. Applying this to any complex $K$ of free abelian groups, we get a natural exact triangle in the derived category -$$ -K \longrightarrow (K \otimes \Bbb Q) \oplus \prod_\ell K^{\wedge,\ell} \longrightarrow \left(\prod_\ell K^{\wedge,\ell}\right) \otimes \Bbb Q \to K[1]. -$$ -If $K \to L$ is a map of two such complexes in the derived category and it induces isomorphisms $K^{\wedge,\ell} \to L^{\wedge,\ell}$ and $K \otimes \Bbb Q \to L \otimes \Bbb Q$ in the derived category, then the map $K \to L$ must then also be an isomorphism in the derived category.<|endoftext|> -TITLE: local property of nonlocal differential operators -QUESTION [6 upvotes]: Motivation: In general, a nonlocal operator acting on the product of two functions doesn't have the product rule as the local operator does. However, the Hilbert transform on the real line has a very nice property -$$H(fg)=gHf+fHg+H(Hf\cdot Hg),$$ -where the remainder (the third term on the RHS) is not just written as a sophisticated commutator. Such an identity is a consequence of the fact that the boundary value of a holomorphic function on the upper half plane must be in the form $f-iHf$. -Now I wonder would it be possible to extend this type of identity to higher dimensions. For example, denote $R_j$ the usual Riesz transform, can we simplify $R_j(fg)-gR_jf-fR_jg$ without using commutator. -Appreciate for your time. - -REPLY [4 votes]: Another proof of the identity for the Hilbert transform can be found by means of Fourier transform. If $\sigma(\xi) = i \operatorname{sign} \xi$, then -$$ \mathcal{F}(H(f g) - f H g - g H f)(\xi) = \int \mathcal{F} f(\eta) \mathcal{F} g(\xi - \eta) (\sigma(\xi) - \sigma(\xi - \eta) - \sigma(\eta)) d\eta . $$ -It turns out that -$$ \sigma(\xi) - \sigma(\xi - \eta) - \sigma(\eta) = \sigma(\xi)\sigma(\xi - \eta) \sigma(\eta) , $$ -and so $H(f g) - f H g - g H f = H(Hf Hg)$. -The same calculation for the Riesz transform, that is, $\sigma(\xi) = \xi_j / |\xi|$, leads to something much more complicated: $\sigma(\xi) - \sigma(\xi - \eta) - \sigma(\eta)$ is not a product of three functions as it was for the Hilbert transform. Therefore, $R_j(f g) - f R_j g - g R_j f$ is not of the form $A(Bf Bg)$ for whatever Fourier multipliers $A$ and $B$. -However, the expression for $\sigma(\xi) - \sigma(\xi - \eta) - \sigma(\eta)$ can be be slightly simplified: if I am not mistaken, -$$ \sigma(\xi) - \sigma(\xi - \eta) - \sigma(\eta) = \sigma(\xi) (1 - (2 + 2 |\xi - \eta|^{-1} |\eta|^{-1} \langle \xi - \eta, \eta\rangle)^{1/2}).$$ -It follows that indeed $R_j(f g) - f R_j g - g R_j f$ is the Riesz transform of something rather regular; namely, it is equal to $R_j(B(f,g))$, where $$B(f, g)(z) = \text{p.v.}\iint f(x) g(y) k(z - x, z - y) dx dy$$ with singular kernel $k$ given by $$\mathcal{F} k(\xi, \eta) = (1 - (2 + 2 |\xi|^{-1} |\eta|^{-1} \langle \xi, \eta\rangle)^{1/2}).$$<|endoftext|> -TITLE: Forcings predicted by core model theory $+$$ZFC$ results proved by the method of core model theory -QUESTION [14 upvotes]: I have two unrelated question. -First question. To motivate the question, let me explain an example. The natural way to force the failure of singular cardinals hypothesis ($SCH$), is to start with a large cardinal $\kappa$, and make it singular while blowing up its power. However, results of core model theory show that if $SCH$ fails at $\kappa,$ then either $\kappa$ is a large cardinal, or it is a singular limit of large cardinals. Motivated by this fact, the long and short extender forcings were developed, showing that the second case can be forced as well. -Now my question is the following: -Question 1. Are there any other example of forcing notions, whose existence is first predicted using core model techniques and then they are discovered? -Second question. Forcing is a powerful tool to prove $ZFC$ results as well, see for example Forcing as a tool to prove theorems, and Examples of ZFC theorems proved via forcing and Proving results about complete Boolean algebras in ZFC using Boolean valued models and Producing finite objects by forcing!. -Surprisingly, one can also use the technique of core model theory to prove $ZFC$ -results. One example that I know, is the following result of Woodin: -Suppose that $V=L[s]$, where $s$ is an $ω$-sequence of ordinals. Then $GCH$ (and in fact much more) holds. See The universe constructed from a sequence of ordinals. -Question 2. Are there any other examples of $ZFC$ results whose proof uses techniques of core model theory. The same question for theories like $ZFC+\phi$, where $\phi$ is the assertion that some large cardinal(s) exist, for which we know a core model exists (so that we can apply the core model techniques). -Remark. I am not interested in results which use some kind of covering in the absence of large cardinals to get some results, like, if there is no measurable cardinal (or even larger cardinals), then square holds at singular cardinals or so on. - -REPLY [4 votes]: Re question 1, how about Bukovsky-Namba forcing? (Wasn't it isolated after Jensen had shown the covering lemma?) -Re question 2, here is an (almost) ZFC example: Assume boldface $(\omega^2+1)-\Pi^1_1$ determinacy. Either lightface $\Pi^1_3$ uniformization is true, or else there is a real $x$ such that $\Sigma^1_3(x)$ uniformization is true. (Every proof of this I know uses Steel's correctness result for the core model below 1 Woodin cardinal.)<|endoftext|> -TITLE: Amenability of $S^{\infty}$ -QUESTION [5 upvotes]: Let $G$ be the group of all permutations of $\mathbb{N}$. If I am not mistaken, this group is denoted by $S^{\infty}$. -Is there a precise locally compact topology on $G$ such that $G$ would be an amenable group? Or is $G$ isomorphic to a dense subgroup of an amenable group? - -REPLY [5 votes]: The answer to this question implies that each locally compact group topology $\tau$ on the permutation group $S^\infty$ is discrete. Since the discrete group $S^\infty$ is known to be non-amenable (it contains a free group with two generators), the locally compact topological group $(S^\infty,\tau)$ is not amenable. -By Corollary in the answer to this question, the permutation group $S^\infty$ is not isomorphic to a dense subgroup of a non-discrete locally compact group. This implies that $S^\infty$ is not isomorphic to a dense subgroup of an amenable locally compact group.<|endoftext|> -TITLE: Is each locally compact group topology on the permutation group discrete? -QUESTION [5 upvotes]: Question. Is each locally compact group topology on the permutation group $S_\omega$ discrete? -Here $S_\omega$ is the group of all bijections of the countable ordinal $\omega$. A group topology on a group $G$ is a topology turning $G$ into a Hausdorff topological group. -This question was motivated by this question of Ali Taghavi. -To my surprise I could neigher find a quick answer nor find a suitable reference. The intuition says that the answer should be affirmative, that is, all locally compact group topologies on $S_\omega$ are discrete. -It is known that $S_\omega$ admits no compact group topology. - -REPLY [9 votes]: Better: any homomorphism $f$ from the infinite symmetric group $S(X)$ ($X$ arbitrary set) to a locally compact group $H$ has a discrete image. -Proof: we can suppose that $f$ has a dense image. -(a) Case when $H$ is totally disconnected. Let $L$ be a compact open subgroup of $H$. Notation: $L^h=h^{-1}Lh$, $h\in H$. Consider $K=f^{-1}(L)$. -Recall that $L$ is commensurated in $H$, in the sense that the function $u_{H,L}(h)=\log([L:L\cap L^h])$ on $H$ takes finite values. Hence $K$ is commensurated in $S(X)$. By a result of Bergman (which easily follows from an earlier 1995 result of Galvin), every subadditive function on $S(X)$ is bounded. By a classical result of Schlichting 1980 (rediscovered by Bergman-Lenstra 1989), this implies that $K$ is groupwise transfixed, in the sense that there exists a normal subgroup $N$ of $S(X)$ that is commensurate to $K$. By Baer's classification of normal subgroups of $S(X)$, normal subgroups of $S(X)$ have no proper subgroup of finite index, with the only exception of the finitary subgroup and its subgroup of index 2. Hence replacing $N$ by a subgroup of index 2 if necessary, we can suppose that $N$ has no proper subgroup of finite index; so $K$ contains $N$ as a subgroup of finite index. Since $N$ has no proper subgroup of finite index, it has no nontrivial homomorphism into any profinite group. So $N$ is contained in the kernel of $f$, and hence $L$ is finite, and hence $H$ is discrete. -(b) General case. By (a), $H/H^\circ$ is discrete, so $H^\circ$ is open. Define $N=f^{-1}(H^\circ)$, then $f(N)$ is dense in $H^\circ$. We conclude by (c) below that $H^\circ=1$, so (a) applies. -(c) Let $N$ be any subquotient of $S(X)$ and $P\neq 1$ a connected locally compact group. Then $N$ has no nontrivial homomorphism $v$ into $P$ with dense image (no continuity assumption). -Proof 1: suppose by contradiction that there's one; modding out the kernel, we can suppose $v$ injective, and modding out $P$, we can suppose that $P$ is Lie, and then that $P$ is either the circle group or a connected simple Lie group with trivial center. The group $P$ being linear, it has all its finitely generated subgroups residually finite (Malcev). So $N$ inherits this property. From Baer's classification of normal subgroups of $S(X)$ (and the simple observation that these are the only subnormal subgroups, see Scott's 1964 book Group Theory), all subnormal subquotients of $S(X)$ contain all countable groups as subgroups, except the finitary subgroup, modulo 1 or modulo the alternating subgroup. We can discard the latter by connectedness. So $N$ is the finitary symmetric group; it is locally finite and not virtually abelian, which is impossible in a linear group in characteristic zero such as $P$. Contradiction. -Proof 2 (without using solution to Hilbert's fifth problem?). The proof by Galvin implying that all subadditive functions on $S(X)$ are bounded adapt without change to all its normal subgroups except those contained in the finitary subgroup. Given this, when $N$ is not contained in the finitary subgroup, and using some word length on $P$ with respect to a compact generating subset, we deduce that $P$ is compact. Then by Peter-Weyl, we can assume that $P$ is Lie compact (and connected). At this point we can conclude as in the first proof by Malcev. An alternative is to use that any infinitely generated field is union of a properly increasing sequence of subfields to deduce that every infinitely generated subgroup of a compact connected Lie group is union of an increasing sequence of proper subgroups ("cofinality $\omega$"), which clearly implies the existence of an unbounded subadditive function. Applying this to the image of $N$ in $P$, we deduce that this image is finitely generated, but again using a word length argument, we deduce that the image is finite, and finitely connectedness forces $P=1$. -It remains the case when $N$ is the finitary subgroup or its alternating subgroup of index 2. In this case I can conclude as in Proof 1 but this makes use of Hilbert V. I have to think if we can conclude by another argument (to discard the possibility of a dense embedding of an infinite alternating group into a connected locally compact group). -References: -G. Bergman and H. Lenstra. Subgroups close to normal subgroups. J. Algebra 127, -80–97 (1989). -YC and Pierre de la Harpe. Metric geometry of locally compact groups. - EMS Tracts in Mathematics, 25. European Mathematical Society (EMS), Zürich, 2016. arxiv link (See Example 4E15 for link between Galvin and boundedness of subadditive maps) -Fred Galvin, Generating countable sets of permutations, J. London Math. Soc. 51 (1995), 230–242 -G. Schlichting. Operationen mit periodischen Stabilisatoren. Arch. Math. 34, 97–99 -(1980).<|endoftext|> -TITLE: Is $Alt_\omega$ a dense subgroup of a non-discrete locally compact topological group? -QUESTION [12 upvotes]: Let $S_\omega$ be the group of bijections of the countable ordinal $\omega:=\{0,1,2,\dots\}$ and $Alt_\omega$ be the subgroup of $S_\omega$ consisting of even permutations of $\omega$ (i.e., the compositions of even number of transpositions). By Onofri-Schreier-Ulam Theorem, $Alt_\omega$ is the smallest non-trivial normal subgroup in the permutation group $S_\omega$. -It is known that -$\bullet$ the group $Alt_\omega$ is not isomorphic to a subgroup of a compact topological group; -$\bullet$ the group $S_\omega$ is not isomorphic to a dense subgroup of a non-discrete locally compact topological groups. -These two results motivate the following - -Problem 1. Is the group $Alt_\omega$ isomorphic to a dense subgroup of a non-discrete locally compact topological group? - -Since the group $Alt_\omega$ is simple, this problem can be reformulated in the following form. - -Problem 2. Let $h:Alt_\omega\to G$ be a homomorphism of $Alt_\omega$ to a locally compact topological group $G$. Is the image $h(Alt_\omega)$ discerete? - -Remark. By the answer of @YCor to this question, -for every homomorphism $h:S_\omega\to G$ to a locally compact topological group $G$ the image $h(S_\omega)$ is discrete. - -REPLY [4 votes]: Corollary 1.5 of this Vissarion Belyaev's paper says that - -A group containing an infinite inert residually finite subgroup - embeds into a non-discrete locally compact group. - -Here, a subgroup $H$ of a group $G$ is called inert if $H\cap g^{-1}Hg$ has a finite index in $H$ for any $g\in G$. -So, in the alternating group $Alt_\omega$, for instance, the (abelian) subgroup generated by the 3-cycles $(1\;2\;3),(4\;5\;6),(7\;8\;9),\dots$ -is inert. infinite, and residually finite . -Thus, the answer to Question 1 is yes (and to Question 2 is no).<|endoftext|> -TITLE: Picard group of a finite type $\mathbb{Z}$-algebra -QUESTION [8 upvotes]: Let $A$ be a finitely generated $\mathbb{Z}$-algebra. Is $\operatorname{Pic}(A)$ finitely generated (as an abelian group)? - -Thoughts: - -We may assume that $A$ is reduced since $\operatorname{Pic}(A) = \operatorname{Pic}(A_{\mathrm{red}})$. -If $A$ is reduced, then the group of units $A^{\times}$ is a finitely generated abelian group, see e.g. [1, Appendix 1, no. 3] or [4, Théorème 1] (which I learned about through this question). -The case $A$ is normal is proved in [3, Chapter 2, Theorem 7.6]. -The following argument is from [2, Lemma 9.6]: Let $B$ be the normalization of $A$, set $X := \operatorname{Spec} A$ and $Y := \operatorname{Spec} B$ and let $\pi : Y \to X$ be the normalization morphism. We have the Leray spectral sequence $$ \mathrm{E}_{2}^{p,q} = \mathrm{H}^{p}(X,\mathbf{R}^{q}\pi_{\ast}\mathbb{G}_{m,Y}) \implies \mathrm{H}^{p+q}(Y,\mathbb{G}_{m,Y}) $$ with differentials $\mathrm{E}_{2}^{p,q} \to \mathrm{E}_{2}^{p+2,q-1}$. Since $\pi$ is a finite morphism (e.g. since $\mathbb{Z}$ is Nagata and [5, 030C]), every invertible sheaf on $Y$ can be trivialized on an open cover obtained as the preimage of an open cover of $X$ (e.g. [5, 0BUT]). Hence $\mathbf{R}^{1}\pi_{\ast}\mathbb{G}_{m,Y} = 0$, so we have $\operatorname{Pic}(Y) \simeq \mathrm{H}^{1}(X,\pi_{\ast}\mathbb{G}_{m,Y})$ from the Leray spectral sequence. Set $Q := \pi_{\ast}\mathbb{G}_{m,Y}/\mathbb{G}_{m,X}$; then the long exact sequence in cohomology associated to the sequence $1 \to \mathbb{G}_{m,X} \to \pi_{\ast}\mathbb{G}_{m,Y} \to Q \to 1$ gives an exact sequence $$ \Gamma(Y,\mathbb{G}_{m,Y}) \to \Gamma(X,Q) \stackrel{\partial}{\to} \operatorname{Pic}(X) \to \operatorname{Pic}(Y) $$ where the first and fourth terms are finitely generated. But what can I say about the sheaf $Q$? I know that it is $0$ on a dense open since $\pi$ is an isomorphism on a dense open (e.g. since $A$ is reduced, the regular locus is an open subset containing the generic points [5, 07R5]). -I should also note that there is a Hartshorne exercise (II, Exercise 6.9) which relates the Picard group of a singular curve (over a field) to that of its normalization. - -References: - -Bass, Introduction to Introduction to Some Methods of Algebraic K-Theory, Number 20 in CBMS Regional Conference Series in Mathematics. American Mathematical Society, 1974. -Jaffe, "Coherent functors, with application to torsion in the -Picard group", Transactions of the American Mathematical Society, vol. 349, no. 2, 1997, pp. 481–527 link -Lang, Fundamentals of Diophantine Geometry, Springer-Verlag (1983) -Samuel, "A propos du théorème des unités", Bulletin des Sciences -Mathématiques, vol. 90, 1966, pp. 89–96 -Stacks Project link - -Keywords: arithmetic scheme, Picard group, finite type $\mathbb{Z}$-algebra - -REPLY [4 votes]: Here I write out some details in R. van Dobben de Bruyn's answer. This argument is from p. 5 of these notes by Konrad Voelkel. - -Let $R$ be a Noetherian normal domain with $\mathrm{Pic}(R) = 0$. Set \begin{align*} A := R[x,y]/(x^{3} = y^{2}) \end{align*} which is the cuspidal cubic over $R$. Then \begin{align*} \operatorname{Pic}(A) \simeq R \end{align*} as abelian groups. - -The normalization of $A$ is \begin{align*} B := R[t] \end{align*} via the $R$-algebra map $A \to B$ sending $(x,y) \mapsto (t^{2},t^{3})$ which identifies $A$ with the subring $R[t^{2},t^{3}]$ of $B$. Let \begin{align*} I := \{a \in A \;:\; aB \subset A\} \end{align*} be the conductor ideal of the inclusion $A \subseteq B$; it is the largest ideal of $B$ contained in $A$. We have \begin{align*} I = \langle x,y \rangle A \end{align*} (for this, use that $A,B$ are $\mathbb{Z}_{\ge 0}$-graded rings and that $A \subset B$ is a graded ring map, hence $I$ is also a graded ideal of $A$; certainly $\langle x,y \rangle A \subset I$ and if $a \in A$ is nonzero then $a \not\in I$ since otherwise $at \in A$). We have a Milnor square $\require{AMScd}$ -\begin{CD} -A @>>> B \\ -@VVV @VVV\\ -A/I @>>> B/I -\end{CD} where $A/I \simeq R$ and $B/I \simeq R[t]/(t^{2})$. By [Wei13, I, Theorem 3.10] we have an exact sequence \begin{align*} (A/I)^{\times} \oplus B^{\times} \stackrel{\alpha}{\to} (B/I)^{\times} \to \operatorname{Pic}(A) \to \operatorname{Pic}(A/I) \oplus \operatorname{Pic}(B) \end{align*} of abelian groups (the Units-Pic sequence). Here $\operatorname{Pic}(A/I) = \operatorname{Pic}(R) = 0$ and $\operatorname{Pic}(B) \simeq \operatorname{Pic}(R) = 0$ where the first isomorphism follows from e.g. Traverso's theorem [Wei13, I, Theorem 3.11]. We have $(A/I)^{\times} = R^{\times}$ and $B^{\times} = R^{\times}$ (since $R$ is reduced) and $(B/I)^{\times} = (R[t]/(t^{2}))^{\times} \simeq R \oplus R^{\times}$ (where the "$R$" in "$R \oplus R^{\times}$" is viewed as an abelian group under addition). Under these identifications, the map $\alpha$ sends $(u_{1},u_{2}) \mapsto (0,u_{1}u_{2}^{-1})$ so $\operatorname{coker} \alpha \simeq R$. -References: -[Wei13] Weibel, The K-book: An Introduction to Algebraic K-theory, volume 145 of Graduate Studies in Mathematics. American Mathematical Society (2013)<|endoftext|> -TITLE: Is a locally finite union of $G_\delta$-sets a $G_\delta$-set? -QUESTION [11 upvotes]: Problem. Let $\mathcal F$ be a locally finite (or even discrete) family of (closed) $G_\delta$-sets in a topological space $X$. Is the union $\cup\mathcal F$ a $G_\delta$-set in $X$? -Remark. The answer to this problem is affirmative in perfect spaces. -A topological space $X$ is perfect if each open subset of $X$ can be written as the countable union of closed sets. I suspect that in general topological spaces a counterexample should exist but I cannot find it, unfortunately. - -Added in Edit. This question has a counterexample "living" in a second-countable Hausdorff space, which is not regular. So, it remains to find a (completely) regular example. - -Added in a Next Edit. A regular counterexample was constructed by Mathieu Baillif in his answer to this question. - -REPLY [3 votes]: I found a simple counterexample, which is however not regular. - -Example. There exists a functionally Hausdorff second-countable space $X$ containing a closed discrete subset $D$, which is not of type $G_\delta$ in $X$. -In this space the countable the family of singleton $\mathcal F=\{\{x\}:x\in D\}$ is a locally finite family of closed $G_\delta$-set whose union $\cup\mathcal F=D$ is not a $G_\delta$-set in $X$. - -Proof. Let $X$ be the real line endowed with the topology $\tau$ consisting of sets $W\subset \mathbb R$ such that for each point $w\in W$ there exists $\varepsilon>0$ such that each point $x\in\mathbb R\setminus \mathbb Q$ with $|x-w|<\varepsilon$ belongs to $W$. It is easy to see that the topological space $X$ is second-countable and functionally Hausdorff but not regular. -The definition of the topology $\tau$ ensures that the countable set $D:=\mathbb Q$ of rational numbers is closed and discrete in $X$. We claim that $D$ is not a $G_\delta$-set in $X$. To derive a contradiction, assume that $D=\bigcap_{n\in\omega}W_n$ for some decreasing sequence $(W_n)_{n\in\omega}$ of open sets in $X$. By the definition of the topology $\tau$, for every $n\in\omega$ and $x\in W_n$ there exists an neighborhood $V_{n,x}$ of $x$ in the Euclidean topology of $\mathbb R$ such that $V_{n,x}\setminus D\subset W_n$. Then $V_n:=\bigcup_{w\in W_n}V_{n,x}$ is an open set such that $D\subset W_n\subset V_n$ and $V_n\setminus D=W_n\setminus D$, which implies that $V_n=D\cup (V_n\setminus D)=D\cup(W_n\setminus D)=W_n$. This means that each set $W_n$ is open in the Euclidean topology of the real line and the set of rational numbers $D=\bigcap_{n\in\omega}W_n$ is of type $G_\delta$ in $\mathbb R$, which contradicts the Baire Theorem. - -Added in Edit. Answering this question, Mathieu Baillif constructed a first-countable zero-dimensional Hausdorff space $X$ of cardinality $|X|=\omega_1$ containing a closed discrete subset $D$, which is not a $G_\delta$-set in $X$. Then $\mathcal F=\{\{x\}:x\in D\}$ is a locally finite family of compact $G_\delta$-sets -in $X$ whose union $\cup\mathcal F$ is not a $G_\delta$ in $X$.<|endoftext|> -TITLE: Is this Mayer-Vietoris sequence motivic? -QUESTION [5 upvotes]: Suppose $Y$ is a variety defined over $\mathbb{Q}$ and $pt$ is a rational point of $Y$. Let $\pi:X \rightarrow Y$ be the blow up of $Y$ at $pt$ and $D$ be the exceptional divisor. For simplicity let's assume both $X$ and $D$ are smooth. Let $CD$ be the cone of $D$ defined to be -\begin{equation} -D \times I/D \times \{0\} -\end{equation} -where $I$ is the unit interval $[0,1]$. Now let $Y'$ be -\begin{equation} -Y'= X \sqcup CD /D \times \{ 1\} -\end{equation} -then the cw complex $Y'$ is homotopic to $Y$, which the morphism $\pi$ is homotopic to the inclusion $i:X \rightarrow Y'$. Then $(CD,X)$ is a cover of $Y'$ while the intersection of $CD$ and $X$ is just $D \times \{ 1\} \simeq D$. The cover $(CD,X)$ of $Y'$ induces a Mayer-Vietoris sequence -\begin{equation} -\cdots \rightarrow H^n(Y',\mathbb{Q}) \xrightarrow{\pi^*} H^n(X,\mathbb{Q}) \xrightarrow{j^*}H^n(D,\mathbb{Q}) \xrightarrow{\delta}H^{n+1}(Y',\mathbb{Q}) \rightarrow \cdots -\end{equation} -where $j$ is the inclusion morphism $D \rightarrow X$. -The morphism $\pi^*$ and $i^*$ is acutually motivic since they are induced by morphisms in the category of $\mathbb{Q}$-varieties. Is the morphism $\delta$ in this long exact sequence motivic? - -REPLY [5 votes]: I'm not sure if that's exactly what you mean by "being motivic", but this long exact sequence comes from a triangle in Voevodsky's category of (integral) motives $DM(\mathbb Q)$. -More generally if $X$ is a $\mathbb Q$-scheme of finite type, $Z\subset X$ a closed subscheme, $\pi : Y \to X$ the blowup at $Z$, and $E=\pi^{-1}(Z)$, then there is a triangle -$$ -M(E) \to M(Y)\oplus M(Z) \to M(X) \to M(E)[1] -$$ -in $DM(\mathbb Q)$. This is proved in Mazza–Voevodsky–Weibel's Lecture notes on motivic cohomology, see equation (14.5.3). It can also be viewed a formal consequence of the six-functor formalism for $DM(-)$, as described by Cisinski and Déglise in their paper Integral mixed motives in equal characteristic. -Added later: Let me sketch the abstract proof, which works in the $\ell$-adic context as well. The proof only uses that $\pi:Y\to X$ is a proper map that restricts to an isomorphism over the complement of $Z$ (an "abstract blow-up"). The triangle comes from a homotopy cocartesian square $1_X=M(Y) \coprod_{M(E)}M(Z)$ in $DM(X)$. Here for $f: X' \to X$ of finite type, we define $M(X')=f_!f^!(1_X)$. Let $i: Z\to X$ be the inclusion and $j: U\to X$ the open complement. Then the pair of functors $(i^!,j^*)$ is conservative. But applying either functor to the given square, using the proper base change theorem, gives a square which is cocartesian for formal reasons (opposite sides are equivalences).<|endoftext|> -TITLE: Is there a first-countable space containing a closed discrete subset which is not $G_\delta$? -QUESTION [7 upvotes]: Being motivated by this problem, I am searching for an example of a first-countable regular topological space $X$ containing a closed discrete subset $D$, which is not $G_\delta$ in $X$. -It is easy to show that such set $D$ cannot be countable. Also the space $X$ cannot be Moore. -On the other hand, there exists a simple example of a non-regular second-countable Hausdorff space, containing a countable closed discrete subsets which is not $G_\delta$. - -REPLY [6 votes]: EDIT: fixed some error in the proof that the diagonal is not a $G_\delta$, and added details. I hope that the proof is now correct. -I think that the following works. First I'll describe a first countable non-regular example, and then explain how it can be modified to obtain a regular one. -Endow $\omega_1$ with the usual order topology. -Take the space $\omega_1\times\omega_1$ with the product topology and refine it by declaring open the sets of the form -$$ U = \{\langle\alpha,\alpha\rangle\}\cup (V-\Delta), $$ -where $V\ni \langle\alpha,\alpha\rangle$ is open in $\omega_1\times\omega_1$ in the usual sense and $\Delta$ is the diagonal, which is thus closed discrete in this topology. Any open set containing $\Delta$ is also open in the product topology, and it is well known that it must then contain $[\alpha,\omega_1) \times [\alpha,\omega_1)$ for some $\alpha$. Hence, any countable intersection of such sets contains $[\beta,\omega_1) \times [\beta,\omega_1)$ for some $\beta$. -This topology is not regular, but we can modify it this way. Only the topology on $\Delta$ will be changed, and -since successor ordinals are isolated in $\omega_1$, we only need to worry about limit ordinals. For each such limit $\alpha\in\omega_1$, fix a sequence of successor ordinals $\alpha_n\nearrow\alpha$. Then a neighborhood of $\langle\alpha,\alpha\rangle$ is given by -$$ - U_{\alpha,k,m} = \{\langle\alpha,\alpha\rangle\} \, \cup \, \bigcup_{n\ge m} [\alpha_{n},\alpha_{n+1}]\times (\alpha_{n+k},\alpha] - \, \cup \, \bigcup_{n\ge m} (\alpha_{n+k},\alpha]\times - [\alpha_{n},\alpha_{n+1}]. -$$ -That is, we take "triangles" (more akin to step pyramids, actually) pointing at $\langle\alpha,\alpha\rangle$ from below and from the side. $\Delta$ is obviously closed discrete since -$U_{\alpha,k,m}\cap\Delta = \{\langle\alpha,\alpha\rangle\}$, and the following holds: -Lemma -If $U\supset\Delta$ is open, then there is some $\beta$ such that $U$ contains the terminal part of $\omega_1\times \{\alpha\}$ whenever $\alpha>\beta$. -By terminal part, I mean $[\gamma,\omega_1)\times \{\alpha\}$ for some $\gamma$. -The proof is by using twice Fodor's Lemma. -First, by definition for each $\alpha$ there is $\beta(\alpha)$ -such that $U$ contains $\{\alpha\}\times[\beta(\alpha),\alpha]$. -Hence by Fodor -there is some $\beta$ such that $U$ contains -$\{\alpha\}\times[\beta,\alpha]$ for $\alpha$ in a stationary subset $E\subset\omega_1$. -If $\alpha>\beta$, $\omega_1\times \{\alpha\}\cap U$ is stationary, so another use of Fodor gives the result. -Corollary -The diagonal $\Delta$ is not a $G_\delta$. -Indeed, a countable collection of open sets containing $\Delta$ will contain the terminal part of $\omega_1\times \{\alpha\}$ if $\alpha$ is big enough. -I hope I did not overlook something, but since this is very similar to the method of "Prüferizing" a surface, which dates back to Rado (1925) and is described in D. Gauld's book on non-metrisable manifolds or in this preprint (free access), I believe that everything works. I am actually fairly convinced that by Prüferizing the diagonal of the square of the longray we can obtain an example which is a topological surface. I have some vague remembrance of an example of this type in a paper of Nyikos, by the way, though I don't remember which.<|endoftext|> -TITLE: The square root of Wilson's theorem when $p\equiv 1 \mod 4$ -QUESTION [20 upvotes]: My question relates, at least superficially, to these old ones: -The value $\pm 1$ for the square root of Wilson's theorem, ((p-1)/2)! mod p -Primes P such that ((P-1)/2)!=1 mod P -When $p\equiv 1 \mod 4$, if $x=((p-1)/2)!$, then $x^2 = -1 \mod p$. -For what primes does $x \in \{1,\ldots,(p-1)/2\}$ (the elements of the set regarded as residues $\mod p$)? -One gets "yes" for $5,13,29,41,53,61,73,89,97,\ldots$ and "no" for - $17,37,101,\ldots$. Despite the slow start for "no", the counts substantially even out, say, when looking at primes up to 100000. Can one prove that the ratio approaches $1/2$? - -REPLY [20 votes]: In the case $p \equiv 1$ mod $4$, the connection is to the real quadratic field ${\mathbb Q}(\sqrt{p})$, whereas the case $p \equiv 3$ mod $4$ is connected to the imaginary quadratic field ${\mathbb Q}(\sqrt{-p})$. -Chowla ("On the class number of real quadratic fields", 1961 PNAS) proves that -$$\left( \frac{p-1}{2} \right)! \equiv (-1)^{\frac{h+1}{2}} \cdot \frac{t}{2} \text{ mod } p,$$ -where $h$ is the class number of ${\mathbb Q}(\sqrt{p})$ and the fundamental unit is $\frac{1}{2}(t + u \sqrt{p}) > 1$ with $t,u \in {\mathbb Z}$. -In fact, 0 < t < 2p. EDIT: that's false. I think I was remembering a converse, that if you find such an element $\frac{1}{2}(t + u \sqrt{p})$ of norm $-1$, and $0 < t < 2p$, then it's a fundamental unit. (This is right, I hope, but not so relevant). See Upper bound on answer for Pell equation for more on bounding $t$. -So it seems like the answer depends on (1) the class number mod 4 and (2) whether $\frac{t}{2} \in \{ 1,2,\ldots,\frac{p-1}{2} \}$ mod $p$. The statistics seem (to me) at least as difficult as the $p \equiv 3$ mod $4$ case.<|endoftext|> -TITLE: Are categories of fibrant objects idempotent complete? -QUESTION [7 upvotes]: If $C$ is a category of fibrant objects, is its associated $\infty$-category idempotent complete, i.e. is it accessible? If this is not always true, besides from the case when it is an $n$-category for finite $n$, when is it true? - -REPLY [12 votes]: If the associated $\infty$-category $C$ is stable, one may consider all possible Verdier quotients, each of which would define a new structure of category of fibrant objects. The fact that all these Verdier quotients are idempotent-complete is equivalent to the vanishing of $K$-theory of $C$ in degree $-1$. Therefore, whenever you start from a singular algebraic variety of positive dimension, some Verdier quotient of the category of perfect complexes on the given variety $X$ gives an example of a category of fibrant objects which is idempotent-complete (as it consists of bounded complexes of vector bundles) but whose localization is not idempotent-complete, because $K_{-1}(X)$ is not zero. -This means that, even if the underlying category is very nice, essentially anything may happen, unless we are considering a Verdier quotient of a category whose $K$-theory vanishes in degree $-1$. This vanishing always occurs in the presence of a bounded $t$-structure. See the paper arXiv:1610.07207 by Antieau, Gepner and Heller. - -REPLY [11 votes]: No, that is not always true. One family of examples is discussed in this answer by Lennart Meier, but it can fail even more subtly. Consider a dual example of the category of finite simplicial sets which is a category of cofibrant objects. It is idempotent complete and weak equivalences are closed under retracts, but its associated $(\infty, 1)$-category is not idempotent complete due to Wall's finiteness obstruction. -A positive result is that it works if limits of towers of fibrations exist and the limit of such tower is a fibration, acyclic if all fibrations in the tower are acyclic. In that case the associated $(\infty, 1)$-category is countably complete. Unfortunately, I don't know any criterion that does not impose infinite limits.<|endoftext|> -TITLE: Bounding Taylor coefficients of $f(z)$ with $f(0)=1$, $f(z)\ne 0$ for $|z|\le 1$ -QUESTION [6 upvotes]: Let $f(z)=1+a_1z+\ldots+a_nz^n+\ldots$ be a complex analytic function defined on the unit disk $|z|\le 1$. Suppose $f(z)\ne 0$ for $|z|< 1$. I would like to know what kind statements one can make concerning bounding the absolute values of $a_n$ from above. As the example of $f(z)=(1+z)^n$ shows individual $a_i$ can be as large as you want, but I wonder still if something non-trivial can be said about the set of possible $a_i$. Maybe there are some classical results on this topic? - -REPLY [5 votes]: The best conjectured in this generality seems to be the conjecture of Lewandowski and Szynal -Lewandowski, Zdzisław; Szynal, Jan, The Landau problem for bounded nonvanishing functions, J. Comput. Appl. Math. 105, No.1-2, 367-369 (1999). ZBL0945.30011. -They conjecture that for any non-vanishing $f$ in the unit disk such that $|f| < 1$ and any $n$ there exist an $L(n)$ such that $|a_0 + \dotsc + a_n| < L(n).$ -As far as I can tell, this is completely open, but some results for small $n$ are known under the additional assumption that the Taylor coefficients are real: -Koulorizos, Philippos; Samaris, Nikolas, The Landau problem for nonvanishing functions with real coefficients, J. Comput. Appl. Math. 139, No.1, 129-139 (2002). ZBL1009.30010.<|endoftext|> -TITLE: Best estimate of the Mertens function without assuming the Riemann Hypothesis -QUESTION [6 upvotes]: I'm searching the best known upper bound for the Mertens function, but without assuming the Riemann hypothesis. -Landau, in 1901, have proved that $M(x)= O(x \exp(-c\sqrt{\ln x})$, but I am unable to find the current estimate in the classical literature. - -REPLY [10 votes]: As Greg Martin said in a comment, the Korobov-Vinogradov zero-free region for $\zeta(s)$ yields -$$M(x)\ll x\exp\bigl(-c(\log x)^{3/5}(\log\log x)^{-1/5}\bigr).$$ -For a reference, see Satz 3 in Section V.5 of Walfisz: Weylsche Exponentialsummen in der neueren Zahlentheorie (VEB Deutscher Verlag der Wissenschaften, Berlin, 1963). -This bound cannot be improved (essentially) without improving the Korobov-Vinogradov zero-free region, see Allison: On obtaining zero-free regions for the zeta-function from estimates of $M(x)$, Proc. Cambridge Philos. Soc. 67 (1970), 333-337.<|endoftext|> -TITLE: Two motivic complexes, compared -QUESTION [8 upvotes]: Bloch defines the motivic complexes $\mathbf{Z}(n)$ in his paper "Algebraic Cycles and Higher K-Theory" (1986). -Some references (that I currently am unable to track down) use $$\check{\mathbf{Z}}(n) := \mathbf{Z}(1)^{\otimes n} = \mathbf{G}_m[-1]^{\otimes n}$$ instead of $\mathbf{Z}(n)$. -What is the relation between $\check{\mathbf{Z}}(n)$ and ${\mathbf{Z}}(n)$? - -REPLY [8 votes]: They are not quasi-isomorphic: your $\check{\mathbf Z}(n)$ is concentrated in a single degree. As Denis points out in the comments, your definition of $\check{\mathbf Z}(n)$ is wrong because you should use the tensor product of sheaves with transfers. In addition, you need to apply Suslin's $\mathbf A^1$-invariantification construction $C_*$ to the result. -Voevodsky's motivic complex $\mathbf Z(n)_V$ is -$$\mathbf Z(n)_V = C_*(\mathbf G_m^{\otimes_{\mathrm{tr}}n})[-n],$$ -where: - -$\mathbf G_m$ is regarded as a presheaf with transfers, i.e., a presheaf on Voevodsky's category $\mathrm{Cor}_k$ of smooth separated $k$-schemes and finite correspondences. The transfers are given by norms of invertible functions. -$\otimes_{\mathrm{tr}}$ is the tensor product of sheaves with transfers, which is the Day convolution of the tensor product $X\otimes Y=X\times Y$ on $\mathrm{Cor}_k$. -$C_*(F)(X)$ is a chain complex concentrated in nonnegative degrees with $C_n(F)(X)=F(X\times \mathbf A^n)$. - -$\mathbf Z(n)_V$ is quasi-isomorphic to Bloch's $\mathbf Z(n)$ as a complex of Zariski sheaves on smooth $k$-schemes. This combines several deep results of Voevodsky, Suslin, and Friedlander. A more or less self-contained proof is in Mazza–Voevodsky-Weibel's Lecture notes on motivic cohomology. -On the other hand, $\mathbf Z(n)_V$ is also a complex of étale sheaves, and if $m$ is prime to the characteristic, then $\mathbf Z/m(n)_V$ is quasi-isomorphic to $\mu_m^{\otimes n}$ as a complex of étale sheaves. This is Theorem 10.3 in the above book. Perhaps this answers your last question. -Edit: Actually the tensor product in the above formula for $\mathbf Z(n)_V$ must be derived, so it's not very explicit. The "official" definition is -$$\mathbf Z(n)_V = C_*(\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\wedge n}))[-n],$$ -where $\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\wedge n})$ means the quotient of the sheaf $\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\times n})$ sending $U$ to $\mathrm{Cor}_k(U,\mathbf G_m^{\times n})$ by the subsheaf generated by $\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\times n-i-1}\times\{1\}\times\mathbf G_m^{\times i})$. The relation with the other formula comes from the fact that $C_*(\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\wedge 1}))\to\mathbf G_m$ is a free resolution of $\mathbf G_m$ as a sheaf with transfers.<|endoftext|> -TITLE: Shortest vectors in a root lattice -QUESTION [7 upvotes]: Let $R$ be a simply-laced root system in a Euclidean vector space $E$, with inner product normalized so that every root has length $\sqrt{2}$. Let $L \subseteq E$ be the lattice spanned by $R$. Is it true that $R$ is the root system of $L$ (i.e., that $R$ is the set of elements of $L$ of length $\sqrt{2}$)? - -REPLY [2 votes]: This statement is true. There is a more general statement behind it which holds for root systems associated to finite, affine or hyperbolic Kac-Moody Lie algebras, cf. "Infinite-dimensional Lie algebras" by V. Kac, Proposition 5.10(a). This proposition states that you can recover all short real roots by taking elements of the root lattice which have minimal positive length.<|endoftext|> -TITLE: What do you call $C$ if $[D,C] = D^\vee \otimes C$ for all $D$? -QUESTION [11 upvotes]: This is different from $C$ being dualizable ($[C,D] = C^\vee \otimes D$). (EDIT: It turns out to be the same -- see Mike Shulman's answer!) But for example, if $C$ is a locally free sheaf of finite rank on a scheme/locally ringed space $X$, then $C$ has this property in the category of quasicoherent sheaves, or in the category of $\mathcal O_X$-modules -- just check locally. -Here I'm working in a symmetric monoidal closed category $\mathcal C$ with monoidal product $\otimes$, unit $I$, internal hom $[-,-]$, and $E^\vee$ denotes the dual $E^\vee = [E,I]$. -I'm really tempted to call such an object $C$ "locally free (of finite rank)", because the condition says that you can understand maps $D \to C$ as long as you internally (i.e. locally) understand maps $D \to I$, i.e. maps into the canonical "free" object. -Perhaps I should say "locally Cauchy-free" instead of "locally free (of finite rank)", since presumably the significance of "locally being able to take finite sums of $I$" is that finite sums of $I$ are $\mathcal C$-enriched absolute colimits (a.k.a. $\mathcal C$-enriched Cauchy colimits) in the additive context. But I don't understand what's going on well enough to firmly draw this connection. - -REPLY [12 votes]: First of all, a nitpick: the condition "$[D,C] = D^\vee\otimes C$" should be stated more precisely as "the canonical map $D^\vee\otimes C \to [D,C]$ is an isomorphism". Now as you mentioned in a comment, it's well-known that the $\forall C$ version of this condition is equivalent to dualizability of $D$, and indeed the $C=D$ case is already equivalent to dualizability. -However, at least in a symmetric monoidal category, it seems to me that the $\forall D$ version is also equivalent to dualizability of $C$. As noted, taking $D=C$ it implies dualizability of $C$. But conversely, using the fact that if $C$ is dualizable then so is $C^\vee$, we have (in the imprecise version) -$$[D,C] = [D,[C^\vee,I]] = [C^\vee,[D,I]] = [C^\vee,D^\vee] = D^\vee \otimes C.$$ -Something analogous might even work in the non-symmetric case, if we kept careful track of the handedness of the duals and internal-homs; I haven't checked. -The connection to your comment about Cauchy-ness is that if $C$ is dualizable, then "copowers by $C$" are also a Cauchy colimit, coinciding with powers by $C^\vee$.<|endoftext|> -TITLE: Blowup formula for motivic cohomology -QUESTION [5 upvotes]: If $X$ is a smooth projective variety over a field, $Z\subset X$ a smooth closed subvariety of codimension $d$, $X'\to X$ the blowup of $X$ along $Z$, there's the blowup formula -$$H^j(X'_{et},\mathbf{Z}(n)) = H^j(X_{et},\mathbf{Z}(n))\oplus\bigoplus_{r=1}^{d-1}H^{j-2r}(Z_{et},\mathbf{Z}(j-r)).$$ -I must be missing something. For $j = 2n+1$ and $n = 1$, we get -$$Br(X') = Br(X) \oplus(things)$$ -and it isn't clear to me the "things" vanish. But we know the cohomological Brauer group is a birational invariant of $X$, so they must vanish. - -What do I not know about $H^{j-2r}(Z_{et},\mathbf{Z}(j-r))$? - -REPLY [2 votes]: There is simply a typo in this formula, the expression -$H^{j - 2r}(Z_{et},j-r)$ should be replaced by $H^{j - 2r}(Z_{et},n-r)$. This is closesly related to the fact that as motives -$M(\mathbb{P}(E)) = \oplus_{r = 0}^{rank(E)} M(X)(r)[2r]$ for a vector bundle $E$ over a smooth variety $E$. In this case the sum in the RHS correspond to the exceptional divisor, which is the projectivisation of the normal bundle for a smooth blow up, the the shift in the cohomology and the weight correspond to the $"(r)"$ and the $"[2r]"$ in the formula for the motive of this vector bundle.<|endoftext|> -TITLE: A question on closed geodesics -QUESTION [6 upvotes]: I heard of a result on Riemannian manifolds but I couldn't find a good reference: - -For every point $p$ on a complete Riemannin manifold there exists a closed geodesic passing through $p$ with length two times the injective radius at $p$. - -Thanks! - -REPLY [14 votes]: The statement is not true. -First, a comment on terminology. There are two related notions for a geodesic that passes twice through a point $p$: a periodic geodesic and a geodesic loop at $p$. The former is smooth at all points, while the latter may be non-smooth at $p$. Sometimes periodic geodesics are called closed geodesics. -By shortening one shows that at any point of a complete Riemannian manifold there is a shortest geodesic loop at that point. Similarly, any compact Riemannian manifold contains a periodic geodesic (but not necessarily through every point). -On a non-compact complete manifold a shortest periodic geodesic through a point may not exist. For example, the surface of revolution obtained by rotating the graph of $z=\frac{1}{x}$ has no periodic geodesics, even though its injectivity radius if finite. (Of course, the standard Euclidean space also has no periodic geodesics but its injectivity radius is infinite). -Another good example is the acute cone of revolution smoothed at the tip. The shortest loop from a point not near the tip goes around the tip point, and is never a periodic geodesic. -Now what is true (in a complete Riemannian manifold) is that the injectivity radius is the minimum of two numbers: - -The half of the length of a shortest geodesic loop at $p$, -The distance from $p$ to nearest conjugate point at $p$. (This number is called the conjugate radius at $p$ and it is bounded below by $\frac{\pi}{\sqrt{k}}$ where $k$ is an upper section curvature bound of the ambient manifold). - -Finally, let me address a possible source of the OP confusion. Recall that the injectivity radius of Riemannian manifold is the infimum of injectivity radii at its points. Klingenberg proved that if the injectivity radius of a compact Riemannian manifold is smaller than $\frac{\pi}{\sqrt{k}}$, then it equals half of the length of a closed periodic geodesic. -Good references are textbooks by P.Petersen and T.Sakai, which are called "Riemannian Geometry".<|endoftext|> -TITLE: Quadratic algebras, quadratic operads, quadratic categories and quantum cohomology -QUESTION [7 upvotes]: Motivated by the quantisation of the symmetric laws in physics, the category of quadratic algebras has been endowed with two tensor products by Manin in his Montreal lectures notes. These products have been extended to the category of quadratic operads by Ginzburg and Kapranov. A long time ago in my master thesis, (I have started to work on this topic in June 1995 and I have defended my master's thesis in September 1995), I have defined the notion of quadratic category which is a category endowed with two tensor products, quadratic algebras and quadratic operads are examples of quadratic categories. More precisely: -A quadratic category $(C,\bullet, \circ)$ is a category $C$ endowed with two tensor products $\bullet$ and $\circ$ such that: -$I_{\circ}$ and $I_{\bullet}$ are the respective neutral elements of $(C,\circ)$ and $(C,\bullet)$. -We denote by $c^{\bullet}:(A\bullet B)\bullet C\rightarrow A\bullet (B\bullet C)$ the associative constraint of $\bullet$. -For every $A\in C$, there exists $A^!$ in $C$, -morphisms: -$b_A:I_{\bullet}\rightarrow A\circ A^!$ -$d_A:A^!\bullet A\rightarrow I_{\circ}$ -two natural morphisms: -$f^1_{A,B,C}:(A\circ B)\bullet C\rightarrow A\circ (B\bullet C)$ -$f^2_{A,B,C}: A\bullet (B\circ C)\rightarrow (A\bullet B)\circ C$ -which verifies: -$f^1_{A\bullet B,C,D}(f^2_{A,B,C}\bullet id_D)=f^2_{A,B,C\bullet D}(Id_A\bullet f^1_{B,C,D})c^{\bullet}_{A,B\circ C,D}$ -$(Id_A\circ f^2_{B,C,D})f^1_{A,B,C\circ D}=c^{\circ}_{A,B\bullet C,D}(f^1_{A,B,C}\circ Id_D)f^2_{A\circ B,C,D}$ -and -$(Id_A\circ d_A)f^1_{A,A^!,A}(b_A\bullet Id_A)=Id_A$ -$(Id_{A^!}\circ d_A)f^2_{A^!,A,A^!}(Id_{A^!}\bullet b_A)=Id_{A^!}$. -We have the following result: -Theorem. -Let $C$ be a quadratic category and $B,D$ two objects of $C$, the functor $A\rightarrow Hom_C(A\bullet B,D)$ is representable by $D\circ B^!$. -Questions. -I would like to know if there exist other examples of such quadratic categories related or not related to the theory of quantum groups ? -In a recent note, Manin studies the interaction between quadratic algebras, quadratic operad, a notion of enriched category due to Kelly and quantum cohomology? Can these relations be interpreted with this framework of quadratic category ? -Reference. -V. Ginzburg, M. Kapranov. Koszul duality for operads. Duke Math 1994. -Yu. Manin Higher structures, quantum group and genus zero operad -https://arxiv.org/pdf/1802.04072.pdf -Yu. Manin. Quantum groups and non–commutative geometry. Publ. de -CRM, Universit´e de Montr´eal (1988), -Tsemo Aristide M\'emoire de D.E.A 1995. - -REPLY [3 votes]: A similar place where two interacting monoidal structures come up is in duoidal categories. -I believe that any star-autonomous category, i.e. a symmetric monoidal closed category with a dualizing object, is an example of a quadratic category. I have only checked a fraction of the coherence diagrams, though. -Examples: - -There are examples coming from linear logic; in deference to that subject where "$!$" means something different, I will use write $A^\ast$ instead of $A^!$. -There are also examples from algebraic geometry: if $X$ is a smooth scheme of pure dimension $d$ over a field $k$, then the derived category of $X$ is an example (because of Serre duality). There are many more examples; see here for some (in that paper, the theory of $\ast$-autonomous categories is essentially rediscovered). -I believe there are many more examples. - -Details: -Let $(\mathcal C,\otimes, I)$ be a symmetric monoidal closed category with dualizing object $D$, and write $[-,-]$ for the internal hom. We may define $I_\bullet = I$, $\bullet = \otimes$, $A^\ast = [A,D]$, $I_\circ= D$, and $A \circ B = [[A,D]\otimes[B,D],D] = (A^\ast \bullet A^\ast)^\ast$. Note that - -There is a canonical isomorphism $A^{\ast\ast} = A$. -There is a canonical isomorphism $Hom(A \otimes B, C^\ast) = Hom(A,(B\otimes C)^\ast)$ - -I believe that $\mathcal C$ is a quadratic category: - -The map $d_A: A^\ast \bullet A \to I_\circ$ is given by the evaluation $[A,D]\otimes A \to D$. -The map $b_A: I_\bullet \to A \circ A^\ast$ is given by the map $I \to [[A,D],[A,D]]$ representing the identity, since $A \circ A^\ast = (A^\ast\otimes A^{\ast\ast})^\ast = (A^\ast \otimes A)^\ast = [[A,D]\otimes A,D] = [[A,D],[A,D]]$. -The map $f^1_{A,B,C}: (A\circ B) \bullet C \to A\circ (B\bullet C)$, i.e. $(A^\ast \otimes B^\ast)^\ast \otimes C \to (A^\ast \otimes (B \otimes C)^\ast)^\ast$ corresponds via (2) to a map $(A^\ast \otimes B^\ast)^\ast \to (C \otimes A^\ast \otimes (B \otimes C)^\ast)^\ast = [C \otimes (B\otimes C)^\ast,A] = [(B\otimes C)^\ast,[C,A]] = [[C,B^\ast],[C,A]]$; the domain can be rewritten as $[B^\ast,A]$. The morphism $[B^\ast,A] \to [[C,B^\ast],[C,A]]$ to which $f^1$ corresponds is the morphism representing postcomposition. -The map $f^2_{A,B,C}$ is defined similarly. - -Proviso: I have only checked the unit equations.<|endoftext|> -TITLE: Unramified non-abelian extension and Galois cohomology -QUESTION [6 upvotes]: Is there an example of a finite Galois extension $E/F$ of number fields, such that $G=\mathrm{Gal}(E/F)$ is non-abelian and the order of the cohomology group $H^1(G,U_E)$ is relatively prime to class number $N$? ($U_E$ denotes the group of units of $E$.) -(Indeed, I think if $E/F$ is an abelian (or at least is a cyclic) unramified (at all finite places) extension, it is not possible for $\# H^1(G,U_E)$ to be relatively prime to the class number $N$. Is that true?) - -REPLY [4 votes]: If I understand your question correctly, every simple unramified extension of a quadratic number field with class number $1$ is an example. Artin constructed the first such extensions, now there are many examples known; see e.g. -"Remark on infinite unramified extensions of number fields with class number one" by D. Brink and the literature there.<|endoftext|> -TITLE: Making sense out of intertwining operators defined by a vector valued integral -QUESTION [5 upvotes]: Let $G$ be the rational points of a connected, reductive group over a $p$-adic field $F$. Let $S$ be a maximal split torus of $G$ with $\Delta$ a set of simple roots corresponding to a minimal parabolic. Let $(\pi,V)$ be an irreducible, admissible representation of a Levi subgroup $M$ of $G$. For $P = MN$ corresponding to some set of simple roots $\theta \subseteq \Delta$, we have the induced representation -$$I(\nu,\pi) = \operatorname{Ind}_{MN}^G \pi \otimes q^{\langle \nu + \rho, H_M(-) \rangle}$$ -where $\nu$ is in the complexified real Lie algebra of $M$, and $\rho$ is half the sum of the roots of $S$ in $N$ counting multiplicity. For $w$ in the Weyl group of $S$, sending $\theta$ to $\theta' \subseteq \Delta$, we get an intertwining operator $A: I(\nu,\pi) \rightarrow I(w(\nu),w(\pi))$ defined by an integral -$$Af(g) = \int\limits_{N_w} f(\dot w^{-1}ng)dn$$ -where $N_w$ is generated by the root subgroups of those roots which are made negative by $w^{-1}$, and $\dot w$ is a nice choice of representative for $w$. -So, I understand formally why this integral intertwines the action of $G$ on each space. What I don't understand and haven't yet found a reference for is how to make sense of the integral itself. It is a vector valued integral, over a function with values in the underlying space of $\pi$. Since we are dealing with smooth representations of $p$-adic groups, this space need not have any topology associated with it. -Usually, vector valued integrals for $p$-adic representations are finite sums, taken over compact sets. I don't believe this should be the case here. If one is stating everything precisely, how should we make sense out of this intertwining operator? - -REPLY [5 votes]: It's not really that the space of locally constant, compactly-supported (complex-valued) functions on a p-adic (or other totally disconnected) group has no topology. Rather, it has a canonical topology with some convenient features. That is, that space is a "strict" (filtered) colimit of finite-dimensional spaces. That is, it is a countable ascending union of finite-dimensional spaces. Finite-dimensional spaces have unique topological vector space topologies. This colimit has a unique (topological vector space) topology. The reason that for many purposes we can ignore the topology is that every linear map from that space to any other topological vector space is continuous: this follows from the corresponding fact for finite-dimensional TVS's and the definition of "colimit". -As in the question, there are times when the topology matters to some degree. Happily, it is quasi-complete (despite not being complete-metric), which is sufficient for various version of vector-valued integrals to make sense, whether Bochner-style integrals or Gelfand-Pettis-style. -As an easier case, it is standard that a compactly-supported, continuous, $V$-valued function (for $V$ locally convex, quasi-complete) $F$ has an integral with expected properties. As with L. Schwartz' treatment of "Schwartz functions" as extending to a suitable one-point compactification, if/when we can compactify the physical set on which we integrate, then we can immediately apply this simplest case of vector-valued integration. -(Of course, the integrals defining these intertwinings only converge for parameters in some cone, and must be meromorphically continued...) -EDIT: to be clear(er), as in comments, no, indeed, the functions in the induced repn are not compactly supported (although compactly supported mod $P$), but they are completely determined by their values on the maximal compact, and this gives the finite-dimensional feature... But/and one easily finds in the literature misleading remarks about spaces of test functions on t.d. groups "having no topology" or "having discrete topology", and I intended partly to address that.<|endoftext|> -TITLE: General position for map from surface to 3-manifold -QUESTION [6 upvotes]: Let f be a smooth map from a (compact,oriented) surface S to a (compact, oriented) 3-manifold M. Suppose that I have an embedded (non-contractible) loop $\gamma$ in my surface $S$, can I find an (immersed) loop $\gamma'$ freely homotopic to $f \circ \gamma$ which is disjoint from $im(S)$? - -REPLY [9 votes]: In general, no you cannot. Consider one dimension down. Take two curves on a torus, intersecting transversely in a point. One of the curves cannot be homotoped to be disjoint from the pair of curves. Now, cross with a circle $S^1$ to get $T^3$. The pair of curves crossed with $S^1$ is an immersed surface (two immersed tori $T^2$). Then either curve lies in the surface, but may not be homotoped to be disjoint from the immersed surface (this may be proved using the intersection product on homology, dual to the cup product). -You might object that the surface is disconnected. To obtain a connected surface, just tube them together to get a homologous surface.<|endoftext|> -TITLE: Homotopy type of the semi-simplicial set of symmetric groups -QUESTION [7 upvotes]: Consider the collection of symmetric groups $\{\Sigma_n\}_{n\geq1}$ as a semi-simplicial set (i.e. a simplicial set without degeneracies) as follows. Consider $i\in\{1,\dots,n+1\}$ and $\pi\in\Sigma_{n+1}$ represented as a sequence $(\pi(1),\dots,\pi(n+1))$, then -$$d_{i-1}(\pi)=(\pi(1)-\epsilon_1,\dots, \widehat{\pi(i)},\dots,\pi(n+1)-\epsilon_{n+1})$$ -where for every $j\in\{1,\dots,n+1\}$ -$$\epsilon_j=\begin{cases} 0 & if\ \ \pi(j)<\pi(i)\\ 1 & if\ \ \pi(j)>\pi(i). -\end{cases}$$ - -I am wondering what it's known about the homotopy type of the geometric realization of this semi-simplicial set? - -REPLY [11 votes]: It is contractible. To see this first observe that it is simply connected. It has 2 arcs $a = (1,2)$ and $b = (2,1)$, however the tirangle $(3,1,2)$ gives us the relation that $ab = b$ and symmetrically $ba = a$ so $\pi_1 = \{1\}$ for this space. To see that the homology vanishes you can use the homotopy operator on the homology complex given by $H((i_1,...,i_{n+1})) = (i_1,...,i_{n+1},n+2)$. -It is pretty clear that $dH \pm Hd = Id$ in this case, just note that all the first face maps just turn formally $n+2$ into $n+1$ (giving precisely the Hd term) while the last face map eliminate the $n+2$-st term, giving back the original chain. -So, this simplicial complex is a simply connected acyclic space, hence contractible.<|endoftext|> -TITLE: Volume ratio of general $\ell_p$ balls and surfaces -QUESTION [6 upvotes]: This question is a generalization of the question Volume ratio of $\ell_1$ balls and $\ell_1$ surfaces -For any $p\in[1,\infty]$ define $\|x\|_p := (|x_1|^p+\cdots+|x_d|^p)^{1/p}$ for $p\in[1,\infty)$ and $\|x\|_\infty := \max_{1\leq i\leq d}|x_i|$ for $p=\infty$. Denote $B_p^d := \{x\in\mathbb R^d: \|x\|_p\leq 1\}$ as the unit $\ell_p$-ball in $d$ dimension, and let $\partial B_p^d := \{x\in\mathbb R^d: \|x\|_p = 1\}$. The quantity of interest is the following "ratio" -$$ -\mathfrak d_{p,d} := \frac{\mathrm{vol}_{d-1}(\partial B_p^d)}{\mathrm{vol}_{d}(B_p^d)}. -$$ -My question is the following: - -For any fixed $p\in[1,\infty]$, how does $\mathfrak d_{p,d}$ asymptotically scale with dimension $d$, as $d\to\infty$? - -For some special cases, both $\mathrm{vol}_d(B_p^d)$ and $\mathrm{vol}_{d-1}(\partial B_p^d)$ have closed form solutions and the ratio $\mathfrak d_{p,d}$ can be calculated explicitly. Below are three examples: - -$p=2$: in this case $\mathrm{vol}_d(B_2^d) = \frac{\pi^{d/2}}{\Gamma(d/2+1)}$, $\mathrm{vol}_{d-1}(\partial B_2^d) = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ and therefore $\mathfrak d_{2,d} = d$; -$p=1$: in this case $\mathrm{vol}_d(B_1^d) = \frac{2^d}{d!}$ and $\mathrm{vol}_{d-1}(\partial B_1^d) = \frac{2^d \sqrt{d}}{(d-1)!}$, and therefore $\mathfrak d_{1,d} = d\sqrt{d}$ -$p=\infty$: in this case $\mathrm{vol}_d(B_\infty^d) = 2^d$ and $\mathrm{vol}_{d-1}(\partial B_\infty^d) = 2^d n$. Therefore $\mathfrak d_{\infty,d} = d$. - -From the above examples, my (very wild) guess is that $\mathfrak d_{p,d} \asymp d^{1+1/p-1/2}$ for $1\leq p\leq 2$ and $\mathfrak d_{p,d} \asymp d$ for $p\geq 2$. But of course I could be very wrong. The challenge for general $p$ is the apparent difficulty in evaluating the volume of $L_p$ sphere areas (e.g., Surface area of an $\ell_p$ unit ball?), but I'm hoping that the (asymptotic) ratio $\mathfrak d_{p,d}$ for general $p$ is potentially easier to evaluate. -Edit: I realized that, because the volume of a unit $\ell_p$ ball does have closed forms for general $p\in[1,\infty]$, the question can be answered if we know the asymptotic dependency of dimension $d$ of the (unit) $\ell_p$-surface area. This question was explicitly mentioned in the comment under the accepted answer of the following question -Surface area of superellipsoid (dice) -Unfortunately, the OP in that question did not pursue this direction. - -REPLY [2 votes]: We have managed to obtain a complete solution to this problem, by applying the divergence theorem and the classical results of Naor and Romik on the affinity between cone and surface measures in $\ell_p$ balls. -The results are different than what I have conjectured in the problem statement. In particular, it is proved that $\mathfrak d_{p,d} \asymp d^{1/2+1/p}$ for all fixed $p<\infty$ and $d\to\infty$. Therefore, there is a phase transition between $p\to\infty$ and $p=\infty$. Such phase transition is because of the asymptotic regime we're considering ($p$ fixed, and $d$ goes to infinity afterwards). In cases where both $p$ and $d$ go to infinity simultaneously, the question remains unsolved. -The proof idea is to first use the divergence theorem to obtain -$$ -d\times \mathrm{vol}_d(\mathbb B_p^d) = \mathrm{vol}_{d-1}(\partial\mathbb B_p^d)\times \int_{\partial\mathbb B_p^d}\frac{\mathrm{d}\sigma_p^{d-1}(v)}{\sqrt{\sum_{i=1}^d|v_i|^{2p-1}}}. -$$ -Here $\sigma_p^{d-1}(v)$ is the surface measure on $\partial\mathbb B_p^{d-1}$, meaning that $\sigma_p^{d-1}(A)=\mathrm{vol}_{d-1}(A)/\mathrm{vol}_{d-1}(\partial\mathbb B_p^d)$ for all measurable $A\subseteq\mathbb B_p^{d-1}$. -The work of Naor and Romik showed that this surface measure is very close (in TV distance) to a cone measure $\gamma_p^{d-1}$, which is equivalently the distribution of $X=Z/\|Z\|_p$ where $Z=(Z_1,\cdots,Z_d)$ consists of i.i.d. random variables distributed according to a generalized Normal distribution $p(\cdot)\propto \exp\{-|\cdot|^p\}$. It is then possible to use Chebyshev's inequality to get an adequate estimate of the integral term above, which yields an asymptotic formula for $\mathfrak d_{p,d}$.<|endoftext|> -TITLE: Intermediate moduli spaces of stable maps -QUESTION [5 upvotes]: In the following paper: -A. Mustata, M. A. Mustata, "Intermediate moduli spaces of stable maps", Invent. math. 167, 47–90 (2007) -the authors introduced a variation on moduli spaces of stable maps denoted by $\overline{M}_{0,0}(\mathbb{P}^r,d,k)$. There exist a natural morphism $\overline{M}_{0,0}(\mathbb{P}^r,d)\rightarrow\overline{M}_{0,0}(\mathbb{P}^r,d,k)$ wich basically contracts the tails mapping with degree less than or equal to $d−k$. -Does there exist a version $\overline{M}_{0,n}(X,d,k)$ of these moduli spaces taking into account also the data of $n$ ordered points on the source curves, and $X$ is an arbitrary projective homogeneous variety? - -REPLY [7 votes]: I am just adding some details to my comments above. Let $k$ be a field. Let $G$ be a simply connected, semisimple algebraic $k$-group; for simplicity, assume that $G$ is split. Let $X$ be a projective $k$-scheme with a transitive, smooth action of $G$. For simplicity, assume that $X$ has a $k$-point with stabilizer $P$ a ("standard", i.e., smooth) parabolic subgroup of $G$. The group of curve classes on $X$ equals $$\text{CH}_1(X)=\text{Hom}_{\mathbb{Z}}(\text{Pic}(X),\mathbb{Z}) = \text{Hom}_{\mathbb{Z}}(\text{Hom}_{k-\text{Gp. Sch.}}(P,\mathbb{G}_m),\mathbb{Z}).$$ For every curve class $\beta\in \text{CH}_1(X)$, denote by $\overline{\mathcal{M}}_{0,n}(X,\beta)$ the $k$-stack parameterizing families of genus-$0$, $n$-pointed stable maps to $X$ with pushforward curve class $\beta$, $$(C,(q_1,\dots,q_n),u:C\to X).$$ This is a smooth, proper algebraic stack with finite diagonal and projective coarse moduli space $\overline{M}_{0,n}(X,\beta)$; it is a Deligne-Mumford stack if the characteristic of $k$ is strictly larger than the degree $\langle c_1(L),\beta \rangle$ for every primitive ample generator $L$ of $\text{Pic}(X)$. -For every geometrically reduced, Cartier divisor $D\subset X$ denote by $U_D\subset \overline{\mathcal{M}}_{0,n}(X,\beta)$ the open substack parameterizing stable maps such that $u^{-1}D$ contains no irreducible component of $C$ and such that the effective Cartier divisor on $C$, $$u^*D + \underline{q}_1+\dots+\underline{q}_n,$$ contains no singular point of $C$ and has multiplicity $\leq 2$ at every point of the support. The complement $B_D$ of $U_D$ has codimension $\leq 2$ at every point. Denote by $m\geq 0$ the intersection number $\langle [D],\beta \rangle$. Considering $u^*D$ as an unordered collection of $m$ points on $C$, the stabilization of $(C,(q_1,\dots,q_m),u^*D)$ gives a point of the quotient $\overline{M}_{0,n+m}/\mathfrak{S}_m$ of $\overline{M}_{0,n+m}$ by the action of the symmetric group $\mathfrak{S}_m$ permuting the last $m$ points. Altogether, the defines a $1$-morphism, $$\rho_D:U_D\to \overline{M}_{0,n+m}/\mathfrak{S}_m.$$ -For every Cartier divisor $A$ on $\overline{M}_{0,n+m}/\mathfrak{S}_m$, the pullback $\rho_D^*A$ extends to all of $\overline{\mathcal{M}}_{0,n}(X,\beta)$ since the complement $B_D$ has codimension $2$, and since the stack is smooth. Also, for every element $g\in G(k)$, there is the translate divisor $g\cdot D\subset X$. For every stable map, there is a dense open subscheme of $G$ parameterizing those $g$ such that the stable map is contained in $U_{g\cdot D}$. Since $G$ is rational, all of the divisor classes $\rho_{g\cdot D}^*A$ are linearly equivalent. Assuming that $A$ itself moves in a basepoint free linear system $(A_t)$, then the pullbacks $\rho_{g\cdot D}^*A_t$ span a basepoint free linear system on $\overline{\mathcal{M}}_{0,n}(X,\beta)$. This linear system defines a contraction of the coarse moduli space, $$\phi_{|A|}:\overline{M}_{0,n}(X,\beta) \to Y_{|A|}.$$ Of course the contraction only depends on the numerical equivalence class of $A$ up to positive multiple, and in fact it only depends on the corresponding contraction, $$\psi_{|A|}:\overline{M}_{0,n+m}/\mathfrak{S}_m \to Z_{|A|}.$$ -These contractions were studied in the articles of Coskun, Harris, and me. -There are many contractions of $\overline{M}_{0,n+m}/\mathfrak{S}_m$, and each of these defines a contraction of $\overline{\mathcal{M}}_{0,n}(X,\beta)$. It is absurd to try to "chart the geography" of all such contractions, but some contractions / basepoint free divisors have many applications. The divisor that has, so far, had the most applications in algebraic geometry is the divisor class identified by Kawamata in his work on the subadjunction formula. In later work, Keel and McKernan proved that this divisor class is basepoint free. This $\mathfrak{S}_{n+m}$-invariant divisor class on $\overline{M}_{0,n+m}$ gives a contraction $\psi$ that contracts all boundary divisor classes except $\Delta_{2,n+m-2}$ (since the target of the contraction has Picard rank $1$, this is enough to uniquely specify the divisor class up to positive multiples). -Already when $X$ equals $\mathbb{P}^r_k$, when $D$ equals a hyperplane, when $m\geq 2$, when $n$ equals $0$, and when $A$ is Kawamata's divisor class, the associated contraction of $\overline{M}_{0,0}(\mathbb{P}^r_k,\beta)$ does not equal the coarse moduli space of a Deligne-Mumford stack equal to the usual Deligne-Mumford stack $\overline{\mathcal{M}}_{0,0}(\mathbb{P}^r_k,\beta)$ over the maximal open $V$ of $Y_{|A|}$ where $\phi_{|A|}$ is an isomorphism. So there is no "modular interpretation" that is a Deligne-Mumford stack. Of course there is an Artin stack whose good moduli space equals $Y_{|A|}$ and restricts to $\overline{\mathcal{M}}_{0,0}(\mathbb{P}^r_k,\beta)$ over $V$; the Artin stack coming from (unparameterized) quasimaps.<|endoftext|> -TITLE: What did Frobenius prove about $M_{12}$? -QUESTION [20 upvotes]: I am interested in this paper which I can't read because it's in German: -Frobenius, G., Über die Charaktere der mehrfach transitiven Gruppen., Berl. Ber. 1904, 558-571 (1904). ZBL35.0154.02. -A free online copy is here. I am specifically interested in the character table of $M_{12}$ which appears at the top of p.568. -I would like to know what, exactly, Frobenius proved with regard to this character table. If I understand correctly, at the time of publication, the existence of $M_{12}$ was still somewhat contentious. Mathieu had written down some permutations that generated a 5-transitive group on 12 letters, but there was some uncertainty as to whether or not these permutations generated all of $A_{12}$. I believe the uncertainty continued until the 1930's, at which point Witt used Steiner systems, to show that $M_{12}$ was really a proper subgroup of $A_{12}$. -In light of this, my guess is that Frobenius proved the following theorem: - -Theorem. Suppose that $G$ is a sharply 5-transitive subgroup of $S_{12}$. Then the character table of $G$ is as follows... - -Note that the theorem as stated does not refer to the specific construction of $M_{12}$ given by Mathieu, but is more a theorem of the kind "Should such a group exist, then...." (a type of theorem that appeared many times in the following 100 years as part of the classification). -However, perhaps I am wrong. So my questions are, specifically: - -Does Frobenius' character table calculation pertain to any sharply 5-transitive subgroup of $S_{12}$, or is it specific to $M_{12}$? -Does Frobenius give a proof of his calculations, or is it more of a sketch? -The same questions as above also apply to $M_{24}$, with obvious edits. - -Edit: Thank you for the useful comments and answers. Perhaps I should add a fourth question to clarify exactly what I am looking for: - -Does Frobenius use any properties of $M_{12}$, apart from sharp 5-transitivity, in his calculation of the character table? - -Edit 2 -- 26 June 2018: My MMath student, Sam Hughes, and I have just uploaded a preprint connected to this question. In it we write down the character table of a sharply 5-transitive subgroup of $A_{12}$ without making any reference to the Mathieu groups. So, even though this is not what Frobenius did (as Frieder Ladisch's answer below makes clear), it is interesting to know that he could have done it if he had wanted to! We have cited Frieder's answer below in the preprint, as it was very helpful for clarifying the history of this work. Thanks again! - -REPLY [17 votes]: It seems to me that Frobenius is using lots of specific facts about the permutation group $M_{12}$ (and $M_{24}$, respectively) here, and that he has no doubts about the existence of these groups. In particular, he is using specific subgroups of $M_{12}$ and $M_{24}$ (and then using induced characters). Most statements are given without proof, and so the reader would have to verify them by whatever tedious calculations (?). -The first paragraph of §5 reads: - -By using the developed theorems, I have calculated the characters of all multiply transitive groups of degree $\leq 24$. Except for the symmetric and alternating groups of the various degrees, no group is known which is more than fivefold transitive, and only two five-transitive groups are known, both discovered by Mathieu, whose characters I want to give here. - -In the next paragraph, Frobenius describes the partition of $M_{12}$ into conjugacy classes, without proofs or references, so as if this is generally known or easy to reproduce. He also explains notation in the character table, e.g., $(6)(3)(2)$ denotes the class of elements with cycle lengths $6,3,2,1$, and $(3)^4$ denotes the class of elements which decompose into $4$ cycles of length $3$. (In the linked scan of the paper, the first column of the table denoting the conjugacy classes is partly hidden in the binding.) The third paragraph of §5 explains more notation (second column contains centralizer orders, first row degrees of characters with superscript if there is more than one of the same degree). -The bottom paragraph on p. 567 reads: - -By the theorems we have proved, $M_{12}$ has the characters $\alpha-1 : 11^{(1)}$, $\frac{1}{2}(\alpha-1)(\alpha-2)-\beta : 55^{(1)}$ and $\frac{1}{2}\alpha(\alpha-3) + \beta : 54$. [Here, $\alpha$ denotes the number of elements fixed and $\beta$ the number of $2$-cycles.] Formula (5), §2 contains three characters of the third dimension of the symmetric group $S_{12}$, and each of these decomposes into two characters of $M_{12}$, namely $99+55^{(3)}$, $120+45$, $144+176$. The first two characters of Formula (6), §2 decompose into $11^{(2)} + 54 + 66 + 144 $ and $ 66 + 120 + 144 $. - -Paragraph following the table on p.568: - -The substitutions of $M_{12}$ fixing a symbol constitute a $4$-transitive group $M_{11}$ of degree $11$ and order $11.10.9.8$. Furthermore, $M_{12}$ contains a subgroup which is isomorphic to $M_{11}$, is $3$-transitive of degree $12$ and has order $12.11.10.6$. So one can also use this group to represent $M_{12}$ as a transitive permutation group on $12$ symbols, and thus one obtains an outer automorphism of $M_{12}$, and thereby the classes $(8)(2)$ and $(8)(4)$ and their squares $(4)^2$ and $(4)^2(2)^2$ are interchanged. By this automorphism, the character $11^{(1)}$ is sent to $11^{(2)}$ and the character $55^{(1)}$ is sent to $55^{(2)}$. The other characters are computed using the subgroup $M_{11}$. - -Next paragraph: - -With benefit, one can also use the following remarkable subgroup of $M_{12}$: Let $(1,2,3,4,5,6)(7,8,9)(10,11)(12)$ be a substitution in the class $(6)(3)(2)$. Then all substitutions $R$ in $M_{12}$, which move the first $6$ (and the last $6$) only among them, constitute a group of order $6!$. Each such substitution $R$ decomposes into two substitutions $R_1$, only permuting the first $6$, and $R_2$, only permuting the last $6$ symbols among themselfs. Both $R_1$ and $R_2$ run through the $6!$ substitutions of the symmetric group $S_6$ of degree $6$, and $R_1$ and $R_2$ correspond by the well known outer autmorphism of this group. In fact, the classes $(6)$, $(3)^2$ and $(2)^3$ of $S_6$ thus correspond to the classes $(3)(2)$, $(3)$ and $(2)$, and their union gives the classes $(6)(3)(2)$, $(3)^3$ and $(2)^4$ of $M_{12}$. The principal character of $S_6$ gives the reducible character $1 + 11^{(1)} + 11^{(2)} +54 +55^{(3)}$ of $M_{12}$, the other linear character of $S_6$ gives the character $11^{(2)} + 55^{(2)} +66$, and thus one gets the two simple characters $55^{(3)}$ and $66$. Then the above formulas give all characters except $16^{(1)}$ and $16^{(2)}$, which can easily be determined from the bilinear relations [=orthogonality relations?]. - -Last paragraph of §5 (p. 569): - -If $\chi(R)$ is a character of $H$, then - $$ \frac{1}{2}(\chi(R)^2 - \chi(R^2)) \:,\quad \frac{1}{2}(\chi(R)^2 + \chi(R^2) ) \tag{1.} -$$ - are linear combinations of characters with positive integer coefficients. If one chooses for $\chi(R)$ the character $16^{(1)}$, then one gets $120$ and $16^{(2)} + 54 + 66$. - -Section 6 on $M_{24}$ is even more sketchy, but has roughly the same structure. Frobenius tells us that he used two other subgroups in addtion to the stabilizer $M_{23}$. One is the setwise stabilizer $M_{16+8}$ of a subset of order $16$, which is the $15+1$-part of an element of cycle structure $15+1+5+3$. The permutations induced on the $16$-subset are said to form the $3$-transitive linear group of order $2^4(2^4-1)(2^4-2)(2^4-2^2)(2^4-2^3)$, and the permutation group induced on the $8$-set is $A_8$. -According to the bottom paragraph on p.~570, on gets another subgroup by taking the stabilizer of a $12$-subset. The stabilizer is isomorphic to $M_{12}$. Each element of the stabilizer decomposes into two permutations $R_1$ and $R_2$ on a $12$-subset and its complement, the correspondence between $R_1$ and $R_2$ is given by the outer automorphism of $M_{12}$ described in §5. Unfortunately, Frobenius gives us no clue how to choose that $12$-subset ("in passender Art")!<|endoftext|> -TITLE: About cartesian closed categories of models of a cartesian theory -QUESTION [7 upvotes]: Let $T$ be a small category, and $\mathrm{Mod}(T)\subset\mathrm{Fun}(T,\textbf{Set})$ the category of cartesian (finite-limit preserving) copresheaves on $T$. If $T$ is a commutative algebraic theory, then $\mathrm{Mod}(T)$ is cartesian closed (see commutative algebraic theory in nLab). -First question: Is the converse implication true? I mean: if $\mathrm{Mod}(T)$ is cartesian closed, can we prove that $T$ commutative? -Furthermore, if $T$ is the full subcategory of the simplicial category $\Delta$ with objects $[0], [1], [2], [3]$ (where $[n]$ is the order $0<1<\dotsm 0$). The category of $T$-algebras is equivalent to the category of $n$-fold cartesian products of sets. - -When $n = 2$, this is the category of rectangular bands. -It should be noted that the first part of Proposition 4.1 may also be found in Proposition 2.3 of Kock's Bilinearity and cartesian closed monads. -In general, we have the following characterisation result for when a (single-sorted) algebraic theory $T$ has a cartesian-closed category of algebras. (Johnstone also discusses a semantic condition for $S$-sorted algebraic theories in Section 9, but does not give the syntactic characterisation.) - -Definition (Johnstone, Section 1). An operation $p$ is strongly non-constant if, whenever we have an identity of the form $p(x_1, \ldots, x_n) = q(v_1(x_{\alpha(1)}), \ldots, v_m(x_{\alpha(m)}))$ and $p$ does not depend on $x_i$, then $q(y_1, \ldots, y_m)$ does not depend on any $y_j$ with $\alpha(j) = i$. - - -Theorem (Johnstone, Theorem 1.2). Let $T$ be a non-degenerate algebraic theory. Then the category of $T$-algebras is cartesian-closed iff every operation is strongly non-constant, and for every $n$-ary operation $p$, there exists an $m$-ary operation $q$, two $m$-tuples of unary operations $u_j, v_j$ ($1 \leq j \leq m$), and a function $\alpha : \underline m \to \underline n$ such that the following identities hold: $$q(u_1(y), \ldots, u_m(y)) = y$$ $$u_j(p(x_1, \ldots, x_n)) = v_j(x_{\alpha(j)}) \qquad (1 \leq j \leq m)$$ - -An elegant characterisation of cartesian-closure for varieties of anomic $S$-sorted theories (i.e. theories with no equations) is provided in Oles's When is a category of many-sorted algebras cartesian closed?. - -Theorem (Oles, Theorems 1 & 7). Let $\Sigma$ be an $S$-sorted signature. The category of $\Sigma$-algebras is cartesian-closed (and furthermore a topos) iff $\Sigma$ is unary (i.e. every operation is unary). - -To answer your first question directly: cartesian-closure of the category of $T$-algebras does not imply that $T$ is commutative. The answer to the second question is affirmative for algebraic theories.<|endoftext|> -TITLE: fibrations of classifying spaces - Leray Hirsch Theorem converse -QUESTION [8 upvotes]: Let $G$ be a topological group and let $H$ be a closed subgroup. Assume that $G \rightarrow G/H$ is a principal $H$-bundle. We have a fibration of classyifing spaces -$$G/H \rightarrow BH \rightarrow BG.$$ -I am interested in the case where the associated spectral sequence degenerates and leads to an isomorphism $H^*(BH) \cong H^*(BG) \otimes H^*(G/H)$. -This holds, for instance, when $G/H$ is contractible or in the case of $G$ a compact connected Lie group, and $H$ the maximal torus on $G$. -Specifically, I am wondering if just assuming that $H^*(BH)$ is a free $H^*(BG)$-module with the structure induced by the inclusion $BH \rightarrow BG$ is enough to talk about the degeneracy of the spectral sequence. If I assume that $G$ is connected , then $BG$ is simply connected and my statement will hold under the Eilenberg-Moore spectral sequence; but I want to consider cases where $G$ is not connected. -EDIT 28/02 -Looking around, I realize that maybe the Leray-Hirsch theorem might play a role here in some specific situations: if the spectral sequence collapses, and $H^*(G/H)$ is a free $R$-module, then $H^*(BH)$ is a free $H^*(BG)$-module. -Conversely, if I assume that $H^*(G/H)$ is a free $R$-module, and $H^*(BH)$ is a free $H^*(BG)$-module, does it follow that the spectral sequence collapses and that -$$H^*(BH) \cong H^*(BG) \otimes H^*(G/H)?$$ - -REPLY [2 votes]: If you work with coefficients in a field $\mathbb{F}$, assume that $H^*(BH;\mathbb{F})$ is a free $H^*(BG;\mathbb{F})$-module, and add the assumption that the Serre spectral sequence has a product structure (i.e. the map -$$H^p(BG;\mathbb{F}) \otimes H^0(BG ; H^q(G/H;\mathbb{F})) \overset{\smile}\longrightarrow H^p(BG ; H^q(G/H; \mathbb{F})) = E_2^{p,q}$$ -is an isomorphism) then the Serre spectral sequence collapses. This can be proved by considering the lowest row a potential differential could start on. (Unfortunately I don't know a reference.) This shows that -$$H^*(BH;\mathbb{F}) \cong H^*(BG;\mathbb{F}) \otimes H^0(BG ; H^*(G/H;\mathbb{F}))$$ -as $H^*(BG;\mathbb{F})$-modules. To get the conclusion you want you must further suppose that $\pi_0(G)$ acts trivially on $H^*(G/H;\mathbb{F})$, but in this case the product structure is automatic. -EDIT: Here is a proof (I will omit coefficients). Suppose that the Serre spectral sequence does not collapse, that the non-zero differential starting on the lowest row starts on the $q$th row, and the longest differential coming out of this row is a $d_r$. -Let $\{\bar{b}_i\}$ be obtained by choosing a homogeneous $H^*(BG)$-module basis for the free module $H^*(BH)$, and discarding those basis elements of degree $\geq q$. Then everything below the $q$th row is a permanent cycle, so the map -$$\mathbb{F} \otimes_{H^*(BG)} H^*(BH) \to H^0(BG ; H^*(G/H))$$ -is surjective in degrees $* < q$. Thus the restrictions $b_i$ of the $\bar{b}_i$ to $H^*(G/H)$ generate $H^0(BG ; H^*(G/H))$ as a $\mathbb{F}$-module in degrees $*r$, but in degee $q+1$ the submodule -$$H^*(BG)\{\bar{b}_i \, \vert \, \vert \bar{b}_i \vert < q+1-r\} \leq H^*(BH)$$ -accounts for all such classes. This gives a nontrivial $H^*(BG)$-linear dependence between the $\bar{b}_i$, which contradicts that these formed part of a $H^*(BG)$-module basis of $H^*(BH)$.<|endoftext|> -TITLE: The Giraud-Benabou construction for splitting fibrations -QUESTION [6 upvotes]: I'm currently reading "Revisiting the categorical interpretation -of dependent type theory" and they give a very terse description of the Giraud-Benabou construction: -For a fibration $p : \mathbb E \to \mathbb B $, the fibration $Rp : R\mathbb E \to \mathbb B$ - is defined by - -Obj $R\mathbb E$ is the set of pairs $(X,\phi)$ where $X \in \mathbb B$ and $\phi : \mathrm{dom}_X \to p$ is a morphism of fibrations from the restriction of the domain fibration to the slice categories $\mathrm{dom}_X: \mathbb B / X \to \mathbb B$. $Rp$ is the first projection. -$\mathrm{Hom}((X, \phi),(Y, \psi))$ consists of pairs $(t,\mu)$ where $t : X \to Y$ and $\mu : \phi \to \psi \circ \mathrm{dom}_t$ over $\mathbb B$. - -They then prove that this is a split fibration in a single line ("for $t:X \to Y$, set $\phi [t] = \phi \circ \mathrm{dom}_X$, ie. $\phi[t](s) = \phi(t \circ s)$ for all $s:Z \to X$, and $t_\phi = (t,id)$. Moreover, $Rp$ is split by the associativity of composition in $\mathbb B$"), and it isn't clear to me exactly what the relation between $Rp$ and $p$ is. -Questions: - -Is there a slightly less terse reference for this that isn't buried somewhere in an almost 500 page book on non-abelian cohomology written in French? -What exactly is the relation between $Rp$ and $p$? -What's the intuition behind this construction? If I were trying to create a split fibration from a fibration, why might I think to do this? -What does this do to the codomain fibration over Set? Does this have a simple description? - -REPLY [4 votes]: Section 2.2 of The local universes model gives a possible answer to (2) and (3): $R$ (or $(-)_*$ in their notation) is a right adjoint to the forgetful functor from split fibrations to non-split fibrations. Their description is no less terse, but perhaps beta-reduced a bit more: - -An object of $T_*$ over $\Gamma\in C$ consists of an object $A$ of $T(Γ)$, together with for each $f : \Gamma'\to \Gamma$ some cartesian lifting $\bar{f} : A_f \to A$, such that $A_{1_\Gamma} = A$, $\bar{1_\Gamma} = 1_A$. - -In the case of the codomain fibration, this means that an object of $T_*$ is a morphism equipped with a chosen pullback square along every morphism into its codomain, such that the chosen pullback along the identity is the identity square.<|endoftext|> -TITLE: Configuration spaces, Ran spaces, free semilattices, Vietoris spaces and power objects -QUESTION [12 upvotes]: These are five important constructions and I would like to know how they are related. -The $n$th unordered configuration space of a space $X$ is -$$ -\operatorname{UConf}_n(X):=\{\text{embeddings of $\{1,...,n\}$ into $X$}\}/(\text{$n$th symmetric group}), -$$ -topologized as a subquotient of $X^n$. -The Ran space of $X$ is the set $\operatorname{Ran}(X)$ of finite subsets of $X$ with the topology generated by sets -$$ -\nabla_{U_1,...,U_n}:=\{S\in\operatorname{Ran}(U_1\cup\cdots\cup U_n)\mid S\cap U_i\ne\varnothing, i=1,...,n\} -$$ -where $U_i$ are disjoint open subsets of $X$. -The free topological semilattice $\operatorname{Sl}(X)$ on $X$ is the value on $X$ of the left adjoint to the forgetful functor from topological semilattices to topological spaces. -The Vietoris space $\mathscr VX$ of $X$ is the set of some (depending on the context) subsets of $X$ topologized by the same kind of $\nabla_{U_1,...,U_n}$ except that they are not required to be disjoint. -Finally, one may choose some nice embedding $I$ of some subcategory of spaces that contains $X$ into a topos in various ways, and consider there the power object $\Omega^{IX}$. Usually it is not in the image of $I$. There are versions like $\operatorname{Sub}_{\mathrm{fin}}(IX)\rightarrowtail\Omega^{IX}$ of objects of finite (say, Kuratowski finite) subobjects of $IX$ which might be. (Note that $\operatorname{Sub}_{\mathrm{fin}}$, with Kuratowski finiteness, is the free internal semilattice functor on any topos whatsoever.) -As a variation on the latter - say, $X$ is a simplicial set; since simplicial sets readily form a topos we have simplicial sets $\operatorname{Sub}_{\mathrm{fin}}(X)\rightarrowtail\Omega^X$. -Questions: - -Is $\operatorname{UConf}_n(X)$ (homeomorphic to) a subspace of $\operatorname{Ran}(X)$? -There is a topology on $\bigcup_n\operatorname{UConf}_n(X)$ with $\{x_1,...,x_n,x_{n+1}\}$ close to $\{x_1,...,x_n\}$ when $x_{n+1}$ is close to $x_n$ in $X$. Is this homeomorphic to $\operatorname{Ran}(X)$? -The same two questions with $\operatorname{Sl}$ in place of $\operatorname{Ran}$. -Is $\operatorname{Ran}(X)$ homeomorphic to $\operatorname{Sl}(X)$? -Are $\operatorname{Ran}(X)$, $\operatorname{UConf}_n(X)$ or $\operatorname{Sl}(X)$ subspaces in $\mathscr VX$ for some nice spaces $X$? -Are there known embeddings of some categories of spaces into toposes such that the image of the embedding is closed under taking $\operatorname{Sub}_{\mathrm{fin}}$? In particular, can $\operatorname{Sub}_{\mathrm{fin}}(IX)$ be isomorphic to $I(\operatorname{Sl}(X))$ for some such $I$? -How does the geometric realization of $\operatorname{Sub}_{\mathrm{fin}}(X)$ relate to $\operatorname{Ran}$, $\operatorname{UConf}_n$, $\operatorname{Sl}$ and $\mathscr V$ of the geometric realization of $X$ for a simplicial set $X$? - -$\ \ \,$0.$\ $Are these considered together and compared somewhere in the literature? - -REPLY [6 votes]: Too long for a comment but it is essentially a comment: -It is easy to see that for a Hausdorff space $X$ the topology on the Ran space coincides with the Vietoris topology and for a non-Hausdorff space $X$ the Ran topology is strictly weaker than the Vietoris topology. -The topology of the free topological semilattice is stronger than the Vietoris topology. For example, for an infinite compact metrizable space $K$ the space $SL(K)$ is a non-metrizable $k_\omega$-space whereas the Vietoris topology is metrizable. So, $SL(K)$ even topologically does not embed into $\mathcal V X$. -By the way, the hyperspace $\mathcal V X$ of non-empty finite subsets of a topological space $X$ endowed with the Vietoris topology coincides with the free Lawson topological semilattice of $X$. -For a Hausdorff space $X$ the configuration space $\mathrm{UConf}_n(X)$ naturally embeds into the free (Lawson) topological semilattice. -For the free Lawson semilattice $\mathcal V X$ this can be shown by comparing the Vietoris (or Ran) topology with the quotient topology on $\mathrm{UConf}_n(X)$. Then combining this with the continuity of the natural maps $\mathrm{UConf}_n(X)\to SL(X)\to \mathcal V X$, we can conclude that $\mathrm{UConf}_n(X)\to SL(X)$ is an embedding, too. -Concerning the literature on the free (locally convex) topological semilattices (at least), you can look at the following papers and references therein: -Banakh and Sakai - Free topological semilattices homeomorphic to $\mathbb R^\infty$ or $\mathbb Q^\infty$ -Banakh, Guran, and Gutik - Free topological inverse semigroups - -As I understand in question (2) on the space $\bigcup_{n\in\mathbb N}\mathrm{UConf}_n(X)$ it is considered the topology of direct limit of the tower $exp_n(X)$ where $exp_n(X)$ is the family of all at most $n$-element subsets of $X$ endowed with the Vietoris (or Ran) topology. This topology is stronger than the topology of $SL(X)$ and I am afraid that two topologies coincide only for $k_\omega$-spaces. For general spaces the operation of taking union is discontinuous with respect to this direct limit topology, so it is not a topological semilattice. This follows from Proposition 4, p. 35, of Banakh, Guran, and Gutik - Free topological inverse semigroups. This proposition says that if for a functionally Hausdorff space $X$ the free topological semilattice $SL(X)$ is a $k$-space, then each closed metrizable subspace of $X$ is locally compact.<|endoftext|> -TITLE: Kazhdan constant and finite index subgroups -QUESTION [7 upvotes]: I am wondering if there is some general relation between Kazhdan constants of a group and it finite index subgroups? -Let $G$ be a finitely generated group with a generating set $\Sigma$ that satisfies Kazhdan property (T) with constant $\kappa(G,\Sigma)$. -1) if $\Gamma$ is a finite index subgroup of $G$, is there a generating set $\Theta$ of $\Gamma$ for which $\kappa(\Gamma,\Theta)$ can be estimated in terms of $\kappa(G,\Sigma)$? -2) if $G$ is a finite index subgroup in $H$, is there a generating set $\Theta$ of $H$ such that $\kappa(H,\Theta)$ can be estimates in terms of $\kappa(G,\Sigma)$? - -REPLY [4 votes]: If $n:=[G:H]$, then $\mathbb C[G] \subset M_n \mathbb C[H]$, where $g \in G$ maps to a permutation matrix decorated with elements from $H$ and the embedding depends essentially only on a choice of a transveral of the quotient map $G \to G/H$. One can arrange things, so that generators $S \subset G$ map to permutations with decorations of length at most $2[G:H]+1$. Those $[G:H]|S|$ decorations generate $H$, this is essentially the content of Schreier's lemma. Let's call this generating set $\Sigma$. We can also arrange that $\mathbb C[H] \subset \mathbb C[G] \subset M_n \mathbb C[H]$ is of the form $h \mapsto h \oplus h^\perp$, where the exact form of $h^{\perp} \in M_{n-1}(\mathbb C[H])$ depends on the situation. -The following proposition is well-known, the book of Bekka-de la Harpe-Valette is an excellent reference for all this. - -Proposition 1.1.9 (Bekka-de la Harpe-Valette) If $(S,\varepsilon)$ is a Kazhdan pair for $G$ and $\delta>0$, then every $(S,\varepsilon\delta)$-invariant vector $\xi$ admits a $G$-invariant vector at distance less than $\delta\|\xi\|$. - -We claim that if $(S,\varepsilon)$ is a Kazhdan pair for $G$, then $(\Sigma,[G:H]^{-1/2}\varepsilon)$ is a Kazhdan pair for $H$. -(1) Suppose that $G$ is a Kazhdan group and $(S,\varepsilon)$ be a Kazhdan pair. Let $\mathcal H$ be a unitary representation of $H$ and $\xi \in \mathcal H$ and assume that $\|h\xi-\xi\|<[G:H]^{-1/2}\varepsilon \|\xi\|$ for all $h \in \Sigma$. Then, we obtain a $G$-representation $\mathcal H^{\oplus n}$ from above and a vector $\eta = (\xi,...,\xi)$ that is $[G:H]^{-1/2}\varepsilon$-fixed by $S$. Thus, there must be a $G$-fixed vector $\eta_0 =(\eta_0^1,...,\eta_0^n)$ at distance less than $[G:H]^{-1/2} \|\eta\|$. It follows that $\eta_0^1$ is non-zero and $H$-fixed. -One can also argue similarly for (2), but it is less clear what the optimal bound would be. -(2) Now suppose that $H$ is a Kazhdan group and let $\mathcal H$ be a unitary $G$-representation and $\xi \in \mathcal H$ with $\|g\xi - \xi\|<\varepsilon \|\xi\|$ for all $g \in S$. It follows that $\|h\xi - \xi\|< (2[G:H]+1)\varepsilon$ and hence, if $\varepsilon>0$ is sufficiently small, there must exist non-zero $H$-invariant vectors nearby. Moreover, the finite group $G/\cap_{g \in G}H^g$ (which is of size at most $[G:H]!$) acts on the space of $H$-invariant vectors. Thus, if $\varepsilon>0$ is small enough, there must exist a $G$-invariant vector. I would guess that the spectral gap of a finite group $L$ is at least of the order $|L|^{-1}$, so that this also gives a quantitative bound.<|endoftext|> -TITLE: linear independence of exponentials -QUESTION [5 upvotes]: Let $X$ be the set of functions $e^{p(x)}$ of the real vector $x$, where $p$ is a multivariate polynomial with $p(0)=0$. -Is any finite subset of $X$ linearly independent? If yes, why? If no, is the answer true for other, restricted choices of $p$? (The answer is yes when the polynomials are restricted to have degree 1.) - -REPLY [6 votes]: Yes. -If you require that no difference $p_j-p_k$ is constant (which follows from -your assumption $p(0)=0$), then $c_1e^{p_1}+\ldots+c_me^{p_m}=0$ implies that all $c_j=0$. -In fact a more general result is true: instead of polynomials one can take any -entire functions. This is called Borel's theorem. -(See, for example, S. Lang, Introduction to complex hyperbplic spaces, Springer 1987, Theorem 1 -in VII, sect 1, p. 186.) -For polynomials in one variable, this is easy: look at the asymptotics, -for example when $z$ tends to infinity on an appropriate ray, and use induction -in degree of the polynomials. To obtain the same for several variables, restrict your identity onto lines in $C^n$.<|endoftext|> -TITLE: Can the subobject classifier be covered by an object in a subtopos? -QUESTION [7 upvotes]: Let $\mathcal{E}$ be a topos and $\Omega$ its subobject classifier. -Is it possible to have a nonidentity local operator (a.k.a Lawvere-Tierney topology) $j\colon\Omega\to\Omega$, a $j$-sheaf $X\in\mathcal{E}_j\subseteq\mathcal{E}$, and an epimorphism $f\colon X\twoheadrightarrow\Omega$? -I'd be interested to see an example. -Thanks! - -REPLY [9 votes]: Consider the Sierpinski topos, i.e., presheaves on the poset $P=\{0<1\}$, and consider the topology in which $1$ is covered by $\{0\}$, i.e., the double-negation topology. Then the presheaf $1+1+1$ (i.e., assign to each node of $P$ a 3-element set and let the transition map be a bijection) is a sheaf. It maps surjectively to $\Omega$ by sending its three points to the three global sections of $\Omega$.<|endoftext|> -TITLE: When is the etale cohomology of a variety simple if its Hodge structure is simple? -QUESTION [5 upvotes]: Suppose $X$ is a smooth projective variety defined over $\mathbb{Q}$, and the pure Hodge structure on $H^{2n}(X)$ is of a very simple form, -\begin{equation} -\mathbb{Q}(-n)^{b^{2n}} -\end{equation} -where $b^{2n}$ is just $\text{dim}\,H^{2n}(X)$. Under what conditions is the etale cohomology $H^{2n}_{et}(X,\mathbb{Q}_{\ell})$ also isomorphic to -\begin{equation} -\mathbb{Q}_{\ell}(-n)^{b^{2n}} -\end{equation} -as representations of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$? -This is however not always true, consider a smooth quadratic surface $S$ of $\mathbb{P}^3_{\mathbb{Q}}$ defined by a quadratic polynomials with determinant say 7. Then the Hodge structure on $H^2(S)\simeq H^2 (\mathbb{P}^1 \times \mathbb{P}^1)$ is $\mathbb{Q}(-1)^{2}$, while $H^{2}_{et}(S,\mathbb{Q}_{\ell})$ is -\begin{equation} -\mathbb{Q}_{\ell}(-1)^{2} \otimes \chi -\end{equation} -where $\chi$ is a Dirichlet character. - -REPLY [5 votes]: Assuming the Hodge conjecture, it's true if and only if the relevant algebraic cycles are defined over $\mathbb{Q}$. -Conjecturally, the cohomologies are different realizations of the same motive. The condition on the Hodge cohomology tells you the motive is a Tate twist of an Artin motive, and you want the Artin motive to be trivial.<|endoftext|> -TITLE: Reference request: The resolvent is analytic in the resolvent set -QUESTION [5 upvotes]: I am busy reading through Taylor's paper Spectral Theory of Closed Distributive Operators. -On page 192, he defines the resolvent and spectrum of $T$: - -Later on in the paragraph, he then proceeds by saying - -When $\rho(T)$ is not empty it is known that $R_\lambda(T)$ is analytic in $\rho(T)$ as a function with values in $[X].$ - -I am, however, having trouble finding a reference which proves that the resolvent is analytic in the resolvent set. Can anyone please point me to a reference in which this is proven? -I did previously ask this question here on Math.SE, but noticed that the answer provided there assumes that the resolvent is continuous in $\lambda$, which is also a result that I do not have. - -REPLY [9 votes]: Although András' comment already answers the question, I think it is worthwile to give a few more details explicitely here, in order to point out that the analyticity is in fact a consequence of the resolvent identity only and has nothing to do with the operator whose resolvent we consider: -Let $X$ denote a complex Banach space, let $[X]$ denote the space of all bounded linear operators on $X$ and let $U \subseteq \mathbb{C}$ be non-empty and open. -Definition. A mapping $\mathcal{R}: U \to [X]$ is called a pseudo-resolvent if it fulfils the so-called resolvent identity -\begin{align*} - \mathcal{R}(\lambda) - \mathcal{R}(\mu) = (\mu - \lambda) \mathcal{R}(\lambda)\mathcal{R}(\mu) -\end{align*} -for all $\lambda, \mu \in U$. -Note that the resolvent of every closed linear operator on $X$ is a pseudo-resolvent, but there are also pseudo-resolvents which are not the resolvent of a closed linear operator (for instance the constant mapping $\lambda \mapsto 0$). -Proposition. Let $\mathcal{R}: U \to [X]$ be a pseudo-resolvent, let $\lambda,\lambda_0 \in U$ and assume that $|\lambda-\lambda_0| < \|\mathcal{R}(\lambda_0)\|^{-1}$ (where we define $0^{-1} := \infty$). Then -\begin{align*} - \mathcal{R}(\lambda) = \sum_{n=0}^\infty (\lambda_0 - \lambda)^n \mathcal{R}(\lambda_0)^{n+1}, -\end{align*} -where the series converges absolutely in $[X]$ (which is endowed with the operator norm). In particular, the mapping $\mathcal{R}: U \to [X]$ is analytic. -Proof. It follows from the resolvent identity that -\begin{align*} - \mathcal{R}(\lambda) \big[\operatorname{id} - (\lambda_0 - \lambda) \mathcal{R}(\lambda_0)\big] = \mathcal{R}(\lambda_0). -\end{align*} -The operator norm of $(\lambda - \lambda_0) \mathcal{R}(\lambda_0)$ is strictly smaller than $1$ by assumption, so we conclude from the Neumann series that the operator in square brackets is invertible and that its inverse is given by $\sum_{n=0}^\infty (\lambda_0 - \lambda)^n \mathcal{R}(\lambda_0)^n$. This proves the assertion. -A few references: -For the case where the pseudo-resolvent is actually a resolvent of a closed operator: - -"K.-J. Engel and R. Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)", Proposition IV.1.3 (as already pointed out by András in the comments) -"W. Arendt, C. Batty, M. Hieber, F. Neubrander: Vector-Valued Laplace Transforms and Cauchy Problems (2011)", Corollary B.3 in the appendix - -For pseudo-resolvents: - -"M. Haase: The Functional Calculus for Sectorial Operators (2006)": in the appendix of this book, pseudo-resolvents are treated as resolvents of so-called multi-valued operators. - -Personal remark concerning the Math.SE post mentioned in the question: -To derive the analyticity of the resolvent as an "immediate" consequence of the resolvent identity without noting that this "proof" requires continuity to be shown first is, at least in my experience, an easily made mistake. I recall falling into this trap myself some time ago.<|endoftext|> -TITLE: Cartography of the duals of GL, PGL, SL, etc -QUESTION [13 upvotes]: A short version of this question could be - -What are the duals of $PGL(2,\mathbf{Q}_p)$, $PGL(2,\mathbf{R})$ and $PGL(2,\mathbf{C})$? - -I should obviously add some precisions. - -there are different notions of duals: admissible, unitary, tempered and generic; all are of interest; -having a parametrization similar to the well-known one for $SL(2, \mathbf{R})$ would be perfect, where principal series, complementary series, discrete series, special representations and parameters involved in their construction appear, as well as the admissible, generic, unitary and tempered parts. Do other drawings of this type exist? even for $SL(2, \mathbf{Q}_p)$? -I stress on $PGL$ because it suits my research purposes, yet having some ideas of the interrelations between $SL$, $PGL$ and $GL$ is probably of interest also. Representation theorists probably know how to switch from one to another, and it could be the reason I always have seen this drawing for $SL(2)$. I would like a reference to be able to do other ones; -the question is about completions of $\mathbf{Q}$, yet more general answers on how to think/visualize representations on every local fields is obviously welcome. I believe it is enough to keep things simple, since the question in mainly about having toy models to think about. - -The aim of all that is to develop some kind of intuition and to have toy models for representation theoretics properties, questions, conjectures, that are as handable and easy to think about as the $SL(2, \mathbf{R})$ case. -Any comment, idea or critics by more enlightened people on this topic to help me formulate the question are obviously welcome. - -REPLY [17 votes]: I'm assuming you're interested in complex representations of these groups. In each case the unitary, tempered and generic duals are contained in the smooth dual, so I'll start by describing that. Let $G$ be any of the groups you're interested in. Actually, most of what I'll say goes through pretty nicely for $G$ the group of points of a connected reductive group over a local field, with some modifications to take the more complicated structure into account. -The first key step in classifying representations into the broad classes you describe is parabolic induction. In $G$ there are certain parabolic subgroups $P$, which have a Levi factorization into a semidirect product $P=MN$, where $M$ is another nice group of the type you're interested in (a Levi subgroup) and $N$ is a normal subgroup. When $G=GL_2$, every parabolic is conjugate to either $GL_2$ itself (which obviously isn't going to help), or to the group $B$ of upper-triangular matrices (the Borel subgroup). The Borel subgroup has a Levi decomposition $B=TU$, where $T$ is the diagonal torus and $U$ is the group of upper-triangular matrices with 1 along the diagonal. In $SL_2$ and $PGL_2$, the same picture is true, but you restrict and project these subgroups, respectively. -Now take $\chi$ an irreducible (so one-dimensional) representation of $T$. You can inflate $\chi$ to $B$ by $B\rightarrow B/U\stackrel{\sim}{\rightarrow}T\stackrel{\chi}{\rightarrow}\Bbb{C}^\times$, and then form the induced representation $Ind_B^G\ \chi$. This is a finite-length representation. Subquotients of this induced representation break into families: - -Finite-dimensional representations. If you're working over a $p$-adic field then these are one-dimensional and factor through $\det$; if not then for each $n\geq 1$ there is a unique family of $n$-dimensional representations, where any two in each family are twists of one another by a character through the determinant. -Principal series representations, which occur when $Ind_B^G \chi$ is irreducible. -If $Ind_B^G\chi$ is reducible, it may or may not be semisimple. If it's semisimple then it splits as a direct sum of two irreducibles, which are called the limit of discrete series. This doesn't happen over a $p$-adic fieldd. Otherwise, it has a unique irreducible subrepresentation, which is one of the finite-dimensional representations. The quotient by this representation is formed of discrete series representations. - -Over the reals or complexes, going through this process gives you all irreducibles. Over a $p$-adic field, there is a family which you don't see in this way: the cuspidal representations. These are much, much harder to construct. -The general picture is similar, but combinatorially it becomes much more complicated as the rank of the group grows. In general there are multiple conjugacy classes of parabolic subgroups, and one has to consider parabolic induction from each of these. Principal series always refers to representations parabolically induced from a Borel subgroup, and discrete series has a more general meaning as well. -Once you know the dual of $GL_2$, you basically know that of the other groups. Given $\pi$ an irreducible representation of $GL_2$, $\pi$ identifies with an irreducible representation of $PGL_2$ if and only if it has trivial central character, and all representations arise in this way. On the other hand, $SL_2$ is a normal subgroup of $GL_2$, and by Clifford theory the restriction of $\pi$ to $SL_2$ splits as a direct sum of pairwise $GL_2$-conjugate irreducible representations (moreover, each appears with multiplicity one). This restriction to $SL_2$ preserves containment in each of the above families of representations.<|endoftext|> -TITLE: Defining computable functions categorically -QUESTION [24 upvotes]: Computable functions may be defined in terms of Turing machines or recursive functions, or some other model of computation. We normally say that the choice doesn't matter, because all models of computation are equivalent. However, something about this bothers me. -Turing machines, for example, operate on strings of tokens, taking strings as input and giving strings as output. If we have another model of computation that also operates on strings then we have no problem saying formally whether it's equivalent to a Turing machine. However, if our model of computation operates on something other than strings then we have to define a mapping from its inputs and outputs to strings, and it seems there is no way to formally define whether that mapping is computable or not. (If I am wrong about this, please correct me.) So whenever we talk about equivalence between Turing machines and some non-string based model of computation, we are invoking an informal step; this seems unsatisfying. -It occurs to me that category theory might offer a way out of this, and I'm wondering if this has been done. For example, suppose we define a category with one object, to be interpreted as the set of all strings on some alphabet. We draw an arrow from this set to itself for every (partial) function that can be computed with a Turing machine, and define composition of arrows as function composition. -It is clear that this forms a category. My question is whether this category by itself (i.e. just the morphisms and their relationships, without reference to the underlying string type) tells us everything we need to know about computable functions. In other words, would it make sense to define Turing equivalence in terms of isomorphism to this category? -What I have in mind here is something like this: suppose we're given a model of computation that computes partial functions from some set $S$ to itself. We can then form a category of all partial functions that can be computed by this system and check whether this category is isomorphic to the similar category for Turing machines as defined above. This is appealing, if it works, because we never needed to define a mapping between the elements of $S$ and the elements of the set of strings. (But I'm worried it might not work, due to the possibility of the category isomorphism itself being uncomputable, in some sense.) -Finally, if this idea doesn't work, is there some way to fix it so that it does? I would be interested to hear about any work on defining or reasoning about the set of computable partial functions from a purely categorical point of view, so that it can be characterised without reference to a particular underlying model of computation. - -REPLY [8 votes]: A more directed answer are Joyal's arithmetic universes. It was once quite difficult to find a description of these, but Maria Maietti has proposed an official description. -Maietti, Maria Emilia, Joyal’s arithmetic universe as list-arithmetic pretopos, Theory Appl. Categ. 24, 39-83 (2010). ZBL1245.03111. -However, for your particular question, I think an arithmetic pretopos suffices. The basic idea is as follows. -First, if $\boldsymbol{1} \xrightarrow{o} \mathbf{N} \xleftarrow{\sigma} \mathbf{N}$ is parameterized natural number object in a a category $\mathcal{A}$ with finite products and disjoint coproducts, then all the $k$-ary primitive recursive functions $f$ are representable in $\mathcal{A}$ in the sense that there is a morphism $\mathbf{N}^k \xrightarrow{\phi} \mathbf{N}$ such that -$$\require{AMScd}\begin{CD} -\boldsymbol{1} @>(\sigma^{n_1}o,\ldots,\sigma^{n_k}o)>> \mathbf{N}^k \\ -@| & @VV{\phi}V \\ -\boldsymbol{1} @>>{\sigma^{f(n_1,\ldots,n_k)}o}> \mathbf{N} -\end{CD}$$ -commutes for all $n_1,\ldots,n_k$. In fact, we can build such a category where all the arrows $\mathbb{N}^k \to \mathbb{N}$ represent primitive recursive functions and nothing more. -Since not all computable functions are primitive recursive, this is not sufficient. To get all computable functions, we need to add unbounded search and then close under composition. That is, given a $k+1$-ary primitive recursive $f$ such that $$(\forall n_1,\ldots,n_k)(\exists m)[f(n_1,\ldots,n_k,m) = 0]$$ the function $$g(n_1,\ldots,n_k) = \min\{m : f(n_1,\ldots,n_k,m) = 0\}$$ is computable. -On the categorical side, this can be achieved by requiring the category to be a pretopos. The image factorization, which comes with pretoposes being regular categories, allows one to prove that the category is closed under unbounded search in the internal sense. Namely, if $G \hookrightarrow \mathbf{N}^k \times \mathbf{N}$ is a subobject such that $G \hookrightarrow \mathbf{N}^k \times \mathbf{N} \to \mathbf{N}^k$ is monic and regular epi, then $G$ is isomorphic the graph of a morphism $\mathbf{N}^k \xrightarrow{\gamma} \mathbf{N}$. One recovers unbounded search by taking $G \hookrightarrow \mathbf{N}^k\times\mathbf{N}$ to be the pullback of $\mathbf{N}^k \times \mathbf{N} \xrightarrow{\phi} \mathbf{N}$ along $\boldsymbol{1} \xrightarrow{o} \mathbf{N},$ where $\phi$ represents the primitive recursive function $$\hat{f}(n_1,\ldots,n_k,m) = \begin{cases} 0 & \text{if $f(n_1,\ldots,n_k,m) = 0$ and $\prod_{i=0}^{m-1} f(n_1,\ldots,n_k,i) \neq 0$,} \\ 1 & \text{otherwise.} \end{cases}$$<|endoftext|> -TITLE: Why the circle for Pontryagin duality? -QUESTION [9 upvotes]: For a locally compact group $G$, we define the Pontryagin dual as $\hat G = Hom(G,\mathbb T)$ where $\mathbb T$ is the circle group and the homomorphisms are continuous group maps. This duality has a lot of nice properties and shows up all over the place so it is probably the right definition. -However, without the benefit of hindsight, why would one choose to define the dual of a group with respect to $\mathbb T$ (instead of some other locally compact group, say). Is there a reason to promote $\mathbb T$ to a special place among all locally compact groups? -A little more broadly, are there other groups $H$ that also lead to a good theory of duality if we define $\hat G_H = Hom(G,H)$? -One possible answer would be to explin the historical context/necessity which led to the definition. But there can also be other motivations and I would be open to both kinds of answers. -On a closely related note, one can also ask a similar question for the Cartier duality in algebraic geometry. Probably, the two will have similar answers. - -REPLY [4 votes]: The group $\mathbb T$ is special. On this site and math.stackexchange the same question has been asked before. -See the second question at When does Pontryagin duality generalize? and Why unitary characters for the dual group in Pontryagin duality if $G$ is not compact? for an explanation. - -REPLY [4 votes]: This is a long comment rather than an answer. Also, I will be very sloppy about the details because I don't have a strong knowledge about them. -I think the fact that Pontryagin duality works with maps $\mathrm{Hom}(\,\_\,,\mathbb{T})$ where $\mathbb{T}=U(1)$ has to do with the following facts: - -Cartier duality, which is a very natural/reasonable thing to do, works with the multiplicative group, wich is $\mathbb{C}^\times$ over the complex numbers. -$\mathbb{T}\subset\mathbb{C}^\times$ is the compact real form of the reductive group $\mathbb{C}^\times$, and the representation theory of a complex reductive algebraic group is "the same" as that of its compact real form. - -[Important edit: as the comment by Yemon Choi reminded me, in what follows the algebraic group $G$ has to be diagonalizable, or at least a version of Cartier duality that I'm aware of is stated for such group schemes in Waterhouse Introduction to affine group schemes. This is very restrictive!] -So what's Cartier duality? Working over $\mathbb{C}$, there is an involution on the category $\mathbf{CHopf}$ of commutative Hopf algebras -$$\mathbf{CHopf}\to\mathbf{CHopf}$$ -$$A:=(V,m,\Delta)\mapsto A^\vee:=(V^\vee,\Delta^\vee,m^\vee)$$ -sending a Hopf algebra $A$ with underlying $\mathbb{C}$-vector space $V$, multiplication $m$ and comultiplication $\Delta$ (and other data that I'm suppressing from the notation) to the Hopf algebra structure $A^\vee$ on the dual vector space $V^\vee=\mathrm{Hom}(V,\mathbb{C})$ with multiplication given by $\Delta^\vee$ and comultiplication by $m^\vee$. The category $\mathbf{aaGrp}$ of affine complex algebraic groups is opposite to the category $\mathbf{CHopf}$ via $A=(V,m,\Delta)\mapsto G:=\mathrm{Spec}(V,m)$ with group operation given by the morphism $\mathrm{Spec}(\Delta):G\times G\to G$. Vice versa, an algebraic group $(G,\mu)$ goes to $V=\mathcal{O}(G)$, the algebra of global functions on $G$, with its natural algebra structure, and comultiplication given by $\mu^{\sharp}:\mathcal{O}(G)\to\mathcal{O}(G\times G)=\mathcal{O}(G)\otimes\mathcal{O}(G)$. -Under this antiequivalence $\mathbf{CHopf}\to\mathbf{aaGrp}$, the involution $A\to A^\vee$ corresponds to $G\mapsto \mathbf{aaGrp}(G,\mathbb{C}^\times)$. -This is to say that "Pontryagin-dualizing" with respect to $\mathbb{T}$ somehow comes from the usual duality of $\mathbb{C}$-vector spaces.<|endoftext|> -TITLE: Example of a locally presentable locally cartesian closed category which is not a topos? -QUESTION [6 upvotes]: The only way I know to get a locally cartesian closed category which is not a topos is to start with a topos and then throw out some objects so that the category is not sufficiently cocomplete to be a topos. Is that the only way there is? -Well, I suppose there's another way: the category of topological spaces and local homeomorphisms is locally cartesian closed. So one can get a similar effect by throwing out morphisms. But in this example the category is not even accessible. -Let's stipulate that the category should also not be a poset. -After all, a topos is just a locally cartesian closed category with a subobject classifier. Is the subobject classifier merely a representability condition, or does it have additional exactness implications beyond local cartesian closure? - -REPLY [9 votes]: Every Grothendieck quasitopos is presentable and locally cartesian closed. These are categories of separated presheaves on a site. The simplest example of a site whose separated presheaves do not form a topos is the one-point topological space: the category of separated presheaf on it is the category of sets with an initial object $0$ freely added. This category doesn't have disjoint coproducts, since $X\times_{X\sqcup Y}Y\simeq \emptyset $ is not initial. More generally, this should work for any site with an initial object that is covered by the empty sieve. -For completeness, let me add the examples discussed in the comments, which are of a different nature: a locale, a.k.a. a $0$-topos, is presentable and locally cartesian closed, but not a topos. -This is an instance of the more general fact that every $n$-topos is presentable and locally cartesian closed, but not an $m$-topos for $m>n$. -Finally, if $k$ is a field, then the ∞-category of motivic spaces over $k$, i.e., $\mathbb A^1$-invariant Nisnevich sheaves on smooth $k$-schemes, is presentable and locally cartesian closed, but is known not to be an ∞-topos: the $0$-truncated group object $a_{Nis}(\mathbb Z[\mathbb A^1-0])$ is an $\mathbb A^1$-invariant Nisnevich sheaf, but it is not equivalent to the loops on its bar construction (so the simplicial colimit in the bar construction is not van Kampen). In light of this, it seems likely that the 1-category of $\mathbb A^1$-invariant Nisnevich sheaves of sets on smooth $k$-schemes (which is locally cartesian closed) is not a topos, but I don't know.<|endoftext|> -TITLE: Looking for generalization of projective model structure -QUESTION [8 upvotes]: If $\mathcal{M}$ is a cofibrantly generated model category and $\mathcal{C}$ is a small category, then we can give $\mathcal{M}^\mathcal{C}$ the projective model structure, in which weak equivalences and fibrations are transformations that are such on each object of $\mathcal{C}$. I recall seeing a generalization in which we take a subset $D$ of objects of $\mathcal{C}$ and say that a transformation $f$ is a weak equivalence or fibration if $f(d)$ is one for every $d\in D$ (and require nothing at the other objects). Am I remembering correctly that this works in this generality, and does anyone have a reference for where this is written down? I'd like to use this but see no need to reinvent the wheel if I can just refer to it. - -REPLY [4 votes]: You might be thinking of the paper "On Modified Reedy and Modified Projective Model Structures" by Mark Johnson. It does exactly what you asked in your question, in Proposition 6.4. Funny you should mention reinventing the wheel; I reproved this result a few years ago in a mathoverflow answer, before I knew about Mark's paper.<|endoftext|> -TITLE: Is a field that never embeds twice in another field necessarily a prime field? -QUESTION [18 upvotes]: Call a field $k$ unrepeatable$^1$ if for every field $L$ there are either zero or one field homomorphisms $k \to L$. Then the prime fields $\mathbb{Q}$ and $\mathbb{F}_p$ for $p$ prime are clearly unrepeatable and it seems very likely to me that those are the only ones. Is that true? -Notice that an unrepeatable field cannot have a non-identity automorphism, and must have a unique embedding into its algebraic closure (so for example $\mathbb{R}$ is not unrepeatable despite having no nontrivial automorphisms). - -$^1$ I made up a term for it because the only name I know for these is the rather unwieldy "subterminal object in the opposite of the category of fields". - -REPLY [15 votes]: It seems that indeed only prime fields are unrepeatable. -Proof: -Let $k$ be unrepeatable and $F\subseteq k$ denote the prime field of $k$. Let $T\subseteq k$ be a transcendence base of $k/F$ and let $G=F(T)$. If $T\neq\emptyset$, then $G/F$ has non-trivial automorphisms (say take one element $t\in T$ to $t+1$). Since $k/G$ is algebraic this extends to a different embedding of $k$ into $\overline G$. Therefore, we see that $k/F$ must be algebraic. -It follows that $F\subseteq k\subseteq \overline{F}$ and since $k$ is unrepeatable (in $\overline F$), $k$ has to be fixed by all automorphisms in the absolute Galois group of $F$. But then $F=k$ and we are done.<|endoftext|> -TITLE: Relative Steenrod's problem -QUESTION [5 upvotes]: Thom's theorem states that for every homology class $\alpha \in H_{*}(X)$ there exists an integer $k = k(\alpha)$ such that the class $k\, \alpha$ comes from the fundamental class of an orientable closed smooth manifold, where $X$ is an arbitrary topological space. To be on the safe side we assume that $X$ is a countable CW-complex. -Given a topological pair $(X,A)$, $A$ is closed, and a homology class $\alpha \in H_{*}(X,A)$. Is it possible to realize $k\, \alpha$ by a map from an orientable manifold with boundary, for $k$ sufficiently large? - -REPLY [9 votes]: Yes. It's quite general. If $h$ is any generalized homology theory (for example, oriented bordism) such that $h_0(point)=\mathbb Z$ and such that $h_n(point)=0$ for all $n<0$ then there is a map from $h$ to ordinary homology inducing an isomorphism $h_0\to H_0$; and after tensoring with $\mathbb Q$ this map $h\to H$ becomes surjective for all spaces and pairs. You can see this using the Atiyah-Hirzebruch spectral sequence with $E^2_{p,q}=H_p(X,A;h_q(point)\otimes \mathbb Q)$, abutting to $h_{p+q}(X,A)\otimes \mathbb Q$. All of the differentials are zero because they are essentially stable operations on rational homology, so the edge $E^2_{p,0}$ survives to $E^\infty$<|endoftext|> -TITLE: Illustrating mathematics with wysiwyg tools -QUESTION [23 upvotes]: What tools are out there for creating mathematical illustrations in a what-you-see-is-what-you-get mode? -Having struggled with tikz for several years, I've found creating figures in Omnigraffle (https://www.omnigroup.com/omnigraffle) to be a liberating experience. (And they have good tech support.) -The webpage http://wiki.illustrating-mathematics.org/wiki/What_tools_do_we_use%3F mentions Inkscape and Illustrator; how do they compare to Omnigraffle for flexibility and ease of use? -If other people know similar systems that they like better, I'd love to know about them. - -REPLY [4 votes]: A free and easy to use software is GeoGebra. After drawing your figure, go to File option and select export as tikz. I personally don't like $\LaTeX$ output of GeoGebra because its generated codes are (long and) not professional, although it draw figures anyway.<|endoftext|> -TITLE: Gödel's speed-up from constructive to classical logic? -QUESTION [11 upvotes]: Gödel's speed-up theorem implies that some proofs can get significantly shortened when allowing extra axioms. There are concrete examples of this phenomenon for instance when moving from Peano arithmetic (PA) to PA + consistency(PA), or when moving from PA to second-order arithmetic, see for instance this question. -However, I am not aware of concrete examples of such a dramatic shortening when moving from constructive logic to classical logic. -So, is there a known example of a statement that is both true in constructive and classical logic, that has a reasonably short proof in classical logic, but such that any proof in constructive logic would be gigantic (and thus not "human-readable")? -Ideally, an example would be a statement that would feel "concrete enough" for the average mathematician, i.e. a statement involving numbers, graphs, algebraic structures, etc. (Friedman's examples on Kruskal's tree theorem fit the bill) -(I would also be interested in an example that has a simple proof in classical logic, that is not yet known to be true in constructive logic but such that a proof, if it exists, must be gigantic.) -Update: There are indeed ways to construct such examples as indicated below, but which feel a bit like cheating as they essentially add the extra axiom in the statement. I am more thinking of some existence theorem of the form -"For every object x satisfying ..., there exists an object y satisfying ...". -Think Friedman's examples, Four-Color Theorem, etc. In other words, an existence theorem where every constructive proof would be gigantic, but where there exists a short nonconstructive proof. Here again, one may hack this by artificially adding something of the form "or (A or not(A))" in the statement, but of course that's not really what I have in mind, nor the sort of statement a mathematician working with classical logic would try to prove on a regular day ;-) - -REPLY [2 votes]: Here is a something which might be an example: -Sometimes there are statements $A$ and $A^*$ which a classical mathematician would consider equivalent and a constructive mathematician would not. An example is the existence of a winning strategy for Hex. -$A(n):$ the game of Hex on an $n \times n$ board will never end in a draw AND there is no winning strategy for the second player (the first could steal it) -$A^*(n)$ There is a winning strategy for the first player in the game of Hex on an $n \times n$ board. -There is a short constructive proof of “$\forall n\,A(n)$ “ -I think a constructive mathematician would accept that $A^*(13)$ is true but consider it possible that any explicit strategy would be enormous. -A classical mathematician would easily accept “ $\forall n\, A^*(n)$ “ That might be unprovable constructively. -——— -A variation: -$B$ no position can end in a draw. -$B^*$ for every position is either a winning strategy for the next player or a winning strategy for the previous player. -From Wikipedia: - -In 1976, Shimon Even and Robert Tarjan proved that determining whether a position in a game of generalized Hex played on arbitrary graphs is a winning position is PSPACE-complete.[12] A strengthening of this result was proved by Reisch by reducing quantified Boolean formula in conjunctive normal form to Hex played on arbitrary planar graphs.[13] In computational complexity theory, it is widely conjectured that PSPACE-complete problems cannot be solved with efficient (polynomial time) algorithms. This result limits the efficiency of the best possible algorithms when considering arbitrary positions on boards of unbounded size, but it doesn't rule out the possibility of a simple winning strategy for the initial position (on boards of unbounded size), or a simple winning strategy for all positions on a board of a particular size.<|endoftext|> -TITLE: Does $\aleph_0$-density of regular open algebra entail existence of countable basis? -QUESTION [8 upvotes]: Suppose that the family $\mathrm{RO}(X)$ of regular open subsets of $(X,\mathscr{O})$ is a basis of $X$. Let the density of $\mathrm{RO}(X)$ (considered as Boolean algebra) be $\aleph_0$. -Does $X$ have to be second-countable? If not, what if we add regularity of $X$ (both $T_1$ and $T_3$ separation axioms)? If answers to both questions are negative, what is the maximal cardinality (relative to $|X|$) of the set of points not covered by dense countable $D\subseteq\mathrm{RO}(X)$? - -REPLY [7 votes]: Joel has already shown you that none of the standard separation axioms is enough to force a positive answer to your question (by the way, the Sorgenfrey line is even perfectly normal). -A notable case for which your question has a positive answer is that of a topological group. -Define the $\pi$-weight of $X$ ($\pi w(X)$) to be the least cardinality of a family $\mathcal{P}$ of non-empty open subsets of $X$ such that for every non-empty open set $U \subset X$ there is $P \in \mathcal{P}$ such that $P \subset U$. It is clear that if $RO(X)$ is a base for $X$ then the density of $RO(X)$ coincides with the $\pi$-weight of $X$. This happens, in particular, if $X$ is a regular space. Now topological groups are regular, and the $\pi$-weight and the weight of a topological group coincide. -(for a proof of that see, for example, Arhangel’skii, Alexander; Tkachenko, Mikhail, Topological groups and related structures, Atlantis Studies in Mathematics 1. Hackensack, NJ: World Scientific; Paris: Atlantis Press (ISBN 978-90-78677-06-2/hbk). xiv, 781 p. (2008). ZBL1323.22001.) -Another case in which $\pi w(X)=w(X)$ is when $X$ is a metric space (because the $\pi$-weight is bounded below by the density and bounded above by the weight, and for metric spaces, the weight and the density coincide).<|endoftext|> -TITLE: A locally presentable locally cartesian closed category that is not a quasitopos -QUESTION [14 upvotes]: This question asks for a locally presentable locally cartesian closed category that is not a topos. All the answers given (at least in the 1-categorical case) are quasitoposes. What is an example of a locally presentable locally cartesian closed 1-category that is not a quasitopos? -Note that there is no difference for this question whether "quasitopos" is meant in the elementary or the Grothendieck sense. By C2.2.13 in Sketches of an Elephant, a locally presentable locally cartesian closed category is a Grothendieck quasitopos iff it is an elementary quasitopos iff it is quasi-effective, i.e. strong equivalence relations are effective. So the question could equivalently be: what is an example of a locally presentable locally cartesian closed category that is not quasi-effective? - -REPLY [8 votes]: Thomas Holder points out to me that this question is answered in Borceux and Pedicchio, A characterization of quasi-toposes, with an example that is reproduced in C4.2.4 of Sketches of an Elephant: the ind-completion of the category $\mathrm{Set}_c$ of finite or countably infinite sets. It is locally finitely presentable essentially by definition, and it is locally cartesian closed because it is a "local exponential ideal" in the presheaf category of $\mathrm{Set}_c$. But it is not a quasitopos because (1) it is balanced, so if it were a quasitopos it would be a topos, and (2) the image of $\mathbb{N}\in \mathrm{Set}_c$ satisfies Freyd's characterization of a natural numbers object, so if it lived in a topos it would be a NNO, but in a cocomplete topos an NNO cannot be finitely presentable.<|endoftext|> -TITLE: Harmonic functions on $(M,g)$ closed, induce an embedding in Euclidean space -QUESTION [7 upvotes]: In Hajime Urakawa's monograph The Spectral Geometry of the Laplacian on page 41, we make an assumption that I can't quite justify on my own. The following is our setup: - -Let $(M^n,g)$ be a closed connected Riemannian manifold, with $\Delta_g$ the Laplacian (a negative operator), and $0=\lambda_0<\lambda_1\le\lambda_2\le\cdots$ the eigenvalues of $-\Delta_g$ with corresponding eigenfunctions $u_0=\hbox{const},u_1,u_2,u_3,\ldots$. Then for $N\gg0$, $$\iota:M\ni x\mapsto(u_1(x),\ldots,u_N(x))\in\mathbb R^N$$ defines an embedding. - -But why is this true? I know that $(u_i)_{i\ge0}$ forms an orthonormal basis for $L^2(M,\mu_g)$, but this doesn't seem to be quite enough to prove that we can have an embedding. Presumably we need to show that we can also approximate functions in $H^1(M,g)$? - -REPLY [8 votes]: You can find a proof of this statement in S Bochner Analytic mappings of compact Riemannian space into Euclidean space Duke Math Journal vol 3 (1937) no 2 pages 339-354. An alternative approach (sketched) is to solve the heat equation for any smooth function as the initial condition using the eigenfunctions of the laplacian . For small time this solution of the heat equation will approximate the initial smooth function in the C^1 topology . Now truncate the eigenfunction expansion of the solution of the heat equation to show that any smooth function can be well approximated by a finite linear combination of eigenfunctions in the C^1 topology .This essentially proves the result.<|endoftext|> -TITLE: Is V, the Universe of Sets, a fixed object? -QUESTION [36 upvotes]: When I first read Set Theory by Jech, I came under the impression that the Universe of Sets, $V$ was a fixed, well defined object like $\pi$ or the Klein four group. However as I have read on, I am beginning to have my doubts. We define -\begin{align} -V_0 & :=\emptyset. \\[10pt] -V_{\beta +1} & :={\mathcal {P}}(V_\beta). \\[10pt] -V_\lambda & :=\bigcup_{\beta <\lambda} V_\beta \text{ for any limit ordinal } \lambda, -\end{align} -and finish by saying -$$V:=\bigcup_{\alpha \in \operatorname{Ord}} V_\alpha.$$ -However this definition seems to create many problems. I can see at least two immediately: First of all, the Power Set operation is not absolute, that is it varies between models of ZFC. Secondly (and more importantly) this definition seems to be completely circular as we do not know a priori what the ordinals actually are. For instance, if we assume some two, mutually inaccessible large cardinals $\kappa , \kappa'$ to exist, and model ZFC as $V_\kappa, V_{\kappa'}$ -respectively, then we get two completely different sets of ordinals! So we seem to be at an impasse: -In order to define the Universe of Sets we must begin with a concept of ordinals, but in order to define the ordinals we need to have a concept of the Universe of Sets! -So my question is to ask: Is this definition circular? The only solution I can think of is that when we define $V$, we implicitly assume a model of ZFC to begin with. Then after constructing the ordinals in this model, we construct $V$ off of them, so to speak. Is this what is being assumed here? - -REPLY [2 votes]: I come a little late, just to clearly state and sum up something that was began to be outlined by some here. -Namely, the syntacticalist view I have that nothing in math or formal sciences in general is a "fixed object" unless its's an actual syntactic object; otherwise it's just a concept in one(s)'s mind(s), possibly formalized into actual syntax and/or conceived from actual syntax. And the concepts themselves are subjectively as diverse as conceivers, only given a same name if those agree enough about their intuitions. -So here, the concepts of π, the Klein Four-Group or the Von Neumann Universe aren't more fixed than by traditional common shared intuition. -But, where the question becomes interesting (to me) and your answer quite true (as your intuition had guessed), is once rephrased given all the above. Sort of: "Has the concept of Von Neumann Universe any concrete syntactical counterpart, an actual avatar, such as π or the Klein Four-Group?" -And the answer indeed is: not really. Yes π has any algorithmic recursive enumeration of its digits in any fixed base. Yes Klein's Group has his Cayley Table. -Though V only has, as said Nate Eldredge in a comment, its defining predicate - as a unary predicate symbol for e.g. ZFC, as a constant symbol of class for e.g. NB. However as this phrase with a hyphen shows, as a defining Formula it really is itself well-"defined" only relatively to some fixed formal system, for which it actually defines something local. Finally, it doesn't alone have an actual independent existence that could unequivocally manifest all its features; only as a string of symbols devoid of most of the answers to the questions one can ask about V as a concept, not at all intrinsically tied to this intuition of ours, enough so to embody it into an actual reality.<|endoftext|> -TITLE: Are $\mathbb C$ , $\mathbb C[X]$ definable in $\mathbb C[[X]]$? -QUESTION [20 upvotes]: Let $L$ be a first-order language and $M$ be an $L$-structure. Let $D \subseteq M^n$ . Let us say $D$ is definable in $M$ if for some finite set (possibly empty) $A=\{a_1,...,a_m\} \subseteq M$ and some formula $\psi[x_1,...,x_n,y_1,...,y_m]$ , $D=\{(b_1,...,b_n)\in M^n : M\vDash \psi[b_1,...,b_n,a_1,...,a_m] \}$. -(i.e. $D$ is definable by a finite set of parameters according to this definition -https://en.wikipedia.org/wiki/Definable_set ). -Now consider the first order theory of commutative rings. Take $M=\mathbb C[[X]]$ (the formal power series ring with complex coefficients) . -My questions are : Is $\mathbb C$ definable in $\mathbb C[[X]]$ ? Is $\mathbb C[X]$ definable in $\mathbb C[[X]]$ - -REPLY [10 votes]: Emil's idea about quantifier elimination is a good one. -The paper "Model Theory of valued fields" by Raf Cluckers cites the thesis "Quelques propriétés des corps valués" by F. Delon, which I wasn't able to find online, for the claim that the field $\mathbb C((t))$ admits quantifier elimination for the language of Macintyre which consists of the language of field theory, a symbol for elements of the valuation ring $\mathbb C[[t]]$, and a symbol for $n$th powers. Thus any set definable in your sense is definable, without quantifiers, in this language. -It is clear that any set definable in this language has only finitely many isolated points and therefore cannot be $\mathbb C$ or $\mathbb C[x]$.<|endoftext|> -TITLE: Did Hilbert laugh? -QUESTION [40 upvotes]: Prof. D. C. McCarty recently gave an interesting -interview (published in January 2015, and easily -found on a large video hosting site), entitled - -What are the limits of mathematical explanation? - -I think this is the best interview with a -philosopher that I have heard or read in a long while, -and hope that Prof. McCarty will give more interviews of -that quality, and I recommend the interview to anyone. -However, there is one problematic minute -during this interview, which is of potential concern to -the mathematical community: during that minute an embellishment is made of one the most often retold legends in the lore of -modern mathematics: Hilbert's public denial that there are unsolvable mathematical questions, his 1930 radio address to that effect (and Gödel's announcement, co-located and simultaneous, of a theorem contradicting Hilbert's belief). -The problematic minute starts at 12 min 1 sec, and ends at 13 min 10 sec, -of the recording that I listened to. I transcribe it here: - -Interviewer: Hilbert said that we will know - everything in mathematics [...] There is no ignorabimus in mathematics. Prof. McCarty: He did say that, and in fact, he did - more than say it: he insisted upon it. He talked about it in his 1901(e1) problems-lecture, and you may have heard this on [...] or - in some other way: his last lecture at Königsberg in 1931.(e2) - He ends the lecture(e3) by talking about the ridiculous - or foolish ignorabimus, and he said that there is no ignorabimus, - and they didn't turn off the recording-device early enough. You can - hear Hilbert's little hideously laugh after he says 'Es gibt kein -ignorabimus.' at the end; and then he goes 'ha ha ha', in a slightly - mean way. It seems to me that Hilbert's utterance there was the last - gasp of a mathematics, and approach to mathematics, that he represented - and was, I think, common in the 19th century. That is, that there aren't - any permanent limits to our mathematical cognition. - -Questions. -1.1. Did Hilbert audibly laugh at the end of his 1930 radio address in Königsberg? -1.2. If so, can the laugh justifiably described as 'hideous' and 'mean'? - -Is there a more extensive recording of Hilbert's address of 8 September 1930 than what can be found here, here, or here (none of which contains any trace of dishonesty or mockery)? -(Not even the file behind the last-mentioned link contains any such trace, which is two seconds longer than e.g. the version which can be found on the largest video hosting service, and still does not contain any laugh? By the way, I did not take the trouble to analyze and compare these audio-files with some specialized software; it is clear that some sort of cutting did occur, but apart from that I don't see what new insight analyzing the available and laughterless recordings would give.) -Independently of what the answer to 1. is, what, if anything, made Prof. McCarty say that Hilbert is on record with a hideous laugh at end of his famous address? - -Remarks. - -I deliberately decided against trying to contact Prof. McCarty privately, -because I think that this question is not injurious (quite the contrary) and because the purpose of this question is to publicly set the historical record straight (to the extent that this is at all possible with such a distant historical event). -Under the 'usual' assumptions about 'history', 'logic' and 'reality', Questions 1.1, 2 and 3 are defined questions with definite answers; in particular, either Hilbert laughed at the end or he didn't (and I think Hilbert would agree with this, given that around the same time he thought he had found proof of the law of exclude middle...). -As regards Question 2, I can think of the following reasons: - -Prof. McCarty was a victim of the well-known psychological mechanism of confabulation -Prof. McCarty had access to an unpublished, uncut version of the radio recording, whereas each publisher of the audio-file either cut the recording to get rid of the 'hideous laugh', or they got an already cut file from some other source -Prof. McCarty read some written account of Hilbert's address, in which a statement about Hilbert laughing at the end is made; if that is so, then of course this raises the further question of what the basis for that hypothetical written account was... - -It may be convenient to readers to have a transcription of Hilbert's address, and a translation; this is what this nice document of J. T. Smith of San Francisco State University offers; I checked the transcription, and found it to be perfect, except for the minuscule detail that "geniesst" must be 'genießt', "grosse" must be 'große', and "heisse" must be 'heiße' (the same orthographical errors are made on the above-mentioned MAA-page, by the way). -I don't think this question is unimportant; other sciences (like chemistry or physics) passionately discuss and document their history and legends too---just think about the vibrant research about the history of physics---and the above-mentioned interview casts one of the most well-known legends of 20th-century mathematics in a new and unbecoming light: Hilbert, against his better knowledge, gleefully giving false hopes to the general public. - -${}$___________________________________ -(e1) By the way, that's a secondary, unimportant mistake in the interview: Hilbert's famous talk at the second ICM in Paris was given in on 8 August 1900, not in 1901. Even the subsequent publication of the talk, both in German and in French traslation, happened already in 1900, so the date 1901 seems completely wrong. -(e2) This is another slight mistake on the part of the interviewee: all sources I know agree that Hilbert's radio address was recorded (and presumably also broadcast---the sources remain silent about that detail) on 8 September 1930, not 1931. -(e3) Presumably, what is meant by 'lecture' here - is just the radio address. - -REPLY [54 votes]: Constance Reid describes the recording on page 196 of her book Hilbert (1969). -The recording was not made directly from the live address. Instead, Hilbert was asked to repeat the conclusion of his speech in the broadcasting studio of a local radio station. This is significant, Hilbert was not speaking in front of an audience, he was speaking into an unfamiliar recording device in the presence of a technician. The faint$^*$ laugh when the technician turned off the machine might just have been a laugh of relief that this was over and had gone well. Hilbert's final sentence was a bit "heavy", if I would be asked to repeat this in front of a recording device I might have felt slightly silly, without the audience, and I might well have laughed to remove the tension. -As reported by Laurent Siebenmann, the laugh was erased in the 1971 restoration process by Springer that transferred the original 1930 recording from a wax cylinder onto vinyl. Attempts to locate the original recording for a CD edition of the speech failed --- it seems the laugh is lost. -$^*$ Reid says the laugh is audible if one listens very carefully.<|endoftext|> -TITLE: What is known about the plethysm $\text{Sym}^d(\bigwedge^3 \mathbb{C}^6)$ -QUESTION [10 upvotes]: What is known about the plethysm $\text{Sym}^d(\bigwedge^3 \mathbb{C}^6)$ as a representation of $\text{GL}(6)$? It is my understanding that this should be multiplicity-free. I tried computing it using the Schur Rings package in Macaulay2 and I cannot see a pattern among the weights that appear. -If a formula is known, a reference would be nice also. Thanks. -EDIT: To save others the work, here is the data for $0 \leq d \leq 5$: -{{0, 0, 0, 0, 0, 0}}, -{{0, 0, 0, 1, 1, 1}}, -{{0, 1, 1, 1, 1, 2}, {0, 0, 0, 2, 2, 2}}, -{{1, 1, 1, 2, 2, 2}, {0, 1, 1, 2, 2, 3}, {0, 0, 0, 3, 3, 3}}, -{{2, 2, 2, 2, 2, 2}, {1, 1, 2, 2, 3, 3}, {1, 1, 1, 3, 3, 3}, {0, 2, 2, 2, 2, 4}, {0, 1, 1, 3, 3, 4}, {0, 0, 0, 4, 4, 4}}, -{{2, 2, 2, 3, 3, 3}, {1, 2, 2, 3, 3, 4}, {1, 1, 2, 3, 4, 4}, {1, 1, 1, 4, 4, 4}, {0, 2, 2, 3, 3, 5}, {0, 1, 1, 4, 4, 5}, {0, 0, 0, 5, 5, 5}}} - -REPLY [3 votes]: $$\bigoplus_{k \ge 0} t^k Sym^{k} V = \frac{1}{(1−tV)(1−t^2 \mathbf{g})(1−t^3 V)(1−t^4)(1−t^4 V_2)},$$ -where $V = \wedge^3 \mathbb{C}^6 = [0,0,1,0,0], V_2 = [0,1,0,1,0]$, and $\mathbf{g} = [1,0,0,0,1].$ -See section 6 of "Series of Lie Groups" by Landsberg and Manivel.<|endoftext|> -TITLE: Holomorphic line bundles on $\mathbb{P}^1$ from gluing data -QUESTION [15 upvotes]: Let $g$ be an $n \times n$ matrix of functions $g_{ij}(z)$ in $\mathbb{C}(z)$. Suppose that the $g_{ij}(z)$ have no poles on the annulus $1-\epsilon < |z| < 1+\epsilon$ and that $\det g(z)$ is nonzero on this annulus. Then I can define a holomorphic vector bundle on $\mathbb{P}^1$ by taking trivial rank $n$ bundles on $\{ |z|<1+\epsilon \}$ and $\{ |z| > 1-\epsilon \}$ and gluing them by $g$. This bundle must be isomorphic to $\bigoplus \mathcal{O}(d_i)$ for some unique $d_1 \geq d_2 \geq \cdots \geq d_n$. - -How do I read off the $d_i$ from the $g_{ij}(t)$? - -For comparison, I know a simple criterion when the annulus is replaced with all of $\mathbb{C} \setminus \{ 0 \}$: $d_1+d_2+\cdots + d_k$ is the minimum, over all $k \times k$ minors of $g$, of the order of vanishing of that minor at $0$. - -REPLY [6 votes]: I am just posting my comment above as an answer. Let $U$ and $V$ be subsets of $\mathbb{CP}^1$, or any other projective algebraic curve $C$, whose complement sets are finite, and such that $\{U,V\}$ is an open cover. Let $g=[g_{i,j}]$ be an $n\times n$ matrix of rational functions on $C$ that are regular on $U\cap V$ and such that $\text{det}(g)$ is nonzero on $U\cap V$. There is an associated locally free $\mathcal{O}_C$-module $\mathcal{E}$ together with isomorphisms, $$\phi:\mathcal{O}_U^{\oplus n} \to \mathcal{E}|_U, \ \ \psi:\mathcal{O}_V^{\oplus n} \to \mathcal{E}|_V,$$ such that the transition function $\psi\circ \phi^{-1}$ on $U\cap V$ is given by the matrix $[g_{i,j}]$. For the reduced divisor $\Delta$ that is the complement of $U\cap V$, for all integers $M\gg 0$, the induced maps, $$ \mathcal{O}_C(-M\underline{\Delta})^{\oplus n}|_U \hookrightarrow \mathcal{O}_U^{\oplus n} \xrightarrow{\phi} \mathcal{E}|_U, \ \ \mathcal{O}_C(-M\underline{\Delta})^{\oplus n}|_V \hookrightarrow \mathcal{O}_V^{\oplus n} \xrightarrow{\psi} \mathcal{E}|_V,$$ extend to morphisms of $\mathcal{O}_C$_modules, $$\phi_M:\mathcal{O}_C(-M\underline{\Delta})^{\oplus n} \to \mathcal{E}, \ \ \psi_M:\mathcal{O}_C(-M\underline{\Delta})^{\oplus n} \to \mathcal{E}.$$ -For me, it is more direct to think about this when I form the "twisted transpose" of these maps, $$phi^\dagger_M:\mathcal{E}^\vee(-M\underline{\Delta}) \to \mathcal{O}_C^{\oplus n}, \ \ \psi^\dagger_M:\mathcal{E}^\vee(-M\underline{\Delta}) \to \mathcal{O}_C^{\oplus n}.$$ Now the problem is to simply understand, for each point $p\in \Delta$, the stalk of $\mathcal{E}(-M\underline{\Delta})$ as a $\mathcal{O}_{C,p}$-submodule of $\mathcal{O}_{C,p}^{\oplus n}$ containing the submodule $(\mathfrak{m}_{C,p}^M)^{\oplus n}$ via $\phi$, resp. via $\psi$. There is a local description of this involving only the integer $M$ and the image of $[g_{i,j}]$ in $\text{Mat}_{n\times n}(\mathcal{O}_{C,p})$. -For every finite subset $\Delta\subset C$, for every integer $M$, for the data at each point $p\in \Delta$ of a $\mathcal{O}_{C,p}$-submodule $E_p$ of $\mathcal{O}_{C,p}^{\oplus n}$ that contains $(\mathfrak{m}_{C,p}^M)^{\oplus n}$, there is a unique $\mathcal{O}_C$-submodule $\mathcal{E}^\vee(-M\underline{\Delta})$ of $\mathcal{O}_C^{\oplus n}$. The problem of understanding this is purely algebraic. As suggested by Sasha, for $C=\mathbb{CP}^1$, we can simply work out which global sections of $\mathcal{O}_C(d)^{\oplus n}$ are contained in the submodule $\mathcal{E}^\vee(-M\underline{\Delta})(d)$ for each integer $d=0,1,\dots,M\text{deg}(\Delta)$ in terms of the vanishing conditions at each point $p\in \Delta$ imposed by the local submodule constraints. This will uniquely determine the splitting type of $\mathcal{E}^\vee(-M\underline{\Delta})$. If you know more about $\Delta$ and the local submodule data, you can say more, e.g., if $\Delta=\{0,\infty\}$ and the submodules are $\mathbb{C}^\times$-invariant then the splitting type is straightforward. For fixed $M$ and $\text{deg}(\Delta)$, there is a parameter space of "local data", and there is a stratification of this parameter space into locally closed subsets on which the splitting type is constant. I doubt we can describe these strata "explicitly" once $M$ and $\text{deg}(\Delta)$ are large. -Historically, I believe that Birkhoff had some papers on this phrased just in terms of matrices whose entries are rational functions. I believe this is how Birkhoff proved what is often called "Grothendieck's splitting lemma" (only for $\textbf{GL}_n$ -- Grothendieck proved his theorem for all semisimple groups). This was also understood by Bertini and del Pezzo via their explicit classification of minimal varieties (excluding Veronese surfaces) as joins of rational curves.<|endoftext|> -TITLE: Order of unipotent matrices over $\mathbb{Z}/q\mathbb{Z}$ -QUESTION [10 upvotes]: Let $q$ be a prime power and let $n\geq2$ be an integer. -Is it known what is the largest order of a unipotent upper-triangular $n\times n$ matrix over the ring $\mathbb{Z}/q\mathbb{Z}$? -I am mostly interested in upper bounds. - -REPLY [10 votes]: This works for any unipotent matrix whose characteristic polynomial is $(x-1)^n$ (including some upper-triangular matrices without ones on the diagonal). By Cayley-Hamilton, the matrix $U$ satisfies $(U-1)^{n}=0$, so we can write $U = 1 + N$ with $N^n=0$. -Then $$ U^{p^k} = (1+N)^{p^k} = \sum_{i=0}^{n-1} {p^k \choose i} N^i$$ so it is sufficient to find $k$ such that ${p^k \choose i }=0\mod q$ for all $i$ from $0$ to $n-1$. -Consider the group action of $\mathbb Z/p^k$ by translation on subsets of $\mathbb Z/p^k$ of size $i$. The order of the stabilizer of any point divides $\gcd ( i,p^k)$, so by the orbit-stabilizer theorem, the binomial coefficient is a multiple of $p^k/\gcd(i,p^k)$. The largest possible value of $\gcd(i,p^k)$ is the largest power of $p$ less than $n$, and we want $p^k$ to be $q$ times it, so it suffices to take -$$p^k = q p^ { \lfloor \log_p (n-1) \rfloor }$$ -and the order of the matrix is at most this. -One can check this is sharp by taking a single unipotent Jordan block, and checking that the lower bound on the $p$-adic valuation of binomial coefficients is sharp. (At this point one needs to use the fact that the order is necessarily a power of $p$ to reduce to the case of a $p^k$th power.)<|endoftext|> -TITLE: Short basis in $\pi_1$ on a hyperbolic surface of bounded diameter -QUESTION [5 upvotes]: First, some terminology. Let $(S,x)$ be a compact surface of genus $g>0$. A standard collection of loops $\gamma_1,\ldots, \gamma_{2g}$ based at $x$ is a collection of loops that cuts $S$ into a topological disk. -Question. Let $(\Sigma,x)$ is a compact hyperbolic surface (without boundary) of genus $g$ and diameter $d$ with a marked point $x$. How to estimate from above in terms of $g$ and $d$ the number $A$ such that there exists a standard collection of geodesic loops on $\Sigma$ based at $x$, all shorter than $A$? -I would be especially grateful for a reference. - -REPLY [3 votes]: I think that $A=2d$ will work, basically by applying Morse theory to the distance function from $x$. Morse theory for distance functions was originally considered by Gromov (and then Cheeger). For the case of surfaces, see a paper of Gershkovich (also see Gershkovich-Rubinstein for a Morse theory of distance functions). -Imagine taking metric balls $B_r(x)$ of radius $r, 0\leq r\leq d$ about $x$, so that $B_d(x)=S$. These will be the image of a hyperbolic disk $D_r$ at $\tilde{x}$ of radius $r$ in the universal cover $\mathbb{H}^2 \overset{\rho}{\to} S$. For $r$ sufficiently close to 0 ($\leq$ the injectivity radius at $x$), $B_r(x)$ will be an embedded image of the disk $\rho(D_r)$. When $r=injrad_x(S)$, there will be some tangencies of pairs of points in $\rho(\partial D_r(x))$. Taking geodesics in $D_r$ from $\tilde{x}$ to these pairs of points in $\partial D_r(x)$, and projecting by $\rho $ to $S$, we see a collection of loops of length $2r$ based at $x$. Now, continue to let $r$ increase. Each time we see a pair of tangencies between pairs of points in $\partial B_r(x)$, create a "dual" loop. Once one reaches $r=d$, we will obtain a collection of loops cutting $S$ up into disks (one may argue this via Morse theory: self-tangencies are the only way to create 1-handles). Then taking a subset of $2g$ of these loops, one obtains a standard collection, each with length $\leq 2d$.<|endoftext|> -TITLE: Number of solutions to polynomial congruences -QUESTION [16 upvotes]: Suppose I have $R$ homogeneous polynomials $F_1, ..., F_R$ with integer coefficients. Let $V$ be the affine variety defined by these polynomials over $\mathbb{C}$. I was wondering if some bound that looks like the following was known? -$$ -\#\{ \mathbf{x} \in (\mathbb{Z}/q \mathbb{Z})^n: F_{i}(\mathbf{x}) \equiv 0 (\text{mod }q) \text{ for each } 1 \leq i \leq R \} \leq C q^{\dim V + \delta} -$$ -for some $\delta > 0$. Here $C>0$ is independent of $q$. Of course if $\delta = n - dim V$ then it's trivial so I was hoping for something smaller. I would greatly appreciate any comments on this. Thank you very much. - -REPLY [3 votes]: The case of $q$ a prime being answered by Igor, let us assume that $q=p^k$ is a power of a prime $p$. In this case there is the following result of Serre, which gives a very partial answer: -Let $d_p$ be the dimension of the $p$-adic analytic space $V$ defined in $\mathbb Z_p^R$ by the equations $F_1=\dots=F_r=0$. Then -$$ -\#\{ \mathbf{x} \in (\mathbb{Z}/q \mathbb{Z})^n: F_{i}(\mathbf{x}) \equiv 0 (\text{mod }q) \text{ for each } 1 \leq i \leq R \} \leq C_p q^{d_p} -$$ -where $C_p$ is some constant depending on $p$ but not on $k$. -The result is Serre's Théorème 8 in his IHES paper on Chebotarev. -Comparing to what you ask, there are two differences: the exponent is $d_p$ instead of $\dim_{\mathbb C} V$ (but they are the same for almost all $p$, aren't they?). And the constant $C$ depends on $p$, though it is the same for all powers $q$ of a given $p$.<|endoftext|> -TITLE: Finitely generated group splitting non-trivially over an infinite virtually cyclic subgroup -QUESTION [7 upvotes]: Let $G$ be a finitely generated group splitting non-trivially over -a infinite virtually cyclic subgroup $V$, namely an amalgamated free product -$$ -A*_V B,\; V \neq A,B. -$$ -Question: can the group $G$ be simple? - -REPLY [6 votes]: The answer is no. As Yves de Cornulier pointed out in a comment, this was proved by Jack Button when $V\cong \mathbb{Z}$. We follow the approach of his proof for the general case. -A subgroup $C$ of a group -$G$ is weakly malnormal if there exists $g \in G$ such that $|C^g ∩ C| < ∞$, where $C^g=gCg^{-1}$. -Suppose that $G = A\ast_V B$ is simple, where $V$ is virtually cyclic, and $V \neq A, B$. By Corollary 2.2 of Minasyan-Osin, if $V$ is weakly malnormal in $G$, then $G$ is either acylindrically hyperbolic or virtually cyclic. In either case, $G$ would not be simple. Acylindrically hyperbolic groups are SQ-universal, hence have many normal subgroups. -Thus, we conclude that $V$ is not weakly malnormal in $G$. -Thus, for every $g\in G$, $|V^g\cap V|=\infty$. But $V$ is virtually cyclic, so $[V:V^g\cap V] < \infty$. Hence we get a homomorphism $\varphi: G \to Comm(V) \cong Comm(\mathbb{Z}) \cong \mathbb{Q}^{\times}$, the abstract commensurator. Since $G$ is simple, $\varphi$ is the trivial homomorphism. Let $Z < V$ be a finite-index copy of $\mathbb{Z}\cong Z$. Then $|Z^g\cap Z|=\infty$, so the homomorphism $g : Z\cap Z^{g^{-1}} \to Z^g\cap Z$, $ a \mapsto gag^{-1}$ must be trivial (since it is trivial as an representative of $Comm(\mathbb{Z})$), which implies that $Z\cap Z^{g^{-1}} = Z^g \cap Z$. Since $G$ is finitely generated, say by $g_1, \ldots, g_n$, there exists a finite-index subgroup $Z^{g_1}\cap \cdots \cap Z^{g_n}\cap Z \leq Z$ which is normalized (in fact, centralized) by each $g_i$, and hence by $G$, a contradiction. - -REPLY [6 votes]: The answer is no: it's not simple. -It's an easy adaptation of the remarks done in Button's paper. -The background: (a) Minasyan-Osin: let $G$ be an amalgam $G=A\ast_C B$ of two groups over a proper subgroup $C$, such that $C$ is weakly malnormal in $B$. Then $A\ast_C B$ is acylindrically hyperbolic. [In case of index (2,2), there's a homomorphism onto the infinite dihedral group.] (b) (Osin) Every finitely generated acylindrically hyperbolic group is SQ-universal (and hence has continuum many normal subgroups). -Given this it is not hard to conclude. Consider an amalgam $A\ast_C B$ with $C$ virtually cyclic. It remains to consider the case when $C$ is weakly malnormal in neither $A$ nor $B$. Since $C$ is infinite virtually cyclic, it follows that it is a commensurated subgroup, in the sense that it is commensurate to all its conjugates, in both $A$ and $B$ (two subgroups of a group are commensurate if their intersection has finite index in both). -Given a group $H$, its abstract commensurator $\mathrm{Comm}(H)$ is the set of isomorphisms $f$ between two finite index subgroups $H_1$, $H_2$, modulo coincidence on a smaller finite index subgroup. If $L$ has finite index in $H$, then the inclusion $L\to H$ induces an isomorphism $\mathrm{Comm}(L)\simeq\mathrm{Comm}(H)$. -Hence for every amalgam $A\ast_C B$ such that $C$ is commensurated in both $A$ and $B$, we have a homomorphism $A\ast_C B\to\mathrm{Comm}(C)$. Here, $C$ being virtually (infinite cyclic), we have $\mathrm{Comm}(C)\simeq\mathbf{Q}^*$. -Hence assuming the amalgam has a finite abelianization, this homomorphism is trivial, which means that deep enough cyclic subgroup $N$ of finite index in $C$ are normal in both $A$ and $B$. Hence $G$ is not simple, and even better, admits the quotient $(A/N)\ast_{C/N}(B/N)$. In particular, unless $C$ has index 2 in both $A$ and $B$, $G$ is SQ-universal. (If the index is twice 2, we deduce that $G$ is virtually $\mathbf{Z}^2$.) -[Actually the latter conclusion (SQ-university except in index 2-2 case) holds if both homomorphisms $A\to\mathrm{Comm}(C)\leftarrow B$ have a finite image. If both have a nontrivial image and at least one has an infinite image, then since both vanish on $C$, we get a free product quotient and get SQ-universality. It remains the case where one has infinite image and the the other is trivial. In this case Button gets SQ-universality with a more complicated argument in the case $C$ is cyclic. I haven't tried to see if it adapts to $C$ virtually cyclic, but in all cases this is not asked by the OP.]<|endoftext|> -TITLE: An equality relation for complex numbers off the nonnegative real axis -QUESTION [5 upvotes]: For every complex number $z$ off the nonnegative real axis there exist positive numbers $p_0,... ,p_n$ such that $\sum_{i=0}^n p_iz^i = 0$. -Finding difficulty in proceeding with the problem. Need some hints. - -REPLY [8 votes]: If $z$ is in the left half plane, there is a quadratic polynomial with non-negative coefficients which has $z$ as a root. If $z$ is not in the left half plane, first find $n$ such that $z^n$ is in the left half plane, then choose a quadratic polynomial in $z^n$.<|endoftext|> -TITLE: Orbits of the function f(x)=2x (mod 1) -QUESTION [5 upvotes]: I am currently studying the dynamics associated with the function $f(x)=2x$ (mod 1). In particular, if we define the orbit of an element $y \in [0,1]$ -$$ orb(y)= \{ f^m(y): m \in \mathbb{Z}\}$$ -it is easy to see, for example, that $orb\big(\frac{1}{2}\big)=\mathbb{Z}\big[\frac{1}{2}\big] \cap [0,1)$. My question concerns what the orbits of elements of the form $\frac{1}{p}$, with $p$ prime would look like. It is easy to see (for example, for $p=3$) that -$$ orb\Big(\frac{1}{3}\Big) \subset \Big(\frac{1}{3} \mathbb{Z} \Big[\frac{1}{2}\Big] \cap [0,1) \Big) \backslash \mathbb{Z} \Big[\frac{1}{2}\Big]$$ -where $\frac{1}{3} \mathbb{Z} \Big[\frac{1}{2}\Big]= \Big\{\frac{m}{3\times 2^n}: m \in \mathbb{Z}, n \in \mathbb{Z}\Big\}.$ My problem is in proving (or disproving) the reverse inclusion. Is there any easy way to see that? - -REPLY [6 votes]: To be precise, $orb\big(\frac{1}{2}\big)=\mathbb{Z}\big[\frac{1}{2}\big] \cap [0,1).$ -Since you say $\mathbb{Z}$ rather than $\mathbb{N}$ you mean the orbit of $y$ to include the solutions $t$ of $f^k(t)=y$ for $k \in \mathbb{Z}.$ -You say that the orbit of $\frac13$ is contained in $$ \Big\{\frac{m}{3\times 2^n}\mid n \in \mathbb{Z},\ 0 \leq m\lt 3\times2^n\text{ and }\gcd(m,3)=1\Big\}$$ and wish to show the reverse inclusion. Induction on $n$ goes pretty easily. -Any rational $r$ can be written uniquely as $r=2^eq$ where $e\in \mathbb{Z}$ and $q=\frac{2a+1}{2b+1}$ has odd numerator and denominator. To keep $r\in [0,1)$ there is an upper bound on $e$ but of course no lower bound.<|endoftext|> -TITLE: What are advantages of chiral algebras over vertex algebras? -QUESTION [13 upvotes]: In Chiral Algebras, Beilinson and Drinfeld introduced the notion of chiral algebra, which based on the pseudo-tensor category of D-modules. On the other hand, There is already a notion of vertex algebras based on formal power series. Chiral algebras on curves are essentially conformal vertex algebras. -My question is: what are advantages of chiral algebras over vertex algebras other than that the definition is kind of shorter? - -REPLY [5 votes]: The main advantage of chiral algebras over vertex algebras is that they admit "very functorial" definitions, and this helps more general concepts and constructions appear naturally. The usual examples involve factorization spaces like the Beilinson-Drinfeld Grassmannian, and applications to the Geometric Langlands program. Another example is the concept of chiral homology, which can be viewed as derived version of a coinvariant construction, but has taken on a new life as factorization homology in the theory of extended TQFTs. -I disagree with the claim that the definition of "chiral algebra" is shorter than that of "vertex algebra". You really need to build up a substantial body of theory even to describe the chiral pseudo-tensor structure on D-modules. Vertex algebras are fundamentally vector spaces with a funny looking multiplication structure. -If you are trying to decide whether you want to invest more time in learning about chiral algebras versus vertex algebras, it may be worth your time to ask which notion has yielded substantial results that interest you. Chiral algebras have been available to the public for about 20 years (since Gaitsgory's IAS notes), and vertex algebras have been around for about 30. That is, neither notion is particularly new.<|endoftext|> -TITLE: Relating two different approaches to the Atiyah-Hirzebruch Spectral Sequence -QUESTION [12 upvotes]: Given a (for simplicity connective) spectrum $E$ and a pointed CW-space $X$ there is "the" (homological) Atiyah-Hirzebruch spectral sequence -$$E_{pq}^2 = \tilde{H}_p( X, \pi_q(E)) \Rightarrow \pi_{p+q}(X \wedge E)$$ -I know of two approaches at arriving at this spectral sequence: - -One is using a CW-structure $\emptyset \subset X_0 \subset \cdots \subset X_s \subset X_{s+1} \subset \cdots $ for $X$. Each pair $X_{s-1} \subset X_s$ gives the long exact sequence -$$ \cdots \rightarrow \tilde H_n (X_{s-1}) \rightarrow \tilde H_n (X_{s}) \rightarrow H_n (X_{s}, X_{s-1}) \rightarrow \tilde H_n (X_{s-1}) \rightarrow \cdots $$ -which we can together assemble into the exact couple $ \bigoplus H_n (X_{s}, X_{s-1}) \rightarrow \bigoplus \tilde H_n (X_{s}) \rightarrow \bigoplus \tilde H_n (X_{s}) \rightarrow \bigoplus H_n (X_{s}, X_{s-1})$. Taking the derived couple and setting $p+q = n, s = q$ gives the above second page of the spectral sequence and with a bit of work one sees that it does converge to the right answer. However, while the second page and the limit do not depend on the chosen CW-Structure for $X$, it is my understanding that the differentials and the induced filtration of $\pi_{p+q}(X \wedge E)$ do. -The second is given by taking the Postnikov tower of $E$ -$$ \cdots P_s E \rightarrow P_{s-1} E \rightarrow \cdots \rightarrow P_1 E \rightarrow P_0E \rightarrow * $$ -The maps $P_s E \rightarrow P_{s-1}E$ have as homotopy fibers shifted Eilenberg Maclane spectra $\Sigma^s H \pi_s(E)$. Smashing with $X$ gives the distinguished triangles -$$ X \wedge \Sigma^s H \pi_s(E) \rightarrow X \wedge P_s E \rightarrow X \wedge P_{s-1} E \rightarrow X \wedge \Sigma^{s+1} H \pi_s(E)$$ -Note that $\pi_n(X \wedge \Sigma^s H \pi_s(E)) = \tilde H_{n-s}(X,\pi_s E) $. Taking the induced long exact sequences in homotopy groups and assembling them together similarly to before we get the exact couple $\bigoplus \tilde H_{n-s}(X,\pi_s E) \rightarrow \bigoplus \pi_n (X \wedge P_s E) \rightarrow \bigoplus \pi_n (X \wedge P_s E) \rightarrow \bigoplus \tilde H_{n-s}(X,\pi_s E) $ -Again setting $n = p+q, s=q$, we already have the second page of the spectral sequence, and using that $E$ is the homotopy limit of the $P_n E$'s gives that the spectral sequence converges to the right terms. - -While it is clear that both spectral sequences compute the same thing, it is not clear to me in what sense they are the same spectral sequence - The Postnikov-tower is unique up to weak equivalence, so the resulting spectral sequence does not depend on this choice, however the choices made in the first case (of a CW-structure for X) seem to make a qualitative difference in how to compute the spectral sequence and I wouldn't expect the filtrations I get in each case to have anything to do with each other. Has this problem already been looked at? I'd be thankful for a guide to the right literature. - -REPLY [18 votes]: For cohomology, this is theorem 3.3 in -Maunder, C.R.F., The spectral sequence of an extraordinary cohomology theory, Proc. Camb. Philos. Soc. 59, 567-574 (1963). ZBL0116.14603. - -Theorem 3.3 If $E_r^{p,q}$ is the spectral sequence associated to a skeletal filtration and $\bar E_r^{p,q}$ is the spectral sequence associated to the Postnikov tower, there exist for all $r\ge 2$ isomorphisms $\phi_r:E_r^{p,q}\to \bar E_r^{p,q}$ such that - $$ \phi_r d_r = d_r\phi_r\,.$$ - -In fact (see ibid., lemma 4.3.1) the proof of the theorem gives an isomorphism of the corresponding exact couples (after they are both indexed to give the $E_2$-page). -I believe the same proof gives the corresponding statement for the homology AHSS.<|endoftext|> -TITLE: Can every curve be made transversal to a foliation by applying a pseudo-Anosov? -QUESTION [5 upvotes]: Let $F$ be a compact oriented surface with a foliation $\cal F$ with $k$-prong singularities only (or, if it helps, assume that $\cal F$ admits an invariant measure). Is it true then there exists a pseudo-Anosov $\phi$ such that for every loop or an arc (with endpoints in $\partial F$) $\gamma$, $\phi^n(\gamma)$ is transversal to $\cal F$ for large $n$? -Here is a rough argument why it is probably true: The $MCG(F)$ action on ${\cal PMF}(F)$ is minimal, cf. Fathi-Laudenbach-Poenaru, Thurston's Work on Surfaces, for closed $F$. (A reference for bounded $F$ would be useful although the original proof seems to work). This implies that the unstable foliations $\cal F^u_\phi$ of pseudo-Anosovs $\phi$ are dense in ${\cal PMF}(F)$. So it is reasonable to expect that there is $\phi$ with all angles between $\cal F^u_\phi$ and $\cal F$ greater than some $\alpha>0.$ (But I don't have a rigorous argument for that.) Now by the work of Thurston (cf. the book above), the angles between $\phi^n(\gamma)$ and $\cal F^u_\phi$ converge to $0$ as $n\to \infty.$ Consequently, $\phi^n(\gamma)$ is transverse to $\cal F$ for large $n$. -EDIT: To address the comment below, let us just assume that $\cal F$ admits an invariant measure of full support. - -REPLY [6 votes]: Given a full support measured foliation $\mathcal F$ and given a pseudo-Anosov $\phi$, what you want will work as long as $\mathcal F^u_\phi$ can be isotoped to be transverse to $\mathcal F$. -The trouble is, this is not always possible. For general choice of $\mathcal F$ and $\phi$, there are a few things one can say. -First, one can say that $\mathcal F$ and $\mathcal F^u_\phi$ can be altered in their Whitehead equivalence classes so that they become transverse. -Second, measured foliations themselves are annoying because of Whitehead equivalence, but you can do away with that annoyance in several ways. The most standard way is to work with measured geodesic laminations: the measured geodesic laminations associated to $\mathcal F$ and $\mathcal F^u_\phi$ will always be transverse, unless they are equal. -Another somewhat less satisfactory way is to work with the "partial measured foliations" that are used in one or two places in FLP. Every measured foliation $\mathcal F$ on $S$ is represented by a partial measured foliation $\mathcal F'$ which is canonical up to isotopy (slice apart all saddle connections, producing a partial measured foliation which has only annulus components and arational components; and then normalize the singularities on the arational components, with no interior saddle connections and with all boundary singularities being 3-pronged). Then one can say that $\mathcal F'$ and $\mathcal F^u_\phi$ will be almost transverse after a homotopy of $\mathcal F^u_\phi$; nonetheless there may still be violations of transversality at singularities, and notice that I wrote "homotopy" instead of "isotopy", because nearly parallel leaf segments of $\mathcal F^u_\phi$ may be forced to coincide in order to achieve transversality with $\mathcal F$. -But to return to your actual question, given $\mathcal F$ does there exist $\phi$ so that $\mathcal F^u_\phi$ can be isotoped to be transverse to $\mathcal F$? The answer is still no, because one can construct a measured foliation $\mathcal F$ which is not transversely recurrent: there is a simple closed curve $\sigma$ which is a concatenation of saddle connections of $\mathcal F$ such that there does not exist a simple closed curve transverse to $\mathcal F$ and having nonempty intersection with $\sigma$. (The reference for this construction, I think, is the Penner-Harer book "Combinatorics of Train Tracks"). -Here's the construction of $\mathcal F$. Choose an essential, separating simple closed curve $c \subset S$, breaking $S$ into a union of two essential subsurfaces $S=S_1 \cup S_2$ with $c = S_1 \cap S_2$. On the surface $S_i$, choose an arational measured foliation $\mathcal F_i$ which is orientable and hence transversely orientable (we are assuming $S$ is oriented). We may, at first, assume that $\mathcal F_i$ has canonical singularities: no interior saddle connections; all boundary singularities 3-pronged. But now alter $\mathcal F_i$ as follows. Pick a boundary saddle connection $\alpha_i$ for $\mathcal F_i$, and alter $\mathcal F_i$ by Whitehead equivalence, collapsing the arc $c - \alpha_i$ to a point $p_i$. Now the only boundary singularity of $\mathcal F_i$ is $p_i$. However, $p_1$ and $p_2$ might not be the same point, but if not then $c$ is a union of two saddle connections with endpoints $p_1,p_2$. Collapse one of those saddle connections to get the final example $\mathcal F$. -Now one can prove that there is no simple closed curve $\gamma$ transverse to $\mathcal F$ that has nontrivial intersection with $c$. If $\gamma$ existed, write it as a concatenation -$$\gamma = \gamma_1 * ... * \gamma_{2K} -$$ -where each $\gamma_{2k+1}$ is contained in $S_1$ transverse to $\mathcal F_1$, and each $\gamma_{2k}$ is contained in $S_2$ transverse to $\mathcal F_2$. The curve $\gamma_1$ has endpoints on $c$ transverse to $\mathcal F$ at each endpoint. At the initial point the orientation on $\gamma_1$ points inwards on $c = \partial S_1$, and at the terminal point it points outward. However, that violates the fact that $\mathcal F_1$ is transversely oriented.<|endoftext|> -TITLE: Characterizing topological spaces $X,Y$ whose function space $C_k(X,Y)$ is Baire -QUESTION [7 upvotes]: I am looking for a characterization of topological spaces $X,Y$ for which the function space $C_k(X,Y)$ is Baire. Here $C_k(X,Y)$ is the space of continuous functions from $X$ to $Y$, endowed with the compact-open topology. -It is well-known that for any complete metric space $Y$ and and compact space $X$ the space $C_k(X,Y)$ is metrizable by the complete metric $d(f,g):=\max_{x\in X}d_Y(f(x),g(x))$. -This implies that for a $k_\omega$-space $X$ and any Polish space $Y$ the function space $C_k(X,Y)$ is Cech-complete and hence Baire. -On the other hand, for any $k$-space $X$ and any Polish space $Y$ the function space $C_k(X,Y)$ is Dieudonne-complete (being homeomorphic to a closed subspace of the Tychonoff product $\prod_{K\in\mathcal K(X)}C_k(K,Y)$ of Cech-complete spaces, where $\mathcal K(X)$ is the family of all compact subsets of $X$). -If $X$ a sequential space, then $C_k(X,Y)$ is Hewitt-complete (being homeomorphic to a closed subspace of the Tychonoff product $\prod_{K\in\mathcal{MK}(X)}C_k(K,Y)$ of Polish spaces, where where $\mathcal{MK}(X)$ is the family of all compact metrizable subsets of $X$). -Nonetheless Hewitt-complete or Dieudonne-complete spaces need not be Baire. -Problem 1. Is there a $k$-space $X$ whose function space $C_k(X,\mathbb R)$ or $C_k(X,\{0,1\})$ is Baire but not Cech-complete? -Problem 2. Let $X$ be a zero-dimensional $k$-space (containing a unique non-isolated point). Are the following two statements equivalent? - -$C_k(X,\mathbb R)$ is Baire; -$C_k(X,\{0,1\})$ is Baire. - -I suspect that such questions should be considered in the literature. Could you give me any references? - -REPLY [7 votes]: The Baireness of $C_k(X)=C_k(X,\mathbb{R})$ has been characterized for some subclasses of $k$-spaces. For example, Gruenhage and Ma proved that for a locally compact or first-countable space $X$, the following are equivalent: - -$C_k(X)$ is Baire. -$X$ has the Moving Off Property. - -Moving Off Property means that if $\mathcal{K}$ is a collection of compact subsets of $X$ such that every compact subset of $X$ is disjoint from some member of $\mathcal{K}$ (such a collection is called "Moving Off"), then $\mathcal{K}$ contains an infinite subcollection with a discrete open expansion. -(1) --> (2) is easily seen to be true, even without the assumption that $X$ is locally compact or first-countable (hint: if $\mathcal{K}$ is a moving off collection then $O_n=\{f \in C_k(X): (\exists K \in \mathcal{K}) (f(K)>n)\}$ is a dense open subset of $C_k(X))$), but it is an open problem whether the above characterization holds for every (completely regular) space $X$. -In the last section of their paper, Gruenhage and Ma offer an example of a locally compact space $X$ such that $C_k(X)$ is Baire but not weakly $\alpha$-favorable. Since every Cech-complete space is weakly $\alpha$-favorable, their example solves your Problem 1.<|endoftext|> -TITLE: Examples of manifolds that do not admit scalar flat metrics -QUESTION [8 upvotes]: The Kazdan-Warner trichotomy states that for $n\ge 3$, a compact $n$-manifold falls into one of three categories: -(A) Every (smooth) function is a scalar curvature. -(B) The manifold is strongly scalar flat. -(C) The manifold only admits scalar curvatures which are negative somewhere. -Of course class (A) is nonempty in all dimensions because it contains $S^n$. Gromov and Lawson showed that (B) contains all tori $T^n$. However, it's not clear to me that (C) is nonempty in all dimensions. Kazdan and Warner (Prescribing Curvatures, Proc. Symp. Pure Math. 27) showed: -Let $M$ be a spin manifold with $\hat A(M)\ne 0$ and $b_1(M)=\dim M$. Then $M$ does not admit a metric of zero scalar curvature. -Consequently, any such manifold must be type (C). They only give the example $T^4\#K3$. Are there examples in dimensions $3$ and $\ge 5$ of type (C) manifolds? Presumably one could use the Kazdan-Warner result above and then apply some knowledge of manifolds with nonzero A-roof genus. They mention Hitchin told them one can strengthen the hypothesis to $b_1(M)\ne 0$. - -REPLY [4 votes]: Christos Mantoulidis showed me how to construct examples in (C) in all dimensions. Namely, if $\Sigma_g^2$ denotes a genus $g$ surface with $g\ge 2$, then $\Sigma_g^2\times T^{n-2}$ does is in class (C). -It does not carry a PSC metric because it is enlargeable (because it carries a metric of nonpositive sectional curvature or because it is a product of enlargeable manifolds). It is not in class (B) because it does not carry a Ricci flat metric. This follows from the general fact that $\mathrm{Ric}\ge 0\implies b_1\le n$. But by the Kuenneth formula, -$$b_1(\Sigma_g^2\times T^{n-2})=b_1(\Sigma_g^2)+b_1(T^{n-2})=2g+n-2,$$ -which is $>n$ when $g\ge 2$. Therefore these manifolds lie in class (C).<|endoftext|> -TITLE: Primes of the form $x^2 + y^2 + 1$ -QUESTION [8 upvotes]: There are infinitely many primes of the form $x^2+y^2+1$, as proved by Bredihin. Motohashi improved the result by showing that there were $\gg x/\log^2 x$ such primes up to $x$. But we expect $\Theta(x/\log^{3/2}x)$ primes up to $x$; has this result been proved? (Failing that, is anything else known about the density since the 70s? I know Sun & Pan proved the Green-Tao theorem for these primes.) - -REPLY [16 votes]: Iwaniec (Acta Arith. 1972) showed that the number of such primes has the order $x/(\log x)^{3/2}$. His result applies more generally to translates of binary quadratic forms. One can also show that short intervals contain the right density of such primes: for example, see the work of Matomaki (Acta Arith. 2007).<|endoftext|> -TITLE: Is a cubic hypersurface determined by its Fano variety of lines? -QUESTION [15 upvotes]: Consider a smooth cubic complex hypersurface $X\subset\mathbf{P}^{n+1}$ of dimension $n\geqslant 3$. The associated Fano variety of lines $F(X)$ is a smooth variety of dimension $2n-4$. Can one recover $X$ from $F(X)$? The answer is positive for $n=3$ (due to Clemens-Griffiths, Tjurin) and $n=4$ (due to Voisin, I think). Are there any conjectures for general $n$? - -REPLY [5 votes]: For what it's worth, I think I now know a complete proof. I learned of this proof from D. Huybrechts's notes in progress http://www.math.uni-bonn.de/people/huybrech/Notes.pdf, proposition 6.21. -Theorem. Let $X$ be a cubic hypersurface of dimension $d\geqslant 3$, with $d\neq 4$. Then $X$ can be recovered from its Fano variety of lines $F(X)$. -Notations: $F=F(X)$, $G=\mathrm{Gr}(2,d+2)$, $\mathscr{O}_F(1)$ and $\mathscr{O}_G(1)$ the Plücker polarisations of $F$ and $G$, respectively. -For future reference, let me state the following results. -Lemma 1. The canonical map $\mathrm{H}^0(G,\mathscr{O}_G(2))\rightarrow\mathrm{H}^0(F,\mathscr{O}_F(2))$ is injective. -Proof. See J. Starr's answer or Altman-Kleiman, Foundations of the theory of Fano schemes, proposition 1.15 (ii). -Lemma 2. $\omega_{F}=\mathscr{O}_{F}(4-d)$. -Proof. See Altman-Kleiman, proposition 1.8. -Lemma 3. If $V$ is a variety such that $\omega_{V}^{\vee}$ is ample, then $\mathrm{Pic}(V)$ is torsion-free. -Proof. Let $L$ be a torsion line bundle on $V$. Then $\mathrm{H}^i(V,L)=0=\mathrm{H}^i(V,\mathscr{O}_V)$ for $i\geqslant 1$ by Kodaira vanishing. As $\chi(L)=\chi(\mathscr{O}_V)$ by Riemann-Roch, we get $\mathrm{H}^0(V,L)=\mathrm{H}^0(V,\mathscr{O}_V)$, -which forces $L$ to be trivial. -Proof of the Theorem. Let $X$ and $X'$ be cubic hypersurfaces and $f:F\xrightarrow{\sim} F'$ an isomorphism between their Fano varieties of lines. By lemma 2 we have the equality -$$(4-d)f^\ast(\mathscr{O}_{F'}(1))=f^\ast(\omega_{F'})=\omega_F=(4-d)\mathscr{O}_{F}(1)$$ -in $\mathrm{Pic}(F)$. For $d\geqslant 5$ lemma 2 shows that we can use lemma 3 to deduce -$$f^\ast(\mathscr{O}_{F'}(1))=\mathscr{O}_{F}(1).$$ -For $d=3$ this is clear. The theorem then follows from proposition 4 of https://arxiv.org/abs/1209.4509. For completeness, I repeat the proof. Let $\mathbf{P}^{d+1}=\mathbf{P}(V)$ be the ambient projective space containing $X$ and $X'$. By lemma 1, if $Q$ is a quadric in $\mathbf{P}(\wedge^2 V)$ containing $F$, then $Q$ also contains $G$. In $\mathbf{P}(\wedge^2 V)$ the Grassmannian $G$ is cut out by quadrics, so in particular $G$ is the intersection of quadrics containing $F$. Hence there is an automorphism $g$ of $G$ taking $F$ to $F'$ via $f$, and $g$ also preserves the Plücker polarisation. By a theorem of Chow on automophisms of Grassmannians (Ann. Math. 2), $g$ is induced by an automorphism $h$ of $\mathbf{P}(V)$. The latter automorphism then induces a (polarisation-preserving) isomorphism $X\xrightarrow{\sim} X'$. (If $L$ is a line on $X$ corresponding to $[L]\in F$, then $h(L)$ is a line on $X'$, and $[h(L)]=f([L])$. Of course $X$ is covered by lines.)<|endoftext|> -TITLE: How to kill a $\Sigma_{n+1}$-correct cardinal softly ($n>1$)? -QUESTION [5 upvotes]: A cardinal $\kappa$ is $\Sigma_n$-correct iff $V_\kappa \prec_n V$. For n>1, how to force a $\Sigma_{n+1}$-correct cardinal to be $\Sigma_{n}$-correct but not $\Sigma_{n+1}$-correct? -For $n=1$, we can force GCH below $\kappa$ and then violate GCH at $\kappa$. -If we assume some large cardinals, there are more partial answers, but the general situation is not clear to me. - -REPLY [8 votes]: This is a very interesting question! -My student Erin Carmody (PhD 2015) had asked this question in -connection with her dissertation, Force to change large cardinal -strength, which contains many -such killing-them-softly results. Her theorem 20 is the -$\Sigma_2$-reflecting cardinal case that you mention. I recall that -at the time we looked into the general $\Sigma_n$-reflecting case, -but it was left open in her dissertation. -How nice finally to have an answer to this. -Theorem. If $\kappa$ is $\Sigma_n$-correct, then there is a -class forcing extension preserving this in which $\kappa$ is not -$\Sigma_{n+1}$-correct. -Proof. Suppose that $\kappa$ is $\Sigma_n$ correct, where -$n>1$. By forcing, if necessary, let me assume that V=HOD holds and -more specifically that there is a $\Delta_2$-definable -well-ordering of the universe. Indeed, let us arrange this -specifically by forcing to code every set into the GCH pattern on a -certain definable sequence of coding points. This forcing preserves -correctness. -Consider the forcing notion $\text{Add}(\kappa,1)$, which is -definable in $V_\kappa$, and consider the dense sets for this -forcing that are $\Sigma_n$-definable in $V_\kappa$. I claim that -there is a subset $s\subset\kappa$ that is $\Sigma_n$ definable in -$V_\kappa$ and generic over $V_\kappa$ with respect to -$\Sigma_n$-definable dense classes for this forcing, meaning that -the initial segments of $s$ meet every $\Sigma_n$-definable dense -subset of this forcing. -If $\kappa$ is inaccessible, then this is easier to see, since in -this case the forcing is $<\kappa$-closed and one can simply meet -the dense sets one by one, using a definable enumeration of them. A -careful version of this argument works even when $\kappa$ is -singular, using the fact that $\kappa$ is $\Sigma_n$-correct. One -simply meets the dense sets one-by-one, and at limits, the -condition produced must be bounded below $\kappa$, since otherwise -one would have a $\Sigma_n$-definable singularization of $\kappa$, -which would violate correctness. -Now let us force with $\text{Add}(\newcommand\Ord{\text{Ord}}\Ord,1)$ to add a generic class -$S\subset\Ord$ extending $s$. (We could also have used a definable -$\Sigma_{n+1}$-generic class, without forcing.) In $V[S]$, let -$\mathbb{P}$ be the class forcing Easton support $\Ord$-iteration -that forces to code $S$ into the GCH pattern at another definable -sequence of ordinals that does not interfere with coding used -above. -Let $V[S][G]$ be the final extension. I claim that $\kappa$ remains -$\Sigma_n$-correct, since if a $\Sigma_n$-statement $\varphi(a)$ is -true in $V[S][G]$ about some parameter $a\in -V_\kappa[s][G_\kappa]$, then this is forced over $V[S][G]$ by some -condition, and by the correctness of $\kappa$ and the -$\Sigma_n$-genericity of $s$, it follows that it holds in -$V_\kappa[s][G_\kappa]$ as desired. Basically, we are lifting the -relation $V_\kappa\prec_{\Sigma_n} V$ to -$V[s][G_\kappa]\prec_{\Sigma_n}V[S][G]$. -But I claim that $\kappa$ is not $\Sigma_{n+1}$-correct in -$V[S][G]$. Note that $V$ is $\Sigma_2$-definable in $V[S][G]$ using -the first sequence of coding points, and $V[S][G]$ thinks that $S$, -which is $\Sigma_2$-definable, is not $\Sigma_n$-definable in $V$, -since in fact it was generic over $V$. But $V_\kappa[s][G_\kappa]$ -thinks that $s$ is $\Sigma_n$-definable in $V_\kappa$. This -violates $\Sigma_{n+1}$-correctness. $\Box$<|endoftext|> -TITLE: Primitivity of subgroups in the Picard groups of anticanonical $K3$ surfaces -QUESTION [9 upvotes]: Let $X$ be a smooth projective threefold with $h^{0,1}(X) = h^{0,2}(X)=0$ that has a smooth anticanonical section $D$. -Then $D$ is necessarily a $K3$ surface. -Consider a subgroup -$$Pic_X(D) = i^*(Pic(X))$$ -of $Pic(D)$, where $i:D \hookrightarrow X$ is the inclusion map. -I tried several examples of $X$, including some blow-ups of toric Fano threefolds and found that $Pic_X(D)$ is a primitive subgroup of $Pic(D)$(i.e. the quotient $Pic(D) / Pic_X(D)$ has no torsion) -Is it generally true or are there any counterexamples? -If $X$ is a smooth Fano threefold, then it is obvious but I am not sure about the general cases. - -REPLY [5 votes]: I think the following may give an example $X$ where the image of the map in question is not primitive. -Let $S$ be the blow-up of $\mathbb{P}^2$ in 9 points that are the intersection of two cubics (so $|{-}K_S|$ is an elliptic pencil). Let $Y \subset \mathbb{P}^1 \times S$ be a general element of $|2H -2K_S|$, for $H$ the hyperplane class of $\mathbb{P}^1$. The linear system is base-point-free, so $Y$ can be taken to be smooth. Let $X$ be the double cover of $\mathbb{P}^1 \times S$ branched over $Y$. -Then anticanonical divisors $D$ of $X$ are double covers of $S$, branched over an element of $|{-}2K_S|$, so $D$ is a K3 with non-symplectic involution whose fixed set is a disjoint union of two elliptic curves. According to Nikulin's classification, the invariant part $N \subset H^2(D)$ (which is generically equal to $\mathrm{Pic}(D)$) is a rank 10 even lattice with signature (1,9), discriminant group $N^*/N \cong (\mathbb{Z}/2)^8$ and alternating discriminant form (in the sense that $t^2 \in \mathbb{Z}$ for every $t \in N^*$). (In other words $N$ is isometric to $2E_8 \oplus U$.) The image $N'$ of the pull-back of the projection $D \to S$ is a sublattice of $N$, with $N' \cong 2H^2(S)$. So $(N')^*/N' \cong (\mathbb{Z}/2)^{10}$, and $N'$ must be an index 2 sublattice of $N$. Concretely, $-K_S$ is a primitive element of $H^2(S)$ whose image in $H^2(D)$ is even. -Meanwhile, I think that one can prove that the pull-back map $H^2(\mathbb{P}^1 \times S) \to H^2(X)$ is an isomorphism. Consider the projectivisation $Z$ of the rank 2 bundle $-K_S \oplus \underline{\mathbb{C}}^2$ over $S$ (so $\pi : Z \to S$ is a $\mathbb{P}^2$-bundle), and let $T$ be the tautological line bundle over $Z$. Then $X$ can be viewed as an element of the linear system $|{-}2T -2 \pi^*(K_S)|$, and I think one can apply a version of the Lefschetz hyperplane theorem (Goretsky-MacPherson or de Cataldo-Migliorini) to show that $H^2(Z) \to H^2(X)$ is an isomorphism. -If correct, this means that the image of $H^2(X) \to H^2(D)$ equals $N'$, so isn't primitive in $H^2(D)$. -Edit: Here is a rather easier example. Fix a line in $\mathbb{P}^3$, and consider a generic quartic K3 $D$ containing that line. Its Picard lattice is generated by the hyperplance class $H$ and by the Poincare dual $L$ to the line, and is isometric to -$$ \begin{pmatrix} 4 & 1 \\ 1 & -2 \end{pmatrix} $$ -Now take a class like $5H+2L$, which is ample. Thus $|5H+2L|$ contains a smooth curve $C$. Let $X$ be the blow-up of $\mathbb{P}^3$ in $C$. The proper transform of $D$ in $X$ is isomorphic to $D$, and is an anticanonical divisor of $X$. The image of $H^2(X) \to H^2(D)$ is generated by $H$ and the Poincare dual to $C$ (i.e. $5H+2L$), so is not primitive.<|endoftext|> -TITLE: Ring structure on K-theory of a quotient of the Fermat quintic -QUESTION [15 upvotes]: Let $Y$ be the Fermat quintic, i.e. $Y \subset \mathbb{C}P^4$ is defined by -$$ -\sum_{i=1}^5 z_i^5 = 0 -$$ -In Section 5.3 of this paper by Volker Braun the author computes the K-groups of a quotient $X = Y/(\mathbb{Z}/5\mathbb{Z})$, where the generator acts by -$$ -[z_1: \dots : z_5] \mapsto [z_1: \alpha z_2 :\alpha^2 z_3 : \alpha^3 z_4 : \alpha^4 z_5] -$$ -with $\alpha = e^{\frac{2 \pi i}{5}}$. This quotient is a non-singular Calabi-Yau manifold and its $K$-groups turn out to be -$$ -K^0(X) \cong \mathbb{Z}^4 \oplus \mathbb{Z}/5\mathbb{Z}\\ -K^1(X) \cong \mathbb{Z}^{44} \oplus \mathbb{Z}/5\mathbb{Z} -$$ -The computation uses the Atiyah-Hirzebruch spectral sequence, which unfortunately does not tell us much about the multiplication on $K^*(X)$. - - -Question: What is the ring structure on $K^*(X)$? - -REPLY [6 votes]: The Atiyah-Hirzebruch spectral sequence does have a multiplicative structure, and I think this can be used to determine the multiplication on K-theory. From the paper of Braun, it follows that the spectral sequence actually degenerates integrally: the only potentially nontrivial differentials are the ones with target $\mathbb{Z}/5\mathbb{Z}\in H^5(X,K^{2q})$, i.e. the differential $d_3^{2,2q}:H^2(X,K^{2q})\to H^5(X,K^{2q-2})$ on the $E_3$ page and the differential $d_5^{0,2q}:H^0(X,K^{2q})\to H^5(X,K^{2q-4})$. By the computations in Braun's paper, these differentials must be trivial because the $\mathbb{Z}/5\mathbb{Z}$ actually appears in the K-theory of $X$. So the spectral sequence must degenerate integrally at the $E_2$-page. Furthermore the extension problems all split, since otherwise there would be no torsion in K-theory. -Now the multiplicative structure of the Atiyah-Hirzebruch spectral sequence states that - -the multiplication on the $E_2$-page is induced from the cup-product on cohomology of $X$. -the multiplication on K-theory is compatible with the relevant filtration for the spectral sequence, and the induced multiplication on the $E_\infty$-page coincides with the multiplicative structure of the spectral sequence. - -Since the filtration on K-theory is split, the multiplication is given by the multiplication on the $E_\infty$-page, which by the degeneration coincides with the multiplication on the $E_2$-page induced from the cup product. All in all, I think the multiplicative structure of the Atiyah-Hirzebruch spectral sequence implies that the multiplication on $K^\ast(X)$ (which actually is just $K^0=H^{ev}$ and $K^1=H^{odd}$) coincides with the cup product on the cohomology of $X$. - -Some more explanations on the relation between multiplication on $K^\ast(X)$ and the $E_\infty$-page are in order. It is true that the relation between multiplication on $K^\ast$ and on the $E_\infty$-page is rather weak: there is a filtration $F^\ast$ on $K^\ast(X)$ which satisfies $F^p\times F^q\subset F^{p+q}$. In particular, even if the filtration splits, the product of elements of $H^p$ and $H^q$ will land in $H^{p+q}\oplus H^{p+q+2}\oplus\cdots \oplus H^{max}$. -I claim that this doesn't happen in the specific case at hand. First of all, we can split off $H^0$ from $K^0$, viewed as K-theory of the point, generated by the trivial line bundle. Then $H^0\times K^\ast(X)\to K^\ast(X)$ will always just be the multiplication coming from the $\mathbb{Z}$-algebra structure. Now for degree reasons, the only product where something strange can happen is the one for $H^2\times H^2\to F^4K^0(X)$. (For instance, multiplication of $H^2$ and $H^3$ lands in the filtration step whose only component is $H^5$ for dimension reasons and therefore this is determined by the cup product on the $E_\infty$-page. Similar arguments for all the other cases.) So the product of two classes from $H^2$ lands a priori in $H^4\oplus H^6$, but then we can check using the Chern character that there is no error term in $H^6$ and the product is really the cup product.<|endoftext|> -TITLE: Correlation of the Möbius function -QUESTION [11 upvotes]: Assume that $$\sum_{n \leq x}\mu(n) \mu(n+1) \gg x.$$ -Question. Does this contradict any well known results, say the prime number theorem or something like that? - -REPLY [27 votes]: The relation you state contradicts the recent result of Terence Tao that -$$ \sum_{n\leq x}\frac{\mu(n)\mu(n+1)}{n}=o(\log x).$$ -Indeed, if $S(x)$ is the sum on your left hand side, we have -$$ \sum_{n\leq x}\frac{\mu(n)\mu(n+1)}{n}= \frac{S(x)}{x}+\int_1^x\frac{S(t)}{t^2}\,dt.$$<|endoftext|> -TITLE: What is this property exhibited by some logical systems? -QUESTION [6 upvotes]: I'm migrating this question from MSE to MO, as in the span of five months, it received 6 upvotes but no answers. If my language needs to be fine-tuned in any way, constructive suggestions and guidance would be greatly appreciated. - -The following property exhibited by some logical systems has captured my attention: -$$\forall X\; ( {\vdash x_1[X]} \implies {\vdash x_2[X]} ) \implies \forall X\; {\vdash (x_1[X]\to x_2[X])},$$ -where $X$ ranges over "ways to fill in the holes in $x_1$ and $x_2$", for any syntactically correct schemas $x_1$ and $x_2$. -In other words, the property states that if $x_2[X]$ is provable whenever $x_1[X]$ is provable, then all instances of the schema ($x_1 \rightarrow x_2$) are also provable. -Some examples off the top of my head of where this property does not hold: - -Classical predicate logic does not have this property because (letting $x_1[P] = P$ and letting $x_2[P] = \forall x.P$), it is necessarily true that $(\vdash P) \implies (\vdash \forall x.P)$, but it is not necessarily true that $\vdash (P \rightarrow \forall x.P)$ -Intuitionistic logic does not have this property because (letting $x_1[A, B, C] = \neg A\to B\lor C$ and letting $x_2[A, B, C] = (\neg A\to B)\lor(\neg A\to C))$, it is necessarily true that $(\vdash\neg A\to B\lor C)\implies(\vdash(\neg A\to B)\lor(\neg A\to C))$, but it is not necessarily true that $\vdash(\neg A\to B\lor C)\rightarrow((\neg A\to B)\lor(\neg A\to C))$. - -My question is 3-fold: - -Does this property have a name? If so, what is it called? -Does classical propositional logic have this property? (I'm assuming it does, but I want to be sure.) What other systems display this property? -Does the presence of this property (or lack thereof) imply any other important properties about the system in question? (I realize that this third part of the question might seem overly broad, but what I really want to know is: is this property important and if so, why? Deep insights appreciated.) - -Partial answers welcomed as well. - -REPLY [10 votes]: This is equivalent to a combination of structual completeness with the deduction theorem. For a start, see Wikipedia.<|endoftext|> -TITLE: MacLane coherence theorem for "monoidal" category without 1 -QUESTION [9 upvotes]: MacLane's coherence theorem for a monoidal category states that once the associators for 4-fold products are compatible (i.e., the pentagon axiom holds), it holds for n-fold products, so I can bracket n-fold products in any way I like. -What happens if I forget the unit, i.e. I consider "semigroup categories" as they are called in the book "Tensor Categories" by Etingof, Gelaki, Nikshych and Ostrik. Is it still true that the pentagon axiom implies that the bracketing of n-fold products does not matter? -The proof of the coherence theorem given in the book relies on the unit. Still, they say that "semigroup categories" categorify semigroups. Then they should better satisfy coherence in my opinion. But I don't know. - -REPLY [4 votes]: The proof of Mac Lane's coherence theorem is fairly ad hoc. However, higher dimensional rewriting provides some general methods to prove coherence-like theorems. Using this, it is very easy to prove the coherence of "semigroup categories". -See for example "Coherence in monoidal track categories", by Y. Guiraud and P. Malbos.<|endoftext|> -TITLE: Forms of automorphism groups of algebraic varieties -QUESTION [5 upvotes]: Let $k$ be an arbitrary field and $X$ be a proper scheme over $k$. Mastumura and Oort proved that the functor $S \longmapsto Aut_{S-sch}(X \times S)$ is representable by a group scheme, locally of finite type, which I denote by $\underline{Aut}(X)$. I am interested in the forms of such a group scheme. More precisely, if $X'$ is a form of $X$ (that is, a (proper) scheme such that $X'_{\overline{k}} \simeq X'_{\overline{k}}$, where $\overline{k}$ is an algebraic closure of $k$ and $X_{\overline{k}}$ stands for the base change $X \times \mathrm{Spec}(\overline{k})$) then by functoriality $\underline{Aut}(X')_{\overline{k}} \simeq \underline{Aut}(X'_{\overline{k}}) \simeq \underline{Aut}(X_{\overline{k}}) \simeq \underline{Aut}(X)_{\overline{k}}$, so that $\underline{Aut}(X')$ is a form of $\underline{Aut}(X)$. - -Question 1: Let $G$ be a form of $\underline{Aut}(X)$. Does there - exist a form $X'$ of $X$ such that $G \simeq \underline{Aut}(X')$? - -Moreover, if $Z$ is a closed subscheme of $X$ then the functor $$S \longmapsto \{g \in \underline{Aut}(X)(S) \mid \text{the automorphism } g \text{ of } X \times S \text{ induces an automorphism of } Z \times S \}$$ is representable by a subgroup scheme of $\underline{Aut}(X)$, which I denote by $\underline{Aut}(X,Z)$ (see Demazure-Gabriel, Théorème II 1.3.6 p165). If $(X',Z')$ is a form of $(X,Z)$ (that is, $Z'$ is a closed subscheme of $X'$ and the isomorphism $X'_{\overline{k}} \simeq X_{\overline{k}}$ induces an isomorphism $Z'_{\overline{k}} \simeq Z_{\overline{k}}$) then as above $\underline{Aut}(X',Z')$ is a form of $\underline{Aut}(X,Z)$. - -Question 2: Let $G$ be a form of $\underline{Aut}(X,Z)$. Does there exist a form $(X',Z')$ of $(X,Z)$ such that $G \simeq -\underline{Aut}(X',Z')$? - -I ask these questions in a quite broad generality. However, in the example I have in mind, $X$ and $Z$ are smooth, projective and geometrically integral, and $\underline{Aut}(X,Z)$ is a smooth connected affine group scheme. I would be happy if the answers were positive with these restrictions. -I suspect that this could be related to cohomology, but I do not want to assume the field $k$ to be perfect so the forms could be trivialised after an inseparable extension. - -REPLY [5 votes]: That is not true. The basic issue is that for every coherent sheaf $\mathcal{F}$ on $X$ that is "intrinsic", and thus admits a linearization of the automorphism group, every cohomology group of $\mathcal{F}$ gives a linear representation of the automorphism group. So for every form of $X$, not only do you get a form of the automorphism group, you get a collection of linear representations for every intrinstic coherent sheaf $\mathcal{F}$. It can easily happen that some form of the automorphism group does not have "enough" linear representations over $k$ to arise from a form of $X$. -Here is a concrete counterexample. Let $(a,b,c,d)$ be positive integers with $a\neq b$, with $c\neq d$, and with the sums $m=a+b$ and $n=c+d$ distinct. Let $A$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[s,t]$. Let $C$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[u,v]$. Let $P$ be a copy of $\mathbb{P}^4$ with homogeneous coordinates $[x_0,x_1,x_2,x_3,x_4]$. Let $Q$ be a copy of $\mathbb{P}^4$ with homogeneous coordinates $[y_0,y_1,y_2,y_3,y_4]$. Denote by $f_{a,b}$, resp. $g_{c,d}$, the closed immersions, $$f_{a,b}:A \to P, \ \ [s,t] \mapsto [s^m,s^{m-1}t,s^at^b,st^{m-1},t^m],$$ $$g_{c,d}:C \to Q, \ \ [u,v] \mapsto [u^n,u^{n-1}v,u^cv^d,uv^{n-1},v^n].$$ Let $R$ be a copy of $\mathbb{P}^{24}$, and denote by $h$ the Segre embedding, $$h:P\times Q \to R,\ \ ([x_i],[y_j]) \mapsto [x_iy_j]_{0\leq i,j\leq 4}.$$ Define $Y$ to be the blowing up of $R$ along the image of $P\times Q$, and define $X$ to be the blowing up of $Y$ along the inverse image of $A\times C$. -The automorphism group of $X$ equals the subgroup of the automorphism group of $Y$ that preserves the inverse image of $A\times C$ as a closed subscheme (rather than "pointwise"). The automorphism group of $Y$ is a semidirect product of $\text{Aut}(P)\times \text{Aut}(Q)$ by a cyclic group $\mathfrak{S}_2$ permuting the factors. The automorphism group of $X$ is the subgroup of this group that preserves $A\times C$ as a closed subscheme of $P\times Q$. Because $m$ does not equal $n$, the cyclic group $\mathfrak{S}_2$ does not preserve $A\times C$. Because $a\neq b$, resp. $c\neq d$, the automorphism group of $A$ in $P$, resp. the automorphism group of $C$ in $Q$, is a copy of the multiplicative group $\mathbb{G}_m$. Thus, the automorphism group of $X$ is isomorphic to $\mathbb{G}_m\times \mathbb{G}_m$. There is a linear representation of this automorphism group on the $k$-vector space of global sections on $X$ of the pullback of $\mathcal{O}(1)$ from $R$. This representation is a direct sum of copies of the characters with weights $(i,j)$ for $$i\in\{ 0,1,a,m-1,m\},\ \ j\in \{ 0,1,c,n-1,n\}.$$ Similarly, the representation on global sections of the pullback of $\mathcal{O}(16) \cong \omega_{R/k}^\vee$ has weights $(i,j)$ ranging from $(0,0)$ to $(16m,16n)$. -The automorphism group of the group scheme $\mathbb{G}_m\times \mathbb{G}_m$ is quite large. For instance, for every separable, degree $2$-field extension $L/k$, the restriction $\text{Res}_{L/k}(\mathbb{G}_{m,L})$ is a group $k$-scheme that is a form of $\mathbb{G}_m\times \mathbb{G}_m$. Since $m$ does not equal $n$, there is no realization of this form arising from a form of $X$: if there were, then the linear representation on the global sections of the twist of the pullback of $\omega_{X/k}^\vee$ would have weights $(i,j)$ that were symmetric under permutation of the two factors of $\mathbb{G}_m\times \mathbb{G}_m$. -There should be similar examples obtained from blowing up finite sets in projective space, but this requires explicit computation of the stabilizer group of the finite point set.<|endoftext|> -TITLE: Symplectic reduction of 4-manifolds with circle actions -QUESTION [8 upvotes]: Let $(M,\omega)$ be a $4$-dimensional closed symplectic manifold. Assume there exists a Hamiltonian $S^1$-action on $M$, let $\mu:M \to \mathbb{R}^*$ be its moment map and let $M_{\text{red}}=\mu^{-1}(r)/S^1$ denote the symplectic reduction/symplectic quotient, where $r$ is a regular value of $\mu$ and where we assume that $S^1$ acts freely on $\mu^{-1}(r)$. For dimensional reasons, $M_{\text{red}}$ is a closed manifold with $\dim M_{\text{red}}=2$, i.e. a closed surface. -Q: (How) can one determine the genus of $M_{\text{red}}$ from the topology of $M$ and the $S^1$-action? - -REPLY [7 votes]: By a result of Hui Li https://arxiv.org/abs/math/0605133, the fundamental groups of $M_{red}$ and $M$ are isomorphic, so one can deduce the genus of $M_{red}$ from the fundamental group of $M$.<|endoftext|> -TITLE: Resource request: Function spaces properties -QUESTION [5 upvotes]: Is there any resource (book, article,...) that presents all the basic function spaces (Metrizable, Normed, Banach, Sobolev, Hilbert, $L^p$, C[a,b], etc.) together with their properties (completeness, separability, reflexivity, compactness, and other special properties maybe...,etc) and diagrams with arrows (which space is also other space> example Normed Space is also a Metrizable space)? -I find it difficult to search anytime properties of these spaces... -Thanks! - -REPLY [7 votes]: Some good references are - -Dunford, Nelson, and Jacob T. Schwartz. Linear Operators Part I: General Theory. Vol. 7. New York: Interscience publishers, 1958. - -(Chapter IV. Special Spaces) - -Johnson, William B., and Joram Lindenstrauss. Handbook of the geometry of Banach spaces. Vol. 1. Elsevier, 2001. - -(The first contribution "Basic concepts in the geometry of Banach spaces" is quite instructive.) - -Triebel, Hans. Theory of function spaces (I-III), Springer - -(As the title says, its more focused on function spaces…) - -REPLY [6 votes]: Terry Tao wrote a blog post describing the following "type diagram of function spaces" a few years ago: - -The axes are labeled by exponents $s$, corresponding to the regularity and $p$ corresponding to the integrability. See the blog post for more details.<|endoftext|> -TITLE: Interesting examples of pro-algebraic completions of groups -QUESTION [9 upvotes]: Fix a field $k$. Given a discrete group $G$, the pro-algebraic completion (or Hochschild-Mostow completion) is the pro-algebraic group $A_{k}(G)$ with is universal with respect to finite dimensional representations of $G$. We usually omit the field $k$ from the notation when it is clear. -I have encountered pro-algebraic completions in a number of places, especially in relation to the schemetization problem, however there are next to no examples of pro-algebraic completions of interesting groups anywhere in the literature. -The only example that I have found is of the pro-algebraic completion of $\mathbb{Z}$, which is -$$A(\mathbb{Z})=\mathbb{G}_{a}(k)\times T\times \widehat{\mathbb{Z}},$$ -where $T$ denotes denotes the pro-torus whose character group is $\mathrm{Hom}(\mathbb{Z},k^{\times})$, and $\widehat{\mathbb{Z}}$ denotes the pro-finite completion of the integers. This example, and its direct generalisation to abelian groups, was found in the following paper of Bass et.al. -https://deepblue.lib.umich.edu/bitstream/handle/2027.42/42819/10711_2004_Article_381723.pdf?sequence=1&isAllowed=y -Can anyone provide any other examples, or references containing examples, of pro-algebraic completions? -Tangentially, does anyone know of any methods to compute pro-algebraic completions? -I am particularly interested in understanding the pro-algebraic completions of the symmetric groups and the free groups. - -REPLY [6 votes]: You ask specifically about the symmetric groups and the free groups. -This is a bit funny, as a finite group forms its own proalgebraic completion (a slight generalization of this fact is Proposition 1 in Bass-Lubotzky-Magid-Mozes), -while the proalgebraic completion of a non-abelian free group is clearly not a manageable object. -Note that it follows that every group which virtually surjects onto a non-abelian free group, e.g a surface group or $\text{SL}_2(\mathbb{Z})$, has a huge completion. -As you mentioned, for f.g abelian groups it should be rather easy to work out the proalgebraic completion. I suppose this is also the case for f.g nilpotent groups: the completion should be the Malcev completion times a protorus times the profinite completion. I would start by thinking explicitly of the Heisenberg group to verify this guess. -As mentioned in Ian Agol's comment, Margulis Super-Rigidity provides a powerful tool for computing proalgebraic completions for higher rank arithmetic groups. -This is also mentioned in the introduction of BLMM. -In particular, when you also have the Congruence Subgroup Property you get very neat results, e.g the -proalgebraic completion of $\text{SL}_n(\mathbb{Z})$ is $\text{SL}_n(\mathbb{C})\times \text{SL}_n(\hat{\mathbb{Z}})$ for $n\geq 3$.<|endoftext|> -TITLE: What are the possible eigenvalues of these matrices? -QUESTION [21 upvotes]: Edit: since we seem a bit deadlocked at this point, let me weaken the question. It's fairly easy to see that the set of 8-tuples of reals which can be the eigenvalues of a matrix of the desired form is closed. We know from jjcale and Caleb Eckhardt that its complement is nonempty. Is its complement dense? That is, would a generic 8-tuple not be the eigenvalues of such a matrix? - -First, here is a baby version of the question, that I already know the answer to. Consider complex Hermitian $4\times 4$ matrices of the form $$\left[\begin{matrix}a I_2&A\cr A^*&b I_2\end{matrix}\right]$$ where $A \in M_2(\mathbb{C})$ and $a,b \in \mathbb{R}$ are arbitrary. Can any four real numbers $\lambda_1 \leq \lambda_2 \leq \lambda_3\leq \lambda_4$ be the eigenvalues of such a matrix, or is there some restriction? Answer: there is a restriction, we must have $\lambda_1 + \lambda_4 = \lambda_2 + \lambda_3$. -The real question is: what are the possible eigenvalues of Hermitian $8\times 8$ matrices of the form $$\left[\begin{array}{c|c}aI_4&A\cr \hline A^*&\begin{matrix}bI_2& B\cr B^*&cI_2\end{matrix}\end{array}\right]$$ with $a,b,c\in\mathbb{R}$, $A \in M_4(\mathbb{C})$, and $B \in M_2(\mathbb{C})$? Can any eight real numbers be the eigenvalues of such a matrix? (I suspect not. If they could, that would tell you that any Hermitian $8\times 8$ matrix is unitarily equivalent to one of this form.) - -REPLY [4 votes]: Let $V$ be the real vector space of the $8\times 8$ matrices of the form given in the question. -Where is an open set in $\mathbb{R}^8$ of possible $8$-tuples of eigenvalues of matrices in $V$. -Proof : -Choose $M_1 \in V$ such that $M_1$ has no degenerated eigenvalues. -Let $v_1,...,v_8$ be an orthonormal base of eigenvectors of $M_1$ . -Then first order perturbation theory tells us that it suffices to show that the map -$V \rightarrow \mathbb{R}^8$, $M \mapsto (v_1^* M v_1,...,v_8^* M v_8)$ has rank $8$. -So I need $7$ additional matrices $M_2,...,M_8$ in $V$ such that the matrix -$X = (v_i^* M_j v_i)_{ij}$ is nonsingular . -Let $f(a,b,c,A,B)$ be the corresponding matrix. -Choose $$ A_1 = \left[\begin{matrix}4&2&3&4\cr5&6&7&8\cr9&10&11&12\cr13&14&15&16\end{matrix}\right] $$ . -Then choose -$M_1 = f(2.7,1,-1,A_1,diag(2,1))$, -$M_2 = f(1,0,0,0,0)$, -$M_3 = f(0,1,0,0,0)$, -$M_4 = f(0,0,1,0,0)$, -$M_5 = f(0,0,0,diag(1,1,1,2),0)$, -$M_6 = f(0,0,0,diag(1,0,0,0),diag(0,1))$, -$M_7 = f(0,0,0,diag(0,1,0,0),diag(1,2))$, -$M_8 = f(0,0,0,diag(0,0,1,0),diag(3,2))$. -This gives $det X = 21.661...$ .<|endoftext|> -TITLE: Does there exist a nonsingular graph for which the determinant of its adjacency matrix remains the same upon deleting a vertex? -QUESTION [8 upvotes]: Consider simple graphs. Any simple graph $G$ is called nonsingular if its $(0,1)$-adjacency matrix $A(G)$ has nonzero determinant. Does there exist any nonsingular simple graph whose determinant value remains the same upon deleting a vertex? - -REPLY [7 votes]: Proposition. The unique smallest graph w.r.t. number of edges of the requested kind is - -. (This graph is isomorphic to graph no. 126 below). - -It has $6$ vertices and $8$ edges. - The determinant of its adjacency matrix is equal to $-4$. Deleting the unique degree-two vertex with two degree-three neighbors (lowermost in picture) leaves a graph whose adjacency matrix has determinant $-4$, too. -Among the 156 isomorphism types of $6$-vertex graphs, the only other graph of the kind requested by the OP is the graph found by Philipp Lampe at 2018-03-05 18:38:11Z, that is, - -, - -(this graph is isomorphic to graph no. 116 below) -and there does not exist any example on five vertices or less. -The above graph has $6$ vertices and $12$ edges. Hence the answer at 2018-03-05 18:38:11Z provided an example of the requested kind with smallest possible number of vertices but missed the example of smallest number of edges (by four edges). ${}\hspace{125pt}$ End of Proposition. -On the content of this answer. -Content. This answer contains -1. a proof of the Proposition (assuming that Sage tells the truth), -2. a correction of two errors in a table in the following published article (which is very relevant to the OP's question): - -[A2012] Alireza Abdollahi, Determinants of adjacency matrices of graphs, Transactions on Combinatorics, Vol. 1, No. 4 (2012), pp. 9-16 - -Notation. For brevity, throughout, 'a.d. (of a graph $G$)' is shorthand for the much longer phrase 'determinant of the adjacency matrix (of a graph $G$)'. (The abbreviation 'a.d.' is for 'adjacency determinant'.) -On the illustrations. Each of the screenshots of the Sage-output below is given in sufficient resolution to make all numbers visible (perhaps you'll have to zoom or click on the picture). -I doctored the pictures of the output in various ways to make them more useful (in particular, the a.d.-value is highlighted in blue), but only to the extent that this was doable in a bearable amount of time. In particular, I did not in any way try to improve upon Sage's graph-drawing. I do not understand the rationale behind the order in which the graphs are listed by Sage, and did not change that either. Not interfering with this ordering makes the data more reproducible. The blue numbers, which I added, give the order in which the graphs were spewn out by Sage. -Details. -I start with item 2. above, because it is useful to have that table available for later double-checking. -On item 2. The incorrect table is this: - - - (source: [A2012; p. 11]; annotations added; the cyan box around the url is in the original) - -The rows marked as wrong are certainly wrong, no matter whether Sage is correct or not. I checked this carefully, and will give reasons below (in short, the rows marked red can be seen to be wrong from the exponents/multiplicities alone, disregarding the values). -The rows marked 'checked with Sage' are probably correct, since I could reproduce each of them with the following Sage code(*): - -g=graphs(9) - v=list(g) -len(v) -results=[] -for i in range(274668): -$\hspace{10pt}$ results.append(v[i].adjacency_matrix().det()) -from collections import defaultdict -histo = defaultdict(int) -for k in results: -$\hspace{10pt}$ histo[k] += 1 -histo - -(here 'histo' is for 'histogram'(**)) whereupon the following output is produced: - -defaultdict(, {0: 133174, -128: 2, 2: 6767, 4: 6950, - 6: 4669, 8: 1566, 10: 1349, 12: 1156, 14: 695, 16: 606, 18: 106, 20: - 297, 22: 173, 24: 240, 26: 95, 28: 91, 30: 61, 32: 46, 34: 5, 36: 32, - 38: 28, 40: 3, 64: 1, 42: 17, 44: 16, 54: 3, -72: 12, 60: 3, -64: 7, - -96: 3, -60: 5, -56: 17, -54: 12, -50: 27, -48: 13, -46: 20, -44: 39, - -42: 47, -40: 103, -38: 52, -36: 110, -34: 128, -32: 593, -30: 199, -28: - 295, -26: 392, -24: 765, -22: 579, -20: 869, -18: 2747, -16: 2247, -14: - 1805, -12: 3062, -10: 4290, -8: 17582, -6: 8531, -4: 14901, -2: 57065}) - -I have annotated the table in various ways; in particular, I have shifted the entries in the row for $n=6$ a little to the left to make room for what I consider to be the correct values. (Incidentally, the only error in [A2012] for this row is the multiplicity of the a.d.-value $4$; the rest of that row is correct. The multiplicity of $4$ is $5$, not $2$ as given in the table.) -How to prove that the rows for $n=5$ and $n=6$ are wrong? This already follows by summing the exponents. Needless to say, for each row, the sum of the exponents must be equal to the number of isomorphism classes (aka number of unlabelled graphs) of graphs with that number of vertices. According to both A000088 and a table on page 105 of [Flajolet-Sedgewick, Analytic Combinatorics, CUP, version 26 June 2009], - -on $3$ vertices, there are $4$ isomorphism classes of graphs, -on $4$ vertices, there are $11$ isomorphism classes of graphs, -on $5$ vertices, there are $34$ isomorphism classes of graphs, -on $6$ vertices, there are $156$ isomorphism classes of graphs, -on $7$ vertices, there are $1044$ isomorphism classes of graphs, -on $8$ vertices, there are $12346$ isomorphism classes of graphs, -on $9$ vertices, there are $274668$ isomorphism classes of graphs, - -while - -in the row for $n=3$, the sum of multiplicities is $3+1=4,$ -which is consistent, -in the row for $n=4$, the sum of multiplicities is $1+7+3=11,$ -which is consistent, -in the row for $n=5$, the sum of multiplicities is $1+25+6+1=33,$ -which contradicts the number $34$ above, -in the row for $n=6$, the sum of multiplicities is $3+5+32+99+10+2+2=153,$ -which contradicts the number $156$ above, -in the row for $n=7$, the sum of multiplicities is $2+2+13+21+20+690+204+40+17+25+5+5 = 1044,$ -which is consistent, -in the row for $n=8$, the sum of multiplicities is $2+2+5+5+7+21+51+43+90+79+128+251+581+813+6551+2416+758+240+73+139+24+23+32+8+1+3=12346,$ -which is consistent, -in the row for $n=9$, the sum of multiplicities is $2+3+12+7+5+17+12+27+13+20+39+47+103+52+110+128+593+199+295+392+765+579+869+2747+2247+1805+3062+4290+17582+8531+14901+57065+133174+6767+6950+4669+1566+1349+1156+695+606+106+297+173+240+95+91+61+46+5+32+28+3+17+16+3+3+1=274668,$ -which is consistent. - -Furthermore, -On item 1. To prove the Proposition, we first note that it is obvious that there does not exist an example on three vertices or less. -Now we show that there is no example on four vertices. -For this, we first note that the only graphs on three vertices without isolated vertices (and hence having any chance of resulting in a nonzero a.d.) are - -the path, with three vertices with a.d. $0$, and -the triangle, with a.d. $2$. - -Hence the only candidates of graphs on four vertices of the kind requested by the OP are those whose a.d. is $2$. Now, the Sage code - -for G in graphs(4): -$\hspace{10pt}$ print([G.adjacency_matrix(),G.adjacency_matrix().det(),G.show()]) - -produced the following output: - - - - - -and this shows that there isn't any four-vertex graph with a.d. $2$. (This is also corroborated by the row for $n=4$ in the table in [A2012; p. 11].) This completes the proof that there is no example on four vertices. -We now show that there is no example on five vertices. Both from the Sage-output above, and from the table in [A2012] we see that the only nonzero a.d.-values on four vertices are $1$ and $-3$. Therefore, the only candidates for an example on five vertices are those graphs on five vertices with an a.d. of either $1$ or $-3$. Now, the Sage code - -for G in graphs(5): -$\hspace{10pt}$ print([G.adjacency_matrix(),G.adjacency_matrix().det(),G.show()]) - -produced the following output: - - - - - - - - - -Both the above, and the table in [A2012; p. 11], strongly suggest that there does not exist any graph on five vertices with a.d. equal to $-3$ or $1$. If this is true (which seems very likely in view of [A2012] reporting to have used $\textsf{GAP}$, a different software from Sage), then this implies that there indeed does not exist an example on five vertices. Needless to say, this is not a mathematical proof; sadly, I cannot imagine how a proof could be given which is better than inspection of all instances, even when using a well-know 'combinatorical' formula for the interpretation determinants found in [H1962; p. 208].That formula still involves both positive and negative summands, which is why I do not think it merits to be considered a genuinely combinatorial interpretation (this is more of an aesthetic judgement, of course). Harary's interpretation so far does not seem to help with the open problem, and it does not offer any computational advantages. -We now prove that on six vertices there are exactly two examples of the kind requested by the OP. From the 34 graphs shown above, it follows that very probably the only nonzero a.d.-values on five vertices are $-4$, $-2$, $2$, $4$. So, the only candidates for examples of the kind requested on six vertices are those with these a.d.-values. Now, the Sage code - -for G in graphs(6): -$\hspace{10pt}$ print([G.adjacency_matrix(),G.adjacency_matrix().det(),G.show()]) - -produced the following output. The same remarks as above apply with regards to the presentation. In particular, my not improving the graph-drawing is more obvious here, since, in several cases such as graphs no. 114, and 184 (and most especially no. 105 and 114, which has some almost invisible edges), Sage's artistic 'choices' were not so sage after all. Most of the time though, Sage did remarkably well in drawing the graphs, and in all cases all information is clear, since I give the adjacency matrices alongside the graphs. I feel that I should justify my not discarding the graphs with an isolated vertex (and hence obviously a vanishing determinant). I did not discard them because this makes the answer more systematic, reproducible and checkable. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -The above strongly suggests that there does not exist a graph on six vertices with a.d.-value $-2$ or $2$. There do exist ten graphs with a.d.-value $-4$ or $4$ though, namely, graphs no. 63, 65, 109, 110, 113, 116, 121, 126, 127, 149. We have to consider each of these in turn, and analyze whether there are vertices whose deletion leaves a graph with the same a.d.-value. -For the graph no. 63, which has a.d. $4$, the deletion of any vertex results in a graph isomorphic to graph no. 30 (in the list of graphs with five vertices), which has a.d. $-2\neq 4$, so 63 is not an example. -For the graph no. 65, which has a.d. $4$, we have to distinguish cases for analyzing the effects of the vertex-deletions. - -when deleting vertex $4$, a graph isomorphic to graph no. 34 results, which has a.d. $-4\neq 4$. -when deleting vertex $0$ or $1$, a graph isomorphic to graph no. 31 results, which has a.d. $-2\neq 4$. -when deleting vertex $2$, a graph isomorphic to graph no. 30 results, which has a.d. $-2\neq 4$. -when deleting vertex $3$ or $5$, a graph isomorphic to no. 32 results, which has a.d. $-2\neq 4$. - -Since the above exhausts all possible vertex-deletions, and the a.d. always changed, the graph no. 65 is not an example. -For the graph no. 109, which has a.d. $-4$, the deletion of any vertex results in a graph isomorphic to no. 16. which has a.d. $0\neq -4$, hence no. 109 is not an example. -For the graph no. 110, which has a.d. $-4$, we again have to distinguish cases: deleting vertex $0$ results in a graph isomorphic to graph no. 22, which has a.d. $2\neq -4$, while deleting vertex $1$ or $2$ results in graph no. 16 with a.d. $0$, and deleting $3$ or $4$ results in no. 31 with a.d. $-2\neq -4$, and finally deleting vertex $5$ results in no. 17 with a.d. $0\neq -4$; this shows that graph no. 110 is not an example either. -For the graph no. 113, which has a.d. $-4$, -deleting $0$ results in graph no. $22$ with a.d. $2\neq -4$, -while deleting $1$ or $2$ results in no. 23 with a.d. $0\neq-4$, while deleting $3$ or $4$ results in no. 32 with a.d. $-2\neq-4$, and finally deleting $5$ results in no. 24. This shows that no. 113 is not an example. -For the graph no. 116, which has a.d. $4$, deleting any of $0$, $1$, or $2$ results in no. 26 with a.d. $-2\neq 4$, -while deleting $3$ or $4$ results in no. 33 with a.d. $-2\neq 4$, but deleting $5$ results in no. 29 with a.d. $4=4$. -Hence no. 116 is an example, namely the example found by Philipp Lampe. In this example, exactly one of the vertices has the required property. -For the graph no. 121, which has a.d. $4$, deleting vertex $0$ or $2$ results in no. 24 with a.d. $-2$, while deleting $1$ results in no. 34 with ad. $-4\neq 4$, while deleting $3$ results in no. 31 with a.d. $-2\neq 4$, while deleting $4$ results in no. 33 with a.d. $-2\neq 4$, -and deleting $5$ results in no. 26 with a.d. $-2\neq 4$. This shows that graph no. 121 is not an example. -For the graph no. 126, which has a.d. $-4$, deleting $0$ or $4$ results in no. 23 with a.d. $0$, while deleting $1$ results in no. 16 with a.d. $0\neq -4$, while deleting $2$ or $5$ results in no. 17 with a.d. $0\neq -4$, but deleting $3$ results in no. 34 with a.d. $-4=-4$. Therefore, graph no. 126 is another example of the required kind. This is the example given at the very beginning of this answer. -For the graph no. 127, which has a.d. $-4$, deleting $0$, $3$ or $4$ results in no. 24 with a.d. $-2\neq-4$, while deleting $1$, $2$ or $5$ results in no. 17 with a.d. $0\neq -4$. -This exhausts all vertices and shows that no. 17 is not an example. -For the graph no. 149, which has a.d. $4$, -and deleting $0$ or $5$ results in no. 26 with a.d. $-2\neq4$, -while deleting $1$, $2$, $3$ or $4$ -results in no. 32 with a.d. $-2\neq 4$. -This shows that no. 149 is not an example. -We have now considered all the candidates, -and found exactly two examples of the kind -requested by the OP, namely, the two graphs -given in the proposition at the beginning of -this answer. This proves the proposition. -Concluding remarks. - -The rather small ratio of $\frac{2}{156} = \frac{1}{78}$ suggests the question of whether the limit - - -$\lim_{n\to\infty}\frac{\text{number of unlabelled graphs on $n$ with at least one vertex of the required kind}}{\text{number of all unlabelled graphs on $n$}}$ - -exists and what it is. Everything other than its being $0$ would seem surprising to me, but I don't know how to prove this. - -Embarassingly, I don't know a construction which would answer the obvious question of - - -whether for every $n\geq 6$ there exists at least one graph on $n$ with the required property (i.e., non-zero determinant of the adjacency matrix, and there is at least one vertex whose deletion leaves a graph with the same determinant of adjacency matrix). - - -Noam Elkies at 2018-03-05 17:23:37Z pointed out the interesting ambiguity in the verbal phrasing of the OP's question, which leaves it unclear what the intended logical quantifier governing the deleted vertex is. Both interpretations are possible. While the OP's accepting the answer at 2018-03-05 18:38:11Z seems to suggest that the OP meant an existential quantifier, the other interpretation is interesting and an open question; i.e., - - -Open problem. Does there exist a finite graph such that the determinant of its adjacency matrix is nonzero and deleting any of its vertices results in a graph whose adjacency matrix has the same value as before? - -Update 13 March, 2018. I am now inclined to think that, contrary to my initial expectations, a graph of the above kind is impossible: In two idle hours I experimentally checked almost all the named graphs at - -http://doc.sagemath.org/html/en/reference/graphs/sage/graphs/graph_generators.html - -and in the case of the graph-constructor allowing parameters, I experimented with many value of these parameters. Moreover, I tested many thousands of circulant graphs (by now, roughly $10^6$ nonisomorphic circulants, on between 11 and 100 vertices, roughly speaking), and for many sizes of 'connection sets', with each 'connection sets' randomly chosen. I also tested random graphs (though there one knows from the get-go that there is no chance, because the vertex-deleted a.d. clearly won't all have the same value). The results was this: - -There are more examples that I care to describe of the a.d. being non-zero while the a.d. after vertex-deletion differs from the previous a.d. by $1$ only. However, amazingly, not a single example of the kind required by the OP was found among thousands of non-isomorphic graphs. This surprised me, because it was contrary to what I had expected. From my experiences with smallish circulants (which came 'close' to what the OP and Noam Elkies asked for), I had expected that randomly mining slightly larger circulants would quickly turn up an example. This was not the case, and I went to circulants on about a hundred vertices, and to intersection sets of size about $5$ (resulting in $10$-regular graphs). - -I am well-aware that circulants are very special graphs and that generalizing from them to all graphs is rather unwarranted. By now, I would be very surprised though, if an affirmative answer of the question in the open problem existed which was a circulant. There seems to be some principle at work which prevents the difference - -$\det(\mathrm{A}(G))-\det(\mathrm{A}(G)|_{(n\setminus\{i\})\times(n\setminus\{i\})})$ - -which for a circulant of course does not depend on $i\in n$, down to $0$, except in the trivial case in which the determinants are both $0$. To repeat myself, this difference can easily brought to $1$ or $-1$ (I have seen hundreds of examples), though I still do not know how to characterize the circulants achieving that. -There might be an appreciable mathematical truth behind these experimental results. Perhaps there is even an understandable reason. I in particular find it intriguing that one can get to within $1$ of what is requested, which is as close as it can get except for an outright solution. -So, unless you cherish the process itself, don't search around in http://doc.sagemath.org/html/en/reference/graphs/sage/graphs/graph_generators.html for an example, for I predict that you won't find any. (This is a rough intuitive judgement of the probability, 'Bayesian' as it were, not an exact frequentist statement.) In my personal opinion it seems more advisable to try to prove the conjecture that such a graph is impossible. -Update March 14, 2018. I think it is not unlikely that someone (like me) finds the article - -[BLP2014] Bapat, R.B.; Lal, A.K.; Pati, S. A formula for all minors of the adjacency matrix and an application. Spec. Matrices 2, No. 1, 89-98 (2014). - -and (like me) initially thinks that this might help with the open problem. I cannot find anything wrong with this interesting article, but I think it might save someone else time to point out that contrary to what the title might suggest, the article [BLP2014] can be completely ignored as far as the open problem is concerned; the reason is that the main focus of [BLP2014] are minors which are indexed by such sets such that the set indexing the 'rows' of the minor is disjoint from the set indexing the 'columns' of the minor, whereas the OP's problem involves minors indexed by sets of the form $(n\setminus\{i\})\times(n\setminus\{i\})$ with $i\in n$, which is about as far from the main purview of [BLP2014] as possible. I am not saying that [BLP2014] had a misleading title: indeed, every kind of minor is taken care of, but for minors of the kind relevant to the present open problem, the formula in [BLP2014] reverts to precisely the formula in [H1962, page 208] (to see this, one has to read until Remark 3.2 in [BLP2014]). -References. -[A2012] Alireza Abdollahi, Determinants of adjacency matrices of graphs, Transactions on Combinatorics, Vol. 1, No. 4 (2012), pp. 9-16 -[B1993] Norman Biggs, Algebraic Graph Theory, Cambridge University Press, 1993, ISBN 9780521458979 -[BLP2014] Bapat, R.B.; Lal, A.K.; Pati, S. A formula for all minors of the adjacency matrix and an application. Spec. Matrices 2, No. 1, 89-98 (2014) -[H1962] Frank Harary, The Determinant of the Adjacency Matrix of a Graph, SIAM Review, Vol. 4, No. 3. (1962), pp. 202-210. -Footnotes. -(*) Incidentally, the case $n=10$ seems to be within the reach of contemporary machines, even with the non-optimized, high-level code I gave above. Yet it is within the reach of the machines available to me. If some readers happens to have both access to a powerful machine/cluster, and has and Sage installed, then running the code I gave with - -'graphs(9)' replaced with 'graphs(10)', - -and - -'range(274668)' replaced with 'range(12005168)' (which is the number of isomorphism classes of graphs on 10 vertices) - -will probably yield the a.d.-values for $n=10$ in a reasonable amount of time and memory. -I would appreciate being sent the output of 'histo' for $n=10$ (e.g. as a comment), and would then add it to this thread. The case $n=11$ is, I guess, out of reach without renting a sizeable amount of time and memory on a supercomputer. -(**) It is worth pointing out that the distribution of a.d.-values is not symmetric around $0$. -This might seem surprising at first thought, but becomes less surprising if one realizes that the usual operations which swap the sign of a determinant, and which often can be used to prove symmetry of vanishing expected values, are not available here: swapping two rows (or columns) does not result in another adjacency-matrix of a graph.<|endoftext|> -TITLE: Tanaka-Meyer formula -QUESTION [6 upvotes]: I have a simple question about Tanaka-Meyer formula, I am having difficulty applying it. Let $X$ be a continous martingale vanishing at zero. From Tanaka-Meyer formula it holds $$d|X_t| = sgn(X_t)dX_t+d \Lambda^X_t(0)$$ -where $\Lambda^X_t(0)$ is the local time accumulated by the process $X$ at the origin. -I am interested in the process $Z = X^2$. From Itô's formula we have -$$dZ_t = 2X_t dX_t + d \langle X \rangle_t.$$ -Since obviously $Z = |Z|$ applying Tanaka-Meyer should have the same differential, however -$$d|Z_t| = 2sgn(Z_t) X_t dX_t + sgn(Z_t)d \langle X \rangle_t + d \Lambda^Z_t(0) ,$$ -and -$$dZ^+_t = 2*\mathbb{1}_{[Z_t>0]} X_t dX_t + \mathbb{1}_{[Z_t>0]}d \langle X \rangle_t + {1 \over 2} d \Lambda^Z_t(0) .$$ -Applying the expected value to processes $|Z|$ and $Z^+$, I have -\begin{align} -\mathbb{E}[\int_0^T sgn(X^2_t)d\langle X \rangle_t +\Lambda^Z_t(0) ] = \mathbb{E}[\int_0^T \mathbb{1}_{[X^2_t>0]}d\langle X \rangle_t +{1 \over 2}\Lambda^Z_t(0)] -\end{align} -which feels like that $\Lambda^Z_t(0) = 0 ~ a.s.$ Where is my reasoning wrong? I have been staring at it longer than I like to admit. - -REPLY [2 votes]: Alternatively, use Corollary 1.9 in Revuz & Yor and $d\langle Z \rangle_t = 4 X_t^2 d \langle X \rangle_t$ (as Eldredge did) to write \begin{align*} -\Lambda_t^Z(0) &= \lim_{\epsilon \downarrow 0} \frac{1}{\epsilon} \int_0^t 1_{\{ 0 \le Z_s < \epsilon\} } d \langle Z \rangle_s \\ -&= \lim_{\epsilon \downarrow 0} \frac{1}{\epsilon} \int_0^t 1_{\{ -\sqrt{\epsilon} < X_s < \sqrt{\epsilon} \}} 4 X_s^2 d \langle X \rangle_s \\ -&\le 4 \lim_{\epsilon \downarrow 0} \int_0^t 1_{ \{-\sqrt{\epsilon} < X_s < \sqrt{\epsilon}\} } d \langle X \rangle_s \\ -&\le 4 ( \Lambda_t^X(0) + \Lambda_t^{-X}(0)) \lim_{\epsilon \downarrow 0} \sqrt{\epsilon} = 0 \quad \text{a.s.} -\end{align*} -which implies $\Lambda_t^Z(0) = 0$ a.s.<|endoftext|> -TITLE: For every prime divisor $p$ of a finite 2-generator group $G$, is there a generating pair containing an element of order divisible by $p$? -QUESTION [10 upvotes]: Let $G$ be a finite 2-generated group. Let $p$ be a prime dividing the order of $G$. Must there exist a generating pair $(g,h)$ of $G$ such that $|g|$ is divisible by $p$? -If not, is this true at least for finite simple groups? - -REPLY [12 votes]: The answer is yes for finite simple groups $G$. Every element of $G$ is an element of a generating pair. This is proved in -Guralnick, Robert, Kantor, William, Probalistic generation of finite simple groups, J. Algebra 234 (2000), p. 743–792. (MR1800754) -Abstract: For each finite simple group $G$ there is a conjugacy class $C_G$ such that each nontrivial element of $G$ generates $G$ together with any of more than $1/10$ of the members of $C_G$. Precise asymptotic results are obtained for the probability implicit in this assertion. Similar results are obtained for almost simple groups. -But the answer in general is no. It is not true for $2$-generated Frobenius groups with a non-cyclic Frobenius complement, because the two generators have to generate the group modulo the Frobenius kernel, and so elements of the kernel do not arise in generating pairs. -For example, $\mathtt{PrimitiveGroup}(9,3)$ in the GAP/Magma databases has order $72$, with Frobenius kernel elementary abelian of order $9$ and complement $Q_8$. All generating pairs consist of two elements of order $4$.<|endoftext|> -TITLE: Computing the relative class group (with Galois action) of relatively large cyclotomic groups -QUESTION [6 upvotes]: For a cyclotomic field $K = \mathbb Q(\zeta_n)$, let $K^+$ be its maximal totally real subfield. We know that $H^+ = Cl(K^+)$ injects into $H = Cl(K)$. I am interested in computing the group $H/H^+$ and the corresponding Galois action. -I found some references for $n$ prime (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.370.4613&rep=rep1&type=pdf) but I am really interested in specific composite n (n = 55,57,63,67, 69, 85...). The corresponding relative class numbers for these values are (10,9,7, 853513, 69, 6205...) -Does anyone know references or suggestions on how to compute the class group? I am specifically interested in figuring out if a specific character occurs in the class group. - -REPLY [2 votes]: Have you had a look at Schoof's paper The structure of the minus class group of abelian number fields, Séminaire de Théorie des Nombres de Paris, 1988--89? Schoof's idea is to conjecture that the Fitting ideal of the minus class group of any abelian complex number field $F$ (meaning the quotient $H_F/H_{F^+}$) should be equal to what he calls a "Stickelberger ideal" generated by Stickelberg elements. As he points out at the end of his introduction, this has somehow been superseeded by Kolyvagin's idea of Euler Systems, but the equality between Fitting and Stickelberger ideal remains open (there is a paper by Kurihara on this, entitled Iwasawa theory, higher Fitting ideals, and Kolyvagin systems of Gauss sum type). -Quite interesting is his Theorem 4.3 saying showing a criterion for freeness of the $\chi$-component of the minus class group.<|endoftext|> -TITLE: Infinite "almost rigid" homogeneous $T_2$-space -QUESTION [6 upvotes]: A topological space $(X,\tau)$ is said to be homogeneous if for all $x,y$ there is a homeomorphism $\varphi:X\to X$ such that $\varphi(x) = y$. -Is there an infinite homogeneous Hausdorff space $(X,\tau)$ such that every continous map $f: X\to X$ is either a homeomorphism, or constant? -EDIT. I forgot to add "infinite" in the original question. - -REPLY [10 votes]: Topological groups are homogeneous. In - -J. van Mill, "A topological group having no homeomorphisms other than translations," Transactions of the AMS 280 (1983), pp. 491-498 (link), - -Jan van Mill constructed an infinite topological group whose only self-homeomorphisms are group translations. Such a space is called "uniquely homogeneous" -- it is homogeneous, but for any pair of points there is exactly one self-homeomorphisms of the space witnessing homogeneity. Jan's group also has the amazing property that removing any point results in a rigid space. -In the same paper (section 4), van Mill shows that, assuming the Continuum Hypothesis, there is a topological group whose only continuous self-maps are either group translations or constant functions. -Thus the answer to your question is "consistently yes, and you can come close in ZFC." I do not know whether anyone else has come along and improved Jan's CH result to a ZFC result (but a quick glance through the papers citing Jan's seems to indicate that no one has).<|endoftext|> -TITLE: Is there a compactification with nontrivial connected remainder? -QUESTION [6 upvotes]: Question: Let $X$ be a continuum and $p \in X$. Under what conditions does there exist a compactification $\gamma (X-p)$ with $\gamma (X-p) - (X-p)$ connected and nondegenerate? -Throughout, $X$ is a continuum. That is to say, a compact connected Hausdorff space. (I'd be interested on any info for non-connected spaces too!) -The story is we want to remove a point $p \in X$ and replace it with a small subcontinuum. Formally we want a space $Y$ and continuous map $Y \to X$ such that all fibres are nontrivial save one, and the nontrivial fibre $F$ is a nondegenerate subcontinuum with void interior. The question is trivial if we allow $F$ to have interior as we can always take a copy of $[0,1]$ and glue it to $p$ at an endpoint. -When $X$ is metric this can always be done. For example we can resolve $X$ at the point $p$ onto the arc: consider the following subspace of $X \times [-1,1]$. -$\displaystyle Y = \Big \{ \Big (x, \sin \Big (\frac{1}{d(x,p)}\Big ) \Big): x \in X-p \Big\} \bigcup \Big ( \{p\} \times [-1,1]\Big )$. -Then the projection map onto the $x$-coordinate satisfies the above condition. -More generally we can resolve onto any locally connected continuum $F$, by recalling $F$ is the image of an arc and constructing a map from $[0,1)$ that visits each point cofinally many times. -When $X$ is not metric we cannot always do this. For example consider the long arc $[0,\omega_1]$. Removing the endpoint $p$ we get the long line whose Stone Čech compactification is known to be the same as the one-point compactification. Since the remainder of the Stone Čech maps onto the remainder of every compactification, we cannot replace $p$ with a nondegenerate continuum, as then that continuum would be the image of a point. -It is known exactly when $X-p$ admits a nontrivial remainder. Namely when $X-p$ has two disjoint closed non-compact subsets. Has it been studied whether there are any conditions for when a nontrivial connected remainder exists? -I know Smith has given a condition for when the Stone Čech remainder is connected (when $p$ fails to be a local cut point) but this is of no immediate use. For example if $X$ is a circle $(X-p)^*$ has exactly two components but the above shows we can always wrap $X-p$ around an arc. In this case we get something like a Warsaw circle. -Edit: You might try to generalise the resolution approach to something involving a gauge $\mathcal G $ of pseudometrics on $X$. But this throws up problems if $\rho(p,x)=0$ for some $x \ne p$ and $\rho \in \mathcal G$. The obvious solution is to replace $\mathcal G$ with a gauge that is positive definite at $p$ meaning $\rho(p,x)=0 \implies p=x$. But this cannot in general be done. In fact if a single $\rho$ is positive definite then $\{p\}$ is immediately a $G_\delta$ set which is not a guarantee for non-metric spaces. - -REPLY [2 votes]: Given a compact Hausdorff space $X$ (not necessarily connected) and a point - $p \in X$, the following are equivalent: - -$X - p$ has a compactification with non-trivial connected remainder. -There is a continuous surjection $q: (X - p)^* \to I$, where $(X-p)^*$ - denotes the Stone-Čech remainder of $X-p$ and $I = [0,1]$. -$X-p$ has a compactification $\gamma(X-p)$ with remainder homeomorphic to $I$. -There is a continuous $f: X-p \to I$ such that every point of $I$ is a - cluster point of $f$ at $p$. (in other words, $\overline{f[U-p]} = I$ - for every neighbourhood $U$ of $p$). - - -Sketch of a proof: -$(1. \implies 2.)$ $(X-p)^*$ can be mapped onto every other - remainder and every non-degenerate continuum can be mapped onto $I$. -$(2. \implies 3.)$ Put $\gamma(X-p) = I \cup_q \beta(X-p)$. Since $X-p$ is open - in $\beta(X-p)$, the inclusion $X-p \to \gamma(X-p)$ is a homeomorphic - embedding. -$(3. \implies 4.)$ The identity mapping of $I$ has a continuous extension - to $f: \gamma(X-p) \to I$. The desired property follows from the density - of $X-p$ in $\gamma(X-p)$ and the correspondence between - the neighbourhoods of $I \subset \gamma(X-p)$ and those of $p \in X$. -$(4. \implies 3. \implies 1.)$ The closure in $X \times I$ of the graph of $f$ - is such a compactification. Clearly $I$ is connected. - -Additional information on condition 1: -A compact Hausdorff space $K$ can be mapped continuously onto $I$ if and -only if $K$ is not scattered. -For sufficiency of this condition we need, by the Tietze-Urysohn extension -theorem, only to see that there is some closed subset that can be mapped -onto $I$. If $K$ is not totally disconnected, we can take this subset to -be any nontrivial component. If $K$ is totally disconnected, then we -can take its perfect kernel, which is zero-dimensional, compact and -dense in itself and by repeated bisection find a mapping onto $\{0,1\}^\omega$, -which maps onto $I$ in a straightforward way. -For necessity, observe that any continous surjection $q: K \to I$ is -proper, which implies that $K$ has a closed subset $L$ such that -$q|_L$ is irreducible. Since $q|_L$ is also closed, it must map every -isolated point of $L$ to an isolated point of $I$, which has no isolated -points. Hence $L$ is dense in itself and $K$ is not scattered. - -Additional information on condition 4: -A easy sufficient, but not necessary, condition for the existence of such -an $f$ is that $X-p$ is not pseudocompact. To see this, suppose -$g$ is an unbounded nonnegative function on $X-p$. Then $g[X-p]$ contains -an unbounded subset $W$ with order type $\omega$. Since $W$ is discrete -and closed in $[0,\infty)$ there is a continuous $h: [0,\infty) \to I$ -such that $h[W]$ is dense in $I$. Putting $f = h \circ g$ gives the -desired result, since if $U$ is a neighbourhood of $p$, $f[U-p]$ contains -all but finitely many points of $h[W]$. -An example of a pseudocompact locally compact Hausdorff space with a -connected Stone-Čech remainder is easily constructed by taking the -product of the long line with a non-degenerate continuum.<|endoftext|> -TITLE: Consistency strength of the existence of a transitive model of $\mathsf{ZFC}^-$ with a $\kappa$-complete ultrafilter -QUESTION [7 upvotes]: Let $\mathsf{ZFC}^-$ be the Zermelo-Fraenkel set theory without power set axiom. -For a transitive model $M$ of $\mathsf{ZFC}^-$ and an cardinal $\kappa\in M$ in the sense of $M$, an unary predicate $U$ over $M$ is an $\kappa$-complete ultrafilter if $M$ thinks it is $\kappa$-complete and $U$ is weakly amenable, that is, for any $F\in {^\kappa}M\cap M$, $\{\xi<\kappa : F(\xi)\in U\}\in M$. -$M$ thinks $\kappa$ is weakly compect. I think in fact we can show $M$ thinks $\kappa$ is Ramsey. However it does not tell about their consistency strength as $\mathsf{ZFC}^-$ is far more weaker than $\mathsf{ZFC}$. -My question is, the exact consistency strength of the existence of the model $M\models \mathsf{ZFC}^-$ with a $\kappa$-complete ultrafilter over $M$ is known? Theorem 1.1 of an article by Gitman and close relation between $\kappa$-complete ultrafilter and elementray embedding suggests its consistency strength may be related to the existence of a weakly compact cardinal. - -REPLY [11 votes]: Let's make some additional assumptions on the ultrafilter $U$. Suppose $M\models{\rm ZFC}^-$ and $\kappa$ is a cardinal in $M$. We say that $U$ is an $M$-ultrafilter if $\langle M,\in,U\rangle$ satisfies that $U$ is a $\kappa$-complete normal ultrafilter on $\kappa$. Because $U$ is only $\kappa$-complete for sequences in $M$ and $M$ might be missing even countable sequences, the ultrapower of $M$ by $U$ may not be well-founded. So let's say that $U$ is good if the ultrapower by $U$ is well-founded. -The consistency of the existence of $M\models{\rm ZFC}^-$ for which there is a good weakly amenable $M$-ultrafilter $U$ on a cardinal $\kappa$ in $M$ is between an ineffable cardinal and an $\omega$-Erdős cardinal, which is much weaker than a Ramsey cardinal. In particular, this assumption is consistent with $V=L$. -Here is a sketch of the argument. Suppose $j:M\to N$ is the ultrapower by $U$ (with $N$ transitive by assumption) with critical point $\kappa$ (by normality). By weak amenability, in $N$, $\kappa$ has the property that every collection of $\kappa$-many subsets of $\kappa$ has a $\kappa$-complete filter. Thus, by elementarity $M$ satisfies that this property holds for cofinally many $\alpha<\kappa$. Now, we argue that each such $\alpha$ is weakly compact in $L_\kappa\models{\rm ZFC}$. Next, we can argue that $\kappa$ has the ineffability property in $N$, and therefore there are cofinally many $\alpha$ that are ineffable in $L_\kappa$. The reason I am going down to $L$ is that we did not assume that $M$ has any powersets and therefore even the notion of inaccessible cardinal does not make sense for $M$. For the upper bound, an $\omega$-Erdős cardinal implies the consistency of what I call a $1$-iterable cardinal $\kappa$ having the property that every $A\subseteq\kappa$ is contained in a model $M\models{\rm ZFC}^-$ of size $\kappa$ with $\kappa\in M$ for which there is a good weakly amenable $M$-ultrafilter on $\kappa$. -If we don't assume that $U$ is normal or that the ultrapower by $U$ has to be well-founded, the consistency strength goes down significantly. At most a weakly compact suffices. If $\kappa$ is weakly compact and $M_0\models{\rm ZFC}^-$ has size $\kappa$, then we can find a $\kappa$-complete ultrafilter $U_0$ for $M_0$ with a well-founded ultrapower. Moreover, if $M_1\models{\rm ZFC}^-$ has size $\kappa$ and extends $M_0$, then we can extend $U$ to a $\kappa$-complete ultrafilter $U_1$ for $M_1$. Thus, in $\omega$-many steps, we can obtain $M=\bigcup_{n\in\omega}M_n$ and a weakly amenable $U$ that is $\kappa$-complete for sequences from $M$. The argument for the "moreover" part can be found here in an article of Peter Holy and Philipp Schlicht. I suspect that you don't even need a weakly compact. (I am assuming $\kappa$ uncountable here).<|endoftext|> -TITLE: Where to publish new mathematical identities? -QUESTION [12 upvotes]: Similar questions have been asked before regarding journals that publish: - -expository work, -recreational mathematics, -computational results, -new proofs of old theorems, -and even math textbooxs. - -However, I did not find any references to journals that publish new mathematical identities. These might include new formulae for $\pi$, $e$, or other known constants that might be useful for researchers in a wide variety of fields. -After briefly looking through papers that contain expositions of new formulae, it would seem that the majority of these papers are given "as-is", i.e., as notes freely available on the webpage of the author. Are there no journals that publish such results? - -REPLY [10 votes]: As Peter Heinig commented, if the mathematics behind the identity is novel and important enough, then you should select a journal like you would select a journal for any other paper—if it's a combinatorial identity, look for a combinatorics journal; if it's a number-theoretic identity, look for a number theory journal, etc. -For the specific case of constants such as $\pi$ and $e$, most novel identities for them are nowadays discovered with significant computer assistance. The journal Experimental Mathematics is one place where such identities have been published, e.g., I'm fond of Jesús Guillera's paper About a New Type of Ramanujan-Type Series, which contains some amazing identities such as the following one due to Gourevitch (which I believe is still open as of this writing): -$$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$ - -REPLY [5 votes]: For new combinatorial identities in the spirit of S. Ramanujan's work, there is the Ramanujan journal. - -REPLY [4 votes]: The American Mathematical Monthly, Mathematics Magazine, and The College Mathematics Journal (all published by the Mathematical Association of America) might be appropriate.<|endoftext|> -TITLE: Supersingularity and Roots of Unity of Zeta Functions -QUESTION [7 upvotes]: Let $C$ be smooth projective curve defined over a finite field $\mathbb{F}_q$. Let $$Z(C,u)=\exp(\sum_{n \ge 1} N_r(C) u^r/r) \in \mathbb{Z}[[u]]$$ be its zeta function, where $N_r$ is the number of $\mathbb{F}_{q^r}$-points on $C$. It is well known (Weil) that $$Z(C,u)=\frac{P_C(u)}{(1-qu)(1-u)},$$ where $P_C$ is a polynomial of degree twice the genus of $C$, that satisfies $P_C(0)=1$ and $P_C(\alpha)=0\implies |\alpha|=q^{-1/2}$. - -What is the standard name for curves $C$ such that the roots of $P_C$ are roots of unity times $q^{-1/2}$? And what is a suitable reference where this name is used? - -Also, - -What is the motivation behind studying such curves? - -I have seen that sometimes such curves are called `supersingular'. I have also seen many other definitions for supersingular curves, not involving the roots of $P_C$. But I have never seen how (and if...) these different definitions are related, which leaves me a bit confused. I list these definitions below. - -Here van der Geer and van der Vlugt define supersingular curves as curves whose jacobians are isogenous to a product of supersingular elliptic curves (over $\overline{\mathbb{F}_q}$). -Wikipedia describes a definition involving Newton polygons. -Rosen (PDF) defines supersingular curves using class numbers. -Here Kodama and Washio mention a definition involving the Hasse-Witt invariant, and also relate this invariant to the roots of $P_C$. - -REPLY [8 votes]: The names supersingular and superspecial appear in the literature in not a very consistent way. -Passing from $\mathbb{F}_q$ to $\mathbb{F}_{q^n}$ changes the roots $\alpha$ to $\alpha^n$. If the roots are roots of unity times $q^{-1/2}$, then passing to a suitable extension, we can assume that they are equal to $q^{-1/2}$ and that $q$ is a square. In this case $P_C(t)= (1-q^{1/2}t)^{2g}$ and the fact that this condition is equivalent to the Jacobian being isogenous to $E^g$, $E$ supersingular with $P_E(t)= (1-q^{1/2}t)^{2}$ follows from Tate's isogeny theorem. -The class number is $P_C(1) = (1-q^{1/2})^{2g}$, which is smallest it can be. Also the number of points $\# C(\mathbb{F}_q) = q + 1 -2gq^{1/2}$ is the smallest it can be. This is a major source of interest in these curves. Also, commonly studied are the maximal curves that attain the Weil bound $\# C(\mathbb{F}_q) = q + 1 +2gq^{1/2}$ and so are as above after extending to the quadratic extension. -The Hasse invariant of these curves is zero because of Manin's theorem relating the characteristic polynomial of the Cartier operator with the reduction modulo the characteristic $p$ of $P_C$. I don't think the converse holds.<|endoftext|> -TITLE: Naturality of Moore-Postnikov systems -QUESTION [9 upvotes]: Where in the literature can I find a naturality statement for Moore-Postnikov towers of maps? Something like the following: -Let $f:X\to A$ and $g:Y\to B$ be maps of connected CW-complexes which both admit a Moore-Postnikov tower of principal fibrations. Then a commuting diagram -$\require{AMScd}$ -\begin{CD} - X @>f>> A\\ - @V \Phi V V @VV \phi V\\ - Y @>>g> B -\end{CD} -(possibly with some extra conditions) induces maps $\Phi_n:X_n\to Y_n$ between the $n$-th stages of the towers of $f$ and $g$, for all $n\ge1$. - -REPLY [9 votes]: Working simplicially (in those days called "semi-simplicially") this is surely due to Moore, with details in unpublished 1956 lecture notes and in John C. Moore, Semi-simplicial complexes and Postnikov systems. 1958 Symposium internacional de topología algebraica International symposium on algebraic topology pp. 232–247. Earlier related work is in the 1954-55 Cartan Seminar "Algebres d'Eilenbeg-Maclane et homotopie". -Taking $f$ and $g$ in the question to be Kan fibrations of simplicial sets, there is a brief treatment on pages 34-35 of my 1967 book "Simplicial objects in algebraic topology"(http://www.math.uchicago.edu/~may/PAPERS1965.html), where I define "the natural Postnikov system of the fibre space $(E,p,B)$". I don't remember whether or not I got that from my adviser's notes cited above, but I imagine I did.<|endoftext|> -TITLE: Are algebraically isomorphic $C^*$-algebras $*$-isomorphic? -QUESTION [19 upvotes]: If A and B are C^*-algebras that are algebraically isomorphic to each other, does -this imply that they are *-isomorphic to each other? - -REPLY [20 votes]: Answering the question in the body of the original post, which seems to be more restricted than the implicit question in the title of the post.... -The answer is YES. See - -L. Terrell Gardner, On isomorphisms of $C^\ast$-algebras. - Amer. J. Math. 87 (1965) 384–396. - MathReview - -Roughly speaking, the proof works by considering the ${\rm C}^*$-algebras $A$ and $B$ as being represented on the GNS spaces $H_A$ and $H_B$ given by all pure states of $A$ and $B$ respectively, and then showing that an algebra isomorphism $A\to B$ can be extended to a spatial isomorphism ${\mathcal B}(H_A)\to {\mathcal B}(H_B)$. -Note that the original question was apparently raised by Sakai, a few years before Gardner's paper.<|endoftext|> -TITLE: Showing an alternating integer series is never $0$ -QUESTION [10 upvotes]: The following series arose in some work related to Hilbert Functions of ideals of points -$$\sum_{k = 0}^m (-1)^k {2m+2 \choose k}[2m(m+1)-k(2m+1)]^{2m-1}.$$ -Experimentally, this series is always negative for $m > 1$ and decreases incredibly quickly. We only need that this is never $0$. We have tried rational roots, taking first differences, but have been unable to make much progress. Any help or suggestions are appreciated. - -REPLY [4 votes]: The first term in the asymptotic expansion for the sum, which I'll call $S_0$, is -$$S_0 \sim -6 * 2^{-(2m+2)}\sqrt{\frac{3}{m(m+1)}} \binom{2m+2}{m+1}(2m-1)!(2m+1)^{2m-1}.$$ -It is explicitly negative and there are no oscillatory terms, so eventually the sum is negative, as conjectured. To complete the proof one would need the second term in the series, show that it is sufficient smaller than the dominant term for a particular $n = n_0$ then check explicitly those values for $n -TITLE: Do commutative rings with "interesting" Jacobson radicals turn up "in nature"? -QUESTION [11 upvotes]: Let $R$ be a commutative ring. Let's say that the Jacobson radical $J(R)$ of $R$ is uninteresting if - -$J(R)$ coincides with the nilradical, or -$J(R)$ is the intersection of a finite number of maximal ideals. - -It seems as if most rings used in algebraic geometry have uninteresting Jacbson radical: - -Every finitely-generated commutative algebra over a field or over a Dedekind domain is Jacobson, so its Jacobson radical coincides with its nilradical, and so is uninteresting by (1). -Every local or semilocal commutative ring has finitely many maximal ideals, and so has uninteresting Jacobson radical by (2). - -For a nonuninteresting example, take the localization $R = \mathbb Z[x]_S$ where $S = \{f(x) \in \mathbb Z[x] \mid f(0) = 1\}$. The maximal ideals of $R$ are of the form $(p,x)$ where $p \in \operatorname{Spec} \mathbb Z$. So the Jacobson radical is $(x)$, which is not uninteresting. -But this example seems rather artificial to me; for example I don't know anywhere a ring like this would show up in algebraic geometry. - -REPLY [3 votes]: Rings similar to the ring $R$ as you consider above, "show up" in real algebraic geometry. If $k:=\mathbb{R}, A:=k[x]$ is the ring of polynomials with real coefficients, and $S\subseteq A$ is the set of polynomials with no real roots ($f(x):=x^2+1$ is such a polynomial), then $B:=S^{-1}A$ may be seen as the global sections of the "structure sheaf" of $X(k):=\mathbb{A}^1_k(k):=Spec(A)(k)$ - the "affine real line" in the sense of real algebraic geometry (this is "vague"). -You may construct a locally ringed space $(X(k), \mathcal{O}_{X(k)})$, where $X(k)$ is the $k$-rational points of $X$ and where $\mathcal{O}:=\mathcal{O}_{X(k)}$ i a sheaf on $X(k)$ with global sections equal to $B$. -Example 1. The rational function $f(x):=\frac{a(x)}{x^2+1}$ is a regular function on $X(k)$ and gives rise to a global section of $\mathcal{O}$ for any polynomial $a(x) \in A$. If $\mathfrak{m}$ is a maximal ideal in the localized ring $B$, it follows the residue field $\kappa(\mathfrak{m})\cong k$ is the field of real numbers. -Hence when you localize at $S$ you have removed all maximal ideals in $A$ with residue field the complex numbers. -Example 2. The global sections $\Gamma(X, \mathcal{O}_X)=A$ is the polynomial ring, hence the real algebraic variety $X(k)$ has "more global sections" than the affine scheme $X$. Again this is "vague" since $X(k)$ is a subspace of $X$. -You may similarly construct the real projective line $\mathbb{P}(1):=\mathbb{P}^1_k(k)$, its structure sheaf $\mathcal{O}_{\mathbb{P}(1)}$ -and the real affine space $\mathbb{A}(n)$ and -you may embed $\mathbb{P}(1)$ as a "closed subvariety" (in the sense of real algebraic geometry) -$$ \phi: \mathbb{P}(1) \rightarrow \mathbb{A}(3)$$ -of affine $3$-space. -The above approach to "real algebraic geometry" uses the classical language of "schemes" and introduce a real algebraic variety using the $k$-rational points of the scheme $\mathbb{A}^1_k$. There are other approaches and you find references at the zentralblatt site under "real algebraic geometry". The apporoach indicated above gives for any scheme $X$ of finite type over $k$, a "real algebraic variety" $(X(k), \mathcal{O}_{X(k)})$, which is a locally ringed space where the structure sheaf $\mathcal{O}_{X(k)}$ has "more" local sections. In the case of projective space $\mathbb{P}^n_k(k)$ you get a space with properties similar to the underlying smooth manifold. You may (similar to the situation with differentiable manifolds and the Whitney embedding theorem) embed $\mathbb{P}^n_k(k)$ into "real affine space" $\mathbb{A}^d_k(k)$. In fact I believe there is an endofunctor $F$ of the category of schemes over $k$ with the property that for any projective scheme $X \subseteq \mathbb{P}^n_k$ it follows the scheme $F(X):=X(k)$ is affine: There is a closed embedding $F(X) \subseteq \mathbb{A}^d_k(k)$. The scheme $F(X)$ is no longer of finite type over $k$. -Example 2. A motivation for doing this is the study of the Weil restriction $W(X)$ of a complex projective manifold $X \subseteq \mathbb{P}^n$. In the case when the Weil restriction $W(X) \subseteq \mathbb{P}^m_k$ is projective over $k$ you get in a functorial way an "affine real algebraic variety" $F(W(X))=Spec(B)$, and any complex holomorphic vector bundle $E$ on $X$ gives a finite rank projective $B$-module $E(k)$. -You may have heard of the "Jouanolou-Thomason trick", which to any quasi projective variety $X$ of finite type over a field $k$ associates an affine variety $Y$ of finite type over $k$ with $K_i(Y)=K_i(X)$. With the above construction you get a similar construction for complex projective varieties which is functorial.<|endoftext|> -TITLE: Nearest matrix orthogonally similar to a given matrix -QUESTION [8 upvotes]: Given $A,B\in\Bbb R^{n\times n}$ is there technique find $$\min_{ T\in O(n,\Bbb R)}\|A-TBT^{-1}\|_F\mbox{ or }\min_{ T\in O(n,\Bbb R)}\|A-TBT^{-1}\|_2$$ within additive approximation error in $\epsilon>0$ in $O\big(\big(\frac{n\cdot\log(\|A\|_2\|B\|_2)}\epsilon\big)^c\big)$ time at fixed $c>0$? - -REPLY [2 votes]: Using the spectral norm, we have the following optimization problem in matrix $\mathrm X \in \mathbb R^{n \times n}$ -$$\begin{array}{ll} \text{minimize} & \| \mathrm A \mathrm X - \mathrm X \mathrm B \|_2\\ \text{subject to} & \mathrm X^\top \mathrm X = \mathrm I_n\end{array}$$ -where matrices $\mathrm A, \mathrm B \in \mathbb R^{n \times n}$ are given. Introducing variable $t \in \mathbb R$ and rewriting in epigraph form, -$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & \| \mathrm A \mathrm X - \mathrm X \mathrm B \|_2 \leq t\\ & \mathrm X^\top \mathrm X = \mathrm I_n\end{array}$$ -Note that $\| \mathrm A \mathrm X - \mathrm X \mathrm B \|_2 \leq t$ is equivalent to -$$t^2 \mathrm I_n - \left( \mathrm A \mathrm X - \mathrm X \mathrm B \right)^\top \left( \mathrm A \mathrm X - \mathrm X \mathrm B \right) \succeq \mathrm O_n$$ -Dividing both sides by $t > 0$, -$$t \, \mathrm I_n - \left( \mathrm A \mathrm X - \mathrm X \mathrm B \right)^\top \left( t \, \mathrm I_n \right)^{-1} \left( \mathrm A \mathrm X - \mathrm X \mathrm B \right) \succeq \mathrm O_n$$ -Using the Schur complement test for positive semidefiniteness, we obtain the following optimization problem in matrix $\mathrm X \in \mathbb R^{n \times n}$ and scalar $t > 0$ -$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & \begin{bmatrix} t \, \mathrm I_n & \mathrm A \mathrm X - \mathrm X \mathrm B\\ \left( \mathrm A \mathrm X - \mathrm X \mathrm B \right)^\top & t \, \mathrm I_n\end{bmatrix} \succeq \mathrm O_{2n}\\ & \mathrm X^\top \mathrm X = \mathrm I_n\end{array}$$ -Relaxing the equality constraint $\mathrm X^\top \mathrm X = \mathrm I_n$, we obtain $\mathrm X^\top \mathrm X \preceq \mathrm I_n$, a linear matrix inequality (LMI) that can be rewritten as follows -$$\begin{bmatrix} \mathrm I_n & \mathrm X\\ \mathrm X^\top & \mathrm I_n\end{bmatrix} \succeq \mathrm O_{2n}$$ -Thus, the relaxation of the original optimization problem is a (convex) semidefinite program (SDP) in matrix $\mathrm X \in \mathbb R^{n \times n}$ and scalar $t > 0$ -$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & \begin{bmatrix} t \, \mathrm I_n & \mathrm A \mathrm X - \mathrm X \mathrm B & & \\ \left( \mathrm A \mathrm X - \mathrm X \mathrm B \right)^\top & t \, \mathrm I_n & & \\ & & \mathrm I_n & \mathrm X\\ & & \mathrm X^\top & \mathrm I_n\end{bmatrix} \succeq \mathrm O_{4n}\end{array}$$ -It remains to be determined whether this relaxation is actually useful.<|endoftext|> -TITLE: Number of zeros of the derivatives of a polynomial -QUESTION [8 upvotes]: What is the maximum total number of zeroes a univariate polynomial $f\in\mathbb{C}[z]$ of degree $d$, together with all of its derivatives, can have at $k$ given points of $\mathbb{C}$? -I am interested in this question for constant $k$, say $k=3$. -One trivial bound is $d \choose 2$, an other is $kd$. -Does there exist an example of a polynomial with more than $k+d$ zeroes in total of it and its derivatives? -In other words, if $p$ is a polynomial and $x_1,\ldots x_k$ are his zeroes then I am asking how many $i,j$ there exist such that $p^{(i)}(x_j)=0$ - -REPLY [7 votes]: The answer is yes. Here is an example for $k=3$. For $m>0$, take $f(z)=z^m(z^2-1)^m$. Then $d=3m$. Of course, $-1$, $0$ and $1$ are roots of $f^{(i)}$ for $0\le i\le m-1$, so we have $3m=d$ roots. However, $f^{(i)}$ is an odd function for $i=m+1,m+3,\ldots, 3m-1$, producing $m$ more roots in $0$. So in this example, the count of roots is at least $\frac{4}{3}d$, which is bigger than $d+3=d+k$ once $m>3$.<|endoftext|> -TITLE: Role of stably free modules in algebraic geometry -QUESTION [7 upvotes]: For any ring $R$, a non-zero module $S$ is stably free if $S\oplus R^a$ is free ($a\geq 1$). This may be an overly vague question, but I am wondering in what contexts do stably free modules arise in algebraic geometry? -I know Serre asked whether projective modules (that turn out to be stably free) over $k[x_1,\ldots, x_n]$ ($k$ field) are free. This, of course, corresponds to asking whether vector bundles over the affine $n$-space $\mathbb{A}^n_k$ are trivial. However, this question was asked back in the 50s (eventually being answered that all projective modules are free!) and I am wondering what role stably free modules now play (or can play) in algebraic geometry. - -REPLY [7 votes]: This is just a long comment. -Stably free modules appear often in many questions regarding number of generators of modules over rings. For example, if $I\subset R$, a polynomial ring over a field in $n$ variables and $I$ an ideal, a question of Murthy is, can one say $I$ is generated by the same number of elements which generates $I/I^2$. Most of the known results use certain stably free module being free. -The universal example of a stably free module is that given by the presentation, -$0\to R\to R^n\to P\to 0$, where $R=k[x_i,y_i, 1\leq i\leq n]/\sum x_iy_i=1$ and the map $R\to R^n$ given by the unimodular row $(x_1,\ldots, x_n)$. If $n\geq 3$, $P$ is stably free and not free. Suslin proved that $(x_1^{a_1},\ldots, x_n^{a_n})$ gives a free module if $(n-1)!$ divides $\prod a_i$. There are topological proofs for the first statement, an algebraic proof of both these (the necessity of the condition of Suslin) uses Grothendieck-Riemann-Roch without denominators (the $(n-1)!$ should be a clue.) Similar principles can be used to show that there are affine varieties (say over complex numbers) of dimension $p+2$, $p$ a prime and stably free modules of rank $p$ which are not free, very far from what you can do with topology, these being all topologically trivial.<|endoftext|> -TITLE: Proof assistant for working in weaker foundations? -QUESTION [32 upvotes]: In some of my works I need to prove some results within the internal logic of categories with not much structures (like pretoposes or even just categories with finite limits). The kind of things I want to prove typically involve manipulating inductive type (like a natural number object, a type of binary tree, some W-types etc...) and prove some of their properties in these relatively weak categories. -Let me give some examples of the sort of results I'm interested in: - -If C is a cartesian category (or category with finite limits) with a parametrized natural number objects $N$ then $N \times N \simeq N$. -If C is a cartesian category with a (parametrized) natural number objects $N$, then $N$ is also a parametrized list object for $N$, and a parametrized finite tree objects. -If C is a pre-topos with parametrized list objects then essentially any kind of finitary free constructions (like free groups, free monoids, free left exact categories etc...) can be performed. -Which of these free construction can already be performed in a cartesian category/ a category with finite limits/an extensive category ? -If C is a a pretopos in which some (non finite) objects are exponentiable, and for which the corresponding W-types exists, then certain free infinitary constructions exists. - -etc... those are only examples, most of them are already well known results, but they are very typical of the kind of things I'm trying to do. Note that I want to work with "internal" proof, and not proving results externally directly in terms of such categories. -I believe that this sort of things is typically a good place to try to use a proof assistant as the proof have to be spelled out in a very high level of details anyway and one can easily makes mistake. -So I'm looking for a proof assistant that would be appropriate for this. -I already started experimenting in Coq. My current strategy is to just be careful about what I'm doing: for example, only use induction principle for propositions without quantifier, or only those authorized by the specific framework I'm working in... And of course not use anythings coming from a library, and only use very explicit tactics to avoid hiding possible problems. So I'm not really using the proof assistant as a definitive witness of validity of a proof, but only as somethings that makes every single step of a proof explicit enough so that I can tell immediately if it makes sense or not in the framework I'm interested in. -This already not too bad, but I was hopping to find a more precise/formal way to do this. - -Is there other proof assistant which have more flexible logical background ? -From my experience (but I'm not very familiar with the precise logical framework that Coq uses, so I can't makes what follows a formal statement, but if someone can confirm it it would be very helpful) everything that I can define in coq can be interpreted in a 'Stack' or 'Sheaves' semantics over the category I'm working in. So the only thing that I need to be careful, is to not apply an induction principle to construct functions into something that is not an object of my category. -So would there be a way in Coq, or in any other proof assistant, to say that I have a nice class of sets (which would be the representables), stable under some operations (corresponding to the structure I'm putting on my category) and that when I define inductive operations or use inductions it has to be restricted to things taking values in a set in this class. But I would still like to be able to use the nice machinery of "match" "induction" "fixpoint" that Coq offers. -Maybe a way to check afterward what kind of induction does a given proof uses ? -Any other suggestions on how to do this sort of things ? - -PS: I was unsure this question was suitable for MO or not. I had the impression it was "mathematical enough" but I do have too admit that a good answer could be something technical about the inner working of some proof assistant. So If you think there is a better place to ask this question, please tell me. - -REPLY [15 votes]: One possibility that's worth thinking about is to use a "meta-proof-assistant" like Twelf, which implements a meta-theoretic logical framework inside of which you can specify any "object language" you like. It's better-adapted to proving meta-theoretic properties about the object-language, but it can also be used to prove theorems in the object language, although it doesn't have nice syntactic sugar like match and fixpoint or support for tactics. -It might also be possible to implement the type theories you're interested in inside of Isabelle or something like it, which is supposed to be a "generic proof assistant", but I'm not very familiar with that. -Regarding Coq, I don't know of an "after the fact" way to check what kind of induction a proof uses. But if you're willing to call induction principles explicitly rather than using match/fix, you could use a "private inductive type" hack similar to the way that we implement HITs in HoTT. That is, postulate your family of "representable" types and its closure conditions, then define your inductive type as a Private Inductive in a module, prove a version of the induction principle that requires the target to be representable, and then export only that restricted induction principle.<|endoftext|> -TITLE: Obstructions to realizing a balanced presentation as a 3-manifold group -QUESTION [8 upvotes]: I am thinking of 3-manifolds as arising from Heegaard splittings which I am thinking about in terms of Heegaard diagrams. I know that 3-manifold groups are rather special in the class of all finitely presentable groups. -I know how to go from a Heegaard diagram ($g$ red curves and $g$ blue curves each forming a cut system on a genus $g$ closed orientable surface) to a presentation for $\pi_1$ - namely I have $g$-generators coming from the red handlebody $g$ relations coming from the disks attached along the blue curves. To find the relations I just orient all of the curves and the surface and for each blue curve I read off the word in the red curves (with the signs of the intersections determining the sign of the words). -I would like to know some explicit balanced group presentations that I can not get in this way - and I would like to see why I can't embed the curves if possible. Of course $< a ,b | [a,b] , [a,b]^2>$ for example fits the bill, since $\mathbb{Z}^2$ is not the fundamental group of a compact 3-manifold. But I would like a more "curves on surfaces" type explanation. -Ideally, I would like some explicit necessary and/or sufficient condition for a word or collection of words to be realizable by embedded curves as above. -Thanks! - -REPLY [3 votes]: $\newcommand{\ZZ}{\mathbb{Z}}$For your first question: the Baumslag-Solitar group $BS = BS(2,3)$ is not residually finite, so cannot embed in a three-manifold group. Thus $BS \times \ZZ$ has a balanced presentation, but is not a three-manifold group. A nice two-generator example is $\ZZ/3\ZZ \times \ZZ$. (For both examples, my proofs rely on geometrization. But see HJRW's comments below.) -For your second question: As Ryan says, the program Heegaard is our friend here. There is extensive documentation explaining the algorithm. It operates on the Whitehead graph of the given presentation. The algorithm performs a sequence Whitehead moves until it either gives up or it makes the Whitehead graph planar. If there is now a good "pairing" of the vertices then the algorithm has produced a Heegaard splitting of the desired three-manifold. -Finally: see page 47, item C.4 of the book 3-manifold groups for a list of interesting references.<|endoftext|> -TITLE: Spectrum of orthogonality graph (2) -QUESTION [9 upvotes]: The orthogonality graph, $\Omega(n)$, has vertex set the set of $\pm 1$ vectors of length $n$, with orthogonal vectors being adjacent. -I am only interested when $4|n$, since otherwise $\Omega(n)$ is empty or bipartite. I am keen to know the spectrum of $\Omega(n)$ - the eigenvalues and their multiplicities. In particular I am seeking the inertia of $\Omega(n)$ - that is the numbers of positive, zero and negative eigenvalues. Many thanks Clive - -REPLY [3 votes]: The graph $\Omega(n)$ is one of the relations of the Hamming association scheme $H(n,2)$, namely, the one corresponding to the Hamming distance $n/2$, see e.g. here. Its eigenvalues are given by the values of the Krawchuk's polynomials $K_k(n/2)$, as $0\leq k\leq n$, see details in my answer to your MO question 295493, -and multiplicities are just $\binom{n}{k}$, $0\leq k\leq n$. -For this one can e.g. compute that if $n$ is divisible by 4 then the number of 0 eigenvalues of $\Omega(n)$ is $2^{n-1}$.<|endoftext|> -TITLE: Rational approximations of $\sqrt{2}$ in $\mathbb{R} \times \mathbb{Q}_7$ -QUESTION [7 upvotes]: Note: this question was updated (2) after GNiklasch's answer was posted, and taking Gro-Tsen's comment into account. The initial question (1) dealt with $\mathbb{Q}_3$. - - Original post (1). Let's try to solve the equation $x^2 - 2 = 0$ with $x = \frac{a}{b} \in \mathbb{Q}$. We can't have $x^2 \neq 2$, so the best we can do is minimize $|x^2 - 2|$. Let's try to find an approximation that works over two different completions. Can we have this? -$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_3 \ll \frac{1}{a} $$ -I'm trying to write an $S$-adic approximate solution over two places $S = \{ 3, \infty\}$ and $x = \frac{a}{b} \mapsto (\frac{a}{b}, \frac{a}{b}) \in \mathbb{Q} \times \mathbb{Q} \subset \mathbb{R} \times \mathbb{Q}_3 $. What are the correct exponents? - -Edit (2). Perhaps i need to find a problem statement that has a solution. gro-tsen suggests I change $p=3$ to $p=7$ so that $\sqrt{2}\in \mathbb{Q}_7$. -$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_7 \ll \frac{1}{a} $$ -possibly I can leave the places the same and change the thing I'm approximating. -$\sqrt{7}\in \mathbb{Q}_3$ so perhaps I can find a rational number $\frac{a}{b}\in \mathbb{Q}$ such that -$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 7 |_3 \ll \frac{1}{a} $$ -Excuse me while I try to state an instance of weak approximation that's not vacuous. - -REPLY [12 votes]: Here is a full solution for the modified problem, inspired by Gro-Tsen's valuable comment. -1. There are infinitely many rational numbers $a/b\in\mathbb{Q}$ in lowest terms such that -$$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{1}{b}\qquad\text{and}\qquad -\left|\frac{a^2}{b^2}-2\right|_7\ll\frac{1}{b}.$$ -To see this, we work in $\mathbb{Z}[\sqrt{2}]$, the ring of integers of $\mathbb{Q}(\sqrt{2})$. In this ring, the rational prime $(7)$ splits as $(3+\sqrt{2})(3-\sqrt{2})$, while the totally positive units are $(3+2\sqrt{2})^m$. For any positive integer $n$, we can choose the positive integer $m$ so that -\begin{align*}a+b\sqrt{2}&=(3+2\sqrt{2})^m(3+\sqrt{2})^n\asymp 7^n,\\ -a-b\sqrt{2}&=(3-2\sqrt{2})^m(3-\sqrt{2})^n\asymp 1.\end{align*} -This is because $a^2-2b^2=7^n$ holds regardless of $m$. By basic arithmetic in $\mathbb{Z}[\sqrt{2}]$, the integers $a$ and $b$ are relatively prime to each other and to $7$ as well. Moreover, $a$ and $b$ are positive and of size $\asymp 7^n$. It follows that -$$\left|\frac{a^2}{b^2}-2\right|_\infty=\frac{7^n}{b^2}\asymp\frac{1}{b} -\qquad\text{and}\qquad -\left|\frac{a^2}{b^2}-2\right|_7=7^{-n}\asymp\frac{1}{b}.$$ -2. By modifying the above argument, we can achieve that -$$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{c}{b}\qquad\text{and}\qquad -\left|\frac{a^2}{b^2}-2\right|_7\ll\frac{c^{-1}}{b}$$ -for any constant $c>0$. This is essentially best possible, because it is straightforward to see that -$$ \left|\frac{a^2}{b^2}-2\right|_\infty\left|\frac{a^2}{b^2}-2\right|_7\geq\frac{1}{|b^2|_\infty|b^2|_7}\geq\frac{1}{b^2}.$$<|endoftext|> -TITLE: An algorithm determining whether two subgroups of a finitely generated free group are automorphic -QUESTION [6 upvotes]: In the book Lyndon, Schupp, Combinatorial Group Theory, P.30 in the edition from 2000 They mention an unpublished work by Waldhausen that is said to give an algorithm to determine whether two subgroups are automorphic given their free generators. I searched for papers written by Waldhausen But I didn't find it. Has Waldhausen or anyone else published a solution to this problem? Where can i find it? This is a copy of a post from StackExchange - -REPLY [7 votes]: According to a comment on the math.SE link, the OP wants to understand the question of deciding when two (finitely generated) subgroups of a given free group $F$ are equivalent by an automorphism of $F$. -I believe that this was first solved by Gersten: - -S. Gersten, On Whitehead’s algorithm, Bull. Am. Math. Soc. 10 (1984) 281– - 284. -Abstract. One can decide effectively when two finitely generated subgroups of a finitely generated free group $F$ are equivalent under an automorphism of $F$. The subgroup of automorphisms of $F$ mapping a given finitely generated subgroup $S$ of $F$ into a conjugate of $S$ is finitely - presented. - -Gersten states that the question asked by the OP was first raised by Whitehead. He also states that he settles this question of Whitehead, and does not cite any papers of Waldhausen. So I guess the paper of Waldhausen never appeared...<|endoftext|> -TITLE: Interesting (combinatorial) actions of the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ -QUESTION [6 upvotes]: I have read many times that it is crucial to understand the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) =: L$, in so far that some have stated (was it Richard Taylor ?) that this understanding is the main goal (or the very definition) of Number Theory (give or take). Anyway, in this question I am concerned with actions of this group. I know that there is, for instance, Grothendieck's "Dessins d'enfants" theory (although Deligne has stated that, at the point of writing and publishing his famous "projective line minus three points paper," the dessins had not helped that understanding very much at all). If one would want to define some action of $L$ on some space/set/geometry/etc., what would make that action interesting ? -Example given: I have never seen descriptions of $L$ on combinatorial geometries such as (axiomatic) projective planes, but (when) would such actions be interesting enough ? What are natural constraints on the geometry acted upon, so that we would have a nice action ? - -REPLY [6 votes]: You should read the paper -I. Bauer, F. Catanese, F. Grunewald: Faithful actions of the absolute Galois group on connected components of moduli spaces, Invent. Math. 199, No. 3, 859-888 (2015). ZBL1318.14034. -Quoting from MathSciNet review: - -The authors study the action of the absolute Galois group $\mathrm{Gal}(\bar{\mathbb{Q}}/ \mathbb{Q})$ on certain hyperelliptic curves, surfaces of general type and moduli spaces of surfaces of general type. -They prove a number of results on faithfulness of the action on moduli spaces and also provide a result on the existence of surfaces $X$ defined over number fields and Galois automorphisms $\sigma$ such that $X$ and its conjugate $X^{\sigma}$ (think of $X^{\sigma}$ as the variety cut out by the equations obtained from applying $\sigma$ to the coefficients of the equations of $X$ in some projective space) have nonisomorphic fundamental groups, in particular, fail to be diffeomorphic. -The existence of such varieties is classical and goes back to J.-P. Serre [C. R. Acad. Sci. Paris 258 (1964), 4194–4196; MR0166197]. The present paper even provides an infinite series of rather explicit examples.<|endoftext|> -TITLE: Is it possible to completely embed complete Heyting Algebras into upsets of a poset? -QUESTION [12 upvotes]: Let $H$ be a Heyting algebra. It is a well-known result that there is a partially ordered set (Kripke frame) X such that there is an embedding of Heyting algebras $f: H \to \mathsf{Up}(X)$, where $\mathsf{Up}(X)$ consists of the up-sets of $X$. Specifically, take $X$ to consist of the prime filters of $H$, ordered by inclusion, and let $f(a) := \{ P \in X | a \in P \}$. -Is it possible to modify the result like this: Let $H$ be a complete Heyting algebra. Does there exists a complete embedding $g: H \to \mathsf{Up}(X)$ for some partially ordered set $X$? By a complete embedding, I mean an embedding preserving arbitrary meets and joins. -The approach above does not seem to work as it might be the case that there are incomplete prime filters on the complete Heyting algebra. - -REPLY [2 votes]: Let me give yet another characterization of these kind of complete Heyting algebras. Although the proof is more involved and set-theoretic in nature, I think the approach is worth since the equivalence is expressed in terms of a distributivity property of the complete Heyting algebra that coincides with complete distributivity in case it is Boolean. It follows, in particular, that any complete Boolean algebra that is not completely distributive (i.e., any Boolean algebra that is not a powerset) is a counterexample. - -Proposition Let $H$ be a complete Heyting algebra. The following are equivalent: -1. H has a complete embedding into Up$(X)$ for some poset $X$. -2. Given a tree of arbitrary height $\gamma$ whose nodes are elements of $H$ such that: -$\bullet$ every node is the join of its immediate successors in the tree, -$\bullet$ every node at a limit level is the meet of its predecessors in the tree, -then whenever a set of nodes $S$ intersects every branch of the tree, the root $r$ is the join $\bigvee_{c \in S}c$. - -Property 2 has been considered in my paper "Infinitary first-order categorical logic" (to appear in the Annals of Pure and Applied Logic) under the name transfinite transitivity (since it can be generalized to a similar property for Grothendieck topologies). It can be seen to imply that $H$ is completely distributive (hence co-Heyting) and in the Boolean case it is precisely equivalent to it, but it is in general strictly stronger than that (as the counterexample of the interval $([0, 1], \leq)$ shows). One way to see the equivalence is via the condition $3$ in Emil's answer. -Proof sketch -Emil's $3$ implies 2. Write the root of the tree as the join of the completely join-irreducible elements below it. Any of these must be below one immediate successor $a$, and hence below one immediate successor of $a$, and so on. One can then define by transfinite recursion a branch of the tree any of whose nodes is above that completely join-irreducible. Since the choice of this latter is arbitrary, we are done. -2. implies Emil's $3$. Let $\delta$ be the cardinality of $H$, and let $\kappa=(2^{\delta})^+$. Consider the canonical well-ordering $f: \kappa \times \kappa \to \kappa$; it has the property $f(\beta, \gamma) \geq \gamma$. For each $a \in H$ let $\mathcal{C}(a)$ be the set of tuples $(b_\alpha)_{\alpha<\lambda}$ of elements of $H$ such that $a = \bigvee_{\alpha<\lambda}b_{\alpha}$. Assume without loss of generality that $\mathcal{C}(a)$ is well-ordered and has order type $\kappa$ (repeating tuples, if needed). Let $a \in H$ be fixed. Then we can build a tree of height $\kappa$ whose nodes are elements of $H$, and having $a$ as a root, by transfinite recursion as follows. Assuming that the tree is defined for all levels $\lambda<\mu$; we show how to define the nodes of level $\mu$. If $\mu$ is a successor ordinal $\mu=\alpha+1$, and $\alpha=f(\beta, \gamma)$, by hypothesis the nodes $\{p_i\}_{i\delta$, it stabilises at some node $c_b$. By construction, this $c_b$ is completely join-irreducible. Indeed, if $c_b=\bigvee_{\alpha<\nu}d_{\alpha}$, at some point of the recursion the successor of $c_b$ at some level will be $c_b \wedge d_{\alpha}$ for some $\alpha<\nu$, which implies that $c_b \leq d_{\alpha}$. -Finally, by 2., $a$ is the join $\bigvee_{b}c_b$, as we wanted to prove.<|endoftext|> -TITLE: Is there a name for this theorem? -QUESTION [6 upvotes]: I wonder if there is a name or reference for the following fact. It is not the proof I am looking for. -Let $s_1, s_2, ...,s_n$ be non-negative real numbers ordered in a non-increasing way. Let $b_1,b_2,...,b_n$ be non-negative real numbers ordered in a non-decreasing way (so opposite of the $s_i$). -Then the average value $$\frac{s_1+s_2+\cdots +s_n}{n}$$ is at least as large as the -weighted average $$\frac{b_1 s_1+\cdots+b_n s_n}{b_1+\cdots+b_n}.$$ -Thanks! - -REPLY [19 votes]: This is known as Chebyshev's Sum Inequality. (I've only ever seen it used in the context of competition math, but Wikipedia gives a reference to Hardy-Littlewood-Polya.) - -REPLY [4 votes]: It is called often called "Chebyshev's other inequality". And this makes more sense than the name "Chebyshev's sum inequality" proposed by Wikipedia, since even Wikepedia knows that there is a version for integrals.<|endoftext|> -TITLE: If $M \otimes -$ is continuous, why is $M$ f.g. projective? Alternative proof -QUESTION [7 upvotes]: Let $R$ be a commutative ring and $M$ be some $R$-module such that $M \otimes -$ is continuous (i.e. preserves all limits). Then one can show that $M$ is f.g. projective. -One way to prove this is to use adjoint functor theorems (anyone you like): We obtain a left adjoint, which has the form $M^* \otimes -$ by Eilenberg-Watts, and then one obtains that $M$ is dualizable, etc. -Is there also a proof which is more direct or elementary? We can also use other characterizations, such as f.g. projective = f.p. flat. It is clear that $M$ is flat (since $M \otimes -$ preserves finite limits), but I do not know how to approach f.p. - -REPLY [9 votes]: I am posting the comment above as one answer. Kernels are examples of limits. So if the functor is continuous, then $M$ is flat. Next, consider the natural transformation $M\otimes_R \text{Hom}_R(-,R)\Rightarrow \text{Hom}_R(-,M)$. Both functors convert colimits to limits, and the natural transformation is an isomorphism on finitely presented objects. Every object is a colimit of finitely presented objects. Thus, the natural transformation is an isomorphism. In particular, consider $\text{Id}_M$ in $\text{Hom}_R(M,M)$. That element is in the image of the natural transformation. Thus, $M$ is finitely presented.<|endoftext|> -TITLE: Space of sections of a fibration under weak homotopy equivalence -QUESTION [5 upvotes]: If I have two (Serre-)fibrations over the same base, and a weak equivalence of the total spaces that is also a map over the base, could I hope that the induced map on the spaces of sections would also be a weak equivalence? -If I restrict to CW-spaces, this is fairly easy, but I don't really know how to handle the space of sections if only weak equivalences are available. - -REPLY [4 votes]: As a complement to Chris Schommer-Pries answer above, the result holds when the base is a CW complex, as can be seen by a (possibly transfinite induction) on the cells. Here's a sketch. -Clearly, the result is true when $B$ is a point. -For a subcomplex $A\subset B$ let $\Gamma(E_i|A)$ be the space of sections of the restriction $E_{i|A} \to A$ for $i=1,2$. Suppose $A\subset C \subset B$, where $C = A\cup D^j$ is the result of attaching a single cell to $A$. Then we have a pullback square -$\require{AMScd}$ -$$ -\begin{CD} -\Gamma(E_i|C) @>>> \Gamma(E_i|A)\\ -@VVV @VVV \\ -\Gamma(E_i|D^j) @>>> \Gamma(E_i|S^{j-1}) -\end{CD}$$ -which is also a homotopy pullback because the horizontal arrows are Serre fibrations. This gives a long exact sequence in homotopy groups -$$ -...\to \pi_n(\Gamma(E_i|C)) \to \pi_n(\Gamma(E_i|D^j)) \oplus -\pi_n(\Gamma(E_i|A)) \to -\pi_n(\Gamma(E_i|S^{j-1}))\to ... -$$ -(where I am being a bit sloppy with the notation for $n \le 1$). -The square for $i=1$ maps to the square for $i=2$. -Inductively, assume the result is true for $A$. The result is clearly true for $S^{j-1}$ and $D^j$ since in these cases where are dealing with a homotopically trivial fibration of the form $F \times X \to X$ where -$X= D^j$ or $S^{j=1}$. Then by the five lemma applied to the long exact sequence, we obtain the result for $C$. This gives the inductive step.<|endoftext|> -TITLE: Herbrand-Ribet and Mazur-Wiles for function fields -QUESTION [6 upvotes]: Is there a version of Herbrand-Ribet or Mazur-Wiles (relating divisibility of class groups to special values of L-functions) for functions fields (over finite fields)? -Probably the proofs would have to be very different since we don't have a nice tool like modular forms to construct representations and extensions with nice properties out of (or do we?). - -REPLY [4 votes]: (I was hoping somebody else would answer this, because function fields are not really my area and I hoped I would learn something from the answer; but nobody seems to be biting, so...) -Iwasawa theory over function fields definitely exists, and in many ways it's easier than number-field Iwasawa theory -- there are more nice tools available, such as the Grothendieck--Lefschetz trace formula, which aren't there in the number field setting. -For instance, here is a paper of Goss and Sinnott from the 1980s which (among many other results) proves an analogue of Herbrand--Ribet for the class groups of function field extensions arising from Drinfeld modules.<|endoftext|> -TITLE: Constructive proof of existence of non-separable normed space -QUESTION [9 upvotes]: I am looking for a constructive proof of one of the following two statements. If they are not constructively provable, I would be very thankful for an explanation as to why that is so (i.e., at which point in a proof must non-constructive means be employed?). - -There exists a normed space X such that for all Y $\subset$ X, if Y is denumerable, then Y is not dense in X. -There exists a normed space X such that for all Y $\subset$ X, if Y is dense in X, then Y is not denumerable. - -I'd consider a proof constructive if it includes no applications of the: - -Law of Excluded Middle: $\phi$ $\lor$ $\neg$$\phi$ -Law of Double Negation Elimination: $\neg$$\neg$$\phi$ $\rightarrow$ $\phi$ -Axiom of Choice or any of its equivalents (Zorn's Lemma, etc.). - -Suggestions would be much appreciated. - -REPLY [3 votes]: Andrew Swan and I proved that in the function realizability topos every metric space is separable (and that every object with decidable equality is countable). Therefore, it is not possible to prove constructively that a non-separable metric space exists. This result strengthens the answer by Matt Frank. -The note is available as arxiv 1804.00427.<|endoftext|> -TITLE: References requestion : Pretopos are algebras for a composed monad? -QUESTION [6 upvotes]: Unless I'm mistaken the "Free completion under finite limits monad" $C \mapsto C^{lex}$ and the "free co-completion monad" $C \mapsto \widehat{C}$ (the categories of small presheaves) satisfies a distributivity law, and the algebra for the composed monad $C \mapsto \widehat{C^{lex}}$ are the infinitary pretopos. -Has this been written out somewhere ? I have a vague memories of seeing something like that some time ago (or maybe it was for completion under co-products and extensive categories), but I have not been able to find it anywhere... -Is there a similar statement for finitary pretoposes ? -I assume $C \mapsto \widehat{C}$ has to be replaced by completion under finite co-products and some co-equalizer, but that seems to be a little bit more tricky. I guess in technical terms what I want to know is: "is it possible to write the free pre-topos monad as the composite of a co-KZ-monad and a KZ-monad satisfying a distributivity law ?" - -REPLY [3 votes]: I think Mike is right: in Garner and Lack, Lex colimits, Proposition 2.3 and the discussion following it show that the free cocompletion 2-monad lifts to the 2-category of finitely complete categories, which is equivalent to the existence of a distributive law between these two 2-monads. Since the 2-monad for finite limits is colax-idempotent and the 2-monad for small colimits is lax-idempotent, one doesn't need to appeal to the theory of fully weak distributive laws between 2-monads for this -- the KZ theory will suffice. -Then Proposition 2.5 shows that the algebras for the composite 2-monad are precisely the infinitary pretoposes.<|endoftext|> -TITLE: Density of intersection with shifted sets -QUESTION [5 upvotes]: Given a subset $S$ of the positive integers $\mathbf{N}$, let $\mathrm{d}^\star(S)$ be its upper asymptotic density, that is, -$$ -\mathrm{d}^\star(S)=\limsup_{n\to \infty}\frac{|S \cap [1,n]|}{n}. -$$ -Also, for each integer $k \ge 0$, set $S+k:=\{s+k:s \in S\}$. - -Question. Do there exist $X,Y \subseteq \mathbf{N}$ such that $\mathrm{d}^\star(X)>0$, $\mathrm{d}^\star(Y)>0$, and - $$ -\mathrm{d}^\star(X \cap (Y+k))=0 \,\,\,\text{ for all }k\ge 0? -$$ - - -Ps. The answer is positive (constructively, modulo my mistakes) for the following finite-analogue: "Do there exist infinite sets $X,Y \subseteq \mathbf{N}$ such that $X \cap (Y+k)$ is finite for all $k\ge 0$?" -To this aim, let $X$ be the set of squares and $Y$ be the set of cubes which are not also squares. Then, by a result of Siegel, the equation $|x-y|=k$ has finitely many solutions with $x \in X$ and $y \in Y$ for every fixed nonzero integer $k$. -Edit: I just discovered an excellent proof of the finite-analogue without translation invariance in Albiac-Kalton "Topics in Banach Space Theory" (lemma 2.5.3), namely: -"There exists an uncountable family $(A_i)$ of infinite sets of $\mathbf{N}$ such that $A_i \cap A_j$ is finite for all $i\neq j$." -Proof:Identify $\mathbf{N}$ with $\mathbf{Q}$ and, for each irrational number $\theta$ let $(a_{\theta,n})$ be a sequence of rationals converging to $\theta$. Then the uncountable family $$\{\{a_{\theta,n}:n\in \mathbf{N}\}: \theta \in \mathbf{R}\setminus \mathbf{Q}\}$$ satisfies the hypothesis. - -REPLY [6 votes]: The answer is "yes". -Let $X = \bigcup [4^n, 2 \cdot 4^n)$ and $Y = \bigcup [2 \cdot 4^n, 3 \cdot 4^n)$. Then $d^\star(X), d^\star(Y) \geqslant \tfrac{1}{4}$, but $X \cap (Y + k)$ is a finite set for every $k \geqslant 0$.<|endoftext|> -TITLE: Can I check the accessibility of a functor on directed colimits of presentable objects? -QUESTION [5 upvotes]: Let $F: \mathcal{K} \to \mathcal{C}$ be a functor between $\lambda$-accessible categories, you can assume $\mathcal{C}$ to be Set if needed. -Is it true that $F$ is $\lambda$-accessible if and only if it preserves $\lambda$-directed colimits of $\lambda$-presentable objects? - -REPLY [6 votes]: Yes, and this doesn't require any assumption on $\mathcal C$. This follows from the following basic fact: if $I$ is filtered and $(A_i)_{i\in I}$ is a diagram of categories with colimit $A$, then the colimit of any functor $F: A\to\mathcal C$ may be computed as -$$ -colim_{a\in A} F(a) = colim_{i\in I}(colim_{a\in A_i}F_i(a)), -$$ -where $F_i$ is the restriction of $F$ to $A_i$ (provided the inner colimits all exist). -Write $\mathcal K=Ind_\lambda(\mathcal K_0)$. Every $X\in \mathcal K$ is then the colimit of the forgetful functor $\mathcal K_0/X \to \mathcal K_0\subset\mathcal K$, and $\mathcal K_0/X$ is $\lambda$-filtered. Let $I$ be $\lambda$-filtered and let $(X_i)_{i\in I}$ be a diagram in $\mathcal K$ with colimit $X$. Then $colim_i \mathcal K_0/X_i = \mathcal K_0/X$ and so - $$ - colim_{i} F(X_i) = colim_{i} (F(colim_{Y\to X_i} Y)) = colim_i (colim_{Y\to X_i} F(Y)) = colim_{Y \to X} F(Y) = F(colim_{Y\to X} Y)=F(X). - $$<|endoftext|> -TITLE: Is there any explicit description of the maximal totally ramified extension of $\mathbb{Q}_p$? -QUESTION [6 upvotes]: It is well known that the maximal unramified extension of $\mathbb{Q}_p$ can be extended by adding the roots of unity of order prime to $p$. Is there any explicit description of the maximal totally ramified extension of $\mathbb{Q}_p$? - -REPLY [10 votes]: A composite of totally ramified extensions need not be totally ramified: -Example 1. (As per LSpice's suggestion) Consider the extensions $\mathbb Q_p(\sqrt{p})$ and $\mathbb Q_p(\sqrt{\varepsilon p})$, where $\varepsilon \in \mathbb Z_p^\times$ is a nonsquare unit (corresponding, for example, to any lift of a generator of $\mathbb F_p^\times$). Then the compositum $\mathbb Q_p(\sqrt{p},\sqrt{\varepsilon p})$ contains the unramified extension $\mathbb Q_p(\sqrt{\varepsilon})$. -Example 2. Consider the extension $\mathbb Q_p \subseteq \mathbb Q_p(\sqrt[n]{p})$, where $n = p^r - 1$. Then the Galois closure of this extension contains all $n$-th roots of unity. But $\mathbb Q_p \subseteq \mathbb Q_p(\zeta_{p^r-1})$ is the unique degree $r$ unramified extension. Thus, we see that the Galois closure of a totally ramified extension need not even be totally ramified. -So there is no such thing as the maximal totally ramified extension. You could in principle still construct some maximal totally ramified extension (i.e. no further extension is totally ramified), but as far as I can tell these fields are not very explicit. By the second example above, it is not a Galois extension of $\mathbb Q_p$. -On the other hand, maximal abelian totally ramified extensions are very well studied: Lubin–Tate theory gives a relatively explicit construction of an abelian field extension $K \to K^{\operatorname{LT}}$ that is linearly disjoint from $K^{\operatorname{ab}}_{\operatorname{nr}}$ such that their compositum equals $K^{\operatorname{ab}}$. Thus, $K^{\operatorname{LT}}$ plays the role of a maximal abelian totally ramified extension in a relatively strong sense. The most explicit case is $K=\mathbb Q_p$; in this case we have $K^{\operatorname{LT}} = \mathbb Q_p(\zeta_{p^\infty})$.<|endoftext|> -TITLE: Fundamental group of a Lie group -QUESTION [10 upvotes]: Is it true that the class of isomorphism classes of fundamental groups of Lie groups coincides with the class of isomorphism classes of finitely generated abelian groups? - -REPLY [23 votes]: Yes, these two classes coincide. One direction splits into two: for a Lie group $G$, - -the fundamental group is abelian: this is already answered here this MO question, and -the fundamental group is finitely generated. This is equivalent (passing to the universal covering of the unit component) to the property that every discrete central subgroup of a connected Lie group is finitely generated, and this is answered here. - -Conversely any finitely generated abelian group is fundamental group of a Lie group. By the classification of finitely generated abelian groups, it is enough to consider cyclic groups (and then use products). Indeed $\mathbf{Z}$ is isomorphic to the fundamental group of the circle group $\mathbf{R}/\mathbf{Z}$, and $\mathbf{Z}/n\mathbf{Z}$ is isomorphic to the fundamental group of $\mathrm{PSL}_n(\mathbf{C})$.<|endoftext|> -TITLE: Number of solutions for the inequality with square roots -QUESTION [6 upvotes]: Let $M$ be some large real number and $\delta>0$. I would like to estimate the number of solutions for the inequality -$$|\sqrt{n_1}+\sqrt{n_2}-\sqrt{n_3}-\sqrt{n_4}|<\delta\sqrt{M},$$ -where $n_i$ are positive integers with $MM/\sqrt{2}$. We conclude that -$$N(M,\delta)>\frac{2\delta}{\sqrt{128}}\cdot\frac{6M^4}{25}>\frac{\delta M^4}{24}.$$ -On the other hand, it is obvious that $N(M,\delta)\geq M^2$, hence we proved for $\delta<1$ that -$$N(M,\delta)>\frac{M^2+\delta M^4}{25}.$$<|endoftext|> -TITLE: Asymptotic distribution of $\lambda_1$ under the $z$-measure for partitions -QUESTION [11 upvotes]: The following question about $z$-measures on Young diagrams came up in some ongoing work with Ofir Gorodetsky. I recall the background and then state our question below in the box. -For parameters $z$ and $z'$, define a measure on $\lambda \vdash n$ with the weights -$$ -M_{n,z,z'}(\lambda):= \frac{(\dim \lambda)^2}{n!\, (z z')_n} \prod_{\square \in \lambda} (z + c(\square))(z' + c(\square)). -$$ -Here $\dim \lambda$ is the number of standard Young tableaux of shape $\lambda$, the product is over all squares $\square$ in the diagram of $\lambda$, and $c(\square) := \mathrm{column}(\square) - \mathrm{row}(\square)$ is the content of $\square$. $(t)_n := t (t+1)\cdots (t + n-1)$. Definitions of the terminology used here can be found in e.g. vol. II of Stanley's Enumerative Combinatorics. Kerov discovered the remarkable fact that these weights induce a probability measure: -$$ -\sum_{\lambda \vdash n} M_{n, z, z'}(\lambda) = 1. -$$ -(At any rate they sum to $1$; they are a probability measure if all of them are non-negative.) Some of the proofs of this fact can be found for instance here or in section 8 here. -Let $\lambda^{(z,z',n)}$ be a random partition of $n$ according to these weights. We are interested in the limiting distribution of $\lambda_1$ under these weights in the case that $z = z' = \alpha > 0$. For $c \in [0,1]$, define -$$ -F_\alpha(c) := \lim_{n\rightarrow\infty}\mathbb{P}(\lambda_1^{(\alpha, \alpha, n)} \leq cn) = \lim_{n\rightarrow\infty}\sum_{\substack{\lambda \vdash n \\ \lambda_1 \leq cn}} M_{n,z,z'}(\lambda). -$$ -This limit is known to exist (and in fact converges to the distribution of $\alpha_1$ of the Thoma complex under the spectral z-measure; see this paper of Borodin). There are a variety of formulas characterizing the spectral z-measure in that paper (e.g. Theorem 2.2.1 for correlation functions of the Thoma point process), but we are interested in positivity of $F_\alpha(c)$ and have not been successful in applying these formulas to this. More exactly, we are interested in the following question: - -Question: Is it true for each $\alpha \in (0,1)$, that $F_\alpha(c) > 0$ for all $c > 0$? - -We are also interested in the behavior of $F_\alpha(c)$ when $\alpha > 1$ (we really understand matters completely only when $\alpha$ is a positive integer), and any information relevant would be interesting, but the question above is the one that is really pressing. -Below are some approximate graphs (generated from $M_{n,\alpha,\alpha}$ with $n=45$) of $F_\alpha(c)$ for $\alpha = .01, .1, .3, .5,$ and $.7$. The graph for $\alpha = .01$ is at the top and $\alpha = .7$ is on the bottom. $c$ of course is the horizontal axis. - -(The apparent changes in convexity for $c$ near $1$ in the bottom two graphs are just a remnant of polynomial interpolation and should be ignored.) - -REPLY [4 votes]: A positive answer to Brad and mine's question is the main result of a recent paper of Olshanski, titled "The Topological Support of the z-Measures on the Thoma Simplex" (link). His proof is based on a previous work of his, as well as a recent paper of Korotkikh. -Interestingly, this question has (conjecturally) a simple number-theoretical interpretation, in terms of the distribution of the $z$th divisor function in short intervals and in arithmetic progressions. See Conjecture 4 in Brad and mine's preprint.<|endoftext|> -TITLE: Reference for flatness in complex-analytic geometry -QUESTION [8 upvotes]: What is a good reference for flat morphisms of complex-analytic spaces? (The book by Grauert and Remmert doesn't treat them). -Topics I'm interested in: openness of flat maps, descent for coherent analytic sheaves. - -REPLY [9 votes]: A reference for the fact that flat maps are open in the complex-analytic category is Theorem 2.12, p. 180 in -C. Bănică, O. Stănăşilă: Algebraic methods in the global theory of complex spaces. -See also my answer to MO question 41158.<|endoftext|> -TITLE: Conditional Expectation for $\sigma$-finite measures -QUESTION [7 upvotes]: Someone knows of some definition or reference of how to define conditional expectation for a measure space with $\sigma$-finite measure. -I think it should be as follows: -Let $(X,\mathcal{B},\nu)$ be a measure space and let $\mathcal{F}\subset\mathcal{B}$ a sub$-\sigma-$algebra, such that $\nu$ is $\sigma-$finite in $\mathcal{F}$. Then for all $f\in L^1(X,\mathcal{B},\nu)$ there exists $g\in L^1(X,\mathcal{F},\nu|_{\mathcal{F}})$ such that -$$\int_{E}fd\nu=\int_Egd\nu|_{\mathcal{F}},\qquad\forall E\in\mathcal{F};$$ -then $g=:\mathbb{E}_{\nu}[f|\mathcal{F}]$ is called the conditional expectation of $f$ given $\mathcal{F}$. -Is this the correct way to define conditional expectation? There is another way to define it without requiring the hypothesis that $\nu$ be $\sigma$-finite in $\mathcal{F}$? - -REPLY [6 votes]: One can define a reasonable notion of conditional expectation -for arbitrary localizable measurable spaces, not necessarily σ-finite. -This is explained in great detail in the answer to -Is there an introduction to probability theory from a structuralist/categorical perspective? -The “pushforward for L_1-spaces” mentioned there is precisely the conditional expectation. -Let me offer a few comments and expand this a little bit. -First, one doesn't really need a measure μ to talk about conditional -expectations, only a measure class [μ], -or equivalently, a σ-ideal N of negligible sets (alias sets of measure 0). -Given a set X with a σ-algebra M of measurable subsets -and a σ-ideal N of negligible subsets, -one can define the set of (finite complex-valued) measures on (X,M,N) -as the set of additive functions M→C that vanish on N. -Infinite measures can be defined using [0,∞] instead of C. -Furthermore, given a faithful finite measure μ on (X,M,N), -one can identify the set of μ-integrable functions f -with the set of finite measures ν via the isomorphism f↦fν -supplied by the Radon-Nikodym theorem. -Given a morphism (X,M,N)→(X',M',N'), one can pushforward a finite -measure on (X,M,N) and get a finite measure on (X',M',N'), -by taking the preimage of a measurable subset of X' and computing -its measure as a subset of X. -This is the conditional expectation. -In particular, in the notation of the original post, -one uses the morphism (X,B,N)→(X,F,N), where N is the σ-ideal -of sets with ν-measure zero. -Although both the domain and codomain have the same underlying set X, -it's best to think of them as different spaces, -and the pushforward can then be thought of as the fiberwise integration. -(For instance, take X=[0,1]×[0,1] with the Borel σ-algebra -and the σ-algebra of “vertical” Borel sets. The resulting morphism -will be isomorphic to the projection [0,1]×[0,1]→[0,1], -and the pushforward will be the fiberwise integration map.) -Let's illustrate the above construction on the example by Iosif Pinelis. -The product of the infinite measure ν and the function f -is the finite measure fν. -Its pushforward along the map (X,B,∅)→(X,F,∅) can be computed as follows. -The codomain is isomorphic to the measurable space consisting of a single point -(which is how one should think about it geometrically). -By definition, measures on (X,F,∅) can be identified with complex numbers, -and the pushforward of a finite measure on (X,B,∅) simply computes -the measure of X. -Thus, in this example the conditional expectation is the measure -on (X,F,∅) that assigns the number 2 (the sum of 1/2^x) to X. -Of course, most analysts are more comfortable with functions rather than measures -and prefer to use the Radon-Nikodym theorem with respect to the pushforward of ν -to convert the pushforward of fν to an integrable function. -However, this is only possible if the pushforward of ν -is a faithful semifinite measure, whereas in this example -it is a purely infinite, nonsemifinite measure. -However, the failure of the Radon-Nikodym theorem doesn't mean -that the conditional expectation doesn't exist, but rather -that it exists only as a finite measure that cannot be -converted to an integrable function. - -REPLY [3 votes]: $\newcommand{\N}{\mathbb N} -\newcommand{\R}{\mathbb R} -\newcommand{\B}{\mathcal B} -\newcommand{\F}{\mathcal F} -\newcommand{\la}{\lambda} -\newcommand{\si}{\sigma} -\newcommand{\Si}{\Sigma} -\renewcommand{\c}{\circ} -\newcommand{\tr}{\operatorname{tr}}$ -The definition you quoted is correct. -However, there can be no reasonable notion of the conditional expectation without the sigma-finiteness condition, even in the discrete setting. E.g., let $X=\N$, $\B=2^\N$, and let $\F$ be any sigma-algebra over $\N$ containing an infinite atom $A\subseteq\N$; for instance, one may take $\F=\{\emptyset,\N\}$, with $A=\N$. Let $\nu$ be the counting measure on $\B=2^\N$, and let $f(x)=1/2^x$ for $x\in\N$. Then $E_\nu f=1\in\R$. -However, on the atom $A$ one cannot reasonably ascribe any value to the conditional expectation $E_\nu(f|\F)$, because such a value (say $v$) could reasonably be only the $\nu$-average of $f$ on $A$. Indeed, if you take $v=0$, this would imply $\int_A f\,d\nu=0$, which is false; if you take $v\ne0$, this would imply $|\int_A f\,d\nu|=|v|\nu(A)=\infty$, which is also false. -The problem here is that, while the measure $\nu$ is sigma-finite, its restriction $\nu|_\F$ to $\F$ is not.<|endoftext|> -TITLE: Etale cohomology of localizations of henselian rings -QUESTION [9 upvotes]: Let $R$ be a ring (say noetherian of finite Krull dimension, possibly with additional hypotheses) henselian along the ideal $(p)$, and let $\hat{R}$ be the $p$-adic completion. Is it true that the étale cohomology of $R[1/p]$ and $\hat{R}[1/p]$ with mod $p$ coefficients coincide? I believe that this should be true (for $K$-theoretic reasons), but I was wondering if there is a direct argument. - -REPLY [13 votes]: TL;DR: Your expectation is right. In fact, there is a third object to compare with $R[1/p]$ and $\hat R[1/p]$, the affinoid rigid space ${\rm Spf}(\hat R)^{\rm rig}$. The cohomology comparison is given by the Gabber-Fujiwara theorem: see Corollary 6.6.4 in [1]. -Let us consider the following general setup: let $(A, I)$ be a henselian couple with $A$ noetherian (i.e. $A$ is henselian along $I$), and let $\hat A$ be the $I$-adic completion. We set $$X={\rm Spec}\, A, \quad \hat X = {\rm Spec}\, \hat A,$$ -$$U= X - V(I), \quad \hat U = \hat X - V(\hat I) \quad \text{ where }\hat I = I\cdot \hat A.$$ -Let $\varepsilon\colon \hat U\rightarrow U$ be the canonical map. -The following is a classical theorem of Elkik. - -Theorem 1 (Cor. p. 579 in [2]). The restriction functor - $$ - \varepsilon^*\colon(\text{finite etale covers of }U)\to (\text{finite etale covers of }\hat U) -$$ - is an equivalence of categories. - -Equivalently, for any finite group $G$, the restriction map $H^1(U, G)\to H^1(\hat U, G)$ is bijective, or $\pi_1(\hat U)\to \pi_1(U)$ is an isomorphism. -The Gabber-Fujiwara theorem (see [1], Cor. 6.6.3) is a far-reaching generalization of this, treating the higher cohomology of arbitrary torsion etale sheaves. Actually, the theorem compares the cohomology of $U$ to the cohomology of the rigid space $U^{\rm rig} = {\rm Spf}(\hat A)^{\rm rig}$. Whatever this object is, it is clear from the notation that $U^{\rm rig}$ depends only on $(\hat A, \hat I)$, and in particular we get the following as a corollary (see [1], Cor.6.6.4): - -Theorem 2 (The formal base change theorem). Let $\mathscr{F}$ be an etale sheaf of sets (resp. torsion groups, resp. torsion abelian groups) on $U$. Then the pull-back map - $$ \varepsilon^*\colon H^q(U, \mathscr{F})\longrightarrow H^q(\hat U, \varepsilon^* \mathscr{F}) $$ - is an isomorphism for $q=0$ (resp. for $q\leq 1$, resp. for $q\geq 0$). - -In particular, at least in good cases (finite cohomological dimension, geometrically unibranch) the Artin-Mazur etale homotopy types of $U$ and $\hat U$ are equivalent. (Disclaimer: as written, [1] treats only abelian cohomology, but I'm sure the nonabelian statements can be easily deduced from Elkik). -Addendum. A forthcoming book by Abbes (a sequel to [3]) is expected to contain a new proof of the Gabber-Fujiwara theorem, based on Gabber's affine analog of the proper base change theorem. -Curiosity. In fact in the $p$-adic situation you describe, both cohomology groups agree with the cohomology of the fundamental group: Theorem 6.7 in here states that for every noetherian $\mathbf{Z}_{(p)}$-algebra $A$ such that $(A, pA)$ is a henselian pair, the scheme $X={\rm Spec}\, A[1/p]$ is a $K(\pi, 1)$. For $\mathbf{F}_p$-coefficients, this can be deduced (via Gabber-Fujiwara) from earlier work of Scholze (Theorem 4.9 in [4]). -[1] Fujiwara, K.: Theory of tubular neighborhood in étale topology. Duke Math. J. 80 (1), 15–57 (1995). -[2] Renee Elkik, Solutions d’equations a coefficients dans un anneau henselien, Ann. Sci. Ecole Norm. Sup. (4) 6 (1973), 553–603 (1974). MR 0345966 (49 #10692) -[3] Abbes, A.: Éléments de géométrie rigide. Volume I. Progress in Mathematics, vol. 286, Birkhäuser/Springer Basel AG, Basel (2010). -[4] Scholze, P.: $p$-adic Hodge theory for rigid analytic varieties. Forum Math. Pi 1, e177 (2013)<|endoftext|> -TITLE: All polynomials are the sum of three others, each of which has only real roots -QUESTION [28 upvotes]: It was asked at the Bulletin of the American Mathematical Society Volume 64, Number 2, 1958, as a Research Problem, if a Hurwitz polynomial with real coefficients (i.e. all of its zeros have negative real parts) can be divided into the arithmetic sum of two or three polynomials, each of which has positive coefficients and only nonpositive real roots. -I would like to know if the following problem is known and how it can be solved: - -Can any polynomial with complex coefficients and degree $n$ be divided into the arithmetic sum of three complex polynomials, each of which has degree at most $n$ and only real roots? - -Any help would be appreciated. - -REPLY [23 votes]: To address the original question about the polynomials with complex coefficients. -Given a polynomial $P\in\mathbb C[x]$ of degree $n$, write $P=Q+iR$ with $Q,R\in\mathbb R[x]$, and fix arbitrarily a polynomial $S\in\mathbb R[x]$ of degree $n$ with all roots real and pairwise distinct. As observed in Eremenko's answer, for any $\varepsilon\ne 0$ sufficiently small in absolute value, the polynomials $Q_\varepsilon:=S+\varepsilon Q$ and $R_\varepsilon:=S+\varepsilon R$ will also have all their roots real, and then -\begin{align*} - P &= Q+iR \\ - &= \varepsilon^{-1}(Q_\varepsilon-S)+i\varepsilon^{-1}(R_\varepsilon-S) \\ - &= \varepsilon^{-1}Q_\varepsilon+i\varepsilon^{-1}R_\varepsilon-(1+i)\varepsilon^{-1}S -\end{align*} -is the required representation of $P$ as a sum of three polynomials of degree at most $n$ with all their roots real.<|endoftext|> -TITLE: A sheaf is a presheaf that preserves small limits -QUESTION [18 upvotes]: There is a common misconception that a sheaf is simply a presheaf that preserves limits. This has been discussed here before many times and I believe I understand it well enough. -However when reading Lurie's DAGVII he goes on to define a sheaf of spectra on an $\infty$-topos $\mathfrak X$ as a presheaf $\mathcal O:{\mathfrak X}^{op}\to \mathsf {Sp}$ which preserves small limits. -Why can the higher analogue of sheaves of rings be defined like this? My guess is that, because it is higher, it sorts out whatever problems you get when defining a normal sheaf like that. But I am seriously clueless on this matter and I would love for some helpful explanations. - -REPLY [20 votes]: This has nothing to do with $\infty$-categories, but with the fact that we look at the full topos and not an arbitrary site of definition: -Theorem: If $T$ is a (Grothendieck) ($1$-)topos, then a "sheaf of set" on $T$ is the same as a functor from $T^{op}$ to $Set$ sending colimits in $T$ to limits in Sets. -Sketches of Proof: Sheaf of sets on T, mean representable so they clearly satisifes this condition. Conversely, if something satisfies this conditons then if you restrict it to a site of definition of T, you get a sheaf on this site, you can then check that your functor is isomorphic to the representable at this sheaf. -You can replace set by group, or any reasonable category (it has to have limits and be well behaved enough so that sheafication exists...) Also this kind of theorem is not true just for $T$ a topos, but in fact for $T$ any presentable category. (this follows from the special adjoint functor theorem... although I think its more like a lemma that you need to prove the special adjoint functor theorem) -But this of course does not work if you replace $T$ by some arbitrary site of definition of $T$.<|endoftext|> -TITLE: $(\infty,2)$-Categorical Analogue of the Local Nature of Equivalences -QUESTION [11 upvotes]: It is well known that, for two functors $F,G : I \to C$ for $I,C$ some $\infty$-categories, the property that a map $\phi: F \to G$ is an equivalence can be checked locally on $I$. Namely, if $\phi(i) : F(i) \to G(i)$ is an equivalence for every $i \in I$, then $\phi$ is an equivalence as a morphism in the -$\infty$-category $Fun(I,C)$. -Is the same true for adjointability? Namely: -Let $I$ be an $\infty$-category, and let $F,G : I \to Cat_\infty$ be two functors into the $(\infty,1)$ category of $\infty$-categories. Let $\phi : F \to G$. -Suppose that, for each $i \in I$, $\phi(i) : F(i) \to G(i)$ admits left adjoint -$L_{\phi(i)}$. Then, there nessecarily exists a map $\psi: G \to F$ restricting to $L_{\phi(i)}$ on every object? Is it essentially unique? does it satisfy some relative version of the property satisfied by left adjoints? -My motivation for believing in it is that the space of choices of the Left adjoint is contractible, so there should be no obstructions to glue them over $I$. - -REPLY [7 votes]: FWIW, there is nothing "truly $\infty$" about this question; the same question can be asked for 1-categories and the answer is the same. In that case it fits into two abstract frameworks: -1: doctrinal adjunction. For a 2-monad $T$ on a 2-category $K$ and a pseudo $T$-algebra map $(g,\bar{g}):A\to B$ whose underlying morphism $g:A\to B$ in $K$ has a left adjoint $f:B\to A$, there is a canonical induced structure of an oplax $T$-algebra map on $f$ (the mate of the pseudo $T$-morphism structure on $g$), and the whole adjunction lifts to the 2-category of $T$-algebras (and pseudo morphisms) if and only if this oplax structure is in fact a pseudo structure. -Now there is a 2-monad $T$ on $\mathrm{Cat}^{\mathrm{ob}(I)}$ whose algebras are functors $I\to \mathrm{Cat}$. The pseudo $T$-morphisms are pseudo natural transformations, and the lax/oplax $T$-morphisms are lax/oplax natural transformations. When doctrinal adjunction is unraveled in the case of this 2-monad, it amounts to exactly the condition mentioned by Denis. -2: property-like structure. You mentioned that the space of adjoints to a given morphism is contractible (if nonempty), i.e. that "having an adjoint" appears to be a mere property of a morphism (rather than structure on it). From this perspective it may be surprising that the adjoints don't fit together. In fact, though, having an adjoint is something in between a "property" and "structure" called a property-like structure: a structure that is unique on objects when it exists, but is not necessarily preserved by morphisms. -One of the simplest examples of property-like structure is "having an identity element" for a semigroup: a semigroup can have at most one identity element, but a semigroup homomorphism need not preserve identities. A more well-known example is "having colimits" for a category: they are unique (up to unique isomorphism) when they exist, but not every functor preserves them (even up to isomorphism). The latter is an example of a special kind of property-like structure called lax-idempotent, in that it is automatically preserved laxly by every morphism (in this case, the comparison map $\mathrm{colim} \circ F \to F \circ \mathrm{colim}$). More precisely, a 2-monad $T$ on a 2-category $K$ is lax-idempotent if every $K$-morphism between $T$-algebras has a unique structure of lax $T$-morphism. -Now there is a 2-monad $T$ on the 2-category $\mathrm{Cat}^{\mathbf{2}}$ whose algebras are functors equipped with a left adjoint, and this 2-monad is lax-idempotent: the unique lax $T$-morphism structure on a commutative square is, again, its mate under the adjunctions. Now your given natural transformation is a functor $I\to \mathrm{Cat}^{\mathbf{2}}$, and the fact that it has adjunctions "pointwise" means that this functor lifts "objectwise" to $T$-algebras. Lax-idempotence of $T$ therefore implies that the functor lifts to the 2-category of $T$-algebras and lax $T$-morphisms; hence it lifts to pseudo $T$-morphisms if and only if all these mates are isomorphisms. - -As far as know, neither of these abstract contexts has yet been worked out in the $\infty$ case. But someone should!<|endoftext|> -TITLE: Finite étale covers of concentrated schemes and extension of base field -QUESTION [6 upvotes]: Let $k'/k$ be an extension of algebraically closed fields of characteristic $0$, and $X$ a concentrated (i.e. quasi-compact and quasi-separated) scheme over $k$. -Question: is the pullback functor from finite étale covers of $X$ to finite étale covers of $X_{k'}$ an equivalence? -If $X$ is connected, then by SGA1, Exposé XIII, Proposition 4.6b we know that $\pi_1(X)=\pi_1(X_{k'})$, and hence the answer is yes. If the connected components of $X$ are open (for example if $X$ is of finite type), then we can reduce ourselves to the connected case, but in general connected components are not open. - -REPLY [5 votes]: Yes, you can reduce to the finite type case by noetherian approximation (Appendix C in Thomason-Trobaugh). Namely, you can write $X=lim_\alpha X_\alpha$ where $X_\alpha$ is of finite type over $k$. Then there is an equivalence of categories $FinEt/X \simeq colim_\alpha FinEt/X_\alpha$.<|endoftext|> -TITLE: A quite puzzling question on Deligne cohomology sheaves and cycle maps -QUESTION [5 upvotes]: Intro. I would be deeply grateful if someone could please clarify the following to me. - -The question. (the main point is (4)) -Let $X$ be a smooth projective variety over $\mathbf{C}$, and $\mathbf{Z}(n)_{\mathcal{D}}$ the Deligne complex on $X$. -We have a triangle: -$$\to\Omega^{ -TITLE: affine vs lipschitz -QUESTION [5 upvotes]: Let $(X,||\cdot||)$ be a normed space where $||\cdot||$ is the sup-norm and let $E$ be a convex and compact subset. Let $f:E\to [0,1]$ be continuous and affine, i.e. for all $x,y\in E$ and all $\lambda \in [0,1]$, $f(\lambda x+(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y)$. -Is $f$ Lipschitz? If not, can you give a simple counter-example? -Technically, I can always modify the metric on $E$ such that $d(x,y)=||x-y||+|f(x)-f(y)|$. Then $f$ is Lipschitz. But this metric is not well-defined over the entire space $X$. I was hoping to show (or not show) that $f$ is Lipschitz under the original sup-norm metric. - -REPLY [8 votes]: For instance, consider the convex subset $E:=\{x\in\ell_\infty: 0\le x_k\le 2^{-k} \text{ for all } k\ge0 \}$ of $\ell_\infty$. By dominated convergence, $E$ is compact and its relative topology coincides with the topology induced by $\ell_1$. -In particular, the affine function $f:E\to [0,1]$ defined by $f(x):=\sum_{k=1}^\infty x_k$ is continuous on $E$. -One has $0\in E$ and $f(0)=0$, while for $u=2^{-n}\sum_{k=1}^n e_k\in E$ one has $\|u\|_\infty=2^{-n} $ and $f(u)=2^{-n}n=n\|u\|_\infty $, so $f$ is not Lipschitz.<|endoftext|> -TITLE: Traces of Sobolev spaces -QUESTION [8 upvotes]: Is there a simple proof of the following fact? - -Theorem. Let $\Omega\subset\mathbb{R}^n$ be a bounded and smooth domain. If $n>2$, then $W^{1,n-1}(\partial\Omega)\subset -W^{1-\frac{1}{n},n}(\partial\Omega)$. That is, there is a bounded - extension operator ${\rm Ext}:W^{1,n-1}(\partial\Omega)\to -W^{1,n}(\Omega)$. - -One can conclude this result from a sequence of results in H. Triebel, -Theory of function spaces. (Reprint of 1983 edition.) -Modern Birkhuser Classics. Birkhauser/Springer Basel AG, Basel, 2010 as follows: -using the following results Triebel's book: -Theorem 2.5.6, Theorem 2.7.1, Proposition 2.3.2.2(8), Theorem 2.5.7 -and 2.5.7(9) (in that order) we obtain -the following relations for function spaces on $\mathbb{R}^{n-1}$: -$$ -W^{1,n-1}(\mathbb{R}^{n-1})= -H^1_{n-1}= -F^1_{n-1,2}\subset -F^{1-\frac{1}{n}}_{n,n}= -B^{1-\frac{1}{n}}_{n,n}= -\Lambda^{1-\frac{1}{n}}_{n,n}= -W^{1-\frac{1}{n},n}(\mathbb{R}^{n-1}). -$$ -I find this proof highly unsatisfactory. -A self contained and elementary (but difficult) proof can also be found in G. Leoni, -A first course in Sobolev spaces. -Graduate Studies in Mathematics, 105. American Mathematical Society, Providence, RI, 2009, see Theorem 14.32, Remark 14.35 and Proposition 14.40. - -REPLY [4 votes]: An elementary proof was shown to me by Jan Malý. It has never been published and with his permission we included it in [1] (Proposition 28). The argument is very elementary (1.5 pages with all details), but still too long to be included here. - -Theorem. For $n\geq 1$ and $p>1$, there is a bounded linear extension operator $$ E:W^{1,p}(\mathbb{R}^{n})\to W^{1,q}\cap - C^\infty(\mathbb{R}^{n+1}_+), \quad \text{where $q=\frac{(n+1)p}{n}$.} - $$ In other words, $W^{1,p}(\mathbb{R}^{n})$ continuously embeds into - the trace space $W^{1-\frac{1}{q},q}(\mathbb{R}^{n})$ of - $W^{1,q}(\mathbb{R}^{n+1}_+)$. - -From this result the following corollary follows right away: - -Corollary. If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ - E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad - \text{where $12$ yields the theorem asked in the question. We need to take $n>2$ since for $n=2$ we have $p=n-1=1$ and the result is false in that case (there are counterexamples, see [1]). -[1] P. Goldstein, P. Hajłasz, -Jacobians of $W^{1,p}$ homeomorphisms, case $p=[n/2]$. - Calc. Var. Partial Differential Equations 58 (2019), no. 4, Art. 122, 28 pp.<|endoftext|> -TITLE: An analogy between the ring of polynomials in two variables and another (commutative?) ring -QUESTION [7 upvotes]: One of the answers to this question says: -"In Serge Lang's Algebra, he says: "One of the most fruitful analogies in mathematics is that between the integers $\mathbb{Z}$ and the ring of polynomials $F[t]$ over a field $F$". He then proves the abc conjecture for polynomials, and for good measure he proves Fermat's Last Theorem for polynomials. In other words, Lang is saying that if something is true for the ring of polynomials, one ought to check if it is true for that rather important ring called the integers. But it turns out that the ring of integers can be rather more troublesome, which may be surprising". - -What happens if we replace $\mathbb{C}[t]$ by $\mathbb{C}[x,y]$? Namely, is there an interesting/a similar analogy between some (commutative?) ring $R$ and $\mathbb{C}[x,y]$? - -Actually, here one can find an example that things over $\mathbb{Z}$ may become more complicated than over other integral domains -($R=k[u^3,v^3,u^2,v^2,uv]$). -Remarks: -(1) Of course one can suggest $R=A_1(\mathbb{C})$, the first Weyl algebra, and the -stable equivalence between the Dixmier and Jacobian Conjectures, but I think I actually prefer that $R$ will be a commutative ring which is 'simpler' than $\mathbb{C}[x,y]$ (Perhaps $R=\mathbb{Z}$ or $R=\mathbb{C}[t]$). -(2) Probably Formanek's paper is close to an answer I am looking for -(with $R=\mathbb{C}[t]$ or $\mathbb{Z}[t]$?). -Thank you very much for any comments/help! -Edit: I think I should have divided my question into two separate questions: -(1) Generalizing Lang's result concerning $R_1=\mathbb{Z}$ and $R_2=\mathbb{C}[x]$ to $R_1[t]$ and $R_2[t]$ (the below answer is nice, except that it is not elaborating what is exactly the generalized result of Lang's result). -(2) Finding connections between the (two-dimensional) Jacobian Conjecture and other theorems in number theory, similarly to what Formanek has done; -perhaps I will ask this in a new question. - -REPLY [6 votes]: This seems like a strange question. Of course there are many analogies between many different commutative rings. However, if you're looking for a similar analogy, you have to replicate the basic features of Lang's analogy - in particular, the fact that the rings have the same dimension. This immediately kills all your suggested answers but $R = \mathbb Z[t]$. -The analogy between $R = \mathbb Z[t]$ and $\mathbb C[x,y]$ is perfectly good generalization of Lang's analogy. Indeed it is almost a special case of Lang's analogy - whenever rings $R_1$ and $R_2$ have similar structure, we should expect that $R_1[t]$ and $R_2[t]$ have similar structure. -Finally I note that for many purposes the analogy between $\mathbb Z$ and $\mathbb C[t]$ is really a composition of two analogies, the first a closer analogy between $\mathbb Z$ and $\mathbb F_q[t]$ and the second an analogy between $\mathbb F_q[t]$ and $\mathbb C[t]$. For some purposes to understand the analogous construction on the other side it is crucial to pass through the middle step and take advantage of the special structures that are only available there (like the global Frobenius).<|endoftext|> -TITLE: Factorization of polynomials into "shortest possible" factors -QUESTION [13 upvotes]: A while ago I asked a question at Mathematica.SE about how to factorize a polynomial into terms with as few monomials as possible each. I now realized that I actually do not know what is rigorous mathematics behind this. -In fact that one was about univariate polynomials, and the same can be asked about several variables. -The question is whether there is some rigorous mathematics (algebra/geometry) behind asking for simultaneously minimizing the number of factors, and the number of monomials in each factor. -For an illustration, here is an example from that question: factorization of -$$1 - q^8 - q^{11} - q^{14} + q^{19} + q^{22} + q^{25} - q^{33}$$into $\mathbb Q$-irreducibles is -\begin{multline*}(1 - q)^3 (1 + q)^2 (1 + q^2) (1 + q^4) (1 - q + q^2 - q^3 + q^4 - q^5 + q^6)\\ -(1 + q + q^2 + q^3 + q^4 + q^5 + q^6)\\ -(1 + q + q^2 + q^3 + q^4 + q^5 + q^6 + q^7 + q^8 + q^9 + q^{10})\end{multline*} -while "my" optimal expression would be$$(1 - q^8) (1 - q^{11}) (1 - q^{14}).$$ -For many variables, and also allowing rational expressions, I would like to recognize in\begin{multline*}\left(x+y^3\right)^2 \left(x^2+y^6\right)^2 \left(x^2-x y^3+y^6\right) \left(x^2+x y^3+y^6\right) \left(x^4+y^{12}\right)\\ \left(x^4-x^2 y^6+y^{12}\right) \left(x^8+y^{24}\right)\end{multline*}that it is equal to$$\frac{\left(x^{12}-y^{36}\right) \left(x^{16}-y^{48}\right)}{\left(x-y^3\right)^2}$$ - -REPLY [7 votes]: This is essentially a question about algebraic circuit complexity. Namely, we can consider formulas which are products of sums of powers of variables (which, using standard notation from circuit complexity, would be called "$\Pi\Sigma\wedge$ formulas"), and you are asking for the smallest expression which evaluates to your given polynomial. -Although I don't know about this particular class of formulas, typically questions of the form "Given a polynomial, what is the smallest formula or circuit in a given class $\mathcal{C}$ that computes it" are at least $\mathsf{NP}$-hard, which means there's unlikely to be an algorithm significantly better than a simple "brute-force" exponential algorithm. (And, even if you think maybe $\mathsf{P}=\mathsf{NP}$, I still wouldn't be holding my breath for a better algorithm any time soon.) -Geometrically, you can view all polynomials that have formulas of the form $\prod_{i=1}^d \sum_{j=1}^{d_i} (\text{powers of variables})$ as the image of a $d$-th Segre re-embedding of $d_i$-th secant varieties of Veronese embeddings. For examples of how this works with other circuit classes (which are better studied, both in complexity theory and in algebraic geometry), e.g. for $\Sigma\Pi\Sigma$ circuits, see Landsberg's article "Geometric complexity theory - an introduction for geometers" or his recent book "Geometry & Complexity Theory".<|endoftext|> -TITLE: Counterexample showing that G-invariant de Rham cohomology different from cohomology of G-invariant sub-complex? -QUESTION [12 upvotes]: If $G$ is a discrete or a Lie Group acting smoothly on a manifold $M$, we can define the algebra of $G$-invariant de Rham classes, $H(M)^G$, and we can also consider the cohomology of the sub-complex of $G$-invariant forms and its cohomology $H(\Omega(M)^G)$ which I note simply $H(M^G)$. -The injection $j:\Omega(M)^G \hookrightarrow \Omega(M)$ passes to cohomology as $j:H(M^G) \to H(M)$, and it is clear that $j(H(M^G)) \subset H(M)^G$, so there is the induced morphism -$$\bar j \colon H(M^G) \to H(M)^G$$ -But $\bar j$ not injective since for $[\alpha]_M^G \in H(M^G)$, $j([\alpha]_M) = 0$ implies only that $\alpha = d\beta$ and $\beta$ has no reason to be $G$-invariant, so no reason why $[\alpha]_M^G$ should be null. -I think that $\bar j$ neither surjective since for $[\alpha]_M \in H(M)^G$, $g \cdot [\alpha]_M = [g \cdot \alpha]_M = [\alpha]_M$ for all $g \in G$ implies only that $g \cdot \alpha = \alpha + d\beta$, so $\alpha$ is not necessarily $G$-invariant. -But is it possible that $[\alpha]_M$ contains another $G$-invariant representant ? -Is there a simple counter-example showing that in general $\bar j$ is neither injective nor surjective ? For injectivity the example in @David's answer is OK, but quid for an example where surjectivité fails ? - -REPLY [17 votes]: If $G$ is compact, the inclusion $H(M^G) \to H(M)^G$ is an isomorphism. The inverse map is defined as follows: Take a class $\omega$ in $H(M)^G$ and lift it to a closed form $\alpha \in \Omega(M)$. Put $\beta = \int_{g \in G} g^{\ast} \alpha$, where the integral is with respect to Haar measure normalized to have volume $1$. Clearly, $\beta \in \Omega(M)^G$. Our lift will map $\omega$ to $[\beta]$. -We must check that $\beta$ is closed, is a de Rham representative of $\omega$, and that its class in $H(M^G)$ is independent of the choice of $\alpha$. Since all $g^{\ast} \alpha$ are closed, so is $\beta$. Since all $g^{\ast} \alpha$ are de Rham representatives of $\omega$, so is $\beta$. Finally, let $\alpha' = \alpha + d \eta$ be another lift of $\omega$. Then -$$\int_{g \in G} g^{\ast} \alpha' = \int_{g \in G} g^{\ast} \alpha + d \int_{g \in G} g^{\ast} \eta$$ -and $\int_{g \in G} g^{\ast} \eta$ is in $\Omega(M)^G$. -When $G$ is not compact, both injectivity and surjectivity can fail. -Failure of injectivity: -Consider $M = \mathbb{R}$ and $G = \mathbb{Z}$ acting by translations. The $1$-form $dx$ is closed and $G$-invariant on $\mathbb{R}$, but has no $G$-invariant integral. So it gives a nonzero class in $H^1(M^G)$, but of course $H^1(M)^G \subseteq H^1(M) = 0$. -Failure of surjectivity Let $M = S^1$. For $\theta \in \mathbb{R}$, define -$$\tilde{\phi}(\theta) = \theta + \tfrac{1}{2} \sin \theta.$$ -Then $\tilde{\phi}: \mathbb{R} \to \mathbb{R}$ descends to a diffeomorphism $\phi: S^1 \to S^1$, with repelling and attracting fixed points at $\theta = 0$ and $\theta = \pi$ respectively. We let $G = \mathbb{Z}$, acting by $\phi$. -The map $\phi$ acts trivially on $H^1(S^1)$. I claim that there is no nonzero smooth $\phi$-invariant $1$-form on $S^1$. Suppose for the sake of contradiction that $\omega$ is a $\phi$ invariant $1$-form. -Let $\omega = g(\theta) d \theta$ by $\phi$ invariant, and suppose for the sake of contradiction that $g(\theta_0) \neq 0$ for some $\theta_0 \not \in \pi \mathbb{Z}$. Then -$$g(\phi^n(\theta_0)) = \prod_{k=0}^{n-1} (\phi')(\phi^k(\theta_0))^{-1} g(\theta_0).$$ -As $k \to \infty$, $\phi^k(\theta_0) \to \pi$ and $\phi'(\phi^k(\theta_0))^{-1} \to 2$. So $g(\phi^n(\theta_0)) \to \infty$ as $n \to \infty$, contradicting that $g(\theta) d \theta$ is supposed to be a continuous $1$-form. -A perhaps nicer failure of surjectivity Take a look at this great post by Robert Bryant. Let $T$ be the torus $S^1 \times S^1$. Let $C_0$ and $C_{\pi}$ be the circles $\{ 0 \} \times S^1$ and $\{ \pi \} \times S^1$. Bryant constructs a flow (in other words, action of $\mathbb{R}$) which takes $C_0$ and $C_1$ to themselves, and where every other orbit approaches $C_0$ as $t \to (-\infty)$, and $C_1$ as $t \to \infty$, with opposite orientations. A picture is worth a thousand words: Here is what the orbits of this flow looks like on one of the two cylinders of $T \setminus (C_0 \cup C_1)$. -I claim that there is no flow invariant $1$-form with $\int_{C_0} \omega \neq 0$. -Let $f(t)$ be one of these nonclosed flows. For $T$ very positive, $f$ takes the interval $(T, T + 2 \pi)$ very close to $C_1$ and, for $T$ very negative, it is very close to $C_0$ with opposite orientation. So, if $\omega$ is a flow invariant continuous $1$-form, then -$$\int_{C_0} \omega = \lim_{T \to - \infty} \int_T^{T+2 \pi} f^{\ast} \omega = \lim_{T \to \infty} \int_T^{T+2 \pi} f^{\ast} \omega = - \int_{C_1} \omega.$$ -On the other hand, $C_0$ and $C_{1}$ are homologous so, if $\omega$ is closed, we have -$$\int_{C_0} \omega = \int_{C_1} \omega.$$ -Combining these, any closed flow-invariant $1$-form has $\int_{C_0} \omega=\int_{C_1} \omega=0$. -One more and I'll shut up Let $C$ be the cylinder $\mathbb{R}^2/\mathbb{Z}$, where we identify $(x,y)$ and $(x+k,y)$ for $k \in \mathbb{Z}$. Let the action of $\mathbb{Z}$ on $C$ be generated by $\phi(x,y) = (x,x+y)$. Let $\gamma$ be the image in $C$ of the line segment $(0,0)$ to $(0,1)$ in $\mathbb{R}^2$. Then $\phi(\gamma)$ is the image of the line segment from $(0,0)$ to $(1,1)$. So, if $\omega$ is a $\phi$-invariant $1$-form, then $\int_{\gamma} \omega = \int_{\phi(\gamma)} \omega$. But $\gamma - \phi(\gamma)$ is a $1$-cycle representing the nontrical class in $H_1(C)$, so we deduce that any $\phi$ invariant $1$-form integrates to $0$ against this class.<|endoftext|> -TITLE: How can a mathematician handle the pressure to discover something new? -QUESTION [211 upvotes]: Suppose I'm an aspiring mathematician-to-be, who started doing research. Although this is really what I love doing, I found that one disturbing point is that there's always the pressure of discovering something new. When I'm doing mathematical research, there's always the fear in the back of my mind, that maybe, I don't get new results. In the past, I could think freely about mathematics, without the pressure, but now that it's "my job", I have these problems. -How to handle this? Since this site is for mathematicians from the graduate level onwards, maybe somebody has a good suggestion. Note that, although this is a mathematics forum, I think this question is appropriate here, since it perfectly matches with the description of the "soft question"-tag. - -REPLY [15 votes]: There is a quotation I once heard that perhaps someone could provide a reference to. It may even have been an MO comment... -The gist of it was that as one gets along in one's mathematical career one's research often starts to feel less like tackling some impossible obstacle than preparing a meal for friends. -As I understand this, we mathematicians find a corner of the subject where we want to resolve some problems that are not so flashy as to draw international attention and gain us a Fields Medal or a tenure-track job at Princeton, but are interesting and important for the area of mathematics we love and are aiding in the growth of. The former sort of goal may drive one's early career, the scratching, biting and clawing that seems necessary to secure the time to think with a decent academic job. Later on, post-tenure, one can try to focus a bit more (with equal or greater intensity) on figuring out what is actually going on with a bit of mathematics. Usually, this corner is interesting to a small community of mathematicians who really will care about the work. It really feels that research is a conversation with these people, and that the work is pursued in a spirit of mutual appreciation of the beauty of that subject and a shared, deep desire for increased clarity...although the realities of promotion and pay still drive the occasional flare up to attempt to become famous... -The fewer of these flare-ups, the better, I think. They are driven by economic forces, rather than honest mathematics. The greatest of us probably think of these sorts of economic forces the least, either because they developed their mathematics early enough to head off these forces early on and now have coveted jobs, or simply they are used to living a spartan lifestyle that shields them from typical economic concerns. -This answer, by the way, is a comment mainly intended for the OP, and may be too riddled with opinion to be appropriate. If down voted sufficiently, I will remove it.<|endoftext|> -TITLE: Combinatorial proof that some model categories are monoidal/enriched? -QUESTION [6 upvotes]: I'm looking for examples of proofs that some Quillen model categories are monoidal, or enriched over an other model category, which are based on explicit computation of the "pushout product" of the generating cofirbations and generating trivial cofibrations, i.e. an explicit combinatorial proofs of a statement like: if $i:A \rightarrow C$ and $j:B \rightarrow D$ are respectively a generating cofibrations and a generating trivial cofibration then the "pushout-product" map: -$$i \otimes' j : C \otimes B \coprod_{A \otimes B} A \otimes D \rightarrow C \otimes D$$ -is a composite of pushout of the generating trivial cofibrations (or maybe a retract of one of these). And same things for pushout-product of generating cofibrations. -I'm specifically interested in $\omega$-combinatorial model category, i.e. locally finitely presentable categories, with the generating cofibration and generating trivial cofibrations being maps between finitely presentable objects. -$\otimes$ above can either bi a monoidal structures or the tensoring for some enrichment. I also accept things that are not quite monoidal structure, as long as they are "bi-closed", so that this sort of condition does implies somethings (for example, the dendroidal "tensor product"), or things that are not quite model structures (like opetopic sets) if you have such combinatorial proof. -I already have some example where the condition are very easy to check: - -The model structure on ordinary category with cartesian product. -Chain complexes. - -In the literature I found such a proof for simplicial sets (both for the model structure for Kan complexes and for quasicategory) -For Kan complex theorem 3.2.3 of Joyal & Tierney notes on simplicial homotopy theory does a special case, but it appears to be enough for deducing the general case by just playing around with the formal properties of the pushout-product. -Similarly, Prop 2.3.2.1 of Lurie's Higher topos theory does a special case for the model structure for quasi-category, which also appears to be sufficient by formal property of the pushout-product. -In both case, the proof of the special case are pure combinatorics. -Cubical sets are also relatively easy to treat, but I still would prefer a reference if it exists. -I would be interested to see other examples of such direct combinatorics proof (like dendroidal sets, cellular sets, complicial sets, other algebraic structure , I assume no such things is known for opetopic sets but I would love to be wrong on that ). -Motivation : I'm finishing a paper about constructing some "weak" (like left & right semi structure) Quillen model structures in constructive mathematics. I have results which says that essentially when these kind of properties are verified then you can get some sort of weak model structure completely constructively. As these properties are purely combinatorial, there proof are generally constructive. I'd like to give as much example as I can, but proving myself that these conditions are indeed constructively valid is out of the scope of the paper if it takes more than five line. I can always invoke principle like Barr's covering theorem to say that constructive proof exists, but having a "clearly constructive proof" in the Litterature that I can quote is considerably better. - -REPLY [2 votes]: This is far from a comprehensive answer, but since you ask for references, here are two. -The result for simplicial sets is obtained by an explicit computation in Appendix H of Joyal's The Theory of Quasi-categories and Its Applications. -For $\Theta_2$-sets, there is Chapter 3 of Oury's Duality for Joyal's Category $\Theta$ and Homotopy Concepts for $\Theta_2$-sets.<|endoftext|> -TITLE: A "strong" Galois-Tukey connection between orders with suborders -QUESTION [5 upvotes]: (Background, may be skipped by the knowledgeable reader: A Galois-Tukey connection between two partial orders $(P,\le)$ and $(Q,\le)$ is a pair of maps $\varphi^+:P\to Q$ and $\varphi^-:Q\to P$ satisfying -$$ \forall p\in P \ \forall q\in Q: \left[ \varphi^-(q)\le p \Rightarrow q\le \varphi^+(p)\right]$$ -for all $p\in P$, $q\in Q$. (Note: Only "$\Rightarrow$", not "$\Leftrightarrow$" as in Galois connections. Also, it's $\le$ on both sides, and the maps are not necessarily monotone.) -If there is such a Galois-Tukey connection, then $cf(P)\ge cf(Q)$, where -$cf(P)$ is the smallest cardinality of a cofinal (or: "dominating") set in $P$. This is easy to see (and well known). END OF BACKGROUND) -I am looking for a reference and/or a name for the following generalisation: -Let $P$, $Q$ be partial orders, $P_0\subseteq P$ and $Q_0\subseteq Q$. -A pair of maps $\varphi^+:P\to Q$ and $\varphi^-: Q_0\to P_0$ is called a BLANK, if we have -$$ \forall p\in P \ \forall q\in Q_0: \left[ \varphi^-(q)\le p \Rightarrow q\le \varphi^+(p)\right]$$ -Is there a better name than "strong GT connection", or perhaps even a well-established one? -My motivation is this: If we define (as in Bartoszyński-Judah 2.1.3.) the cardinal $cf(P_0, P)$ as the smallest size of a subset $D$ of $P$ which "dominates" $P_0$ (i.e., $\forall p_0\in P_0\ \exists d\in D: p_0\le d$), and dually the notion $add(P_0,P)$, then it is easy to see that the existence of a strong GT function as above will imply $add(P_0,P)\le add(Q_0,Q)$ and $cf(P_0,P)\ge cf(Q_0,Q)$. - -Edit: Thanks to Peter Vojtas and Andreas Blass for pointing out that - -a (generalized) Galois-Tukey relation (or "morphism") does not have to be between two orders; any two relations (subsets of $P_0\times P$ and $Q_0\times Q$, -where no inclusion relation is required between $P_0$ and $P$, or between $Q_0$ and $Q$) will do; -Letting $\le'_P$, $\le'_Q$ be the restrictions of $\le_P$, $\le_Q$ to $P_0\times P$ and $Q_0\times Q$, respectively, my "strong GT relation" is just the usual generalized GT relation/morphism between $\le'_P$ and $\le'_Q$. Both relations happen to be partial orders, but that is irrelevant. - -REPLY [2 votes]: As indicated in Peter Vojtas's answer, this notion is a special case of what he called generalized Galois-Tukey connections (in the paper he linked to) and what I later called morphisms (in my chapter of the Handbook of Set Theory). Specifically, what you describe is a morphism from $(P_0,P,\leq)$ to $(Q_0,Q,\leq)$. (In the terminology of generalized G-T connections, "from" and "to" would be interchanged.) In general, the domain and codomain of a morphism could be any triples of the form $(A_-,A_+,R)$ where $R\subseteq A_-\times A_+$, i.e., any two sets and a relation between them. What's special in your situation seems to be just that (1) $P_0\subseteq P$, (2) the $\leq$ relation that you actually use (between $P_0$ and $P$) is the restriction of a partial order on all of $P$, and (3,4) the same for $Q_0$ and $Q$.<|endoftext|> -TITLE: Linking topological spheres -QUESTION [14 upvotes]: Is there a simple proof of the fact that: - -If $A\subset S^3$ is homeomorphic to $S^1$, then there is a circle $B$ - embedded into $S^3\setminus A$ that such that the circles $A$ and $B$ - are linked with the linking number $1$? - -If $A$ is smoothly embedded, than it is easy. The problem is that the general topological embedding can be very complicated and I do not see a simple geometric explanation of this fact. I know that $H_1(S^3\setminus A)=\mathbb Z$ (Corollary 1.29 in Vick's Homology Theory) so the generator of the homology group should be a circle. -Edit 1: -This is actually true as John Klein pointed out in his answer below, but I do not know the answer to the higher dimensional version of the problem stated below. Only a partial answer is given in comments below. -The question can be then generalized to higher dimensional spheres. - -If $A\subset S^n$ is homeomorphic to $S^k$, there should be a - topological sphere in $S^n\setminus A$ of dimension $n-k-1$ that links - $A$ with the linking number $1$. - -Edit 2. A counterexample is given below in my answer. - -REPLY [3 votes]: Let me add another answer. This answer stems from the fruitful discussion on MathOverflow. - -Theorem. There is a topological embedding $\iota:\mathbb{S}^1\to\mathbb{S}^5$ such that $\pi_3(\mathbb{S}^5\setminus\iota(\mathbb{S}^1))=0$. -Therefore, no -$3$-sphere can be linked with $\iota(\mathbb{S}^1)$. - -Proof. -It is well known that there are $3$-dimensional integer homology spheres whose universal cover is $\mathbb{R}^3$. For example, there are particular constructions in [1,2,5] of hyperbolic integer homology spheres. Note that the universal cover of a hyperbolic $3$-manifold is the hyperbolic space that is homeomorphic to $\mathbb{R}^3$. -Other examples are listed in Homology sphere with $\mathbb{R}^3$ as the universal cover. -Let $\mathcal M$ be such an integer homology sphere. Since the universal cover of $\mathcal M$ is contractible, $\pi_3(\mathcal M)=0$. According to the celebrated theorem of Cannon and Edwards [3,4], the double suspension of an integer homology sphere is homeomorphic to a topological sphere. Let $h:S^2{\mathcal M}\to\mathbb{S}^5$ be such a homeomorphism. -$\mathcal M$ is a deformation retract of the complement of the vertices of the suspension $S\mathcal M$. -Therefore, $\mathcal M$ is also a deformation retract of the complement of the suspension of the vertices in $S^2\mathcal M$. Denote the suspension of the vertices by $X$, so $\mathcal M$ is a deformation retract of $S^2{\mathcal M}\setminus X$ and hence $\pi_3(S^2\mathcal{M}\setminus X)=0$. -$X$ is homeomorphic to $\mathbb{S}^1$. If $g:\mathbb{S}^1\to X$ is a homeomorphism, then $\iota=h\circ g:\mathbb{S}^1\to\mathbb{S}^5$ is a topological embedding and clearly $\pi_3(\mathbb{S}^5\setminus\iota(\mathbb{S}^1))=\pi_3(\mathbb{S}^5\setminus h(X))=\pi_3(S^2{\mathcal M}\setminus X)=0$. -The proof is complete. $\Box$ -Edit. While the result is a straightforward consequence of known results, I decided to publish it as a short note that I dedicated to MathOverflow [6]. -[1] J. Baldwin, J., S. Sivek, -Stein fillings and $SU(2)$ representations. Geom. Topol. 22 (2018), 4307-4380. -[2] J. Brock, N. M. Dunfield, -Injectivity radii of hyperbolic integer homology $3$-spheres. -Geom. Topol. 19 (2015), 497-523. -[3] J. W. Cannon, Shrinking cell-like decompositions of manifolds. Codimension three. Ann. of Math. 110 (1979), 83-112. -[4] R. D. Edwards, The topology of manifolds and cell-like maps. Proceedings of the International Congress of Mathematicians (Helsinki, 1978), pp. 111–127, Acad. Sci. Fennica, Helsinki, 1980. -[5] J. Hom, T. Lidman, -A note on surgery obstructions and hyperbolic integer homology spheres. -Proc. Amer. Math. Soc. 146 (2018), 1363-1365. -[6] P. Hajłasz, Linking topological spheres. Atti Accad. Naz. Lincei Rend. Lincei Mat. Appl. 30 (2019), 907-909. -https://arxiv.org/abs/1906.01771<|endoftext|> -TITLE: Spectral sequence in Betti cohomology -QUESTION [5 upvotes]: Let $X$ be a smooth projective algebraic variety over the complex numbers, and let us name -$$f : X_{\rm an}\to X_{\rm Zar}$$ -the morphism of sites induced by sending a Zariski open $U\subset X$ to $U^{\rm an}$. -Consider the constant sheaf $\mathbf{Q}(1)$ on $X_{\rm an}$ whose value group is $2\pi i\cdot\mathbf{Q}$ and let $\mathbf{Q}(n) := \mathbf{Q}(1)^{\otimes n}$. -Does the Leray spectral sequence -$$H^p(X_{\rm Zar}, R^qf_*\mathbf{Q}(n))\Rightarrow H^{p+q}(X_{\rm an}, \mathbf{Q}(n))$$ -degenerate at the $E_2$-page? That is, are all the differentials in the $E_2$ page of the spectral sequence, zero? -This would follow if $$Rf_*\mathbf{Q}(n) = \bigoplus_pR^pf_*\mathbf{Q}(n)[-p]$$ -in the derived category of abelian Zariski sheaves on $X$, so I'm wondering: is some Hard Lefschetz Theorem for the sheaves $R^pf_*\mathbf{Q}(n)$ true? -(This question originated from clever remarks by the user Merlin, on a previous question of mine. I expect the answer to be negative, but if so, I would like to understand why exactly.) - -REPLY [2 votes]: No; this spectral sequence usually does not degenerate. It appears that the group in the left hand side are often infinite dimensional. In particular, the $E_2^{p,p}$-term is the group of codimension $p$ cycles modulo algebraic equivalence (see Corollary 7.4 of the paper Bloch, S., & Ogus, A. (1974). Gersten’s conjecture and the homology of schemes. Ann. Sci. École Norm. Sup.(4), 7(181-201), 12); hence the edge map to $H^{2p}(X,\mathbb{Q})$ is often not injective (starting from $p=2$; see Corollary 7.5 of loc. cit.).<|endoftext|> -TITLE: Simplification of integral on the sphere -QUESTION [5 upvotes]: In the article: https://arxiv.org/abs/0906.3217 the authors prove in Lemma 1 a formula which helps compute more easily the integral of the Hessian of a function defined on $\Bbb{S}^2$. More precisely, if $h : \Bbb{S}^2 \to \Bbb{R}$ is a $C^2$ function, $Hess(h)(X,Y) = \langle \nabla_X \nabla h,Y\rangle$ is the Hessian of $H$ and $H(h)$ is the determinant of the Hessian, then -$$ \int_{\Bbb{S}^2} H(h) dA = \frac{1}{2} \int_{\Bbb{S}^2} |\nabla h|^2 dA$$ -where $dA$ is the area element on $\Bbb{S}^2$. This formula is helpful and simplifies some aspects of my numerical computations. Here and in the following, $\nabla$ and $\Delta$ represent the tangential gradient and Laplace-Beltrami operator on $\Bbb{S}^2$. -The above formula is proved in connection with bodies of constant width, but in the proof in the article they don't seem to use this fact. In my numerical experiments the formula gives the expected result in the general case. However, I work in the case where $H(h)>0$ (I don't know if this is relevant or not...) - -I was wondering if it is possible to obtain a similar simplification for the integral $\int_{\Bbb{S}^2} h H(h) dA$? More precisely, is it possible to obtain something of the form - $$ \int_{\Bbb{S}^2} h H(h) dA = \int_{\Bbb{S}^2} \mathcal{F}(h,\nabla h,\Delta h)dA $$ - where $\mathcal{F}$ is "simple" (polynomial)? - -REPLY [6 votes]: You can also derive the lemma from the Bochner formula, which in general dimensions can be written -$$ \frac{1}{2}\Delta\lvert\nabla u\rvert^2 = \lvert\nabla^2u\rvert^2 - (\Delta u)^2 + \delta\left((\Delta u)\,du\right) + \mathrm{Ric}(\nabla u,\nabla u) . $$ -If $(M,g)$ is a Riemannian surface with Gauss curvature $K$, then $\mathrm{Ric}=Kg$ and the determinant of the Hessian is $H(u)=\frac{1}{2}\left((\Delta u)^2-\lvert\nabla^2u\rvert^2\right)$. Multiplying the Bochner formula by $u$ and integrating by parts gives -$$ \int u\,H(u)\,dA = \frac{1}{4}\int \left(2Ku - 3\Delta u\right)\lvert\nabla u\rvert^2\,dA . $$ -In your situation, $K=1$.<|endoftext|> -TITLE: Closed, sum-free form for the $n$-th derivative of $\operatorname{arcsinh}(\frac1x)$ in $x=1$ -QUESTION [6 upvotes]: During research involving the Born–Jordan quantization I came across the expression -$$ -\frac{d^k}{dx^k}\operatorname{arcsinh}\Big(\frac1x\Big)\Big|_{x=1}\tag1 -$$ -for $k\in\mathbb N_0$. It is not too hard to write this expression as a sum -$$ -(1)=\frac{\sqrt{2}}{2^k}\sum_{j=0}^{k-1} a_j^k\tag{2a} -$$ -for any $k\in\mathbb N$ where $(a_j^k)_{j\in\mathbb Z,k\in\mathbb N}$ is a recursive sequence of integers given by -$$ -a_j^k:=\begin{cases} a_0^1=-1&\\a_j^k=0&\text{if }j<0\text{ or }j\geq k\\ a_j^{k+1}=a_j^k(2j-k)+a_{j-1}^k(2j-3k-1)&\text{else}\end{cases},\tag{2b} -$$ -(basically a modified version of Pascal's triangle). Unfortunately, I so far was not able to find a closed (sum-free) form of $(1)$ / $(2a)$ for arbitrary $k\in\mathbb N_0$. -This recursive sequence is nice for explicit calculations (especially speeds up things for larger $n$) - but I'm interested how $(1)$, or rather of the arising matrix elements -$$ -M_{nn}:=2\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{2^k}{k!}\Big(\frac{d^k}{dx^k}\operatorname{arcsinh}\Big(\frac1x\Big)\Big|_{x=1}\Big)\tag3 -$$ -behave for large $n$. - -Explicit calculations (up until $n=200$) suggest that $M_{nn}\overset{n\to\infty}\longrightarrow0$ with $M_{nn}=\mathcal O(\frac1n)$ for $n\to\infty$. What could be an approach to potentially prove this? What is more realistic: trying to find some bound for $(3)$ or trying to find a closed, sum-free form for $(1)$, respectively $(2a)$ / $(2b)$? - -Being fairly new here I hope this "question" (or rather problem) is suitable for mathoverflow. If it is not, feel free to tell me so I can outsource it to math.stackexchange. Thanks in advance for any answer or comment! - -REPLY [4 votes]: The numbers $c_k = \frac{1}{k!} \left.\left(\frac{d}{dx}\right)^k\operatorname{arcsinh}\frac1x\right|_{x=1}$ are the coefficients in the expansion: -$$\operatorname{arcsinh}\frac1x = c_0 + c_1(x-1) + c_2(x-1)^2 + \dots.$$ -It follows that $2^kc_k$ is the coefficient of $t^k$ in $\operatorname{arcsinh}\frac1{1+2t}$. -Now, -$$M_{nn} = 2\sum_{k=0}^n \binom{n}{k} 2^kc_k = 2\cdot [t^n]\ (1+t)^n \operatorname{arcsinh}\frac1{1+2t}.$$ -Using Lagrange inversion, we get the generating function for $M_{nn}$: -$$\sum_{n\geq 0} M_{nn} t^n = \frac{2}{1-t} \operatorname{arcsinh}\frac{1-t}{1+t}.$$ -From here the asymptotic for $M_{nn}$ can be obtained using the standard tools (e.g., see the answer from GH from MO). - -To get an explicit formula for $\left.\left(\frac{d}{dx}\right)^k\operatorname{arcsinh}\frac1x\right|_{x=1}$ (and thus for $c_k$), we notice that -$$\left(\frac{d}{dx}\right)^k\operatorname{arcsinh}\frac1x = - \left(\frac{d}{dx}\right)^{k-1} (x^2+x^4)^{-\frac12}.$$ -To expand the last expression, one can use Faà di Bruno's formula: -$$\left(\frac{d}{dx}\right)^{k-1} (x^2+x^4)^{-\frac12}$$ -$$ = \sum_{i=1}^{k-1} \binom{-\frac{1}{2}}{i} i! (x^2+x^4)^{-\frac12-i} B_{k-1,i}(2x+4x^3,2+12x^2,24x,24,0,0,\dots,0),$$ -where $B_{k-1,j}$ are Bell polynomials. -Evaluating at $x=1$, for $k>0$, we get -$$\left.\left(\frac{d}{dx}\right)^k\operatorname{arcsinh}\frac1x\right|_{x=1} = --\sum_{i=1}^{k-1} \binom{-\frac{1}{2}}{i} i! 2^{-\frac12-i} B_{k-1,i}(6,14,24,24,0,0,\dots,0)$$ -$$ = -\frac{(k-1)!}{\sqrt{2}}\sum_{j_1+2j_2+3j_3+4j_4=k-1} \frac{(-1)^{j_1+j_2+j_3+j_4}(2(j_1+j_2+j_3+j_4))!}{(j_1+j_2+j_3+j_4)!j_1!j_2!j_3!j_4!} 2^{-3(j_1+j_2+j_3+j_4)} 6^{j_1}7^{j_2}4^{j_3}$$ -$$=-\frac{(k-1)!}{\sqrt{2}}\sum_{j_1+2j_2+3j_3+4j_4=k-1} \frac{(-1)^{j_1+j_2+j_3+j_4} (2(j_1+j_2+j_3+j_4))!}{(j_1+j_2+j_3+j_4)!j_1!j_2!j_3!j_4!} 2^{-2j_1-3j_2-j_3-3j_4} 3^{j_1}7^{j_2}.$$<|endoftext|> -TITLE: Why is the definition of entropy solution necessary to prove uniqueness for hyperbolic conservation laws? -QUESTION [5 upvotes]: I'm aware that there are a lot of counterexamples to show that distributional solutions for hyperbolic (scalar) conservation laws are not unique. -However, I'd like to ask: - - -Conceptually, at which point of a proof of uniqueness is the definition of distributional solution not enough to go on? - -Why is the definition of entropy solution useful in the proof of uniqueness for hyperbolic conservation laws? - -REPLY [3 votes]: Unless an evolution PDEs be linear, a uniqueness proof is always nonlinear in essence: you prove that some distance $d(u(t),v(t))$ between two solutions is bounded in terms of $d(u(0),v(0))$. To carry out the proof, you need to be able to compute the time derivative of $d(u,v)$, and for this you need the chain rule. It turns out that the solutions of hyperbolic conservation laws display sharp discontinuities, and therefore the chain rule does not apply. -The entropy condition is a differential inequality that involves such a nonlinear quantity, typically $d(u,k)$ where $k$ is a constant state. This immediately yiedls uniqueness if the initial data is constant. When it is not, you can still conclude uniqueness if the solution is Lipschitz continuous, by a so-called weak-strong argument (similar to that for the Navier-Stokes equation). If the equation is scalar, Kruzkhov used the fact that the distance $|v-u|$ is a convex entropy with respect to both arguments and proved uniqueness of weak entropy solutions. For systems, the situation is much more complicated, and totally open in several space dimensions.<|endoftext|> -TITLE: When did people start thinking of elliptic curves as groups? -QUESTION [28 upvotes]: I have been reading some old papers of Cassels and Selmer from around 1950, and they talk about generators of rational solutions to elliptic curves, in the sense of Mordell–Weil, but do not appear to use the word group. (Edit: Taking another look, at least some of Cassels' papers from this period do use the word group.) -Weil - L'arithmétique sur les courbes algébriques (footnote 1, p. 281) says: - -Afin de réserver le mot de groupe au sens qu'il a pris depuis Galois, je parlerai toujours de systèmes de points, bien qu'on air l'habitude en géométrie algebrique de parler de groupes de points sur une courbe. - -Question: When did it become common parlance to call elliptic curves groups? - -REPLY [18 votes]: The first mathematician who talked about groups of points on elliptic curves (in the sense of Galois, i.e., in the modern sense of the word group) was -Juel [Ueber die Parameterbestimmung von Punkten auf Curven zweiter -und dritter Ordnung. Eine geometrische Einleitung in die Theorie der -logarithmischen und elliptischen Funktionen, Math. Ann. 47 (1896), 72-104]. -Poincare, in his important article referred to by ThiKu, does not use the concept of groups (this is the point of Schappacher's article). Even Mordell proved his theorem, namely that the group of rational points on an elliptic curve is finitely generated, without using the notion of a group: this was only done by Weil. -For a long time, most people interested in elliptic curves regarded them as -a variety of diophantine equations, and preferred thinking about secant and -tangent methods instead of group operations. Only when it became clear in -the 1960s that the rank of an elliptic curve could be computed by applying -all kinds of homomorphisms and determining the orders of kernels and images of such homomorphisms, the group theoretic point of view became indispensible.<|endoftext|> -TITLE: Periods in the trivial extension algebra of the incidence algebra of the divisor lattice -QUESTION [5 upvotes]: Definition of $C_L$ for people who like number theory: -Let $m$ be a number with prime factorisation $m=p_1^{n_1} ... p_r^{n_r}$ with $n_i>0$. Define $I_m$ to be the incidence algebra of the divisor lattice of $m$. Up to isomorphism this just depence on the $n_i$ and not on the primes so lets define with $L=[n_1+1,...,n_r+1]$, $I_L$ as the incidence algebra of the divisor lattice of a number $m$ with prime factorisation $m=p_1^{n_1} ... p_r^{n_r}$. -Define $C_L$ as the trivial extension algebra of $I_L$. For the definition of trivial extension algebra see for example https://math.stackexchange.com/questions/229412/trivial-extension-of-an-algebra (together with a proof that this is always a Frobenius algebras) or the textbook on Frobenius algebras I by Skowronski and Yamagata. -Definition of $C_L$ for people who like Nakayama algebras: -For a natural number $n \geq 2$ denote by $A_n$ the hereditary Nakayama algebra with dimension $n$ given by quiver and relations over a field $K$ (that is the Nakayama algebra with Kupisch series $[n,n-1,...,2,1]$). -For a list of integers $L=[n_1,...,n_r]$. Define the algebra $C_L$ as the trivial extension of the algebra $A_{n_1} \otimes_K \cdots \otimes_K A_{n_r}$. -Questions: -Is $C_L$ isomorphic to a known/studied algebra from the literature? -Experiments with small lists suggest that the algebra is always periodic or at least all simple modules are periodic with the same period. -(Recall that a module $M$ is called periodic with period $n$ when $n$ is the smallest integer such that $\Omega^n(M) \cong M$. An algebra is called periodic in case the regular module is periodic as a bimodule.) -The period seems to behave in a strange way. What could it be (in case such a finite periodic exists)? -Here some examples, where after the list $L$ the number is the period of a simple module of $C_L$. Sadly even with a computer it takes long to do such calculations. -$L=[t]: 2t$ for $t \geq 1$ (this is well known) -$L=[2,2]: 5$ -$L=[3,2]: 22$ -$L=[4,2]:29$ -$L=[5,2]: 12$ (this was surprising) -$L=[6,2]: 43$ -$L=[4,3]:42$ -$L=[3,3]: 8$ -$L=[2,2,2]: 6$ -$L=[3,2,2]: 26$ -Guesses what the period might be for general $L$ are also welcome. I should say that it is not too good tested yet as even for those small lists it takes several minutes to do the calculations with a computer. -More general, one might ask what those periods are (in case they exist) for the trivial extension algebra of tensor products of hereditary path algebras of Dynkin type. For example the trivial extension of $k D_4 \otimes_K k A_2$ has the property that all simple modules have period equal to 6. - -REPLY [5 votes]: Calculating periods of simple modules for the trivial extension algebra $TA$ can be reduced to a calculation with $A$-modules (at least if $A$ has finite global dimension), which is much easier. -The starting point is a result of Happel, that for a finite-dimensional algebra $A$ with finite global dimension, the bounded derived category $D^b(\text{mod }A)$ is equivalent as a triangulated category to the stable module category $\underline{{\text{mod}}}\hat{A}$ of the repetitive algebra $\hat{A}$ of $A$, or equivalently the stable category $\underline{\text{mod}}_{\mathbb{Z} }TA$, of $\mathbb{Z}$-graded modules for the trivial extension algebra $TA=A\ltimes DA$, regarded as a graded algebra with $A$ in degree zero and $DA$ in degree one. -This equivalence sends an $A$-module $M$, regarded as an object of $D^b(\text{mod }A)$ concentrated in degree zero, to the graded $TA$-module with $M$ as degree zero component, and other degree components zero, with $DA$ acting as zero. -The shift functor in $D^b(\text{mod }A)$ is the usual "shift to the left" of complexes, and the shift functor in $\underline{\text{mod}}_{\mathbb{Z} }TA$ is the cosyzygy functor $N\mapsto\Omega^{-1} N$, which commutes with "forgetting the grading". -The "shift grading" functor $N\mapsto N(1)$ for $\underline{\text{mod}}_{\mathbb{Z} }TA$ corresponds to the functor $M\mapsto FM:=\mathbf{L}\nu M[1]$ for $D^b(\text{mod }A)$, where $\mathbf{L}\nu$ is the derived Nakayama functor (take a projective resolution and apply $-\otimes_ADA$). -So to calculate the period of an $A$-module $S$, considered as a $TA$ module, we can repeatedly apply $F$ to $S$, considered as an object of $D^b(\text{mod }A)$, until the first time we get a shift $S[p]$ of $S$, and then $p$ is the period of $S$. -Now let's apply this to the algebra $I_L=A_{n_1}\otimes_K\dots\otimes_K A_{n_r}$. -For the hereditary algebra $A_n$, let $X(n,1),X(n,2),\dots,X(n,n)$ be the simple modules, in the natural order where $X(n,1)$ is the injective simple and $X(n,n)$ is the projective simple, and let $X(n,0)$ be the $n$-dimensional projective-injective module. -The derived Nakayama functor for $A_n$ satisfies $\mathbf{L}\nu X(n,i)=X(n,i+1)[1]$ for $00$ for which $F^lX(i_1,\dots,i_r)$ is a shift of $X(i_1,\dots,i_r)$ is $\operatorname{lcm}(n_1+1,\dots,n_r+1)$, and for each $j$, two in every $n_j+1$ of the modules in the $F$-orbit of $X(i_1,\dots,i_r)$ have $i_j\in\{0,n_j\}$, so the total shift of $F^lX(i_1,\dots,i_r)$, and hence the period of the corresponding module for the trivial extension algebra, is -$$\operatorname{lcm}(n_1+1,\dots,n_r+1)\left(r+1-2\sum_{j=1}^r\frac{1}{n_j+1}\right).$$ -Note that this only calculates the periods of simple modules (and shows that they all have the same period). Probably the trivial extension algebra itself has the same period, but I don't see how to prove that using this method.<|endoftext|> -TITLE: Are holomorphic vector bundles over Kähler manifolds Kähler -QUESTION [5 upvotes]: Let $X$ be a Kähler manifold and $E\to X$ a holomorphic vector bundle. Is there a Kähler structure on $E$ compatible with its complex structure? - -REPLY [6 votes]: Proposition 3.18 of Voisin's Hodge Theory and Complex Algebraic Geometry I says that, if $X$ is compact Kahler and $E$ is a holomorphic vector bundle over $X$, then $\mathbb{P}(E)$ is Kahler. Since $E$ embeds as an open submanifold of $\mathbb{P}(E \oplus \mathbb{C})$, this establishes your result for $X$ compact, and I think her proof could be simplified if you just want the vector bundle version and not the projective bundle version. But it looks to me like she actually is using compactness in a nontrivial way.<|endoftext|> -TITLE: Does the lattice of all topologies embed into the lattice of $T_1$-topologies? -QUESTION [9 upvotes]: Let $\kappa$ be an infinite cardinal, and let $\text{Top}(\kappa)$ be the lattice of all topologies on $\kappa$, ordered by $\subseteq$. Let $\text{Top}^{T_1}(\kappa)$ be the lattice of all $T_1$-topologies on $\kappa$. -Is there an injective lattice homomorphism $\varphi: \text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa)$? - -REPLY [5 votes]: This is a long comment about Bjørn's answer rather than an answer itself. -Hartmanis proves -Thm 1 If $\kappa>2$, then $\text{Top}(\kappa)$ is a simple lattice. -Thm 2 If $\kappa$ is infinite, then -(1) For any finite subset $F\subseteq \kappa$, the restriction map -$\rho_F:\tau\mapsto \tau|_{\kappa-F}$ is a nonconstant, noninjective, complete lattice homomorphism from -$\text{Top}^{T_1}(\kappa)$ to $\text{Top}^{T_1}(\kappa-F)\;(\cong\text{Top}^{T_1}(\kappa))$. -(2) For any finite subset $F\subseteq \kappa$, $\ker(\rho_F)$ is a proper, complete congruence on $\text{Top}^{T_1}(\kappa)$. Every proper, complete congruence has this form. If $F\neq G$ are distinct finite subsets of $\kappa$, then $\ker(\rho_F)\neq\ker(\rho_G)$. \\\ -Theorem 1 implies that if $\varphi:\text{Top}(\kappa)\to\text{Top}^{T_1}(\kappa)$ is an embedding, then for any congruence $\theta$ on $\text{Top}^{T_1}(\kappa)$ one must have either -Case 1. the image of $\varphi$ is contained in a single $\theta$-class, or -Case 2. the image of $\varphi$ is contained in a $\theta$-transversal. -Otherwise $\varphi^{-1}(\theta)$ would be a nontrivial proper congruence of a simple lattice. Thus each congruence $\theta$ on $\text{Top}^{T_1}(\kappa)$ restricts the possibilities for $\varphi$. -Theorem 2 implies that there are lots of $\theta$'s to choose from, namely all of the form $\ker(\rho_F)$. -Thms 1&2 imply that there are lots of restrictions $\varphi$ must satisfy, but there are not enough restrictions to rule out the existence of $\varphi$. In my answer on this page I construct $\varphi$ whose image is contained in a $\theta$-transversal for every $\theta$ of the form $\ker(\rho_F)$.<|endoftext|> -TITLE: Analog of Tennenbaum's theorem for EFA -QUESTION [8 upvotes]: EFA can prove the exponential function to be total, but it cannot prove the superexponential function to be total. Is there an analog of Tennenbaum's theorem (which states the PA has no recursive non-standard models) for EFA which states that EFA has no "sub-superexponential" non-standard models? (Here "sub-superexponential" should mean something like "only finitely iterated exponential" space complexity.) Can one say even more, like that EFA has no primitive recursive non-standard models? - -REPLY [12 votes]: Tennenbaum’s theorem applies unrestricted to EFA, and even to the much weaker theory $IE_1$ (Robinson’s arithmetic + induction for bounded existential formulas): - -$IE_1$ has no recursive nonstandard models (Wilmers [1]). - -On the other hand, $\mathit{IOpen}$ (induction for quantifier-free formulas) does have recursive nonstandard models (Shepherdson [2]), and the same also holds for extensions of $\mathit{IOpen}$ by certain “algebraic” axioms: Berarducci and Otero [3] prove this for $\mathit{IOpen}$ + “normality” (= integral closure), and Mohsenipour [4] sketches a proof for $\mathit{IOpen}+\mathit{GCD}$. -In fact, there seems to be a kind of dividing line between $\mathit{IOpen}$ and its mild extensions on the one side, and $IE_1$ or theories of similar strength on the other side, with several results strengthening Tennenbaum’s theorem to the effect that nonstandard models of theories behind the line are very complicated: - -Every nonstandard model of $IE_1$ has a nonstandard cut that is a model of Peano arithmetic (Paris [5]). More generally, let $T$ be any $\Sigma_1$-sound recursively axiomatizable extension of $I\Sigma_1$. Then every nonstandard model of $IE_1$ has a nonstandard cut that is a model of $T$ ([5] + McAloon [6]). -The additive reduct $(M,+,{\le})$ of any model of $IE_1$ is recursively saturated (Wilmers [1]). -The real closure of [the fraction field of] a nonstandard model is recursively saturated. This is not actually known to hold for $IE_1$, but it is true for “unbounded” models of $IE_2$, and models of $\mathit{PV}_1$ or $\Sigma^b_1\text-\mathit{LLLLLIND}$ (J and Kołodziejczyk [7]). - -References: -[1] George Wilmers, Bounded existential induction, Journal of Symbolic Logic 50 (1985), no. 1, pp. 72–90. -[2] John C. Shepherdson, A non-standard model for a free variable fragment of number theory, Bulletin de l’Académie Polonaise des Sciences, Série des Sciences Mathématiques, Astronomiques et Physiques 12 (1964), no. 2, pp. 79–86. -[3] Alessandro Berarducci and Margarita Otero, A recursive nonstandard model of normal open induction, Journal of Symbolic Logic 61 (1996), no. 4, pp. 1228–1241. -[4] Shahram Mohsenipour, A recursive nonstandard model for open induction with GCD property and confinal primes, Logic in Tehran (A. Enayat, I. Kalantari, M. Moniri, eds.), Lecture Notes in Logic, no. 26, Association for Symbolic Logic, 2006, pp. 227–238. -[5] Jeff B. Paris, O struktuře modelů omezené $E_1$-indukce, Časopis pro pěstování matematiky 109 (1984), no. 4, pp. 372–379 (in Czech). -[6] Kenneth McAloon, On the complexity of models of arithmetic, Journal of Symbolic Logic 47 (1982), no. 2, pp. 403–415. -[7] Emil Jeřábek and Leszek Kołodziejczyk, Real closures of models of weak arithmetic, Archive for Mathematical Logic 52 (2013), no. 1–2, pp. 143–157.<|endoftext|> -TITLE: Fedosov vs. Kontsevich deformation quantization : a beginner survey -QUESTION [8 upvotes]: I'm a condensed matter physicist who tries to understand the details of deformation quantization. -In my self-made training, I've found two huge pieces of work, namely - -Fedosov, B. V. (1994). "A simple geometrical construction of deformation quantization". Journal of Differential Geometry, 40 : 213–238. -Kontsevich, M. (2003). "Deformation Quantization of Poisson Manifolds". Letters in Mathematical Physics, 66 : 157-216. - -[remarks : Fedosov's work seems to be also available with details in a book Deformation quantization and index theory. Are the two references overlapping ? -- I've found several documents from Kontsevich having similar titles, from 1997 to 2003, but I've no access to the reference of 2003, is the arXiv version the same as the final one ?] -My problem is that these works are really deep, long and difficult to me, so before continuing reading them, I'd like to understand whether these two works are equivalent or not, if they overlap somehow, and which kind of problem these works solved. If the answers could be without too much details for a physicist I'd really appreciate continuing learning this interesting topic. - -REPLY [6 votes]: Fedosov's work seems to be also available with details in a book Deformation quantization and index theory. Are the two references overlapping ? - -Yes, indeed. The book contains strictly more than the contents of the paper from 1994. - -I've found several documents from Kontsevich having similar titles, from 1997 to 2003, but I've no access to the reference of 2003, is the arXiv version the same as the final one ? - -Again, yes. The preprint from 1997 is essentially the same as the paper published in LMP in 2003. - -I'd like to understand whether these two works are equivalent or not, if they overlap somehow, and which kind of problem these works solved. - -As it is explained in the comment by Bertram Arnold, Fedosov's quantization works for symplectic manifolds only (and can actually be generalized to regular Poisson manifolds) while Kontsevich's quantization works for general Poisson manifolds (even the ones having non-regular symplectic foliation). -In the symplectic case, the local deformation quantization is known to exist, and is essentially unique (it's the Moyal-Weyl star-product). So, the deformation quantization problems is a globalization problem, that is solved by Fedosov using methods from formal geometry and homological algebra. -In the general Poisson case, even finding a local formula is a difficult task. This is actually the most difficult part of Kontsevich's theorem. Once this is done, one proves that the local formula satisfy certain property that allows to run a globalization precedure that is similar to the one used by Fedosov. -Hence, Kontsevich's result definitely encompasses the one of Fedosov. - -which kind of problem these works solved - -Fedosov's work solves the problem of finding a star-product quantizing the Poisson bracket for functions on a symplectic manifold. -Kontsevich's work solves the far more general problem of finding a star-product quantizing the Poisson bracket for functions on a Poisson manifold. -Concerning the path integral approach to deformation quantization (work of Cattaneo--Felder), that is mentionned in Bertram Arnold's comment, Kontsevich's approach relies on a Sigma model with 2d source/worldsheet (the so-called Poisson sigma-model), while Fedosov's procedure can be unerstood as relying on a sigma model with 1d source/worldsheet (see e.g. Cattaneo's Part III of Déformation, quantification, théorie de Lie -in English despite the french title-, Chapters 10-11, or the more recent work of Grady--Li--Li https://arxiv.org/pdf/1507.01812.pdf).<|endoftext|> -TITLE: Bhargava's work on the BSD conjecture -QUESTION [19 upvotes]: How much would Bhargava's results on BSD improve if finiteness of the Tate-Shafarevich group, or at least its $\ell$-primary torsion for every $\ell$, was known? Would they improve to the point of showing $100$% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture? (which, of course, would still not prove the BSD conjecture) -I've just attended a very nice seminar talk about the topic, and I'm curious to get some expert info. - -REPLY [8 votes]: This was meant to be a comment, but it won't fit, so here we go. -Just a few naive observations. -For completeness: - -Weak BSD For any global field $K$ and any abelian variety $A$ defined over $K$, we have: - $$\text{ord}_{s=1}L(A/K,s) = \text{rk}_{\mathbf{Z}}A(K).$$ -BSD For any global field $K$ and any abelian variety $A$ defined over $K$, we have: -(1) $\text{Sha}(A/K)$ is a finite abelian group. -(2) $\rho :=\text{ord}_{s=1}L(A/K,s) = \text{rk}_{\mathbf{Z}}A(K).$ -(3) $\lim_{s\to 1}L(A/K,s)(s-1)^{-\rho} = \frac{R_A\cdot\Omega_A\cdot\#\text{Sha}(A/K)}{|\Delta_K|^{1/2}\cdot \# A(K)_{\rm tor}\cdot\# A^{\vee}(K)_{\rm tor}}.$ -Super-weak BSD The proportion of abelian varieties defined over $K$ that satisfy Weak BSD above, is 100%. - -If $K$ is the global function field of a smooth projective geometrically irreducible curve over $\mathbf{F}_q$, then it is known by work of Tate, Milne, Bauer, Schneider, Kato, Trihan, that Weak BSD is equivalent to BSD, and BSD is in turn equivalent to finiteness of the $\ell$-primary torsion in the Tate-Shafarevich group for some prime $\ell$ (allowed to be the characteristic of $K$). -If $K$ is a number field, the above stream of equivalences is still expected to be true, but not known yet, likely because there's still no robust cohomological method to put us in the position to mimic the flat cohomology/crystalline-syntomic cohomology methods employed in the positive characteristic case, which Iwasawa theory for abelian varieties was partially meant for. Let us grant for a moment the following: - -Expectation Let $K$ be a number field. For any abelian variety $A$ defined over $K$, the following are equivalent: -(1) $\text{Sha}(A/K)$ is finite. -(2) For all primes $\ell$, the $\ell$-primary torsion subgroup in $\text{Sha}(A/K)$ is finite. -(3) For some prime $\ell$, the $\ell$-primary torsion subgroup in $\text{Sha}(A/K)$ is finite. -(4) Weak BSD is true for $A$. -(5) Strong BSD is true for $A$. - -This expectation, a Theorem if $K$ were a positive characteristic global function field, reveals the problem is really about showing finiteness of $\text{Sha}(A/K)(\ell)$. -Over number fields, it is not even known that $\text{Sha}(A/K)(\ell)$ vanishes for almost all primes $\ell$. -The takeaway from the Bhargava-Skinner-Zhang paper, and from the answers and comments here, is that knowledge of finiteness $\text{Sha}(A/K)(\ell)$ does not actually help much or at all to improve their progress on Super-weak BSD ($\text{Sha}$-finiteness enters through the parity conjecture, a theorem unconditionally, and definitely a much weaker statement than finiteness of $\text{Sha}(A/K)(\ell)$), which to me just means such methods fail to get to the point of the problem itself, and will not solve it. - -In other words, I don't see any trace of the ability to produce interesting algebraic cycles on abelian varieties, into them, and according to the Expectation above, true in char $p$ though open in char $0$ so far, this should be the whole point. - -Regardless, clearly Super-weak BSD, which is what the discussion in the question, comments, and around the Bhargava-Skinner-Zhang paper, were about, is not equivalent to weak BSD under any circumstances, no matter that $K = \mathbf{Q}$ and $A$ is $1$-dimensional. -Showing 100% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture does not show weak BSD (as the OP took care to make clear in his/her question). Such result, if ever available, should be regarded as being motivational only. -EDIT: I should also add that several of the averages (both unconditional and conjectural) that are key to the Bhargava-Skinner-Zhang methods, fail in char $p$, while I'd regard a method towards BSD to be "promising", if it were able to make progress or settle its char $p$ counterpart, first.<|endoftext|> -TITLE: Relative/acylindrical hyperbolicity of free-by-cyclic groups -QUESTION [7 upvotes]: Is this statement true? - -Let $\mathbb{F}$ denotes a finitely generated free group, $\Phi$ an automorphism of $\mathbb{F}$ and $\varphi$ its image in $\mathrm{Out}(\mathbb{F})$. -If $\varphi$ is polynomially growing and of infinite order, then the semidirect product $\mathbb{F}\rtimes_\Phi\mathbb{Z}$ -is acylindrically hyperbolic -but not relatively hyperbolic. - -If there is anything unclear about notation, please tell me. This statement is closely related to Problem 8.2 in A. Minasyan and D. Osin. Acylindrical hyperbolicity of groups acting on trees. -math. Annalen, 362:1055–1105, 2015 (arXiv link). This problem asks which mapping tori of injective endomorphisms of finitely generated free -groups are acylindrically hyperbolic. - -REPLY [5 votes]: In addition to Henry's answer, I would like to mention that the acylindrical hyperbolicity of free-by-cyclic groups is fully characterised in the recent preprint Acylindrical hyperbolicity of automorphism groups of infinitely-ended groups, allowing us to remove the word virtually in Button and Kropholler's statement. Actually, we have the more general proposition: -Theorem: Fix a group $H$ and a morphism $\varphi : H \to \mathrm{Aut}(\mathbb{F}_n)$ where $n \geq 2$. The semidirect product $\mathbb{F}_n \rtimes_\varphi H$ is acylindrically hyperbolic if and only if -$$\mathrm{ker} \left( H \overset{\varphi}{\to} \mathrm{Aut}(G) \to \mathrm{Out}(G) \right)$$ -is a finite subgroup of $H$. -In the particular case where $H= \mathbb{Z}$, we find that $\mathbb{F}_n \rtimes_\Phi \mathbb{Z}$ is acylindrically hyperbolic if and only if $\Phi$ has infinite order in $\mathrm{Out}(\mathbb{F}_n)$. -Such a behaviour seems to be quite common. The same statements hold for infinitely-ended groups, non-elementary hyperbolic groups, most non-elementary relatively hyperbolic groups (including toral relatively hyperbolic groups), irreducible right-angled Artin/Coxeter groups (and more generally, most graph products of groups; see this preprint).<|endoftext|> -TITLE: Differential algebraic geometry vs Diffiety theory -QUESTION [11 upvotes]: Algebraic geometry is said to be useful to study not only specific solutions of polynomial equations but to understand the intrinsic properties of the totality of solutions of a system of equations. -However, varieties (which are the central object of study in algebraic geometry) are generalisations of solution sets of polynomial equations and thus do not seem suitable for investigating (intrinsic properties of solutions of a system of) differential equations. -But precisely the investigation of differential equations in such an abstract manner would probably enable great applications in physics and computer science. -While looking for approaches that generalise the notions of algebraic geometry to the study of differential equations I basically came across the following two (if someone knows more, please let me know as well): - -Differential algebraic geometry -Diffiety Theory - -However, the information provided on the wiki-sides are not very helpful. Furthermore, I only found this post on mathoverflow about diffieties which is not directly addressing my question. I found a short introduction to the idea of a diffiety on the ncat-lab but it does not contrast the idea and its applications. -My questions are: - -Does anyone know both approaches well enough to compare their methods and applications? If so, please share this knowledge. -If I want to learn one of these approaches, which one should I choose? -And what would be good books to start? - -PS: I found a link to a so-called diffiety institute which provides access to a lot of writings on diffiety theory. And here are some more information on differential algebra. -EDIT: The diffiety institute site is not being updated anymore except for the library but I found a more up-to-date website carrying on the promotion of the idea of a diffiety: The Levi-Civita Institute. - -REPLY [6 votes]: Disclaimer: I don't know much about this, and I hope more knowledgeable people weigh in. -You might want to take a look at Ayoub's differential Galois theory for schemes and the foliated topology (see preprint). -If we are interested in solutions of a single polynomial equation in one variable (over a field and its algebraic extensions), the relevant part of algebra is Galois theory. There is a well-known version of Galois theory for differential equations over (differential) fields, also called Picard-Vessiot theory. -From the point of view of algebraic geometry, Galois theory is the study of the etale site of the spectrum of a field. Thus a natural version of your question (though maybe not exactly what you wanted) asks for a version of differential Galois theory for varieties. -Since the extension of the classical Galois theory to varieties (or schemes) relies on the etale topology, a natural lead is to look for a topology with an analogous relation to Picard-Vessiot theory and its natural extension to varieties. As far as I understand, this is what Ayoub's foliated topology does (among other things). -A somewhat relevant quote from Ayoub (p. 4) - -En un sens, la topologie étale est une approximation profinie de la topologie transcendante. Un slogan que j’aimerai proposer est le suivant: la topologie feuilletée est une approximation pro-algébrique de la topologie transcendante!<|endoftext|> -TITLE: On critical reviews of Hawking's lecture "Gödel and the end of the universe" -QUESTION [38 upvotes]: The search for a neat Theory of Everything (ToE) which unifies the entire set of fundamental forces of the universe (as well as the rules which govern dark energy, dark matter and anti-matter realms) has been the subject of a long-standing all-out effort of physicists since the early 20th century. Some complicated theories such as Quantum Field Theory and M-Theory have been developed along these lines. However, the ultimate theory of everything still seems far out of reach and highly controversial.(A related debate: 1, 2). -The difficulty of the hopeless situation brought some physicists, such as Hawking, on the verge of total disappointment. Their general idea was that maybe such an ultimate theory is not only out of reach (with respect to our current knowledge of the universe) but also fundamentally non-existent. In his Gödel and the End of the Universe lecture Hawking stated: - -Some people will be very disappointed if there is not an ultimate -theory that can be formulated as a finite number of principles. I used -to belong to that camp, but I have changed my mind. I'm now glad that -our search for understanding will never come to an end, and that we -will always have the challenge of new discovery. Without it, we would -stagnate. Gödel’s theorem ensured there would always be a job for -mathematicians. I think M-theory will do the same for physicists. I'm -sure Dirac would have approved. - -A glimpse of Hawking's lecture makes it clear that the argument which he uses for refuting the possibility of achieving the Theory of Everything in his lecture, is loosely (inspired by and) based on Gödel's incompleteness theorems in mathematics. Not to mention that Hawking is not the only person who brought up such an argument against ToE using Gödel's theorems. For a fairly complete list see here and here. Also, some arguments of the same nature (using large cardinal axioms) could be found in this related MathOverflow post. -Hawking's view also shares some points with Lucas-Penrose's argument against AI using Gödel's incompleteness theorems, indicating that human mind is not a Turing machine (computer) and so the Computational Theory of Mind's hope for constructing an ultimate machine that has the same cognitive abilities as humans will fail eventually. -There have been a lot of criticism against Lucas-Penrose's argument as well as the presumptions of Computational Theory of Mind. Here, I would like to ask about the possible critical reviews on Hawking's relatively new idea. - -Question: Is there any critical review of Hawking's argument against the Theory of Everything in his "Gödel and the End of the Universe" lecture, illustrating whether it is a valid conclusion of Gödel's incompleteness theorem in theoretical physics or just yet another philosophical abuse of mathematical theorems out of the context? -Articles and lectures by researchers of various background including mathematicians, physicists, and philosophers are welcome. - -REPLY [7 votes]: Based on some of the comments I have seen, some commentators are finding Hawking's arguments vague or unconvincing. These are legitimate criticisms. But I think he has been partially, if not completely vindicated, by two recent advances that were not fully developed at the time the paper was published. - -The undecidability of the spectral gap problem. - -Hawking states: if there are mathematical results that can not be proved, there are physical problems that can not be predicted. -Since Hawking wrote this article, Cubitt, Perez-Garcia and Wolf showed that the spectral gap is undecidable by proving that the spectral gap decision problem is equivalent to the halting problem. And the halting problem is the theoretical computational equivalent of Godel's first theorem. This result does indeed suggest that some problems regarding the in quantum physics cannot be predicted, and many of the most challenging and long-standing open problems in theoretical physics concern the spectral gap, - -David Wolpert's inference engine proof - -Hawking states: "But we are not angels, who view the universe from the outside. Instead, we and our models are both part of the universe we are describing. Thus a physical theory is self referencing, like in Godel’s theorem." -The Wolpert "inference engine" proof is , similar to the results of Gödel’s incompleteness theorem and Turing’s halting problem, and, indeed, even the the Spectral Gap problem. It relies on a variant of the liar’s paradox—ask Laplace’s demon to predict the following yes/no fact about the future state of the universe: “Will the universe not be one in which your answer to this question is yes?” For the demon, seeking a true yes/no answer is like trying to determine the truth of “This statement is false.” Knowing the exact current state of the entire universe, knowing all the laws governing the universe and having unlimited computing power is no help to the demon in saying truthfully what its answer will be. -Philippe M. Binder, a physicist at the University of Hawaii at Hilo, suggests that Wolpert's theory implies researchers seeking unified laws cannot hope for anything better than a “theory of ALMOST everything.”<|endoftext|> -TITLE: Advanced software for OEIS? -QUESTION [27 upvotes]: Is there (if not, why?) a software where I can input a sequence of integers, like into the OEIS, and then it makes some simple transformations on it to check whether the sequence can be obtained from some other sequence(s)? -For example, if I enter 2, 4, 6, 8, 10, then currently OEIS returns A005843: The nonnegative even numbers (which starts with 0, but OEIS can search for subsequences). -But suppose that A005843 is not in the database yet. -Then OEIS will not return anything. -Instead, a more intelligent search software could return A000027: The positive integers. -I know that this raises several questions, like what transformations, which sequences to display first etc., but the feature could be quite useful. -Even better, if the software could do more complicated things, like check whether my sequence is the sum of two OEIS entries. -ps. My motivation came from the sequence 2, 4, 9, 16, 27, 38 currently missing from OEIS, which was posed on this Hungarian puzzle page. -(The puzzle has already expired, so feel free to discuss.) - -REPLY [15 votes]: There has been some previous discussion on the OEIS mailing list about similar topics. For instance, about the sum of two sequences. (If I recall correctly, there was a university project that performed a more in depth search to find new combinations for existing sequences, but I can't seem to find the details now...). -For more of an answer to your question, I would check out this thread: http://list.seqfan.eu/pipermail/seqfan/2015-February/014455.html where a user links to some code (on github) which is currently implemented on a website http://www.sequenceboss.org/ (was a bit slow on first load for me). -Plugging in the above numbers gives me: - -Result -SequenceBoss thinks the sequence -$$(a_n)_{n≥1}=2,4,9,16,27,38,…$$ -is generated by -$$a_n=(−1+n)^2+\text{prime}(n)$$ -If true, the sequence continues -$$2,4,9,16,27,38,53,68,87,110,131,…$$ -http://sequenceboss.org/?q=2%2C+4%2C+9%2C+16%2C+27%2C+38<|endoftext|> -TITLE: Is Yetter's invariant multiplicative under connected sum? -QUESTION [8 upvotes]: Classical formulation -Consider the (untwisted) Dijkgraaf-Witten invariant, defined for an oriented, connected, closed manifold $M$ and a finite group $G$: -$$DW_G(M) := \lvert \operatorname{Hom}(\pi_1(M), G )\rvert $$ -We're counting group homomorphisms from the fundamental group of the manifold to $G$. I am multiplying the more common definition by the size of the group, in order to simplify the following formulas. (Some are of the opinion that each homomorphism should be weighted by the inverse size of its automorphism group. I think this is not so, see e.g. this article. Instead, I think you can also sum over equivalence classes of $G$-bundles, which correspond to conjugacy classes of homeomorphisms, and there you need the weights.) -Now this invariant satisfies a simple identity, where $\#$ denotes connected sum: -$$DW_G(M_1 \# M_2) = \lvert \operatorname{Hom}(\pi_1(M_1 \# M_2), G )\rvert = \lvert \operatorname{Hom}(\pi_1(M_1) * \pi_1(M_2), G )\rvert = \lvert \operatorname{Hom}(\pi_1(M_1)) \rvert \cdot \lvert \operatorname{Hom} (\pi_1(M_2), G )\rvert = DW_G(M_1) \cdot DW_G(M_2) $$ -The Dijkgraaf-Witten invariant is multiplicative under direct sum. -We've used the free product of groups, denoted as $*$. We know from the Seifert-van-Kampen theorem that $\pi_1(M_1 \# M_2) = \pi_1(M_1) * \pi_1(M_2)$. -Homotopy theory formulation -We can also write the Dijkgraaf-Witten invariant as the number of homotopy classes into the classifying space of $G$: -$$DW_G(M) = \lvert [M, BG] \rvert$$ -Indeed, instead of $BG$ we could insert any (sufficiently finite) homotopy 1-type. This calls for a generalisation, known as the Yetter invariant. -An aside: The twisted invariant -Let $[M]$ be the fundamental class of the oriented manifold $M$, $\omega \in H^{\operatorname{dim} M}(G, k^\times)$ a group cohomology element (with values in the unit group of a field) and $c\colon M \to B(\pi_1(M))$ the canonical map. Then the twisted Dijkgraaf-Witten invariant is defined as: -$$DW_G^\omega(M) := \sum_{\phi\colon \pi_1(M) \to G} \langle \phi^*(\omega), c_*([M]) \rangle $$ -The inner product $\langle - , - \rangle$ comes from Poincaré duality. -It is easy to see that if $\omega$ is the trivial cocycle, the sum ranges over $1$, and we recover the original formula. -Yetter invariant -David Yetter, and then later Tim Porter, defined, for a homotopy 2-type $\mathcal{T}$, the following invariant: -$$Y_\mathcal{T}(M) = \lvert [M, \mathcal{T}] \rvert$$ -I think the original definition is more hands-on, in terms of crossed modules. -Edit: I seem to have omitted the correct groupoid cardinalities here. See Arun Debray's answer further below for the correct version, or this article. -Since there are some kinds of higher formulations of Seifert-van-Kampen (as discussed in this question) suitable for this situation, I'm asking the following question: -Question: Is the Yetter invariant multiplicative under connected sum? I.e. does the following hold: -$$Y_\mathcal{T}(M_1 \# M_2) = Y_\mathcal{T}(M_1) \cdot Y_\mathcal{T}(M_2)$$ - -REPLY [3 votes]: As Kevin Walker pointed out in a comment, Dijkgraaf-Witten invariants are weighted by $1/\mathrm{Stab}(\rho)$. In -the same way, the Yetter invariant for $\mathcal T$ and $M$ is generally defined such that it's weighted using the -2-groupoid cardinality of $\pi_{\le 2}\mathrm{Map}(M, \mathcal T)$, so that the invariant is -$$\sum_{[f\colon M\to\mathcal T]} \frac{|\pi_2(\mathrm{Map}(M, \mathcal T), f)|}{|\pi_1(\mathrm{Map}(M, \mathcal T), -f)|}.$$ -If we use this normalization, the Yetter invariants are the partition functions of a TQFT $Z_{\mathcal T}$, usually -called the Yetter model. In this case, a different MathOverflow -answer by Kevin Walker tells us that $Z_{\mathcal T}$ is multiplicative -under connect sum iff - -$\dim Z_{\mathcal T}(S^{n-1}) = 1$, and -$Z_{\mathcal T}(S^n) = 1$. - -The state space $Z_{\mathcal T}(M^{n-1}) := \mathbb C[[M, \mathcal T]]$, so for $n = 3$, the first property doesn't -hold: $\dim Z_{\mathcal T}(S^2) = |\pi_2(\mathcal T)|$. A similar problem occurs for $n = 2$. -If $n > 3$, then $[S^{n-1}, \mathcal T] = 0$, so the first property holds. The second property does not quite hold: -$[S^n, \mathcal T] = 0$, but we have to calculate the weighting. Since $\mathrm{Map}(S^n, \mathcal -T)\simeq\mathrm{Map}(\mathrm{pt}, \mathcal T)\cong \mathcal T$, -$$Z_{\mathcal T}(S^n) = \frac{|\pi_2(\mathcal T)|}{|\pi_1(\mathcal T)|},$$ -which is frequently not equal to 1. -I don't know about the unweighted version you mentioned, since it doesn't come from a TQFT as far as I know.<|endoftext|> -TITLE: Affine Paved Affine Varieties; not Affine Space -QUESTION [11 upvotes]: Let $V$ be an algebraic variety. If there is a finite ascending chain of Zariski closed sets $\emptyset=V_0\subset V_1\subset \cdots \subset V_n=V$ such that $V_i-V_{i-1}$ is a fintie disjoint union of copies of affine space $\mathbb{A}^i$ we say $V$ is affine paved (so $V$ is "algebraically cellular"). -Note: there are non-equivalent variations of this definition (see here). -One can deduce that an affine paved variety (over $\mathbb{C}$) has no odd cohomology and its even cohomology is free abelian. -Examples: - -Finite disjoint unions of affine space are affine paved. Let's call these examples "trivial." -Projective space is affine paved. -The Bruhat cells in a flag variety show there are interesting projective examples. - - -Question: Are there non-trivial affine paved affine varieties? - -This very well might be a silly question. Perhaps the affine cone over an affine paved projective variety always works? It doesn't seem clear to me, and I figured someone out there might have thought about such examples before (Google doesn't seem to know any). - -REPLY [14 votes]: One example is $uw = v(v+1)$. The equations $u=v=0$ cut out an $\mathbb{A}^1$, and the complement is isomorphic to $\mathbb{A}^2$ by the map $(x,y) \mapsto (y, xy-1, x(xy-1))$. Note that the hypersurface $u=v=0$ is not principal; by dhy's comment, it can't be. -More generally, any of the Danielewski surfaces $u w^k = \prod_{i=1}^r (v-\alpha_i)$, with the $\alpha_i$ distinct, should be paveable with one $\mathbb{A}^2$ and $(r-1)$ copies of $\mathbb{A}^1$. -I wrote a related blogpost. Much good discussion in the comments!<|endoftext|> -TITLE: Is every total computable function definable by a normalizing lambda term? -QUESTION [8 upvotes]: $\newcommand{\nat}{\mathbb{N}}$ -$\newcommand{\then}{\ \Longrightarrow\ }$ -A partial function $f : \mathbb{N} \to \mathbb{N}$ is said to be $\lambda$-definable if there is a term $F \in \Lambda$ such that -$$ \forall n \in \nat: \quad f(m)\!\downarrow = n \then -F c_n =_\beta c_m $$ -where $=_\beta$ is beta conversion, and $c_n = \lambda f s. f^n(s)$ is Church's encoding of numerals in $\Lambda$. -It is well-known that every partial computable function is $\lambda$-definable. However, whenever a particular recursive definition of the function $f$ makes use of the $\mu$-operator (minimization), then the $\lambda$-term corresponding to that definition will contain a fixed point combinator, hence not be normalizing in general. -On the other hand, large classes of total recursive functions can be $\lambda$-defined by normalizing terms. For example, Girard showed that every function which is provably total in higher-order arithmetic can be defined by a term typable in the system $F\omega$, and every term in this system is normalizing. - -REPLY [6 votes]: Yes, you can actually encode all computable functions by normalizable terms since it is always possible to turn a term into a normal form which computes the same function on Church's numerals. We just need to put $n (\lambda x. x)$ in front of every lambda in your term, where $n$ is the argument of the function. Formally, define function $r$ as follows: -\begin{align*} -r(n, x) & = x \\ -r(n, \lambda x. t) & = n (\lambda x. x) (\lambda x. r(n, t)) \\ -r(n, f a) & = r(n, f) r(n, a) -\end{align*} -Now, if we have a term $F$, then we can define $F'$ as $\lambda n. r(n, F n)$. Clearly, $F'$ is a normal form and, for every natural number $n$, terms $F c_n$ and $F' c_n$ are $\beta$-equivalent.<|endoftext|> -TITLE: Factor traces of the Temperley-Lieb algebra -QUESTION [7 upvotes]: Given $\delta\in\mathbb C$, let $A(\delta)$ denote the complex unital $*$-algebra generated by an identity $1$ and selfadjoint elements $e_k$, $k\in\mathbb N$, satisfying $e_k^2=\delta e_k$, $e_ke_l=e_le_k$ for $|k-l|\geq2$ and $e_ke_{k\pm1}e_k=e_k$ (This puts well-known constraints on the possible values of $\delta$). I wonder if it is known what all the extremal traces (factor traces) of $A(\delta)$ are, i.e. the positive normalized functionals $\tau:A(\delta)\to\mathbb C$ such that $\tau(ab)=\tau(ba)$ for all $a,b\in A(\delta)$ and having the property that $\tau$ cannot be written as a non-trivial convex combination of other traces. -Markov traces are known to be factor traces. But are there additional factor traces on $A(\delta)$, and can they be specified explicitly? Any reference to the literature would be welcome. - -REPLY [2 votes]: For $\delta < 2$, I think you can read this off from Jones's "index for subfactors" paper that the "finite depth"-ness forces that the Markov trace is the only trace for each allowed $\delta$. -For $\delta \ge 2$, again from Jones's paper the C$^*$-envelope of $A(\delta)$ is independent of $\delta$, and is in fact a quotient of $C^*(S_\infty)$ (which is easy to see for $\delta=2$ from the Schur-Weyl duality). The factor traces on the latter are classified by the Thoma parameter $0 \le \alpha_n$, $0 \le \beta_n$, $\sum_n \alpha_n + \beta_n \le 1$ ($1 \le n < \infty$), and those induced from $A(\delta)$ correspond to $\alpha_1 + \alpha_2 = 1$ (Wassermann's PhD thesis is the original reference I think?). These correspond to the Markov traces again. -So at the end of the day, the Markov traces are the only factor traces if I'm not mistaken.<|endoftext|> -TITLE: Does the Day convolution induce the structure of a bimonoidal category on $Fun(C,D)$? -QUESTION [6 upvotes]: Let $\mathcal{C}$ and $\mathcal{D}$ be symmetric monoidal categories and assume that the symmetric monoidal product $\otimes_{\mathcal{D}}$ on $\mathcal{D}$ preserves colimits in both variables. -Then the Day convolution $*$ defines a symmetric monoidal structure on the category of functors $Fun(\mathcal{C},\mathcal{D})$. -(The obligatory nlab link and here a treatment for $\infty$-categories. In fact nlab only considers the case that $\mathcal{D}$ is the category we're enriched in, so $\mathit{Set}$ for the purpose of this question.) -There is also a (much easier to define) symmetric monoidal structure on $Fun(\mathcal{C},\mathcal{D})$, given by 'point-wise multiplication' using $\otimes_\mathcal{D}$. (This does not use the symmetric monoidal structure on $\mathcal{C}$.) - -I have two questions about these structures: - - -Does $\otimes_{\mathcal{D}}$ distribute over $*$, i.e. do we have $F\otimes (G* H) \cong (F\otimes G) * (F \otimes H)$ ? - - -More precisely, does $Fun(\mathcal{C},\mathcal{D})$ have the structure of a bimonoidal category/rig category, where $*$ is '$+$' and $\otimes$ is '$\times$'? -If not, are they compatible in some other sense? - - -Is the Day convolution of two symmetric monoidal functors again symmetric monoidal? - - -This is easy for the other product $\otimes_{\mathcal{D}}$, which in fact induces a symmetric monoidal structure on the category of symmetric monoidal functors $Fun^{\otimes}(\mathcal{C}, \mathcal{D})$. -Combined the question would be whether this extends to a bimonoidal structure using the Day convolution. - -I believe both of the answers to be yes when $\mathcal{C}$ is the category of oriented $1$-bordisms between $0$-manifolds (aka the free symmetric monoidal category on one dualisable object). So I was wondering if this in general gives a bimonoidal structure on the category of TFTs. This would be nice because the definition of the direct sum of two TFTs is kind of awkward and written out it looks a lot like the Day convolution. (One defines $(F\oplus G)(M) := F(M) \oplus G(M)$ for connected objects $M$ and then uses the fact that every manifold decomposes uniquely as the disjoint union of its connected components.) -On the other hand it might only be this nice property of the bordism category, namely having a 'unique prime decomposition', that makes the result work in this case. - -Edit: There was a mistake concerning which way the distributivity is supposed to go, I hope this is more clear now. - -REPLY [5 votes]: I don't know in general whether $\otimes_D$ distributes over $\ast$. However, I think it is unlikely, because there is in general a different compatibility between them: they are a duoidal category. This is stated without proof on the nLab page, probably as a generalization from the special case $[V_\kappa,V]$ discussed in arXiv:1507.08710.<|endoftext|> -TITLE: Eigenvalue and eigenvector of ergodic Markov operator for continuous space Markov chain -QUESTION [5 upvotes]: As we know that the transition matrix $P$ of a Markov chain with finite space is a stochastic matrix, and from Perron-Frobenius Theorem, we know that the spectral radius of the matrix $P$ is $1$, and the eigenvalue $1$ is simple. I was wondering if we consider ergodic Markov process with continuous spaces (from which we know that there exists a unique invariant measure), what would be the eigenvalue and eigenfunction of the corresponding Markov operator (sometimes called Perron-Frobenius operator)? Is the eigenvalue guaranteed to have module smaller than $1$? Is the eigenvalue simple as well? - -REPLY [3 votes]: With respect to slightly modified versions of your second and third questions (Is the second eigenvalue guaranteed to have module smaller than 1? Is the first eigenvalue simple as well?), the answers are both Yes, if the transition matrix is Harris recurrent and the associated operator self-adjoint and compact. -The case of MC on continuous state spaces $S$ is different from the case of finite state space. First of all, the transition matrix is replaced by a transition kernel $P(x,A)$=probability of reaching $A\subset S$ from the state $x.$ In this case, if there is an invariant distribution $\pi$ and $P$ is $\pi$ irreducible, then $\pi$ is the unique invariant distribution of $P,$ moreover, if $P$ is aperiodic, then for $\pi$-almost all $x$ we have the convergence of $P^n(x,A)$ (as $n\to \infty$) to $\pi(A)$ in the Total Variation Norm. -One can recover a similar version of the finite state space case only when $P$ is Harris recurrent, as the convergence occurs for all $x.$ Under this assumption one can do the following: - -Define the Hilbert space $L^2(\pi)=\{f: \int |f|^2 d\pi<\infty \}$ with the natural inner product. -If $L^2(\pi)$ is separable and it has an orthonormal basis, one can define a linear operator on $L^2(\pi)$ by $Af(x):=\int p(x,y)f(y)dy,$ where $p(x,\cdot)$ is the transition density for $P.$ - -This operator $A$ will be analogous to your transition matrix $P$ for the finite state case: $1$ is an eigenvalue with eigenfunction $f(y)\equiv 1$ (and it is the unique largest eigenvalue). If you make extra assumptions on your operator $A$ you can get more information of the other eigenvalues, for example, if $A$ is self-adjoint and compact then the eigenvalues are in $[0,1],$ there are at most countably many eigenvalues, and the unique possible accumulation point is the origin. You can even study the spectral gap by approximating the second eigenvalue. The importance of the spectral gap in this setting is similar to the finite space case, measuring the efficiency of algorithms for approximating integrals (with all the applications that this may have). It is usually a hard problem and there is a lot of interest for practical Monte Carlo Markov chains. -You may be also interested in the following references: - -I. Kontoyiannis and S.P. Meyn. Geometric -ergodicity and the spectral gap of non-reversible Markov chains. -Probability Theory and Related Fields. (2012). -M. Ahues, A. Largillier and B. Limaye. Spectral -Computations for Bounded Operators. (2001). -G. Helmberg. Introduction to Spectral Theory in Hilbert Space. Elsevier. (2014). -Q. Qin, J.P. Hobert and K. Khare. Estimating the spectral gap of a trace-class Markov operator, arXiv<|endoftext|> -TITLE: If a measure $\mu$ and Lebesgue measure $\lambda$ are singular, is the derivative of $\mu$ with respect to $\lambda$ $\infty$, $\mu$-a.e.? -QUESTION [7 upvotes]: If a positive Radon measure $\mu$ and the Lebesgue measure $\lambda$ are singular, can we show that the derivative of $\mu$ with respect to $\lambda$ is $\infty$, $\mu$-a.e.? Namely, can one show that -$$\frac{\mu(B(x,r))}{\lambda(B(x,r))}\to\infty \quad \mbox{ as } \ r\to0, \quad \mu\mbox{-a.e.?}$$ -Does this result also hold for any pair of singular Radon measures $\mu\,\bot\,\lambda$? -Actually, a result I can find in [Thm 1.29, Evans-Gariepy, Measure Theory and Fine Properties of Functions] is that, for any Radon measures $\mu$ and $\lambda$, -$$\lim_{r\to0}\frac{\mu(B(x,r))}{\lambda(B(x,r))}<\infty \quad \mbox{ as } \quad r\to0, \quad \lambda\mbox{-a.e.}$$ - -REPLY [6 votes]: Let A be such that $\mu(A) = 1, \lambda(A) = 0$. There is an open set $O$ with $ A \subset O, \lambda(A) < \epsilon$. Suppose for convenience that $$\liminf\frac{\mu(B(x,r))}{\lambda(B(x,r))}< a $$ on $A$, otherwise replace $A$ with the set on which it is true. For each x in A pick $r_x$ so that $$B(x, r_x) \subset O \quad \text{ and } \quad \frac {\mu(B(x,r_{3x}))}{\lambda(B(x,r_{3x}))}< a .$$ By the Vitali covering lemma there is a disjoint subcollection $J$ with $$A \subset \cup_J B(x, r_{3x})$$ But then $$1 = \mu(A) \le \sum_J\mu(B(x,r_{3x})) \le a \sum_J \lambda(B(x,r_{3x} )) \le a \cdot 3 \cdot \epsilon $$ and since $\epsilon$ is at your disposal, this is a contradiction. I've written this for 1 dimension but I think it applies mutatis mutandis to all.<|endoftext|> -TITLE: Reference request: cohomology of Eilenberg Maclane spaces with $p$-local groups -QUESTION [7 upvotes]: In Rudyak's 'On Thom Spectra, Orientability, and Cobordism', the following fact is used: - -Let $\mathbb{Z}_{(p)}$ be $\mathbb{Z}$ localized at the ideal $(p)$. Let $\pi,\tau$ be two cyclic $\mathbb{Z}_{(p)}$-modules, that is, $\pi,\tau=\mathbb{Z}/p^m$ or $\mathbb{Z}_{(p)}$. Then the stable cohomology groups of the Eilenberg Maclane spaces $\varinjlim H^{i+N}(K(\pi,N);\tau)$ have exponent $p$, i.e. $px=0$ for every element $x$, for $i>0$. - -Rudyak says this fact is well-known and references Cartan's Seminaire in 1959. But I have skimmed through the articles in both Seminaire 1958-1959 and Seminaire 1959-1960, yet have not found this fact proved anywhere. Now it is perfectly possible that I have simply overlooked a proposition, since I can barely read French, but I would like to ask if anyone can provide a more specific reference that proves the above fact? -Alternatively, it would be appreciated if anyone can provide a sketch of a proof. Since Rudyak references Cartan Seminaire, it is likely that a proof using Serre's spectral sequence, which is heavily used in the Seminaire, can be constructed. Serre's mod $\mathcal{C}$ theory does not apply here since the groups with exponent $p$ do not form a Serre class. I believe a proof for the case $\pi=\mathbb{Z}/p$ would require a close inspection on how the order of the group dies down. And I don't have any idea how to start with $\pi=\mathbb{Z}_{(p)}$. -Any help is appreciated! - -REPLY [9 votes]: Here is a sketch proof. -Step 1: For sensible spaces or spectra (connected, finite type) $X$, $H^*(X;\tau) $ will have exponent $p$ for all the coefficient groups $\tau$ you list exactly when the image of the Bockstein $\beta: H^*(X;\mathbb Z/p) \rightarrow H^{*+1}(X;\mathbb Z/p)$ equals the kernel of $\beta$. This can be proved by fooling around with the various coefficient sequences defining the Bockstein. Bill Browder developed this idea into the `Bockstein spectral sequence', which he made good use of. -Step 2: So now we need to look at $H^*(H\pi;\mathbb Z/p)$, viewed graded vector space with the Bockstein action, and we want to show that this module is `$\beta$--acylic'. The Kunneth theorem applies, and we can reduce to the cases when $\pi = \mathbb Z$ or $\pi=\mathbb Z/p^m$. -Now we use the calculations of Serre, et. al. First of all, $H^*(H\mathbb Z/p;\mathbb Z/p) = A$, the mod $p$ Steenrod algebra. When written with Serre's admissible basis, one can see that $A$ is $\beta$--acyclic. (In fact, $A$ is forced by general Hopf algebra theory to be free as a module over the exterior algebra on $\beta$.) -Then $H^*(H\mathbb Z;\mathbb Z/p) = A/A\beta$, which one checks is also $\beta$--acyclic ($\beta$ acts on the left). Finally, if $m>1$, then $H^*(H\mathbb Z/p^m;\mathbb Z/p) = A/A\beta \oplus \Sigma A/A\beta$, and so is also $\beta$--acyclic. -So this is what Rudyak meant. -The canonical mod 2 reference is Serre, Jean-Pierre Groupes d'homotopie et classes de groupes abéliens. (French) Ann. of Math. (2) 58, (1953). 258–294. Odd primes would have been later, but not by much. -A little bit more added later $\dots$ -The cofibration sequence $H\mathbb Z \rightarrow H\mathbb Z \rightarrow H\mathbb Z/p$ induces a short exact sequence $0 \rightarrow \Sigma H^*(H\mathbb Z; \mathbb Z/p) \rightarrow A \rightarrow H^*(H\mathbb Z; \mathbb Z/p) \rightarrow 0$. It is not hard to see that the short exact sequence $0 \rightarrow im \beta \rightarrow A \rightarrow coker \beta \rightarrow 0$ maps to this (with the identity in the middle slot). The fact that $A$ is $\beta$--acyclic ($im \beta = ker \beta$) implies that $im \beta$ and $coker \beta$ are the `same size', as graded vector spaces (up to a suspension). This forces the map between the short exact sequences to be an isomorphism, and so one has $H^*(H\mathbb Z; \mathbb Z/p) = coker \beta = A/A\beta$.<|endoftext|> -TITLE: Geometric interpretation of algebraic tangent cone -QUESTION [13 upvotes]: Suppose $(A,\mathfrak m)$ is a Neotherian local $k$-algebra with residue field $k$. Then, we define (the coordinate ring of) its algebraic tangent cone to be the $k$-algebra $A_c = \sum_{i\ge 0} \mathfrak m^i/\mathfrak m^{i+1}$. -On the other hand, we also have the Zariski tangent space $(\mathfrak m/\mathfrak m^2)^\vee$, whose coordinate ring is the symmetric algebra $A_t = \sum_{i\ge 0} \text{Sym}^i(\mathfrak m/\mathfrak m^2)$. There is an evident surjective map of $k$-algebras $A_t\to A_c$, which gives an embedding of the algebraic tangent cone into the Zariski tangent space. -Griffiths-Harris, in their book, say that the tangent cone to a variety (over $\mathbb C$) at a point is the collection of tangent vectors obtainable as velocities of analytic arcs through that point lying on the variety. A naïve attempt at formalizing this in this algebraic setting could be the following. -A $k$-algebra map $A\to k[t]/t^2$ is the same as a Zariski tangent vector, and we can ask if this can be lifted (non-uniquely) to a $k$-algebra map $A\to k[[t]]$ (roughly this corresponds to going from a vector to a "formal path" tangent to it). We could then ask if the image of the embedding $\text{Spec } A_c\to\text{Spec }A_t = (\mathfrak m/\mathfrak m^2)^\vee$ corresponds exactly to those algebra maps to $k[t]/t^2$ which admit lifts to $k[[t]]$. -This is not the correct characterization as the example of $\mathbb C[[x,y]]/(y^2-x^3)$ shows, since any two formal power series $f,g$ in the variable $t$ with $g^2 = f^3$ must be divisible by $t^2,t^3$ respectively. Is there a way then to modify the above question so that the answer provides a nice characterization of the image of the tangent cone inside the Zariski tangent space? - -REPLY [5 votes]: You may wish to read the discussion on pages 106-108 of Eisenbud/Harris "The Geometry of Schemes", where they explicitly compute the tangent cone of $(y^2-x^3)$ as a degeneration of the tangent cones in the family $(y^2 - tx^2 - x^3)_{t \in \mathbb{A}^1}$. -In a direction close to what @JasonStarr said, though one requiring a background in jet schemes, you could look at Example 5.13 in Ein/Mustață "Jet Schemes and Singularities" where they describe some features of the arc space of the cusp that you seem to be curious about.<|endoftext|> -TITLE: Tensor product of coaugmented conilpotent coalgebras -QUESTION [5 upvotes]: Let $\mathbb{K}$ be a field of char. 0. -Let $\mathrm{A}, \mathrm{B}$ be conilpotent cocommutative coaugmented counital dg-coalgebras over $\mathbb{K}$ -(i.e. that their corresponding cokernel of their coaugmentation is a conilpotent cocommutative non-counital dg-coalgebra). -Is it always true that their tensorproduct $\mathrm{A} \otimes_{ \mathbb{K} } \mathrm{B}$ is conilpotent as a coaugmented counital dg-coalgebra? -In other words lifts the tensor product $- \otimes_{ \mathbb{K}} -$ of the category of dg-$\mathbb{K}$-vector spaces to the category of cocommutative coaugmented counital dg-coalgebras over $\mathbb{K}$? -From the very definition of conilpotency it follows that the tensorproduct $\mathrm{A} \otimes_{ \mathbb{K} } \mathrm{B}$ of two cocommutative non-counital dg-coalgebras $\mathrm{A}, \mathrm{B}$ over $\mathbb{K}$ is conilpotent as a non-counital dg-coalgebra if $\mathrm{A}$ or $ \mathrm{B}$ are conilpotent as non-counital dg-coalgebras. -But taking the cokernel of the coaugmentation only defines an oplax symmetric monoidal functor from cocommutative coaugmented counital dg-coalgebras to cocommutative non-counital dg-coalgebras. -However it follows from this that if we endow the category of cocommutative coaugmented counital dg-coalgebras with the smash-product $- \wedge_{ \mathbb{K} } -$ over $\mathbb{K}$, then $\mathrm{A} \wedge_{ \mathbb{K} } \mathrm{B}$ -is conilpotent as a coaugmented counital dg-coalgebra if $\mathrm{A}$ or $ \mathrm{B}$ are conilpotent as coaugmented counital dg-coalgebras. -But I want to consider the symmetric monoidal structure on the category of cocommutative coaugmented counital dg-coalgebras given by $- \otimes_{ \mathbb{K} } -$ to describe conilpotent cocommutative bialgebras as monoids in the category of conilpotent cocommutative coaugmented counital dg-coalgebras. - -REPLY [6 votes]: Yes, it is true. Moreover, neither cocommutativity nor characteristic $0$ are needed as conditions. The tensor product of two conilpotent coassociative dg-coalgebras over any field $k$ is a conilpotent coassociative dg-coalgebra over $k$. -The definition of a conilpotent dg-coalgebra includes two conditions: the dg-coalgebra $(C,d)$ should be coaugmented as a dg-coalgebra, that is, the coaugmentation $\gamma\colon k\to C$ should be a morphism of dg-coalgebras (which means the equation $d\circ\gamma=0$), and the underlying graded coalgebra should be conilpotent (which means that the cokernel of its coaugmentation $C/\gamma(k)$ should be a conilpotent coalgebra without counit). -Checking that the tensor product of two coaugmented dg-coalgebras is a coaugmented dg-coalgebra is quite easy, so let me skip it. Then the question reduces to showing that the tensor product of two conilpotent coalgebras is a conilpotent coalgebra. -Any (counital or noncounital) coassociative coalgebra is the union of its finite-dimensional subcoalgebras. In particular, a coagumented coalgebra is the union of its finite-dimensional coaugmented subcoalgebras. A finite-dimensional coalgebra is the dual vector space to a finite-dimensional algebra. In particular, a finite-dimensional coaugmented coalgebra is the dual vector space to a finite-dimensional augmented algebra. -A (counital or noncounital) coalgebra is conilpotent if and only if all its finite-dimensional (coaugmented) subcoalgebras are, and if and only if every its element is contained in a conilpotent finite-dimensional subcoalgebra. A finite-dimensional coalgebra $W$ without counit is conilpotent if and only if its dual finite-dimensional algebra $U=W^*$ satisfies $U^n=0$ for $n$ large enough. A finite-dimensional coaugmented coalgebra $C$ is conilpotent if and only if the augmentation ideal $A_+$ of its dual finite-dimensional algebra $A=C^*$ satisfies $A_+^n=0$ for $n$ large enough. -Let $A$ and $B$ be two augmented associative algebras such that $A_+^m=0$ and $B_+^n=0$. The tensor product $A\otimes_k B$ is an augmented associative algebra with the augmentation ideal $$(A\otimes_kB)_+=A_+\otimes B+A\otimes B_+=(A_+\otimes k) \oplus (A_+\otimes B_+) \oplus (k\otimes B_+) \subset A\otimes_kB.$$ Now you can easily see that $(A\otimes_kB)_+^{m+n-1}=0$. -As the duality (anti-equivalence of categories) between finite-dimensional algebras and finite-dimensional coalgebras takes tensor products to tensor products, and as the tensor product of directed unions is the union of tensor products, the desired assertion follows.<|endoftext|> -TITLE: Sum-regular $\{0,1\}$-matrices -QUESTION [11 upvotes]: Let $n\in\mathbb{N}$ be a positive integer. We say that an $n\times n$-matrix $A$ with all entries in $\{0,1\}$ is $k$-regular for some $k\in \{0,\ldots,n\}$ if the sum of every row and the sum of every column of $A$ equals $k$. Let $M(n, k)$ be the number of $k$-regular $n\times n$-matrices with all entries in $\{0,1\}$. -It's easy to see that $M(n,1)=n!$. Moreover, a symmetry argument shows that $M(n,k) = M(n, n-k)$ for all $k\in \{0,\ldots,n\}$. -Question. Given $n>1$, is it true that for all $k\in\{0,\ldots, n\}$ we have $M(n,k)\leq M(n, \lfloor\frac{n}{2}\rfloor)$? - -REPLY [9 votes]: I can prove a related result. Perhaps someone can -modify the proof to solve Dominic's problem. -I use multivariate notation such as $x^\alpha=x_1^{\alpha_1}\cdots -x_m^{\alpha_m}$, where $\alpha=(\alpha_1,\dots,\alpha_m)$. Let -$\alpha\in \{0,1,2,\dots\}^m$ and $\beta\in\{0,1,2,\dots\}^n$. Let $f(\alpha,\beta)$ be the -number of $m\times n$ matrices with entries $0,1,2$, -and with each entry equal to 1 colored either red or blue, with row -sum vector $\alpha$ and column sum vector $\beta$. -Theorem. - $$ f(\alpha,\beta) \leq f((n,n,\dots,n),(m,m,\dots,m)). $$ -Proof. Let $g(\alpha,\beta)$ be the -number of $m\times n$ matrices with entries $-2,0,2$, -and with each entry equal to 0 colored either red or blue, with row -sum vector $\alpha$ and column sum vector $\beta$. By dividing each -entry of such a matrix by 2 and then adding 1, it is clear that - $$ f(\alpha,\beta)=g\left(2\alpha-2(n,\dots,n), - 2\beta-2(m,\dots,m)\right). $$ -Hence we want to show that - $$ g(\alpha,\beta) \leq g((0,0,\dots,0),(0,0,\dots,0)). $$ -We have for fixed $m,n$ that - $$ \sum_{\alpha,\beta} g(\alpha,\beta)x^\alpha y^\beta = - \prod_{r=1}^m\prod_{s=1}^n (x_r^{-1}y_s^{-1}+x_ry_s)^2. $$ -Since for any integer $k$ we have $\int_0^{2\pi}e^{ikx}dx = 1$ if $k= - 0$ and otherwise is $0$, it follows that - $$ g(\alpha,\beta) = \frac{1}{(2\pi)^{m+n}} - \int_0^{2\pi}\cdots \int_0^{2\pi} e^{-i(\alpha_1 \theta_1+\cdots+ - \alpha_m\theta_m+\beta_1\psi_1+\cdots+\beta_n\psi_n)}\\ - \prod_{r=1}^m\prod_{s=1}^n (e^{-i(\theta_r+\psi_s)} - +e^{i(\theta_r+\psi_s)})^2\,d\theta\,d\psi. $$ -Now $(e^{-i(\theta_r+\psi_s)}+e^{i(\theta_r+\psi_s)})^2$ is a -nonnegative real number. Hence by the triangle inequality, - \begin{eqnarray*} g(\alpha,\beta) & \leq & - \frac{1}{(2\pi)^{m+n}} - \int_0^{2\pi}\cdots \int_0^{2\pi} - \prod_{r,s=1}^n (e^{-i(\theta_r+\psi_s)} - +e^{i(\theta_r+\psi_s)})^2\,d\theta\,d\psi\\ & = & - g((0,\dots,0),(0,\dots,0)).\ \Box \end{eqnarray*}<|endoftext|> -TITLE: Parabolic subgroups of relatively hyperbolic and CAT(0) groups -QUESTION [10 upvotes]: Let $G$ be a finitely generated group. We say that $G$ is CAT(0) if it acts properly and co-compactly by isometries on a CAT(0) space. -We say it is hyperbolic relative to a collection $\Omega$ of subgroups if it acts properly by isometries on a Gromov-hyperbolic space $X$ such that limit points are either conical or bounded parabolic and the stabilizers of the parabolic points are precisely the elements of $\Omega$. Those subgroups are called (maximal) parabolic subgroups or peripheral subgroups. -Question: Assume that $G$ is both CAT(0) and relatively hyperbolic. Are the peripheral subgroups CAT(0) themselves ? - -REPLY [9 votes]: Peripheral subgroups of relatively hyperbolic CAT(0) groups are indeed CAT(0) themselves. In fact, more is true: Morse subgroups of CAT(0) groups are CAT(0) themselves. -Definition. Given a finitely generated group $G$, $H \subset G$ is a Morse subgroup if, for every $A>0$ and $B \geq 0$, there exists some constant $K \geq 0$ such that any $(A,B)$-quasigeodesic between any two points of $H$ stays in the $K$-neighborhood of $H$ (in some fixed Cayley graph constructed from a finite generating set; the definition does not depend on this choice). -In his paper Quasi-convexity of hyperbolically embedded subgroups, Sisto proved that hyperbolically embedded subgroups are Morse subgroups, so in particular peripheral subgroups of relatively hyperbolic groups are Morse subgroups. -Now, I claim that Morse subgroups in CAT(0) groups are always convex-compact. (I already proved this statement for groups acting geometrically on CAT(0) cube complexes in my article Hyperbolicities in CAT(0) cube complexes (Proposition 4.2).) -Proposition: Let $X$ be a complete CAT(0) space and $S \subset X$ a Morse subset. The Hausdorff distance between $S$ and its convex hull is finite. -Proof. Let $x$ be a point in the convex hull of $S$, and fix a point $y \in S$. The first observation is that, for every $\epsilon>0$, there exists some $z \in S$ whose projection onto $[x,y]$, say $p$, is at distance less than $\epsilon$ from $x$. Otherwise, if $m$ denotes the point of $[x,y]$ at distance $\epsilon$ from $x$ and $\pi : X \to [x,y]$ the projection onto $[x,y]$, then $\pi^{-1}([m,y])$ would be a convex subspace(*) containing $S$ but not $x$, contradicting the fact that $x$ belongs to the convex hull of $S$. -Now, I claim that $[y,p] \cup [p,z]$ is a $(\sqrt{2},0)$-quasigeodesic. The only point to verify is that, if $a \in [p,y]$ and $b \in [p,z]$, then $d(a,b) \geq \frac{1}{\sqrt{2}} (d(a,p)+d(p,z))$. -Consider a comparison triangle $\Delta = \Delta(\bar{a},\bar{b},\bar{p})$ for $[a,p]\cup[p,b] \cup [a,b]$. Notice that $\angle_{\bar{p}}(\bar{a},\bar{b}) \geq \angle_p(a,b) \geq \pi/2$. Therefore, $d(\bar{a},\bar{b})$ greater or equal to the length of the hypothenuse of a right-angled triangle whose small sides have lengths $\alpha := d(\bar{a},\bar{p})$ and $\beta :=d(\bar{b},\bar{p})$. One has -$$\begin{array}{lcl} d(a,b) & = & d(\bar{a},\bar{b}) \geq \sqrt{\alpha^2+\beta^2} = (\alpha + \beta) \cdot \frac{\sqrt{1+(\alpha/\beta)^2}}{1+ \alpha/\beta} \\ & \geq & \frac{1}{\sqrt{2}} (\alpha+\beta) = \frac{1}{\sqrt{2}} ( d(\bar{a},\bar{p})+d(\bar{b},\bar{p}) \\ & \geq & \frac{1}{\sqrt{2}} ( d(a,p)+d(b,p) \end{array}.$$ -Because $S$ is Morse, there exists a universal constant $K$ such that $d(p,S) \leq K$. Consequently, $d(x,S) \leq d(x,p)+d(p,S) \leq K+ \epsilon$. -So the conlusion is: if $K$ is such that any $(\sqrt{2},0)$-quasigeodesic between any two points of $S$ stays in the $K$-neighborhood of $S$, then the Hausdorff distance between $S$ and its convex hull is at most $K$. $\square$ -Corollary: Let $G$ be a group acting geometrically on a CAT(0) space $X$. If $H \subset G$ is a Morse subgroup, then $H$ is convex-cocompact, ie., there exists a convex subspace $Y \subset X$ on which $H$ acts cocompactly. -Proof. The map $\left\{ \begin{array}{ccc} G & \to & X \\ g & \mapsto & g \cdot x_0 \end{array} \right.$, where $x_0 \in X$ is a fixed basepoint, is a quasi-isometry, so the orbit $H \cdot x_0$ is a Morse subspace. It follows from the previous proposition that $H$ acts cocompactly on the convex hull of $H \cdot x_0$. $\square$ -(*) The assertion is not true. The pre-image of a single point under a projection onto a geodesic is not necessarily convex in a CAT(0) space. So the proof above is not correct. I do not erase the argument in case it would lead to a correct proof.<|endoftext|> -TITLE: Cardinality of the image of a polynomial modulo $p^n$ -QUESTION [7 upvotes]: Let $f \in \mathbb{Z}[x]$ be a nonconstant polynomial and let $p$ be a prime number. I'm looking for results about $$N_f(p^k) := \#\{(f(n) \bmod p^k) : n \in \mathbb{Z}\},$$ as $k \to +\infty$, where $(a \bmod b)$ is the remainder of $a$ divided by $b$. -If $N_f(p) = p$ and $f^\prime(x) \not\equiv 0\bmod p$, for all $x \in \mathbb{Z}$, then using Hensel's lemma it can be proved that $N_f(p^k) = p^k$ for all $k \geq 1$. However, I did not find more general result. -Thank you for any suggestion. - -REPLY [4 votes]: At this question, it was proven that for all $f$, there exists a rational number $\delta$ such that $N_f(p^k)\sim \delta\cdot p^k$ as $k\rightarrow\infty$.<|endoftext|> -TITLE: How to characterize "matching-transitive" regular graphs? -QUESTION [9 upvotes]: I am interested in regular graphs $G$ such that for each pair of 1-factors (=perfect matchings) $F$ and $F'$ there is an automorphism of $G$ that takes $F$ to $F'$. Let's call this property matching transitivity. -Those graphs should have a fairly big automorphism group. I am wondering whether it is possible to characterize them. -Trivial examples of matching-transitive graph are $K_{2n}$ or the complete bipartite graph $K_{n,n}$. -EDITED after Brendan McKay's comment: -If we remove successively 1-factors from such a graph, it is clear not true that this property is maintained at each stage. -The question is inspired by the existence of graphs whose 2-factors are isomorphic (as graphs), i.e. all have the same partition $\pi$ as cycle type. For a discussion and examples of such graphs for various cycle types see here and here. All those graphs are cubic, and BTW I don't think that such a graph can be $k$-regular with $k>3$. (Is there an easy argument for that?) -Some of those graphs have lots of symmetries, others a rather small automorphism group, like for instance, the graph of type $(5,11)$ given in this answer, which has a unique triangle and automorphism group $S_3$, but not less than $12$ different 1-factors (so it cannot be matching-transitive). On the other hand, the Heawood graph ($|Aut(G)|=336$) is, and so is the Coxeter Graph (same automorphism group) as outlined here. -EDIT: As mentioned in another comment, there are many graphs with a unique perfect matching, which is obviously not what I am after. So I'll add the (somewhat mild) condition that each edge should belong to some perfect matching. Such a graph is called $1$-extendable (cf. page 113 in [László Lovász, Michael D. Plummer, Matching Theory, Annals of Discrete Mathematics 29, North-Holland, 1986, ISBN: 0 444 87916 1]). -So: - -For given $n$ and $k$, can anything be said about the matching-transitivity of a connected $1$-extendable $k$-regular graph on $n$ vertices in terms of the size (or structure) of its automorphism group? - -Note that both Heawood and Coxeter Graph are indeed $1$-extendable, and so is the $(5,11)$ graph mentioned before. There is possibly more hope now that, roughy speaking, the bigger the automorphism group, the bigger the chance for the graph to be matching-transitive. - -REPLY [3 votes]: One infinite family is the lexicographic product (composition) $C_s[\bar K_t]$, where $s$ is odd and $t$ is even. Basically you take the cycle $C_s$ and expand each vertex into a cluster of $t$ independent vertices. Here is $C_5[\bar K_2]$ with a representative perfect matching. In general, any perfect matching has $t/2$ edges between each consecutive cluster of vertices. Any two perfect matchings are related by an automorphism that permutes the vertices in each cluster. - -Now I can report on some computations. It turned out to be trickier that expected because of graphs with extremely large groups and matching counts. For the record I'll give the method. -$G$ is a graph with $n$ vertices and automorphism group $\varGamma$. "Matching" means "perfect matching", and "matching-transitive" means "$\varGamma$ acts transitively on the set of all matchings". -$G$ is matching-transitive iff each of its components is matching-transitive, so let's assume $G$ is connected. -Let $P_1,\ldots,P_k$ be the orbits of the action of $\varGamma$ on the -pairs of edges of $G$. The "profile" of a matching $M$ is the list $m_1,\ldots,m_k$ where for each $j$ exactly $m_j$ of the $\binom{n/2}{2}$ pairs of edges of $M$ lie in $P_j$. Clearly, every matching in a matching-transitive graph has the same profile (but the converse might not be true). - -Compute $\varGamma$. -Start a scan of all matchings, halting when any of these conditions become true: (A) two different profiles have been seen; (B) the number of matchings exceeds $|\varGamma|$; (C) the number of matchings exceeds some huge number chosen to prevent memory overflow; (D) the search completes without (A)-(C) happening. -In the case of (A) or (B), $G$ is not matching-transitive. -In the case of (C), we failed; think of something else to do. -In the case of (D), apply $\varGamma$ to the list of all matchings to test if they are all equivalent. - -Case (C) can happen with a matching-transitive graph having a huge number of matchings. I excluded known examples: complete graphs, complete bipartite graphs, $C_s[\bar K_t]$, where $s$ is odd and $t$ is even (defined above). With those cases out of the picture, I have not encountered case (C). -In the case of vertex-transitive graphs, the only examples I found were: - -complete graphs, balanced complete bipartite graphs, $C_s[\bar K_t]$, where $s$ is odd and $t$ is even -Petersen graph, Coxeter graph, Heawood graph - -These tests did not require that every edge is contained in some matching, but that is true for all the examples. The graphs tested were: - -All transitive graphs to 46 vertices -All cubic transitive graphs to 1280 vertices -All quartic symmetric (arc-transitive) graphs up to 640 vertices - -The problem for large graphs is that the my crude backtrack for listing all the matchings starts to take too long. However, all the large cubic or quartic graphs except $C_s[\bar K_2]$ could be eliminated by a different method: use a heuristic to find a hamiltonian cycle and check whether the two (or three, for cubic graphs) matchings implied by the cycle are equivalent. Sometimes this required testing multiple hamiltonian cycles. -For connected regular graphs that are not vertex-transitive, there are too many trivial cases with very few matchings, so I added the criterion that every edge must be contained in a matching. For connected non-transitive cubic graphs, here are the counts starting at $n=6$: 0, 1, 1, 2, 3, 4, 8, 8, 14, 24. For connected non-transitive quartic graphs, the counts starting at $n=8$ are: 0, 1, 2, 6, 5. I didn't attempt to make any sense out of them.<|endoftext|> -TITLE: Group rings over central products -QUESTION [5 upvotes]: I have a proof of the following result but I was wondering if anyone had a reference for it. I have asked on math.stackexchange here but didn't receive any replies. -Let $G$ a finite group given by the (inner) central product of two subgroups $H$ and $K$ over $M$ (I am using the definition of central product given in Gorenstein's "Finite groups"; in particular $G=HK$, $H \cap K=M \subset Z(G)$ and $H$ centralizes $K$). If R is a commutative ring, then $R[G]=R[H] \otimes_{R[M]} R[K]$. - -REPLY [6 votes]: I don't have a reference, but the proof is routine so that the result should be well-known. Also note that $H, K$ don't need be finite. -For a proof first note that if $H, K$ are subgroups of $G$ and $M$ is a subgroup of $H \cap K$ than $$\mu: R[H] \times {R[K]} \to R[G],\,(h,k) \mapsto h \cdot k$$ -is bilinear such that $\mu(hm,k)=\mu(h,mk)$. Hence $\mu$ induces a homomorphism -$$\mu: R[H] \otimes_{R[M]} R[K]\to R[G],\,h \otimes k \mapsto hk.$$ -Now let $G$ be the central product of $H,K$, i.e. (1) $G=HK$, (2) elements of $H,K$ commute, (3) $M= H\cap K$ is central in $G$. -By (1) $\mu$ is surjective and by (2) $\mu$ is an $R$-algebra homomorphism. Define a map -$$\varphi: G \to R[H] \otimes_{R[M]} R[K],\,hk \mapsto h \otimes k.$$ -By (3) $\varphi$ is well-defined (i.e. doesn't depend on the choice of $h,k$). Extend $\varphi$ linearly to $R[G]$. Than by definition, $\varphi \circ \mu = \text{id}$, i.e. $\mu$ is injective and hence an isomorphism of $R$-algebras. q.e.d.<|endoftext|> -TITLE: Is an Eulerian lattice shellable? -QUESTION [6 upvotes]: The notion of Eulerian lattice generalizes the notion of face lattice of a convex polytope. -(Bruggesse-Mani): The boundary complex of a convex polytope is shellable. -(Björner-Wachs): A poset is said to be shellable if its order complex is shellable. A bounded poset is shellable if and only if its proper part is shellable. -(Wachs): The order complex of the face poset of a simplicial complex is its barycentric subdivision. -Pre-question: Is the shellability kept by taking the barycentric subdivision? -If so, the face lattice of a convex polytope is shellable. This leads to: -Question: Is an Eulerian lattice shellable? - -REPLY [9 votes]: The simplest example of a nonshellable (even non-Cohen Macaulay) Eulerian lattice is the disjoint union of two boolean algebras of rank three with their bottom elements identified and their top elements identified. The total number of elements is 14.<|endoftext|> -TITLE: Countable $\mathbf\Sigma^1_2$ equivalence relations -QUESTION [9 upvotes]: This question is meant to be viewed under moderate large cardinal hypotheses, e.g., enough to ensure $\aleph_1^{L[x]}<\aleph_1$ for all reals $x$. -In analogy with the (well-developed) theory of countable Borel equivalence relations, what can be said about countable $\mathbf\Sigma^1_2$ (or $\mathbf\Delta^1_3$, whatever make the most sense) equivalence relations? -A natural example of such an equivalence relation is relative constructability, i.e., $x\equiv_c y$ if and only if $L[x]=L[y]$ for reals $x$ and $y$. -Specifically, I'm looking for: - -An analogue of the Feldman-Moore theorem on countable equivalence relations being induced by appropriately-measurable actions of countable groups. -Analogues of the Silver and Glimm-Effros dichotomies. -A theory of hyperfiniteness (i.e., equivalence relations which are a increasing union of $\mathbf\Sigma^1_2$ finite equivalence relations), and its relationship to appropriately-measurable $\mathbb{Z}$-actions. - -Are there any references which address these issues? -I suspect both that many arguments in the Borel case can be adapted (assuming appropriate uniformization results under large cardinals), but also that I might be blind to possible pitfalls... - -REPLY [2 votes]: We will show the generalization of the Feldman-Moore theorem on countable equivalence relations to the context of thin $\kappa$-Suslin equivalence relations where $\kappa$ is any infinite cardinal. Under further hypotheses, that is determinacy axioms, then every $\Sigma^1_{2n+2}$ set of reals is $\delta^1_{2n+1}$-Suslin and every thin equivalence relation will be a countable equivalence relation by the perfect set theorem for all sets under $AD$. Just in $ZFC$, the pointclass $\Sigma^1_2$ is $\delta^1_1$-Suslin where $\delta^1_1=\aleph_1$ but the perfect set theorem at this level requires a bit more. -We will thus show the following: let $X$ be a Hausdorff space and assume $E\subseteq X\times X$ is a thin $\kappa$-Suslin equivalence relation (in the non $AD$ context we may just say $E$ is an equivalence relation of size $\leq \kappa$). Call a group thin if it does not contain a perfect set of elements. Then there exists a $\kappa$-Suslin action of a thin group $F$ on $X$ such that $E$ is obtained as an orbit equivalence relation, that is we have that $$E=\{(x,y)\in X\times X: Fx=Fy\}.$$ Notice that the second set is always $\kappa$-Suslin by definition. Furthermore the group $F$ is generated by the set $$\{f_k:k\in \omega\}$$ such that $f_k$ has order 2, $(f_k)^2=e$ for all $k\in \omega$ and $$E(x,y) \text{ holds if and only if either } x=y, \text{ or } f_k(x)=y$$ for some $k\in \omega$. -We now start the proof. By Caicedo, Clemens, Conley and Miller we have the generalization of the Luzin-Novikov uniformization theorem to the $\kappa$-Suslin sets. More specifically: If $\kappa$ is an infinite cardinal, if $X$ is a Hausdorff space and if $A\subseteq X^{\omega}$ is $\kappa$-Suslin then one of the following holds: -1) The set $A$ is the union of $\kappa$-many graph intersecting sets which are $\kappa^+$-Borel when considered as subsets of $A$, -2) The set $A$ has a pairwise disjoint perfect subset. -Two sequences $x$ and $y$ are graph disjoint if $x(n)=y(n)$ for all $n\in \omega$. A set $A$ is called graph intersecting if no two sequences in $A$ are graph disjoint. The Luzin-Novikov uniformization theorem follows as a special case of the Caicedo-Clemens-Conley-Miller theorem by considering constant sequences. -Next, since the equivalence relation $E\subseteq X\times X$ has $\leq \kappa$-size sections then by the Caicedo-Clemens-Conley-Miller theorem we have $E=\bigcup_{\alpha\in \kappa} Y_{\alpha}$ where for every $\alpha\in \kappa$, $Y_{\alpha}\subseteq X\times X$ is the graph of a $\kappa^+$-Borel function. We define $$Y_{\alpha}^{-1}=\{(x,y)\in X\times X: (y,x)\in Y_{\alpha}\}$$ By symmetry of $E$ we obtain that $$E=\bigcup_{\alpha<\kappa} Y_{\alpha}$$ Define next the following sets: $$Y_{\alpha,\beta}=Y_{\alpha}\cap Y^{-1}_{\beta}$$ We also must have that $$E=\bigcup_{\alpha,\beta<\kappa} Y_{\alpha,\beta}$$ We may assume for the remainder of the proof that $X$ is the unit interval $I$. -Let $J,K\subseteq I$ be two disjoint intervals with rational endpoints. Since $J$ and $K$ are disjoint we must have $$J,K\subseteq (I\times I)\setminus\{(x,x):x\in I\}$$ For all $\alpha,\beta<\kappa$, define also the set $$A_{\alpha,\beta}(J,K)$$ by $$A_{\alpha,\beta}(J,K)=proj(\{(x,y)\in Y_{\alpha,\beta}: x\in J \wedge y\in K\})$$ where we take the projection onto the first coordinate. Associated to each such set $A_{\alpha\beta}(J,K)$ we have a map $$f_{\alpha,\beta}(J,K):A_{\alpha,\beta}(J,K) \to X$$ such that the graph $$\Gamma_f=\{(x,y)\in Y_{\alpha,\beta}: x\in Z\}$$ By definition of the set $A$ we must have $$f(A)\cap A=\emptyset$$ and $$E(x,f(x))$$ since $Y_{\alpha,\beta}\subseteq E$ and $(x,f(x)\in Y_{\alpha,\beta}$ for all $x\in A$. In addition notice that $f$ is an injective function. -We thus define a $\kappa$-Suslin bijection $$f': f(A)\cup A\to f(A)\cup A$$ by letting $f(f(x))=x$ and finally define an automorphism of the space $X$ fixing everything outside of $A\cup f(A)$, that is $f'(x)=x$ where $x\in X\setminus (A\cup f(A))$. -By definition of the intervals $J$ and $K$ there are only $\aleph_0$ many such automorphisms $f$ of $X$. Let $F$ be the group generated by the set $$\{f_k:k\in \omega\}$$ $F$ satisfies the theorem and the $\kappa$-Suslin equivalence relation $E$ comes as an orbit equivalence relation. -There are probably more results that can be generalized along these lines and more natural equivalence relations that could be studied by looking at inner models with Woodin cardinals and the $\mathcal{Q}$-degrees. Here's a reference of the Caicedo-Clemens-Clinton-Miller article I'll stop here, this is already too long.<|endoftext|> -TITLE: simple conjecture on palindromes in base 10 -QUESTION [8 upvotes]: The conjecture says that for any a, b belong to the the set of non-negative integers ($a$ and $b$ are not necessarily distinct), taking any natural value of $c$; we have always that $$(10^c-1) \cdot \frac{10^a+10^{2a}+1}{3} \cdot \frac{10^b+10^{2b}+1}{3}$$ is palindromic in base $10$. This conjecture was experimented well. -Example: taking $a=2,b=3,c=5$, -we will get $$3367 \cdot 333667 \cdot 99999=112344555443211$$ which is palindromic in base $10$, -my question is to prove or disprove this conjecture. -Note that the conjecture above was proved before editing . -another conjecture that I ask to prove or disprove it is that the sequence of the numbers of the form (10^a+10^2a+1)/3 is the maximally dense sequence in base 10 with the palindromic products property as described above . - -REPLY [16 votes]: Using -$$\frac{10^c-1}{9}=\sum_{m=0}^{c-1} 10^m,$$ -the product in question equals -$$\sum_{n=0}^{2a+2b+c-1}r(n)10^n,$$ -where $r(n)$ is the number of times $n$ occurs among the numbers (counted with multiplicity) -$$\begin{matrix} -m+2a+2b,&m+2a+b,&m+2a,\\ -m+a+2b,&m+a+b,&m+a,\\ -m+2b,&m+b,&m,\\ -\end{matrix}$$ -for some $m\in\{0,1,\dots,c-1\}$. For a given $n$, each entry in the above $3\times 3$ grid equals $n$ for at most one value of $m$, hence $r(n)\in\{0,1,\dots,9\}$ is the $n$-th decimal digit of the product in question. In addition, subtracting the above grid from $2a+2b+c-1$ yields the same grid reversed (with $m$ replaced by $c-1-m$), hence $r(2a+2b+c-1-n)=r(n)$ follows as well. This shows that the product in question is indeed palyndromic in base $10$.<|endoftext|> -TITLE: Automorphism groups of Hirzebruch surfaces -QUESTION [6 upvotes]: The Hirzebruch surfaces are the $\mathbb{P}^1$ bundles $\mathbb{F_n}$ ($n\geqslant 0$) which can be obtained projectivizing the rank $2$ vector bundles $\mathcal{O}_{\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1}(n)$. Those surfaces are famous because they exhaust ($n\neq 1$) the minimal rational (smooth, projective) surfaces other than $\mathbb{P}^2$. The most familiar examples are $\mathbb{F_0}\simeq \mathbb{P}^1\times\mathbb{P}^1$ and $\mathbb{F_1}$ which is isomorphic to $\mathbb{P}^2$ blown up at a point. -Other features of Hirzebruch surfaces are apparently well-known, such as its intersection theory. -My question is about automorphism groups of Hirzebruch surfaces. For the sake of comparation, the automorphism group of $\mathbb{P}^2$ is isomorphic to $PGL(3,\mathbb{C})$ and I manage to find a very nice description of $Aut(\mathbb{F_0})$ in this answer (Automorphisms of a smooth quadric surface $Q\subset\mathbb{P}^{3}$), since a quadric smooth projective surface $Q\subset \mathbb{P}^3$ is isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$. -Do we have a similar description of $Aut(\mathbb{F_n})$ for positive $n$? - -REPLY [10 votes]: For $n \ge 2$ the surface $\mathbb{F}_n$ is the blowup of the weighted projective plane $\mathbb{P}(1,1,n)$ at its singular point. Because of that -$$ -Aut(\mathbb{F}_n) \cong Aut(\mathbb{P}(1,1,n)). -$$ -The automorphism group of $\mathbb{P}(w_1,w_2,\dots,w_m)$ is the group of non-degenerate $m$-by-$m$ matrices with the $(i,j)$-entry being a polynomial of degree $w_i - w_j$, up to simultaneous rescaling.<|endoftext|> -TITLE: "Inner product" between prime factorizations -QUESTION [14 upvotes]: Let $x \in \mathbb{Q}^+$. Then $x$ can be expressed uniquely as a product over primes: -$$x = \prod_{p \text{ prime}} p^{\nu_p(x)}$$ -where $\nu_p(x)$ is the p-adic valuation of $x$. Let $\nu(x) = \langle \nu_2(x), \nu_3(x), \ldots \rangle \in \mathbb{Z}^{\oplus \mathbb{N}}$. Then $\nu(x) + \nu(y) = \nu(xy)$ and $a \nu(x) = \nu(x^a)$. Thus the positive rationals can be viewed as a module where addition and multiplication of two module elements correspond to multiplication and exponentiation of their corresponding rationals. Suppose we define the following "inner product": -$$\nu(x) \cdot \nu(y) = \sum_{p \text{ prime}} \nu_p(x) \nu_p(y)$$ -This gives rise to a notion of orthogonality, angles, lengths, and volumes between positive rationals. My question is this: Do this "inner product" and the aforementioned notions have any significance, from the perspective of number theory? Have they been studied before? -For example: if $x,y \in \mathbb{Z}^+$ then $\nu(x) \cdot \nu(y) =0$ iff $x$ and $y$ are coprime. Furthermore, if $x \in \mathbb{Z}^+$ then $\lVert \nu(x) \rVert_1 = \Omega(x)$ and $\lVert \nu(x) \rVert_0 = \omega(x)$. - -REPLY [4 votes]: You can find the same idea from Graham, Knuth, and Patashnik's book Concrete Mathematics on page 115 and some related results.<|endoftext|> -TITLE: Is it always possible to calculate the limit of an elementary function? -QUESTION [33 upvotes]: I already asked this question on https://math.stackexchange.com/questions/2691331/is-it-always-possible-to-calculate-the-limit-of-an-elementary-function but as I received no answer; maybe it is not as obvious as I originally thought. -The precise formulation of my question is: define an "strong elementary function" by only admitting rational for the "constant function" in the usual definition of "elementary" function (see for example: https://en.wikipedia.org/wiki/Elementary_function). Let $a$ be an "elementary real" if the constant function $f(x)=a$ is a strong elementary function. With this definition some non rational reals are elementary (for example $\pi=4⋅arctan(1)$); but there are reals that are not elementary. Now let $f(x)$ be a strong elementary function defined in an open interval of an elementary real $a$ with the possible exception of $a$. Suppose that -$lim_{x\rightarrow a} f(x)$ -exists. Is this limit necessarily an elementary real? -The idea behind this question is this: it seems that "limits" can always be calculated with some simple tricks (Hospital rule, etc...) in elementary calculus; but is there a general argument that shows that it is always possible? (the precise formulation of the question does not ask for an algorithm, I expect a positive answer to be constructive, but that is not entirely clear). -Update: clarification of the notion of elementary functions. -Definition -By function $\mathbb{R}\rightarrow\mathbb{R}$; I mean a partial function. -The class of (strong) elementary functions is the smallest class of functions such that: - -the constant function $f(x)=1$ is elementary. -if $f$ and $g$ are elementary; so is $f+g$; $f-g$; $f\cdot g$; $f/g$. (the domain of $f/g$ is ${\rm dom}(f)\cap {\rm dom}(g)\cap \{x \ | \ g(x)\neq 0\}$. -if $n$ is a natural number; $f(x)=x^n$ and $f(x)=\sqrt[n]{x}$ are elementary; the domain of the later is $\mathbb{R}^+$. -$\sin$, $\cos$, $\tan$ are elementary -$\arcsin$, $\arccos$ and $\arctan$ are elementary. ($\arcsin:[-1\ 1]\rightarrow[-\pi\ \pi]$; $\arccos(x)={\pi\over 2}-\arcsin(x)$; $\arctan: ]-\infty\ +\infty[\rightarrow]-\pi\ \pi[$) -$\exp$ is elementary. -$\ln$ is elementary. ($\ln: \mathbb{R}^+\rightarrow \mathbb{R}$). -if $f$ and $g$ are elementary; so is $f\circ g$; the domain of the latter is $\{x\ | \ x\in{\rm dom}(g) \wedge g(x)\in{\rm dom}(f)\}$. - -I have not tried to avoid redundancy but -note that the point 3 is not redundant because of the domain of the functions considered; for example $f(x)=x^2$ is defined on $\mathbb{R}$ but $f(x)=\exp(2\cdot\ln(x))$ is defined on $\mathbb{R}_0^+$. Also $\sqrt[n]{x}$ is defined for $x=0$ but not $\exp(\frac{\ln(x)}{n})$. -I think this is the class of functions we consider in the Risch algorithm (https://en.wikipedia.org/wiki/Risch_algorithm) except that I do not take all constant functions as elementary; that would obviously make no sense for my question. -I hope I have not missed something obvious. I do not think a small modification of my definition will make any difference; if it is it would be interesting to discus. - -REPLY [6 votes]: In fact, zero-equivalence for combinations of polynomial and sine functions is undecidable, which means that it is undecidable whether the limit is zero in that setting (This goes back to at least Paul Wang's 1974 paper The undecidability of the existence of zeros of real elementary functions).<|endoftext|> -TITLE: One colored infinity operads via symmetric sequences? -QUESTION [8 upvotes]: The question -One standard approach to the theory of 1-colored (symmetric) operads in classical 1-categorical theory is via monoids in symmetric sequences with respect to the composition product. Has this been worked out for $\infty$-operads in the sense of Higher Algebra? -More details -More concretely, I am not even interested in the composition product, but in something that seems rather obvious but still requires a proof. For a 1-colored $\infty$-operad $\mathcal{P}$ (my operads are actually reduced) we can associate a symmetric sequence of spaces $\{\mathcal{P}(n)\}_{n\ge0}$ of $n$-arry operations. From this sequence, one can construct an endofunctor of spaces -$$T_{\mathcal{P}}(X) = \coprod_{n\ge0}(\mathcal{P}(n)\times X^n)//\Sigma_n$$ -which is the monad obtained from the free forgetful adjunction $\mathcal{S}\leftrightarrows Alg_{\mathcal{P}}(\mathcal{S})$ (so far, everything can be extracted from HA). -Now let $\mathcal{P}\to \mathcal{Q}$ be a map of 1-colored $\infty$- operads. This induces a forgetful functor of algebras in spaces $Alg_{\mathcal{Q}}(\mathcal{S}) \to Alg_{\mathcal{P}}(\mathcal{S})$ from which we get by abstract nonsense a canonical map $T_{\mathcal{P}}(X)\to T_{\mathcal{Q}}(X)$. I need to know that this map is indeed the obvious thing one would expect. That is, that it is induced from the map of symmetric sequences $\{\mathcal{P}(n)\}_{n\ge0} \to \{\mathcal{Q}(n)\}_{n\ge0}$ which is itself induced from the map of operads. -This is definitely true, but I am looking for a rigorous (and preferably short) argument for it. My hope is that a comprehensive treatment of 1-colored $\infty$-operads in terms of symmetric sequences will include enough stuff to deduce this from. -Remarks on the literature: -In Higher Algebra section 6.3 the approach via symmetric sequences is mentioned, but not developed formally. This whole thing (and much more) follows morally from this paper which identifies 1-colored $\infty$-operads with analytic $\infty$-monads, but the identification is not so transparent (to me), especially when working with $\infty$-operads using the HA model. There are model category approaches to this, but again the comparisons are usually not so simple and as a first choice I would prefer to stay in the quasi-category model. - -REPLY [2 votes]: As far as I have understood your question, you ask the following question: -(When I write category in the following, I always mean $\infty$-category.) -Let $\mathcal{C}$ be a symmetric monoidal category compatible with small colimits (i.e. $\mathcal{C}$ admits small colimits that are preserved by the tensorproduct in each variable). -Denote $\mathrm{Sym}(\mathcal{C})$ the category of symmetric sequences in $\mathcal{C}$ viewed as a monoidal category with the composition product. -An associative algebra $\mathcal{O} $ in $\mathrm{Sym}(\mathcal{C})$ gives rise to a monad $\mathrm{T}_{ \mathcal{O} }, $ i.e. an associative algebra in $\mathrm{Fun}( \mathcal{C}, \mathcal{C})$ (endofunctors of $\mathcal{C}$) -via a monoidal functor $\mathrm{Sym}(\mathcal{C}) \to \mathrm{Fun}( \mathcal{C}, \mathcal{C})$ that sends $ \mathcal{O} $ to $ \mathrm{T}_{\mathcal{O}},$ -where $ \mathrm{T}_{\mathcal{O}}(X) := \coprod_{n\ge0}(\mathcal{O}(n)\otimes \mathrm{X}^n)//\Sigma_n.$ -Given a monad $\mathrm{T} $ denote $ \mathrm{Alg}_{ \mathrm{T}} ( \mathcal{C} )$ its category of algebras. -Given a monoid $\mathcal{O} $ in $\mathrm{Sym}(\mathcal{C})$ its category $ \mathrm{Alg}_{ \mathcal{O} } ( \mathcal{C} )$ is the category of left modules over $ \mathcal{O} $ in $\mathrm{Sym}(\mathcal{C})$ on objects of $\mathrm{Sym}(\mathcal{C})$ that belong to $\mathcal{C} \subset \mathrm{Sym}(\mathcal{C}).$ -You ask how one can show the existence of an equivalence -$ \mathrm{Alg}_{ \mathcal{O} } ( \mathcal{C} )\simeq \mathrm{Alg}_{ \mathrm{T}_{ \mathcal{O} } } ( \mathcal{C} )$ over $ \mathcal{C} $ natural in $\mathcal{O} \in \mathrm{Alg}(\mathrm{Sym}(\mathcal{C})) . $ -To answer this question, some preparations: -Denote $\Sigma \simeq \coprod_{ \mathrm{n} \geq 0 } \mathrm{B}( \Sigma_{\mathrm{n}} )$ the category of finite sets and isomorphisms. -Then one can define $\mathrm{Sym}(\mathcal{C}) = \mathrm{Fun}(\Sigma, \mathcal{C}) \simeq \prod_{ \mathrm{n} \geq 0 } \mathrm{Fun}(\mathrm{B}( \Sigma_{\mathrm{n}}), \mathcal{C}). $ -We have a small colimits preserving fully faithful functor $\iota: \mathcal{C} \subset \prod_{ \mathrm{n} \geq 0 } \mathrm{Fun}(\mathrm{B}( \Sigma_{\mathrm{n}}), \mathcal{C})$ that is the identity on the first factor and the constant functor with image the initial object of $ \mathcal{C} $ on every other factor. -As next I define the composition product via the Day-convolution on $\mathrm{Fun}(\Sigma, \mathcal{C}):$ -We endow $\Sigma$ with the symmetric monoidal structure given by the coproduct of finite sets. -Then $\Sigma$ is the free symmetric monoidal category on the contractible category, i.e. for every symmetric monoidal category $ \mathcal{E} $ -evaluation at the set with one element $1 \in \Sigma $ yields an equivalence -$$ \mathrm{Fun}^\otimes (\Sigma, \mathcal{E}) \simeq \mathcal{E}, $$ -where $\mathrm{Fun}^\otimes (\Sigma, \mathcal{E}) $ denotes the category of symmetric monoidal functors $\Sigma \to \mathcal{E}.$ -We endow the functor-category $\mathrm{Sym}(\mathcal{C}) = \mathrm{Fun}(\Sigma, \mathcal{C}) $ with the Day-convolution symmetric monoidal structure. -The Day-convolution symmetric monoidal structure on $ \mathrm{Fun}(\Sigma, \mathcal{C}) $ restricts to the full subcategory $\iota: \mathcal{C} \subset \mathrm{Fun}(\Sigma, \mathcal{C}) $ so that $\iota$ gets a symmetric monoidal functor. -We have a unique small colimits preserving symmetric monoidal functor $\mathcal{S} \to \mathcal{C} $ starting at the category of spaces $\mathcal{S}$ (endowed with the cartesian symmetric monoidal structure) that sends the contractible space to the tensorunit $ \mathbb{1}_{ \mathcal{C}} $ of $ \mathcal{C} $ and gives rise to a symmetric monoidal functor $ \mathrm{Fun}(\Sigma, \mathcal{S}) \to \mathrm{Fun}(\Sigma, \mathcal{C})$ on Day-convolutions. -Composing with the symmetric monoidal Yoneda-embedding -$ \Sigma \simeq \Sigma^\mathrm{op} \subset \mathrm{Fun}(\Sigma, \mathcal{S}) $ we get a symmetric monoidal functor $\phi: \Sigma \to \mathrm{Fun}(\Sigma, \mathcal{C}) $ -that sends $\mathrm{n} \in \Sigma $ to the symmetric sequence concentrated in degree $\mathrm{n} $ with value $\Sigma_{\mathrm{n}} \times \mathbb{1}_{ \mathcal{C}}. $ -$ \phi $ satisfies the following universal property: -The Day-convolution $ \mathrm{Fun}(\Sigma, \mathcal{C})$ is a symmetric monoidal category compatible with small colimits and we have a -small colimits preserving symmetric monoidal functor $ \iota: \mathcal{C} \to \mathrm{Fun}(\Sigma, \mathcal{C}). $ -For every symmetric monoidal category $ \mathcal{D} $ compatible with small colimits equipped with a small colimits preserving symmetric monoidal functor $ \mathcal{C} \to \mathcal{D} $ -composition with $\phi: \Sigma \to \mathrm{Fun}(\Sigma, \mathcal{C}) $ yields an equivalence $$ \mathrm{Fun}^\otimes_{ \mathcal{C} } (\mathrm{Fun}(\Sigma, \mathcal{C}) , \mathcal{D}) \to \mathrm{Fun}^\otimes (\Sigma, \mathcal{D})\simeq \mathcal{D}, $$ -where $ \mathrm{Fun}^\otimes_{ \mathcal{C} } (\mathrm{Fun}(\Sigma, \mathcal{C}) , \mathcal{D}) $ denotes the category of symmetric monoidal functors $\mathrm{Fun}(\Sigma, \mathcal{C}) \to \mathcal{D}$ compatible with the symmetric monoidal functors from $\mathcal{C}.$ -So the equivalence $\mathrm{Fun}^\otimes_{ \mathcal{C} } (\mathrm{Fun}(\Sigma, \mathcal{C}) , \mathcal{D}) \to \mathrm{Fun}^\otimes (\Sigma, \mathcal{D})\simeq \mathcal{D}$ evaluates at $\phi(1)\in \mathrm{Fun}(\Sigma, \mathcal{C})$, i.e. at the symmetric sequence concentrated in degree $1 $ with value $ \mathbb{1}_{ \mathcal{C}}. $ -Let $\mathrm{Z} \in \mathcal{D} $ corresponding to $ \bar{\mathrm{Z}} \in \mathrm{Fun}^\otimes_{ \mathcal{C} } (\mathrm{Fun}(\Sigma, \mathcal{C}) , \mathcal{D}) .$ -Then for every $\mathrm{X} \in \mathrm{Fun}(\Sigma, \mathcal{C})$ we have a canonical equivalence $$\bar{\mathrm{Z}}( \mathrm{X} ) \simeq \coprod_{ \mathrm{n} \geq 0 } \iota(\mathrm{X}_{ \mathrm{n}}) \otimes_{\Sigma_{\mathrm{n}} } \mathrm{Z}^{ \otimes \mathrm{n}}.$$ -Taking $\mathcal{D}= \mathrm{Fun}(\Sigma, \mathcal{C})$ we get a canonical equivalence $$ \mathrm{Fun}(\Sigma, \mathcal{C}) \simeq \mathrm{Fun}^\otimes_{ \mathcal{C} } (\mathrm{Fun}(\Sigma, \mathcal{C}) , \mathrm{Fun}(\Sigma, \mathcal{C})).$$ -The category $\mathrm{Fun}(\Sigma, \mathcal{C}) \simeq \mathrm{Fun}^\otimes_{ \mathcal{C} } (\mathrm{Fun}(\Sigma, \mathcal{C}) , \mathrm{Fun}(\Sigma, \mathcal{C})) $ carries a monoidal structure given by composition. -Given a monoidal category $\mathcal{B} $ denote $\mathcal{B}_\mathrm{rev }$ -the reverse monoidal structure on $ \mathcal{B} $ with $\mathrm{X} \otimes_{ \mathcal{B}_\mathrm{rev } } \mathrm{Y} = \mathrm{Y} \otimes_{ \mathcal{B} } \mathrm{X}.$ -One can define the composition product on $\mathrm{Fun}(\Sigma, \mathcal{C})$ to be the reverse monoidal structure of the monoidal structure on $ \mathrm{Fun}(\Sigma, \mathcal{C}) \simeq \mathrm{Fun}^\otimes_{ \mathcal{C} } (\mathrm{Fun}(\Sigma, \mathcal{C}) , \mathrm{Fun}(\Sigma, \mathcal{C})) $ given by composition. -Denote $\mathrm{Sym}(\mathcal{C})$ the category $\mathrm{Fun}(\Sigma, \mathcal{C})$ endowed with the composition product. -For $\mathrm{X}, \mathrm{Z} \in \mathrm{Sym}(\mathcal{C})$ their composition product $\mathrm{X} \circ \mathrm{Z} $ is by definition -$$ (\bar{\mathrm{Z}} \circ \bar{\mathrm{X}}) (\phi(1)) \simeq \bar{\mathrm{Z}} (\mathrm{X}) \simeq \coprod_{ \mathrm{n} \geq 0 } \iota(\mathrm{X}_{ \mathrm{n}}) \otimes_{\Sigma_{\mathrm{n}} } \mathrm{Z}^{ \otimes \mathrm{n}}. $$ -Let $\mathrm{Y} \in \mathcal{C} $. Then $\mathrm{X} \circ \iota(\mathrm{Y} ) \simeq \coprod_{ \mathrm{n} \geq 0 } \iota(\mathrm{X}_{ \mathrm{n}}) \otimes_{\Sigma_{\mathrm{n}} } \iota(\mathrm{Y})^{ \otimes \mathrm{n}} \simeq \iota(\coprod_{ \mathrm{n} \geq 0 } \mathrm{X}_{ \mathrm{n}} \otimes_{\Sigma_{\mathrm{n}} } \mathrm{Y}^{ \otimes \mathrm{n}}). $ -As a monoidal category $\mathrm{Sym}(\mathcal{C})$ acts on itself from the left. -So by the line above the left action of $\mathrm{Sym}(\mathcal{C})$ on $\mathrm{Sym}(\mathcal{C})$ restricts to a left action of $\mathrm{Sym}(\mathcal{C})$ on $ \mathcal{C} \subset \mathrm{Sym}(\mathcal{C})$. -Given a monoid $\mathcal{O}$ of $\mathrm{Sym}(\mathcal{C})$ denote $\mathrm{Alg}_{\mathcal{O}}(\mathcal{C}):= \mathrm{LMod}_{ \mathcal{O} } ( \mathcal{C}) $ the category of left modules in $\mathcal{C}$ over $\mathcal{O}$ with respect to the left action of $\mathrm{Sym}(\mathcal{C})$ on $ \mathcal{C}. $ -By the theory of endomorphism objects Higher Algebra 4.7.2 the category $\mathrm{Fun}( \mathcal{C}, \mathcal{C}) $ admits a monoidal structure given by composition that acts on $ \mathcal{C}$ in a universal way: -This implies that the left action of -$\mathrm{Sym}(\mathcal{C})$ on $ \mathcal{C} $ is the pullback of the -universal left action of $\mathrm{Fun}( \mathcal{C}, \mathcal{C})$ on $ \mathcal{C}$ -along a unique monoidal functor $\mathrm{T}: \mathrm{Sym}(\mathcal{C}) \to \mathrm{Fun}( \mathcal{C}, \mathcal{C})$. -For every $\mathrm{Y} \in \mathcal{C} $ and $\mathcal{O} \in \mathrm{Sym}(\mathcal{C})$ one has $\mathrm{T}(\mathcal{O} )( \mathrm{Y}) \simeq \coprod_{ \mathrm{n} \geq 0 } \mathcal{O}_{ \mathrm{n}} \otimes_{\Sigma_{\mathrm{n}} } \mathrm{Y}^{ \otimes \mathrm{n}}. $ -So $\mathrm{T}: \mathrm{Sym}(\mathcal{C}) \to \mathrm{Fun}( \mathcal{C}, \mathcal{C})$ induces a functor $\mathrm{Alg}(\mathrm{Sym}(\mathcal{C})) \to \mathrm{Alg}(\mathrm{Fun}( \mathcal{C}, \mathcal{C}))$ that sends an associative algebra in $ \mathrm{Sym}(\mathcal{C})$ to its associated monad on $\mathcal{C}$. -As the left action of -$\mathrm{Sym}(\mathcal{C})$ on $ \mathcal{C} $ is the pullback of the -universal left action of $\mathrm{Fun}( \mathcal{C}, \mathcal{C})$ on $ \mathcal{C}$ along $\mathrm{T}: \mathrm{Sym}(\mathcal{C}) \to \mathrm{Fun}( \mathcal{C}, \mathcal{C})$, -the map $$ \Phi: \mathrm{LMod}( \mathcal{C}) \to \mathrm{Alg}(\mathrm{Sym}(\mathcal{C})) \times \mathcal{C} $$ of cartesian fibrations over $\mathrm{Alg}(\mathrm{Sym}(\mathcal{C}))$ is the pullback of the map -$$ \Psi: \mathrm{LMod}( \mathcal{C}) \to \mathrm{Alg}(\mathrm{Fun}( \mathcal{C}, \mathcal{C})) \times \mathcal{C} $$ of cartesian fibrations over $\mathrm{Alg}(\mathrm{Fun}( \mathcal{C}, \mathcal{C})) $ along the functor -$ \mathrm{Alg}(\mathrm{Sym}(\mathcal{C})) \to \mathrm{Alg}(\mathrm{Fun}( \mathcal{C}, \mathcal{C}))$. -Denote ${\mathrm{Cat}_\infty}_{/ \mathcal{C} }$ the category of small $(\infty$-) categories over $\mathcal{C}. $ -$\Psi$ classifies a functor $\alpha: \mathrm{Alg}(\mathrm{Fun}( \mathcal{C}, \mathcal{C})) \to {\mathrm{Cat}_\infty}_{/ \mathcal{C} }$ that sends a monad to its category of algebras. -$\Phi $ classifies a functor $\beta: \mathrm{Alg}(\mathrm{Sym}(\mathcal{C})) \to {\mathrm{Cat}_\infty}_{/ \mathcal{C} }$ that sends an associative algebra in $ \mathrm{Sym}(\mathcal{C})$ to its category of algebras. -As $ \Phi $ is the pullback of $ \Psi$ along the functor -$ \mathrm{Alg}(\mathrm{Sym}(\mathcal{C})) \to \mathrm{Alg}(\mathrm{Fun}( \mathcal{C}, \mathcal{C}))$, the functor $\beta$ factors as -$$\mathrm{Alg}(\mathrm{Sym}(\mathcal{C})) \to \mathrm{Alg}(\mathrm{Fun}( \mathcal{C}, \mathcal{C})) \xrightarrow{\alpha} {\mathrm{Cat}_\infty}_{/ \mathcal{C} },$$ which provides a canonical equivalence -$$ \mathrm{Alg}_{ \mathcal{O} } ( \mathcal{C} )\simeq \mathrm{Alg}_{ \mathrm{T}( \mathcal{O} ) } ( \mathcal{C} ) $$ over $ \mathcal{C} $ natural in $\mathcal{O} \in \mathrm{Alg}(\mathrm{Sym}(\mathcal{C})) . $<|endoftext|> -TITLE: A finite group that splits and does not split -QUESTION [5 upvotes]: Is there an example of a finite group $A$ that acts on a finite group $C$ irreducibly (that is, $C$ has no proper nontrivial $A$-invariant subgroup) -such that there exists an epimorphism $$\tau \colon A \ltimes C \to A$$ which does not split? - -REPLY [5 votes]: I don't believe this is possible. Let $G$ be a minimal counterexample. Then $G = AC$ with $C \unlhd G$ and $A$ a complement of $C$, and there exists $D \unlhd G$ with $G/D \cong A$, but $D$ has no complement if $G$. Note that $G/D \cong G/C \Rightarrow |C|=|D|$. Since $A$ acts irreducibly on $C$, we have $C \cap D = 1$. -We prove first that $D \le A$. Since $C \cap D = 1$, we have $CD \cap A \cong D$, and $CD \cap A$ is a subgroup of $C \times D$ of the form $\{(\phi(d),d) : d \in D \}$, where $\phi:D \to C$ is a homomorphism. -But $CD \cap A \unlhd A$, and for $a \in A$ (with notation $g^h=h^{-1}gh$) we have $(\phi(d),d)^a = (\phi(d)^a, d^a)$, but this must also equal $(\phi(d^a),d^a)$, so $\phi(d^a) = \phi(d)^a$ and hence ${\rm Im}(\phi)$ is $A$-invariant so, by assumption, it must be $C$ or $1$. -If ${\rm Im}(\phi) = C$ then $\ker(\phi)=1$, so $D \cap A = 1$. But then $A$ is a complement of $D$ in $G$, contrary to assumption. So $ {\rm Im}(\phi) = 1$ and $D \le A$ as claimed. -So now $G/D$ is a split extension of $C$ by $A/D$ and, since $D$ centralizes $C$, $A/D$ must act irreducibly on $C$. -If $D$ had a complement $B$ in $A$, then $CB$ would be a complement of $D$ in $G$, contrary to assumption. So $G/C \cong A$ is a nonsplit extension of $D$ by $A/D$. But we are assuming that $G/C \cong G/D$, so $G/C$ is a smaller counterexample than $G$.<|endoftext|> -TITLE: Integral inequality involving an analytic function -QUESTION [7 upvotes]: I have been trying to prove for any $\delta>0,$ $$ -\int_0^{2\pi}\left|1+ e^{i\theta}f(e^{i\theta})\right|^{\delta}d\theta\leq \int_0^{2\pi}\left|1+e^{i\theta}\right|^{\delta}d\theta -$$ for any analytic function $f(z)$ in $|z|\leq 1$ with $|f(e^{i\theta})|\leq 1.$ Can anyone help me in this? - -REPLY [10 votes]: Let $F(z)=1+z$ and $\phi(z)=zf(z)$. Then $\phi$ is an analytic function mapping -the unit disk into itself, and $\phi(0)=0$, and your inequality becomes the subordination inequality -$$\int_0^{2\pi}|F\circ\phi(e^{i\theta})|^\delta d\theta\leq\int_0^{2\pi}|F(e^{i\theta})|^\delta d\theta,$$ -see, for example, -J. E. Littlewood, Lectures on the Theory of Functions. Oxford University Press, 1944, Theorem 210 on p. 164. -Here is a short proof of the subordination inequality. Let $u$ be subharmonic in the disk, $v$ the least harmonic majorant (=harmonic function matching $u$ on the unit circle), and let $\phi$ be analytic, mapping the unit disk into itself, $\phi(0)=0$. Denote by $I$ the average over the unit circle. -Then: -$$I(u\circ \phi)\leq I(v\circ\phi)=v\circ\phi(0)=v(0)=I(v)=I(u),$$ -where we used the average property of harmonic functions twice: once for $v$ and another for $v\circ\phi$. -In our application $u=|F|^\delta$.<|endoftext|> -TITLE: Can we glue characteristic 0 and characteristic p representations of a finite group given equality of (Brauer) characters? -QUESTION [11 upvotes]: Suppose I have a prime $p$ and a finite group $G$ together with representations $\sigma: G \to GL_n(\mathbb{Q}_p)$ and $\pi: G \to GL_n(\mathbb{F}_p)$. My question is: - -When does there exist a representation $\rho: G \to GL_n(\mathbb{Z}_p)$ such that $\sigma$ is isomorphic to the base change of $\rho$ to $\mathbb{Q}_p$, and $\pi$ is isomorphic to the base change of $\rho$ to $\mathbb{F}_p$? - -One necessary condition is that for any $p$-regular element $g \in G_{reg}$, the trace of $\sigma(g)$ is equal to the Brauer character of $\pi(g)$. However, I can't tell if this is sufficient or not - the results in Serre's Linear Representations book only give me existence theorems up to semisimplification. I tried to cook up a suitable $G$-stable $\mathbb{Z}_p$-lattice using affine buildings, but I got hopelessly lost. - -REPLY [13 votes]: The condition on Brauer characters is not sufficient. -Let $G$ be a $p$-group, $\pi$ any nontrivial representation over $\mathbb{F}_p$, and $\sigma$ the trivial representation over $\mathbb{Q}_p$ of the same degree as $\pi$. Then a representation over $\mathbb{Z}_p$ whose base change to $\mathbb{Q}_p$ is isomorphic to $\sigma$ must also be trivial, so its base change to $\mathbb{F}_p$ can’t be isomorphic to $\pi$. -But the condition on Brauer characters is trivially satisfied.<|endoftext|> -TITLE: Inertia of a class of Cayley graphs -QUESTION [5 upvotes]: Let $H^n_2(d)$ be the Cayley graph with vertex set $\{0,1\}^n$ where two strings form an edge iff they have Hamming distance at least $d$. What is the inertia of these graphs, that is, the numbers of positive, negative and zero eigenvalues? Thank you. - -REPLY [5 votes]: Such graphs are studied a lot in coding theory and in the theory of association schemes. In paricular the eigenvalues can be explicitly written down in terms of Krawchuk polynomials $K_k(u)$. -You can find this e.g. in Bannai & Ito "Association Schemes I": https://www.amazon.com/Algebraic-Combinatorics-Association-Schemes-Mathematics/dp/0805304908 -I never saw inertia of these graphs mentioned anywhere, but it looks easy to figure out. -To give you an idea, the eigenvalues of the relations of the Hamming scheme $H(n,2)$ (that is, the $n+1$ adjacency matrices $A_j$ of graphs $\Gamma_j$ ($0\leq j\leq n$) partitioning $K_{2^n}$, so that two vertices of $\Gamma_j$ are adjacent if the Hamming distance between them is $j$, may be compactly presented (discounting multiplicities) as an $(n+1)\times (n+1)$ matrix $P$, -with $j$-column containing the eigenvalues of $A_j$, and $j$-th row containing the eigenvalues corresponding to the $j$-th common eigenspace of $A_j$'s (the latter works as $A_j$'s pairwise commute). -In general, the entries of $P$ are given by -$$ -\newcommand{\kk}{k} -\newcommand{\uu}{u} -P_{ku}=K_{\kk}(\uu)=\binom{n}{\kk}\sum_{j=0}^k -\frac{(-\kk)_j(-\uu)_j 2^j}{(-n)_j j!}=\binom{n}{\kk}\sum_{j=0}^k (-1)^j 2^j -\frac{\binom{u}{j}\binom{k}{j}}{\binom{n}{j}} -$$ -To get the eigenvalues of $H_2^n(d)$ you just need to sum up the last $n-d+1$ columns of $P$. The multiplicities are given by the binomial coefficients $\binom{n}{j}$, $0\leq j\leq n$, also this is what you get in the 1st row of $P$. - -Here are the experimental results; I coded up this in Sagemath and computed the number $p(d)$ of positive eigenvalues of $H_2^{2d}(d)$ for -$1\leq d\leq 18$. Here is what I got -[1,6,36,136,496,2016,8256,32896,130816,523776,2098176, - 8390656,33550336,134209536,536887296,2147516416,8589869056,34359607296] - -While the usual interface does not produce a match, Superseeker does tell me a lot, e.g. it says that -Guesss (a program by Harm Derksen) suggests that the generating function $F(x)$ -may satisfy the following algebraic or differential equation: -$$x^2+1/8 x+1/16+(x^3-1/4 x^2+1/4 x-1/16)F(x) = 0$$ -as well as that the exponential generating function for this sequence is -$$2 \exp(4 x) - \sin(2 x) - \cos(2 x).$$ -Thus, conjecturally, this is the answer for the number $p(d)$ of positive eigenvalues; there are no 0 eigenvalues, and the number of the negative ones is just $2^{2d}-p(d).$ - -But in fact it's much easier to state, namely, conjecturally -$$ -\frac{2^{2d}-2p(d)}{2^d}= -\begin{cases} -1 &\text{if $d\mod 4\in\{1,2\}$ and}\\ --1 &\text{otherwise.} -\end{cases} -$$<|endoftext|> -TITLE: Critical points of Dirichlet L functions -QUESTION [11 upvotes]: Let $L(s,\chi)$ denote a Dirichlet $L$-function for a real-valued non-principal -character $\chi$. This has limiting value $L(\infty,\chi) = 1$ and we are interested in how this limit is approached through real values of $s$, $1 < s < \infty$ . -For example, let $\chi_4$ be the non-trivial character $(\mathrm{mod}\ 4)$ taking values $(1,0,-1,0)$ at $(1,2,3,4)$. We have $L(1,\chi_4) < 1$ and, by an alternating series argument, $L(s,\chi_4)$ is monotonically increasing for $s > 1$. -Next consider $\chi_7$ $(\mathrm{mod}\ 7)$ taking values $(1,1,-1,1,-1,-1,0)$ at $(1,2,3,4,5,6,7)$. Note that $L(1,\chi_7) > 1$. However, $L(s,\chi_7)$ is not monotone decreasing for $s > 1$, but first increases with a peak near $s = 1.1$ before decreasing. What can one say in general? -Questions: 1) Are there infinitely many $\chi$ (real-valued as above) for which -$L(s,\chi)$ has no critical points on the interval $1 < s < \infty$ (i.e. monotonic)? -2) Is it possible for $L(s,\chi)$ to have more than one critical point in $(1,\infty)$? If so, is there any upper bound on the number of such critical points which holds for all $\chi$? -3) Are there any books or papers which display graphs of $L$-functions on the real line? -Thanks - -REPLY [8 votes]: I can answer 2): if q=857, the L-function has 2 critical points. I would guess -the number of critical points is unbounded. I do not know -if this is the smallest example (draw on the same graph the L-function -for q=17 and for q=857 starting at t=2, say (the critical points are smaller): -the graphs almost coincide; Andrew Granville confirms that this is is a case -of what he calls "pretentious characters", this is why I tried 857 immediately. -UPDATE: I have run a program for $q$ from $-10000$ to $10000$: there are 27 positive -q and 27 negative q for which there are 2 critical points (unfortunately none -with more than 2), the first positive $q$ being 593, 857, 1697,..., and the -first negative $q$ being $-548$, $-632$, $-872$, ... -I should also have answered question 3): in Pari/GP the command -L=lfuninit(D,[5,5,1]);ploth(t=1,10,lfun(L,t)) -plots the $L$-function of a real character $\chi_D$ from $1$ to $10$ for instance.<|endoftext|> -TITLE: Do $2^{n-1}\equiv1\pmod n$ and $(n-1)/2$ prime imply $n$ prime? -QUESTION [5 upvotes]: Do $2^{n-1}\equiv1\pmod n$ and $(n-1)/2$ prime imply $n$ prime? -Equivalently: Does $n$ being a Fermat pseudoprimes to base 2 (OEIS A001567) imply that $(n-1)/2$ is composite? That holds for all $n<2^{64}$, based on Jan Feitsma's table. -Motivation is a simplification in the search of safe primes as used in cryptography. - -Progress so far: Pocklington's theorem states that if $q>\sqrt n-1$ is a prime dividing $n-1$, and $a^{n-1}\equiv1\pmod n$, then $n$ is prime or $\gcd(a^{(n-1)/q},n)\ne1$. Applying this for $a=2$, it comes that any counterexample $n$ to the propositions would be a multiple of $3$. -The question then boils down to: do $6k+1$ prime imply $4^{6k+1}\not\equiv1\pmod{4k+1}$ ? - -REPLY [19 votes]: We have $2^{\varphi(n)}\equiv 1 \pmod n$, thus $2^k\equiv 1$, where $k=\text{gcd}(\varphi(n),n-1)$. Note that $k$ is even, since both $n-1$ and $\varphi(n)$ are even. If $n=2p+1$ for prime $p=(n-1)/2$, then even divisor of $n-1$ is either 2 or $2p$. If $k=2$, we get $n|2^2-1=3$; if $k=2p=n-1$, we get $\varphi(n)\geqslant n-1$, thus $n$ is prime.<|endoftext|> -TITLE: Representing field elements in a computer -QUESTION [11 upvotes]: I'm wondering if there is existing terminology to describe fields $F$ with the properties below. I don't have a completely precise description of the concept I have in mind, but hopefully this will be enough. - -Every element of $F$ requires only a finite amount of information to describe (so that it can be stored in a computer in some way). -There exist (finite) algorithms which can be used implement the field operations of $F$. - -For example, the field $\mathbb Q$ (and any finite extension of it) has these properties, but the fields $\mathbb R$ and $\mathbb Q_p$ (where $p$ is prime) do not. I would appreciate any help in formulating my question more precisely, and any references which might discuss this concept. - -REPLY [16 votes]: You are looking at a computable field (if your focus is on the field), or a computable presentation of a field (if your focus is on the details of how elements and operations are coded). These objects are studied in computable structure theory.<|endoftext|> -TITLE: Rational map given by pfaffians -QUESTION [5 upvotes]: Consider a general skew-symmetric $(n+1)\times (n+1)$ matrix $Z$, and let su map $Z$ to the point of $\mathbb{P}^n$ determined by $[pf_0(Z):\dots:pf_n(Z)]$ where the $pf_i(Z)$ are the principal Pfaffians of order $n$ of $Z$. -Is the closure of the image in $\mathbb{P}^n$ of this rational map a known variety? - -REPLY [4 votes]: At least in characteristic zero, the image of this map is all of $\mathbb P^n$. In finite characteristic, I am not sufficiently familiar with Pfaffians to say whether the argument carries through. -Indeed, start by noting that the map is only well-defined for $Z$ of rank $n$, since otherwise all principal Pfaffians of rank $n$ vanish. For $Z$ of maximal rank, consider instead the map that sends $Z$ to $\left[(-1)^0pf_0(Z):\dotsm:(-1)^npf_n(Z)\right]$, which agrees with your map up to an automorphism of $\mathbb P^n$. This map sends $Z$ to its kernel, and every line arises as the kernel of some skew-symmetric matrix.<|endoftext|> -TITLE: Symmetric C* property -QUESTION [6 upvotes]: This is a cross-post from math.stackexchange.com -What can be said about elements $a$ of a C*-algebra which fulfil the 'symmetric C* property' -$$\| a^\ast a+ aa^\ast\|=2\| a\|^2$$ -I'd guess that this is not a general property, but I don't have a good idea which elements (except for self-adjoint and normal ones, of course) should satisfy this. - -REPLY [7 votes]: This is hardly a complete answer. -As you say, all normal operators $a \in B(\mathcal H)$ satisfy this equality. Moreover, so does any element $c = a\oplus b \in B(\mathcal H\oplus \mathcal H)$ where $a$ is normal and $\|b\| \leq \|a\|$. This is because -$$ \|c^*c + cc^*\| = \max\{\|a^*a + aa^*\|, \|b^*b + bb^*\|\} = 2\|a\| = 2\|c\|. $$ -On the other hand, this is certainly not a general property. For instance, if $a = \left[\begin{matrix} 0 & 1\\ 0&0\end{matrix}\right]$ then $$\|a^*a + a^*a\| = \left\|\left[\begin{matrix} 1 & 0\\ 0&1\end{matrix}\right]\right\| = 1 \quad \textrm{but} \quad 2\|a\|^2 = 2.$$ -And on the third hand, $a = \left[\begin{matrix} 0 & 1 & 0 \\ 0& 0& 1\\ 0&0&0\end{matrix}\right]$ satisfies this equality. This makes me doubt that there will be any nice characterization.<|endoftext|> -TITLE: Are two triangles with equal corresponding medians, congruent? -QUESTION [9 upvotes]: Is the hyperbolic or spherical analogy of the following Euclidean fact, true? - -Two triangles with equal corresponding medians are congruent. - -More precisely: Assume that $\Delta ABC$ and $ \Delta A'B'C'$ are two triangles in the hyperbolic space $\mathbb{H}^2$ or elliptic space $\mathbb{S}^2$ such that $$AM_1=A'M_1',\; BM_2=B'M_2',\;CM_3=C'M_3'$$ where $M_i (M_i')$ in $XM_i(X'M_i')$ is the mid point of the edge opposite to the vertex $X(X')$, respectively. -Under this condition, is there an isometry of the corresponding space carrying the first triangle to the second one? - -REPLY [7 votes]: This is not true for $S^2$. Namely, one can construct two convex non-isometric spherical triangles such that all medians have length $\frac{3\pi}{4}-\varepsilon$. -1) The first triangle is the standard equilateral triangle. Clearly medians of such triangles all have the same length and their length vary from $0$ to $\pi$. -2) The second triangle will be a perturbation of the following one. Take on $S^2$ two opposite points $A$ and $B$ and join them by two geodesics (of length $\pi$) that bound a sector with angle $\frac{3\pi}{4}$. Call one of these geodesics $AB$. And choose $C$ as the mid point of the other geodesic. It is easy to see that for this degenerate triangle all medians have length $\frac{3\pi}{4}$. Now, by perturbing slightly $A$, $B$ and $C$ one can decrease the lengths of all medians by the same amount. -Note. It is important in the second construction that by varying $A$ and $B$ as little as we want, we can achieve that the median starting from $C$ takes any length in $(0,\pi)$. However the length of two other medians stay close to $\frac{3\pi}{4}$. -PS, hyperbolic case. Concerning the hyperbolic case, it looks like the answer is positive. -If I would like to prove it, I would do as follows. -i) Consider the map from the convex cone in $\mathbb R_{\ge 0}^3$ consisting of triples $(a,b,c)$ satisfying the (non-strict) triangle inequality to itself: lengths of sides $\to$ lengths of medians. -ii) Realise that this map if proper (preimage of compact is compact). And it is an isomorphism on the boundary on the cone. -iii) The map is a "diffeo" close to the point $(0,0,0)$ - because it is so for Euclidean triangles. -iv) This is the most complicated bit -- show that the map has non-vanishing differential. This is where one needs to make a calculation. But it looks very plausible. -v) If all the above holds then the map is an isomorphism. QED<|endoftext|> -TITLE: Can we estimate the error $\left| \frac{1}{N^2} \sum f ( \{ \sqrt{2} m + \sqrt{3} n \} ) - \int_0^1 f(x) \, dx \right|$? -QUESTION [7 upvotes]: As a computer experiment I did a few Riemannian sums to see if I could quantify the density statement $\overline{\mathbb{Q}(\sqrt{2}, \sqrt{3})} = \mathbb{R}$ : -$$ \Big| \frac{1}{N^2} \sum_{0 \leq m,n \leq N} \{ \sqrt{2} m + \sqrt{3} n \}^5 - \frac{1}{6} \Big| \stackrel{?}{<} \frac{1}{N^2} $$ -A log-plot shows the correct exponent is a bit less than $2$. Is it a Hausdorff dimension of some kind? - -The general quantitative statement looks like some error term to an average: -$$ \Big| \frac{1}{N^2} \sum_{0 \leq m,n \leq N} f \big( \{ \sqrt{2} m + \sqrt{3} n \} \big) - \int_0^1 f(x) \, dx \Big| \stackrel{?}{\ll} \frac{1}{N^2} $$ -This is certainly false... what might the a good exponent be? The statement could be more general - I have some kind of totally real number field - but then we get a worse exponent. - -REPLY [5 votes]: From Koksma-Hlawka inequality (see here), the following holds for functions with bounded variation in Hardy-Krause sense. -If $f:[0,1]\rightarrow\mathbb{R}$ and $g:[0,1]\times [0,1]\rightarrow\mathbb{R}$ defined by $g(x,y)=f(\{x+y\})$ is bounded variation in Hardy-Krause sense, then $$ -\left|\frac1{N^2} \sum_{0\leq m,n \leq N} f(\{\sqrt 2 m + \sqrt 3 n\})-\int_0^1 f(x)dx\right|\leq C V(g) D_{N^2} -$$ -where $C>0$ is absolute, $V(g)$ is the Hardy-Krause variation of $g$, and $D_{N^2}$ is the discrepancy of the double sequence $\{(\{\sqrt 2 m\},\{\sqrt 3 n\} ) \ | \ 0\leq m,n\leq N\}$. -As @GH from MO suggested, Erdos-Turan-Koksma inequality gives an upper bound of $D_{N^2}$. The crucial estimate here is the bound of $||x||$ the distance between $x$ and its nearest integer, for certain numbers $x$. -Let $1\leq H\leq N^2$, we have -$$\begin{align} -D_{N^2}&\ll \frac1H + \sum_{h_1 -TITLE: Variants of reflection principle -QUESTION [5 upvotes]: This question concerns two definitions of the reflection principle. One of them known to be a consequence of the other one. I would like to understand if the reverse is true. -Let us state the first definition of reflection. -Definition(Reflection): A stationary set $S$ in $[H_\lambda]^\omega$ reflects at $X\subseteq H_\lambda$ if $S\cap [X]^\omega$ is stationary in $[X]^\omega$. -The following is the definition of reflection principle which widely is used in literature. See Jech's book, for example. - -Definition(Reflection principle) We say ${\rm RP(\lambda)}$ holds if every stationary set of $[H_\lambda]^\omega$ reflects at a set of size $\aleph_1$. We also let ${\rm RP}$ denote ${\rm RP}(\lambda)$ for all regular $\lambda\geq\aleph_2$. - -Now we define another definition of reflection: -Definition(Reflection$^*$): A stationary set $S$ in $[H_\lambda]^\omega$ reflects if there is a continuous $\in$-chain $\langle M_\alpha:\alpha<\omega_1\rangle$ of countable elementary substructures of $H_\lambda$ such that $\{\alpha<\omega_1:M_\alpha\in S\}$ is stationary in $\omega_1$. -Now using this definition we can build a principle: - -Definition(Reflection principle$^*$) We say ${\rm RP}^*(\lambda)$ holds if every stationary subset of $[H_\lambda]^\omega$ reflects. We also let $RP^*$ denote ${\rm RP}^*(\lambda)$ for all $\lambda\geq\aleph_2$. - -The following theorem is easy. -Theorem: ${\rm RP}^*$ implies ${\rm RP}$. -Both of these definitions are used and known in literature as reflection principle. Note that ${\rm SRP}$ (Strong reflection principle) implies ${\rm RP}^*$. I would like to know if they are not equivalent. - -Question: Is there any model of ${\rm RP}+\neg {\rm RP}^* $? - -REPLY [8 votes]: The answer is "yes". The principle you call $\text{RP}^*$ is called "reflection to internally club sets" in the literature; as far as I know this terminology first appeared in Foreman-Todorcevic's ``A new Lowenheim-Skolem Theorem". See their paper for the definition of internally stationary, internallly club, and internally approachable. -The main result of Krueger's ``Internal approachability and reflection" gives a model where RP holds---in fact in his model, every stationary subset of any $[H_\lambda]^\omega$ reflects to an internally stationary set of size $\omega_1$---yet there is a stationary set of countable models that doesn't reflect to any internally club set of size $\omega_1$. He doesn't state the main result that way (he states that RP on internally approachable sets fails), but if you look at the proof you'll see that his stationary set that fails to reflect to an internally approachable set, in fact fails to reflect to an internally club set (of size $\omega_1$). -Note that if $\vec{M}$ is a continuous $\in$-chain as in your definition of Reflection*, then its union is an internally club set (i.e. if $W$ is its union, then $W \cap [W]^\omega$ contains a club). And it is straightforward to show that if $S$ is a stationary subset of $[H_\lambda]^\omega$, $S$ reflects to an internally club set of size $\omega_1$ if and only if there exists a continuous $\in$-chain $\vec{M}$ such that $M_\alpha \in S$ for stationarily many $\alpha < \omega_1$.<|endoftext|> -TITLE: Is K theory ever trivial because of the ring, and not because of the kinds of modules we look at? -QUESTION [11 upvotes]: Let $\mathcal C$ be some Waldhausen category; we know that the K-theory $K(\mathcal C)$ might be trivial if $\mathcal C$ contains objects that are too ``big'' in some sense---for instance, via the Eilenberg swindle. -(Example: An infinite-dimensional vector space is isomorphic to a direct sum of itself and a finite-dimensional vector space, so any finite-dimensional and any infinite-dimensional vector space have the same K-theory class; on the other hand, any infinite-dimensional vector space represents an idempotent element in the $K_0$ group, but the only idempotent in a group is the identity, so must represent zero.) -So I have two related questions: -(1) In known examples where the K-theory of $\mathcal C$ is trivial, does it tend to be because of Eilenberg-swindle-type reasons? -(1)(a) Specializing to the categories of sheaves on a variety, or on a non-commutative space, I understand that adjectives like quasi-coherent sheaves, or perfect sheaves, or coherent sheaves, or Ind-coherent sheaves, are important to distinguish how big the objects in our categories are. If you would be able to clarify the size or `infinite-dimensionality' of the objects in each such category, I would be very grateful! For instance, should I just always assume that the stable oo-category of Ind-coherent sheaves always has trivial K-theory? -(2) Are there examples of when the underlying `ring' is so big that the K-theory has to be trivial? For instance, if I have some non-compact space with gigantic $\pi_1$, is it true that if I looked at its based loop space and modules over the based loop space, its K-theory will be trivial? What if I only look at perfect modules over such an algebra? (Here, you can take singular chains to think about based loops as an algebra in chain complexes, or you can think of it as an $E_1$ ring spectrum. I am interested in either answer.) - -REPLY [10 votes]: If $R\cong R\oplus R$ as an $R$-bimodule, then for any module $R$-module $N$, we have -$$N\cong N\otimes R\cong N\otimes (R\oplus R)\cong (N\otimes R)\oplus -(N\otimes R)\cong N\oplus N$$ -so that every $[N]$ is trivial in $K$-theory. -For a concrete example, let $S$ be any ring and $R$ the ring of column-finite countable-by-countable matrices over $S$.<|endoftext|> -TITLE: Example of a (presentable $k$-linear $\infty$-)category which is dualizable but not compactly generated? -QUESTION [8 upvotes]: Is there an example of a presentable, stable, $k$-linear $\infty$-category which is dualizable but not compactly generated, where $k$ has characteristic zero, and which is $\text{QCoh}(X)$ (by which I mean the derived dg category of quasicoherent sheaves on $X$) for some prestack $X$? -Or, perhaps by removing some of the conditions above, e.g. being $\text{QCoh}(X)$, $k$ having characteristic zero, stability. - -REPLY [11 votes]: If $X$ is a locally compact topological space, then -$\mathrm{Shv}(X, \mathrm{Mod}_{k} )$ is a presentable $k$-linear stable $\infty$-category which is dualizable (in fact, self-dual), but is rarely compactly generated (for example, this fails for $X = \mathbf{R}$).<|endoftext|> -TITLE: Natural model for genus $6$ curves -QUESTION [11 upvotes]: By Brill-Noether theory, the generic genus $6$ curve is birational to a sextic plane curve in $\mathbb{P}^2$. I was wondering if there is a direct/natural construction of this birational map. In other words, - - -do we know a divisor $D$ on the curve inducing this map? Is it given by a complete system, or by some natural subspace thereof? - - -This question came up while talking with other people, we were trying to recollect the beautiful picture relating genus $6$ curves, del Pezzo surface, and plane sextics. - -REPLY [12 votes]: On the canonical model $C_{10} \subset \mathbb{P}^5$ of a smooth curve of genus six there exist five special $g_4^1$, obtained as follows: we take an arbitrary point on the curve, and the remaining three are determined so that the four points lie on a plane. We can also see these five $g^1_4$ as cut on the $C_{10}$ by the five pencils of conics of the Del Pezzo surface containing it. -Projecting the canonical model $C_{10}$ from one set of four coplanar points belonging to one of the $g_4^1$, we obtain a birational plane - model $X_6 \subset \mathbb{P}^2$, namely a sextic with four nodes. The five $g_4^1$ are cut on $X_{6}$ by the lines through a node and by the conics through the four nodes. -For more details you can look at the classical book -H. F. Baker, Principles of Geometry, Vol. 5, Note II page 97.<|endoftext|> -TITLE: Formally real fields with unique non-Archimedean ordering -QUESTION [7 upvotes]: My question is rather simple. Do there exist a formally real field that admits a unique ordering (so sums of squares are the positive elements) and such that this ordering is not archimedean? -Oh, I have forgotten to add that the field I am looking for cannot be euclidean (in particular it cannot be real closed)! -The question I have in mind is the following. I proved many years ago that if K is a field that admits a unique ordering and this ordering is archimedean, then every automorphism f of K(x), where x is an indeterminate, maps K onto K. I wonder if archimedeanity can be dropped in this statement, and I realize that I know very few fields K whose unique ordering is not archimedean; essentially, real closed fields that cannot be embedded into the reals. And for such fields I know that f(K) = K!! - -REPLY [2 votes]: In addition to the above very nice model-theoretic construction, it seems interesting to describe such fields Galois-theoretically. Here is one such construction: -By a result of Ershov, the free profinite product of absolute Galois groups of fields is again an absolute Galois group of some field, and in fact such a field can have an arbitrarily large cardinality. -Let $K$ be a field of cardinality $>\aleph$ whose absolute Galois group is the free profinite product of a group of order $2$ and of some absolute Galois group of a non-ordered field (e.g. of $\hat{\mathbb{Z}}$). By a result of Herfort and Ribes https://eudml.org/doc/152727, every torsion element in a free profinite product is conjugate to an element of one of the free factors. The Artin-Schreier theorem therefore implies that $K$ has exactly one ordering. Since a field with an Archimedean ordering embeds in $\mathbb{R}$, the ordering on $K$ is non-Archimedean.<|endoftext|> -TITLE: Must a bending of the cylinder leave the bases planar? -QUESTION [8 upvotes]: Set $M=\{(\cos(\theta),\sin(\theta),z):\theta\in[0,2\pi],z\in[0,1]\}$. A bending of $M$ is a smooth map $\Gamma:M\times [0,1]\rightarrow \mathbb{R}^3$ such that -1) $\Gamma[M\times\{t\}]$ is a submanifold with boundary of $\mathbb{R}^3$ (I call it $M_t$) -2) For every $m\in M$, we have $\Gamma(m,0)=m$ -3) For every $t\in[0,1]$, the map $m\rightarrow \Gamma(m,t)$ is an isometry between $M$ and $M_t$ -Question: Must any bending of $M$ leave the bases planar? -The only smooth bendings I can think of so far are those which change the shape of the circular bases to elipses -Thank you - -REPLY [3 votes]: Take a geodesic $C$ on a smooth surface $S$. Through each point of $C$ draw a tangent to $S$ perpendicularly to $C$. The union of these tangents will be a developable surface, and $C$ is a geodesic on it. In particular, if $C$ is a closed geodesic, the union of tangent segments of length $\epsilon$ centered at the points of $C$ will be isometric to a cylinder. Now, if $C$ is non-planar, this cylinder will have non-planar bases. -In order to construct a continuous deformation of a straight cylinder into a cylinder with non-planar bases, we need a family $(S(t), C(t))$ with, say $S(t)$ a sphere and $C(t)$ its equator (and $C(t)$ non-planar for some $t$). I am quite sure such a family exists, but have no concrete example.<|endoftext|> -TITLE: Is an ∞-topos of local homotopy dimension $\leq n$ of homotopy dimension $\leq n$? -QUESTION [5 upvotes]: [All references are wrt to Lurie's "Higher Topos Theory" in its latest online available version (March 10, 2012)] -Definition 7.2.1.8: -An ∞-topos $X$ is locally of homotopy dimension $\leq n$ if there exists a collection $\{U_\alpha\}$ of objects of $X$ which generate $X$ under colimits, such that each $X_{/U_\alpha}$ is of homotopy dimension $\leq n$. -Definition 7.2.1.1: -An ∞-topos $X$ has homotopy dimension $\leq n$ if enery $n$-connective object $U \in X$ admits a global section $1_X \to U$ ($1_X$ being the final object of $X$). -Question: Assume that $X$ is locally of homotopy dimension $\leq n$. Is $X$ of homotopy dimension $\leq n$? -My idea to prove this is using the collection of all objects in the first definition. However, I do not get a grasp on any connection between the connectivity of objects of $X$ and the connectivity of objects of some slice ∞-topos $X_{/U}$. -What I actually want to show is that an ∞-topoi which is of homotopy dimension $\leq n$ for some integer $n$ is hypercomplete. Corollary 7.2.1.12 does this for ∞-topoi which are locally of finite homotopy dimension. Its proof lies on Proposition 7.2.1.10 where I do not understand a crucial point ($\phi$ determines a point of $F(X(0))$), so I am unable to adopt this proposition to ∞-topoi of finite homotopy dimension. - -REPLY [3 votes]: As a counterexample to the statement at the end of the question, the $\infty$-topos of parameterized spectra is even of homotopy dimension $\le -1$, but not hypercomplete--as Mathieu Anel pointed out to me when I asked him a similar question.<|endoftext|> -TITLE: Would mathematics be different if not written one-dimensionally? -QUESTION [7 upvotes]: Mathematics is written one-dimensionally, using symbols that make sense when put together on a line. The 2d sheets of paper that we use don't have enough room to write mathematics two-dimensionally. As a result, most symbols in mathematics are operations involving two objects (or one object acting on another object) and many mathematical concepts seem to depend on this fact (non-commutativity, left/right action). - -Question: Would mathematics be different for 4d beings writing on 3d sheets of paper? For example, can it make sense (for them) to extend the notion of left/right action to up/down? - -REPLY [9 votes]: This should be a comment, but I do not know how to put an image there. Some commutative diagrams in homological algebra are in fact depicted "3-dimensionally" (or, at least, using perspective in order to simulate 3-dimensionality).<|endoftext|> -TITLE: Continuous images of $\beta \mathbb{N} \setminus\mathbb{N}$ -QUESTION [15 upvotes]: Let $\beta \mathbb{N}$ denote the Stone-Cech compatification of the natural numbers and $\beta \mathbb{N} \setminus\mathbb{N}$ -denote the reminder of this compactification. I wonder if there is a characterization in ZFC of the continuous images of $\beta \mathbb{N} \setminus\mathbb{N}$. I mean: Which compact Hausdorff spaces are continuous images of $\beta \mathbb{N} \setminus\mathbb{N}$? - -REPLY [23 votes]: This is a great question. There has been quite a bit of work done to figure out what the continuous images of $\beta \mathbb N \setminus \mathbb N$ are. I'll do my best to summarize some of that work here. -In some sense the answer to your question is yes: a space $X$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$ if and only if it is the remainder $X = \gamma \mathbb N \setminus \mathbb N$ of some compactification $\gamma \mathbb N$ of $\mathbb N$. -This answer is a bit unsatisfying, though, because what one really wants is an internal characterization of the images of $\beta \mathbb N \setminus \mathbb N$, some recognizable property of $X$ that tells us straightaway whether $X$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$ or not. -There is no complete characterization of this kind that I am aware of. However, there are plenty of theorems that give sufficient conditions: - -$\bullet$ Every separable compact Hausdorff space is a continuous image of $\beta \mathbb N \setminus \mathbb N$ -$\bullet$ Every compact Hausdorff space of weight$^1$ $\leq\! \aleph_1$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$ (Parovicenko, 1963)$^2$ -$\bullet$ Every compact Hausdorff space of weight $<\! \mathfrak{p}$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$ (van Douwen and Przymusinski, 1980)$^3$ -$\bullet$ Every perfectly normal$^4$ compact space is a continuous image of $\beta \mathbb N \setminus \mathbb N$ (Przymusinki, 1982) - -Notice that Parovicenko's theorem implies that, under ZFC+CH, a compact Hausdorff space is a continuous image of $\beta \mathbb N \setminus \mathbb N$ if and only if it has weight $\leq\! \aleph_1$. This gives you a very nice characterization of the continuous images of $\beta \mathbb N \setminus \mathbb N$ under ZFC+CH, but you may already know this -- you asked for a ZFC characterization. -Just to give you an idea of how hard it is, in general, to decide whether something is a continuous image of $\beta \mathbb N \setminus \mathbb N$, let me point out that it was an open question for years whether (in ZFC) every first countable compact Hausdorff space is a continuous image of $\beta \mathbb N \setminus \mathbb N$. This was finally solved by Murray Bell in a 1990 paper, where he showed that (in the Cohen model) there is a first countable space that is not a continuous image of $\beta \mathbb N \setminus \mathbb N$. -Along similar lines, there is also an interesting paper of Dow-Hart from 1999 entitled "$\omega^*$ has almost no continuous images" where they show . . . well, they show that many things you might suspect are continuous images of $\beta \mathbb N \setminus \mathbb N$ just aren't (assuming OCA). Work of Ilijas Farah indicates that OCA+MA is something of an optimal hypothesis for ensuring $\beta \mathbb N \setminus \mathbb N$ has as few continuous images as possible (see his book Analytic Quotients for the full story, or his much more accessible article "The fourth head of $\beta \mathbb N$ for a taste). -Finally, let me recommend Jan van Mill's survey article on $\beta \mathbb N$ from the Handbook of set-theoretic topology (link), which looks at this question from several angles and gives a lot of good information on the topic. - - -Recall that the weight of $X$ is the smallest cardinality of a basis for $X$. - -The $\aleph_1$ in this result is optimal. Kunen showed in his thesis that in the Cohen model, the weight-$\aleph_2$ space $\omega_2+1$ is not a continuous image of $\beta \mathbb N \setminus \mathbb N$. - -Again, the cardinality bound is optimal. It is consistent with $\mathfrak{p} = \mathfrak{c}$ that there is a compact Hausdorff space of weight $\mathfrak{c}$ that is not a continuous image of $\beta \mathbb N \setminus \mathbb N$. A construction is outlined in van Mill's survey article on $\beta \mathbb N$ from the Handbook of set-theoretic topology. - -Recall that a perfectly normal space is one in which every closed subset is a $G_\delta$.<|endoftext|> -TITLE: Is $Spin(N)$ a subgroup of $SU(N)$ -QUESTION [8 upvotes]: Is $Spin(N)$ a subgroup of $SU(N)$? If so, how can we embed $Spin(N)$ into $SU(N)$? I would love to find a representation where both $Spin(N)$ and $SU(N)$ act faithfully and see explicitly how the embedding is done. For example, $SO(N)$ can be embedded easily in $SU(N)$ by consider the fundamental representation. - -REPLY [25 votes]: $\def\Spin{\mathrm{Spin}}$The spin group has a representation $S$ called the spin representation, which has dimension $2^m$ for $N=2m$ or $2m+1$. It is irreducible for $N$ odd and splits into two complex conjugate representations $S_+ \oplus S_-$ for $N$ even, with each of $S_{\pm}$ being of dimension $2^{m-1}$. These are the smallest representations of $\Spin(N)$ which do not factor through $SO(N)$. So, once the power of $2$ exceeds $N$, the spin group will not embed in $SU(N)$. -This leaves us to analyze $1 \leq N \leq 6$, and $N=8$. Each of these has some special form. -$\Spin(1)$ is the trivial group. Clearly, this embeds in $SU(1)$. -$\Spin(2)$ is the circle group $\mathrm{R}/\mathrm{Z}$. Note that $SO(2)$ and $\mathrm{Spin}(2)$ are isomorphic as abstract groups, but the map $\mathrm{Spin}(2) \to SO(2)$ is the double cover. Clearly, this embeds in $SU(2)$. -When $N=3$, the spin representation $S$ is two dimensional and gives an isomorphism $\Spin(3) \cong SU(2)$. So, in particular, $\Spin(3)$ embeds in $SU(3)$. -When $N=4$, the representations $S_+$ and $S_-$ are two dimensional, giving an isomorphism $\Spin(4) \cong SU(2) \times SU(2)$. Conveniently, this embeds into $SU(4)$ via the representation $\mathbb{C}^2 \boxtimes 1 \oplus 1 \boxtimes \mathbb{C}^2$. We cannot use just one of the two $SU(2)$ representations, as the individual projections have kernels. Particle physicists call working with $\Spin(4)$ in the form $SU(2) \times SU(2)$ the "spinor helicity formalism". If you talk to the sort of particle physicists who care what the signature of spacetime is, you'll want to know that $\Spin(3,1) \cong SL_2(\mathbb{C})$. -When $N=5$, $\dim S = 4$. Indeed, this $4$-dimensional vector space can be identified with a $2$-dimensional quaternionic vector space and $\Spin(5)$ is the group of $2 \times 2$ unitary quaternion matrices. So $\Spin(5)$ embeds into $SU(4)$ (and also into $SU(5)$.) -When $N=6$, we have $\dim S_+ = 4$ and gives an isomorphism $\Spin(6) \cong SU(4)$. So $\Spin(6)$ embeds into $SU(6)$ with lots of room. -When $N=7$, we have $\dim S = 2^3 > 7$, so $\Spin(7)$ does not embed in $SU(7)$. It does embed in $SU(8)$, though. -When $N=8$, we have $\dim S_+ = \dim S_- = 8$. Unfortunately, the corresponding maps $\Spin(8) \to SU(8)$ have kernel. Indeed, there are three maps $\Spin(8) \to SO(8)$, related by the triality symmetry of $\Spin(8)$, and each of them has a central two element group as kernel. So there are interesting maps $\Spin(8) \to SU(8)$, but no embeddings. The composition $\Spin(7) \to \Spin(8) \to SO(8)$ for one of the nononbvious maps $\Spin(8) \to SO(8)$ is the embedding of $\Spin(7)$ I referenced above. -Once $N \geq 9$, the spin representations have dimension much larger than $N$.<|endoftext|> -TITLE: Is the "Ramond sector" invariant of a 3-framed lattice always divisible by 24? -QUESTION [10 upvotes]: For the purposes of this question, a rank-$r$ (integral) lattice is a full-rank discrete subgroup $L \subset \mathbb R^r$ such that $\langle \ell, \ell' \rangle \in \mathbb Z$ for all $\ell \in L$. It is even if $\ell^2 = \langle \ell,\ell\rangle \in 2\mathbb Z$ for all $\ell \in L$ and odd if there is some $\ell$ such that $\ell^2$ is odd. Define $L^* = \{ v\in \mathbb R^n \text{ s.t. } \langle v,\ell \rangle \in \mathbb Z \, \forall \ell \in L\}$. $L$ is unimodular if $L = L^*$. -Let $L$ be an odd unimodular lattice, and define $L_{ev} = \{\ell \in L \text{ s.t. } \ell^2 \in 2\mathbb Z\}$. Then $L_{ev}$ has index $2$ in $L$, and so index $4$ in $L_{ev}^*$. The cosets of $L_{ev}$ in $L_{ev}^*$ are $L_{ev}$, $L_{odd} := L \smallsetminus L_{ev}$, and two more that I will arbitrarily name $L_+$ and $L_-$. Vectors in $L_\pm$ might have fractional length. (The structure of the group $L_{ev}^*/L_{ev}$ and the lengths of vectors in $L_\pm$ depend on the rank $r$ mod $8$.) -Given a discrete subset $S \subset \mathbb R^r$, the Theta series of $S$ is the $q$-series -$ \Theta_S(q) = \sum_{s\in S} q^{s^2/2}. $ -Suppose that $L$ is odd and unimodular. I find myself interested in the following function: -$$ Z_{RR}(L) := \Theta_{L_+}(q) - \Theta_{L_-}(q)$$ -The letters "$RR$" stand for "Ramond-Ramond", because this this shows up as an RR-sector partition function of a spin conformal field theory built from $L$ (times a power of $\eta$). I'm pretty sure that $Z_{RR}(L)$ is automatically a (level-1) modular form of weight $r$. [Edit: $r/2$.] -Let $\sqrt3\mathbb Z \subset \mathbb R$ denote the (non-unimodular) odd lattice of vectors $\ell$ such that $\ell^2 \in 3\mathbb Z$. -A unimodular lattice $L$ of rank $r$ is 3-framed if it contains a copy of $(\sqrt3\mathbb Z)^r$. 3-framed lattices are in natural bijection with self-dual ternary codes; Harada and Munemasa used this bijection, together with Borcherds' classification of unimodular lattices of rank $\leq 25$, to classify self-dual ternary codes of rank $24$. -Recall the discriminant $\Delta(q) = q \prod_{n=1}^\infty (1-q^n)^{24}$. It is a modular form of weight $12$. I'm pretty sure that -Proposition: If $L$ is 3-framed and unimodular, then $Z_{RR}(L) \in \mathbb Z[\Delta]$. [Edit: $\mathbb Z[\Delta^{1/2}]$.] -I.e. I am claiming that $Z_{RR}(L)$ vanishes unless $r = 12k$, in which case $\Delta^{-k}Z_{RR}(L) \in \mathbb Z$. -By simply checking against Borcherds' classification, I know that the following conjecture is true in ranks $r = 12$ and $24$: -Conjecture: Suppose $L$ is a 3-framed unimodular lattice of rank $r = 12k$. Then the integer $\Delta^{-k}Z_{RR}(L)$ is divisible by $24$. [Edit: $\Delta^{-k/2}$.] -Is the conjecture true in general? Is it known? -[Edit: And please continue to point out errors in the comments — I'm sure there are more.] - -REPLY [3 votes]: The discussion in the comments establishes the conjecture when $r$ is divisible by $24$. When $r$ is merely divisible by $12$, the comments establish that $\Delta^{-r/24} Z_{RR}$ is divisible by $12$. (The latter case follows from the former because $Z_{RR}$ is easily seen to be multiplicative.) -Specifically, when $r$ is divisible by $24$, $Z_{RR}$ is a difference of $\Theta$ series of two even unimodular lattices of rank $r$. Theorem 12.1 of https://math.berkeley.edu/~reb/papers/on2/on2.pdf establishes that every $\Theta$ series of an even unimodular lattice is a polynomial of $\Delta$ and $c_4$, and the coefficient of $\Delta^{r/24}$ is divisible by $24$. -This answer is community wiki in case someone wants to expand it.<|endoftext|> -TITLE: Software for computing equivariants -QUESTION [6 upvotes]: If $\Gamma$ is a finite group with action on two vector spaces $\mathbb R^n$ and $\mathbb R^m$ denoted by $\gamma_n$ and $\gamma_m$ respectively, the fundamental equivariants are the polynomials $f: \mathbb R^n \rightarrow \mathbb R^m$ that commute with $\Gamma$, i.e. $$\gamma_m \cdot f(x) = f(\gamma_n^{-1} \cdot x) \quad \forall x \in \mathbb R^n.$$ -It it well known that this set is a finitely generated free module over the ring of primary invariants. -Is there any software available to compute a set of generators of the module of equivariants? -An algorithm is described in page 12 of this paper "Zeros of equivariant vector fields..." by P. Worfolk (citeseerx link, unrestricted) for instance in pseudo code, but surprisingly I was not able to find any implementation online in Singular, Magma or similar. - -REPLY [2 votes]: The canonical reference appears to be Karin Gatermann's book: -Gatermann, Karin, Computer algebra methods for equivariant dynamical systems, Lecture Notes in Mathematics. 1728. Berlin: Springer. xv, 153 p. (2000). ZBL0944.65131. -In the book she frequently alludes to the Maple package symmetry which she uses to compute invariants and equivariants.<|endoftext|> -TITLE: Complexity of $\Vdash_{\mathbb{P}}$ when $\mathbb{P}$ is a class definable iteration -QUESTION [10 upvotes]: Let $\mathbb{P}$ some class forcing iteration which is absolute enough. For instance, suppose that $\mathbb{P}$ is a $\Delta_1$-definable class iteration. Normally, this is the case for iteration of standard forcings like Cohen, collapses and so on. -My question is basically technical and its related with the degree of complexity of the forcing relation $\Vdash_{\mathbb{P}}$. It is well-known that for set forcing notions $\mathbb{P}$ the forcing relation $\Vdash_{\mathbb{P}}$ for $\Sigma_n$-formulae ($n\geq 1$) is $\Sigma_n$-definable with $\mathbb{P}$ as a parameter. On the contrary, if $\mathbb{P}$ is a class this is not clear at all. Notice that in this context the quantifiers over conditions are unbounded so probably the complexity increases. -My question is the following: -Question: If $\mathbb{P}$ is a $\Delta_1$-definable class forcing iteration (suppose if you like that the iterates are Cohen forcings of whatever standard), then the forcing relation $\Vdash_{\mathbb{P}}$ for $\Sigma_n$-formulae is also $\Sigma_n$-definable? If not, could you give a bound for its complexity? -Thanks for your help! - -REPLY [2 votes]: There are several technical subtleties to consider when we try to define the forcing relation $\Vdash$ in the context of class forcing. First we need to distinguish between definable class forcing, i.e. in the context of $ZFC$-set theory and class forcing in the context of $KM$-set theory. Next for definable class forcing, there is the issue that the forcing relation $\Vdash$ might not even be definable in the ground model $M$. In fact both the definability and truth lemma can fail for class forcing notions -For example one may try to define the following: -$p \Vdash \tau_1\in\tau_2 \leftrightarrow \{p':\exists (\tau_3,p'')\in \tau_2 \text{ such that } p'\leq p'', p'\Vdash \tau_2=\tau_3\}$ is dense below $p$. -But the logical complexity of such a statement depends on the ranks of the names and since the forcing is a proper class then the complexity of the statement cannot be bounded and thus determined. -Thus to determine whether a class forcing notion is definable, we will have to involve restrictions on the class forcing notions $\mathbb{P}$. One such property to impose on class forcing is the property of pretameness defined by Friedman: -A class forcing $\mathbb{P}$ is pretame if whenever we have an $(M,A)$-definable (where $A\subseteq M$) sequence of dense classes $\langle D_{\alpha}: \alpha\in X\rangle$ and $X\in M$, $p\in \mathbb{P}$, then there is a condition $q\leq p$ and a new sequence $\langle E_{\alpha}: \alpha\in X\rangle\in M$ with $E_{\alpha}\subseteq D_{\alpha}$ and $E_{\alpha}$ is predense below $q$ for every $\alpha\in X$. -In his book on Fine Structure and Class Forcing, Friedman shows that for any formula $\varphi$ the forcing relation is now $(M,A)$-definable. In addition the truth lemma also holds in this case. You will definitely find more information in that book.<|endoftext|> -TITLE: Thurston's "tinker toy" problem -QUESTION [37 upvotes]: In the article "On Being Thurstonized" by Benson Farb (located here), a particular result of Thurston is mentioned. -Namely, suppose a "tinker toy" $T$ is a contraption consisting of a multitude of rods. They can either be bolted to a table, or attached to each other along hinges. The configuration space $C(T)$ consists of all possible configurations of $T$. For example, if $T$ is a single segment, then $C(T)$ is a circle. -The claim is then that any compact, smooth manifold can be obtained as a component of some $C(T)$. -I've thought about it a bit (though perhaps not enough), and this theorem escapes me completely. Are there any notes or explanations out there that describe the proof of this theorem? - -REPLY [34 votes]: I was Thurston's (undergraduate) student in the mid 1980s, when he was thinking about linkages. Here's how Thurston explained the proof to me. -By a theorem of Nash (see https://en.wikipedia.org/wiki/Nash_functions), any smooth manifold is diffeomorphic to a solution space of a set of real polynomial equations. So now all one needs to do is devise planar linkages which implement addition and multiplication of real numbers, and hook them together in a way that mirrors the algebraic equations from Nash's theorem. -I never worked through the details of the above sketch myself. - -[Added: Igor Rivin's answer overlaps with mine, and has additional details.] - -REPLY [25 votes]: The result comes by way of Nash's theorem which states that every smooth manifold is a component of a real algebraic variety. -Nash, John, Real algebraic manifolds, Ann. Math. (2) 56, 405-421 (1952). ZBL0048.38501. -The game is now to represent a real algebraic variety as a configuration space of a linkage. This is a very old game, going back to Kempe 1876 - the so called Kempe Universality Theorem which states that any real plane algebraic curve can be thus represented. The result of Kapovich and Millson in Andy Putman's answer is a direct descendant of Kempe's result. As far as I know, Thurston did not prove this result, but did give it to Bill Goldman as a Senior Thesis problem. I was under the impression that Goldman solved the problem, but maybe it was a restricted version. - -REPLY [24 votes]: Sometimes the Kempe Universality Theorem -(see Igor Rivin's answer) -is expressed as: There is a linkage -that signs your name. - -          - - -As an indication of how difficult it is to achieve this in practice, here is a little -linkage that when driven by $b$ rotating on the pinned circle center $a$, -draws a rough approximation to a handwritten letter J (purple), drawn out -by joint $z$. - -          - - -          - -Fig. 2.16 in How To Fold It. -Linkage designed by Don Shimamoto. -Computations in Cinderella.<|endoftext|> -TITLE: Reflexive subspaces of bidual Banach spaces -QUESTION [7 upvotes]: The answer to the question is almost surely negative (as almost always in Banach space theory) but I cannot find a relevant example. -Is there an example of an infinite-dimensional Banach space $X$ such that $X^{**}$ does not contain infinite-dimensional reflexive subspaces? -Note that such a space must be HI-saturated. - -REPLY [7 votes]: The answer is that there is indeed an example of such space. This is established in Theorem 6.27 of: -Argyros, Spiros A.; Arvanitakis, Alexander D.; Tolias, Andreas G. Saturated extensions, the attractors method and hereditarily James tree spaces. Methods in Banach space theory, 1–90, London Math. Soc. Lecture Note Ser., 337, Cambridge Univ. Press, Cambridge, 2006. -This book chapter is freely available online at https://arxiv.org/pdf/0807.2392.pdf -Briefly, Theorem 6.27 establishes the existence of an infinite dimensional Banach space $\mathfrak{X}_{\mathcal{F}_s'}$ whose dual does not contain any infinite dimensional reflexive subspace. Since $\mathfrak{X}_{\mathcal{F}_s'}$ admits a boundedly complete basis, denoted $(e_n)$, $\mathfrak{X}_{\mathcal{F}_s'}$ is isomorphic to a dual space. In particular, the closed linear span in $\mathfrak{X}_{\mathcal{F}_s'}^\ast$ of the coordinate functionals $(e_n^\ast)$ biorthogonal to $(e_n)$ provides an example of such a space as requested in OP's question.<|endoftext|> -TITLE: Example of a prime action on a compact Hausdorff Space -QUESTION [5 upvotes]: Suppose that a discrete group $\Gamma$ acts on a compact Hausdorff space $X$ via homeomorphisms. This action induces an action on $C(X)$, the space of all continuous functions from $X$ to $\mathbb{C}$, by $s.f(x)=f(s^{-1}x)$. The action is said to be 'prime' if $C(X)$ doesn't admit any invariant unital $C^*$-subalgebra other than $\mathbb{C}$ and $C(X)$. -There is an equivalent condition for the same which is : $C(X)$ doesn't admit any invariant $C^*$ -sub-algebra other than $\mathbb{C}$ and $C(X)$ iff the only $\Gamma$-equivariant surjective continuous map from $X$ to a compact Hausdorff $\Gamma$-space $Y$, when $Y$ is not a singleton, is one-to-one. -I wonder if there are any examples of such an action for $X$ which doesn't have discrete topology. I couldn't find any example of such an action. It would be great if there any research papers/books which mention of such an action. -Thanks for the help!! - -REPLY [2 votes]: Thompson's group $V$, which is a finitely presented infinite simple group, consists of all homeomorphisms of the Cantor set $\{0,1\}^\mathbb N$ that can be described the following way. A prefix code is a collection of finite words none of which is a prefix of another. A finite prefix code $C$ is maximal if it is not contained in another prefix code. Equivalently, it is maximal if removing the vertices of the binary tree corresponding to $C$ disconnects it. -An element of Thompson's group is given by specifying two maximal prefix codes $C_1=\{u_1,\ldots, u_n\}$ and $C_2=\{v_1,\ldots, v_n\}$ of the same size and a permutation $\sigma\in S_n$. The action of the corresponding element $f_{C_1,C_2,\sigma}$ on $x\in \{0,1\}^n$ is as follows. There is a unique factorization $x=u_iz$ with $u_i\in C_1$. Then $f_{C_1,C_2,\sigma}(x) = v_{\sigma(i)}z$. Note that the prefix codes $C_1,C_2$ are not uniquely determined by the element of $V$. -Let $X=\{0,1\}^\mathbb N$ and let $R$ be an equivalence relation such that $Y=X/R$ is compact Hausdorff and the projection $X\to Y$ is $V$-equivariant. Suppose that $R$ is not the equality relation and has more than one equivalence class. Note that $R$ must be closed as a subset of $X\times X$ by Hausdorffness. Since the complement of $R$ in $X\times X$ is non-empty and open, it follows that there are finite words $u,v$ such that $ux$ and $vy$ are never equivalent for any infinite words $x,y$. By extending the shorter of the two words, we may assume $|u|=|v|=m$ (where $|\cdot|$ is the length). -Since $R$ is not the equality relation we can find two distinct infinite words $x_1,x_2$ which are equivalent under $R$. I claim we can find an element $g\in V$ such that $g(x_1)\in u\{0,1\}^\mathbb N$ and $g(x_2)\in v\{0,1\}^\mathbb N$. This will contradict that $X\to Y$ is $V$-equivariant. -Let $n$ be the length of the longest common prefix of $x_1,x_2$. Let $k>\max\{n,m\}$ and let $C$ be the maximal prefix code consisting of all words of length $k$. Then $x_1,x_2$ have different prefixes $p,q$ of length $k$. Also, since $k>m$, we can find $p',q'\in C$ such that $u$ is a prefix of $p'$ and $v$ is a prefix of $q'$. There is an element $g=f_{C,C,\sigma}\in V$ such that $g$ takes all infinite words $px$ to $p'x$ and $qx$ to $q'x$. It follows that $g(x_1)$ and $g(x_2)$ are not equivalent under $R$, a contradiction.<|endoftext|> -TITLE: Graph automorphism group -QUESTION [12 upvotes]: Let $A_w$ denote such set of positive integer $n$ that: for any two permutations $\pi_0,\pi_1\in S_n$, if $\pi_1$ is not a power of $\pi_0$, then there exists a (labeled non oriented) graph $G$ of size $n$ such that $\pi_0\in Aut(G)$ and $\pi_1\notin Aut(G)$. -Let $A_s$ denote such set of positive integer $n$ that: for any permutation $\pi \in S_n$, there exists a (labeled non oriented) graph $G$ of size $n$ with $Aut(G) = \langle\pi\rangle$. (Pointed by @Ycor.) -Q1: Is $A_w$ or even $A_s$ infinite? -Q2: Does $A_w$ or even $A_s$ contain almost all positive integers? -Q3: What are the smallest elements in these two sets? -Is there any known results? (Papers on this subjects are welcomed.) - -REPLY [4 votes]: Peter and YCor already gave a counterexample, so this answer is just some additional commentary. I'll ignore loops for simplicity. If $\varGamma$ is a permutation group on $\lbrace 1,\ldots, n \rbrace$, then the 2-closure $\varGamma_2$ of $\varGamma$ is the largest subgroup of $S_n$ such that the orbits of $\varGamma_2$ on the set of unordered pairs $\lbrace v,w\rbrace$ are the same as the orbits of $\varGamma$ on the set of unordered pairs. Clearly $\varGamma\le\varGamma_2$. (Google "2-closure" and "orbital" for tons of references and more information.) -Now, the key observation is that every automorphism group of a graph is 2-closed; i.e. it is equal to its own 2-closure. So to find counterexamples to your question just find any $\pi_0$ such that $\langle \pi_0\rangle$ is not 2-closed. YCor's example is a permutation with one cycle: for $n\ge 3$ and $\pi_0=(1\,2\,\ldots\,n)$, the 2-closure of $\langle \pi_0\rangle$ also contains an additional $n$ elements of order 2 (think of the reflections of a regular polygon about a line through the centre). -(Incidentally, many sources define "2-closed" using ordered pairs rather than unordered pairs. For undirected graphs we need unordered pairs.) -ADDED. -I'll start but not yet complete a full description of the set $F(n)$ defined by Peter. To recap: $F(n)$ is the set of all pairs $(\pi_0,\pi_1)$ of permutations in $S_n$ such that every $n$-vertex graph $G$ that has $\pi_0$ as an automorphism also has $\pi_1$ as an automorphism. Describing $F(n)$ is equivalent to describing the groups $\Gamma_{\pi_0}=\lbrace \pi_1 \mid (\pi_0,\pi_1)\in F(n)\rbrace$. ($\Gamma_{\pi_0}$ is a group because the intersection of groups is a group.) -The cases $n=1$ and $n=2$ are trivial and so is the case $n\ge 3, \pi_0=(1)$; I'll leave them for the reader. Assume that $n\ge 3 $ and that $\pi_0$ is not the identity. -Two obvious facts, the first by definition the second by symmetry: (1) $\langle \pi_0\rangle\le \Gamma_{\pi_0}$. (2) For any $\gamma\in S_n$, $\Gamma_{\pi_0^\gamma}=\Gamma_{\pi_0}^\gamma$, where exponentiation denotes conjugacy.Fact (2) says that we only need be concerned with the cycle structure of $\pi_0$. -So take $\pi_0\in S_n$ and let the cycles $C_1,C_2,\ldots,C_k$ of $\pi_0$ have lengths $\ell_1\ge \ell_2\ge\cdots\ge\ell_k$. -For $1\le j\le k$, let $D_j$ be the unique dihedral group of degree $\ell_j$ that contains $C_j$. (Think of $C_j$ as the rotational symmetries of a regular $\ell_j$-gon and $D_j$ as the full symmetry group including reflections.) For small orbits with $\ell_j\le 2$, take $D_j=C_j$. Now define $\Gamma^*_{\pi_0}= D_1\oplus \cdots \oplus D_k$. By a $t$-gon we mean a single vertex if $t=1$, a single edge if $t=2$ and a (graph) cycle of length $t$ if $t\ge 3$. If we insert a $t$-gon into a $t$-cycle of $\pi_0$, we'll assume that the $t$-cycle is an automorphism of the $t$-gon. -Lemma 1. $~\Gamma_{\pi_0}\le \Gamma^*_{\pi_0}$. -Proof. For any $\ell_j\ge 2$, take a graph with an $\ell_j$-gon in $C_j$ and isolated vertices elsewhere. The automorphism group fixes $C_j$ setwise, so $\Gamma_{\pi_0}$ does too. If there are any 1-cycles, connect them into a path and connect one end of the path to some $t$-cycle ($t\ge 2$) into which insert a $t$-gon. No more edges. The automorphism group fixes each of the 1-cycles, so $\Gamma_{\pi_0}$ does too. Thus we have proved that $\Gamma_{\pi_0}$ fixes the cycle partition of $\pi_0$ cell-wise. Now take a graph in which $C_j$ contains an $\ell_j$-gon for every $j$ and no extra edges. The subgroup of its automorphism group that fixes the cycle partition of $\pi_0$ cell-wise is $\Gamma^*_{\pi_0}$. This completes the proof. -$\square$ -To complete the determination of $\Gamma_{\pi_0}$, we have to determine which elements of $\Gamma^*_{\pi_0}$ preserve all possible joins between different cycles. This is slightly complicated because it depends on arithmetic properties: the number of edges between $C_i$ and $C_j$ is divisible by the least common multiple of $\ell_i$ and $\ell_j$. On the bright side, the solution for two cycles should imply the solution for any number of cycles. I need to think of how to do it smoothly.<|endoftext|> -TITLE: Neighborhood fingerprint of a graph -QUESTION [8 upvotes]: Let $G=(V,E)$ be a finite, simple, undirected graph. For $v\in V$ we set $N_0(v) = N(v) = $ $\{w\in V: \{v,w\} \in E\}$ and for $k\in \omega$ let $$N_{k+1}(v) = N_k(v) \cup \bigcup\big\{N(z): z\in N_k(v)\big\}.$$ -We define the neighborhood fingerprint of $G$ as $F_G: V\times \omega \to \omega$ defined by $(v,k) \mapsto |N_k(v)|.$ -It is clear that if two graphs $G,H$ are isomorphic, then there is a bijection $\varphi:V(G)\to V(H)$ such that for all $v\in V(H)$ and all $k\in \omega$ we have $F_G(v,k) = F_H(\varphi(v), k)$. -Does the converse hold? More precisely, if there is a bijection $\varphi$ as described above, are the graphs $G,H$ isomorphic? - -REPLY [5 votes]: Answer. Arguably the simplest infinite set of counterexamples to "Does the converse hold?" are provided by the Möbius ladder's vis-à-vis the prism graphs. Any pair of such graphs (of equal orders) bear stronger similarity to one another than an equal neighborhood fingerprint: all subgraphs induced by balls of radius $r$ are isomorphic (hence the OP's hypothesis holds for such pairs), unless $r\geq\lfloor\frac{\text{number of spokes}}{2}\rfloor$, i.e., unless the balls are large enough to engulf the whole graph and detect the non-isomorphy. -This answers the question. -Remark. While we are at it: the above example can be seen as a reason why the (in)famous Reconstruction Conjecture is so incredible: the condition that all non-trivial balls are isomorphic seems very strong---at first sight perhaps even stronger than the hypothesis in the Reconstruction Conjecture. In particular, the largest balls almost 'look' like the vertex-deleted subgraphs featuring in the Reconstruction Conjecture; and yet, the conclusion that the graphs would have to be isomorphic is false.<|endoftext|> -TITLE: $1$ as difference of composites with same number of prime factors -QUESTION [12 upvotes]: I noticed and found only first three cases: -We can write $1$ as difference of two composites that have one prime factor $$3^2-2^3=1$$ -and as difference of two composites that have two prime factors $$3\cdot 5 - 7\cdot 2 = 1$$ -and as difference of two composites that have three prime factors $$2^2 \cdot 3^2 \cdot 43-7 \cdot 13 \cdot 17=1$$ -I believe that this holds for every $k \in \mathbb N$, that is, that for every $k \in \mathbb N$ there exist composites $a_k$ and $b_k$ that have exactly $k$ prime factors and are such that we have $a_k-b_k=1$. - -Is my belief true? Is this known? What is known about all of this and similar problems? Can someone find solutions for some larger $k$´s? - -Examples exist at least for $k=1,2,...11$ by this list . - -REPLY [12 votes]: Erdos mentions in his book "Topics in the Theory of Numbers" the following: - -"It is stil unkown if the equation $\omega(n+1)=\omega(n)$ has - infinitely many solutions...It is known that $|\omega(n+1)-\omega(n)|$ - takes a certain value infinitely often". - -Here $\omega(x)$ denotes the number of prime factors of $x$. -So, I suppose this is still open. I am not sure if there is an update on this problem.<|endoftext|> -TITLE: Is this morphism of regular local rings regular? -QUESTION [6 upvotes]: Let $R$ be a regular, local $\mathbb{Q}$-algebra with a regular system of parameters $x_1, \dotsc, x_n$, and let -$$f \colon \mathbb{Q}[X_1, \dotsc, X_n]_{(X_1, \dotsc, X_n)} \rightarrow R$$ -be the map given by $X_i \mapsto x_i$. Then $f$ is flat (for instance, by Bourbaki, cf. EGA III, 0.10.2.2). -Is $f$ a regular morphism, that is, are the (geometric) fibers of $f$ regular? Certainly, the closed fiber is regular, but how about the others? -The answer should be yes, and I would appreciate an argument for this. In the case when $R$ is excellent, the positive answer seems to be a special case of EGA IV, 7.9.8, but a less contrived argument (and one that would not use an additional excellence assumption) would be greatly appreciated. - -REPLY [3 votes]: You do not need $R$ excellent. Since $\mathbb{Q}[X_1, \dotsc, X_n]_{(X_1, \dotsc, X_n)}$ is excellent it follows from M. André, Localisation de la lissité formelle.<|endoftext|> -TITLE: Is there a universal way to force the Axiom of Choice to be true? -QUESTION [15 upvotes]: Given a model of set theory $V$ there are various ways to construct a model in which the Axiom of Choice holds, such as Gödel's constructible universe $L^V$ or by using forcing*. I'm wondering if any of these constructions have a nice universal property in the sense of category theory. -In particular are there any adjoints to the inclusion 2-functor $\mathcal{ZFC}\to\mathcal{ZF}$? -(Where $\mathcal{ZF}$ is the 2-category whose objects are either - -Models of ZF -Toposes - -and whose morphisms are either - -Geometric morphisms -Logical functors -Elementary embeddings** - -and where $\mathcal{ZFC}$ is the full subcategory on the objects that obey Choice.) -I think that the concept of the "minimal model" might allow one to construct a right adjoint which is similar in character to the constructable universe. Forcing goes "the other way" by adding sets rather than restricting them, so I suspect it might give a left adjoint. -*Or not? See comments. -**Asaf Karagila notes that there can't exist an elementary embedding between a model where Choice holds and one where it doesn't. So there can't be an adjunction in this case, because its unit or counit would sometimes have to be such a morphism. But perhaps there's some other kind of morphism between models of ZF that does allow an adjunction? - -REPLY [4 votes]: Following up on my remarks in the comments, allow me to answer from -the perspective of model-theoretic interpretations of theories. I -view interpretations of theories as providing particularly strong -forms of the desired functors between these categories. -Specifically, an interpretation of one theory $S$ in another -theory $T$, is a uniform way of defining a model of $S$ inside any -model of $T$. Given a model of the latter theory $M\models T$, one -defines a domain $N$ and functions and relations on this domain, -such that with this structure it becomes a model of the first -theory $N\models S$. -Two theories are mutually interpretable, if each of them is -interpreted in the other. So in any model of the one theory -$M\models T$ you can define model of the other theory $N\models S$, -and inside $N$ you can define a model $\bar M\models T$ of the -first theory again. -The thing to notice is that with mutual interpretability, there is -no insistence that these interpretations are inverse of each other, -and it could be that $M$ and $\bar M$ are not much related. Perhaps -this makes them rather un-adjoint-like. -A much stronger notion, therefore, imposes the uniform inverse -requirement. Specifically, theories $S$ and $T$ are -bi-interpretable if they are mutually interpretable in such a way -that the interpreted model $\bar M$ arising in $N$ is isomorphic to -$M$ and furthermore, isomorphic by an isomorphism that is definable -in $M$, and vice versa in the other direction. -The relevance of this for your question is that ZF and ZFC are not -bi-interpretable. -Theorem. Distinct extensions of ZF are never bi-interpretable. -Thus, one cannot transform ZF models and ZFC models into one -another in such a way that they form a bi-interpretation, and I -take this to be a kind of negative answer to a strong version of -the question. -I recently made a blog post providing a proof and further -discussion of this theorem and related matters: - - -Different set - theories are never - bi-interpretable. - - -(The theorem follows from results of Albert Visser in his 2006 -paper, "Categories of theories and interpretations." In addition, -there is a nice automorphism group model-theoretic argument of Ali -Enayat showing for the specific case of ZF and ZFC that they are -not bi-interpretable. Follow the link at my blog.)<|endoftext|> -TITLE: The largest disk contained by a 'product' of two simply connected plane regions with unit conformal radii -QUESTION [5 upvotes]: Consider a pair of holomorphic functions $f,g \in \mathcal{O}(\Delta)$ on the complex unit disk $\Delta = \{|z| < 1\}$ that both satisfy $f(0) = g(0) = 0$ and $f'(0) = g'(0) = 1$. Does the domain -$$ -f(\Delta) \cdot g(\Delta) := \{f(z)g(w) \mid z, w \in \Delta \} \subset \mathbb{C} -$$ -contain the unit disk $\Delta$, with equality if and only if $f(\Delta) = g(\Delta) = \Delta$? -In this setting, it may be useful to recall Bloch's theorem, which states that $f(\Delta)$ and $g(\Delta)$ contain disks of a fixed (absolute) radius; however those disks need not of course be centered at the origin. - -REPLY [5 votes]: Let's start with the simple reduction. Notice that $f(\Delta)$ and $g(\Delta)$ are connected open sets containing small disks near the origin, so if one of them is unbounded, $f(\Delta)g(\Delta)=\mathbb C$. -Let $a\in\Delta\setminus\{0\}$ (we certainly have $0=f(0)g(0)$, so the origin is never problematic) be not in $f(\Delta)g(\Delta)$. Consider $f(\Delta)\ni 0$ and $(a/g)(\Delta)\ni\infty$. Those are connected but not necessarily simply connected disjoint open sets. Let $\Omega$ be the union of $f(\Delta)$ and all bounded connected components of $\mathbb C\setminus f(\Delta)$ (or, which is the same, the complement of the connected component of $\mathbb C\setminus f(\Delta)$ containing $\infty$). If $(a/g)(\Delta)$ intersects $\Omega$, it must intersect $f(\Delta)$ as well (you cannot reach a bounded connected component of a complement of an open set by a path (actually even an open sausage, if you want) from $\infty$ without crossing the set itself. Now replace $f$ with the conformal mapping $\varphi$ from $\Delta$ to $\Omega$ with $\varphi(0)=f(0)$, $\Phi=\varphi'(0)>0$. Then, by the Schwarz lemma, $\Phi\ge f'(0)=1$, so if we consider $\widetilde f=\Phi^{-1}\varphi$, we will have $\Phi^{-1}a\notin \widetilde f(\Delta)g(\Delta)$ and $\widetilde f$ is now univalent. Similarly we can make $g$ univalent. -Now comes the main -Lemma: Let $f$ be a (bounded and, if you want, analytic up to the boundary) univalent function such that $f(0)=0, f'(0)=1$. Let $A$ be the area on $\mathbb C\setminus\{0\}$ given by $dA(z)=|z|^{-2}dm_2(z)$, which is invariant under $z\mapsto az (a\ne 0)$ and $z\mapsto z^{-1}$. Then -$$ -A(f(\Delta\setminus r\Delta))\ge 2\pi\log\frac{1}{r}+o(1)\text{ as }r\to 0^+\,. -$$ -Proof: -Let $S=\Delta\setminus r\Delta$. We have (since $f$ is univalent) -$$ -A(f(S))\times 2\pi\log\frac 1r=\left[\int_S\frac{|f'|^2}{|f|^2}\,dm_2\right]\left[\int_S\frac{1}{|z|^2}\,dm_2\right] -\\ -\ge -\left[\int_S\frac{|f'|}{|f|}\,\frac{dm_2(z)}{|z|}\right]^2=I^2\,. -$$ -Note that -$$ -I=\int_{[0,2\pi]}d\theta\int_r^1d\rho \frac{|f'(\rho e^{i\theta})|}{|f(\rho e^{i\theta})|}\ge \int_{[0,2\pi]}d\theta\int_r^1 d\rho \frac d{d\rho}(\log|f(\rho e^{i\theta})|) -\\ -= -\int_{[0,2\pi]}d\theta\log|f(e^{i\theta})|-\int_{[0,2\pi]}d\theta\log|f(re^{i\theta})|=I_1-I_2\,. -$$ -However, under our assumptions we have $I_1=0$ while $I_2=2\pi\log r+O(r)$ as $r\to 0^+$ whence the lemma. -Now life gets easy. Notice that for sufficiently small $r>0$ both $f(S)$ and $(a/g)(S)$ lie in the annulus $\{w\in\mathbb C\,:\,(1-o(1))r\le|w|\le (1+o(1))|a|r^{-1}$. (we use both the boundedness and the univalence properties here). But the invariant area of this annulus is only $4\pi\log\frac 1r+\log|a|+o(1)$, so the images must overlap somewhere, thus finishing the story. -Cute question, by the way :-)<|endoftext|> -TITLE: A combinatorial question on complete graphs and polynomials -QUESTION [5 upvotes]: Let $K_n$ denote the complete graph on $n$ vertices. Let us denote the vertices simply by $1,\ldots,n$. Suppose that, for each edge $ij$, with $1\leq i -TITLE: Abel and Galois (and Arnold) -QUESTION [42 upvotes]: Question Is there a connection between Abel and Galois theories of polynomial equations? -Recall that for every polynomial $p(x)\in \mathbb{Q}[x]$ (say, without the free coefficient), Abel considered the monodromy group of the Riemann surface of the analytic function $w(z)$ defined by $p(w(z))+z=0$.. There is an expression of $w(z)$ in radicals if the monodromy group is solvable (is it an "if and only if" statement?). -On the other hand, for every polynomial $p(x)\in \mathbb{Q}[x]$, Galois considered the automorphism group (the Galois group) of the splitting field of $p$. The roots of $p(x)$ are expressed in radicals if and only if the Galois group is solvable. -The question asks whether there are any known connections between the monodromy group of $w(z)$ and the Galois group of $p(x)+z$ (considered as polynomial in $x$). -I am pretty sure this is well known. I just cannot find it in the literature. -Update 1. What I called "Abel's proof" of Abels' theorem is in fact Arnold's proof written by Alexeev (an English translation can be found here). Abel's proof was based on different ideas, see this text. So some instances of the word "Abel" above should be replaced by "Arnold". Added "Arnold" to the title. -Update 2. I found a very nice book by Askold Khovanskii, -"Topological Galois Theory", where Arnold's proof and its strong generalizations to other types of equations, including differential equations, are explained in detail. Highly recommended. - -REPLY [26 votes]: The action of the monodromy group of $w(z)$ on the fiber $p^{-1}(a)$ for a non-critical value $a$ of $p$ (that is $|p^{-1}(a)|=\deg p$) is the same as the action of the Galois group of $p(x)+z$ over $\mathbb C(z)$ on the roots of $p(x)+z$ in some splitting field. One can see this by comparing each of these groups with the deck transformation group of the cover $X\to\mathbb P^1\mathbb C$, where $X$ is the normal hull of the cover $P^1\mathbb C\to P^1\mathbb C$, $x\mapsto p(x)$. (Remark: Though it doesn't make a difference, it is slightly more convenient to work with $p(x)-z$ instead of $p(x)+z$.)<|endoftext|> -TITLE: Bruhat-Tits building of $SL_n(\mathbb{Q})$, hyperbolic isometries and its axis -QUESTION [6 upvotes]: Consider $G=SL_n(\mathbb{Q})$ and $p$ a prime integer. Associated to $G$ and $p$ we have its Bruhat-Tits building $\Delta$. -It is well known that $\Delta$ can be provided with a canonical $CAT(0)$ metric and an action of $G$ by isometries. Hence every element in $G$ determines an isometry that it is either: -Elliptic: it has a fixed point, or -Hyperbolic: it has an axis, i.e. a geodesic $L$ on which acts as a nontrivial translation. -Associated to $G$ there is a system of apartments $\mathcal{A}=\{g\Sigma : \Sigma \text{ the fundamental apartment}\}$ of $\Delta$ which is not the full system of apartments. -My question is the following: -Suppose $\alpha\in G$ is a hyperbolic isometry of $\Delta$ and $L$ is an axis of $\alpha$, is there an apartment in $\mathcal{A}$ containing $L$? -There is certainly an apartment in the full system of apartments of $\Delta$ that contains $L$. -Thank you all. - -REPLY [2 votes]: If my understanding of the question is correct then the answer is "not necessarily". -I understand that $\Delta$ is the Bruhat-Tits building of $H=\text{SL}_n(\mathbb{Q}_p)$, while $G=\text{SL}_n(\mathbb{Q}) -TITLE: Mathematical theory of aesthetics -QUESTION [20 upvotes]: The notion of beauty has historically led many mathematicians to fruitful work. Yet, I have yet to find a mathematical text which has attempted to elucidate what exactly makes certain geometric figures aesthetically pleasing and others less so. Naturally, some would mention the properties of elegance, symmetry and surprise but I think these constitute basic ideas and not a well-developed thesis. -In this spirit, I would like to know whether there are any references to mathematicians who have developed a mathematical theory of aesthetics as well as algorithms(if possible) for discovering aesthetically pleasing mathematical structures. -To give precise examples of mathematical objects which are generally considered aesthetic, I would include: - -Mandelbrot set -Golden ratio -Short proofs of seemingly-complex statements(ex. Proofs from the Book) - -I think the last example is particularly useful as Jürgen Schmidhuber, a famous computer scientist and AI researcher, has attempted to derive a measure of beauty using Kolmogorov Complexity in his series of articles titled 'Low Complexity Art'. Meanwhile, I find the following research directions initiated by computer scientists particularly fruitful: - -Bayesian Surprise attracts Human Attention -Curiosity and Fine Arts -Low Complexity Art -Novelty Search and the Problem with Objectives - -Note: From a scientific perspective, researchers on linguistic and cultural evolution such as Pierre Oudeyer have identified phenomena which are both diverse and universal. Diversity is what makes our cultures different and universality enables geographically-isolated cultures to understand one another. In particular, many aesthetics have emerged independently in geographically isolated cultures especially in cultures which developed in similar environments. Basically, I believe that if we take into account what scientists have learned from the fields of cultural -and linguistic evolution, embodied cognition, and natural selection I think we could find an accurate mathematical basis for aesthetics which would also be scientifically relevant. - -REPLY [2 votes]: What about this paper where the aesthetics of fractal dimension is measured. The peak seem to be (according to the paper) near Hausdorff dimension 1.5.<|endoftext|> -TITLE: Reference for Weyl's law for higher order operators on closed Riemannian manifolds -QUESTION [5 upvotes]: I am looking at page 32 (beginning of Chapter 5) here. We are given a formally self-adjoint, metrically defined differential operator $A$ on $(M^n,g)$ of order $2l$ with positive definite leading symbol, such that $A_{\tilde{g}}=c^{-2l}A_g$ whenever $\bar{g}=c^2g$ is a change of metric for some $c>0$ constant. We are told that the asymptotic behaviour of eigenvalues of $A$ satisfies Weyl's formula -\begin{equation}\tag{1} -\lambda_j\sim C(g,A)\,j^\frac{2l}{n}~~~\mathrm{as}~j\rightarrow\infty, -\end{equation} -but I am struggling to find a reference for this result. It is also stated here without proof, but even in the suggested references I still can't find anything. -Lots of the literature (for instance this paper of Gilkey) also only states (1) for (higher order) Laplace-type operators, i.e. operators for which the leading symbol is (a power of) the metric. Am I correct in saying the conditions above are more general? Thanks in advance. -EDIT: by the conditions above I mean in particular the scaling property $A_{\tilde{g}}=c^{-2l}A_g$. Does this imply, or is it implied by, $A$ having its leading symbol being a power of the metric? I guess this is a second question; answers to either would be welcomed. - -REPLY [2 votes]: One possible reference is Seeley's paper on Complex powers of Elliptic Operators, where Seeley did it for the Laplacian (page 6). But the discussion carries over to all elliptic $\Psi DO$s without much difficulty. For a "modern" expository article, see this. Seeley's argument would imply a term of type $j^{2l/n}$, just as the article claimed. A standard textbook is Shubin's book, where the result is proved in page 130.<|endoftext|> -TITLE: A necessary and sufficient condition for a space curve to lie on a ellipsoid -QUESTION [11 upvotes]: Any (arc-length parametrized) space curve is uniquely determined (up to rigid motion) by its curvature  and its torsion. -For instance we know that a necessary and sufficient condition for a space curve to lie on a sphere is -$R^2+(R')^2T^2=const$, where $R=1/\kappa$, $T=1/\tau$, and $R'$ is the derivative of $R$ relative to $s$. -I want to know if there is a necessary and sufficient condition for a space curve to lie on a ellipsoid (in terms of its curvature and torsion). - -REPLY [8 votes]: Robert describes the differential equations which one can write in terms of $\tau$ and $\kappa$, and the inherent limitations in this local approach. But maybe one can find more reasonable or useful conditions in terms of integral equations, and the whole problem could be more interesting or natural if we consider closed curves. In other words, a global approach could be more enlightening. -For instance, a necessary condition for a closed curve to lie on a sphere is that $\int \tau=0$, e.g. see p.171 of Millman and Parker, which incidentally turns out to characterize spheres. Furthermore, any closed curve lying on a convex surface must have at least $4$ points where $\tau=0$, which is a generalization of the classical four vertex theorem due to Sedykh; see also this paper for another proof, and this paper for a generalization. Another necessary condition for a curve to lie on an ellipsoid is that it have a pair of parallel tangent lines, which turns out to characterize ellipsoids, as described in this paper with Bruce Solomon. -It would be interesting to find more simple or nice necessary conditions for a closed curve to lie on an ellipsoid, and I think it is possible that a collection of these may turn out to be sufficient as well. -Addendum 1: I found a nice paper which seems to be relevant: -Kreyszig, Erwin; Pendl, Alois -Spherical curves and their analogues in affine differential geometry. -Proc. Amer. Math. Soc. 48 (1975), 423–428. -In this paper the authors define a curve to be spherical in the affine sense if all its normal planes pass through a common point. If I am not mistaken these include curves which lie on ellipsoids, but other curves as well. At any rate, they obtain a very nice characterization for these "affine spherical curves": -$$ -\left(\frac{1}{\widetilde\tau}\right)''+\frac{\widetilde\kappa}{\widetilde\tau}=0, -$$ -where $\widetilde\kappa$, $\widetilde\tau$ are the affine curvature and torsion and diferrentiation is with respect to affine arc length. -Addendum 2: Maybe a necessary condition could be that $\int\widetilde\tau=0$, but this is just a quick guess.<|endoftext|> -TITLE: Generating prime knots (in order) -QUESTION [6 upvotes]: In this really cool paper https://arxiv.org/abs/1612.03368, A. Malyutin shows that the probability that a random prime knot of up to $N$ crossings (as $N$ goes to infinity) is not generically hyperbolic (mod some "standard conjectures"). Now, the question is, how would one go about generating prime knots, either in order of crossing number (preferred) or in some randomized way? It is an interesting question of which conjectures are actually not correct (since, just like HRJW in the comments, I believe that a random prime knot should be hyperbolic). - -REPLY [4 votes]: I'm not sure this is what you are looking for, but Adams et al. (https://arxiv.org/abs/1208.5742) show how every knot is obtained from a permutation in $S_n$. We sampled random permutations and used SnapPy to determine the resulting knot types, and they seem to be generically hyperbolic (https://arxiv.org/abs/1711.10470, Figure 14).<|endoftext|> -TITLE: Are there Type III codes with small but nonzero "index"? -QUESTION [8 upvotes]: Recall that a Type III code of rank $r$ is a linear subspace $C \subset \mathbb F_3^r$ which is self-dual for the standard inner product. (These occur only when $r$ is divisible by $4$.) Elements of $C$ are called code words. The Hamming weight of a code word is its number of non-zero entries. I will call a code word maximal if its Hamming weight is $r$. (Maximal codewords clearly only occur when $r$ is divisible by $12$.) -The set of maximal codewords can be partitioned into two subsets as follows: two maximal words $w_1,w_2$ are in the same subset if $w_1-w_2$ has even Hamming weight, and are in opposite subsets if $w_1-w_2$ has odd Hamming weight. Call these subsets $M_+$ and $M_-$. -I suggest the following definition: -Definition: The index of $C$ is $|M_+| - |M_-|$. Of course, this is only defined up to sign. I could take its absolute value if I cared. The name is because this is related to the "supersymmetric index" of a certain supersymmetric field theory. -The discussion in the comments of this question implies the following when $r$ is divisible by $24$: -Proposition: The index of any Type III code of rank $r=24k$ is divisible by $24$. -Conjecture: This is true also when $r \equiv 12 \mod 24$. (It is trivially true for $r \equiv \pm 4 \mod 12$, as then $M_\pm$ are empty.) -Moreover, I suspect that both $|M_+|$ and $|M_-|$ are necessarily divisible by $24$. -Example: The Ternary Golay code, with rank $12$ has index $24$. It is the unique rank-$12$ code with non-zero index. -Example: The complete classification of Type III codes of rank $24$ is known. Assuming I read it correctly, there are precisely five codes of rank $24$ and index $24$. - -Question: In higher rank, do there exist Type III codes with index exactly $24$? For example, what about rank $36$? - -At best, there would be some general algorithm that produces a code with index $24$ for each rank $r = 12k$. -By the way, I know how to prove: -Proposition: If the code $C$ contains words of Hamming weight $3$, then its index vanishes. -So if you are looking for such a code, you know not to look at such codes. -It's pretty easy to show: -Lemma: The index multiplies when you take the direct sum of codes. -Since $24^2$ is pretty big, you probably will want to work with indecomposable codes. - -REPLY [4 votes]: Index $24$ isn't hard for length $36$. -For example, the Type III code with generator matrix -+ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 + - + 0 - - + 0 + + - + + - 0 - -0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 - 0 - + - - 0 0 - 0 0 0 0 - + - - - -0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 - + 0 + - - - + + - + 0 - 0 - + -0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 + 0 + - + 0 0 + 0 + 0 - - + - -0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 - + 0 0 - 0 - + 0 + 0 - - 0 + 0 + - -0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 + - 0 - - 0 + + 0 + 0 + 0 - 0 - + 0 -0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 - + 0 0 - + 0 - - 0 0 - + - 0 0 - + -0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 + + - - - - - + 0 + - - - 0 + - 0 -0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 + + - - + + 0 + - 0 + 0 0 + + + + - -0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 - + + + 0 0 - + + - + - - 0 + 0 - - -0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 + + 0 0 0 - + + + + + 0 - 0 - 0 0 + -0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 + 0 0 0 - - + 0 0 - 0 + - - + 0 - + -0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 + 0 - 0 0 - + - - 0 - + + - 0 0 + -0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 + 0 - + - - - - 0 0 - 0 - - - - - - -0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 - 0 - - - 0 + + 0 + - 0 - 0 - + 0 0 -0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 + - - + 0 0 - 0 - + - + - - - + + 0 -0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 - 0 + - 0 - 0 + 0 - - 0 + 0 0 - - - -0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 + + + + - + 0 - - 0 + 0 + - - - + - -has index $24$. (The total number of maximal codewords is $520$; of -$70$ "random" Type III codes I tried, $26$ had index $24$ -- which indeed -was the most common index -- and all had index divisible by $24$, -while the number of maximal codewords was always a multiple of $8$ -but not necessarily $24$.)<|endoftext|> -TITLE: Why the unreasonable applicability of complex numbers in physics/engineering? -QUESTION [10 upvotes]: After years of using complex numbers in every kind of analysis of physical and electrical engineering problems I am starting to wonder: why is this particular algebra so effective in modelling the world? Because from a purely mathematical point of view $\mathbb{C}$ is no more than the field $\mathbb{R^2}$ with the operations given by: -$$(x,y) + (a,b) = (x+a, y+b)$$ -$$(x,y) * (a,b) = (x*a - y*b,y*a + x*b)$$ -Is there a physical motivation for these definitions which might help explain why this particular algebra is so powerful in modelling physical systems? -PS: My friend suggests that perhaps this algebra can be derived to be the algebra that is satisfied by Fourier coefficients. And that it is some kind of implicit Fourier analysis which is actually modelling the physical systems. Can that be true? -EDIT: This question is different from the one about demystifying complex numbers. That one asks for examples for the usefulness of complex numbers. I know they are useful. I would like to know why they are useful if that is an answerable question. - -REPLY [9 votes]: I think this question can be answered without appealing to quantum mechanics (and the question is more general than quantum mechanics). -In classical physics, quantum physics, electrical engineering, chemistry, etc. you're often describing your the system at hand with differential equations. When solving differential equations, complex numbers pop up all over the place. As alluded to in the quote above, a straightforward solution to many differential equations often makes use of complex analysis, so there's no surprise that they are used and are useful if the tools we're using come from complex analysis. -However, we don't have to limit ourselves to just differential equations. Pick a random polynomial and solve for all the roots. You're far more likely to end up with complex roots than purely real roots. If we describe the world around us in mathematical terms and everything from simple polynomials to differential equations give complex answers more often than purely real answers, it not surprising (to me at least) that complex numbers are going to show up everywhere. -A more boring answer is that complex numbers usually show up whenever phase is important (quantum mechanics, wave mechanics, electrical circuits, etc.) You don't strictly need to use complex numbers, but it sure makes the notation a lot more compact and intuitive. So, we use them because they're convenient.<|endoftext|> -TITLE: Classification of singularities of plane curves of fixed degree (reference request) -QUESTION [8 upvotes]: We know the answers to some questions like What is the maximal number of singularities of (reduced) plane curves of degree $d$? for general $d$ (in this case $\tfrac{1}{2}d(d-1)$, obtained by $d$ lines in general position). On the other hand, more subtle questions like Given $d$, what is the largest $n$ such that some plane curve of degree $d$ has a singular point of type $A_n$ can be really hard. Nonetheless, in small degrees, complete lists can be given and in very small degrees, the lists are actually rather short and obvious. -I would really appreciate if someone could give an up to date list of the major results in the field. -Since this is probably too much asked for, here are some more, rather concrete questions: - -I am aware of Chung-Man Hui's '79 thesis Plane quartic curves classifying singularities of plane curves of degree four. Is there something similar for quintics or even sextics? -Wall's book Singular Points of Plane Curves (most notably sections 7.5-7.7) contains relevant information. Is the information I can find there (more or less) complete and up to date? -Back to one of the questions above, out of curiosity, for which degrees do we know the maximal $n$ such that some curve of the given degree has an $A_n$-singularity? We have: -$$\begin{array}{lccccc} -d & 2 & 3 & 4 & 5 & 6 & \cdots \\ -\text{max. }A_n\text{ on a }C_d & 1 & 3^{1)} & 7^{2)} & 12^{3)} & 19^{3)} & ? -\end{array}$$ -1) On an irreducible cubic, there is at worst an $A_2$. -2) On an irreducible quartic, there is at worst an $A_6$. -3) Added after JNS' answer; here, the maximal $A_n$ is attained by irreducible curves. - -REPLY [6 votes]: Here is what is known for quintics and sextics. Starting from $d=7$, I do not know the answer to your question. - -Quintics: The maximal $A_n$ on a $C_5$ is $n=12$. For example, the curve given by the zero set of $$ -(y^2-xz)^2(\frac{1}{4}x+y+z)-x^2(y^2-xz)(x+2y)+x^5 -$$ has an $A_{12}$-singularity at $(0:0:1)$, c.f. Wall, C. (1996). Highly singular quintic curves. Mathematical Proceedings of the Cambridge Philosophical Society, 119(2), 257-277. He gives a classification of quintics with ''large'' Milnor number. -Sextics: The maximal $A_n$ on a $C_6$ is $n=19$. This was proved by Yang, who gave a complete classification of sextics with simple singularities. -See: Yang, Jin-Gen. Sextic curves with simple singularities. Tohoku Math. J. (2) 48 (1996), no. 2, 203--227. - -As a side note, you have observed that the maximal $A_n$-singularities for $d=2,3,4$ occur only for reducible curves. However, for $d=5,6$ the maximal singularities are achieved only by irreducible curves.<|endoftext|> -TITLE: When is $max Spec R$ homotopy equivalent with $Spec R$ (with Zariski topology)? -QUESTION [9 upvotes]: A commutative ring with unity is called pm-ring if every prime ideal is contained in a unique maximal ideal. In [dMO71], it is shown that pm-rings are characterized by the fact that $\operatorname{max Spec} R$ (the set of all maximal ideals under Zariski subspace topology) is a retract of $\operatorname{Spec} R$, and in this case, the unique retraction is given by $u: \operatorname{Spec} R \to \operatorname{max Spec} R$ , $u(P)$ is the unique maximal ideal containing $P\in \operatorname{Spec} R$ . -It can moreover be shown that for pm-rings, this retract is actually also a deformation retract , because $H : \operatorname{Spec} R \times [0,1] \to \operatorname{Spec} R$ given by $H(P,t)=P, \forall t \in [0,1)$ and $H(P,1)=u(P)$ is continuous, so gives a homotopy between $i\circ u$ and $Id_{\operatorname{Spec} R}$, where $i:\operatorname{max Spec} R \to \operatorname{Spec} R$ - is the inclusion map. -So the questions I want to ask are the following : -(1) Can we characterize (possibly algebraic characterization) commutative rings (with unity) $R$ such that $\operatorname{max Spec} R$ is homotopy equivalent with $\operatorname{Spec} R$ ? -(2) Can we characterize commutative rings (with unity) $R$, such that $i : \operatorname {max} \operatorname {Spec} R \to \operatorname {Spec} R$ is a homotopy equivalence i.e. there exists a map $g : \operatorname {Spec} R \to \operatorname {max} \operatorname {Spec} R$ such that $i\circ g$ and $g \circ i$ are homotopic to the respective identity maps ? -As noted, pm-rings are definitely in both the class, but what are all such rings ? Even if we can't say what are all such rings, can we atleast find class of rings for each case (1) and (2) which are not necessarily pm-rings ? - -References. -[dMO71] De Marco, Giuseppe; Orsatti, Adalberto, Commutative rings in which every prime ideal is contained in a unique maximal ideal, Proc. Am. Math. Soc. 30, 459-466 (1971). ZBL0207.05001. - -REPLY [2 votes]: This is not a full answer, but some examples to show that the situation is a bit tricky. (This answer was worked out together with Dmitrii Pirozhkov.) -Lemma. Let $R$ be a domain. Then $X = \operatorname{Spec} R$ is contractible. -Proof. In fact, the inclusion of the generic point $\eta$ is a deformation retract, by the map -\begin{align*} -[0,1] \times X &\to X \\ -(t,\mathfrak p) &\mapsto \left\{\begin{array}{ll}\mathfrak p, & t = 0, \\ \eta, & t > 0.\end{array}\right. -\end{align*} -(One easily shows that this map is continuous.) $\square$ -Thus, for a domain, the question is when $\operatorname{Specmax} R$ is contractible. Even for simple examples like $R = k[t]$, the result depends on the base field $k$: -Example. Let $R$ be a $1$-dimensional Noetherian domain. Then $X = \operatorname{Specmax} R$ is a cofinite topological space. But the homotopy type depends on the cardinality of $X$: - -If $|X| = 1$, then $R$ is local and $X$ is evidently contractible (since it is a point). -If $X$ is countable but has more than $1$ element, then $X$ is not path connected (see e.g. this MO post), so in particular it is not contractible. -If $|X| \geq |\mathbb R|$, then we can choose a bijection $\phi \colon (0,1) \times X \to X$. Extend $\phi$ to a map $\phi \colon [0,1] \times X \to X$ by setting $\phi(0,-) = \operatorname{id}$ and $\phi(1,-) = \mathfrak m_0$ a constant map. Then each fibre of $\phi$ is closed, hence $\phi$ is continuous. Thus, $\phi$ is a homotopy from the identity to a constant map, so $X$ is contractible.<|endoftext|> -TITLE: Are indecomposable representations of a finite group of Lie type absolutely indecomposable? -QUESTION [6 upvotes]: Let $G = G(K)$ be a Chevalley group over an algebraically closed field $K$ of characteristic $p > 0$. Consider the finite group $G(q) = G(\mathbb{F}_q)$. (For example, if $G = \operatorname{SL}_n(K)$ then $G(q) = \operatorname{SL}_n(\mathbb{F}_q)$). -It was proven by Steinberg that every irreducible $\mathbb{F}_q[G(q)]$-module is absolutely irreducible, and correspond to $q$-restricted dominant weights (i.e. weights $\lambda$ such that $0 \leq \langle \lambda, \alpha \rangle < q$ for all simple roots $\alpha$). -Consider an indecomposable $\mathbb{F}_q[G(q)]$-module $V$. Is $V \otimes_{\mathbb{F}_q} K$ an indecomposable $K[G(q)]$-module? -EDIT: In case this is not true, I also would be interested if what I am asking for is true under some mild restrictions on $q$ and/or $G$. - -REPLY [9 votes]: This is rarely the case for all $V$. $KG(q)$ would need to have finite representation type, so $G(q)$ would need to have a cyclic Sylow $p$-subgroup. Otherwise a counterexample can be found by taking an indecomposable module over some field extension of $\mathbb{F}_q$, that is not defined over $\mathbb{F}_q$ and regarding it as a module for $\mathbb{F}_qG$. -Suppose $KG$ has infinite representation type, where $K$ is the algebraic closure of $\mathbb{F}_q$. -Then for some $d$ there are infinitely many isomorphism classes of $d$-dimensional indecomposable $KG$-modules. Since there are only finitely many $d$-dimensional $\mathbb{F}_qG$-modules up to isomorphism, there must be a $d$-dimensional $KG$-module $M$ that is not defined over $\mathbb{F}_q$: i.e., such that there is no $\mathbb{F}_qG$-module $N$ such that $M\cong N\otimes_{\mathbb{F}_q}K$. -The structure constants of $M$ only involve finitely many elements of $K$, and so $M$ is defined over some finite extension $k$ of $\mathbb{F}_q$: i.e., $M\cong L\otimes_kK$ for some $kG$-module $L$, where $L$ is indecomposable and can't be defined over $\mathbb{F}_q$. -Let $L'$ be $L$ regarded as an $\mathbb{F}_qG$-module (so $L'$ is $[k:\mathbb{F}_q]d$-dimensional). -Then $L'\otimes_{\mathbb{F}_q}k$ is a direct sum of $[k:\mathbb{F}_q]$ copies of $L$, so there can be no indecomposable direct summand $N$ of $L'$ with $N\otimes_{\mathbb{F}_q}k$ indecomposable.<|endoftext|> -TITLE: Existence of spectral gap -QUESTION [7 upvotes]: I would like to start by saying that any comment or idea is highly appreciated. -Let us observe that for Hilbert-Schmidt operators $H_1,H_2$ on an infinite-dimensional separable complex Hilbert space $H$ and a bounded self-adjoint operator $T: H \rightarrow H$ -$$\operatorname{Tr} \left(H_1 ([T,H_2])^*\right) = \operatorname{Tr} \left([T,H_1] H_2^* \right).$$ -In other words, the operator $S_T(H_1):=[T,H_1]$ is self-adjoint with respect to the Hilbert-Schmidt inner product. -Then, we can look at $0 \le S_T^2(H_1) = [T,[T,H]]$ -and see that this operator is positive, as it is a square. -Now, I would like to ask the following questions: -(i)Is there a self-adjoint bounded $T$ such that $S_T^2$ has a spectral gap? (FALSE! See answer by Mateusz Wasilewski.) -(ii)Is there a finite collection of self-adjoint bounded $T_1,...,T_n$ such that $S_{T_1}^2 + ... + S_{T_n}^2$ has a spectral gap? -(iii) Is there an infinite collection of positive bounded $(T_n)$ such that $\sum_n T_n$ converges pointwise such that $\sum_n S_{T_n}^2$ has a spectral gap? -I say $S_T^2$ has a spectral gap if -$$\operatorname{Tr}(S_T^2(H_1)H_1^*) \ge \lambda \operatorname{Tr}(H_1H_1^*)$$ -for some positive $\lambda>0$. -It looks to me that somebody with a operator algebra background may be more familiar with this problem. I assume the three questions are increasingly likely to be true? - -REPLY [7 votes]: The answer to (ii) is positive. Here is a construction, which relies on graphs with spectral gap. For simplicity I write things for trees, but one can probably do the same for more general graphs with spectral gap. -Start with $X_n$ an $n \geq 3$-regular tree, and label each edge with a label in $\{1,\dots,n\}$ in such a way that no edge with the same label share a vertex. Define $\sigma_i$ the permutation of $X_n$ permuting the endpoints of every edge labelled $i$, and $T_i \colon \xi \in \ell_2(X_n) \mapsto \xi_i \circ \sigma_i$. If you prefer groups, $X_n$ is the Cayley graph of $G_n$, the free product of $n$ copies of $(\mathbf Z/2\mathbf Z)$, $\sigma_i$ are the standard generators of $G_n$ and $T_i =\lambda(\sigma_i)$ for the left-regular representation. -Then $S_{T_i}^2(H) = 2(H-T_i H T_i)$ and $\sum_{i=1}^n S_{T_i}^2(H) = 2n(H-\frac{1}{n} \sum_{i}^n T_i H T_i^*)$ (I write $T_i^*$ eventhough $T_i$ is self-adjoint). So it suffices to show that the map $H \mapsto \frac{1}{n} \sum_{i}^n T_i H T_i^*$ has norm $<1$. But identifying the space of Hilbert-Schmidt operators on $H$ with $H \otimes \overline H$, this operators is $\frac{1}{n} \sum_{i}^n T_i \otimes T_i^*$, which (by Fell's absorption principle) has the same norm as $\frac{1}{n} \sum_{i}^n T_i$, the operator of the Random walk on $X_n$. By a famous computation by Kesten, this norm is $\frac{2 \sqrt{n-1}}{n}$. This gives spectral gap $\lambda = 2n-4\sqrt{n-1}$. -A last remark: this method cannot be applied to $n=2$ (because it constructs self-adjoint unitaries $T_1,\dots,T_n$, and for $2$ unitaries there cannot be spectral gap). I am sure that (ii) has already positive answer for $n=2$, but this would require a different construction.<|endoftext|> -TITLE: Sets $A$ stable under $(x,f(x))\mapsto x+f(x)$ -QUESTION [5 upvotes]: Let $A$ be a finite set of real numbers or integers. We know how to characterize, broadly speaking, sets $A$ such that $A+A$ is not much larger than $A$ (Freiman's theorem). I have a question that feels somewhat related but may or may not be. -What are the sets $A$ such that there is a bijection $f:A\mapsto A$ for which $$\left|\{x+f(x): x\in A\}\cap A\right|\geq (1-\epsilon) |A|,$$ -where $\epsilon$ is small? In particular, if $A$ is an interval in the integers (say, $A =\{1,\dotsc,n\}$, or $A=\{-n,-n+1,\dotsc,n\}$), is there such a bijection $f$? -The related question of whether there are such $f$ altogether (with $A$ being initially unspecified) is easily settled. See below. The answer does imply that there are very many sets $A$ for which there are such $f$, but it is unclear to me whether there is a good characterization of such sets $A$ (and whether, say, $A$ can be an interval in the integers). - -Require that $f$ be not just a bijection but (seen as a permutation) a long cycle. (This is not a very restrictive assumption; in fact, it is very easy to see that, if $f$ does not have many short cycles, it can be easily modified at $\epsilon |A|$ places so that it satisfies the assumption.) So, let us order the elements of $A$ (which is still an unspecified set with $n$ elements) in the following way: -$$x_1,\; x_2 = f(x_1),\; x_3 = f(x_2),\;\dotsc,\; x_n = f(x_{n-1}).$$ -Then there is a permutation $\pi$ of $\{1,2,\dotsc,n\}$ such that $$x_i+f(x_i) = x_{\pi(i)}$$ for almost all $1\leq i\leq n$ (meaning: $(1-\epsilon) n$ values of $1\leq i\leq n$). In other words $$x_i + x_{i+1} = x_{\pi(i)}$$ for almost all $1\leq i\leq n$. -Now we see that, for any permutation $\pi$, we have a system of $(1-\epsilon) n$ linear equations in $n$ variables, and thus we must have a solution - an $(\epsilon n)$-dimensional space of solutions, in fact. Checking that there are few repetitions among the variables $x_i$ should take some work, but that should be generically the case. - -REPLY [7 votes]: I'll address the special cases $A = [-n, n]$ and $A = [n]$. -For $A = [-n, n]$ we can always find a bijection where all the sums are distinct and in the correct range: map $[-n,0]$ to $[0,n]$ and $[1,n]$ to $[-n, -1]$, preserving order in each case. -For $A = [n]$ the maximum size of the intersection is probably $\lfloor \frac{2n-1} 3 \rfloor$ (which holds for $n \leq 14$ by exhaustive search). When $n = 3k-1$ this is $2k-1$, which can be achieved by mapping $[k]$ to $[k, 2k-1]$ and $[k+1,2k-1]$ to $[k-1]$, preserving order in each case. I'll show that $2k-1$ is also an upper bound for the size of the intersection when $n=3k-1$. -I'll restate the problem as follows. Let $T = \{(x, y, z) \in [n]^3 : x + y = z) \}$. We seek to bound the size of a subset $S$ of $T$ such that the projections of $S$ onto the $x$, $y$ and $z$ coordinates are injective. -For each $j \in [3k-1]$, write - -$X_j = \{(x,y,z) \in S : x \leq j)\}$. -$Y_j = \{(x,y,z) \in S : y \leq j\}$. -$Z_j = \{(x,y,z) \in S : z \geq 3k-j\}$. - -Observe that $|X_j|, |Y_j|, |Z_j| \leq j$. -Let $W$ be the multiset union of $X_1, \ldots, X_{2k-1}, Y_1, \ldots, Y_{2k-1}, Z_1, \ldots, Z_{2k-1}$. I claim that this contains at least $3k$ copies of every element of $T$. Indeed, if $x, y \leq 2k$ and $z \geq k$, then $(x,y,z)$ is in $X_x, X_{x+1}, \ldots, X_{2k-1}, Y_y, Y_{y+1}, \ldots, Y_{2k-1}, Z_{3k-z}, Z_{3k-z-1}, \ldots, Z_{2k-1}$, which is a total of $(2k-x) + (2k-y) + (z-k) = 3k$ sets. -The other three cases ($x > 2k$, $y > 2k$, $z < k$) are mutually exclusive. If $x > 2k$, then $(x,y,z)$ is in $Y_y, Y_{y+1}, \ldots, Y_{2k-1}, Z_{3k-z}, Z_{3k-z-1}, \ldots, Z_{2k-1}$, which is a total of $(2k-y) + (z-k) = k+x > 3k$ sets. Similarly, if $y > 2k$, then $(x,y,z)$ is in $k + y > 3k$ sets. Finally, if $z < k$, then $(x,y,z)$ is in $X_x, X_{x+1}, \ldots, X_{2k-1}, Y_y, Y_{y+1}, \ldots, Y_{2k-1}$, which is a total of $(2k-x) + (2k-y) = 4k-z > 3k$ sets. -It follows that -$$ -3k|S| \leq \sum_{j=1}^{2k-1} (|X_j| + |Y_j| + |Z_j|) \leq 3 \sum_{j=1}^{2k-1} j = \frac {3(2k)(2k-1)} 2, -$$ -and therefore $|S| \leq 2k-1$, as claimed. - -We can also read structural information about the sets $S$ achieving this upper bound out of the proof. If $|S| = 2k-1$ then the inequalities in the preceding display must be equality; for example, none of the $(x,y,z) \in T$ that appear in more than $3k$ of the $X_j, Y_j, Z_j$ can be in $S$.<|endoftext|> -TITLE: What does the flow of the principal symbol of the differential operator tell us about the PDE? -QUESTION [8 upvotes]: Disclaimer: Let me apologize in advance for asking this slightly vague question -Let $M$ be a manifold and let $P$ be a partial differential operator acting on $C^{\infty}(M)$. Associated to $P$ there's the obvious linear PDE: -$$P(f)=0$$ -Naturally associated with $P$ we also have a hamiltonian system $$\Phi= (T^*M, \omega,H=\sigma(P))$$ where the symplectic form $\omega$ is the standard one and principal symbol $\sigma(P)$ of $P$ is taken as the hamiltonian. - -Question: What does the dynamics of $\Phi$ tell us about the original differential equation? - -Obviously since we are only considering the principal symbol we won't get a terrible amount of information. On the other hand we are not taking only the zero locus $\{ \sigma(P)=0\}$ (AKA the characteristic variety / set ) so one might hope that we can at least find some information in $\Phi$ which is not already present in the geometry of $\{\sigma(P)=0\}$. - -REPLY [5 votes]: It goes something like this (I can't promise that what I've written below is completely correct. It is only to help you read the rigorous details in more definitive reference): -The initial observation is that the smoothness of a distribution $f$ on $\mathbb{R}^n$ can be measured using the decay rate of its Fourier transform $\hat{f}(\xi)$ as a function of $|\xi|$. Microlocal analysis means to localize this idea in the cotangent bundle. -The wavefront set $WF(u) \subset T^*M$ of a distribution $u$ on $M$ has the property that -$$ WF_x(U) = WF(u)\cap T^*_xM $$ -is a closed conic subset of $T^*_xM$. -First, it suffices to restrict to a compactly supported distribution $u$ on a coordinate chart $O \subset M$. We can therefore assume everything is on $\mathbb{R}^n$. Then $WF_x(u)$ at $x \in \mathbb{R}^n$ is a closed cone in $\mathbb{R}^n\backslash\{0\}$, which should be viewed as $T^*_x\mathbb{R}^n\backslash\{0\}$. In particular, $\xi_0 \notin WF_x(u)$ if and only if for any neighborhood $N$ of $x$ there exists $\chi \in C^\infty_0(N)$ such that the Fourier transform $\widehat{\chi u}$ decays rapidly (faster than polynomial decay) in a conical neighborhood of $\xi_0$. -Let $\sigma$ be the symbol of a real linear differential operator $P$ and $\Sigma = \sigma^{-1}(0)$ its characteristic variety. We say that $P$ is of real principal type if the Hamiltonian vector field of $\sigma$ restricted to $\Sigma$ is everywherer nonzero. -Hormander's original propagation of singularities theorem (which has many antecedents and by now many generalizations) says that if $P$ is a real differential operator of real principal type and $u$ is a distribution on $M$ such that $Pu \in C^\infty(M)$, then $WF(u) \subset \Sigma$ and is invariant under the null bicharacteristic flow, which is the Hamiltonian flow of $\sigma$ restricted to $\Sigma$.<|endoftext|> -TITLE: Homotopy invariant structure: Stasheff versus Segal -QUESTION [5 upvotes]: To describe homotopy invariant algebraic structures on spaces, there are different approaches. - -The Stasheff / Boardman–Vogt / May approach, where operations and equations are replaced by spaces of operations, witnessing higher homotopy coherence. -The Segal approach, where the structure maps that would be isomorphisms (in the strict version) are merely required to be weak equivalences. - -In the introduction to [1], Badzioch writes: "Also, $A_{\infty}$-spaces can be viewed as homotopy algebras over the algebraic theory $T_{\mathrm{Mon}}$ such that the corresponding strict algebras describe monoids." He then points out that his main result specialized to the case of monoids yields an equivalence of homotopy theories $$h\mathrm{Alg}_{\mathrm{Mon}} \simeq \mathrm{Alg}_{\mathrm{Mon}}$$ between homotopy monoids à la Segal and (strict) topological monoids. Combined with the equivalence $$A_{\infty}-\mathrm{Spaces} \simeq h\mathrm{Alg}_{\mathrm{Mon}}$$ which he alluded to, this provides an alternate proof of the known equivalence $A_{\infty}-\mathrm{Spaces} \simeq \mathrm{Alg}_{\mathrm{Mon}}$ explained beautifully here. - -Question. Is there a good reference explaining the equivalence - $$A_{\infty}-\mathrm{Spaces} \simeq h\mathrm{Alg}_{\mathrm{Mon}}$$ - between $A_{\infty}$-spaces and homotopy monoids in spaces à la Segal? (Without going through strict topological monoids.) - -For this example, I can imagine an equivalence constructed more or less by hand. A functor in the Segal-to-Stasheff direction would involve some choices. -References to more general results of the form "homotopy algebras à la Stasheff $\simeq$ homotopy algebras à la Segal" would be welcome. -Here are some references I've looked at. - -Other papers by Badzioch as well as papers by Bergner [2][3] contain related material, but I couldn't find the answer to the specific question above. -I believe that Balzin has worked on the distinction between the Stasheff approach and the Segal approach, more specifically some advantages of the Segal approach in some situations. -These notes by Leinster look at the problem from a (higher) categorical perspective. Section 3.4 suggests that the answer is not obvious. - -[1] Badzioch, Bernard, Algebraic theories in homotopy theory, Ann. Math. (2) 155, No.3, 895-913 (2002). ZBL1028.18001. -[2] Bergner, Julia E., Simplicial monoids and Segal categories., Davydov, Alexei (ed.) et al., Categories in algebra, geometry and mathematical physics. Conference and workshop in honor of Ross Street’s 60th birthday, Sydney and Canberra, Australia, July 11--16/July 18--21, 2005. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-3970-6/pbk). Contemporary Mathematics 431, 59-83 (2007). ZBL1134.18006. -[3] Bergner, Julia E., Rigidification of algebras over multi-sorted theories, Algebr. Geom. Topol. 6, 1925-1955 (2006). ZBL1125.18003. - -REPLY [7 votes]: As requested, my comments in the form of an answer: -At the end of Categories and Cohomology Theories Segal gives a fairly detailed sketch of how to compare these theories in the harder, $\mathbb{E}_{\infty}$ case. The same sketch works in the $\mathbb{A}_{\infty}$-case. Later, May-Thomason elaborated on Segal's remarks and gave an axiomatic treatment of infinite loop space machines. -Thomason, later still, then wrote down a comprehensive treatment of the single loop space results, which should in particular provide a published reference for what you have in mind. -That said, the functor going in one direction is easy to write down: associated to an $\mathbb{A}_{\infty}$-space $X$ one has the simplicial space $[n] \mapsto \mathbb{A}_{\infty}(n) \times X^{\times n}$, i.e. the bar construction. This is evidently a $\Delta$-space, or homotopy monoid, in the sense of Segal. From here one could construct an inverse a la Segal, May-Thomason, and Thomason, or else proceed in any number of ways...<|endoftext|> -TITLE: Generators for commutator subgroup of surface group -QUESTION [7 upvotes]: Let $\pi_1(\Sigma_g) = \langle\text{$x_1,\ldots,x_{2g}$ $|$ $[x_1,x_2]\cdots[x_{2g-1},x_{2g}]$}\rangle$ be a surface group. Can anyone tell me an explicit free basis for the commutator subgroup of $\pi_1(\Sigma_g)$? I would prefer one consisting of conjugates of the elementary commutators $[x_i,x_j]$. - -REPLY [7 votes]: Before I answer this old question, a confession: the user "Linda" is actually me! I asked this basically to make sure that something I proved was not already known. This is such a classical subject that you can never be sure! -Anyway, I finally got around to writing up a paper that among other things answers this question. It can be downloaded here. Theorem B of it shows that the commutator subgroup of a surface group is freely generated by the set -$$\{\text{$[x_i,x_j]^{x_i^{k_i} \cdots x_{2g}^{k_{2g}}}$ $|$ $1 \leq i -TITLE: Inequality for the inner product in the probabilistic simplex -QUESTION [8 upvotes]: Let $x, y \in \mathbb{R}^{n}_{\geq 0}$ satisfy $\sum_i x_i = \sum_i y_i = 1$ and let $c \geq 1$. I am trying to find out whether the following inequality holds: -\begin{equation} -\left(\sum_i x_i y_i\right)^{2c-1} \leq \sum_i(x_i y_i)^c -\end{equation} -It seems like there should be a simple proof using standard inequalities, but it eludes me. I need the inequality to hold as part of a larger proof dealing with isoperimetric properties of the boolean hypercube. A proof or a counterexample would be much appreciated! - -REPLY [7 votes]: Let me explain how this may be solved without tricks, this may help for other similar issues. -1) The inequality must hold true for $\sum x_i\leqslant 1$, $\sum y_i\leqslant 1$ (instead of equalities). Indeed, if we replace $x_i$ to $tx_i$, $t>1$, we get a stronger inequality, and this reduces to the case $\sum x_i=1$, analogously to $\sum y_i=1$. -2) Denote $z_i=x_iy_i$. What can be said about these numbers if we know only that $\sum x_i\leqslant 1,\sum y_i\leqslant 1$? For sure, we know that $\sum \sqrt{z_i}=\sum\sqrt{x_iy_i}\leqslant \sum \frac{x_i+y_i}2=1$, but actually nothing more: if $\sum \sqrt{z_i}\leqslant 1$, then it may occur $x_i=y_i=\sqrt{z_i}$. Note that even when we do not see an inequality for $z$'s immediately, we could come to it by minimizing $\sum y_i=\sum z_i/x_i$ subject to $\sum x_i\leqslant 1$. -3) Thus the problem restates as $\sum \sqrt{z_i}\leqslant 1$, we need to prove $(\sum z_i)^{2c-1}\leqslant \sum z_i^c$. Now make the inequality homogeneous, that is, write a (stronger) homogeneous in $z$'s inequality $$\left(\sum z_i\right)^{2c-1}\leqslant \left(\sum z_i^c\right)\left(\sum \sqrt{z_i}\right)^{2c-2}.$$ -Ok, after this standard preparation we get a pure Hölder inequality -$$\left(\sum a_i\right)^\alpha\left(\sum b_i\right)^\beta\geqslant -\left(\sum a_i^{\frac{\alpha}{\alpha+\beta}} b_i^{\frac{\beta}{\alpha+\beta}}\right)^{\alpha+\beta} -$$ -for $\alpha=1,\beta=2c-2$, $a_i=z_i^c,b_i=\sqrt{z_i}$.<|endoftext|> -TITLE: What is "topology in dimension 3.5"? -QUESTION [7 upvotes]: I've noticed a couple of conference titles which reference something called -"topology in dimension 3.5," such as this one and this one. This subject seems quite mysterious to me — it looks like it's part of geometric topology, but I'm guessing it's not about 3.5-dimensional manifolds, so what kinds of things are studied in 3.5-dimensional topology? - -REPLY [4 votes]: cw answer: As mentioned in the comments, this just refers to the relations between 3-dimensional and 4-dimensional topology.<|endoftext|> -TITLE: Arithmetic representation stability and Galois action -QUESTION [10 upvotes]: I am thinking about representation stability phenomena (as considered by Church, Farb, Ellenberg and others) in arithmetic settings where Galois action enters the picture, and am curious about their relevance to learning about Galois action. -To describe a concrete problem which mimics one considered by them in topological setting, start with an arithmetic scheme X over a number field K. Consider the sequence of its configuration spaces $C_n(X) = \{(p_1,..., p_n) : p_i'\mbox{s are distinct geometric points of }X\}$. -Symmetric group $S_n$ acts on $C_n(X)$ in the obvious way, as does the absolute Galois group of $K$, $G_K$, and the two actions commute. The actions are inherited by (etale) cohomology groups $H^i(C_n(X))$. Thus, we are in the Church-Farb setting of configuration spaces of manifolds except that the FI-module formed by consistent $S_n$-representations $H^i(C_n(X))$ is here also a Galois module. -My question is: in cases of X and K where the $S_n$-sequence is representation stable in their sense, what if anything can we learn about the Galois action on X from the fact of stability? What if we take $K$ to be a $p$-adic field instead of a number field? - -REPLY [4 votes]: I strongly doubt that one can say anything in particular about the Galois action on $X$ from the fact that representation stability holds for the configuration spaces of points on $X$. -Church's original proof of representation stability for configuration spaces of points on oriented manifolds used very little "manifold-machinery". In fact the only place being a manifold was used was to write down the Cohen-Taylor spectral sequence computing the cohomology of the configuration space. As noted by Totaro, the Cohen-Taylor spectral sequence is the Leray spectral sequence for the inclusion of the configuration space into X^n. From this one can rather easily deduce that the Cohen-Taylor spectral sequence exists in the world of algebraic geometry and has the same formal properties, and you can run Church's proof with no modications. So you know e.g. that if X is any smooth connected variety over a field $k$, then $H^\ast_{\textrm{\'et}}(C_n(X) \otimes \overline k,\mathbb Z_\ell)$ is a representation stable sequence of Galois modules.<|endoftext|> -TITLE: A log structure on the moduli space of curves -QUESTION [7 upvotes]: Let $M_{g, n}$ be the moduli space of curves of genus $g$ with $n$ marked points. Let $M_{g, \vec{n}}$ be the moduli space of marked curves with a choice of a (possibly zero) tangent vector at each marked point. There is a map $i:M_{g, n}\to M_{g, \vec{n}}$ (the zero section), and there is a log structure on $M_{g, \vec{n}}$ associated to the normal crossings divisor which is the complement to the locus where all specified tangent vectors are nonzero. Let $M_{g, n}^{log}$ be the pullback of this log structure on $M_{g, \vec{n}}$ to $M_{g, n}$ along $i.$ -Question: what is $M_{g, n}^{log}$ called? Who has studied it? At one point I heard the term "log moduli space of curves with log structure" floating around, though I've never seen it explicitly defined. Is my $M_{g, n}^{log}$ the (restriction from $\bar{M}_{g,n}$ to $M_{g,n}$) of this log moduli space? - -REPLY [5 votes]: As Piotr Achinger suggested in a comment, your log moduli space is the direct product of $M_{g,n}$ with the log point $\operatorname{Spec}(\mathbb{N}^n \to \mathbb{C})$ given by the monoid map $(x_1,\ldots,x_n) \mapsto 0$. The main reason is that the monoid $\mathbb{N}^n$ only has automorphisms given by reordering basis elements, and in this case, the basis elements are attached to marked points that have a specified order. That is, the local system is trivial for rather elementary reasons.<|endoftext|> -TITLE: wrong formula for Atkin--Lehner operators? -QUESTION [20 upvotes]: I am really embarrassed to ask this question since I am supposed to be an expert on modular forms, but here goes anyway since I have been stuck on this -for several days. -Let $q$ be a prime divisor of $N$ such that $q^2\nmid N$, and let $\chi$ be some Dirichlet character modulo $N$. If $F$ is a normalized eigenform (newform of course) in -$S_k(\Gamma_0(N),\chi)$ with Fourier coefficients $a(n)$, it is known from -Atkin--Lehner--Li that if $W_q$ denotes an Atkin--Lehner involution then -$F$ is an eigenform for the action of $W_q$ with eigenvalue $-a(q)/q^{k/2-1}$, and also that $a(q)^2=\chi(q)q^{k/2-1}$. -I am implementing this in Pari/GP and found a contradiction. Most probably -my implementation is wrong, but please follow the following simple example -(I know that David Loeffler had a ticket about a similar problem in Sage, but that was solved). -Consider the space $S_3(\Gamma_0(12),\chi_{-4})$, where as usual $\chi_{-4}$ -is the odd character mod 4. It has a single eigenfunction defined over -$\mathbb Q(\sqrt{-3})$, together with its conjugate. -I can choose $W_3=[3,-1;12,-3]$, and $\tau=i=\sqrt{-1}$, so that -$W_3(\tau)=(3i-1)/(12i-3)$. I then compute $1000$ Fourier coefficients -of $F$, and the value $F(i)$ and $F|_3W_3(i)$. Instead of finding -$-a(3)/3^{1/2}$ as ratio, I find $+a(3)/3^{1/2}$ (equal to $i$ and $-i$, but in the wrong order or equivalently with the wrong sign). -I did many other experiments, both in odd and even weights, and with nonreal characters, and my -implementation tells me that the eigenvalue should be -$-a(q)\overline{\chi(q)}/q^{k/2-1}$. -Now I went through Winnie Li's proof which I essentially reproduced in my book with F. Stromberg, and it seems correct. So I am stuck. Can someone at least check the above example, or tell me what I am doing wrong ? Note that -I am not mixing the two possible embeddings of my form: both give me the wrong sign. -ADDED March 27: I am very surprised not to have any (positive or negative) -answer. Since the error is systematic, let me give another example in -even weight with numerical values. Consider $S_2(\Gamma_0(35),\chi_5)$. -It is of dimension $2$ generated by an eigenform and its conjugate. One -specific embedding of the eigenform is (sorry for the GP notation) -$$F(\tau)=q + 2*I*q^2 - I*q^3 - 2*q^4 + (-2 - I)*q^5 + 2*q^6 - I*q^7 + 2*q^9 + O(q^{10}).$$ -I take a random point, say $\tau=I/2+1/Pi$. Simply by summing sufficiently -many terms I find $F(\tau)=-0.015178...+0.03676...*I$. -I choose as Atkin--Lehner matrix for the prime $q=7$ the matrix -$W_7=[7,-3;35,-14]$ which has the required properties. I compute -$F|_2W_7(\tau)$ (here I use 2000 terms since the imaginary part is small) -and I find that it is equal to $0.03676+0.015178...*I$, so the eigenvalue -is $-I$. However, Atkin--Lehner-Li tells me it should be equal to $-a(7)$, -which as one sees from above is equal to $+I$, contradiction. -Can someone please check this in Sage, Magma, or for that matter Pari/GP ? - -REPLY [7 votes]: Let's examine the case $N=12$, weight $3$, with central character $\chi_{-4}$. According to the LMFDB, there are two newforms $f$ in this space. I will take it as numerically-verified that for both of these forms, $f|W_3$ is proportional to $f$. -Theorem 2.1 of Atkin-Li (Twists of newforms and pseudo-eigenvalues of W-operators. -Invent. Math. 48 (1978), no. 3, 221–243) then states -$f|W_3 = \lambda f,$ -where $$\lambda = -\frac{|a_f(3)|}{a_f(3)}.$$ -Here the minus sign comes from their convention that the Gauss sum of a trivial character is $-1$, under the conditions at hand. Here I multiplied by $|a_f(3)|$ because the pseudo-eigenvalue has absolute value $1$. Indeed, the earlier Proposition 1.1 from that paper shows $\lambda^2 = -1$, so the only question is the sign of $\lambda$. Now the above formula for $\lambda$ becomes $\lambda = - \frac{\overline{a_f(3)}}{|a_f(3)|} = \frac{a_f(3)}{|a_f(3)|}$, since $\lambda = \pm i$. Isn't this the numerical value that is claimed in the original question? -added later: I consulted Theorem 3 of Li's 1975 Math. Annalen paper (Newforms and functional equations). I will continue the discussion of this example. Li defines a version of the Atkin-Lehner operator, denoted $V_3^M$ (I will drop the superscript in what follows). However, I think that her choice of $V_3$ is inconsistent with $W_3$. According to Li's definition, we may choose $V_3 = \begin{pmatrix} -3 & -1 \\ 12 & 3 \end{pmatrix}$. She requires the lower-right entry to be $3$. With $W_3$ given as in the original question, we have $$V_3 W_3^{-1} = \begin{pmatrix} 7 & -2 \\ -24 & 7 \end{pmatrix}.$$ -Since $\chi_{-4}(7) = -1$, we have $$f|V_3 = - f |W_3.$$ -It should be mentioned that the definition of the Atkin-Lehner operator in the Atkin-Li paper has slightly different congruence conditions. In particular, I checked that the choice of $W_3$ in the original question is consistent with the definition of the Atkin-Lehner definition. -Unless I made some mistakes, I believe the above discussion clears up the problem, at least in this level $12$ example.<|endoftext|> -TITLE: Does Beilinson's conjecture on values L-functions work for smooth projective varieties over a number field -QUESTION [5 upvotes]: In Nekovar's introductory paper "Beilinson's Conjecture" -http://math.stanford.edu/~conrad/BSDseminar/refs/BeilinsonintroII.pdf -The conjecture is formulated for smooth projective varieties over $\mathbb{Q}$. However all the statements and proofs seem to equally work even for smooth projective varieties defined over an arbitrary number field $L$. Is this true? Are there any differences between the case over $\mathbb{Q}$ and number fields $L$? - -REPLY [3 votes]: In addition to François's answer, I'll address the second question. - -Are there any differences between the case over $\mathbb Q$ and a number fields $L$? - -The main difference - which can be dealt with but should not be forgotten nevertheless - is that there are now many embeddings of the field of definitions into $\mathbb C$, and that some but not necessarily all of them might be real embeddings. -Since Beilinson's conjecture makes crucial use of the Hodge structure on the singular cohomology $H^i((X\times_{L,\sigma}\mathbb C)(\mathbb C),\mathbb Q)$ of the variety $X$ (with $\sigma:L\hookrightarrow\mathbb C$ an embedding) and more precisely of the Hodge structure on $H^i((X\times_{L,\sigma}\mathbb C)(\mathbb C),\mathbb Q)\otimes_{\mathbb Q}\mathbb C$ (which is always an $\mathbb R$-Hodge structure, but not always an $\mathbb R$-Hodge structure over $\mathbb R$) one must be careful when generalizing from $\mathbb Q$ to $L$: should one fix an embedding? consider them all at once? does the conjecture depend on this choice? what if some embeddings are real and some complex? etc. -Incidentally, the same difficulty arises for the Bloch-Kato conjectures predicting the exact value of special values of $L$-function with respect to the $p$-adic étale realization (in that case, there might be many primes above $p$ in $L$ and the $D_{\operatorname{dR}}$-module appearing in the conjecture might depend on the choice of the prime).<|endoftext|> -TITLE: Misunderstanding of Hodge conjecture -QUESTION [5 upvotes]: I ask a question on math stack exchange about Hodge conjecture and have not got any reply or comment. -https://math.stackexchange.com/questions/2704988/clarify-hodge-conjecture -so I decide to post this question on MO. Let's focus on smooth projective varieties defined over $\mathbb{Q}$. The Hodge conjecture could also be formulated as the Hodge realisation functor $\text{R}$ being full-faithful, -\begin{equation} -\text{R}:\textbf{M}_{\text{num}}(\mathbb{Q},\mathbb{Q}) \rightarrow \textbf{HS}_{\mathbb{Q}} -\end{equation} -where $\textbf{M}_{\text{num}}(\mathbb{Q},\mathbb{Q}) $ is the category of motives defined over numerical equivalence and $\textbf{HS}_{\mathbb{Q}}$ is the abelian category of pure Hodge structures over $\mathbb{Q}$. Its equivalence to the usual statement is explained carefully in the post -https://math.stackexchange.com/questions/2249701/why-is-the-hodge-conjecture-equivalent-to-the-assertion-that-mathcalr-ma -However a naive question has puzzled me. Suppose $\chi:(\mathbb{Z}/N\mathbb{Z})^\times \rightarrow \mathbb{Q}$ is a real non-trivial Dirichlet character, i.e. it takes values in $\{ -1,1\}$, then by decomposing the Artin motive $h^0(\text{Spec}\,\mathbb{Q}(\zeta_N))$ there is a pure motive $\chi$ which is a direct summand of $h^0(\text{Spec}\,\mathbb{Q}(\zeta_N))$. Its construction could also be found in section 1.1 of the paper -https://arxiv.org/pdf/math/0101071.pdf -The Hodge realistion of the Artin motive $h^0(\text{Spec}\,\mathbb{Q}(\zeta_N))$ is just $\mathbb{Q}(0)^m$, where $m=[\mathbb{Q}(\zeta_N):\mathbb{Q}]$, so as a direct summand, the Hodge realisation of $\chi$ should be $\mathbb{Q}(0)$ as its is one dimensional. If the functor $\text{R}$ is full-faithful, this would imply the motive $\chi$ is isomorphic to the Tate motive $\mathbb{Q}(0)$. -However for a prime $\ell \neq N$, the $\ell$-adic representation of $\chi$ is just the Galois representation associated with the Dirichlet character $\chi$, which is non-trivial. So the motive $\chi$ cannot be isomorphic to $\mathbb{Q}(0)$. -There must be something that I haven't understood correctly, could some one point out the naive mistakes that I have made? - -REPLY [10 votes]: The Hodge conjecture implies that the functor -$R\colon M_{num}(k,\mathbb{Q})→HS(\mathbb{Q})$ is fully faithful when $k$ is the algebraic closure of $\mathbb{Q}$, not $\mathbb{Q}$ itself, as your example illustrates.<|endoftext|> -TITLE: Some binomial coefficient determinants -QUESTION [27 upvotes]: It is well known that for $n>0$ -$$d(n)=\det\left(\binom{2i+2j+1}{i+j}\right)_{i,j=0}^{n-1}=1.$$ -Computer experiments suggest that more generally -$$d(n,k)=\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,j=0}^{(2k+1)n-1}=(2n+1)^{k}.$$ -Has anyone an idea how to prove this for general $k$? - -REPLY [5 votes]: Johann Cigler and I have posted a solution on arXiv: -"An interesting class of Hankel determinants", arXiv:1807.08330. -Let $d_r(N)=\det\left({2i+2j+r\choose i+j}\right)_{i,j=0}^{N-1}$. We show that for $k,n\ge 1$, -\begin{align} -&d_{2k+1}((2k+1)n)=d_{2k+1}((2k+1)n+1)=(2n+1)^k,\\ -&d_{2k+1}((2k+1)n+k+1)=(-1)^{k+1\choose 2}4^k(n+1)^k,\\ -&d_{2k}(2kn)=d_{2k}(2kn+1)=(-1)^{kn},\\ -&d_{2k}(2kn+k)=-d_{2k}(2kn+k+1)=(-1)^{kn+{k\choose 2}}4^{k-1}(n+1)^{k-1}. -\end{align}<|endoftext|> -TITLE: Continuum Hypothesis and the fact that every co-finite topological space, with uncountable underlying set , is contractible -QUESTION [14 upvotes]: Let $X$ be a co-finite topological space. If $|X| \ge 2^{\aleph_0}=\mathfrak c$, then $X$ is contractible (https://en.wikipedia.org/wiki/Contractible_space) . Indeed, there is a bijection $f: X \times (0,1) \to X$; fix a point $a \in X$; define $H: X \times [0,1] \to X$ as $H(x,0)=x, \forall x \in X; H(x,1)=a,\forall x \in X ; H(x,t)=f(x,t), \forall (x,t)\in X \times (0,1)$. Then $H$ is continuous, hence a homotopy between $Id_X$ and the constant map on $X$ which takes everything to $a$. This shows that, under Continuum Hypothesis, any co-finite topological space $X$ with uncountable $X$, is contractible. -My question is : If every co-finite topological space with uncountable underlying set is contractible , then is it true that Continuum Hypothesis holds ? Or atleast, can we say that the fact "every co-finite topological space with uncountable underlying set is contractible" cannot be proved without Continuum Hypothesis ? - -REPLY [12 votes]: Nice question! -I claim that this property does not necessarily imply CH. As Todd -guessed in his comment, the answer is related to certain cardinal -characteristics of the continuum. -Specifically, let us define the closed-partition number to be the size $\kappa$ of the smallest nontrivial partition of the unit interval $[0,1]$ into closed sets. That is, $\kappa$ is the smallest size of a set $I$, such that the unit interval admits a partition $$[0,1]=\bigsqcup_{i\in I}C_i$$ into at least two pairwise disjoint nonempty closed sets $C_i$. -It is a -standard exercise to show that $\kappa$ is uncountable (see this -nice explanation of Timothy -Gowers); in other words, the unit interval is not a nontrivial union of countably many disjoint nonempty closed sets. A somewhat more -refined observation is that $\text{cov}(\mathcal{M})\leq\kappa$, as -explained in this MO answer of Andreas -Blass. A further refined -observation is that $\frak{d}\leq\kappa$, made by Taras Banakh in a -comment on Andreas's answer. I'm not sure if this $\kappa$ already -has a name or if it is provably equal to one of the well-known -cardinal invariants. -Meanwhile, Arnie Miller proved in - -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. - -that it is consistent with ZFC that $\kappa=\omega_1$, even while -CH fails. So the unit interval can be the disjoint union of -$\omega_1$ many nonempty closed sets, even when CH fails. This -situation will be the key to answering your question. -Let's begin with the following observation. -Observation. If $X$ is a $T_1$ path-connected space with at -least two points, then $X$ has size at least the closed-partition number $\kappa$. -Proof. If $f:[0,1]\to X$ is a path between two distinct points, -then the sets $C_x=\{t\mid f(t)=x\}$ are disjoint closed sets, -whose union is $[0,1]$. So $X$ must have size at least $\kappa$. -$\Box$. -One can now characterize exactly which cofinite spaces are contractible. -Theorem. Suppose that $X$ is a cofinite space with at least two -points. Then the following are equivalent. - -$X$ is contractible. -$X$ is path connected. -$X$ has size at least $\kappa$, the closed-partition number. - -Proof. Clearly every contractible space is path connected. And -we proved in the observation that every path-connected $T_1$ space (and the cofinite -topology is $T_1$) has size at least $\kappa$. -What remains is to prove that every cofinite space $X$ of size at -least $\kappa$ is contractible. Fix a closed particition -$[0,1]=\sqcup_{i\in I} C_i$, where $I$ has size $\kappa$. Also fix -distinct points $x_i\in X$ and another distinct point $a\in X$. -Define a map $H:X\times[0,1]\to X$ as follows. This is an analogue -to the contraction defined in the OP under CH. Namely, let -$H(x,0)=x$ and $H(x,1)=a$; and for other values of $t$, let -$H(x,t)=x_i$, where $t\in C_i$. -I claim that this map is continuous and hence is a contraction of -the space. To see that it is continuous, consider any open set in -$X$, which is the complement of a finite set in $X$. The preimage -of any point $x\in X$ not of the form $x_i$ or $a$ is just the -point $(x,0)$, which is closed. The pre-image of $a$ is -$X\times\{1\}\cup\{(a,0)\}$, which also is closed. And the -pre-image of $x_i$ is $X\times C_i$, plus the point $(x_i,0)$, and -this also is a closed set. So the preimage of any closed set in $X$ -is a finite union of closed sets and hence is closed. And so the -map $H$ is continuous, and therefore $X$ is contractible, as -desired. $\Box$ -Corollary. If ZFC is consistent, then it is consistent with ZFC -that every uncountable cofinite space is contractible, yet CH -fails. -Proof. Miller provides a model where $\kappa=\omega_1$, yet CH -fails. In this model, every uncountable cofinite space has size at -least $\kappa$, and so all of them are contractible, yet CH fails. -$\Box$.<|endoftext|> -TITLE: Algebraically closed field of cardinality greater than $\mathfrak c$ -QUESTION [10 upvotes]: It is a 20th century result that there exists only one algebraically closed field for a given characteristic $p$ and cardinality $\kappa>\aleph_0$, up to isomorphism. Is there a better way to imagine such fields, other than adjoining $\kappa$ transcendental elements to $\mathbb Q$ or $\mathbb F_p$ and taking the algebraic closure? -In characteristic 0, do these fields enjoy some nice topological properties? Can you maybe do analysis on them? -I have not found much about fields of cardinality greater than $\mathfrak c$, not even about the "next simplest case" $2^\mathfrak c$. Do you know of an references? - -REPLY [2 votes]: In the caracteristic zero case, I suggest one way to look at this is to choose a prescribed real closed subfield $R_{\kappa}$ of said algebraically closed field $F_{\kappa}$ of cardinal $\kappa$ that one finds particularly interesting, so that geometric intuition (for instance basically that of a plane) or "analytic" intuition may apply to $F_{\kappa}$. -Of course each $R_{\kappa}$ appears "as an isomorphic copy" in each algebraically closed field of cardinal $\kappa$, but not all of them are isomorphic, so this is a means to distinguish ordered pairs $(F_{\kappa},R_{\kappa})$ between them. -Each choice naturally transfers some structure from $R_{\kappa}$ to $F_{\kappa}=R_{\kappa}\oplus R_{\kappa}.\sqrt{-1}$, be it an absolute value $F_{\kappa}\rightarrow R_{\kappa}^{\geq 0}$ (and thus a nice sequential uniform structure), a valuation extending that of $R_{\kappa}$, a Hahn series structure, a transfer principle, order related saturation properties, an exponential map... -That said I don't think there is a distinguished real closed subfield of $F_{\kappa}$ in a field-theoretic standpoint in general.<|endoftext|> -TITLE: baricentric distributions -QUESTION [5 upvotes]: Let $\mathcal{P}$ be the convex hull of a point set $p_1, \dotsc, p_n$ (for simplicity, assume that no $p \in P$ lies in the convex hull of the other points.) -Now, pick a point uniformly at random from the simplex $x_1 + \dotsc + x_n = 1, x_i \geq 0,$ and take $\phi=\sum x_i p_i.$ The point $\phi$ will be contained in $\mathcal{P}$ and in the simple case where $\mathcal{P}$ is a triangle, $\phi$ is uniformly distributed. In general, however, it is not: The two graphics are for the case $\mathcal{P}$ is a square. The first is a simple plot of a million points sampled from the distribution - the distribution looks like it is trying to be supported on the quadrangle spanned by the midpoints of the sides of the square. - -The second is the density histogram: - -which seems to indicate that the distribution has a bit of a peak at the center of gravity. -Any deep thoughts? - -REPLY [2 votes]: The problem seems to lie in the way you sample the unit simplex. -Let $(U_i)_{1\leq i\leq n}$ be a sequence of i.i.d. uniform random variables, $S=\sum U_i$ and define $X_i=U_i/S.$ Then non-intuitively, $(X_1,\dots,X_n)$ is not uniformly distributed over the unit simplex. -The following plot shows 300000 points on the unit square with barycentric coordinates generated using this method: -First method -The method can easily be fixed: if the $U_i$s defined above are i.i.d. exponential random variables, then $(X_1,\dots,X_n)$ is uniformly distributed over the simplex. Again, the following plot shows 300000 points generated according to this second method: -Second method -It should be noted that uniformly distributed $X_i$s doesn't imply uniformly distributed $\phi,$ as said by Mateusz Kwaśnicki in a comment.<|endoftext|> -TITLE: The first unstable homotopy group of $Sp(n)$ -QUESTION [20 upvotes]: Thanks to the fibrations -\begin{align*} -SO(n) \to SO(n+1) &\to S^n\\ -SU(n) \to SU(n+1) &\to S^{2n+1}\\ -Sp(n) \to Sp(n+1) &\to S^{4n+3} -\end{align*} -we know that -\begin{align*} -\pi_i(SO(n)) \cong \pi_i(SO(n+1)) \cong \pi_i(SO), \quad i &\leq n-2\\ -\pi_i(SU(n)) \cong \pi_i(SU(n+1)) \cong \pi_i(SU), \quad i &\leq 2n - 1 = (2n+1) - 2\\ -\pi_i(Sp(n)) \cong \pi_i(Sp(n+1)) \cong \pi_i(Sp), \quad i &\leq 4n+1 = (4n + 3) - 2. -\end{align*} -These values of $i$ are known as the stable range. So the first unstable groups are $\pi_{n-1}(SO(n))$, $\pi_{2n}(SU(n))$, and $\pi_{4n+2}(Sp(n))$ respectively. -I was able to find $\pi_{n-1}(SO(n))$ for $1 \leq n \leq 16$ by combining the tables on the nLab page for the orthogonal group and appendix A, section 6, part VII of the Encyclopedic Dictionary of Mathematics. The groups are -$$0, \mathbb{Z}, 0, \mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z}_2, \mathbb{Z}, 0, \mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z}_2\oplus\mathbb{Z}_2, \mathbb{Z}\oplus\mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z}_2, \mathbb{Z}, \mathbb{Z}_2, \mathbb{Z}\oplus\mathbb{Z}.$$ -There doesn't seem to be any pattern here, so I guess that there is no general result for $\pi_{n-1}(SO(n))$. (Feel free to correct me if I'm wrong.) I just noticed that every second term contains a copy of $\mathbb{Z}$, while every fourth term contains two copies. -The case of $SU(n)$ is completely different: in The space of loops on a Lie group, Bott proved, among other things, that $\pi_{2n}(SU(n)) \cong \mathbb{Z}_{n!}$, see Theorem 5. -Again consulting the Encyclopedic Dictionary of Mathematics, I was able to find $\pi_{4n+2}(Sp(n))$ for $n = 1, 2, 3$. The groups are $\mathbb{Z}_{12}$, $\mathbb{Z}_{120}$, and $\mathbb{Z}_{10080}$. This seems to suggest that this case is more similar to $SU(n)$ than $SO(n)$, so one might hope there is a Bott-type result. - -Is there an analogue of Bott's result for $Sp(n)$? That is, is there some increasing function $f : \mathbb{N} \to \mathbb{N}$ such that $\pi_{4n+2}(Sp(n)) \cong \mathbb{Z}_{f(n)}$? - -OEIS has no sequences beginning $12, 120, 10080$, so I have no guess what $f(n)$ could be. It is interesting to note that $12 \mid 120$ and $120 \mid 10080$ which is another similarity with the $SU(n)$ case. -Of course, three groups is not much to go on, so this may be a completely misguided guess. Some questions that would be nice to answer before seriously hoping for such a result are: - -Is $\pi_{4n+2}(Sp(n))$ always cyclic? -Is $\pi_{4n+2}(Sp(n))$ always finite? -Is $|\pi_{4n+2}(Sp(n))|$ increasing in $n$? - -Any information regarding these three questions would also be interesting to know. -Falling short of answering any of these questions, have any more of these groups (namely $\pi_{18}(Sp(4)), \pi_{22}(Sp(5)), \dots$) been computed? - -Update: I added the sequence $|\pi_{4n+2}(Sp(n))|$ to the OEIS: A301898. -Also, the answer to the question I asked was also in the Encyclopedic Dictionary of Mathematics on page 1746. - -REPLY [27 votes]: The answer appears to be in the paper Homotopy groups of symplectic groups by Mimura and Toda. They claim the calculation was already in a paper of Harris, but that was stated in terms of a symmetric space and it's not immediately obvious to me how to translate into information about the groups. -They state that the group is $\mathbb Z_{(2n+1)!}$ if $n$ is even and $\mathbb Z_{(2n+1)! \cdot 2}$ if $n$ is odd, which agrees with your data. - -REPLY [24 votes]: The first unstable homotopy groups of $SO(n)$ are actually 8-periodic (except for some junk at the beginning). Some more unstable homotopy groups of $SO(n)$ can be found in: - -M. Kervaire. Some nonstable homotopy groups of Lie groups. Illinois J. Math. 4 (1960), 161-169. (link to journal website) - -The 8-periodicity for the orthogonal group comes about as follows: the relevant piece of the stabilization sequence is -$$ -\pi_n S^n\to \pi_{n-1}SO(n)\to \pi_{n-1}SO(n+1)\to 0. -$$ -The unstable homotopy groups $\pi_{n-1}SO(n)$ are then direct sums of the stable stuff from $\pi_{n-1}SO(\infty)$ plus a cyclic quotient of $\pi_n S^n\cong \mathbb{Z}$. The 8-periodicity effectively comes from the stable summand (check the list of homotopy groups of the infinite orthogonal group). The cyclic quotient of $\pi_n S^n$ is only 2-periodic, alternating between $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$; I think this basically comes from the corresponding Euler class of the sphere alternating between 2 and 0. -The description of the unstable homotopy of the symplectic groups given in Will Sawin's answer can also be found in - -B. Harris. Some calculations of homotopy groups of symmetric spaces. Trans. Amer. Math. Soc. 106 (1963), 174-184. (link to journal website)<|endoftext|> -TITLE: Lower bound for the fractional part of $(4/3)^n$ -QUESTION [7 upvotes]: My son, who is 16, is doing some independent research. A lower bound depending on $n$ for $\left\{ \left( \frac{4}{3} \right)^n \right\}=\left( \frac{4}{3} \right)^n-\left\lfloor \left(\frac{4}{3} \right)^n \right\rfloor$, the fractional part of $\left( \frac{4}{3} \right)^n$, might help him improve his results. Note that $\frac{1}{3^n}$ is an obvious bound. Is there a better bound known? - -REPLY [9 votes]: It may be interesting to note that, subject to the ABC conjecture, you have the fantastically good estimate - $$ \left\{ \left( \frac43 \right)^n\right\} \gg_\delta \delta^n,\quad \delta\in(0,1). $$ -The proof goes as follows. -Let - $$ 4^n = 3^nk+r,\quad 00$ we have then -\begin{align*} - 2^{-d}4^n &\ll_\varepsilon \left( {\rm rad}(2^{-d}4^n\cdot3^nk_0\cdot r_0) \right)^{1+\varepsilon} \\ - &\le\ \ \ (6k_0r_0)^{1+\varepsilon} \\ - &\ll_\varepsilon (kr)^{1+\varepsilon}\cdot 2^{-d}. -\end{align*} -In view of $k<(4/3)^n$, this implies - $$ 4^n \ll_{\varepsilon} \left(\frac43\right)^{(1+\varepsilon)n} r^{1+\varepsilon} $$ -and, as a result, - $$ \left\{ \left( \frac43 \right)^n \right\} = \frac r{3^n} \gg_{\varepsilon} 4^{-\varepsilon n/(1+\varepsilon)}. $$ -The assertion follows by choosing $\varepsilon$ to satisfy $4^{-\varepsilon/(1+\varepsilon)}=\delta$.<|endoftext|> -TITLE: Quotable equivalents of Martin's axiom -QUESTION [8 upvotes]: I am seeking "quotable equivalents" for MA (Martin's axiom). For the continuum hypothesis, examples of such statements are as follows. -(a) (Sierpinski) The (xy) plane can be covered by countably many $x \mapsto y$ and $y \mapsto x$ functions. -(b) (Zoli) The set of transcendental reals is a union of countably many transcendence bases for $\mathbb{R}$. -(c) (Erdős) There is an uncountable family of analytic functions on $\mathbb{C}$ that takes only countably many values at each complex number. -(d) (Freiling) There is a function $F$ from $\mathbb{R}$ to the family of countable subsets of $\mathbb{R}$ such that for every $x, y \in \mathbb{R}$, either $x \in F(y)$ or $y \in F(x)$. -Since each one of (a)-(d) refers to a "countable/uncountable" dichotomy, it would be reasonable to have statements with a "continuum/smaller than continuum" dichotomy. - -REPLY [8 votes]: Martin's axiom is equivalent to the assertion that $$H_{\frak{c}}\prec_{\Sigma_1} V[G]$$ -for all c.c.c. forcing extensions $V[G]$. In other words, $H_{\frak{c}}$ is existentially closed in all c.c.c. forcing extensions. -This is proved in Bagaria, Joan, A characterization of Martin’s axiom in terms of absoluteness, J. Symb. Log. 62, No. 2, 366-372 (1997). ZBL0883.03039. The characterization also appears independently, attributed to J. Stavi, in Stavi, Jonathan; Väänänen, Jouko, Reflection principles for the continuum, Zhang, Yi (ed.), Logic and algebra. Providence, RI: American Mathematical Society (AMS). Contemp. Math. 302, 59-84 (2002). ZBL1013.03059. -This characterization leads naturally to the resurrection axioms, which I explored with Thomas Johnstone in several articles. The resurrection axiom for a class of forcing $\Gamma$ is the assertion that for every $\Gamma$ extension $V[g]$ there is further $\Gamma$ forcing $V[g][h]$ such that -$$H_{\frak{c}}\prec H_{\frak{c}}^{V[g][h]}.$$ -Thus, one attains full elementarity, at the cost of further forcing.<|endoftext|> -TITLE: Operator that commutes with projections -QUESTION [11 upvotes]: We investigate the Hilbert space $\ell^2(\mathbb{N}_0)$ with standard orthonormal basis vectors $e_n:=(0,...,0,1,0,...).$ -Consider the family of self-adjoint rank $1$ projections $P_n\bullet:= \langle \bullet,e_n \rangle e_n.$ -Take any $n\in\mathbb{N}_0$. My question is this: Does there exist a bounded linear operator $T$ on $\ell^2(\mathbb{N}_0)$ that commutes with all $P_m$ for $m \neq n$ but not with $P_n$? - -REPLY [17 votes]: Maybe a quicker way to see this is, if $TP_m = P_mT$ for all $m \neq n$ then $TP = PT$ where $P =\sum_{m\neq n} P_m = I - P_n$. Since $T$ commutes with $I$, it must therefore commute with $P_n$. - -REPLY [8 votes]: The answer is no. Indeed, without loss of generality (wlog), $n=0$, so that $TP_m=P_mT$ for all $m=1,2,\dots$ (assuming here that $\mathbb{N}_0=\{0,1,\dots\}$). Since $P_0=I-\sum_{m\ge1}P_m$ and $T I=IT$, it follows that $TP_0=P_0T$. (Here, of course, $I$ is the identity operator and the operator $S:=\sum_{m\ge1}P_m$ is defined by the condition that $Sx:=\sum_{m\ge1}P_m x$ for all $x\in\ell^2(\mathbb{N}_0)$.)<|endoftext|> -TITLE: Deformations of Calabi-Yau manifolds -QUESTION [14 upvotes]: Let $X$ be a compact complex smooth manifold with holomorphically trivial canonical class. - -It is true that any (sufficiently small?) deformation of the complex structure of $X$ also has holomorphically trivial canonical class? - -REPLY [21 votes]: The answer in general is no. Nakamura has constructed here (pp.90, 96-99, solvmanifolds of type III-(3b)) an example of a compact complex (non-Kähler) manifold $M$ with $TM$ holomorphically trivial (so in particular $K_M$ is holomorphically trivial) which has arbitrarily small deformations $M_t$ with negative Kodaira dimension. -On the other hand, if $M$ is compact Kähler with $K_M$ holomorphically trivial, then its sufficiently small deformations $M_t$ are Kähler (Kodaira-Spencer) and still have $K_{M_t}$ holomorphically trivial. Indeed, we have $c_1(K_{M_t})=0$ in $H^2(M_t,\mathbb{Z})$ (a topological condition), so by the Calabi-Yau theorem they admit Ricci-flat Kähler metrics $g_t$. On the other hand $\dim H^0(M_t,K_{M_t})=h^{n,0}(M_t)$ is locally constant hence equal to $1$, so you have a nontrivial holomorphic section $\Omega_t$ of $K_{M_t}$. A Bochner formula gives $\Delta_{g_t}|\Omega_t|^2_{g_t}=|\nabla \Omega_t|^2_{g_t},$ which can be integrated on $M_t$ to see that $\Omega_t$ is parallel with respect to $g_t$, hence it must be nowhere vanishing. This gives you a holomorphic trivialization of $K_{M_t}$. -Prompted by Piotr Achinger's comment below, let me also note the following generalization. If $M$ is compact complex with Hodge-de Rham (aka Frölicher) spectral sequence degenerating at $E_1$ (this is true for all compact Kähler manifolds), and with $K_M$ holomorphically trivial, then its sufficiently small deformations $M_t$ also have spectral sequence degenerating at $E_1$ (again by Kodaira-Spencer), and $K_{M_t}$ holomorphically trivial. Indeed, it is well-known that the degeneration at $E_1$ is equivalent to the equality -$$b_k(M)=\sum_{p+q=k}h^{p,q}(M),$$ -for all $k$, so in particular the Hodge numbers $h^{p,q}(M_t)$ are locally constant. As above, you get a nontrivial holomorphic section $\Omega_t$ of $K_{M_t}$. Now we don't have Ricci-flat Kähler metrics anymore, but by Proposition 1.6 and 1.1 of this paper, we can find Hermitian metrics $g_t$ on $M_t$ whose first Chern form (i.e. Chern-Ricci curvature) vanishes, and then the Bochner formula goes through essentially as before ($\nabla$ now being the Chern connection of $g_t$), see Lemma 2.1 in that paper.<|endoftext|> -TITLE: Unique smooth structure on 3-manifolds -QUESTION [12 upvotes]: Do you know a good reference for the existence and uniqueness of a smooth structure on $3$-manifolds? -As far as I understand topological $3$-manifolds admit a unique smooth structure. -I could find the following references for this result: -It follows from Hauptvermutung for $3$-manifolds (Theorems 3 and 4 in [2]) and from the fact that a combinatorial $3$-manifold has a unique smoothing (see Theorem 4.2 in [1]). -However, I am not quite satisfied with this answer since it requires a good understanding of what is written in these papers. For someone like me, who does not know geometric topology well, it would be better to have a reference to an explicit statement. -[1] M. W. Hirsch, B. Mazur, -Smoothings of piecewise linear manifolds. -Annals of Mathematics Studies, No. 80. -Princeton University Press, Princeton, N. J.; University of Tokyo Press, Tokyo, 1974. -[2] E. E. Moise, -Affine structures in 3-manifolds. V. -The triangulation theorem and Hauptvermutung. -Ann. of Math. 56 (1952), 96-114. -There is a related post, but I do not find the answer posted there satisfactory. Smooth structures on closed $3$-manifolds are unique up to diffeomorphism? -I think Moise does not talk about smooth structures only about triangulations. - -REPLY [16 votes]: An alternative to Moise's paper for the existence and uniqueness of piecewise linear (PL) structures on topological 3-manifolds is the paper "The triangulation of 3-manifolds" by A.J.S. Hamilton in Quart. J. Math. Oxford (2), 27 (1976), 63-70. The result is stated as Theorem 2 there and proved in the rest of the paper using the famous Kirby torus trick together with several basic results about PL 3-manifolds. -For the existence and uniqueness of smooth structures on PL 3-manifolds there is a nice exposition in section 3.10 of Thurston's book "Three-Dimensional Geometry and Topology".<|endoftext|> -TITLE: On Erdos-Kakutani like Equivalents of the Failure of Continuum Hypothesis -QUESTION [19 upvotes]: Among all mysterious equivalents of the Continuum Hypothesis and its negation, there is an algebraic combinatorial equivalent of the $\neg CH$ by Erdos and Kakutani (MR0008136) as follows: -Definition. A linear homogeneous equation $a_1x_1 + a_2x_2 + … + a_nx_n = 0$ with real coefficients is called $\aleph_0$-regular if every coloring of the real numbers by $\aleph_0$-many colors has a monochromatic solution to the above equation in distinct $x_i$. -Remark. Not all linear homogeneous equations with real coefficients are $\aleph_0$-regular. An example is Schur's equation: $x_1 + x_2-x_3=0$. - -Theorem 1. (Erdos - Kakutani) The followings are equivalent: -(a) $2^{\aleph_0}>\aleph_1$ -(b) The equation $x_1+x_2 - x_3 - x_4=0$ is $\aleph_0$-regular. - -In the other words, not only there must be many reals if one can always find a monochromatic solution for such an equation with respect to every coloring, but also if there are so many reals then such a solution actually exists for every $\aleph_0$-coloring. -Inspired by some results of Komjath, Fox (MR2360680) has proved the following generalization of Erdos - Kakutani theorem: - -Theorem 2. (Fox) The followings are equivalent: -(a) $2^{\aleph_0}>\aleph_n$ -(b) The equation $x_1+nx_2 - x_3 - ... - x_{n+3}=0$ is $\aleph_0$-regular. - -My questions are in two different directions. First, note that there are proper class many instances of the failure of $CH$ in the form of $2^{\aleph_0}>\aleph_\alpha$ while the number of linear homogeneous equations $a_1x_1 + a_2x_2 + … + a_nx_n = 0$ with real coefficients is limited. Thus there must be a place where the correlation between the size of continuum and $\aleph_0$-regularity of linear homogeneous equations with real coefficients breaks! So there must be a minimum $\alpha$ where there is NO Erdos-Kakutani-like equivalent of the $2^{\aleph_0}>\aleph_\alpha$. How far is it? - -Question 1. What is the minimum ordinal $\alpha$ such that there is NO linear homogeneous equation $a_1x_1 + a_2x_2 + … + a_nx_n = 0$ with real coefficients such that: -$$2^{\aleph_0}>\aleph_\alpha\Leftrightarrow a_1x_1 + a_2x_2 + … + a_nx_n = 0 ~ \text{is}~\aleph_0 \text{-regular}$$ - -Fox and Erdos-Kakutani theorems indicate that the existence of monochromatic solutions for certain linear equations over real numbers contains valuable information about the size of the continuum. What about arbitrary equations, particularly the simple non-linear ones? And what sort of information do they contain? Can their $\aleph_0$-regularity impose upper bounds on the size of the continuum rather than lower bounds? - -Question 2. Is there any non-linear version of Erdos-Kakutani's theorem? A non-linear equation $p(x_1,...x_n)=0$ whose $\aleph_0$-regularity is equivalent to a statement about the size of the continuum? - -REPLY [2 votes]: During a personal communication, Schmerl has suggested the results in the following papers which are closely related to the answer to question 2. I add them here for the sake of those who are interested: -Ref: -(1) James H. Schmerl, Avoidable algebraic subsets of Euclidean space, Trans. -Amer. Math. Soc. 352 (2000), 2479-2489. -(2) James H. Schmerl, Chromatic numbers of algebraic hypergraphs, Combinatorica 37 (2017), no. 5, 1011–1026. -(3) James H. Schmerl, Deciding the Chromatic Numbers of Algebraic Hypergrahs, (2016). - -Here are some definitions and results from the above papers. You may find further information in the mentioned references: -Definition. A polynomial $p(x_0, x_1,\cdots , x_{k-1})$ over the reals $\mathbb{R}$ is $(k, n)$-ary if each $x_i$ is an $n$-tuple of variables. We say that a $(k, n)$-ary polynomial $p(x_0, x_1,\cdots , x_{k-1})$ is avoidable if the points of $\mathbb{R}^n$ can be colored with countably many colors such that whenever $a_0, a_1,\cdots , a_{k-1}\in \mathbb{R}^n$ are distinct and $p(a_0, a_1,\cdots , a_{k-1})=0$, then there are $i < j < k$ such that the points $a_i, a_j$ are differently colored. The polynomial is unavoidable if it is not avoidable. -Similarly, for an infinite cardinal $\kappa$, the $(k, n)$-ary polynomial $p(x_0, x_1,\cdots , x_{k-1})$ is $\kappa$-avoidable -if the points of $\mathbb{R}^n$ can be colored using $\kappa$ colors such that whenever $a_0, a_1,\cdots , a_{k-1}\in \mathbb{R}^n$ are distinct and $p(a_0, a_1,\cdots , a_{k-1})=0$, then there are $i < j < k$ such that the points $a_i, a_j$ are differently colored. A polynomial is $\kappa$-unavoidable if it is not $\kappa$-avoidable. -In fact, $p(x_0, x_1,\cdots , x_{k-1})$ is avoidable if the chromatic number $\chi (H)$ of its zero hypergraph $H$ is countable, and it is $\kappa$-avoidable if $\chi (H)\leq\kappa$. -Example 1. For each $k<\omega$ and ordinal $\alpha$, the polynomial, $x_0+\cdots+x_{k}-x_{k+1}-kx_{k+2}$, in Fox' theorem is a $(k + 3, 1)$-ary polynomial -which is $\aleph_{\alpha}$-avoidable if and only if $2^{\aleph_0}\leq \aleph_{\alpha+k}$. -Example 2. The $(3, n)$-ary polynomial $p(x, y, z)=||x-y||^2-||y-z||^2$, for $2\leq n<\omega$ where $||.||$ denotes the Euclidean norm in $\mathbb{R}^n$, is avoidable. -Theorem 1. For each $n < w$ there is a countable partition of $\mathbb{R}^n$ which avoids every avoidable $(-, n)$-ary polynomial over $\mathbb{Q}$. -Theorem 2. Assuming $2^{\aleph_0}\geq \aleph_{m}$, every avoidable polynomial is $m$-avoidable. -Theorem 3. Assuming $2^{\aleph_0}\leq \aleph_{m}$, every $m$-avoidable polynomial is avoidable. -Corollary. The followings hold: -(a) Assuming $2^{\aleph_0}\geq \aleph_{m}$, a polynomial is avoidable iff it is $m$-avoidable. -(b) Assuming $2^{\aleph_0}\leq \aleph_{m}$, a polynomial is avoidable iff it is $\omega$-avoidable.<|endoftext|> -TITLE: Is there a well-defined notion of "pitch shift" without time dilation? -QUESTION [6 upvotes]: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous periodic function. If we replace $f$ by $$g(t) := f(\frac{1}{\lambda}t)$$ for some $\lambda > 0$, this new function $g$ has period $\lambda$ times the period of $f$, and$\mathbf{{}^1}$ $$\hat{g}(s) = \lambda \hat{f}(\lambda s)$$. -That is $f$ has been pitch-shifted down by $\log_2 \lambda$ octaves, but also time-dilated by a factor of $\lambda$. -My question is: is there mathematically well-defined operation of pitch shift without time dilation? -I strongly suspect the answer is "no", because any continuous linear operation $T_{\lambda}$ which correctly pitch-shifts down by $\log_2 \lambda$ octaves all functions of the form $f(t) = sin(\omega t)$, i.e., for which $T_{\lambda}(sin( \omega t)) = sin(\frac{\omega}{\lambda} t)$, will necessarily have $T_{\lambda}(f(t)) = f(\frac{1}{\lambda}t)$ for every continuous periodic function $f$. -This doesn't immediately rule out a non-linear (or possibly non-continuous) operation, so I'm wondering if there is anything in the literature about such an operation, or a "no-go" theorem showing that the notion of pitch-shift without time-dilation is incoherent, period. -Most papers about this subject are written by engineers, who really only care about performing this operation approximately to give more-or-less the correct psycho-acoustic effect. E.g. this paper using wavelets, which I take to be roughly state-of-the-art. Other past approaches have been even worse: e.g. breaking up a signal into short time intervals doing a naive pitch-shift with time dilation (possibly repeating or clipping), and then assembling them back together, introducing discontinuities when moving from one time interval to the next, which are then smoothed over by some filter. -My motivation here is to be able to have an ideal standard by which to judge the degree of accuracy of a pitch-shift algorithm. If there is none, then the question of the accuracy of a pitch-shift algorithm is an empirical question about subjective experience rather than a mathematical one. - -$\mathbf{{}^1}$ The Fourier transform will have to be a distribution rather than a function. - -REPLY [3 votes]: Let us model the signal $f$ as follows: -\begin{equation} - f(t)=\Re\sum_{j=1}^n c_j(t)K((t-t_j)/\tau_j)e^{i\omega b_j t} -\end{equation} -for $t\in[0,T]$, -where $n$ is not very large, $\omega$ is a large positive real number, $b_j\asymp 1$ ($b_j\in\mathbb R$), $K$ is a kernel function (such as $K(x)=e^{-x^2}$), the $t_j$'s are time moments in the interval $[0,T]$, $\tau_j\asymp1$ ($\tau_j\in\mathbb R$), and the $c_j$'s are complex-valued "amplitude" functions, which are slowly varying as compared with the harmonics $e^{i\omega b_j \cdot}$, except maybe only one of the harmonics. -I think what human beings may be doing is as follows: Locally, over time intervals that are small enough for the amplitudes to noticeably change and yet much greater that $2\pi/\omega$, using measurements of $f(t)$ at sufficiently many time moments $t$, they obtain estimates $\hat c_j$, $\hat K$, $\hat t_j$, $\hat\tau_j$, $\hat\omega$, $\hat b_j$ of $c_j$, $K$, $t_j$, $\tau_j$, $\omega$, $b_j$ and then pitch-shift, to get the transformed signal over the same time interval $t\in[0,T]$, of the same tempo, with as many "notes": -\begin{equation} - \hat f_\lambda(t)=\Re\sum_{j=1}^n \hat c_j(t)\hat K((t-\hat t_j)/\hat \tau_j)e^{i\hat \omega \hat b_j t/\lambda}. -\end{equation} -This does not look pretty, and yet it seems a reasonable way for the brain to act. -Here is an illustration of this pitch-shifting, with $\omega=30$ and $\lambda=2$:<|endoftext|> -TITLE: Jones-Wenzl-type projectors for Brauer algebras -QUESTION [6 upvotes]: Jones-Wenzl projectors are an explicit combinatorial description of the central projectors of the Temperley-Lieb algebra. -They also describe very explicitly the failure of certain representations to exist for $q$ a root of unity and generate the negligible ideal (the ideal of elements $a$ such that $\operatorname{tr}(ab) = 0$ for every $b$.) -Is there a similar combinatorial description of such projectors for the Brauer algebra? -Here the Brauer algebra is drawn with the same diagrams as Temperley-Lieb, but the strands are allowed to cross. -It is a graphical description of the decomposition of tensor powers ofan irrep of a classical orthogonal or symplectic algebra. -I think the answer is "no" at least for all of the properties I mentioned above, but I'd love to be surprised. -I also don't think I need all the properties above: just knowing some elements of the negligible ideal (for whichever value of the loop parameter and whatever number of strands) would be useful. -I think this question is related to mine, except that I looked up the book mentioned and didn't find it helpful. -The paper "On central idempotents in the Brauer algebra" shows how to compute some of the idempotents, but non-recursively and without a formula for the trace, so it may be that the question I'm asking doesn't even have a partial answer. - -REPLY [2 votes]: Via email, Noah Snyder pointed me to the papers Tuba and Wenzl - On braided tensor categories of type BCD and Lehrer and Zhang - The second fundamental theorem of invariant theory for the orthogonal group. I wound up going in a different direction than this question was suggesting, so I did not read the papers in detail and try to translate their results to more diagrammatic language, but it seems like that should be possible. -In particular, Section 7.9 of the first paper (by Tuba and Wenzl) has the type of recursive formula I was looking for.<|endoftext|> -TITLE: Liftable rational varieties -QUESTION [9 upvotes]: Is there an example of a rational smooth projective variety over a perfect field of characteristic $p$, that is not liftable to characteristic zero? - -REPLY [11 votes]: Two such examples were given by Achinger and Zdanowicz [AZ17], both of which satisfy a whole bunch of other good properties (e.g. their classes in the Grothendieck ring of varieties are polynomials in the Lefschetz motive $\mathbb L = [\mathbb A^1]$). The easiest one to state is probably the following: -Example. (Achinger–Zdanowicz [AZ17, §4]) Let $k = \bar{\mathbb F}_p$, let $P = \mathbb P^3_k$, and construct $X$ from $P$ by first blowing up all $\mathbb F_p$-points of $P$, and then blowing up all strict transforms of lines defined over $\mathbb F_p$. Then $X$ cannot be lifted to any ring $A$ in which $pA \neq 0$. -The proof is relatively elementary, and relies on matroid theory and elementary projective geometry, as well as a little deformation theory. - -References. -[AZ17] Achinger, Piotr; Zdanowicz, Maciej, Some elementary examples of non-liftable varieties. Proc. Amer. Math. Soc. 145 (2017). ZBL06769127.<|endoftext|> -TITLE: Ramsey Number R(3,3,4) -QUESTION [15 upvotes]: How much is known about the Ramsey number R(3,3,4)? There is a trivial upper bound of 34, but are any tighter bounds known? - -REPLY [19 votes]: According to the reference 1 below to earlier work of the authors in this paper here, it is known that -$$ -30\leq R(3,3,4)\leq 31 -$$ -Edit: A more recent paper, by Codish, Frank, Itzhakov and Miller, available here has shown that $R(3,3,4)=30.$ Thanks to @Julian. - -Piwakowski, K., and Radziszowski, S. P., Journal of Combinatorial Mathematics and Combinatorial Computing, 27:135-141, 1998.<|endoftext|> -TITLE: $p$-adic lifts of tropical varieties -QUESTION [7 upvotes]: What is currently known about lifts of tropical varieties to varieties over $\mathbb{Q}_p$ or its extensions? Starting with an appropriate rational polyhedral cone complex what are the obstructions to its deformation to a (toric or log-smooth) variety over a $p$-adic field classified by? In which major cases are they known to vanish? What are the corresponding deformation spaces? -Summary and references would be really appreciated. - -REPLY [3 votes]: There are moduli spaces of tropical lifts. They satisfy "Murphy's law": Any behavior which can happen on a scheme of finite type can happen on them. See Katz and Payne "Realization spaces for tropical fans" for the details.<|endoftext|> -TITLE: Eulerian ordering of the integers modulo n -QUESTION [5 upvotes]: Let $n>1$ be an integer. Consider the set $C_n := \{0,1, \dots , n-1\}$. -An Eulerian ordering of $C_n$ is an ordering $r_1, \dots, r_n$ of its elements such that: -$$\forall i \le n \ \forall j1$ square-free? -It is checked by Sage for $n\le 500$ (see below). -If yes, let $n>1$ be a square-free integer: -Stronger question: Can any partial Eulerian ordering of $C_n$ ($r_1, \dots, r_s$ with $s -TITLE: Tropicalization of perfectoid spaces and their tilts -QUESTION [19 upvotes]: Does tropicalization exist in the world of perfectoid spaces? Since it does for Huber's adic spaces, I thought it might for perfectoid spaces too, yet I can't find any explicit references so far. -For concreteness let $K$ be the completion of $\mathbb{Q}_p(p^{\frac{1}{p^\infty}})$, and $X$ a perfectoid space over $K$. Can one tropicalize $X$ and its tilt $X^\flat$? If so, how are the two tropicalizations related? Are they isomorphic (as rational polyhedral cone complexes)? -Further, what is the information tropicalization would retain in this setting? Is there a shadow left of the Galois action on $X$ and $X^\flat$? - -REPLY [10 votes]: Sorry for the late answer! -Tropicalization works for all analytic adic spaces $X$ (in particular, for perfectoid spaces). One minor variation on the usual story is that adic spaces have their valuations in abstract totally ordered abelian groups. On analytic adic spaces, these valuations always have unique rank $1$ generalizations, taking values in $\mathbb R_{>0}$ up to isomorphism. In other words, the value group is only defined up to rescaling. If you fix a nonarchimedean base field $K$, you can get rid of this ambiguity by fixing the norm on $K$. More generally, you can fix a global topologically nilpotent unit $\varpi\in \mathcal O_X^\times(X)$ and normalize the norm of $\varpi$. But that's actually just a special case of the following: -For any analytic adic space $X$ and global units $f_0,\ldots,f_n\in\mathcal O_X^\times(X)$, such that the locus $\{|f_0|=\ldots=|f_n|=1\}$ is empty, one can define a tropicalization map -$$|X|\to \mathbb P^n(\mathbb R)$$ -taking any $x\in |X|$ with rank $1$-generalization $\tilde{x}$ to -$$[\log |f_0(\tilde{x})|:\ldots:\log |f_n(\tilde{x})|].$$ -If $X$ is affinoid (or Stein, in a suitable sense), one can recover the maximal Hausdorff quotient of $|X|$ (i.e., "the Berkovich space") as the inverse limit of the images of all tropicalization maps (for varying $f_i$'s). -In the case of perfectoid spaces, if $X$ is perfectoid with tilt $X^\flat$, and one chooses $f_0^\flat,\ldots,f_n^\flat\in \mathcal O_{X^\flat}^\times(X^\flat)$ with $f_i=(f_i^\flat)^\sharp\in \mathcal O_X^\times(X)$, then $|X|\cong |X^\flat|$ and the corresponding tropicalization maps agree. -About the Galois action: Everything here is functorial, so you can get Galois actions, I guess. -Unfortunately, I don't really know of any published work combining perfectoid and tropical geometry, but I agree that this is a natural idea. In fact, when I tried to prove the full weight-monodromy conjecture using perfectoid spaces, the remaining step was to find algebraic varieties of the correct dimension in certain open subsets of $|\mathbb P^n_{K^\flat}|$ (as an adic space). This can (essentially) be phrased as a problem of finding algebraic varieties with given tropicalization, and fits a line of research whether tropical varieties can be realized by actual varieties. Note that the tropicalization can be realized by an algebraic variety over $\mathbb P^n_K$ by assumption, so whatever combinatorial conditions one might put on the tropicalization, they ought to be satisfied. Unfortunately, I could never really make progress on these questions.<|endoftext|> -TITLE: Fields of rationality as a notion of automorphic size -QUESTION [5 upvotes]: I want to interpret the degree of the field of rationality of an automorphic form as a notion of size, analogously to the conductor, and this question is about the possible obstructions to do so. The results I know are limited to the case of Hecke cusp forms, I hense recall them and raise the natural questions and results to Maass forms and automorphic representations. -1. The case of Hecke cusp forms. Let $f$ be a cuspidal Hecke eigenform. Its field of rationality $\mathbf{Q}(f)$ is the extension of $\mathbf{Q}$ generated by its Fourier coefficients, that is to say -$$\mathbf{Q}(f) = \mathbf{Q}(a_1(f), a_2(f), a_3(f), \ldots). \qquad (1)$$ -It is known that the degree of the field of rationality grows with the level (in the sense that the proportion of $f$ with field of rationality of bounded degree converges to zero when the level grows). -2. The case of automorphic representations. I am wondering about what can be said of an analogous notion for automorphic representations. Let $\pi_v$ be an admissible representation of $GL(2, F_v)$ for a local field $F_v$, its field of rationality is defined as -$$\mathbf{Q}(\pi_v) = \{ \sigma \in Aut(\mathbf{C}) \ : \ {}^\sigma \pi \simeq \pi \}.$$ -For an automorphic representation $\pi$ of $GL(2, \mathbf{A})$, decomposed by Flath theorem as $\pi = \otimes_v \pi_v$, the field of rationality of $\pi$ is -$$\mathbf{Q}(\pi) = \prod_v \mathbf{Q}(\pi_v). \qquad (2)$$ -3. The case of Maass forms. These two notion agree in the case of Hecke cusp forms. I wonder what can be said in the case of Maass forms: a field of rationality can be defined by its coefficients as in (1), and also by the attached representation as in (2). - -(A) Do the two notions agree in the case of Maass forms? - -4. A kind of automorphic size. As already stated, in the case of cusp forms the degree of the field of rationality essentially grows with the level (Serre, Shin-Templier, Binder for references). This endows -$$d(\pi) = [\mathbf{Q}(\pi) : \mathbf{Q}],$$ -with a size flavor. However, here are some natural questions in this direction: - -(B) Is there any result in the weight aspect? -(C) Is $d(\pi)$ always finite? -(D) Is there any infinite family with constant degree of field of rationality? What can be said about such families? - -I apologize for these maybe loosely related questions, it appeared to me that they are relevant in this spirit of "size", and I wished to asked them together. - -REPLY [3 votes]: I haven't thought about the case of Maass forms, but I can tell you what is expected about the weight. -Maeda's conjecture implies that all cuspidal eigenforms in $S_k(1)$ are conjugate, so $\bar d(\pi)$ should be $\dim S_k(1)$, where $\bar d$ denotes the degree of the Galois closure. In particular, the answer to (D) should be no for level 1, but I don't think anyone knows how to prove such a statement. See also this question: modular eigenforms with integral coefficients [Maeda's Conjecture] -A generalization of Maeda's conjecture was proposed by Tsaknias for more general level. For instance, say $N$ is squarefree. Then if you look at newforms of level $N$, you can distinguish Galois orbits by looking at Atkin-Lehner sign patterns (of which there are $2^m$, $m =$ number of primes dividing $N$). The conjecture is that for $k$ large all forms with the same Atkin-Lehner sign pattern are conjugate.<|endoftext|> -TITLE: Do you know important theorems that remain unknown? -QUESTION [170 upvotes]: Do you know of any very important theorems that remain unknown? I mean results that could easily make into textbooks or research monographs, but almost -nobody knows about them. If you provide an answer, please: - -State only one theorem per answer. When people will vote on your answer they will vote on a particular theorem. -Provide a careful statement and all necessary definitions so that a well educated graduate student -working in a related area would understand it. -Provide references to the original paper. -Provide references to more recent and related work. -Just make your answer useful so other people in the mathematical community can use it right away. -Add comments: how you discovered it, why it is important etc. -Please, make sure that your answer is written at least as carefully as mine. I did invest quite a lot of time writing my answers. - -As an example I will provide three answers to this question. I discovered these results while searching for papers related to the questions I was working on. - -REPLY [24 votes]: Ionin–Pestov theorem is not very well known, but it deserves to be included in standard introductory texts on differential geometry of curves. -It gives the simplest meaningful example of a local-to-global theorem which is what differential geometry is about. -Theorem. Assume that a plane region $F$ is bounded by a simple loop with curvature at most $1$. Then $F$ contains a unit disc. -The original reference: - -Пестов, Г. Г., Ионин В. К. О наибольшем круге, вложенном в замкнутую кривую // Доклады АН СССР. — 1959. — Т. 127, № 6. - -We used it in our textbook What is differential geometry.<|endoftext|> -TITLE: Discriminant of numerator of inverse logarithmic derivative operator iteration -QUESTION [7 upvotes]: Let $T:\mathbb Q(x)\to \mathbb Q(x)$ be the operator of inverse logarithmic derivative, i.e. $$Tf=\frac{f}{f'}.$$ Define $$p_n(x)=T^n\left(x-\frac{x^2}{2}\right).$$ Let $f_n(x) \in \mathbb Z[x]$ be the numerator of $p_n(x)$, taken with positive leading term. The sequence $D_n$ of discriminants of our polynomials starts with -$$\begin{aligned} -D_0&=4,\\ D_1&=4,\\ D_2&=4,\\ D_3&=-256=2^8,\\ D_4&=2^{32}5^2,\\ D_5&=2^{128}5^6331^2, \\D_6&=2^{512}5^{20}331^646599695357^2.\\ -\end{aligned} -$$ -Is it true that we always have $D_n=\pm a_n^2$ for some integer $a_n$? What else can be said about the arithmetic properties of $a_n$? For example, is it true that for all $n \geq 2$ we have $\nu_2(D_n)=2^{2n-3}$? - -REPLY [10 votes]: First, let's make the substitution $x\to 1-x$, so the function $x-\frac12x^2$ -becomes $\frac12(1-x^2)$. The key here is that it is an even function, and the -discriminant of even and odd polynomials are essentially squares. More precisely, -$$\begin{aligned} - \text{Disc}\bigl( A(x^2) \bigr) &= \pm\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ - &= \pm 2^{\deg(A)} \cdot A(0)\cdot \text{Disc}\bigl( A(x) \bigr)^2 -\end{aligned} -$$ -and -$$\begin{aligned} - \text{Disc}\bigl( xA(x^2) \bigr) &= \pm\text{Resultant}\bigl(xA(x^2),A(x^2)+2x^2A'(x^2)\bigr)\\ - &= \pm A(0)\text{Resultant}\bigl(A(x^2),2x^2A'(x^2)\bigr)\\ - &= \pm 2^{\deg(A)} \cdot A(0)^3\cdot \text{Disc}\bigl( A(x) \bigr)^2.\\ -\end{aligned} -$$ -Next, define three operators on the ring of rational functions: -$$ T(f)=\frac{f}{f'},\quad D(f)=\frac{f'}{f},\quad I(f)=\frac{1}{f}. $$ -The question asks about iterates of $T$, but note that $T=I\circ D$ and -$D\circ I=-D$, so -$$ T^n(f) = (I\circ D)^n(f) = I\circ(D\circ I)^{n-1}\circ D(f) = (-1)^{n-1}I\circ D^n. $$ -So up to sign, the numerator and denominator of $T^n(f)$ are the reverse of those for $D^n(f)$, -so it's enough to look at the logarithmic derivative $D$. Finally, we note that -if we start with an even function, then $Df$ is odd, and likewise if we start with an odd function, -then $Df$ is odd. Hence if $f$ is an even (or odd) rational function, then for all $n\ge1$ we -have $D^nf=A_n/B_n$ with one of $A_n$ and $B_n$ odd and the other even. -From above, the absolute values of the discriminants of $A_n$ and $B_n$ are squares, -up to the power of $2$ and the $A(0)$ term. But it's easy to check that for the particular -function $\frac12(1-x^2)$, the power of $2$ will be even and the $A(0)$ term will be $\pm1$.<|endoftext|> -TITLE: surface with rational curve in the double locus -QUESTION [5 upvotes]: I am interested in the existence of a surface $X$ over $\mathbb{C}$ with the following properties (or a reason for why one cannot exist): - -$X$ is slc (and not-normal) -There is rational curve $C \subset X$ contained in the double locus (i.e. non-normal locus) -The cotangent $\Omega^1_Y$ is ample, where $Y \to X$ is the normalization - -Note that this last condition implies that $Y$ has no rational or elliptic curves. In particular, since the pre-image of the double locus under the normalization is a 2-to-1 cover branched over the pinch points, -Riemann-Hurwitz gives that $C$ contains at least 6 pinch points. -The only examples I know of non-normal surfaces with a rational curve in the double locus containing $> 2$ pinch points have non-positive Kodaira dimension. - -REPLY [2 votes]: Here is a possible example building up on what I think was the idea of @Sasha. -Take $Y$ to be a non-isotrivial fibration of genus $g_1 \geq 3$ curves over a genus $g_2 \geq 2$ curve that contains a closed fiber $C'$ which is hyperelliptic. Such surfaces (Kodaira fibrations) have ample cotangent. -Consider the surface $X$ obtained by the pushout of $Y$ given by the quotient by the involution, i.e. -$\require{AMScd}$ -\begin{CD} - C' @>\sigma>> \mathbb{P}^1\\ - @V V V @VV V\\ - Y @>>> X -\end{CD} -This is an slc surface $X$, and $Y$ has a finite map to $X$ that is birational and finite. Moreover one can show that $Y \to X$ is the normalization. -By construction of the pushout the image of the hyperlliptic fiber $C'$ is a $\mathbb{P}^1$ inside the double locus of $X$. -Follow up question: Is this surface contained in the closure of the main component of the moduli space of surfaces of general type (i.e. is $X$ a smoothable surface in the sense of KSBA)?<|endoftext|> -TITLE: Every odd integer greater than $1$ is of the form $a+b$ with $a^2+b^2$ being prime -QUESTION [8 upvotes]: Let $m$ be an odd integer greater than $1$. Is it true that there are positive $a, b$ such that $m=a+b$ and $a^2+b^2$ is a prime number? -It seems that for every odd $m$ there are many $(a,b)\in \mathbb{N}^2$ which sum to $m$ and their sum of squares give a prime number but I don't see how to prove this. -Or, equivalently, is it true that if we list all primes $p=a^2+b^2\equiv 1\pmod{4}$ then $a+b$ covers every odd integer? - -REPLY [5 votes]: The question is not new. It is originally due to Ming-Zhi Zhang. One may consult https://oeis.org/A036468 .<|endoftext|> -TITLE: Is the intersection of two Caccioppoli (i.e. finite perimeter) sets Caccioppoli? -QUESTION [14 upvotes]: Recall that we say that a bounded measurable set $S\subset\mathbb R^n$ is said to be Caccioppoli if the indicator function $1_S$ is BV, and we set -$$ -\operatorname{perim}(S)=\| \nabla 1_S\|_{TV} -$$ -where $\|\cdot\|_{TV}$ denotes the total variation. So, if $S$ and $T$ are Caccioppoli sets, is it known whether $S\cap T$ is Caccioppoli? - -REPLY [5 votes]: Why not indulge in some overkill: - -Theorem (cf. Federer, Theorem 4.5.11). Let $\newcommand{\RR}{\mathbb{R}}A\subset\RR^n$ be measurable with finite n-dimensional Lebesgue measure. Then $A$ has finite perimeter iff $\mathcal{H}^{n-1}(\partial^* A)<\infty$. - -Here $\mathcal{H}^s$ denotes the $s$-dimensional Hausdorff measure and $\partial^*A\subset\RR^n$ the measure-theoretic boundary of $A$. The latter is determined by -$$ -x\in\RR^n\setminus\partial^*A \quad\Longleftrightarrow\quad \lim_{r\to 0+}\frac{\mathcal{H}^n(A\cap B(x,r))}{\mathcal{H}^n(B(x,r))}\in\{0,1\}. -$$ -In other words, $x\in\partial^*A$ iff $A$ does not have Lebesgue density $0$ or $1$ at $x$. -By Federer's theorem it suffices to show that $\partial^*(S\cap T)\subset\partial^*S\cup\partial^*T$, or equivalently, $$(\RR^n\setminus\partial^*S)\cap(\RR^n\setminus\partial^*T)\subset\RR^n\setminus\partial^*(S\cap T).$$ -The rest is elementary. Let $x$ be an element of the left hand side. Clearly, if $S$ or $T$ has density $0$ at $x$, then $S\cap T$ has density $0$ at $x$. So suppose both $S$ and $T$ have density $1$ at $x$. But as -$$(S\cap B(x,r))\setminus (B(x,r)\setminus T) = S\cap T\cap B(x,r),$$ -also $S\cap T$ has density $1$ at $z$. So $x$ is an element of the right hand side. - -REPLY [4 votes]: The intersection of two Caccioppoli sets is again a Caccioppoli set: as Piotr Hajlasz points out, a Caccioppoli set is simply a set of finite perimeter therefore the (standard) proof I report below is based on the following, slightly more general, definition of the (relative) perimeter of a set. -Definition. Let $S$ a Lebesgue measurable set in $\mathbb{R}^n$. For any open subset $\Omega\subseteq\mathbb{R}^n$ the perimeter of $S$ in $\Omega$, denoted as $P(S,\Omega)$, is the variation of $\chi_S$ in $\Omega$ i.e. -\begin{split} -P(S,\Omega)&=\sup\left\{\int_S \mathrm{div}\varphi\,\mathrm{d}x\,:\,\varphi\in [C_c^1(\Omega)]^n, \|\varphi\|_\infty\leq1\right\}\\ -& =\| \nabla (\chi_{S\cap\Omega})\|_{TV}=TV(S,\Omega) -\end{split} -The perimeter so defined is a lower-semicontinuous set function which is additive respect to the argument $\Omega$. -Theorem. Let $S$ and $T$ two Lebesgue measurable sets in $\mathbb{R}^n$: then, for any open set $\Omega\subseteq\mathbb{R}^n$ -$$ -P(S\cup T,\Omega)+P(S\cap T,\Omega)\leq P(S,\Omega) + P(T,\Omega) -$$ -Proof. We can choose two smooth sequences of functions $\{s_k\}, \{t_k\}$ in $C^\infty(\Omega)$ converging respectively to $\chi_S$ and $\chi_T$ in $L^1_{loc}(\Omega)$, for example by using partitions of unity, and such that -$$ -0\leq s_k\leq 1\quad 0\leq t_k \leq 1, -$$ -and -$$ -\lim_{k\to\infty}\int_\Omega |\nabla s_k|\mathrm{d}x=P(S,\Omega)\quad -\lim_{k\to\infty}\int_\Omega |\nabla t_k|\mathrm{d}x=P(T,\Omega) -$$ -Now since $s_kt_k\to\chi_{S\cap T}$ and $s_k+t_k-s_kt_k\to\chi_{S\cup T}$, the theorem follows by passing to the limit for $k\to\infty$ in the following elementary inequality -$$ -\int_\Omega |\nabla (s_kt_k)|\mathrm{d}x+\int_\Omega |\nabla (s_k+t_k-s_kt_k)|\mathrm{d}x \leq -\int_\Omega |\nabla s_k|\mathrm{d}x+ -\int_\Omega |\nabla t_k|\mathrm{d}x\quad\blacksquare -$$ -An immediate consequence of this theorem is the following corollary, which includes the sought for answer to the posed question. -Corollary. If $S$ and $T$ are Caccioppoli sets, so are $S\cup T$ and $S \cap T$. -Notes - -The proof is taken almost verbatim from the wonderful book of Ambrosio, Fusco and Pallara ([1], §3.3 p. 144): De Giorgi, Colombini and Piccinini ([2], §1.3 pp.18-19) prove the same result in a different way and assuming from the start that $S$ and $T$ are Caccioppoli sets. -A direct proof of the result $P(S),P(T)<\infty\Rightarrow P(S\cap T)<\infty$ is presented by Gurtin, Williams and Ziemer ([3], p. 6) and is based on lattice and measure theoretical considerations: in fact, it is worth noting that this property of Caccioppoli sets make them very important in the formal definition of the concept of body in the modern rational continuum mechanics. -It is perhaps also worth to point out that the property of BV functions used by Piotr Hajlasz is in reality their original definition as conceived by Lamberto Cesari back in 1936. - -[1] Ambrosio, Luigi; Fusco, Nicola; Pallara, Diego (2000). Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs. New York and Oxford: The Clarendon Press/Oxford University Press, New York, pp. xviii+434, ISBN 0-19-850245-1, MR1857292, Zbl 0957.49001. -[2] De Giorgi, Ennio; Colombini, Ferruccio; Piccinini, Livio (1972), Frontiere orientate di misura minima e questioni collegate [Oriented boundaries of minimal measure and related questions], Quaderni (in Italian), Pisa: Edizioni della Normale, pp. 180, MR0493669, Zbl 0296.49031. -[3] Gurtin, M. E.; Williams, W. O. & Ziemer, W. P., "Geometric measure theory and the axioms of continuum thermodynamics", Archive for Rational Mechanics and Analysis, 1986, 92, 1-22, MR816619, Zbl 0599.73002.<|endoftext|> -TITLE: What is the (genuine) name for the Gutik hedgehog? -QUESTION [9 upvotes]: Working with non-regular topological semigroups, my collegue Oleg Gutik discovered a special space $H$ which we named Gutik's hedgehog. It is homeomorphic to the space -$$H:=\{(0,0)\}\cup\{(\tfrac1n,0):n\in\mathbb N\}\cup\{(\tfrac1n,\tfrac1{nm}):n,m\in\mathbb N\},$$ endowed with the topology $\tau$ consisting of sets $U\subset H$ satisfying the following two conditions: -(1) if $(\frac1n,0)\in U$ for some $n\in\mathbb N$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $k\ge m$; -(2) if $(0,0)\in U$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $n\ge m$ and all $k\in\mathbb N$. -It turns out that Gutik's hedgehog is a test space for regularity in the class of first-countable Hausdorff spaces. -Theorem. A first-countable Hausdorff space is regular if and only if it contains no topological copies of the Gutik hedgehog. -Because of this fundamental role in testing regularity, I admit that Gutik's hedgehog is known in topology under some different name. I would be grateful for any information in this respect. -Remark 1. The Gutik's hedgehog resembles (but is not equal to) the non-regular space of Smirnov, see Example 64 in "Counterexamples in Topology". - -REPLY [3 votes]: Searching Steen & Seebach, Counterexamples in Topology for "Hausdorff, separable, not regular" I found this example. Is that it? -This is Example 79, page 97. - -Irregular Lattice Topology - Let $A = \{(i,k) : 0 < i,k \in \mathbb{N}\}$, $B = \{(i,0) : i \geq 0\}$ and $X = A \cup B$. Declare each point of $A$ to be open. Let the set of all $U_n((i,0)) = \{(i,k) : k=0$ or $k \geq n\}$ form a local basis at any point with $i \neq 0$ and the set of all $V_n = \{(i,k) : i=k=0$ or $i,k\geq n\}$ be a local basis at $(0,0)$.<|endoftext|> -TITLE: Minimal ideals of the ring of continuous functions -QUESTION [5 upvotes]: A minimal ideal of a commutative ring $R$ is a nonzero  ideal which contains no other nonzero ideal.  -Let $X $ be a completely regular topological space and $C (X) $ the ring of all real valued continuous functions over $X $. Is there any characterization for minimal ideals of $C (X) $? - -REPLY [7 votes]: Let us first observe that an ideal $I$ of a commutative ring $R$ with identity is minimal in OP's sense if and only if $I$ is a non-zero simple module over $R$. From now on, we will favour this terminology. -Note also that a topological space $X$ with trivial topology, i.e., the open sets of $X$ are $X$ and $\emptyset$, is completely regular for a trivial reason. -Let $C(X)$ be the ring of continuous real-valued functions on $X$. If $X$ is a topological space with trivial topology, then the ring $C(X)$ is isomorphic to $\mathbb{R}$ and hence is a simple ring. -This leads to - - -Claim. Let $X$ be a completely regular topological space. - Then $I \subseteq C(X)$ is a non-zero simple ideal if and only if $I$ is the set of continuous functions $f$ such that either $f = 0$ or $\{ x \in X \,|\, f(x) \neq 0 \} = Y$ where $Y = Y(I) \subseteq X$ is an open subspace depending only on $I$ and whose induced topology is trivial. In particular, any non-zero simple ideal of $C(X)$, if it exists, is isomorphic to $\mathbb{R}$. - - -If $X$ is a completely regular topological space and $Y$ is an open subspace of $X$ with trivial induced topology, then $Y$ is also closed. -In particular, if $x \in X$ is an isolated point, then $\{x\}$ is closed and the ideal consisting of the continuous functions $f$ such that $f(y) = 0$ for every $y \neq x$ is simple. I am indebted to Zach Teitler who mentioned this fact first. - - -Proof. - Let $R = C(X)$ and let $Y \subseteq X$ be a topological subspace whose induced topology is trivial. Then any $f \in R$ is constant on $Y$ and since $Y$ is both closed and open, any continuous function $f$ defined on $Y$ can be continuously extended to $X$ by setting $f(x) = 0$ for $x \notin Y$. Thus the ideal consisting of all functions which are zero on $X \setminus Y$ is a simple ideal. - Consider now a simple ideal $I = Rf$ with $f \neq 0$ and let $Y = \{ x \in X \,|\, f(x) \neq 0 \}$. Since $Rf = Rf^2$, the function $1/f$ defined over $Y$ can be continuously extended to $X$. From this we infer that $Y$ is closed. Indeed, if there is $x \in \overline{Y} \setminus Y$, then the continuous extension of $1/f$ cannot be bounded on an open subset containing $x$, a contradiction. Now let $g \in R$ be such that its restriction to $Y$ is not zero. Since $Rfg = Rf = I$, we deduce that $g$ has no zero in $Y$. Let us prove that $g$ is constant. Reasonning by contradiction, we consider $x,y \in Y$ such that $g(x) \neq g(y)$. Let $F$ be the pre-image under $g$ of a closed segment in $\mathbb{R}$ that contains $g(x)$ but not $g(y)$. Then $F$ is a closed set that contains $x$ but not $y$. Since $X$ is completely regular, we can find a function $h \in R$ such that $h$ vanishes on $F$ and $h(y) = 1$. This is impossible since we have established before that $h$ cannot have any zero in $Y$. Hence every $g \in R$ is constant over $Y$. Since $Y$ is both closed and open, every continuous function on $Y$ extends continuously to $X$. Therefore any continuous function on $Y$ is constant. As $Y$ is also a completely regular topological space, its topology is trivial.<|endoftext|> -TITLE: Coordinates of the Weyl vector of $E_8$ (and the 135 classes of $W(E_8)/W(D_8)$) -QUESTION [15 upvotes]: Consider the root system of $E_8$, written in its standard "even" coordinate system: i.e., it is the set of all $240$ vectors in $\mathbb{R}^8$ which whose coordinates are either all integers or all integers-plus-a-half, have an even sum, and whose sum of squares is $2$. -For any choice of a set of positive roots, the half-sum of the positive roots is known as the corresponding "Weyl vector". For example, if we take the positive roots to be those whose last nonzero coordinate is positive, the Weyl vector is $(0,1,2,3,4,5,6,23)$. Since the Weyl vector lies in the interior of the Weyl chamber, the Weyl group acts freely on its orbit, and there are $\#W(E_8) = 2^{14} \cdot 3^5 \cdot 5^2 \cdot 7 = 696\,729\,600$ Weyl vectors, exactly one for each choice of positive roots. -My question is essentially whether we can describe this set of $696\,729\,600$ Weyl vectors in a simple way through its coordinates. -One obvious reduction is that the Weyl group of $D_8$, which is a subgroup (of order $8!\times 2^7 = 2^{14}\cdot 3^2\cdot 5\cdot 7 = 5\,160\,960$) of that of $E_8$ acts by permuting the $8$ coordinates in any way and changing the sign of an even number of them (see remarks below). So all that need to be described are the $696\,729\,600 / 5\,160\,960 = 3^3\cdot 5 = 135$ orbits of Weyl vectors modulo this action. It's not difficult to list them explicitly, e.g., a simple computation gives me: -(0, 1, 2, 3, 4, 5, 6, 23) -(-1/2, 3/2, 5/2, 7/2, 9/2, 11/2, 13/2, 45/2) -(0, 1, 3, 4, 5, 6, 7, 22) -(1/2, 3/2, 5/2, 9/2, 11/2, 13/2, 15/2, 43/2) -(0, 1, 2, 5, 6, 7, 8, 21) -(1, 2, 3, 4, 6, 7, 8, 21) -(1/2, 3/2, 5/2, 9/2, 13/2, 15/2, 17/2, 41/2) -(3/2, 5/2, 7/2, 9/2, 11/2, 15/2, 17/2, 41/2) -(0, 1, 3, 4, 7, 8, 9, 20) -(1, 2, 3, 5, 6, 8, 9, 20) -(2, 3, 4, 5, 6, 7, 9, 20) -(-1/2, 3/2, 5/2, 7/2, 15/2, 17/2, 19/2, 39/2) -(1/2, 3/2, 7/2, 9/2, 13/2, 17/2, 19/2, 39/2) -(3/2, 5/2, 7/2, 11/2, 13/2, 15/2, 19/2, 39/2) -(5/2, 7/2, 9/2, 11/2, 13/2, 15/2, 17/2, 39/2) -(0, 1, 2, 3, 8, 9, 10, 19) -(0, 2, 3, 4, 7, 9, 10, 19) -(0, 1, 4, 5, 6, 9, 10, 19) -(1, 2, 4, 5, 7, 8, 10, 19) -(2, 3, 4, 6, 7, 8, 9, 19) -(1/2, 3/2, 5/2, 7/2, 15/2, 19/2, 21/2, 37/2) -(-1/2, 3/2, 7/2, 9/2, 13/2, 19/2, 21/2, 37/2) -(1/2, 5/2, 7/2, 9/2, 15/2, 17/2, 21/2, 37/2) -(1/2, 3/2, 9/2, 11/2, 13/2, 17/2, 21/2, 37/2) -(3/2, 5/2, 9/2, 11/2, 15/2, 17/2, 19/2, 37/2) -(0, 1, 3, 4, 7, 10, 11, 18) -(-1, 2, 3, 5, 6, 10, 11, 18) -(1, 2, 3, 4, 8, 9, 11, 18) -(0, 2, 4, 5, 7, 9, 11, 18) -(0, 1, 5, 6, 7, 8, 11, 18) -(1, 3, 4, 5, 8, 9, 10, 18) -(1, 2, 5, 6, 7, 9, 10, 18) -(-1/2, 3/2, 5/2, 9/2, 13/2, 21/2, 23/2, 35/2) -(-3/2, 5/2, 7/2, 9/2, 11/2, 21/2, 23/2, 35/2) -(1/2, 3/2, 7/2, 9/2, 15/2, 19/2, 23/2, 35/2) -(-1/2, 5/2, 7/2, 11/2, 13/2, 19/2, 23/2, 35/2) -(-1/2, 3/2, 9/2, 11/2, 15/2, 17/2, 23/2, 35/2) -(3/2, 5/2, 7/2, 9/2, 17/2, 19/2, 21/2, 35/2) -(1/2, 5/2, 9/2, 11/2, 15/2, 19/2, 21/2, 35/2) -(1/2, 3/2, 11/2, 13/2, 15/2, 17/2, 21/2, 35/2) -(0, 1, 2, 5, 6, 11, 12, 17) -(-1, 2, 3, 4, 6, 11, 12, 17) -(0, 2, 3, 5, 7, 10, 12, 17) -(-1, 3, 4, 5, 6, 10, 12, 17) -(0, 1, 4, 5, 8, 9, 12, 17) -(-1, 2, 4, 6, 7, 9, 12, 17) -(1, 2, 4, 5, 8, 10, 11, 17) -(0, 3, 4, 6, 7, 10, 11, 17) -(0, 2, 5, 6, 8, 9, 11, 17) -(0, 1, 6, 7, 8, 9, 10, 17) -(-1/2, 3/2, 5/2, 9/2, 11/2, 23/2, 25/2, 33/2) -(1/2, 3/2, 5/2, 11/2, 13/2, 21/2, 25/2, 33/2) -(-1/2, 5/2, 7/2, 9/2, 13/2, 21/2, 25/2, 33/2) -(-1/2, 3/2, 7/2, 11/2, 15/2, 19/2, 25/2, 33/2) -(-3/2, 5/2, 9/2, 11/2, 13/2, 19/2, 25/2, 33/2) -(-3/2, 5/2, 7/2, 13/2, 15/2, 17/2, 25/2, 33/2) -(1/2, 5/2, 7/2, 11/2, 15/2, 21/2, 23/2, 33/2) -(-1/2, 7/2, 9/2, 11/2, 13/2, 21/2, 23/2, 33/2) -(1/2, 3/2, 9/2, 11/2, 17/2, 19/2, 23/2, 33/2) -(-1/2, 5/2, 9/2, 13/2, 15/2, 19/2, 23/2, 33/2) -(-1/2, 3/2, 11/2, 13/2, 17/2, 19/2, 21/2, 33/2) -(0, 1, 3, 4, 5, 12, 13, 16) -(0, 2, 3, 5, 6, 11, 13, 16) -(0, 1, 3, 6, 7, 10, 13, 16) -(-1, 2, 4, 5, 7, 10, 13, 16) -(-1, 2, 3, 6, 8, 9, 13, 16) -(-2, 3, 4, 6, 7, 9, 13, 16) -(1, 2, 3, 6, 7, 11, 12, 16) -(0, 3, 4, 5, 7, 11, 12, 16) -(0, 2, 4, 6, 8, 10, 12, 16) -(-1, 3, 5, 6, 7, 10, 12, 16) -(-1, 3, 4, 7, 8, 9, 12, 16) -(0, 1, 5, 6, 9, 10, 11, 16) -(-1, 2, 5, 7, 8, 10, 11, 16) -(1/2, 3/2, 5/2, 7/2, 9/2, 25/2, 27/2, 31/2) -(1/2, 3/2, 7/2, 9/2, 11/2, 23/2, 27/2, 31/2) -(-1/2, 3/2, 7/2, 11/2, 13/2, 21/2, 27/2, 31/2) -(-1/2, 3/2, 5/2, 13/2, 15/2, 19/2, 27/2, 31/2) -(-3/2, 5/2, 7/2, 11/2, 15/2, 19/2, 27/2, 31/2) -(-5/2, 7/2, 9/2, 11/2, 15/2, 17/2, 27/2, 31/2) -(1/2, 5/2, 7/2, 11/2, 13/2, 23/2, 25/2, 31/2) -(1/2, 3/2, 7/2, 13/2, 15/2, 21/2, 25/2, 31/2) -(-1/2, 5/2, 9/2, 11/2, 15/2, 21/2, 25/2, 31/2) -(-1/2, 5/2, 7/2, 13/2, 17/2, 19/2, 25/2, 31/2) -(-3/2, 7/2, 9/2, 13/2, 15/2, 19/2, 25/2, 31/2) -(-1/2, 3/2, 9/2, 13/2, 17/2, 21/2, 23/2, 31/2) -(-3/2, 5/2, 11/2, 13/2, 15/2, 21/2, 23/2, 31/2) -(-3/2, 5/2, 9/2, 15/2, 17/2, 19/2, 23/2, 31/2) -(0, 1, 2, 3, 4, 13, 14, 15) -(1, 2, 3, 4, 5, 12, 14, 15) -(0, 1, 4, 5, 6, 11, 14, 15) -(-1, 2, 3, 6, 7, 10, 14, 15) -(0, 1, 2, 7, 8, 9, 14, 15) -(-2, 3, 4, 5, 8, 9, 14, 15) -(-3, 4, 5, 6, 7, 8, 14, 15) -(1, 2, 4, 5, 6, 12, 13, 15) -(0, 2, 4, 6, 7, 11, 13, 15) -(0, 2, 3, 7, 8, 10, 13, 15) -(-1, 3, 4, 6, 8, 10, 13, 15) -(-2, 4, 5, 6, 8, 9, 13, 15) -(0, 1, 4, 7, 8, 11, 12, 15) -(-1, 2, 5, 6, 8, 11, 12, 15) -(-1, 2, 4, 7, 9, 10, 12, 15) -(-2, 3, 5, 7, 8, 10, 12, 15) -(-2, 3, 4, 8, 9, 10, 11, 15) -(3/2, 5/2, 7/2, 9/2, 11/2, 25/2, 27/2, 29/2) -(1/2, 3/2, 9/2, 11/2, 13/2, 23/2, 27/2, 29/2) -(-1/2, 5/2, 7/2, 13/2, 15/2, 21/2, 27/2, 29/2) -(1/2, 3/2, 5/2, 15/2, 17/2, 19/2, 27/2, 29/2) -(-3/2, 7/2, 9/2, 11/2, 17/2, 19/2, 27/2, 29/2) -(-5/2, 9/2, 11/2, 13/2, 15/2, 17/2, 27/2, 29/2) -(-1/2, 3/2, 9/2, 13/2, 15/2, 23/2, 25/2, 29/2) -(-1/2, 3/2, 7/2, 15/2, 17/2, 21/2, 25/2, 29/2) -(-3/2, 5/2, 9/2, 13/2, 17/2, 21/2, 25/2, 29/2) -(-5/2, 7/2, 11/2, 13/2, 17/2, 19/2, 25/2, 29/2) -(-3/2, 5/2, 7/2, 15/2, 19/2, 21/2, 23/2, 29/2) -(-5/2, 7/2, 9/2, 15/2, 17/2, 21/2, 23/2, 29/2) -(0, 1, 5, 6, 7, 12, 13, 14) -(-1, 2, 4, 7, 8, 11, 13, 14) -(0, 1, 3, 8, 9, 10, 13, 14) -(-2, 3, 5, 6, 9, 10, 13, 14) -(-3, 4, 6, 7, 8, 9, 13, 14) -(-1, 2, 3, 8, 9, 11, 12, 14) -(-2, 3, 4, 7, 9, 11, 12, 14) -(-3, 4, 5, 7, 9, 10, 12, 14) -(-3/2, 5/2, 7/2, 15/2, 17/2, 23/2, 25/2, 27/2) -(-1/2, 3/2, 5/2, 17/2, 19/2, 21/2, 25/2, 27/2) -(-5/2, 7/2, 9/2, 13/2, 19/2, 21/2, 25/2, 27/2) -(-7/2, 9/2, 11/2, 15/2, 17/2, 19/2, 25/2, 27/2) -(-7/2, 9/2, 11/2, 13/2, 19/2, 21/2, 23/2, 27/2) -(0, 1, 2, 9, 10, 11, 12, 13) -(-3, 4, 5, 6, 10, 11, 12, 13) -(-4, 5, 6, 7, 9, 10, 12, 13) -(-9/2, 11/2, 13/2, 15/2, 17/2, 21/2, 23/2, 25/2) -(-5, 6, 7, 8, 9, 10, 11, 12) - -(Again, any element of this list is defined only up to permutation of the coordinates and an even number of sign changes: here I've sorted the coordinates in absolute value and written the minus sign, if necessary, on the first coordinate, but the representatives in question might not be the best.) -I can see no clear pattern in this list. Maybe I'm looking at it in all the wrong way. -Question: How can we describe this set simply? -(A followup question might be whether we can easily multiply two elements of $W(E_8)$ represented as transformations on such Weyl vectors. But the first step is, of course, to recognize them.) -Remarks: - -The embedding of $W(D_8)$ into $W(E_8)$ comes from the fact that the root system $D_8$ is a closed subsystem of $E_8$: as such, it can be described by Borel-de Siebenthal theory: take the extended (i.e., affine) Dynkin diagram of $E_8$ and remove node number $1$ in Bourbaki's numbering of the node, this leaves us with the Dynkin diagram of $D_8$. But more concretely, in the chosen coordinate system, the root system of $D_8$ consists of those $112$ roots having shape $(\pm1,\pm1,0,0,0,0,0,0)$ (for some choice of the two signs and some permutation of the coordinates) inside that of $E_8$ which additionally consists of the $128$ roots having shape $(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2})$ (for some choice of the signs such that an even number are minus). -The motivation of the problem is to understand $W(E_8)$ better and see how its elements can be represented (also, now that I experimentally found a list of numbers, I'm naturally inclined to try to find patterns in it…). To perhaps better explain why I think this is a natural question, consider the analogous case of $A_n$ in the standard system of $n+1$ coordinates all integer with sum zero: the Weyl vector is $(0,1,2,\ldots,n)$ minus whatever constant is necessary to make it sum to zero (viz., $\frac{n-1}{2}$); its orbit under $W(A_n) \cong \mathfrak{S}_{n+1}$ consists of all permutations of $(0,1,2,\ldots,n)$ (minus constant), these are very easy to recognize, and it is fairly natural to represent an element $w$ of $\mathfrak{S}_{n+1}$ by the corresponding Weyl vector $\rho_w = w\cdot\rho_0$ (where $\rho_0$ is a fixed Weyl vector, say the one I just wrote); in fact, trying to compute $w$ (as product of Coxeter generators) from $\rho_w$ is essentially a sorting algorithm. The case of the other classical root systems is similarly easy; so I thought it natural to try to look at the exceptional root systems, and, of these, $E_8$ seems the most interesting because there is a coordinate system that is really pleasant (because of the relation with $D_8$). - -Edit (2019-06-17): To answer a question by André Henriques in the comments, the Sage code used to produce the list above is here (there are a few comments in the code explaining how it works). The code also produces a representation of the same list as the graph of the decomposition of the Weyl chamber of $W(D_8)$ as $W(E_8)$-cells (with adjacency being given by the $W(E_8)$-reflection hyperplanes, colored according to the nodes of the $E_8$ Dynkin diagram), and it looks like this: - -REPLY [2 votes]: Concerning the first question in the header (and some of your preparatory remarks), it's useful to keep in mind the Planche VII for $E_8$ at the end of Chapters 4-6 of Bourbaki's treatise Groupes et algebres de Lie (English translation, Springer). For example, (VII) gives explicit coordinates for $2\rho$ in terms of both the standard basis of $\mathbb{R}^8$ and the simple roots. There is a lot of useful information in these planches, though I wish they had used some of the blank space to fill in all of the positive roots arranged by height (as T.A. Springer does in his IHES paper for exceptional types). It would also be helpful to include a planche for the "exceptional" type $D_4$. -Concerning your other more specialized question about Weyl group orbits, it might help to give some motivation. Aside from that, it is probably not easy to compute these orbits or interpret the results in this case.<|endoftext|> -TITLE: Definition of asymptotic frequency -QUESTION [5 upvotes]: Consider a sequence $(x_n)_{n\in\mathbb{N}}$ of 0s and 1s. The asymptotic frequency of 1s in $x$ is usually defined as: -$$f=\lim_{n\to+\infty} \frac{1}{n}{\sum_{i=0}^{n-1} x_i}$$ -when this limit exists. But sometimes the limit does not exist yet it sounds "reasonable" to say that the asymptotic frequency still exists. Is there a general way to define asymptotic frequency so that it applies to a much broader class of sequences?. -For simplicity, we can focus only on defining $f=0$: how to define there are "infinitely more" 0s than 1s in the sequence. Maybe this could involve probability theory, maybe not. Of course the definition can't be invariant under any permutation since there is nothing to say in terms of cardinal except there are as many 1s as 0s: countably many. -The tricky example I have in mind is successive groups of length $2^k$. In each group the values are equal: all 0s or all 1s. There are infinitely many groups with value 1 but these become more and more sparse for example with frequency $1/k$. It looks like it: - -1-00-1111-0000000- ....... groups of length $2^k$: most are 0, with - sometimes a group of 1s. - -If you compress visually each group into a single digit, it looks like: - -101000010000000000000100000000000000000000000000000000000000000000001.... - -The limit does not exist as the average is greater than 1/2 infinitely many times. Yet, this example can be thought as the realization of a random sequence $(X_n)_{n\in\mathbb{N}}$ that tends to 0 in probability. Hence it could make sense saying the asymptotic frequency is 0. - -REPLY [2 votes]: $\newcommand{\ep}{\epsilon} -\newcommand{\ga}{\gamma} -\newcommand{\la}{\lambda} -\newcommand{\Si}{\Sigma} -\newcommand{\R}{\mathbb{R}} -\newcommand{\E}{\operatorname{\mathsf E}} -\newcommand{\PP}{\operatorname{\mathsf P}}$ -I don't think such a "reasonable" much broader definition is possible. The reason is that the relative frequency $f_n:=s_n/n$ (where $s_n:=\sum_1^n x_i$) varies very slowly in $n$ when $n$ is large. So, it will take $f_n$ a very long time to significantly depart from whatever value it has attained. -To express this thought rigorously, let us now indeed "focus only on defining $f=0$", as you suggested; for any other value of $f$, we can reason quite similarly. Take some $\ep\in(0,1/2)$ and suppose that $f_M\le\ep$ and $f_N\ge2\ep$ for some natural $M$ and $N>M$; this situation would "reasonably" occur if a "generalized limit frequency" is $0$, but $f_n\not\to0$. Then $s_M\le M\ep$, $s_N\ge2N\ep$, and hence -\begin{equation} - N-M\ge s_N-s_M\ge(2N-M)\ep, -\end{equation} -which implies $N\ge\frac{1-\ep}{1-2\ep}\,M$, and so, -\begin{equation} - s_N-s_M\ge\frac\ep{1-2\ep}\,M. -\end{equation} -Note that $s_N-s_M$ is the number of $1$'s in the sequence $(x_n)$ for natural $n\in(M,N]$. -Thus, this number of $1$'s -- after the time $M$ (at which $f_M$ was $\ep$-close to $0$) -- needed to get a relative frequency at least $2\ep$ is at least proportional to the time $M$. -So, for the $f_n$'s, it is impossible get a picture like the one below, where the intervals of significantly non-zero values of $f_n$ become (relatively) negligibly short in time. Therefore, I don't think one can reasonably say that such a sequence has a negligible amount of $1$'s. - -For such a picture to be possible, $f_n$ would need to grow arbitrarily fast over some intervals -- but it cannot do that, because $x_n\in\{0,1\}$ for all $n$ and hence -$$|f_{n+1}-f_n|=\Big|\frac{nf_n+x_{n+1}}{n+1}-f_n\Big|=\Big|\frac{x_{n+1}-f_n}{n+1}\Big| \le\frac1{n+1}.$$<|endoftext|> -TITLE: Are the rationals definable in any number field? -QUESTION [16 upvotes]: Let $K$ be a number field. Is it necessarily true that $\mathbb{Q}$ is a first-order definable subset of $K$? Equivalently (since in any number field, its ring of integers is a definable subset), is $\mathbb{Z}$ necessarily a definable subset of $K$, or of $\cal{O}_K$? -Edit: And if not, is $\mathbb{Q}$ always interpretable in $K$? - -REPLY [12 votes]: According to R. S. Rumely, Undecidability and Definability for the theory of global fields, AMS Trans., 262, pp. 195-217, prime subfield is always definable in global field, and in number case, you can define $\Bbb N$.<|endoftext|> -TITLE: Which edition of Philosophiae Naturalis Principia Mathematica of Isaac Newton would you recommend to me? -QUESTION [14 upvotes]: I'm searching for a good edition of Philosophiae Naturalis Principia Mathematica of Isaac Newton in English. Which edition of the Principia can you suggest me? If it's possible, cheap and similar to original with the original proofs of Isaac Newton. -I'm sorry for my poor English, I'm just learning it. - -REPLY [4 votes]: This 2003 edition by Densmore & Donahue has the following book review - -Makes the great adventure of Principia available not only to modern - scholars of history of science, but also to nonspecialist - undergraduate students of humanities. It moves carefully from Newton's - definitions and axioms through the essential propositions, as Newton - himself identified them, to the establishment of universal gravitation - and elliptical orbits. The guidebook unfolds what is implicit in - Newton's words as he himself would have filled in the steps and - completes the argument in ways that are authentic and not - anachronistic, exactly following Newton's thinking rather than - substituting tools of modern calculus or the formulations of modern - physics. It is Newton in his own terms. This is a wonderful book. - ―Richard S. Westfall<|endoftext|> -TITLE: The "Johnson polychora" -QUESTION [9 upvotes]: Firstly, a definition: - -A convex polyhedron, whose faces are regular polygons (2D polytopes). - -This includes the 92 Johnson solids, 13 Archimedean solids, 5 Platonic solids and two infinite familes - prisms and antiprisms. -However, I see two ways of extending this definition to 4D: - -1) A convex polychoron (4-polytope), whose 2D faces are regular polygons, ie. its 3D faces satisfy the first definiton. -2) A convex polychoron, whose 3D faces are regular polyhedra (3D polytopes). - -The list I gave above is known to be complete. But is there (at least partial) progress on the 4D case? The second definition is much more restrictive, so maybe a complete classification of polychora, which satisfy 2) exists. Can you point me to some reference, please? - -REPLY [4 votes]: Even so the above cited webpages indeed are being hosted by Anton Sherwood, everything what belongs under https://bendwavy.org/klitzing/home.htm is my own content, he just provides webspace to me. -The set of convex polytopes, where their facets are bound to be regular, indeed are the polytopes already investigated by Gerd and esp. his wife Roswitha Blind, cf. https://bendwavy.org/klitzing/explain/johnson.htm#blind. - What they proved could be enlisted nearly fully as this $complete$ listing - -the bipyramid of the tetrahedron, and similarily the bipyramid on every higher dimensional regular simplex -the pyramid on the octahedron, and similarily the pyramid on every higher dimensional regular crosspolytope -the pyramid on the icosahedron, and the bipyramid on the icosahedron -the adjoin of an octahedral pyramid and the rectified 5-cell (aka apiculated rectified 5-cell) -millions of more or less symmetrical edge-facetings of the 600-cell (using tetrahedra and icosahedra for cells), e.g. the snub 24-cell is one of those. - -The set of convex polytopes, where only its polygonal faces are bound to be regular, are known as CRF (convex, regular faced polytopes). This set is much vaster and by no means fully classified. Even so under the restriction to 4D cases only, a large list of examples is being provided on my wbsite, cf. https://bendwavy.org/klitzing/explain/johnson.htm#crf. (Directly above the table even a downloadable spreadsheet with the full so far known listing is available.) -The picture displayed above not only is my proprietary, it moreover displays the first example historically being known what later became known as the set of scalifomrs, cf. https://bendwavy.org/klitzing/explain/scaliform.htm (symmetry acts transitively on the vertex set, edges are all of the same size, polygonal faces are regular). The displayed picture e.g. has for facets a total of 2 truncated tetrahedra, 8 trigonal cupola, and 6 tetrahedra (the latters acting as digonal antiprisms). - ---- rk<|endoftext|> -TITLE: Two smooth tangent almost complex curves in a $4$-manifold -QUESTION [10 upvotes]: I would like to know if following is correct. -Statement. Suppose we have a smooth (i.e., $C^\infty$) almost complex structure on $\mathbb R^4$ and $C_1, C_2$ are two $J$-holomorphic curves passing through $(0,0)$, tangent at $(0,0)$ and regular at $(0,0)$. Then there exist $C^{\infty}$ smooth complex coordinates $(z,w)$ such that $C_1$ is locally given by $w=0$ and $C_2$ by $w=z^n$. -PS. I think it would be enough for me to know that $C_2$ can be given by $w=z^n+O|z^{n+1}|$. However everything must be $C^{\infty}$ - the coordinates, and the $O|z^{n+1}|$ term. If, on the other hand one can not expect to have a $C^{\infty}$ diffeo, what is the best one can expect? - -REPLY [2 votes]: It turns out that it is nice to read books. The answer to the weaker version of my question with $O|z|^{n+1}$ term is contained on page 17 of McDuff-Salamon book [MS] (no need of Micallef-White!): -https://people.math.ethz.ch/~salamon/PREPRINTS/jholsm.pdf -Proof. In the proof of Lemma 2.2.3 of [MS] one uses coordinates in $\mathbb C^2$ such that $C_1$ is given by $w=0$ and the almost complex structure $J$ along the line $(z,0)$ is the standard one. Then it is explained that the almost complex map $z\to \mathbb C^2$ corresponding to $C_2$ is given by -$$z\to (p(z)+O(|z^{n+1}|), az^n+O(|z^{n+1}|))$$ -where $p(z)$ is a polynomial of order at most $n$, $a\ne 0$. In our case of course $p'(0)\ne 0$. It is now clear that in these coordinates $C_2$ is as need. QED. -Comment. The above proof is elementary and does not use Micallef-White. Similarly to Micallef-White's, statement it can be used to answer the original question with a $C^1$-smooth change of coordinates (instead of $C^{\infty}$). Indeed, after a smooth reparameterization in $z$ and scaling in $w$ the above map for $C_2$ looks as -$$z\to (z, z^n+O(|z^{n+1}|)).$$ -Denote the second term by $f(z)$. Then the map -$(z,w)\to (z,w-(f(z)-z^n)/z^n)$ is $C^1$ and it sends the couple $C_1,C_2$ to the couple $(w=0, w=z^n)$. -I wonder still if one can make this last change of coordinates $C^{\infty}$...<|endoftext|> -TITLE: Why do these two Monster-related calculations yield $163$? -QUESTION [17 upvotes]: Fact 1: (1979, Conway and Norton)$^{1}$ -"There are $194-22-9=\color{blue}{163\,}$ $\mathbb{Z}$-independent McKay-Thompson series for the Monster." -Note: There are 194 (linear) irreducible representations of $\mathbb{M}$, hence 194 conjugacy classes. Of these, there are 22 that are complex quadratic valued. Of the remaining, there are 9 linear dependencies.$^{2}$ - -Fact 2: (2004, C. Cummins)$^{3}$ -"The genus 0 moonshine groups have $132 + 1 + 4 + 5 + 13 + 1 + 7=\color{blue}{163}$ equivalence classes with period 1." -Note: There are exactly $6486$ genus $0$ moonshine groups, but only $371$ equivalence classes. Of these, $310\,$ have a rational representative. (Curiously, $\color{red}{310} =163+67+43+19+11+7$.) -\begin{array}{|c|r|c|c|c|c|c|c|c|c|c|c|} -\hline -\text{Period}& & 15^* & 11^*& 7^*& 3^*& 2^*& i& i\,2& 2 & 5 &\color{brown}{\text{Total}}\\ -\hline -1&132& 0& 1& 4& 5& 0& 13& 1& 0& 7 & \color{blue}{163}\\ -2&120& 0& 0& 2& 1& 7& 4& 2& 2& 2& 140\\ -3& 26& 1& 0& 1& 4& 0& 2& 0& 0& 0& 34\\ -4& 16& 0& 0& 0& 0& 2& 0& 0& 0& 0& 18\\ -6& 10& 0& 0& 0& 0& 0& 0& 0& 0& 0& 10\\ -8& 3& 0& 0& 0& 0& 0& 0& 0& 0& 0& 3\\ -12& 2& 0& 0& 0& 0& 0& 0& 0& 0& 0& 2\\ -24& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\\ -\hline -\color{brown}{\text{Total}}&\color{red}{310}& 1& 1& 7& 10& 9& 19& 3& 2& 9& \color{brown}{371}\\ -\hline -\end{array} - -Question: While the value $163$ was calculated differently in Facts 1 and 2, are they just different ways of saying the same thing, and that one implies the other? - -References: - -Monstrous Moonshine, by J.H. Conway & S.P. Norton. -Sporadic and Exceptional, by Yang-Hui He & John McKay. -Congruence Subgroups of Groups Commensurable with PSL (2, Z) of Genus 0 and 1, by C. J. Cummins. -Tables by C. J. Cummins. - -REPLY [8 votes]: If there is a connection, I would expect it to be related to the fact that the $j$ invariant is connected to both Moonshine and to class field theory. Note that 7, 11, 19, 43, 67, and 163 are all the absolute values of odd discriminants with class number 1. (Strangely, 3 is missing from the list! Otherwise the list is complete.)<|endoftext|> -TITLE: When is the image of a regular map an algebraic variety? -QUESTION [5 upvotes]: Let $f:X\rightarrow Y$ be a regular map of smooth connected algebraic varieties (say over an algebraically closed field). I know that the image $f(X)$ is only a constructible set, in general, but I am interested in conditions that ensure $f(X)$ being an algebraic variety. -A precise question: suppose the differential $df$ has a constant rank. Is $f(X)$ an algebraic variety (or a locally closed subvariety of $Y$)? - -REPLY [6 votes]: I am correcting the first sentence of the comment. -Even if the rank of $df$ is constant, the image may be only constructible. Let $X$ be the complement of the $s$-axis in the $(s,t)$-affine plane, $X=\{(s,t): t\neq 0\}.$ Let $Y$ be the affine $3$-space with coordinates $(u,v,w).$ Let $f$ be the function $f(s,t)=(s^2-1,s(s^2-1),s+t)$. The image of $f$ is the disjoint union of two locally closed subsets whose union is not locally closed. The first locally closed set is $\text{Zero}(v^2-u^2(u+1))\setminus \text{Zero}(u-(w^2-1),v-w(w^2-1))$. The second locally closed set is $\text{Zero}(u,v)$. The union of these two locally closed sets is not locally closed. It is only constructible.<|endoftext|> -TITLE: $\infty$-operads and $E_\infty$-algebras -QUESTION [15 upvotes]: I work in algebraic geometry. Lately, the answer to most of my questions seems to be "you should read Lurie's Higher Algebra." I took this advice seriously, however it turned out not to be an easy task. -From what I gather, I should learn how to work with $E_\infty$-rings. I understand the definition of a symmetric monoidal $\infty$-category as a certain coCartesian fibration $p\colon C^\otimes \to N({\rm Fin}_*)$, and that an $E_\infty$-ring in a stable symmetric monoidal $\infty$-category $C$ should be a certain section of $p$. -In Higher Algebra, $E_\infty$-algebras are a special case of algebras over $\infty$-operads, where the operad ${\rm Comm}$ is just the identity $N({\rm Fin}_*)\to N({\rm Fin}_*)$. Lurie treats general $\infty$-operads in detail in two chapters of about 150 pages each, and in later chapters he specializes to the case of the operads $E_k$. Since I seem to be interested in the final operad ${\rm Comm}$ only, it is difficult for me to motivate myself to read through the chapters about the general theory of $\infty$-operads. - -Question. As an algebraic geometer interested in making the step from commutative rings to $E_\infty$-rings, should I care about $E_k$-rings for $k<\infty$, or algebras over more general $\infty$-operads? If so, why? - -REPLY [11 votes]: Here are two examples where $E_k$-algebras show up in algebraic geometry, for $1 -TITLE: Classification of oriented vector bundles of rank 5 over closed oriented 5-manifolds -QUESTION [10 upvotes]: I am looking for a complete classification in terms of characteristic classes and "computable" (preferably geometric) invariants. There is this work where the authors classify oriented vector bundles for manifolds with $w_2(M)\neq0$ and a condition on $H^4(M;\mathbb Z)$. But this is a purely homotopy theoretic approach which does not take into account, that the underlying space is a manifold. -Is there anything else in the literature about this topic? -Edit: Found something. - -REPLY [4 votes]: I don't know how to answer this question completely, but I'll make some observations. -The oriented vector bundles over $M$ are classified by homotopy classes of maps $f:M\to \tilde{G}_5$, the oriented Grassmannian, denoted $\pi_0 ([M, \tilde{G}_5])$, where $[X,Y]$ denotes the space of mappings from $X$ to $Y$. Let $E\to M$ be a 5-dimensional vector bundle over $M$. Then the Euler class $e(E)=0$, and hence there is a global section. So $E=F\oplus \varepsilon_1$, where $\varepsilon_1=M\times \mathbb{R}$ is the trivial bundle, and $F\to M$ is a 4-dimensional oriented vector bundle. Hence $\pi_0 [M,\tilde{G}_4]\to \pi_0 [M,\tilde{G}_5]$ is onto. I'm not sure if this helps, but it is related to the observation made in the paper that you linked to that in the case of spin 5-manifolds, this bundle is quaternionic. -A second observation is that $M$ admits a cell structure with one $5$-cell. The 4-skeleton $M^4\subset M$ is a cofibration (note that up to homotopy, $M^4 \simeq M - pt.$, so is well defined up to homotopy). Hence the map $[M, \tilde{G}_5]\to [M^4,\tilde{G}_5]$ is a fibration, with fiber $[(S^5,\ast),(\tilde{G}_5,\ast)]$. The fiber has two components, in bijection with the components of $[S^5,\tilde{G}_5]$ from the homotopy exact sequence and the fact that $\tilde{G}_5$ is simply-connected for the fibration $[(S^5,\ast),(\tilde{G}_5,\ast)]\to [S^5,\tilde{G}_5]\to \tilde{G}_5$. -A third observation is that by cellular approximation, $\pi_0[M^4,\tilde{G}_5] = \pi_0[M^4, \tilde{G}_5^5]$, where $\tilde{G}_5^5$ is the 5-skeleton. This is because any map $f:M^4 \to \tilde{G}_5$ may be homotoped to $f: M^4\to \tilde{G}_5^4$. Moreover, homotopies between such maps may be homotoped to $\tilde{G}_5^5$, since $M\times [0,1]$ is 5-dimensional. -A fourth observation is that $\tilde{G}_5^5 \cong \tilde{G}_\infty^5$, where $\tilde{G}_\infty$ is the infinite oriented Grassmannian obtained as the nested union $\tilde{G}_1\subset \tilde{G}_2 \subset \cdots$, where each inclusion is induced by stabilization. This can be proved using the cell structure on Grassmannians given by Schubert cells, but I suspect there may be a more direct approach. -Then $\pi_0 [M^4,\tilde{G}_5]=\pi_0[M^4,\tilde{G}_5^5]=\pi_0[M^4,\tilde{G}_\infty^5] = \pi_0[M^4,\tilde{G}_\infty] =\pi_0 [M^4,BSO]\leq \tilde{KO}(M^4)$, the real reduced K-theory of $M^4$. The image of $\pi_0 [M^4,\tilde{G}_\infty] \to \pi_0[M^4,G_\infty]=\tilde{KO}(M^4)$ should correspond to the components where $im \{\pi_1(M^4)\to \pi_1(G_\infty)\}=0$, where $\pi_1(G_\infty)=\mathbb{Z}/2$. In principle, this should be computable from the Atiyah-Hirzebruch spectral sequence and the cohomology of $M^4$, which is determined by the cohomology of $M$. -A fifth observation is that with the fibration $[(S^5,\ast),(\tilde{G}_5,\ast)]\to [M, \tilde{G}_5] \to [M^4,\tilde{G}_5]$, -the components in the image will correspond to maps where $im\{\pi_0 [M^4,\tilde{G}_5]\to \pi_0 [S^4,\tilde{G}_5]\} =0$, where $S^5\subset M^4$ is the attaching map of $\partial B^5$, the 5-handle. Again, this should be determined by the map $\tilde{KO}(M^4) \to \tilde{KO}(S^4)$. -Now for the above fibration, the fiber has two components, so the number of components should be at most twice the number of components of the base (determined by the above K-theory computations). I'm not sure how to determine the image $\pi_1 [M^4,\tilde{G}_5]\to \pi_0 [(S^5,\ast),(\tilde{G}_5,\ast)]$. -The upshot is that for each 5-dimensional vector bundle over $M$, one may modify it by "inserting" a copy of the tangent bundle to $S^5$ (this can be descibed in terms of clutching functions of the attaching map of the 5-cell). Then the above question reduces to whether this changes the vector bundle up to isomorphism or keeps it the same? I think the above description classifies them up to this operation.<|endoftext|> -TITLE: When does the enveloping algebra functor lift to the category of bialgebras? -QUESTION [7 upvotes]: Let $\mathrm{Ass} $ denote the operad, whose algebras are associative unital algebras, considered as a dg-operad. -Denote $\mathrm{Ch} $ the category of chain complexes over a commutative ring $\mathrm{R} $ and denote $\mathrm{Bialg}_{ } (\mathrm{Ch} )$ the category of cocommutative (counital and unital) bialgebras in $\mathrm{Ch}. $ -Let $\mathcal{O}$ be a dg-operad and $\phi: \mathcal{O} \to \mathrm{Ass} $ a map of dg-operads. $\phi$ induces a free-forgetful adjunction $ \mathcal{U} : \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch} ) \rightleftarrows \mathrm{Alg}_{ \mathrm{Ass} } ( \mathrm{Ch} ).$ -When does the left adjoint $ \mathcal{U} : \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch}) \to \mathrm{Alg}_{ \mathrm{Ass} } (\mathrm{Ch} ) $ -lift to a functor $ \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch} ) \to \mathrm{Bialg}_{ } (\mathrm{Ch} )? $ -For $\mathcal{O}$ the Lie operad there is such a lift. -For $\mathcal{O}$ the Lie operad the left adjoint $\mathcal{U}$ is the enveloping algebra. For every Lie-algebra $\mathrm{X}$ the bialgebra structure on $\mathcal{U}(\mathrm{X})$ is encoded in a symmetric monoidal structure on the category $\mathrm{LMod}_{ \mathcal{U}(\mathrm{X}) } (\mathrm{Ch} )$ (being the category of representations of $\mathrm{X}$) lifting the symmetric monoidal structure on $\mathrm{Ch} .$ -The category of representations of $\mathrm{X}$ can also be described as the category $\mathrm{Mod}^{\mathcal{O}}_{ \mathrm{X} } (\mathrm{Ch} ) $ -of $\mathrm{X}$-modules over the Lie-operad. -This leads to my 2. question: For which dg-operads $ \mathcal{O}$ and -$ \mathcal{O}$-algebras $\mathrm{X}$ does the symmetric monoidal structure on $\mathrm{Ch} $ lift to a symmetric monoidal structure on $\mathrm{Mod}^{\mathcal{O}}_{ \mathrm{X} } (\mathrm{Ch} ) ?$ -More generally given a dg-operad $\mathcal{O}$ and a $\mathcal{O}$-algebra $\mathrm{X}$ there is a enveloping associative algebra $\mathcal{U}_{ \mathcal{O}}( \mathrm{X} ) $ with the property that we have a canonical equivalence -$$\mathrm{LMod}_{\mathcal{U}_{ \mathcal{O}}( \mathrm{X} ) } (\mathrm{Ch} ) \simeq \mathrm{Mod}^{\mathcal{O}}_{ \mathrm{X} } (\mathrm{Ch} ) .$$ -As a cocommutative bialgebra structure on an associative algebra corresponds to a symmetric monoidal structure on its category of left modules lifting the symmetric monoidal structure on $ \mathrm{Ch} $, my 2. question is equivalent to the following: -For which dg-operads $ \mathcal{O}$ does the enveloping associative algebra functor $ \mathcal{U} : \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch}) \to \mathrm{Alg}_{ \mathrm{Ass} } (\mathrm{Ch} ) $ -lift to a functor $ \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch} ) \to \mathrm{Bialg}_{ } (\mathrm{Ch} )? $ - -REPLY [2 votes]: I was going to post this as a comment but rambled on too long. I haven't check the details of the following. -To every operad $O$ there's an algebra $A(O)$ in the category of right $O$-modules created by marking one of the inputs as special. The algebra structure is given by composing two elements of $A(O)$ along this special input and the right $O$-module structure given by composing with an element of $O$ in any non-special input. If this algebra is a bialgebra in the category of right $O$-modules and furthermore $A(O)$ is a free right $O$-module then I believe the enveloping algebra functor extends to the category of bialgebras. This is because $A(O)\circ_O\mathfrak{g}$ is the enveloping algebra of an $O$-algebra $\mathfrak{g}$. -The reason for asking that $A(O)$ is free is so that the functor $(-)\circ_O \mathfrak{g}$ from right O-modules to vector spaces is monoidal. However on second thoughts the functor might be monoidal without this condition. I would however be (pleasantly) surprised if you could find an operad $O$ whose algebra $A(O)$ was a bialgebra and not free as a right $O$-module. -Having said that I'm now doubting that $A(Lie)$ is free as a right $Lie$-module if we work over the integers. It is free if you work over rationals or with shuffle collections, but perhaps not with symmetric collections if you can't make the symmetric group actions respect the generators.<|endoftext|> -TITLE: A condition for a dga to be minimal -QUESTION [5 upvotes]: I'm reading a book "Complex Geometry" by Daniel Huybrechts. In this book he says that a simply connected dga satisfying some conditions must be minimal. (p.147, Remark 3.A.13) I tried to prove this statement but eventually failed. It seems that there is a counter example. -The statement in the book is as follows: - -If $ (M=\bigoplus _{i\ge0} M^i ,d) $ is a simply connected dga over a field $ k $ satisfying -1) $ M^0 =k $ -2) $ M^+ := \bigoplus _{i>0} M^i $ is free, i.e. there exist homogeneous elements $ x_1 , x_2 , ... $ of positive degree $d_1 , d_2 , ...$, such that $M^+ =S^+ $ (the positive part of the (super)symmetric algebra of the vector space generated by $x_1 , x_2 , ...$. -and such that $d(M)\subset M^+ \cdot M^+$, then $M$ is minimal. In fact, 1) and 2) suffice to deduce $M^1 =0$ as in the proof of the previous lemma. - -And the counterexample I found is as follows: - -Let $M=S^*$ with $x, y$ and $z$ homogeneous elements of degree $1$. -Set $dx=yz, dy=zx$ and $dz=xy$. -Using Leibniz rule and $x^2 = y^2 =z^2 = 0$ by graded commutativity, we can show that $d$ satisfies $d^2 = 0$. -e.g. $d(dx)=d(yz)=(dy)z-ydz=zxz-yxy=-xzz+xyy=0$ -Moreover, since $d(ax+by+cz)=ayz+bzx+cxy=0$ implies $a=b=c=0$, we also know that $H^1(M,d)=0$, i.e. $(M,d)$ is simply connected. -Therefore, $(M,d)$ satisfies all assumptions in the statement but fails to have $M^1 =0$. -Moreover, $(M,d)$ is not minimal as $dx, dy, dz \ne0$ - -Question : Is this indeed a counterexample and the statement is wrong? Or is there something I am confused or missing? -And then what is the importance of this remark? It seems that this is not used in the rest of the book. - -REPLY [6 votes]: You are correct and the statement, as you cite it, is wrong. The counterexample you describe is well-known, and is usually cited to demonstrate precisely this failure.<|endoftext|> -TITLE: Why only $\bar\partial$ but not $\partial$ in Dolbeault cohomology -QUESTION [15 upvotes]: While I learn about $\partial$ and $\bar{\partial}$ operators, I had some questions about the reason why people prefer $\bar\partial$ over $\partial$. Specifically, - -When defining Dolbeault cohomology, one uses $\bar{\partial}$ but not $\partial$. I wonder whether there happens any problem if one define a cohomology by $\partial$. Or is this because it isn't interesting? -Any textbook on complex variables say only about $\bar{\partial}$-Poincare lemma. Is there a version with $\partial$? If not, where the difference between two operators fundamentally comes? -For a holomorphic vector bundle $E$, we define the operator $\bar{\partial}_E$ only. Again why don't we define $\partial_E$? -Let $(E,h)$ be an hermitian holomorphic vector bundle on a compact hermitian manifold $(X,g)$. When we show that the operator $\bar\partial_E^*:=-\bar*_{E^*}\circ\bar\partial_{E^*}\circ\bar*_E$ on $A^{p,q}(X,E)$ is adjoint to $\bar\partial_E$, one uses $$\int_X\bar\partial(\alpha\wedge\bar*_E\beta)=\int_X d(\alpha\wedge\bar*_E\beta)$$ for $\alpha\in A^{p,q}(X,E)$ and $\beta\in A^{p,q+1}(X,E)$(c.f. "complex geometry" by Huybrechts, p.170). Here how we know $$\int_X\partial(\alpha\wedge\bar*_E\beta)=0$$? Again, the two operators appear to have different rules. - -REPLY [3 votes]: If $X$ is a complex manifold and $E\to X$ is a holomorphic vector bundle, only $\bar\partial_E$ can be defined naturally, i.e., it depends only on the complex structures of $X$ and $E$. The $\partial_E$ operator cannot be defined intrinsically. -If $E$ has a flat connection, you can also define the $\partial$-operator. This is the case when $E$ is trivial, and then the $\partial$ operator you mention is the one we all know. The flat connection is $d$. -In general, the operators $\bar\partial$ and $\partial$ have different roles, even if $\partial$ is well-defined. -To answer your questions: - -$\partial$ is not defined, therefore it cannot be used to define some cohomology. If you have a holomorphic vector bundle over $X$, then the $\bar\partial$ complex give you some cohomology which is isomorphic to the Cech cohomology, and this a very deep result. -Sure, the $\bar\partial$-Poincare lemma is a local statement, and it can be conjugated to get a statement about the $\partial$ operator. No problem here. -Because $\partial_E$ cannot be defined naturally. You need some other conditions (like flatness or a Hermitian metric) in order to define the $\partial$ operator. - -"Again why don't we define $\partial_E$?" Because it doesn't exist. - -You already have some metric since you can define the star operator. The short answer is "for bidegree reasons"<|endoftext|> -TITLE: Does knight behave like a king in his infinite odyssey? -QUESTION [46 upvotes]: The Knight's Tour is a well-known mathematical chess problem. There is an extensive amount of research concerning this question in two/higher dimensional finite boards. Here, I would like to tackle this problem on the unbounded infinite board, $\mathbb{Z}\times\mathbb{Z}$. -The first issue is how to construct a knight's tour on the infinite plane. An intuitive way is to build it locally by mimicking the king's spiral tour in $\mathbb{Z}\times\mathbb{Z}$ board as follows: - -Roughly speaking, a knight may follow the king's pattern by completing a knight's tour on an $n\times n$ block and then moving to the next block in the spiral order. For instance, in the below picture each square represents a certain $n\times n$ part of the $\mathbb{Z}\times\mathbb{Z}$ plane and the blue path is a knight tour performed in that particular block. It is connected to a point in the neighbor blocks with a knight move. - -Of course, the existence of such a nice natural number $n$ should be checked. However, it sounds reasonable to expect that for a sufficiently large $n$ (which there are so many open knight tours with various beginning and ending points) there are some tours which can serve as the building blocks of the knight's king-like spiral. (Actually only a few straight and right-oriented $n\times n$ tours with appropriate beginning and ending points are needed. The other blocks will be redundant up to isomorphism). -Here, the first question arises: - -Question 1. Is there (a concrete example of) a knight's tour on the infinite plane? -In the special case, does the already described strategy for building a knight's tour locally actually work? If yes, what is the minimum required $n$? - -Assuming an affirmative answer to the above question, the next step is to speculate on the general structure of all possible knight tours on the infinite plane: - -Question 2. Do all infinite knight's tours on the infinite plane arise as a combination of local finite knight's tours on a grid of blocks of the same size (whose local knight's tours are attached to their neighbor counterparts in a king's move fashion)? -Precisely, is it true that for any given knight's tour $t$ on the infinite plane $\mathbb{Z}\times\mathbb{Z}$, there is a (probably large) natural number $n$ and a partition $p$ of the plane into $n\times n$ blocks in a grid such that the restriction of $t$ into any block in $p$ is an $n\times n$ knight's tour in that particular block itself? We call the partition $p$ a localization of the knight's tour $t$ into a grid of $n\times n$ blocks. -If no, what is an example of an infinite knight's tour which can't be seen as a combination of local knight's tours in a grid of finite blocks of the same size (no matter what the resolution of the gird is)? -If yes, what is: -$$min\{n_t|~t~\text{is a knight's tour of $\mathbb{Z}\times\mathbb{Z}$ & there is a localization of}~t~\text{into a grid of}~n\times n~\text{blocks}\}$$ -In better words, what is the finest possible resolution of a grid for a knight's tour of the infinite plane, $\mathbb{Z}\times\mathbb{Z}$? - -As a remark it is worth mentioning that not all partitions of the infinite plane into $n\times n$ blocks are in the form of a perfect grid. For example, see the left picture in the following diagram which is a partition of the plane via $n\times n$ blocks different from a grid of $n\times n$ blocks (right): - -So, for the sake of completeness, one may reduce the grid partition condition in question 2 to merely a partition of the infinite plane into $n\times n$ blocks: - -Question 3. What is the answer to question 2 if we merely consider partitions of the infinite plane into $n\times n$ blocks, rather than those partitions which arrange blocks in a perfect grid? - -The question could be asked in a more general sense as well: - -Question 4. Is the knight's tour on the infinite plane always decomposable into finite knight's tours? Precisely, is it true that for every knight's tour $t$ on the infinite plane there is a partition $p$ of the plane into finite squares (not necessarily consisting of blocks in a grid or of the same size) such that the restriction of $t$ to any block in $p$ is a finite knight's tour as well? - - -Update. Thanks to Joel's answer and Eric's counterexample, the questions 1 and 2-4 have been answered positively and negatively respectively. Thus, we know that there are open knight's tours of the unbounded infinite plane, $\mathbb{Z}\times\mathbb{Z}$, of different nature, namely those which are formed of local knight's tours of finite planes and those total tours which don't arise this way. In this sense the answer to the question in the title is: "Sometimes but not always!" -Remark. As an additional piece of information, it is worth mentioning that there are some simple infinite sub-spaces of $\mathbb{Z}\times\mathbb{Z}$-plane which don't admit a knight's tour while some others do. For example, it is easy to see that one can obtain a knight's tour for $\mathbb{N}\times\mathbb{N}$ sub-plane through a local block by block construction via Joel's $5\times 5$ blocks introduced in his answer. However, it is intuitively clear that the infinite strip (i.e. $[-n,+n]\times\mathbb{Z}$ for some $n$) doesn't admit a knight's tour because the knight in such a sub-space must cover both infinite sides of the strip in a back and forth movement and so it should pass the central part of the strip infinitely many times which is impossible simply because after finitely many passages there will be no empty room left in the central region. Anyway, it sounds an interesting question to classify all infinite sub-spaces of $\mathbb{Z}\times\mathbb{Z}$-plane which admit a knight's tour. The same has already been done for all finite rectangle-shape sub-spaces of $\mathbb{Z}\times\mathbb{Z}$-plane, namely $m\times n$-boards. - -REPLY [3 votes]: I've pictured the start of a more-or-less random-5-by-5 Hamkins-variant infinite Knight's tour construction. It begins at the center and proceeds to the SE corner of the 25-by-25 board.<|endoftext|> -TITLE: Difference of adjacent dominant weights is a root? -QUESTION [5 upvotes]: The basic set-up here makes sense in the theory of abstract root systems if one brings (integral) weights into the picture, but it may be more natural to think about the classical characteristic 0 theory of finite-dimensional representations of a semisimple Lie algebra. (This question was recently raised by a colleague, who had observed that it was true in type $A$ and was tempted to go on in case-by-case fashion.) -First fix a simple system of roots in an irreducible root system, hence notions of positive roots and dominant weights. Recall the usual partial ordering of weights: $\mu \leq \lambda$ means that $\lambda -\mu$ is a sum of positive roots. There is some evidence that the following statement is always true: -$(*)$ Suppose $\lambda, \mu$ are dominant weights. If $\lambda > \mu$ and there is no intermediate dominant weight, then $\lambda - \mu$ is a single root (not necessarily simple, of course). - -Is $(*)$ true for all root systems, and if so is there a reference? - -[UPDATE] Jantzen's recent short paper here has an affirmative case-by-case answer in 2.1-2.2. Although his paper deals with modular representations of semisimple algebraic groups, this part of the proof only concerns root-and-weight computations. (However, he does rely on the classification of irreducible root systems, so it's still a question whether there is a uniform proof.) -P.S. The argument due to Steinberg in Stembridge's 1998 paper seems to answer this last question, as pointed out by Sam Hopkins. - -REPLY [4 votes]: A more precise description of positive roots $\gamma$ such that $\mu=\lambda-\gamma$ can be found in the paper -http://iopscience.iop.org/article/10.1070/SM1988v061n01ABEH003200/meta -See, in particular, Proposition 1 and Lemma 3.<|endoftext|> -TITLE: Some Elementary Schubert Calculus Calculations -QUESTION [6 upvotes]: Here are some simple geometry problems I am unable to resolve to my satisfaction. I asked the question on Math Stack (https://math.stackexchange.com/questions/2713754/a-problem-in-elementary-combinatorial-space-geometry) but it has received no interest, so I ask here in a different format. -Let $\delta$ denote the configuration of two intersecting lines, in other words, a conic with a double point, in three dimensional projective space. -Let $\nu$ denote the condition that one of the lines in $\delta$ intersects some given line. -Let $\mu$ denote the condition that the plane of $\delta$ passes through a given point. -Let $\rho$ denote the condition that the point of intersection of $\delta$ lies on a given plane. -Prove: -$$\delta\mu^2\nu^4\rho=17$$ -$$\delta\mu\nu^4\rho^2=17$$ -$$\delta\mu\nu^6=70$$ -$$\delta\nu^6\rho=70$$ -$$\delta\mu\nu^5\rho=50$$ -The first and second are the most baffling to me. The third and fourth, I have an idea, but I wish I had a better one. -Regarding the first, it means that the number of $\delta$ whose plane passes through two given points( $\mu^2$) and thus passes through a fixed line, and such that the lines of $\delta$ intersect $4$ given lines is $17$. -This can be analyzed as follows -${\bf Case 1}$ One of the lines of $\delta$ intersects three of the given lines. Then it intersects, these three lines and the axis of $\mu^2$, as ${\bf there \ are \ two \ lines \ intersecting \ four \ given \ lines \ in \ space}$ and there are -$\binom{4}{3}=4$ such choices there are $8$ lines. The second line is then uniquely determined. -${\bf Case 2}$ This is the real problem. Each line of $\delta$ intersects $2$ of the given lines, as there are $\frac{1}{2}\binom{4}{2}=3$ such partitions -each must have $3$ solutions, provided $17$ is the right answer. -How to obtain this last calculation is the real problem. -Well I have further thoughts, but I'll wait to see if any interest in this question. Am I missing something obvious ? - -REPLY [4 votes]: I would like to propose a slightly more elementary derivation of $17$ in the language that you used on math.stackexchange. Indeed $17=2\cdot 4+3\cdot 3$. -So our hero: $l_1,l_2,l_3,l_4$ four lines, $l$ is one more line and $P$ is a plane. We are looking for coupes $l',l''$ so that $l'\cap l''\in P$, $l\subset P(l_1,l_2)$, and $l'\cup l''$ intersect all four $l_i$. -In your answer on math.stack you have explained why $2\cdot 4$ and why $3$. So only the last $3$ should be explained. Here is a claim: -There exists exactly $3=4-1$ couples $l',l''$ where $l'$ intersects $l_1,l_2$ and $l''$ intersects $l_3, l_4$. -Proof. Note that the family of lines that intersect $3$ lines $l,l_1,l_2$ form a quadratic surface in $P^3$. Denote this surface $Q_{12}$. Denote by $Q_{34}$ the surface corresponding to lines $l, l_3,l_4$. Note now that $Q_{12}$ and $Q_{34}$ both intersect $P$ in conics. Denote these conics $C_{12}$ and $C_{34}$. By construction both $C_{12}$ and $C_{34}$ contain the intersection point $l\cap P$. And additionally to $l\cap P$ they have $3=4-1$ intersection points. Now, for each of these $3$ points we can pass two desired lines $l'$ and $l''$.<|endoftext|> -TITLE: Is $\mathbb{Q}$ the continuous image of a Golomb-like space, or vice versa? -QUESTION [5 upvotes]: The set $${\cal B} = \big\{\emptyset\big\}\cup\big\{\{a + bn: n\in\omega\}: a\in\omega, b\in(\omega\setminus\{0\})\big\}$$ is a basis for a topology $\tau$ on $\omega$. Is there a surjective continuous map from $\mathbb{Q}$ with the Euclidean topology onto $(\omega,\tau)$, or the other way round, or in neither direction? -Note. These two spaces are not homeomorphic, since $\mathbb{Q}$ is homogeneous, and $(\omega,\tau)$ is not. - -REPLY [6 votes]: Your topology is by definition the profinite topology of $\mathbf{Z}$, restricted to the nonnegative numbers $\mathbf{N}$. Hence it's a nonempty countable metrizable space. By a classical theorem of Sierpiński (Dasgupta - Countable metric spaces without isolated points, pdf), this implies that it is homeomorphic to $\mathbf{Q}$, in contradiction with your claim. -Golomb's space is defined as yours, but restricting to positive numbers and to coprime arithmetic progressions: precisely this subtlety drastically changes the topology (which is then connected—removing 0 is necessary to have it Hausdorff).<|endoftext|> -TITLE: Matrix trace & norm -QUESTION [14 upvotes]: For any nonnegative semidefinite matrix $A$ and any matrix $B$, we have -$$\mbox{tr} (AB) \le \mbox{tr} (A) \, \|B\|$$ -where $\mbox{tr}(\cdot)$ is the trace and $\|\cdot\|$ is the operator norm. How to prove this? - -REPLY [9 votes]: We can work in an orthonormal basis in which $A$ is diagonal. In that case the diagonal entries of $AB$ are $a_{ii}b_{ii}$, so -$$ -tr(AB) \leq|tr(AB)|=\left|\sum a_{ii}b_{ii}\right| \leq \sum a_{ii}|b_{ii}| \leq \sum a_{ii} \|B\|=tr(A)\|B\|. -$$<|endoftext|> -TITLE: Decomposition of tensor power of symmetric square -QUESTION [7 upvotes]: Studying some representation theory I came up with the following problem. -We work over a field of characteristic $0$. Let $V$ be the standard representation of $\mathrm{GL}_n$ and let $W$ be the representation $(\mathrm{Sym}^2(V))^{\otimes n}$. -Is it possible to describe the weights of the irreducible components of $W$ (after choosing the usual Borel and so on)? In particular I would like to know if the representation $\det(V)^{\otimes 2}$ appears in $W$. - -REPLY [9 votes]: By Pieri's formula, a partition with $2n$ elements in $n$ rows, corresponding to a representation of $GL_n$, occurs in this representation with multiplicity equal to the number of ways of obtaining that partition by starting with the empty partition and $n$ times adding two elements, no two in the same column. -For the determinant squared, which corresponds to a partition with $2$ columns of length $n$, this occurs with multiplicity exactly one.<|endoftext|> -TITLE: Does annular Khovanov homology detect the unknot (in annulus)? -QUESTION [7 upvotes]: Recently Kronheimer and Mrowka showed that Khovanov homology detects the unknot. It's still not known if the Jones polynomial detects the unknot. -Does annular Khovanov homology detect the unknot in an an annulus? -More generally, does annular Jones polynomial detect the unknot in an annulus (perhaps there's a counterexample?) - -REPLY [4 votes]: The full answer is obtained in our recent arxiv preprint here. See Theorem 1.3. It shows that the annular Khovanov homology detects the unlink in the thickened annulus. -The proof relies on a spectral sequence relating the annular Khovanov homology to the annular instanton homology, sutured instanton Floer homology for sutured manifolds with tangles developed in this preprint and Batson-Seed's unlink detection result for Khovanov homology.<|endoftext|> -TITLE: What is the intuition behind Almgren's frequency function? -QUESTION [12 upvotes]: It is by now well-known that for a harmonic function $u : B_1^n(0) \to \mathbb{R}$, the ratio -$$ -N(r) := \frac{r\int_{B_r(0)}|\nabla u|^2}{\int_{\partial B_r(0)} u^2} -$$ -is a non-decreasing function of $r$. -This and analogous expressions have been used in many different fundamental works on elliptic equations, but to me, it has always seemed like a mystery and as a result I have trouble even remembering this expression and usually have to look it up. -What's the intuition behind the definition of $N$? What does this ratio of energy in the ball and L^2 norm on the sphere even represent? - -REPLY [18 votes]: Observe that if $u$ is homogeneous of degree $\alpha$, then $N(r) \equiv \alpha$. (After integrating by parts, the numerator becomes $r \int_{\partial B_r} u u_{\nu} = \alpha \int_{\partial B_r} u^2$.) In addition, harmonic functions that are homogeneous of degree $\alpha$ tend to oscillate over distance $\sim \alpha^{-1}$ on $\partial B_1$ (think $r^{\alpha}\cos(\alpha\theta)$ in $\mathbb{R}^2$). Roughly, the monotonicity of $N$ says that the dominant frequency represented in $u$ at scale $r$ is increasing. -Some intuition comes from the expansion of $u$ in harmonic polynomials. For $r$ small, $N(r)$ only sees the lowest frequency (the first term in the expansion dominates), and as $r$ grows, the higher frequencies become more and more relevant. One proof of monotonicity comes directly from the expansion. In the context of unique continuation, monotonicity says "if the dominant frequency at $r = 0$ is $\infty$, then the same is true at all larger scales, so $u$ has to vanish in $B_1$." -Another perspective (which I find more general and illuminating) is variational. Note that $N$ is invariant under transformations that preserve harmonicity (that is, under $w \rightarrow \tilde{w} := Aw(rx)$ we have $N_w(r) = N_{\tilde{w}}(1)$). In particular, $N_w$ is constant when $w$ is homogeneous of any degree. To show monotonicity of $N_u$ when $u$ is harmonic, a useful idea is to compare $u$ with the function $w$ that is homogeneous of degree $\gamma:= N_u(1)$ (the dominant frequency of $u$ in $B_1$) and has the same boundary values as $u$ on $\partial B_1$. Roughly, since $u$ is a minimizer of the Dirichlet energy, $N_u$ has to grow faster than $N_w$ (which is constant) at $r = 1$. More precisely, the condition $N_u'(1) \geq 0$ becomes -$$\int_{\partial B_1} (|\nabla u|^2 - 2\gamma u u_{\nu}) - (n-2) \int_{B_1} |\nabla u|^2 \geq 0.$$ -The second term is larger than the corresponding quantity for $w$ by energy minimality. The first term is -$$\int_{\partial B_1} |\nabla_{\partial B_1} w|^2 + (u_{\nu} - \gamma u)^2 - \gamma^2w^2 = \int_{\partial B_1} |\nabla w|^2 - 2\gamma ww_{\nu} + (u_{\nu} - \gamma u)^2$$ -which is larger than the corresponding quantity for $w$, completing the proof. -In my view, monotonicity formulae in variational problems (free boundary problems, minimal surfaces, etc.) come from playing with quantities that are invariant under the natural scaling of the problem, and involve the energy. (This is the way I remember the Almgren quantity). The monotonicity of these quantities on minimizers comes from comparison with the scaling-invariant objects with the same boundary data. The clearest example is minimal surfaces: if $0$ is in a surface of dimension $n-1$, the quantity ($A(r)$:= area in $B_r$)/$r^{n-1}$ is dilation-invariant and constant on cones. The monotonicity on a minimal surface comes from comparing with the cone over the intersection of the surface with $\partial B_1$. Indeed, the area of the cone in $B_r$ grows more slowly at $r = 1$ since it crosses $\partial B_1$ orthogonally, and its area in $B_1$ is larger than that of the minimal surface, so $A'(1) - (n-1)A(1)$ for the minimal surface is larger than the corresponding quantity for the cone, which vanishes. -I hope this helps!<|endoftext|> -TITLE: Questions about categorification (& combinatorial simplification of the Russian approach to Lusztig's conjectures, in zero & positive characteristic) -QUESTION [12 upvotes]: This is a question about the proofs of Kazhdan-Lusztig's conjectures for category $\mathcal{O}$ using higher representation theory (avoiding Beilinson-Bernstein's geometric localization theory). -Using Bernstein-Frenkel-Khovanov and generalizations (Sussan, Stroppel-Mazorchuk, etc), the Kazhdan-Lusztig conjectures in type A are equivalent to the following statement: the classes of the simple modules in their categorification correspond to a "dual canonical basis" in a tensor product representation of $\mathfrak{sl}_k$ (for appropriately chosen $k$). -Using Losev-Webster and Webster, that statement can be deduced. They construct a theory of tensor product categorifications in type A (i.e. existence + uniqueness). -Can [Losev-Webster] be simplified, so that the connection to Soergel's J.AMS paper (www.ams.org/jams/1990-03-02/S0894-0347-1990-1029692-5/) becomes clear? It would be interesting to construct a theory of tensor product categorifications, with the input data being a Dynkin diagram (i.e. a simple Lie algebra). - -REPLY [4 votes]: Concerning your other question (4), it's probably too early to say what will eventually happen with Lusztig's various conjectures in prime characteristic. By now there is a lot of literature, following Williamson's cautionary observations about Lusztig's Santa Cruz conjecture in 1979 modeled on the statement of the Kazhdan-Lusztig conjecture. The modular situation turns out to be even more complicated than expected. See for example the papers and preprints listed on Achar's and Williamson's homepages, such as here or here. -While the earlier work by Andersen-Jantzen-Soergel proved Lusztig's first conjecture for all "sufficiently large" primes $p$, the explicit bound on $p$ found afterward by Fiebig is extremely large. There remains a serious problem about "intermediate" primes, and no firm conjecture about what happens when $p$ is smaller than the Coxeter number. While the current work encourages hopes for a solution, the end result is likely to be far more complicated than the Kazhdan-Lusztig conjecture. It's definitely worthwhile to seek alternative approaches but probably premature to expect a simple answer involving just the Kazhdan-Lusztig polnomials for an affine Weyl group.<|endoftext|> -TITLE: Simple $C^*$ algebras with invariant subspace property -QUESTION [7 upvotes]: Edit: According to the valuable comment of Yemon Choi I revise the question by replacing "faithful" with "irreducible". -We say that a $C^*$ algebra $A$ satisfies the invariant subspace property if there exist an irreducible representation $\phi: A \to B(H)$, for some Hilbert space $H$, such that $\forall a \in A,\; \phi(a)$ has a nontrivial closed invariant subspace of $H$. - -What is an example of a simple $C^*$ algebra with this property not isomorphic to the matrix algebra or the algebra of compact operators? - -In particular, does $C^*_{red}(F_2)$ satisfy this property? - -REPLY [2 votes]: Any unital C*-algebra has this property. This follows from these two facts: -Fix $a\in A$. -(1) If $a-\lambda\cdot 1$ is not left invertible then there is an irreducible representation $\pi$ such that $\lambda$ is an eigenvalue of $\pi(a)$. -(2) The set of $\lambda$ such that $a-\lambda \cdot 1$ is not left invertible contains the boundary of the spectrum of $a$. In particular, this set is nonempty. -Proof of (1): Assume first that $a$ is not left invertible, so that $\overline{Aa}$ is a proper left ideal in $A$. Choose a maximal left ideal $L$ containing $\overline{Aa}$. Then there exists a pure state $\phi$ such that -$$ -L=\{x\in A:\phi(x^*x)=0\}. -$$ -See Theorem 2.9.5 in Dixmier's "C*-algebras". Now do the GNS construction with this state. We get that $\pi_{\phi}(a)\xi_\phi=0$ (because $a\cdot 1\in L$). So $\pi_\phi(a)$ has kernel. Now, in general, if $a-\lambda\cdot 1$ is not left invertible then we apply the same argument to $a-\lambda \cdot 1$. -Notice that, conversely, if $\lambda$ in the spectrum of $a$ can be realized as an eigenvalue in some representation, then $a-\lambda \cdot 1$ cannot be left invertible. -Proof of (2): (This is well known.) Suppose that $a$ is left invertible and in the closure of the set of invertible elements. Say $ba=1$ and $c_n\to a$, with $c_n$ invertible. Then $bc_n\to 1$. So $bc_n$ is invertible for large $n$. Since $c_n$ is invertible, we get that $b$ is invertible, whence $a$ is invertible. -Somewhat digressing into the question of realizing the full spectrum as eigenvalues: -(1) If $A$ is finite (i.e., the unit is a finite projection) then left invertible implies invertible so the full spectrum can be realized as eigenvalues. (Pointed out by Yemon in the comments.) -(2) If $a$ is normal, then again left invertible implies invertible, so the same works. This case is a theorem in Pedersen's book "C*-algebras and their automorhism groups".<|endoftext|> -TITLE: Are open orientable 3-manifolds parallelizable via obstruction theory? -QUESTION [5 upvotes]: In the case of orientable closed $3$-manifolds, we have 4 ingredients that ensure parallelizability: -1a) Closed smooth $n$-manifolds have homotopy type of a $CW$-complex -1b) Closed smooth $n$-manifolds are Poincare duality spaces. -2) First Stiefel-Whitney class is $0$, from orientability -3) $BSU(2)$ is $3$-connected -1b) gives us Poincare duality and all of the familiar machinery of Steenrod squares and Wu's classes, which, combined with 2), provides for the vanishing of higher classes. -This, in turn, allows us to construct a map to $BSU(2)$. Throwing in 1a), we have that this map is null-homotopic QED. -For open 3-manifolds, I don't see why the computations with characteristic classes should still be valid (as 1b doesn't hold), but one can still show that the map to $BSU(2)$ is null-homotopic, provided it exists. -I am aware of Whitehead's paper https://www.sciencedirect.com/science/article/pii/B9780080098722500260, but I haven't read it, and I don't know if he uses obstruction theory, or something else. -Can the classical proof using obstruction theory still be salvaged in open case? -EDIT: Per Andy Putman's comment, I've replaced the false statement in 1b). - -REPLY [11 votes]: I think you are saying: for a closed $3$-manifold, vanishing $w_1$ implies vanishing $w_2$ by Wu's relations, but is this still true if the manifold is not closed? The answer is yes. -For a compact manifold with boundary, you can form the double (union of two copies of $M$ along $\partial M$). This will again be orientable, therefore parallelizable, therefore so is $M$. -For a noncompact manifold without boundary, you can prove the vanishing of $w_2$ by observing that $w_2$ restricts to zero on each compact set (because of the previous paragraph). This forces $w_2$ to be zero. -(This implication -- a cohomology class must be zero if it restricts to zero on every compact set -- is valid for homology with field coefficients, but not for integral cohomology.)<|endoftext|> -TITLE: Is measure preserving function almost surjective? -QUESTION [16 upvotes]: Let $F:[0,1]\to[0,1]$ be a Lebesgue measure preserving function. Is $F$ almost surjective, i.e., the image of $F$ has interior measure one? -This question is motivated by the following observation. If $F$ satisfies the above and $X$ is uniformly distributed over $[0,1]$, then $F(X)\sim Unif[0,1]$, i.e., $F(X)$ seems to appear almost everywhere in [0,1]. I believe the answer is no, but a counterexample is nontrivial. -A relating post is https://math.stackexchange.com/questions/2612075/is-every-measure-preserving-function-almost-surjective but the counter example therein does not apply to my question. (The function constructed there has different domain and range.) - -REPLY [3 votes]: A few remarks (where $m$ is the Lebesgue measure): - -If $F[0,1]$ is a Lebesgue measurable set, then $F[0,1]$ has Lebesgue measure equal to one (you can read Fedor Petrov's comment for this post). -Because of Birkhoff's Ergodic Theorem: For every $[a,b]\subset [0,1]$ $$ -\int_0^1\left(\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{[a,b]}(F^i (x))\right)dm=m([a,b]). -$$ -If $F[0,1]$ is a Lebesgue measurable set and $F$ is also Ergodic, then for $m$-a.e. $x$ $$ -1=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{F[0,1]}(F^i (x))=m(F[0,1]), -$$ -which is another proof that $m(F[0,1])=1.$ -If $F[0,1]$ is a Lebesgue measurable set, then $$ -1=\int_0^1\left(\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{F[0,1]}(F^i (x))\right)dm=m(F[0,1]), -$$ -which is another proof that $m(F[0,1])=1.$ -If $T$ is ergodic and $K\subset [0,1]\setminus F[0,1]$ is a Lebesgue measurable set, then $m(K)=0.$ By contradiction, we would obtain that for $m$-a.e. $x$ $$ -0=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{K}(F^i (x))=\mathcal{Leb}(K)\neq 0. -$$ -If $K\subset [0,1]\setminus F[0,1]$ is a Lebesgue measurable set, then $m(K)=0.$ By contradiction, we would obtain that -$$ -0=\int_0^1\left(\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{K}(F^i (x))\right)d m=m(K)\neq 0. -$$ -Let $\mathcal{A}=\{K:K\subset F[0,1], K \mbox{ measurable }\}$ and $\mathcal{B}=\{K:K\subset [0,1]\setminus F[0,1], K \mbox{ measurable }\}.$ Then $\sup_{K\in \mathcal{A}} m(K)=1.$ By contradiction, if not, by sigma aditivity -$$ -1=m[0,1]=\sup_{K\in \mathcal{A}} m(K)+\sup_{K\in \mathcal{B}} m(K)=\sup_{K\in \mathcal{A}} m(K)<1, -$$ -which is another proof that the image of $F$ has inner measure one.<|endoftext|> -TITLE: The space of contractible loops of a finite dimensional $K(\pi,1)$ -QUESTION [5 upvotes]: Let $X$ be a finite dimensional $K(\pi,1)$ manifold. Is it true that the space contractible loops of this manifold can be contracted to the space of constant loops on $X$? What if $X$ is a finite dimensional $CW$-complex? -I understand that the answer is positive if $X$ is a negatively curved manifold, since in this case there is a geometric argument. - -REPLY [10 votes]: The statement is true for a $K(\pi,1)$ but not true for other $X$. Finite dimensionality is not relevant. -Here's a sketch: let $X = K(\pi,1)$ and I will assume $X$ has the homotopy type of a CW complex. Let $\Omega_0 X$ be the space of contractible based loops in $X$. It's easy to show that the homotopy groups of this space are trivial in every degree (in degree 0 by definition and in higher degrees since $X$ is a $K(\pi,1)$ space, using the fact that $\pi_j(\Omega_0X) = \pi_{j+1}(X)$ for $j >0$). -Let $L_0 X$ be the space of contractible unbased loops. Then the sequence -$$ -\Omega_0 X \to L_0 X \to X -$$ -is a fibration, where the second map is evaluation at the basepoint of the circle. Since the fiber has trivial homotopy groups, it follows that the map $L_0X \to X$ is a weak homotopy equivalence. It follows that the section $X\to L_0X$ given by the constant loops is also a weak homotopy equivalence. It's therefore a homotopy equivalence since the spaces in question have the homotopy type of a CW complex.<|endoftext|> -TITLE: Modular forms, Maass forms and Automorphic representations -QUESTION [5 upvotes]: I am beginning to learn about automorphic forms, and stay perplex concerning the two languages of "forms" versus "representations" often used at the same time. As far as I understand, - -a modular/Maass form generates an automorphic representation (by considering the space of its right translations) -conversely, an automorphic representation gives a unique automorphic form (as a newvector, i.e. a nontrivial element in $\pi^K$ which is one dimensional for a suitable (?) compact subgroup $K$) - - -I would like to understand, where does the distinction between modular - form and Maass form appear in the representation language? - -Thanks in advance for any reference or explanation - -REPLY [7 votes]: The archimedean component of the representation determines the type of the underlying automorphic form. Loosely speaking, - -if $\pi_\infty$ is a discrete series (of index $k$), the underlying form is an holomorphic cusp form (of weight $k$, and level the arithmetic conductor of $\pi_f$) -if $\pi_\infty$ is a principal or complementary series (of index $\lambda$), the underlying form is a Hecke-Maass cusp form (of eigenvalue $\lambda$) - -The correspondence is detailed and proved in Gelbart, Automorphic forms on Adele Groups, 5.C "Some explicit features of the correspondence between cusp forms and representations". -Maybe you should look at the more form-casted answer given here, and the associated reference in Bump.<|endoftext|> -TITLE: References on Gerbes -QUESTION [15 upvotes]: I am looking for some references related to gerbes and their differential geometry. Almost every article I have seen that is related to gerbes there is a common reference that is Giraud's book Cohomologie non-abelienne. For me, it is not readable as I can not read french. -Only references I am familiar with are - -https://arxiv.org/abs/math/0212266 Introduction to the language of stacks and gerbes -https://arxiv.org/abs/math/0106083 Differential Geometry of Gerbes -https://arxiv.org/abs/math/0611317 Notes on 1- and 2-gerbes - -Out of these, only first article is more or less readable. The other two by Lawrence Breen are really not readable for me. -Some excerpt of comments answering such a request were - -"The standard reference is Giraud's book Cohomologie non-abelienne. This book is unreadable in the strongest possible meaning of the word unreadable." -"I find most of the contemporary articles in this area, which are often nonsystematic in terminology and notation, plus wave hands and use jargon on most issues, much less readable than Giraud's book." -"See Giraud's book on nonabelian cohomology." -"I don't think telling someone to see a dense 470 page book in French on non-abelian cohomology is a helpful comment." - -I am getting demotivated and irritated by lack of notes on gerbes and even in Mathoverflow there are not so much to see. Is this out of fashion now? Are there any one else who read/work on these? Iam not looking for something in Physics perspective. - -REPLY [4 votes]: The following reference might be helpful for you: -Hitchin, Lectures on Special Lagrangian submanifolds, $\S1$.<|endoftext|> -TITLE: Rate of convergence of $\frac{1}{\sqrt{n\ln n}}(\sum_{k=1}^n 1/\sqrt{X_k}-2n)$, $X_i$ i.i.d. uniform on $[0,1]$? -QUESTION [27 upvotes]: Let $(X_n)$ be a sequence of i.i.d. random variables uniformly distributed in $[0,1]$; and, for $n\geq 1$, set -$$ -S_n = \sum_{k=1}^n \frac{1}{\sqrt{X_k}}\,. -$$ -It follows from the generalized central limit theorem (as in [1] and [2, Theorem 3.1]) that -$$ -\frac{S_n-2n}{\sqrt{n\ln n}} -$$ -converges in law to a Gaussian distribution. However, in this case Berry—Esseen fails to give any convergence rate, as the summands are not square integrable. -Moreover, Hall's results for this sort of sum do not apply. (In [3], this would correspond to $\alpha=2$, while the results hold for $0<\alpha<2$.) - -What is the rate of convergence of $\frac{S_n-2n}{\sqrt{n\ln n}}$ to Gaussian? - - -[1] Shapiro, Jesse M., “Domains of Attraction for Reciprocals of Powers of Random Variables.” SIAM Journal on Applied Mathematics, vol. 29, no. 4, 1975, pp. 734–739. JSTOR, www.jstor.org/stable/2100234. -[2] Christopher S. Withers and Saralees Nadarajah, Stable Laws for Sums of Reciprocals. December 2011. -[3] Hall, P. (1981), On the Rate of Convergence to a Stable Law. Journal of the London Mathematical Society, s2-23: 179-192. doi:10.1112/jlms/s2-23.1.179 - -REPLY [12 votes]: $\newcommand{\de}{\delta} -\newcommand{\De}{\Delta} -\newcommand{\ep}{\epsilon} -\newcommand{\ga}{\gamma} -\newcommand{\Ga}{\Gamma} -\newcommand{\la}{\lambda} -\newcommand{\Si}{\Sigma} -\newcommand{\thh}{\theta} -\newcommand{\R}{\mathbb{R}} -\newcommand{\E}{\operatorname{\mathsf E}} -\newcommand{\PP}{\operatorname{\mathsf P}}$ -Let $V_k:=1/\sqrt{X_k}$, $b_n:=\sqrt{n\ln n}$, -\begin{equation*} - Z_n:=\frac{S_n-2n}{\sqrt{n\ln n}}=\frac1{b_n}\sum_1^n (V_k-\E V_1), -\end{equation*} -\begin{equation*} - \de_n:=\sup_{x\in\R}|\De_n(x)|, -\end{equation*} -where -\begin{equation*} - \De_n:=F_n-G,\quad -F_n(x):= P(Z_n1\}$, where $I$ denotes the indicator. -Note that $|e^{iu}-1-iu|\le2|u|$ and $|e^{iu}-1-iu+u^2/2|\le|u|^3/6$ for real $u$. -So, for the characteristic function (c.f.) $f_V$ of $V_k$ and $|t|\le1$ we have -\begin{multline*} - \frac12\,f_V(t)=\int_1^\infty\frac{e^{itx}}{x^3}\,dx - =\int_1^\infty\frac{1+itx}{x^3}\,dx - -\int_1^{1/|t|}\frac{t^2x^2/2}{x^3}\,dx \\ - +\int_1^{1/|t|}\frac{e^{itx}-1-itx+t^2x^2/2}{x^3}\,dx - +\int_{1/|t|}^\infty\frac{e^{itx}-1-itx}{x^3}\,dx \\ - =\frac12+it-\frac{t^2}2\,\ln\frac1{|t|} - +c\int_1^{1/|t|}\frac{|t|^3x^3}{x^3}\,dx - +c\int_{1/|t|}^\infty\frac{|t|x}{x^3}\,dx \\ - =\frac12+it-\frac{t^2}2\,\ln\frac1{|t|}+ct^2. -\end{multline*} -So, for $|t|\le1$ we have $\ln f_V(t)=2it-t^2\,\ln\frac1{|t|}+ct^2$ and hence for the characteristic function $f_n:=f_{Z_n}$ of $Z_n$ we have -\begin{multline*} - \ln f_n(t)=-i2nt/b_n+n\ln f_V(t/b_n) - =-\frac{t^2}{\ln n}\,\ln\frac{\sqrt{n\ln n}}{|t|}+c\frac{t^2}{\ln n} \\ - =-\frac{t^2}2-\frac{t^2}{\ln n}\,\Big(\frac12\,\ln\ln n-\ln|t|\Big)+c\frac{t^2}{\ln n} \\ - =-\frac{t^2}2+\frac{t^2}{\ln n}\,\ln|t|+c\frac{t^2}{\ln n}\,\ln\ln n \tag{2} -\end{multline*} -for $|t|\le b_n=\sqrt{n\ln n}$. So, with $\ep_n$ as in (1), -\begin{equation*} - \ln f_n(t)= - \begin{cases} - -\frac{t^2}2+c\ep_n|t| & \text{ if }|t|\le1 \\ - -\frac{t^2}2+c\ep_n t^2\ & \text{ if }1\le|t|\le\ln n, - \end{cases} -\end{equation*} -whence -\begin{multline*} - \int_{|t|<\ln n}\frac{|f_n(t)-e^{-t^2/2}|}{|t|}\,dt \\ - \le \int_{|t|<1}\frac{|e^{c\ep_n|t|}-1|}{|t|}\,dt - +\int_{\R}\frac{|e^{-(1-2c\ep_n)t^2/2}-e^{-t^2/2}|}{|t|}\,dt \le c\ep_n; -\end{multline*} -the latter integral was bounded using the identity $\int_0^\infty\frac{e^{-at^2/2}-e^{-t^2/2}}t\,dt=\ln(1/\sqrt a)$ for $a>0$. -By the Esseen smoothing inequality (see e.g. formula (6.4)), -\begin{equation*} - \de_n\le c\int_{|t|<\ln n}\frac{|f_n(t)-e^{-t^2/2}|}{|t|}\,dt +c/\ln n. -\end{equation*} -Now the second inequality in (1) immediately follows. -It remains to prove the first inequality in (1). For real $t$ and real $A>0$, -\begin{multline*} - \int_0^\infty e^{itx}d\De_n(x)=\int_0^A e^{itx}d\De_n(x)+c(1-F_n(A))+c(1-G(A)) \\ - =c\De_n(A)+c\De_n(0)-it\int_0^A e^{itx}\De_n(x)dx+2c(1-G(A))+c\de_n \\ - =c\de_n+c|t|A\de_n+ce^{-A^2/2} - =c\de_n+c|t|\de_n\sqrt{\ln\frac1{\de_n}} -\end{multline*} -if $A=\sqrt{2\ln\frac1{\de_n}}$. -Similarly estimating $\int_{-\infty}^0 e^{itx}d\De_n(x)$, we have -\begin{equation*} - f_n(t)-e^{-t^2/2}=\int_{-\infty}^\infty e^{itx}d\De_n(x)=c\de_n+c|t|\de_n\sqrt{\ln\frac1{\de_n}}. -\end{equation*} -Letting $t=1$ here and in (2), we see from the second line in (2) that -$$\de_n\sqrt{\ln\frac1{\de_n}}\gg\frac{\ln\ln n}{\ln n},$$ -whence the first inequality in (1) immediately follows. -It appears that similar techniques should work for a somewhat wide class of distributions, like the ones referenced by the OP.<|endoftext|> -TITLE: What are zeta functions good for? -QUESTION [14 upvotes]: I know a couple of answers to the above question: - -They can be used for point counting over finite fields/estimating the distribution of primes in characteristic 0. -There are various conjectures/results relating the special values of L-functions with other stuff in the vein of the class number formula/Birch Swinnerton-Dyer conjecture, Iwasawa theory on the other hand. - -What else can we do (conjecturally or otherwise) with zeta functions? I am interested in connections of the zeta functions to objects that have nothing to do with zeta functions as such (but are still of interest to arithmetic geometers and other mathematicians). My interests and background are definitely very algebraic so I have almost no idea about what results on the analytic side imply. - -REPLY [3 votes]: Depending on what kind of zeta functions you want, the Selberg zeta function allows you to relate lengths of closed geodesics to eigenvalues of the Laplacian. In particular, you can use the Selberg zeta function in combination with a trace formula to prove the prime geodesic theorem for compact Riemann surfaces and get Weyl's law. This also leads to construct isospectral manifolds. -Similarly, for graphs one can look at the analogous Ihara zeta function to relate lengths of "geodesics" to certain spectral quantities. In particular, one can get a characterization of Ramanujan graphs in terms of the Ihara zeta function. There are also numerous variants to count different things in graphs (Bartholdi zeta function, path zeta functions), and I have a conjecture with Christina Durfee that zeta functions are better at distinguishing graphs spectrally than the usual (adjacency matrix or Laplacian) spectra considered.<|endoftext|> -TITLE: Solve this Diophantine equation $(2^x-1)(3^y-1)=2z^2$ -QUESTION [16 upvotes]: Find the positive integers $(2^x-1)(3^y-1)=2z^2$ have three solutions -$$(1,1,1),(1,2,2),(1,5,11)$$I already know $(2^x-1)(3^y-1)=z^2$ has no solution. See: P.G.Walsh December 2006 [On Diophantine equations of the form] paper but there is a factor of $2$ that seems complicated, and I didn't know anyone had studied this before. If so, please help me with the article or link, thanks. - -REPLY [4 votes]: I don't have a solution but since we're facing a narrow margins situation I'll answer. I followed the approach I guessed Gerhard Paseman's comment was talking about but there was an unforeseen complication. -If $\gcd(2^x-1,3^y-1)=1$ we have -$$2^x-1=p^2$$ -$$3^y-1=2q^2$$ -for integers $p,q$. We can factor the first one in $\mathbb{Z}[i]$ -$$(-i)^x(1+i)^{2x}=(p+i)(p-i)$$ -the two factors on the right hand side have the same modulus so they both must be a unit times $1+i$ -$$p+i=u_1(1+i)^x$$ -$$p-i=u_2(1+i)^x.$$ -Then -$$2i=(u_1-u_2)(1+i)^x$$ -we have unique factorization so the power of $1+i$ on the right hand side has to match the one on the LHS so $x\leq 2$. $x=2$ doesn't work mod 3 so we have $x=1$. -Now splitting into three cases for each residue of $y\mod 3$: - -$y=3w$ - -Then -$$w^3=2z^2+1$$ -and to transform into an elliptic curve $w=w_1/2$ and $z=z_1/4$ -$$z_1^2=w_1^3-8.$$ -This is a Mordell curve so we can just look up the results. Bennet and Ghadermarzi have a table with all these solutions and looking up this and the other two cases, which turn out to have coefficients $-72$ and $-648$, we find that the three are the only solutions. -Now if $\gcd(2^x-1,3^y-1)\neq1$. I couldn't solve this case, but when $\gcd(x,y)=1$ I found some really weird behavior which made me think solving it in general might be harder than expected because it involves a relationship between $\text{ord}_p(2)$ and $\text{ord}_p(3)$. If a prime $p$ divides both sides of the gcd we have -$$2^x\equiv 3^y\equiv 1\mod p$$ -$$\gcd(\text{ord}_p(2),\text{ord}_p(3)) = 1$$ -This is relatively fast to calculate on a computer so I was able to find that $p=683,599479$ are the only possibilities under $10^9$. This is now OEIS sequence A344202, where you'll find 6 more terms which means mathematica has a faster way of calculating these than I do. -I feel like the $\gcd(x,y)\neq 1$ case might simplify to the previous one, but ruling out the few primes that satisfy the condition about the multiplicative orders of two and three will be difficult. For both of the primes I found we have $\text{ord}_p(2)\text{ord}_p(3)=p-1$ and there exists an unique element $r$ in $\mathbb{Z}_p$ so that $r^{\text{ord}_p(2)}\equiv 3$ and $r^{\text{ord}_p(3)}\equiv 2$. For $p=683$ it was $r\equiv 218$ and for $p=599479$, $r\equiv 45077$.<|endoftext|> -TITLE: An estimate on deviation of two smooth tangent $J$-holomorphic curves -QUESTION [5 upvotes]: Take $\mathbb C^2$ with coordinates $(z,w)$. Suppose that $J$ is a $C^{\infty}$ almost complex structure on $\mathbb C^2$ such that the line $w=0$ is $J$-holomorphic and $J(0,0)$ is given by $(z,w)\to (iz,iw)$. -Suppose that there is a smooth function $f(z):\mathbb D\to \mathbb C$ defined on a unit disk $\mathbb D$ such that the graph $(z,f(z))\subset \mathbb C^2$ is a $J$-holomorphic subvariety of $(\mathbb C^2, J)$ (we don't assume that $z\to (z,f(z))$ is a $J$-holomorphic map). -Question. Is it true that there is $n$ and a constant $c\in(0,1)$ such that $c|z^n|<|f(z)|<\frac{1}{c} |z^n|$ for sufficiently small $z$? -Remark. The statement is trivial in the case when $J$ coincides with the standard complex structure on $\mathbb C^2$. This question is a follow up to Two smooth tangent almost complex curves in a $4$-manifold - -REPLY [7 votes]: Yes, this is true. In fact, a more precise statement holds: Unless $f$ vanishes identically on $\mathbb{D}$, there is an integer $n$ and a nonzero complex number $a$ such that $f(z) = a\,z^n + f_{n+1}$, where $f_{n+1}$ is a smooth function on the disk that vanishes to order $n{+}1$ at $z=0$. -The proof follows immediately from a Taylor series expansion argument, plus a fact about solutions of elliptic equations with the same symbol as the Cauchy-Riemann equations. -Here is the argument: Let $\Omega$ be a nonzero complex-valued $2$-form on $\mathbb{C}^2$ such that $\Omega$ is of type $(2,0)$ with respect to $J$. Thus, a $2$-dimensional submanifold $\Sigma$ of $\mathbb{C}^2$ is $J$-holomorphic if and only if $\Sigma^*(\Omega)=0$. (I use the notation $\Sigma^*$ to denote the operation of pulling back forms to $\Sigma$.) This $\Omega$ is uniquely defined up to multiplication by a nonzero complex function. Because you have assumed that $J(0,0)$ is the standard complex multiplication, it follows that, up to a nonzero multiple, $\Omega$ is equal to $\mathrm{d}z\wedge\mathrm{d}w$ plus a term that vanishes at $(z,w)=(0,0)$. In particular, we can take $\Omega$ to have the form -$$ -\Omega = (\mathrm{d}z -a_1\,\mathrm{d}\bar z - a_2\,\mathrm{d}\bar w)\wedge -(\mathrm{d}w -b_1\,\mathrm{d}\bar z - b_2\,\mathrm{d}\bar w) -$$ -where $a_1,a_2,b_1,b_2$ all vanish at $(z,w)=(0,0)$. Letting $L\subset\mathbb{C}^2$ be the line defined by $w=0$, we see that, -because $w=0$ is a $J$-holomorphic curve, we must have -$$ -0 = L^*\Omega = (\mathrm{d}z -L^*a_1\,\mathrm{d}\bar z )\wedge -( -L^*b_1\,\mathrm{d}\bar z ) = -(L^*b_1)\,\mathrm{d}z\wedge\mathrm{d}\bar z. -$$ -Consequently, $b_1$ vanishes on $L$, so it must be of the form $b_1 = c_1\,w + c_2\,\bar w$ for some smooth functions $c_1$ and $c_2$. -Now, suppose that a $J$-holomorphic curve $C\subset\mathbb{C}^2$ is defined by an equation $w = f = f(z,\bar z)$ in a neighborhood of $(z,w) = (0,0)$, and that $f(0,0)=0$. If $f$ vanishes to infinite order at $z=0$ (i.e., its Taylor series at $z=0$ vanishes identically), then elliptic theory tells us that $f$ must vanish identically in a neighborhood of $z=0$. Let's assume that this is not the case, so that there is an $n\ge 1$ such that $f$ vanishes to order $n$ at $z=0$ but not to order $n{+}1$. Write $f = f_n + f_{n+1}$ where $f_n$ is the order $n$ Taylor polynomial of $f$ at $z=0$ and $f_{n+1}$ vanishes to order $n{+}1$. Now $C^*(\Omega)=0$, and we can compute its $(n{-}1)$-th order Taylor expansion as -$$ -0 \equiv C^*(\Omega) \equiv (\mathrm{d}z)\wedge(\mathrm{d}f_n ) -\equiv \frac{\partial f_{n}}{\partial\bar z}\,\mathrm{d}z\wedge\mathrm{d}\bar z, -$$ -where the equivalences mean that we ignore terms that vanish to order at least $n$ at the origin. (This uses the fact that $a_1$, $a_2$, and $b_2$ vanish to order at least $1$ at the origin while $C^*b_1 = C^*\!c_1\,f + C^*\!c_2\,\bar f$ vanishes to order at least $n$, while $\mathrm{d}f$ and $\mathrm{d}\bar f$ vanish to order at least $n{-}1$.) It follows that $\partial f_{n}/{\partial\bar z} = 0$ and, hence, that $f_n = a\,z^n$ -for some constant $a$, which is nonzero, since $f_n$ is nonzero.<|endoftext|> -TITLE: What’s the etiquette on using diagrams that need color to be understood? -QUESTION [83 upvotes]: I’m working on a paper that makes heavy use of colorful diagrams to supplement the text. For most of these it would probably not be possible to create grayscale versions that convey the same information as effectively. I’m a bit worried about this because (1) I imagine that some people like to print out papers to read them but these people might not want/be able to print in color, and (2) some readers may be colorblind. -What are the expectations on an author in my situation? Will it be considered rude to leave as-is, so long as the diagrams are not technically necessary to verify the arguments? Am I expected to include a description in the captions (“this region is red, this region is blue...”)? Or most stringently, would I be expected to post a “colorblind version” somewhere that tries to recreate the diagrams in grayscale as best as I can? -I’m interested in all opinions, but especially those of people who would have trouble with color for whatever reason. - -REPLY [3 votes]: Of course going through existing coloured diagrams and redesigning them in a monochrome (or grey-scale) colour scheme would be a lot of work, and depending on the complexity of the diagram the diagram may actually benefit from the use of colour. -But just because no other answer has mentioned the benefits of black & white (or greyscale) diagrams: - -there are no issues for people with colour vision deficiency -many journals do not print in colour (unless you are prepared to pay for colour printing), so library copies will usually be black & white -some researchers will print the electronic version (or view it on an ebook reader) and having diagrams compatible with standard black & white laser printers is very convenient (not only for the printed journal copy). - - -Even though colours can be helpful, every diagram can be redesigned in black & white and there are several techniques for distinguishing elements as you would with colour: -lines - -style (solid, dotted, dashed, ...) -thickness - -objects - -shape (circle, square, triangle, diamond, cross, ...) -filled vs. unfilled - -areas - -pattern (plain, striped, dotted, checkered, ...) -direction of pattern<|endoftext|> -TITLE: Is there a generalization (surely there is) of this simple combinatorial identity? -QUESTION [6 upvotes]: I was just doing some algebra on a paper and obtained: $$\sum_{l=0}^{n-1} {{n+l} \choose l}={2n \choose {n+1}}$$ -Are there some generalizations of this identity? -One possible generalization would be $$F_m(n)=\sum_{l=0}^{n-1} {{n+l} \choose l}^{m}$$ but it is not in my reach to obtain what would $F_m(n)$ equal to for every $m \in \mathbb N$ -Other possible generalization could be of the form $$G_k(n)=\sum_{l=0}^{n-1} k^l \cdot{{n+l} \choose l}$$ -I am interested in any generalization(s), not neccessarily of the forms I mentioned (I could mention some other forms but so could you so there is no need to do that). -You do not need to prove in an answer a generalization that you mention but it would be nice if you would point me to a direction where that generalization is mentioned and proven. -I asked a same question on MSE and received neither a comment nor an answer so I deleted that question there and decided to ask it here, although this is a low-level question for MO. - -REPLY [9 votes]: What about the generalised Vandermonde identity: -$$ -{ n_1+\dots +n_p \choose m }= \sum_{k_1+\cdots +k_p = m} {n_1\choose k_1} {n_2\choose k_2} \cdots {n_p\choose k_p} -$$ -See also Hockey-stick identity and Identities involving binomial coefficients. - -REPLY [6 votes]: Since $\binom{n+l}{l} = \binom{n+l}n$, we have -$$\sum_{l=0}^{n-1} \binom{n+l}{l} = \sum_{i=n}^{2n-1} \binom{i}{n}.$$ -Here the lower and upper summation bounds as well as the lower index of binomial coefficients depend on $n$. This is a particular case of a more general formula, where all three entities are independent: -$$\sum_{i=n}^m \binom{i}{k} = \binom{m+1}{k+1} - \binom{n}{k+1}.$$ - -REPLY [5 votes]: Formula 4.2.5.37 on page 503 in Prudnikov, Brychkov, Marichev is -\begin{equation} - \sum_{k=0}^n\binom{a+k}a\binom{b-k}{b-n}=\binom{a+b+1}n. -\end{equation} -Substituting here $a=n+1$ and $b=n$, and then replacing $n$ by $n-1$, we get your identity. -The book Prudnikov, Brychkov, Marichev is in Russian, but this should be no problem here.<|endoftext|> -TITLE: First order decidability of limit of gradient flow? -QUESTION [11 upvotes]: Let $f: \mathbb{R}^n\to\mathbb{R}$ be a polynomial function, and let $p$ be a critical point. Consider the ascending manifold $A_p$ consisting of all points whose limit under the gradient flow of $f$ is equal to $p$. Is $A_p$ a semialgebraic set? If not, is there an additional assumption on $f$ that makes it so? -This is an offshoot of a previous question of mine that did not get much attention. - -REPLY [6 votes]: Overview: The boundaries of the basins of attraction are lower dimensional stable manifolds. In two dimensions, they are the arcs flowing from repelling fixed points to saddle points. I expect that these are basically never analytic near the repelling fixed point. More specifically, if $f(x,y) = \alpha x^2 + \beta y^2 + (\mbox{higher order terms})$, then $dx/dt \approx 2 \alpha x$ and $dy/dt \approx 2 \beta y$, so the solution to the differential equation is roughly $x \approx \exp(2 \alpha t)$ and $y \approx \exp(2 \beta t)$, giving $x \approx y^{\beta/\alpha}$. If $\beta/\alpha$ is irrational, then this can't be algebraic. -Here is a case where I can make this analysis precise. Let -$$f(x,y) = a x^2 (1-x^2/(2p^2)) + b y^2 (1-y^2/(2 q^2)) - c x^2 y^2$$ -with $a$, $b>0$, $a/b$ irrational and $c > \max(a/p^2, b/q^2)$. Then $f$ has a repelling fixed point at $(0,0)$ (if Morse flow goes up hill), has attracting fixed points at $(\pm p, 0)$ and $(0, \pm q)$, and has a saddle point in each quadrant. The $x$ and $y$-axes are flow lines, so none of the other flow lines can cross them. I claim that none of the other flow lines through $(0,0)$ are analytic at $(0,0)$. I'll analyze the case of a flow line through the first quadrant, so $x(t)$, $y(t)>0$. -The Morse flow equation is -$$\begin{bmatrix} dy/dt \\ dx/dt \end{bmatrix} = \begin{bmatrix} \partial f/\partial y \\ \partial f/\partial x \end{bmatrix} = \begin{bmatrix} 2y (b-by^2/q^2 - c x^2) \\ 2x (a - bx^2/p^2 - c y^2) \\ \end{bmatrix}. $$ -This means that -$$\frac{d \log y/dt}{d \log x/dt} = \frac{y^{-1} d y/dt}{x^{-1} d x/dt} = \frac{b-by^2/q^2 - c x^2}{a - ax^2/p^2 - c y^2}.$$ -So, by L'Hospital, on any flow line where $x$ and $y \to 0^+$, we have -$$\frac{\log y}{\log x} \to \frac{b}{a}.$$ -If the flow line were an algebraic arc, then $\tfrac{\log y}{\log x}$ would approach a rational limit, a contradiction. -In particular, the flow from $(0,0)$ to the saddle point in the first quadrant is the boundary between the basins of attraction of $(p,0)$ and $(0,q)$, and is not algebraic. -Again, I don't think this example is special. I think it is just unusual in that I can carefully analyze the behavior as $(x,y) \to (0,0)$, since $x$ and $y$ divide $\tfrac{\partial f}{\partial x}$ and $\tfrac{\partial f}{\partial y}$. -Here is a picture. I took $p=1$, $q=1.1$ and $(a,b,c) = (0.2633, 0.4733, 1)$. Those values were chosen to make the attracting fixed points look like $-c(x^2+y^2)$, so the flow comes into them without any funny nodes.<|endoftext|> -TITLE: How can I prove that $(n-1)$-dimensional manifold is not contained in a $(n-2)$-dimensional affine variety? -QUESTION [5 upvotes]: I am having trouble proving the following statement, which I think is true (and possibly very basic). Let $M$ be a real differentiable manifold of dimension $(n-1)$ sitting inside $\mathbb{R}^n$. Let $W $ be an algebraic set defined by homogeneous forms in $\mathbb{R}[x_1, ..., x_n]$ for which $W \subseteq \mathbb{A}_{\mathbb{\mathbb{C}}}^n$ has dimension $n-2$ (as an algebraic set in $\mathbb{C}^n$). I want to prove that -$$ -M \not \subseteq W \cap \mathbb{R}^n. -$$ -I think one reason why I am having trouble proving this (even though I suspect it might be quite basic) may be that I don't have a good understanding on how the dimension of the manifold and dimension of algebraic sets relate to one another. I would greatly appreciate any comments or references. Thank you. - -REPLY [4 votes]: You seem to be asking whether it is possible that the topological dimension of the set of real points of an algebraic variety is greater than its algebraic dimension (in your case, the former is $n-1$ and the latter is $n-2.$) It seems to be a standard fact that this is impossible (inequality in the opposite dimension is quite possible, as in the standard example of $x^2+y^2 = -1.$) The standard reference seems to be: -Basu, Saugata; Pollack, Richard; Roy, Marie-Françoise, Algorithms in real algebraic geometry, Algorithms and Computation in Mathematics 10. Berlin: Springer (ISBN 3-540-33098-4/hbk). x, 662 p. (2006). ZBL1102.14041.<|endoftext|> -TITLE: Interesting examples of vector bundles on hyperkahler varieties -QUESTION [7 upvotes]: I'm looking for a few concrete examples of vector bundles on hyperkahler varieties of dimension $\ge 4$. Here are a few examples I know already: - -For $X$= the Hilbert scheme of points $S^{[n]}$ on a K3 surface with a line bundle $L$, one has the -tautological bundle $L^{[n]}$, which has rank $n$. -For $X$=the variety of lines of a cubic fourfold, one has the -restriction of the universal bundles on the ambient Grassmannian. -Similarly for other hyperkahler varieties embedded in Grassmannians. - -Are there other natural examples of interesting vector bundles like this? (I'm mostly interested in hyperkahler manifolds of $K3^{[n]}$-type, but also other cases are interesting). - -REPLY [5 votes]: Let $S$ be a K3 surface, and consider a moduli space of sheaves $M$ where you've cooked it up so that stable = semi-stable and there are only vector bundles, no torsion-free sheaves. Suppose moreover that it's a fine moduli space, so there's a universal bundle $U$ on $M \times S$. I think Mukai vector $(5,1,0)$ on a quartic surface of Picard rank 1 should do the trick, but no promises. -Then the "wrong-way slices" $U|_{M \times \text{pt}}$ are very interesting bundles on $M$. Probably they're stable, and $S$ is a component of the moduli space of stable bundles on $M$ via this construction, but no one knows how to prove it. - -REPLY [4 votes]: On a hyperkähler manifold $M$ with a circle action which fixes just one complex structure $I$, and rotates $J$ and $K$, and whose Kähler form $\omega_I$ is integral, there exists a hyperholomorphic line bundle $$\mathcal L\to M,$$ -i.e., it admits a connection such that its $(0,1)$-part with respect to any of the complex structures in the 2-sphere generated by $I,J,K$ is a holomorphic structure. It is used in the c-map construction to obtain a correspondence between those hyperkähler manifolds and certain quaternionic Kähler manifolds. -The construction has its origin in physics, was carried out by Haydys (J. Geom. Phys. 58 (2008)), and is nicely explained in Hitchin's paper "On the hyperkähler/quaternion Kähler correspondence".<|endoftext|> -TITLE: Lower bound on the eigenvalues of the Laplacian -QUESTION [7 upvotes]: I am looking for a graph for which $2 d_{i} < \mu_{i}$, for some index $i$, where $\mu_{1} \leq \mu_{2} \leq \dots\leq \mu_{n}$ are the eigenvalues of the Laplacian matrix $L(G)$ and $d_{1} \leq d_{2} \leq \dots \leq d_{n}$ are the node degrees. -According to the literature and existing upper/lower bounds on the eigenvalues of the Laplacian as a function of node degrees, it seems -there is a graph with this property. However, I was not able to find such a graph by generating all graph with $4, 5, \dots, 10$ vertices. -Any help or advice for a possible hypothesis confirmation or rejection would be appreciated. -Related references: - -Miriam Farber, Ido Kaminer, Upper bound for the Laplacian eigenvalues of a graph, June 2011. -A. E. Brouwer, W. H. Haemers, A lower bound for the Laplacian eigenvalues of a graph—proof of a conjecture by Guo, February 2008. - -REPLY [4 votes]: This is an expanded solution based on the comment of @mostafa. -Such graphs dose not exist. -Let $L$ be the Laplacian of the graph. Suppose that diagonal elements of $L$ are sorted -sequence of degrees $d_1\leq \ldots \leq d_n$. - By Min-Max theorem, we have -$$\mu_i= \min_{U<\mathbb R^n,\dim(U)=i} \big\{ \max_{x \in U, \|x\|=1} x^TL x \big\} $$ -So, if there exists an $i$-dimensional subspace of $\mathbb R^n$, say $U$, such that -$$ -\max_{x \in U, \|x\|=1} x^TL x \leq 2d_i -$$ -then we have $\mu_i \leq 2d_i$. -Let $U := \langle e_1,\ldots,e_i\rangle$, where $e_k$ is $k$-th element of the standard base of $\mathbb R^n$, for $k=1,\ldots,i$. Suppose that $x\in U$ and $\|x\|=1$. Note that only first $i$ elements of $x$ are nonzero. we have -$$ -x^TL x = \sum_{ab \in E(G),a -TITLE: Periodicity isomorphism in KR theory -QUESTION [6 upvotes]: I am currently working my way through Atiyah's paper "On K-Theory and Reality" and can't get my head around a remark stated in the paper. -In the third section $KR$-Theory is related to regular $K$-theory using so called "coefficient theories", i.e. functors $X \mapsto KR(X \times S^{p,0})$. In Proposition 3.1 Atiyah proves periodicity isomorphisms in the cases $p=1,2,4$ using the maps $\mu_p \colon X \times S^{p,0} \times \mathbb{R}^{0,p} \to X \times S^{p,0} \times \mathbb{R}^{p,0}$, $(x,s,u) \mapsto (x,s,su)$ and Bott periodicity $\beta^p \colon KR \to KR^{p,p}$. -Afterwards he claims, that these isomorphisms $$ \gamma_p = \mu_p^* \circ \beta^p \colon KR(X \times S^{p,0}) \longrightarrow KR^{-2p}(X \times S^{p,0})$$ are given by multiplication by $c_p = \gamma_p(1) = \mu^*_p(b^p \cdot 1)$ for $1 \in KR(S^{p,0})$ and $b= [H] -1$ the Bott element, since $\mu^*_p$ and $\beta^p$ are $KR(X)$-module homomorphisms. -I am struggling with this claim: Both maps are clearly $KR(X)$-module homomorphisms but to prove this claim I think one also needs to know that - -the external product map $KR(X) \otimes KR(S^{p,0}) \to KR(X \times S^{p,0})$ is surjective and -$KR(S^{p,0}) \cong \mathbb Z$ generated by $1 \in KR(S^{p,0})$ - -Especially the first point troubles me. Or am I missing something here? - -REPLY [2 votes]: As it turns out my confusing resulted from a – supposed – typo: The isomorphisms $\mu^*_p$ and $\beta^p$ are in fact $KR(X \times S^{p,0})$-linear (and not just $KR(X)$-linear), which is quite easy to see by following the definitions and obviously implies the statement about $\mu^*_p \circ \beta^p$.<|endoftext|> -TITLE: About the existence of characters on $B(X)$ -QUESTION [13 upvotes]: Let $X$ be a Banach space. Let $B(X)$ be the space of all bounded linear operators on $X$. Does $B(X)$ have an empty character space for any $X$? -I know the proof of the fact that $M_n(\mathbb{C})$ has no characters and also $B(H)$, where $H$ is a Hilbert space. - -REPLY [3 votes]: I think that this question is a suitable reason to remind everyone about a very interesting (but not so well known) unpublished paper C.J. Read, "Different forms of the approximation property", Unpublished manuscript, approximate date: 1989. In this paper Read shows the existence of Banach spaces with "approximate characters". The techniques used by Read is of the same type as the techniques of Mankiewicz in the paper mentioned in the answer of Tomek Kania. The techniques was introduced by Efim Gluskin in his papers in 1981. -(I posted Read's paper on my web page because one of my papers relies heavily on Read's construction, which is not available anywhere.)<|endoftext|> -TITLE: Is the existence of double complement of a set provable in Intuitionistic ZF? -QUESTION [11 upvotes]: In Powell's article [1] he introduces the axiom of double complement, which says a double complement $\{x : \lnot\lnot(x\in A)\}$ is a set for any set $A$. -I can't find similar axiom from other references, even in the Friedman's article [2] on double negation interpretation over set theories. Hence it is natural to ask the relation between his axiom and other axioms of IZF. -(Note: Powell consider the axiom of collection rather than the replacement in his article, but I will consider full IZF.) -I have made some attempts on this problem: if $f(\beta):=\sup\{\alpha\in\mathrm{On} : \lnot\lnot(\alpha<\beta)\}$ exists for each ordinal $\beta$, then the axiom of double complement holds: then $\{x\in V_{f(\operatorname{rank}(A))} : \lnot\lnot(x\in A)\}$ would be the double complement of $A$. (Here $V_\alpha := \bigcup_{\beta\in\alpha} \mathcal{P}(V_\beta)$ is a von Neumann hierarchy. Axiom of power set is necessary in my argument.) However checking $f(\beta)$ is a set is at least as hard as checking the axiom of double complement. -Forcing or realizability seems not helpful to me. This is because we need to generate a set whose double complement is proper class to prove the independence of the axiom of double complement, and it seems to need a proper-class sized name. However both methods just deal with set-sized names. -My question is: Is the axiom of double complement provable from full IZF? If not, is it indenpendent from IZF? Is the axiom of double complement related to the law of excluded middle? I would appreciate any help. -(Added in Jan 04, 2019: Realizability can be used to prove some non-classical principles are compatible with the axiom of double complement. In fact, if $V$ is a model of ZFC and $\mathcal{A}$ is a pca then the realizability model $V(\mathcal{A})$ validates the axiom of double complement.) -(Added in Jan 06, 2019) I recheck details of the proof of the above statement and I found my proof does not work. Beeson states we can prove the consistency of IZF + Double complement + Church's thesis via realizability in his book Foundations of Constructive Mathematics without proof. I do not know it really holds. - -References -[1] Powell, William C. "Extending Gödel's negative interpretation to ZF." The Journal of Symbolic Logic 40.2 (1975): 221-229. -[2] Friedman, Harvey. "The consistency of classical set theory relative to a set theory with intuitionistic logic 1." The Journal of Symbolic Logic 38.2 (1973): 315-319. - -REPLY [3 votes]: This is not a full answer, but I think it is worth to mention: we can prove the theory CZF + Full separation + ￢Double complement is consistent! -The proof goes as follows: consider the Lubarsky's first model in 1. Let $\check{0}(\kappa)=\varnothing$ and $\check{1}(\kappa) = \{\check{0}\upharpoonright (\mathrm{On}\setminus\kappa)\}$. They will behave as 0 and 1 of the model. Furthermore, for each ordinal $\kappa$ define -$$\check{1}_\kappa (\lambda ) = \begin{cases}\varnothing & \text{if }\kappa<\lambda\text{ and} \\ \{\check{0}\upharpoonright (\mathrm{On}\setminus\lambda)\} & \text{otherwise.}\end{cases}$$ -I will claim that the element $x\in M_0$, defined by $x(\nu)=\{\check{1}\upharpoonright (\mathrm{On}\setminus\nu)\}$ for every ordinal $\nu$, has no double complement. First, it is tedious to check -$$\kappa \vDash \lnot\lnot (\check{1}_\nu \in x) \iff \forall \lambda \ge \kappa \exists \mu\ge\lambda : \mu\models \check{1}_\nu \in x$$ -and the later statement holds: take any $\mu>\max(\lambda, \nu)$. Hence if $y\in M_0$ is a double complement of $x$, then it must contain every $\check{1}_\nu$, which is impossible. (In fact, we need to prove any $y\in M_\xi$ for any $\xi$ cannot be a double complement of $x$. However, its proof is not too different from my proof.) -However, my proof (if correct) has some unsatisfactory points. First, it requires the consistency of ZFC. Lubarsky proved that CZF + Full separation is equiconsistent with the second-order arithmetic [2]. I wonder CZF + Full separation + ￢Double complement is equiconsistent with Second order arithmetic. Kripke models seems not adequate to derive such kind of equiconsistency result (unless we form a Kripke model of CZF over CZF.) -Second, it does not settle my original question: what happenes if we assume the axiom of power set? - -(Added in Jan 24, 2019) I found that V. H. Hahanyan makes a bunch of result on the axiom of double complement over IZF (example). But most of his article is written in Russian so it is not easily accessible to me. - -References -(1) Lubarsky, Robert S. "Independence results around constructive ZF." Annals of Pure and Applied Logic 132.2-3 (2005): 209-225. -(2) Lubarsky, Robert S. "CZF and Second Order Arithmetic -." Annals of Pure and Applied Logic 141.1-2 (2006): 29-34.<|endoftext|> -TITLE: Universal enveloping algebra and the algebra of invariant differential operators -QUESTION [6 upvotes]: Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra. Then $\mathfrak{g}$ may be interpreted as the Lie algebra of right (equivalently left) invariant vector fields. Let $\mathcal{U}(\mathfrak{g})$ be its enveloping algebra. - -Why $\mathcal{U}(\mathfrak{g})$ may be interpreted as the algebra of all right invariant differential operators? - -In other words: as there are no algebra relations within $\mathcal{U}(\mathfrak{g})$ except $XY-YX=[X,Y]$ the same should be true for invariant differential operators: however, is it obvious? - -REPLY [5 votes]: From the Lie algebra inclusion $\mathfrak{g}\rightarrow \Gamma(T_G)$ of right invariant vector fields we get a map $U(\mathfrak{g})\rightarrow\Gamma(\mathcal{D}_G).$ This induces a map $U(\mathfrak{g})\otimes\mathcal{O}_G\rightarrow\mathcal{D}_G$ of $\mathcal{O}_G$-modules. We have a structure of $G$-equivariant sheaf on both sides, with respect to the right translation action of $G$ on itself. (The $G$-equivariant structure on the left hand side comes from tensoring up the $G$-equivariant structure on $\mathcal{O}_G,$ i.e., the $G$ action on the $U(\mathfrak{g})$ part is trivial.) Our map is $G$-equivariant, and if we know it is an isomorphism, we will get an isomorphism $\Gamma(U(\mathfrak{g})\otimes\mathcal{O}_G)^G\rightarrow\Gamma(\mathcal{D}_G)^G,$ or $U(g)\rightarrow\Gamma(\mathcal{D}_G)^G$, as desired. -Now let us show that $U(\mathfrak{g})\otimes\mathcal{O}_G\rightarrow\mathcal{D}_G$ is an isomorphism. We have a canonical filtration $F_i\mathcal{D}_G$ on $\mathcal{D}_G$ by order of differential operators. On the left hand side, the PBW filtration $F_iU(\mathfrak{g})$ induces a filtration $F_iU(\mathfrak{g})\otimes\mathcal{O}_G$. -As our map sends $F_iU(\mathfrak{g})\otimes\mathcal{O}_G$ to $F_i\mathcal{D}_G$ and both filtrations are exhaustive, it suffices to check that the map on associated gradeds is an isomorphism. The associated graded of the left hand side is $\operatorname{Sym}^{\bullet}\mathfrak{g}\otimes\mathcal{O}_G.$ -For the right hand side, we know that $F_i\mathcal{D}_G/F_{i-1}\mathcal{D}_G$ can be canonically identified with $\operatorname{Sym}^iT_G.$ The inclusion $\mathfrak{g}\rightarrow \Gamma(T_G)$ gives rise to a map $\mathfrak{g}\otimes\mathcal{O}_G\rightarrow T_G$ which can be seen to be an isomorphism. This identifies the associated graded of the right hand side with $\operatorname{Sym}^{\bullet}\mathfrak{g}\otimes\mathcal{O}_G,$ and the map on associated gradeds becomes the identity.<|endoftext|> -TITLE: Convex hull of the Stiefel manifold with non-negativity constraints -QUESTION [11 upvotes]: Consider the Stiefel manifold -$$\mathrm{St}(n,k) :=\{X \in \mathbb{R}^{n\times k} : X^TX = I_k\},$$ -where $I_k$ is the $k$-dimensional identity matrix. It is well known that -$$\mathrm{conv} \left( \mathrm{St}(n,k) \right) = \{X \in \mathbb{R}^{n\times k} : \|X\|_2 \leq 1\}$$ -where $\|\cdot\|_2 $ is induced $2$-norm. -Question: Is there a characterization for the convex hull of the Stiefel manifold with non-negativity constraints: -$$\mathrm{conv}(\mathrm{St}(n,k) \cap \mathbb{R}^{n\times k}_+)$$ -where $\mathbb{R}^{n \times k}_+$ is the set of all $n \times k$ matrices with non-negative elements? - -REPLY [2 votes]: @Arian has supplied the solution for $k = 1$. It is a known result that for $k = n$, the desired convex hull is the set of all $n$ by $n$ doubly stochastic matrices. -I offer this conjecture for the solution for general $k$, as the following extension of the answer by @Arian .The remainder of this post is conjecture: -$$\mathrm{conv}(\mathrm{St}(n,k) \cap \mathbb{R}^{n\times k}_+) = \{X\in\mathbb{R}^{n\times k}_+:\|X\|_2 \le 1, \Sigma_{j=1}^n X_{ij} \ge 1\ \forall i\}$$ -$\{X\in\mathbb{R}^{n\times k}_+:\|X\|_2 \le 1, \Sigma_{j=1}^n X_{ij} \ge 1\ \forall i\}$ implies $\|X\|_2 \ge \sqrt{\frac{k}{n}}$ and $\sqrt{n-k+1} \ge \Sigma_{j=1}^n X_{ij}, \forall i$. -So putting it altogether, we can state -$\mathrm{conv}(\mathrm{St}(n,k) \cap \mathbb{R}^{n\times k}_+) =$ $$\{X\in\mathbb{R}^{n\times k}_+: \sqrt{\frac{k}{n}} \le \|X\|_2 \le 1, \sqrt{n-k+1} \ge \Sigma_{j=1}^n X_{ij} \ge 1\ \forall i\}$$ -When $k = n$, $\|X\|_2 = 1, \Sigma_{j=1}^n X_{ij} = 1\ \forall i$ must hold, and this reduces exactly to the set of all $n$ by $n$ doubly stochastic matrices, thereby matching the known result for square $X$.<|endoftext|> -TITLE: Tietze transformations for sites of toposes -QUESTION [7 upvotes]: I have read that people think of a site as a presentation of the corresponding sheaf topos. For instance, on page 7 of this text by Caramello: as Grothendieck observed himself, a site of definition for a given topos can be seen as a sort of presentation of it by generators and relations (one can think of the objects of the category underlying the site as defining the generators, and of the arrows and covering sieves as defining the relations). -Now, for presentations of monoids (or groups), Tietze transformations are a very powerful tool. Namely, any two presentations of a given monoid can be related by a sequence of the following transformations: - -add or remove a definable generator, -add or remove a derivable relation. - -(You can think of finite presentations, although larger ones can be handled by considering transfinite sequences of transformations). -Thus my question: are there similar "Tietze transformations" on sites, such that any two sites having equivalent categories of sheaves can be related by Tietze transformations? - -REPLY [6 votes]: There are two observation to be made that limit a little this kind of analogy: -1) Site are indeed in some sense presentations, but an infinity theory (I mean with operation of infinite arity) something like the theory of ininitary pretopos, i.e. categories satisfying all of Giraud axioms except being presentable, but including being co-complete. -Because of that you can expect to describe the transformation that you need to go from a site to an other by a finite sequence move involving a finite number of objects. -2) Site are presentations, but more in the sense of "generators + ideal", not "generators" + "generator of an ideal". So "adding definable relation" does not really makes sense, essentially any valid covering relation is already part of your topology. -But you could consider the notion of topology generated by some class of covering of course. -Once this said, Grothendieck Comparison lemma (see https://ncatlab.org/nlab/show/comparison+lemma) can be used to do something relatively close to what you have in mind: -It says something which can roughly be translated by "if I add a bunch of new object to my site that can be covered by the previous one then I don't change the topos of sheaves " (This is very rough, I refer you to the precise statement of theorem 3.2 on the nLab page). -Any equivalence between toposes of sheaves on two different sites can always be obtained by two application of this lemma (one going up and the other going down). Essentially, by taking a subcategory of the topos of sheaves large enough to contains the two sites that you have.<|endoftext|> -TITLE: Is every graph an isomorphic factor of some complete graph? -QUESTION [10 upvotes]: Q1: -For any given finite simple graph G with e edges, does there always exist an $n$ such that the edges of $K_n$ can be partitioned into $\frac{\binom{n}{2}}{e}$ edge-disjoint copies of $G$? If so, can any upper bounds be placed on the minimal required $n$? -Q2: -Similarly for digraphs (with twice as many copies)? -Q3: -For any loopless multidigraph with maximum edge multiplicity $m$, it seems clear that for some $M>m$, an $M$-complete multidigraph can be exactly partitioned into isomorphic copies of it, even without increasing $n$: if nothing smaller, one can always overlay $n!$ copies of the original multidigraph with each permutation of the vertices. Can this partition always be accomplished with $M=m$, increasing $n$ instead? If not, can tighter bounds be placed on $M$? -Q4: Does allowing loops (and adjusting edge-counts appropriately) fundamentally change any of the above? - -REPLY [11 votes]: Q1: yes, this is a theorem by Wilson; see the first paragraph here: https://arxiv.org/abs/1604.07282 -Edit: perhaps the book Decomposition of graphs by J. Bosak might be helpful (the preview on google books is quite limited).<|endoftext|> -TITLE: The Axiom of Determinacy and the Banach-Mazur game -QUESTION [11 upvotes]: The Wikipedia article on the Axiom of Determinacy (AD) claims: - -Equivalent to the axiom of determinacy is the statement that for every subspace X of the real numbers, the Banach–Mazur game BM(X) is determined. - -Is this claim true? -AD is usually stated for the Gale-Stewart game $G(S)$ with payoff set $S\subseteq \omega^\omega$, in which the players build a sequence in $\omega^\omega$ by alternately choosing integers, and Player I wins if this sequence is in $S$. -In the Banach-Mazur game $BM(X)$ with payoff set $X\subseteq \mathbb{R}$, the players build a descending sequence $U_0\supseteq U_1\supseteq U_2\supseteq \dots$ by alternately choosing nonempty open sets, and Player II wins if $\bigcap_{n\in \omega}U_n\subseteq X$. -AD implies the determinacy of $BM(X)$ for all $X\subseteq \mathbb{R}$, since $BM(X)$ can be replaced by an equivalent game in which the players are additionally required to play open intervals with rational endpoints, which can be coded by natural numbers. -After a bit of Googling, I couldn't find any information about the converse (aside from the claim on Wikipedia). Of course, this might just be because it's very obvious. But on the other hand, the fact that $BM(B)$ is determined when $B$ is Borel (or even just has the Baire property) is an easy theorem, while the corresponding fact for $G(B)$ is a very hard theorem ("Borel determinacy"). So I can believe that the converse might be false. -Edit: In light of the answer to this question, I've edited the Wikipedia page, replacing the (false) equivalence with the (true) implication: - -The axiom of determinacy implies that for every subspace X of the real numbers, the Banach–Mazur game BM(X) is determined. - -REPLY [13 votes]: The claim is false. The Banach-Mazur game, also known as the $**$-game shows (and is equivalent to) that every set of reals has the Baire property. What is true, as you've noted, is that if one has a pointclass $\Gamma$ which is adequate and closed under Borel ($\Delta^1_1$) substitutions then we have $$Det(\Gamma) \rightarrow\forall A\in \Gamma, G^{**}(A)\text{ is determined}$$ This can be shown by noting that the payoff set of the Banach-Mazur Game $G^{**}(A)$ is in $\Gamma$ -However the Baire property alone is not enough to imply determinacy for all sets of reals. The Baire is much weaker in consistency strength than the assertation that the axiom of determinacy holds in $L(\mathbb{R})$ for example. By Work of Shelah, the assertation that the Baire property holds for all sets of reals ordinal definable from a real has same consistency strength as $ZFC$. In other words, one may "take away" the inaccessible Solovay used in his construction. However, by work of Woodin the theory $ZF+AD$ has same consistency strength as $ZFC+\omega$-many Woodin cardinals. -On the other hand, $AD$ is actually equivalent to Turing determinacy in $L(\mathbb{R})$, or also equivalent to cofinally many below $\Theta$ cardinals having the strong partition property.<|endoftext|> -TITLE: Kernel of evaluation map into field of quotients -QUESTION [6 upvotes]: Let $R$ be an integral domain and for $a \in R$ denote by $\text{eval}_a: R[X] \to R$ evaluation at $a$. It's well-known (and easy to see) that -$$\ker(\text{eval}_a)=(X-a).$$ -The next more complicated thing in this setting is to evaluate at an element $q$ of the quotient field $K$ of $R$: $\text{eval}_q: R[X] \to K,\,f \mapsto f(q)$. - -Question 1: Is there an explicit description of the generators of $\ker(\text{eval}_q)$ ? - -In particular, I wonder if - -$$\ker(\text{eval}_q) =(\,bX-a \mid q=\frac{a}{b};\,a,b \in R\,)\qquad ? $$ - -I could solve the following special cases: - -If $q=a\in R$ then $\ker(\text{eval}_q)=(X-a)$ -If $q=1/b$ then $\ker(\text{eval}_q)=(bX-1)$ -If $R$ is a GCD and $q=\frac{a}{b}$ with $a,b$ coprime then $\ker(\text{eval}_q)=(bX-a)$ - -According to 3. I wonder, if the GCD assumption is really needed: - -Question 2: If $a, b\in R$ are coprime, i.e. $(a,b)=R$, is $\ker(\text{eval}_q)=(bX-a)$ for $q=\frac{a}{b}$ ? - -$$$$ -For a proof of 3. note that $bX-a\in R[X]$ is irreducible and hence prime (since $R$ is GCD, $R[X]$ is also GCD and irreducible elements in a GCD are prime). If $f \in R[X]$ annulates $q$, write $f=(X-q)h$ for some $h \in K[X]$. By clearing denominators, there is $r \in R$ and $\tilde{h}\in R[X]$ such that $rf =(bX-a)\tilde{h} \in (bX-a)$. Since $(bX-a)$ is prime and $r \not\in (bX-a)$ we finally obtain $f \in (bX-a)$. -Remark: I have asked the question on math.SE but didn't get any reply: https://math.stackexchange.com/questions/2718227/kernel-of-evaluation-map-into-field-of-quotients - -REPLY [5 votes]: The following claim characterizes circumstances under which $\ker(\text{eval}_q)$ for $q \in K = \text{Frac}(R)$ is a principal ideal of $R[X]$. In particular, this answers question $2$ in the positive and provides a recipe to construct non-principal ideals of the form $\ker(\text{eval}_q)$. - -Claim 1. Let $R$ be an integral domain and let $q = \frac{a}{b} \in \text{Frac}(R)$ . Then the following are equivalent: -$(i)$ The ideal of denominators of $q$, that is $\mathfrak{d} \Doteq \{d \in R \,\vert\, dq \in R\}$, is the principal ideal of $R$ generated by $b$. -$(ii)$ For every $c \in R$, if $b$ divides $ac$ then $b$ divides $c$, in other words $Ra \cap Rb = Rab$. -$(iii)$ The ideal $\ker(\text{eval}_q)$ is the principal ideal of $R[X]$ generated by $bX - a$. - -Edit: This characterization is well-known under a slightly different form, see Addendum at the bottom. -Note that if $a, b \in R$ are co-prime in the sense that $Ra + Rb = R$, then $(ii)$ is satisfied. If $R$ is a Prüfer domain, e.g., a Dedekind domain, the condition $(ii)$ is equivalent to $Ra + Rb = R$. The condition $\text{gcd}(a, b) = 1$ doesn't imply $(ii)$ in general, but If $R$ is a pre-Schreier domain, then $(ii)$ is equivalent to $\text{gcd}(a, b) = 1$. You can find in [1] an example of a Schreier domain which is not a GCD domain. - -Proof of Claim 1. $(i) \Rightarrow (ii)$ is immediate. Let us show that $(ii) \Rightarrow (iii)$. To do so, let us consider $f(X) = \sum_{i = 0}^n a_i X^i \in \ker(\text{eval}_q)$ with $n > 0$. There is $g(X) = \sum_{i = 0}^{n - 1} b_i X^i \in K[X]$ such that $f(X) = (bX - a)g(X)$. We deduce from the latter identity that $b_{n - 1 - i} = \frac{ab_{n - i} + a_{n - i}}{b}$ for every $0 \le i \le n - 1$, agreeing that $b_n = 0$. The identity $f(q) = 0$ can be re-written as $$a_n a^n + a_{n- 1}a^{n - 1}b + \cdots + a_1 ab^{n - 1} + a_0b^n = 0.$$ Hence it follows from $(ii)$ that $b$ divides $a_n$, so that the above equality is equivalent to - $$ -a^{n - 1}(ab_{n - 1} + a_{n - 1}) + a_{n - 2}a^{n - 2}b + \cdots + a_1 ab^{n - 2} + a_0b^{n - 1 } = 0 -$$ - where $b_{n - 1} = \frac{a_n}{b} \in R$. Substituting $ab_{n - 1} + a_{n - 1}$ with $bb_{n -2}$, dividing the left-hand side by $b$ and using $(ii)$ again yields $b_{n - 2} \in R$. By repeating this process we eventually obtain that $g(X) \in R[X]$. Therefore $f(X) \in R[X](bX - a)$, which establishes $(iii)$. - We will complete the proof of Claim $1$ by showing that $\neg (i) \Rightarrow \neg (iii)$. By hypothesis we can find $d \in \mathfrak{d}$ such that $b$ doesn't divide $d$. As result $dX - dq \in \ker(\text{eval}_q) \setminus R[X](bX - a)$. - -The next claim underlines the special rôle of GCD domains with respect to OP's questions. - -Claim 2. - Let $R$ be an integral domain. Then the following are equivalent: -$(i)$ For every $q \in \text{Frac}(R)$, the ideal of denominators of $q$ is a principal ideal of $R$. -$(ii)$ $R$ is a GCD domain. - -If $q = \frac{a}{b}$, then $\mathfrak{d} = \frac{1}{a}(Ra \cap Rb)$. Thus the ideal of denominators of $q$ is principal if and only $Ra \cap Rb$ is principal, that is $\text{lcm}(a,b)$ exists. In this case, we know that $\text{gcd}(a, b)$ exists and is such that $\text{lcm}(a,b)\text{gcd}(a, b) = ab$ up to multiplication by a unit. - -Proof of Claim 2. Assertion $(i)$ is equivalent to the fact that $\text{lcm}(a,b)$ exists for every $a, b \in R$. The latter is equivalent to the fact that $\text{gcd}(a,b)$ exists for every $a, b \in R$. - -The following consequence is immediate. - -Corollary. Let $R$ be an integral domain. Then the following are equivalent: -$(i)$ For every $q \in \text{Frac}(R)$, the ideal $\ker(\text{eval}_q)$ is a principal ideal of $R[X]$. -$(ii)$ $R$ is a GCD domain. - -In general, we observe that $\{ bX - a \,\vert \, q = \frac{a}{b} \} = \{ dX - dq \,\vert \, d \in \mathfrak{d} \setminus \{0\}\}$ is the set of polynomials in $\ker(\text{eval}_q)$ with degree $1$. The question as to whether the latter set generates $\ker(\text{eval}_q)$ was answered in the negative by a user44191's comment: this doesn't hold when $R = \mathbb{Z}[\sqrt{5}]$ and $q = \frac{1 + \sqrt{5}}{2} \in \text{Frac}(R)$. Clearly, the ideal $\mathfrak{d}$ of denominators of $q$ contains $2$ and $1 - \sqrt{5}$. It is easy to check that the ideal $\mathfrak{m} = R \cdot 2 + R \cdot (1 - \sqrt{5})$ is a non-principal maximal ideal of $R$ of index $2$. Thus $\mathfrak{d} = \mathfrak{m}$ is not principal. As $X^2 - X -1 \in \ker(\text{eval}_q)$, any polynomial in $\ker(\text{eval}_q)$ is of the form $f(X)(X^2 - X - 1) + g(X)$ with $f(X), g(X) \in R[X]$, $g(X) = 0$ or $\deg(g(X)) \le 1$. Therefore $$\ker(\text{eval}_q) = R[X]((1 - \sqrt{5})X + 2) + R[X]((2X - (1 + \sqrt{5})X) + R[X](X^2 - X - 1).$$ -But we can get a positive answer for a class of rings which are not necessarily Noetherian. A domain $R$ is called a locally GCD domain if every localization $R_{\mathfrak{m}}$ of $R$ at a maximal ideal $\mathfrak{m}$ of $R$ is a GCD domain. - -Claim 3. Let $R$ be a locally GCD domain, e.g., a Prüfer domain, and let $q \in \text{Frac}(R)$. Then $\ker(\text{eval}_q)$ is generated as an ideal of $R[X]$ by the set $\{ bX - a \,\vert \, q = \frac{a}{b} \}$. - -I ignore if the converse of Claim 3 holds true. - -Proof of Claim 3. Let $i \ge 1$ and let $\mathfrak{c}_i$ be the ideal of $R$ generated the leading coefficients of the polynomials in $\ker(\text{eval}_q)$ with degree $i$. We have in particular $\mathfrak{c}_1 = \mathfrak{d}$, the ideal of denominators of $q$. We observe first that $\ker(\text{eval}_q)$ is generated by polynomials of degree at most $n \ge 1$ if and only if $\mathfrak{c}_i = \mathfrak{c}_n$ for every $i > n$. We shall establish that $\mathfrak{c}_i = \mathfrak{c}_1$ for every $i > 1$. Let $\mathfrak{m}$ be a maximal ideal of $R$ and let $\ker(\text{eval}_q)_{\mathfrak{m}}$ be the kernel of the evaluation map $f \mapsto f(q)$ from $R_{\mathfrak{m}}[X]$ to $\text{Frac}(R)$. Since $\mathfrak{c}_i = \left( \frac{b}{a^i} (Ra + Rb)^{i - 1} \right)\cap R$, the ideal generated by the leading coefficients of the polynomials in $\ker(\text{eval}_q)_{\mathfrak{m}}$ with degree $i$ is $\mathfrak{c}_i R_{\mathfrak{m}}$. As $R_{\mathfrak{m}}$ is a GCD domain by hypothesis, we deduce from Claim 2, that $\mathfrak{c}_i R_{\mathfrak{m}} = \mathfrak{c}_1 R_{\mathfrak{m}}$ for every $i > 1$. As this holds for every maximal ideal $\mathfrak{m}$, we deduce that $\mathfrak{c}_i = \mathfrak{c}_1 $ holds for every $i > 1$. - -Addendum. -I discovered that Claim 1 above is known under the following form: - -Claim 4 [2, Exercise 17.2]. Let $R$ be a commutative integral domain and $a \in R, b \in R \setminus \{0\}$ and let $q = \frac{a}{b} \in F(R)$. Then the following are equivalent: - -$(i)$ The sequence $(b, a)$ is a regular. -$(ii)$ The cohomology group $H^1(K(b,a))$ of the Koszul complex of $(b, a)$ is trivial, that is $(Rb: Ra)/Rb = \{0\}$. -$(iii)$ The polynomial $(bX - a)$ is a prime element of $R[X]$. -$(iv)$ The ideal $\ker(\text{eval}_q)$ is the principal ideal of $R[X]$ generated by $bX - a$. - - -A sequence $(a_1, \dots, a_n)$ of elements in a ring $R$ is said to be regular if for each $i$ the element $a_i$ is a regular element of $R/(Ra_1 + \cdots Ra_{i - 1})$. Given two ideals $I, J$ of $R$, we used above the following notation: $(I : J) \Doteq \{ r \in R \,\vert\, rJ \subseteq I \}$. - -[1] P. M. Cohn, "Bezout rings and their subrings", 1968. -[2] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 1995.<|endoftext|> -TITLE: Asymptotics of unrooted labeled forests -QUESTION [5 upvotes]: It is well known that the number of unrooted labeled trees on vertex set -$[n]={1,2,...,n}$ is $n^{n-2}$. Let $U(z)$ be the exponential generating function of the sequence of these numbers. Then $F(z)=\text{exp}(U(z))$ is the exponential generating function for the number of unrooted forests of $[n]$. -My question is, how do you prove that $n![z^n]F(z) \sim e^{1/2} n^{n-2}$. -In other words, that is the asymptotic number of unrooted forests on $[n]$. I found this result on page 406 of the Flajolet-Sedgewick book, without proof. -If that helps, $U(z)= T(z) - T(z)^2 /2$ , where $T(z)$ is the exponential generating function for the numbers of rooted trees. But while $T(z)$ satisfies the relation $T=z \ \text{exp}(T)$, I could not find such a relation for F. - -REPLY [6 votes]: The idea in the book by Flajolet-Sedgewick is to use singularity analysis. The generating function $F(z)=\exp(U(z))$ with $U=T-T^2/2$ inherits the dominant singularity $e^{-1}$ of $T(z)=-W(-z)$, where $W$ is the Lambert $W$ function. From the local expansion of $T(z)$ -$$T(z)=1-\sqrt{2}\sqrt{1-ze}+\frac{2}{3}(1-ze)-\frac{11\sqrt{2}}{36}(1-ze)^{3/2}+O((1-ze)^2),\quad z\rightarrow e^{-1}$$ -follows that of $F$: -$$F(z)=e^{1/2}\left(1-(1-ze)+\frac{2\sqrt{2}}{3}(1-ze)^{3/2}+O((1-ze)^2)\right).$$ -From there, singularity analysis is straightforward. The asymptotic behaviour of the coefficients comes from the term in $(1-ze)^{3/2}$, leading to -$$[z^n]F(z)\sim \frac{e^{1/2}}{\sqrt{2\pi}}n^{-5/2}e^{n}.$$ -That coefficient is the number of unrooted labeled trees divided by $n!$, so that multiplying by Stirling's formula gives the desired $e^{1/2}n^{n-2}$.<|endoftext|> -TITLE: Collection of Mathematical Constants -QUESTION [10 upvotes]: I recently stumbled over the large collection of mathematical constants of Mauro Fiorentini; it is in Italian and appears to be something in the vein of the famous OEIS, however maintained by a single person in isolation. -Question: -what is known about the author of, and the motivation for, that collection of constants and, is something similar available from other sources (the link leads to the first of several hundred tables of constants)? - -REPLY [5 votes]: Steven Finch used to have a site full of essays about his favourite constants including a table, though the contents page is actually more interesting -He then turned this into a book Mathematical Constants (Encyclopedia of Mathematics and its Applications 94, CUP, 2003), formalising the essays with an expanded table on pages 543-566. The website then shrank before disappearing<|endoftext|> -TITLE: When to publish minor results? -QUESTION [35 upvotes]: One of the main things I enjoy about MO and MSE is the chance to solve somewhat difficult problems. On a couple of occasions, I've found proofs of results (posed as questions mostly on MSE) which seem original and not entirely trivial. The examples I'm most fond of are short but tricky, not in any way revolutionary, but have a satisfying "aha!" at a crucial point. -Disappointingly, these questions seem to always disappear into the mists of time. While my answers to "soft questions" often get enormously upvoted and see lots of views, those answers to serious questions which actually have some original intellectual content often go largely unnoticed. Once they fall off the top of the question list, they essentially are never seen again. -Perhaps this fate is inevitable, but it seems a shame to let a cool trick die. This has got me wondering whether to try to publish any of these. I'm not a professional mathematician, so this likely would not materially benefit me in any way. The only goal would be to put the results out in a place where they might be of use. The options I know of are: - -Drop it on arxiv. -Publish in the usual way, in a topical journal. -Publish a note in an expository journal like American Mathematical Monthly. -Submit as a problem to a journal that publishes problems. - -I've published in the Monthly before, and it seems a good place for small results of broad interest. However, it's not without cost (submitting takes work), it's not totally clear that results published in the Monthly "do any more good" than those rotting in the bowels of MSE, and moreover not all minor results are of sufficiently broad interest for an expository journal to be a viable option anyway. -So my questions are: - -What's the best way(s) (in the sense of "being maximally useful to those who may find them interesting") to disseminate minor mathematical results? -Is it worth the effort? Or are all minor discoveries bound to languish in obscurity until such time that they are rediscovered and applied to a major result? -Are there any good rules of thumb to determine what minor results are worthy of dissemination (if any)? - - -A couple afterthoughts: - -I've noticed stackoverflow doesn't have the same problem with questions not having any later use, presumably because programming questions are more likely to be shared by many people. -I can give examples of the minor results that inspired this question if requested. -I think this question should be community wiki, but apparently I'm not allowed to designate it as such myself. - -REPLY [14 votes]: If you're not interested in self-promotion but are purely thinking about how best to disseminate your results, then I think that we're at a point in history where the question you should be asking isn't "Where should I publish this?" but "What can I do to maximize the probability that when people in the future are in need of this idea, their search engines will find it for them?" -One of the motivations for the Tricki was to serve as a repository for "useful tricks." The Tricki has had limited success IMO but I believe it is a step in the right direction. -I looked at the two examples you provided and it seemed to me that in both cases, even if you went to the trouble of formally publishing the results, the chances of someone discovering your paper with a search engine are not as high as one would like, because there's no obvious keyword or search term that they would use. Thus, the work required to publish the result doesn't seem quite worth it, unless you also find some way to solve the problem of making your work discoverable by search engines. -Of course I'm making an assumption here—that for example you would want your answer to fedja's question regarding a proof of Enflo's theorem to be discoverable by more people than just the small number of people who are interested in different proofs of Enflo's theorem. Under that assumption, I think that it might be fruitful to ask yourself, "Are there other results similar to mine that are scattered across the literature and the Internet, and if so, what is a good organizing principle that would make it easier for people to navigate this corner of the mathematical universe?" Obviously, this takes way more work than simply publishing your result, but the benefits will be substantial if you succeed.<|endoftext|> -TITLE: Finite dimensional irreducible representations of quasisplit p-adic groups -QUESTION [5 upvotes]: For split groups over a $p$-adic field, every irreducible smooth (complex) representation is either infinite-dimensional or one-dimensional. Is it true for quasisplit groups that split over an unramified extension, or quasisplit groups in general? Unfortunately I don't have a good feel for examples, since the books I consult for learning the representation theory of $p$-adic groups usually stick to groups like $GL_n$. - -REPLY [6 votes]: See Prop. 3.9 of http://math.stanford.edu/~conrad/JLseminar/Notes/L2.pdf for an optimal affirmative answer (no quasi-split condition needed: any connected reductive group over any non-archimedean local field, with the minimal necessary isotropicity hypotheses).<|endoftext|> -TITLE: Bounds on Tamagawa numbers of reductive groups -QUESTION [10 upvotes]: Let $G$ be a reductive algebraic group over a number field $k$. Weil's conjecture on Tamagawa numbers (now a theorem) tells us that the Tamagawa number $\tau(G)$ of $G$ is 1 if $G$ is semisimple and simply-connected. Are there known bounds for $\tau(G)$ for the general case? -Presumably one might be able to use the formula $\tau(G) = |\text{Pic}(G)|/|\text{Sha}(G)|$ and try to estimate the numerator, but I am not familiar with the literature in this respect. - -REPLY [4 votes]: Yes, the formula is correct, see Sansuc, J.-J. Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80, (10.1.2). -In the extreme cases (when G is semisimple or a torus) there are formulas for ${\rm Pic}(G)$ in Lemma 6.9 of Sansuc's paper. -Namely, when $G$ is semisimple with fundamental group $B=\pi_1(G_\bar k)$, we have -$${\rm Pic}(G)=X^*(B)^\Gamma,$$ -where $X^*$ denotes the character group and $\Gamma={\rm Gal}(\bar k/k)$. -It follows that the order of the Picard group is bounded by the order of the fundamental group of $G_{\bar k}$. -When $G$ is a torus, we have -$${\rm Pic}(G)=H^1(k, X^*(G)).$$ -For a general reductive group $G$, set -$G^{\rm ss}=[G,G]$ and $G^{\rm tor}=G/G^{\rm ss}$, -then for the semisimple part of $G$ we obtain $$\#{\rm Pic}(G^{\rm ss})\le\#\pi_1(G^{\rm ss}_{\bar k})$$ -and for the toric part we obtain -$$\#{\rm Pic}(G^{\rm tor})=\# H^1(k,X^*(G)).$$ -From the short exact sequence -$$ 1\to G^{\rm ss}\to G\to G^{\rm tor}\to 1$$ -we obtain an exact sequence -$${\rm Pic}\ G^{\rm tor}\to{\rm Pic}\ G\to {\rm Pic}\ G^{\rm ss},$$ -see Sansuc's paper, Corollary 6.11. We obtain the following bound: -$$\#\tau(G)\ \le\ \#{\rm Pic}(G)\ \le \# H^1(k,X^*(G))\cdot \#\pi_1(G^{\rm ss}_{\bar k}).$$<|endoftext|> -TITLE: Lifting a representation from a discrete subgroup -QUESTION [7 upvotes]: Let $\pi_0$ be an irreducible representation of a discrete subgroup $\Gamma$ of the reductive group $G$. My question in general is - -What can be said of representations $\pi$ of $G$ extending $\pi_0$, in the sense that $\pi$ acts on $\Gamma$ as $\pi_0$? - -I would like to know technical possibilities in this direction, even assuming freely further properties on $\pi_0$. What if in particular if $\pi_0$ is finite dimensional? -The base case $\Gamma = SL(2, \mathbb{Z})$ and $G=SL(2, \mathbb{Q})$ is already of interest for me. - -REPLY [5 votes]: If $\Gamma$ is a lattice in a connected semisimple $\mathbb{R}$-algebraic group $G$ without compact factors, then the Borel density theorem says that $\Gamma$ is Zariski dense in $G$. This implies that there can be at most one extension of a representation of $\Gamma$ to a rational representation of $G$. -Of course, extensions need not exist. For finite-dimensional representations, however, you can often get information using the Margulis superrigidity theorem. This says that if $\Gamma$ is a lattice in a semisimple Lie group $G$ satisfying some technical assumptions and if $V$ is a finite-dimensional representation of $\Gamma$, then there exists a finite-index subgroup $\Gamma'$ of $\Gamma$ such that the action of $\Gamma'$ on $V$ extends to a rational representation of $G$. -Passing to a finite-index subgroup is necessary here. For instance, if the action of $\Gamma$ on $V$ factors through a finite group, then you first have to pass to a subgroup $\Gamma'$ that acts trivially on $V$ (and the extension to $G$ is then the trivial action). -The most important technical assumption in the superrigidity theorem is that $G$ is of higher rank. It applies to $\text{SL}(n,\mathbb{Z})$ in $\text{SL}(n,\mathbb{R})$ for $n \geq 3$, but not for $n=2$.<|endoftext|> -TITLE: generators for the handlebody group of genus two -QUESTION [5 upvotes]: Is the handlebody group of genus two surface generated by Dehn twists along properly embedded disks and annuli? -Are there alternative ways to describe a set of generators that are conceptually simple (not necessarily finite but conceptually simple)? For this part, really I'm asking the question for the handlebody subgroup of the mapping class group so I am considering only the restriction of the homeomorphisms to the boundary of the handlebody. Therefore a description in terms of Dehn twists that are tractable is appreciated. -Context: Define a handlebody $B$ as a regular neighbourhood of a graph embedded in $\mathbb{R}^3$. The genus of a handlebody is defined as the rank of its fundamental group. The boundary of a genus $g$ handlebody is a surface of genus $g$ then. Let the handlebody group of genus $g$ be the group of homeomorphisms of the handlebody up to isotopy. For any properly embedded disk $D$ or annulus $A$ in $B$, one can define a positive Dehn twist along $D$ or $A$ in the standard way by either twisting once counter-clockwise around the center of a regular neighborhood of $D$ or twisting along the $S^1$ direction in a regular neighborhood of $A$. - -REPLY [5 votes]: As stated in Ian Agol's answer, the mapping class group of a handlebody $B$ maps onto $Out(\pi_1B)$. This is easy to show by lifting known generators for the automorphism group of a free group. Twists along disks lie in the kernel of the surjection to $Out(\pi_1B)$, and it is a theorem of Luft (Math. Ann. 234 (1978) 279-292) that this kernel is generated by twists along disks. (McCullough showed this kernel is not finitely generated when the genus of $B$ is at least two.) -Twists along annuli need not lie in the kernel of the map to $Out(\pi_1B)$, and in fact the subgroup they generate maps onto a subgroup of $Out(\pi_1B)$ of index at most two. For example, if one considers $B$ as obtained by attaching a 1-handle $H$ to a handlebody $B'$ of genus one less than the genus of $B$, then the homeomorphism of $B$ obtained by dragging one end of $H$ around an embedded loop in $\partial B'$ disjoint from the disk where the other end of $H$ attaches can be realized by a twist along an annulus. The annulus lies in $B'$ and is isotopic to an annulus in $\partial B'$ which is a neighborhood of the loop used for the dragging operation, with one end of $H$ lying in the interior of the annulus in $\partial B'$ and the other end lying outside this annulus. The induced automorphism of $\pi_1B$ fixes $\pi_1B'$ and takes the generator corresponding to $H$ to the product of itself with the element of $\pi_1B'$ represented by the embedded loop. Such automorphisms generate an index two subgroup of $Out(\pi_1B)$, the subgroup mapping to automorphisms of $H_1(B)$ of positive determinant. To get the rest of $Out(\pi_1B)$ one can take a half-twist along a disk in $B$ splitting off a solid torus.<|endoftext|> -TITLE: What is the scope of validity of Kunneth formula for de Rham? -QUESTION [7 upvotes]: In books like Bott-Tu or all pdf texts I have found on internet, the Kunneth formula for manifolds $M$ and $N$ and their de Rham cohomology -$$ H^{\bullet}_{dR}(M \times N) \simeq H^{\bullet}_{dR}(M) \otimes H^{\bullet}_{dR}(N)$$ -is proved under various finiteness hypothesis : one of the two manifolds is compact, or with finite dimensional de Rham spaces, or admitting a finite good cover. -On another side, for singular cohomology of topological spaces $X$ and $Y$ and a PID $A$ (let's say $Z$) there is the following more general Kunneth formula : -$$H^n(X \times Y;A) \simeq (\sum_{i+j=n} H^i(X;A)\otimes H^j(Y;A))\oplus(\sum_{p+q=n+1} Tor(H^p(X;A),H^q(Y;A))$$ -If $X$ and $Y$ are manifolds and we take $A \equiv \mathbb{R}$, the $H^p(X;\mathbb{R})$ ans $H^q(Y;\mathbb{R})$ are vector spaces, so the $Tor$ part is null, and using the de Rham theorem we end up with a Kunneth formula for de Rham without any finitenes hypothesis. -But Bott-Tu p 108 give an explicit counterexample to the Kunneth formula when both manifolds have infinite dimensional cohomology, and write "that some sort of finiteness hypothesis is necessary for Kunneth and Leray-Hirsch to hold". -So what is wrong with the Kunneth formula for de Rham "deduced" from kunneth for singular above ? -And what is the real scope of Kunneth formula for de Rham i.e. the minimal hypothesis for the formula to hold ? - -REPLY [7 votes]: Let me convert my comment into an answer and add something: -I think the assumptions (good cover / compact) you mention are simply there to provide a simpler proof. If it's true for singular cohomology, then it's true for de Rham cohomology, by de Rham's theorem. I'm pretty sure the most general statement you can get for de Rham's cohomology is that you just want $H^k(Y)$ to be finite-dimensional for all $k$. CW is not even necessary (thanks to CW approximations), and "free" is automatic over a field (well, if you assume the axiom of choice I guess). -Note that this finiteness assumption is also necessary for general spaces, the Künneth formula you wrote is only valid when at least one of the two spaces have finite-type cohomology. -There is a Künneth formula for homology, which has no assumptions about finiteness, but if you pass to cohomology without finiteness assumptions, you run into problem. The actual statement for cohomology is $H^n(X \times Y; F) \cong \bigoplus_{i+j=n} H^i(X; H^j(Y;F))$ for a field $F$. But if $H^j(Y;F)$ is not finitely generated, you don't necessarily have $H^i(X; H^j(Y;F)) \cong H^i(X;F) \otimes_F H^j(Y;F)$. So there is no contradiction here.<|endoftext|> -TITLE: Is the Ford-Fulkerson algorithm a tropical rational function? -QUESTION [24 upvotes]: The Ford-Fulkerson algorithm -Let me recall the standard scenario of flow optimization (for integer flows at least): -Let $\mathbb{N} = \left\{0,1,2,\ldots\right\}$. Consider a digraph $D$ with vertex set $V$ and arc set $A$. Fix two distinct vertices $s$ and $t$ of $D$, which we call the source and the sink (despite not requiring them to satisfy anything specific). A capacity function means a map $c : A \to \mathbb{N}$, which is thought of as assigning to each arc of $D$ the maximum amount of whatever it carries. Given a capacity function $c$, a $c$-flow (or integer $c$-flow) means a map $f : A \to \mathbb{N}$ satisfying the following two properties: - -the capacity constraints: $0 \leq f\left(a\right) \leq c\left(a\right)$ for any arc $a \in A$; -the conservation constraints: $\sum\limits_{\text{arcs } a \text{ having source } v} f\left(a\right) = \sum\limits_{\text{arcs } a \text{ having target } v} f\left(a\right)$ for any vertex $v \notin \left\{s,t\right\}$. - -The value of a $c$-flow is defined to be the number \begin{equation} -\sum\limits_{\text{arcs } a \text{ having source } s} f\left(a\right) - \sum\limits_{\text{arcs } a \text{ having target } s} f\left(a\right) \\ -= \sum\limits_{\text{arcs } a \text{ having target } t} f\left(a\right) - \sum\limits_{\text{arcs } a \text{ having source } t} f\left(a\right) . -\end{equation} -The Ford-Fulkerson algorithm (see, e.g., Section 8.2 in Jeremy L. Martin's Lecture Notes on Algebraic Combinatorics, or Section 4.3 in Lex Schrijver's A Course in Combinatorial Optimization, or Timon Thalwitzer's Max-Flow Min-Cut, or Section 1.8 in my Spring 2017 Math 5707 Lecture 16) constructs a $c$-flow with the maximum value. The algorithm proceeds by starting with the identically-$0$ flow and progressively improving it by finding a path in a certain (changing) digraph and modifying the flow along that path. The algorithm is not deterministic (it relies on finding a path in a digraph), but there are various ways of making it deterministic (such as taking the lexicographically smallest path, but there are also smarter ways; Section 4.4 in Schrijver's notes shows one). -The question - -Question 1. Is there any deterministic version of the Ford-Fulkerson algorithm that yields a maximum-value $c$-flow as a "tropical rational map" of $c$ ? - -What do I mean by "tropical rational map"? For each $i$ and $k$, we let $\pi_i$ denote the map $\mathbb{N}^k \to \mathbb{N}$ that sends each $k$-tuple to its $i$-th entry. A tropical rational function is a map $\mathbb{N}^k \to \mathbb{N}$ that can be formed from the coordinate projections maps $\pi_i$ using addition, subtraction, minimum and maximum. (For instance, $\mathbb{N}^5 \to \mathbb{N},\ \left(a,b,c,d,e\right)\mapsto \min\left\{a,b\right\} + \max\left\{c,d\right\} - e$ is a tropical rational function... a rather important one in fact. So is $\mathbb{N}^2 \to \mathbb{N},\ \left(a,b\right) \mapsto \left|a+b\right| + \left|a-b\right|$.) A tropical rational map is a map $g : \mathbb{N}^k \to \mathbb{N}^\ell$ such that $\pi_i \circ g$ is a tropical rational function for each $i \in \ell$. -Any deterministic version of the Ford-Fulkerson algorithm can be regarded as a map $\mathbb{N}^A \to \mathbb{N}^A$ that sends any capacity function $c$ to a $c$-flow $f$. A priori, there is no reason for such a map even to be continuous (e.g., if it chooses between two parallel edges based on which one has bigger capacity, then it won't be continuous). Even for "reasonable" deterministic versions of the Ford-Fulkerson algorithm, I am not sure if they yield a tropical rational function (where we identify $\mathbb{N}^A$ with $\mathbb{N}^k$ for $k = \left|A\right|$). Indeed, if we allow capacities and flows to have non-integer values, then a "stupid" choice of augmenting paths in the Ford-Fulkerson algorithm leads to a never-terminating execution which doesn't even converge to a maximum-value flow; this doesn't happen with properly chosen augmenting paths, however. -Motivation: The Hillman-Grassl algorithm -The following probably will mostly only make sense to the enumerative and algebraic combinatorialists. Here is why I suspect the Ford-Fulkerson algorithm to have a tropical-rational-function avatar. -The Hillman-Grassl corresondence (a bijection between reverse plane partitions of a given shape and arbitrary arrays of this shape; see, e.g., Section 5 of Alejandro Morales, Igor Pak, Greta Panova, Hook formulas for skew shapes I. q-analogues and bijections for a definition) is an algorithm that progressively decreases the entries of array by following something like "anti-augmenting paths". From this point of view, it is eerily similar to Ford-Fulkerson, apart from the fact that it keeps decreasing rather than increasing numbers. (It differs from Ford-Fulkerson also in what it records -- it's as if you wouldn't care about the maximum-value flow, but instead care about the augmenting paths you've used to build it.) - -Question 2. Can we view the Hillman-Grassl correspondence as a Ford-Fulkerson algorithm for some directed graph, made deterministic according to some path-choosing rule? - -Anyway, here is the motivation for the above question: - -Proposition 3. The Hillman-Grassl correspondence is a tropical rational map. - -Here is an outline of a proof of Proposition 3 using a lot of handwaving (I don't really want to claim I'm sure of it). Recall how the Hillman-Grassl correspondence "disassembles" a reverse plane partition $\pi$ -- it works in multiple steps, where each step constructs a lattice path with north and east steps. In the form in which it is usually stated, the Hillman-Grassl algorithm decrements all entries of $\pi$ on this path by $1$; but we can modify this to make it decrement them by $r$ instead, where $r$ is the largest number they can be decremented by without breaking the reverse-plane-partition condition (explicitly: $r$ is the smallest difference of the form $\pi_{i,j} - \pi_{i-1,j}$ where $\left(i,j\right) \to \left(i,j+1\right)$ is an eastward step of the path). (This just corresponds to taking $r$ consecutive steps of the Hillman-Grassl algorithm.) Now, the crucial point is that there is a specific total order on all relevant lattice paths ("relevant" means that the path begins on a southern edge of the partition, and ends on an eastern one) such that the Hillman-Grassl algorithm uses the lattice paths in this order. (This is stronger than Lemma 4.2.3 in Bruce E. Sagan, The Symmetric Group, 2nd edition 2001 (errata); that lemma just says that the hooks that span the lattice paths appear in a specific order; but the rest is easy.) Thus, we can replace the algorithm by a "blind" version ("blind" in the sense that the lattice paths no longer depend on the entries of the reverse plane partition), which tries out all relevant lattice paths in this total order (there are only finitely many), decrementing each of them as much as possible. (Paths that don't appear in the original Hillman-Grassl algorithm will appear in this "blind" version, but the entries on them get decremented by $0$, so they don't make a difference.) So the "blind" version describes a tropical rational map. - -Question 4. Is the inverse of the Hillman-Grassl correspondence (i.e., the map back from arrays to reverse plane partitions) a tropical rational map as well? - -Note that I am aware of an "alternative" to the Hillman-Grassl correspondence (found by Pak, Hopkins and Sulzgruber independently), which is manifestly a tropical rational function (in both directions), obtained essentially by composing toggles. But Hillman-Grassl itself is not clearly of this form, and even the Greene-Kleitman invariants allegedly proven by Gansner in his thesis don't make that clear. - -REPLY [2 votes]: In the comments I suggested that Hillman-Grassl may be related to a "column insertion" version of RSK. Based on a few examples (larger than $2\times 2$) I checked, I am now less sure that there is a direct connection. However, answering your question four, I think that "inverse Hillman-Grassl correspondence", as you call it, (i.e., the map from fillings of $\lambda$ with arbitrary nonnegative integers to reverse plane partitions of shape $\lambda$) can be realized as a tropical rational map. -The reason is: I think the description (which you mentioned) of Hillman-Grassl that appears as Theorem 3.3 in the paper "The Hillman-Grassl correspondence and the enumeration of reverse plane partitions -" of Gansner (https://www.sciencedirect.com/science/article/pii/0097316581900418) explains exactly how to do this. Namely, I think to compute the quantity $a_k(M_l)$, in the language of that theorem, we can do the following. Let's call a modified A-chain in $M_l$ a sequence of boxes $u_1,u_2,\ldots,u_m$ in the shape of $M_l$ which does not repeat any boxes and such that $u_{i+1}$ is weakly southwest of $u_i$ for all $i=1,\ldots,m-1$. Associate to such a chain the weight which is simply the sum of the $M_l$ entries of the boxes it visits. Then $a_k(M_l)$ should be the maximum over all collections of $k$ disjoint modified A-chains of the sum of the weights of these chains. This fact, together with the Theorem 3.3, explains how to write the entries of the output reverse plane partition entries as tropical rational functions in the input entries. -EDIT: -Darij, in case you are still interested in this question- I believe this recent preprint of Garver, Patrias, and Thomas https://arxiv.org/abs/1812.08345 not only shows that the Hillman-Grassl correspondence is a PL-function, but in fact gives a "toggle" description of the map, akin to the "toggle" description of PL-RSK. In fact, they show that these two maps (Hillman-Grassl and RSK) are essentially built out of the same toggles, but correspond to two different orientations of the Type A Dynkin diagram.<|endoftext|> -TITLE: How to compute the group cohomology of $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ with coefficient in a trivial module? -QUESTION [6 upvotes]: The group cohomology of cyclic groups can be computed easily due to the periodity. Now how can one compute the group cohomology $H^r(\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z},M)$? As least in some special case, e.g. $M$ has trivial action, or even $M=\mathbb{Z}$? I would like to know the case that $M$ is not $\mathbb{Z}$. -Furthermore, can one compute $H^r(\mathbb{Z}/n\mathbb{Z}\times\cdots\times \mathbb{Z}/n\mathbb{Z},M)$? - -REPLY [5 votes]: If $M$ is torsion-free (as a $\mathbb{Z}-$module), then the Kunneth formula still holds. If $M = \mathbb{Z}$ with trivial group action, then it holds by the corresponding result for the CW-complex category and abstract nonsense; more generally, it holds for chain complexes under some fairly mild restrictions (including having finite homology in each dimension, but that's clear here.)<|endoftext|> -TITLE: Functions that are approximately differentiable a.e -QUESTION [8 upvotes]: The classical definition of an approximately differentiable function is as follows: - -Definition. -Let $f:E\to\mathbb{R}$ be a measurable function defined on a measurable set $E\subset\mathbb{R}^n$. We say that $f$ is - approximately differentiable at $x\in E$ if there is a linear function - $L:\mathbb{R}^n\to\mathbb{R}$ such that for any $\varepsilon>0$ the - set $$ \left\{ y\in E:\, \frac{|f(y)-f(x)-L(y-x)|}{|y-x|} <\varepsilon - \right\} \quad \text{has $x$ as a density point.} $$ - -It turns out that this definition is equivalent to the one described in the following result. - -Theorem. A measurable function $f:E\to\mathbb{R}$ defined in a measurable set $E\subset\mathbb{R}^n$ is approximately differentiable - at $x\in E$ if and only if there is a measurable set $E_x\subset E$ - and a linear function $L:\mathbb{R}^n\to\mathbb{R}$ such that $x$ is a - density point of $E_x$ and $$ \lim_{E_x\ni y\to x} - \frac{|f(y)-f(x)-L(y-x)|}{|y-x|} = 0. $$ - -Do you know any reference for the proof of this result? Since we could not find a good reference, we proved it in the appendix of the paper listed below, but I am pretty sure it can be found somewhere else. -P. Goldstein, P. Hajłasz, A measure and orientation preserving homeomorphism with approximate Jacobian equal −1 almost everywhere. -Arch. Ration. Mech. Anal. 225 (2017), 65–88. -In my opinion it is a folklore result that most of the people working with approximately differentiable functions know. The characterization given in the theorem is much easier to use than the original definition. - -REPLY [3 votes]: The equivalence between the two definitions is in fact really straightforward. Call $E_x^\varepsilon$ the set appearing in the first definition. Clearly the second definition implies the first one, as for $r$ small enough (depending on $\varepsilon$) $E_x\cap B_r(x)\subseteq E_x^\varepsilon$. -Conversely, let $k_0:=0$ and take, for each integer $j>0$, the smallest integer $k_j>k_{j-1}$ such that -$$|E_x^{1/j}\cap(B_r(x)\setminus B_{r/2}(x))|\ge(1-1/j)|B_r(x)\setminus B_{r/2}(x)|$$ -for all $r\le 2^{-k_j}$. Notice that you can pick -$$E_x:=\bigcup_{j=1}^\infty\bigcup_{k=k_j}^{k_{j+1}-1}E_x^{1/j}\cap (B_{2^{-k}}(x)\setminus B_{2^{-k-1}}(x)).$$<|endoftext|> -TITLE: How many disjoint compact sets are needed to form a connected compactum? -QUESTION [5 upvotes]: Let's assume all spaces are metrizable. For each connected compact space $X$ let $\mathscr K(X)$ be the set of all partitions of $X$ into non-empty compact sets, excluding the trivial partition $\{X\}$. Let $$\mathfrak \kappa=\min\{|\mathcal K|:(\exists \text{ a connected compactum }X)(\mathcal K\in \mathscr K(X))\}.$$ -We know $$\aleph_0<\kappa \leq |\mathbb R|;$$ -the first inequality is due to Sierpinski (Theorem 6.1.27 in Engelking's Topology), and the second is true because every connected space has a partition into $\mathbb R$-many singletons. -Is it necessarily true that $\kappa=|\mathbb R|$? Or is this axiom-dependent? -EDIT: What about just for the space $X=[0,1]$? Is the minimum cardinality of a compact partition of $X$ necessarily $|\mathbb R|?$ - -REPLY [5 votes]: It is indeed axiom independent. In the Solovay random model (adding $\aleph_2$ random reals over a model of ZFC+GCH) you have $2^\omega=\aleph_2$ and there exists a $\aleph_1$ partition of $[0,1]$ into nowhere dense closed sets (hence compact). The idea is to define such a collection in the ground model and use the key fact that no Cohen real is added in random forcing to capture each new real using some nowhere dense closed set. In this way, no new (random) real escapes from the partition we fixed to begin with in the gound model. I thought it was folklore but actually found that Jacques Stern has an article about this.<|endoftext|> -TITLE: Using Dunwoody's results on cohomological dimension to learn about a von Neumann regular group ring -QUESTION [7 upvotes]: Just recently I've stumbled across Warren Dicks' book Groups, trees and projective modules (1980) and I was pretty stunned. I know nothing of group cohomology, but I gather the "tree" component is a special case of space studied in general group cohomology. -The results that caught my attention were: - -The augmentation ideal of $R[G]$ is projective iff $G$ has cohomological dimension at most $1$, and it has cohomological dimension exactly $1$ iff it is an infinite group of cohomological dimension $1$. -The augmentation ideal of $R[G]$ is projective iff $G$ is the fundamental group of a graph of finite groups having order invertible in $R$. - -For me this resonated with two other results I know well about group rings: - -(Renault) $R[G]$ is right self-injective iff $R$ is right self-injective and $G$ is finite. - -and - -(Connell) $R[G]$ is von Neumann regular iff 1) $R$ is von Neumann regular; 2) $G$ is locally finite; and 3) the order of every finite subgroup of $G$ is invertible in $R$. - - -Suppose from here on that $R[G]$ is von Neumann regular, $G$ is (at most) countable, and you don't know Connell's proof about regular group rings. -Then the augmentation ideal is countably generated, and by a result of Kaplansky the augmentation ideal must be projective. Apparently now $G$ is "the fundamental group of a graph of finite groups having order invertible in $R$." -First question: is it obvious somehow that $G$ is locally finite and that the groups which are vertices of the graph of groups tell us about the finite subgroups of $G$ and the invertibility of their orders? (I.e., can you recover Connell's theorem from Dunwoody's theorem?) -Second question: now additionally assume $R[G]$ is right self-injective and you don't know Renault's result. Can right self-injectivity be interpreted in this context to explain why $G$ is finite, i.e. $G$ has cohomological dimension $0$? (I.e., can you recover Renault's theorem in the special case of VNR right self-injective rings from Dunwoody's theorem by explaining why injectivity of $R[G]$ decides that $G$ is finite?) -I'm just probing around here seeing if I can get some connection between the two disciplines. - -REPLY [2 votes]: I am not sure exactly what constitutes an answer to question 1, but here is a proof using the Dunwoody-Dicks stuff that if $G$ is a countable locally finite group, each of whose finite subgroups has order invertible in $R$, then the augmentation ideal of $RG$ is projective (recovering Kaplansky's result without using Connell). -Since $G$ is countable, we can write $G=\bigcup_{n\geq } G_n$ where $\{1\}=G_1\subset G_2\subset \cdots$ is a countable chain of finite subgroups (an infinite chain if $G$ is infinite, which is the interesting case) with orders invertible in $R$. -Let $T$ be the tree with vertex set $\coprod_{n\geq 1}G/G_n$ and we connect $gG_n$ to $gG_{n+1}$ by an edge for $g\in G$. Then $G$ acts on the tree $T$ in an obvious way, and the vertex stabilizers are conjugates of the $G_n$, and hence finite with orders invertible in $R$. Thus $G$ has a projective augmentation module by the result from Dunwoody-Dicks you cited. -The corresponding graph of groups is just a one-sided infinite ray with $G_i$ at vertex $i$ and edge $i$ and the edge groups include in the natural way ($G_i$ is mapped to $G_i$ by the identity and to $G_{i+1}$ by the inclusion) so the expression as a fundamental group of a graph of groups is the direct limit. -Having said that, this is most likely almost the same as Kaplansky's proof (I've never seen it) written in a geometric way. Going from an action on a tree with finite stabilizers of order invertible in $R$ to projective augmentation module is the easy direction. One just needs that the permutation module $R[G/H]$ is projective when $H$ is finite with order invertible in $R$, which is obvious since it is isomorphic to $RGe$ with $e$ the idempotent $e=\frac{1}{|H|}\sum_{h\in H}h$. Then one can use that the augmented simplicial chain complex of a tree is exact. -Going from projective augmentation module to the action on the tree is the hard direction and uses the Almost Stability Theorem, which is a fancy version of Stallings Ends Theorem. - -REPLY [2 votes]: I'm adding a new answer since my other answer was the converse. -Question 1: -I can `simplify' Connell's argument using Dunwoody-Dicks. Assume that $RG$ is von Neumann regular. Note that if $g\in G$, then $(1-g)r(1-g)=1-g$ for some $r\in R$. So $(1-g)(1- r(1-g))=0$. As $r(1-g)$ is in the augmentation ideal, it follows that $1-r(1-g)\neq 0$ and so $1-g$ is a left zero divisor. This implies $g$ has finite order by standard group ring arguments (in order for $a=ga$ we must have $g$ permutes the finite support of $a$ and hence has finite order). So $G$ is torsion. -If $G$ is countable, it has projective augmentation ideal (since countably generated left ideals in a von Neumann regular ring are projective) and so $G$ acts on a tree with finite stabilizers with order invertible $R$ by Dunwoody-Dicks and because the augmentation ideal is projective. Thus each finitely generated subgroup of $H$ also acts on a tree with finite stabilizers of order invertible in $R$ and we prove that $H$ is finite. But any finitely generated torsion group acing on a tree has a global fixed point (see Serre's book, where it is shown any finitely generated group of elliptic automorphisms has a global fixed point) and so $H$ is finite with order invertible in $R$. Alternatively, a finitely generated group acting on a tree with finite stabilizers has a finite index free subgroup by Bass-Serre theory, which must be trivial in our case since $H$ is a torsion group. So $H$ is finite and hence is a subgroup of a vertex stabilizer so its order is invertible in $R$. Thus $G$ is locally finite.<|endoftext|> -TITLE: Is $\eta(\tau)^2$ a modular form of weight 1 on $\Gamma(12)$? -QUESTION [15 upvotes]: As we know, the Dedekind eta function $\eta(\tau)$ acquires a phase $\exp(2\pi i/24)$ under the modular transformation: $\tau \rightarrow \tau+1$. Therefore $\eta(\tau)^2$ is invariant under $\tau \rightarrow \tau+12$. Here comes the question, is $\eta(\tau)^2$ invariant on $\Gamma(12)$, where $\Gamma(12)$ denotes the principal subgroup? - -REPLY [14 votes]: Yes, it is. There are standard criteria (stated for example in Ken Ono's book "Web of Modularity" - Theorems 1.64 and 1.65) that indicate when an eta quotient is modular on $\Gamma_{0}(N)$, and these show that $\eta(12\tau)^{2}$ is a modular form of weight $1$ on $\Gamma_{0}(144)$ with Nebentypus $\chi_{-4}$ (with order of vanishing $1$ at every cusp, somewhat surprisingly). This function $\chi_{-4}$ is the Dirichlet character where $\chi_{-4}(n) = 0$ if $n$ is even, $\chi_{-4}(n) = (-1)^{(n-1)/2}$ if $n$ is odd. -For $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, define $g(\tau) | M = \det(M)^{1/2} (c\tau+d)^{-1} g\left(\frac{a\tau+b}{c\tau+d}\right)$ to be the weight $1$ "slash" operator. Saying that $g(\tau) = \eta(12\tau)^{2}$ has character $\chi_{-4}$ means that if $M \in \Gamma_{0}(144)$ we have that -$g | M = \chi_{-4}(d) g$. -If we let $f(\tau) = \eta(\tau)^{2}$, then $f | \begin{bmatrix} 12 & 0 \\ 0 & 1 \end{bmatrix}$ transforms nicely under $\Gamma_{0}(144)$, and hence $f(z)$ transforms nicely under $\begin{bmatrix} 12 & 0 \\ 0 & 1 \end{bmatrix} \Gamma_{0}(144) \begin{bmatrix} 1/12 & 0 \\ 0 & 1 \end{bmatrix}$. In particular, if $M = \begin{bmatrix} a & b \\ 144c & d \end{bmatrix} \in \Gamma_{0}(144)$, we get that $g | M = \chi_{-4}(d) g$ and this gives that -$$ - f | \begin{bmatrix} a & 12b \\ 12c & d \end{bmatrix} = \chi_{-4}(d) f. -$$ -In particular, for any $N = \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix} \in {\rm SL}_{2}(\mathbb{Z})$ with $a' \equiv d' \equiv 1 \pmod{4}$ and $b' \equiv c' \equiv 0 \pmod{12}$, we have that $f | N = f$. The set of such $N$ is a group which has $\Gamma(12)$ as an index $4$ subgroup.<|endoftext|> -TITLE: Non-separated étale algebraic spaces -QUESTION [11 upvotes]: Let $f: X \to S$ be a morphism of algebraic spaces, where $S$ is a scheme. If $f$ is separated and étale then Knutson's criterion says that $X$ is actually a scheme. -I have a some closely related questions. - - -Why does one require separatedness in Knutson's result? What is an example of an étale cover of a scheme which is not a scheme? - - -I cannot visualise at all what a non-separated étale morphism could look like. - - -What is an example of a non-separated étale morphism $f: X \to S$ with $X$ an algebraic space? Can there even exist such a morphism with $X$ a scheme? - -REPLY [20 votes]: To answer question 2 (with $X$ a scheme), just take for $S$ the affine line over a field $k$, for $X$ the usual ``$S$ with the origin doubled'' (two copies of $S$ glued along $S\smallsetminus\{0\}$) and for $f$ the obvious projection (which is the identity on each copy). -For question 1, the best example I know is a ``Galois twist'' of the above, where $f$ is still an isomorphism above $S\smallsetminus\{0\}$ but $f^{-1}(0)$ is the spectrum of a $\mathbb{Z}/2$-extension $k'$ of $k$, instead of two $k$-rational points. -To achieve this, start with the constant $S$-group scheme $G:=(\mathbb{Z}/2\mathbb{Z})_S$. Remove the nontrivial point of $G$ above $0\in S$: you get an open subgroup $H$ of $G$. Fixing a $\mathbb{Z}/2$-extension $k'$ of $k$, the $S$-scheme $S':=S\times_kk'$ has a natural action of $G$, induced by the Galois action on $k'$, and the quotient $S'/G$ is of course $S$. -Now the required example is $X:=S'/H$, with the natural projection on $S$. Clearly the $H$-action defines an étale equivalence relation, so $X$ is an algebraic space. Next, if you extend the scalars to $k'$, you get (the $k'$-version of) the preceding example. To see that $X$ is not a scheme, look at the only point $x$ of $X$ above $0$ (with residue field $k'$). If $U$ is any open subspace of $X$ containing $x$, then $U\times_kk'$ must contain both ``funny'' points of $X\times_kk'$, so $U$ cannot be affine (not even separated). -EDIT to answer Daniel's comments: - -Call $X_0$ the first example (standard affine line with double origin). Then the second example $X$ can actually be obtained as a Galois twist of $X_0$: the group $\Gamma:=\mathbb{Z}/2\mathbb{Z}\cong\mathrm{Gal}(k'/k)$ acts on $X_0$ by swapping the two copies of $S$, and then $X$ is just the corresponding Galois twist, i.e. the contracted product $X_0\times_k^\Gamma\mathrm{Spec}(k')$. -Even a twist of a separated scheme need not be a scheme. Hironaka has constructed proper, non-projective $k$-varieties, and you can easily tune the construction to find such a variety $Z$ having an involution $\sigma$ such that there is an orbit $Y$ which is not contained in any affine open set. Then the twist $Z':=Z\times_k^\Gamma\mathrm{Spec}(k')$ is an algebraic space but not a scheme, since the twist of $Y$ is a point of $Z'$ which cannot have an affine neighbourhood. (By the same argument, the quotient $Z/<\sigma>$ is not a scheme, and neither is the symmetric square of $Z$ over $k$).<|endoftext|> -TITLE: Kovari-Sos-Turan theorem -QUESTION [6 upvotes]: Let $r \leq s$ be fixed natural numbers. Then by the Kővári–Sós–Turán theorem, any graph on $n$ vertices with at least $cn^{2-\frac{1}{r}}$ edges contains a complete bipartite subgraph $K_{r,s}$ for a constant $c.$ -I was wondering if we can say that if we have at least $cn^{2-\frac{1}{r}}+1$ edges, then there are at least $c'n$ complete bipartite subgraphs $K_{r,s}$, for a constant $c'$? -For example, if a graph has at least $\frac{n^2}{4}+1$ edges, then it has at least $n$ triangles! -I'm pretty sure the answer to my question should be Yes, but I do not know how to prove it. - -REPLY [2 votes]: Let's look at the case $r = 2$ and choose to interpret the question as asking how many $K_{2,2}$ or $C_4$'s one has in a graph with $n$ vertices and ex$(n, C_4)$ + 1 edges, where ex$(n, C_4)$ is the largest number of edges in an $n$-vertex graph with no copy of $C_4$. This is known to be approximately $\frac{1}{2} n^{3/2}$. -The exact value of ex$(n, C_4)$ is only known for certain values of $n$, but for those cases the lower bound constructions come from projective planes. The spirit of the extremal examples is that each vertex has neighbourhood of order $\Theta(\sqrt{n})$ and these neighbourhoods cover the pairs of vertices in the graph, with (almost) every pair appearing in exactly one neighbourhood (if a pair were in two neighbourhoods, we would have a $C_4$). -What happens if we add an extra edge to such a construction? Suppose the new edge is $uv$. Then, for every vertex $w \in N(u)$, the edge $vw$, which was already covered by a neighbourhood, is covered by $N(u)$ and so forms a $C_4$. Similarly, for every $w \in N(v)$, the edge $uw$ produces a $C_4$. Overall, we produce $O(\sqrt{n})$ new copies of $C_4$, different to the expectation of $\Omega(n)$ expressed in the question. I suspect that $\Omega(\sqrt{n})$ is the correct answer in this case.<|endoftext|> -TITLE: Which affiliation to use when publishing, when invited professor at second university -QUESTION [7 upvotes]: I am a PhD student at one university and an invited professor at a second, i.e. I do not have a permanent position in the second one. Now I need to indicate an affiliation in a journal paper but I do not know whether to indicate or not to indicate a university where I am an invited professor. - - -What’s the etiquette for the affiliation to indicate in such a case - to indicate both or only the first one? - -REPLY [21 votes]: If you get support (financial and moral) from both places, you should list them both. - -REPLY [9 votes]: You can write: -University XXX (your permanent address) -Current address: -University YYY (your current address) -If it happens that the paper is going to be published after you are back at your home university, then you can remove the current address (when you receive proofs for final corrections) and leave only the address of the home university. However, you should put in the acknowledgements something like that: (This is an example of an actual acknowledgement from a published paper) -Final version was prepared when the last author was staying -at Mathematisches Institut der Universitat Bonn; he gratefully acknowledges the -generosity of Humboldt Foundation and hospitality of his colleagues from Bonn. - -REPLY [6 votes]: I think you should list the institutions that supported any substantial part of the research presented in your paper. As for your other affiliations, I think you may want to list some or all or none of them.<|endoftext|> -TITLE: Expected value of determinant of simple infinite random matrix -QUESTION [16 upvotes]: Suppose we have a matrix $A \in \mathbb{R}^{n\times n}$ where -$$A_{ij} = \begin{cases} 1 & \text{with probability} \quad p\\ 0 &\text{with probability} \quad1-p\end{cases}$$ -I would like to know the following expected value -$$\lim_{n \rightarrow \infty} \mathbb E(| \det (A) |)$$ -i.e., the asymptotic behavior as $n$ becomes large. - -What I tried so far -It feels like this has been done already, but after searching for quite a while I performed simulations and it looks like -$$\mathbb E(|\det(A)|) \propto e^{n f(p) }$$ -where $f$ is some function of the probability $p$. -$p$." /> -I would be very happy if someone knows the result or a good reference where I could look it up. -Edit: I changed the figure and added plots of the solutions -$$ \text{log} \mathbb E(|\det(A)|) \propto \frac{n}{2} \text{log} \frac{n p (1-p)}{e}$$ -given by Richard Stanley and RaphaelB4 which coincide up to multiplicative terms if the factorial in Richard Stanleys solution is replaced by stirlings formula. - -REPLY [6 votes]: Here is a solution to Hipstpaka's question about $\det(A)^2$. I don't -have enough space to answer in a comment. I don't know where a -statement appears in the literature, but the proof uses a standard -technique discussed for instance in Enumerative Combinatorics, -vol. 2, Exercise 5.64. -Write $\varepsilon_w$ for the sign of the permutation $w\in S_n$. -Then - \begin{eqnarray*} \sum_A \det(A)^2 & = & - \sum_{i,j=1}^n\sum_{a_{ij}=0,1} \left(\sum_{w\in S_n} - \varepsilon_w a_{1,w(1)}\cdots a_{n,w(n)}\right)^2\\ & = & - \sum_{i,j=1}^n\sum_{a_{ij}=0,1} \sum_{u,v\in S_n} - \varepsilon_u\varepsilon_v a_{1,u(1)}\cdots a_{n,u(n)} - a_{1,v(1)}\cdots a_{n,v(n)}. \end{eqnarray*} -Let $f(u,v)$ be the number of distinct variables occurring among -$a_{1,u(1)},\dots, a_{n,u(n)},a_{1,v(1)},\dots, a_{n,v(n)}$. The -product $a_{1,u(1)}\cdots a_{n,u(n)}a_{1,v(1)}\cdots a_{n,v(n)}$ is 1 -with probability $p^{f(u,v)}$ and is otherwise 0. Moreover, $f(u,v)= -2n-\mathrm{fix}(uv^{-1})$, where $\mathrm{fix}(uv^{-1})$ denotes the -number of fixed points of $uv^{-1}$. If $E$ denotes expectation, then -we get - $$ E(\det(A)^2) = \sum_{u,v\in S_n}\varepsilon_u\varepsilon_v - p^{2n-\mathrm{fix}(uv^{-1})}. $$ -Setting $w=uv^{-1}$ and noting that -$\varepsilon_u\varepsilon_{wu^{-1}} = \varepsilon_w$, we get - \begin{eqnarray*} E(\det(A)^2) & = & \sum_{w\in S_n} - p^{2n-\mathrm{fix(w)}} - \sum_u\varepsilon_u\varepsilon_{wu^{-1}}\\ & = & - n!p^{2n}\sum_{w\in S_n}p^{-\mathrm{fix(w)}}\varepsilon_w. - \end{eqnarray*} -Let $g(n)=\sum_{w\in S_n}p^{-\mathrm{fix(w)}}\varepsilon_w$. By -standard generating function techniques (Enumerative Combinatorics, -vol. 2, Section 5.2) we have - \begin{eqnarray*} \sum_{n\geq 0} g(n)\frac{x^n}{n!} & = & - \exp\left( \frac 1p x-\frac{x^2}{2}+\frac{x^3}{3} - -\frac{x^4}{4}+\cdots\right)\\ & = & - (1+x)\exp \left( \frac 1p-1\right)x. \end{eqnarray*} -It is now routine to extract the coefficient of $x^n$ and then compute - $$ E(\det(A)^2)= n!\,p^n(p-1)^{n-1}(1+(n-1)p). $$ -In general we don't have $E(\det(A)^2)=E(|\det(A)|)^2$, so this does not directly answer the original question. -For a related question (and answer) see Expected determinant of a random NxN matrix.<|endoftext|> -TITLE: Ultrafilters and diagonal arguments -QUESTION [22 upvotes]: Is there a diagonal argument to show that if $x$ is infinite then ${\cal P}(x)$ (the power set of $x$) is smaller than $\beta x$ (the set of ultrafilters on $x$)? -(Added later. I tried commenting but it wouldn't let me!) -My reason for interest in this was roughly as follows. With AC one can prove that an infinite set $X$ has $2^{2^{|X|}}$ ultrafilters. If choice fails very badly then there might only be $|X|$-many. I was hoping that there might be a diagonal construction lurking in the background which can feed off the choice principles one supplies. As my correspondents know, i spend a lot of time thinking about Quine's NF, and one question there is: how many ultrafilters are there on the universe? If every ultrafilter is principal then very few, yes. But if there are nonprincipal ufs at all, is there some elementary argument available to show that there must be as many of them as there are sets? If there is such a diagonal construction i'd like to know. It could be useful paedogogically, too. - -REPLY [15 votes]: Even without amorphous sets, this cannot happen. -Suppose there are no free ultrafilters on some infinite set $x$, then $\beta x$ has a natural bijection with $x$. Which is strictly smaller than $\mathcal P(x)$, as you very well know. -And of course, this is consistent for $x=\omega$, and for $x$ being any infinite set.<|endoftext|> -TITLE: A generalization of Erdős-Ko-Rado theorem -QUESTION [8 upvotes]: Is there any result known about the following generalization of the Erdős-Ko-Rado theorem? -Let $n, k, r, s$ be positive integers. We call a family $\mathcal{F}$ of $k$-element subsets of $\{1,\ldots, n\}$, an $(r, s)$-intersection family if among every $r$ elements of $\mathcal{F}$ at least two have an $s$-element intersection. What is the maximum size of such a family? - -REPLY [6 votes]: The case $s=1$ is Erdős hypergraph mathcing conjecture from -Paul Erdős (1965). A problem on independent $r$-tuples. Ann. Univ. Sci. Budapest. Eötvös Sect. Math. 8 (1965), 93–95. -users.renyi.hu/~p_erdos/1965-01.pdf -A recent paper about it is -Peter Frankl (2017) Proof of the Erdős matching conjecture in a new range, Israel Journal of Mathematics 222(1), pp 421-430 -doi:10.1007/s11856-017-1595-7 -The generalization you are asking about is studied in -Christos Pelekis, Israel Rocha (2017), A generalization of Erdős' matching conjecture. arxiv:1710.04633 -In the notation of this paper you are asking for the maximum number of edges in a $k$-uniform hypergraph, such that its $s$-matching number is strictly less than $r$. Their abstract says that they identify a collection of candidate solutions, and show that it contains the optimum when $n\geqslant 4k\binom{k}{s}r$.<|endoftext|> -TITLE: Research topics in distribution theory -QUESTION [17 upvotes]: The theory of distributions is very interesting, and I have noticed that it has many applications especially with regard to PDEs. But what are the research topics in this theory? also in terms of functional analysis - -REPLY [18 votes]: While I do not know much about current development of the general theory of distributions, I can say something about the current research topics in a special class of distributions, the theory of Sobolev spaces. -Theory of Sobolev spaces was one of the greatest discoveries in -the XXth century mathematics. This theory is the most important single -tool in studying nonlinear partial differential equations, both in its -theoretical aspects and numerical implementation. -Although the theory of Sobolev spaces has been created in the late -thirties, in recent years, there have been major breakthroughs in the theory, by -expanding the applications to new areas of pure mathematics -like analysis on metric spaces, geometric group theory or algebraic -topology as well as to areas in applied mathematics, like for example to non-convex -calculus of variations. -I will list some of the active research areas, just to set an example, but the list is far from being complete. I will focus mostly on the areas that I am familiar with. These areas are not directly related to partial differnetial equations, where the apllications are well known. For each topic I will provide just one reference as otherwise I would have to put hundreds. It will then be easy to search MathSciNet to find other relevant references. - -Sobolev spaces on irregular domains. The classical embedding and extension theorems assume that the boundary of a domain is quite regular. However, a substantial effort has been put in extending these results to domains whose boundary might be fractal. It seems that the first important paper in that direction was [6]. -Sobolev mappings between manifolds. This class of mappings appear in a natural way in the study of geometric variational problems for mappings between manifolds. Like for example, the theory of harmonic mappings. One of the early problems was the question whether smooth mappings are dense. That led to a very fruitful research showing deep connections to algebraic topology. See for example [4]. -Theory of quasiconformal and quasiregular mappings. Quasiconformal mappings are homeomorphisms that distort balls in a certain controlled way. Quasiregular mappings are roughly speaking quasiconformal mappings that can have branching set where they are not one-to-one. Just like conformal mappings versus holomorphic functions. One of the main tools in study of such mappings is the theory of Sobolev spaces. Applications of the theory include conformal parametrizations of surfaces, dynamical systems and rigidity results like the Mostow rigidity theorem. For a recent impressive result see [3]. -Mappings of finite distortion. The theory of quasiregular maps led to this theory. It has important applications in the non-linear elasticity. -For basic results, see [7]. -Non-linear elasticity. An approach to non-linear elasticity proposed by J.Ball [1], led to new questions in the analysis and geometry of Sobolev mappings. This development is related to the theory of quasiregular mappings and mappings of finite distortion discussed above. -Variable exponent Sobolev spaces. This is the extension of the theory to the case in which the $L^p$ spaces are replaced with $L^{p(x)}$. That is, the exponent of integrability is a function [2]. -At last, but not least: Analysis on metric spaces. Quite surprisingly, the first order analysis involving derivatives can be extended to metric spaces equipped with a measure. This is a very active research area that has already its MSC classification 30L. Sobolev spaces on metric spaces play an important role in the development of this theory [5]. - -[1] J. M. Ball, Convexity conditions and existence theorems in nonlinear elasticity. Arch. Rational Mech. Anal. 63 (1976/77), 337–403. -[2] L. Diening, P. Harjulehto, P. Hästö, M. Růžička, Lebesgue and Sobolev spaces with variable exponents. Lecture Notes in Mathematics, 2017. Springer, Heidelberg, 2011. -[3] D. Drasin, P. Pankka, -Sharpness of Rickman's Picard theorem in all dimensions. -Acta Math. 214 (2015), 209–306. -[4] F. Hang, F. Lin, Topology of Sobolev mappings. II. Acta Math. 191 (2003), 55–107. -[5] J. Heinonen, P. Koskela, N. Shanmugalingam, J. T. Tyson, Sobolev spaces on metric measure spaces. An approach based on upper gradients. New Mathematical Monographs, 27. Cambridge University Press, Cambridge, 2015. -[6] P. W. Jones, Quasiconformal mappings and extendability of functions in Sobolev spaces. Acta Math. 147 (1981), 71–88. -[7] S. Hencl, Stanislav; P. Koskela, Lectures on mappings of finite distortion. Lecture Notes in Mathematics, 2096. Springer, Cham, 2014.<|endoftext|> -TITLE: The "Spaces of Schwartz distributions are finite dimensional" challenge -QUESTION [8 upvotes]: The more I study Schwartz distributions and the corresponding spaces, the more the latter look "finite dimensional" to me. Of course they are not finite dimensional in the technical sense but they are essentially finite dimensional in the sense that "theorems which are true for $\mathbb{R}^n$ should be true for them too". -I would like to get a better hold on what "essentially finite dimensional" should mean. So I propose this question/challenge (maybe this should be made community wiki?). - -Question: Can you find examples of theorems which only use the topological vector space structure of $V$ and are true when $V$ is finite dimensional, typically false when $V$ is an infinite dimensional Banach space, but true again when $V$ is $\mathcal{S}'(\mathbb{R}^n)$ or $\mathcal{D}'(\mathbb{R}^n)$ (with the strong topology). - -Here are some example to get started. - -A subset $A$ of $V$ is compact iff it is closed and bounded. -$V$ is reflexive. -A sequence of Borel probability measures on $V$ converges weakly iff the characteristic functions $V'\rightarrow\mathbb{C}$ converge pointwise to a function that is continuous at the origin. - -I would prefer examples of properties/theorems which are not entirely a consequence of being nuclear. - -Motivation: -I don't yet have a definition of "essentially finite dimensional" (EFD) which is a fluctuating concept to me. -Of course EFD should be of the form "nuclear+some mystery ingredients". I am hoping the answers to the question as I asked it would tell me more about the optimal mix of mystery ingredients rather than the nuclear part. One might think that what I am after is a phrase-definition like "nuclear, barrelled and quasi-complete" but it is not quite that as I try to explain in what follows. -The definition of finite dimensional vector space is that $V$ is isomorphic to $\mathbb{R}^n$ for some $n$. This involves a concrete model as a space of functions on a discrete set, here the finite set $\{1,2,\ldots,n\}$. I suppose EFD could be defined similarly, as follows. Let $\Omega$ be a set of "pre-seminorms" on $\mathbb{R}^{\mathbb{N}}$. By "pre-seminorm" I mean the same definition as seminorm but allowing the value $\infty$. -Then define $V=\{x\in \mathbb{R}^{\mathbb{N}} | \forall \omega\in\Omega, \omega(x)<\infty\}$ with the topology defined by the restrictions of the elements of $\Omega$ which then become true seminorms. I suppose one could restrict to $l^1$ pre-seminorms of the form -$$ -\omega(x)=\sum_{n\in \mathbb{N}}\omega_n\ |x_n| -$$ -with weights $\omega_n\ge 0$. What I would ultimately like is an explicit property of $\Omega$ which makes the corresponding space as close as possible to a finite dimensional space. The part of that property concerning nuclearity was worked out at long time ago by Grothendieck for echelon spaces and Pietsch in general. The property should be "The Pietsch criterion+???". -Anyway, all I asked is examples of theorems, the more surprising the better, and the further away from just being nuclear the better too. - -Motivation for the motivation: I have been thinking about how one could improve existing presentations of the theory of distributions, see e.g. -Nice applications for Schwartz distributions One approach I am exploring is to use sequence space representations and I would eventually like to narrow down the possibilities to a class of $\Omega$'s which is easy to define and could serve as "optimal" class of concrete models for EFDs. - -REPLY [14 votes]: Like others have pointed out the key, concept is that of nuclear spaces and a good presentation can be found in volume 4 of Gelfand's Generalized Functions. -Kolmogorov has introduced a concept of dimension of a functional space and the nuclear spaces have finite Kolmogorov dimension; see Chap I. Sec. 3.8 of the above reference. For example, in this section it is proved the space of real analytic functions on $\mathbb{R}^n$ has Kolmogorov dimension $n$. As an aside, this concept of dimension has lead to proofs of many generalizations of Hilbert 13th problem. -The above reference also contains a beautiful presentation of the Schwartz kernel theorem. I recommend this source enthusiastically.<|endoftext|> -TITLE: Homomorphism to multiplier algebra of groupoid $C^\ast$-algebra -QUESTION [7 upvotes]: If I have a functor $X\to Y$ between topological groupoids with appropriate Haar measures, such that $X_0 \to Y_0$ is injective and a homeomorphism onto its image, then I should have (or rather, I hope to have) a non-degenerate $\ast$-homomorphism $C^\ast(X) \to M(C^\ast(Y))$, where the latter is the multiplier algebra of $C^\ast(Y)$. I've been hunting for a description of this map but have been unsuccessful, can anyone point me to a treatment? -Edit In fact I think I only need the case that $X_0 \to Y_0$ is in fact a homeomorphism, since I can factor the general case that I'm interested in into that case, followed by something I think I understand. - -Addendum: in the special case that the groupoids have only one object, and so can be identified with groups, I think what I'm after is the (extension of the) pushforward map that takes an integrable function on the domain and gives back a measure on the codomain. But I'm aware that generalising to groupoids sometimes isn't quite so straightforward. - -REPLY [2 votes]: This is not a complete answer, but that was way too long for a comment: -First I started almost sure that this sort of things would be in the literature, but it seems harder than I thought to find something talking about this. -A pretty good lead is the paper "A universal property for groupoid C* algebras,I" by Alcides Buss, Rohit Holkar and Ralf Meyer which might answer your question with a little bit of work. -Their main theorem (3.23) says that given a $C^*$-algebra $D$ and a Hilbert $D$-module $F$, representations of $C^*(G)$ on $F$ corresponds to a notion of "representations of $G$ on $F$". This is an extention of the theory of integration and desintegration of representations with value in Hilbert space to representation with value in a Hilbert module. -As morphisms to the multiplier algebra $A \rightarrow M(D)$ are exactly the same as representations of $A$ on the Hilbert $D$-module $D$. This tells you precisely what you need to have in order to get such a morphisms. -I believe this is a good approach to that problem as this universal property is the simplest way to construct the map you are looking for in the case of group: -The universal property of the maximal group C* algebra $C^*(G)$ is that a morphisms from $C^*(G)$ to $M(A)$ is the same as unitary action of $G$ by multiplier on $A$. In particular, any morphism from $G$ to $H$ induced a unitary action by multiplier of $G$ on $C^*(H)$ by simply restricting the action of $H$ and this corresponds to a morphisms $C^*(G) \rightarrow M(C^*(H))$. -Now I haven't read their paper in much details yet so I'm not totally sure how their notion of 'representation' would behave with respect to composition of morphisms of groupoids, but it should anyway give you a good understanding of when this kind of morphisms exists or not. -I'm especially a little worried in general about a possible condition of "compatiblity" of the Haar measure on the source and the target which might makes the results false in general, but true in lots of cases (like étale groupoids, Lie groupoids, or groups, where the choice of a Haar measure is not really a problem)<|endoftext|> -TITLE: Perfect $p$-group with nontrivial center -QUESTION [5 upvotes]: For a $p$-group $G$, which we know has a nontrivial center, can we have $G=G'$? -Obviously not when $G$ is finite. But the question makes sense for infinite groups (here $p$-group means that each element has finite order, which is a power of some given prime number $p$). - -REPLY [2 votes]: You do not need a Tarski monster. Just take the free group $\langle a,b\rangle$ impose two relations of the form $a=w, b=w'$ satisfying a small cancellation condition, say, $C'(.000001)$ where $w,w'$ are products of commutators. The factor-group $G$ is hyperbolic (small cancellation) and perfect. Apply the method of Olshanskii from "Periodic quotient groups of hyperbolic groups" to obtain an infinite quotient $H$ of $G$ of prime exponent $p\gg 1$. Then the same proof as in Ashmanov-Olshanskii (it is in Olshanskii's book "Geometry of defining relations") proves that the Schur multiplier of $H$ is the free Abelian group of countable rank. Then take the universal central extension of $H$. -Update: If a finite exponent is not needed, a periodic 2-generated $p$-group example can be constructed as a lacunary hyperbolic group, see my answer to this question.<|endoftext|> -TITLE: Picture of 18 smooth reflexive polytopes of dimension 3 -QUESTION [8 upvotes]: It was proven by Batyrev in 1981 -http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=im&paperid=1581&option_lang=eng -that there exist exactly 18 smooth toric Fano three-folds. I would like to know if there is some place where the pictures of the corresponding 18 simple reflexive polytopes are presented. -More precisely, I would like to see the moment polytopes, i.e. the polytopes that are dual to 18 reflexive polytopes depicted in the reference given below by David. These polytopes should be Delzant, i.e., each vertex can be send by an element of $SL(3,\mathbb Z)$ to the standard vertex given by equations $x_i\ge 0$. - -REPLY [6 votes]: The article that was linked by David G. Stork actually contains information about how one can get these pictures. Namely, the SageMath computer algebra system contain a database of all reflexive polytopes and the figure from the article includes their position in this database (off by one). You can do much more with these objects in Sage than just getting their polar dual. Thus I can give you easily more than just a picture, I can give you a 3D model that you can rotate and zoom in your browser. Just paste the following code into SageMathCell and hit Evaluate. -# change the number from 1 to 18 -n = 3 -polytopes_list = [1, 5, 6, 7, 8, 25, 26, 27, 28, 29, 30, 31, 82, 83, 84, 85, 219, 220] - -P = ReflexivePolytope(3, polytopes_list[n-1]-1) -dP = P.polar() -dP.plot3d()<|endoftext|> -TITLE: Do all Fano threefolds have effective $c_2$? -QUESTION [9 upvotes]: Let $X$ be a smooth complex projective Fano threefold. Then the class $c_1(X)$ can be realised as an effective divisor in $X$. It is it true that the class $c_2(X)$ can be realised as an effective curve? -(Note that $c_3$ cannot, since the Euler characteristics of the cubic Fano $3$-fold is negative) - -REPLY [9 votes]: By a theorem of Miyaoka (Theorem 6.1 in Y. Miyaoka `The Chern classes and Kodaira dimension of an algebraic variety', 1987), we have that for a Fano variety $X$, $c_2(X)\cdot H\ge 0$ for any ample divisor $H$. By duality of the cone of curves and the nef cone, it follows that $c_2(X)$ is at least pseudoeffective (i.e., a limit of effective classes). On the other hand, since $X$ is Fano, the Cone theorem shows that the cone of curves is rational polyhedral, spanned by finitely many classes of rational curves, so $c_2(X)$ is in fact effective. -EDIT: As C. Jiang correctly points out, to apply Miyaoka, one needs also the fact that the tangent bundle of a Fano 3-fold is generically nef. This was proved by Kollár--Miyaoka--Mori--Takagi, and later Peternell. -See also Theorem 1.2 in Q. Xie `On Pseudo-Effectivity of the Second Chern Classes for Terminal Threefolds'.<|endoftext|> -TITLE: A game on integers -QUESTION [37 upvotes]: $A$ and $B$ take turns to pick integers: $A$ picks one integer and then $B$ picks $k > 1$ integers ($k$ being fixed). A player cannot pick a number that his opponent has picked. If $A$ has $5$ integers that form an arithmetic sequence, then $A$ wins. $B$'s goal is to prevent that from happening. Who has a winning strategy (after finite number of turns)? Does it depend on $k$? - -REPLY [4 votes]: The minimalist strategy for $A$ of "don't think, just pick the smallest unchosen non-negative integer!" has a simple counter strategy for $B$ which is interesting to analyze as to how far it gets before it eventually fails. I wonder how good it is. -The number of turns for $A$ before winning increases exponentially with $k$ (provided $A$ uses this minimalist strategy and $B$ uses the one I will describe.) This is true also for thee term arithmetic progressions. -Consider the game where $A$ needs to get an arithmetic progression of length $m$ to win and $B$ gets $k$ choices each turn. Then I think my strategy for $B$ matched with the minimalist strategy for $A$ will have $A$ win on a turn no smaller than $k^c$ where $$c=\log_{\frac{m}{m-1}}(m-1).$$ -A better strategy (for three term sequences) for $A$ would win in $k+2$ turns (this could easily be improved to $\frac{k}3$ turns.) Roughly, on the first $k$ turns $A$ selects available even numbers $a_1,\cdots,a_{k}$ then, whatever choices $B$ has made so far, it is possible to pick an even $a_{k+1}$ so that none of $\frac{a_{k+1}-a_i}2$ have been chosen yet. $B$ must leave one of those unchosen on her turn so $A$ picks that next turn and wins. -How to extend this to longer progressions is not clear however that argument shows that wlog $A$ gets three free moves on the first turn (we could arrange that the entire progression determined by $a_i,a_{k+1}$ is free.) -Below is a strategy for $B$ if she knows for sure that $A$ will blindly follow the minimalist strategy. The moves are all known in advance so it is just a question of when $A$ wins. For ease of analysis I will allow $k \gt 0$ to be any real number and interpret this as: on turn $t$ if $B$ has previously chosen $m$ integers she has $\lfloor kt \rfloor-m$ moves available and can make none or some or all and hold the rest for future turns. -The strategy is that she will pick in order the non-negative integers with a $4$ in their base $5$ expansion and let $A$ work up through the rest one at a time. This blocks any $5$ term progressions as long as $B$ stays ahead. However $B$ will save her moves and only pick enough numbers that $A$ will not win on the next turn (which may turn out to take more moves than she has). -When $B$ loses it will because $A$ was able to play a number in some interval $[5^j-5^{j-1},5^j-1].$ This will happen in the event that $$\frac{5^{j-1}-4^{j-1}}{4^{j-1}} \leq k \lt \frac{5^{j}-4^{j}}{4^{j}}.$$ The exact winning move depends on where $k$ falls in that range. -For example, If $k \geq \frac{61}{64} =0.953125$ then $A$ at some point picks $125$ without having won. No problems occur until after the turn when $A$ picks $468=3333_5.$ At this point anything in $[469,624]$ would be a winner for $A$ on the next move. But $B$ has moves in reserve, certainly enough to choose all of $[629,499].$ However to be able to go on and choose all of $[500,634]$ requires $k \geq \frac{369}{256}=1.44140625.$ If so, then everything is good until at least $2499.$<|endoftext|> -TITLE: Cohomology of $ko,tmf,MSpin,MString$ with coefficients $\mathbb{Z}/p$ for odd primes $p$ -QUESTION [13 upvotes]: It is well-known that $$H^*(ko,\mathbb{Z}/2)=\mathcal{A}\otimes_{\mathcal{A}(1)}\mathbb{Z}/2$$ -$$H^*(tmf,\mathbb{Z}/2)=\mathcal{A}\otimes_{\mathcal{A}(2)}\mathbb{Z}/2$$ -where $\mathcal{A}$ is the mod 2 Steenrod algebra. -$H^*(MSpin,\mathbb{Z}/2)$ and $H^*(MString,\mathbb{Z}/2)$ are closely related to the above because of the Atiyah-Bott-Shapiro orientation and Witten genus. -I find in Adams and Priddy's Uniqueness of BSO: -$$H^*(ko,\mathbb{Z}/p)=\bigoplus_{s=0}^{\frac{p-3}{2}}\Sigma^{4s}\mathcal{A}_p/(\mathcal{A}_pQ_0+\mathcal{A}_pQ_1)=\bigoplus_{s=0}^{\frac{p-3}{2}}\Sigma^{4s}\mathcal{A}_p\otimes_{E(Q_0,Q_1)}\mathbb{Z}/p$$ -where $\mathcal{A}_p$ is the mod $p$ Steenrod algebra for odd primes $p$ -and $Q_0=\beta,Q_1=P^1\beta-\beta P^1$. -I want to know what is $H^*(MSpin,\mathbb{Z}/p)$, $H^*(tmf,\mathbb{Z}/p)$ and $H^*(MString,\mathbb{Z}/p)$ for odd primes $p$. -Any references and partial answers are appreciated. -Edit: Big thanks to Mark Behrens for his answer. As a complement, I find -$$H^*(MSpin,\mathbb{Z}/p)=H^*(MSO,\mathbb{Z}/p)$$ for odd primes $p$ and the homology of $MSO$ at odd primes as a comodule over the dual Steenrod algebra is Lemma 20.38 of Switzer's textbook. - -REPLY [20 votes]: [Some folks started emailing me about this so I supposed I should click on the link and post what I knew...] -The homology of tmf at all primes as a comodule over the dual steenrod algebra is Theorem 21.5 of Charles Rezk's tmf notes -https://faculty.math.illinois.edu/~rezk/512-spr2001-notes.pdf -The spaces Bspin, BString are all spaces of the spectrum ko. At odd primes ko is a summand of ku, and the mod p cohomology of these spaces was determined by Bill Singer "Connective fiberings over BU and U". By the Thom isomorphism theorem, this gives an answer for MSpin and MString, but maybe not the desired answer because the isomorphism is not an isomorphism of modules over the steenrod algebra. -The more refined computation of H^*MO at odd primes was done by Rosen in his Northwestern Thesis - this computation is redone by Hovey and Ravenel in their paper "THE 7-CONNECTED COBORDISM RING AT p = 3" (despite the title it does the cohomology for all odd primes p). -Hope this helps...<|endoftext|> -TITLE: Infinitely many integer solutions to $X^4+Y^4-18Z^4= -16$ -QUESTION [13 upvotes]: We found infinitely many integer solutions to -$$X^4+Y^4-18Z^4= -16 \qquad (1)$$. -The interesting part in this diophantine equation is the sum of -the reciprocals of the degrees is $3/4 < 1$, which is related to -Vojta's more general abc conjecture. - -Q1 Is this result trivial or known? - -Consider the diophantine equation -$$ aX^n+bY^m+cZ^l=d \qquad (2)$$ -where $1/n+1/m+1/l < 1$. Solution of (2) is trivial if almost always -$d \in \{aX_0^n,bY_0^m,cZ_0^l\}$ and the sum of the other two -monomials vanishes. - -Q2 Is there (2) with infinitely many integer non-trivial solutions, - except for generalization of (1) with $n=m=l=4$? - -REPLY [26 votes]: I do not know if the result is trivial or known, but your result is fun -and follows from the identity -$$(x-1)^4+(x+1)^4+16=2(x^2+3)^2$$ -so that you find infinitely many solutions by solving the Pell equation -$x^2-3z^2=-3$, so $x+z\sqrt{3}=\pm\sqrt3(2+\sqrt3)^n$. -You can of course replace $x-1$ and $x+1$ by $ax+b$ and $ax-b$ and find -infinitely many solutions to similar equations.<|endoftext|> -TITLE: Does the nerve have a right adjoint, sometimes? -QUESTION [8 upvotes]: Let $F : G\to PA$ be a functor ($G$ a small category, $PA = [A°,Set]$ the presheaf category of $A$). Let $N_F$ be the functor $PA\to PG$ defined by left Kan-extending the Yoneda embedding $y_G : G\to PG$ along $F$: - -The following procedure seems to exhibit a right adjoint for $N_F$; this is strange, as $N_F$ is in general only a right adjoint. Can you spot the error (if any) in my argument? - -Consider the extended diagram - - where the functor $PG \to PA$ is the left extension of $y_A$ along $N_F y_A$, that can easily shown to be equal to the left extension of the identity along $N_F$. And this latter functor shall be the right adjoint to $N_F$: counit and unit are determined by the suitable universal properties. - -Many things can be true at this point: - -The functor does not exist. -The functor exists, but fails to be a right adjoint to $N_F$ (the extension is not absolute/not preserved by $N_F$). -The functor is indeed a right adjoint to $N_F$; I didn't know it had one in the special case where its domain is itself a free cocompletion. - -Note that taking the right extension of $y_A$ along $N_Fy_A$ does not give a left adjoint to $N_F$: if such a left adjoint exists, it is unique and must exhbit the universal property of the left extension of $F$ along $y_A$: -$$ -\text{Lan}_{y_A}F \dashv \text{Lan}_Fy_A -$$ -(that's an instance of the $F$-nerve and $F$-oidal realization yoga). Indeed, a key step in the above argument is that $\text{Lan}_{N_F y_A}y_A\cong \text{Lan}_{N_F}(\text{Lan}_{y_A}y_A) \cong \text{Lan}_{N_F}1$, as $y_A$ is dense. But it's not codense: in fact, $\varphi \to \text{Ran}_{y_A}(y_A)(\varphi)$ is the $\varphi$-component of the unit of Isbell adjunction. - -REPLY [4 votes]: To answer the title question, but not the question about your argument.... -$PG$ and $PA$ are locally presentable categories, so $N_F : PA \to PG$ has a right adjoint if and only if it preserves small colimits. -From the formula -$$ N_F(Q)(g) \cong \hom_{PA}(F(g), Q) $$ -the condition that $N_F$ preserves colimits is equivalent to natural isomorphisms -$$ \hom_{PA}(F(g), \operatorname{colim}_j Q_j) -\cong \operatorname{colim}_j N_F(Q_j)(g) -\cong \operatorname{colim}_j \hom_{PA}(F(g), Q_j) $$ -In other words, $F(g)$ must be a tiny object for every $g$. -But in a presheaf category, the tiny objects are precisely the retracts of representables. If the domain category is idempotent complete, this simplifies to the tiny objects being precisely the representables. -In other words, $N_F$ has a right adjoint if and only if $F$ can be written as a composite of a functor $G \to \operatorname{Idem}(A)$ followed by the yoneda embedding.<|endoftext|> -TITLE: Group (Co)Homology of Symmetric Group -QUESTION [6 upvotes]: The question concerns the group homology or group cohomology of symmetric groups. - -The entries in groupprops.subwiki.org and in this MO post show the results for the symmetric group S$_4$. - -groupprops.subwiki.org, $H_q(\text{S}_4,\mathbb{Z})$: - -this MO post, $H_k(\text{S}_{n=4},\mathbb{Z})$: - - -If the groupprops.subwiki.org has the correct result, is it correct to say that the torsion parts of Group (Co)Homology of Symmetric Group are related by: - -$$H^q(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_q(\text{S}_4,\mathbb{Z})?$$ -Thus, -$$H^1(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_1(\text{S}_4,\mathbb{Z})=\mathbb{Z}_2?$$ -$$H^2(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_2(\text{S}_4,\mathbb{Z})=\mathbb{Z}_2?$$ -$$H^3(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_3(\text{S}_4,\mathbb{Z})=\mathbb{Z}_2\oplus \mathbb{Z}_4\oplus \mathbb{Z}_3?$$ -$$H^4(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_4(\text{S}_4,\mathbb{Z})=\mathbb{Z}_2?$$ - -More generally, do we have a precise relation between the group cohomology and group homology of different coefficients: - -$$H^q(G,\mathbb{R}/\mathbb{Z}) \text { and } H_p(G,\mathbb{Z}),$$ -say for any finite group $G$, and here in particular $\text{S}_4$? (Here $\mathbb{R}/\mathbb{Z}=S^1=$U(1).) - -REPLY [5 votes]: First of all, let $X$ be a space (with usual nice properties). Then -as ${\mathbb Q}/{\mathbb Z}$ is injective, the universal coefficient theorem for cohomology gives an isomorphism -$$Hom (H_*(X,{\mathbb Z}) ,{\mathbb Q}/{\mathbb Z}) -\cong H^*(X,{\mathbb Q}/{\mathbb Z}). $$ -Now, when $X$ is the classifying space of a finite group $BG$, in positive dimension, $A=H_*(X,{\mathbb Z})$ is finite abelian group, thus there is non-canonical isomorphism between $A$ and $Hom(A,{\mathbb Q}/{\mathbb Z})$. -Thus the answer to the third question (which includes the second one) is "yes".<|endoftext|> -TITLE: Flows in Hilbert spaces -QUESTION [7 upvotes]: Let $\varphi: [0,T] \rightarrow H$ be a Hilbert space valued $C^1$-function. Let $H = X \oplus X^{\perp}$ such that $\varphi(0) \in X$ and the implication $\varphi(t) \in X \Rightarrow \varphi'(t) \in X$ holds. -I ask: How can I show that $\varphi$ stays in $X$? It sounds natural and I guess it is true, but I fail to see how it can be shown. - -REPLY [7 votes]: Consider $\phi(t) = (1-e^{-1/t^2}, t)\in \mathbb R^2$. It satisfies your assumptions for $X= 0\times \mathbb R$, but it does not stay in $X$.<|endoftext|> -TITLE: Are (semi)simple Lie groups some sort of "homotopy quotient groups" of their maximal tori? -QUESTION [24 upvotes]: Warning: non-specialist writing, some rubbish possible. -The formula $h^*(BG)\cong h^*(BT)^W$ valid for complex oriented cohomology of the classifying space of a compact Lie group $G$ with maximal torus $T$ and Weyl group $W$ suggests that it might come from some sort of equivalence like $G\sim T///W$ where "///" stands for some hypothetical homotopy quotient, like homotopy colimit but not in spaces but rather some sort of equivariant homotopy colimit in the category of topological groups or maybe rather in the category of homotopy associative H-spaces and homotopy coherent homomorphisms, or maybe some sort of stacky quotient instead. Is there such a thing? -(Slightly later: comments by Ben Wieland show that what I wrote is not clear enough. Let me try to explain at least approximately: by $T$///$W$ I mean a universal object among topological group homomorphisms $f:T\to G$ together with homotopies between $f$ and $fw$ for each $w\in W$, where $w:T\to T$ is the action map. Presumably these homotopies must satisfy coherence conditions, and maybe I should require them to induce homomorphisms $T\to G^{[0,1]}$ or something similar, here I am not sure.) -My motivation (not counting that it might be useful to know such thing if it is true) comes from the fascinating paper "Root systems and elliptic curves" by Looijenga (Invent. Math. 38 (1976), pp. 17-32) where he constructs a "third invariant theory" using the Weyl group action on, in place of the Cartan subalgebra or maximal torus that are used for the "first", resp. "second" invariant theory, the abelian variety $A:=Q^\vee\otimes E$ where $Q^\vee$ is the lattice generated by the dual root system and $E$ is an elliptic curve. (Invariants are with respect to the $W$-action on sections of the line bundle over $A$ corresponding to the divisor formed by the fixed hypertori of $W$-reflections, and are expressed using certain theta functions.) -I wonder if some similar "homotopy quotient group" $A///W$ can be formed to yield some new kind of group-like object. -However I must admit that before this irresponsible "jump from second to third" it would be more appropriate to first understand the "first" case -- whether the corresponding Lie algebra $\mathfrak g$ might be realized as some sort of "homotopy quotient Lie algebra" $\mathfrak h///W$ of the Cartan subalgebra by the Weyl group... - -REPLY [13 votes]: Here's the "answer" that I started writing, then put away for a while. The short answer is: although "T//W" is not the same as G, they "look sort of the same" from the point of view of certain generalized cohomology theories. If you replace the role of the maximal torus with "arbitrary abelian subgroups", this "looking the same" can be given homotopical meaning. Finally, this data that your generalized cohomology (such as elliptic cohomology) "sees" can be encoded (according to Lurie) in terms of objects of derived algebraic geometry, which for elliptic cohomology you can think of as the "A///W" that you want to have. -... -As pointed out in the comments, $h^*(BG)\neq h^*(BT)^W$ for many compact simply connected $G$ and generalized cohomology theories (though equality does hold for $G=U(n)$ or $SU(n)$ and $h^*$ complex oriented). -If you try to naively make a "homotopical" version of this, by talking about classifying spaces, it gets even worse: in general $BG \neq BT_{hW}$, where the right-hand side is the homotopy invariant quotient of $BT$ by the action of the Weyl group (in fact, $BT_{hW}= B(NT)$, the classifying space of the normalizer.) -What is true is if the appropriate set of primes is inverted in your cohomology theory, then you get an equivalence: e.g., if $|W|$ is inverted in $h^*$, then $h^*(BG)=h^*(NT)$ for connected $G$ (as Saal Hardali points out in the comments). But it's not true integrally. -(It is true rationally. Which you can then reinterpret as a statement of Chern-Weil theory: $H^*(BG;\mathbb{R}) = (\operatorname{Sym} \mathfrak{g}^*)^G$; i.e., seen by $\mathbb{R}$-cohomology, $BG$ "looks like" $\mathfrak{g}//G$.) -Another thing to do is to generalize from "maximal torus" to "arbitrary abelian subgroup". So let $\mathcal{A}(G)$ be the category of "abelian subgroups of $G$": really, the category whose objects are $G/A$, the $G$-orbits with abelian isotropy group $A$, and whose maps are the space of $G$-equivariant maps between such orbits. Note that for connected $G$ this category contains the object $G/T$ with $\mathrm{Aut}(G/T)=W$, so it really is a generalization of "$W$ acting on $T$". -Then we obtain a "better" approximation map: -$$ \mathrm{hcolim}_{G/A\in \mathcal{A}(G)} BA = \mathrm{hcolim}_{G/A\in \mathcal{A}(G)} (G/A)_{hG}\to BG,$$ -which actually makes sense for all compact Lie group $G$ (including finite groups). This approximation actually turns out to be an equivalence (though not for deep reasons). Note: $\mathrm{hcolim}$ is homotopy colimit, and $(-)_{hG}$ is homotopy orbits. -Side note: the above "formula" implies a spectral sequence for any cohomology theory $h^*$ for computing $h^*(BG)$. This spectral sequence has as its "edge" map: -$$ -h^*(BG) \to \operatorname{lim}_{G/A\in \mathcal{A}(G)} h^*(BA). -$$ -Hopkins, Kuhn, and Ravenel proved that for a finite $G$, and a complex oriented $h^*$, this edge map becomes an isomorphism after inverting $|G|$. This is one version of their famous "character theorem". -There are analogues of this that apply equivariant cohomology theories other than "Borel equivariant cohomology", in which case things are more subtle. For instance, consider equivariant $K$-theory. If for a $G$-space $X$ I let $K_G(X)$ denote the spectrum whose homotopy groups are the $G$-equivariant $K$-theory of $X$, then there is supposed to be an equivalence -$$ -K_G(*) \approx \mathrm{hlim}_{G/A\in \mathcal{A}(G)} K_G(G/A) =\mathrm{hlim}_{G/A\in \mathcal{A}(G)} K_A(*). -$$ -Here $G$ is any compact Lie group, including finite ones (in which case it is a kind of Artin induction theorem). (This is basically asserted in Lurie's "Survey of Elliptic Cohomology", though I haven't seen a proof written out in full anywhere. $\mathrm{hlim}$ is homotopy limit.) -In other words, equivariant $K$-theory "sees" $G$ (more precisely, $(*//G)$) as indistinguishable from a derived colimit of abelian subgroups $A$ (more precisely, $(*//A)$). This is reminiscent of and related to (but distinct from and more complicated than) the observation that $RG=RT^G$ for representation rings of compact connected groups (because $\pi_0 K_G(*)=RG$). -Another equivariant cohomology theory you might try is elliptic cohomology. If you had a "naturally occuring" example of an equivariant elliptic cohomology, you would conjecture that it satisfies an analogue of the above equivalence. Or, if you're Jacob Lurie, you construct equivariant elliptic cohomology for arbitrary abelian compact Lie groups, then extend it to arbitrary compact Lie groups by imposing an equivalence as above. -Thus, Lurie associates to each abelian compact Lie group $A$ an object $M_A$ in derived algebraic geometry, so that: - -$M_{U(1)}$ is a "derived elliptic curve" $E$, -$M_{A\times B}=M_A\times M_B$, so in particular -$M_{U(1)\otimes Q} = E\otimes Q$ for a lattice $Q$, and -$M_{Z/n}= $ kernel of $[n]:E\to E$. - -Then for an arbitrary compact Lie group $G$ you should set $M_G$ to be some sort of colimit of a functor on $\mathcal{A}(G)$ that you can construct which sends $G/A\mapsto M_A$. (Apparently it's delicate to know exactly what sort of geometric object $M_G$ will be, but for the purposes of extracting an equivariant cohomology theory out of this business it's probably not too important. I'll pretend it is some kind of geometric object for the purposes of this discussion.) -For $G$ connected, you should think of $M_G$ as a derived and souped-up version of $E\otimes Q^{\vee}//W$ (souped-up because we use all abelian subgroups of $G$ and not just the torus with its Weyl group action). -As a bonus, Lurie can extend this set-up to Lie 2-groups $\widetilde{G}$ which are extensions of the form $1\to \mathbb{B}U(1)\to \widetilde{G}\to G\to 1$, associated to some cohomology class $c\in H^4(BG;\mathbb{Z})$. -Then $M_{\widetilde{G}}$ is (I think, and modulo things that I haven't said quite correctely) supposed to be the principal space of a line bundle over $M_G$, whose sections are to be (presumably) described by the theta-functions that Looijenga writes down in that paper you mention. (As an aside, I have no idea what motivates Looijenga to do whatever it is he does in that paper.) -Unfortunately, I don't know much more, since most of this is only described in Lurie's "Survey" mentioned above, which is only an announcement. (Versions of this story were perceived much earlier, most notably by Grojnowski in "Delocalized equivariant elliptic cohomology".)<|endoftext|> -TITLE: How do (co)limits in posets of subobjects relate to (co)limits in ambient category? -QUESTION [5 upvotes]: Sorry if this is elementary. -Let $C$ be a category, $X$ an object in $C$ and let $S(X)$ denote the poset of subobjects of $X$. -According to the nlab entry, if $C$ has all limits and co-limits, so does $S(X)$. It's a little unclear to me why this is true, and in particular how to construct a limit in $S(X)$ using limits in $C$. -For example, in the category $Mod_R$ of modules over a ring $R$, $S(X)$ is the poset of submodules of $X$, and the limit and colimit are respectively the intersection and "sum" of submodules. It's not clear to me immediately how these relate to limits and colimits in $Mod_R$. -Hence the title of the question: -How do (co)limits in poset of subobjects relate to (co)limits in ambient category? - -REPLY [4 votes]: To summarize what's been said in comments: - -If $C$ has small limits, then so does $S(X)$: they are inherited from $C/X$, which inherits them from $C$ by adding $X$ as a terminal object in the diagram being taken a limit of. -In particular, therefore, if $C$ is well-powered, so that $S(X)$ is small, then it is a small complete lattice and hence also has all colimits, constructed by the adjoint functor theorem. However, they are not a priori related to colimits in $C$, and not well-behaved (e.g. not stable under pullback). -If $C$ has a factorization system where the right class is the subobject inclusions, then joins in $S(X)$ can be constructed as the images of coproducts in $C$. -If $C$ is a pretopos, then finite joins in $S(X)$ can be constructed as pushouts under intersections, as described at the nLab page.<|endoftext|> -TITLE: Minimal length presentations of cyclic groups -QUESTION [12 upvotes]: By the length of a finite presentation I mean the sum of the lengths of the relators. I am interested in knowing what the minimal length of a presentation of $\mathbb{Z}/n\mathbb{Z}$. I'm even more interested in knowing the minimal length of a balanced presentation $\mathbb{Z}/n\mathbb{Z}$. -For concreteness, I am really interested in the particular case of $\mathbb{Z}/173\mathbb{Z}$. For example this has a presentation of length 24 given by $$ - is this a minimal length presentation? Is this presentation minimal length among balanced presentations? - -REPLY [7 votes]: I can beat that by one. -$$C_{173} \cong \langle a,b,c,d,e \mid a^4=b, b^2=c,c^4=d,d^4=e, abcde=1 \rangle.$$ -It might be hard to prove minimality. I am sure the best you can do is $O(\log n)$, but it could be interesting to see what constant you could get. The approach of using squares of successive generators together with the binary expansion of $n$ gives a bound of $4 \log_2 n$, which could be reduced to about $3.5 (\log_2 n+1)$ by using positives and negatives in the binary expansion of $n$. -But I think using cubes and the ternary expansion of $n$ is better in general. That seems to give a bound of $5 (\log_3 n +1)\sim 3.155 \log_2 n+5$.<|endoftext|> -TITLE: When does the forgetful functor from algebras over a monad commute with homotopy geometric realizations? -QUESTION [10 upvotes]: Let $\mathcal{C}$ be a combinatorial model category and $\mathrm{T}$ a monad on -$\mathcal{C}.$ -Assume that the model structure on $\mathcal{C}$ lifts to a model structure on -the category of $\mathrm{T}$-algebras $\mathrm{Alg}_{\mathrm{T} }(\mathcal{C}),$ -where the weak equivalences and fibrations are those of $\mathcal{C}.$ -Assume that $\mathrm{T}: \mathcal{C} \to \mathcal{C} $ preserves homotopy colimits indexed by $\Delta^{\mathrm{op}}.$ -Does the forgetful functor $\mathrm{Alg}_{\mathrm{T} }(\mathcal{C}) \to \mathcal{C}$ preserve homotopy colimits indexed by $\Delta^{\mathrm{op}}?$ -More generally one can ask the question replacing $\Delta^{\mathrm{op}}$ by an arbitrary category. -Remark: -If $\mathrm{T}: \mathcal{C} \to \mathcal{C} $ preserves -colimits indexed by some category $\mathrm{K}, $ -the forgetful functor $\mathrm{Alg}_{\mathrm{T} }(\mathcal{C}) \to \mathcal{C}$ preserves colimits indexed by $\mathrm{K}.$ -This also holds for $\infty$-categories with the appropriate notion of monad and algebras over a monad. - -REPLY [2 votes]: Denote by U: Alg_T(C)→C and Free: C→Alg_T(C) the adjoint functors -between Alg_T(C) and C. -Suppose U(j) is a cofibration in C for all j, -where j is a cobase change of Free(i) in Alg_T(C), -where i is a generating cofibration in C. -In this case the preservation of sifted homotopy colimits follows from the preservation of sifted colimits by the forgetful functor and the fact that sifted homotopy colimits can be computed by replacing the diagram by a weakly equivalent projectively cofibrant diagram. -This follows from the key fact that the forgetful functor -from sifted diagrams of T-algebras in C to sifted diagrams in C -preserves projectively cofibrant diagrams. -Indeed, the forgetful functor preserves sifted colimits -and sends cobase changes of free morphisms on generating projective -cofibrations to projective cofibrations by assumption on T. -This condition is satisfied in many situations of interest, -e.g., when T is induced by a colored operad in a symmetroidal model category, as explained in Theorem 6.6 of arXiv:1410.5675. -The model categories of simplicial sets, simplicial symmetric spectra, -and chain complexes in characteristic 0 are symmetroidal. -If we replace symmetric operads with nonsymmetric operads, then -a tractable monoidal model category will suffice, which includes -almost all important examples.<|endoftext|> -TITLE: Does the axiom of choice follow from the statement "Every simple undirected graph is either connected, or its complement is connected"? -QUESTION [10 upvotes]: Using the Well-Ordering Principle, which is equivalent to the Axiom of Choice, it can be proved that - - -(S): for every simple, undirected graph $G$, finite or infinite, either $G$ or its complement $\bar{G}$ is connected. - - -Does (S) imply (AC)? - -REPLY [27 votes]: (S) is a theorem of ZF. -Proof: Let $G$ be a graph, and let $v$ be a vertex of $G$. Define -$$P_v = \{w \,:\, \text{there is a path from } v \text{ to } w\}.$$ -If $P_v$ is the vertex set of $G$, then $G$ is connected. If not, then $\overline{G}$ (the complement of $G$) contains the complete bipartite graph on $P_v$ and $V(G) \setminus P_v$. But then it's easy to see that any two vertices in $\overline{G}$ are connected by a path of length $1$ or $2$.<|endoftext|> -TITLE: Number of collinear ways to fill a grid -QUESTION [22 upvotes]: A way to fill a finite grid (one box after the other) is called collinear if every newly filled box (the first excepted) is vertically or horizontally collinear with a previously filled box. See the following example: - -Let $g(m,n)$ be the number of collinear ways to fill a $m$-by-$n$ grid. Note that $g(m,n) = g(n,m)$. -Question: What is an explicit formula for $g(m,n)$? -Conjecture (user44191): $g(m,n)=m!n!(mn)!/(m+n-1)!$. -Definition (user44191): Let $g(m,n,i)$ be the number of collinear ways to fill $i$ boxes in a $m$-by-$n$ grid such that every row and every column contain at least one filled box. -Remark: $g(m,n) = g(m,n,mn)$. -Proposition (user44191): Here is a recursive formula for $g(m,n,i)$: - -$g(1,1,1) = 1$. -If $m=0$ or $n=0$ or $ i< \min(m,n)$, then $g(m,n,i) = 0$. -$g(m,n,i+1)=(mn-i) g(m,n,i) + mn g(m-1,n,i) + mn g(m,n-1,i).$ - -Proof: The two first points are obvious. We consider the number of collinear ways to fill $i+1$ boxes in a $m$-by-$n$ grid such that every row and every column contain at least one filled box. -There are three cases, corresponding to the three components of the recursive formula: - -The last filled box is not the only filled box in its row and not the only filled box in its column. -The last filled box is the only filled box in its row. -The last filled box is the only filled box in its column. - -By the collinear assumption, 2. does not overlap 3. $\square$ -One way to answer the question is to prove the conjecture using the above recursive formula. -We checked the conjecture for $1\le m \le n \le 5$, using the recursive formula (see below). -Remark: This question admits an extension to higher dimensional grids. -Remark: This question was inspired by that one. - -Sage program -# %attach SAGE/grid.sage - -from sage.all import * - -import copy - -def grid(m,n,j): - if [m,n,j]==[1,1,1]: - return 1 - elif j < min(m,n) or m==0 or n==0: - return 0 - else: - i=j-1 - return (m*n-i)*grid(m,n,i) + m*n*grid(m-1,n,i) + m*n*grid(m,n-1,i) - -def IsFormulaCorrect(m,n): - return grid(m,n,m*n)==factorial(m)*factorial(n)*factorial(m*n)/factorial(m+n-1) - -def CheckFormula(M,N): - for m in range(1,M+1): - for n in range(M,N+1): - if not IsFormulaCorrect(m,n): - return False - return True - -Computation -sage: CheckFormula(5,5) -True - -REPLY [2 votes]: Here's an elementary solution to the following equivalent problem: An $n \times m$ matrix is nice if it contains every integer from $1$ to $mn$ exactly once and $1$ is the only entry which is the smallest both in its row and in its column. Prove that the number of $n \times m$ nice matrices is $(nm)!n!m!/(n+m-1)!$. -For ease of reading, we'll switch the roles of $m$ and $n$ in the problem statement. -After making it continuous, taking complements, and flipping signs, the problem is equivalent to the following: For $1 \le i \le m$, $1 \le j \le n$, define a value independently, identically, and at random $x_{ij} \in U[0,1]$, where $U[0,1]$ is the uniform distribution on $[0,1]$. A pair $(i,j)$ is good if both $x_{ij} \ge x_{iv}$ for all $v$ and $x_{ij} \ge x_{uj}$ for all $u$. Then the probability that there are at least two good pairs $(i,j)$ is exactly $1 - \frac{m! n!}{(m+n-1)!}$. -Let $g$ denote the number of good pairs. Observe that, by a generalized form of the Principle of Inclusion-Exclusion, $$\text{Pr}[g \ge 2] = \sum_{(i_1,j_1),(i_2,j_2)\text{ distinct}}\frac{1}{2!}\text{Pr}[(i_1,j_1),(i_2,j_2)\text{ both good}]$$ $$-\sum_{(i_1,j_1),(i_2,j_2),(i_3,j_3)\text{ pairwise distinct}}\frac{2}{3!}\text{Pr}[(i_1,j_1),(i_2,j_2),(i_3,j_3)\text{ all good}]$$ $$+\sum_{(i_1,j_1),(i_2,j_2),(i_3,j_3),(i_4,j_4)\text{ pairwise distinct}}\frac{3}{4!}\text{Pr}[(i_1,j_1),(i_2,j_2),(i_3,j_3),(i_4,j_4)\text{ all good}]$$ $$\pm\cdots$$ $$=\sum_{k \ge 1}\sum_{(i_1,j_1),(i_2,j_2),\cdots,(i_k,j_k)\text{ pairwise distinct}} \frac{(-1)^k (k-1)}{k!}\text{Pr}[(i_1,j_1),(i_2,j_2),\cdots,(i_k,j_k)\text{ all good}].$$ (Indeed, one can check that $\sum_{k\ge 1}\binom{g}{k}(-1)^k(k-1)$ is $0$ for $g=0,1$, and is $1$ for $g \ge 2$.) -Note that if any two of $(i_1,j_1),(i_2,j_2),\cdots,(i_k,j_k)$ share a row or share a column, then $\text{Pr}[(i_1,j_1),(i_2,j_2),\cdots,(i_k,j_k)\text{ all good}] = 0$. Thus we can rewrite the sum $$\text{Pr}[g \ge 2] = \sum_{k \ge 1}\sum_{(i_1,j_1),(i_2,j_2),\cdots,(i_k,j_k)\text{ pairwise distinct rows, pairwise distinct columns}} \frac{(-1)^k (k-1)}{k!}\text{Pr}[(i_1,j_1),(i_2,j_2),\cdots,(i_k,j_k)\text{ all good}]$$ $$=\sum_{k \ge 1}\sum_{\begin{array}{c} i_1,i_2,\cdots, i_k \text{ pairwise distinct} \\ j_1,j_2,\cdots, j_k \text{ pairwise distinct}\end{array}} \frac{(-1)^k (k-1)}{k!}\text{Pr}[(i_1,j_1),(i_2,j_2),\cdots,(i_k,j_k)\text{ all good}]$$ $$=\sum_{k=1}^{\min (m,n)} (-1)^k (k-1) \binom{m}{k} \binom{n}{k} k! \text{ Pr}[(1,1),(2,2),\cdots,(k,k)\text{ all good}]$$ $$=\sum_{k=1}^{\min (m,n)} (-1)^k (k-1) \binom{m}{k} \binom{n}{k} k! p_k,$$ where $p_k$ is the probability that $(1,1),(2,2),\cdots, (k,k)$ are all good. -Lemma 1. $p_k = \frac{(m+n-k-1)!}{(m+n-1)!}$ for all $1 \le k \le \min (m,n)$ -Proof. Without loss of generality assume $\min (m,n) = m$. We'll calculate $q_k = \frac{1}{k!}{p_k}$, which is the probability that $(1,1),(2,2),\cdots, (k,k)$ are all good and $x_{11} \le x_{22} \le \cdots \le x_{kk}$. Equivalently, $q_k$ is the probability that, for each $1 \le i \le k$, $x_{ii}$ is the maximum of the $m+n-2i+1$ numbers $x_{mi},x_{(m-1)i}, \cdots, x_{ii},x_{i(i+1)}, x_{i(i+2)}, \cdots, x_{in}$, and $x_{11} \le x_{22} \le \cdots \le x_{kk}$. -The probability distribution of the maximum of $w$ independently identically distributed uniform random variables over $[0,1]$ is $A_w(t) = \text{Pr}[\max = t] = wt^{w-1}$. Thus, $q_k$ is $\prod_{i=1}^k \frac{1}{m+n-2i+1}$ multiplied by the probability that $z_1 \le z_2 \le \cdots \le z_k$, where each $z_i$ is distributed according to the function $A_{m+n-2i+1}$. We can write $$q_k = \prod_{i=1}^k \frac{1}{m+n-2i+1} \int_{z_1 \le z_2 \le \cdots \le z_k} A_{m+n-1}(z_1)A_{m+n-3}(z_2)\cdots A_{m+n-2k+1}(z_k) $$ $$= \prod_{i=1}^k \frac{1}{m+n-2i+1} \int_{z_1 \le z_2 \le \cdots \le z_k} (m+n-1)z_1^{m+n-2} (m+n-3) z_2^{m+n-4} \cdots (m+n-2k+1) z_k^{m+n-2k}$$ $$= \int_{z_1 \le z_2 \le \cdots \le z_k} z_1^{m+n-2} z_2^{m+n-4} \cdots z_k^{m+n-2k}$$ $$= \int_{z_2 \le z_3 \le \cdots \le z_k}\frac{1}{m+n-1} (z_2^{m+n-1}) z_2^{m+n-4} z_3^{m+n-6} \cdots z_k^{m+n-2k} $$ $$= \int_{z_2 \le z_3 \le \cdots \le z_k}\frac{1}{m+n-1} z_2^{2(m+n)-5} z_3^{m+n-6} \cdots z_k^{m+n-2k}$$ $$= \int_{z_3 \le z_4 \le \cdots \le z_k} \frac{1}{m+n-1} \frac{1/2}{m+n-2} (z_3^{2(m+n)-4})z_3^{m+n-6}z_4^{m+n-8} \cdots z_k^{m+n-2k}$$ $$= \cdots$$ $$= \frac{1}{m+n-1}\frac{1/2}{m+n-2}\frac{1/3}{m+n-3} \cdots \frac{1/k}{m+n-k}.$$ -Thus $p_k = k!q_k = \frac{1}{m+n-1}\frac{1}{m+n-2}\frac{1}{m+n-3}\cdots \frac{1}{m+n-k} = \frac{(m+n-k-1)!}{(m+n-1)!}$ as desired. $\blacksquare$ -Our desired probability is then $$\text{Pr}[g \ge 2] = \sum_{k=1}^{\min (m,n)} (-1)^k (k-1) \binom{m}{k} \binom{n}{k} k! p_k$$ $$=\sum_{k=1}^{\min (m,n)} (-1)^k (k-1) \frac{1}{k!} \frac{m!n!}{(m-k)!(n-k)!} \frac{(m+n-k-1)!}{(m+n-1)!}$$ $$=\frac{m!n!}{(m+n-1)!}\sum_{k=1}^{\min (m,n)} (-1)^k(k-1)\frac{1}{k!} \frac{(m+n-k-1)!}{(m-k)!(n-k)!}$$ -We'll use the method of Snake-Oil for evaluating combinatorial sums. Declare formal variables $a,b$ and consider the formal power series $f(a,b) = \sum_{m,n \ge 1}\sum_{k=1}^{\min (m,n)} (-1)^k(k-1)\frac{1}{k!} \frac{(m+n-k-1)!}{(m-k)!(n-k)!} a^m b^n$. Our desired probability is then $\frac{m!n!}{(m+n-1)!}$ times the coefficient of $a^mb^n$ in this power series. However, we can switch the order of summation: $$f(a,b) = \sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!} \sum_{n \ge k} \frac{1}{(n-k)!} b^n \sum_{m \ge k} \frac{(m+n-k-1)!}{(m-k)!} a^m.$$ Using the fact that $\binom{c}{d} = (-1)^d \binom{-c+d-1}{d}$ and the Generalized Binomial Theorem, we have $$f(a,b) = \sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!} \sum_{n \ge k} \frac{1}{(n-k)!} b^n \sum_{m \ge k} (n-1)! \binom{m+n-k-1}{m-k} a^m$$ $$=\sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!} \sum_{n \ge k} \frac{(n-1)!}{(n-k)!} b^n \sum_{m \ge k} \binom{-n}{m-k}(-1)^{m-k} a^m$$ $$=\sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!} \sum_{n \ge k} \frac{(n-1)!}{(n-k)!} b^n a^k (1-a)^{-n}$$ $$=\sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!} a^k \sum_{n \ge k} \frac{(n-1)!}{(n-k)!} \left(\frac{b}{1-a}\right)^n $$ $$=\sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!} (k-1)! a^k \sum_{n \ge k}\binom{n-1}{n-k}\left(\frac{b}{1-a}\right)^n$$ $$=\sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!} (k-1)! a^k \sum_{n \ge k}\binom{-k}{n-k}(-1)^{n-k}\left(\frac{b}{1-a}\right)^n$$ $$=\sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!} (k-1)! a^k \left(\frac{b}{1-a}\right)^k \left(1-\frac{b}{1-a}\right)^{-k}$$ $$=\sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!} (k-1)! a^k \left( \frac{b}{1-a-b} \right)^k$$ $$=\sum_{k \ge 1} (-1)^k (k-1)\frac{1}{k!}(k-1)!\left(\frac{ab}{1-a-b}\right)^k$$ $$=\sum_{k \ge 1} \frac{k-1}{k} \left(\frac{-ab}{1-a-b}\right)^k $$ $$= \frac{\frac{-ab}{1-a-b}}{1-\left(\frac{-ab}{1-a-b}\right)} + \log \left(1 - \left(\frac{-ab}{1-a-b}\right)\right)$$ $$=\frac{-ab}{1-a-b+ab} + \log\left(1 + \frac{ab}{1-a-b}\right)$$ $$=\frac{-ab}{(1-a)(1-b)} + \log \left( \frac{1-a-b+ab}{1-a-b} \right)$$ $$= -\frac{a}{1-a}\frac{b}{1-b} + \log(1-a) + \log(1-b) - \log(1-(a+b))$$ $$=-(a+a^2+a^3+\cdots)(b+b^2+b^3+\cdots) - (a + \frac{1}{2}a^2 + \frac{1}{3}a^3 + \cdots) - (b + \frac{1}{2}b^2 + \frac{1}{3}b^3 + \cdots) + ((a+b) + \frac{1}{2}(a+b)^2 + \frac{1}{3}(a+b)^3 + \cdots)$$ It is readily seen that for $m,n \ge 1$, the coefficient of $a^mb^n$ is $\frac{1}{m+n}\binom{m+n}{m} - 1 = \frac{(m+n-1)!}{m!n!} - 1$, so $\text{Pr}[g \ge 2]$ is equal to $\frac{m!n!}{(m+n-1)!}$ times this coefficient, which is $\boxed{1 - \frac{m!n!}{(m+n-1)!}}$ as desired. $\blacksquare$<|endoftext|> -TITLE: Connection on a Principal bundle and transition functions, as in Hitchin's notes -QUESTION [5 upvotes]: This is along the lines of this question - -Gerbes are not just topological objects: we can do differential geometry with them too. We shall next describe what a connection on a gerbe is. -To begin with, let’s look at a connection on a line bundle which is given -by transition functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow S^1\subseteq \mathbb{C}^*$. A connection on consists of $1$ forms $A_\alpha$ defined on $U_\alpha$ such that on a twofold intersection $U_\alpha\cap U_\beta$ we have $iA_\alpha-iA_\beta=g_{\alpha\beta}^{-1}dg_{\alpha\beta}$ - -I know what is a connection $1$ form (on principal bundle) but not as in above version. I am trying to relate what I know with what is given here. -Let $(P,M,\pi)$ be a principal $G$ bundle with $\mathfrak{g}=T_eG$ and transition functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow G$. - -Definition : A connection form on $P$ is a $\mathfrak{g}$ valued $1$ -form $\omega$ on $P$ such that - -$\omega(p)(A^*(p))=A$ for all $A\in \mathfrak{g}$ and $p\in P$ -$(\delta_g^*\omega)(p)(v)=Ad_{g^{-1}}(\omega(p)(v))$ for all $p\in P,g\in G$ and $v\in T_pP$. - - -Given a connection $1$ form $\omega$ on $P$ I am trying to associate a collection of $1$ forms $\{A_\alpha\}$ with some compatability conditions (here $A_\alpha$ is a $1$ form on $U_\alpha$). -Given local trivialization $\psi_\alpha$, we have a section of $\pi$ namely $\sigma_\alpha:U_\alpha\rightarrow P$ defined as $\sigma_\alpha(x)=\psi_\alpha^{-1}(x,e)$. Given local trivialization $\psi_\beta$, we have a section of $\pi$ namely $\sigma_\beta:U_\beta\rightarrow P$ defined as $\sigma_\beta(x)=\psi_\beta^{-1}(x,e)$. -Suppose $x\in U_\alpha\cap U_\beta$ then, we have $\sigma_\alpha(x)\in \pi^{-1}(x)$ and $\sigma_\beta(x)\in \pi^{-1}(x)$. Thus, there exists $g\in G$ (depending on $x$) such that $\sigma_\beta(x)=\sigma_\alpha(x)g$. Given $x\in U_\alpha\cap U_\beta$ there is an obvious choice for an element of $G$ namely $g_{\alpha\beta}(x)$. I see that we have $\sigma_\beta(x)=\sigma_\alpha(x)g_{\alpha\beta}(x)$ for all $x\in U_{\alpha}\cap U_\beta$ i.e., $\sigma_\beta=\sigma_\alpha g_{\alpha\beta}$. -Given a $1$ form $\omega$ on $P$ and we can pull back $\omega$ to $U_\alpha$ under $\sigma_\alpha$ to get $1$ form $\omega_\alpha=\sigma_\alpha^*\omega$ on $U_\alpha$ similarly we can pull back to $U_\beta$ to get $1$ form $\omega_\beta=\sigma_\beta^*\omega$ on $U_\beta$. -As $\sigma_\alpha$ and $\sigma_\beta$ are related by $\sigma_\alpha=\sigma_\beta g_{\alpha\beta}$, one can expect that $\omega_\alpha$ and $\omega_\beta$ are related some how. Given $g_{\alpha\beta}:U_{\alpha\beta}\rightarrow G$ we can produce a $1$ form on $U_\alpha\beta$ as pull back of $\theta$ on $G$ i.e., the canonical $1$ form on $G$ which is a left invariant $1$ form determined by $\theta(e)(A)=A$ for all $A\in \mathfrak{g}$. Let us denote pull back of $\theta$ to $U_\alpha\cap U_\beta$ by $\theta_{\alpha\beta}$. Then, I am expecting some compatibility relation between $1$ forms $\omega_\alpha,\omega_\beta$ and $\theta_{\alpha\beta}$ that should come from -$\sigma_\alpha=\sigma_\beta g_{\alpha\beta}$. -EDIT : I could see in Kobayshi and Nomizu that these local connection forms $\sigma_\alpha$ are related by -$$ \sigma_\beta(p)(X_p)=ad(g_{\alpha\beta}(p)^{-1})(\sigma_\alpha(p)(X_p))+\theta_{\alpha\beta}(p)(X_p) \text{ on } U_\alpha\cap U_\beta.$$ -In short, we have -$$\sigma_\beta=ad(g_{\alpha\beta}^{-1})\sigma_\alpha+\theta_{\alpha\beta} \text{ on } U_\alpha\cap U_\beta$$ -I see that people use $g_{\alpha\beta}^{-1}dg_{\alpha\beta}$ (notation of Maurer Cartan differential of a map $g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow G$) to denote what we have called as $\theta_{\alpha\beta}$. So, in this notation, we have -$$\sigma_\beta=ad(g_{\alpha\beta}^{-1})\sigma_\alpha+g_{\alpha\beta}^{-1}d g_{\alpha\beta}\text{ on } U_\alpha\cap U_\beta.$$ -Question : I am trying to understand how $\sigma_\beta=ad(g_{\alpha\beta}^{-1})\sigma_\alpha+g_{\alpha\beta}^{-1}d g_{\alpha\beta}\text{ on } U_\alpha\cap U_\beta$ on principal bundles has its similar part as $iA_\alpha-iA_\beta=g_{\alpha\beta}^{-1}dg_{\alpha\beta}$ in case of line bundles. -Any suggestion/reference on how to see this is welcome. - -REPLY [4 votes]: A principal $G$-bundle $P\to M$ can be described by an open cover $(U_\alpha)$ of $M$ and a cocycle $g_{\beta\alpha}: U_\alpha\cap U_\beta\to G$. The total space is the quotient of the disjoint union of the spaces $U_\alpha\times G$ via the equivalence relation -$$ U_\alpha\times g\ni (x, g')\sim (y, g'')\in U_\beta\times G $$ -iff -$$ x=y,\;\;g''=g_{\beta\alpha}(x)\cdot g'. $$ -A section of a $G$ bundle is then a collection of smooth maps -$$\sigma_\alpha: U_\alpha\to G $$ -satisfying the gluing conditions -$$ \sigma_\beta(x)=g_{\beta\alpha}(x)\cdot \sigma_\alpha(x),\;\;\forall x\in U_\alpha\cap U_\beta. $$<|endoftext|> -TITLE: Does $SU(2)\cong Sp(1)\subset SO(5)$ factor through $Spin(5)\cong Sp(2)$ as the standard embedding $Sp(1) \to Sp(2)$? -QUESTION [6 upvotes]: $SU(2)$ can be seen as a subgroup of $SO(5)$ through the following chain of subgroups -$$ -SU(2) \subset SO(4) \subset SO(5). -$$ -If we identify $SU(2)\cong Sp(1)$, does the inclusion $Sp(1) \to SO(5)$ factor through $Spin(5) \cong Sp(2) \to SO(5)$ as the standard embedding of $Sp(1)$ in $Sp(2)$. I understand, that there is a lift of $Sp(1) \to SO(5)$ to $Sp(1) \to Sp(2)$, but I can not see that this is the standard embedding. -I already asked this also at mathstackex. - -REPLY [5 votes]: Yes. The rep of $Sp(2)$ on $\mathbb{C}^5$ that induces the covering map is the complement to the line spanned by the symplectic form in $\bigwedge{}^{2}\mathbb{C}^4$. What you are asking is whether there is another line in this space which is invariant under $Sp(1)$, that is, if the action of $Sp(1)$ on the wedge square $\bigwedge{}^{2}\mathbb{C}^4$ fixes a 2-d subspace. Since the representation on $ \mathbb{C}^4$ of $Sp(1)\cong SU(2)$ is the defining rep on $\mathbb{C}^2$ and two trivials, you get two invariant 2-forms from the symplectic form on $\mathbb{C}^2$ and wedging the two trivials.<|endoftext|> -TITLE: The arithmetic progression game and its variations: can you find optimal play? -QUESTION [19 upvotes]: Consider the arithmetic progression game, a two-player game of -perfect information, in which the players take turns playing -natural numbers, or finite sets of natural numbers, all distinct, -and the first player wins if there is a sufficiently long -arithmetic progression amongst her numbers. This question grows out -of Haoran Chen's question, A game on the -integers, and my answer -there. -We have a natural parameterized family of such games, depending on -the number of integers each player is allowed to play on each move -and the desired length of the winning arithmetic progression. -Specifically, for positive-valued functions $f,g$ on the natural -numbers and natural number $n$, let $G(f,g,n)$ be the game in which - -Player A is allowed to play $f(i)$ many numbers on move $i$. -Player B is allowed to play $g(i)$ many numbers on move $i$. -Player A wins if she can form an arithmetic progression of length $n$ using her numbers. - -All the integers played must be distinct. -Generalizing, let $G(f,g,\omega)$ be the game where player A aims -in infinite play to create an infinite arithmetic progression in -her set, and let $G(f,g,{<}\omega)$ be the game in which she wins -by creating a set with arbitrarily large finite arithmetic -progressions. -The original game at the other question was $G(1,k,5)$, with $k$ -constant. -It is not difficult to see that all the games $G(f,g,n)$ are -determined, meaning that one of the players has a winning strategy. -After all, for finite $n$ the game $G(f,g,n)$ is an open game and -therefore determined; and the infinitary games $G(f,g,\omega)$ and -$G(f,g,<\omega)$ have winning conditions of complexity $\Sigma^0_2$ -and $\Pi^0_2$, respectively, and all such games are determined -because they are very low in the Borel hierarchy. -In my answer at the other question, I had made several -observations. - -Player A wins the game $G(1,k,n)$, where player A plays one -number on each move and player B plays at most -a constant $k$ many numbers. The winning -strategy for A is simply to play the smallest available number, -which will ensure that set A has positive density, with proportion $1/(k+1)$ in any initial segment, -and all such sets with positive density contain -arbitrarily long arithmetic progressions by Szemerédi's -theorem. -Thus, actually, player A wins $G(f,k,<\omega)$ for any $f$ and any -$k$, and all with the same strategy. -A generalization of this argument adapts to show that if $g=O(f)$, then player A wins $G(f,g,n)$ and indeed $G(f,g,<\omega)$, since this hypothesis ensures that she can play a set with positive density, which will therefore have arbitrarily long finite arithmetic progressions. -Meanwhile, a diagonal argument shows that player B wins $G(f,1,\omega)$ for any $f$, the game where -player A is aiming to produce an infinite arithmetic progression, -since player B can block the $n^{th}$ progression with a single number at step $n$. -Thus, player B also wins $G(f,g,\omega)$ for any functions $f$ and $g$. -Meanwhile, player B also wins $G(1,<\omega,3)$, since at each move player B -can play all the numbers up to double the largest current number. -This will prevent A from making three numbers in arithmetic -progression. -Indeed, a more refined version of this argument shows that -player B wins $G(1,i-1,3)$, using the function $g(i)=i-1$, since at move -$i$, player A has added one new number, which creates $i-1$ pairs -with earlier numbers, and player B can block each of those pairs from -continuing to an arithmetic progression of length $3$ by playing -$i-1$ many blocking moves. - -I have a number of questions. -Question 1. How quickly can player A win the game $G(1,k,n)$, -where $k$ is constant? Is the -always-play-the-smallest-available-number strategy in any sense -close to (or far from) optimal? -Question 2. Does the game $G(1,k,n)$ have finite game value? -That is, can we bound the length of play in advance, where the -bound depends only on $k$ and $n$ and not on the particular moves -of player B? -I presume the answer is yes, but note that not every open game has -a finite game value. -Question 3. For which functions $g$ does player A win -$G(1,g,n)$? -Above I argued that player A wins $G(1,g,3)$ for constant $g$, and -player B wins for $g(i)=i-1$. For $n=3$, that bound seems likely -close to optimal, since if player B ever fails to block a number, -then player A can win. But of course, perhaps some of the numbers -to be blocked had already been blocked at earlier stages of play. -So the answer seems to be somewhere between the constant functions -and the predecessor function. -Question 4. Can one show that every unbounded function $g$ -enables player B to win $G(1,g,n)$ for some sufficiently large $n$? -Question 5. How much of the analysis carries over to the -corresponding geometric progression games? That is, the games -where player A aims to create geometric integer progressions. -I don't know if there is a geometric progression analogue of -Szemerédi's theorem, but meanwhile, some of the other arguments do -generalize. - -REPLY [3 votes]: For $G(1,f,3)$, a more optimal strategy for A would be to choose the next available odd power of $2$, unless player A can complete a three term arithmetic progression on this turn. -The reason to choose odd powers of $2$ is that this prevents any "collisions" in the sense that on any round $n$ of the game the set of midpoints between two of A's choices has maximal size $\binom{n}{2}$, and the third elements of an arithmetic progression starting with two of A's choices has maximal size $\binom{n}{2}$. Furthermore, the odd power requirement ensures that these two sets don't overlap. -[Details: If $2^x < 2^y$ are two of player A's choices, the two arithmetic progressions are $2^x,2^{x-1}+2^{y-1},2^y$ and $2^x,2^y,2^{y+1}-2^x$. In either case, it's easy to recover $x$ and $y$ from the third element, whether $2^{x-1}+2^{y-1}$ or $2^{y+1}-2^x$. The restriction to odd $x$ ensures that the binary representation of $2^{x-1}+2^{y-1}$ never has consecutive set bits. On the other hand, the set bits of the binary representation of $2^{y+1}-2^x$ are all contiguous.] -In order for B to win against this strategy, player B must cover all of these third elements at each round. This is only possible if $\sum_{i=1}^n f(i) \geq 2\binom{n}{2}$ for each $n$. Otherwise player A's power of $2$ strategy will necessarily win as soon as B's choices fail to cover all of A's $2\binom{n}{2}$ ways to complete a three element arithmetic progression. -By skipping some odd powers of $2$ so that B cannot preemptively cover any of A's ways to finish a three term arithmetic progression, we see that A can win $G(1,f,3)$ unless $f(n) \geq 2n-2$ for every $n$. -This strategy doesn't require A to know anything about $f$. If A can additionally use information about $f$ then one can devise a winning strategy for A in $G(1,f,3)$ unless $f(n) \geq 3n-3$ for every $n$. The idea is to first choose an astronomical number $\omega$ and rather than playing odd powers of $2$, A chooses numbers of the form $\omega + 2^x$ where $x$ is odd. This way, because $\omega$ is very large, three term arithmetic progressions can also be completed below $\omega$ in addition to the two ways to complete them above $\omega$. Then there are the maximum $3\binom{n}{2}$ ways for player A to complete a three term at round $n$. Note that this requires that $\omega$ to be very large, but a suitable $\omega$ can be calculated using $f$ in such a way that $A$ wins on the first round where $f(n) < 3n-3$. -This is optimal. If $f(n)\geq 3n-3$ for every $n$, then B has a winning strategy by covering all of A's possible ways to complete a three term arithmetic progression at each step. - -The games $G(r-2,f,r)$ admit a similar analysis. This time, player A's strategy is to pick sufficiently spread out powers of $(r-1)!$ shifted by an astronomical number $\omega$, unless player A can grab the remaining $r-2$ elements of an arithmetic progression containing two previous picks. -Let's look at $r=4$ for a concrete example. Then all of A's picks are powers of $6$. Given two such numbers $\omega+6^i < \omega+6^j$ there are $6 = \binom{4}{2}$ ways to fit these numbers in a four term arithmetic progression: - -$\omega+6^i,\omega+6^j,\omega+2\cdot6^j-6^i,\omega+3\cdot6^j-3\cdot6^i$ -$\omega+2\cdot6^i-6^j,\omega+6^i,\omega+6^j,\omega+2\cdot6^j-6^i$ -$\omega+3\cdot6^i-2\cdot6^j,\omega+2\cdot6^i-6^j,\omega+6^i,\omega+6^j$ -$\omega+6^i,\omega+6^i/2+6^j/2,\omega+6^j,\omega-6^i/2+3\cdot6^j/2$ -$\omega+3\cdot6^i/2-6^j/2,\omega+6^i,\omega+6^i/2+6^j/2,\omega+6^j$ -$\omega+6^i,\omega+2\cdot6^i/3+6^j/3,\omega+6^i/3+2\cdot6^j/3,\omega+6^j$ - -Excluding $\omega+6^i,\omega+6^j$, the union of these arithmetic progressions consists of the nine numbers $$F(i,j) = \left\{ -\begin{split} -&\omega+3\cdot6^i-2\cdot6^j,\quad & -&\omega+2\cdot6^i-6^j,\quad & -&\omega+3\cdot6^i/2-6^j/2, \\ -&\omega+2\cdot6^i/3+6^j/3,\quad& -&\omega+6^i/2+6^j/2,\quad& -&\omega+6^i/3+2\cdot6^j/3, \\ -&\omega-6^i/2+3\cdot6^j/2,\quad& -&\omega+2\cdot6^j-6^i,\quad& -&\omega+3\cdot6^j-2\cdot6^i -\end{split}\right\}.$$ The "spread" requirements on picks by A should be that any two such sets are disjoint from each other, and disjoint from B's previous picks. -By inspection, in order to block all six arithmetic progressions containing $\omega+6^i$ and $\omega+6^j$, player B must pick at least four of the nine numbers from $F(i,j)$. This means that at round $n$, we must have $\sum_{i=1}^n f(i) \geq 4\binom{2n}{2}$. In fact, we must have $f(n) \geq 16n-12$ for every $n$ or else A's strategy will win at the first $n\geq1$ such that $f(n)<16n-12$. -Similar calculations for $r>4$ show that there is always a linear function $an+b$, where $a$ and $b$ depend on $r$, such that player A's strategy will win at the first $n$ such that $f(n) -TITLE: Is there a sensible notion of a winding number of a closed curve in $\mathbb{R}^n$, $n\geq 3$, with respect to a point not lying on it? -QUESTION [6 upvotes]: I have been browsing "Topological Degree Theory and Applications" by Cho, Chen and O'Regan as well as "Mapping Degree Theory" by Outerelo and Ruiz, but I have not been able to quite answer myself the following question: - -Let $\gamma:\mathbb{S}^1\to\mathbb{R}^n$, $n\geq 3$, be a closed (rectifiable, piece-wise smooth, smooth, if necessary) curve and let $p\in\mathbb{R}^n\setminus\gamma(\mathbb{S}^1)$ be a point not lying on $\gamma$. Do we have a good notion of a winding number $w(\gamma,a)$ that generalizes the case $n=2$ ? - -Remarks: - -If $\gamma$ is smooth with $D\gamma$ of rank $1$ everywhere, then $\gamma(\mathbb{S}^1)$ is a submanifold ($\gamma$ is injective by definition and $\mathbb{S}^1$ is compact). In particular, it is a connected manifold, and we have a well-defined topological degree $\deg(\gamma)$ since it does not depend on a choice of a regular value of $\gamma$ on the image. This is one straight-forward generalization of the case $n=2$ for curves in higher dimensions. There are probably other approaches as well. -Outerelo and Ruiz introduce $w(f,a)$ for continuous $f$ defined only on hypersurfaces $X:=\partial\bar{U}\subseteq\mathbb{R}^n$ (for some $U$ open and bounded). In particular, the construction does not seem to work for curves for $n\geq 3$. More generally, the topological degree appears to be defined only for maps whose domain and codomain are of equal dimension. (In this interpretation and modulo technical details, the winding number in the plane is the degree of a map defined on the unit disk which restricts on the boundary to our curve.) - -I am hoping that (1.) could somehow be "relativized" to include an external vantage point $p$. - -REPLY [7 votes]: It's natural to assume that the winding number is an integer. Nevertheless, this number should behave in a continuous way (see what follows). Consider a curve in $\ \mathbb R^n,\ (n\ge3).\ $ Consider points $p\ q\in\mathbb R^n\ $ outside the curve. Then there exists continuous $ f:[0;1]\rightarrow\mathbb R^n\ $ which is outside the curve, and is connecting $\ p\ q.\ $ The winding number should be constant for points $\ f(t),\ $ hence it should be the same for $\ p\ $ and $\ q.\ $ Thus, under the given understanding, a winding number would be useless.<|endoftext|> -TITLE: etale topology of local schemes -QUESTION [13 upvotes]: My apologies if this is too elementary, but since I have seen similar questions here I offer it. -After years doing almost exclusively classical complex geometry, using analysis and topology, I am trying to learn schemes, starting from the “red book”, which I have browsed for years, but vowing now not to skip anything. After spending 6 enjoyable months on chapter I, varieties, which I already “knew”, I am up to section II.2, the definition of preschemes, and became puzzled by essentially the first example, “Ex. F bis”. Mumford says the punctured local scheme spec(C[[X1,...,Xn]]) - m, where m is the unique maximal ideal, has "topological properties identical to those of the ordinary 2n-1 sphere", with reference to “chapter 8”. -Since when n=1 this scheme has a one point underlying topological space I assumed this was a misprint, until I thought about it in the context of “K-valued points” discussed later. For n=1 this scheme, spec(C((X))), apparently admits maps of every degree n, coming from the field extensions C((X^1/n)) of fractional Laurent series, which mimic finite covering spaces of the circle. Further googling turns up statements that the “etale cohomology” of this scheme resembles topological cohomology of the 2n-1 sphere, where etale cohomology is apparently ordinary derived functor sheaf cohomology, applied to the category of all etale maps rather than that of just open immersions. -I have found comparison theorems for etale cohomology, e.g. in answers to questions on this site, that the etale cohomology of a variety over the complex numbers, at least for finite coefficient groups, is the ordinary topological cohomology of the underlying topological space, but I have not found comparison theorems that I can see apply to the local analytic case in the example above. -So I have this question, and a guess as to what such theorems might say. -Suppose we are given a polynomial, or a convergent power series f in n+1 complex variables X0,...,Xn, which defines an isolated singularity at the origin, and we mod out the power series ring by the corresponding principal ideal (f), obtaining the analytic local ring R of the singularity. Then I believe the C algebra structure of this ring determines the singularity up to local analytic isomorphism.hence the algebraic invariants of that ring should also determine local topological properties of the singularity. -In particular, if we intersect the variety defined by f with a small 2n+1 sphere centered at the origin, we get a smooth 2n-1 real dimensional manifold K, the boundary of the Milnor fiber of the singularity. Then the topological properties of the punctured local scheme S = spec(R) - m, where m is the maximal ideal, should reflect those of the manifold K. In Mumford’s example, we have f = X0, R = C[[X1,...,Xn]], and K = the 2n-1 sphere in the hyperplane X0 =0. -In Singularities of Complex Hypersurfaces, Milnor proves this manifold K is n-2 connected at least when f is a polynomial defining an isolated singularity, and I ask if this is reflected in the etale cohomology of the scheme S. In particular -1) does the etale fundamental group of S vanish for n ≥ 3? -2) does the etale cohomology of S vanish in degrees 1 through n-2? -3) does the “top” degree etale cohomology of S, in degree 2n-1, not vanish? -If so where can one find such local comparison theorems? What about results for isolated complete intersection singularities, or more general punctured local spectra?" - -REPLY [6 votes]: As an addition to Will Sawin's excellent answer: I was wondering about the same thing when I was writing my PhD thesis, and I managed to prove that over $\mathbb{C}$, the etale homotopy type (as defined by Artin and Mazur) of the "algebraic Milnor fiber" (either using henselization or completion) agrees with the homotopy type of the classical Milnor fiber up to profinite completion. In particular, the scheme you mention does not just share the properties of the sphere, it in some precise sense "is" the sphere. -The result can be found in Chapter 4 of my thesis. This is really nothing fancy: it is deduced from the comparison theorems for the fundamental group and cohomology from SGA1 and SGA4 mentioned by Will and a formal argument due to Artin and Mazur, which is probably why I never decided to publish that part. For the comparison of henselization and completion, you need a bit more work, but there are stronger results of this kind in the literature, due to Fujiwara and Gabber.<|endoftext|> -TITLE: Combinatorial curvature of real projective plane -QUESTION [6 upvotes]: There is a notion of combinatorial curvature due to Forman, see here (published paper) or here (preprint). I checked for a couple of small triangulations of $\mathbb{RP}^2$ (6-vertex, 7-vertex, 9-vertex) and the 1-curvature seemed to be non-positive for any edge in these. Is there any triangulation which has at least some positive curvature? (My expectation is no: subdivisions seem to have more and more negative combinatorial curvature, so positive curvature triangulations should be small.) If there is no triangulation of $\mathbb{RP}^2$ with some positive curvature, what would be the explanation? - -REPLY [2 votes]: Take any triangulation and any pair of adjacent faces $ABC$ and $BCD$. Now subdivide the shared edge $BC$ and both faces into three edges $BE$, $EF$, $FC$ and six faces $ABE$, $AEF$, $AFC$, $BED$, $EFD$, $FCD$. The edge $EF$ has $1$-curvature $2+2-2=2$.<|endoftext|> -TITLE: Is the quotient of a toric variety by a finite group still toric -QUESTION [14 upvotes]: I have asked this question on Math.StackExchange, but haven't got any reply. -Suppose $X$ is a toric variety with fan $\Sigma$, and the lattice of one-parameter subgroups of its torus is $N$. Suppose $\phi$ is a representation of finite group $G$ in $N \otimes_{\mathbb{Z}} \mathbb{R}$, -\begin{equation} -\phi:G \rightarrow \text{GL}( N \otimes_{\mathbb{Z}} \mathbb{R}) -\end{equation} -which sends the lattice $N$ to $N$, and the cones of $\Sigma$ to cones of $\Sigma$, therefore each $\phi(g)$ defines a toric automorphism of $X$. -Question 1: Is the quotient variety $X/G$ (which exists from GIT) still a toric variety? Is the quotient morphism $X \rightarrow X/G$ toric? -Question 2: If so, how to construct the fan of $X/G$ and the toric morphism $X \rightarrow X/G$? -Any references on the two questions? - -REPLY [36 votes]: No. If $X$ is the 47-dimensional torus and $G=\mathbb{Z}/47\mathbb{Z}$ acting by the permutation representation, then $X/G$ is not a rational variety (R. Swan, Inv. math. 7, 148-158 (1969)), therefore not toric.<|endoftext|> -TITLE: Hopf Subalgebras of Quantized Algebras -QUESTION [7 upvotes]: As is well known, quantized enveloping algebras $U_q(\frak{g})$ admit far fewer sub-Hopf algebras than classical enveloping algebras $U(\frak{g})$. As one can check directly, for appropriate subsets of the (simple root) standard generators $E_i,F_i,K_i$, a Hopf subalgebras are seen to be generated. However, I do not know of sub-Hopf algebras which are not generated by the standard generators. Are there examples of objects? -Moreover, since the notion of a Hopf sub-algebra is so restrictive for $U_q(\frak{g})$, is it naive to expect some kind of classification? - -REPLY [4 votes]: Since the OP is asking for examples of - -sub-Hopf algebras which are not generated by the standard generators - -i.e. the Chevalley generators (which are actually the generators of the Cartan–Weyl basis with a different normalization), satisfying the Chevalley–Serre Relations, maybe the following method might appear useful for constructing such examples: -The quantum universal enveloping (super)algebras $U_q(g)$, where $g$ is a simple Lie (super)algebra, are known to have alternative—isomorphic—descriptions in terms of $q$-deformed paraparticle (parabosonic or parafermionic) algebras: such descriptions are called pre-oscillator or paraparticle realizations in the mathematical physics literature. -(They extend the corresponding oscillator realizations of the undeformed cases, where UEA of Lie (super)algebras are shown to be isomorphic or homomorphic to the usual (undeformed) parabosonic or parafermionic algebras, or homomorphic to Weyl and Clifford algebras, i.e., the CCR and the CAR relations of quantum mechanics). -See for example Palev - A superalgebra morphism of $U_q(osp(1/2N))$ onto the deformed oscillator superalgebra $W_q(N)$ written in terms of $q$-deformed parabosonic generators or Palev - Quantization of $U_q[so(2n + 1)]$ with deformed para-Fermi operators written in terms of $q$-deformed parafermionic generators. -In the paper Palev and Van der Jeugt - The quantum superalgebra $U_q\big( osp(1/2n)\big)$: deformed para-Bose operators and root of unity representations, the authors use deformed realizations to connect the rep theory of $U_q\big( osp(1/2n)\big)$ with the deformed paraboson Fock spaces. They describe the isomorphism between the $q$-deformed UEA of the Lie superalgebra $osp(1/2n)$ and the $q$-deformed parabosonic algebra: see p. 2608, relations (2.12) and their converse: (2.13), (2.14), (2.15) for the undeformed case and p. 2609, relation (3.3), (3.4) and their converse in p. 2610, rel. (3.5), for the deformed case. -These realizations are then used (see p. 2611, Proposition 4) to provide a description of the sub-hopf algebra -$$ -U_q\big(gl(n)\big)\subset U_q\big(osp(1/2n)\big) -$$ -in terms of deformed parabosonic generators. The $U_q\big(gl(n)\big)$ sub-hopf algebra is described in terms of a suitable subset of deformed parabosonic generators and relations (which are not linear combinations of the Cartan-Weyl generators). -Similar methods, utilizing bosonic, fermionic or paraparticle generators, can be used to describe—and maybe used as a tool at some classification attempt—other hopf subalgebras of various $q$-deformed UEAs. There are lots of works in a similar spirit in the mathematical physics literature—see also the references in the cited articles. -I hope the above might be of some interest for the purposes of the OP.<|endoftext|> -TITLE: Conjugacy classes in $GL_{n}(Z / pZ)$ -QUESTION [5 upvotes]: Let $p$ be a prime number and $G=GL_n ( \mathbb{Z} / p \mathbb{Z} -)$. Consider the set $U$ of upper-triangular matrices of $G$ -having entries of $1$ on the diagonal. The cardinality of $U$ is $p^{\frac {n(n-1)} 2}$ and $U$ is a subgroup of $G$, in particular $U$ is a -Sylow $p$-subgroup of $G$. It is well known that the Sylow $p$-subgroups of a -group $G$ are conjugate, and every $p$-subgroup $H$ of $G$ is contained in -some Sylow $p$-subgroup of $G$. Then there exists $g\in G$ such that $H\leq -gUg^{-1}$, which allows us to compute the number of conjugacy classes of -elementary abelian subgroups of rank $2$ ($H=( \mathbb{Z} / p \mathbb{Z} ) ^2$) in the Sylow $p$-subgroup $U$. Any help would be appreciated so much. Thank you all. - -REPLY [2 votes]: I would say the problem is open. -Consider the case where $p>n$, i.e. every element has order $p$ in $U$. -The number of such subgroups is $(|U|·k(U)-1-\frac{|U|-1}{p-1}(p^2-1))\frac{1}{(p^2-1)(p^2-p)}$, where $k(U)$ is the number of conjugacy classes of $U$. -Let me explain the formula. The number of commuting pairs in any finite group G is $|G|⋅k(G)$. Such pairs may - -Generate a trivial group (only 1 choice) -Generate a group of order $p$. There're $\frac{|U|-1}{p-1}$ groups of order $p$ in $U$, and for each group, there're $p^2-1$ choices to generate the group with $2$ elements. -Generate a group of order $p^2$. The group is surely rank $2$, and has $(p^2-1)(p^2-p)$ generating pairs. Dividing the number of such pairs in $U$ by $(p^2-1)(p^2-p)$ gives the answer. - -So, the question is equivalent with determining the number of conjugacy classes of $U$, which is the Higman Conjecture.<|endoftext|> -TITLE: Szemerédi–Trotter theorem -QUESTION [7 upvotes]: Szemerédi–Trotter theorem asserts that given $n$ points and $m$ lines in the plane, the number of incidences (i.e., the number of point-line pairs, such that the point lies on the line) is: -$O((mn)^{\frac{2}{3}}+m+n).$ -I was wondering how many different proofs are known for this theorem? Is there any survey or references for these proofs? I know there are two proofs here https://www.cs.princeton.edu/~zdvir/papers/Dvir-survey.pdf - -REPLY [2 votes]: I can think of at least three proofs: - -The original cell decomposition proof of Szemerédi-Trotter -The proof via the Crossing Lemma given by Székely -The proof using the polynomial ham sandwich theorem - -Adam Sheffer has an early draft of a book on incidence theory and the polynomial method, which mentions all three proofs.<|endoftext|> -TITLE: Hecke operators for hermitian modular forms of general level -QUESTION [5 upvotes]: It has bug me for a while that I don't have a good understanding of the theory of Hecke operators. For elliptic modular forms, it was explained in Koblitz's book that they arose from viewing the modular forms as function on modular points (lattices in $\mathbb{C}$, possibly with additional structures) but I feel this is very particular to elliptic modular forms as there doesn't seem to be a similar interpretation for other kind of modular forms such as Siegel modular forms. For Siegel modular forms of general level $\Gamma = \Gamma^{(2)}(N)$, it was defined by Andrianov in his book Modular Forms and Hecke Operators: He proved that the commeasurator of $\Gamma^{(2)}(N)$ in $G = GSp_4(\mathbb{Q})^+$ is the whole group $G$ (Lemma 3.1) and commented that - -Using Lemma 3.1 as a point of departure, one could determine the Hecke ring - of the pair $(\Gamma, G)$ and then consider its representations on spaces of modular forms for the group $\Gamma$. However, the structure of the Hecke rings that arise is in general unknown, and one does not yet have a concrete general theory of Hecke operators. -Because our constructions are not meant as an end in themselves, but rather as a means for studying Diophantine problems in number theory, we shall simplify the situation by, in the first place, limiting ourselves to the types of congruence subgroups that arise in arithmetic, and, in the second place, considering certain subrings of the Hecke ring of the pair $(\Gamma, G)$, rather than the entire Hecke ring. - -(I changed Andrianov's $K$ to $\Gamma$ and $S$ to $G$.) -The subring Andrianov mentioned is defined on page 124, namely the ring of Hecke operator associated to the Shimura pair $(\Gamma, S(\Gamma))$ for certain subgroup $S(\Gamma)$ of $G$ defined in equation (3.5). -Now, I have no idea how Andrianov came up with that $S(\Gamma)$. Is there a systematic explanation of the ring? I am aware that for level 1, the theory seems to be explainable from the representation theoretic perspective and Satake isomorphism. -I would imagine that a similar theory could be developed for Hermitian modular forms (as in Ikeda's paper https://www.math.kyoto-u.ac.jp/~ikeda/hermitian) but I am not so sure what one should change from the Siegel case. What should be "the" Hecke algebra for Hermitian modular forms of level $N$? (The group $U(2,2)$ I think does not split so I don't know how Satake isomorphism will work here.) - -REPLY [4 votes]: It's best to think of these things adelically. Suppose you have a reductive group $G$ over a number field $F$, and you set -$$\mathcal{G} = G(\mathbf{A}_{F, f}) = \sideset{}{'}\prod_{\text{$v$ finite place of $F$}} G(F_v).$$ -For any open compact $U \subset \mathcal{G}$ you can form the Hecke algebra $\mathbf{C}[U \backslash \mathcal{G} / U]$ of compactly supported, bi-$U$-invariant functions on $\mathcal{G}$, and this will act on the $U$-invariants of any $\mathcal{G}$-representation (in particular, on spaces of automorphic forms for $G$ of level $U$). -Now, if $U$ is of the form $\prod_v U_v$, you get a formula -$$\mathbf{C}[U \backslash \mathcal{G} / U] = \sideset{}{'}\bigotimes_v \mathbf{C}[U_v \backslash G(F_v) / U_v],$$ -so it suffices to understand each of the local Hecke algebras -- this is the advantage of the adelic approach. -For all but finitely many $v$, the group $G(F_v)$ is unramified (= quasi-split and splits over an unramified extension) and $U_v$ is a hyperspecial maximal compact. In this case, the local Hecke algebra is commutative, and its structure is described by the Satake isomorphism (which works fine in this degree of generality -- it's not necessary to assume that $G$ is split). In particular, in your $U(2, 2)$ setting, for places $v$ split in the quadratic field defining the unitary group, we have $G(F_v) \cong \operatorname{GL}_4(F_v)$ and the spherical Hecke algebra is the same as that of $GL_4$, which is a polynomial ring $\mathbf{C}[X_1, X_2, X_3, X_4^{\pm 1}]$. -The problem is what to do at the bad places. There is, essentially, no hope of describing $\mathbf{C}[U_v \backslash G(F_v) / U_v]$ in any concrete way for a totally general $U_v$. Ideally, one would want to define some collection of "nice" open subgroups of $G(F_v)$, and some "nice" (ideally commutative) subalgebras of the Hecke algebras of these subgroups, which would still be rich enough that the actions of these subalgebras on smooth representations encode the "interesting" information. For $G = \mathrm{GL}_2$ we understand this very well, thanks to Atkin, Lehner, and Casselman. However, it's not clear what the "right" approach is for general $G$. -For $G = \operatorname{GSp}(4)$ several approaches have been tried. Andrianov's approach is to use subgroups $U_v$ that look like "block upper-triangular" matrices, with the bottom right 2x2 block congruent to 0 modulo powers of $v$. Then you can look for a subalgebra of the Hecke algebra that somehow "comes from" the $GL_2$ blocks along the diagonal, and that is Andrianov's $S(\Gamma)$. However, there are other approaches; Roberts and Schmidt have worked out a theory of newforms, oldforms, and Hecke operators using the "paramodular subgroups", which has essentially no overlap with Andrianov's setup. -There are generalisations for generic representations of $GL(n)$, and a few other groups of small rank, e.g. $GSp(4)$ and $U(2, 1)$; see this question. However, there's nothing like a general theory for arbitrary $G$.<|endoftext|> -TITLE: Is there a field with finitely many abelian extensions, that is neither separably closed nor real closed? -QUESTION [5 upvotes]: If $K$ has only finitely many Galois extensions, then $K$ must be either separably closed or real closed. Are there any other fields whose abelianizations are finite extensions (i.e. whose absolute Galois groups have finite abelianizations)? - -REPLY [4 votes]: Here is another example, which also answers the "followup question": -Let $K$ be the field of Laurent series over $\mathbb{R}$. Its absolute Galois group is the infinite profinite dihedral group $\hat{\mathbb{Z}}\rtimes(\mathbb{Z}/2\mathbb{Z})$, where the action is by inversion. This group is the free profinite product of two groups of order $2$. Its abelianization is therefore $(\mathbb{Z}/2\mathbb{Z})^2$. In particular, $K$ has the desired property.<|endoftext|> -TITLE: fixed points of quadratic iteration -QUESTION [5 upvotes]: Consider the well-known iteration $f:z\to z^2 + c,$ and consider the values of $c$ for which $0$ is a periodic point. Experiment shows that most such values of $c$ (about $480$ out of $512$ for period $10$) lie on the real axis, and those that don't lie near the cusp part of the Mandelbrot set cardioid part, along the boundary (the cardioid itself). I assume all this is well-known, but not to me, so any enlightenment/references welcome. - -REPLY [5 votes]: The number of real centers of hyperbolic components of exact period $n$ is OEIS A000048. This isn't bad to work out from the combinatorics of kneading sequences which Rivin links to sources for. A number of references linked by OEIS imply this, although I oddly can't find one which explicitly states it and gives a proof. I know this because Sarah Koch, Dylan Thurston and I were computing these numbers earlier this year and I worked out the combinatorics, which I'm willing to write up if you care enough. -As OEIS says, this gives the formula -$$\frac{1}{2n} \sum_{d|n,\ d \ \mbox{odd}} \mu(d) 2^{n/d} \approx \frac{2^n}{2n}.$$ -By contrast, the number of $c$ of exact period $n$ is -$$\sum_{d|n} \mu(d) 2^{n/d} \approx 2^n.$$ -So the fraction of real roots is roughly $\tfrac{1}{2n}$. - -REPLY [4 votes]: Upon investigation: - -Mathematica is screwing up. Using it's own function CountRoots[], which counts real roots (exactly), the number of real roots, while quite far from zero (there are $30$ for $n=9,$ for example) is not the majority of the fixed points. -The parameter values corresponding to the periodic points are equidistributed on the boundary of $M$ with respect to harmonic measure. References are: - -Ahlfors, Lars V., Conformal invariants. Topics in geometric function theory, McGraw-Hill Series in Higher Mathematics. New York etc.: McGraw-Hill Book Company. VII, 157 p. $ 10.95 (1973). ZBL0272.30012. -and -Brolin, H., Invariant sets under iteration of rational functions, Ark. Mat. 6, 103-144 (1965). ZBL0127.03401. - -Real fixed points correspond to the fixed points of the quadratic iteration on the intervals - a beautiful subject, one of the most beautiful papers on which is - -Milnor, John; Thurston, William, On iterated maps of the interval, Dynamical systems, Proc. Spec. Year, College Park/Maryland, Lect. Notes Math. 1342, 465-563 (1988). ZBL0664.58015. -(many thanks to Curt McMullen for the refs).<|endoftext|> -TITLE: The statement that $A \ge B$ implies $A^{-1} \le B^{-1}$ is still true for matrices? -QUESTION [5 upvotes]: Problem: -Suppose we have two real, symmetric and positive definite square matrices $A$ and $B$, i.e., -$$A_{ij}, B_{ij}\in \mathbb{R}$$ -$$A^T=A$$ -$$B^T=B$$ -$$x^TAx>0 \forall x$$ -$$x^TBx>0 \forall x$$ -If $A \ge B$, i.e., $A-B$ is semi-postive definite ($x^T(A-B)x \ge0,\,\forall x $), -then is the statement that $A^{-1} \le B^{-1}$ true, i.e. $A^{-1}-B^{-1}\ge0$ ($x^T(B^{-1}-A^{-1})x\ge0,\,\forall x $) -Remarks: -Obviously if $A$ and $B$ can be diagonalized simultaneously with the same similarity transformation, then the statement is true. -What about the general case? I tried some numerical examples, it seems the statement is true. But I don't know how prove it. -I would appreciate if anyone can give a proof or point out any reference that has the solution of the above problem. -Thanks in advance! - -REPLY [21 votes]: This is a well-known fact. A simple proof : setting $y=B^{1/2}x$, we have $\|y\|^2\le y^TB^{-1/2}AB^{-1/2}y$, that is $I_n\le B^{-1/2}AB^{-1/2}$. The eigenvalues of the latter symmetric matrix are thus $\ge1$. Its inverse $B^{1/2}A^{-1}B^{1/2}$ has eigenvalues $\le1$, that is $B^{1/2}A^{-1}B^{1/2}\le I_n$. This gives $z^TB^{1/2}A^{-1}B^{1/2}z\le\|z\|^2$. Setting $w=B^{1/2}z$, this writes $w^TA^{-1}z\le z^TB^{-1}z$, that is $A^{-1}\le B^{-1}$. -More generally, Loewner theory tells you what are the operator-monotone functions. By definition, a numerical function $f:I\rightarrow{\mathbb R}$ ($I$ an interval) is monotone operator if whenever $A,B$ are real symmetric matrices with spectra included in $I$, the inequality $A\le B$ implies $f(A)\le f(B)$. The beautiful theorem is that $f$ is operator monotone if and only if it admits a holomorphic extension to the upper half plane $\Im z>0$, with values in the upper half plane (of course this extension is unique) ; such a holomorphic function is called a Pick function. In particular, operator monotone functions are analytic ! For instance, if $\alpha>0$, the map $A\mapsto A^\alpha$ is monotone operator over $(0,+\infty)$ if and only if $\alpha\le1$ !! -W. F. Donoghue dedicated a full book to Loewner theorey. See also R. Bhatia's book.<|endoftext|> -TITLE: Inverse of a small submatrix -QUESTION [11 upvotes]: Let $A$ be a large matrix (say, $1000 \times 1000$), and let $\mathcal I = \{2,3,5\}$ be a set of row/column indices. Let $(A^{-1})_{\cal I \times I}$ denote the submatrix of $A^{-1}$ that consists of the $\{2,3,5\}$ rows and columns of $A^{-1}$. -Is there an efficient way of computing the following $3 \times 3$ matrix inverse $((A^{-1})_{\cal I \times \cal I})^{-1}$ without inverting the large matrix $A$? - -REPLY [11 votes]: One way to go about this is as follows: -For $i,j \in \mathcal{I}$ Compute $e_i^TA^{-1}e_j$ by using the approach based on Gaussian quadrature; see for instance, a precise algorithm and analysis in our paper "Gauss quadrature for matrix inverse forms with applications." -Now that you have $[A^{-1}]_{\mathcal{I},\mathcal{I}}$, getting its inverse is an easy matter since $|\mathcal{I}|$ is small.<|endoftext|> -TITLE: Rational homology spheres and geometric properties of the Wu manifold -QUESTION [6 upvotes]: I am interested in simply connected rational homology spheres. The first such example is in dimension 5 and it is the Wu manifold $SU(3)/SO(3)$. You can find a discussion about simply connected rational spheres in: -Simply-connected rational homology spheres. -I need to understand the geometric properties of this manifold in order to construct mappings with certain properties. So my questions are: -1. Why is the Wu manifold important except for being a simply connected rational homology sphere? -2. Where can I read about the geometric properties of the Wu manifold? -I am interested more in geometric properties of $SU(3)/SO(3)$ rather than its applications to algebraic topology since, it seems, I have to construct certain mappings explicitly without using algebraic topology. I will describe my problem later in another post. - -REPLY [6 votes]: Another feature of the Wu manifold which makes it an important example is that it is not spin${}^c$. In contrast, every closed orientable manifold of dimension at most four is spin${}^c$. -One geometric property of the Wu manifold is that it admits a metric of positive scalar curvature. Gromov and Lawson showed that a simply connected non-spin manifold of dimension at least five always admits a metric of positive scalar curvature.<|endoftext|> -TITLE: A flatness result of Fiedorwicz for amalgamated free products of monoids in connection with classifying spaces of monoids -QUESTION [10 upvotes]: In Lemma 5.2(a) of Z. Fiedorowicz, Classifying Spaces of Topological Monoids and Categories American Journal of Mathematics Vol. 106, No. 2 (Apr., 1984), pp. 301-350 the author proves the following. - -Lemma 5.2(a) Suppose that $\{M_i\}_{i\in I}$ is a collection of monoids with a common submonoid $W$ such that the monoid ring $\mathbb ZM_i$ is flat as a left $\mathbb ZW$-module for each $i\in I$. Let $M$ be the amalgamated free product (or pushout) of the $M_i$ over $W$. Then $\mathbb ZM$ is flat as a left $\mathbb ZM_j$-module for each $j\in I$. - -His proof is one line, but I cannot understand it. I would very much love to use this result. I'm a bit nervous about it because the proof he sketches would seem to work in the category of rings, or more generally $k$-algebras over a commutative ring with unit $k$, but papers I saw of Cohn on amalgamated free products of rings seem to involve much more complicated colimits diagrams than Fiedorwicz is using. Moreover, Warren Dicks pointed out to me an example of an amalgamated free product of $k$-algebras, with $k$ a field, where the factors are flat over the subring but the amalgamated free product is not flat over the factors. He derives it from the paper https://arxiv.org/abs/math/0205034. -I would be happy for a proof I can understand in the case of two factors. I did try to contact the author, but he is retired and didn't respond. I now reproduce his proof. - -Proof. This is immediate because $\mathbb ZM$ is the direct limit of -$$\mathbb ZM_j\otimes_{\mathbb ZW} \mathbb ZM_{i_1}\otimes_{\mathbb ZW} \mathbb ZM_{i_2}\otimes_{\mathbb ZW}\cdots \otimes_{\mathbb ZW}\mathbb ZM_{i_k}$$ $i_1,\ldots, i_k\in I$ as a left $\mathbb ZM_j$-module and hence is flat over $\mathbb ZM_j$. - -What I don't understand is exactly what is the directed (or filtered) diagram that this direct limit is indexed over (i.e., what are the maps). Each of the terms is flat, so if $\mathbb ZM$ is such a direct limit, then I am fine with the proof. There is a map from each of these iterated tensors to $\mathbb ZM$ induced by multiplication (and the ``inclusions'' into the pushout) but I don't see how to organize this into a direct limit (=filtered colimit). - -REPLY [6 votes]: This appears to be false to me (unless maybe there are commutativity hypotheses), unless I've made a mistake below. (This is based on an example that I saw Andrew Ranicki give, a number of years ago, about non-exactness of Cohn localization.) -Consider the diagram of monoids -$$ -\Bbb Z \leftarrow \Bbb N \rightarrow \Bbb N \ast \Bbb N -$$ -where the right-hand arrow is the inclusion of the first factor into the free product. On taking monoid algebras, this is a diagram of rings -$$ -\Bbb Z[x^{\pm 1}] \leftarrow \Bbb Z[x] \rightarrow \Bbb Z\langle x,y\rangle. -$$ -The left factor is clearly flat over $\Bbb Z[x]$ and the right factor is free on the basis $$\{1\} \cup\{y\cdot w \mid w \text{ is a monomial in }x,y\}.$$ -Thus both factors are flat left modules. The monoid algebra on the pushout $M$ is the ring $\Bbb Z\langle x^{\pm 1}, y\rangle$ by a universal property argument. -However, consider the exact sequence of left $\Bbb Z\langle x,y\rangle$-modules -$$ -0 \to \Bbb Z\langle x,y\rangle \oplus \Bbb Z\langle x,y\rangle \xrightarrow{[x,y]} \Bbb Z\langle x,y\rangle \to \Bbb Z \to 0, -$$ -where the first map sends $(a,b)$ to $ax+by$ and the second sends $x$ and $y$ to zero. Tensoring on the left with $\Bbb Z\langle x^{\pm 1}, y\rangle$ gives the sequence -$$ -0 \to \Bbb Z\langle x^{\pm 1},y\rangle \oplus \Bbb Z\langle x^{\pm 1},y\rangle \xrightarrow{[x,y]} \Bbb Z\langle x^{\pm 1}, y\rangle \to 0 \to 0. -$$ -This is not left exact, because the first factor of the direct sum is taken isomorphically to the middle term by invertibility of $x$.<|endoftext|> -TITLE: Endomorphism ring of simple ordinary abelian variety -QUESTION [9 upvotes]: Is there an example of an ordinary and simple abelian variety $A$ over an algebraically closed field $K$ (of characteristic $p>0$) such that ${\rm End}(A)$ is not commutative? Note that the answer is no if $K=\overline{\bf F}_p$ (in that case ${\rm End}(A)_{\bf Q}$ is a CM field). My question is for other fields. - -REPLY [6 votes]: Let $D$ be a non-split quaternion algebra over $\mathbb{Q}$, split at $p$ and $\infty$. Let $\mathcal{O}$ be a maximal order (I think these are all conjugate). Then $\mathcal{O}^1$, the multiplicative group of norm 1 elements, embeds into $SL_2(\mathbb{R})$ by picking an isomorphism $\iota:D\otimes \mathbb{R}\cong M_{2\times 2}(\mathbb{R})$. In particular, $\mathcal{O}^1$ acts on the upper half plane $\mathbb{H}$. It turns out that this action is properly discontinuous and cocompact; form the quotient $\mathcal{X}=[\mathbb{H}/\mathcal{O}^1]$. This (Deligne-Mumford stack) is the moduli space of principally polarized abelian surfaces with multiplication by $\mathcal{O}$; these abelian surfaces are also sometimes called "fake elliptic curves (with multiplication by $\mathcal{O}$)". -It turns out this Shimura curve has a good integral model, and in particular may be reduced modulo $p$. Let $\pi:A\rightarrow X$ denote the reduction mod $p$ together with the universal fake elliptic curve and let $K:=\mathbb{F}_p(X)$ denote the function field of $X$. -First of all, $A$ is generically ordinary. Here is one way to see this: $A[p^{\infty}]$ has an action by $\mathcal{O}\otimes\mathbb{Z}_p\cong M_{2\times 2}(\mathbb{Z}_p)$. By projecting onto idempotents, we see that $A[p^{\infty}]\cong \mathcal{G}\oplus \mathcal{G}$ where $\mathcal{G}$ is a height 2, dimension 1 $p$-divisible group. Moreover, $\mathcal{G}$ is everywhere versally deformed along $X$ (this is essentially Serre-Tate). On the other hand, there are only two possible Newton Polygons by the numerical constraints on $\mathcal{G}$: ordinary and supersingular. Therefore $\mathcal{G}$ and hence $A$ is generically ordinary. -Now, $A_{K}$ is an ordinary abelian surface over $K$ with multiplication by $\mathcal{O}$. We just need to check that $A_{K}$ is geometrically simple. If it weren't, there would be a finite extension $L/K$ such that $A_L$ is isogenous to $E\times E'$. By Ogg-Shafarevich, $E$ and $E'$ both have good reduction over the smooth projective curve $Y$ corresponding to $L$. This implies that $E$ and $E'$ are isotrivial over $Y$, which is a contradiction (e.g. by monodromy).<|endoftext|> -TITLE: Branching laws for $SO(n)$ -QUESTION [7 upvotes]: The branching laws for the $SO(n-1)$ as a subgroup of $SO(n)$ are well known and easy to find. See for example the Wikipedia article: -https://en.wikipedia.org/wiki/Restricted_representation#CITEREFMurnaghan1938 -I am having trouble finding the next most complicated example, i.e. -$$ -SO(n-2) \times SO(2) \subseteq SO(n). -$$ -Where can a description of these branching laws be found? -Edit: In particular I would like to know examples of representations of $SO(n-2) \times SO(2)$ which appear with multiplicity $1$ in any branching. - -REPLY [2 votes]: check Eastwood-Wolf, branchig of ...., Arxive 0812.0822 math[RT] in this paper you find who to compute branching laws useing LiE.<|endoftext|> -TITLE: sum of character product over derangements -QUESTION [10 upvotes]: It is widely known that -$$ \frac{1}{n!}\sum_{\pi\in S_n}\chi_\lambda(\pi)\chi_\mu(\pi)=\delta_{\lambda,\mu},$$ -where $S_n$ is the permutation group and $\chi$ are its irreducible characters. -In exercise 7.63 of his classic book Enumerative Combinatorics, Richard Stanley computes explicitly the value of -$$\sum_{\pi\in D_n}\chi_\lambda(\pi),$$ -where $\lambda$ is a hook and $D_n$ is the set of derangements (permutations without fixed points). -I would like to know the value of $$ \sum_{\pi\in D_n}\chi_\lambda(\pi)\chi_\mu(\pi),$$ -at least when $\lambda$ and/or $\mu$ is a hook. Is anything known about this sum? (It is a generalization of the previous one, to which it reduces when $\mu=(n)$). - -REPLY [8 votes]: Using standard symmetric function notation, we have - \begin{eqnarray*} \sum_{n\geq 0}\sum_{\lambda,\mu\vdash n} - \frac{1}{n!}\left(\sum_{\pi\in D_n}\chi_\lambda(\pi)\chi_\mu(\pi)\right) - s_\lambda(x)s_\mu(y) & = & \sum_{n\geq 0}\frac{1}{n!} - \sum_{\pi\in S_n}\left.p_{\rho(\pi)}(x)p_{\rho(\pi)}(y)\right|_{p_1(x)=0}\\ - & = & \sum_\nu \left.s_\nu(x)s_\nu(y)\right|_{p_1(x)=0}\\ - & = & e^{-p_1(x)p_1(y)}\sum_\nu s_\nu(x)s_\nu(y), - \end{eqnarray*} -since - $$\sum_\nu s_\nu(x)s_\nu(y)=\exp \sum_{m\geq 1}\frac - {p_m(x)p_m(y)}{m}. $$ -Thus your sum is obtained by expanding $e^{-p_1(x)p_1(y)}\sum_\nu - s_\nu(x)s_\nu(y)$ in terms of Schur functions and taking - $n!$ times the coefficient of $s_\lambda(x)s_\mu(y)$. To do this - expansion you could write - $$ e^{-p_1(x)p_1(y)} = \sum_{m\geq 0} - (-1)^m\frac{s_1(x)^ms_1(y)^m}{m!} $$ -and iteratively apply Pieri's formula for multiplying a Schur -function by $s_1$.<|endoftext|> -TITLE: Number of irreducible polynomials of degree $r$ in $F_2[x]$ -QUESTION [6 upvotes]: This is admittedly, probably an easy question for the right person here, but I cannot seem to track down an answer. The question itself may not be hard, but the answer is crucial to a math paper I am writing (and I don't know enough number theory). - -True or false: There are at least exp$(r)$ irreducible polynomials in $\mathbb{F}_2[x]$ of degree $r$ or less. [I don't care about the hidden constants in the exp$(r)$-notation; could be $2^{\frac{r}{100}}$ or it could be $\frac{2^r}{r}$.] - -If True a reference would be terrific. -I am aware that there are roughly $n/\log n$ primes of size $n$ or less for $n$ large, I am trying to adapt that to a polynomial ring where the coefficients are in a characteristic 2 field. - -REPLY [13 votes]: There is an old and well-known "prime number theorem" for irreducible polynomials in $\mathbb F_p[x]$ whose proof is much easier than the prime number theorem for $\mathbb Z$. You can find it in many textbooks, for example Ireland-Rosen A Classical Introduction to Number Theory Chapter 7, Section 2. -Theorem The number of monic irreducible degree $n$ polynomials in $\mathbb F_p[x]$ is -$$ \frac{1}{n}\sum_{d\mid n} \mu(n/d)p^d. $$ -It's not hard to see that the $p^n$ term in the sum dominates, so you get $O(p^n/n)$ as desired. The proof is not hard, one first shows that if we let $F_d(x)$ be the product of the monic irreducible polynomials of degree $d$, then -$$ \prod_{d\mid n} F_d(x) = x^{p^n}-x. $$<|endoftext|> -TITLE: A reference to a characterization of metric spaces admitting an isometric embedding into a Hilbert space -QUESTION [14 upvotes]: I am looking for a reference to the bipartite version of the Schoenberg's criterion of embeddability into a Hilbert space. The Schoenberg criterion is formulated as Proposition 8.5(ii) of the book "Geometric Nonlinear Functional Analysis" by Benyamini and Lindenstrauss: -Theorem (Schoenberg). A metric space $(X,d)$ admits an isometric embedding into a Hilbert space if and only if the function $d^2$ is negative definite in the sense that $\sum d^2(x_i,x_j)c_i\bar c_j\le 0$ for all $x_1,\dots,x_n\in X$ and all complex scalars $c_1,\dots,c_n$ satisfying $\sum c_j=0$. -This characterization has a bipartite version: -Theorem (??). A metric space $(X,d)$ admits an isometric embedding into a Hilbert space if and only if -$$\sum_{i -TITLE: An inequality involving a sum of power terms -QUESTION [5 upvotes]: I am currently working in a problem in Information Theory and I came across a difficult inequality. After many attemps, I simplified the inequality, which now looks at follows. -Consider a positive integer $x \in \{1,2,3,...\}$ and a real number $q \in [1,x]$. Prove that -\begin{equation} -\sum_{m=1}^{x} \frac{m}{\min(m+q-1,x)} \binom{x}{m} \left( \frac{x-q}{x+q}\right)^m \leq \frac{x-q+2}{x+q} \left[ \left(\frac{2x}{x+q}\right)^x-1 \right] -\end{equation} -Note that the LHS and the RHS are equal for $q=1$ and $q=x$. Moreover, please note that -\begin{equation} -\sum_{m=1}^{x} \binom{x}{m} \left( \frac{x-q}{x+q}\right)^m =\left[ \left(\frac{2x}{x+q}\right)^x-1 \right]. -\end{equation} -At this point, I got stuck. I tried by induction over $x$ first, but I wasn't able to go anywhere as the computations become untractable. I then tried to consider the derivative with respect to $q$, but again I wasn't able to reach any conclusion. -I am used to deal for my research with this kind of inequalities and I usually sort them out by applying in some way Cauchy-Schwarz, Jensen, AM-GM or any power mean inequality or by using some real analysis technique with convex/concave functions. However, in this case, all the classical techniques seem to fail. The main issue seems to be given by the power terms $\left( \frac{x-q}{x+q}\right)^m$, which make the inequality difficult. -If someone can help me by providing new directions/techniques to look at, pointing me out similar inequalities in the literature, or give me new any idea to solve it, I would be grateful. - -REPLY [6 votes]: $\newcommand{\al}{\alpha} -\newcommand{\de}{\delta} -\newcommand{\De}{\Delta} -\newcommand{\ep}{\varepsilon} -\newcommand{\ga}{\gamma} -\newcommand{\Ga}{\Gamma} -\newcommand{\la}{\lambda} -\newcommand{\Si}{\Sigma} -\newcommand{\thh}{\theta} -\newcommand{\R}{\mathbb{R}} -\newcommand{\E}{\operatorname{\mathsf E}} -\newcommand{\PP}{\operatorname{\mathsf P}} -\newcommand{\EE}{\mathcal E} -\newcommand{\F}{\mathcal F} -\newcommand{\I}{\mathcal I} -\newcommand{\x}{\mathbf x} -\newcommand{\size}{\text{size}} -\newcommand{\pow}{\text{power}} -\newcommand{\st}{\text{stupid}}$ -This conjecture is true. Of course, the hardest part of this problem is the presence of the $\min$. So, the crucial point in the proof is the following upper bound: - -Lemma 1. - \begin{equation*} - \frac1{\min(m+q-1,x)}\le\frac Am+\frac Bx, -\end{equation*} - where - \begin{equation*} - A:=\frac{(x-q+1)^2}{(x+q-1)^2},\quad B:=1-A=\frac{4 (q-1) x}{(x+q-1)^2}, \tag{3} -\end{equation*} - \begin{equation*} - 10$. So, $\de(x-2/5,x)>0$ for $x\ge2$. -Recalling that $\de(q)=\de(q,x)$ is decreasing in $q\in[1,x-2/5]$, we see that $\de(q,x)>0$ for $x\ge2$ and $q\in[1,x-2/5]$. -Thus, we get the inequality in question for $x\ge2$ and $q\in[1,x-2/5]$. The case $x=1$ is trivial. -So, it remains to consider the case when $x\ge2$ and $q\in(x-2/5,x]$. This case is much easier than the one considered, because in this case the $\min$ does not cause trouble: indeed, for $q\in(x-2/5,x]$ and $m=2,\dots,x$ we have $\min(m+q-1,x)=x$, a constant. So, in this case the difference between the left- and right-hand sides of the inequality in question times $2 q x (x+q)$ equals $2 x (x^2 - q (x - 2)) - q ((x - q)^2 + 4 x)r$, with $r$ as before. Hence, the inequality in question can be rewritten here as -\begin{equation} - \de(q):=\de(q,x):=\ln r-\ln\frac{2 x (x^2 - q (x - 2))} {q r ((x - q)^2 + 4 x)}\ge0; \tag{7} -\end{equation} -here we use the same notation, $\de(q)=\de(q,x)$, for an expression (somewhat similar to but) different from the one in (5). Here, we have -\begin{multline*} - q (x + q) (x^2 - q (x - 2)) ((x - q)^2 + 4 x)\de'(q) \\ - =q^3 (11 - 6 x) x^2 + x^4 (4 + x) + q x^3 (4 - 11 x - 2 x^2) \\ - + - 2 q^4 (2 - 3 x + x^2) + q^2 x^2 (-20 + 5 x + 6 x^2)<0 -\end{multline*} -for all $q\in[1,x]$, so that $\de(q,x)$ is decreasing in $q\in[1,x]$. -Also, $\de(x,x)=0$ and hence $\de(q,x)>0$ for all $q\in[1,x]$. -Thus, the inequality in question is completely proved.<|endoftext|> -TITLE: A characterization of metric spaces admitting a bi-Lipschitz embedding into a Hilbert space? -QUESTION [8 upvotes]: Theorem (??) derived in this MO-post from Schoenberg's theorem yeilds a "bipartite" characterization of metric spaces that admit an isometric embedding into a Hilbert space. This Theorem (??) implies the following necessary condition of bi-Lipschitz embeddability into a Hilbert space. -Theorem. If a metric space $X$ admits a bi-Lipschitz embedding to a Hilbert space, then there exists a positive real constant $L$ such that the inequality -$$\sum_{i -TITLE: A difficult determinant -QUESTION [13 upvotes]: (EDIT: I have removed the denominators I had in a previous version as they were superfluous) -The $N\times N$ determinant -$$D(a,\vec{b})=\det\left((2N+a+b_j-i-j)!\right)$$ -has the nice form -$$D(a,\vec{b})=\prod_{j=1}^N(N+a+b_j-j)!\prod_{i=j+1}^N(b_j-b_i-j+i).$$ -Since from the definition it is clear that $D(a,\vec{b})$ is antisymmetric in the variables $x_j=(N+b_j-j)$, it should be proportional to the Vandermonde of the $x$. -I would like to know if the generalization where $a$ is allowed to vary with $i$ has a nice expression as well, -$$D(\vec{a},\vec{b})=\det\left((2N+a_i+b_j-i-j)!\right)=?$$ -This is antisymmetric in both $x_j=(N+b_j-j)$ and $y_i=(N+a_i-i)$, so it should be proportional to both Vandermondes of the $x$ and the $y$... -I know Krattenthaler has this great paper about determinants, but I was not able to find help there. - -REPLY [5 votes]: I have found a solution myself, at least in the case when $a$ and $b$ are partitions. -The determinant can be written as -$$ D=\det((x_i+y_j)!)=\det\left( \int z^{x_i+y_j}e^{-z}dz\right)$$ -We resort to the Andreief identity, -$$ \int dz \det(f_{i}(z_{j}))\det(g_{i}(z_{j}))=N! -\det\left(\int f_{i}(z)g_{j}(z)dz\right).$$ -This is usually used from left to right, but I used in reverse to write -$$D=\frac{1}{N!}\int dz \det(z_{j}^{x_j})\det(z_{j}^{y_i})\prod_{i=1}^Ne^{-z_i}$$ -Now, if $x_i=a_i-i+N$ and $y_i=b_i-i+N$ and if $a\vdash n$ and $b\vdash m$ are partitions, then $\det(z_{j}^{x_j})\det(z_{j}^{y_i})=(V(z))^2s_a(z)s_b(z)$, where $V$ is the Vandermonde and $s$ are the Schur functions. Then -$$D=\frac{1}{N!}\int dz (V(z))^2s_a(z)s_b(z)\prod_{i=1}^Ne^{-z_i}$$ -The Littlewood-Richardson coefficients are defined by $s_as_b=\sum_{\rho\vdash n+m} c^\rho_{ab}s_\rho$. Using them we have -$$D=\frac{1}{N!}\sum_{\rho\vdash n+m} c^\rho_{ab}\int dz (V(z))^2s_\rho(z)\prod_{i=1}^Ne^{-z_i}.$$ -This is an integral of the Selberg type, and the result is known: -$$\int dz (V(z))^2s_\rho(z)\prod_{i=1}^Ne^{-z_i}=N!\frac{d_\rho}{(n+m)!}\prod_{j=1}^N (\rho_i+N-i)!^2 \quad (\ell(\rho)\le N),$$ where $d_\rho$ is the dimension of the irreducible representation of the permutation group labeled by $\rho$. -Therefore, -$$D=\frac{1}{(n+m)!}\sum_{\substack{\rho\vdash n+m\\\ell(\rho)\le N}} d_\rho c^\rho_{ab}\prod_{j=1}^N (\rho_i+N-i)!^2.$$ -Curiously, there are no Vandermondes in this solution.<|endoftext|> -TITLE: Automorphisms of power set lattice mod finite -QUESTION [8 upvotes]: Let $N$ be a countably infinite set and let $\mathcal P$ denote power set. -I get that the automorphisms of $(\mathcal P(N),\subseteq)$ are all induced by permutations of $N$. -But what can be said about automorphisms of $\mathcal P(N)$ mod finite? That is, mod out by the equivalence relation $A\sim B\iff A$ and $B$ differ only finitely. - -REPLY [12 votes]: This is known as Rudin-Shelah problem. Note that, by Stone duality, this is equivalent to determine the self-homeomorphism group of the Stone-Cech boundary of $N$. Notably, consider the group induced by bijections between two cofinite subsets of $N$ (modulo cofinite coincidence). It maps homomorphically injectively into the automorphism group of $A=\mathcal{P}(N)/\mathrm{fin}$. The problem is whether this map $\Phi$ is surjective, i.e., is every automorphism induced by a bijection between two cofinite subsets. -Using basic model theory Rudin (1956) proved that the continuum hypothesis (CH) implies that the automorphism group of $A$ has cardinal $2^{2^{\aleph_0}}$, so $\Phi$ is far from surjective. Shelah (1982) showed the consistency of ZFC + ($\Phi$ is surjective). -See van Douwen's 1990 (posthumous) paper: "The automorphism group of $\mathcal{P}(\omega)/\mathrm{fin}$ need not be simple". In notes of this article, it is mentioned that under CH, this group (thus of cardinal $2^{2^{\aleph_0}}$) is simple, and attributed to Rubin-Štěpánek (reference Handbook of boolean algebras, vol 2, 1989) .<|endoftext|> -TITLE: Moments of maximum of independent Gaussian random variables -QUESTION [5 upvotes]: Let $X = (X_1, \ldots, X_d) \in \mathbb{R}^d$ be a mean-zero Gaussian random vector with identity covariance matrix. Are there upper bounds for -$$E \left(\|X\|_{\infty}^k \right)$$ for $k=1, \ldots, 6$ ? It is easy to see for example for $k=1$, the upper bounded is of the order $\sqrt{2\log(2d)}$. In particular for lower order moments (like when $k=6$) can one still obtain upper bounds that are logarithmic in $d$ ?. - -REPLY [6 votes]: $\newcommand{\de}{\delta} -\newcommand{\De}{\Delta} -\newcommand{\ep}{\epsilon} -\newcommand{\ga}{\gamma} -\newcommand{\Ga}{\Gamma} -\newcommand{\la}{\lambda} -\newcommand{\Si}{\Sigma} -\newcommand{\thh}{\theta} -\newcommand{\R}{\mathbb{R}} -\newcommand{\E}{\operatorname{\mathsf E}} -\newcommand{\PP}{\operatorname{\mathsf P}}$ -Let $M:=\|X\|_{\infty}$. Then for $x>0$ -\begin{equation} - \PP(M>x)\le d\,\PP(|X_1|>x)=2d\, G(x)<2d\, f(x)/x, -\end{equation} -where $G:=1-F$, $F$ is the standard normal cdf, and $f$ is the standard normal cdf. Also, trivially $\PP(M>x)\le1$. So, -\begin{equation} - \E M^k=\int_0^\infty kx^{k-1} \PP(M>x)dx\le I_1+2kdI_2, -\end{equation} -where $x_d:=\sqrt{2\ln d}$, -\begin{equation} - I_1:=\int_0^{x_d} kx^{k-1}dx=x_d^k=(2\ln d)^{k/2}, -\end{equation} -\begin{equation} - I_2:=\int_{x_d}^\infty x^{k-2}f(x)\,dx - \sim x_d^{k-3}f(x_d)\ll(\ln d)^{(k-3)/2}/d=o((\ln d)^{k/2}/d), -\end{equation} -where the asymptotic equivalence $\sim$ follows by l'Hospital's rule. So, -\begin{equation} - \E M^k\le(2\ln d)^{k/2}(1+o(1)); -\end{equation} -the convergence everywhere here is as $d\to\infty$. - -This asymptotic bound is in fact the best possible one. Indeed, take any $c\in(0,2)$. Then for $y=y_{d,c}:=\sqrt{c\ln d}$ we have -\begin{multline} - \PP(M>y)=1-(1-G(y))^d\ge1-\exp\{-d\,G(y)\} \\ - =1-\exp\Big\{-d\,\exp\Big(-\frac{y^2}{2+o(1)}\Big)\Big\} - =1-\exp\big\{-d^{1-c/2+o(1)}\big\}\to1 -\end{multline} -and hence $\PP(M>x)\to1$ uniformly over all $x\in[0,y]$. So, -\begin{multline} - \E M^k=\int_0^\infty kx^{k-1} \PP(M>x)dx\ge \int_0^y kx^{k-1}\,dx\,(1-o(1)) \\ - =y^k(1-o(1))=(c\ln d)^{k/2}(1+o(1)). -\end{multline} -Taking now $c$ to be arbitrarily close to $2$, we conclude that -\begin{equation} - \E M^k=(2\ln d)^{k/2}(1+o(1)). -\end{equation}<|endoftext|> -TITLE: Differential operators and quasi-finite morphisms -QUESTION [8 upvotes]: Let $ X, Y $ be smooth affine varieties over $ \mathbb C $. Let $ T : X \rightarrow Y $ be a dominant quasi-finite morphism and let $ T^\# : \mathbb C[Y] \rightarrow \mathbb C[X] $ be the resulting map on coordinate rings. -($ T $ being quasi-finite means that it can be factored as an open embedding followed by a finite morphism. It also means that $T^\# $ is injective and every element of $\mathbb C[X] $ is algebraic over $ \mathbb C[Y] $.) -Let $ D(X) $ denote the ring of differential operators on $ X $. Is the following statement is true? - -Let $ d \in D(X) $. Suppose that $ d(T^\#(f)) = 0 $ for all $ f \in \mathbb C[Y] $. Then $ d = 0 $. - -REPLY [5 votes]: Edit: I think skipped a step in the original argument - it is not immediately clear to me that $d$ commutes with $\mathbb C[Y]$ in $D(X)$ unless $d$ is a derivation. I have added an inductive argument for this below. - -I believe this is true. Here is a sketch of an argument. -We will prove this by induction on the degree of the differential operator $d$. It is certainly true for differential operators of degree 0. -Suppose $d\in D(X)$ is of degree $\leq k$, and $d(f)=0$ for all $f\in \mathbb C[Y]$ (which I identify with a subring of $\mathbb C[X]$). -Lemma: We have $[d,f]=0$ for all $f\in \mathbb C[Y]$. -Proof of lemma: The differential operator $[d,f]$ is of order $\leq k-1$, and $[d,f](g)=0$ for all $g\in \mathbb C[Y]$. The lemma now follows from the inductive hypothesis. -It remains to show that any $\mathbb C[Y]$-linear differential operator $d\in D(X)$ is zero. -Consider the finite field extension $$K := \mathbb{C}(Y) \subseteq L := \mathbb{C}(X)$$ -Note that $D(X)$ injects in to $D(L/\mathbb C)$ and $d$ lands in the subring $D(L/K)$. In other words $d$ may be considered as an element of the ring $D(L/K)$ (here I am using Grothendieck's definition of differential operators for a $K$-algebra). -I claim that $D(L/K)=L$, and thus $d=0$. Indeed, $L$ is a smooth $K$-algebra of dimension $0$ (as we are in characteristic $0$), and thus: - -The ring of $D(L/K)$ is generated by derivations $Der_K(L,L)$ and $L$. -$Der_K(L,L)=0$.<|endoftext|> -TITLE: In what ways is ZF (without Choice) "somewhat constructive" -QUESTION [23 upvotes]: Let me summarize what I think I understand about constructivism: -"Constructive mathematics" is generally understood to mean a variety of theories formulated in intuitionist logic (i.e., not assuming the law of excluded middle, $\neg\neg A\Rightarrow A$ (LEM)) so that, broadly speaking, in order to prove $A \lor B$ one must prove either $A$ or $B$ ("disjunction property") and, in order to prove $\exists x.A(x)$ one must "construct" some $x$ and prove $A(x)$ for it ("existential property"). There are a number of systems of constructive mathematics, and variations on the requirements (I don't claim to know the details), but, as far as I understand it, all of them operate in the absence of the LEM which is considered one of, if not "the", root cause of non-constructivism. So constructive mathematics in this sense represents a fairly radical departure from classical mathematics, as the abandonment of the LEM represents a considerable break from accustomed practice; the upside, of course, is that a constructive proof will provide some benefits that a classical proof will not (e.g., extracting algorithms from proofs or something of the sort). -(I hope I didn't write too much nonsense here. Onward to $\mathsf{ZF}$.) -On the other hand, consider $\mathsf{ZF}$, i.e. Zermelo-Fraenkel set theory without the axiom of choice, formulated in classical logic. This also represents a departure from the more habitual $\mathsf{ZFC}$, although not as radical as the abandonment of the LEM. Some of the differences (with classical $\mathsf{ZFC}$), however, are vaguely of the same flavour: e.g., in (some varieties of) constructive mathematics one cannot prove the existence of discontinuous real functions, while in $\mathsf{ZF}$ one cannot prove the existence of non-measurable real functions; in constructive mathematics, the powerset of a finite set might fail to be finite, while in $\mathsf{ZF}$ the powerset of a well-orderable set might fail to be well-orderable; in both cases, one need be careful about what "finite" means (although the subtleties are considerable greater in constructive mathematics as in $\mathsf{ZF}$). This leads me to wonder whether there is a pattern of similarity. -Very often, we ("we" being classical mathematicians) say that we cannot exhibit some object because its construction requires the axiom of choice: e.g., I cannot exhibit a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space or a non-principal ultrafilter on $\mathbb{N}$ or this kind of things. -But conversely, we will say that we can exhibit a bijection between $X$ and $Y$ given injections $X\to Y$ and $Y\to X$, because the proof of the Schröder-Bernstein theorem does not use the axiom of choice (even though it is probably not constructive); and, while hardline constructive mathematicians will disagree, I think most classical mathematicians will agree that it does, indeed, "construct" a bijection. Similarly, this paper, which shows in $\mathsf{ZF}$ that if there is a bijection between $n\times X$ and $n\times Y$ (where $n$ is a natural number) then there is one between $X$ and $Y$ does it by "exhibiting" a kind of algorithm which it claims is somehow canonical ("pan galactic division"). This is all reminiscent of the ideas of constructivism, even though it is not fully constructive. -(This is far too long, but I feel it is necessary to illustrate where I'm getting at.) -All of the above leads me to ask whether the similarity/analogy between $\mathsf{ZF}$ and constructive mathematics can be made precise. Or, a little less vaguely: - -Question(s): Is there some well-defined sense in which $\mathsf{ZF}$ (in classical logic, but as opposed to $\mathsf{ZFC}$) can be said to be "somewhat constructive"? E.g., does it satisfy some very weak version of the existential property? Can we formalize the intuitive idea that proving the existence of an object in the absence of Choice "exhibits" the object in a stronger sense than proving its existence in $\mathsf{ZFC}$ (as the case of Schröder-Bernstein illustrates)? - -A followup or related question might be whether there exist theories defined in classical logic that have an even stronger "constructive flavour" than $\mathsf{ZF}$, i.e., if the above question makes sense, that are "fairly constructive" without going all the way to abandoning the LEM. -Additional comments: One (to me) confusing fact is that constructive theories often incorporate some variant of the axiom of choice (see, e.g., here); the relation between these constructive axioms of choice and the axiom of choice from $\mathsf{ZFC}$ (which, as I explain above, seems to be anti-constructive) is unclear to me: so maybe I'm missing a big part of the picture. And at the other extreme, $V=L$, which seems to be a kind of super axiom of choice, makes it possible to "exhibit" some things which could not be exhibited without it, e.g., a well-ordering of $\mathbb{R}$ (or just about anything: take the $<_L$-least element such that ). - -REPLY [14 votes]: As I have said in a comment, Levy proves a weak form of the existential property for $ZF$ and $\Pi_2$ sentences. He also proves that his results are best possible. Let me state a simple fact that is how I like to think about this existential property. If $ZF$ proves a $\Pi_2$ sentence $\forall x\exists y A$ then $ZF$ proves that $\forall x\exists y\in L(x); A$. That is, for every $\Pi_2$ existential $ZF$-theorem there is a relatively constructible witness. This is easy to prove.<|endoftext|> -TITLE: Torsors over complete local fields -QUESTION [6 upvotes]: Let $G$ be a linear algebraic group scheme, and let $R$ be a complete discrete valuation ring, with quotient field $K$ and residue field $k$. -If $T$ is an $R$-torsor, it yields by base change a $k$-torsor $T_k$. -Apparently, we have the following theorem: -Thm. If $G$ is smooth and $T,T'$ are two $R$-torsors, then $T_k\simeq T'_k$ implies that $T\simeq T'$. -This is proved in SGA3,p. 401,prop. 8.1. In fact, more is proved in SGA3 (it says that any $k$-torsor can be lifted in a unique way to an $R$-torsor, up to isomorphism), but I only need this particular case. -Unfortunately, I do not have access to SGA3, and I'm not really comfortable with the different topologies, so here is a couple of questions: - -Is the topology used in the theorem the étale topology ? -Would it be true if $G$ is not necessarily smooth but if we use torsors for the fppf topology ? - -Typically, I wonder what would happen for $G=\mu_2$ defined over a field of characteristic $2$ ? -Thanks a lot! - -REPLY [6 votes]: The isomorphism classes of torsors for $\mu_{n}$ over $R$ and $k$ are -classified by the fppf cohomology groups $H^{1}(R,\mu_{n})$ and $H^{1}% -(k,\mu_{n})$. From the Kummer sequence, one sees that $H^{1}(R,\mu_{n})\simeq -R^{\times}/R^{\times n}$ and $H^{1}(k,\mu_{n})\simeq k^{\times}/k^{\times n}$. -If $n$ is prime to the residue characteristic (so $\mu_{n}$ is etale), -then $R^{\times}/R^{\times n}\rightarrow k^{\times}/k^{\times n}$ is a -bijective, but if $n$ equals the residue characteristic then it is not injective; -this means that there are nonisomorphic torsors over $R$ that become -isomorphic over $k$. (Yes, a key point is that for smooth groups, the torsors are classified by the etale cohomology groups.)<|endoftext|> -TITLE: Intersection of two generic extensions -QUESTION [8 upvotes]: It is well known that the intersection of two models of ZFC does not have to be a model of ZFC (or even ZF). Now what if we restrict ourselves to models $M[G]$, $M[H]$ which are generic over $M$ for the same poset $\mathbb{P}$? What about their intersection. It is not too difficult to show that if $G \times H$ is $\mathbb{P} \times \mathbb{P}$ generic over $M$, then $M[G] \cap M[H] = M$ is a model of ZFC. But if $G$ and $H$ are arbitrary then this need not be the case and I have an example below. But this example looks fairly complicated and I am wondering whether this is already the case for Cohen forcing, i.e: -Can we find two Cohen reals $c_0$, $c_1$ over $M$ so that $M[c_0] \cap M[c_1]$ is not a model of ZF? -The simplest reason for $M[c_0] \cap M[c_1]$ not satisfying ZF would be a failure of power set in the form that $M[c_0] \cap M[c_1]$ contains no set of all reals but I see no obvious way to do this. If $c_0$, $c_1$ are Sacks reals over $M$ then $M[c_0]$ either contains the same reals as $ M[c_1]$ or they only share those of $M$ so this sort of failure is impossible in this case. -The following is an example where this is true: -Let $\mathbb{P}$ be $\prod_{n \in \omega} \mathbb{C}(\omega_n)$. This is the standard Easton product for adding one $\omega_n$-Cohen real for every $n$. Let $A \subseteq \omega$ be some very bad real that codes, say, a well order of $M$'s height. Now I claim that we can find two generics $\bar x = \langle x_n \rangle_{n \in \omega}$, $\bar y = \langle y_n \rangle_{n \in \omega}$ so that $x_n = y_n$ exactly when $n \in A$ and for any $n \notin A$, $M[\bar x]$ and $M[\bar y]$ have no common $\mathbb{C}({\omega_n})$ generics over $M$. Then if $M[\bar x] \cap M[\bar y]$ was a model of ZF, we could decode $A$. -How to construct $\bar x, \bar y$? Let $\langle D_i \rangle_{i \in \omega}$ enumerate all dense subsets of $\mathbb{P}$ in $M$ and $\langle (\tau^0_i, \tau^1_i, n_i,) \rangle_{i \in \omega}$ enumerate all triples where $\tau^0_i$, $\tau^1_i$ are $\mathbb{C}({\omega_{n_i}})$ names in $M$ for $n_i \notin A$. Now construct $\langle p_i \rangle$, $\langle q_i \rangle$ decreasing sequences in $\mathbb{P}$ as follows: We assume that $p_i,q_i$ agree on coordinates in $A$. Extend $p_i$ to $p'$ and $q_i$ to $q'$ so that $p',q' \in D_i$ and $p',q'$ still agree on coordinates in $A$ (first extend $p_i$ to $p_i'' \in D_i$ , then let $q''$ extend $q_i$ by agreeing with $p''$ on places where $q_i$ and $p_i$ already agreed, then extend $q''$ to $q' \in D_i$ and then $p''$ to $p'$... ). -Now note that $\prod_{n \neq n_i} \mathbb{C}(\omega_n)$ does not add any $\mathbb{C}({\omega_{n_i}})$ generic over $M$. This means that if $\tau$ is a $\mathbb{P}$ name for a $\mathbb{C}(\omega_{n_i})$ generic and $r \in \mathbb{P}$, then $r$ can be extended to $r'$ on coordinates different than $n_i$ so that there are $\sigma \perp \sigma' \in \mathbb{C}({\omega_{n_i}})$ so that $\sigma \subseteq \tau$ or $\sigma' \subseteq \tau$ can be forced by extending the $n_i$'th coordinate (and only this coordinate). Now using this extend $p'$ to $p_{i+1}$ and $q'$ to $q_{i+1}$ again so that they agree on coordinates in $A$ but $\tau^0_i[\bar x]$ and $\tau^1_i [\bar y]$ will disagree. - -REPLY [11 votes]: Let $M$ be a countable transitive model. Fix $a\subseteq \omega$ such that $a$ is not generic over $M$. We plan to construct a pair of $M$-generic Cohen reals $G_0$ and $G_1$ such that $P(\omega)\cap M[G_0]\cap M[G_1]$ encodes $a$ and hence does not belong to $M[G_0]$ or $M[G_1]$ let alone $M[G_0]\cap M[G_1]$. -Since adding $\omega$ Cohen reals is the same as adding a single one, we might as well instead construct a pair of $M$-generics $G_0,G_1\subseteq \text{Add}(\omega,\omega)$ with the property described above. -We build $M$-generic sets $G_0,G_1\subseteq \text{Add}(\omega,\omega)$ such that if $\langle g^0_n : n < \omega\rangle$ and $\langle g^1_n : n < \omega\rangle$ are the associated reals then - -For all $n\in a$, $g^0_n = g^1_n$. -For all $n\notin a$, $g^0_n\times - G_1$ is generic for $\text{Add}(\omega,1)\times - \text{Add}(\omega,\omega)$. -For all $n\notin a$, $g^1_n\times - G_0$ is generic for $\text{Add}(\omega,1)\times - \text{Add}(\omega,\omega)$. - -It follows that $S = P(\omega)\cap M[G_0]\cap M[G_1]\notin M[G_0]$, since $a = \{n : g^0_n\in S\}$ can be computed from $S$ and $G_0$ and $a\notin M[G_0]$. In particular $S\notin M[G_0]\cap M[G_1]$, so $M[G_0]\cap M[G_1]$ does not satisfy the Powerset Axiom. -The generics are constructed by building two decreasing sequences of $\text{Add}(\omega,\omega)$ conditions $\langle p^i : i < \omega\rangle$ and $\langle q^i : i < \omega\rangle$ converging to $G_0$ and $G_1$ respectively, inductively meeting the countably many relevant dense sets while maintaining at each stage $i$ that $p^i_n = q^i_n$ for $n\in a$ by copying the information on the coordinates in $a$ every time one extends a condition. This can be done because the steps enforcing genericity in $\text{Add}(\omega,1)\times\text{Add}(\omega,\omega)$ only add information on the coordinates in $a$ to one of the sequences.<|endoftext|> -TITLE: Non embedding of $Y\times Y$ into $\mathbb{R}^3$ -QUESTION [31 upvotes]: I know that this is a well known result, but where can I find a proof? I am also interested to see more general non-embedding results of this type. - -Theorem. Let $Y$ be the union of two segments with one segment being - attached in the middle of the other one. Then - there is no topological embedding of the space $Y\times Y$ - into $\mathbb{R}^3$. - -It was a homework problem in my undergraduate (sophomore) topology class taught by Professor Karol Sieklucki at University of Warsaw! - -REPLY [7 votes]: I will show that $Y^k$ cannot embed in $\mathbb{R}^p$ when $p < 2k$. -Write $\mathrm{Conf}(2,X) = \{(x,x') \in X^2 \, | \; x \neq x'\}$ for the deleted diagonal of $X$, and note that any injection $Y \hookrightarrow X$ induces an $S_2$-equivariant map $\mathrm{Conf}(2,Y) \to \mathrm{Conf}(2, X)$. -We need the following corollary of Theorem 1.8 from my paper "Configuration space in a product". - -If an embedding $A \subseteq B$ induces a homotopy equivalence of pairs $(A^2, \mathrm{Conf}(2,A)) \simeq (B^2, \mathrm{Conf}(2,B))$, then it also induces a homotopy equivalence $\mathrm{Conf}(2,A^k) \simeq \mathrm{Conf}(2,B^k)$ for all $k \geq 0$. - -Here is a short proof of the corollary in the case $k=2$. -Two points in $A^2$ are distinct if and only if they are distinct in their first coordinate, or their second coordinate, or both. In other words, $\mathrm{Conf}(2,A^2)$ is covered by the two open subsets $A \times \mathrm{Conf}(2,A)$ and $\mathrm{Conf}(2,A) \times A$, and the intersection of these sets is $\mathrm{Conf}(2,A) \times \mathrm{Conf}(2,A)$. As a consequence, $\mathrm{Conf}(2,A^2)$ is the homotopy pushout of a diagram that depends only on the pair $(A^2, \mathrm{Conf}(2,A))$. By our assumption, the inclusion $A \subseteq B$ induces a pointwise homotopy equivalence on these pushout diagrams, and hence on homotopy pushouts. This concludes the proof for $k=2$; the case of general $k$ requires a larger homotopy colimit, but is otherwise similar. -"Reordering the cars in the driveway twice" gives an equivalence $\mathrm{Conf}(2,Y) \simeq \mathrm{Conf}(2, \mathbb{R}^2)$, (even when restricted to the driveway and a little part of the street, two cars may wind around each other), leading to an equivalence of pairs -$$ -(Y^2, \mathrm{Conf}(2,Y)) \simeq (\mathbb{R}^4, \mathrm{Conf}(2,\mathbb{R}^2)) -$$ -induced by the usual inclusion $Y \subset \mathbb{R}^2$. By the corollary, $\mathrm{Conf}(2, Y^k)$ is homotopy equivalent to $\mathrm{Conf}(2, \mathbb{R}^{2k})$, and moreover, this map is $S_2$ equivariant. -Since $\mathrm{Conf}(2, \mathbb{R}^{p}) \simeq_{S_2} (S^{p-1}, \tau)$, where $\tau$ denotes the antipodal action, and similarly $\mathrm{Conf}(2,Y) \simeq_{S_2} \mathrm{Conf}(2, \mathbb{R}^{2k}) \simeq_{S_2} (S^{2k-1}, \tau)$, any embedding $Y^{k} \subseteq \mathbb{R}^p$ induces an $S_2$-map -$$ -(S^{2k-1}, \tau) \to (S^{p-1}, \tau), -$$ -which is impossible for $p<2k$ by the Borsuk-Ulam theorem.<|endoftext|> -TITLE: Does the existence of a unique chromatic (possibly transfinite) number for every (possibly non-finite) simple graph imply the axiom of choice? -QUESTION [6 upvotes]: Assuming the axiom of choice I can write for any cardinal number $\kappa$ and any simple graph $G$ that a function $f$ is a $\kappa\text{-coloring}$ of $G$ if and only if the cardinality of the image of $f$ is equal to $\kappa$ and that: -$$\forall u,v\in V(G)\left[\{u,v\}\in E(G)\implies f(u)\neq f(v)\right]$$ -Now letting $f=\text{id}_{V(G)}$ one sees every simple graph $G$ has a $|V(G)|\text{-coloring}$. Which proves there exists a cardinal number capable of coloring any simple graph, now by the axiom of choice I know that the cardinals are well ordered which means that there must exist a smallest such cardinal that colors any simple graph, thus one can always unambiguously define the chromatic number of a simple graph as the smallest cardinal number which colors it but only if we assume the axiom of choice. So I'm curious if the seemingly stronger converse of this proposition holds, or if not then what can be said on the manner and would really appreciate any references or relevant comments. - -REPLY [14 votes]: It seems that your question has a positive answer, as shown by Galvin and Komjáth in their paper - -Galvin, F.; Komjáth, P., Graph colorings and the axiom of choice, Period. Math. Hung. 22, No.1, 71-75 (1991). ZBL0748.05056.<|endoftext|> -TITLE: The division problem for tempered functions -QUESTION [5 upvotes]: It is well known (see for example S Łojasiewicz, Sur le problème de la division, Studia Math. 8 (1959), 87–136.) that any linear partial differential operator with constant coefficients is surjective on $\mathcal{S}'(\mathbb{R}^n).$ (The space of tempered distributions.) -A function $f \in C_{\infty}(\mathbb{R}^n,\mathbb{C})$ is tempered if it can be controlled (as well as all its derivatives) by a polynomial when $|x|\to +\infty$. Of course, the distribution associated to a tempered function is a tempered distribution. Let us note $T(\mathbb{R}^n)$ the space of tempered functions. - -Is every linear partial differential operator with constant coefficients surjective on $T(\mathbb{R}^n)$ ? - -Of course if $f \in T(\mathbb{R}^n)$ and if $P$ is such an operator, I can find $v\in \mathcal{S}'(\mathbb{R}^n)$ such that $P.v = u_f$, but nothing tells me that $v$ is associated to a tempered function. -Thanks for any help. - -REPLY [7 votes]: Your space $T(\mathbb R^d)$ is what Laurent Schwartz introduced as $\mathscr O_M$ because these slowly increasing smooth functions (each partial derivative is bounded by a polynomial whose degree depends on the derivative) act as opérateurs de multiplication on $\mathscr S'$ (L. Schwartz, Théorie des distributions (1966), page 246). -Surjectivity of linear partial differential operators with constant coefficients is discussed in the article Surjectivity of differential operators and the division problem in certain function and distribution spaces of Julian Larcher in Journal of Mathematical Analysis and Applications 409, p. 91-99. -For example, it is proved there that every hypoelliptic operator is surjective (this not completely trivial: If $P(D)v=f\in \mathscr O_M$ with $v\in\mathscr S'$ then $v$ is smooth and a tempered distribution but a priori need not be a smooth tempered function -- this is a good reason to avoid the term tempered function). Generalizing an example of M.S. Baouendi from 1965, Larcher shows that in $\mathbb R^2$ the operator $\partial_x\partial_y+c$ is surjective if and only if $c$ is real. There is also a conjecture about a characterization of surjectivity on $\mathscr O_M$ but this is an open problem.<|endoftext|> -TITLE: Definability in the field of reals with a predicate for some powers of two -QUESTION [14 upvotes]: In "The field of reals with a predicate for the powers of two", Van den Dries has proved that the set of integers is not definable in $(\mathbb{R}, +,\cdot, \leq, 0, 1, 2^{\mathbb{Z}})$, where - $2^{\mathbb{Z}}=\{2^n: n \in \mathbb{Z} \}$. -Question. Is there a subset $S$ of $2^{\mathbb{Z}}$ such that $\mathbb{Z}$ is definable in $(\mathbb{R}, +,\cdot, \leq, 0, 1, S)?$ - -REPLY [8 votes]: The answer is no, due to Friedman and Miller, Expansions of o-minimal structures by sparse sets, Fundamenta Mathematicae, 1(167), 55-64. Thanks to Erik Walsberg for providing the reference. -The main result of the paper is the following remarkable theorem: - -Let $\mathfrak{R}$ be an o-minimal expansion of $(\mathbb{R}, <, +)$. Let $E \subseteq \mathbb{R}$ be such that, for every $m\in \mathbb{N}$ and $f: \mathbb{R}^m \to \mathbb{R}$ definable in $\mathfrak{R}$, the closure - of $f(E^m)$ is a finite union of discrete sets. Then every subset of $\mathbb{R}$ definable in $(\mathfrak{R},E)^\#$ either has interior or is a finite union of discrete sets. - -Here $(\mathfrak{R},E)^\#$ is the structure obtained by adding to $\mathfrak{R}$ predicates picking out every subset of every cartesian power $E^k$ of $E$. -Moreover, they show that the same is true if "a finite union of discrete sets" is replaced by "nowhere dense", "null", "countable", or "discrete". -As an example, Friedman and Miller observe that the work of van den Dries implies that the theorem applies to $E = 2^{\mathbb{Z}}$. So you can expand the real field by all subsets of $2^\mathbb{Z}$ at once, without defining $\mathbb{Z}$ (since defining $\mathbb{Z}$ implies defining $\mathbb{Q}$ which has empty interior but is not a finite union of discrete sets). -Note, however, that there are subsets $X\subseteq 2^{\mathbb{Z}}$ such that $(\mathbb{R},+,\cdot,\leq,0,1,X)$ interprets $(\mathbb{Z},+,\cdot)$. For example, let $X = \{2^{n^2}\mid n\in \mathbb{N}\}$. Since every natural number is the sum of four squares, $X^4 = 2^\mathbb{N}$, and by taking quotients, we can define $2^\mathbb{Z}$. Then $(2^\mathbb{Z},\cdot,X)$ is isomorphic to $(\mathbb{Z},+,S)$, where $S$ is a predicate picking out the perfect squares, and multiplication on $\mathbb{Z}$ is definable in $(\mathbb{Z},+,S)$.<|endoftext|> -TITLE: One point compactification of $(\mathbb{C}^{\ast})^n$ -QUESTION [5 upvotes]: I would like to know if there is a closed form formula for the homotopy type of $\widehat{(\mathbb{C^{\ast}})^n}$? For example, it is not difficult to see that $\widehat{\mathbb{C^{\ast}}}$ has the homotopy type of $S^1\vee S^2$. -My guess is that the formula, for general $n$, should look something like this -$$ \bigvee_{i=0}^n \left(\bigvee_{\binom{n}{i}} S^{n+i}\right). $$ -The hunch is based on the fact that the compactly supported cohomology of the $n$-torus has the homology of the above wedge of spheres. However, I am not sure how I can prove this. -Note: I did post the same question on Math StackExchange; here is the link. - -REPLY [12 votes]: If $\widehat X$ is the 1-point compactification of $X$, then there is a homeomorphism (for, say, locally compact Hausdorff spaces) -$$ -\widehat{X \times Y} \cong \widehat X \wedge \widehat Y -$$ -with the smash product. Moreover, the smash product preserves homotopy equivalences for well-pointed spaces, which $\widehat{\Bbb C^*}$ and $S^1 \vee S^2$ both are. -Therefore, because the smash product distributes over the wedge product, there is a homotopy equivalence -$$ -\begin{align*} -\widehat{(\Bbb C^*)^n} &\simeq (S^1 \vee S^2)^{\wedge n}\\ -&\cong \bigvee_{i_k \in \{1,2\}} S^{i_1} \wedge S^{i_2} \wedge \dots \wedge S^{i_n}\\ -&\cong \bigvee_{i=0}^{n} \bigvee_{\binom{n}{i}} S^{n+i} -\end{align*} -$$ - as desired.<|endoftext|> -TITLE: Looking for a tractable algorithm or formula for the determinant of a tensor -QUESTION [5 upvotes]: It is possible to define the determinant of a tensor. -We think of a tensor as a collection of numbers but this collection easily extends to a proper multilinear map. -If $T:\{1,....,n\}^m\to \mathbb C$ then one can define -$$\operatorname{Det}T:=\sum_{\sigma_2,...,\sigma_m\in S_n}\left(\left[\prod_{i=2}^m\operatorname{Sign}\sigma_i\right]\left[\prod_{k=1}^n T(k,\sigma_2(k),...,\sigma_m(k))\right]\right).$$ -This determinant satisfies some rules that are similar to the normal determinant. For example, if we add a slice to a different slice then the determinant remains the same (for matrices, if we add one column to another one then the determinant does not change). If we view map $T$ as a tensor then the value of the determinant is also invariant under a change of basis (if the transition matrix has determinant one). -For fixed $m$, the naive algorithm to determine the number runs in $O(n!^{m-1})$ (ie not very efficiently). It would be very helpful to have an efficient algorithm to make some small calculations, or even knowing the computational complexity of calculating this number would be very helpful (preferably polynomial, of course). If there is any literature on this determinant (I have not been able to find any on this) then this would also be hugely welcomed. - -REPLY [3 votes]: This is the Cayley hyperdeterminant, see [1,2], which is believed to be an NP-hard computation [3]. - -F. Gherardelli, Osservazioni sugli iperdeterminanti, Istit. -Lombardo Accad. Sci. Lett. Rend. A 127, 107 (1993). -The Cayley -Determinant of the Determinant Tensor and the Alon Tarsi -Conjecture (1997). -On the hardness of the noncommutative determinant (2009).<|endoftext|> -TITLE: Are there any books/articles that apply abstract coordinate free differential geometry to basic thermodynamics? -QUESTION [10 upvotes]: The mathematical structure of thermodynamics by Peter Salamon (pdf) would be an example, but i would like a more abstract natural formulation of application of differential geometry or even geometric algebra to for example Maxwell relations in thermodynamics that does not use coordinates. - -REPLY [3 votes]: In this context, one should probably mention "Contact Geometry: the Geometrical Method of Gibbs's Thermodynamics" by V.I. Arnold (Proc. of the Gibbs Symposium,Yale University, 1989, 163-179).<|endoftext|> -TITLE: Proving convergence of sum over $\mathbb{Z}^n$ -QUESTION [14 upvotes]: In my research, I am trying to use the following construction by Benson Farb and John Franks, which proves that for all $n$, the group of $n\times n$ matrices with 1's on the diagonal, 0's above the diagonal and integer entries below embeds as a subgroup of $C^1(S^1)$. -http://www.math.uchicago.edu/~farb/papers/nilpotent.pdf -The construction hinges on defining interval lengths with the following sums: -Let $K>0$ and $B_K : \mathbb{Z}^n \to \mathbb{R}$ be defined by -\begin{align*} -B_K(q_1,q_2,\ldots,q_n) &= K + \sum_{j=1}^{n} q^{2n-2j+2}_j\\ -&=q^{2n}_1+q^{2n-2}_2+\cdots+q^4_{n-1}+q^2_n +K -\end{align*} -and let $S_K$ be defined by -$$ S_K = \sum_{(q_1,q_2,\ldots,q_n)\in\mathbb{Z}^n} \frac{1}{B_K(q_1,q_2,\ldots,q_n)}.$$ -The authors off-handedly say the sum defining $S_K$ converges by the integral and comparison tests. When I first saw the sum, I was like "Pfff of course, I'll just do an inductive comparison to the harmonic series. Easy squeezy lemons." -Two days later and I have no idea how to prove convergence. -I've tried induction on $n$. Clearly converges for $n=1$, assume true for $n-1$ and write -\begin{align*} S_K =\sum_{q_1\in\mathbb{Z}}\ \sum_{(q_2, q_3,\ldots,q_n)\in\mathbb{Z}^{n-1}} \frac{1}{q^{2n}_1+q^{2n-2}_2+\cdots+q^4_{n-1}+q^2_n +K}\end{align*} -but I cannot figure out a way to extract a $q_1$ term to prepare for the second summation. -I've also tried to sum one $q_i$ at a time, performing the integral test at each step, i.e. -$$S_K=\sum_{(q_1, q_2,\ldots,q_{n-1})\in\mathbb{Z}^{n-1}}\ \sum_{q_n\in\mathbb{Z}} \frac{1}{A + q^2_n}$$ -where $A$ is all of the other terms. Then, by the integral test -$$S_K \leq \sum_{(q_1, q_2,\ldots,q_{n-1})}\frac{1+\frac{\pi}{2}\sqrt{A}}{A}$$ -but this square root stirs up trouble for me because I'm forced to halve the exponents of all the terms in $A$ for my next approximation. This works for $n=2$ and $n=3$, but I can't push it further. -I'm close to foaming at the mouth, so if y'all have any input it would be greatly appreciated. - -REPLY [6 votes]: The authors just posted a correction to the paper on arXiv, where they say: -"In an earlier version of this paper we used $2n − 2j + 2$ as the exponent in the definition of BK instead of $4n − 2j + 2.$ As a result (as was pointed out to us) the series SK might not converge. However, the only properties we use of BK are that SK converges and that the ratio of the value of BK at certain points limits to 1."<|endoftext|> -TITLE: Non embedding of the Heisenberg group -QUESTION [11 upvotes]: It is well known that Heisenberg groups cannot be bi-Lipschitz embedded into Euclidean spaces. A standard proof uses the fact that a Lipschitz mapping from a Heisenberg group into a Euclidean space is almost everywhere Pansu differentiable. Do you know proofs of this result that do not use Pansu differentiability? I have heard that there are such proofs, but I do not know where to find them. -EDIT: I like all three answers of rob, Robert Young and YCor so I cannot accept any of them since I would like to accept each of them. - -REPLY [3 votes]: See -Li, Sean Markov convexity and nonembeddability of the Heisenberg group. Ann. Inst. Fourier (Grenoble) 66 (2016), no. 4, 1615–1651.<|endoftext|> -TITLE: What are the applications of the Atiyah-Bott Yang Mills paper? -QUESTION [23 upvotes]: I recently finished a seminar going through Atiyah and Bott's paper ''The Yang-Mills Equations over Riemann surfaces''. The ideas going into the proof were surprising and very beautiful to me. -However, beyond its proof's beauty, I'm having trouble seeing the use of what I've just read. For instance, as I understand it the main result of the paper is an inductive formula for the cohomology of the space $\mathcal{C}(n,k)$ (the holomorphic vector bundles of rank $n$ and Chern class $k$ over Riemann surface $M$). This makes what the $\mathcal{C}$ look like a little clearer to me, but I've heard that if $g(M)\ne 0,1$ no very explicit of the $\mathcal{C}$ are known, so the only application I can think of (helping obtain an explicit description of the $\mathcal{C}$) seems not to have worked yet. -That naive train of thought lead me to ask: - -What subsequent mathematics has heavily used the results of the - Atiyah-Bott paper? Or, more petulantly, what's the point of the - result? - -(I know that there was a lot of activity on the Yang-Mills ideas which appear in the proof by Donaldson etc., but I'm asking about more direct applications as opposed to something like that.) - -REPLY [6 votes]: With 562 citations on Mathscinet, it's hard to summarize all of the applications of this influential paper! One important one was the extension of the Atiyah-Bott results to the setting of parabolic bundles and also to bundles over 2-dimensional orbifolds. The latter, coupled with calculations by Fintushel and Stern, permits one to calculate the Instanton Floer homology of Brieskorn homology spheres. These are Seifert-fibered spaces with 3 or more fibers; once you get up to 5 fibers you need to calculate the homology of the space of SU(2) representations of some 2-dimensional orbifolds. -I wouldn't underestimate the broader influence as well; the use of equivariant Morse theory for one, and also the many ideas that were used in Donaldson's work on 4-manifold invariants.<|endoftext|> -TITLE: Two questions about basic sequences -QUESTION [8 upvotes]: Suppose $(x_n)$ and $(y_n)$ are two basic sequences in a separable Banach space $X$ such that $\overline{span}\{(x_n), (y_n)\}=X$. Can we always pass to subsequences $(x_{n_k})$ and $(y_{n_k})$ such that $\overline{span}\{(x_{n_k}), (y_{n_k})\}\neq X$? - -I had the impression that this must be "obviously true", and perhaps still is embarrassingly easy, but I just can't see the argument. - -A somewhat related question, to which I do not know the answer either: Can any separable Banach space be written as the closed span of finitely many basic sequences? - -REPLY [10 votes]: I would suggest to get a positive answer to the second question as follows: Let $\{x_i\}$ be a basic sequence in $X$ (existing by Mazur's result) and let $\{z_i\}$ be a rapidly converging to zero sequence in $X$ with dense linear span. Let $\{y_i\}$ be given by $y_i=x_i+z_i$. The sequence $\{y_i\}$ is basic by the result of Krein-Milman-Rutman. The linear span of the union is dense because it contains $\{z_i\}$. -The answer to the first question is negative. It is done by using the construction for the second question, and in addition picking $\{z_i\}$ in such a way that each of it subsequences has a dense linear span in $X$. Such constructions are known, one of them: Consider a sequence $\{u_k\}$ of normalized vectors with dense linear span and let $z_i=\sum_{k=1}^\infty\left(\tau_i\right)^ku_k$, where $\{\tau_i\}$ is a rapdly converging to $0$ sequence of real numbers. Any subsequence of $\{z_i\}$ has a dense linear span because otherwise there is $x^* \in X^*$, $x^*\ne 0$, such that the analytic (in $\lambda$) function $\sum_{k=1}^\infty\lambda^kx^*(u_k)$ would be a nonzero function with infinitely many zeros in $[0,1]$.<|endoftext|> -TITLE: Dynamical system and omega limit set -QUESTION [5 upvotes]: Can Omega limit sets of dynamical systems be connected but not road connected? -In the process of reading Wiggins, we have encountered the definition and properties of Omega limit sets for autonomous equations in a finite dimensional space. The connectivity of Omega limit set of a point is described in nature, but its path connectivity is not discussed. -It is because the Omega limit set itself may not satisfy the road connectivity. Or is it because we can't use the Omega limit set's road connectivity in the research process? - -REPLY [10 votes]: Yes, the omega-limit set of a dynamical system can be connected but not path-connected. In fact this is the case for hyperbolic flows with an attractor, e.g. the Lorenz attractor, the Plykin attractor, -some uniformly hyperbolic attractors -etc. -For these systems, the omega-limit set is the attractor. This attractor contains a dense unstable leaf, hence is the closure of a path-connected set, and thus connected. On the other hand, a path in the attractor must stay in an unstable leaf, because the attractor is a Cantor set transversally to the unstable leaves. Since the attractor is not a single unstable leaf, there are points on the attractor that cannot be connected by a path. Actually, each unstable leaf is of empty interior in the attractor.<|endoftext|> -TITLE: Minimum number of operations necessary to arrive at any configuration -QUESTION [6 upvotes]: Let $k \geq 2$ and $N_1, N_2, ..., N_k$ be positive integers. -Let $S=\{(a_1,a_2,...,a_k) \in \mathbb{Z}^k:1 \leq a_i \leq N_i\}$ and $A=\{1,2,...,\prod_{i=1}^{k} N_{i}\}$. -Given a bijective map $f:S \to A$ we define a change as the operation of choosing any two $s_1,s_2 \in S$, such that $s_1$ and $s_2$ differ in only one coordinate, and then we interchange the values of $f(s_1)$ and $f(s_2)$. -Consider the set $F$ whose elements are all bijective maps $f:S \to A$. -Is it possible to find the exact value of $M$ (as a function of $k$ and $N_1,N_2,...,N_k$) such that for any $f,g \in F$ after at most $M$ changes beginning from $f$ the result is $g$? -Does anyone know if this result holds for specific values of $k$, such $k=2$ or $k=3$? -Any help or reference would be appreciated. - -REPLY [3 votes]: As said in the comments above, it's easier to think of this as a word length problem on $\mathfrak{S}_S$, the symmetric group of $S$, in terms of certain transpositions (which we will also call "changes"). -For $k = 2$, the formula is $2 N_1 N_2 - N_1 - N_2$. More generally: -Lemma: There is an element $\sigma \in \mathfrak{S}_S$ with minimal length exactly $|S| \sum_i (1 - \frac{1}{N_i})$ in terms of changes. -Let $\sigma \in \mathfrak{S}_S$ be the element that takes each $(a_1, a_2, ..., a_k)$ to $(a_1 + 1 (mod N_1), a_2 + 1 (mod N_2), ..., a_k + 1 (mod N_k))$. The minimum length expression of $\sigma$ in terms of changes has length $|S| \sum_i (1 - \frac{1}{N_i})$. The proof is similar to the proof that the long cycle in $\mathfrak{S}_n$ has minimum length $n - 1$ in terms of transpositions: each change can only increase one coordinate of the output for one input; the expression above is the number of increased coordinates in $\sigma$, and as such the minimal length is at least the above expression. -In more detail: For any $\sigma \in \mathfrak{S}_S$, let $m(\sigma) = \sum_{s \in S} |\{1 \leq i \leq k|\sigma(s)_i > s_i\}|$. -Claim: for any $\sigma \in \mathfrak{S}_S$ and change $\tau$, $m(\tau \sigma) \leq m(\sigma) + 1$. -Proof: The change only affects the sum for two elements $s, s'$ in $S$. Furthermore, it only affects one coordinate. Finally, for one of $s, s'$, that coordinate is decreased. Therefore, $m$ can increase by at most $1$. -Induction then shows that the minimum length of any $\sigma \in \mathfrak{S}_S$ is at least $m(\sigma)$. It's not hard to see that for our chosen $\sigma$, $m(\sigma) = |S| \sum_i (1 - \frac{1}{N_i})$, and we are done. $\square$ -As a side note, this bound is sharp, that is, there is a way to express this $\sigma$ with that length: we can focus on one coordinate at a time. For each "axis" in the direction of some coordinate $i$, correcting that coordinate is equivalent to a long cycle on the $N_i$ elements, which therefore takes $N_i - 1$ changes. We therefore need $\frac{|S|}{N_i} (N_i - 1) = |S| (1 - \frac{1}{N_i})$ changes to correct the $i$th coordinate, which we can then leave alone. That totals to the above expression. -Lemma: For $k = 2$, any element $\sigma \in \mathfrak{S}_S$ can be written using at most $2 N_1 N_2 - N_1 - N_2$ changes. -Proof: When $N_1 = 1$, this says that any element of $\mathfrak{S}_{N_2}$ can be written using at most $N_2 - 1$ transpositions, which is already known to be true. As such, we have our base case. -The essential idea is that we consider the elements of $S$ as boxes in a rectangle; we start with the boxes lexicographically ordered, and we want to end up in some arbitrary reordering; we're allowed "moves" that switch two boxes either in the same row or column. The proof is done by correctly "filling" the last row, and bounding the number of moves we take to do so. We therefore only care about the boxes that will eventually go into that last row. There are $N_1$ of these. -We start by making sure there is one of these boxes in each column. We do this by repeating the following step: if there is an empty column, move the box that is supposed to be in that column to that column. This can only be done finitely many times (and we will later discuss how many times). -Once that is done, each column has exactly one box, by pigeonhole principle. Move every box to the last row. This takes at most $N_1$ changes. -Finally, as all boxes are in the last row, they can be moved to the correct position through changes that switch last-row boxes. -Claim: This takes at most $2 N_1 - 1$ changes. -Proof: First, we note that the first step doesn't affect at least one element. That is because at least one element is moved into the last empty column; as it doesn't leave an empty column, there must be some element there that wasn't moved. -The set of columns that were never empty is then also the set of columns affected by the third step. This set is nonempty, so (by the usual argument for word length with respect to arbitrary transpositions), the maximum number of transpositions in the third step is at most the size of the set minus 1. -As such, the total number changes in the first and third step is at most the number of columns that were ever empty plus the number of columns that were never empty minus 1, for a total of $N_1 - 1$. The second step takes at most $N_1$ steps, so the total number of steps is at most $2 N_1 - 1$. -Combining this with the base case, the induction is clear, and we are done. $\square$ -I think that for $k > 3$, the above argument can be generalized, with enough care, to show that the lower bound in the first lemma is sharp, but don't immediately see it. The proof here essentially relied on the fact that we can choose any order in which to fix the elements of the last row, which is no longer true in the higher-dimensional case. I think a general proof has a good chance of working, using the facts that: -1) $M(N_1, N_2, ..., N_{k - 1}, 1) = M(N_1, N_2, ..., N_{k-1})$, and -2) $M(N_1, N_2, ..., N_k + 1) - M(N_1, N_2, ..., N_k) = N_1 N_2 ... N_{k - 1} + M(N_1, N_2, ..., N_{k-1})$ -The first term should represent the number of moves in the $k$ direction that may be necessary to get everything to the last hyperplane (as in the second step in the above proof); the second term should represent the number of moves to rearrange everything in the other directions - though, as here, some of those rearrangements may have to happen outside of that plane.<|endoftext|> -TITLE: How many homotopy types of lens spaces L(p,q) if the given integer p is not prime? -QUESTION [9 upvotes]: There is a theorem of Whitehead that lens spaces $L(p,q)$ and $L(p,q')$ are of the same homotopy type iff $\pm qq'≡ m^2 (\mathrm{mod}\ p)$ for some $m$. As a consequence, for a given $p$, there is only one homotopy type of $L(p,q)$ if $p=4k+3$ is prime, and two if $p=4k+1$ is prime (see Rolfsen's book "Knots and Links"). What if $p$ is not prime? Is it true that there are also two homotopy types? If not, is there any counterexample? - -REPLY [17 votes]: No; in fact, there can be arbitrarily many homotopy types. -The theorem you quote says that the number of homotopy types, for a given $p$, is the same as the size of the following quotient group: $(\Bbb Z/p\Bbb Z)^\times$, the group of units in the ring $\Bbb Z/p\Bbb Z$, modded out by the subgroup generated by its squares and $-1$. By the Chinese remainder theorem, we can calculate this size on the prime-power factors of $p$ and multiply all the answers together. -Moreover, it's easy (given known elementary number theory statements related to primitive roots, which are already used in the prime-$p$ result you mention), to show that if $p=\ell^r$ is a power of the prime $\ell$ ($r\ge1$), then the size of this quotient is $1$ if $\ell\equiv3\pmod4$, is $2$ if $\ell\equiv1\pmod4$, is $1$ if $\ell=2$ and $1\le r\le2$, and is $2$ if $\ell=2$ and $r\ge3$. -Consequently, define $f(p) = \#\{\ell\mid p\colon \ell \text{ is prime, } \ell\equiv1\pmod 4\}$, and define $g(p) = f(p)+1$ if $8\mid p$ and $g(p) = f(p)$ if $8\nmid p$. (Here $a\mid b$ means "$a$ divides $b$".) Then the number of homotopy types for a given $p$ is exactly $2^{g(p)}$. -The smallest $p$ for which there are more than $2$ homotopy types is $p=40$; there are $4$ homotopy types, with the corresponding equivalence classes of $q$ represented by: -\begin{align*} -&\{1,9,31,39\}\\ -&\{11,19,21,29\}\\ -&\{3,13,27,37\}\\ -&\{7,17,23,33\} -\end{align*} -The smallest $p$ for which there are more than $1{,}000$ homotopy types is $p=8\cdot5\cdot13\cdot17\cdot29\cdot37\cdot41\cdot53\cdot61\cdot73 = 91{,}783{,}456{,}403{,}080$.<|endoftext|> -TITLE: Longest path through hypercube corners -QUESTION [12 upvotes]: Is the longest Hamiltonian path through the $2^d$ unit hypercube vertices known, -where path length is measured by Euclidean distance in $\mathbb{R}^d$? -The unit hypercube spans from $(0,0,\ldots,0)$ to $(1,1,\ldots,1)$. -For example, for the cube in $\mathbb{R}^3$, I believe the longest path -has length -$3\sqrt{2}+4\sqrt{3} \approx 11.17$, -avoiding all edges of length $1$, and using all $4$ of the long diagonals -and $3$ short diagonals: - -          - - -          - -Path: $(1,7,2,8,3,5,4,6)$. - - -This likely has been studied, in which case pointers would be appreciated. -If exact values are not known, bounds would be useful. - -REPLY [3 votes]: We can in fact, a corrollary from MTyson's post is that we can construct a graph $G$ isomorphic to a hypercube $H$ (where the vertices are the $d$-bit strings and two vertices are adjacent in $H$ if they differ in precisely one bit) such that every edge in $G$ has length at least $\sqrt{d-1}$ in this metric, and $G$ admits a Hamiltonian circuit where half the edges have length $\sqrt{d}$. -Vertex $(u_1,u_2,\ldots, u_d)$ is adjacent in $G$ to $(1-u_1,1-u_2,\ldots, 1-u_{i-1}, u_i, 1-u_{i+1}, \ldots, 1-u_d)$ for all $i=1,2,\ldots, d-1$, and then vertex $(u_1,u_2, \ldots, u_d)$ is also adjacent in $G$ to its opposite $\bar{u}= (1-u_1,1-u_2, \ldots, 1-u_d)$. -Then $G$ is indeed isomorphic to a hypercube $H$ and has all edges of length at least $\sqrt{d-1}$. Furthermore, as $H$ admits a Hamiltonian circuit that has all edges of the form $\{(u_1, u_2, \ldots, u_{d-1}, u_d), (u_1, u_2, \ldots, u_{d-1}, 1-u_d)\}$ (i.e., the vertices differ in precisely their last bit), it follows that $G$ admits a Hamiltonian circuit that has all edges $\{u,\bar{u}\}$. -One can also check that no Hamiltonian cycle can have longer length than -$2^{d-1}(\sqrt{d} + \sqrt{d-1})$. Indeed, for each vertex $u$, there is only one other vertex that is $\sqrt{d}$ distance from $u$. So this implies that the edges $\{e = \{u,v\}; ||u-v||_2 = \sqrt{d} \}$ form a matching. Thus only half of the edges $e$ in a Hamiltonian cycle can have length $\sqrt{d}$, the remaining must have length no more than $\sqrt{d-1}$, giving the upper bound.<|endoftext|> -TITLE: Some more binomial coefficient determinants -QUESTION [13 upvotes]: The setup is similar to this question, but generalizes the size of the Hankel matrix. We'll define -$$d(n,k,r):=\det\left(\binom{2i+2j+k+r}{i+j}\right)_{i,j=0}^{kn-1}.$$ -Edit: Thanks to Johann Cigler for checking, which made me discover that the originally given formula $d(n,k,r):=\det\left(\binom{2i+2j+k}{i+j}\right)_{i,j=0}^{kn-1+r}$ was not the one I used in my implementation. Sorry for that! So in fact my generalization only concerns matrix sizes that are multiples of $k$. Now I am not sure if the picture becomes more coherent when including other matrix sizes as well, but it may be wotrh giving it a look. -The original question is the special case when $k$ is odd and $r=0$ and conjectures $d(n,k,0)=(2n+1)^\frac{k-1}{2}$. (For even $k$, we have $d(n,k,0)=(-1)^{ n\frac k2}$ according to Krattenthaler.) -Computer experiments suggest that in the more general case, there is still a certain proportion of those determinants which are, up to sign, exact $\bigl[\frac{k+r-1}2\bigr]^\text{th}$ powers (including $\pm1$), while the others are not only no powers but usually have quite large prime factors (e.g. see the two grey entries in the $k=4$ column just below). In the following I will ignore those other values and only consider those which are powers. It may not be a surprise that they all come in intriguing patterns. -E.g. for $r=1$, the table of $d(n,k,1)$ starts with $$\begin{matrix} -n \backslash k &2&3&4&5&6&7&8&9&10&11 \\ \hline \ -0 \quad| &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\ \ -1 \quad|&-4 & 8 & \color{grey} {\scriptstyle {(-61)}} & & & & & \\ \ -2 \quad| &3& -16& -8^2 & -24^2 & & & & \\ \ -3 \quad|&\color{red}5 &1 & \color{grey}{\scriptstyle {(5\cdot139)}} & -36^2 & 12^3 & -48^3 & & \\ \ -4 \quad|&\color{red}{-12} & 1 & 7^2 & & & 64^3 & 16^4 & 80^4 \\ \ -5 \quad|&\color{red}7 &32 & \color{blue}{9^2} & 1 & & & & 100^4& -20^5 & 120^5 \\ \ -6\quad|&9 &-40 & & -1 & 11^3 & & & & & -144^5 \\ \ -7 \quad|&-20 &1 & \color{blue}{-24^2} & & \color{red}{13^3} & 1 & & \\ \ -8 \quad|&11 &1 & & 84^2 & & 1 & 15^4 & \\ \ -9 \quad|&13 & 56 & \color{blue}{15^2} & 96^2 & & & 17^4 & 1 \\ -10\quad|&-28 & -64& 17^2& & \color{red}{36^3} & & & -1 & 19^5 \\ -11\quad|&15 & 1 & & & & -160^3 & & & 21^5 & 1 & \\ -12 \quad|&17 & 1 & -40^2 & & & 176^3 & & & & 1 & \\ -13 \quad|&-36 & 80 & & & \color{red}{23^3} & & 48^4 & \\ -\end{matrix} $$ -These first entries largely suffice to conjecture the general patterns of $d(n,k,1)$ for even and odd $k$. -For $r=2,3,...$ the patterns are not too different from the ones encountered for $r=1$. (See at the end.) -Instead of $d(n,k,r)$, wherever it is up to sign such a $\bigl[\frac{k+r-1}2\bigr]^\text{th}$ power, let us consider the root $e(n,k,r)$ defined by $$e(n,k,r):=\sqrt[{\bigl[\frac{k+r-1}2 \bigr]}]{|d(n,k,r)|} \in\mathbb N$$ -If $r$ and $k$ are fixed, the $e(n+\lambda(k+r),k,r)_{\lambda\in\mathbb N_0}$ always seem to form an arithmetic sequence. For $k+r$ even, its difference is either $2k(r+k)$ or $0$ (in the latter case, we have the constant sequence $1,1,...$), i.e. the first conjecture will be - -$$e(n+k+r,k,r)-e(n,k,r)=\begin{cases}2k(k+r)\\ 0\end{cases}.$$ - -It seems much harder to formalize where the $\pm1$'s occur (and where the others), at least for odd $r\ge3$. OTOH, the pattern seems clear if $r$ is even: - -For even $r$, we have $|d(n,k,r)|=1$ iff $k$ is also even and $n=\lambda\frac{k+r}{\gcd(k,r)}$, where $\lambda \in\mathbb N_0$. In this case, more precisely, $$d(n,k,r)= (-1)^{ n\frac k2(\frac r2-1)}.$$ - -Further, if $k$ and $r$ are coprime and moreover $(k+r)$ is odd, the $k^\text{th}$ column consists of equidistant triples (three of them are colored in the above table) such that the middle value of the root is the sum of the two others, e.g. from $d(10,4,1)$ and $d(12,4,1)$ we can guess $d(14,4,1)=23^2$. BTW, the middle value is always $\equiv0\pmod4$. So for these columns, the roots come in exactly $3$ arithmetic sequences. - -There must be a deeper reason for this "sum behavior". Any idea what is happening here? - -The density (i.e. the proportion of $\bigl[\frac{k+r-1}2\bigr]^\text{th}$ powers among all values) of a given column is for coprime $k,r$ $$ \begin{cases}\frac3{k+r} & \text{if}\ k\not\equiv r\pmod 2\\ \frac4{k+r} & \text{if} \ k\equiv r\pmod 2\text{ i.e. both odd}\end{cases}$$ and a certain multiple of that if $k,r$ are not coprime. -The first entries of $e(n,k,r)$ for $r=2,...,8$ are displayed below, with a "$-$" sign meaning that the corresponding power $d(n,k,r)$ bears a negative sign. (So in fact, compared with the $d(n,k,1)$ table above I have just omitted the exponents $\bigl[\frac{k+r-1}2\bigr]$ for better lisibility.) Sadly I couldn't get the entries of the columns to flush right, and I did not want to do that manually. -r=2: -n\k 0 1 2 3 4 5 6 7 8 9 10 -0 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -8 -4 12 -2 1 -4 1 3 32 8 40 -3 1 3 -16 1 5 -72 -4 1 3 1 36 1 7 -5 1 -8 -24 7 16 1 9 -6 1 5 1 -16 1 80 1 -7 1 5 -32 9 60 11 120 -8 1 -12 1 1 -28 -9 1 7 -40 1 15 -192 -10 1 7 1 13 84 15 128 1 -11 1 -16 -48 -28 36 19 -12 1 9 1 15 1 36 1 200 1 -13 1 9 -56 108 21 - -r=3: -n\k 0 1 2 3 4 5 6 7 8 9 -0 1 1 1 1 1 1 1 1 1 1 -1 1 1 -12 4 -16 -2 1 -8 -1 3 -48 -3 1 8 3 24 5 1 -4 1 1 -8 1 48 1 -5 1 1 5 -36 1 -80 -6 1 -16 -1 9 -120 -7 1 16 48 9 11 -8 1 1 7 1 20 1 120 1 -9 1 1 -16 -60 11 -96 13 -28 -10 1 -24 9 -1 -1 -192 -11 1 24 72 17 -12 1 1 1 128 17 1 -13 1 1 11 -84 1 -1 -14 1 -32 -24 -1 17 -264 -15 1 32 13 96 36 21 220 - -r=4: -n\k 0 1 2 3 4 5 6 7 8 9 -0 1 1 1 1 1 1 1 1 1 1 -1 1 1 16 4 -20 -2 1 12 1 3 -3 1 -4 -1 32 1 5 -4 1 1 -12 -5 1 3 -24 5 48 -1 -6 1 3 1 12 1 80 1 -7 1 7 64 -8 1 -8 36 1 11 -24 -9 1 -1 80 11 1 -10 1 5 1 24 1 -11 1 5 -48 96 13 -140 15 -12 1 1 11 1 1 -13 1 -12 24 112 19 - -r=5: -n\k 0 1 2 3 4 5 6 7 8 9 -0 1 1 1 1 1 1 1 1 1 1 -1 1 1 20 -4 24 -2 1 4 -1 3 -3 1 -12 1 -40 -4 1 12 3 32 1 84 -5 1 60 7 -6 1 -1 -1 -96 -7 1 -1 5 -48 7 -80 1 -140 -8 1 1 16 1 -9 1 24 12 9 100 -24 -10 1 -24 -1 -11 1 7 1 -120 13 -1 -12 1 1 80 1 -28 1 -13 1 1 140 15 192 17 - -r=6: -n\k 0 1 2 3 4 5 6 7 8 9 10 -0 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -24 -4 28 -2 1 16 1 3 -3 1 3 -48 -4 1 4 1 40 1 96 -5 1 1 -72 7 -6 1 32 5 1 -7 1 3 -96 1 -8 1 3 1 1 140 1 -9 1 7 80 9 -120 -10 1 48 1 -20 1 13 -11 1 8 11 -144 -12 1 1 9 1 256 -13 1 -168 15 - -r=7: -n\k 0 1 2 3 4 5 6 7 8 9 10 11 -0 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -28 4 -32 -2 1 -20 -1 3 -3 1 3 56 -4 1 16 1 -5 1 -16 3 -40 1 -84 -1 -6 1 -72 -1 16 -7 1 8 1 -12 112 -8 1 1 1 160 -9 1 1 5 -140 1 -216 -10 1 -1 -1 -11 1 9 120 11 168 -12 1 32 80 1 -24 1 15 -13 1 -32 13 -196 - -r=8: -n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 -0 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 32 4 -36 -2 1 -4 1 3 -3 1 -20 1 64 -4 1 3 -8 1 -5 1 4 -1 96 1 -6 1 1 84 1 -7 1 -1 128 9 -8 1 40 7 1 -9 1 3 1 160 -1 -10 1 3 1 1 216 1 -11 1 7 192 -12 1 1 1 -13 1 -60 -16 11 -168 13 224 -32 - -REPLY [5 votes]: Johann Cigler and I have posted a proof of many of these observations on arXiv: -"An interesting class of Hankel determinants", arXiv:1807.08330. -Let $d_r(N)=\det\left({2i+2j+r\choose i+j}\right)_{i,j=0}^{N-1}$. We show that for $k,n\ge 1$, -\begin{align} -&d_{2k+1}((2k+1)n)=d_{2k+1}((2k+1)n+1)=(2n+1)^k,\\ -&d_{2k+1}((2k+1)n+k+1)=(-1)^{k+1\choose 2}4^k(n+1)^k,\\ -&d_{2k}(2kn)=d_{2k}(2kn+1)=(-1)^{kn},\\ -&d_{2k}(2kn+k)=-d_{2k}(2kn+k+1)=(-1)^{kn+{k\choose 2}}4^{k-1}(n+1)^{k-1}. -\end{align} -We do not prove the formulas for $d_r(rn-1)+d_r(rn+2)$. Possibly these could be proven by modifying the first or last step of the inductions in Lemmas 7.2 or 7.5.<|endoftext|> -TITLE: Degree of hypersurfaces and flag varieties -QUESTION [6 upvotes]: In the book Discriminants, Resultants, and Multidimensional Determinants of Andrei Zelevinsky,M.M. Kapranov and Izrail' Moiseevič Gel'fand, the authors give the following definition of degree of a hypersurface in a Grassmannian. - -As they say, in generale a hypersurfaces in a projective variety is not given by the vanishing of a polynomial in its coordinate ring, but for Grassmannians this is true, since its coordinate ring is a UFD, therefore every height-one prime is principal by Krull Theorem. -However I'm stuck on the definition of degree of a hypersurface in a Grassmannian. To be more precise...I wuold prove that this definition is well posed, as in the case of projective hypersurfaces: -The set of pencils $P_{NM}$ is parametrized by the flag variety $\mathcal{Fl}(n;k-1,k+1):=\lbrace N\subset M \: : \: \dim N=k-1, \dim M=k+1 \rbrace$. I would check that there exists a nonempty open subset $U$ of $\mathcal{Fl}(n;k-1,k+1)$ such that the intersection number of $Z$ with any pencil $P_{NM}$ in $U$ is equal to the maximal number of intersection points of $Z$ with a pencil $P\not\subset Z$. -Any help? - -REPLY [2 votes]: I am a novice, but just to start you off, I would ask myself what happens to the number of intersections when you deform slightly such a pencil? It seems if there are any tangential intersections or intersections through a singularity, that the number should go up. Thus if we start from a pencil with a maximal number of intersections, it seems that these are all transversal, and hence the number should remain constant under small deformation. Since all such pencils form a single "family", the generic one should have the maximal number of intersections. -I would also try to gain insight from an example, such as the grassmann of lines in P^3. In general there is a standard (Plucker) embedding of the grassmann into projective space, and a hypersurface of degree one in the grassmannian should be one that is cut by a hyperplane in that embedding. Maybe you can use your ufd arguments to argue that every hypersurface is cut by some projective hypersurface for that embedding. It would seem to follow at least that every such pencil does meet every hypersurface. -But I would start thinking about lines in P^3 and note that there it seems the only obvious hypersurface defined by a Schubert condition, is the lines that meet a given line. Then it seems this set has indeed degree one by your definition, since fixing a general point and general plane through that point, exactly one line in the given plane and through the given point, also meets the given line. -Generalizing from this example should give a Schubert cycle - hypersurface in a general grassmannian which does meet your pencil in exactly one point. This suggests that your pencil is the right curve to measure degree. After seeing that every hypersurfce must meet your pencils, I would see if I can show that some such pencil is not contained in a given hypersurface. In the example case, take any line not representing a point in the given hypersurface and just choose a point on it and a plane containing it, to define a pencil not lying wholly in the hypersurface. Thus if every pencil meets a given hypersurface, and some pencil meets it a finite number of times, and the family of all pencils is irreducible, then it seems a generic pencil meets it the maximal number of times.<|endoftext|> -TITLE: Is there a general way to turn a 2-monad into a lax-idempotent (a.k.a. KZ) 2-monad? -QUESTION [8 upvotes]: Often a 2-monad is best replaced with a KZ monad. For example: - -$Fun(B,Cat)$ is 2-monadic over $Cat/Ob B$, but KZ over $Cat/B$. -$SymMonCat$ is 2-monadic over $Cat$, but KZ over $Cat/Fin_\ast$. -The category of Beck-Chevalley bifibrations over $B$ is 2-monadic over $Cat/B$, but KZ over $Cat/Span(B)$ (where it consists of the cocartesian fibrations). -(This one is a bit of a stretch.) Finitely-cocomplete categories are 2-monadic over pre-derivators (i.e. functors $FinCat^{coop} \to Cat$), but the KZ forgetful functor to $Cat$ is often easier to use. - -In case you've never heard of lax-idempotent (KZ) monads, I've included some background and expanded on the above examples at the end. -(N.B. I am blithely ignoring strictness considerations in this question.) -The pattern: The general pattern I am getting at above is that a 2-monad on one 2-category $C$ can often fruitfully be replaced with a lax-idempotent 2-monad on a differerent 2-category $C'$, with the same algebras. Thus one trades objects of $C$ with certain structure for objects of $C'$ with certain properties. One wins if an object of $C'$ is not much harder to specify than an object of $C$, and the properties in question are easier to check than the structure in question is to specify. Sometimes it's not even feasible to describe the original 2-monad directly, but it is nonetheless one can approach its algebras via the alternative description. -Question: Is there something systematic going on here? For example, is there some kind of adjunction between lax-idempotent 2-monads and all 2-monads, and we sometimes get lucky and find that it induces an equivalence of categories of algebras? -I think that the work of Cruttwell and Shulman, and also of Hermida is somehow relevant here. But it seems that passing from a categorical structure to the "virtual" version gives slightly different KZ replacements than the ones I've listed above. For example, Cruttwell and Shulman's theory takes the 2-monad for symmetric monoidal categories, and yields the adjunction between symmetric monoidal categories and symmetric multicategories -- not the adjunction between symmetric monoidal categories and Gamma-categories. And in their framework, one still needs to work directly with the original 2-monad. -Clubs also seem relevant, but I'm not sure they cover all the cases above. - -Here is some background explanation. -What is a lax-idempotent 2-Monad? -There are a number of different ways to approach lax-idempotent 2-monads (sometimes called K(ock-)Z(oberlein) monads). For a flavor, here are some equivalent definitions. A lax-idempotent 2-monad $(T,\eta,\mu)$ on a 2-category $C$ is a 2-monad such that... - -... there are canonically split adjunctions $\eta_{Tc} \dashv \mu_c \dashv T \eta_c$ for each $c \in C$. -... the simplicial object $T^\bullet : \Delta^{op} \to Fun(C,C)$ extends to a 2-functor $\Delta^{op} \to Fun(C,C)$ (where $\Delta^{op}$ is considered as a full 2-subcategory of $Cat^{op}$). -... a $T$-algebra $m : Tc \to c$ is equivalent to a split adjunction $m \dashv \eta_c$. -... One need not even specify the 2-monad structure explicitly. It suffices to give an assignment on objects $c \mapsto Tc$ and unit cells $\eta_c: c \to Tc$ satisfying a certain property -- see Marmolejo and Wood. - -As usual in 2-category theory, there are 3 dual notions: (co)lax-idempotent 2-(co)monads. -Examples of lax-idempotent 2-monads: -The key fact, which can be seen from (3) above, is that an object $c \in C$ can admit a $T$-algebra structure in at most one way. That is, lax-idempotent 2-monads encode property-like structure. Replacing a structure with a property is often a good idea in higher category theory, so there are many familiar examples: - -The "free colimit completion" 2-monad on $Cat$ is lax-idempotent. Likewise, the "free finite coproduct completion", or any other completion under a class of colimits. - -We also have the examples alluded to above: - -Let $B$ be a category. The unstraightening functor $Fun(B,Cat) \to Cat/B$, or equivalently (by the Grothdieck construction) the forgetful functor $OpFib(B) \to Cat/B$ is 2-monadic, inducing a lax-idempotent 2-monad on $Cat/B$, which sends $p : E \to B$ to $B \downarrow p$. -The functor $SymMonCat \to Fun(Fin_\ast, Cat)$ sending a symmetric monoidal category to the associated $\Gamma$-category is 2-monadic, inducing a lax-idempotent 2-monad on $Fun(Fin_\ast, Cat)$. -Composing the above two examples, one sees (I think -- I should be careful, since monadic functors are not stable under composition!) that $SymMonCat \to Cat/Fin_\ast$ is another example. -Let $B$ be a category with pullbacks. Let $BC(B) \subseteq Cat/B$ be the subcategory of functors which are simultaneously cartesian and cocartesian fibrations, and satisfy the Beck-Chevalley condition for all pullback squares. The forgetful functor is 2-monadic, but not lax-idempotent (in fact, it can be obtained from a distributive law between the 2-monad for cocartesian fibrations mentioned above, and the dual one for cartesian fibrations). However, we can encode the same data in terms of a fibration over $Span(B)$, and thus the forgetful functor $BC(B) \to Span(B)$ is lax-idempotent by the fibration example above. Variations are possible with different kinds of "Burnside categories" as the Barwick school likes to call them. - -REPLY [2 votes]: I don't expect you'll be able to find one context that includes all such examples of "lax-idempotent replacements". But I can think of a couple other general examples that apply whenever your 2-monad equips categories with "algebraic structure". - -If $T$ is the Lawvere theory for the algebraic structure, i.e. the free category-with-products containing a model, then algebras are product-preserving functors $T\to \mathrm{Cat}$, and product-preservation is a lax-idempotent (even pseudo-idempotent) property. -If $T$ is the free multicategory (or cartesian multicategory, etc.) containing a model of the algebraic structure, then algebras are multicategory functors $T\to \mathrm{Cat}$, which can be identified with opfibrations of multicategories over $T$, again a lax-idempotent property on multicategories over $T$.<|endoftext|> -TITLE: Distance from nonnegativity of some orthonormal vectors -QUESTION [8 upvotes]: Suppose that $1 < k < n$. Does there exist a constant $\beta > 0$, such that for every $k$ orthonormal vectors $f_1,\ldots,f_k \in \mathbb R^n$, -there exist $k$ orthonormal vectors with nonnegative elements, $x_1,\ldots,x_k\in \mathbb R_+^n$, such that -$$\sum_{i=1}^k \|x_i - f_i\|^2_2 \leq \beta \sum_{i=1}^k\|f_i^-\|_2^2$$ -where $f_i^- := \max\{-f_i,0\}$ is the negative part of the vectors $f_i$? - -In another way, I am interested in the estimation of the distance of a $n\times k$ dimensional matrix $F$ whose columns are orthonormal from the set of $n\times k$ matrices whose columns are nonnegative and orthonormal, i.e., -$$ -\mathrm{dist}(F;St_+(n,k)) -$$ -Where $St_+(n,k)$ is the set of $n\times k$ matrices whose columns are nonnegative and orthonormal. If we drop orthonormal condition and compute $\mathrm{dist}(F;\mathbb{R}^{n\times k }_+)$, we obtain $\|F^-\|$ as a lower bound for the above distance. In this term, my question is as follows: Is a multiple of $\|F^-\|$ an upper bound for the above distance, i.e., - -Is there a constant $c>0$, such that - $$ -\|F^-\| \leq \mathrm{dist}(F;St_+(n,k)) \leq c \|F^-\| -$$ - for every $n\times k$ dimentional matrix $F$ whose columns are orthonormal? - -In the special case $k=1$, the above statement is true, with $c = 2$. -I'm interested in the special case of small values of $k$, such as $k=2$. Experimentally, -for $k>1$ and random matrices $F$ and by using Frobenius norm, I get an upper bound for $\mathrm{dist}(F; St_+(n,k))$ by alternating projection to nonnegative matrices and orthonormal matrices. I guess that the above statement is true for $c \approx 2$, $(\beta \approx 4)$. - -REPLY [3 votes]: $\newcommand{\de}{\delta} -\newcommand{\De}{\Delta} -\newcommand{\ep}{\varepsilon} -\newcommand{\ga}{\gamma} -\newcommand{\Ga}{\Gamma} -\newcommand{\la}{\lambda} -\newcommand{\Si}{\Sigma} -\newcommand{\thh}{\theta} -\newcommand{\R}{\mathbb{R}} -\newcommand{\E}{\operatorname{\mathsf E}} -\newcommand{\PP}{\operatorname{\mathsf P}} -\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$ -The following two simple lemmas are crucial. - -Lemma 1. For any nonnegative numbers $a_1,\dots,a_k$, - \begin{equation*} - \sum_1^k a_i^2-\max_1^ka_i^2\le\sum_{1\le i0$.)<|endoftext|> -TITLE: Question on Witten’s paper “Supersymmetry and Morse theory” -QUESTION [6 upvotes]: EDIT. I am trying to read the article “Supersymmetry and Morse theory” by E. Witten (JDG 17 (1982)). -This well known article applies some tools developed by physicists (e.g. path integrals) to topology of manifolds. -I am a mathematician, but I am familiar with some necessary physical ideas, although probably on more elementary level than necessary to understand the paper. To be more specific, I cannot understand the computation of matrix elements of the (twisted) de Rham differential $d_t$ on p. 672-673. The exposition there is too concise for me. - -Is there a more detailed exposition of Witten’s paper today? However mathematical rigor is not so important for me. I would like to understand the physicists tools and ideas. - -REPLY [6 votes]: You can find much more on the specific family $d_t$ if you search for the key phrase "Witten deformation"; I would try to give some specific references here but I am a little puzzled by the statement that you are "NOT looking for a mathematically more rigorous exposition". Could you clarify; does this rule out e.g. anything with definitions, theorems and proofs? I guess the Helffer-Sjöstrand theory is not what you're after? -Perhaps Section 5 of this paper of Rogers, which uses (stochastic calculus) path integral techniques to compute those matrix elements could be helpful? -More generally this part of the computation has been generalized / abstracted as an instance of localization in equivariant cohomology. About this, there has been much written in both theoretical physics and mathematics; one physics source that I found inspiring (although also with few details) is Part II of this review of Cordes, Moore and Ramgoolam. Chapter 12 is all about SUSY QM, though let me warn that it's likely not readable on its own. It essentially specializes the "symmetries, fields, equations" viewpoint developed in chapters 9, 10, 11 to this example. Appendix 4 of Chapter 12 gives a brief resummation of Witten's paper on Morse theory that you cite.<|endoftext|> -TITLE: Historically, which came first: the Lie algebras or their classification? -QUESTION [14 upvotes]: The classification of the complex simple Lie algebras by their Dynkin diagrams gives rise to five exceptional complex simple Lie algebras: $F_4, G_2, E_6, E_7$ and $E_8$. -I am trying to find out whether the classification was discovered first (attributed to Wilhelm Killing [1888-1890]), or whether some/all of the exceptional complex simple Lie algebras were discovered before the classification. -It's said here that Killing discovered $G_2$ in 1887, which would mean that $G_2$ at least came before the classification. I suspect that, since this discovery came only very shortly before his discovery of the classification, this means that the others were discovered as a consequence of the classification, but I'm hoping for clarification. - -REPLY [20 votes]: The classification came first. As Killing says in his introduction (translation by Coleman (1989)): - -For each $l$ there are four structures supplemented for $l = 2, 4, 6, 7, 8$ by exceptional simple groups. For these exceptional groups I have various results that are not in fully developed form; I hope later to be able to exhibit these groups in simple form and therefore am not communicating the representations for them that have been found so far. - -More details in Hawkins (cited in your linked reference) (1982, p. 156): - -The discovery that type $G_2$ actually exists seems to have transformed Killing’s attitude towards the possible existence of further new simple groups. (...) after this discovery, when he obtained the $\smash{a_{ij}}$ and associated root systems for the exceptional types $E$ and $F$ and the “new” general type $\smash{C_l}$, Killing maintained that simple groups for these types exist even though he never managed to carry his calculations far enough to be able to write down multiplication tables for them. - -and in Cartan (1894, pp. 94-95): - -As to the determination of simple group structures, Mr. Killing is content to show that to types A), B), D) there correspond long known simple groups, namely the general projective group of $l$-dimensional space and the projective groups of a nondegenerate quadric in spaces of $2l$ and $2l-1$ dimensions. But he doesn’t give the explicit calculations allowing to show that they are the only ones, and as to the other integer systems, he only gives vague indications on the corresponding structures, except however for type C).<|endoftext|> -TITLE: Number theory question from Homotopy groups of spheres -QUESTION [11 upvotes]: Let $n$ be some integer. -Is it true that there exists odd prime $p$ such that -$4n = (p-1) \cdot k$, -where $k$ is an integer coprime with $p$? -This question asked Roman Mikhailov. This is corresponds with Homotopy groups of spheres. Unfortunately I do not know the details. - -REPLY [6 votes]: Probably yes. The question is equivalent to: for any (nonzero) integer $n$, there exists an odd prime $p$ not dividing $n$ such that $p-1$ divides $4n$. -The divisors of $4$ are $1,2,4$; of these, $2,4$ are one less than odd primes $3,5$. So if $3\nmid n$ then choose $p=3$, and if $5\nmid n$ then choose $p=5$. -So we may suppose that $3,5$ both divide $n$. The divisors of $4\cdot3\cdot5=60$ that are one less than odd primes are $2, 4, 6, 10, 12, 30, 60$. So if $7\nmid n$ then choose $p=7$, if $11\nmid n$ then choose $p=11$, if $13\nmid n$ then choose $p=13$, if $31\nmid n$ then choose $p=31$, and if $61\nmid n$ then choose $p=61$. -So we may suppose that $3,5,7,11,13,31,61$ all divide $n$. The divisors of $4\cdot3\cdot5\cdot7\cdot11\cdot13\cdot31\cdot61$ that are one less than odd primes are $2, 4, 6, 10, 12, 22, 28, 30, 42, 52, 60, 66, 70, 78, 130, 156, 210, \ -310, 330, 366, 372, 420, 462, 546, 660, 682, 732, 858, 910, 1092, -1302, 1612, 1708, 1830, 1860, 2002, 2310, 2730, 2860, 4026, 4092, -4270, 4620, 4758, 6006, 8052, 8580, 8866, 13420, 14322, 16926, 18910, -20020, 24180, 25620, 28182, 28210, 41602, 47580, 47740, 53196, 55510, -56730, 79422, 84630, 88660, 93940, 104676, 132370, 186186, 294996, -624030, 794220, 873642, 930930, 1221220, 1248060, 1474980, 1622478, -1831830, 1861860, 2912140, 3244956, 3663660, 11357346$; and so it continues.... -All we need to prove that the answer to your question is "yes" is for this iterative procedure to contain infinitely many primes. Although that might be hard to prove, it would be incredible if the procedure halted.<|endoftext|> -TITLE: What "metatheory" did early set theory/logic researchers use to prove semantic results? -QUESTION [25 upvotes]: Things like the first-order completeness theorem and the Löwenheim-Skolem theorem are considered foundational in mathematical logic. -The modern approach seems to be, usually, to interpret a "model" specifically as a set in some other (typically first-order) "set metatheory." So when we talk about a model of PA, for instance, we typically mean that we are formalizing it as a subtheory of something like first-order ZFC. Models of ZFC can be formalized in stronger set theories, such as those obtained by adding large cardinals, etc. -But when Godel proved that a first-order sentence has a finite proof if and only if it holds in every "model" -- what was he talking about? Likewise, how can we understand the Löwenheim-Skolem theorem if models didn't even exist at the time? -It is clear that these researchers were not talking about using first-order ZFC as a metatheory, as that theory didn't even gain popularity until after Cohen's work on forcing in the 60s. Likewise, NBG set theory had not yet been formalized. And yet, they were obviously talking about something. Did they have a different notion of semantics than the modern set-theoretic one? -In closing, two questions: - -In general, how did early researchers (let's say pre-Cohen) formalize semantic concepts such as these? -Have any of these original works been translated into English, just to see directly how they treated semantics? - -REPLY [9 votes]: To address the main question about Gödel's original proof of the completeness theorem, which as Timothy Chow explains can be found in Gödel's Collected Works, he is using the notion of “Erfüllungssystem”, which means something like "satisfactory system". It is a structure intended to satisfy a formula $\phi$ that is not refutable (i.e., such that $\neg \phi$ is not provable), whose underlying set is the natural numbers. He then extends his theorem for countable theories proving the compactness theorem. -The structure is built as what we would now call a filtered colimit of a countable chain of finite structures. This is done through the use of König's lemma, by first building a finitely branching tree of finite structures with the property that the there are structures at any finite level and the satisfaction in a given structure at level $n$ restricts to the satisfaction at the immediate predecessor. Then there is a cofinal branch through which he builds the model of $\phi$ on the structure given by the natural numbers. -It is worth mentioning that as an immediate consequence of his proof he gets the downward Löwenheim-Skolem theorem, since he builds a model on the structure whose underlying set is the natural numbers for any countable set of satisfiable formulas. Löwenheim's original "proof" had actually some flaws which later Skolem fixed. - -REPLY [8 votes]: As Andreas Blass pointed out, the meta-theory can be ordinary mathematics, at least in theory. In practice, without an explicit meta-theory, authority figures decide what is allowed, and what not. Tarski (like Cantor before him) learned this lesson the hard way, as can be read in accounts of Tarski's theorem about choice from 1924: - -... when he tried to publish the theorem in Comptes Rendus de l'Académie des Sciences Paris, Fréchet and Lebesgue refused to present it. Fréchet wrote that an implication between two well known propositions is not a new result. Lebesgue wrote that an implication between two false propositions is of no interest. - -It is no surprise that the modern notion of model and meta-theory are due to Tarski (and his colleague Robert Vaught) from 1956. But Tarksi already presented a "non-modern" notion of meta-theory in 1933, see the SEP entry on Tarski's Truth Definitions. -The issue is a little bit complex, but here is my own summary of the difference between those two notions for a start: - -If I understood it correctly, for the 1933 version, the model (i.e. the algebraic structure about which we talk) is part of the meta-language and not mentioned separately. The assignment of objects to variables on the other hand is what can satisfy a given formula. A formula is (defined to be) true if it is satisfied by all possible assignments of objects to variables. -The 1956 version is treated less explicitly in the linked SEP entry, but it is hinted at that the model is no longer an implicit part of the meta-language, but an explicit object from set-theory. A model can satisfy a given formula (or sentence), similar to how an "assignment of objects to variables" could satisfy a given formula for the 1933 version. But the text also hints that the 1956 now relies stronger on an underlying set-theory, while the 1933 explicitly tried to minimize "the set-theoretic requirements of the truth definition".<|endoftext|> -TITLE: Lickorish-Wallace theorem for torsion spin$^c$ 3-manifolds? -QUESTION [7 upvotes]: The Lickorish-Wallace theorem tells us that any closed 3-manifold $Y$ is an integer link surgery on $S^3$, which yields an oriented cobordism between $S^3$ and $Y$. Filling out the $S^3$ by a 4-ball $B^4$, we obtain a compact 4-manifold bounding $Y$, and as a corollary we have that the 3rd oriented cobordism group $\Omega^{SO}_3=0$. -Now think of spin$^c$ analogue of this situation. We start with a spin$^c$ 3-manifold $(Y,\mathfrak{t})$. The Lickorish-Wallace theorem applied to the underlying manifold $Y$ gives a cobordism $W$ which only consists of 2-handles. To fill out the $S^3$-side of $W$ by a 4-ball, we should impose the condition that the $S^3$ is endowed with the (unique) torsion spin$^c$ structure $\mathfrak{s}_0$. Thus, a temporary version for spin$^c$ Lickorish-Wallace theorem is the existence of a spin$^c$ structure on $W$ which extends $\mathfrak{t}$ and restricts to the torsion spin$^c$ structure on $S^3$. But this cannot be true in full generality, as already seen from the case when $W$ is the product cobordism. (In this case, $\mathfrak{t}$ should be the torsion spin$^c$ structure.) -But (it seems) it is a forklore that a sufficient condition to make the theorem work is requiring that $\mathfrak{t}$ is torsion. For clarity, let me state the modified theorem again: -For a link surgery cobordism $W$ from $(S^3,\mathfrak{s}_0)$ to a closed 3-manifold $Y$ and a torsion spin$^c$ structure $\mathfrak{t}$ on $Y$, we can always find a spin$^c$ structure which restricts to $\mathfrak{s}_0$ on $S^3$ and $\mathfrak{t}$ on $Y$ (so that $(Y,\mathfrak{t})$ represent 0 in $\Omega^{Spin^c}_3$). -Question. How we can prove this theorem? This theorem appears in many places, e.g. it is a key step in defining absolute $\mathbb{Q}$-gradings in Heegaard Floer homology. But those literature do not include the proof. I guess a few obstruction theory would work, but it is mysterious to me how to exploit the torsion condition to extend the spin$^c$ structure. - -REPLY [5 votes]: First, let me remark that $S^3$ has a unique spin$^c$ structure $\mathfrak{t}_0$, which is also torsion. (I decided to call it $\mathfrak{t}_0$, because I prefer to use $\mathfrak{t}$ for spin$^c$ structures on 3-manifolds, and $\mathfrak{s}$ for spin$^c$ structures on 4-manifolds.) So, you don't need to specify that $\mathfrak{t}_0$ is the unique torsion spin$^c$ structure, and this probably will remove the source of your confusion. -In general, spin$^c$ structures on 3- and 4-manifolds are an affine space (a torsor, some say) over $H^2$; more precisely, for a 3-manifold $Y$ (respectively, a 4-manifold $W$) there is a free and transitive action of $H^2(Y;\mathbb{Z})$ on $\rm{Spin}^c(Y)$ (resp., of $H^2(W;\mathbb{Z})$ on $\rm{Spin}^c(W)$)). I'll denote the action with a + (probably not the best choice, but it works). -Moreover, the action is compatible with the restriction map of spin$^c$ and the restriction map on cohomology. That is, if $\iota: Y \hookrightarrow W$, $\alpha\in H^2(W;\mathbb{Z})$, and $\mathfrak{s}$ is a spin$^c$ structure on $W$, then $\iota^*\alpha + \iota^*\mathfrak{s} = \iota^*(\alpha + \mathfrak{s})$. -In your case, $W$ is a 2-handlebody, so $\iota^*: H^2(W;\mathbb{Z}) \to H^2(Y;\mathbb{Z})$ is onto, since the next term in the long exact sequence for the pair is $H^3(W,Y;\mathbb{Z}) \cong H_1(W,S^3) \cong H_1(W) = 0$, where the first equality is Poincaré-Lefschetz duality, the second is excision, and the third is the absence of 1-handles. -Combining the action and the surjectivity of $\iota^*: H^2(W;\mathbb{Z}) \to H^2(Y;\mathbb{Z})$, you get that every spin$^c$ structure $\mathfrak{t}$ on $Y$ extends to some spin$^c$ structure on $W$. (The restriction on the $S^3$ boundary is unique, so there is no problem there.) -This actually proves the stronger version of the statement that you wanted, without the restriction on being torsion. (So your confusion was, in a sense, justified.)<|endoftext|> -TITLE: What is so special about set theory anyway? -QUESTION [17 upvotes]: (Later edit - tried to clarify a couple of vague places concerning interpretations of theories that became evident in comments (thanks to Andrej Bauer, Mauro ALLEGRANZA and Emil Jeřábek). (To closers and downvoters: may I humbly direct your attention to the soft-question tag down there...)) -This is an ideal question: I don't even know what I am asking -It has been inspired by reading What "metatheory" did early set theory/logic researchers use to prove semantic results? -Specifically by - -The modern approach seems to be, usually, to interpret a "model" specifically as a set in some other (typically first-order) "set metatheory." So when we talk about a model of PA, for instance, we typically mean that we are formalizing it as a subtheory of something like first-order ZFC. Models of ZFC can be formalized in stronger set theories, such as those obtained by adding large cardinals, and etc. - -Indeed. It makes me wonder: there are all kinds of theories; there is a very general notion of interpretation of one theory in another; nothing about sets comes into play so far. Then all of a sudden, when one says "model", the first thing comes to mind is "a set with blablabla...". Why? -I mean, is there a rigorous way to distinguish some theories among all others by some formal property which (a) guarantees that any sensible notion of model is subsumed by interpretability in a theory with this property and (b) every theory with this property is equivalent (in some rigorous sense) to "a theory of sets"? What does this latter even mean? -Are there any other such "all-encompassing" classes of theories? This last question might seem senseless since any two "all-encompassing" theories must be mutually interpretable and thus be equivalent to each other (I think, at least in some sense?). But still maybe there is some notion of "meta-encompassing" that makes some "all-encompassings" more "all-encompassing" than others...? - -REPLY [13 votes]: I agree with other respondents that it is unlikely that one will be able to come up with some kind of formal argument that distinguishes set theory from other "mathematics-complete" systems (to use Mike Shulman's term, which I like!), because mathematicians are so good at rephrasing one language in terms of another. There is surely also some degree of what one might call "historical accident" involved; we instinctively think of a model as a set because that's how we were taught, and how our teachers were taught, etc. -That said, I think a case can be made that set theory has some psychological advantages when it comes to addressing certain questions, e.g., is mathematics consistent? and exactly which assumptions are used to derive which theorems? Today, most mathematicians take a rather breezy attitude towards the consistency question, assuming that it's all been sorted out by logicians and that it's no concern of theirs. If you take consistency seriously, however, then it is vitally important to try to build everything from the ground up, one small step at a time, in as simple and clear a manner as possible. As long as you limit yourself to finitary mathematics, several alternatives to set theory are available (arithmetic, syntax, type theory, ...), but for infinitary reasoning, set theory seems to be the psychologically best choice for most people. -Even for finitary reasoning, set theory has the advantage of being a very flexible and fine-grained tool. In reverse mathematics, the standard approach is to consider subsystems of second-order arithmetic. Arithmetic is certainly a natural foundation for finitary mathematics, but it quickly becomes an irresistible temptation to introduce set-theoretic reasoning because it makes it so easy to introduce any new concept or axiom that you might come up with. That is why "second-order" arithmetic becomes the de facto foundation. -There are other goals you might have for a "foundation for mathematics" for which set theory is arguably not the optimal approach. For example, maybe you have developed some conception of the mathematical universe, and you want some formal theory that directly captures the structures and concepts you have in mind. Then category theory or homotopy type theory might be more attractive. Still, I think set theory is hard to beat as a way to "get off the ground" so to speak. To fully appreciate the merits of category theory or homotopy type theory requires a lot of prior mathematical knowledge and experience, and I think that most people would have trouble acquiring all that background knowledge without ever appealing to any set-theoretic concepts along the way.<|endoftext|> -TITLE: Who was in the Fields committee for ICM 1962 (the first appointed by IMU)? -QUESTION [9 upvotes]: Traditionally, at the presentation of the Fields medals at the ICM opening ceremony, the composition of the Fields medal committee is disclosed. This information can be found in the first volume of the respective ICM Proceedings (for older ICM, somewhat hidden in the presentation speech of the chair). That is, except for the 1962 ICM at Stockholm, the first for which the Fields Committee was appointed via IMU). It is only clear that Nevanlinna (who gave the presentation speech) was the chair: - -After an interval, Professor Nevanlinna in his capacity of chairman of - the Fields Medals Committee, read the report of the Committee - which was as follows: [..] The problem of suggesting names for the - award of the Fields medals has, since the last congress, been - entrusted to the International Mathematical Union. To prepare the - names at this Congress in Stockholm, the Union appointed a Fields - medals committee. This committee has the honour to make known its - decision here. - -Some more details are given in the Secretary's report: - -The scientific programme was drawn up in close cooperation - with the International Mathematical Union, which for this - purpose nominated a Consultative Committee with Professor - G. de Rham as chairman. The Swedish representatives in charge - of the scientific programme were Professors L. Gårding, L. - Carleson and L. Hörmander. Thus a first meeting was held in - Zurich in November 1960 followed by a meeting in Düsseldorf in - January 1961. As a result of these two meetings the - "International Fields Committee", that elects the two prize - winners, was constituted. Furthermore a list was made of - speakers to be invited to deliver one-hour addresses on chosen - topics in different fields of mathematics. - -From this passage, one might guess that there was an overlap of the Fields and the Program Committee (which would be a historical exception), but I couldn't find more explicit information in the volume. Does someone know more about the composition of the committee? - -REPLY [3 votes]: Fields Medal Prize Committee 1962 - -R. Nevanlinna (chair) -P.S. Aleksandrov -E. Artin -S.-S. Chern -C. Chevalley -H. Whitney -K. Yosida - -Only Aleksandrov was also in the IMU consultative committee, the other members were not. For the record, the 1962 consultative committee was chaired by de Rham, with IMU members Aleksandrov, Chandrasekharan, Eckmann, Hodge, Hopf, Montgomery, and Morse, and local Swedish members Carleson, Frostman, Gårding, Hörmander, and Pleijel.<|endoftext|> -TITLE: Is the intersection of two subgroups, defined below, always trivial? -QUESTION [25 upvotes]: Suppose, $G = \mathbb{Z} \ast H$, where $H$ is an arbitrary group. Suppose, $g \in G$ and $g \notin \langle\langle H \rangle \rangle $. -Is $\langle\langle g \rangle \rangle \cap H$ always trivial? -($\ast$ stands for free product, and $\langle \langle \dots \rangle \rangle$ stands for normal closure) -Yesterday, I have asked this question on math.stackexchange.com and was advised to re-ask it there: -Is the intersection of two subgroups, defined below, always trivial? v2.0 -Any help will be appreciated. - -REPLY [2 votes]: As pointed out, the Kervaire conjecture requires an assumption about the exponent sum on $g$. So, in fact, the answer to the original question is"no". -Take $H$ be the cyclic group of order 6 generated by $a$. Take $\mathbb{Z}$ to be generated by $t$. Lastly, take $g=a^2ta^3t^{-1}$. With these assumptions, we get $H\subset \langle\langle g\rangle\rangle$. -This is most easily seen by looking in the quotient group $G=(H*\mathbb{Z})/\langle\langle g\rangle\rangle$. Since $g=1$, we have $a^2=ta^3t^{-1}$. -Now, since $a^6=1$ we have $a^4=(a^2)^2=(ta^3t^{-1})^2=1$. -Also, $1=(a^2)^3=(ta^3t^{-1})^3=ta^3t^{-1}$ whence $a^3=1$. So $a=1$. It follows that $H\subseteq \langle\langle g\rangle\rangle$. \ No newline at end of file